Nonlinear Partial Differential Equations II

Summary of the course Nonlinear partial diﬀerential equations II Stefan M¨uller Bonn University Summer term 2019

This is only a summary of the main results and arguments discussed in class and not a complete set of lecture notes. These notes can thus not replace the careful study of the literature. In particular, these notes only contain selected proofs. As discussed in class, among others the following books are recommended

[Ev] L.C. Evans, Partial diﬀerential equations, Amer. Math. Soc., 1998 (2nd edition 2010).

[St70] E.M. Stein, Singular integrals and diﬀerentiability properties of functions, Princeton Univ. Press, 1970.

These notes are based on the books mentioned above and further sources which are not always mentioned speciﬁcally. These notes are only for the use of the students in the class ’Nonlinear partial diﬀerential equations II’ at Bonn University, Summer term 2019.

Please send comments and corrections to [email protected].

1 [July 12, 2019] Contents

4 Parabolic equations 4 4.1 Linear equations ...... 4 4.1.1 Existence by discretization in space ...... 6 4.1.2 Higher regularity ...... 11 4.2 Integration theory and Lp spaces for Banach space-valued functions ...... 19 4.2.1 Simple and measurable functions ...... 19 4.2.2 The Bochner–Lebesgue integral ...... 20 4.2.3 Lebesgue and Lp-spaces ...... 23 4.3 Sobolev functions with values in Banach spaces ...... 26 4.4 Maximum principles and the Harnack inequality ...... 36 4.5 Other approaches to regularity ...... 45 4.5.1 Equations with measurable coeﬃcients ...... 45 4.5.2 Campanato, Schauder and Lp theory ...... 45 4.6 Time discrete approximation for nonlinear equations . . . . . 47

5 Hyperbolic equations 53 5.1 Second order hyperbolic equations ...... 53 5.1.1 Definition, existence and uniqueness of weak solutions 53 5.1.2 Regularity ...... 59 5.1.3 Finite speed of propagation ...... 62 5.2 First order hyperbolic systems ...... 63 5.2.1 Definitions ...... 63 5.2.2 Symmetric first order systems ...... 64 5.3 Local existence in time existence for nonlinear hyperbolic systems with smooth initial data ...... 72 5.3.1 Main result ...... 73 5.3.2 Refinements ...... 83 5.3.3 Examples ...... 91

6 Hyperbolic conservation laws, weak solutions and the Lax entropy condition 95 6.1 Weak solutions and the Rankine-Hugoniot jump condition . . 95 6.2 Riemann’s problem ...... 100 6.2.1 Simple waves and rarefaction waves ...... 101 6.2.2 Shock waves ...... 104 6.2.3 Local solution to Riemann’s problem ...... 106

7 Fourier methods 109 7.1 Examples ...... 110 7.1.1 The Fourier transform ...... 110 7.1.2 The Fourier transform on tempered distributions . . . 111

2 [July 12, 2019] 7.1.3 Relations with PDE ...... 113 7.2 Singular integrals, Lp estimates and multiplier theorems . . . 120 7.3 Strichartz estimates and the nonlinear Schr¨odingerequation . 130 7.4 Restriction theorems, oscillatory integrals and Strichartz estimates...... 143 7.4.1 Setting of the problem and connection with PDE . . . 143 7.4.2 Curved hypersurfaces and the Tomas-Stein theorem for the sphere ...... 144 7.5 A quick look back ...... 157

8 Epilogue: some highlights in PDE 158

3 [July 12, 2019] [1.4. 2019, Lecture 1]

4 Parabolic equations

4.1 Linear equations This subsection is taken from [Ev], Chapters 7.1.1 and 7.1.2. I occasionally slighty relaxed the regularity assumptions on f to emphasize the idea of maximal regularity. The existence of weak solutions, e.g., is proved for f ∈ L2(0,T ; H−1(Ω)) rather than f ∈ L2(0,T ; L2(Ω)). n Let Ω ⊂ R be open and bounded and set

ΩT = (0,T ] × Ω for some T > 0. We will study the initial boundary-value problem

∂tu + Lu = f in ΩT , u = 0 on ∂Ω × [0,T ], (4.1) u = g on {0} × Ω, where f :ΩT → R and g :Ω → R are given and u : ΩT → R is the unknown, u = u(t, x). The letter L denotes a second order partial-diﬀerential operator, having either divergence form

n n X ij X i Lu = − ∂i(a (t, x)∂ju) + b (t, x)∂iu + c(t, x)u i,j=1 i=1 or else nondivergence form

n n X ij X i Lu = − a (t, x)∂i∂ju + b (t, x)∂iu + c(t, x)u i,j=1 i=1 for given coeﬃcients aij, bi, c.

Deﬁnition 4.1. We say that the partial diﬀerential operator ∂t + L is parabolic if L is elliptic, i.e, if there exists a θ > 0 such that

X ij 2 n a (t, x)ξiξj ≥ θ|ξ| ∀ξ ∈ R , ∀(t, x) ∈ ΩT .

Remark. For simplicity we have considered only scalar equations. Most results, however also hold for parabolic systems (i.e., systems for which L is strongly elliptic) in divergence form as long as the coeﬃcients are continuous (see below).

4 [July 12, 2019] Discuss: function of (t, x) vs. map from a time interval to a function space. t 7→ u(·, t). Brieﬂy discuss L1(I,X) (and Lp(I,X)). Measurable, hence norm measurable; can deﬁne L1 (semi)norm. Main point if X is separable then simple functions are dense.

We write H1(Ω) as shorthand for W 1,2(Ω) and we denote by H−1(Ω) 1 the dual space of H0 (Ω). From now on we will assume that ij ∞ i ∞ ∞ 2 −1 2 a ∈ L (ΩT ), b ∈ L (ΩT ), c ∈ L (ΩT ), f ∈ L (0,T ; H (Ω)), g ∈ L (Ω). (4.2) We will also assume that

aji = aij ∀i, j = 1, . . . , n. (4.3)

Until further notice we will assume that L is in divergence form and we deﬁne the quadratic form associated to L by

Z N X ij X B[u, v, t] := a (t, ·)∂ju∂iv + bi(t, ·)∂iu v + c(t, ·)uv. (4.4) Ω i,j i=1

If ∂t + L is parabolic and (4.2) holds then there exist β > 0 and γ ≥ 0 such that 2 2 B[u, u, t] ≥ βkuk 1 − γkuk 2 . (4.5) H0 (Ω) L (Ω) This follows from the estimate

X i ε 2 1 2 k b ∂ u uk 1 ≤ kbk ∞ ( k∇uk + kuk ) i L L 2 L2 2ε L2 i which implies that

θ 2 2 2 2 B[u, u, t] ≥ k∇ukL2 − CkukL2 ≥ βkukH1 − γkukL2 2 0 where we used the Poincaréinequality in the last step. Note that (4.5) also holds for systems if either a) L is strongly elliptic and the coefficients ij m×m a ∈ R are continuous or b) L is super-strongly elliptic. To define a weak solution we multiply the equation by a test function v and integrate by parts. As common for time dependent equations we often view the function u on ΩT as a map from (0,T ] to a function space. If X is a Banach space we say that a map u : (0,T ) → X has a weak derivative v if Z Z 0 ∞ ϕ (t)u(t) dt = − ϕ(t)v dt ∀ϕ ∈ Cc (I). I I In this case we write u0 = v. See Subsection 4.3 for further results on weak derivatives and Sobolev spaces of Banach-valued maps.

5 [July 12, 2019] Deﬁnition 4.2. We say that

2 1 u ∈ L (0,T ; H0 (Ω)) (4.6) with weak derivative u0 ∈ L2(0,T ; H−1(Ω)) (4.7) is a weak solution of (4.1) if

0 1 (i) hu (t), vi + B[u(t), v, t] = hf(t), vi ∀v ∈ H0 (Ω) and a.e. t ∈ [0,T ]; (ii) u(0) = g.

Remark. 1. Conditions (4.6) and (4.7) imply that u has a unique representative in C0([0,T ]; L2(Ω)) (see Theorem 4.33 below). Condition (ii) is required for this representative. 2. For brevity we sometimes write u0, u and f instead of u(t), u0(t) and f(t) in (i).

4.1.1 Existence by discretization in space Motivation Let u be suﬃciently regular weak solution and we take v(t, ·) = d 2 u(t, ·) in the deﬁnition of the weak solution. Using the identity dt ku(t)kL2 = 2(u, u0) = 2hu0, ui we get

d 2 1 2 β 2 ku(t)kL2 + B[u, u, t] = hf, vi ≤ kfkH−1 + kukH1 . dt 2β 2 0

Now by (4.5)

d 1 2 β 2 2 1 2 ku(t)kL2 + kukH1 ≤ γkukL2 + kfkH−1 . dt 2 2 0 2β

It follows from Gronwall’s inequality (see Lemma 4.5 below) that

max ku(t)kL2 ≤ CT (kfkL2(0,T ;H−1) + kgkL2 ) t∈[0,T ] and hence kuk 2 1 ≤ CT (kfk 2 −1 + kgk 2 ). L (0,T ;H0 ) L (0,T ;H ) L

Estimates for a ﬁnite dimensional approximation We now show that the formal reasoning outlined above can be used to rigorously prove bounds for ﬁnite dimensional approximations. Then one can use weak convergence (and linearity of the equation) to pass to the limit thus proving the existence of a weak solution. We assume that

∞ 2 {w}k=1 is an orthonormal basis of L (Ω),

6 [July 12, 2019] 2 i.e., (wk, wl)L2 = δkl and span {wk : k = 1,..., ∞} is dense in L , and ∞ 1 {w}k=1 is an orthogonal basis in H0 (Ω).

One may take, e.g., wk as the set of (weak) eigenfunctions of −∆. We now consider ﬁnite dimensional approximations (or Galerkin approximations) of the form

X k um(t, ·) = dm(t)wk(·). k≤m

k We obtain a system of ODEs for the coeﬃcients dm(t) by using the weak for of the equation for all test functions wj with j ≤ m. Since the wk are orthonormal in L2 this yields the following system of ODEs

k 0 dm (t) + B[um, wk, t] = (f, wk) for a.e. t ∈ (0,T ] ∀ k ≤ m k (4.8) dm(0) = (g, wk) ∀ k ≤ m Proposition 4.3. For each m the equation (4.8) has a unique solution with 1 m 1,2 m 1,2 −1 dm = (dm, . . . , dm) ∈ W (0,T ; R ). If, in addition, f ∈ W (0,T ; H (Ω)) ij i 2,2 m and if the coeﬃcients a , b and c do not depend on t then dm ∈ W (0,T ; R ). Proof. This follows from standard ODE theory. Details (not discussed in class): the equation can be written explicitly as the m dimensional system of ODEs

0 dm (t) + Adm(t) = fm(t) for a.e. t ∈ (0,T ], dm(0) = gm m 2 m k where gm ∈ R , fm ∈ L (0,T ; R ), and where the elements Aj (t) of the ∞ m×m matrix A(t) are given by B[wj, wk, t]. Hence A(t) ∈ L (0,T, R ). Now 1,2 m 0 m dm is a solution in W (0,T ; R ) if and only if dm ∈ C (0,T ; R ) and the integral equation Z t Z t dm(t) = A(s)dm(s) ds + fm(s) ds =: N(dm)(t) (4.9) 0 0 0 m holds. The map N is a contraction on C (0,T ; R ) (equipped with the −Mt norm supt∈[0,T ] e |u(t)| with M = M(m) suﬃciently large) and hence has a unique ﬁxed point. 1,2 m If ,in addition, A(s) is independent of s and fm ∈ W (0,T ; R ) then 2,2 m it follows from (4.9) that dm ∈ W (0,T ; R ).

Theorem 4.4. Assume that ∂t + L is parabolic and (4.2) as well as (4.3) hold. Then there exists a constant C which depends on Ω, T and the coeﬃ- cients aij, bi, c such that 0 max kum(t)kL2(Ω) + kumkL2(0,T ;H1(Ω)) + kumkL2(0,T ;H−1(Ω)) t∈[0,T ] 0 ≤C kfkL2(0;T ;H−1(Ω)) + kgkL2(Ω) . (4.10)

7 [July 12, 2019] Remark. 1. The constant depends on the coefficients only through the constants β and γ in the (4.5). 1 2. The assumption that the function wk are orthogonal in H0 is only used to 0 2 −1 prove (4.11) below which in turn implies the bound for um in L (0,T ; H ). This bound for can be derived in the same way as long as L2 orthogonal projection Pm onto Xm = span{wk : k ≤ m} satisfies the kPmvk 1 ≤ H0 1 Ckvk 1 for all v ∈ H and all m ∈ . This observation is useful for H0 0 N constructing numerical approximation (using, e.g., hat functions for n = 1 or finite elements for n ≥ 1).

k Proof. Multiply (4.8) by dm(t) and sum over k. Since wk is an orthonormal basis this gives

1 2 β 2 1 2 ∂tkumkL2 + B[um, um, t] = (f, um) ≤ kumkH1 + kfkH−1 . 2 2 0 2β

Using (4.5) we get

1 2 β 2 2 1 2 ∂tkum(t)kL2 + kumkH1 ≤ γkumkL2 + kf(t)kH−1 . 2 2 0 2β Thus Gronwall’s inequality gives

2 kum(t)kL2 ≤ CT (kum(0)kL2 + kfkL2(0,T ;H−1)).

2 Since the wk are orthonormal in L we get

kum(0)kL2 ≤ kgkL2 .

Thus it only remains to prove the estimate for u0. Here we will use the 1 fact that the wk are also orthogonal in H0 which has not been used so far. Note ﬁrst that by deﬁnition

0 0 ku (t)k −1 = sup{hu (t), wi : kwk 1 ≤ 1}. m H (Ω) m H0 (Ω)

Let Xm be the ﬁnite dimensional space spanned by w1, . . . , wm,

Xm := span{wk : 1 ≤ k ≤ m}.

2 2 Let Pm denote the L orthogonal projection from L (Ω) onto Xm. The projection is characterized by the conditions

Pmh ∈ Xm, (Pmh, wj)L2 = (h, wj) ∀j ≤ m.

1 Since the functions wj are also an orthogonal basis in H0 (Ω) it follows that

1 (Pmh − h, wl) 1 = 0 ∀l > m, ∀h ∈ H (Ω). H0 (Ω) 0

8 [July 12, 2019] 2 2 2 Thus khk 1 = kPmhk 1 + kh − Pmhk 1 and in particular H0 (Ω) H0 (Ω) H0 (Ω)

kPmhk 1 ≤ khk 1 . (4.11) H0 (Ω) H0 (Ω)

2 Now since the wk are an orthonormal basis in L m 0 X k 0 0 hum(t), wi = (dm) (t)(wk, w)L2 = hum(t),Pmwi k=0 and

0 hum(t),Pmwi = −B[um(t),Pmw, t] + hf, Pmwi ≤Ckumk 1 kPmwk 1 + kfk −1 kPmwk 1 . H0 (Ω) H0 (Ω) H (Ω) H0 (Ω) In connection with (4.11) we deduce that

0 ku (t)k ≤ Ckumk 1 + kfk −1 . m H0 (Ω) H (Ω)

0 2 1 Thus the bound for um follows from the bound for um in L (0,T,H0 (Ω)). 1,1 Lemma 4.5. Let α ∈ [0, ∞). Assume that a, b : [0,T ] → R with a ∈ W , a ≥ 0, b ≥ 0, b ∈ L1 and a0 ≤ αa + b. Then αt a(t) ≤ e kbkL1((0,t)) Proof. The functiona ˜ = e−αta satisﬁesa ˜0 ≤ e−αtb. Thus Z t −αs a˜(t) ≤ e b(s) ds ≤ kbkL1 . 0

[1.4. 2019, Lecture 1] [5.4. 2019, Lecture 2]

Passage to the limit

Theorem 4.6. Assume that ∂t + L is parabolic and (4.2) as well as (4.3) hold. Then there exists one and only one weak solution u of (4.1). Moreover u satisﬁes

0 max ku(t)kL2(Ω) + kukL2(0,T ;H1(Ω)) + ku kL2(0,T ;H−1(Ω)) t∈[0,T ] 0 ≤C kfkL2(0;T ;L2(Ω) + kgkL2(Ω) . (4.12) where C depends on Ω, T and the constants β and γ in (4.5)

9 [July 12, 2019] Proof. Fix T > 0. Let um be the solutions for (4.8). It follows from the bounds (4.10) that there exist a subsequence such that (see Theorem 4.35 below)

2 1 uml * u in L (0,T,H0 (Ω)), (4.13) u0 * h in L2(0,T,H−1(Ω)). (4.14) ml It follows from the deﬁnition of weak derivatives that u ∈ W 1,2(0,T,H−1(Ω)) and u0 = h. 1 Assume that k ≤ ml, let ψ ∈ C ([0,T ]) with ψ(T ) = 0 multiply (4.8) by ψ and integrate by parts. This yields

Z T Z T Z T 0 −(g, wk)ψ(0)− (uml , wk)ψ dt+ B(uml , wk, t)ψ(t) dt = (f, wk)ψ dt. 0 0 0 Taking the limit l → ∞ we get

Z T Z T Z T 0 −(g, wk)ψ(0)− (u, wk)ψ dt+ B(u, wk, t)ψ(t) dt = (f, wk)ψ dt ∀k ∈ N. 0 0 0

1 2 1 Since the wk form a basis of H0 (Ω) and since u ∈ L (0,T ; H0 (Ω)) it follows that Z T Z T Z T 0 1 −(g, v)ψ(0)− hu, viψ dt+ B(u, v, t)ψ(t) dt = (f, v)ψ dt ∀v ∈ H0 (Ω). 0 0 0 We know that u ∈ W 1,2(0,T ; H−1(Ω)). In particular u ∈ C0([0,T ]; H−1(Ω)) (see Theorem 4.32 below). Thus t 7→ hu(t), vi is in W 1,2(0,T ) ∩ C0([0,T ]) 1 for each v ∈ H0 (Ω) and integration by parts yields (recall that ψ(T ) = 0) Z T −(g, v)ψ(0) + hu(0), viψ(0) + hu0, vi + B[u, v, t] − hf, vi ψ(t) dt = 0. 0 (4.15) ∞ Taking ψ ∈ Cc ([0,T ]) we see that (i) in Definition 4.2 holds. Finally since ψ(0) is arbitrary we get u(0) = g. Thus u is a weak solution in the sense of Definition 4.2. 2 1 0 2 −1 The bounds for u ∈ L (0,T ; H0 (Ω)) and u ∈ L (0,T ; H (Ω)) follow from weak lower semicontinuity of the norm, see Theorem 4.35. Finally the bound for maxt∈[0,T ] ku(t)kL2(Ω) can also be deduced from weak lower semicontinuity or from Theorem 4.33 (iv). To show uniqueness assume that u1 and u2 are weak solutions for given f and g. Then u = u1 − u2 is a weak solution for f = 0 and g = 0. Taking v = u in the definition of the weak solution we get (using Theorem 4.33)

1 d kuk2 + B[u, u, t] = 0 2 dt L2(Ω)

10 [July 12, 2019] which implies that 1 d 2 kuk 2 ≤ γkuk . 2 dt L (Ω) L2(Ω) Since u(0) = 0 it follows from Gronwall’s inequality that u(t) = 0 for all t ∈ [0,T ].

4.1.2 Higher regularity Motivation We now aim to show that if f and g are more regular than u is more regular. The guiding principle is ’maximal regularity’. We seek to prove that if ∂tu + Lu = f, u(0, ·) = g then both ∂tu and Lu are ’as regular as f’ (provided that the initial data g are regular enough). Then one can use elliptic regularity theory (see NPDE I) to show subsequently that all second spatial derivatives of u are ’as good as f’. The existence theory corresponds to the choice f ∈ L2(H−1) and gives 2 −1 2 −1 2 1 ∂tu ∈ L (H ) and ∆u ∈ L (H ) which then implies that u ∈ L (H0 ). Now we look for similar estimates if f ∈ L2(L2) or if we have additional information on ∂tf. To see what estimates we can expect we ignore for the moment all tech- nicalities and look at the simplest equation

∂tu − ∆u = f, u(0, ·) = g.

There are two simple things we can do the with this equation.

• Multiply the equation by u and integrate in Ω.

• Multiply the equation by ∂tu and integrate in Ω. Multiplication by u was used in the proof that weak solutions exist and yields 1 2 2 ∂ kuk + k∇uk = (f, u) ≤ Ck∇uk 2 kfk −1 2 t L2 L2 L H where we used the Poincar´einequality in the last step. Thus we obtain the bound

kuk ∞ 2 + kuk 2 1 ≤ CT (kfk 2 −1 + kgk 2 ). (4.16) L (0,T ;L ) L (0,T ;H0 ) L (0,T ;H ) L

Multiplication by ∂tu gives after integration by parts 1 1 1 k∂ uk2 + ∂ k∇uk2 = (f, ∂ u) ≤ kfk2 + k∂ uk2 . t L2 2 t L2 t 2 L2 2 t L2

11 [July 12, 2019] This leads to the estimate

k∂tuk 2 2 + kuk ∞ 1 ≤ CT (kfk 2 2 + kgk 1 ). L (0,T,L ) L (0,T ;H0 ) L (0,T ;L ) H0

Rewriting the equation as −∆u = −∂tu + f and using elliptic regularity we also get kuk 2 2 ≤ CT (kfk 2 2 + kgk 1 ). L (0,T ;H ) L (0,T ;L ) H0

If we have information on ∂tf we can get further bounds by observing that u˜ := ∂tu satisﬁes the equation

∂tu˜ − ∆˜u = ∂tf, u˜(0) = ∂tu(0) = ∆u(0) + f(0) = ∆g + f(0).

Now applying (4.16) tou ˜ we get

k∂tuk ∞ 2 + k∂tuk 2 1 ≤ CT (k∂tfk 2 −1 + k∆g + f(0)k 2 ) L (0,T,L ) L (0,T ;H0 ) L (0,T ;H ) L (4.17) ∞ 2 and rewriting the equation as −∆u = f − ∂tu we get u ∈ L (0,T ; H ).

Rigorous results Theorem 4.7. Assume that L is parabolic and (4.2) as well as (4.3) hold. Assume further that the coeﬃcients

aij, bi and c are independent of t and belong to C∞(Ω)¯ and ∂Ω ∈ C∞.

Let u be a weak solution of

∂tu + Lu = f in ΩT , u = 0 on ∂Ω × [0,T ], u = g on {0} × Ω.

(i) Assume that 1 2 2 g ∈ H0 (Ω), f ∈ L (0,T ; L (Ω)). Then

2 2 ∞ 1 0 2 2 u ∈ L (0,T ; H (Ω))∩L (0,T ; H0 (Ω)), u ∈ L (0,T ; L (Ω)) (4.18) and

0 kuk 2 2 + kuk ∞ 1 + ku k 2 2 L (0,T ;H (Ω)) L (0,T ;H0 (Ω)) L (0,T ;L (Ω)) ≤C kfk 2 2 + kgk 1 (4.19) L (0,T ;L (Ω)) H0 where C depends only on Ω, T and the coeﬃcients.

12 [July 12, 2019] (ii) If, in addition, g ∈ H2(Ω), f 0 ∈ L2(0,T ; H−1(Ω)), f ∈ L∞(0,T,L2(Ω)) then ∞ 2 0 ∞ 2 2 1 u ∈ L (0,T ; H (Ω)), u ∈ L (0,T ; L (Ω)) ∩ L (0,T ; H0 (Ω)), u00 ∈ L2(0,T ; H−1(Ω)) and we have the estimate 0 0 00 kuk ∞ 2 + ku k ∞ 2 + ku k 2 1 + ku k 2 −1 L (0,T ;H (Ω)) L (0,T ;L (Ω)) L (0,T ;H0 (Ω)) L (0,T ;H (Ω)) ≤C kfkH1(0,T ;H−1(Ω)) + kfkL∞(0,T,L2(Ω)) + kgkH2(Ω) (4.20) where C depends only on Ω, T and the coeﬃcients.

0 2 −1 2 1 Remark. If in (ii) we assume that f ∈ L (0,T ; H (Ω)) and f ∈ L (0,T,H0 (Ω)) then automatically f ∈ C0([0,T ]; L2(Ω). Moreover, using elliptic regularity we also get u ∈ L2(0,T,H3(Ω)) with an estimate for the corresponding norm. Proof.

0 Step 1. Testing the discrete equation with um. k 0 Multiply (4.8) by dm (t) and sum over k to get

0 0 0 1 2 1 0 2 ku (t)k 2 +B[u (t), u (t)] = hf(t), u (t)i ≤ kf(t)k + ku (t)k . m L (Ω) m m m 2 L2(Ω) 2 m L2(Ω)

Now write B = B1 + B2 where n n X ij X i B1[u, v] := a (x)∂ju ∂iv, B2[u, v] := b (x)∂iu v + c(x)uv. i,j=1 i=1 Since the coeﬃcients aij are independent of t and symmetric we get 1 d B [u (t), u0 (t)] = B[u (t), u (t)]. 1 m m 2 dt m m Moreover we have 2 C 2 B2[u, v] ≤ εkvkL2 + kukH1 . ε 0 1 Applying this with ε = 4 we get

1 0 2 1 d 1 2 2 kum(t)kL2(Ω) + B1[um(t), um(t)] ≤ kf(t)kL2(Ω) + Ckum(t)kH1(Ω). 4 2 dt 2 0 (4.21) 2 By parabolicity and the Poincar´einequality we have kvk 1 ≤ CB1[v, v]. H0 Thus (4.21) and Gronwall’s inequality imply that

max B1[um(t), um(t)] ≤ CT (kfkL2(0,T,L2(Ω) + B1[um(0), um(0)]). t∈[0,T ]

13 [July 12, 2019] Step 2. 2 We have um(0) = Pmg where Pm is the L projection to span {wk : 1 ≤ k ≤ m}. Thus by (4.11)

B1[um(0), um(0)] ≤ Ckum(0)k 1 ≤ Ckgk 1 . H0 (Ω) H0 (Ω) Hence it follows from Step 1 and Gronwall’s inequality that

max kum(t)kH1(Ω) ≤ C max B1[um(t), um(t)] ≤ CT (kfkL2(0,T,L2(Ω)+kgkH1 ). t∈[0,T ] 0 t∈[0,T ] 0

Integrating (4.21) in time we also deduce that

0 2 2 kumk 2 2 ≤ CT (kfkL2(0,T,L2(Ω)) + kgk 1 ). L (0,T ;L (Ω) H0

∗ ∞ 1 0 Thus for a subsequence um * v in L (0,T ; H0 (Ω)) and um * w in L2(0,T ; L2(Ω)). We have already seen in the existence proof that (for a fur- 2 1 0 0 2 −1 ther subsequence) um * u in L (0,T ; H0 (Ω)) and um * u in L (0,T ; H (Ω)). Thus v = u and w = u0 and the corresponding bounds for u and u0 follow from weak lower semicontinuity of the norm, see Theorem 4.35.

Step 3. Bound for kukL2(0,T,H2(Ω) by elliptic regularity. We rewrite the weak form of the equation as

0 1 B[u(t), v] = hf(t) − u (t), vi ∀v ∈ H0 for a.e. t ∈ [0,T ].

This is the weak form of the equation

Lu(t) = f(t) − u0(t) and elliptic regularity yields

0 kukH2(Ω) ≤ C(kf(t)kL2(Ω) + ku (t)kL2(Ω)).

Taking the square and integrating in t we obtain the desired bound. This ﬁnishes the proof of (i).

Step 4. Bounds for u0. Assume now in addition g ∈ H2(Ω), f 0 ∈ L2(0,T ; H−1(Ω)). Let um be the solution of the discrete system and set

0 u˜m = um.

k 2,2 By Proposition 4.3 we have dm ∈ W (0,T ). Hence we can diﬀerentiate the discrete equation with respect to t and we get

0 0 (˜um, wk) + B[˜um, wk] = hf , wki.

14 [July 12, 2019] k 0 Multiply by (dm) and sum over k to get

0 0 (˜um, u˜m) + B[˜um, u˜m] = hf , u˜mi.

Using Gronwall’s inequality we get

Z T 0 2 0 2 0 2 0 2 sup kum(t)kL2(Ω) + kumkH1(Ω) ≤ C(kum(0)kL2(Ω) + kf kL2(0,T ;H−1(Ω)) t∈[0,T ] 0 0 (4.22)

[5.4. 2019, Lecture 2] [8.4. 2019, Lecture 3]

k 2,2 1 Since dm ∈ W (0,T ) ,→ C ([0,T ]) we can pass to the limit t ↓ 0 in the k 0 equation of dm (t) and we get

0 (um(0), wk) = (f(0), wk) + B(um(0), wk).

k 0 and multiplication by (dm) (0) and summation yields

0 2 0 0 kum(0)kL2(Ω) = (f(0), um(0)) + B[um(0), um(0)] 0 ≤(kfkL∞(0,T ;L2(Ω)) + kum(0)kH2 ) kum(0)kL2(Ω). (4.23)

0 Thus the desired estimate for um follows from (4.22) if we can show that

kum(0)kH2(Ω) ≤ CkgkH2(Ω)

(see the next step). Then the estimate for u0 follows by passing to the weak limit and lower semicontinuity of the norm (see Theorem 4.35). There is one subtlety in the last inequality in (4.23). In principle we only now that kf(t)k ≤ kfkL∞(0,T ;L2(Ω)) for a.e. t but we would like to use this bound for a speciﬁc t namely t = 0. To see that the estimate 0 is nonetheless correct let a(t) := |(f(t), um(0))|. Then a is continuous on 0 1 [0,T ] because um(0) ∈ H0 as ﬁnite linear combination of the wk and f ∈ 1,2 −1 0 W (0,T,H (Ω)). Moreover a(t) ≤ kfkL∞(0,T,L2(Ω)kum(0)kL2 (Ω)) for a.e. t ∈ (0,T ). Since a is continuous this bound holds for every t and in particular for t = 0.

Step 5. Bound for kum(0)kH2 . 2 We have um(0) = Pmg where Pm is the L orthogonal projection onto span {wk : 1 ≤ k ≤ m}. We thus have to show that

2 1 kPmgkH2(Ω) ≤ CkgkH2(Ω) ∀g ∈ H (Ω) ∩ H0 (Ω).

By elliptic regularity k|gk| = k∆gkL2(Ω)

15 [July 12, 2019] 2 1 is an equivalent norm on H (Ω) ∩ H0 (Ω). It thus suﬃces to show that

k∆PmgkL2(Ω) ≤ k∆gkL2(Ω). (4.24)

1 Now we strongly use that the wk ∈ H0 (Ω) are eigenfunctions of −∆ and 2 hence in H (Ω) (by elliptic regularity). Moreover −∆wk = λkwk implies 1 2 that ∆wk ∈ H0 (Ω) ∩ H (Ω). Hence by Lemma 4.8

2 (∆Pmg, ∆Pmg) = (Pmg, ∆ Pmg) (4.25)

2 2 2 Now ∆ wk = λkwk and hence ∆ Pmg ∈ Xm. Thus by the deﬁnition of the projection

2 2 (Pmg, ∆ Pmg) = (g, ∆ Pmg) = (∆g, ∆Pmg) ≤ k∆gkL2(Ω)k∆PmgkL2(Ω) and combining this with (4.25) we get (4.24).

Step 6. Bound for u00. 00 0 0 We bounded um =u ˜m by the same argument we used to bound um in the proof the apriori estimate, Theorem 4.4. As in Step 4 we have

0 0 (˜um(t), wk) = −B[˜um(t), wk] + hf (t), wki.

2 for a.e. t ∈ [0,T ]. Recall that Pm is the L orthogonal projection onto 1 span {wk : 1 ≤ k ≤ m}. Then for v ∈ H0 (Ω)

0 0 0 (˜um(t), v) = (˜um(t),Pmv) = −B[˜um(t),Pmv] + hf (t),Pmvi 0 ≤(Cku˜m(t)k 1 + kf (t)k −1 ) kPmvk 1 . H0 (Ω) H (Ω) H0 (Ω)

Since kPmvk 1 ≤ kvk 1 (see 4.11) we get H0 H0

0 0 0 ku˜ (t)k −1 = sup{(˜u (t), v): kvk 1 ≤ 1} ≤ Cku˜m(t)k 1 +kf (t)k −1 . m H m H0 (Ω) H0 (Ω) H (Ω)

00 0 0 Hence the bounds for um =u ˜m follow from those for um. Passage to the weak limit and lower semicontinuity of the norm yields the bound for u00 in L2(0,T ; H−1(Ω)).

Step 7. Bounds for u by elliptic regularity. As in Step 3 elliptic regularity gives

0 ku(t)kH2(Ω) ≤ C(kf(t)kL2(Ω) + ku kL2(Ω)).

Thus u ∈ L∞(0,T,H2(Ω)) and the bound in the corresponding norm follows from the bound for u0 in L∞(0,T ; L2(Ω)).

16 [July 12, 2019] 1 2 Lemma 4.8. (i) If f ∈ H0 (Ω) and g ∈ H (Ω) then

(f, −∆g)L2 = (∇f, ∇g)L2 .

1 2 (ii) If f, g ∈ H0 (Ω) ∩ H (Ω) then

(f, −∆g)L2 = (∇f, ∇g)L2 = (−∆f, g)L2 .

∞ Proof. (i) follows from the deﬁnition of the weak derivative if f ∈ Cc (Ω). The general case follows by density. The second identity in (ii) follows by exchanging the roles of f and g in (i).

Theorem 4.9 (Estimates for higher derivatives and compatibility conditions). Let m = 0. Let the assumptions of Theorem 4.7 (i) hold and assume that in addition dkf g ∈ H2m+1(U), ∈ L2(0,T,H2m−2(Ω)), for k = 0, ..., m. dt Suppose also that the following mth-order compatibility conditions hold

1 gk ∈ H0 (Ω) for k = 0, ..., m where dk−1f g = g, g = f(0) − Lg , g := (0) − Lg . 0 1 0 k dtk−1 k−1 Then dku ∈ L2(0,T ; H2m+2−2k(Ω)) for k = 0, ..., m + 1. dtk and we have the estimate

m+1 k m k ! X d u X d f k ≤ C k + kgkH2m+1(Ω) dt 2m+2−2k dt 2m−2k k=0 H (Ω) k=0 H (Ω) where C depends only on m, Ω, T and the coeﬃcients of L.

Sketch. This is proved by induction over m. The case m = 0 is covered by Theorem 4.7(i). Induction step: Assume that the result holds for m and that f and g satisfy the assumptions for m + 1. The main point is to prove thatu ˜ := u0 is a weak solution of ˜ ∂tu˜ + Lu˜ = f in ΩT , u˜ = 0 on ∂Ω × [0,T ], (4.26) u˜ =g ˜ on {0} × Ω. where f˜ = ∂tf, g˜ = f(0, ·) − Lg.

17 [July 12, 2019] To show this note that by assumption f ∈ L2(0,T ; H2(Ω))∩H1(0,T ; L2(Ω)) 3 1 1 1 and g ∈ H (Ω)∩H0 (Ω). Hence by Theorem 4.7(ii) we have u ∈ H (0,T ; H0 (Ω)) 0 1 −1 1 andu ˜ = u ∈ H (0,T ; H (Ω)). It follows that for v ∈ H0 the functions

t 7→ hu,˜ vi, t 7→ B[u(t), v] and t 7→ hf(t), vi are in W 1,2(0,T ) and diﬀerentiating the weak form of the equation for u we get hu˜0(t), vi + B[˜u, v] = hf 0(t), vi for a.e. t ∈ [0,T ]. Moreover since H1(0,T ; X) ,→ C0([0,T ]; X) we can pass to the limit t ↓ 0 in the equation hu˜(t), vi + B[u(t), v] = hf(t), vi and we get

1 hu˜(0), vi + B[g, v] = hf(0), vi ∀v ∈ H0 (Ω).

2 1 Since g ∈ H (Ω) ∩ H0 (Ω) this implies that

u˜(0) = −Lg + f(0) =g. ˜

This shows thatu ˜ is a weak solution of (4.26) Now one can check easily that f˜ andg ˜ satisfy the assumptions of the theorem for m. Hence, by induction assumption,u ˜ satisﬁes the conclusion for m. It follows that dku dk−1u = ∈ L2(0,T ; H2m+2−2k(Ω)) for k = 1, ..., m + 1. dtk dtk−1 The remaining claim u ∈ L2(0,T ; H2m+2(Ω) follows by elliptic regularity since B[u(t), v] = hf(t) − u0(t), vi which is the weak form of Lu(t) = f(t) − u0(t). The corresponding bounds follow in the same way by induction.

Corollary 4.10 (C∞ smoothness for parabolic equations). Assume

∞ ∞ g ∈ C (Ω)¯ , f ∈ C (ΩT ).

th and the m -order compatibility conditions hold for all m ∈ N. Then

∞ u ∈ C (ΩT ).

Proof. Apply Theorem 4.9 for all m ∈ N and use elliptic regularity. [8.4. 2019, Lecture 3] [12.4. 2019, Lecture 4]

18 [July 12, 2019] 4.2 Integration theory and Lp spaces for Banach space-valued functions This subsection was not discussed in class. Let I ⊂ R be an interval and let X be a Banach space. We want to define an integral (called the Bochner-Lebesgue integral) for maps f : I → X and the spaces L1(I; X) and Lp(I; X). There are two natural approaches. The first proceeds like for real-valued functions. One defines an integral for Pn simple functions f = i=1 aiχEi (with Ei ⊂ R measurable) and then passes to completion. This leads to the notion of strongly measurable functions. The second approaches is based on the idea of using ’coordinates’. More precisely we say that f is weakly measurable if for every T ∈ X0 (the space of bounded linear functions from X to R) the map T ◦ f : I → R is measurable. If the real-valued map t 7→ kf(t)k is integrable (or has an integrable R majorant) then the maps T ◦ f are integrable and T 7→ I T ◦ f dx is a 0 R bounded linear map from X to R. Hence the integral f dx can be defined as an element of the bidual space (X0)0. If simple functions are dense then the integral is actually an element of X since X can be viewed a closed subspace of (X0)0. The Dunford-Pettis Theorem provides a natural connection between the two approaches. If X is separable (or, more generally, if there exists a null set N such that the range f(I \ N) is separable) then every weakly measurable function is strongly measurable. The content of this section is mainly based on the textbook [AE09] of Amann and Escher and on the textbook [Ev] of Evans. For the rest of this subsection let • T > 0 and

1 • X = (X, k · kX ) be a real Banach space .

4.2.1 Simple and measurable functions Deﬁnition 4.11 (Almost everywhere). Let P be a property that is either true or false for the points t ∈ [0,T ]. Then P holds for almost every t ∈ [0,T ], if there exists a null set N such that P (t) is true for all t ∈ N c. Alternatively, in this case one says that P holds almost everywhere in [0,T ]. Abbreviatory notation in common use is P holds for a. e. t ∈ [0,T ] and P holds a. e. in [0,T ], respectively. Deﬁnition 4.12 (Simple function). A function s : [0,T ] → X is called simple if n X s (t) = χIi (t) xi, t ∈ [0,T ] , i=1 1Most of the presented results are also true for complex Banach spaces but for conve- nience the presentation is restricted to the real case.

19 [July 12, 2019] where Ii, i = 1, . . . , n, are Lebesgue measurable disjoint subsets of [0,T ] and xi ∈ X for i = 1, . . . , n. The set of all simple functions s : [0,T ] → X is denoted by SF ([0,T ]; X).

Deﬁnition 4.13 (Strongly measurable function). A function u : [0,T ] → X is called strongly measurable, or measurable for short, if there exists a sequence (sk) ∈ SF ([0,T ]; X) of simple functions such that

sk (t) −→ u (t) for a. e. t ∈ [0,T ] as k −→ ∞.

The set of all (strongly) measurable functions u : [0,T ] → X is denoted by L0 ([0,T ]; X). Deﬁnition 4.14 (Weakly measurable function). Let X be a real Banach space and X∗ the corresponding dual space. A function u : [0,T ] → X is called weakly measurable if for each u∗ ∈ X∗ the mapping t 7→ hu∗, u (t)i is measurable.

Deﬁnition 4.15 (Separable space). X is called separable if it contains a countable and dense subset U.

For finite-dimensional or separable Banach spaces the definitions of strong and weak measurability coincide, but for non-separable Banach spaces a weakly measurable function is not necessary strongly measurable. In this case of a non-separable Banach space there is one additional requirement on the function u, that is introduced in the following definition.

Deﬁnition 4.16 (Almost separably valued function). A function u : [0,T ] → X is called almost separably valued if there exists a null set N ⊂ [0,T ] such that u (N c) is separable.

The next theorem gives a characterization of strongly measurable functions.

Theorem 4.17 (Dunford–Pettis). A function u : [0,T ] → X is (strongly) measurable if and only if it is weakly measurable and almost separably valued.

Another formulation of the Dunford–Pettis theorem is that for a separable Banach space the properties of strong and weak measurability are equivalent. Thus, in the case of a separable Banach space the veriﬁcation of strong measurability of a Banach space-valued function u : [0,T ] → X is reduced to the veriﬁcation of measurability of a real-valued function.

4.2.2 The Bochner–Lebesgue integral It is the purpose of this paragraph to deﬁne the so-called Bochner–Lebesgue integral. As a basis the integral of a simple function is deﬁned previously.

20 [July 12, 2019] Deﬁnition 4.18 (Integral of simple functions). For a simple function s ∈ SF ([0,T ]; X), the integral of s over [0,T ] is given by

n n Z T Z T X X s (t) dt = χIi (t) xi dt := |Ii| xi, 0 0 i=1 i=1 where Ii, i = 1, . . . , n, are Lebesgue measurable subsets of [0,T ] as in Deﬁ- nition 4.12 and |Ii| denotes the Lebesgue measure of Ii. The above deﬁned integral for simple functions has the expected properties, i. e. it is linear, continuous, and, in the case X = R, monotone. Now consider the space SF ([0,T ]; X) of simple functions endowed with the seminorm Z T ksk1 := kskX dt, s ∈ SF ([0,T ] ,X) . 0

In general, k · k1 is not a norm, since for a null set N= 6 ∅ one has χN 6= 0 but kχN k1 = 0. A Cauchy sequence in SF ([0,T ]; X) with respect to k·k1 is called an L1- Cauchy sequence. The idea is to introduce the Bochner–Lebesgue integral by completing the space SF ([0,T ]; X) of simple functions with respect to k · k1. Deﬁnition 4.19 (Integrable function). A function u : [0,T ] → X is called Bochner–Lebesgue integrable, or integrable for short, if there exists an L1-Cauchy sequence (sk) ∈ SF ([0,T ] ,X) with

lim ksk (t) − u (t)kX = 0 a. e. in [0,T ] k→∞ The space consisting of all (Bochner–Lebesgue) integrable functions is denoted by L1 ([0,T ]; X). To be able to state a reasonable deﬁnition of the Bochner–Lebesgue integral some further considerations are required. For an L1-Cauchy sequence (sk) of simple functions converging to an integrable function u a. e. in [0,T ], R T the sequence 0 sk dt of integrals converges in X. If one wants to deﬁne the limit of the sequence of integrals as the Bochner–Lebesgue integral one must previously ensure that it is independent of the sequence of approximating simple functions. These considerations are summarized in the following theorem.

21 [July 12, 2019] Theorem 4.20. Suppose (sk) and (rk) are L1-Cauchy sequences converging R T R T a. e. to the same function. Then limk 0 sk dt and limk 0 rk dt exist in X and the equation

Z T Z T lim sk (t) dt = lim rk (t) dt k→∞ 0 k→∞ 0 holds.

With these preliminaries one can deﬁne the Bochner–Lebesgue integral.

Deﬁnition 4.21 (Bochner–Lebesgue integral). Let (sk) ∈ SF ([0,T ]; X) be an L1-Cauchy sequence with sk → u ∈ L1 ([0,T ]; X) a. e. Then the Bochner–Lebesgue integral of u is deﬁned as

Z T Z T u (t) dt = lim sk (t) dt. 0 k→∞ 0

For an integrable function u ∈ L1 ([0,T ]; X) its norm kukX belongs to L1 ([0,T ]; R). Additionally, if (sk) is an L1-Cauchy sequence converging to u a. e., then Z T Z T ku (t)kX dt = lim ksk (t)kX dt 0 k→∞ 0 holds. Furthermore, if one deﬁnes

Z T kuk1 := ku (t)kX dt 0 for a function u ∈ L1 ([0,T ]; X), then k · k1 is a seminorm on L1 ([0,T ]; X). It is called L1-seminorm. One can show that SF ([0,T ]; X) is dense in L1 ([0,T ]; X) and that L1 ([0,T ] ,X) is a complete space with respect to the L1-seminorm (see Corollary 2.9 and Theorem 2.10 in the book by Amann and Escher) As expected, the Bochner–Lebesgue integral is a linear and continuous map Z T · dt : L1 ([0,T ]; X) → X. 0 The next theorem gives a characterization of (Bochner–Lebesgue) integrable functions.

Theorem 4.22 (Bochner). For a function u ∈ L0 ([0,T ]; X) the following statements are equivalent:

(i) u ∈ L1 ([0,T ]; X),

(ii) kukX ∈ L1 ([0,T ]; R),

22 [July 12, 2019] R T (iii) 0 kukX dt < ∞. If one of the above conditions (i)–(iii) is satisﬁed, then

Z T

u (t) dt ≤ ku (t)k1 < ∞. 0 X

4.2.3 Lebesgue and Lp-spaces

n p Let Ω ⊂ R be open. Fubini’s theorem suggest that the space L ([0,T ] × Ω; R) p p can be identiﬁed with L ([0,T ]; L (Ω; R)) for 1 ≤ p < ∞. This is indeed the case (see Theorem below ), but there is no corresponding result for p = ∞). For the proofs of the results stated below see Chapter 10 in the book by Amann and Escher.

Deﬁnition 4.23 (Essentially bounded function, essential supremum). A function u ∈ L0 ([0,T ]; X) is called essentially bounded if there exists a constant α ≥ 0 such that the set {kukX > α} is a Lebesgue null set. In this case the expression

kuk∞ := ess sup ku (t)kX := inf {α ≥ 0; |{kukX > α}| = 0} t∈[0,T ] is called the essential supremum of u.

Deﬁnition 4.24 (Lebesgue space). Let u ∈ L0 ([0,T ]; X). For 0 ≤ p < ∞ deﬁne Z T 1/p p kukLp([0,T ];X) = kukp := kukX dt , 0 1/p where ∞ := ∞ and kukL∞([0,T ];X) = kuk∞. Then for 0 ≤ p ≤ ∞ the set n o Lp ([0,T ]; X) := u ∈ L0 ([0,T ]; X); kukp < ∞ is called Lebesgue space over [0,T ].

One can now introduce Lp-spaces of functions u : [0,T ] → X taking values in a Banach space.

23 [July 12, 2019] p Deﬁnition 4.25 (L -space). Let N := {u ∈ L0 ([0,T ]; X); u = 0 a. e.}. Then for 0 ≤ p ≤ ∞ the quotient space denoted by

p L ([0,T ]; X) := Lp ([0,T ]; X) /N is called the Lp-space over [0,T ]. Now, some frequently used inequalities for Lebesgue spaces are presented.

Theorem 4.26 (H¨older’sinequality). For 1 ≤ p ≤ ∞ let u ∈ Lp ([0,T ]; X) 0 and v ∈ Lp0 ([0,T ]; X), where p denotes the dual exponent of p. Then uv ∈ L1 ([0,T ]; X) and the following relation holds:

kuvk1 ≤ kukp kvkp0 . A consequence of H¨older’sinequality is the fact that the triangle inequality holds in Lebesgue spaces.

Theorem 4.27 (Minkowski inequality). Let u, v ∈ Lp ([0,T ]; X) for 1 ≤ p ≤ ∞. Then u + v belongs to Lp ([0,T ]; X) and

ku + vkp ≤ kukp + kvkp . For the sake of completeness some important results concerning Lp- spaces that are frequently used in this thesis are presented in the following remark.

Remark 4.28. (1) For 0 ≤ p ≤ ∞ the set N := {u ∈ L0 ([0,T ]; X); u = 0 a. e.} p is a vector subspace of Lp ([0,T ]; X). Hence, the quotient spaces L ([0,T ]; X) are well-deﬁned real vector spaces.

(2) The elements in Lp ([0,T ]; X) are not functions on [0,T ] but equivalence classes of Lp ([0,T ]; X) with respect to N . However, for [u] ∈ L0 ([0,T ]; X) and representatives u, v of [u] there holds u−v ∈ N , i. e. two functions u and v can be identiﬁed, if u equals v a. e. in [0,T ].

p (3) For 1 ≤ p ≤ ∞ and ||| · |||p : L ([0,T ]; X) → R≥0, [u] 7→ kukp, the relations

||| [u] |||p = kukp = 0 ⇐⇒ u = 0 a. e. ⇐⇒ [u] = 0

p hold. This implies that ||| · |||p deﬁnes a norm on L ([0,T ]; X) and thus, Lp ([0,T ]; X) is a normed vector space.

(4) As usual, the following notation is used henceforth: For 1 ≤ p ≤ ∞ an equivalence class [u] = u + N ∈ Lp ([0,T ]; X) is denoted by u and the corresponding norm ||| · |||p is denoted by k · kp or k · kLp([0,T ];X), respectively.

24 [July 12, 2019] (5) For 1 ≤ p ≤ ∞ there holds

p n 0 o L ([0,T ]; X) = u ∈ L ([0,T ]; X); kukp < ∞ .

(6) For every 1 ≤ p ≤ ∞ the space Lp ([0,T ]; X) is a Banach space.

(7) If H is a (real) Hilbert space, then L2 ([0,T ]; H) is a Hilbert space with respect to the inner product

Z T 2 2 (·|·)2 : L ([0,T ]; H) × L ([0,T ]; H) → R, (u, v) 7→ (u|v)H dt. 0

2 (8) In particular, the space L ([0,T ]; R) is a Hilbert space with respect to the inner product Z T (u|v)2 = uv dt. 0 (9) For 1 ≤ p < ∞ one has SF ([0,T ]; X) ,→ Lp ([0,T ]; X) and SF ([0,T ]; X) is dense in Lp ([0,T ]; X).

(10) Lq ([0,T ]; X) ,→ Lp ([0,T ]; X) for 1 ≤ p < q ≤ ∞, Lq ([0,T ]; X) is dense in Lp ([0,T ]; X) and

1/p−1/q q kukp ≤ T kukq , u ∈ L ([0,T ]; X) . (4.27)

(11) For 1 ≤ p < ∞ there holds C ([0,T ]; X) ,→ Lp ([0,T ]; X) and C ([0,T ]; X) is dense in Lp ([0,T ]; X). For the case p = ∞ one has C ([0,T ]; X) ,→ L∞ ([0,T ]; X), but the embedding is not dense.

(12) For 0 ≤ p ≤ ∞ the spaces Lp ([0,T ]; X) ,Lp ((0,T ]; X) ,Lp ([0,T ); X) and Lp ((0,T ); X) are isomorph (cf. [AE09, Remark 5.2]).

p The next theorem gives a characterization of the space L ([0,T ] × Ω; R).

25 [July 12, 2019] Theorem 4.29. Let 1 ≤ p < ∞. Then the map

p p p L ([0,T ] × Ω; R) → L ([0,T ]; L (Ω; R)) , u 7→ (t 7→ u (t, ·)) , is an isometric isomorphism.

Due to this result, for 1 ≤ p < ∞, one can identify the Banach spaces p p p L ([0,T ] × Ω; R) and L ([0,T ]; L (Ω; R)) in virtue of the above isometric isomorphism. There is a diﬀerent situation in the case p = ∞.

Remark 4.30. The assertion of Theorem 4.29 is not true for p = ∞;

∞ ∞ ∞ L ([0,T ] × Ω; R) 6' L ([0,T ]; L (Ω; R)) .

In the case p = ∞ the proof of Theorem 4.29 fails – not because of the boundedness but because of the measurability (cf. [AE09, Remark 6.23]).

4.3 Sobolev functions with values in Banach spaces Here we summarize sum useful facts on Sobolev spaces of functions which take values in Banach spaces. Except for the results on weak and strong compactness, Theorem 4.35 and Theorem 4.37, the following material is taken from [Ev], Section 5.9.2, with a small modification in the proof of Theorems 4.32 and 4.33 suggested by Evans. Instead of extension by zero one extends by reflection. Let I ⊂ R be an open interval. Let X be a Banach space. 1 Definition 4.31. (i) We say that v ∈ Lloc(I; X) is the weak derivative 1 of u ∈ Lloc(I; X) if Z Z 0 ∞ ϕ (t)u(t) dt = − ϕ(t)v dt ∀ϕ ∈ Cc (I). (4.28) I I In this case we write u0 = v.

(ii) The Sobolev space W 1,p(I; X) consists of all functions u ∈ Lp(I; X) such that the weak derivative u0 exists and belongs to Lp(I; X). We set

( p 0 p 1/p ku(t)kX + ku (t)kX dt if 1 ≤ p < ∞, kuk 1,p := W (I;X) 0 esssupt∈I (ku(t)kX + ku (t)kX ) .

As usual we write H1(I; X) = W 1,2(I; X). If I = (a, b) we write Lp(a, b; X) as a shorthand for Lp((a, b); X).

It is easy to say by approximation that we obtain an equivalent deﬁnition ∞ 1 if in (4.28) we replace ϕ ∈ Cc (I) by ϕ ∈ Cc . Similarly one see that by

26 [July 12, 2019] approximation that if u and v are in L1(I,X) and I is compact then (4.28) is equivalent to Z Z 0 1 ¯ ϕ (t)u(t) dt = − ϕ(t)v dt for all ϕ ∈ C (I) with ϕ|∂I = 0. I I As for scalar-valued function each function in u ∈ W 1,p(I; X) has a continuous representative.

Theorem 4.32. Let u ∈ W 1,p(I,X) with weak derivative u0. Then there exists u¯ ∈ C0(I,X¯ ) such that u(t) =u ¯(t) for a.e. t ∈ I and Z t u¯(t) − u¯(s) = u0(τ) dτ s for every s, t ∈ I¯.

Remark. Usually one does not distinguish between u andu ¯.

Proof. The proof is the same as for X = R. If I 6= (−∞, ∞) extend u by reflection at the endpoints to a slightly larger interval I˜. Then one can check directly from the definition of the weak derivative that the extended function 1,p 0 2 still belongs to W (I,X˜ ) and the u is extended antisymmetrically Let ψk be the standard mollifiers, i.e., ψ ∈ C∞(−1, 1), R ψ = 1, ψ (t) = kψ(kt). c R k Let uk = ψk ∗ u. Then

p 0 0 p uk → u in L (I; X), uk → u in L (I; X)

1 In particular there exists a subsequence ukj and an L null set N such that

ukj (t) → u(t) for all t ∈ I \ N.

1 Now uk ∈ C (I; X) and thus for s, t ∈ I

Z t 0 uk(t) − uk(s) = uk(τ) dτ. s

0 0 p 1 Since uk → u in L ((s, t); X) and hence in L ((s, t); X) we get Z t u(t) − u(s) = u0(τ) dτ for all s, t ∈ I \ N. s

2 ∞ Indeed it suﬃces to consider test functions ϕ ∈ Cc which are supported in a small neighbourhood of the boundary point a If ϕ is symmetric that both sides in the deﬁnition of the weak derivative vanish. If ϕ is antisymmetric then ϕ(a) = 0 and thus ϕ = 0 on ∂I so that the desired identity holds.

27 [July 12, 2019] Let fix s ∈ I \ N and defineu ¯ by Z t u¯(t) := u(s) + u0(τ) dτ for t ∈ I¯. s Thenu ¯(t) = u(t) for all t ∈ I \ N. Morever it follows from the definition thatu ¯ ∈ C0(I¯; X).

For parabolic equations the following fact is very useful.

2 1 0 2 −1 Theorem 4.33. Suppose that u ∈ L (I; H0 (Ω)) and u ∈ L (I; H (Ω)). Then (i) u ∈ C0(I; L2(Ω)) (after possible modiﬁcation of a set of measure zero);

(ii) The map 2 t 7→ ku(t)kL2 is weakly diﬀerentiable with d ku(t)k2 = 2hu0(t), u(t)i for a.e. t ∈ I. dt L2

(iii)

0 2 0 2 2 max ku(t) − u(t )kL2 ≤ 8 ku kL2(I;H1(Ω)) + kukL2(I;H−1(Ω) . (4.29) t,t0∈I¯ 0

(iv) 0 max ku(t)kL2 ≤ C(T ) kukL2(I;H1(Ω)) + ku kL2(I;H−1(Ω)) . t∈I¯ 0

Remark. This is one of the simplest example of a trace theorem. Given n some information on u on the upper half-space (0, ∞) × R we can assign n in a unique way values for u on the boundary {0} × R in a suitable space. 2 n Note that the conclusion u(0, ·) ∈ L (R ) in the above theorem is optimal. 2 n For any g ∈ L (R ) the existence of weak solutions of the parabolic equation n ∂tu − ∆u = 0 in (0, ∞) × R with u(0, ·) = g provides an extension u of g with the required regularity. Note that the boundary values u(0, ·) lie in a space which is ’intermediate’ be- 2 0 2 tween the spaces X0 and X1 where u ∈ L (0, ∞; X0) and u ∈ L (0, ∞; X1). This is no coincidence and there is a natural connection between trace theorems, parabolic equations and real interpolation, see, e.g., [Li69] and [Ta07].

Idea of proof. By Theorem 4.32 the we may assume that u ∈ C0(I¯; H−1). Extend u again by reﬂection to a slightly larger interval I˜ ⊃ I¯ and set uk = ψk ∗ u as in the proof of Theorem 4.32. Then

2 1 0 −1 0 0 2 −1 uk → u in L (I; H0 ), uk → u in C (I; H ),, uk → u in L (I; H ).

28 [July 12, 2019] 1 Choose a subsequence (not relabelled) such that uk(t) → u(t) in H0 (and hence in L2) for a.e. t ∈ I. We have

d 2 0 0 0 0 ku (t)−u (t)k = 2(u (t)−u (t), u (t)−u (t)) 2 = 2hu (t)−u (t), u (t)−u (t)i. dt j k L2 j k j k L j k j k

2 Now choose s ∈ I such that uk(s) → u(s) in L and use the identity above 2 2 to estimate kuj(t) − uk(t)kL2 − kuj(s) − uk(s)kL2 . This yields Z 2 0 0 lim sup sup kuj(t) − uk(t)kL2 ≤ lim sup 2|huj(t) − uk(t), uj(t) − uk(t)i| dt j,k→∞ t∈I j,k→∞ I Z 0 0 2 2 ≤ lim sup kuj − ukkH−1 + kuj − ukkH1 dt = 0. j,k→∞ I 0

0 2 0 2 Thus uj is a Cauchy sequence in C (I¯; L ) and hence converges in C (I¯; L ) 0 2 −1 to a function v ∈ C (I; L ). But we already know that uj(t) → u(t) in H for every t ∈ I¯. Hence v = u in I and u ∈ C0(I; L2). To prove (ii) start from the identity

Z t 2 2 0 kuk(t)kL2 − kuk(s)kL2 = 2 huk(τ), uk(τ)i dτ for s, t ∈ I s and pass to the limit to get

Z t 2 2 0 ku(t)kL2 − ku(s)kL2 = 2 hu (τ), u(τ)i dτ. (4.30) s Since the integrand on the right hand side is in L1(I) this implies (ii). (iii) and (iv): Homework. Hint: (iv) follows form (iii).

The following generalisation is useful to establish higher regularity of solutions to parabolic equations.

∞ Theorem 4.34. Let k ∈ N. Assume that ∂Ω ∈ C . Suppose that u ∈ L2(I; Hk+2(Ω)) and u0 ∈ L2(I; Hk(Ω)). Then

u ∈ C0(I; Hk+1(Ω)).

j Proof. Homework. Hint: consider ﬁrst the result for space H0(Ω) and then use that the exists a bounded extension operator E : Hj(Ω) → Hj(Ω)˜ for some Ω˜ ⊃ (Ω).

Compactness Weak* compactness in Lp0 (I; X0) follows simply by com- 0 p bining weak*compactness in X and weak*or weak compactness in L (I; R), working initially on a countable and dense subset of test functions.

29 [July 12, 2019] Theorem 4.35 (Weak* compactness in Lp0 (I; X0)). Let X be a separable Banach space. Let 1 ≤ p < ∞ and let p0 ∈ (1, ∞] denote the dual exponent. Assume that

kukkLp0 (I;X0) ≤ M ∀k ∈ N. p0 0 Then there exists a subsequence ukl and u ∈ L (I; X ) such that such that Z Z p lim hf, ukl i dt = hf, ui dt ∀f ∈ L (I; X). (4.31) l→∞ I I Moreover

kuk p0 0 ≤ lim inf kukk p0 . (4.32) L (I;X ) k→∞ L (I;X) If, in addition, X is reﬂexive then the roles of X and X0 can be interchanged.

Notation We denote the convergence in (4.31) by ∗ p0 0 ukl * u in L (I,X ). If X is reﬂexive and 1 < p < ∞ we also write p0 0 ukl * u in L (I,X ).

Remark. This notation is justiﬁed by the following result. Let A be a R 0 measurable set. Let Tf (u) = Ahf, ui dt. If X and X are separable one can show that for 1 ≤ p < ∞ the map f 7→ Tf is an isometric isomorphism from Lp0 (A; X0) to (Lp(A; X))0 (see [Ed65], Theorem 8.18.2 for p = 1 and Theo- rem 8.20.3 for 1 < p < ∞)3. Thus the convergence in (4.31) is equivalent weak* convergence in Lp0 (I; X0) and (4.32) just expresses the weak lower semicontinuity of the norm. If X is reﬂexive and 1 < p < ∞ then weak* and weak convergence are equivalent. In concrete application we rarely needs that Lp0 (I; X0) is (isometrically isomorphic to) the dual of Lp(I; X). The existence of a subsequence for which (4.31) holds, the uniqueness of the limit u if (4.31) holds and the lower semicontinuity estimate are usually enough.

Proof. Set q = p0. By choosing a subsequence (not relabeled) we may assume that

L := lim kukkLq(I;X0) = lim inf kukkLq(I;X0). k→∞ k→∞ Choosing a further subsequence we may assume that q kukkX0 * v in L (I). (4.33) 3Edwards does not assume separability and shows that in this more general setting the dual of Lp(A, X) is isometrically isomorphic to the space of weakly measurable functions p0 0 Lw (A, X ).

30 [July 12, 2019] Step 1. The key point is to show that there exists u ∈ Lq(I,X0) and a subsequence such that

q hukl , hi * hu, hi in L (I) ∀h ∈ X. (4.34) This is proved by the usual diagonalization argument, working ﬁrst on a ˜ ˜ dense set in X. Let D ⊂ X be countable and dense. Let D = spanQ D be Pk ˜ set the of all ﬁnite (!) linear combination i=1 αigi with gi ∈ D and α ∈ Q. Then D is also countable and dense. For each g ∈ D the functions

Ug,k(t) := huk(t), gi are bounded in Lq(I) and hence contain a weakly convergent subsequence (in the following weak convergence should always be replaced by weak* convergence if q = ∞). Since D is countable the usual diagonalization argument implies that there exists a single subsequence ukl and functions q Ug ∈ L (I) (deﬁned for g ∈ D) such that

q hukl , gi *Ug in L (I) ∀g ∈ D. (4.35)

Since the left hand side is Q-linear in g we have

Ug+h = Ug + Uh,Uαg = αUg a.e. ∀g, h ∈ D, ∀α ∈ Q. (4.36)

The estimate |Ug,k(t)| ≤ kuk(t)kX0 kgkX implies that

|Ug(t)| ≤ v(t)kgkX a.e. ∀g ∈ D. (4.37)

The null set in (4.36) and (4.37) might depend on g, h and α, but since D × D × Q is countable there exists a single nullset N such that these identities hold for all t∈ / N, all g, h ∈ D and all α ∈ Q. Fix t∈ / N. Then

Thus the map g 7→ Ug(t) is Lipschitz as a map from D to R with Lipschitz constant v(t). Since D is dense in X this map as a unique Lipschitz extension to X. Since the map is Q-linear on D it follows that the extension is a linear map on X, thus an element of X0 which we denote by u(t). Then

hu(t), gi = Ug(t) ∀g ∈ D, t∈ / N and it follows from (4.37) that

ku(t)kX0 ≤ v(t) ∀t∈ / N. (4.38) Thus

khu, g − hikLq(I) ≤ kvkLq(I) kg − hkX ∀g, h ∈ X, k ∈ N.

31 [July 12, 2019] Together with the estimate

q khukl , gi − hukl , hikL (I) ≤ Mkg − hkX ∀g, h ∈ X, l ∈ N, the convergence (4.35) for g ∈ D and the density of D we easily deduce (4.34). Moreover (4.38), (4.33) and weak lower semicontinuity of the norm under weak convergence in Lq(I) imply

kukLq(I;X0) ≤ kvkLq(I) ≤ L.

Step 2. Convergence for f ∈ Lp(I,X). p We reduce this to weak convergence for L functions with values in R by using simple test functions. Denote space of simply functions with values in X by

( k ) X 1 S := χAi (t)fi : fi ∈ X,Ai ⊂ I measurable, L (Ai) < ∞, k ∈ N . i=1 If f in S then

k Z X Z Tf (ukl ) := hukl , fi dt = ukl dt, fi . I i=0 Ai Now by Step 1 Z Z Z

hukl , fi dt = hukl , fiiχAi dt → hu, fiiχAi . Ai I I Hence

Tf (ukl ) → Tf (u) ∀f ∈ S. We have

q p p |Tf (ukl ) − Tg(ukl )| ≤ kukl kL (I;X) kf − gkL (I;X) ≤ Mkf − gkL (I;X)

p and the same bound holds for u instead of ukl . Since S is dense in L (I; X) the assertion follows.

Step 3. Reflexive X. By definition the reflexivity the canonical injection i : X → X00 given by hi(x),T i = hT, xi for all T ∈ X0 is an isometric isomorphism. Thus separability of X implies that X00 is separable. It follows that X0 is separable.

Now assume that kukkLp0 (I;X) and letu ˜k = i ◦ uk. Then ku˜kkLp0 (I;X00) =

32 [July 12, 2019] 0 kukkLp0 (I;X). Application of the theorem with X replaced by Y = X shows p0 00 that there exists a subsequenceu ˜kl andu ˜ ∈ L (I; X ) such that Z Z Z p 0 hf, ukl i dt = hu˜kl , fi dt → hu,˜ fi dt ∀f ∈ L (I,X ). I I I Now deﬁne u(t) = i−1u˜(t). Then hu,˜ fi = hf, ui and thus Z Z p 0 hf, ukl i dt → hf, ui dt ∀f ∈ L (I,X ). I I This is the desired assertion.

[12.4. 2019, Lecture 4] [15.4. 2019, Lecture 5]

We now turn to strong compactness and ﬁrst recall the Arzela-Ascoli theorem which was proved in the course PDE and Functional Analysis.

Theorem 4.36. Let K be a compact subset of a metric space and let X be a Banach space. Then A ⊂ C0(K; X) is compact in C0(K; X) if and only if the following two conditions are sat- isﬁed

(i) For each t ∈ K the set At := {u(t): u ∈ A} is compact in X; (ii) the functions in A are equicontinuous, i.e.,

∀ε > 0 ∃δ > 0 ∀u ∈ A d(t, s) < δ =⇒ ku(t) − u(s)kX < ε.

Thus two properties are need for compactness in C0(K; X): compactness of u(t) in X for each t and uniform control of u(t) − u(s) if t and s are close. We now turn to compactness in Lp(I; Y ). A very useful observation is that the uniform control of u(t)−u(s) can be in a (much) weaker space Z. This is crucial for parabolic equations where we have control of the time derivative only in some weak space (usually a negative Sobolev space).

Theorem 4.37 (Aubin-Lions lemma). Let X,Y,Z be Banach spaces and assume that X,→ Y is compact and Y,→ Z is continuous. Let I ⊂ R be a bounded interval and let

p 1,q 0 A := {u ∈ L (I; X) ∩ W (I,Z): kfkLp(I,X) + kf kLq(I;Z) ≤ M}.

(i) Assume that 1 ≤ p < ∞ and q ≥ 1. Then

A is compact in Lp(I; Y ).

33 [July 12, 2019] (ii) Assume that p = ∞ and q > 1. Then A is compact in C0(I¯; Y ).

Proof. The proof of assertion (i) is Homework, the proof of assertion (ii) is similar and is given below. The strategy is the same as in the proof of the Frechet-Kolmogorov compactness criterion for real-valued Lp functions. Hints: First show by contradiction

∀δ > 0 ∃Cδ > 0 kakY ≤ δkakX + CδkakZ . (4.39)

Then extend the functions u ∈ A by reflection at the end points and define Aε := {ϕε ∗ u : u ∈ A} ∞ where ϕ ∈ Cc (−ε, ε) is the standard mollifier. We view the functions in A as maps from I to X. If suffices to show that

0 p (i) Aε is a compact subset of C (I¯; Y ) (and hence of L (I,Y )).

p (ii) ∀η > 0 ∃ε > 0 A ⊂ Bη(Aε) in L (I,Y ). More explicitly the last inclusion is equivalent to

∀f ∈ A ∃g ∈ Aε kg − fkLp(I;Y ) < ε.

To prove (i) show that all functions v ∈ Aε satisfy.

0 kvkC0(I¯;X) ≤ CεM and kv kC0(I¯;Z) ≤ CεM.

This implies that

kv(t) − v(s)kY ≤ δCεM + CδCεM |t − s|.

Deduce that the functions in Aε are equicontinuous and apply the Arzela- Ascoli theorem. 0 To show (ii) set h(t) := ku (t)kZ show that

Z t+|τ| ku(t + τ) − u(t)kZ ≤ h(s) ds, t−|τ| deduce that Z t+ε k(ϕε ∗ u)(t) − u(t)kZ ≤ C h(s) ds t−ε and conclude that

p p p kϕε ∗ u − ukLp(I,Z) ≤ CεkhkL1 ≤ CεM .

34 [July 12, 2019] Now combine this with (4.39) to show that for each η > 0 there exists δ(η) > 0 and ε(η) > 0 such that

kϕε ∗ u − ukLp(I,Y ) < η ∀u ∈ A. To prove (ii) letu ¯ : I¯ → Z denote the continuous representative of 1,q u ∈ W (I; Z). We ﬁrst show that if εk → 0 then ϕεk ∗ u is a Cauchy 0 ¯ 1 sequence in C (I; Y ). For α = 1 − q we have Z t+ε 0 α k(ϕε ∗ u)(t) − u(t)kZ ≤ C ku (s)kZ ds ≤ Cε M t−ε and k(ϕε ∗ u)(t)kX ≤ M. Thus (4.39) implies that for 0 < ε0 < ε α ¯ k(ϕε ∗ u)(t) − (ϕε0 ∗ u)(t)kY ≤ 2δM + CδCε ∀t ∈ I, δ > 0. 0 ¯ This implies that ϕεk ∗ u is a Cauchy sequence in C (I,Y ) if εk → 0. Hence there exists v ∈ C0(I¯; Y ) such that 0 ¯ ϕεk ∗ u → v in C (I; Y ). On the otheru ¯ ∈ C0(I¯; Z) implies that 0 ¯ ϕεk ∗ u → u¯ in C (I; Z). 0 0 Thusu ¯ = v ∈ C (I¯; Y ). Setting ε = εk and taking the limit k → ∞ we get the uniform bound α sup k(ϕε ∗ u)(t) − u¯(t)kY ≤ 2δM + CδCI ε ∀t ∈ I,¯ δ > 0. (4.40) u∈A Similarly we get α k(ϕε ∗ u)(t) − (ϕε ∗ u)(s)kZ ≤ |t − s| M, k(ϕε ∗ u)(t) − (ϕε ∗ u)(s)kX ≤ 2M and thus α k(ϕε ∗ u)(t) − ϕε ∗ u)(s)kY ≤ 2δM + CδM|t − s| ∀t ∈ I,¯ δ > 0. 0 ¯ Since ϕεk ∗ u → u¯ in C (I; Y ) it follows that α ku¯(t) − u¯(s)kY ≤ 2δM + CδM|t − s| ∀t ∈ I,¯ δ > 0. This shows that the family A is equicontinuous in C0(I¯; Y ). It only remains to show that At := {u¯(t): u ∈ A} is compact in Y . Now ε At := {(ϕε ∗ u)(t): u ∈ A} is compact in Y since kϕε ∗ u)(t)kX ≤ M for all u ∈ A. Finally by (4.40) for every η > 0 ε there exists an ε > 0 such that At ⊂ Bη(At ) in Y . Thus At is compact in Y .

35 [July 12, 2019] 4.4 Maximum principles and the Harnack inequality In this subsection, which is taken from Evans, Chapter 7.1.4, we consider an elliptic operator L in non divergence form with symmetric coeﬃcients ij ji a = a . As before, we set ΩT := (0,T ] × Ω, and set

ΓT := ΩT \ ΩT . (4.41)

Note that ΓT contains {0} × Ω but not {T } × Ω. In analogy to the elliptic case, we will prove weak and strong maximum principles for solutions to parabolic equations.

Weak maximum principle Set

2 2 C1 (ΩT ) := {u :ΩT → R : u, Dxu, Dxu, ut ∈ C(ΩT )}. (4.42)

A function u is called a subsolution (respectively supersolution) if

ut + Lu ≤ 0 (respectively ut + Lu ≥ 0) in ΩT .

ij i 2 Theorem 4.38. Suppose that a , b , c ∈ C(ΩT ). Assume that u ∈ C1 (ΩT )∩ C(ΩT ) is a subsolution, i.e.,

ut + Lu ≤ 0 in ΩT .

(i) If c ≡ 0, then max u = max u. ΩT ΓT

(ii) If c ≥ 0, then max u ≤ max u+. ΩT ΓT

Proof. The proof carries over essentially verbatim from the elliptic case (see NPDE I, Theorem 2.93).

(i) Suppose that c ≡ 0, and consider ﬁrst the case that ut + Lu < 0 in ΩT . Assume for the sake of contradiction that there is (x0, t0) ∈ ΩT such that u(x , t ) = max u. There are two possibilities: 0 0 ΩT If 0 < t0 < T , then (x0, t0) is an interior point of ΩT , and thus

ut(x0, t0) = 0.

Otherwise, t0 = T . Then we have ut(x0, t0) ≥ 0. In both cases,

2 Dxu(x0, t0) = 0 and Dxu(x0, t0) ≤ 0. (4.43)

36 [July 12, 2019] Consequently (see the footnote in NPDE I, page 183)

n X ij (∂t + L)u(x0, t0) = − a uxixj (x0, t0) ≥ 0, i,j=1

which is a contradiction to the strict inequality ut + Lu < 0 in ΩT . Consider now the case of general u with ut + Lu ≤ 0 in ΩT . For ε > 0 ε ε ε set u (x, y) := u(x, y) − εt. Then ∂ut + Lu < 0, and by the above argument max uε = max uε. Letting ε → 0+ we obtain the claim ΩT ΓT by pointwise convergence uε → u.

(ii) Suppose now that c ≥ 0. Suppose that u attains a positive maximum + at (x0, t0) ∈ ΩT with u(x0, t0) > maxΓT u ≥ 0. Suppose ﬁrst that ∂tu + Lu < 0. Then cu(x0, t0) ≥ 0, and we derive a contradiction as in part (i). The general case follows again by approximating u ε ε by u . Note that if ε > 0 is chosen small enough, then u (x0, t0) > + ε + ε maxΓT u ≥ maxΓT (u ) . Hence u attains a local maximum at some point (x1, t1) ∈ ΩT which yields a contradiction.

Remark 4.39. Note that part (ii) can also be reduced to (i) as was done in the elliptic case (see NPDE I, Corollary 2.94) with a slightly more careful argument.

Corollary 4.40. Under the assumptions of Theorem 4.38, if ut + Lu = 0 in ΩT , then max |u| = max |u|. ΩT ΓT

Proof. Exercise. Observe that one inequality ia trivial since ΓT ⊂ ΩT .

Harnack’s inequality Theorem 4.41. Suppose that bi ≡ c ≡ 0, and let aij ∈ C∞, aij = aji. Suppose V ⊂⊂ U is connected. For each 0 < t1 < t2 ≤ T , there exists a ij 2 constant C = C(V, t1, t2, a ) with the following property: If u ∈ C1 (ΩT ) satisﬁes

ut + Lu = 0 and u ≥ 0 in ΩT , (4.44) then sup u(·, t1) ≤ C inf u(·, t2). V V

37 [July 12, 2019] Remark 4.42. (i) For solutions to elliptic equations Lu = 0 in Ω with mild assumptions on the coeﬃcients, the Harnack inequality states that if u ≥ 0 then for every set V ⊂⊂ Ω,

sup u ≤ C inf u, V V i.e., it compares function values of u in a ﬁxed set V . In the parabolic case, the Harnack inequality compares function values in a set V with function values in the same set but at a later time.

(ii) The classical proof by Moser (see [Mo64]) follows essentially ideas of the proof of the Harnack inequality for elliptic equations discussed in NPDE I (see Theorem 2.45) but is much more involved. It requires much less regularity assumptions on the coeﬃcients and the solution, and can then be used to show local H¨olderregularity of solutions to parabolic equations.

Proof. (Theorem 4.41) For the following computations we assume that u is smooth enough. Without loss of generality we may assume that inf u > 0 in + ΩT (otherwise consider u + ε and let ε → 0 at the end), and that V ⊂⊂ Ω is an open ball (otherwise use a covering argument). We consider

v := log u in ΩT . (4.45)

We aim at proving the diﬀerential inequality

2 vt ≥ α|∇v| − β, (4.46) with constants α, β independent of v, which implies the assertion by integration (see Step 7). Note that (4.46) reads in terms of u

u |∇u|2 t ≥ α − β. u u2 Step 1: Setting up a formulary. We will need to compute various derivatives that we collect here for later reference: We have u v = xi , (4.47) xi u ux x u − ux ux v = i j i j , (4.48) xixj u2 and thus by (4.44),

n n ut 1 X X v = = aiju = aij(v + v v ) = w +w, ˜ (4.49) t u u xixj xixj xi xj i,j=1 i,j=1

38 [July 12, 2019] where n n X ij X ij w := a vxixj , w˜ := a vxi vxj . (4.50) i,j=1 i,j=1

By symmetry of the coeﬃcients aij,

n n X X w˜ = aij v v + 2 aijv v , (4.51) xk xk xi xj xixk xj i,j=1 i,j=1 n n X X w = aij v + aijv . (4.52) xk xk xixj xixj xk i,j=1 i,j=1

From now on, we will denote by R1 a term which may change from expression to expression with the following property: For every ε > 0, there exists Cε such that

2 2 2 |R1| ≤ ε|D v| + Cε|Dv| + Cε. (4.53)

Note that we can absorb any term of the form vxi vxj , vxixj or (by Young’s inequality) vxixj vxk into R1. We have

n X ij ij w˜xkxl = 2a vxixlxk vxj + 2a vxixl vxj xk + R1. (4.54) i,j=1

Further, by (4.49),

n X kl kl kl kl vtxi = axi vxkxl + a vxkxlxi + axi vxk vxl + 2a vxkxi vxl (4.55) k,l=1

Step 2: Diﬀerential inequality for w. By (4.49) and (4.54),

n X ij ij vxkxlt = wxkxl + (2a vxixlxk vxj + 2a vxixl vxj xk ) + R1, (4.56) i,j=1 and thus, using (4.52) and (4.56)

n X kl kl wt = a vxkxlt + at vxkxl k,l=1 n n n X kl X ij X ij kl = a wxkxl + 2 a vxj wxi + 2 a a vxixk vxj xl + R1. k,l=1 i,j=1 i,j,k,l=1 (4.57)

39 [July 12, 2019] Set n X kl bk := −2 a vxl . (4.58) l=1 Then, choosing ε > 0 small enough to absorb ε|D2v|2, we have by (4.57), (4.53) and the parabolicity condition,

n n n X kl X k X ij kl wt − a wxkxl + b wxk ≥ 2 a a vxixk vxj xl − |R1| k,l=1 k=1 i,j,k,l=1 ≥ θ˜2|D2v|2 − C|Dv|. (4.59)

[15.4. 2019, Lecture 5] [26.4. 2019, Lecture 6]

Step 3: Diﬀerential inequality for w˜. We proceed similarly forw ˜. We have n n n (4.50) X ij X ij X ij w˜t = at vxi vxj + a vxitvxj + a vxi vxj t i,j=1 i,j=1 i,j=1 (4.55) X ij kl = R1 + 2 a a vxkxlxi vxj + vxkxi vxl vxj i,j,k,l

(4.54) X kl X ij kl X ij kl = R1 + a w˜xkxl − 2 a a vxixl vxj xk + 2 a a vxkxi vxl vxj k,l i,j,k,l i,j,k,l

(4.51) X kl X ij kl X kl = R1 + a w˜xkxl − 2 a a vxixl vxj xk + a w˜xk vxl , k,l i,j,k,l k,l which implies by (4.58), (4.53), and the parabolicity condition,

n n X kl X k 2 2 2 w˜t − a w˜xkxl + b w˜xk ≥ −C|D v| − C|Dv| − C (4.60) k,l=1 k=1 with constants independent of v (or w or u). Step 4: Diﬀerential inequality for w + κw˜. We set

wˆ := w + κw.˜ (4.61)

If κ > 0 is chosen small enough, then by (4.59) and (4.60)

n n X X θ2 wˆ − aklwˆ + bkwˆ ≥ |D2v|2 − C|Dv|2 − C. (4.62) t xkxl xk 2 k,l=1 k=1

∞ Step 5. Choose a cut-oﬀ function ψ ∈ Cc (ΩT ) such that 0 ≤ ψ ≤ 1, ψ = 0 4 on ΓT , and ψ ≡ 1 on V ×[t1, t2]. We claim that ψ wˆ+µt ≥ 0 in ΩT . Assume

40 [July 12, 2019] for the sake of contradiction that ψ4wˆ + µt attains a negative minimum at (x0, t0) ∈ ΩT . Note that 0 > wˆ(x0, t0) = [w + κw˜](x0, t0) implies by (4.50) and parabolicity

2 X i,j |D v(x0, t0)| ≥ −C a vxi,xj (x0, t0) i,j X ij 2 ≥ Cκ a vxi vxj (x0, t0) ≥ C|Dv| (x0, t0), i,j and thus in particular

2 |wˆ(x0, t0)| ≤ C|D v|(x0, t0). (4.63)

Further, as in the proof of Theorem 4.38 (note that the assumption 4 ψ wˆ(x0, t0) + µt(x0, t0) < 0 implies that ψ(x0, t0) 6= 0, and we may divide 3 by ψ (x0, t0))

4ψxk wˆ(x0, t0) + ψwˆxk (x0, t0) = 0, (4.64)

2 4 and (since Dx(ψ wˆ + µt)(x0, t0) ≥ 0)

4 X kl 4 0 ≥ (ψ wˆ + µt)t − a (ψ wˆ + µt)xkxl (x0, t0) k,l 4 X kl ≥ µ + ψ wˆt − a wˆkl (x0, t0) − R2 (4.65) k,l where by (4.64)

2 X kl 2 2 |R2| ≤ [Cψ |wˆ| + |2 a 4ψ ψxl (ψwˆk)](x0, t0) ≤ Cψ |wˆ|(x0, t0). (4.66) k,l

Further, by (4.58) and (4.64), we have

n X X 1 bkwˆ (x , t ) = aklv wˆ ψ (x , t ), xk 0 0 xl xk ψ 0 0 k=1 k,l and thus by (4.62) ! θ˜2 0 ≥ µ + ψ4 |D2v|2 − C|Dv| − C − R, (4.67) 2 where we estimate for arbitrary δ > 0, using (4.63) and Young’s inequality

2 3 |R| ≤ C[ψ |wˆ| + ψ |Dv||wˆ|](x0, t0) 2 2 3 2 4 2 2 ≤ C[ψ |D v| + ψ |Dv||D v|) ≤ δψ |D v| + Cδ. (4.68)

41 [July 12, 2019] Note that here the power ψ4 was useful. Inserting (4.68) into (4.67), we ﬁnd choosing δ > 0 small enough,

4˜2 2 2 0 ≥ µ + ψ θ 4|D v| − Cδ − Cδ, which is a contradiction if we choose µ large enough (independently of v). 4 Step 6: Proof of (4.46). From Step 5 we know that ψ wˆ + µt ≥ 0 in ΩT , and thus (since ψ ≡ 1 in V × [t1, t2]) in particularw ˆ + µt ≥ 0 in V × [t1, t2]. Hence, by (4.50) and parabolicity, there are constants α, β independent of v such that

vt = w +w ˜ =w ˆ + (1 − κ)w ˜ ≥ −µt + (1 − κ)w ˜ X ij 2 = −µt + (1 − κ) a vxi vxj ≥ −µt + α|Dv| 2 ≥ α|Dv| − β in V × [t1, t2]

Step 7: Conclusion. For ﬁxed x1, x2 ∈ V we have (recall that V is a ball and in particular convex)

Z 1 v(x2, t2) − v(x1, t1) = Dv · (x2 − x1) + vt(t2 − t1) ds 0 (4.46) Z 1 2 ≥ −|Dv||x2 − x1| + (t2 − t1)[α|Dv| − β] ds ≥ −γ 0 with γ depending only on α, β, |x2 − x1| and t2 − t1. Since v = log u we deduce that −γ u(x2, t2) ≥ e u(x1, t1)∀x1, x2 ∈ V.

−k2t Example. Let V = B(0,R) and Ω = B(0, 2R). Take u = e cos kx1 π where k = 4R . Then u is a solution of the heat equation, u ≥ 0 in [0, ∞) × B(0, 2R) and

2 2 π t2−t1 π t2−t1 sup u(t1, x) ≥ e 16 R2 sup u(t2, x) ≥ e 16 R2 inf u(t2, x). x∈B(0,R) x∈B(0,R) x∈B(0,R)

This shows that for V = B(0,R) and L = −∆ the constant in Harnack’s 2 π t2−t1 inequality has to be at least e 16 R2 .

A look back at the proof of the Harnack inequality. Let us retrace the steps of the proof in simplest case ut − ∆u = 0. We set again v = log u and we have to show that

∀L > 0 ∃γL > 0 vt − L|∇v| + γL ≥ 0.

42 [July 12, 2019] This implies v(t , x ) − v(t , x ) ≥ −γ (t − t ) if L = |x2−x1| (see Step 7). 2 2 1 1 L 2 1 t2−t1 A suﬃcient condition is

2 vt ≥ α|∇v| − β in [t1, t2] × V (4.69) for some α > 0. For the heat equation Step 1 reduces to the calculations v v ut = e vt and ∇u = e ∇v which give

2 vt = ∆v + |∇v| = w +w. ˜

The key identity for Steps 2 and 3 is

∆|∇v|2 = 2|∇2v|2 + 2∇v · ∇∆v.

This gives immediately

2 2 wt − ∆w − 2∇v · ∇w = 2|∇ v| . (4.70)

We also get w˜t = 2∇v · ∇vt = 2∇v · ∇∆v + 2∇v · ∇w.˜ Together with the key identity this gives

2 2 w˜t − ∆w ˜ − 2∇v · ∇w˜ = −2|∇ v| .

1 To show (4.69) with α = 2 and it suﬃces show that 1 wˆ := ∆v + |∇v|2 ≥ −β in [t , t ] × V 2 1 2 The identities for w andw ˜ give

2 2 wˆt − ∆w ˆt − 2∇v · ∇wˆ = |∇ v| and the lower bound forw ˆ follows essentially from the maximum principle applied to W = ψ4wˆ + µt where ψ is the cut-oﬀ function in Step 5. If W ≥ 0 in [t1, t2] × V then the desired inequality holds with β = µt2. Assume for a contradiction that W has a negative minimum at (t0, x0). Then at (t0, x0) we have Wt = ∇W = 0 and ∆W ≥ 0 and thus

0 ≥ (∂tW − ∆W − 2∇v · ∇W )(t0, x0) 4 ≥ µ + ψ (∂twˆ − ∆w − 2∇v · ∇w)(t0, x0) − error 4 2 2 = µ + ψ |D v(t0, x0)| − error.

43 [July 12, 2019] 3 4 3 Here the error involves terms of the form ψ ∂tψ w, ∆ψ w, ψ ∇ψ · ∇w and ∇v · ∇ψ w. Finally we use that the assumption W (t0, x0) < 0 implies that wˆ(t0, x0) < 0 and hence

2 2 |∇v(t0, x0)| ≤ 2|∇ v(t0, x0)| and |w(t0, x0)| ≤ |∆v(t0, x0)|.

Finally the condtion ∇W (t0, x0) = 0 gives a bound of ψ∇w in terms of |w∇ψ|. Since the ﬁrst and second derivatives of ψ are bounded we get

2 2 3 2 |error| ≤ C|ψ| |∇ v(t0, x0)| + Cψ |ψ||∇v| |∇ v(t0, x0)| 2 2 3 2 3/2 ≤C|ψ| |∇ v(t0, x0)| + Cψ |∇ v(t0, x0)| .

1 2 1 2 3 4/3 4/3 1 −4 4 Using Young’s inequalities ab ≤ 2 δa + 2δ b and ab ≤ 4 δ |a| + 4 δ b we get a contradiction if µ is chosen suﬃciently large.

Theorem 4.43 (Strong maximum principle for c = 0). Let

X ij L = a (t, x)∂iu∂ju i,j

ij ij ij ∞ and assume that a = a and the a ∈ C (ΩT ) and ∂t + Lu is parabolic. 1,2 Suppose that Ω is connected and u ∈ C (ΩT ) ∩ C(ΩT ). (i) If ∂t + Lu ≤ 0 in ΩT

and u attains its maximum over Ω¯ T at a point (t0, x0) ∈ ΩT , then

u is constant on Ωt0 = (0, t0] × Ω.

(ii) If ∂t + Lu ≥ 0 in ΩT

and u attains its minmum over Ω¯ T at a point (t0, x0) ∈ ΩT , then

u is constant on Ωt0 = (0, t0] × Ω.

Proof. See Evans, Section 7.1.4.

Similarly one can show that if c ≥ 0, ∂tu + Lu + cu ≤ 0 and u has a nonnegative at (t0, x0) then u is constant on Ωt0 (see [Ev], Section 7.1.4, Theorem 12).

Remark 4.44. This shows that parabolic equations have inﬁnite speed of propagation. Indeed, if u(0, ·) ≥ 0 in Ω and V = {x : u(0, x) > 0} is a non-empty subset of Ω, then u(t, ·) > 0 in Ω for all t > 0.

Proof. Exercise, see Homework 9.

44 [July 12, 2019] 4.5 Other approaches to regularity 4.5.1 Equations with measurable coefficients Consider a weak solution of the equation n X ij n ∂tu − ∂i(a (t, x)∂ju) = 0 in (0, ∞) × R i,j=1 where the coefficients aij satisfy n 2 X ij 2 θ|ξ| ≤ a (t, x)ξiξj ≤ Θ|ξ| , with 0 < θ ≤ Θ < ∞. (4.71) i,j=1 The fundamental result of Nash is that bounded solutions are weakly Höldercontinuous. Theorem 4.45 (Nash [Na58]). There exist constants A > 0 and α ∈ (0, 1), depending only on θ, Θ and n such that for all t2 ≥ t1 > t0 > 0 " α α/(2α+2)# |x1 − x2| t2 − t1 |u(t1, x1)−u(t2, x2)| ≤ A sup |u(t0, ·)| √ + . t1 − t0 t1 − t0 Nash’s proof is based on a careful analysis of the parabolic Green’s func- n tion (the analogue of the heat kernel for the heat equation in R ). For a four page summary which outlines the main ideas see [Na57]. Later Moser [Mo64] gave a different proof by proving first a Harnack inequality by his iteration technique analogous to (but technically much more demanding than) the elliptic case.

4.5.2 Campanato, Schauder and Lp theory The energy estimates in Sections 4.1.1 and 4.1.2 can be localized by using cut-oﬀ functions on parabolic cylinders

2 2 (t0 − R , t0 + R ) × B(x0,R) and based on this one derives estimates in parabolic Campanato spaces analogous to the Campanato estimates for elliptic equations and systems discussed in NPDE I. Again as in the elliptic case one can use the relation between Campanato spaces and H¨olderspaces as well as interpolation between BMO and L2 to obtain Schauder and Lp estimates (see [Ca66b] for scalar equations and Schlag [Sc96] for systems). For simplicity we consider only zero initial conditions. Consider thus the initial boundary value problem

ij i ∂tu − a (t, x)∂i∂ju + b (t, x)∂iu + c(t, x)u = f in ΩT , (4.72)

u = 0 on ΓT (4.73)

45 [July 12, 2019] and assume that aij = aji and (4.71) For µ ∈ (0, 1) we deﬁne the Schauder seminorm by |u(t, x) − u(t0, x0)| [g]µ,ΩT := sup 0 0 0 µ 0 µ/2 (t,x),(t ,x )∈ΩT |x − x | + |t − t | and we deﬁne the Schauder space by

µ/2,µ C (ΩT ) := {g :ΩT → R :[g]µ,ΩT < ∞}. Note that functions in that space are uniformly continuous and have a unique extension to {0} × Ω. We write [g] = sup |g| (or esssup for L∞ func- 0,ΩT ΩT tions). Theorem 4.46 (Schauder estimates for parabolic equations). Let µ ∈ (0, 1). Assume that µ/2,µ 2,µ a, b, c, f ∈ C (ΩT ), ∂Ω ∈ C . Assume in addition that f satisﬁes the compatibility condition

f(0, x) = 0 ∀x ∈ ∂Ω.

1,2 2 Let u ∈ C (ΩT ) be a solution of (4.72), (4.73). Then ∇ u and ∂tu are in µ/2,µ C (ΩT ) and

2 [∇ u]µ,ΩT + [∂tu]µ,ΩT ≤ C (|u|0,ΩT + fµ,ΩT ) where C depends on θ, Θ, µ, n, Ω and the H¨older norm of the coeﬃcients. Theorem 4.47 (Lp estimates for parabolic equations). Assume that 1 < p < ∞ and ∞ 2 a ∈ C(ΩT ), b, c ∈ L (ΩT ), ∂Ω ∈ C , 1,2 Let u ∈ C (ΩT ) be a solution of (4.72), (4.73).Then 2 p p p p p [D u]L (ΩT ) + [∂tu]L (ΩT ) ≤ C0 |u|L (ΩT ) + fL (ΩT ) ≤ C1(T )kfkL (ΩT ) where the constants depend on p, n, θ, Θ, Ω, |a|0,ΩT , |b|0,ΩT and |c|0,ΩT and the modulus of continuity of a. The Schauder estimates can alternatively be proved by a blow-up argument which we sketched for the elliptic case in NPDE I (see [Si]; the reason for introducing the seemingly complex notation at the beginning of the paper is that this notation allows one to treat the elliptic and the parabolic case simultaneously). n p For the inhomogeneuous heat equation in R with zero initial data the L estimates can also be obtained from the Marcinkiewicz multiplier theorem. Indeed if we extend u(t, x) and f(t, x) by 0 for t < 0 then u is a solution of

n ∂tu − ∆u = f in R × R .

46 [July 12, 2019] n+1 Let Fu(τ, ξ) denote the Fourier transform in R . Then iτ F(∂ u)(τ, ξ) = Ff(τ, ξ) t iτ + |ξ|2

iτ and we have seen that the multiplier m(τ, ξ) = iτ+|ξ|2 satisﬁes the assumptions of the Marcinkiewicz multiplier theorem. Thus

k∂tukLp(Rn+1) ≤ CkfkLp(Rn+1) and hence k∆ukLp(Rn+1) ≤ CkfkLp(Rn+1)

p ˙ p n Combined with elliptic regularity one obtains estimates in L (R; L2(R )). By suitable localization and perturbation arguments one can also obtain local Lp estimates and estimates for equations with variable coeﬃcients (com- pare the discussion for elliptic equations in NPDE I). p q Finally one can also prove estimates for ∂tu and ∆u in L (I; L (Ω)) provided that the right hand side is in the same space. In simple cases these estimates can be derived from the Lp(Lp) estimates by a Banach valued version of the Calderon-Zygmund singular integral estimates, see [BCP62], Theorem 1 for the precise formulation. For recent very general results on Lp(Lq) estimates, see [DHP07]. A classical reference for existence and regularity results on parabolic equations is the book by Ladyshenskaja, Solon- nikov and Uraltseva [LSU68]. [26.4. 2019, Lecture 6] [29.4. 2019, Lecture 7]

4.6 Time discrete approximation for nonlinear equations Forward vs. backward time discretization for ODEs

0 u (t) = Au(t), u(0) = u0.

Let h > 0. The derivative can be approximated by a forward time diﬀerence

u(t + h) − u(t) (∂hu)(t) = t h or a backward time diﬀerence u(t) − u(t − h) (∂−hu)(t) = t h Thus for t = kh we can iteratively deﬁne approximate solutions by

h + + + ∂t u (kh) = Au (kh), k = 1, 2, . . . , u (0) = u0 and −h − − − ∂t u (kh) = Au (kh) k = 1, 2, . . . , u (0) = u0

47 [July 12, 2019] The iteration for u+ is explicit, i.e., just requires an application of the matrix A while the deﬁnition of u− involves the application of the inverse (Id − hA)−1. Both schemes converge for h → 0. If A has large negative eigenvalues λ and −hλ > 1 then the u+ has wild solutions which have nothing to do with the continuous solution while u− converges nicely.

Nonlinear parabolic equations The following exposition is a much wa- n tered down version of [AL83]. Let Ω ⊂ R be an open bounded and connected domain with Lipschitz boundary. We consider the equation

∂tu − div a(∇u) = f in (0,T ) × Ω, u = 0 on (0,T ) × ∂Ω, (4.74) u(0, ·) = u0

We make the following assumptions

n n a : R → R is continuous, (4.75)

2 n (a(p1)−a(p2), p1 −p2) ≥ θ|p1 −p2| ∀p1, p2 ∈ R and some θ ≥ 0, (4.76) n |a(p)| ≤ C(1 + |p|) ∀p ∈ R and some C > 0, (4.77) 2 n (a(p), p) ≥ c|p| − C ∀p ∈ R and some c, C > 0, (4.78) 2 2 −1 u0 ∈ L (Ω), f ∈ L (0,T ; H (Ω)). (4.79)

Note that if (4.76) holds with θ > 0, then (4.78) follows by taking p2 = 0 in (4.76) and using Young’s inequality.

Deﬁnition 4.48. A function u is a weak solution of (4.74) if

2 1 0 2 −1 u ∈ L (0,T ; H0 (Ω)), u ∈ L (0,T ; H ),

0 1 hu (t), vi + (a(∇u(t)), ∇v) = hf(t), vi ∀v ∈ H0 (Ω) and a.e. t ∈ [0,T ], and u(0) = u0. Theorem 4.49. Assume (4.75)–(4.79). Then a weak solution of (4.74) exists.

Lemma 4.50 (Solution of the problem for one time step). Assume that g ∈ H−1(Ω) and let (4.75)–(4.78) hold. Let h > 0. Then there exists one 1 and only one U ∈ H0 (Ω) such that

1 (U, v) + h(a(∇U), v) = hg, vi ∀v ∈ H0 (Ω).

48 [July 12, 2019] Proof. This can be proved by Galerkin approximation and the Minty-Browder trick exactly as the corresponding result in NPDE I without the term (U, v). The existence of solutions of the ﬁnite dimensional problem follows from the coercivity of a and the Leray-Schauder theorem.

T Proof of Theorem 4.49. Let K ∈ N \{0} and let h = K . In the following we consider k ∈ {0, 1,...,K}. We extend f by 0 to (−∞,T ] and the extension by a function from (−h, T ] to H−1(Ω) which is piecewise constant 1 Z kh fh(t) = f(s) ds for t ∈ ((k − 1)h, kh]. h (k−1)h Note that by Jensen’s inequality

kfh(t)kL2(−h,T ;H−1(Ω)) ≤ kfkL2(0,T ;H−1(Ω)) (4.80) 1 and we deﬁne an approximate solution uh :(−h, T ] → H0 (Ω) which is constant on each interval ((k − 1)h, kh] inductively as follows. First set

uh(t) = u0h for t ∈ (−h, 0].

Now assume that uh is deﬁned on (−h, (k − 1)h]. Then deﬁne uh(t) as the solution of −h (∂t uh(t), v) + (a(∇uh(t)), ∇v) = hfh(t), vi for t ∈ ((k − 1)h, kh]. (4.81)

By Lemma 4.50 this problem does indeed have a unique solution uh(t). Here 1 u0h ∈ H0 (Ω) is chosen such that 2 u0h → u0 in L (Ω), hku0hk 1 → 0 as h → 0. H0 (Ω)

Step 1. Energy estimate Test the equation with uh(t). For the time continuous equation we used the fact that 1 ∂ u u = ∂ u2. t t 2 For the time discretized equation the main observation is 1 1 1 1 1 (a − b)a = a2 − b2 + (a − b)2 ≥ a2 − b2. 2 2 2 2 2 This implies that 1 (∂−hu , u ) ≥ ∂−h ku k2. t h h t 2 h Here it is crucial that we used backward time diﬀerences and not forward time diﬀerences. Now integration in t from 0 to T , (4.78) and (4.80) give after a short calculation the energy estimate

kuhk 2 1 ≤ C. L (0,T ;H0 (Ω)) Hence for a subsequence (which we do not relabel in order not to overload the notation) 2 1 uh * u¯ in L (0,T ; H0 (Ω)).

49 [July 12, 2019] Step 2. Compactness in time. Letu ˜h be the piecewise aﬃne function which agrees with uh at all points 1 1 kh. Thenu ˜h is in H (0,T ; H0 (Ω)) and u (kh) − u ((k − 1)h) ∂ u˜ (t) = h h = ∂−hu (t) for all t ∈ ((k − 1)h, kh). t h h t h It follows from (4.81), (4.77) and (4.80)

k∂tu˜hkL2(0,T ;H−1(Ω)) ≤ C.

Now by convexity of the norm and of the function s 7→ s2 we get

2 2 h 2 ku˜kL2(0,T ;H1(Ω)) ≤ kukL2(0,T ;H1(Ω)) + ku0hkH1 . 0 0 2

2 Since hku0hkH1 → 0 as h → 0 we can apply the Aubin-Lions lemma to get (for a further subsequence)

2 2 u˜h → u in L (0,T ; L (Ω)).

It is easy to see (homework) that this implies that

2 2 uh → u in L (0,T ; L (Ω)).

1 −1 Hence u =u ¯. From the bound on ∂tu˜h we also get thatu ¯ ∈ H (0,T ; H (Ω)) and −h 2 −1 ∂t uh = ∂tu˜h * ∂tu¯ in L (0,T ; H (Ω)). In particularu ¯ has a unique representative in C0([0,T ]; H−1(Ω)) and this representative satisﬁes u¯(0) = u0. (4.82) 1 Indeed for each v ∈ H0 (Ω) the functions t 7→ hu˜h(t), vi are uniformly H¨older −1 continuous and thusu ˜h(t) * u¯(t) in H (Ω) for all t, and in particular for 2 t = 0. Sinceu ˜h(0) = u0h → u0 in L (Ω) this proves (4.82).

Step 3 Compactness in space/ Minty-Browder trick. For θ > 0 we will show strong convergence

2 1 uh → u¯ in L (0,T ; H0 (Ω)).

Using this we can easily pass to the limit to show thatu ¯ is a weak solution. To prove compactness we ﬁrst test the equation with uh − u¯ and integrate in time to get

Z T −h (∂t uh, uh − u¯) + (a(∇uh), ∇uh − ∇u¯) − hfh, uh − u¯i dt = 0. (4.83) 0

50 [July 12, 2019] Now we use

−h 0 2 −1 2 −1 ∂t u¯ → u¯ in L (0,T ; H ), fh → f in L (0,T ; H )

2 1 (see homework). Since uh − u¯ * 0 in L (0,T,H0 (Ω)) this yields Z T −h lim (∂t u,¯ uh − u¯) + (a(∇u¯), ∇uh − ∇u¯) − hfh, uh − u¯i dt = 0. h→0 0 Subtracting this from (4.83) and using monotonicty of a and the relation (see Step 1) 1 (∂−h(u − u¯), u − u¯) ≥ ∂−h ku − u¯k2 t h h t 2 h L2 and (4.82). we easily see that

Z T 2 1 2 lim sup θ k∇uh − ∇u¯kL2 dt ≤ lim sup ku0h − u0kL2 = 0 h→0 0 h→0 2 and this is the desired assertion. If θ = 0 one can pass to the limit using the Minity-Browder trick (homework).

Degenerate parabolic equations, parabolic-elliptic equations and free boundary value problems As shown in the paper by Alt and Luckhaus [AL83] the arguments above can be reﬁned to give existence (and further properties) of weak solution to equations of the form

∂tb(u) − div a(b(u), ∇u) = f(b(u)) where b is monotone, but not necessarily strictly monotone. In particular when b ≡ 0 the equation is no longer parabolic but elliptic. Their approach also allows one to study degenerate parabolic equations such as the porous medium equation m ∂tv − ∆v = 0 which can be brought into the above framework (for v ≥ 0) by setting u = vm and b(s) = s1/m. By a limiting procedure one can also consider functions b which have jumps. This is relevant for free boundary value problems which arise in the theory of phase transitions such as the Stefan problem.

51 [July 12, 2019] Maximal monotone operators The monotonicity of the operator u 7→ Au := −div a(∇u) plays a crucial role in the argument. Indeed one can build a very general theory for solutions of the inclusion

∂tu ∈ A(u) where A is a so called maximal monotone operator (the value of A(u) may be a set rather than just a point. Maximality essentially means that the graph {(u, v): u ∈ D(A), v ∈ A(u)} is closed). A typical example is obtained as follows. Let H be a Hilbert space, let W : H → [0, ∞] be convex and let A(u) be the set of subgradients of W at u, i.e., A(u) := {v : H(u + u0) − H(u) ≥ (u, v) ∀u0 ∈ H}. Taking W to be zero on a convex closed set and ∞ elsewhere one can handle equations with constraints in a natural way. For a beautiful exposition of the theory see Brezis’ book [Br73]. A short discussion of linear maximal monotone operators can be found in Chapter 7 of [Br11] (for the French original see [Br83]).

To add just a little word of warning note that while the theory of monotone operators is very beautiful a number of very important PDE (like the Navier-Stokes equations or problems in elasticity) do not ﬁt into this framework. [29.4. 2019, Lecture 7] [10.5. 2019, Lecture 8]

52 [July 12, 2019] 5 Hyperbolic equations

5.1 Second order hyperbolic equations This subsection is essentially taken from [Ev, Section 7.2]. We use the n notation of Section 4.1. Let Ω ⊂ R be open and bounded, T > 0 and set ΩT := Ω × (0,T ]. For given f :ΩT → R and g, h :Ω → R, we seek solutions u :ΩT → R of the initial/boundary value problem

utt + Lu = f in ΩT u = 0 on ∂Ω × [0,T ]

u = g, ut = h on Ω × {0}. (5.1) We assume that L is in divergence form, i.e.,

n n X ij X i Lu = − a (x, t)ux + b (x, t)ux + c(x, t)u. i xj i i,j=1 i=1

2 Deﬁnition 5.1. The operator ∂t +L is called (uniformly) hyperbolic if there exists θ > 0 such that n X ij 2 n a (x, t)ξiξj ≥ θ|ξ| ∀(x, t) ∈ ΩT and ∀ξ ∈ R . (5.2) i,j=1 Note that the wave equation is a special case of (5.1) for the choice L = −∆.

5.1.1 Deﬁnition, existence and uniqueness of weak solutions In this subsection, make the following assumptions:

aij = aji, (5.3) ij i 1 a , b , c ∈ C (ΩT ) for i, j = 1, . . . , n, (5.4) 2 1 2 f ∈ L (ΩT ), g ∈ H0 (Ω), and h ∈ L (Ω). (5.5)

As in (4.4) we deﬁne the bilinear form associated to L, Z n " n # X ij X B[u, v; t] := a (t, ·)∂ju∂iv + bi(t, ·)∂iu v + c(t, ·)uv dx. Ω i,j=1 i=1 Note that for general b 6= 0 the form B is not symmertric in u and v. We denote the leading order term by Z n X ij A[u, v; t] := a (t, ·)∂ju∂iv. (5.6) Ω i,j=1

53 [July 12, 2019] By the assumptions on the coeﬃcients (5.5), (5.4) and (5.2) there are con- 1 stants C, γ > 0 such that for every v, w ∈ H0 (Ω) and all 0 < t ≤ T ,

|B[v, w; t]| ≤ Ckvk 1 kwk 1 H0 (Ω) H0 (Ω) |(B − A)[v, w; t]| ≤ Ckvk 1 kwk 2 (“continuity”) (5.7) H0 (Ω) L (Ω) θ 2 2 B[v, v; t] ≥ kvkH1(Ω) − γkvkL2 2 0 ˜ 2 A[v, v; t] ≥ θkvk 1 , (“coercivity”). (5.8) H0 (Ω) Further,

∂B ∂A [v, w; t] , [v, w; t] ≤ CkvkH1(Ω)kwkH1(Ω), (5.9) ∂t ∂t 0 0 where Z n " n # X ij X i ∂tB[u, v; t] := at (t, ·)∂ju∂iv + bt(t, ·)∂iu v + ct(t, ·)uv dx, Ω i,j=1 i=1 and similarly for A. To obtain the weak form of the equation, we again 1 interpret L and f as linear functionals on H0 , which means that we consider −1 u with ∂ttu(t) ∈ H . Roughly speaking, we multiply by a test function and integrate by parts to ﬁnd the following characterization.

Deﬁnition 5.2. A function

2 1 u ∈ L (0,T ; H0 (Ω)) (5.10) with

u0 ∈ L2(0,T ; L2(Ω)), and u00 ∈ L2(0,T ; H−1(Ω)) (5.11) is called a weak solution of the hyperbolic initial/boundary value problem 5.1 if

00 1 (i) hu , vi + B[u, v; t] = (f, v) for each v ∈ H0 (Ω) and a.e. 0 ≤ t ≤ T , (ii) u(0) = g and u0(0) = h.

Remark 5.3. Conditions (5.10) and (5.11) imply that u has a representative u¯ for which u¯ ∈ C([0,T ]; L2(Ω)) and u¯0 ∈ C([0,T ]; H−1(Ω)). Condition (ii) is required for this representative.

We aim at proving existence and uniqueness of weak solutions. For that, we proceed along the lines of Section 4.1 and employ a Galerkin approximation.

54 [July 12, 2019] Theorem 5.4. Under the assumptions (5.3), (5.4) and (5.5), there exists one and only one weak solution to (5.1). The rest of this subsection is devoted to the proof of Theorem 5.4. We show existence ﬁrst. For that, we proceed along the lines of Section 4.1 and ∞ discretize in space. Choose smooth functions {wk}k=1 such that {wk} is an 1 2 orthogonal basis of H0 (Ω) and an orthonormal basis of L (Ω). For ﬁxed m ∈ N we set m X k um(t) := dm(t)wk. (5.12) k=1 Following the lines of Section 4.1, we split the proof in several steps.

Existence of approximate solutions.

Theorem 5.5. For each m ∈ N there exists one and only one um of the form (5.12) such that k k0 dm(0) = (g, wk) and dm(0) = (h, wk) ∀k = 1, . . . , m 00 (um, wk) + B[um, wk; t] = (f, wk) for 0 ≤ t ≤ T and k = 1, . . . , m. (5.13) Proof. Similarly to the parabolic case, (5.13) is equivalent to an m×m linear system of second order ordinary diﬀerential equations. Since the coeﬃcients are in L∞(0,T ) and the right hand side is in L2, the assertion follows from standard ODE theory.

Energy estimates. Lemma 5.6. There exists C > 0 depending only on Ω, T and the coeﬃcients of L such that for all m ∈ N 0 00 max kum(t)kH1(Ω) + ku (t)kL2(Ω) + kumkL2(0,T ;H−1(Ω) 0≤t≤T 0 ≤ C kfk 2 2 + kgk 1 + khk 2 . L (0,T ;L (Ω)) H0 (Ω) L (Ω) 0 Proof. We test the deﬁning equation with um, i.e., we multiply (5.13) with k0 dm(t) and sum over k. Thus 00 0 0 0 (um, um) +B[um, um; t] = (f, um) for a.e. 0 ≤ t ≤ T. (5.14) | {z } = 1 d ku0 k2 2 dt m L2(Ω) 0 0 0 We decompose B[um, um; t] = A[um, um; t] + (B − A)[um, um, t] and use symmetry of A to obtain 0 0 0 B[um, um; t] = A[um, um; t] + (B − A)[um, um, t] 1 d 1 ∂ = A[u (t), u (t); t] − A[u , u ; t] + (B − A)[u , u0 , t]. 2 dt m m 2 ∂t m m m m

55 [July 12, 2019] We combine this with (5.14), and estimate (recall (5.9) and use Young’s inequality)

1 d ku0 k2 + A(u (t), u (t); t) 2 dt m L2(Ω) m m 1 ∂ = A(u , u ; t) − (B − A)(u , u0 , t) + (f, u0 ) 2 ∂t m m m m m 0 2 2 2 ≤ C kumk 2 + kumk 1 + kfk 2 . (5.15) L (Ω) H0 (Ω) L (Ω) Using the coercivity condition (5.8), we deduce

d ku0 k2 + A[u (t), u (t); t] ≤ dt m L2(Ω) m m 0 2 2 ≤ C kumkL2(Ω) + A[um(t), um(t); t] + kfkL2(Ω) . (5.16)

We now apply Gronwall’s lemma to the function

0 2 η(t) := kum(t)kL2(Ω) + A[um(t), um(t); t].

2 2 For that, we note that (recall that kum(0)k 1 ≤ kgk 1 ) H0 (Ω) H0 (Ω)

0 2 η(0) = kum(0)kL2(Ω) + A(um(0), um(0); 0) ≤ 2 2 ≤ C khk 2 + kgk 1 . (5.17) L (Ω) H0 (Ω) Thus, by (5.16), (5.17) and Gronwall’s lemma, it follows that

Z t 0 2 C1t 2 kum(t)kL2(Ω) + A(um(t), um(t); t) ≤ e η(0) + C2 kf(s)kL2(Ω) ds 0 2 2 2 ≤ C khk 2 + kgk 1 + kfk 2 2 . L (Ω) H0 (Ω) L (0,T ;L (Ω)) Using again the coercivity (5.8), and recalling that 0 ≤ t ≤ T was arbitrary, we conclude the proof of the ﬁrst two estimates, i.e.,

0 max kum(t)kH1(Ω) + ku (t)kL2(Ω) ≤ 0≤t≤T 0 ≤ C kfk 2 2 + kgk 1 + khk 2 . (5.18) L (0,T ;L (Ω)) H0 (Ω) L (Ω) It remains to show that

00 ku k 2 −1 ≤ C kfk 2 2 + kgk 1 + khk 2 . m L (0,T ;H (Ω)) L (0,T ;L (Ω)) H0 (Ω) L (Ω)

By deﬁnition of the dual norm and in view of (5.18), it suﬃces to show that

00 1 |hu , vi| ≤ C kfk 2 + kumk 1 ∀v ∈ H (Ω) with kvk 1 ≤ 1. m L (Ω) H0 (Ω) 0 H0 (Ω)

56 [July 12, 2019] This follows from the variational equality as the estimate for u0 in the 2 parabolic case. Precisely, denote by Pm the L -orthogonal projection onto 1 span {wk : 1 ≤ k ≤ m}. Since {wk} are also H0 -orthogonal, we have

1 kPmvk 1 ≤ kvk 1 ∀v ∈ H (Ω). H0 (Ω) H0 (Ω) 0

2 1 Now, since {wk} is an orthonormal basis in L , we have for every v ∈ H0 (Ω) with kvk 1 ≤ 1, H0 (Ω)

00 00 00 hum(t), vi = (um(t), v) = (um(t),Pmv) = (f, Pmv) − B[um(t),Pmv(t); t] ≤ C kfk 2 + kumk 1 kPmvk 1 . L (Ω) H0 (Ω) H0 (Ω) | {z } ≤1

This concludes the proof of the energy estimates.

Passage to the limit. By the energy estimates, the sequence of ap- 2 1 0 proximate solutions m 7→ um is bounded in L (0,T ; H0 (Ω)) with m 7→ um 2 2 00 2 −1 bounded in L (0,T ; L (Ω)) and m 7→ um bounded in L (0,T ; H (Ω)). Since the latter three spaces are reﬂexive, and by deﬁnition of the weak 2 1 0 2 2 derivatives, there is u ∈ L (0,T ; H0 (Ω) with u ∈ L (0,T ; L (Ω)) and u00 ∈ L2(0,T ; H−1(Ω)), and a subsequence such that

2 1 uml * u weakly in L (0,T ; H0 (Ω)) u0 * u0 weakly in L2(0,T ; L2(Ω)) ml u00 * u00 weakly(*) in L2(0,T ; H−1(Ω)). ml

We aim at proving that u is a weak solution of (5.1). Fix N ∈ N, choose ml ≥ N, and test the Galerkin equation for uml with

N X k v(t, ·) := d (t)wk(·), (5.19) k=1 where dk ∈ C1([0,T ]). Integration in t yields

Z T Z T hu00 , vi + B[u , v, t] dt = (f, v) dt. ml ml 0 0 We pass to the limit l → ∞. For the ﬁrst term, we use Theorem 4.35, for the second one that B is continuous, and obtain

Z T Z T hu00, vi + B[u, v, t] dt = (f, v) dt. (5.20) 0 0

57 [July 12, 2019] 2 1 Since functions of the form (5.19) are dense in L (0,T ; H0 (Ω)), the identity 2 1 (5.20) holds for all v ∈ L (0,T ; H0 (Ω)), and in particular hu00, vi + B[u, v; t] = (f, v) for a.e. 0 ≤ t ≤ T.

2 1 It remains to verify the initial conditions. Let now v ∈ C ([0,T ]; H0 (Ω)) of the form v(x, t) = φ(t)w(x) with v(T ) = v0(T ) = 0, and consider

Z T hu00, vi + B[u, v; t]. 0 First, similarly to the parabolic case, integration by parts applied to the ﬁrst term twice yields Z T Z T hu00, vi + B[u, v; t] = hv00, ui + B[u, v; t] dt + (u(0), v0(0)) − (u0(0), v(0))(5.21). 0 0 Second, doing the same on the level of the approximations, we ﬁnd Z T Z T hu00, vi + B[u, v; t] = lim hu00 , vi + B[u , v; t] ml ml 0 l→∞ 0 Z T = lim hv00, u i + B[u , v; t] dt + (u (0), v0(0)) − (u0 (0), v(0)) ml ml ml ml l→∞ 0 Z T = hv00, ui + B[u, v; t] dt + (g, v0(0)) − (h, v(0)). (5.22) 0 Since v(0) and v0(0) are arbitrary, we deduce from (5.21) and (5.22) that

u(0) = g and u0(0) = h.

2 1 Remark 5.7. Since for X = L or X = H0 , the set 2 {v ∈ L (0,T ; X) : ess sup kv(t)kX ≤ C} t is a bounded and convex subset of L2(0,T ; X), the energy estimates im- ∞ 1 0 ∞ 2 00 ply that in fact u ∈ L (0,T ; H0 (Ω)), u ∈ L (0,T ; L (Ω)) and u ∈ L2(0,T ; H−1(Ω)). Uniqueness. To show uniqueness of weak solutions to the linear equation (5.2) it suﬃces to show that the only weak solution for data f ≡ g ≡ h ≡ 0 is u ≡ 0. Remark 5.8. Uniqueness of weak solutions would be easy to obtain if we could use v := u0(t) as a test function (which we cannot because u0(t) is in 2 1 general only L but not H0 ). Assume that u is solution to the homogeneous wave equation

0 utt − ∆u = 0, with u(0) = 0, u (0) = 0.

58 [July 12, 2019] If u was suﬃciently regular, we could test the equation with u0 and integrate by parts to ﬁnd conservation of the energy, i.e., 1 d 0 = ku k2 + k∇uk2 . 2 dt t L2(Ω) L2(Ω) Since the energy vanishes at t = 0, we would get u ≡ 0, which would imply uniqueness of weak solutions.

The rigorous proof was not discussed in detail on June 16. For ﬁxed 0 ≤ s ≤ T , we set ( R s u(τ) dτ if t ≤ s v(t) := t 0 otherwise.

1 0 Then v(t) ∈ H0 (Ω) for each 0 ≤ t ≤ T , v (t) = −u(t) for t < s, and u0(0) = v(s) = 0. Thus we obtain from the equation, integrating by parts, that Z s 0 = hu00, vi + B[u, v; t] dt 0 Z s = −(u0, v0) + B[u, v; t] dt 0 Z s = hu0, ui − B[v0, v; t] dt. 0 Note that in case of the wave equation this equality can be rewritten as Z s 1 d 2 2 0 = kukL2(Ω) − k∇vkL2(Ω) 2 0 dt 1 = ku(s)k2 + k∇v(0)k2 , 2 L2(Ω) L2(Ω) and thus u ≡ 0. In the case of general hyperbolic equations, the t-dependence of the coeﬃcients and the non-symmetric part of B require more work. [10.5. 2019, Lecture 8] [13.5. 2019, Lecture 9]

5.1.2 Regularity Even though solutions of parabolic and of hyperbolic equations usually be- have very diﬀerent, the L2 regularity theory is very similar. The reason is that the theory is essentially based on energy estimates, obtained by multiplying the equation (or time derivatives of the equation) by ∂tu or higher time derivatives of u.

59 [July 12, 2019] To simplify the discussion we make the following additional assumptions for the rest of this subsubsection.

∂Ω ∈ C∞,

ij i ∞ a , b , c ∈ C (ΩT ), aij, bi and c are independent of t. Recall that we also assume that aij = aji. Moreover as in the parabolic case we let wk be the eigenfunctions of −∆ corresponding to the eigenvalues λ1 ≤ λ2 ≤ .... Theorem 5.9 (Improved regularity). (i) Assume that

1 2 2 2 g ∈ H0 (Ω), h ∈ L (Ω), f ∈ L (0,T ; L (Ω)). Let u be a weak solution of (5.1). Then in fact

∞ 1 0 ∞ 2 u ∈ L (0,T ; H0 (Ω)), u ∈ L (0,T ; L (Ω)) and

0 ess sup ku(t)kH1(Ω) + ku (t)kL2(Ω) t∈[0,T ] 0 ≤C kgk 1 + khk 2 + kfk 2 2 H0 (Ω) L (Ω) L (0,T ;L (Ω))

(ii) If, in addition,

2 1 0 2 2 g ∈ H (Ω), h ∈ H0 (Ω), f ∈ L (0,T ; L (Ω)). then

∞ 2 0 ∞ 1 00 2 2 u ∈ L (0,T ; H (Ω)), u ∈ L (0,T ; H0 (Ω)), u ∈ L (0,T ; L (Ω)) and

0 0 ess sup ku(t)kH2(Ω) + ku (t)kH1(Ω) + ku (t)kL2(Ω) t∈[0,T ] 0 ≤C kgk 2 + khk 1 + kfk 1,2 2 H (Ω) H0 (Ω) W (0,T ;L (Ω))

Remark. In (i) it actually suffices that f ∈ L1(0,T ; L2(Ω)). For the estimates of u, u0 and u00 in (ii) it suffices that, in addition f 0 ∈ L1(0,T : L2(Ω)). 0 0 To see this one use the estimate (um(t), f(t)) ≤ kum(t)kL2 kf(t)kL2 before applying Gronwall’s inequality and one applies a variant of Gronwall’s inequality to solutions of the differential inequality η0 ≤ Cη + h(t)η1/2 to estimate η(t) in terms of η(0) and khkL1(0,T ).

60 [July 12, 2019] Proof. (i) In the proof of Theorem 5.4 we have already shown that estimate

0 sup kum(t)kH1(Ω) + kum(t)kL2(Ω) ≤ C kgkH1(Ω) + khkL2(Ω) + kfkL2(0,T ;L2(Ω)) t∈[0,T ] 0 0

Passing to the limit m → ∞ we get assertion (i). 0 (ii) Diﬀerentiate the identity for um and writeu ˜m = um. This yields

00 0 (˜um, wk) + B[˜um, wk] = (f , wk) for 1 ≤ k ≤ m.

k 00 Multiplying by dm and summing over k we get

00 0 0 0 0 (˜um, u˜m) + B[˜um, u˜m] = (f , u˜m)

As in the proof of the energy estimate we deduce that d ku˜0 k2 + A[˜u , u˜ ] ≤ ku˜0 k2 + A[˜u , u˜ ] + kf 0k2 dt m L2(Ω) m m m L2(Ω) m m L2(Ω) (5.23) 2 Now let Pm denote the L orthogonal projection onto span {w1, . . . , wm}. Then

0 u˜m(0) = um(0) = Pmh, 0 00 u˜m(0) = um(0) = Pm(−Lum(0) + f(0)) = Pm(−LPmg − f(0)).

2 1 2 1 Since Pm is bounded on L , H0 and H ∩H0 (see the discussion of parabolic equations) we get

0 ku˜m(0)k 1 ≤ Ckhk 1 , ku˜ (0)k 2 ≤ Ckgk 2 + kf(0)k 2 . H0 (Ω) H0 (Ω) m L (Ω) H (Ω) L (Ω) By the Sobolev embedding theorem in one dimension we have

0 kf(0)k ≤ C(kfkL2(0,T ;L2(Ω)) + kf kL2(0,T ;L2(Ω))).

Thus (5.23) and the Gronwall inequality yield

00 2 0 2 sup kum(t)kL2(Ω) + k∇umkH1(Ω) t∈[0,T ] 0 0 2 ≤ sup ku˜m(t)kL2(Ω) + A[˜um(t), u˜m(t)] t∈[0,T ] ≤C kgk 2 + khk 1 + kfk 1,2 2 H (Ω) H0 (Ω) W (0,T ;L (Ω))

Passing to the limit m → ∞ we obtain the bounds for u00 and u0. The bound for u then follows from the equation Lu = −u00+f and elliptic regularity.

Higher regularity can then be obtained by induction as in the parabolic case.

61 [July 12, 2019] Theorem 5.10 (Higher regularity). Let m ∈ N. Assume that

g ∈ Hm+1(Ω), h ∈ Hm(Ω),

dk ∈ L2(0,T ; Hm−k(Ω)) for k = 0, 1, . . . , m. dtk Suppose also that the m-th order compatibility conditions hold.

1 g0, h1, g2, h3, . . . , gm ∈ H0 (Ω) if m = 2l, l ∈ N,

1 g0, h1, g2, h3, . . . , hm ∈ H0 (Ω) if m = 2l + 1, l ∈ N, where d2l−2 g := g, g := (0, ·) − Lg , 0 2l dt2l−2 2l−2 d2l−1 h := h, h := (0, ·) − Lh . 1 2l+1 dt2l−1 2l−1 Then dku ∈ L∞(0,T ; Hm+1−k(Ω)) for k = 0, 1, . . . , m + 1, dtk and we have the the corresponding estimate.

Theorem 5.11 (C∞ regularity). Assume that

∞ ∞ g, h ∈ C (Ω), f ∈ C (ΩT ) and the m-th order compatibility conditions hold for all m ∈ N. Then the hyperbolic initial-boundary value problem has a unique solution

∞ u ∈ C (ΩT ).

Proof. Apply theorem (5.10) for all m ∈ N.

5.1.3 Finite speed of propagation See [Ev], Chapters 2.4.3b and 7.2.4.

Proposition 5.12 (Finite speed of propagation for the wave equation). 2 n Suppose that u ∈ C (R × (0, ∞)) solves the wave equation utt − ∆u = 0 in n n R × (0, ∞). Let x0 ∈ R and t0 > 0, and set

C := {(x, t) : 0 ≤ t ≤ t0, |x − x0| ≤ t0 − t}.

If u ≡ 0 and ut ≡ 0 on B(x0, t0) × {t = 0}, then u ≡ 0 in C. Proof. See Satz 3.9 of Einf¨uhrungin die partiellen Diﬀerentialgleichungen.

62 [July 12, 2019] Theorem 5.13 (Finite speed of propagation for variable coefficients). As- sume that u is a smooth solution of the hyperbolic equation n X ij n utt − a (x)uxixj = 0 in R × (0, ∞), i,j=1 with coefficients independent of time, smooth and symmetric, i.e., aij = aji. n n Let (x0, t0) ∈ R × (0, ∞). Suppose that q : R → R satisfies n X ij n a qxi qxj = 1 in R \{x0}, i,j=1 n q > 0 in R \{x0}, q(x0) = 0, n and q smooth in R \{x0}. Define

K := {(x, t): q(x) < t0 − t}, and for t > 0 set Kt := {x : q(x) < t0 − t}.

If u ≡ 0 and ut ≡ 0 on K0, then u ≡ 0 in K.

Proof. Exercise. See Homework 6.

n Remark 5.14. Note that by deﬁnition Dq 6= 0 in R \{x0}, and thus ∂Kt is a smooth (n − 1)-dimensional submanifold for 0 ≤ t < t0.

5.2 First order hyperbolic systems This subsection is taken from [Ev], Chapter 7.3.

5.2.1 Deﬁnitions

n m We look for maps u : [0, ∞) × R → R which satisfy the ﬁrst order system of equations j X n ∂tu + Bj∂xj u = f in (0, ∞) × R (5.24) i=1 and the initial condition

n u(0, ·) = g on R . (5.25) Here the unknown is u and the functions

n m n m n m×m f : [0, ∞) × R → R , g : R → R and B : [0, ∞) × R → R are given.

63 [July 12, 2019] n Notation For each y ∈ R set n X n B(t, x; y) := yjBj(t, x), (t, x) ∈ [0, ∞) × R . j=1 Deﬁnition 5.15 (Hyperbolic system). The system of PDE (5.24) is called hyperbolic if the m × m matrix B(t, x; y) is diagonalizable (over R) for each n x, y ∈ R and each t ∈ [0, ∞). In other words (5.24) is hyperbolic if for each x, y, t the matrix B(t, x; y) has m real eigenvalues

λ1(t, x; y) ≤ . . . λm(t, x; y) and the corresponding eigenvectors ri(t, x; y) with i = 1, . . . , m form a basis m of R . Deﬁnition 5.16. (i) We say that (5.24) is a symmetric hyperbolic system if Bj(t, x) is a symmetric matrix for each t, x and each j = 1, . . . , n. n (ii) We say that (5.24) is a strictly hyperbolic system if for each x, y ∈ R with y 6= 0 and each t ∈ [0, ∞) the matrix B(t, x; y) has m distinct real eigenvalues λ1 < . . . < . . . [13.5. 2019, Lecture 9] [17.5. 2019, Lecture 10]

5.2.2 Symmetric first order systems In my view the heart of the matter is the following. If we view the equation n X ∂tu = − Bj∂xj u + f j=1 as an abstract ODE n 0 X u = Au + f, with A = Bj∂xj j=1 then we cannot use the usual Picard-Lindelöfargument to prove existence because A is not bounded in L2. What saves us is the fact that ∂j is a skew-symmetric operator (defined 2 on a dense subspace in L ) and hence for symmetric Bj the operator Bj∂j is skew-symmetric up to a bounded operator Z 1 (Bju, u) = (∂xj Bju, u) + ∂xj (Bu · u) dx, Rn 2 | {z } =0

64 [July 12, 2019] see Step 2 in the proof of Theorem 5.19 for a rigorous statement and justiﬁca- tion. Hence when testing the equation with u the operator A behaves like a bounded operator and hence we can get an estimate for ku(t)kL2 by a Gron- wall inequality. The term ∂xj Bj arises as the commutator of the symmetric multiplication operator Bj and the skew-symmetric operator ∂xj . Indeed if A and B are bounded operators, A is symmetric and B is skew-symmetric then then twice the symmetric part of the product AB is given by

AB + (AB)T = AB + BT AT = AB − BA = [A, B].

One cannot naively extend such abstract calculations to unbounded operators but they often provide a good guide for a rigorous proof. From now on we assume that

n the matrices Bj(t, x) are symmetric for all (t, x) ∈ [0, ∞) × R and all j = 1, . . . n.

We also assume that

2 n m×m Bj ∈ C ([0, ∞) × R ; R ) and 2 M := sup sup |Bj| + |Dt,xBj| + |Dt,xBj| < ∞ j [0,∞)×Rn and 1 n m 1 n m f ∈ H ((0,T ) × R ; R ), g ∈ H (R ; R ). Note that

1 n m 2 1 n m 0 2 2 n m f ∈ H ((0,T )×R ; R ) ⇔ f ∈ L (0,T,H (R ; R )) and f ∈ L (0,T,L (R ; R )).

To simplify the notation we deﬁne the bilinear form

n Z X B[u, v; t] := Bj(·, t)∂xj u · v dx n R j=1

1 n m for t ∈ [0,T ] and u, v ∈ H (R ; R ). Deﬁnition 5.17. We say that

2 1 n m 0 2 2 n m u ∈ L (0,T ; H (R ; R )) with u ∈ L (0,T ; L (R ; R )) is a weak solution of the initial value problem (5.24)– (5.25) provided

0 1 n m (i) (u , v) + B(u, v; t) = (f, v) for each v ∈ H (R ; R ) and a.e. t ∈ [0,T ] and

(ii) u(0) = g

65 [July 12, 2019] 2 n m Here and afterwards ( , ) denotes the scalar product in L (R ; R ). 2 n m Note that in view of Theorem 4.32 we have u ∈ C([0,T ]; L (R ; R ) and hence the initial condition (ii) makes sense. To show existence we approximate (5.24)– (5.25) be the parabolic initial- value problem

ε ε Pn ε n ∂tu − ε∆u + j=1 Bj∂xj u = f in (0,T ] × R ε ε n (5.26) u (0, ·) = g on R for 0 < ε ≤ 1 and gε = ηε ∗ g where ηε is the standard molliﬁer. We ﬁrst show that for each ε > 0 has a unique solution (which converges to zero as |x| → ∞ in an appropriate sense). Then we establish uniform estimates for the uε and show that their weak limit is a weak solution of the hyperbolic initial-value problem (5.24)–(5.25).

Theorem 5.18 (Existence of approximate solutions). For each ε > 0 there exists a unique solution uε of (5.26) with

ε 2 3 n m ε0 2 1 n m u ∈ L (0,T ; H (R ; R )), u ∈ L (0,T ; H (R ; R ).

∞ 1 n m Proof. This is proved by a ﬁxed point argument in the space X = L (0,T ; H (R ; R )) using the estimates for the heat kernel Φε (which is the distributional solution of ∂u − ε∆u = 0 and u(0, ·) = δ0) In particular we use the estimate

1 kΦ (t − s, ·) ∗ h(s, ·)k 1 n ≤ C(ε) + 1 kh(s, ·)k 2 n . ε H (R ) (t − s)1/2 L (R )

For the details see [Ev], Section 7.3.2, Theorem 1. The regularity of uε0 follows from energy estimates completely analogous to those in Theorem 4.7 (ii)4 The the regularity of uε follows from elliptic regularity.

Theorem 5.19 (Energy estimates). The exists a constant C, depending only on n, T and the bound M on the coeﬃcients Bj, such that

ε ε 0 max ku (t)kH1( n; m)+ku (t) kL2(0,T ;L2( n; m)) ≤ kgkH1( n; m)+kfkH1((0,T )× n; m) t∈[0,T ] R R R R R R R R for each ε ∈ (0, 1].

4 ∞ n ∞ n Assume that f ∈ Cc ([0,T ] × R ) and g ∈ C (R ). Then the initial-value problem ∂tu − ε∆u = f, u(0, ·) = g has a unique solution and u and all time derivatives belong 2 m n to L (0,T ; H (R )) for all m ∈ N as can be easily seen by the representation of u using the heat kernel. Now diﬀerentiate the equation with respect to xk and multiply by 2 2 n 2 ∂ ∂ u. This gives the bound for ∂ ∂ u in L (0,T ; L ( )) in terms of k∇ gk 2 n and t xk k t R L (R ) kfk 2 1 n . L (0,T ;H (R ))

66 [July 12, 2019] ε Proof. Step 1. We ﬁrst aim for an estimate for ku (t)kL2 by Gronwall’s inequality and compute

n d 1 X kuεk2 = (uε, uε0) = (uε, f − B ∂ uε + ε∆uε). dt 2 L2(Rn;Rm) j xj j=1

Now 1 1 |(uε, f)| ≤ kuεk2 + kfk2 2 L2(Rn;Rm) 2 L2(Rn;Rm) and (uε, ε∆uε) = −ε(∇uε, ∇uε) ≤ 0.

Step 2. Estimate for the term with spatial derivatives. ∞ n m Let v ∈ Cc (R ; R ). Then

 n  n X X Z v, Bj∂xj v = (Bj∂xj v) · v dx n j=1 j=1 R n n 1 X Z 1 X Z = ∂xj ((Bjv) · v) dx − (∂xj Bj)v · v) dx 2 n 2 n j=1 R j=1 R where we used the symmetry of Bj in the last identity. Since v has compact support it follows that   n n Z X 1 X 2 v, Bj∂xj v ≤ (∂xj Bj)v · v) dx ≤ nMkvkL2( n; m). 2 n R R j=1 j=1 R By approximation we therefore have   n ε X ε ε 2 u , Bj∂xj u  ≤ Cku k 2 n m . L (R ;R ) j=1 Using the estimates in Step 1 we get

d 1 kuεk2 ≤ C kuεk2 + kfk2 . dt 2 L2(Rn;Rm) L2(Rn;Rm) L2(Rn;Rm)

By Gronwall’s inequality we deduce

ε 2 2 2 max ku (t)kL2( n; m) ≤ C kgkL2( n; m) + kfkL2(0,T );L2( n; m)) . t∈[0,T ] R R R R R R

ε since kg kL2(Rn;Rm) ≤ kgkL2(Rn;Rm).

67 [July 12, 2019] Step 3. We derive estimates for ∇xu by diﬀerentiating the equation with respect to x.

Step 4. We derive estimates for ∂tu by diﬀerentiating the equation with respect to t.

Theorem 5.20 (Existence of weak solutions). Under the assumptions stated the initial-value problem has a weak solution.

Sketch of proof. By the energy estimates there exists a subsequence εk → 0 such that

ε 2 1 n m u k * u in L (0,T ; H (R ; R )), ε 0 0 2 2 n m u k * u in L (0,T ; L (R ; R )).

1 1 n m Now we have for every v ∈ C ([0,T ]; H (R ; R )) Z T Z T ε0 ε (u , v) + ε(Dxu ,Dxv) + B(u, v, t) dt = (f, v) dt. 0 0

Passing to the limit εk → 0 we get Z T Z T (u0, v) + B[u, v, t] dt = (f, v) dt. 0 0 1 n m 1 With the choice v(t) = ϕ(t)w where v ∈ H (R ; R ) and ϕ ∈ C ([0,T )) we get

0 1 n m (u (t), w) + B[u(t), w, t] = (f, w) ∀w ∈ H (R ; R ) and a.e. t ∈ [0,T ]. (5.27) R T ε0 To see that u(0) = g assume that v(T ) = 0, integrate the term 0 (u , v) dt 1 n m by parts, pass to the limit εk → 0 and use that gεk → g in H (R ; R ) to deduce that Z T Z T −(g, v(0)) + −(u, v0) + B(u, v, t) dt = (f, v) dt. 0 0

R T 0 Integrating 0 −(u, v ) dt by parts and using (5.27) we get −(g, v(0)) + 1 n m 1 (u(0), v(0)) = 0. Since v(0) ∈ H (R ; R ) is arbitrary and since H is dense in L2 it follows that u(0) = g.

We now prove uniqueness of weak solution, even including lower order terms. We say that u is a weak solution of the initial value problem

∂ u + Pn B ∂ u + Du = f in (0,T ] × n, t j=1 j t R (5.28) u(0, ·) = g

68 [July 12, 2019] if u is as in Deﬁnition 5.17 with condition (i) replaced by (u0, v) + B[u, v, t] + (Du, v) = (f, v). . Theorem 5.21 (Uniqueness of weak solutions). Assume that B ∈ C1([0,T ]× n m×m 0 n m×m R ; Rsym ), D ∈ C ([0,T ] × R ; R ) with T M1 := kdiv BkL∞ + kD + D kL∞ < ∞. 2 2 n m 2 n Let f ∈ L (0,T ; L (R ; R )), g ∈ L (R ). Then there exists at most one weak solution of the initial value problem (5.28). Moreover

1 (M1+1)T sup ku(t)kL2 ≤ e 2 (kgkL2 + kfkL2(0,T ;L2)). (5.29) t∈[0,T ] Proof. It suffices to prove that every weak solution satisfies (5.29) Applying this estimate to the difference of two solutions (which is a weak solution with f = 0 and g = 0) we obtain the uniqueness result. To prove (5.29) first 2 1,1 note that by Theorem 4.33 the map t 7→ ku(t)kL2 is in W (0,T ) and for a.e. t its derivative is 2(u, u0). In Step 2 in the proof of Theorem 5.19 we have shown that 1 X 1 B[v, v, t] = − ( ∂ B v, v) = − (div B v, v) ∀v ∈ H1. 2 xj 2 j Thus, for a.e. t, d ku(t)k2 = 2(u, u0) = −2B[u, u, t] − 2(Du, u) + 2(f, u) dt L2 T 2 2 2 ≤ kdiv B − (D + D )kL∞ ku(t)kL2 + kf(t)kL2 + ku(t)kL2 . The assertion follows from the Gronwall inequality. [17.5. 2019, Lecture 10] [20.5. 2019, Lecture 11]

Corollary 5.22 (Smooth solutions). Assume that 2 ∞ n m×m Bj ∈ C ([0,T ]; C (R ; Rsym )) with j k sup |Dt DxB| ≤ Cj,k for all j ≤ 2 and k ∈ N [0,T ]×Rn and 1 l n l n f ∈ C ([0,T ]; H (R )), g ∈ H (R ) for all l ∈ N. Then the unique weak solution of the initial value problem (5.28) satisifes 2 l n m u ∈ C ([0,T ],H (R ; R )) for all l ∈ N. 2 ∞ n m In particular u ∈ C ([0,T ]; C (R ; R )).

69 [July 12, 2019] Sketch of proof. This was not discussed in class We consider again solutions uε of the parabolic regularisation

n ε ε X ε ε ∂tu − ε∆u + Bj∂xj u = f, u (0, ·) = g. j=1

We have already seen that there exists a solution uε ∈ L2(0,T ; H3) with uε0 ∈ L2(0,T ; H1) and that uε converges weakly in L2(0,T,H1) to a weak solution of n X ∂tu + Bj∂xj u = f, u(0, ·) = g. j=1 In view of the regularity of uε, B, f and g we may diﬀerentiate the equa- ε tion for u with respect to xk and we obtain the following system for the i,ε i,ε derivatives vk = ∂xk u

n ε ε X ε ε ε ∂tv − ε∆v + Bj∂xj v + Lv = Dxf, v (0, ·) = Dxg j=1

i Pn i where (Lv)k = j=1(∂xk Bj)vj. One can now derive a priori estimates for vε in the same way as for uε (the extra term Lv poses on diﬃculty in applying Gronwall’s inequality) and we conclude that vε converges weakly in L2(0,T,H1) to a weak solution of

n X ∂tv + Bj∂xj v + Lv = Dxf, v(0, ·) = Dxg. j=1

s By induction one sees that for each s ∈ N the collection Dxu of all spatial derivatives of order s satisﬁes is a weak solution of n s X s s s ˜ s ∂tDxu + Bj∂xj Dxu + LDxu = Dxf + Lu, v(0, ·) = Dxg, j=1

s where L(Dxu) is a linear combination of the partial derivatives of order s ˜ with coeﬃcients ∂xk Bj and where Lu is a linear combination of terms of the form ∂αB ∂βu with |β| ≤ s − 1 and |α| + |β| = s + 1. Inductive application 2 of the estimate (5.29) to the equations for u, Dxu, Dxu, etc. shows that

2 s u ∈ L (0,T ; H ) for all s ∈ N.

It follows from the equation for u and the regularity of the coefficients that 2 s 0 s ∂tu ∈ L (0,T ; H ) for all s and hence u ∈ C ([0,T ]; H ). Using the equation again we see first that u ∈ C1([0,T ]; Hs) and finally that u ∈ C2([0,T ]; Hs).

70 [July 12, 2019] Remark 5.23. The results on existence and uniqueness can be extended to systems of the form

n X B0∂tu + Bj∂xj u + Du = f j=1 provided that B0 is symmetric and we assume in addition that there exists α > 0 such that

2 n m (B0(t, x)ξ, ξ) ≥ α|ξ| ∀(t, x) ∈ [0,T ] × R , ξ ∈ R and

2 n 2 B0 ∈ C ([0,T ] × R ) with sup |B0| + |Dt,xB0| + |Dt,xB0| < ∞, [0,T ]×Rn

1 n D ∈ C ([0,T ] × R ) with sup |D| + |Dt,xD| < ∞. [0,T ]×Rn Indeed existence of solutions of the parabolic regularisation can be proved, e.g., by discretization in time and the apriori estimates can be derived in the d same way as in the case B0 = Id by computing dt (B0u, u) (see also below). The passage to the limit can then be carried out as before. The apriori estimate (5.29) takes the form

−1/2 Mt sup ku(t)kL2 ≤ α e (sup |B0|kgkL2 + kfkL2(0,T ;L2)). (5.30) t∈[0,T ] where now n 1 T X M := (kdiv Bk ∞ + kD + D k ∞ + 1), div B := ∂ B + ∂ B . 2α L L t 0 xj j j=1

d The estimate (5.30) can be easily veriﬁed by calculating dt (B0u, u) and using Gronwall’s inequality. d (B u, u) = ((∂ B )u, u) − 2B[u, u, t] − 2(Du, u) + 2(f, u) dt 0 t 0 = (div B − (D + DT )u, u) + 2(f, u) ≤ 2αM(u, u) + kfk2 2 ≤ 2M(B0u, u) + kfk .

Gronwall yields

2Mt 2 (B0u, u) ≤ e ((Bg, g) + kfkL2(0,T ;L2)). or 1 (u, u) ≤ e2Mt(sup |B | kgk2 + kfk2 ). α 0 L2(0;TL2)

71 [July 12, 2019] Remark. (DiPerna-Lions theory) The fact that the apriori estimate for u involves only div B and not the full derivative of DxB allows one to deﬁne a good notion of solutions for ODEs X˙ = b(X) for a.e. initial value for vectorﬁelds which are not Lipschitz but only satisfy the conditions b ∈ W 1,1 and div B ∈ L∞, see [DPL89]. This is one of three crucial ingredients in the proof of the global existence of weak solutions of the Boltzmann equation by DiPerna and P.L. Lions. For recent improvements and a number of interesting open questions see [Am04].

5.3 Local existence in time existence for nonlinear hyperbolic systems with smooth initial data This section is based on [Ma84]. Let m ∞ m G ⊂ R open, Fj ∈ C (G; R ), j = 1, . . . , n. We seek a map n m u : [0,T ] × R → R which solves the initial value problem

n X n ∂tu + ∂xj Fj(u) = 0, in (0,T ) × R , (5.31) j=1 n u(0, ·) = g in R (5.32)

n m where the initial data g : R → R are given. Equation (5.31) is called a conservation law. Set ˜ ∞ m×m Aj := (∇Fj) ∈ C (G, R ) 1 m Then u ∈ C ([0,T ] × R ; G) is a solution of if and only if

n X ˜ n ∂tu + Aj(u)∂xj u = 0, in (0,T ) × R ,. (5.33) j=1

In order to be able to apply the theory of symmetric hyperbolic systems we impose the following symmetrization condition.

Deﬁnition 5.24. We say that the conservation law (5.31) satisﬁes condition ∞ m×m (S) if t here exists A0 : C (G; R ) and α > 0 such that

2 1 2 m (i) α|ξ| ≤ A0(y)ξ · ξ ≤ α |ξ| ∀ξ ∈ R , y ∈ G,

(ii) A0A˜j is symmetric for all y ∈ G. | {z } =:Aj (y)

72 [July 12, 2019] If condition (S) holds then (5.31) is equivalent to the nonlinear symmetric hyperbolic system

n X n A0(u)∂tu + Aj(u)∂xj u = 0, in (0,T ) × R ,. (5.34) j=1

5.3.1 Main result Theorem 5.25 (Lax/ Majda). Assume that condition (S) holds. Assume n G1 ⊂⊂ G is open, s > 2 + 1 and

s n n g ∈ H (R ), g(x) ∈ G1 ∀x ∈ R .

Let G2 be open with G1 ⊂⊂ G2 ⊂⊂ G. Then there exists a T > 0 (which may depend on Aj, G1, G2, g, s, m and n) such that the initial-value problem 1 n (5.34), (5.32) has a unique classical solution u ∈ C ([0,T ] × R with

n ∞ s n m 1,∞ s−1 n m u(t, x) ∈ G2 ∀(t, x) ∈ [0,T ]×R , u ∈ L ([0,T ]; H (R ; R ))∩W ([0,T ]; H (R ; R )).

Remark. Various reﬁnements are possible:

(i) The existence time depends on g only throught kgkHs ,

0 s n m 1 s−1 n m (ii) u ∈ C ([0,T ]; H (R ; R )) ∩ C ([0,T ]; H (R ; R )), (iii) One can characterize the maximal existence time

T = sup{T : ∃ solution u ∈ C0([0,T ]; Hs)}, see Section 5.3.2.

Proof of Theorem 5.25. The assertion is proved by the method of successive approximation (see Step 2.), i.e., a ﬁxed-point argument. There is an interesting twist, however, since one cannot directly prove contraction in the natural space. Instead one shows boundedness in the natural space and contraction in a weaker space (see Steps 3 and 4). This suﬃces to conclude. For simplicity we assume that

m G = R , G1 = B(0, 1), i.e, the initial data satisfy |g| < 1,

G2 = B(0, 2), i.e, we seek a solution with |u| < 2,

A0 = Id .

73 [July 12, 2019] Step 1. Smoothing of initial data. Set x g = η ∗ g, η (x) = ε−nη( ) ε ε ε ε ∞ s where η ∈ Cc (B(0, 1)) is the standard molliﬁer. If g ∈ H then

lim kηεu − ukHs = 0, kgεu − ukL2 ≤ CεkukH1 . ε→0 Set −k k εk = 2 ε0, g = ηεk .

Step 2. Iteration scheme Deﬁne a sequence of approximate solutions as follows (i) u0(t, x) = g0(x) ∀t ∈ [0, ∞);

(ii) uk+1 is the solution of the linear symmetric hyperbolic initial-value problem ∂ uk+1 + Pn A (uk)∂ uk+1 = 0 in (0,T ) × n, t j=1 j xj k+1 R (5.35) uk+1(0, ·) = gk.

Note that such a solution exists for suﬃciently small Tk+1 by the existence and regularity theory in the previous subsection. One may take Tk+1 as the largest time such that the condition uk+1(t, x) ∈ B(0, 2) holds for all n t < Tk+1 and all x ∈ R . We need to show in the course of the proof that there exists a T∗ > 0 such that Tk+1 ≥ T∗ for all k.

Step 3. Bound in high norms.

Lemma 5.26. There exist constants L1 > 0, L2, T∗ > 0 such that k+1 0 max ku (t) − g kHs ≤ L1, t∈[0,T∗] k+1 0 max ku (t) − g kC0 ≤ 1, t∈[0,T∗] k+1 max k∂tu kHs−1 ≤ L2. t∈[0,T∗]

Step 4. Contraction in low norms.

P Lemma 5.27. There exist T∗∗ ∈ (0,T∗], θ ∈ (0, 1) and βk with k |βk| < ∞ such that

k+1 k k k−1 max ku (t) − u (t)kL2 ≤ θ max ku (t) − u (t)kL2 + βk. (5.36) t∈[0,T∗] t∈[0,T∗]

74 [July 12, 2019] Step 5. Conclusion of the argument. We pass to the limit by interpolating between the convergence in low norms and the boundedness in high norms. Set T = T∗∗. From Steps 3 and 4 we get5 k 0 2 k u → u in C ([0,T ]; L ), max ku (t)kHs ≤ L. t∈[0,T∗] s n m Now (see below for a proof) for any function v ∈ H (R ; R )

1−λ s λ n/2 kvk 0 ≤ Ckvk kD vk with λ = , (5.37) C L2 L2 s 1−µ s µ 1 + n/2 kD vk 0 ≤ Ckvk kD vk with µ = (5.38) x C L2 L2 s Thus uk is a Cauchy sequence in C0([0,T ]; C1) and

uk → u in C0([0,T ]; C1).

From the equation for uk+1 it follows that

k 0 0 ∂tu → ∂tu in C ([0,T ]; C ).

1 n Thus u ∈ C ([0,T ] × R ) and u solves the equation (5.34). Finally the ∞ s ∞ s−1 bounds for u in L (0,T ; H ) and ∂tu in L (0,T ; H ) in follow from Step 3 by letting k → ∞ and using to weak* convergence in L∞(0,T ; X). To prove (5.37) we use Fourier transform and the Cauchy-Schwarz inequality in Fourier space Z (2π)n sup |v| ≤ |vˆ| dξ Rn Rn Z Z = |vˆ| dξ + |ξ|−s |ξ|s|vˆ| dξ B(0,R) Rn\B(0,R) n/2 n/2−s s ≤CR kvkL2 + CR kD vkL2 .

s s This holds for every R > 0. Now optimize in R by taking R = kD vkL2 /kvkL2 . The estimate (5.38) is derived in the same way using that Z (2π)n sup |∇v| ≤ |ξvˆ| dξ Rn Rn n/2+1 n/2+1−s s ≤CR kvkL2 + CR kD vkL2 .

5To see that uk is a Cauchy sequence in C0([0,T ]; L2) one can, e.g., argue as follows. l+1 l l 1 0 Pl l−m First one shows by induction that ku − u k ≤ θ ku − u k + m=1 θ βm The last P∞ term in the estimate can be rewritten as m=1 χm≤lθl−m|βm Summing this term in l from 1 to L one gets

∞ L ∞ X X l−m X 1 1 X χ θ |β | ≤ |β | ≤ B, where B = |β | < ∞. l≥m m 1 − θ m 1 − θ m m=1 l=1 m=1 m P∞ l+1 l −1 1 0 Hence l=1 ku − u k ≤ (1 − θ) (ku − u k + B).

75 [July 12, 2019] [20.5. 2019, Lecture 11] [24.5. 2019, Lecture 12]

Step 6. Uniqueness. This follows easily by writing down the equation for the diﬀerence of two solutions, noting that the Aj are Lipschitz on B(0, 2) and using the Gronwall estimate (5.29) for the linear equation for a suﬃciently short time interval [0,T1]. 0 1 n Details: Assume that u andu ˜ are solutions in C ([0,T ],C (R )) ∩ 1 0 n C ([0,T ]; C (R )) with sup |u| ≤ 2 and sup |u˜| ≤ 2. Set w =u ˜ − u. Then n n X X ∂tw + Aj(u)∂xj w = f, with f = (Aj(˜u) − Aj(u)) ∂xj u,˜ j=1 j=1 w(0, ·) = 0.

Set Bj = Aj ◦ u. Then M := sup |div B| < ∞. Moreover

|Aj(˜u) − Aj(u)| ≤ sup |DA| |w| ≤ C|w|. B(0,2)

Let 0 < T1 ≤ T . Then by the Gronwall estimate (5.29) for the linear equation

1 1 (M+1)T1 (M+1)T1 1/2 sup kw(t)kL2 ≤ e 2 kfkL2(0,T ;L2) ≤ Ce 2 T1 sup kw(t)kL2 . t∈[0,T1] t∈[0,T1]

Thus for T1 suﬃciently small we get

sup kw(t)kL2 = 0. t∈[0,T1]

Now we can repeat the argument on [kT1, (k +1)T1] as long as (k +1)T1 ≤ T and finally on [KT1,T ]. Proof of Lemma 5.26. The key idea to get Hs estimates is to take s derivatives of the equation for uk+1 with respect to x. This yields an equation for each derivative ∂αuk+1 to which we can apply the theory linear symmetric systems. Since differentiation and multiplication by a coefficient do not commute exactly an additional term arises, but this term is of lower order. k We first note that for each k there exists times Tk > 0 such that u ∈ 2 l n C ([0,Tk; H ) for all l ∈ N and |u| ≤ 2 in [0,Tk] × R . This is easily seen by 0 0 induction. The assertion clearly holds for u ≡ g and every T0 > 0. If the k assertion holds for k then by Corollary 5.22 (applied with Bj = Aj ◦ u ) the k+1 2 l solution u exists on [0,Tk] and belongs to C ([0,Tk]; H ) for all l ∈ N. In particular we have Z t k+1 k+1 k+1 k+1 k+1 ku (t) − g kC0 ≤ ku (t) − g kHs−1 ≤ k∂tu kHs−1 dt ≤ Ckt. 0

76 [July 12, 2019] k+1 It follows that ku (t)kC0 ≤ 2. Thus the desired assertion holds for k + 1.

Now we will prove a more quantitative statement by induction. In particular we will show that the times Tk are uniformly bounded from below.

−1 Claim. Let L ≤ min(CS , 1) where CS is the Sobolev constant CS = supf6=0 kfkC1 /kfkHs . Then there exist an ε0 > 0 and T∗ > 0 such that

k 0 max ku (t) − g kHs ≤ L ∀k ∈ N (5.39) t∈[0,T∗] k 0 max ku (t) − g kC0 ≤ 1 (5.40) t∈[0,T∗]

We prove the claim by induction. The case k = 0 is trivial u0 − g0 = 0. Assume that the estimates hold for k. Note that (5.39) implies (5.40) and k (5.40) implies that sup |u (t)| ≤ 2 for t ≤ T∗. It thus suﬃces to show that (5.39) holds for k + 1. We will show below that on (0,T ∗]

d k+1 0 2 k 2 k+1 0 2 k s 2 0 2 ku −g k s ≤ C(1+ku (t)k s ) ku (t)−g k s +C(1+ku (t)k s ) kg k . dt H H H H Hs+1 (5.41) Together with Gronwall’s inequality this implies that

k+1 0 MT∗ k+1 0 1/2 0 sup ku − g kHs ≤ e (kg − g kHs + CMT∗ kg kHs+1 ) t∈[0,T∗] where k s M = C(1 + sup ku (t)kHs ). t∈[0,T∗] 0 k 0 Since kg kHs ≤ kgkHs and ku (t) − g kHs ≤ L ≤ 1 by induction assumption we get s M ≤ C(1 + (1 + kgkHs ) ) =: M0 Now we have

k+1 0 k+1 0 kg − g kHs ≤ kg − gkHs + kg − gkHs ≤ 2ω(ε0) where ω(ε) := sup kηδ ∗ g − gkHs δ∈(0,ε)

Recall from Step 1 that limε→0 ω(ε) = 0. Moreover

0 C kg kHs+1 ≤ kgkHs . ε0 Now choose

77 [July 12, 2019] L (i) ε0 so small that ω(ε0) ≤ 4 ,

1/2 C L (ii) T > 0 so small that T M kgk s ≤ , 1 1 ε0 0 H 4 (iii) T = min( ln 2 ,T ) ∗ M0 1 Then k+1 0 L L sup ku − g kHs ≤ 2( + ) ≤ L. t∈[0,T∗] 4 4 This ﬁnishes the proof of the claim modulo the proof of (5.41). To prove (5.41) we simplify the notation and set

k+1 0 k v := u − g , u := u ,Bj = Aj ◦ u

Then n X ∂tv + Bj∂xj v = H, j=1 v(0, ·) = gk+1 − g0 where n X 0 H := − (Aj ◦ u)∂xj g . j=1 0 l k Note that g = ηε0 ◦ g ∈ H for all l ∈ N. Since u is also smooth it follows 0 l that H ∈ C ([0,Tk]; H ) for all l ∈ N. We have

DxBj = (DAj) ◦ u Dxu.

n Since u(t, x) ∈ B(0, 2) for t ≤ T∗ and since Aj is smooth in R it follows that sup |DxBj| ≤ C sup |Dxu| ≤ C max ku(t)kHs < ∞ t∈[0,T∗]

Multiplying the equation for ∂tv by v we get d kvk2 ≤ (sup |div B| + 1) kvk2 + kHk2 dt L2 L2 L2 2 2 0 2 ≤(C max ku(t)kHs + 1) kvkL2 + C(|A(0)| + max ku(t)kHs ) kg kHs . t∈[0,T∗] t∈[0,T∗]

To get estimates for spatial derivatives of v we diﬀerentiate the equation. This gives

n α X α ∂t∂ v + Bj ∂xj ∂ v = −Fα + Hα (5.42) j=1 ∂αv(0, ·) = ∂α(gk+1 − g0)

78 [July 12, 2019] where n α X 0 Hα = −∂ (Aj ◦ u)∂xj g , j=1 α α Fα = ∂ ((Aj ◦ u)∂xj v) − (Aj ◦ u)∂ ∂xj v.

The main point is that Fα has the structure of a commutator and thus the L2 norm of F α can be estimated using only |α| derivatives of v and not |α| + 1 derivatives. In fact the ’calculus inequalities’ in Lemma 5.28 below and the estimate al ≤ 1 + as for a ≥ 0 and 0 ≤ l ≤ s give the key estimates s kFα(t)kL2 ≤ C(1 + ku(t)kHs ) kv(t)kHs ∀|α| ≤ s (5.43) and s 0 kHα(t)kL2 ≤ C(1 + ku(t)kHs ) kg kHs+1 (5.44) Multiplying the equation (5.42) by ∂αv and summing over all |α| ≤ s we get the desired energy estimate

d 2 s 2 2 s 2 0 2 kvk s ≤ C(1 + ku(t)k s ) kv(t)k s + C(1 + ku(t)k s ) kg k . (5.45) dt H H H H Hs+1 This completes the proof of the claim. k+1 To ﬁnish the proof of Lemma 5.26 we only need to estimate ∂tu 0 s−1 in C ([0,T∗],H ). This estimate follows directly from the equation for k+1 ∂tu and the calculus inequalities in Lemma 5.28 below.

k+1 k k Proof of Lemma 5.27. Let w = u −u and recall that Bj = Aj ◦u Then n X X k k−1 k ∂tw + Bj∂xj w = f := (Aj ◦ u − Aj ◦ u )∂xj u , j=1 j=1 w(0) = gk+1 − gk.

For t ∈ [0,T∗] the right hand side is bounded by k k−1 k 0 k k−1 kf(t)kL2 ≤ sup |DA| k(u −u )(t)kL2 kDxu (t)kC0 ≤ C(L+kg kHs )k(u −u )(t)kL2 B(0,2) where L is the constant in (5.39). Moreover k |DxB| ≤ sup |DA| sup |Dxu k ≤ CL. B(0,2)

Thus (5.29) implies that for all T ∈ (0,T∗] k+1 k sup ku − u (t)kL2 = sup kw(t)kL2 t∈[0,T ] t∈[0,T ] (CL+1)T k+1 k ≤e (kg − g kL2 + kfkL2(0,T ;L2)) (CL+1)T −k 1/2 k k−1 ≤e (C2 kgkH1 + CT (L + kgkHs ) sup k(u − u )(t)kL2 . t∈[0,T ]

(CL+1)T∗∗ 1/2 Thus (5.27) holds if we choose T∗∗ such that e T∗∗ ≤ θ.

79 [July 12, 2019] It remains to show the key estimates (5.43) and (5.44). The main point is to ﬁnd conditions which guarantee that

s s f, g ∈ H =⇒ fg ∈ H and kfgkHs ≤ CkfkHs kgkHs s s s u ∈ H , a ∈ C =⇒ a ◦ f ∈ H and ka ◦ ukHs bounded in terms of kukHs .

To illustrate the main idea let us ﬁrst look at the cases s = 1 and s = 2. For s = 1 we have

k∂j(fg)kL2 = kf∂jgkL2 + ∂jfgkL2 ≤ kfkL∞ kDfkL2 + kDfkL2 kgkL∞

n This is bounded by CkfkH1 kgkH1 if 2 < s = 1, i.e., if n = 1. For s = 2 we get

∂k∂j(fg) = f∂k∂jg + ∂jf ∂kg + ∂kf ∂jg + ∂k∂jf g.

2 The ﬁrst term can be estimated as before by kfkL∞ kD gkL2 and hence by n kfkH2 kgkH2 is 2 < s = 2 (i.e. n ≤ 3). The last term can be estimate analogously. For the middle term we have two options. First can use the Sobolev embedding theorem. For n ≤ 3 this gives

k∂jfkLp ≤ kfkH2 if 2 ≤ p ≤ 6.

Thus k∂jf ∂kgkL2 ≤ k∂jfkL4 k∂kgkL4 ≤ kfkH2 kgkH2 . The second option is to use the interpolation estimate

1/2 2 1/2 k∂jfkL4 ≤ kfkL∞ kD fkL2 . From this argument one obtains even the slightly stronger estimate

2 2 2 kD (fg)k ≤ C(kfkL∞ kD gkL2 + kD fkL2 kgkL∞ ).

∞ To prove the interpolation estimate note that for f ∈ Cc Z Z Z |Df|4 dx = |Df|2Df · Df dx = − div (|Df|2Df) f dx. Rn Rn Rn

Thus Z Z 4 2 2 |Df| dx ≤ C |Df| |D f| dx kfkL∞ Rn Rn and the desired estimate follows from the Cauchy-Schwarz inequality. ????? Use (5.38) ? The estimate for the composition are similar. By the chain rule one has

∂j(a ◦ u) = (Da ◦ u)(∂ju)

80 [July 12, 2019] and 2 ∂k∂j(a ◦ u) = (D a ◦ u)(∂ju, ∂ku) + (Da ◦ u)(∂k∂ju) 2 n where we view D a as a bilinear form. If s > 2 then kukL∞ ≤ CkukHs so that Da ◦ u and D2a ◦ u are bounded because the derivatives of a are uniformly bounded on bounded sets. Then 2 k(D a ◦ u)(∂ju, ∂ku)kL2 ≤ Ck∂jukL4 k∂kukL4 and we can conclude the argument as before. We now state the general result and we ﬁrst focus on an estimate by s s kfkHs kgkHs . For a reﬁned bound in terms of kfkL∞ kD gkL2 +kD fkL2 kgkL∞ see Lemma 5.34 below. Lemma 5.28 (Calculus inequalities, basic version). Let Dlf denote the collection of all partial derivatives of order l. n s n l n (i) If s > 2 , l ≤ s, f ∈ H (R ), g ∈ H (R ) then l kD (fg)kL2 ≤ Cl,skfkHs kgkHl (5.46)

n s n l−1 n (ii) If s > 2 + 1, l ≤ s, f ∈ H (R ), g ∈ H (R ) then l l kD (fg) − fD gkL2 ≤ Cl,skfkHs kgkHl−1 (5.47)

n l m d l n m (iii) If s > 2 , l ≤ s, a ∈ C (R ; R ), u ∈ H (R ; R ) then l r r kD (a ◦ u)kL2 ≤Cl,m,d max sup k(D a) ◦ uk kukHs (5.48) 1≤r≤l (5.49)

[24.5. 2019, Lecture 12] [27.5. 2019a, Lecture 13]

Proof. This argument was only sketched in class. (i) Let |α| = l. By the product rule we have

α X β γ ∂ (fg) = cβ,γ∂ f ∂ g. β+γ=α Given (β, γ) with β + γ = α we say that a pair (p, q) is admissible if β γ k∂ fkLp ≤ CkfkHs , k∂ gkLq ≤ kgkHl . Note that the condition β + γ = α implies that γ |β| n ∂ g ∈ H (R ) . We have

1 1 1 β γ β γ (p, q) admissible and + = =⇒ k∂ f ∂ gk 2 ≤ k∂ fk p k∂ gk q ≤ kfk s kgk l . p q 2 L L L H H By the Sobolev embedding we have the following case distinctions.

81 [July 12, 2019] n (i) If |β| < s − 2 then p = ∞, q = 2 is admissible. n (ii) If |β| = s − 2 then |β| ≥ 1 and hence any (p, q) with p ∈ [2, ∞) and 2n q ∈ [2, n−2 ] is admissible (q ∈ [2, ∞) if n = 2). n (iii) If |β| > 2 then p = 2, q = ∞ is admissible. n (iv) If |β| = 2 then |β| < s and hence any (p, q) with q ∈ [2, ∞) and 2n p ∈ [2, n−2 ] is admissible (p ∈ [2, ∞) if n = 2). n n (v) If |β| > s − 2 and |β| < 2 then (p, q) is admissible if p ∈ [2, pβ] and 1 1 s−|β| 1 1 |β| q ∈ [2, qβ] with = − and = − pβ 2 n qβ 2 n

Now in all cases but the last it is obvious that there is an admissible pair 1 1 1 with p + q = 2 . In the last case we compute 1 1 s 1 + = 1 − < . pβ qβ n 2

1 1 1 Hence there exists an admissible pair with p + q = 2 . (ii) This is similar to (i). The main diﬀerence is that the term f∂αg does not appear. (iii) For a derivative of order l we get

l α X X r β1 βr ∂ (a ◦ u) = cβ1,...,βr (D a ◦ u)(∂ u, . . . , ∂ u) r=1 β1+...+βr=α, |βj |≥1

We say that βj pj is admissible if k∂ ukLpj ≤ CkukHs . By the Sobolev imbedding theorem

j n (i) If |β | < s − 2 then pj is admissible if pj ∈ [2, ∞] j n (ii) If |β | = s − 2 then pj is admissible if pj ∈ [2, ∞)

j (iii) If |βj| > s− n then p is admissible if p ∈ [2, r ] where 1 = 1 − s−|β | . 2 j j j rj 2 n

We need to show that there is a choice of admissible pj such that

X 1 1 = . p 2 j j

Then by H¨older’sinequality

β1 βr β1 β1 r k∂ u . . . ∂ ukL2 ≤ k∂ ukLp1 ... k∂ ukLpr ≤ CkukHs .

j n If |β | ≤ s − 2 for all indices j we can take pj = 2r.

82 [July 12, 2019] j n If |β | > s − 2 for all indices j then

X 1 X 1 s |βj| 1 s l 1 = − + ≤ − + ≤ rj 2 n n 2 n n 2 j j | {z } <0

P j since β = α and |α| = l. Hence we can chose pj ∈ [2, rj] such that P 1 = 1 . pj 2 j n If |β | > s − 2 for at least one and at most r − 1 indices j then

X |βj| l − 1 ≤ n n j n j,|β |>s− 2 and thus

X 1 X 1 s |βj| 1 s l − 1 1 1 = ( − ) + ≤ ( − ) + ≤ − . r 2 n n 2 n n 2 n j n j j n j,|β |>s− 2 j,|β |>s− 2

1 j n P 1 1 If we take pj = when |β | ≤ s − then j n ≤ . Hence there nr 2 j,|β |≤s− 2 pj n exist admissible p such that P 1 = 1 . j j pj 2

5.3.2 Reﬁnements Corollary 5.29. The time existence T in Theorem 5.25 can be chosen in such a way that it depends on g only the kgkHs . Sketch of proof. We argue like in the proof of Lemma 5.26 to prove the following claim by induction.

Claim. Let L = 2kgkHs . Then there exists T∗ > 0 such that

k s (i) supt∈[0,T∗] ku (t)kH ≤ L for all k ∈ N,

k+1 k+1 (ii) ku − g kC0 ≤ 1 for all k ∈ N. Since u0 ≡ g0 the assertion holds for k = 0. Assume that the assertion holds for k. Now we proceed as in the proof of Lemma 5.26 but we estimate k+1 k+1 0 ku (t)kHs rather then ku − g kHs . Then the term Hα does not appear and we get d k+1 2 k s k+1 2 ku k s ≤ C(1 + ku (t)k )ku k s . dt H H Thus k+1 MT k+1 MT sup ku (t)kHs ≤ e kg kHs ≤ e kgkHs t∈[0,T∗]

83 [July 12, 2019] where M = C(1 + Ls). Taking

ln 2 T ≤ ∗ M we get k sup ku (t)kHs ≤ L. t∈[0,T∗] k+1 k+1 k+1 To estimate ku (t) − g kC0 we use the equation for u and the ’calculus inequalities’ to derive a bound

k+1 sup k∂tu (t)kHs−1 ≤ C(L) t∈[0,T∗] where L 7→ C(L) is an increasing function and C(L) < ∞ for all L > 0. Using the embedding Hs−1 ,→ C0 this implies that

k+1 k+1 k+1 k+1 sup ku (t) − g kC0 ≤ CSku − g kHs−1 ≤ T∗CSC(L). t∈[0,T∗]

Hence we if require in addition that 1 T∗ ≤ CSC(L) then kuk+1 − gk+1k ≤ 1 and the proof of the claim is ﬁnished.

Corollary 5.30. The solution belongs to C0([0,T ],Hs) ∩ C1([0,T ],Hs−1).

Sketch of proof. It suﬃces to show that u ∈ C0([0,T ],Hs). The diﬀerentia- bility of u and the assertion of ∂tu then follow from the equation (and the chain rule in connection with the calculus inequalities).

Step 1. Right continuity at t = 0: limt↓0 ku(t) − gkHs = 0. Let h ∈ Hs+1 and estimate ku(t) − h Hs| similar to the argument in the proof of Lemma 5.26. This shows that

s+1 lim sup ku(t) − hkHs ≤ kg − hkHs ∀h ∈ H . t↓0

Since Hs+1 is dense in Hs this yields the assertion.

s Step 2. Right continuity that t0 ∈ [0,T ): limt↓t0 ku(t) − u(t0)kH = 0. s Since u(t0) ∈ H there exists a T˜ > 0 and a classical solution w of the problem

∂ w + Pn A (w)∂ w = 0, in (0, T˜) × n, t j=1 j xj R . (5.50) w(0, ·) = u(t0).

84 [July 12, 2019] and by Step 1 lim kw(τ) − u(t0)kHs = 0. τ↓0

On the other hand the function τ 7→ u(τ + t0) is also a classical solution n of (5.50) (on [0,T − t0] × R ). By uniqueness w(τ) = u(t0 + τ) and right continuity at t0 follows.

s Step 3. Left continuity at t0 ∈ (0,T ]: limt↑t0 ku(t) − u(t0)kH = 0. Let w be the solution of

∂ w − Pn A (w)∂ w = 0, in (0, T˜) × n, t j=1 j xj R . (5.51) w(0, ·) = u(t0).

By Step 1 limτ↓0 kw(τ) − u(t0)kHs = 0. Now τ 7→ u(t0 − τ) is a classical solution of the same initial value problem. Thus w(τ) = u(t0 − τ) which yields the assertion.

[27.5. 2019a, Lecture 13] [27.5. 2019b, Lecture 14]

Theorem 5.31. Let u be the solution constructed in the proof of Theo- rem 5.25. Let M = sup |Dxu|. (5.52) [0,T ]×Rn

Then the exists a constant C = C(s, G2) such that

CMT sup ku(t)kHs ≤ e ku(0)kHs . (5.53) t∈[0,T ]

Proof. We have for |α| ≤ s

α k+1 X k α ∂t∂ u + (Aj ◦ u )∂xj ∂ u = −Fα j

k k where Fα is the commutator deﬁned above. Setting Bj = Aj ◦ u be obtain as before k k sup |div B | ≤ C1M where k Mk := sup |Dxu |. [0,T ]×Rn Testing the equation with ∂αu we get

α k+1 2 α k+1 2 α k+1 ∂tk∂ u kL2 ≤ C1Mk∂ u kL2 + kFαkL2 k∂ u kL2 .

85 [July 12, 2019] Now using the reﬁned calculus estimates in Lemma 5.34 we get

X k k+1 |α| k k+1 kFαkL2 ≤ kD(Aj ◦ u )kL∞ ku kHs + kD (Aj ◦ u )kL2 kDu kL∞ . j

Thus k+1 k kFαkL2 ≤ CMkku kHs + ku kHs Mk+1. Hence

k+1 2 k+1 2 k+1 2 k k+1 ∂tku kHs ≤ C1Mkku kHs +C2Mkku kHs +C2Mk+1ku kHs ku kHs . It follows that

k+1 k ∂tkukHs ≤ (C1 + C2)Mkku kHs + C2Mk+1ku kHs .

Now if we could pass to the limit k → ∞ we would get the corresponding estimate for ku(t)kHs and the assertion would follow by Gronwall’s inequality. 2 s We not know, however, whether uk converges strongly to u in L ([0,T ]; H ). And hence we cannot immediately pass to the limit. From the bounds in L∞([0,T ],Hs)∩W 1,∞([0,T ],Hs−1), the convergence in L∞([0,T ]; L2) and the interpolation inequalities we can deduces at least that k 0 1 n m u → u in C ([0,T ]; C (R ; R ). This implies that Mk → M. Set k k h (t) = ku (t)kHs and 0 M = (C1 + C2)M + 1.

Then we have for k ≥ k0

(hk+1)0(t) ≤ M 0hk+1(t) + M 0hk(t).

Since the sequence uk is bounded in C0([0,T ]; Hs) it follows that the sequence hk is bounded in C0[0,T ]. Let

h(t) := lim sup hk(t). k→∞

For ﬁxed t there thus exists a subsequence that hkj (t) → h(t). Since −hk is bounded from below we can apply Fatou’s lemma along the subsequence kj − 1 and we get Z t Z t Z t lim sup hkj −1(τ) dτ ≤ lim sup hkj −1(τ) dτ ≤ h(τ) dτ. j→∞ 0 0 j→∞ 0

86 [July 12, 2019] Thus Z t 0 h(t) ≤ kgkHs + 2M h(τ) dτ 0 for all t ∈ [0,T ]. From this we easily conclude that

2M 0t h(t) ≤ e kgkHs (5.54) which together with the lower semicontinuity of the Hs norm yields the desired assertion. 2M 0t To see that (5.54) holds let A > kgkHs . Then h(t) ≤ e A on a maximal interval [0, τ¯) (since h(t) ≤ kgkHs + 2MkhkL∞ t). Ifτ ¯ 6= T then

τ¯ 0 Z 0 0 0 e2M τ¯A = h(τ) < A + 2M 0 e2M τ A dτ = A + (e2M τ¯ − 1)A = e2M τ¯A, 0 a contradiction. If τ = T then we deduce that h(t) ≤ e2M 0tA for all A > kgkHs and all t ∈ [0,T ]. This yields (5.54)

1 n Corollary 5.32. Let u ∈ C ([0,T ) × R ) be a classical solution. Let s > n 2 + 1. Assume that G2 ⊂⊂ G,

s n m u(0, ·) ∈ H (R ; R ), (5.55)

n u(t, x) ∈ G2 ∀t ∈ [0,T ) × R (5.56) and M = sup |Dxu| < ∞. (5.57) [0,T )×Rn 0 s n m 1 s−1 n m Then u ∈ C ([0,T ); H (R ; R )) ∩ C ([0; T ); H (R ; R )

CMT sup ku(t)kHs ≤ e ku(0)kHs . (5.58) t∈[0,T ]

Sketch of proof. For a short time interval this follows directly from the The- orem 5.31 and the uniqueness result in Theorem 5.25. Let Tˆ be the supremum over all times such the assertion is true in [0, T˜). Then there exists and ε > 0 such that Theorem 5.31 can be applied on an interval of length at least 2ε with initial value u(Tˆ − ε). Thus the solution can be extended to [0, max(Tˆ + ε, T )) and the bounds hold on that interval. This contradicts the deﬁnition of Tˆ.

Theorem 5.33. If (0, T ) is a maximal interval of existence for solutions in 0 s n m C ([0,T ); H (R ; R ) and T < ∞ then one of the following two conditions must be satisﬁed

(i) lim supt↑T k∂tu(t)kL∞ + kDxu(t)kL∞ = ∞;

87 [July 12, 2019] n (ii) for every compact set K ⊂ G there exists a times tj ↑ T and xj ∈ R such that u(tj, xj) ∈/ K for all j ∈ N. Sketch of proof. We argue by contradiction. Let

T = sup{T : ∃ solution u ∈ C0([0,T ]; Hs)}.

Assume that T¯ < ∞ and assume that

lim sup k∂tu(t)kL∞ + kDxu(t)kL∞ ≤ M (5.59) t↑T and

m n there exists a compact set K ∈ R such u(t, x) ∈ K if (t, x) ∈ [0, T¯) × R .

We ﬁrst derive a contradiction if (5.60) is replaced by the stronger condition lim sup ku(t)kHs ≤ M (5.60) t↑T In this case we can use the usual continuation argument. By the existence result, Theorem 5.25 and Corollary 5.29 there exists T˜ > 0 (which depends on M and K) such that for initial data g with kgkHs ≤ M and g(x) ∈ K m s 1 s−1 for all x ∈ R the is a solution wg ∈ C([0,T∗]; H ) ∩ C ([0,T∗]; H ) of the ¯ 1 ˜ initial value problem. Let g = u(T − 2 T ) and deﬁne ( u(t) if t < T¯ − 1 T˜, u˜(t) = 2 ¯ 1 ˜ ¯ 1 ˜ ¯ 1 ˜ w(t − (T − 2 T )) if T − 2 T ≤ t ≤ T + 2 T .

0 ¯ 1 ˜ s Thenu ˜ is a solution in u ∈ C ([0, T + 2 T ]; H ) and this contradicts the deﬁnition of T¯. To reach a contradiction assuming only the weaker condition (5.59) if suﬃces to apply Corollary 5.32.

Lemma 5.34 (Calculus inequalities, reﬁned version). Let s ∈ N, s ≥ 1 and denote by Ds the collection of all partial derivatives of order s.

s n ∞ (i) If f, g ∈ H (R ) ∩ L then

s s s kD (fg)kL2 ≤ Cs(kfkL∞ kD gkL2 + kgkL∞ kD fkL2 ) (5.61)

s n ∞ s−1 ∞ (ii) If f ∈ H (R ), Df ∈ L , g ∈ H ∩ L then

s s s−1 s kD (fg) − fD gkL2 ≤ Cs(kDfkL∞ kD gkL2 + kgkL∞ kD fkL2 ) (5.62)

88 [July 12, 2019] (iii) If a ∈ C then

s s r r−1 kD (a ◦ u)kL2 ≤ Cs,mkD ukL2 max sup k(D a) ◦ uk kuk ∞ . (5.63) 1≤r≤s L

The proof is very similar to the proof of Lemma 5.28 and uses the following interpolation estimates. Lemma 5.35 (Special case of Gagliardo-Nirenberg interpolation estimates). Assume that p, q, r ∈ [2, ∞]. Let k, l ∈ N and assume that 1 l − k 1 k 1 0 < k < l, = + . q l q l r Then l−k k k l l l kD uk q n ≤ kuk kD uk L (R ) Lp(Rn) Lr(Rn) where C depends on p, q, r, k, l and n. Proof. This was not discussed in class. We only consider the case q < ∞. The case q = ∞ (which implies p = r = ∞) can easily be obtained by a slight variation of the argument. We proceed by induction over s.

Step 1. l = 2. We need to show that

1/2 2 1/2 1 1 1 1 1 kDuk q ≤ kuk kD uk , where = + . (5.64) L Lp L2 q 2 p 2 r

∞ n By density it suﬃces to show the result for u ∈ Cc (R ). Now Z Z Z |Du|q dx = Du · |Du|q−2Du dx = − udiv (|Du|q−2Du) dx Rn Rn Rn Now by the chain rule (here we use q ≥ 2)

|div (|Du|q−2Du)| ≤ |Du|q−2 |D2u|.

(with the usual interpretation that |Du|q−2 = 1 if q = 2). For q = 2 we immediately obtain the desired estimate kDukL2 ≤ kuk k∆ukL2 . For q > 2 we note that 1 q − 2 1 2 1 1 + + = 1 − + + = 1. p q r q p r q Thus we can apply H¨older’s inequality with exponents p, q−2 and r and we get Z Z q q−2 2 q−2 2 |Du| dx ≤ C |u| |Du| |D u| dx ≤ CkukLp kDukLq kD ukLr . Rn Rn q−2 Dividing by kDukLq we get (5.64).

89 [July 12, 2019] Step 2. Induction step. Assume that the result holds for all (k, l) with 0 < k < l ≤ L. We will show the estimate for 0 < k < l with l = L + 1. Assume ﬁrst k < L. Then we can estimate Dku in terms of u and DLu. On the other hand by setting v = Dku and using the induction assumption we can bound DLu = DL−kv in terms of v and DL+1−kv = DL+1u. The combination of these two estimate will give the result. We begin with the second estimate and deﬁner ˜ by 1 1 1 L − k 1 = + . r˜ L + 1 − k q L + 1 − k r Then by the induction assumption

1 L−k L−k L+1−k L+1−k L+1−k kD vkLr˜ ≤ CkvkLq kD vkLr . Setting v = Dku this yields

1 L−k L k L+1−k L+1 L+1−k kD ukLr˜ ≤ CkD ukLq kD ukLr . Now a short calculation shows that L − k 1 k 1 1 + = L p L r˜ q Thus L−k k k L L L q kD ukL ≤ CkukLp kD ukLr˜. Inserting the estimate for DLu and taking the inequality to the power L(L+ 1 − k) we get

k L(L+1−k) a k b L+1 c kD ukLq ≤ kukLp kD ukLq kD ukLr . with a = (L − k)(L + 1 − k), b = k, c = (L − k)k Thus

a + c = (L − k)(L + 1), a + b + c = L2 + L − kL = L(L + 1 − k) and hence L+1−k k k L+1 L+1 L+1 kD ukLq ≤ kukLp kD ukLr . If k = L one uses the induction assumption for v = Du to obtain an estimate for DLu in terms of Du and DL+1u. Then one estimates Du in terms of u and DLu.

90 [July 12, 2019] Proof of Lemma 5.34. Let us just consider (i), the other estimates being similar. As before α X β γ ∂ (fg) = cβ,γ∂ f ∂ g β+γ=α Set 2s 2s 1 m = , m = with the convention = ∞. β |β| γ |γ| 0 Then 1 1 1 + = if |α| = s mβ mγ 2 and deﬁning |β| s − |γ| λ = = s s we get

1−λ 1−λ β γ β m γ m λ s s λ k∂ f ∂ gkL2 ≤ k∂ fkL β k∂ gkL γ ≤ CkfkL∞ kD fkL2 kgkL∞ kD gkL2 .

1 p 1 p0 1 1 Using Young’s inequality ab ≤ p a + p0 b with p = λ and p0 = 1 − λ we conclude that

β γ s s k∂ f ∂ gkL2 ≤ C (λkfkL∞ kD gkL2 + (1 − λ)kgkL∞ kD fkL2 ) .

5.3.3 Examples The Euler equations for an isentropic compressible fluid The equation for an ideal isentropic fluid without viscosity or heat conduction follow from the conversation of mass and linear momentum. The ρ denote the density of the fluid, v its velocity and p the pressure. For every test volume n Ut ⊂ R which moves with the fluid we have d Z ρ dx = 0 conservation of mass dt Ut Z Z d n−1 ρv dx = force on Ut = −pν dH conservation of momentum dt Ut ∂Ut where ν denotes the outer normal of Ut. This gives Z Z n−1 0 = ∂tρ dx + ρ v · ν dH Ut ∂Ut Z Z Z n−1 n−1 −pν dH = ∂tρv dx + ρv v · ν dH ∂Ut Ut ∂Ut If v, ρ and p are sufficiently regular we can use Gauss’ theorem to convert the boundary integrals into volume integrals. Using the fact that Ut was an

91 [July 12, 2019] arbitrary (suﬃciently regular) test volume we obtain the following system of PDEs

∂tρ + div (ρv) = 0,

∂t(ρv) + div (ρv ⊗ v) + ∇p = 0.

These are (n + 1) PDEs for the (n + 2) variables ρ, v1, . . . , vn, p. These equations hold for every ideal fluid. The properties of a specific fluid (or gas) are characterized by a specific relation (also known as a constitutive equation) between pressure and density

p =p ˆ(ρ) described by the constitutive functionp ˆ : (0, ∞) → (0, ∞). Thus the Euler equations for an ideal compressible ﬂuid are given by

∂tρ + div (ρv) = 0,

∂t(ρv) + div (ρv ⊗ v) + ∇pˆ(ρ) = 0.

For further details on the derivation of the equation see the lecture notes to PDE and Modeling (summer term 2013). Set u ρv u = ρv, U = = . ρ ρ Then the Euler equations can be written as system of conservation laws

n X ∂tU + Fj(U) = 0 j=1 with 1 ρ uju +p ˆ(ρ)ej Fj(U) = for j = 1, . . . , n. uj For ease of notation we can consider the case of one space dimension, n = 1. Then ∂tU + ∂xF (U) = 0 where u2/ρ +p ˆ(ρ) F (U) = . u Thus ∂tU + A(U)∂xU = 0 with 2 ! 2u − u +p ˆ0(ρ) A(U) = DF (U) = ρ ρ2 . 1 0

92 [July 12, 2019] The matrix A(U) is not symmetric but the system can be symmetrized by using a diagonal matrix A0

1 0 A = . 0 0 µ

Indeed

2 ! ! 2u u 0 2u µ ρ − ρ2 +p ˆ (ρ) T ρ A0A = ,A A0 = u2 0 . µ 0 − ρ2 +p ˆ (ρ) 0

Thus u2 A A = AT A ⇐⇒ µ = − +p ˆ0(ρ) 0 0 ρ2 and hence the one-dimensional compressible Euler equation satisﬁes the condition (S) if and only if

pˆ0(ρ) > 0 and |v| < ppˆ0(ρ).

Indeed one can easily check that the system is hyperbolic (i.e., A is real diagonazible) if and only if the above condition holds. This condition has a very natural interpretation. Linearizing the 1d Eu- ler equations around ρ = const ≡ ρ0 and v = 0 one sees that the linearization satisﬁes the linear system

0 ∂tρ˜ + ρ0∂xv˜ = 0, ρ0∂tv˜ +p ˆ (ρ0)∂xρ˜ = 0.

Deriving the ﬁrst equation with respect to t and the second with respect to x we get the linear wave equation

2 0 2 ∂t ρ˜ − pˆ (ρ0)∂xρ˜ = 0.

Thus ppˆ0(ρ) is the speed of propagation in the linearized equation for (in- ﬁnitesimally) small density variations, i.e., the speed of sound. Hence the hyperbolicity condition of the 1d Euler equation is simply that the speed v of the ﬂuid should be less then the (local) speed of sound. The quotient of this two quantities is also known as the Mach number M. [27.5. 2019b, Lecture 14] [31.5. 2019, Lecture 15]

Elastodynamics This example was not discussed in class. Let ϕ : [0,T ]× n n R → R and view ϕ(t, ·) as the deformation of an elastic body whose n reference conﬁguration ﬁlls R . Then

2 ρR ∂t ϕ = div T,

93 [July 12, 2019] where ρR denotes the density in the reference conﬁguration and where for a hyperelastic material the Piola-Kirchhoﬀ stress tensor T has the form

m×m T = Tˆ(∇ϕ) = DW (∇ϕ) where W : R → R.

Thus ϕ satisﬁes the nonlinear wave equation

2 ρR ∂t ϕ = div (DW (∇ϕ)).

Assume for simplicity that the reference density is constant and

ρR ≡ 1.

Then the nonlinear wave equation can be written as a ﬁrst order conservation law for the the variable U = (∇ϕ, ∂tϕ). For ease of notation we do this only for n = 1 and set

0 w = ∂xϕ, v = ∂tϕ, p(w) := W (w).

Then ∂tw − ∂xv = 0, ∂tv − ∂x(p(w)) = 0. This system is often called the p-system. It can be written as

∂tU + ∂xF (U) = 0 with w w v U = ,F ( ) = − . v v p(w) Thus 0 1 A(U) = DF (U) = − . p0(w) 0 The system can be symmetrized with the help of the diagonal matrix

λ 0 A = − . 0 0 1

Indeed 0 λ 0 p0(w) A A = − ,AT A = − 0 p0(w) 0 0 λ 0 0 Thus A0 symmetrizes the system if λ(U) = p (w) and

The p-system satisﬁes condition (S) ⇔ p is strictly monotone ⇔ W is strictly convex ⇔ the p-system is hyperbolic.

94 [July 12, 2019] 6 Hyperbolic conservation laws, weak solutions and the Lax entropy condition

6.1 Weak solutions and the Rankine-Hugoniot jump condition In the previous section we have seen that the initial value problem for hyperbolic conservation laws which satisfy the symmetrisability condition (S) posseses a smooth solution on some time interval [0,T ), provided the initial data are sufficiently regular. We will see shortly that in general solutions can become discontinuous in finite time. Thus it is natural to look for weak solutions. In contrast to elliptic equations, however, hyperbolic conservation laws admit many different weak solutions including ’unphysical’ solutions. Thus one needs to impose additional conditions on weak solutions (called ad- missibility criteria) to have a good theory. Even though conservation laws have been of great interest since the 19th century (in particular in view of applications in gas dynamics) and the formation of singularities was well understood the problem of finding a good notion of solution (even in one space dimension) was only solved in the 1950’s, independently by Peter Lax and Olga Oleinik. A comprehensive treatment of hyperbolic conservation laws can be found in Dafermos’ beautiful book6. This section is an introduction to the problem and follows closely [Ev], Chapter 11. Many interesting and deep aspects are not covered, including the existence of solutions by the iterative solution of Riemann problems (Glimm’s random choice scheme), the theory of solutions in the space of functions of bounded variations or the stability of entropy solutions under weak convergence [Ta79, Ta83, DP85] . Let us begin with a specific example of a scalar conservation law, the initial value problem for Burgers equation7.

 u2 ∂ u + ∂ = 0 in × (0, ∞), t x 2 R (6.1) u(0, ·) = g in R. Assume that ∞ g ∈ Cc (R), g = 0 on [1, ∞), g = 1 on [−1, 0]. (6.2) 6Here is a quote from the review of the first edition by D. Serre: ’This book has several qualities which make it the best book ever written on the subject, in the reviewer’s opinion. First, the style is so clear that some parts, especially the first few, can be read like a novel. Next, the choice of topics is so clever that the book will survive at least several decades, or even forever. At a finer level, Dafermos has chosen as often as possible to present theorems with physically motivated assumptions. This will be strongly appreciated by people looking for references. Last, the historical notes which follow each chapter, and the exhaustive bibliography, form a priceless database which does not exist elsewhere.’ 7The following material is partly taken from the lecture Introduction to PDE, WS 2012

95 [July 12, 2019] By Theorem 5.25 there exists a T > 0 such that (6.1) has a unique classical 1 n solution u ∈ C ([0,T ] × R ). We now show that T cannot be bigger than 1. 1 n Theorem 6.1. Assume that u ∈ C ([0,T ] × R ) is a classical solution of (6.1) and that u0 satisiﬁes (6.2). Then T < 1. Proof. Assume that T ≥ 1. The key idea is to look at the characteristics of the equation and to show that the characteristics must intersect at a time t ≤ 1. A C1 curve t 7→ (t, γ(t)) is a characteristic if u(t, γ(t)) = const. If u is C1 then t 7→ (t, γ(t)) is characteristic if

0 ∂tu(t, γ(t)) + γ (t)∂xu(t, γ(t)) = 0. On the other hand if u is a C1 solution of the Burgers equation then

∂tu(t, γ(t)) + u(t, γ(t))∂xu(t, γ(t)) = 0. Hence t 7→ (t, γ(t)) is a characteristic if γ0(t) = u(t, γ(t)) = const. Taking γ(t) = x0 + u0(x0)t we get

u(t, x0 + u0(x0)t) = u(x0).

Applying this with x0 = −1 we get

u(1, 0) = u0(−1) = 1 while application with x0 = 0 gives

u(1, 0) = u0(0) = 0. This is a contradiction.

Example. Let  1 if x ≤ 0  g(x) = 1 − x if x ∈ (0, 1) (6.3) 0 if x ≥ 1 . Then u(t, γ(t)) = g(0, γ(0)) with γ0(t) = g(γ(0)) gives  1 falls x ≤ t, t ∈ [0, 1) 1 − x u(x, t) = falls 0 ≤ t ≤ x < 1 (6.4)  1 − t 0 falls x ≥ 1, t ∈ [0, 1) . Then for t < 1 the function u is piecewise aﬃne in x and solves the equation a.e. At t = 1 the function u has a discontinuity. Since we cannot expect global C1 solution we deﬁne weak (or distributional or integral) solutions.

96 [July 12, 2019] m m Deﬁnition 6.2. Let F : R → R be continuous. We say that u ∈ ∞ m L ((0, ∞) × R; R ) is a weak solution of initial value problem

∂tu + ∂xF (u) = 0 on [0, ∞) × R, u(0, ·) = g on R. if Z Z ∞ u·∂tv +F (u)·∂xv dt dx+ g ·v = 0 ∀v ∈ Cc ([0, ∞)×R). (6.5) (0,∞)×R R Note that (6.5) incorporates both the weak form of the equation and the initial condition.

Theorem 6.3 (Rankine-Hugoniot jump condition for piecewise C1 solu- ∞ tions). Let u ∈ L (R × (0, ∞)) be a weak solution. Let ω ⊂ (0, ∞) × R be 1 open, let Ω ⊂ (0, ∞) × R be an open set with C boundary. If there exist two 1 1 functions v1 ∈ C (ω ∩ Ω) and v2 ∈ C (ω \ Ω) such that ( v in ω ∩ Ω, u = 1 v2 in ω \ Ω then ∂tu + ∂xF (u) = 0 in ω \ ∂Ω (6.6) and v − v 1 2 · ν = 0 on ∂Ω ∩ ω. (6.7) F (v1) − F (v2)

∞ Proof. To derive (6.6) it suﬃces to use test functions v ∈ Cc (ω ∩ Ω) and ∞ v ∈ Cc (ω \ Ω). Finally (6.7) follows from the divergence theorem.

Example. Consider the Burgers equation with initial data ( 0 if x ≤ 0 g(x) = (6.8) 1 if x > 0 .

For a ∈ (0, 1) deﬁne ( 0 falls x ≤ at ua(x, t) = (6.9) 1 falls x > at

Let Ω = {(x, t): x < at}, u1 = 0, u2 = 1. The tangent√ vector to the curve t, at is (1, a) and hence the unit normal is ν = (−a, 1)/ 1 + a2. Hence (6.7) yields −1 −a · = 0 . (6.10) −1/2 1

97 [July 12, 2019] 1 Thus ua is a weak solution if and only if a = 2 . This solution is known is a 1 shock. The shock (i.e. the discontinuity of u) moves with speed 2 . Now deﬁne  0 if x ≤ 0  u∗(x, t) = x/t if 0 < x < t (6.11) 1 if x ≥ t .

The function u∗ is continuous and hence (6.7) holds trivially. One easily sees that u2 ∂ + ∂ = ∂ + u∂ u = 0 in U = {(x, t) : 0 < x < t}. t x 2 t x

Hence u∗ is another weak solution (known as a rarefaction wave). We continue to follow [Ev], Chapter 11. Consider a semilinear hyperbolic system in nondivergence form

ut + B(u)ux = 0 in R × (0, ∞), (6.12)

m m×m m where B : R → R is given, and u : R × (0, ∞) → R is wanted. If B = DF , then for smooth functions, (6.12) is equivalent to the conservation law ut + F (u)x = 0. Deﬁnition 6.4. The equation (6.12) is called strictly hyperbolic if for each m z ∈ R , the eigenvalues of B(z) are real and distinct. For the remainder of this subsection, we consider only strictly hyperbolic systems.

Notation. Denote the eigenvalues of B(z) by

λ1(z) < λ2(z) < ··· < λm(z).

For k = 1, . . . , m let rk(z) be an eigenvector of B(z) to the eigenvalue λk(z), i.e., B(z)rk(z) = λk(z)rk(z). T For k = 1, . . . , m let `k(z) be an eigenvector of B (z) to the eigenvalue λk(z) (since B is quadratic, B and BT have the same spectrum), i.e.,

T B(z) `k(z) = λk(z)`k(z).

One often identiﬁes ` with the row vector `T , for which

T `(z)B(z) = λk(z)`(z).

Thus {rk(z)} are called right eigenvectors of B(z), and {`k(z)} left eigenvectors of B(z). We collect some properties in the next lemma.

98 [July 12, 2019] m Lemma 6.5. (i) By strict hyperbolicity, for every z ∈ R , {rk(z): k = m 1, . . . , m} (and similarly {`k(z): k = 1, . . . , m}) span R .

(ii) `j(z) · rk(z) = 0 if j 6= k. (iii) Invariance of hyperbolicity under change of coordinates. Let u be a m m smooth solution of (6.12). Let Φ: R → R be a smooth diﬀeomor- phism with inverse Ψ := Φ−1. Then u˜ := Φ(u) is a solution to ˜ ˜ u˜t + B(˜u)˜ux = 0 in R × (0, ∞) with B(z) := DΦ(Ψ(z))B(Ψ(z))DΨ(z). (6.13) The system (6.13) is strictly hyperbolic. Proof. (i) by deﬁnition (ii) We compute

λk(z)(`j(z) · rk(z)) = `j(z) · (B(z)rk(z)) = T = (B (z)`j(z)) · rk(z) = λj(z)(`j(z) · rk(z)),

and thus the assertion follows since λj 6= λk if j 6= k. (iii) follows by direct computations, using chain rule. Details can be found in [Ev, Theorem 1, Section 11.1.2].

Theorem 6.6 (Dependence of eigenvalues and eigenvectors on parameters). m m×m Assume that B : R → R is smooth and that B at each point has m real and distinct eigenvalues. Then for k = 1, . . . , m, m (i) λk(z) depend smoothly on z ∈ R ,

(ii) rk(z) and `k(z) can be chosen to depend smoothly on z and being nor- malized, i.e., |rk(z)|, |`k(z)| = 1.

Proof. We only sketch the proof here. For details see [Ev, Theorem 2, m Section 11.2.1]. Fix z0 ∈ R , k ∈ {1, . . . , m} and rk(z0) with |rk(z0)| = 1. First show that in a neighborhood of z0, we ﬁnd smooth functions λk(z) and rk(z) with the desired properties. To do so, apply the implicit function theorem to the smooth function Φ(r, λ, z) := (B(z)r − λr, |r|2). ∂Φ(r,λ,z) Invertibility of ∂(r,λ) (rk(z0), λk(z0), z0) follows from strict hyperbolicity. To show global existence of λk and rk with the desired properties, set

Rmax := sup{r > 0 : there exist smooth λk(z), rk(z) on B(0, r)}.

Assume for the sake of contradiction that Rmax < ∞, cover ∂B(0,Rmax) by ﬁnitely many balls into which λk and rk can be extended smoothly as sketched above. This yields a contradiction to the deﬁnition of Rmax.

99 [July 12, 2019] Remark 6.7. (i) Note that in the above theorem, the eigenspaces are also globally provided with an orientation.

(ii) The proof relies on the fact that the eigenspaces are one-dimensional. An analogue statement is not true for general smooth functions B : m m×m R → R , consider e.g., 2 2 1 cos( ) sin( ) − 2 z z B(z) := e z 2 2 ,B(0) := 0 sin( z ) − cos( z ) for z → 0.

Example. Recall the p-system

vt − wx = 0

wt − p(v)x = 0.

This can be written in the form (6.12) with

0 −1 B(z) := 0 . −p (z1) 0

0 1/2 The eigenvalues of B(z) are ±(p (z1)) , and they are real and distinct if p0 > 0. Recall that the wave equation

utt − (p(ux))x = 0 in R × (0, ∞) can be written as p-system with v := ux and w := ut. The strict hyperbolicity condition then states that p(ux) is an increasing function of ux. If m = 1, this arises as balance law of adiabatic thermoelasticity in Lagrangian coordinates, assuming no body force and zero heat source. Here, v = ux stands for the (scalar valued) deformation gradient, and if the specimen is an elastic bar undergoing longitudinal oscillation, v measures the strain, and p(v) = p(ux) is the stress, which then has to be an increasing function of the strain in the case of strict hyperbolicity. For details see [Da00]. [31.5. 2019, Lecture 15] [17.6. 2019a, Lecture 16]

6.2 Riemann’s problem Consider the system of conservation laws with piecewise constant initial data

ut + F (u)x = 0 in R × (0, ∞) (6.14) ( u if x < 0 u(x, 0) = g(x) = ` (6.15) ur if x > 0.

100 [July 12, 2019] We aim to prove existence of solutions to (6.14) & (6.15) if ul and ur are close. The solution will be built of three types of constructions, namely constant states u ≡ c, shock waves and rarefaction waves. Combined with Glimm’s random choice method and rather delicate estimate this existence result for the Riemann problem can be used to prove existence of weak solutions to the conservation law for general initial data with small total variation, see [Da00]. As preparational step, we consider the constructions of rarefaction waves and shock waves separately.

6.2.1 Simple waves and rarefaction waves

2 The scalar valued case. Let m = 1, let F : R → R be C and uniformly convex. Assume that u` < ur. We aim to show that there exists a rarefaction wave solution, as was computed explicitly for Burger’s equation in the last lecture. For that, we observe that the equation and the initial condition are invariant under the rescaling x → αx and t → αt. This motivates to study solutions to (6.14) of the form x u(x, t) = v( ). t Inserting the ansatz into the equation, we find x x x x 1 x 1 x x 0 = v0 + F 0v v0 = v0 F 0v − . t t2 t t t t t t t 0 x x Thus, the equation is satisfied if F (v( t )) = t . Note that by the assumption of uniform convexity, F 0 is strictly increasing and onto, and therefore invertible. Thus, by the above computation, the function x x u(x, t) = v( ) = (F 0)−1( ) t t is a solution to (6.14). Consequently, the function  u if x < F 0(u )  ` t ` 0 −1 x 0 x 0 u(x, t) = (F ) ( t ) if F (u`) < t < F (ur) (6.16)  x 0 ur if t > F (ur) is a continuous weak solution since it is continuous in R × (0, ∞) and is a classical solution in every domain of definition.

Systems of conservation laws. We now consider the vectorial case, which we will reduce to the scalar valued case above. For that, we make a simple wave ansatz, i.e., we look for solutions to (6.14) & (6.15) of the form

u(x, t) = v(w(x, t)), (6.17)

101 [July 12, 2019] m with unknowns w : R × (0, ∞) → R and v : R → R . Inserting the ansatz (6.17) into (6.14), we ﬁnd

v˙(w)wt + DF (v(w))v ˙(w)wx = 0. (6.18) Set B := DF and use the notation introduced above for the eigensystems of B. If for some k = 1, . . . , m we have

v˙(s) = rk(v(s)) and (6.19)

wt + λk(v(w))wx = 0, (6.20) then this also yields a solution to (6.18) since then

v˙(w)wt + DF (v(w))v ˙(w)wx = rk(v(w))wt + B(v(w))rk(v(w)) wx = 0. | {z } =λk(v(w))rk(v(w)) We therefore from now on search for solutions to (6.20) & (6.19). The associated u = v(w) is then called a k-simple wave. Note that (6.19) is an ODE for v, where the function rk on the RHS is smooth. Further, if we ﬁx a solution v to (6.19), then (6.20) is a conservation law for the scalar valued function w. Precisely, given a solution v to (6.19), we rewrite (6.20) as Z s wt + Fk(w)x = 0 with Fk(s) := λk(v(t)) dt. (6.21) 0

We search for conditions to ensure that Fk is convex (or concave), which 0 implies that Fk is injective. For that, we compute derivatives, i.e., 0 Fk(s) = λk(v(s)), (6.22) and thus 00 Fk (s) = Dλk(v(s)) · v˙(s) = Dλk(v(s)) · rk(v(s)). (6.23) m Hence, Fk is convex if Dλk(z) · rk(z) > 0 for all z ∈ R , and concave m if Dλk(z) · rk(z) < 0 for all z ∈ R . We also note that Fk is linear if Dλk(z) · rk(z) = 0, which motivates the following deﬁnition.

Deﬁnition 6.8. The pair (λk(z), rk(z)) is called genuinely nonlinear if m Dλk(z) · rk(z) 6= 0 ∀z ∈ R .

p 0 p 0 T Example. For the p-system, we have λ1 = p (z1) and r1k(−1, p (z1)) , p 0 p 0 T and λ2 = − p (z1) and r2k(1, p (z1)) . Thus, the characteristic fields are genuinely nonlinear if p00 6= 0 everywhere. We also introduce some notation for trajectories of solutions to the ODE (6.19). m th Definition 6.9. For a state z0 ∈ R , we define the k -rarefaction curve m Rk(z0) to be the path in R of the solution to the ODE (6.19), which passes through z0.

102 [July 12, 2019] Notation. If (λk, rk) is genuinely nonlinear, set

+ Rk (z0) := {z ∈ Rk(z0): λk(z) > λk(z0)} and

− Rk (z0) := {z ∈ Rk(z0): λk(z) < λk(z0)}. Note that then + − Rk(z0) = Rk (z0) ∪ {z0} ∪ Rk (z0). Theorem 6.10 (Existence of k-rarefaction waves). Suppose that for some k ∈ {1, . . . , m}

(i) (λk, rk) is genuinely nonlinear

+ (ii) ur ∈ Rk (u`). Then there exists a continuous weak solution u of Riemann’s problem (6.14)&(6.15), which is a k-simple wave constant along lines through the origin.

Remark 6.11. The solution u is called (centered) k-rarefaction wave.

Proof. We employ the strategy outlined above. We choose a solution v to (6.19) which passes through u` (and thus by (ii) also through ur). Pick w` and wr ∈ R such that u` = v(w`) and ur = v(wr). Suppose that w` < wr, the other case can be treated similarly, using concavity instead of convexity. Consider the scalar valued conservation law (6.21) with piecewise constant initial condition ( w if x < 0 w(x, 0) = g(x) = ` wr if x > 0. 0 0 By (ii), we have λk(ur) > λk(u`), and by (6.22), Fk(wr) > Fk(w`). Hence, by 0 (i), we conclude that Fk is strictly convex, and thus, Fk is injective, and the 0 0 0 whole interval [Fk(w`),Fk(wr)] is contained in the range of Fk. Therefore, a continuous weak solution is given by a rarefaction wave connecting w` and wr, i.e.,  w if x < F 0 (w )  ` t k ` 0 −1 x 0 x 0 w(x, t) = (Fk) ( t ) if Fk(w`) < t < Fk(wr)  x 0 wr if t > Fk(wr). Thus, u(x, t) = v(w(x, t)) is a solution with the desired properties.

[17.6. 2019a, Lecture 16] [17.6. 2019b, Lecture 17]

103 [July 12, 2019] 6.2.2 Shock waves We now consider solutions to (6.14) & (6.15), for which the initial states are joined not by a rarefaction wave but by a shock. In view of the Rankine- Hugoniot condition (see Theorem 6.3) we need

F (u`) − F (ur) = σ(u` − ur) for some σ ∈ R.

Given a state z0 we ﬁrst characterize all states which satisfy the above compatibility condition.

m Deﬁnition 6.12. For z0 ∈ R deﬁne the shock set

m S(z0) := {z ∈ R : ∃σ = σ(z, z0) s.t. F (z) − F (z0) = σ(z − z0)}.

m Theorem 6.13 (Structure of the shock set). Fix z0 ∈ R . There is an open neighborhood of z0 in which S(z0) consists of m smooth curves Sk(z0), k = 1, . . . , m with the following properties:

(i) Sk(z0) passes through z0 with tangent rk(z0)

(ii) limz→z0, z∈Sk(z0) σ(z, z0) = λk(z0)

λk(z)+λk(z0) 2 (iii) σ(z, z0) = 2 + O(|z − z0| ) as z → z0, z ∈ Sk(z0). m m m×m Proof. We ’linearize’and set B : R × R → R , Z 1 B(z, z0) := DF (z0 + t(z − z0)) dt. 0 Then

B(z, z0)(z − z0) = F (z) − F (z0), and B(z, z) = DF (z).

Observe that the arguments below exploit the fact that B (and thus also its eigensystem) is symmetric in z and z0 (see [Da00]). Note that by deﬁnition z ∈ S(z0) if and only if

∃σ ∈ R :(B(z, z0) − σ Id )(z − z0) = 0. (6.24)

By strict hyperbolicity, B(z0) = B(z0, z0) has m real and distinct eigenvalues, and thus, so does B(z, z0) for z close to z0. To see this, one can use for example the fact that the characteristic polynomial of B(z0) has m disjoint real roots and therefore each root is simple. Thus the implicit function theorem implies that a small (real) perturbation of the characteristic polynomial also has m disjoint real roots. ˆ We denote the ordered eigenvalues of B(z, z0) by λk(z) and the associated ˆ eigenfunctions according to Theorem 6.6 byr ˆk(z) and `k(z), k = 1, . . . , m, and the eigenvalues and -functions of DF by λk, rk and `k, respectively.

104 [July 12, 2019] ˆ Then λk(z0) = λk(z0) etc. The eigenvector equation (6.24) for B(z, z0) is ˆ satisﬁed if and only if σ = λk(z) and (z − z0)krˆk(z) for some k = 1, . . . , m, the latter condition being equivalent to

(z − z0) · `ˆj(z) = 0, for all j 6= k. (6.25) This is a system of (m − 1) equations for the m unknown components of z, while in the definition of S(z0) we have to solve a system of m equations for the (m+1) unknown (z, σ). Clearly, z = z0 is a solution to (6.25). We use the inverse function theorem to solve (6.25) in a neighborhood of z0. Precisely, m m−1 note that solutions of (6.25) corresponds to zeros of Φ : R → R given by ˆ ˆ Φk(z) := (..., `k−1(z) · (z − z0), `k+1(z) · (z − z0),... ). ˆ The differential DΦk(z0) has full rank m − 1 since the vectors `j(z0) = `j(z0) are linearly independent. Thus in a neighborhood of z0 the solutions to (6.25) lie on a smooth curve Sk(z0) which can be parametrized by φk : m ˙ R → R such that φk(0) = z0, |φk(t)| = 1 and Φk(φk(t)) = 0 for all t close to 0. At z0, by definition, the curve is orthogonal to all `j(z0), j 6= k, and thus its tangent is parallel to rk(z0). Up to reparametrization of φk, the curve Sk(z0) has tangent rk(z0) at z0. This shows (i). On each branch Sk(z0), we have by construction

F (φk(t)) − F (z0) = σ(φk(t), z0)(φk(t) − z0) (6.26) with a smooth function σ. Differentiating with respect to t and evaluating at t = 0 yields DF (z0)rk(z0) = σ(z0, z0)rk(z0), which implies σ(z0, z0) = λk(z0), and thus (ii). Set σ(t) := σ(φk(t), z0). The first derivative of (6.26) gives 0 0 0 DF (φk(t))φk(t) = σ (t)(φk(t) − z0) + σ(t)φk(t). Differentiating again and evaluating at t = 0 we get

0 2 00 2σ (0) rk(z0) − D F (z0)(rk(z0), rk(z0)) = (DF (z0) − λk(z0) Id )φk(0).(6.27)

On the other, hand, let ψk be a unit speed parametrization of the k- rarefaction function and consider the eigenvector equation

DF (ψk(t))˜rk(t) = λk(t)˜rk(t), wherer ˜k(t) = rk(ψ(t)). Diﬀerentiate with respect to t and set t = 0 to obtain

2 0 0 D F (z0)(rk(z0), rk(z0)) − λk(0)rk(z0) = −(DF (z0) − λk(z0) Id )˜rk(0)(6.28).

105 [July 12, 2019] m Now we use that for every v ∈ R

T lk(z0)·(DF (z0)−λk(z0) Id )v = ((DF ) (z0)−λk(z0) Id )lk(z0), v) = (0, v) = 0.

Thus adding (6.27) and (6.28), and taking the scalar product with `k(z0) we get 0 0 2σ (0) = λk(0), which implies 2 2σ(t) − 2σ(0) = λk(t) − λk(0) + O(t ).

Using σ(0) = λk(0), we obtain (iii).

6.2.3 Local solution to Riemann’s problem

Suppose ul and ur can be joined by a k-shock, i.e., ur ∈ Sk(ul). Then, in view of Theorem 6.13 (iii), if ul is close to ur then either

λk(ur) < σ(ur, ul) < λk(ul), (6.29) or all inequalities reversed. For physical reasons we consider only shocks satisfying (6.29). This condition is called Lax entropy condition. We give some rough motivation for it in the following remark.

Remark 6.14. (i) For nice solutions of scalar-valued conservation laws, 0 0 the condition F (ul) > σ > F (ur) ensures that if one moves back- wards in time along characteristics, one does not encounter lines of discontinuity for u (see [Ev, Section 3.4.1 b)]).

(ii) Hence, roughly speaking, for k-shocks that satisfy the condition (6.29), information from the past propagating along k-characteristics is ab- sorbed in the k-shock, while shocks that satisfy the reverse inequalities become sources of new information which is then carried on along k- characteristics (see [Da00, Chapter 8.3]). For the gas dynamic equations, the Lax entropy condition is related to the fact that the entropy increases across the shock (see [Sm83]).

(iii) Roughly speaking, (6.29) is a stability condition in the sense that shocks are unstable if a smoother solution, e.g., a rarefaction wave, exists that connects the two states.

± In analogy to the deﬁnitions of Sk , we introduce the following notation.

Deﬁnition 6.15. Let (λk, rk) be genuinely nonlinear. Set

+ Sk (z0) := {z ∈ Sk(z0): λk(z0) < σ(z, z0) < λk(z)}, − Sk (z0) := {z ∈ Sk(z): λk(z0) < σ(z, z0) < λk(z0)}.

106 [July 12, 2019] − + Note that Sk = Sk ∪ {z0} ∪ Sk . We set

+ − Tk(z0) := Rk (z0) ∪ {z0} ∪ Sk (z0). (6.30)

1 Remark 6.16. (i) By Theorem 6.13, Tk(z0) is a C -curve.

(ii) If x ∈ Tk(y), then y and x can be joined by a shock wave or a rarefaction wave.

Theorem 6.17. Suppose that (λk, rk) is genuinely nonlinear for all k, and let ul be given. There is an open neighborhood N of ul such that for every ur ∈ N there exists a weak solution to Riemann’s problem which is constant along lines through the origin.

Proof. We show existence of a solution using the inverse function theorem. m For a state z ∈ R , i ∈ {1, . . . , m}, and ε ∈ R with |ε| small, set

i m τε(z) ∈ R to be the state characterized by the following properties:

i (i) τε(z) ∈ Ti(z), i + i − (ii) if ε > 0, then τε(z) ∈ Ri (z); and if ε < 0, then τε ∈ Si (z), i (iii) the distance from z to τε(z) along the curve Ti(z) equals ε. i By construction, τε(z) can be joined to z by a shock or a rarefaction wave. m Let N(0) ⊂ R be a suﬃciently small neighborhood of 0, and deﬁne Φ : m N(0) → R by

m m−1 1 Φ(ε) := τεm ◦ τεm−1 ◦ · · · ◦ τε1 (ul), ε = (ε1, . . . , εm).

1 m Note that Φ ∈ C (N(0), R ), Φ(0) = ul, and by deﬁnition of R(z) and Theorem 6.13, i τεi (u) = u + εiri(u) + o(εi), which implies DΦ(0) = (r1(ul), . . . , rm(ul)).

Since {rj(ul)} are linearly independent, DΦ(0) has full rank, and the inverse function theorem guarantees that there is a neighborhood of ul such that for every ur in this neighborhood there exists a unique t in a neighborhood of 0 such that Φ(t) = ur. Note that Φ(t) yields a construction manual for a solution to the Riemann problem if the shock waves and rarefaction waves used to connect to intermediate states do not intersect, i.e., the wave

107 [July 12, 2019] speeds of distinct characteristic families are strictly separated (cf. [Da00, Theorem9.3.1]). For that, label the intermediate states

z := τ k ◦ · · · ◦ τ 1 (u ). k εk ε1 l

Recall that a k-rarefaction wave joining zk−1 and zk in particular takes values x x zk−1 if t < λk(zk−1), and zk is λk(zk) < t . If zk−1 and zk are close to ul, then λk(zk−1) ≈ λk(ul) ≈ λk(zk). Similarly, if zk−1 and zk are joined by a k- x x shock, then it takes values zk−1 if t < σk(zk, zk−1), and zk if σ(zk, zk−1) < t . Since σ(ul, ul) = λk(ul), if the intermediate states are close to ul, then σ(zk, zk−1) ≈ λk(ul). By strict hyperbolicity, λ1(ul) < ··· < λm(ul), and thus, the shock waves and rarefaction waves connecting the intermediate states do not intersect if the neighborhoods are chosen suﬃciently small.

[17.6. 2019b, Lecture 17] [24.6. 2019a, Lecture 18]

108 [July 12, 2019] 7 Fourier methods

In NPDE I we studied nonlinear elliptic equations and systems of the form

−div A∇u = −div f and −div A∇ = g and a central theme were optimal regularity results of the form

f ∈ X =⇒ ∇u ∈ X g ∈ X =⇒ ∇2u ∈ X, where X is a function space. Very roughly we used three key ideas

(i) Multiplication of the equation by u (’energy estimates’) lead to L2 estimates

(ii) Scaling and decay estimates for the homogeneous equation lead to estimates in the Campanato space L2,λ. In particular we obtained estimates in the spaces

• C0,α = L2,n+2α for α ∈ (0, 1) (Schauder estimates) • BMO = L2,n

(iii) Interpolation between L2 and BMO leads to Lp estimates.

One important general observation was that the key point is to understand equations with constant coefficients. Continuous coefficients or coefficients in Ck,α can then be handled by perturbation arguments. In this section we will study a new approach to optimal regularity for n equations in R with constant coefficients. The key observation is that the Fourier transform turns differentiation into multiplication. Hence the solu- n tion of linear PDEs with constant coefficients in R can be formally reduced to a combination of Fourier transform, division and inverse Fourier transform. In view of Plancherel’s theorem this gives immediately L2 estimates for elliptic equations. In this chapter we discuss when these estimates can be extended to Lp estimates and we also go beyond elliptic equations. An excellent reference is [St70]. The basic properties of the Fourier transform were discussed in the course Analysis III. Other references for the Fourier transform include Chapter 7.1 of [Ho83] and Chapter 7 in [Ru73] (note that Rudin uses a definition of the Fourier transform which differs by a factor (2π)−n/2 and a different definition of convolution which also differs by a factor (2π)−n/2).

109 [July 12, 2019] 7.1 Examples 7.1.1 The Fourier transform

1 n Deﬁnition 7.1. For a function u ∈ L (R ; C) the Fourier transform is deﬁned by Z uˆ(ξ) = Fu(ξ) := u(x)e−ix·ξ dx. (7.1) Rn We recall four key properties of the Fourier transform.

(i) 1 F can be extended to a bijective isometry from L2( n; ) to (2π)n/2 R C itself. Thus Z dξ Z FuFv n = uv dx. (7.2) Rn (2π) Rn (ii) The Fourier transform turns diﬀerentiation into multiplication

∂dju = iξju,ˆ ixdju = −∂ξj uˆ (7.3)

(iii) The Fourier transform turns convolution into multiplication

−n u[∗ v =u ˆv,ˆ uvc = (2π) uˆ ∗ vˆ (7.4)

(iv) The inverse of the Fourier transform is given by Z −1 ix·ξ dξ (F v)(x) = v(ξ)e n . (7.5) Rn (2π)

Here the ﬁrst identity in (7.3) holds initially for u ∈ W 1,1 and can be extended to W 1,2 by using (i). The second identity in (7.3) holds if (1+|x|)u ∈ L1. In this caseu ˆ is in C1. From (iv) we immediately obtain the following useful relations

Fg = (2π)n(F −1g¯), (7.6) 1 1 F −1f = F(Rf) = R(Ff) (7.7) (2π)n (2π)n where Rf(x) = f(−x). Thus in particular F 2 = (2π)nR,RFR = F. (7.8)

110 [July 12, 2019] The Fourier transform has the following scaling properties.

F(u(· − a))(ξ) = e−ia·ξFu(ξ) (7.9) ξ F( u(s·) )(ξ) = s−n(Fv) . (7.10) | {z } s L∞ scaling | {z } L1 scaling ξ F( snu(s·) )(ξ) = (Fv) . (7.11) | {z } s L1 scaling | {z } L∞ scaling To memorize the second and third scaling relation it suﬃces to keep in mind that both F and F −1 map L1 to L∞. A very useful fact is that the Fourier transform maps Gaussians into Gaussians. 2 2 1 − |·| − |ξ| F e 2 (ξ) = e 2 (7.12) (2π)n/2 Using the scaling relation we deduce that

n/2 2 ! 2 a − a|·| − |ξ| F e 2 (ξ) = e 2a (7.13) (2π)n/2 for a > 0. A short argument (see Homework 1) show that (7.13) also holds for a ∈ C as long as Re a > 0. If A is a symmetric positive deﬁnite n × n matrix then 1/2 ! −1 (det A) − (A·,·) − (A ξ,ξ) F e 2 (ξ) = e 2 (2π)n/2

7.1.2 The Fourier transform on tempered distributions The Fourier transform can be extended to (tempered) distributions as follows. The Schwartz class of rapidly decaying smooth functions is deﬁned as

∞ n β α S := {ψ ∈ C (R ; C) : sup |x ∂ ψ(x)| < ∞ for all multiindices α, β} x (7.14) On S one deﬁnes seminorms and a metric as follows

β α X −|α|−|β| [ϕ − ψ]α,β [ψ]α,β = sup |x ∂ ψ(x)|, d(ϕ, ψ) = 2 . x 1 + [ϕ − ψ]α,β α,β Note that the sum converges since ∞ ∞ n ∞ X X X Y X X 2−|α| = ... 2−α1 ... 2−αn = 2−αj = 2n, 2−|α|−|β| = 4n. n n α∈N α1=0 αn=0 j=1 αj =0 α,β∈N

111 [July 12, 2019] With this metric S is complete (thus S is a Frechet space). Using (7.3) and (7.5) one easily sees that F maps S bijectively to itself. Moreover it is easy to check that F is continous map from S to itself (see Homework 1). Elements in the dual S0 are called tempered distributions. Note that ∞ Cc ⊂ S and hence every tempered distribution is a distribution. The reason ∞ to introduce S is that Cc is not mapped by F to itself. For T ∈ S0 one now deﬁnes the Fourier transform by duality

(FT )(ϕ) := T (Fϕ) (7.15)

If f ∈ L1 and T (ϕ) = R fϕ dx then f Rn

FTf = TFf , (7.16) i.e., the Fourier transform on distributions is consistent with the usual definition if the distribution comes from an L1 function. To prove (7.16) note that application of (7.6) with g = h¯ yields Fh¯ = (2π)n(F −1h). Taking g = h¯ in Plancherel’s identity we thus get Z Z fh dx = Ff F −1h dξ. (7.17) Rn Rn For ϕ ∈ S and h = Fϕ we therefore obtain Z Z −1 (FTf )(ϕ) = Tf (h) = fh dx = Ff F h dξ = TFf (ϕ) Rn Rn which proves (7.16). Using again that Fϕ = (2π)n(F −1ϕ¯) we can easily extend the Plancherel identity (7.2) to tempered distributions 1 FT (Fϕ) = FT (F −1ϕ¯) = T (ϕ ¯). (7.18) (2π)n

Note that diﬀerentiation and multiplication by a smooth function with polyomial growth induce continuous maps on S. Hence for T ∈ S we can deﬁne as usual

(∂αT )(ψ) = (−1)|α|T (∂αψ), (fT )(ψ) = T (fψ).

It follows that (7.3) also holds for tempered distributions

F(∂αT ) = (iξ)αFT, F((−ix)αT ) = ∂α(FT ). (7.19)

Finally we deﬁne ϕ ∗ T for ϕ ∈ S and T ∈ S0 and verify that (7.4) still holds, i.e.,

F(ϕ ∗ T ) = Fϕ FT, F(ϕT ) = (2π)−nFϕ ∗ FT. (7.20)

112 [July 12, 2019] First note that for f ∈ L1, g, h ∈ L1 ∩ L∞ we have Z Z (f ∗ g)h dx = f(Rg ∗ h) dx, where Rg(x) = g(−x). Rn Rn Thus we deﬁne

(ϕ ∗ T )(ψ) := T ((Rϕ) ∗ ψ), ∀ T ∈ S0, ϕ, ψ ∈ S.

1 −1 This ensures that ϕ ∗ Tf = Tf∗ϕ for f ∈ L . Using the identities F = (2π)−nFR, F 2 = (2π)nR and RFR = F (see (7.8)) we get

(F(ϕ ∗ T ))(ψ) = (ϕ ∗ T )(Fψ) = T (Rϕ ∗ Fψ) =T (F −1(FRϕ F 2ψ) = T (FR(FRϕ Rψ)) =FT (RFRϕ ψ) = FT (Fϕ ψ) = (Fϕ FT )(ψ) and this proves the ﬁrst identity in (7.20). To obtain the second identity one can argue similarly or one can apply F to the ﬁrst identity and use the relation F 2 = (2π)nR where RT (φ) := T (Rϕ). The following identities are often useful

n Fδ0 = 1, F1 = (2π) δ0 (7.21)

7.1.3 Relations with PDE Example 1: The Poisson equation Assume that n −∆u = f in R . If u and f are suﬃciently smooth and integrable this is equivalent to

|ξ|2uˆ(ξ) = fˆ(ξ).

Since ∂\j∂ku = −ξjξkuˆ we have −ξ ξ ∂\∂ u = j k fˆ(ξ) (7.22) j k |ξ|2 | {z } m(ξ) ’Fourier multiplier’ Now an optimal regularity estimate in L2 follows from the boundedness of m and Plancherel’s identity. 1 1 1 k∂ ∂ uk2 = k∂\∂ uk2 = kmfˆk2 ≤ kfˆk2 = kfk2 . j k L2 (2π)n j k L2 (2π)n L2 (2π)n L2 L2 Using the fact that the Fourier transform converts convolution in multiplication we can at least formally rewrite the relation (7.22) as

−1 ∂j∂ku = K ∗ f, where K = F m.

113 [July 12, 2019] Since the solution u can be represented in terms of the Greens function we get formally that (for n ≥ 3) c K(x) = ∂ ∂ n . j k |x|n−2

Thus K is homogeneous of degree −n and therefore just fails to be in L1. The convolution with K is called a singular integral and one of our tasks is to give a meaning to such singular integrals (and to analyse under which conditions they map Lp to Lp). First we state the L2 estimate explicitly.

∞ n Proposition 7.2. Let m ∈ L (R ) and deﬁne the operator T by

T u := F −1(mFu).

2 n Then T maps L (R ) to itself and

kT ukL2 ≤ kmkL∞ kukL2

A key question is under which additional conditions on m the operator T maps Lp to Lp. We ﬁrst look at some other examples.

Example 2: The heat equation. Consider a solution of

n ∂tu − ∆u = 0 in (0, ∞) × R n u(0, x) = u0(x) ∀x ∈ R

For each t ≥ 0 letu ˆ(t, ·) denote the Fourier transform of u(t, ·), i.e., Z uˆ(t, ξ) = u(x)e−iξ·x dx. Rn Then

2 ∂tuˆ(t, ξ) + |ξ| uˆ(t, ξ) = 0 uˆ(0, ξ) = uc0(ξ). Thus the PDE has been transformed into an inﬁnite family of ODEs. Their solution is given by −t|ξ|2 uˆ(t, ξ) = e uc0(ξ). The Fourier multiplier 2 m(ξ) = e−t|ξ| is of rapid decay for large ξ (’high frequencies’). This leads to a strong smoothing of the initial data for positive time. A simple way to see this

114 [July 12, 2019] 2 (for u0 ∈ L ) is the following calculation for which we use the multiindex notation α α1 αn ∂ = ∂x1 . . . ∂xn , |α| = α1 + ... + αn. Z Z α 2 1 α 2 |∂ u| (t, x) dx = n |ξ uˆ(t, ξ)| dξ Rn (2π) Rn 1 Z 2 ≤ |ξ|2|α|e−2t|ξ| |u (ξ)|2 dξ n c0 (2π) Rn 1 Z 2 = t−|α| (t|ξ|2)|α|e−2t|ξ| |u (ξ)|2 dξ n c0 (2π) Rn | {z } ≤C|α| −|α| 2 ≤C|α|t ku0kL2 .

k,2 n It follows that u(t, ·) is bounded in all Sobolev spaces W (R ) and hence smooth. To obtain a representation of the solution in real space we can use the formula f[∗ g = fˆgˆ to conclude that

−1 −t|·|2 u(t, ·) = Kt ∗ u0,Kt = F e .

From the formula for the Fourier transform of Gaussians (7.12) we get −n/2 2 K 1 (x) = (2π) exp(−|x| /2) and the scaling relation (7.10) implies that 2

2 1 − |x| K (x) = e 4t . (7.23) t (4πt)n/2

Example 3: The Schr¨odingerequation. Consider a solution of

n i∂tu − ∆u = 0 in (0, ∞) × R (7.24) n u(0, x) = u0(x) ∀x ∈ R and let againu ˆ(t, ξ) denote the Fourier transform of u(t, ·) at ﬁxed t. We obtain again an inﬁnite system of ODEs which has the solution

it|ξ|2 uˆ(t, ξ) = e uc0(ξ). Now the Fourier multiplier

2 m(ξ) = eit|ξ| satisﬁes |m| = 1. Thus ku(t, ·)kL2 = ku0kL2 . (7.25) For this Fourier multiplier there are no Lp estimates for any p 6= 2. Thus it p is not possible to bound the L norm of u(t, ·) in terms of ku0kLp .

115 [July 12, 2019] Some interesting estimates can, however, be obtained from the real space representation of the solution in the form

u(t, ·) = Lt ∗ u0. (7.26)

Completely formally the Schr¨odingerequation can be obtained from the heat equation by replacing t by t/i. Making this change in the kernel Kt suggests that the kernel Lt is given by

n/2 2 i − i|x| L = e 4t . t (4πt)n/2 A short calculation shows that indeed

i∂tLt − ∆Lt = 0

2 2 Moreover one can show that Lt ∗ u0 → u0 in L as t → 0 if u0 ∈ L . To do ∞ n so one can ﬁrst assume that u0 ∈ Cc (R ) and approximate Lt by

n/2 2 ε (i + ε) − (i+ε)|x| L = e 4t t (4πt)n/2

(for related issues see Homework 1). Thus (7.26) really deﬁnes a solution of the Schr¨odingerequation. From this representation and the fact that −n/2 |Lt| ≤ (4πt) one obtains the algebraic decay estimate 1 ku(t, ·)k ∞ ≤ ku k 1 . (7.27) L (4πt)n/2 0 L Interesting further estimates can be obtained by interpolating between this estimate and the L2 estimate (7.25). Alternatively one can look at the inhomogeneous problem

n i∂tu − ∆u = f in R × R and take the Fourier transform in space and time, given by Z (Fu)(τ, ξ) = u(t, x)e−i(tτ+x·ξ) dx dt. R×Rn This yields i(iτ) + |ξ|2 Fu = Ff (7.28) The Fourier multiplier m(τ, ξ) = −τ + |ξ|2 is not elliptic. Indeed it vanishes on the lower dimensional singular set

S = {(τ, ξ): τ = |ξ|2}.

The fact that S is curved is of crucial importance for more subtle estimates.

116 [July 12, 2019] Example 4: General scalar elliptic equations. Consider the diﬀerential operator

X α Lu = aα∂ u |α|≤m with constant coeﬃcients aα. There u and the aα may take values in R or C. The symbol of L is deﬁned as

X α σ(ξ) = aα(iξ) |α|≤m and the principal symbol is

X α σm(ξ) = aα(iξ) . |α|=m

Different authors use different conventions whether to use ξ or iξ in the definition of the symbol.

Deﬁnition 7.3. The operator L is called elliptic if

n σm(ξ) 6= 0 ∀ξ ∈ R \{0}.

Remark. One easily sees that L is elliptic if and only if there exists a c > 0 such that m n |σm(ξ)| ≥ c|ξ| ∀ξ ∈ R . (7.29) n−1 Indeed, since σm is m-homogeneous it suﬃces to consider ξ ∈ S and we may take c = minSn−1 |σm(ξ)|. If the aα are real then L can only be elliptic −m if m is even. Indeed, if m is odd then i σm is an odd real-valued function on Sn−1 and thus by the intermediate value theorem must have a zero on Sn−1. The remainder of this subsubsection was not discussed in detail in class.

2 n Theorem 7.4. Let L be elliptic, f ∈ L (R ) and σ = σm. Let

m,2 n α 2 n Xm := {u ∈ Wloc (R ): ∂ u ∈ L (R ) if |α| = m} Then

(i) There exists a constant C which only depends on σm, m and n with the following property. If u ∈ W m,2 and Lu = f then

X α 2 2 k∂ ukL2 ≤ CkfkL2 . (7.30) |α|=m

117 [July 12, 2019] 2 (ii) For each f ∈ L there exists a u ∈ Xm which satisﬁes Lu = f. More- over u is unique up to the addition of a polynomial of degree ≤ m − 1. In addition u satisﬁes (7.30).

Proof. (i) The Fourier transforms of u and all its partial derivatives up to 2 order m are in L and we have σmFu = Ff. Thus for |α| = m

α 2 α 2 1 α 2 1 (iξ) k∂ ukL2 = n k(iξ) FukL2 = n Ff (2π) (2π) σm(ξ) L2 1 ≤ CkFfk2 = Ckfk2 . (2π)n L2 L2

m Here we used that kσm(ξ)k ≥ c|ξ| with c > 0. (ii) The uniqueness result follows from (i) applied to the difference of two −1 1 solutions. A natural approach to show existence is to define u = F σ Ff. 1 It is, however, not obvious from this definition that u is in Lloc or even that 1 the inverse Fourier transform of σ Ff is defined. It is easier to start from the m-th derivatives of the putative solutions u. If u is a solution these derivatives satisfy

(iξi1 ) ... (iξim ) F(∂i1 . . . ∂im u) = Ff. σm(ξ) To turn this approach into a rigorous proof we deﬁne nm functions by

−1 (iξi1 ) ... (iξim ) vi1...im := F Ff. σm(ξ) Then Plancherel’s theorem and the lower bound |σ(ξ)| ≥ c|ξ|m imply that 2 n m,2 n vi1...im ∈ L (R ). It remains to show that there exists a u ∈ Wloc (R ) such that

∂i1 . . . ∂im u = vi1...im .

This follows from Lemma 7.5 below. Indeed, symmetry of the vi1...im with respect to permutation of the indices is clear. To verify (7.31) we use Plancherel’s identity. We have Z Z dξ vi1...im−1j ∂kϕ dx = Fvi1...im−1j F∂kϕ n Rn Rn (2π) Z (iξi1 ) ... (iξim−1 )(iξj) dξ = Ff(ξ) (iξk)Fϕ(ξ) n . Rn σm(ξ) (2π) The last expression is symmetric in j and k and hence (7.31) holds. Now it is easy to verify that u indeed solves the equation. We have

X α X Lu = aα∂ u = aαvα α α

118 [July 12, 2019] since aα = 0 if |α|= 6 m. Thus

X X (iξ)α FLu = a Fv = a Ff = Ff. α α α σ (ξ) α m Thus Lu = f.

m Lemma 7.5. Let m ∈ N\{0} and assume that the n functions vi1...im with 1 n ij ∈ {1, . . . , n} belong to Lloc(R ) and satisfy

(i) viπ(1)...iπ(m) = vi1...im for all permutations π;

(ii) ∂kvi1...im−1j = ∂jvi1...im−1k in the sense of distributions, i.e., Z ∞ n vi1...im−1j ∂kϕ − vi1...im−1k ∂jϕ dx = 0 ∀ϕ ∈ Cc (R ). (7.31) Rn

m,1 n Then there exists u ∈ Wloc (R ) such that

∂i1 . . . ∂im u = vi1...im

p n in the sense of weak derivatives. If, in addition, vi1...im ∈ Lloc(R ) for some m,p n p ∈ (1, ∞], then u ∈ Wloc (R ). Here the v may be real or complex-valued.

Proof. Homework 1. Hint: First consider m = 1 and then do the general case by induction. For ε m = 1 deﬁne the regularized functions vj = ρε ∗ v and show that these functions still satisfy (ii). Then the classical result from Analysis II shows ε 1 ε ε that there exists a u ∈ C such that ∂ju = vj . By subtracting a constant if needed we may assume that Z uε dx = 0. B(0,1)

ε Thus the (generalized) Poincar´einequality gives estimates for ku kL1(B(0,R)) ε δ ε 1 and ku − u kL1(B(0,R)). Conclude that u → u in Lloc as ε → 0. For the assertion about Lp also use the Poincar´einequality.

2 n Corollary 7.6. Let L be elliptic, f ∈ L (R ). Then there exists a constant C which only depends on σ and m with the following property. If u ∈ 2 n Xm ∩ L (R ) and Lu = f then

X α 2 2 2 k∂ ukL2 ≤ C kfkL2 + kukL2 . |α|=m

119 [July 12, 2019] Proof. Homework. 2 n m,2 n Hint: By Fourier transform one easily sees that Xm ∩ L (R ) = W (R ). Show ﬁrst that c |(σ − σ )(ξ)| ≤ |ξ|m + C m 2 P α and bring the term |α|

7.2 Singular integrals, Lp estimates and multiplier theorems This subsection follows closely [St70]. We look for criteria on the multiplier m which guarantee that the operator T deﬁned by

T f := F −1mFf

p n is a bounded operator from L (R ) to itself. For p = 2 a necessary and sufficent criterion is that m ∈ L∞. To derive a criterion for p 6= 2 we first note that at least formally T f = K ∗ f where K is the tempered distribution K = F −1m (note that every L∞ function defines a tempered distribution since S ⊂ L1). We first show a result which involves more general kernels. Theorem 7.7 (Lp estimates for Calderon-Zygmund kernels). Suppose that 2 n T is is a linear map which maps functions in L (R ) with compact support n to measurable functions on R and satisfies 2 n (i) kT fkL2 ≤ AkfkL2 for all f ∈ L (R ) with compact support; n n (ii) there exists a measurable function K on R × R such that Z T f(x) = K(x, y)f(y) dy for all x∈ / supp f Rn 2 n and all f ∈ L (R ) with compact support. Moreover Z |K(x, y) − K(x, y¯)| dx ≤ A ∀y ∈ B(¯y, r). (7.32) Rn\B(¯y,2r)

Then kT fkLp ≤ ApkfkLp ∀p ∈ (1, 2] 2 n for all f ∈ L (R ) with compact support and thus T can be extended to a p n bounded operator from L (R ) to itself. The constant Ap depends only on A, p and n.

120 [July 12, 2019] Proof. Step 1. Marcinkiewicz interpolation In view of the Marcinkiewicz interpolation theorem (see NPDE I) it suﬃces to show that T maps L1 into weak-L1, more precisely that there exists a constant A1 such that

n A1 L {x : |T f(x)| > t} ≤ kfk 1 . (7.33) t L More precisely, since T is originally only defined on L2 functions with compact support we will first show (7.33) for L2 functions with compact support and then use this estimate and assumption (i) to show that T can be extended to a linear operator T¯ on L1+L2 such that T¯ satisfies the counterpart of (7.33) for all f ∈ L1. Then one can apply the Marcinkiewicz interpolation theorem to T¯. Step 2. Calderon Zygmund decompostion To prove the weak-L1 estimate we divide f into a good part g and a bad part b which is supported on a well-controlled set. n Lemma 7.8 (Calderon-Zygmund decomposition II). Let Q0 ⊂ R be a 8 1 cube . Assume that f ∈ L (Q0) with Z – |f| dx ≤ s.

Then there exist at most countably many closed subcubes Ik with disjoint interior and a decomposition

f = b + g such that

(i) |g| ≤ 2ns a.e.;

(ii) Z X n+1 b = bk, supp bk ⊂ Ik, – |bk| dx ≤ 2 s, k Ik Z bk dx = 0; (7.34) Ik

(iii) P Ln(I ) ≤ 1 R |f|; k s Q0 (iv) R |g| dx ≤ R |f| dx, R |b| dx ≤ 2 R |f| dx. Q0 Q0 Q0 Q0

8As in NPDE I by cube we mean a cube parallel to the coordinate axis unless something else is state explicitly

121 [July 12, 2019] Remark. The same result holds with the same proof if f takes values in a normed space X provided that |f| is replaced by kfkX .

Proof of Lemma 7.8. This can be proved by dyadic decomposition of Q0 as in the proof of the Calderon-Zygmund decomposition in NPDE I. Alterna- tively one can use that result (applied to h = |f|) and deﬁne

X Z Z g = χ S f + χ – f, b = χ (f − – f) Q0\ k Ik Ik k Ik k Ik Ik Then the properties above follow easily from the result in NPDE I.

Note that t t {|T f| > t} ⊂ {|T g| > } ∪ {|T b| > }. (7.35) 2 2 Step 3. Estimate for T g. Here and in the next step we assume that f ∈ L2 has compact support. We ﬁx t > 0 and assume that the Q is so large that it contains supp f and that Z 0 t ≥ – |f| dx. We extend g and b by zero outside Q0.

Q0 Since g ∈ L∞ we can use the L2 estimate for g. We have Z n t 4 2 L ({|T g| ≥ } ≤ 2 |T g| dx 2 t Rn Z Z n 2 4 2 2 4 2 n 4 2 A ≤ 2 A |g| dx ≤ 2 A 2 t |g| dx ≤ kfkL1 . t Rn t Rn t Step 4. Estimate of T b. Lety ¯k be the center of the cube Ik. Let Bk = B(¯yk, rk) be the smallest ball ∗ ∗ which contains the interior of Ik and let Bk = B(¯yk, 2rk). For x ∈ Bk we have Z Z T bk(x) = K(x, y)bk(y) dy = (K(x, y) − K(x, y¯k))bk(y) dy Ik Ik R where we used that bk dx = 0. By Fubini’s theorem and the assumption Ik Z |T bk(x)| dx n ∗ R \Bk Z Z ≤ |K(x, y) − K(x, y¯k)| dx |bk(y)| dy n ∗ Ik R \Bk | {z } ≤A Z ≤A |bk| dy. Ik

122 [July 12, 2019] Since the functions bk have disjoint support we get from Lemma 7.8(iv) Z X Z |T b| dx ≤ A |bk| dy ≤ 2AkfkL1 . n S ∗ R \ k Bk k Ik

n ∗ n/2 n Finally we note that L (Bk) = n ωnL (Ik) where ωn is the volume of the n unit ball in R . Thus Z n t n [ ∗ 2 L {|T b| > } ≤ L ( Bk) + |T b| dx 2 t n S ∗ k R \ k Bk n/2 n [ 4A ≤ n ω L ( I ) + kfk 1 n k t L C ≤ kfk 1 t L where we used Lemma 7.8(iii) in the last step. Step 5. Extension of T to L1 + L2 with a weak weak-L1 bound. This was not discussed in class Let

2 n X = {f ∈ L (R ): f has compact support}. 1 1 We want to extend T to an operator T1 on L such that the weak-L estimate still holds. It follows from Steps 3 and 4 and (7.35) that

n C L ({|T f| > t} ≤ kfk 1 ∀f ∈ X. t L Since X is dense in L1 we would like to conclude that the same estimate holds for all f in L1. A slight subtlety is that the expression A N(g) := inf{A > 0 : Ln{|g| > t} ≤ ∀t > 0} t is only a quasi-norm and not a norm because it only satisﬁes a weakened triangle inequality of the form

N(f + g) ≤ 2(N(f) + N(g)).

This subtlety can be easily circumvented in the following way. We first 1 1 show that if f ∈ L , hk ∈ X and hk → f in L as k → ∞ then T hk converge to a measurable function F which only depends on f and not on the particular sequence hk. This implies in particular that F = T f if f ∈ X (in this case one may take hk = f. We then define an operator T1 by T1f := F . It follows easily that linearity of T on X implies linearity of T1 1 1 on L . We finally show that the weak L bounds extends from T to T1. 1 Let f ∈ L . Then there exist gk ∈ X such that

−k kf − gkk ≤ 2 kfk.

123 [July 12, 2019] Thus n C −k+1 C L {|T g − T g | > t} ≤ kg − g k 1 ≤ 2 . k+1 k t k+1 k L t Hence the functions T gk converge in measure to a limit F . Now consider 1 hk ∈ X with hk → f in L as k → ∞. Then n n n L {|F − T hk| > t} ≤ L {|F − T gk| > t/2} + L {|T gk − T hk| > t/2} 2C ≤Ln{|F − T g | > t/2} + kg − h k. k t k k Now the ﬁrst term goes to zero by the deﬁnition of F and the second term 1 goes to zero since gk − hk → 0 in L . Thus T hk converges in measure to F . Moreover Ln({|F | > t}) n n ≤L ({|F − T gk| > t/2}) + L ({|T gk| > t/2})

n C ≤L ({|F − T g | > t/2}) + kg k 1 k t k L n C C ≤L ({|F − T g | > t/2}) + kfk 1 + kg − fk 1 . k t L t k L Now the ﬁrst and third term in the last line go to zero as k → ∞ and thus

n n C L ({|T f| > t}) = L ({|F | > t}) ≤ kg − fk 1 . 1 t k L Finally by assumption (i) the linear operator T also has a unique exten- 2 sion to a bounded linear operator T2 from L to itself. By uniqueness of the 1 2 1 2 extensions T1 (to L ) and T2 (to L ) we see that T1 = T2 on L ∩ L . Now for any function f of the form f = g + h with g ∈ L1 and h in L2 we deﬁne

T¯ f = T1g + T2h. 1 2 0 0 The fact that T1 = T2 on L ∩ L guarantees that T1g + T2h = T1g + T2h if g + h = g0 + h0, i.e., T¯ depends only on f and not on the particular decomposition of f into an L1 function and an L2 function. Now T¯ satisfies the assumptions of the Marcinkiewicz interpolation theorem. Theorem 7.9 (Mihlin-Hörmandermultiplier theorem). Assume that m ∈ l n C (R \{0}) and |m| ≤ A and |∂αm(ξ)| ≤ A|ξ|−|α| ∀|α| ≤ l n where l is the smallest integer with l > 2 . Then the operator T := F −1mF 2 n which is originally defined on L (R ) can be extended to a bounded operator p n from L (R ) to itself for 1 < p < ∞ and the norm Ap of the extended operator depends only on A, p and n.

124 [July 12, 2019] The proof will be based on a dyadic decomposition of m. We ﬁrst state a corresponding partition of unity result. Proposition 7.10 (Dyadic partition of unity). There exists a function ϕ ∈ ∞ 1 Cc (B(0, 4) \ B(0, 2 )) such that ∞ X ϕ(2−jξ) = 1 ∀ξ 6= 0. j=−∞ We may take ϕ radially symmetric.

∞ 1 Proof. Let ψ ∈ Cc (B(0, 4) \ B(0, 2 )) with ψ = 1 on B(0, 2) \ B(0, 1) and ψ ≥ 0. Set ψ(ξ) ϕ(ξ) = P∞ −k k=−∞ ψ(2 ξ) Note that the denominator is ﬁnite since at most three terms in the sum are non-zero. Moreover the denominator is bounded from below by 1 for each ξ. If ψ is radially symmetric then so is ϕ.

∞ 3 Alternative proof. Let ψ ∈ Cc (B(0, 2)) with ψ = 1 on B(0, 2 ). Set ϕ(ξ) = ∞ 1 ψ(ξ) − ψ(2ξ). Then ϕ ∈ Cc (B(0, 2) \ B(0, 2 )) and ϕ(2−jξ) = ψ(2−jξ) − ψ(2−(j−1)ξ)

Thus K X ϕ(2−jξ) = ψ(2−K ξ) − ψ(2K+1ξ) j=−K The right hand side equals 1 in B(0, 2K ) \ B(0, 2−K ) and taking K → ∞ we get the assertion.

Proof of Theorem 7.9. The proof follows [St93], pp. 245–247. It suﬃces to prove the result for 1 < p ≤ 2. Then the general case follows by duality. Step 1. Dyadic decomposition ∞ 1 Let ϕ ∈ Cc (B(0, 4) \ B(0, 2 )) be as in Proposition 7.10 and set −j −1 −1 mj(ξ) := ϕ(2 ξ)m(ξ),Kj := F mj,Tj := F mjF.

1 n ∞ n ∞ n 2 n Then mj ∈ L (R ) ∩ L (R ) and hence Kj ∈ L (R ) ∩ L (R ). Moreover P∞ n Pk j=−∞ mj = m on R \{0} and j=−k |mj| ≤ 3|m|. Thus it follows from 2 n the dominated convergence theorem that for all f ∈ L (R )

k k X X 2 n T f := lim Tjf = lim Kj ∗ f in L (R ) k→∞ k→∞ j=−k j=−k

125 [July 12, 2019] 2 n Note that the convolution Kj ∗f is well deﬁned since Kj and f are in L (R ). We will see below that the limit

k X 2 n K := lim Kj exists in Lloc(R \{0}). (7.36) k→∞ j=−k

∞ n The convergence holds even in C (A) if A is a compact subset of R \{0} but we will not use this. 2 n Thus for f ∈ L (R ) with compact support and x∈ / supp f we have Z T f(x) = K(x − y)f(y) dy. Rn In view of Theorem 7.7 it thus suﬃces to show that (7.32) holds for the kernel K˜ (x, y) = K(x − y). Since in this case we may take without loss of generalityy ¯ = 0 the required estimate reduces to Z |K(x − y) − K(x)| dx ≤ A0 ∀y ∈ B(0, r). (7.37) Rn\B(0,2r)

2 Step 2. Weighted L estimates for Kj Using Plancherel’s identity and the fact that Fourier transform and inverse Fourier transform convert diﬀerentiation into multiplication we get for |γ| ≤ l Z Z γ 2 1 γ 2 |(−ix) Kj(x)| = n |∂ mj(ξ)| dξ (7.38) n (2π) j 1 j R B(0,4·2 )\B(0, 2 2 ) Now by the Leibniz rule and the assumption ∂βm(ξ) ≤ C|ξ|−|β| and the fact j 1 j that mj is supported on B(0, 4 · 2 ) \ B(0, 2 2 )

γ X α −j β −j|α| −j|β| −j|γ| ∂ξ mj(ξ) = cαβ ∂ξ ϕ(2 ξ) ∂ξ m(ξ) ≤ C2 2 = C2 . α+β=γ

Summing (7.38) over all multiindices γ with |γ| = M we thus get for 0 ≤ M ≤ l Z 2M 2 −2jM jn |x| |Kj(x)| dx ≤ C2 2 . (7.39) Rn If we use this estimate with M = 0 for j ≤ 0 and M = l for j > 0 we see that (7.36) holds. 1 Step 3. L estimates for Kj and proof of (7.37) We now use the Cauchy-Schwarz inequality to turn the weighted L2 estimates into L1 estimates. Applying (7.39) with M = 0 we get Z n/2 jn/2 |Kj| dx ≤ Ca 2 . B(0,a)

126 [July 12, 2019] On the other hand application of (7.39) with M = l yields Z Z −l l |Kj| dx = |x| |x| |Kj| dx Rn\B(0,a) Rn\B(0,a) ≤Can/2−l 2−jl 2jn/2 = Can/2−l 2j(n/2−l). (7.40) Combining these estimates for a = 2−j we deduce that Z |Kj| dx ≤ C. Rn Similarly one shows that Z |∇Kj| dx ≤ C2j. Rn R 1 Since K(x − y) − K(x) = 0 −∇K(x − ty) · y dt this implies that Z j |Kj(x − y) − Kj(x)| ≤ C|y|2 . Rn Assume that |y| ≤ r and use the last estimate if 2j ≤ r−1 and (7.40) if 2j > r−1. This yields ∞ X Z |Kj(x − y) − Kj(x)| dx n j=−∞ R \B(0,2r) X X ≤ Cr2j + Crn/2−l2j(n/2−l) {j:2j ≤r−1} {j:2j >r−1} ≤Crr−1 + Crn/2−lr−(n/2−l) ≤ C. This implies (7.37).

[24.6. 2019b, Lecture 19] [1.7. 2019a, Lecture 20]

Remark. It can be easily seen from the proof that the pointwise assumption on the derivatives can be replaced by the following L2 version Z sup R−n−2α |∂αm|2 dξ ≤ C ∀|α| ≤ l. R B(0,2R)\B(0,R) The Marcinkiewicz multiplier theorem provides an even weaker condition which only involves estimates of the form Z k ∂ m sup ≤ A xk+1,...,xn R ∂x1 . . . ∂xk k for all dyadic rectangles R ⊂ R and each 1 ≤ k ≤ n (see Theorem 7.11 for a precise statement).

127 [July 12, 2019] Examples

(i) Elliptic equations of order k without lower order terms. In this case

ikξ . . . ξ m(ξ) = j1 jk σk(ξ)

n−1 Thus m(ξ) = m(ξ/|ξ|) and m is smooth on S since σk does not vanish on Sn−1. Hence m satisﬁes the assumptions of the Mihlin- H¨ormandermultiplier. It follows that in Theorem 7.4 one can replace L2 by Lp with p ∈ (1, ∞). In combination with the estimate (Home- work)

|α| α k − k∂ ukLp ≤ Cp,α εk∇ ukLp + ε k−|α| kukLp for |α| < k, 1 < p < ∞

one sees that also Corollary 7.6 holds for Lp with 1 < p < ∞.

(ii) The Riesz transform. The Riesz transform Rj is the operator T given by the multiplier iξj/|ξ|. Since the multiplier is also homogeneous of n−1 degree zero and smooth on S the operator Rj is a bounded operator p n from L (R ) to itself for p ∈ (1, ∞). Symbolically one sometimes writes −1/2 Rj = ∂j(−∆) . P 2 Note that j Rj = −Id and thus

n X f = − Rj(Rjf). j=1

It thus follows that for p ∈ (1, ∞)

α p n −1 k p n ∂ f ∈ L (R ) ∀ |α| = k ⇐⇒ F |ξ| Ff ∈ L (R ) (7.41)

(iii) m(ξ) = h(|ξ|) with h ∈ Cl(0, ∞), |h(k)(t)| ≤ At−k for t > 0 and 0 ≤ k ≤ l. In particular the assumptions are satisﬁed for

|ξ|r m(ξ) = if r ≤ s (1 + |ξ|2)s/2

(Homework). For a diﬀerent argument that m is a good Fourier multiplier for r = s see [St70], p. 133, Lemma 2.

(iv) m(ξ) = |ξ|is

128 [July 12, 2019] Example ii) motivates the following deﬁnition of the fractional Sobolev p spaces Ls

p n 0 −1 2 s/2 p n Ls(R ) := {f ∈ S : F (1 + |ξ| ) Ff ∈ L (R )}, for s ∈ R. For s ≥ 0 one obtains an equivalent deﬁnition if one starts from f ∈ Lp 0 p k,p rather than f ∈ S . It follows from (7.41) and Example iii) that Lk = W for k ∈ N and 1 < p < ∞.

An example of a multiplier which does not satisfy the assumptions of the Mihlin-H¨ormandertheorem is the multiplier

2 m(ξ) = ei|ξ| which appears in the Schr¨odingerequation.

Scaling To explore the strengths and weaknesses of the Mihlin-Hörmander multiplier theorem we consider the effect of scaling on the multiplier. Define x S f(x) := f( ) λ λ Then by (7.10)

n −1 n −1 FSλ = λ S 1 F, F λ S 1 = SλF λ λ Moreover for the product of two functions we have the relation

m (S 1 g) = S 1 [(Sλm) g] λ λ

−1 Thus writing Tm = F mF we get

−1 TmSλ = SλTSλm or TSλm = Sλ TmSλ. −n/p p Since λ Sλ is an isometry on L it follows that

p p p p kTSλmkL(L ,L ) = kTmkL(L ,L ) One can easily check that the conditions in the Mihlin-H¨ormandertheorem are invariant under the transformation m 7→ Sλm. Now consider the anisotropic scaling operator x x (Sf)(x) = f( 1 ,..., n ) λ1 λn Then it is easy to see that

n Y −1 FS = λj S F j=1

129 [July 12, 2019] and thus

−1 TSm = S TmS, so that kTSmkL(Lp,Lp) = kTmkL(Lp,Lp). Note, however, that the assumptions of the Mihlin-H¨ormandertheorem are not invariant under anisotropic scaling. The assumptions of the following Marcinkiewicz multiplier theorem are invariant under anisotropic rescaling. We say that I is a dyadic interval if I is of the form I = [2j, 2j+1] or j+1 j k I = [−2 , −2 ] with j ∈ Z, and we say that R ⊂ R is a dyadic rectangle if R = I1 × ... × Ik where each interval Ii is a dyadic interval. n n Theorem 7.11. Assume that m is a C function in each ’octant’ {x ∈ R : σjxj > 0} where σj = ±1 and suppose (i) |m| ≤ A;

k (ii) for each 1 ≤ k ≤ n and each dyadic rectangle R ⊂ R Z k ∂ m sup dx1 . . . dxk ≤ A xk+1,...,xn R ∂1 . . . ∂k where the ’sup’ sign is omitted if k = n; (iii) the condition analogous to (ii) holds for every one of the n! permutations of the variables x1, . . . , xn −1 p n Then T = F mF can be extended to a bounded operator from L (R ) to itself for p ∈ (1, ∞) and the operator norm Ap of T depends only on A, p and n. Proof. See [St70], pp. 109–112.

Example. Consider the multiplier

iξ1 m(ξ) = Pn 2 iξ1 + j=2 |ξj| which appears in connection with maximal regularity estimates for the heat equation in space-time. This multiplier satisﬁes the assumption of the Marcinkiewicz multiplier theorem, but does not satisfy the assumptions of the Mihlin-H¨ormandermultiplier theorem (Homework).

7.3 Strichartz estimates and the nonlinear Schr¨odingerequation This subsection also closely follows [Tao99]. We would like to study nonlinear Schr¨odinger equations like

2 m i∂tu − ∆u = |u| u on (0, ∞) × R , m u(0, ·) = f on {0} × R .

130 [July 12, 2019] To do so we will rewrite the equation as an integral equation and then use a fixed point argument. The key ingredient are space-time estimates for the linear equation known as Strichartz estimates, see Theorem 7.12 below. To rewrite the equation as an integral equation we first recall the varia- k k tions of constants formula from the theory of linear ODE. Let A : R → R k be linear and let F : [0, ∞) → R be continuous. Then a function y ∈ 1 k C ([0, ∞); R ) is a solution of y0 = Ay + F on (0, ∞), y(0) = f. if and only if Z t y(t) = etAf + e(t−s)AF (s) ds. 0 Here etA = P (tA)k/k! is the matrix exponential function. More ab- k∈N stractly etAf = z(t) where z(t) is the unique solution of the homogeneous ODE z0 = Az, z(0) = f. For the homogeneous Schrödingerequation we already know the solution of the homogeneous equation. If f ∈ L2 and

m i∂tu − ∆u = 0 on (0, ∞) × R , m u(0, ·) = f on {0} × R then u(t) = S(t)f where it|ξ|2 Fm(S(t)f)(ξ) = e Fmf(ξ) (7.42) m 1 where Fm denotes the Fourier transform in R . If, in addition, f ∈ L we can also represent S(t) in real space

Z m/2 2 i − i|x−y| (S(t)f)(x) = e 4t f(y) dy (7.43) m/2 Rm (4πt) We thus say that u is a mild solution of the inhomogeneous Schrödinger equation if Z t u(t) = S(t)f + S(t − s)(−iF (s)) ds. (7.44) 0 Here the factor (−i) in the integral arises from the fact that we have to multiply the equation by −i to bring it to the form ∂tu = −i∆u − iF . It is easy to see that for sufficiently regular data with sufficient decay at ∞ a mild solution is also a classical solution and vice versa (Homework). To emphasize the analogy with the ODE case one often use the notation

S(t) = e−it∆.

131 [July 12, 2019] The following theorem contains the key space-time estimates for the linear Schr¨odingerequation. Theorem 7.12 (Strichartz estimates for Schr¨odinger). Let n = m + 1, n m R+ = (0, ∞) × R and m + 2 n + 1 p = 2 = 2 . m + 4 n + 3 2 m p n Then for all f ∈ L (R ) and all F ∈ L (R+). −it∆ ke fk p0 n ≤ CkfkL2( m), (7.45) L (R+) R Z ∞ it∆ e F (t) dt ≤ CkF k p n , (7.46) L (R+) 0 L2(Rm) Z t −i(t−s)∆ e F (s) ds ≤ CkF k p n . (7.47) L (R+) 0 p0 n L (R+)

Notation. The symbol t in (7.45) is to be understood as follows. Deﬁne −it∆ u(t, x) := (e f)(x). Then kuk p0 n ≤ CkfkL2( m). A similar remark L (R+) R applies to (7.47).

Remark. 1. It is no coincidence that the exponent p agrees with the optimal exponent in the Tomas-Stein restriction theorem which appears below. Indeed will see at the end of Section 7.4.1 that a space-time estimate like (7.45) is equivalent to an L2 − Lp0 extension estimate for the paraboloid 2 m n p 2 {(|ξ| , ξ): ξ ∈ R } ⊂ R and hence to an L − L restriction estimate. 2. It is easy to see by scaling that the above estimates can only hold for m+2 −it∆ p = 2 m+4 . Consider, e.g., (7.45). If u(t, x) = (e f)(x) is the solution 2 for initial datum f then uλ(t, x) = u(t/λ , x/λ) is the solution for initial 0 datum fλ(x) = f(x/λ). Thus (7.45) can only hold if (m + 2)/p = m/2 or 2/p = (m + 4)/(m + 2). 3. Everything will be reduced to the proof of (7.47). This is an application of the so-called T ∗T trick, see the remark after the end of the proof for further discussion. For the proof of Theorem 7.12 we will use the one-dimensional version of the following result on ’fractional integration’ which will be proved below. Lemma 7.13 (Fractional integration). Suppose that α ∈ (0, n), 1 1 α n = + − 1 and p ∈ 1, . q p n n − α

p n Assume that f ∈ L (R ). Then 1 the function y 7→ |f(x − y)| is integrable for a.e. x ∈ n (7.48) |y|α R

132 [July 12, 2019] and 1 k ∗ fk q n ≤ C kfk p n (7.49) | · |α L (R ) α,p,n L (R )

Remark. This would follow by the usual convolution estimate if the kernel |·|−α was in the space Ln/α. The point is that this is not true but the kernel is in the Marcinkiewicz space weak-Ln/α = Ln/α,∞.

Proof of Theorem 7.12. Step 1. Proof of (7.47). Set Z t −i(t−s)∆ ˜ h(t) = e F (s) ds , h(s) = kF (s)kLp . 0 0 Lp (Rm) We have to show that ˜ khkLp0 (0,∞) ≤ CkhkLp(0,∞) We know that

−it∆ −it∆ C ke fk 2 = kfk 2 , ke fk ∞ ≤ kfk 1 . L L L tm/2 L

2 Interpolating these inequalities with θ = m+2 we get

−it∆ C ke fk 0 ≤ kfk p Lp tθm/2 L where 1 1 1 1 1 m + 4 m + 2 = (1 − θ) + θ1 = (1 + ) = , p = 2 . p 2 2 m + 2 2 m + 2 m + 4 Thus Z ∞ C |h(t)| ≤ h˜(s) ds. mθ/2 0 |t − s| Now 1 1 2 2 mθ − = 1 − = − = − 1. p0 p p m + 2 2 Thus the desired estimate follows by the lemma on fractional integration lemma (see Lemma 7.13) in dimension 1. Step 2. (7.47) =⇒ (7.46). This is essentially the T ∗T trick. The estimate (7.46) is equivalent to

Z ∞ Z ∞ it∆ is∆ 2 Re e F (t) dt, e F (s) ds ≤ CkF k p n . L (R+) 0 0 L2(Rm)

133 [July 12, 2019] ∞ n We will first show this identity for more regular F , e.g. F ∈ Cc (R+) or 1 2 m at least F ∈ L ((0, ∞); L (R )). Then the general case follows by density. Note that the scalar product on the left hand side is already real, but inserting the real part will be useful shortly. We rewrite this as Z ∞ Z ∞ it∆ is∆ 2 Re e F (t), e F (s) ds dt ≤ CkF k p n . L2( m) L (R ) 0 0 R + Since it∆ is∆ is∆ it∆ e F (t), e F (s) = (e F (s), e F (t)) 2 m L2(Rm) L (R ) the integrand is symmetric in s and t. It suffices to consider the double integral for s ≤ t and the estimate is equivalent to Z ∞ Z t it∆ is∆ 2 2 Re e F (t), e F (s) ds dt ≤ CkF k p n . L2( m) L (R ) 0 0 R + it∆ −it∆ Using the identity (e g, h)L2(Rm) = (g, e h)L2(Rm) (which can be easily proved by Fourier transform) this can be rewritten as Z ∞ Z t −i(t−s)∆ 2 2 Re F (t), e F (s) ds dt ≤ CkF k p n . L (R+) 0 0 L2(Rm) Using Hölder’sinequality we easily deduce this estimate from (7.47). Step 3. (7.45) ⇐⇒ (7.46) by duality. Estimate (7.46) is equivalent to the estimate

Z ∞ it∆ 2 p e F (t) dt, f ≤ CkF k p n kfk 2 m ∀ f ∈ L ,F ∈ L . L (R+) L (R ) 0 L2(Rm) (7.50) Now we use again that

it∆ −it∆ (e g, h)L2(Rm) = (g, e h)L2(Rm). Thus (7.50) is equivalent to

Z −it∆ 2 p F (t, x)e f(x) dx dt ≤ CkF kLp( n ) kfkL2( m) ∀ f ∈ L ,F ∈ L . n R+ R R+ (7.51) Now (7.45) implies (7.51). Conversely (7.45) follows from (7.51) by taking the supremum over all F ∈ Lp.

Remark. The proof is just another variant of the T ∗T trick. Set Z ∞ TF := eis∆F (s) ds, 0 Z t (AF )(t) = e−i(t−s)∆F (s) ds. 0

134 [July 12, 2019] Then

(T ∗f)(t) = e−it∆f, Z ∞ (A∗F )(t) = e−i(t−s)∆F (s) ds. t and Z ∞ (T ∗TF )(t) = e−i(t−s)∆F (s) ds = (AF + A∗F )(t) 0 Step 3 is just the argument that T is bounded if and only if T ∗ is bounded. Step 2 is just the implication A bounded =⇒ T bounded which is based on the identies

(TF,TG) = (T ∗TF,G) = (AF, G) + (A∗F,G) = (AF, G) + (F, AG) which for G = F reduces to (TF,TF ) = 2 Re(AF, F ). This identity also shows that T bounded =⇒ A + A∗ bounded. This, however, does in general not imply that A is bounded. Thus the convolution operator A+A∗ is easier to analyse then the operator A which is sometimes called the ’retarded’ operator (because only times with t − s > 0 are considered).

Remark. Even though the proof of Theorem 7.12 is short the result is truly remarkable in view of the fact that every ﬁxed t 6= 0 the operator U(t) = e−it∆ is not bounded from L2 to any Lq with q 6= 2. Indeed if U(t): L2 → Lq was bounded then U(t)U ∗(t) would be bounded from Lq0 → Lq. But this is impossible since U(t)U ∗(t) = Id . Note also that U(t) cannot be bounded from Lr0 to Lr for any 1 ≤ r < 2. Otherwise the Hausdorﬀ-Young theorem would imply that FU(t) is bounded from Lr0 to Lr0 and it would follow that F is bounded from Lr0 to Lr0 . By scaling this is only possible for r0 = 2

Remark. This remark was not discussed in class. The argument can easily be extended to spaces with diﬀerent integrability exponents in time and space. In particular we have

Z t −it∆ −i(t−s)∆ 0 0 e f Lq(Lr) ≤ CkfkL2 , e F (s) ds ≤ CkF kLq (Lr ) 0 Lq(Lr) as long as r ≥ 2 and q > 2 and 1 1 1 1 2( − ) + m( − ) = 1. 2 q 2 r

1 1 This formula is easy to remember: 2 − q is one half of the gain in regularity in q0 q 1 1 time (from L to L ) and 2 − m is one half the gain of regularity in each space

135 [July 12, 2019] variable. Now the total gain is 2 (since we have second order equation in space) and by scaling one time dimension counts like two space dimensions. 1 1 The restriction q > 2 is equivalent to m( 2 − r ) < 1 which arises from the application of the fractional integration theorem. For r = q = p0 one 0 0 2(m+2) recovers the condition (m + 2)/p = m/2 or p = m which is equivalent 2(m+2) to p = m+4 . [1.7. 2019a, Lecture 20] [1.7. 2019b, Lecture 21]

Proof of Lemma 7.13. We may assume that kfkLp = 1. Let Mf denote the Hardy-Littlewood maximal function of f. The key idea is to estimate the nonlocal expression R |x − y|−αf(y) dy in terms of local expressions in Mf. p Since 1 < p < ∞ we have Mf ∈ L and kMfkLp ≤ CpkfkLp . In particular n Mf(x) < ∞ for a.e. x ∈ R . We have Z 1 n−α α |f(x − y)| dy ≤ CR Mf(x) B(0,R)\B(0,R/2) |y| Summing over B(0, 2−kB) \ B(0, 2−k+1R) we get Z 1 n−α α |f(x − y)| ≤ CR Mf(x) B(0,R) |y|

One the other hand Young’s inequality and the assumption kfkLp = 1 imply that 0 Z Z !1/p 1 1 n/p0−α α |f(x − y)| dy ≤ αp0 ≤ CR n |y| n |y| R \B(0,R) R \B(0,R) Here we used that 1 n − α α 1 α > = 1 − and hence < p n n p0 n 0 n so that p α > n. Combining the estimates for B(0,R) and R \ B(0,R) we see that (7.49) holds since Mf < ∞ a.e. Moreover

Z 1 n n−α n−α− p α f(x − y) dy ≤ CR Mf(x) + CR (7.52) Rn |y| for all R > 0. Taking R−n/p = Mf(x) and using that n − α n − α p = p = 1 − n/p n q we see that the right hand side of (7.52) is bounded by C(Mf(x))p/q.

This implies (7.49) since kMfkLp ≤ CpkfkLp = Cp.

136 [July 12, 2019] 2 m Lemma 7.14 (Energy identity for mild solutions). Let f ∈ L (R ), F ∈ p n m+2 L (R+) with p = 2 m+4 and let Z t u(t) := e−it∆f − i e−i(t−s)∆F (s) ds. 0 Then for t > s ≥ 0 Z 2 2 ku(t)kL2 − ku(s)kL2 = 2 Re(−iF u¯) dτ dx. (7.53) (s,t)×Rm

∞ n Proof. Homework. Hint: ﬁrst show the result for nicer F , e.g., F ∈ Cc (R+) by arguing as in Step 2 of the proof of Theorem 7.12. Alternatively one can use that for nice f and F the function u is a classical solution of the linear d Schr¨odingerequation and compute dt (u(t), u(t)). Then use the estimates in p n Theorem 7.12 to pass to obtain the result for F ∈ L (R+).

We now consider the cubic Schr¨odingerequation in two space dimensions

2 2 i∂tu − ∆u = |u| u in (0, ∞) × R , (7.54) u(0, ·) = f. (7.55)

The cubic nonlinearity is critical in two space dimensions. One can see this in the following way. On the one hand by Lemma 7.14 for (mild) 2 4 solutions of the equation we have ku(t)kL2 = kfkL2 since |u| uu¯ = |u| is real. On the other hand the equation has a scaling invariance. If u is a solution −1 2 with initial datum f then uλ(t, x) = λ u(t/λ , x/λ) is a solution with initial −1 datum fλ(x) = λ f(x/λ). Non criticality is reﬂected by the fact that in two dimensions is that kfλkL2 = kfkL2 , i.e., scaling preserves the norm which is conserved. We say that u is a mild solution of the cubic Schr¨odingerequation if

u = N(u) (7.56) where Z t N(u) = e−it∆f − i e−i(t−s)∆(|u|2u)(s) ds. 0 Note that for m = 2 the exponents in Theorem 7.12 are 4 p = , p0 = 4. 3

Thus u ∈ Lp0 , implies |u|2u ∈ Lp (this is another sign that the cubic nonlinearity is critical in two space dimensions).

137 [July 12, 2019] Theorem 7.15 (Existence, uniqueness and scattering for small data). There exists ε0 > 0 and M > 0 with the following properties.

(i) If kfkL2(R2) ≤ ε0 then the cubic Schr¨odingerequation (7.54), (7.55) has a unique mild solution and

kuk 4 3 ≤ M, ku(t)k 2 2 = kfk 2 2 ∀t > 0. L (R+) L (R ) L (R ) Moreover the solution depends continuously on f, i.e., the map f 7→ u 2 2 4 3 is continuous as a map from B(0, ε0) ⊂ L (R ) to L (R+). 2 2 (ii) There exists f+ ∈ L (R ) such that

−it∆ lim ku(t) − e f+kL2( 2) = 0. (7.57) t→∞ R

Remark. The property (7.57) is known as scattering. It states that for t → ∞ the solution of the nonlinear cubic Schr¨odingerequation with initial datum f behaves like the solution of the linear homogeneous equation with another initial datum f+.

4 3 Proof. Let X = L (R+). To prove (i) we use the Banach ﬁxed point theorem. It suﬃces to show that for some closed ball B(0,M) ⊂ X

(i) N maps B(0,M) to itself and

(ii) N is a contraction, i.e, 1 kN(v) − N(u)k 4 ≤ kv − uk 4 ∀u, v ∈ B(0,M). L 2 L

We begin with the contraction property. Assume that u, v ∈ B(0,M) We have9 2 2 3 2 2 |v| v − |u| u ≤ (|u| + |v| )|v − u| 2 3 1 1 Since 4 = 2 + 4 H¨older’sinequality gives

kghkL4/3 ≤ kgkL2 khkL4 Thus

2 2 3 2 2 2 k|v| v − |u| uk 4/3 ≤ (kuk + kvk )kv − uk 4 ≤ 3M kv − uk 4 . L 2 L4 L4 L L Let C0 be the constant in (7.47). Then

0 2 kN(v) − N(u)kL4 ≤ 3C M kv − ukL4 .

9 2 0 2 Proof: let ut = (1 − t)u + tv, h(t) = |ut| ut. The |h (t)| ≤ 3|ut| |v − u| ≤ 3((1 − 2 2 R 1 0 t)|u| + t|v| )|v − u|. Use that h(1) − h(0) = 0 h (t) dt.

138 [July 12, 2019] Thus the contraction property holds if 1 M 2 ≤ . 6C0 To prove (i) it now suﬃces to show that M kN(0)k 4 ≤ . L 2

1 1 00 since kN(v) − N(0)k ≤ 2 kvk ≤ 2 M. Let C denote the constant in (7.45). Then 00 00 kN(0)kL4 ≤ C kfkL2 ≤ C ε0. Hence both conditions (i) and (ii) are satisﬁed if M 1 ε = and M 2 ≤ . (7.58) 0 2C00 6C0 This implies all the statements in assertion (i) of the theorem except the 2 identity ku(t)kL2 = kfkL2 . This identity follows from (7.53) since Re −i|u| uu¯ = Re −i|u|4 = 0. To prove scattering we apply the inverse of the solution operator S(t) for the linear homogeneous equation at time u(t) and deﬁne

v(t) = eit∆u(t).

Then by the deﬁnition of a mild solution Z t v(t) = f − i eis∆|u(s)|2u(s) ds. 0 This suggests to deﬁne Z ∞ is∆ 2 f+ = f − i e (|u| u)(s) ds. 0 Then Z ∞ is∆ 2 v(t) − f+ = i e (|u| u)(s) ds. t Since S(t) = e−it∆ is an isometry on L2 the estimate (7.46) yields

−it∆ 2 ku(t) − e f+kL2 = kv(t) − f+kL2 ≤ Ckχ(t,∞)|u| ukL4/3 . Now the right hand side goes to zero as t → ∞ by the monotone convergence theorem.

Theorem 7.16 (Local-in-time existence for large data). Let f ∈ L2. Then there exists a T > 0 such that the cubic Schr¨odingerequation has a mild 4 2 solution in L ((0,T ) × R ).

139 [July 12, 2019] Proof. We argue as before but all spaces and norms are now restricted to 2 (0,T ) × R where T > 0 will be chosen later. As before we choose 1 M 2 ≤ 6C0 where C0 is the constant in (7.47). Then N is a contraction on B(0,M) ⊂ 4 2 X = L ((0,T ) × R ). The main point is to show that M kN(0)k ≤ . X 2 To see this note that by (7.45) Z ∞ 4 kN(0)(t)kL4 dt < ∞. 0 Hence there exists a T > 0 such that Z T 4 4 M kN(0)(t)kL4 dt < 4 . 0 2

The rest of this subsection was not discussed in class Once we have understood the most diﬃcult critical nonlinearity we can also establish results for the Schr¨odingerequation with subcritical nonlinearity h(u) of weaker growth. We seek a mild solution of

2 i∂tu − ∆u = h(u) on (0, ∞) × R , (7.59) u(0, ·) = f, (7.60) i.e., a solution of u = N(u) where now Z t (N(u))(t) = e−it∆f + e−i(t−s)∆h(u(s)) ds. 0 We always assume that h(0) = 0. We begin with the simplest case that h is Lipschitz

|h(a) − h(b)| ≤ L|a − b| ∀a, b ∈ C. In this case the subtle Strichartz estimates are not needed and it suﬃces to R t −i(t−s)∆ the simple ’energy’ estimate for v(t) = 0 e F (s)) ds. We have

kvkL∞(L2) ≤ kF kL1(L2). (7.61)

140 [July 12, 2019] This estimate follows directly from Jensen’s inequality. Indeed Z t Z t −i(t−s)∆ kv(t)kL2 ≤ ke F (s)kL2 ds = kF (s)kL2 ds. 0 0 Alternatively one can use the energy identitiy (7.53) for mild solutions with f = 0, take the supremum over t on the left hand side estimate the right hand side by kukL∞(L2) kF kL1(L2) (one can ﬁrst considers more regular F and then pass to the limit). Now it follows directly from (7.61) that N is a contraction in X = L∞((0,T ); L2) for small T . Indeed

kN(u) − N(v)kX ≤ Lku − vkL1((0,T );L2) ≤ LT ku − vkX . To treat more general subcritical nonlinearities we work simultaneously in the energy space L∞(L2) and in L4(L4) use both the energy estimate (7.61) and the Strichartz estimate (7.47). In addition we need two estimates which intertwine the energy picture and the Strichartz picture. For brevity ∞ 2 ∞ 2 2 we write L (L ) for L (0,T ; L (R )) etc. Lemma 7.17. Let Z t v(t) = e−i(t−s)∆F (s) ds. 0 Then

4/3 4/3 kvkL∞(L2) ≤ CkF kL4/3(L4/3) ∀F ∈ L (L ) (7.62) 1 2 kvkL4(L4) ≤ CkF kL1(L2), ∀F ∈ L (L ). (7.63)

Proof. Homework. Sketch: For the ﬁrst estimate use that kv(t)kL2 = it∆ ke v(t)kL2 and apply (7.46). For the second estimate use duality. Show ﬁrst that Z ∞ Z Z ∞ Z ∞ v(t, x)G(t, x) dx dt = eis∆F (s), eit∆G(t) dt ds 0 R2 0 s L2 and then use again (7.46).

Theorem 7.18 (Existence for the subcritical Schr¨odinger equation). As- sume that α ∈ (0, 2) and that h : C → C satisﬁes α h(0) = 0 and |h(a) − h(b)| ≤ L(1 + (|a| + |b|) ) |a − b| ∀ a, b ∈ C. Then the following assertions hold.

2 2 (i) For each E0 > 0 there exists a T > 0 such that for every f ∈ L (R ) with kfkL2 ≤ E0 the nonlinear Schr¨odingerequation (7.59), (7.60) has 4 4 ∞ 2 a unique mild solution in X = XT = L ((0,T ); L ) ∩ L ((0,T ); L ) and the solution depends continuously on f.

141 [July 12, 2019] (ii) If in addition h(a)¯a is real for all a ∈ C then the equation has a global 2 mild solution (i.e., T = ∞) for all f ∈ L and ku(t)kL2 = kfkL2 for all t > 0.

Proof. Homework. Hints: to prove the ﬁrst assertion introduce the norm

kukX := max(kukL4((0,T );L4), kukL∞((0,T );L2)) and note that by the ﬁrst Strichartz estimate (7.45)

kN(0)kX ≤ CkfkL2 ≤ CE0.

Let M = 2CE0. It only remains to show that N is a contraction in B(0,M). To see this let k ≥ 1 and deﬁne

F1 = χ(h(u) − h(v)),F2 = (1 − χ)(h(u) − h(v)) where χ = χ{|u|+|v|≤k},

Z t −i(t−s)∆ wj(t) := e Fj(s) ds. 0 Note that

α α−2 2 |F1| ≤ L(1 + k )|u − v|, |F2| ≤ 2Lk (|u| + |v|) |u − v| and show that

α α−2 2 kF1kL1(L2) ≤ TL(1 + k )ku − vkX , kF2kL4/3(L4/3) ≤ 8Lk M ku − vkX .

Now use (7.61), (7.63) as well as (7.62) and (7.47) and choose ﬁrst k large and then T small to achieve the contraction. To prove (ii) use the energy identity (7.53) and the standard continuation argument. Let

∗ T = sup{T > 0 : ∃ mild solution in XT }.

∗ 1 If T < ∞, let E0 = kfk and δ = 2 T (E0) where T (E0) is the minimal ∗ existence time in (i). By the energy identity ku(T − δ)kL2 = kfkL2 = E0. Hence there exists a mild solution v on [0, 2δ) with v(0) = u(T ∗ − δ). Deﬁne ( u(t) if t ≤ T ∗ − δ, u˜(t) = v(t − (T ∗ − δ)) if t > T ∗ − δ.

Then one can easily check from the deﬁnition of a mild solution thatu ˜ is a mild solution on [0,T ∗ + δ). This contradicts the deﬁnition of T ∗.

142 [July 12, 2019] Extension to m 6= 2 All the results above can be extended to arbitrary space dimension m with the following modiﬁcations (i) The critical nonlinearity is h(u) = |u|4/mu;

4 (ii) The subcritical case is characterized by the condition α ∈ (0, m ). [1.7. 2019b, Lecture 21] [8.7. 2019a, Lecture 22]

7.4 Restriction theorems, oscillatory integrals and Strichartz estimates. This subsection closely follows lecture notes by T. Tao [Tao99]

7.4.1 Setting of the problem and connection with PDE

n Let S ⊂ R be a hypersurface, equipped with a measure ω. We consider the following two problems.

Restriction problem. For which p, q do we have an estimate

kFfkLq(S,ω) ≤ CkfkLp(Rn) ?

Extension problem. For which p, q do we have an estimate

kF(gω)k 0 ≤ Ckgk 0 ? Lp (Rn) Lq (S,ω) Here p0 and q0 are the dual exponents of p and q, respectively.

By a duality argument one easily sees that the two problems are equivalent.

Connection with PDE. Let n = m + 1 and consider the Schr¨odinger equation

m i∂tu − ∆u = 0 in R × R , m u(0, ·) = u0 in R .

m Let Fm be the Fourier transform in the x variable, i.e., in R . We have seen that it|ξ|2 Fmu(t, ξ) = e Fmu0(ξ) −1 Using Fm we can write the solution u as

1 Z 2 u(x, t) = ei(t|ξ| +x·ξ)F u (ξ). m/2 m 0 (2π) Rm

143 [July 12, 2019] n We now express u as the inverse Fourier transform of a measure in R = m+1 R . Let S be the hypersurface S = {(τ, ξ): τ = |ξ|2},

n m m let π(τ, ξ) = ξ be the projection from R = R × R to R . Consider on S the pull-back measure ω = π∗Lm, i.e., ω(A) := Lm(π(A ∩ S)). This can be rewritten as Z Z m m 2 χA dω = L {ξ ∈ R :(R × {ξ}) ∩ (A ∩ S) 6= ∅} = χA(|ξ| , ξ) dξ. Rn Rm By the usual approximation arguments one sees that Z Z 2 n h dω = h(|ξ| , ξ) dξ ∀ h ∈ Cc(R ). (7.64) Rn Rm Alternatively one can use the Riesz-Radon theorem to use (7.64) to deﬁne a measure ω. Indeed the right hand side of (7.64) deﬁnes a positive, bounded n and linear functional on Cc(R ). Hence there exists a unique Radon measure n ω on R such that the identity in (7.64) holds. Let i(tτ+x·ξ) hx,t(τ, ξ) := e , g(τ, ξ) := Fmu0(ξ). Then 1 Z 1 Z u(x, t) = h (|ξ|2, ξ)g(|ξ|2, ξ) dξ = h g dω m/2 x,t m/2 x,t (2π) Rm (2π) Rn and thus u(x, t) = (2π)1/2F −1(gω). By Plancherel’s theorem we have Z Z Z 2 2 2 m/2 2 |g| dω = g(|ξ| , ξ) dξ = |Fu0(ξ)| dξ = (2π) ku0kL2 . Rn Rm Rm p0 2 Thus a space-time estimate for u in L for initial data u0 ∈ L is equivalent to the restriction problem with q0 = 2 (and hence q = 2).

7.4.2 Curved hypersurfaces and the Tomas-Stein theorem for the sphere We now return to the restriction and extension problem for a hypersurface S with measure ω. Since x 7→ eix·ξ is continuous and bounded we have the trivial L1 − L∞ estimate

kFfkL∞(S,ω) ≤ kfkL1 .

144 [July 12, 2019] 2 2 It is also easy to show that there is no L − L estimate. Indeed let gk be a sequence such that kgkkL2(S,ω) = 1 and kgkkL2(Rn) → 0. Now take −1 fk = F gk and use Plancherel’s identity. Another simple observation is that there is no nontrivial Lp − Lq restriction estimate for ﬂat hypersurfaces. Indeed let S = {x : xn = 0} be equipped n−1 n−1 ∞ with the Lebesgue measure L . Let g ∈ S(R ) and let h ∈ Cc (R) with Fh(0) = R h = 1. Consider R x f (x) = g(x , . . . , x ) λ−1h( ) λ 1 n−1 λ Then Ff(ξ) = Fg(ξ1, . . . , ξn−1) Fh(λξn) and

1 −1+ p kFfkLq(S,ω) = kFgkLq(Rn−1), kfkLp(Rn) = kgkLp(Rn−1)λ khkLp(R). Taking λ → ∞ we see that there can be no restriction estimate for p > 1. As we have seen a restriction estimate is equivalent to an extension estimate kF(fω)k 0 ≤ Ckfk 0 . For the case of a ﬂat hypersurface Lp (Rn) Lq (S,ω) we can write

n−1 F(fH ∠{xn = 0})(ξ) Z Pn−1 −i xj ξj = f(x1, . . . xn−1)e j=1 dx1 . . . dxn−1 Rn−1 =(Ff)(ξ1, . . . , ξn−1).

In particular the Fourier transform is independent of ξn and hence cannot p0 n 0 be in any of the spaces L (R ) for p < ∞. A key observation is that for genuinely curved surfaces the Fourier transform of the surface measure always decays at inﬁnity. The simplest example is the sphere and we will show the following result. Theorem 7.19. Let ω by the n−1 dimensional Hausdorﬀ measure restricted to the unit sphere Sn−1. Then there exists a constant such that for |ξ| → ∞ ! Cei|ξ| (Fω)(ξ) = Re + O(|ξ|−(n+1)/2) as |ξ| → ∞ |ξ|(n−1)/2 with C = 2(−2πi)(n−1)/2 Notation. The statement a(ξ) = O(|ξ|α) as ξ → ∞ means that there exist constants C and R such that |a(ξ)| ≤ C|ξ|α if |ξ| ≥ R.

The main ingredient in the proof of Theorem 7.19 are estimates for oscillatory integrals known as the principle of instationary phase and the more subtle principle of stationary phase.

145 [July 12, 2019] Theorem 7.20 (Principle of instationary phase). Let M be a smooth mani- ∞ ∞ fold, let ψ ∈ Cc (M), ϕ ∈ C (M). Assume that

∇ϕ 6= 0 on supp ψ.

Then for all N ∈ N we have Z iλϕ(x) −N e ψ(x) dvolM = O(λ ). (7.65) M

Proof. The idea is that for large λ the function eiλϕ(x) oscillates rapidly in at least one direction and that one can use integration by parts. To make this rigorous we first localize the problem and then make a change of variables so that the oscillations are in one coordinate direction. For each point x0 ∈ supp ψ there exists a ball B(x0, ρ) such that B(x0, ρ)∩ M is in the image of a single chart. Since the set supp ψ is compact it can be covered by the images of finitely many such charts. Hence it suffices to n prove the result for a single chart, or equivalently for M = R . n n Assume thus that M = R . We first show that for each a ∈ R there ∞ exists an open ball B(a, r) such that the assertion holds for ψ ∈ Cc (B(a, r)). The full assertion can then be proved using a smooth partition of unity since supp ψ is compact by assumption. After a rotation of coordinates we may assume that ∂nϕ(a) 6= 0. Define a change of variables by

η(x1, . . . , xn) = (x1, . . . , xn−1, ϕ(x) − ϕ(a) + an)

Then det ∇η(a) = ∂nϕ(a) 6= 0.

Thus by the inverse function theorem there exists r > 0 such that η|B(a,r) ∞ is invertible with smooth inverse. Hence for ψ ∈ Cc (B(a, r)) the change of variables x = η−1(y) yields Z eiλϕ(x)ψ(x) dx B(a,r) Z = eiλ(yn+ϕ(a)−an)ψ(η−1(y)) | det(∇η−1)(y)| dy η(B(a,r)) Z =eiλ(ϕ(a)−an) eiλyn ψ˜(y) dy η(B(a,r)) where ψ˜(y) = ψ(η−1(y))| det(∇η−1)(y)|. ˜ ∞ ˜ Thus ψ ∈ Cc (η(B(a, r)) and we may extend ψ by zero to a function in ∞ n Cc (R ). Successive integration by parts in the integral with respect to yn

146 [July 12, 2019] yields Z Z 1 Z 1 iλyn ˜ iλyn ˜ iλyn N ˜ e ψ(y) dy = e ∂nψ(y) dy = N e ∂n ψ(y) dy Rn Rn −iλ Rn (−iλ) and the assertion follows.

[8.7. 2019a, Lecture 22] [8.7. 2019b, Lecture 23]

∞ n Theorem 7.21 (Principle of stationary phase). Let ϕ ∈ C (R ). Let n x0 ∈ R and assume that ϕ has a non-degenerate critical point at x0, i.e., 2 ∇ϕ(x0) = 0, det ∇ ϕ(x0) 6= 0.

∞ Then there exists a δ > 0 with the following property. If ψ ∈ Cc (B(x0, δ)) 2 then with H(x0) := ∇ ϕ(x0), Z 1/2 n/2 iλϕ(x) n/2 i sgn(H(x0))π/4 iϕ(x0) −1 | det H(x0)| λ e ψ(x) dx = (2π) e e ψ(x0)+O(λ ) Rn (7.66) as λ → ∞, where sgn(H(x0)) denotes the signature of H(x0), i.e., sgn(H(x0)) := #{positive eigenvalues of H(x0)}−#{negative eigenvalues of H(x0)}.

Remark. To remember the prefactors one can start from the fact that for a symmetric positive deﬁnite matrix A one has

−n/2 1/2 − 1 (λAx,x) (2π) det(λA) e 2 → δ0 as λ → ∞

2 and then replace A by −i∇ ϕ(x0).

Proof. We may assume wlog that x0 = 0 and ϕ(x0) = 0.

1 2 Step 1 n = 1 and ϕ(x) = ± 2 x . 1 2 We ﬁrst assume ϕ(x) = 2 x . By (7.13) we have for a ∈ C with Re a > 0

1/2 − a |·|2 (2π) − 1 |ξ|2 F(e 2 )(ξ) = e 2a a1/2 Applying this with a = −(i − ε)λ, using the Plancherel formula in the form Z Z fh dx = Ff F −1h dξ R R (see (7.17)) and taking the limit ε ↓ 0 we get Z Z 1/2 iλ |x|2 1/2 iπ/4 − i |ξ|2 −1 λ e 2 ψ(x) dx = (2π) e e 2λ F ψ(ξ)dξ. R Rn

147 [July 12, 2019] Now Z Z (Ff)(0) = f and with f = F −1ψ we get ψ(0) = F −1ψ. R R iα Combining this with the estimate |e − 1| ≤ |α| for α ∈ R we get Z Z − i |ξ|2 −1 − i |ξ|2 −1 e 2λ F ψ dξ − ψ(0) ≤ e 2λ − 1 F ψ(ξ) dξ R R 1 Z ≤ |ξ|2|F −1ψ(ξ)| dξ ≤ Cλ−1 2λ R ∞ k −1 2 since ψ ∈ Cc (R) implies that | · | F ψ ∈ L (R) for all k ∈ N and hence by Cauchy-Schwarz Z Z 1/2 Z 1/2 2 −1 1 2 4 −1 2 |ξ| |F ψ(ξ)| dξ ≤ 2 dξ (1 + |ξ| )|ξ| |F ψ| dξ < ∞. R R 1 + |ξ| R 1 2 If ϕ(x) = − 2 x one takes a = λ(i + ε), considers again the limit ε ↓ 0 and obtains Z Z 1/2 − iλ |x|2 1/2 −iπ/4 i |ξ|2 −1 λ e 2 ψ(x) dx = (2π) e e 2λ F ψ(ξ)dξ. R Rn The remainder of the proof is as before.

1 2 2 2 2 Step 2 n ≥ 1 and ϕ(x) = 2 (x1 + . . . xk) − (xk+1 + ... + xn) . If ψ(x) = ψ1(x1) . . . ψn(xn) this follows from Step 1 and Fubini’s theorem. Since linear combinations of such functions are dense (see Lemma 7.23) the assertion follows.

Step 3 General case. We reduce this to Step 2 by a change of variables. By the Morse lemma (see Lemma 7.22 below) there exists an open neighbourhood V of 0 and a n smooth invertible map η : V → η(V ) ⊂ R such that 1 ϕ˜(y) := ϕ(η(y)) = (y2 + ... + y2) − (y2 + ... + y2) . 2 1 k k+1 n 2 n n We view ∇ ϕ(0) as a linear map from R to R . Since ∇ϕ(0) = 0 we have ∇2ϕ˜(0) = (∇η(0))T ∇2ϕ(0) ∇η(0) an thus det ∇2ϕ˜(0) = det ∇2ϕ(0)(det ∇η(0))2. Choose δ > 0 so small that B(0, δ) ⊂ η(V ). Since supp ψ ∈ B(0, δ) the change of variables x = η(y) yields Z Z Z eiλϕ(x)ψ(x) dx = eiλϕ˜(y)ψ(η(y)) | det ∇η(y)| dy = eiλϕ˜(y)ψ˜(y) | det ∇η(0)| dy Rn Rn Rn

n ∞ Lemma 7.22 (Morse lemma). Let U ⊂ R be open, let f ∈ C (U) and let x0 ∈ U be nondegenerate critical point of f. Let k be the number of positive eigenvalues of ∇2f. Then there exists a neighbourhood V of 0 and a smooth n and invertible change of variables η : V → R with η(0) = x0 such that 1 f(η(y)) = f(x ) + (y2 + ... + y2) − (y2 + ... + y2) . 0 2 1 k k+1 n Proof. See [Mi63], Lemma 2.2.

∞ n Lemma 7.23 (Density of tensor products in Cc (R )). For functions ψ1, . . . , ψn : R → K with K = R or K = C deﬁne the tensor product by n Y (ψ1 ⊗ ... ⊗ ψn)(x1, . . . xn) = ψj(xj). j=1 Let D be the set of ﬁnite linear combinations of tensor products

∞ D = span(ψ1 ⊗ ... ⊗ ψn : ψj ∈ Cc (R; K)).

∞ n ∞ n Then D is dense in Cc (R , K), i.e., for each ψ ∈ Cc (R ; K) there exists a n compact set K ⊂ R and ηk ∈ D such that α α n supp ηk ⊂ K, ∂ ηk → ∂ ψ uniformly ∀α ∈ N . Proof of Theorem 7.19. Let ω be the n − 1 dimensional Hausdorff measure n−1 n restricted to S . Since ω is invariant under rotations in R it is easy to see that the Fourier transform Fω is rotationally symmetric. Thus it suffices to compute Z Z −ix·λen n−1 −iλxn n−1 Fω(λen) = e dH = e dH Sn−1 Sn−1 Now we first apply the instationary phase theorem with M = Sn−1 and ϕ(x) = −xn. The only critical points of ϕ are the north pole en and the south pole −en. Let 0 < δ 1 and consider smoth cut-off functions ψ± : Sn−1 → [0, 1] with

∞ n−1 ψ± ∈ Cc (B(±en, δ)) ∩ S , ψ± = 1 in B(±en, δ/2)

149 [July 12, 2019] Then for all N ∈ N Z −iλxn n−1 −N e (1 − ψ+(x) − ψ−(x)) dH = O(λ ). Sn−1 Next we use the stationary phase theorem to estimate Z −iλxn n−1 e ψ+(x) dH . Sn−1 We have

n−1 p 2 supp ψ+ ⊂ graph f, where f : B(0, δ) ⊂ R → R, f(y) = 1 − |y| .

Thus Z Z −iλxn n−1 −iλf(y) p 2 e ψ+(x)H = e (ψ+ ◦ f)(y) 1 + |∇f(y)| dy Sn−1 B(0,δ) | {z } =ψ˜+(y)

Now the only critical point of f is y = 0 and

2 ∇ f(0) = −Id , ψ˜+(0) = ψ+(0) = 1.

Thus the stationary phase theorem (with n replaced by n − 1) yields Z (n−1)/2 −iλxn n−1 (n−1)/2 −iλ −1 λ e ψ+(x) dH = (2πi) e + O(λ ) Sn−1 Similarly one shows that Z (n−1)/2 −iλxn n−1 (n−1)/2 iλ −1 λ e ψ−(x) dH = (−2πi) e + O(λ ). Sn−1 This ﬁnishes the proof.

Theorem 7.24 (Restriction theorem for Sn−1, ﬁrst version). Assume that 4n p = (3n + 1)

Then kFfkL2(Sn−1) ≤ CkfkLp Proof. Let ω be the restriction of Hn−1 to Sn−1.

150 [July 12, 2019] Step 1. Reformulation as an Lp − Lp0 estimate. We want to prove Z 2 FfFf dω ≤ CkfkLp ∀f ∈ S. Sn−1 We view (Ff) ω as a tempered distribution. Then the desired estimate is 2 (Ff ω)(Ff) ≤ CkfkLp Since the Fourier transform intertwines convolution and multiplication (see (7.20)) we have 1 Ff ω = F(f ∗ F −1ω) = F(f ∗ Fω) (2π)n where the last identity follows from (7.7) and the fact that ω is invariant under the reﬂection x 7→ −x. Thus by Plancherel’s identity (7.18) Z 1 ¯ ¯ (Ff ω)(Ff) = n F(f ∗ Fω)(Ff) = (f ∗ Fω)(f) = (f ∗ Fω)f dx (2π) Rn It therefore suﬃces to show that

kf ∗ FωkLp0 ≤ CkfkLp ∀f ∈ S.

Step 2. Fractional integration. By Theorem 7.19 |F −1ω|(ξ) ≤ C|ξ|−(n−1)/2 (7.67) for large |ξ|. Since |F −1ω(ξ)| ≤ Hn−1(Sn−1) for all ξ the estimate (7.67) holds for all ξ. We have 1 1 2 6n + 2 2n − 2 (n − 1)/2 − = 1 − = 1 − = − 1 = − 1. p0 p p 4n 4n n Thus the fractional integration lemma (see Lemma 7.13) with α = (n − 1)/2 shows that

kf ∗ (Fω)kLp0 ≤ CkfkLp . This ﬁnishes the proof.

Remark. Step 1 is an example of the T ∗T method which is based on the following observation. A linear operator T : Lp → L2 is bounded if and only if T ∗T is bounded from Lp to Lp0 . Here T ∗ is deﬁned by (T ∗f, g) = (f, T g) for f ∈ L2 and g ∈ Lp. This equivalence follows from the estimates ∗ ∗ 2 (T f, T f) = (T T f, f) ≤ kT T kL(Lp,Lp0 )kfkLp and ∗ 2 (T T f, g) = (T f, T g) ≤ kT kL(Lp,L2) kfkLp kgkLp by taking the supremum over all functions f and g with kfkLp = kgkLp = 1. Finally we state without proof the optimal restriction estimate.

151 [July 12, 2019] Theorem 7.25 (Tomas-Stein restriction theorem). Assume that n + 1 p ≤ 2 n + 3 Then

kFfkL2(Sn−1) ≤ CkfkLp(Rn). n+1 Proof. We will prove the result for p < 2 n+3 and sketch the argument for n+1 the end point case p = 2 n+3 . As before we denote by ω the Hn−1 measure restricted to Sn−1. By Step 1 in the proof of Theorem 7.24 it suﬃces to show that

kFω ∗ fkLp0 ≤ CkfkLp

Step 1. Dyadic decomposition. We break up the kernel Fω into pieces which are supported on annuli of k ∞ 1 size 2 . As shown in Proposition 7.10 there exists ϕ ∈ Cc (B(0, 4)\B(0, 2 )) such that

X n −k ϕk = 1 in R \{0} where ϕk(x) = ϕ(2 x). k∈Z Set ∞ X ψ0 = 1 − ϕk k=1 and deﬁne

T0f = (ψ0Fω) ∗ f, Tkf = (ϕkFω) ∗ f for k ≥ 1.

Now ∞ ∞ X X kFω ∗ fkLp0 = k TkfkLp0 ≤ kTkfkLp0 . (7.68) k=0 k=0 Thus we are done if we can show that

−εk kTkfkLp0 ≤ 2 CkfkLp .

The estimate for T0 is easy. Indeed supp ψ0 ⊂ B(0, 4) since supp ϕk ⊂ B(0, 4) for all k ≤ 0. Moreover in B(0, 4) we have ψ0 = 1 − ϕ1 − ϕ2 − ϕ3 ∞ ∞ n and thus ψ ∈ Cc (B(0, 4)). Moreover Fω ∈ C (R ) since ω has compact ∞ support (use the second identity in (7.3)). Hence T0 is convolution by a Cc function and hence the estimate for T0 holds for all p ≤ 2.

152 [July 12, 2019] p p0 Step 2. L − L estimates for Tk. In general Lp − Lp0 estimates are diﬃcult to prove directly but here we can obtain them by interpolating between to simply estimates. First we have the obvious L1 − L∞ estimate

−k(n−1)/2 kTkfkL∞ ≤ kϕkFωkL∞ kfkL1 ≤ C2 kfkL1 .

−k−1 Here we used Theorem 7.19 and the fact that ϕk = 0 on B(0, 2 ). Second we have an L2 − L2 estimate by Fourier transform. Indeed

FTkf = F(ϕk Fω)Ff and thus kTkfkL2 ≤ kF(ϕk Fω)kL∞ kfkL2 Now −n 2 F(ϕk Fω) = (2π) (Fϕk ∗ (F) ω) = Fϕk ∗ ω where we used that ω is symmetric and hence F 2ω = (2π)nω. Now

kn k (Fϕk)(ξ) = 2 (Fϕ)(2 n).

Set η = Fϕ. Then η ∈ S Z kn k (Fϕk ∗ ω)(x) = 2 η(2 (x − y)) dω(y) Sn−1

Dividing the integral into rings |x−y| ≤ 2−k and 2−k+j−1 < |x−y| ≤ 2−k+j for j ≥ 0 and observing that these sets intersect Sn−1 is a set of Hn−1 measure ≤ C2(−k+j)(n−1) we see that

 ∞  kn −k(n−1) X (−k+j)(n−1) |(Fϕk ∗ ω)(x)| ≤ 2 C 2 + 2 sup η (7.69) n j j=1 R \B(0,2 )

−Nj Since η ∈ S we have supRn\B(0,2j ) η ≤ CN 2 for any N. Hence the sum over j converges an we get

k |(Fϕk ∗ ω)(x)| ≤ 2 C.

Now the Riesz-Thorin theorem yields

−θk(n−1)/2+(1−θ)k 1 1 kT fk 0 ≤ C2 kfk p , where = θ1 + (1 − θ) . k Lp L p 2 The exponent on the right hand side vanishes if θ = θ∗ where n − 1 2 θ∗ + (1 − θ∗) = 1 or θ∗ = . 2 n + 1

153 [July 12, 2019] This corresponds to 1 1 1 1 1 n + 3 = = + θ∗ = p p∗ 2 2 2 n + 1 or n + 1 p∗ = 2 n + 3 For p < p∗ we have θ > θ∗ and hence

−εk kTkfkLp0 ≤ C2 kfkLp . This ﬁnishes the proof for p < p∗.

Step 3. Idea of proof for p = p∗. This was not discussed in class. The main point is to avoid the use of the triangle inequality in (7.68). To do so one shows the estimates ∞ X [ n−1 +it]k 2 2 f ∗ (ϕkFω) ≤ CkfkL1 k=1 L∞

∞ X [−1+it]k 2 f ∗ (ϕkFω) ≤ CkfkL2 k=1 L2 Let ∞ X [z n−1 +(1−z)(−1)]k) T (z)f := 2 2 f ∗ (ϕkFω) k=1 Then ∞ ∗ X T (θ )f := f ∗ (ϕkFω) k=1 and T (1 + is) maps L1 to L∞ while T (is) maps L2 to L2. Thus one can use complex interpolation (with θ = θ∗) to get the Lp − Lp0 estimate for T (θ∗) with p = p∗ (see Theorem 7.26 below). The L1 − L∞ estimate is easy to prove. Indeed for each ξ at most three of the functions ϕk do not vanish at ξ. From this and the decay of Fω one easily deduces that for all t ∈ R ∞ X n−1 k k 2 2 ϕkFωkL∞ ≤ C k=1 and this gives the ﬁrst estimate. The proof of the L2 − L2 estimate requires a reﬁnement of the argument in (7.69) which uses radial symmetry of ϕ and the extra condition Z ϕˆ(x0, 0) dx0 = 0, Rn−1 see [Tao99], Lecture 2, for the details.

154 [July 12, 2019] Theorem 7.26 (Riesz-Thorin theorem for an analytic family of operators). Let S = (0, 1) × R ⊂ C. Assume that for z ∈ S¯ the linear map T (z) maps (L1 ∩ L∞)(E) into (L1 + L∞)(E0) and Z z 7→ T f g dx is analytic in S ∀f ∈ (L1 ∩ L∞), g ∈ (L1 ∩ L∞). E0 Z z 7→ T f g dx is continuous in S¯ ∀f ∈ (L1 ∩ L∞), g ∈ (L1 ∩ L∞). E0

Let p0, p1, q0, q1 ∈ [1, ∞], θ ∈ (0, 1) and assume that

kT (it)fkLq0 ≤ k0kfkLp0 ,

kT (1 + it)fkLq1 ≤ k1kfkLp1 .

Then 1−θ kT (θ)f)kLqθ ≤ k0 k1θkfkLpθ provided 1 1 1 1 1 1 = (1 − θ) + θ and = (1 − θ) + θ pθ p0 p1 qθ q0 q1 with the usual convention 1/∞ = 0. Proof. This is proved exactly like the Riesz-Thorin theorem (where T is independent of z). The details of the proof were not discussed in class. One considers functions of the form

N N X X f = α χE , g = β χ 0 k k k Ek k=1 k=1

0 for disjoint sets Ek and Ek. By scaling one may assume that

kfk pθ = kgk 0 = 1. L L(qθ) Then one deﬁnes

0 pθ (qθ) f g 0 f(z) = |f| pz , g(z) = |g| (qz) , |f| |g|

S 0 S 0 (with f = 0 on E \ k Ek and g = 0 on E \ k Ek) and applies the three line theorem to the analytic function Z z−1 −z h(z) = k0 k1 T (z)f(z) g(z)dx.

Details: note that

p0 p1 pθ kf(is)kLp0 = kf(1 + is)kLp1 = kfkLpθ = 1,

155 [July 12, 2019] and similarly

0 0 kg(is)kL(q0) = kg(1 + is)kL(q1) = 1. Thus by assumption |h(z)| ≤ 1 on ∂S. Hence the three line theorem gives Z θ−1 −θ k k T (θ)f g = |h(θ)| ≤ 1. 0 1 Since functions f and g of the above form are dense in the unit sphere of 0 Lpθ and L(qθ) this implies the assertion. Lemma 7.27 (Necessary conditions for restriction estimates). The restriction estimate

kFfkLq(Sn−1,ω) ≤ CkfkLp(Rn) can only hold if n + 1 2n p0 ≥ q and p < n − 1 n + 1 In particular the Tomas-Stein restriction is optimal for q = 2. Idea of proof. This was not discussed in class. The main idea is to use an anisotropic scaling argument. More explicitly one considers f such that Ff is of the form ξ0 ξ − 1 Ff(ξ0, ξ ) = g h n for 0 < λ 1. n λ λ2 If h(t) = 1 for |t| ≤ 1 one can arrange g such that Ff = g on Sn−1 and one can easily determine the scaling norms of Ff and f with respect to λ. To derive the second restriction on p one uses that the restriction estimate is equivalent to the extension estimate

kF(gω)k 0 ≤ kgk 0 . Lp (Rn) Lq (Sn−1) Now one takes g = 1 on Sn−1 and uses Theorem 7.19

Very brief comments on the literature In [St77] Strichartz systemati- cally obtained Lp estimates for the Schr¨odingerequation, the wave equation and the Klein-Gordon equation through the corresponding restriction/ extension theorems for quadratic surfaces using again explicit calculations of Fourier transforms (to get the decay), now combined with complex interpolation (similar to the argument for the endpoint case in the Tomas-Stein theorem) and the T ∗T trick. For the wave equation Strichartz obtained the estimate in an earlier pa- 2 per [Str70] kuk p m+1 ≤ k(∂ − ∆)uk 0 using explicit calculation L (R ) t Lp (Rm+1) for the Fourier transform. The estimates for the wave equation were opti- mized by Ginibre and Velo [GV95] using an approach which merges Fourier methods (in particular stationary phase estimates) and operator theoretic methods (like the a more systematic use of the T ∗T trick). For further de- velopments and applications to nonlinear PDE see, e.g., [KT98, So95, SS98].

156 [July 12, 2019] 7.5 A quick look back Some central ingredients and ideas

• Scaling (everywhere); anisotropic scaling arises in the Marcinkiewicz multiplier theorem, limits on the exponents in restriction theorems, estimates for parabolic equations and the Schr¨odingerequation

• Dyadic decomposition (on each block the action of multiplication or diﬀerentiation operators is very simple)

• Calderon-Zygmund decomposition (another case of dyadic composition); for those who know probability: this is an important special case of a martingale

• Weak type-estimates

• Interpolation (Marcinkiewicz - real interpolation; Riesz-Thorin - complex interpolation)

• Oscillatory integrals, stationary phase estimates (’Fourier picture’)

• T ∗T trick (’operator picture’)

For elliptic equations isotropic scaling, CZ decomposition and (Marcinkiewicz) interpolation are suﬃcient and one obtains results for all p ∈ (1, ∞) (as well as in borderline spaces such as BMO and H1.) For optimal results for parabolic equation one needs anisotropic scaling (or other ideas like function space valued multipliers, see later). Again results typically hold for all p ∈ (1, ∞). For hyperbolic and dispersive equations one needs estimates for oscillatory integrals, restriction theorems, etc. This is more delicate and here there are typically restrictions on the values p. [8.7. 2019b, Lecture 23] [12.7. 2019, Lecture 24]

157 [July 12, 2019] 8 Epilogue: some highlights in PDE

This is a completely personal selection which leaves out many important contributions. As all such lists it says much more about the writer than about the subject.

(i) 1912 Distribution of large eigenvalues (Weyl); this generated a large and ongoing activity: spectral theory of operators in quantum mechanics, can one hear the shape of a drum?, heat kernel asymptotics ↔ geometry

(ii) 1933 Leray-Schauder degree and applications to PDEs in ﬂuid mechanics

(iii) 1934 Global weak solutions of the Navier-Stokes equation (Leray): ’Sobolev spaces’, weak solution, weak convergence, partial regularity, conjecture about self-similar singularities (resolved only in 1994 in the negative)

(iv) 1930’s Existence of minimal surfaces (FM Douglas 1936)

(v) Late 1940’s Distributions and their Fourier transform (FM L. Schwartz 1950)

(vi) Late 1930’s–50’s Sobolev spaces, Lax-Milgram, weak solution of elliptic PDEs, higher regularity

(vii) 1950’s and 60’s Theory of general linear PDE (FM H¨ormander1962, hypoellipticity, wave front set, pseudodiﬀerential operators, parametrix, beginning of microlocal analysis)

(viii) Mid 1950’s Admissibilty criteria for hyperbolic conservation laws (Lax, Oleinik)

(ix) 1956 Nash’s embedding theorem: a) totally unexepect low regularity solutions in low codimension, b) smooth solution in higher codimension. Huge impact: Ad a) Gromov’s theory of convex integration, recent applications to the Euler equations and turbulence; Ad b) Nash-Moser hard implicit function theorem.

(x) 1957/58 Regularity of divergence form elliptic and parabolic PDE with L∞ coeﬀ. (Nash, De Giorgi, early 60’s: new proof by Moser iteration); for nondivergence from late 70’s early 80’s (Evans, Krylov), applications to fully nonlinear elliptic PDE.

158 [July 12, 2019] (xi) 1960’s and 70’s Functional analytic approaches: evolution equations, (Kato, ...), semigroups (Hille-Yoshida theorem), commutators; interpolation, traces spaces, . . . (Lions-Magenes, Peetre, Triebel); problems in quantum mechanics including scattering, mostly for single particle systems [RS72+]

(xii) 1966 Arnold: solutions incompressible Euler equation are geodesics in the group of volume preserving diﬀeomorphisms (functional analytic version in [EM70])

(xiii) 1970’s Several highlights in harmonic analysis (e.g. real variable theory of Hardy space, Feﬀerman-Stein, FM Feﬀerman 1978).

(xiv) 1960’s to 80’s Geometric PDE I: Existence and regularity of minimal surfaces and harmonic maps in arbitrary dimensions (Schoen- Uhlenbeck (Abel Preis 2019), Hildebrandt, . . . ). Geometric measure theory for generalized geometric objects: sets of finite perimeter, varifolds, currents (De Giorgi rectifibility thm. 1957, Federer-Fleming closure thm. 1960, Allard regularity thm. 1971, Alm- green’s big regularity paper 1984 (new proof by Delellis-Spadaro 2013). Monotonicity formula, tangent cones, connections to gauge theory in physics and topology in particular through the Yang-Mills equations, FM S.T. Yau 1982 (positive mass conjecture in general relativity, Calabi-Yau conjecture), mean curvature flow (Huisken, . . . ).

(xv) Late 60’s- Topological methods: mountain pass lemma, Morse theory, gradient ﬂows index linking number, Floer homology, . . . [St08]

(xvi) 1970’s: Free boundary value problems (e.g., the form of a water wave, a fluid jet or a freezing crystal): flatness implies smoothness (Caffarelli), monotonicity formula (Alt-Cafarelli-Friedman [ACF84])

(xvii) 1970’s– Strichartz estimates for dispersive equations

(xviii) Mid 1970’s Systematic study of oscillation and concentration eﬀects in PDE: compensated compactness (Murat-Tartar), later: H-measures (Tartar), concentration compactness (P.L. Lions), subtle integrability properties of compensated compactness quantities and Hardy spaces.

(xix) 1977 Existence of minimizers in nonlinear elasticity (J.M. Ball, build- ing on work of Morrey on quasiconvexity).

(xx) 1980’s Viscosity solutions for ﬁrst order Hamilton-Jacobi equations and second order PDE (Crandall-Evans-Lions, Ishii, . . . , [CIL92])

(xxi) Late 1980’s Global weak solution of the Boltzmann equation (Di Perna-Lions, FM P.L. Lions, 1994), uses among others: velocity aver-

159 [July 12, 2019] aging for linear kinetic equations, renormalized solutions and DiPerna- Lions ﬂows for ODE

(xxii) 1980’s- Geometric PDE II: Viscosity solutions for the mean curvature flow and other flows (Evans-Spruck, Chen-Giga-Goto), Hawking- Penrose conjecture (Huisken-Ilmanen), Ricci flow (Hamilton, Yau, Perel- man’s proof of the Poincaréconjecture (FM 2010, rejected) [MT07]), geometric wave equations (existence and blow-up, bounded curvature conjecture for the Einstein equations, see 2010’s)

(xxiii) Late 1990’s Dissipative PDE as gradient ﬂows in the Wasserstein metric, Otto calculus, connections with optimal transport (Villani, McCann), new view on Sobolev and many other functional inequalities (FM Figalli 2018) deﬁnition of Ricci curvature in metric-measure spaces and compactness (Sturm, Lott-Villani)

(xxiv) 2000– Convergence to equilibrium for large solutions of the Boltz- mann equation under some regularity assumptions (Villani), Con- vergence Boltzmann to incompressible Navier-Stokes (Golse, Saint- Raymond), Landau damping (Mouhut-Villani), FM Villani 2010

(xxv) 2010’s Systematic study of irregular solutions of the Euler equations (DeLellis-Szekelyhidi, with precursors in the work of Scheﬀer and Schnirelman), proof of the Onsager conjecture for Euler equation, non-uniqueness for the Navier-Stokes equation; ’thin’ free boundary value problems (Caﬀarelli, FM Figalli 2018); Einstein equations (L2 curvature conjecture; Klainermann, Rodniansky, Szeftel; linear stability of Schwarzschild black holes; M. Dafermos, Holzegel, Rodnianski).

Many important results are missing from this list including results on: the interaction of PDE theory and effizient simulation (Cea’s lemma, Babuska- Brezzi condition, multigrid methods, boundary elements, H-matrices, . . . ), global analysis (e.g., the Atiyay-Singer index theorem), holomorphic functions in several variables (∂ problem), Hodge theory, degenerate and subel- liptic equations, Cauchy-Kowalevskaya theorems, harmonic measure, inverse problems, semiclassical limits, Carleman estimates and unique continuation, nodal properties of solutions, completely integrable equations and inverse scattering, dynamical systems methods, semiclassical methods, many-body quantum mechanics, equations of mixed type, stochastic pde, . . . Also putting a time to the emergence of a certain method is very tricky. Many methods and results have precursors (e.g., topologically methods cer- tainly enter the field in a profound way already with the Leray-Schauder degree in 1933) and have a strong influence on later work.

Two observations:

160 [July 12, 2019] • Inspiration often came from the study of fundamental equations in physics or geometry

• The historic development was often much less systematic than a modern presentation might suggest. E.g., Leray’s work on weak solutions of the nonlinear time-dependent Navier-Stokes equations came much earlier than the systematic development of weak solutions of the linear elliptic equations.

161 [July 12, 2019] References

[AE09] H. Amann, J. Escher, Analysis III. Translated from the 2001 Ger- man original, Birkh¨auser,2009.

[BL76] J. Bergh and J. L¨ofstr¨om,Interpolation spaces, Grundlehren No. 233, Springer, 1976.

[Br73] H. Brezis, Op´erateursmaximaux monotones et semi-groupe de con- tractions dans les espaces de Hilbert, North-Holland, 1973.

[Br83] H. Brezis, Analyse functionelle, Th´eorieet applications, Masson, 1983.

[Br11] H. Brezis, Functional analysis, Sobolev spaces and partial diﬀeren- tial equations, Springer, 2011.

[Da00] C. Dafermos, Hyperbolic conservation laws in continuum physics, Grundlehren, Springer, 2000, 3rd ed., 2010.

[Ev] L.C. Evans, Partial diﬀerential equations, Amer. Math. Soc., 1998.

[Ho83] L. H¨ormander,The analysis of linear partial diﬀerential operators I, Springer, 1983.

[LSU68] O.A. Ladyˇzenskaja, V.A. Solonnikov, N.N. Uralceva, Linear and quasilinear equations of parabolic type (translated from the Rus- sian), Am. Math. Soc., 1968.

[Li69] J.L. Lions, Quelques méthodes de résolutionsdes problèmesaux limites non linéaires,Dunod Gauthier-Villars, 1969.

[Ma84] A. Majda, Compressible ﬂuid ﬂow and systems of conservation laws in several space variables, Springer, 1984.

[Ru73] W. Rudin, Functional analysis, McGraw Hill, 1973 (reprinted by Tata McGraw Hill 1974).

[So95] C.D. Sogge, Lectures on nonlinear wave equations, International Press, 1995.

[St70] E.M. Stein, Singular integrals and diﬀerentiability properties of functions, Princeton Univ. Press, 1970.

[St93] E.M. Stein, Harmonic analysis - Real-variable methods, orthogo- nality, and oscillatory integrals.

[SS98] J. Shatah, M. Struwe, Geometric wave equations, Courant Lecture Notes in Mathematics, New York University, 1998.

162 [July 12, 2019] [Sm83] J. Smoller, Shock waves and reaction-diﬀusion equations, Springer, 1983.

[St93] E.M. Stein, Harmonic analysis - Real-variable methods, orthogo- nality, and oscillatory integrals, Princeton Univ. Press, 1993.

[St08] M. Struwe, Variational methods, Springer, 4th edition, 2008.

[Tao99] T. Tao, Lecture notes for 254b, 1999, http://www.math.ucla.edu/~tao/254b.1.99s/.

[Ta07] L. Tartar, An introduction to Sobolev spaces and interpolation spaces, Lecture Notes of UMI vol. 3, Springer, 2007.

[Tr83] H. Triebel, Theory of function spaces, Birkh¨auser,1983.

References

[ACF84] H.W. Alt, L.A. Caﬀarelli, A. Friedman, Variational problems with two phases and their free boundaries, Trans. AMS 282 (1984),

[AL83] H.W. Alt,, S. Luckhaus, Quasilinear elliptic-parabolic diﬀerential equations, Math. Z. 183 (1983), 311–341.

[Am04] L. Ambrosio, Transport equation and Cauchy problem for BV vector ﬁelds, Invent. Math. 158 (2004), 227–260.

[BCP62] A. Benedek, A.P. Calderon, R. Panzone, Convolution operators on Banach space valued functions, Proc. Natl. Acad. Sci. 48 (1962), 356–365.

[Ca66b] S. Campanato, Equazioni paraboliche del secondo ordine e spazi L2,θ(Ω, δ), Ann. Mat. Pura Appl. 73 (1966), 55–102.

[CIL92] M.G. Crandall, H. Ishii, P.-L- Lions, User’s guide to viscosity solutions of second order partial diﬀerential equations, Bull. AMS 27 (1992), 1–67.

[DP85] R. DiPerna, Compensated compactness and general systems of conservation laws, Trans. AMS 292 (1985), 383–420.

[DPL89] R. DiPerna, P.L. Lions, Ordinary diﬀerential equations, transport theory and Sobolev spaces, Invent. Math. 98 (1989), 511–547.

[DHP07] R. Denk, M. Hieber, J. Pr¨uss,Optimal Lp − Lq estimates for parabolic boundary value problems with inhomogeneous data, Math. Z. 257 (2007), 193–224.

163 [July 12, 2019] [EM70] D.G. Ebin, J. Marsden, Groups of diﬀeomorphisms and the motion of an incompressible ﬂuid, Ann. Math. 92 (1970), 102–163.

[Ed65] R.E. Edwards, Functional analysis, Holt, Rinehart and Winston, 1965, reprinted by Dover, 1995.

[GV95] J. Ginebre, G. Velo, Generalized Strichartz estimates for the wave equation, J. Funct. Anal. 133 (1995), 50–68.

[KT98] M. Keel, T. Tao, Endpoint Strichartz estimates, Amer. Math. J. 120 (1998), 955–980.

[Mi87] A. Mielke, Uber¨ maximale Lp RegularitätfürDifferentialgleichun- gen in Banach und Hilberträumen,Math. Ann. 277 (1987), 121– 133.

[Mi63] J. Milnor, Morse theory, Annals of Mathematical Studies 51, Princeton University Press, 1963.

[MT07] J. Morgan, G. Tian, Ricci ﬂow and the Poincar´econjecture, Clay Math. Monographs, Amer. Math. Soc, 2007.

[Mo64] J. Moser, A Harnack inequality for parabolic diﬀerential equations, Comm. Pure Appl. Math. 17(1) (1964), 101–134.

[Na57] J. Nash, Parabolic equations, Proc. Natl. Acad. Sci. 43 (1957), 754–758.

[Na58] J. Nash, Continuity of solutions of parabolic and elliptic equations, Amer. J. Math. 80 (1958), 931–954.

[RS72+] M. Reed, B. Simon, Methods of modern mathematical physics I - IV, Academic Press, 1972, 1975, 1978, 1979.

[Sc96] W. Schlag, Schauder and Lp estimates for parabolic systems via Campanato spaces, Comm. PDE 21 (1996), 1141–1175.

[Si] L. Simon, Schauder estimates by scaling, Calc. Var. 5 (1997), 391– 407.

[Str70] R.S. Strichartz, A priori estimates for the wave equation and some applications, J. Funct. Anal. 5 (1970), 218–235.

[St77] R.S. Strichartz, Restrictions of Fourier transforms to quadratic surfaces and decay of solutions to wave equations, Duke Math. J. 44 (1977), 705–714.

164 [July 12, 2019] [Ta79] L. Tartar, Compensated compactness and applications to partial diﬀerential equations, Research Notes in Math., Nonlinear Analysis and Mechanics: Heriot-Watt Symposium, Vol. 4 (R. J. Knops, ed.), Pitman Press, 1979.

[Ta83] L. Tartar, The compensated compactness method applied to systems of conservation laws, Systems of Nonlinear Partial Diﬀerential Equations (J. M. Ball, ed), NATO ASI Series, C. Reidel, 1983.

165 [July 12, 2019]