Advanced Partial Differential Equations Prof. Dr. Thomas

Advanced Partial Diﬀerential Equations Prof. Dr. Thomas Sørensen summer term 2015

Marcel Schaub

July 2, 2015

1 Contents

0 Recall PDE 1 & Motivation 3 0.1 Recall PDE 1 ...... 3

1 Weak derivatives and Sobolev spaces 7 1.1 Sobolev spaces ...... 8 1.2 Approximation by smooth functions ...... 11 1.3 Extension of Sobolev functions ...... 13 1.4 Traces ...... 15 1.5 Sobolev inequalities ...... 17

2 Linear 2nd order elliptic PDE 25 2.1 Linear 2nd order elliptic partial diﬀerential operators ...... 25 2.2 Weak solutions ...... 26 2.3 Existence via Lax-Milgram ...... 28 2.4 Inhomogeneous bounday value problems ...... 35 2.5 The space H−1(U) ...... 36 2.6 Regularity of weak solutions ...... 39

A Tutorials 58 A.1 Tutorial 1: Review of Integration ...... 58 A.2 Tutorial 2 ...... 59 A.3 Tutorial 3: Norms ...... 61 A.4 Tutorial 4 ...... 62 A.5 Tutorial 6 (Sheet 7) ...... 65 A.6 Tutorial 7 ...... 65 A.7 Tutorial 9 ...... 67 A.8 Tutorium 11 ...... 67

B Solutions of the problem sheets 70 B.1 Solution to Sheet 1 ...... 70 B.2 Solution to Sheet 2 ...... 71 B.3 Solution to Problem Sheet 3 ...... 73 B.4 Solution to Problem Sheet 4 ...... 76 B.5 Solution to Exercise Sheet 5 ...... 77 B.6 Solution to Exercise Sheet 7 ...... 81 B.7 Solution to problem sheet 8 ...... 84 B.8 Solution to Exercise Sheet 9 ...... 87

2 0 Recall PDE 1 & Motivation

0.1 Recall PDE 1 We mainly studied linear 2nd order equations – speciﬁcally, elliptic, parabolic and hyperbolic equations. Concretely: • The Laplace equation ∆u = 0 (elliptic) • The Poisson equation −∆u = f (elliptic)

• The Heat equation ut − ∆u = 0, ut − ∆u = f (parabolic)

• The Wave equation utt − ∆u = 0, utt − ∆u = f (hyperbolic) We studied (“main motivation; goal”) well-posedness (à la Hadamard) 1. Existence of solutions 2. Uniqueness of solutions 3. Continuous dependence on the data. Also “properties of solutions” were important (regularity, estimates, mean value formulae, infinite/finite propagation speed and “techniques”). Recall that on existence, two types of results exist (1) Write down explicit solution formula (i.e. derive it – somehow – then prove that it does produce a solution). We did this “several times”, but in very specific cases and n 2 n very limited! For example for the Laplace equation −∆u = f in R , for f ∈ Cc (R ). n We studied harmonic polynomials, i.e. ∆u = 0 in R with u a polynomial. Then we studied the problem in the half-space

( n ∆u = 0 in R+ n u = g on ∂R+ n−1 ∞ n−1 for g ∈ C(R ) ∩ L (R ). The same for a Ball and g ∈ C(∂B(0, r)). Or the heat equation

( n ut − ∆u = 0 in R × (0, ∞) n u = g on R × {t = 0} n ∞ n for g ∈ C(R ) ∩ L (R ). Or ( n ut − ∆u = f in R × (0, ∞) n u = 0 on R × {t = 0} and combinations. Also the wave equation was studied ( utt − uxx = 0 in R × (0, ∞) u = g, ut = h on R × {t = 0}

and on R+ × (0, ∞)  utt − uxx = 0 in + × (0, ∞)  R u = g, ut = h on R+ × {t = 0}  u = 0 on {x = 0} × (0, ∞)

and for dimensions n > 2 odd/even. I.e.

3 (a) Equations are very speciﬁc (not general elliptic/parabolic/hyperbolic) n (b) The geometry is very speciﬁc (all of space (R ), half-space, ball). (2) One exception: Perron’s Method for ( ∆u = 0 in U u = g on ∂U

for g ∈ C(∂U) guaranteed existence for every g ∈ C(∂U) provided some (complicated) condition on ∂U is satisﬁed (certain simple geometric conditions (on ∂U) can assure that this holds). More general geometry – but still a very speciﬁc equation. Recall that in fact we are interested in (homogeneous/inhomogeneous) Lu = 0, Lu = f (elliptic) n in U ⊆ R or ut − Lu = 0, ut − Lu = f (parabolic) in U × [0,T ] or utt − Lu = 0, utt − Lu = f (hyperbolic) in U × [0,T ] with

n n X X Lu = aij(x)uxixj + bi(x)uxi + c(x)u(x) i,j=1 i=1

and (maybe) conditions on aij, bj, c. Hence, in PDE 2, we shall concentrate on methods to existence of solutions – mostly (only...) illustrated on elliptic linear 2nd order equations Lu = f. Possibly we also discuss parabolic/hyperbolic equations. Learning techniques, methods, and ideas that can often be applied also for other equations – diﬀerent order, and more or less nonlinear.

Two main ideas (1) Extend the concept of a “solution” (from “classical solutions”, as in PDE 1)

(2) Use some Functional Analysis.

Today (and next times), we will start by saying something (more or less vague) about these two ideas: “why & how”. Depending on where/with whom you did PDE 1, you might know some of this. There is no hope to solve “every” PDE by ﬁnding a solution. Therefore, we need something more systematic/general, which enters Functional Analysis. Functional Analysis (linear!) is the study of

(1) linear (inﬁnite dimensional) spaces, with a topology (i.e. notion of convergence/approximation): Banach spaces, Hilbert spaces, Fréchet-spaces, locally convex topological vector spaces.

(a) In general (i.e. all Banach spaces) (b) Speciﬁc examples (i.e. Lp(Ω), `p).

(2) Linear maps (mostly continuous linear maps) between such spaces: T : X −→ Y .

The idea of using Functional Analysis is to reformulate the question “Does a solution to F (...) = 0 (PDE) exist?” as “Does the map F : X −→ Y vanish somewhere?” m n m n Note that the study of f : R −→ R, f : R −→ R (1) (nonlinear) and A: R −→ R (2) (linear) in (1) is Analysis I-III and (2) is Linear Algebra. This is also motivated by questions about solutions to equations (so is “all” of abstract algebra). Hence, the “goal” is to reformulate questions in PDE via FA. Nevertheless, this is not an FA-course! We will need

4 certain tools which should be known from FA 1. If need be, we can do some in the Tutorials. We need to ﬁx appropriate vector spaces of functions (“function spaces”). A natural choice is Ck(U) or Ck(U) with supremum-norm (sup-norm). kfk = P kDαfk . Ck(U) |α|6k L∞(U) We get a Banach space (i.e. complete). However, the relevant operators (relevant for PDE) T : X −→ Y turn out not to be “well-behaved” in this setup (maybe more precise later). One possible replacement is Hölder-spaces. n Deﬁnition 0.1. Let U ⊆ R open, 0 < γ 6 1. γ (i) If u: U −→ R and there is a constant C > 0 such that |u(x) − u(y)| 6 C|x − y| for every x, y ∈ U then u is Hölder-continuous with exponent γ.

(ii) If u: U −→ R is bounded and continuous, we let kukC(U) := supx∈U |u(x)| and if u is also Hölder-continuous with exponent γ, we let |u(x) − u(y)| [u]C0,γ (U) := sup γ x,y∈U |x − y| x6=y

(the γ’th Hölder semi-norm) and kukC0,γ (U) := kukC(U) + [u]C0,γ (U) (the γ’th Hölder- norm of u).

k,γ (iii) The Hölder-space C (U), k ∈ N0, γ ∈ (0, 1] is deﬁned as k,γ k C (U) := {u: U −→ R | u ∈ C (U), kukCk,γ (U) < ∞}

where the norm k · kCk,γ (U) is

X α X α kukCk,γ (U) := kD ukC(U) + [D u]C0,γ (U |α|6k |α|=k

k,γ Theorem 0.2. (C (U), k · kCk,γ (U)) is a Banach space. Furthermore, for many problems in PDE (especially elliptic), these spaces are “good” for studying PDE’s (i.e. the operators T : X −→ Y one needs to study have “good” properties). For example one can prove “Elliptic Schauder-estimates”. However, this is often very, very hard work – this is not easy to explain, but basically, the reason is the sup-norm is hard to deal/work with – norms given by integrals are much, much “better”. So the idea is to take Ck(U) – or Ck,γ(U) – and use an integral norm – say, Lp-norm (p ∈ [1, ∞)): P α P α kD uk p instead of kD uk . The problem is: With these norms, we |α|6k L (Ω) |α|6k C(U) do not get a complete space (i.e. Ck(U), or Ck,γ(U) with these norms are not Banach spaces). The solution could be to abstractly “complete” the space (always possible!) but this is not a good solution if we have no concrete representation (of added objects). It turns out that one does have one. Q is not complete – R is – and “added” objects are still numbers. Riemann integrable functions on [0, 1] – R[0, 1] – with L1-norm – are not complete. But L1[0, 1] is complete – the “added” objects are still “functions”. This leads to the deﬁnition of “weak derivatives” and “Sobolev spaces” (next chapter). Note that this leads to proofs of existence of solutions to PDE’s which might be one of the “new added functions” – called “weak solutions” – and not a “true” classical solution. Secondly (often relatively independent) is afterwards to study regularity: Prove that this weak solution is in fact a true Ck-function (and that solves equation as a classical solution). I.e. Sobolev spaces are a tool to prove existence of some kind of solutions (“generalised solutions”), but the real interest is in classical solutions (always) regularity studies. We will study “weak derivatives”, “Sobolev spaces”, and “weak solutions”.; These concepts are very important in many PDE contexts (also other than studied in PDE 2). However

5 (1) There are also other concepts of “generalised derivatives” (in distribution theory)‘.

(2) For other types of equations, other types of “generalised solutions” might be relevant. For example, “entropy solutions”, “kinetic solutions”, “viscosity solutions”, “shock waves”.

(always “+ regularity studies”).

Note. (1) M. Hairer got the Fields medal (2014) for inventing and studying a new concept of “generalised solution” for certain types of PDE.

(2) Nirenberg & Nash got the Abel Price 2015 for studying regularity theory of certain nonlinear elliptic equations from geometry.

I.e. the programme for PDE 2:

• Introduce “weak derivatives” and the appropriate function spaces (“Sobolev spaces”) and study their properties.

• Prove existence of “weak solutions” (to linear elliptic 2nd order PDE’s).

• Prove regularity: That in fact are classical solutions.

6 1 Weak derivatives and Sobolev spaces

n Motivation (Integration by parts). Recall that for U ⊆ R open, ∞ ∞ Cc (U) := {φ: U −→ R | φ ∈ C (U), supp(φ) ⊆ U compact} 1 ∞ (“compact” in the relative topology in U!). Let u ∈ C (U), φ ∈ Cc (U) (φ is called “test-function”). By integration by parts:

u(x)φxi (x)dx = − uxi (x)φ(x)dx i = 1, . . . , n (∗) ˆU ˆU ∞ k (no boundary terms since φ ∈ Cc (U)!). Generally, for k ∈ N and u ∈ C (U) and using (∗) k times, k = |α|, we get

u(x)Dαφ(x)dx = (−1)|α| (Dαu)(x)φ(x)dx (∗∗) Û Û The right side makes sense if u ∈ Ck(U) (a little less would do...). However, the left side α 1 1 makes sense if uD ϕ ∈ L (U). This always holds if u ∈ Lloc(U) where 1 1 Lloc(U) := {u: U −→ R | u|K ∈ L (K) ∀K ⊆ U compact} ∞ α ∞ k since φ ∈ Cc (U) and so D φ ∈ Cc (U). Hence, if u∈ / C the equality (∗∗) a priori makes no sense. But we can use it as a motivation to define the “weak derivative” Dαu. 1 n th Definition 1.1. Let u, v ∈ Lloc(U), α ∈ N0 . We call v an α weak derivative of u iff v and u satisfy

α |α| ∞ u(x)D φ(x)dx = (−1) v(x)φ(x)dx φ ∈ Cc (U) () Û Û In this case, we write Dαu := v. 1 th α Remark. I.e. given u ∈ Lloc(U), u has an α weak derivative (called D u) iff there is 1 v ∈ Lloc(U) such that () holds. 1 Lemma 1.2 (Uniqueness of weak derivatives). Let u ∈ Lloc(U), α a multiindex. If u has th 1 α α an α weak derivative, then it is unique: If v, v˜ ∈ Lloc(U) satisfy v = D u, v˜ = D u (i.e. () holds for both v and v˜), then v =v ˜ a.e. in U. α α Proof. If v = D u, v˜ = D u, then, by ()

α |α| |α| ∞ uD u dx = (−1) vφ dx = (−1) vφ˜ dx ∀φ ∈ Cc (U) Û Û Û Hence, (v(x) − v˜(x))φ(x) dx = 0 Û ∞ for every φ ∈ Cc (U) and so, by the fundamental lemma of the calculus of variations B.1.2 v − v˜ = 0 a.e. in U.

n 1 Example 1.1. Let n = 1, U = (0, 2) ⊆ R = R = R and let ( ( x 0 < x 1 1 0 < x 1 u(x) := 6 v(x) := 6 1 1 < x < 2 0 1 < x < 2

Then u0 = v in the weak sense (Exercise 1 on Sheet 1).

7 Example 1.2. Let n = 1, U = (0, 2) ( x 0 < x 1 u(x) := 6 2 1 < x < 2

Then u0 does not exist in the weak sense (i.e. u has no weak derivatives) (Exercise 1 on Sheet 1). Hence, it is not enough that “u0 exists a.e. (in the classical sense) and this is an 1 Lloc-function”.

1.1 Sobolev spaces k,p Deﬁnition 1.3. For p ∈ [1, ∞], k ∈ N0, the Sobolev space W (U) is deﬁned by ( ) u ∈ L1 (U), ∀α ∈ n, |α| k : Dαu exists k,p loc N0 6 W (U) = u: U −→ R α p in the weak sense and D u ∈ L (U)

k k,2 k Remarks. (1) For p = 2, we write H (U) := W (U), k ∈ N0, since – see later – H (U) is a Hilbert-space.

0 2 0,p p k,p m,p (2) H (U) = L (U) and W (U) = L (U). Also, W (U) ⊆ W (U) for k > m. (3) As usual, when treating Lp-spaces, we identify functions in W k,p(U) which agree a.e. in U.

(4) The notation varies a lot in the literature.

Definition 1.4 (Sobolev norms). For u ∈ W k,p(U) we define its W k,p-norm as  1/p  X α p  |(D u)(x)| dx p ∈ [1, ∞)  Û |α|6k kukW k,p(U) := X α  ess sup |D u| p = ∞  U |α|6k

1/p P α p α Remarks. (a) I.e. kuk k,p = kD uk , p ∈ [1, ∞) where D u is the W (U) |α|6k Lp(U) αth weak derivative of u.

n (b) Introduce an ordering on the set Ak := {α ∈ N0 | |α| 6 k} and letting nk := |Ak| we see that n α o kukW k,p(U) = kD ukLp(U) α∈A n k `p(R k ) q n n One can use other ` (R k )-norms, since all norms on R k are equivalent (this gives k,p α equivalent norms on W (U), e.g. for p = ∞: max|α|6k kD ukLp(U) or for p = 1: P α kD uk p . |α|6k L (U) k,p k,p Deﬁnition 1.5 (Convergence in Sobolev spaces). Let (um)m∈N ⊆ W (U), u ∈ W (U). k,p n→∞ k,p (i) We say um converges to u in W (U), written um −−−→ u in W (U), iﬀ

lim kum − ukW k,p(U) = 0 m→0

8 n→∞ k,p n→∞ k,p (ii) We write um −−−→ u in Wloc (U) iﬀ whenever V b U, then um −−−→ u in W (V ).

Recall that V b U iﬀ V ⊆ U (V open) and V ⊆ U is compact (in the relative topology). k·k k,p ∞ W k,p(U) Deﬁnition 1.6. W0 (U) := Cc (U) .

k,p k,p ∞ ∞ k,p I.e. W0 (U) is the closure (in the W (U)-norm) of Cc (U). Note that Cc (U) ⊆ W (U) ∞ k,p for all k, p. Note also that this means that Cc (U) is dense in W0 (U) (by definition). k,p ∞ Hence, u ∈ W0 (U) iff there is a sequence (um)m ⊆ Cc (U): limm→∞ kum −ukW k,p(U) = 0. k,p k,p k,p In particular W0 (U) ⊆ W (U). Later we shall see: One can interpret W0 as “those k,p α functions u ∈ W (U) such that D u = 0 on ∂U for all |α| 6 k − 1”. Note the relation to k boundary values – in particular, to Dirichlet problem). Again, the notation is H0 (U) := k,2 W0 (U), p = 2. 1,p Remark. For n = 1, U ⊆ R an open interval u ∈ W (U) iff there is v absolutely continuous with

(a) u = v a.e

(b) v0 ∈ Lp(U).

(v0 exists as a classical derivative a.e. when v is absolutely continuous). Hence, in Di- mension 1 (but only here!) one has a “simple characterization” via absolutely continuous functions and classical derivatives). Reference: “A First Course in Sobolev Spaces”, Leoni, AMS, GSM-series.

n −a Example 1.3. Let U = B1(0) ⊆ R the unit ball, a > 0 and u(x) = |x| for x ∈ U, x 6= 0. 1,p n−p 1,p Then u ∈ W iﬀ a < p (Exercise 1 Sheet 1). Especially p > n implies u∈ / W (U).

Example 1.4. Let (vk)k∈N ⊆ B1(0) be countable and dense. Let ∞ X 1 u(x) = |x − v |−a 2k k k=1

1,p n−p x ∈ U. Then (!) u ∈ W (U) for a < p but u is unbounded on each open subset of U! We will prove certain elementary properties of weak derivatives

• some of which (all?!) are trivial for “real” (classical) derivatives.

• Here, we have to always work with the deﬁnition of weak derivatives.

k,p Theorem 1.7 (Properties of weak derivatives). Let u, v ∈ W (U) and |α| 6 k. Then α k−|α| β α (i) D u ∈ W (U) and for all multiindices α, β with |α|+|β| 6 k we have D (D u) = Dα(Dβu) = Dα+βu.

k,p α α α (ii) For all λ, µ ∈ R: λu + µv ∈ W (U) and D (λu + µv) = λD u + µD v for all |α| 6 k. (iii) If V ⊆ U is an open subset, then u ∈ W k,p(V ) (see also 1.5 (ii)).

9 ∞ k,p (iv) For any ζ ∈ Cc (U), we have ζu ∈ W (U) and Leibniz’s formula X α Dα(ζu) = (Dβζ)(Dα−βu) β β6α α α! n holds with β := β!(α−β)! for α, β ∈ N0 , α! := α1!α2! ··· αn! and α 6 β ⇐⇒ αi 6 βi for all i = 1, . . . , n. For (iv), compare with Ex 1, Sheet 1, PDG 1 (WS 13/14). Proof. Exercise 2 on Sheet 1.

n Theorem 1.8 (Sobolev spaces as function spaces). Let U ⊆ R be open. Let k ∈ N0, k,p p ∈ [1, ∞]. Then (W (U), k · kW k,p(I)) is a Banach space. k,p Proof. That W (U) is a linear space is Theorem 1.7 (ii). We ﬁrst check that k · kW k,p(U) k,p is a norm. Let λ ∈ R, u ∈ W (I), then clearly kλukW k,p(U) = |λ| · kukW k,p(U) (use Theorem 1.7 (ii)). Also, kukW k,p(U) = 0 then kukLp(U) = 0 then u = 0 a.e. in U. The triangle inequality: Let u, v ∈ W k,p(U), p ∈ [1, ∞) (exercise: do p = ∞!). Minkowski’s inequality gives: 1/p X α α p ku + vkW k,p(U) = kD u + D vkLp(U) |α|6k 1 p /p X α α = kD ukLp(U) + kD vkLp(U) |α|6k p α = kD ukLp(U) + kD vkLp(U) p n α∈Ak α∈Ak ` (R k ) p α kD ukLp(U) + kD vkLp(U) 6 p n p n α∈Ak ` (R k ) α∈Ak ` (R k ) 1/p 1/p X α p X α p = kD ukLp(U) + kD vkLp(U) |α|6k |α|6k

= kukW k,p(U) + kvkW k,p(U) k,p It remains to prove completeness. Let (um)m∈N ⊆ W (U) be Cauchy (in k · kW k,p(U)). α p p Then for all α with |α| 6 k: (D um)m∈N ⊆ L (U) is Cauchy. As L is complete, there is p α m→∞ p m→∞ uα ∈ L (U) such that D um −−−−→ uα in L (U). In particular um −−−−→ u(0,0,...,0) =: u p k,p α 1 in L (U). Claim: u ∈ W (U) and D u = uα for all |α| 6 k. Firstly, u ∈ Lloc(U) by ∞ Hölders inequality – exercise. Fix φ ∈ Cc (U), then α α |α| α uD φ dx = lim umD φ dx = lim (−1) (D um)φ dx Û m→∞ Û m→∞ Û |α| = (−1) uαφ dx Û p α since um → u in L and Hölder. Hence uα = D u (by definition and uniqueness) – and p k,p α m→∞ α since uα ∈ L (U) for every |α| 6 k, we have u ∈ W (U). Hence, D um −−−−→ D u in p L (U) for all |α| 6 k, so we have that

lim kum − uk k,p = 0 m→∞ W (U) k,p k,p I.e. um → u in W (U), so every Cauchy sequence in W (U) is convergent (in norm in W k,p(U)), to an element in W k,p(U).

10 1.2 Approximation by smooth functions Approximation: To “compute” in Sobolev spaces, it turns out to be practical to resort to approximation by smooth functions. Note that C∞(U) ⊆ W k,p(U) for all k, p, when U is bounded. Fix k ∈ N, p ∈ [1, ∞]. Recall for ε > 0: Uε := {x ∈ U | dist(x, ∂U) > ε} and let ∞ n n η be a molliﬁer (see also Sheet 1), i.e. η ∈ Cc (R ), 0 6 η 6 1, η ≡ 0 on R \ B1(0) and η = 1. Rn ´ Theorem 1.9 (Local approximation by smooth functions). Let u ∈ W k,p(U) for some ε −n x p ∈ [1, ∞) and let u := ηε ∗ u in Uε with ηε(x) = ε η( ε ). Then ε ∞ (i) u ∈ C (Uε) for all ε > 0

ε ε→0 k,p (ii) u −−−→ u in Wloc (U). Proof. (i) Exercise 4 on Sheet 1.

α ε α α ε th (ii) We claim that D u = ηε ∗ D u in Uε. Note that “D u ” is the α partial derivative ∞ ε α th of the C -function u , whereas “ηε ∗ D u” is the ε-molliﬁcation of the α weak derivative of u. Let x ∈ Uε. Then

(Dαuε)(x) = Dα η (x − y)u(y) dy = (Dαη )(x − y)u(y) dy ˆ ε ˆ x ε = (−1)|α| (Dαη )(x − y)u(y) dy ˆ y ε α α = ηε(x − y)D u(y) dy = [ηε ∗ D u](x) ˆU

∞ Note: For fixed x ∈ Uε, the function φ(y) := ηε(x − y) is in Cc (U). Hence, by the th definition of the α weak derivative (of u). Now let V ⊆ U open wih V b U (i.e. V ⊆ U, V compact). Hence (see Ex 4(c), Sheet 1), Dαuε −−−→ε→0 Dαu in Lp(V ) for all α, with |α| k. Hence (think!) kuε −ukp = P kDαuε −Dαukp −−−→ε→0 0 6 W k,p(V ) |α|6k Lp(V ) since there are finitely many terms in the sum.

k,p k,p The next goal is to not only approximate in Wloc (U) by smooth functions but in W (U). However, by functions in C∞(U) – not in C∞(U) – see later.

n Theorem 1.10 (Global approximation by smooth functions). Let U ⊆ R be open and k,p bounded. Let u ∈ W (U) for some p ∈ [1, ∞). Then there is a sequence (um)m∈N ⊆ ∞ k,p m→∞ k,p C (U) ∩ W (U) such that um −−−−→ u in W (U).

∞ Remark. This does not say um ∈ C (U) – ! – see later.

1 Proof. Let, for i ∈ N, Ui := {x ∈ U | dist(x, ∂U) > i }, whence Ui is open. Then S∞ U = i=1 Ui. Let Vi := Ui+3 \ U i+1 (open), i ∈ N. Choose any open V0 b U so that S∞ U = i=0 Vi. Let {ζi}i∈N be a smooth partition of unity subordinate to the open sets ∞ P∞ {Vi}i∈N (of U), i.e. ζi ∈ Cc (Vi), 0 6 ζi 6 1, i=1 ζi = 1 (construction: see B.2.4). Let k,p k,p u ∈ W (U). By Theorem 1.7 (iv), ζiu ∈ W (U) and supp(ζiu) ⊆ Vi. Fix δ > 0. Then i i δ choose εi > 0 so small that, with u := ηεi ∗ (ζiu), we have ku − ζiukW k,p(U) 6 2i+1 , i ∈ N i P∞ i and supp u ⊆ Wi with Wi := Ui+4 \ U i ⊇ Vi. Let v := i=0 u , then v is welldeﬁned and ∞ C (U), since for all sets V b U at most ﬁnitely many terms in the sum are non-zero on

11 P∞ P∞ P∞ V . Recall that i=0 ζi = 1 on U so u = 1 · u = ( i=0 ζi) · u = i=0(ζiu). Hence, for V U b ∞ ∞ X i X 1 ku − vk k,p ku − ζ uk k,p δ = δ W (V ) 6 i W (U) 6 2i+1 i=0 i=0

Hence kv − ukW k,p(V ) 6 δ for all V b U. Take the supremum of V ’s with V b U to get ∞ k,p kv − ukW k,p(U) 6 δ, v ∈ C (U) ∩ W (U). Recall that in 1.10, we approximate with C∞(U)-functions – these might not be in C∞(U). It may not have a continuous extension to U (or, the derivatives have not). To assume approximations with C∞(U)-functions, one needs ∂U (the boundary of U) to be suﬃciently “nice”.

Theorem 1.11 (Global approximation by functions smooth up to the boundary). Let n 1 k,p U ⊆ R open and bounded, ∂U ∈ C . Then for every u ∈ W (U), k ∈ N0, p ∈ [1, ∞) ∞ n→∞ k,p there is a sequence (um)m ⊆ C (U) such that um −−−→ u in W (U). Remark. It will be “normal” that if we want “some” result to hold “up to the boundary” then we need some “niceness” of the boundary (here C1 assumed). It is also “normal” that the proof gets a bit more “technical”.

0 1 n−1 Proof. Fix x ∈ ∂U. By definition, since ∂U is C , there exists r > 0 and γ : R −→ R 1 0 0 (a C -function) such that U ∩ Br(x ) = {x ∈ Br(x ) | xn > γ(x1, . . . , xn−1)} (possibly after relabeling the coordinates). i.e. “∂U is locally the graph of a C1-function”. Let 0 0 ε V := U ∩ B r (x ) ⊆ U ∩ Br(x ). Define the “shifted point” x := x + λεen for x ∈ V (en 2 the unit vector in xn-direction). Note that there is λ > 0 (fixed sufficiently large) such ε 0 that Bε(x ) ⊆ U ∩ Br(x ) for all x ∈ V and small ε > 0, i.e.

ε 0 ∃λ > 0 ∃ε0 > 0 ∀x ∈ V ∀ε 6 ε0 : Bε(x ) ⊆ U ∩ Br(x )

ε ε ε 1 ε ∞ Let uε(x) := u(x ). Then let v := ηε ∗ u (ηε standard mollifer) . We have v ∈ C (U). ε ε→0 k,p Claim: v −−−→ u in W (V ). For, let α be a multiindex with |α| 6 k. Then

α ε α α ε α α α kD v − D ukLp(V ) 6 kD v − D uεkLp(V ) + kD uε − D ukLp(V ) Now

α α ε→0 p •k D uε −D ukLp(V ) −−−→ 0 since (see B.3.1) translation is continuous in the L -norm.

α ε α ε→0 •k D v − D uεkLp(V ) −−−→ 0 by arguments like in the proof of Theorem 1.9.

0 0 Let δ > 0. As ∂U is compact (closed and bounded) we have the existence of {x1, . . . , xN } ⊆ 0 N ∂U and {r1, . . . , rN } ⊆ + such that {B ri (x )} cover ∂U. For each i ∈ {1,...,N} do as R 2 i i=1 0 ∞ SN 0 above to get sets Vi = U∩B ri (x ) and functions vi ∈ C (V i) such that ∂U ⊆ B ri (x ) 2 i i=1 2 i SN k,p and kvi −ukW (Vi) 6 δ. Finally take an open set V0 b U such that U ⊆ i=0 Vi. Theorem ∞ k,p 1.9 gives v0 ∈ C (V 0) with kv0 − ukW (V0) 6 δ. Then let {ζi}i∈N be a smooth partition

1The idea is: If x was too close to ∂U, we could not mollify – we have moved “up” away from ∂U – so note that at xε, there is space to mollify.

12 0 0 PN of unity subordinate to {V0,B r1 (x1),...,B rN (x )}. Let v := ζivi (“pasting the vi’s 2 2 N i=0 ∞ together via partition of unity”). Then v ∈ C (U) and (by Theorem 1.7) for |α| 6 k

N α α X α α p p kD v − D ukL (U) 6 kD (ζivi) − D (ζiu)kL (Vi) i=0 N X k,p 6 C kvi − ukW (Vi) (∗) i=0 6 C(N + 1)δ For (∗) see Tutorial Sheet 3. Check carefully the above, to see this is enough to prove: ∞ ∀ε > 0 ∃v ∈ C (U): kv − ukW k,p(U) < ε. Remarks. (1)“ ∂U ∈ C0” should be enough.

(2) For a cutted disk, the result is not true (see book by Maz’ya).

1.3 Extension of Sobolev functions The next “themes” will also depend on ∂U (here we need ∂U ∈ C1)

(1) Extending W k,p-functions to larger sets.

(2) Restricting W k,p(U) (to ∂U).

We start with (1), the problem to extend u by setting u ≡ 0 outside U could give “bad derivatives” (not in W k,p in larger set).

n Theorem 1.12 (Extension Theorem). Let p ∈ [1, ∞] and U ⊆ R open and bounded with 1 ∂U ∈ C . Let V open and bounded with U b V . Then there exists a linear and bounded 1,p 1,p n 1,p operator E : W (U) −→ W (R ) such that for all u ∈ W (U): (i) Eu = u a.e. in U.

n (ii) Eu has support within V (i.e. Eu = 0 a.e. on R \ V ).

(iii) kEukW 1,p(Rn) 6 CkukW 1,p(U) with C > 0 depending only on p, U, V (i.e. not on u). n We call Eu an extension of u to R . 0 Proof. Fix some x ∈ ∂U. Assume first that ∂U is flat near x0, lying in the (hyper)plane 0 0 + {xn = 0} (near x ). Then there exists a Ball B(x , r) ≡ B such that B := B ∩ {xn > − n 1 0} ⊆ U and B := B ∩ {xn 6 0} ⊆ R \ U. Assume additionally that u ∈ C (U). Then define ( u(x) x ∈ B+ u(x) := xn − −3u(x1, . . . , xn−1, −xn) + 4u(x1, . . . , xn−1, − 2 ) x ∈ B u is called a higher-order reflection of u from B+ to B−. We claim u ∈ C1(B). Let − + − + th u := u|B− and u := u|B+ . First we show (u )xn = (u )xn on {xn = 0} (n partial − xn derivative). By definition, uxn (x) = 3uxn (x1, . . . , xn−1, −xn) − 2uxn (x1, . . . , xn−1, − 2 ) + 1 and hence ((u )xn = uxn – also on ∂U since u ∈ C (U)) if we put xn ≡ 0:

− + uxn = uxn {xn=0} {xn=0}

13 + − Also (insert xn = 0 in the deﬁnition of u): u = u on {xn = 0} (i.e. on B ∩ {xn = 0}). + − Hence, all other derivatives of u , u are along {xn = 0}, so

− + uxi = uxi i = 1, . . . , n − 1 {xn=0} {xn=0} From this follows

Dαu− = Dαu+ i = 1, . . . , n − 1 {xn=0} {xn=0}

1 for all α with |α| 6 1 hence u ∈ C (B). Also using these calculations, we get a constant C > 0 such that kukW 1,p(B) 6 CkukW 1,p(B+) with C independent of u but on B and on p. Recall that we assumed ∂U to be ﬂat near x0 (and that u ∈ C1(U)...). If this was not the case, we need to “straighten out ∂U”. Assume ∂U is Ck at x0 ∈ ∂U (use: k = 1). Then n−1 k there exists r > 0 and γ : R −→ R is a C -function such that

0 0 U ∩ B(x , r) = {x ∈ Br(x ) | xn > γ(x1, . . . , xn−1)}

(possibly after relabeling the coordinates). Then deﬁne

i yi = xi =: Φ (x) i = 1, . . . , n − 1 n yn = xn − γ(x1, . . . , xn−1) =: Φ (x)

n n so y = Φ(x), Φ: R −→ R

i xi = yi =: Ψ (y) i = 1, . . . , n − 1 n xn = yn + γ(y1, . . . , yn−1) =: Ψ (y) so x = Ψ(y). Then Φ = Ψ−1 and both Φ and Ψ are Ck-maps. The map x 7−→ Φ(x) = y “straightens out ∂U near x0”. Note that det DΦ = det DΨ = 1. Let u˜(y) := u(Ψ(y)) and choose a ball as before (around y0 where the boundary is flat). Do as above to get an extension of u˜ from B+ to a function u˜ defined on the whole ball B with u˜ ∈ 1 C (B) and ku˜kW 1,p(B) 6 Cku˜kW 1,p(B+). Let W := Ψ(B) (open set since Ψ and Φ are diffeomorphisms). Converting back to the x-variables, we get an extension u of u to W by u(x) := u˜(Φ(x)), x ∈ W , with, by change of variables and det DΦ = det DΨ = 1,

kukW 1,p(W ) 6 CkukW 1,p(U) Now we go from “local to global”: We need a partition of unity argument to glue everything 0 0 together. ∂U is compact, so there exist {x1, . . . , xN } ⊆ ∂U, open sets {W1,...,WN } and 0 SN extensions ui of u to Wi (each constructed as above with xi ∈ Wi) such that ∂U ⊆ i=1 Wi N 1,p 1,p S and kuikW (Wi) 6 CkukW (U). Finally we take W0 b U so that U ⊆ i=0 Wi and let N PN {ζi}i=0 be an associated partition of unity. Let u := i=0 ζiui, with u0 := u. Using SN ∂U ⊆ i=1 Wi and Theorem 1.7 we get

kukW 1,p(Rn) 6 CkukW 1,p(U) (∗) for some constant, depending on U, p and n but not on u. We can also arrange for supp u ⊆ V , U b V by replacing Wi by Wfi := Wi ∩ V , we have

N N [ [ supp u ⊆ supp ζi ⊆ Wfi ⊆ V i=0 i=0

14 Write Eu := u. Note that u 7−→ Eu is linear. Recall that we assumed above u ∈ C1(U). 1,p Let now u ∈ W (U), p ∈ [1, ∞) and choose (by Theorem 1.11) a sequence (um)m∈N ⊆ ∞ m→∞ 1,p ∞ C (U) with um −−−−→ u in W (U). Since E is linear and (∗) holds for C (U)-functions. For every m, `:

kEum − Eu`kW 1,p(U) 6 Ckum − u`kW 1,p(U)

1,p n 1,p n Hence, (Eum)m ⊂ W (R ) is Cauchy and so, by Theorem 1.8, there exists u ∈ W (R ) m→∞ 1,p n 1,p such that Eum −−−−→ u in W (R ). Set Eu := u. Then E is deﬁned on W (U) and this deﬁnition does not depend on the chosen sequence (um)m. E is also linear and bounded.

Remarks. (i) Assume ∂U ∈ C2 (not only C1). Checking the proof of 1.12 one sees: 2,p 2,p n The operator E is in fact a bounded linear operator E : W (U) −→ W (R ). One needs to observe that the “higher order reﬂection” ( u(x) x ∈ B+ u(x) := xn −3u(x1, . . . , xn−1, −xn) + 4u(x1, . . . , xn−1, − 2 ) x ∈ B−

2 2 2,p is not (in general) C (if u is C ), but it is W (B) and one has a bound kukW 2,p(B) 6 CkukW 2,p(B+).

∞ k,p (ii) Even if ∂U ∈ C the reﬂection above does not give an extension for W (U), k > 2. One needs a more involved “higher order reﬂection” than the above (see for example Haroske & Triebel p.75 - 77).

1.4 Traces Next we discuss “traces”, which in this context means “restriction” to a lower-dimensional submanifold/hypersurface, typically (here only) ∂U.

n Theorem 1.13 (Trace Theorem). Let p ∈ [1, ∞) and let U ⊆ R open and bounded. Assume ∂U is C1. Then there exists a bounded linear operator T : W 1,p(U) −→ Lp(∂U) such that

1,p (i) T u = u|∂U if u ∈ W (U) ∩ C(U).

1,p (ii) kT ukLp(∂U) 6 CkukW 1,p(U) for all u ∈ W (U) with C > 0 depending only on p and U.

We call T u the trace of u on ∂U.

Remarks. (a) Point (i) is why it is a “trace”: T is the restriction to the boundary ∂U when this should make sense (i.e. if u is also continuous). The theorem gives a way to assign an Lp-function on ∂U to a general W 1,p(U)-function in a way which is continuous in the appropriate norms.

(b) Point (ii) is just restating “bounded linear operator”.

We proceed as in the proof of Theorem 1.12 – we do it for suﬃciently regular u and conclude by approximation (i.e. density-argument). For regular u, do a partition of unity-argument and for the local problem, straighten the boundary. Finally do the “straight-boundary- case”. As in the proof of 1.12, we start with the end.

15 1 0 0 Proof. Assume u ∈ C (U), x ∈ ∂U and ∂U is ﬂat near x , lying in {xn = 0}. As earlier, 0 + − choose a ball Br(x ) ≡ B such that B := B ∩ {xn > 0} ⊆ U and B := B ∩ {xn 6 n 0 0} ⊆ \ U and let Bˆ be the ball with the same center and half the radius, Bˆ := B r (x ). R 2 ∞ ˆ Let ζ ∈ Cc (B) be such that 0 6 ζ 6 1 and ζ ≡ 1 on B (Exercise 4 Sheet 2). Let ˆ ˆ 0 n−1 1 Γ = ∂U ∩ B ⊆ B. Let x = (x1, . . . , xn−1) ∈ R = {xn = 0}. Then, since u ∈ C (U), we 0 have u|∂U ∈ C (∂U) (even better...) and

p 0 p 0 p 0 ! p |u| dx = ζ(x)|u| dx 6 ζ(x)|u| dx = − (ζ|u| )xn dx () + ˆΓ ˆΓ ˆ{xn=0} ˆB because by partial integration in xn:

p p p p 0 − 1(ζ|u| )xn dx = 1 · ζ|u| dS + 1xn ζ|u| dx = ζ|u| dx + + + ˆB ˆ∂B ˆB ˆ{xn=0}

Hence, () is equal to

p p−1 0 p p−1 − (|u| ζxn + p|u| sgn(u)uxn ) dx 6 C (|u| + |u| |Du|) ˆB+ ˆB+ p (p−1) p−1 p p 0 p−1 0 |u| |u| + |Du| = C |u| (|u| + |Du|) 6 C p−1 + ˆ + ˆ + p B B p C0 = (|u|p + |Du|p) dx =: C (|u|p + |Du|p) dx p ˆB+ ˆB+ by Young’s inequality (see A.2.2). Now if x0 ∈ ∂U, but ∂U is not ﬂat near x0, we straigthen out the boundary to get the setting above. Applying the estimate above and changing variables we get

p p p |u| dS 6 C (|u| + |Du| )dx ˆΓ ˆU 0 0 0 with Γ ⊆ ∂U some subset containing x . By compactness of ∂U there are x1, . . . , xN ∈ ∂U, SN and open subsets Γi ⊆ ∂U, i = 1,...,N such that ∂U = i=1 Γi and

kukLp(Γ) 6 CkukW 1,p(U) where C > 0 depends on p, Γi but not on u. Then

N X p p 1,p kukL (∂U) 6 kukL (Γi) 6 CekukW (U) i=1 SN since ∂U = i=1 Γi with Ce := max16i6N C(Γi) · N. Setting T u := u|∂U which makes sense for u ∈ C1(U) we have

1 kT ukLp(∂U) 6 CkukW 1,p(U) ∀u ∈ C (U) (∆)

1,p ∞ 1 m→∞ Assume now u ∈ W (U) and take a sequence (um)m ⊆ C (U) ⊆ C (U) with um −−−−→ u 1,p in W (U). Then (∆) implies kT um − T u`kLp(∂U) 6 Ckum − u`kW 1,p(U) so (T um)m is p Cauchy in L (∂U) which is Banach. Let T u := limm→∞ T um. As in the proof of Theorem 1.12 (see also Tut 3, ex 5) this implies T : W 1,p(U) −→ Lp(∂U) is a well-deﬁned, linear and bounded operator. Assume ﬁnally u ∈ W 1,p(U) ∩ C(U). Note that the functions ∞ 1,p constructed in Theorem 1.11, (um)m ⊆ C (U), um → u in W (U) converge uniformly 1,p to u on U (in particular on ∂U). Hence, T u = u|∂U for u ∈ W (U) ∩ C(U) (since true ∞ for um ∈ C (U) and T um = um|∂U ).

16 1 1 ap bq Lemma (Young’s inequality). Let 1 < p, q < ∞, p + q = 1. Then ab 6 p + q (a, b > 0). Recall that eventually, we want to study PDE – for example boundary value problems (BVP) à la ( −∆u = f in U u = 0 on ∂U so we need to unterstand exactly when this happens (i.e. T u = 0).

1,p n Theorem 1.14 (Trace-zero functions in W ). Let U ⊆ R open and bounded with ∂U ∈ 1 1,p 1,p C . Assume u ∈ W (U). Then u ∈ W0 (U) iﬀ T u = 0.

W 1,p(U) 1,p ∞ 1,p Remark 1.15. Recall W0 (U) = Cc (U) , i.e. u ∈ W0 (U) iﬀ there is a sequence ∞ 1,p (um)m ⊆ Cc (U) such that um → u in W (U) (Deﬁnition 1.6). 1,p Proof (Only one way - the simple way!). Assume u ∈ W0 (U). Then there is a sequence ∞ 1,p ∞ (um)m∈N ⊆ Cc (U), um → u in W (U). For um ∈ Cc (U) we have T um = 0 for m ∈ N 1,p p since T v = v|∂U , v ∈ C(U). Since T : W (U) −→ L (U) is a linear bounded operator, it 1,p p is continuous, so um → u in W (U) implies T um → T u in L (∂U). Recapitulation: We have by now

• Deﬁned weak derivatives

• Deﬁned Sobolev spaces (Banach!)

• Studied approximation via C∞-functions

• Studied extensions (E)

• Studied traces (T ; restriction to the boundary)

1.5 Sobolev inequalities Next, we are going to focus on Sobolev inequalities and Sobolev embeddings. These topics are

(1) Extremely important

(2) Extremely interesting (!)

(3) Vast!

Hence, we will not cover everything – in particular, we will not give all proofs. We will do give some motivation: What is this about and why it is useful? Morally: What do the theorems say and how do we roughly remember the result: Statements of all versions, the proofs of some (the!) inequalities. If wanted, we can do more on that in the tutorials.

1,p p Motivation. (A) Recall that W (U) & L (U) (U open and bounded, k ≡ 1, p ∈ [1, ∞]. I.e. u ∈ W 1,p(U) implies u ∈ Lp(U). Of course u ∈ Lp(U) then u ∈ Lq(U) for all q 6 p. Hence, the moral (although not clear why) is “the bigger the p, the better” (at least “more”). Sobolev inequalities come in now: If u is not just in Lp(U), but in u ∈ W 1,p(U) (i.e. Du ∈ Lp(U), then in fact u ∈ Lq(U) for some q > p! I.e. “W 1,p(U) ⊆ Lq(U) for some q > p”.

17 1,p (B) Moreover, recall Exercise 3 Sheet 2. Let u ∈ W ([a, b]), p ∈ (1, ∞) and [a, b] ⊆ R. 1 Then α := 1 − p > 0 and

0 kukC0,α([a,b]) 6 |u(x0)| + Cku kLp([a,b]) i.e. if the right hand side is ﬁnite, then the left hand side is also ﬁnite. If we had 0 1,p kukC0,α([a,b]) 6 C(kukLp([a,b]) + ku kLp([a,b])) (∆) then u ∈ W ([a, b]) implies u ∈ C0,α([a, b]) (Hölder continuous). Then in particular:

(a) u is continuous on [a, b] (b) u ∈ Lq([a, b]) for all q ∈ [1, ∞].

Statements like “W 1,p(U) ⊆ Lq(U) for some q > p” and “u ∈ W 1,p([a, b]) =⇒ u ∈ C0,α([a, b])” are called “embedding results” (Sobolev embedding) since it “embeds one space in another”. Inequalities like (∆) are called Sobolev inequalities (in general – (∆) has a more speciﬁc name). The connection between “embeddings” and “inequalities”:

(a) From the inequality (∆) we can “read”: If u ∈ W 1,p([a, b]), then u ∈ C0,α([a, b]), i.e. W 1,p(U) ⊆ C0,α(U), U = [a, b]. (b) But, more importantly! It is a “continuous identiﬁcation”. One normally proves inequalities like (∆) (i) For every nice function (say, C∞) (ii) Then (like for E in Theorem 1.12 and T in Theorem 1.14 one has that the identity operator has a unique bounded extension (i.e. one identiﬁes u ∈ W 1,p(U) with some (!) u ∈ C0,α(U) – recall that a priori u ∈ W 1,p(U) could be changed on a set of measure 0...).

1.5.1 Gagliardo-Nirenberg-Sobolev inequality The discussion above was for W 1,p(U) (k = 1). In general, we have Theorems on W k,p(U), n U ⊆ R , so there are three parameters: k, p, n. Hence, the difficulty is to distin- guish/remember the slightly different looking results (as above Lq or C0,α) depending on the relation between k, p, n: Hard (but necessary!!!) to remember. The first aim is to understand where this relation comes via “The Mother of all Sobolev inequalities” (for k = 1).

n Deﬁnition 1.16. Let n ∈ N (as in R ). For 1 6 p < n (i.e. p ∈ [1, n)), the Sobolev conjugate of p (or, “the critical Sobolev exponent” – see later) is np p∗ := n − p

1 1 1 ∗ Note that p∗ = p − n > 0 and p > p (“the Sobolev gap”). Theorem 1.17 (Gagliardo-Nirenberg-Sobolev inequality). Assume 1 6 p < n, i.e. p ∈ ∗ np [1, n). Let p = n−p be the Sobolev conjugate of p. Then there exists a constant C = C(p, n) such that

1 n kuk p∗ n CkDuk p n ∀u ∈ C ( ) (∗) L (R ) 6 L (R ) c R Remarks 1.18. (a) We need u to have compact support, otherwise the inequality cannot hold (take u ≡ 1). But the constant does not depend on this compact support (size for example).

18 (b) If an inequality like (∗) holds for some q, i.e. kukLq(Rn) 6 CkDukLp(Rn), then it can only hold for for one speciﬁc q, namely q = p∗ (the claim of 1.17 is that it indeed does hold for p∗). Where does p∗ come from? Assume

kukLq(Rn) 6 CkDukLp(Rn) (∗∗)

∞ n ∞ n for some C > 0, some q with 1 6 q < ∞ and all u ∈ Cc (R ). Let u ∈ Cc (R ) n ∞ n ﬁxed and deﬁne, for all λ > 0, uλ(x) := u(λx), x ∈ R . Then uλ ∈ Cc (R ) and (∗∗) implies

kuλkLq(Rn) 6 CkDuλkLp(Rn) (∗ ∗ ∗) But 1 |u |q dx = |u(λx)|q dx = |u(x)|q dx ˆ λ ˆ λn ˆ Rn Rn Rn whereas the right hand side is λp |Du |p dx = λp |(Du)(λx)|p dx = |(Du)(y)|p dy ˆ λ ˆ λn ˆ Rn Rn Rn Inserting in (∗ ∗ ∗) gives

1 λ kukLq( n) kDukLp( n) λn/q R 6 λn/p R so

n n 1− p + q kukLq(Rn) 6 Cλ kDukLp(Rn) () n n Assume now 1 − p + q 6= 0. n n ∞ n • If 1 − p + q > 0, let λ & 0, then (from ()) kukLq(Rn) = 0 for every u ∈ Cc (R ) . n n ∞ n • If 1 − p + q < 0 we let λ % ∞ then kukLq(Rn) = 0 for all u ∈ Cc (R ) . n n np ∗ So if (∗∗) holds, then 1 − p + q = 0, i.e. q = n−p = p ! ∗ n Proof. First assume p = 1, whence p = n−1 . Then

xi u(x) = u (x , . . . , x , y , x , x ) dy i = 1, . . . , n ˆ xi 1 i−1 i i+1 n i −∞ So ∞ |u(x)| 6 |Du(x1, . . . , yi, . . . , xn)| dyi i = 1, . . . , n ˆ−∞ Hence, using

n ∞ 1 n Y n−1 |u(x)| n−1 |Du(x , . . . , y , . . . , x )| dy 6 ˆ 1 i n i i=1 −∞

19 we integrate with respect to x1: ∞ ∞ n ∞ 1 n Y n−1 |u(x)| n−1 dx1 6 |Du| dyi dx1 ˆ ˆ ˆ−∞ −∞ −∞ i=1 1 ∞ n 1 ∞ n−1 Y ∞ n−1 = |Du| dy1 |Du| dyi dx1 ˆ−∞ ˆ ˆ−∞ −∞ i=2 1 n 1 ∞ n−1 Y ∞ ∞ n−1 |Du| dy |Du| dx dy ( ) 6 ˆ 1 ˆ ˆ 1 i −∞ i=2 −∞ −∞ by the “generalised Hölder inequality”. If p ∈ [1, ∞], Pm 1 = 1. Then |u ··· u |dx i i=1 pi U 1 m 6 m Q p ´ k=1 kukkL k (U). Now integrate () with respect to x2:

∞ 1 ∞ n n ∞ n−1 1 n−1 Y n−1 |u| dx1dx2 6 |Du| dx1dy2 Ii dx2 ˆ ˆ−∞ ˆ −∞ −∞ i=1,i6=2 ∞ ∞ ∞ with I1 := −∞ |Du|dy1, Ii := −∞ −∞ |Du|dx1dyi, i = 3, 4, . . . , n. Use again generalised Hölder to get´ ´ ´

∞ 1 1 n ∞ ∞ n−1 ∞ n−1 |u| n−1 dx1 dx2 6 |Du| dx1 dx2 · |Du| dy1 dx2 ˆ ˆ−∞ ˆ−∞ ˆ−∞ −∞ n 1 Y ∞ ∞ ∞ n−1 |Du| dx dx dy ˆ ˆ ˆ 1 2 i i=3 −∞ −∞ −∞

Continue to integrate (and use gereralised Hölder with respect to x3, . . . , xn) to get n ∞ ∞ 1 n n Y n−1 n−1 |u(x)| n−1 dx ··· |Du| dx , ··· dy ··· dx = |Du(x)| dx ˆ 6 ˆ ˆ 1 i n ˆ i=1 Rn −∞ −∞ Rn i.e.

kuk p∗ n kDuk 1 n ( ) L (R ) 6 L (R ) ∗ n γ p = n−1 , hence the claim for p = 1. Assume 1 < p < n to v := |u| with γ > 1 (we will choose γ later). Then

n−1 n−1 n n n γn γ γ−1 |v| n−1 dx = |u| n−1 dx 6 D|v| = γ |u| |Du| dx ˆRn ˆRn ˆ ˆ Rn Rn p−1 1 p p p (γ−1) p 6 γ |u| p−1 dx |Du| dx (∗) ˆRn ˆRn Choose now γ so that the two u-integrals (left & right side) involve the same power of u, γn p p(n−1) γn p i.e. n−1 = (γ − 1) p−1 . That is γ := n−p > 1 since 1 < p < n. Then n−1 = (γ − 1) p−1 = pn ∗ n−p = p . I.e. (∗) becomes

n−1 p−1 1 n p p p∗ p∗ p |u| dx 6 γ |u| |Du| dx ˆRn ˆRn ˆRn

20 n−1 p−1 np−p−np+n n−p 1 Dividing leads n − p = np = np = p∗ . Hence we have

p(n − 1) kuk p∗ n kDuk p n L (R ) 6 n − p L (R )

We get the same estimate for bounded domains via extension – now for W 1,p-functions.

1,p n n Theorem 1.19 (Estimates for W (U), p ∈ [1, n), U ⊆ R ). Let U ⊆ R be open and 1 1,p ∗ np p∗ bounded with ∂U ∈ C . Let p ∈ [1, n) and u ∈ W (U). Let p = n−p . Then u ∈ L (U) and there exists C = C(p, n, U) > 0 such that

kukLp∗ (U) 6 CkukW 1,p(U) (∆)

Remark. I.e. “W 1,p(U) ⊆ Lp∗ (U)” and the embedding is continuous by (∆).

1 1,p n Proof. By Theoem 1.12 (Extension, recall that ∂U is C ) there exists u := Eu ∈ W (R ) with u = u in U, u has compact support and kukW 1,p(Rn) 6 CkukW 1,p(U). From Theorem n 1.9 (local approximation by smooth functions, and u has compact support in R ) there ∞ n 1,p n is (um)m∈N ⊆ Cc (R ), um → u in W (R ). By Theorem 1.17, we just proved kum − p∗ n u k p∗ n CkDu −Du k p n for all `, m 1, hence u → u in L ( ). By Theorem ` L (R ) 6 m ` L (R ) > m R 1.17, ku k p∗ n CkDu k p n . So, kuk p∗ n CkDuk p n . Combining this with m L (R ) 6 m L (R ) L (R ) 6 L (R ) the extension

kuk p∗ = kuk p∗ kuk p∗ n CkDuk p n Ckuk 1,p n C˜kuk 1,p L (U) L (U) 6 L (R ) 6 L (R ) 6 W (R ) 6 W (U) Summary: “Mother of all Sobolev inequalities” + Extension implies the result.

1.5.2 Poincaré inequality

1,p n Theorem 1.20 (Estimates for W0 (U), p ∈ [1, n), Poincaré inequality). Let U ⊆ R be 1,p open and bounded. Let u ∈ W0 (U) for p ∈ [1, n). Then

kukLq(U) 6 CkDukLp(U) (∗) for all q ∈ [1, p∗].

In particular (check! exercise!) for every p ∈ [1, ∞]: kukLp(U) 6 CkDukLp(U) (∗∗). Remark. (i) I.e. for all p ∈ [1, n) and q ∈ [1, p∗] there is a constant C(p, q, n, U) > 0 such that (∗) holds for u ∈ W 1,p(U).

∗ np (ii) Recall p = n−p .

1,p 1,p (iii) It says W0 (U) (not W (U)). Also, no condition on ∂U. (iv) On the right side of (∗): Not the W 1,p-norm, only the Lp-norm of Du.

(v) For all q ∈ [1, p∗] not only q = p∗ (but: How deep?!)

(vi) The inequality (∗∗) is called (a) Poincaré inequality.

1,p As a consequence: On W0 (U), if U is bounded: kukW 1,p(U) and kDukLp(U) are both norms and they are equivalent.

21 1,p ∞ Proof. Let u ∈ W0 (U), so by deﬁnition there is a sequence (um)m ⊆ Cc (U) such that 1,p n ∞ n um → u in W (U). Extend each um by 0 on R \ U (then um ∈ Cc (R )) and then by Theorem 1.17 and arguments as above (in proof of 1.19), we get kukLp∗ (U) 6 CkDukLp(U). n Since λ (U) < ∞ (ﬁnite Lebesgue measure) we get (from Hölder), kukLq(V ) 6 CkukLp(U) if q ∈ [1, p∗].

Remark. Also in Theorem 1.19 we have (since U is bounded!) u ∈ Lq(U) for all q ∈ [1, p∗] not only q = p∗. However, Theorem 1.17 only holds for q = p∗ for reasons of scaling – this breaks down if U is bounded. On the other hand, the argument why 1.19 holds for q ∈ [1, p∗] uses that U has ﬁnite Lebesgue measure (see also Exercise 1 Tut 4).

What happens if the initial p is larger (i.e. p > n)? This should be “better”... ∗ np 1,n Note. p = n−p % ∞ as p % n. So, for p = n, we could expect/hope that: u ∈ W (U) ∞ n implies u ∈ L (U). But, this is wrong if n > 1. Example: Let U = B1(0) ⊆ R , u(x) = log(log(1 + 1/|x|)). This case is in fact much more complicated.

1.5.3 Morrey’s inequality

I.e. we study the case n < p 6 ∞ (i.e. p ∈ (n, ∞]). As in Exercise 3 Sheet 2, we get Hölder-continuity (not only Lq).

Theorem 1.21 (Morrey’s inequality). Assume n < p 6 ∞, i.e. p ∈ (n, ∞]. Then there exists a constant C = C(p, n) such that

kukC0,γ (Rn) 6 C · kukW 1,p(Rn)

1 n 1,p n n for all u ∈ C (R ) ∩ W (R ), where γ := 1 − p ∈ (0, 1]. Proof. No proof here (in Tut...)! The statement is exam material!

Deﬁnition 1.22. We call u∗ a version of a given function u iﬀ u = u∗ almost everywhere.

1,p n Theorem 1.23 (Estimates for W , p ∈ (n, ∞]). Let U ⊆ R be open and bounded, with 1 1,p ∂U ∈ C . Assume n < p 6 ∞, i.e. p ∈ (n, ∞] and let u ∈ W (U). Then u has a version ∗ 0,γ n u ∈ C (U) for γ = 1 − p ∈ (0, 1] with the estimate

∗ ku kC0,γ (U) 6 CkukW 1,p(U) for some constant C = C(p, n, U).

Remarks 1.24. (a) No proof here (uses extension: Theorem 1.12) Statement is exam material!

(b) We shall, as a consequence of 1.23, always identify any u ∈ W 1,p(U), p > n with its continuous version (1.23 in particular says such one exists).

1,p n We have now Sobolev inequalities and embedding for W (U) and U ⊆ R bounded with ∂U ∈ C1 for all p. The larger the p, the better the result (check, how this behaves in the k,p n dimension n, for ﬁxed p). Now, we want to study W (U), U ⊆ R , k ∈ N. Expectation: “The bigger the k, the better”.

22 n Theorem 1.25 (General Sobolev inequalities). Let U ⊆ R be open and bounded with ∂U ∈ C1. Assume u ∈ W k,p(U).

n q 1 1 k np (i) If k < p , i.e. kp ∈ [1, n), then u ∈ L (U) where q = p − n , i.e. q = n−kp and

kukLq(U) 6 CkukW k,p(U) for some C = C(k, p, n, U) > 0.

n n k−1−b p c,γ (ii) If k > p , i.e. kp ∈ (n, ∞] then u ∈ C (U) where  n n n 1 − − b c ∈/ N γ = p p p n any positive number < 1 p ∈ N

and

kuk n Ckuk k,p Ck−1−b p c,γ (U) 6 W (U)

for some C = C(k, p, n, γ, U) > 0.

Remarks 1.26. (a) No proof (see Evans) (statement: Exam!).

n (b) Note that the dividing point between (i) and (ii) is k = p (or kp = n) – i.e. kp < n or kp > n. Just as for W 1,p it was p < n or p > n. Also, in part (i), the critical q ∗ ∗ np (sometimes still denoted p = p (p, k, n)) is q = n−kp . This is an essential number to remember.

(1) Gives the critical q in Lq. (2) Decides where the borderline is (n − kp > 0 (i) or n − kp < 0 (ii)).

m,γ m n (c) Part (ii) says u ∈ C (U) (i.e. u ∈ C in particular) for m := k − 1 − b p c < k and n n n n if p is not integer then γ = 1 − p − b p c ∈ (0, 1). If p is integer, any γ ∈ (0, 1) but not γ = 1!

(d) Clearly, for n, p fixed and k increasing, we get better & better results. In particular: If we want to prove u ∈ Cm for some (given m) then it is enough to prove (for fixed n, p) that u ∈ W k,p for some (sufficiently large and computable) k.

1.5.4 Compact embeddings – Rellich-Kondrachov The above – Sobolev inequalities and embeddings – gives the relationship between k, n, p for having continuous embedding of Sobolev spaces in “nice” spaces – Lq and Cm,γ. In particular for all p ∈ [1, n) and q ∈ [1, p∗] we have W 1,p(U) ,→ Lq(U) (continuous embedding). However, one can often do better!

Definition 1.27. Let X,Y be Banach spaces, X ⊆ Y with norms k · kX , k · kY . (i) We say that X is continuously embedded in Y , written X,→ Y iff there is C > 0 such that kukY 6 CkukX for all u ∈ X. (ii) We say that X is compactly embedded in Y iff X,→ Y and each bounded sequence in X is pre-compact in Y . We write X b Y .

23 Remarks 1.28. (a) X,→ Y implies that whenever (um)m∈N ⊆ X convergent/Cauchy in k · kX it follows that (um)m ⊆ Y is convergent/Cauchy in k · kY .

(b) X b Y implies much stronger: If (um)m ⊆ X is bounded in X, i.e. kumkX 6 X for all m ∈ N and some C > 0, then there is a subsequence (umj )j∈N which converges in Y to

some limit in Y , i.e. there is an element u ∈ Y such that limj→∞ kumj −ukY = 0. It is the “closest” one gets to a substitute for Bolzano-Weierstrass (“Any bounded sequence n in R has a convergent subsequence”). Any sequence which is bounded in X has a subsequence which converges in Y (but not necessarily in X).

Hence, in general, when studying function spaces (Cm,γ,W k,p, etc.) compact embeddings are very desirable!

n Theorem 1.29 (Rellich–Kondrachov/Kondrashov/...). Let U ⊆ R be open and bounded, 1 1,p q ∗ ∗ np with ∂U ∈ C . Let p ∈ [1, n). Then W (U) b L (U) for every q ∈ [1, p ) where p = n−p . Remarks. (1) The known, continuous embedding is also compact except when q = p∗ (the “critical Sobolev-exponent”).

(2) No proof here (see Evans!)

(3) Of course, one can write down similar statements for W k,p, p > 1 and one also has information on compact embeddings in the case of Hölder spaces (see exercises). We shall not need this here – but it is very interesting & useful!

The ﬁeld of “Sobolev spaces” is huge – we shall need a bit more later, but for now we are done!

Recapitulation • Weak derivatives

• Sobolev spaces (Banach spaces via derivatives and integral norms).

∞ ∞ • Approximation via C and Cc functions (density results). • Extensions and traces (restrictions)

• Sobolev inequalities & embeddings (continuous & compact)

These themes also appear in other setups/generalisations:

k,p • Fractional Sobolev spaces (W (U) for k ∈ R+). Sobolev spaces on manifolds (etc.) • Weighted Sobolev spaces (for example change Lebesgue measure in Lp w(x)dx, w > 0). ; • Variable power p = p(x) ( |u(x)|p(x) dx). ´ For more, see list of literature.

24 2 Linear 2nd order elliptic PDE

2.1 Linear 2nd order elliptic partial diﬀerential operators Recall that the overall goal is to study elliptic PDE and boundary value problems, for example −∆u = f in U and u = g on ∂U. The main problem is existence here – or, more generally ( Lu = f in U (∆) u = g on ∂U

n where U ⊆ R is open and bounded and u: U −→ R, u = u(x) is the unknown. f : U −→ R, g : ∂U −→ R is the given (data) and L is a linear second order partial diﬀerential operator having for u ∈ C2(U) either the form

n n X ij X i Lu = − (a (x)uxi )xj + b (x)uxi + c(x)u i,j=1 i=1 = − div(A(x)∇u) + b(x)∇u + c(x)u = − div(A(x)Du) + b(x) · Du + c(x)u ()

ij for A(x) = (a (x))ij or

n n X ij X i Lu = − a (x)uxixj + b (u)uxi + c(x)u () i,j=1 i=1 for given coeﬃcients (data) aij, bi, c, i, j = 1, . . . , n. We say Lu = f is in divergence form if it is like () and in non-divergence form if it is like (). We shall most often discuss the case g ≡ 0 – which is called Dirichlet’s boundary condition.

ij 1 Remark. Why are () and () diﬀerent?!? They are not – provided a ∈ C because we can take () and diﬀerentiate: n n X ij X ˜ Lu = a (x)uxixj + bi(x)uxi + c(x)u () i,j i=1

˜ i Pn ij with bi := b − j=1 axj (x), i = 1, . . . , n and this (i.e. ()) is in non-divergence form. However, there are two reasons why we look at both (divergence & non-divergence form) seperately

(1) Diﬀerent methods are best for diﬀerent forms: The divergence form is best treated with “energy methods” whereas the non-divergence form can be treated better by the “Maximum Principle”.

(2) The above argument does not hold if we cannot diﬀerentiate aij (ex. if aij ∈ C0 – or only aij ∈ L∞). But the methods we develope below can be used also in this case (and, cases where only aij ∈ L∞ – for example – do occur in particular in intermediate steps when studying nonlinear equations (semi-linear, quasi-linear) elliptic problems). Pn ij Pn ij For example: i,j=1 a (x, u, Du)uxixj , i,j=1(a (x, u, Du)uxi )xj (quasi-linear). We shall always assume the symmetry condition aij = aji, i, j = 1, . . . , n, i.e. A(x) = ij (a (x))ij is symmetric for all x ∈ U.

25 nd Definition 2.1. Let L be a linear 2 order partial differential operator as in () or (). We say that L is (uniformly) elliptic in U iff there is θ > 0 such that

n X ij 2 hA(x)ξ, ξi = a (x)ξiξj > θ|ξ| = θhξ, ξi i,j=1

n for a.e. x ∈ U, ξ ∈ R , i.e. as matrices A(x) > θ1Rn for a.e. x ∈ U.

Remark. This implies that all eigenvalues λi = λi(x) of the symmetric matrix A(x) are bigger or equal θ for a.e. x ∈ U (in fact, it is equivalent).

ij i Example. Let a = δij, b ≡ 0, c ≡ 0, L = −∆. We will see: Solutions for Lu = 0 (or Lu = f) often share many properties with harmonic functions (already seen...). Except: Now we do not have an explicit formula for solutions of Lu = 0 in U and u = g on ∂U. Recall that the plan/idea is (1) To prove the existence of weak solutions (i.e. “generalised” solutions, Sobolev-space solutions).

(2) Then study regularity (i.e. W k,p, Cm,γ, Ck-properties). The hope/aim is to prove that the solutions are in fact classical solutions (i.e. u ∈ C2, (Lu)(x) = f(x) for all x ∈ U). Compare Weyl’s Lemma, Theorem 2.28 in PDG 1. Note. This setup/scheme is very general. For other types of equations, one introduces (maybe...) other types of “generalised solutions” – part (2) is then still the second part/step.

2.2 Weak solutions Consider ( Lu = f in U (∗) u = 0 on ∂U with L in divergence form, i.e.

n n X ij X i Lu = − (a (x)uxi )xj + b (x)uxi + c(x)u i,j=1 i=1 with aji = aij, i, j = 1, . . . , n. Assume aij, bi, c ∈ L∞(U), i, j = 1, . . . , n in all that follows. Later we possibly assume more. Let f ∈ L2(U). Motivation. If u solves (∗) and if u is smooth (say, C∞) then

∞ (a) Multiply Lu = f with a test function v ∈ Cc (U): vLu = vf (pointwise for all x ∈ U). (b) Integrate over U:

vLu dx = vf dx ˆU ˆU i.e. n n X X −v (aij(x)u ) + v bi(x)u + vc(x)u dx = vf dx ˆ xi xj xi ˆ U i,j=1 i=1 U

26 (c) Integrate by parts with respect to xj in the ﬁrst term). Notice that v ≡ 0 on ∂U so the boundary terms vanish. So we get

n n X X aij(x)u v + bi(x)u v + cuv = fv (∗∗) ˆ xi xj xi ˆ U i,j=1 i=1 U

(d) Note two things:

∞ 1 (1) Computed for v ∈ Cc (U) – by approximation this will also hold for v ∈ H0 (U) = 1,2 W (U) (p = 2). This involves v, vxj . 1 nd (2) (∗∗) makes sense also if only u ∈ H (U), for, we have u, uxi – but no 2 order derivatives on u; note how this – from partial integration – was a consequence of L being in divergence form (Exercise: Try to do the same if L is in non-divergence 1 1 form). Note that we want u ∈ H0 (U) (not only H (U)) to reﬂect the boundary 1 condition “u = 0 on ∂U” in (∗) – recall Theorem 1.15: T u = 0 iﬀ u ∈ H0 (U) for u ∈ H1(U).

(∗∗) is the so-called “weak” (or “variational”) formulation of (∗).

Deﬁnition 2.2. Let L be a 2nd order elliptic partial diﬀerential operator in divergence form (as in ()). (i) The bilinear form B[·, ·] associated with L is

n n X X B[u, v] := aij(x)u v + bi(x)u v + cuv dx ˆ xi xj xi U i,j=1 i=1

= hA(x)Du, DviRn + hb(x), DuiRn v + cuv dx ˆU

1 for u, v ∈ H0 (U). 1 (ii) We call u ∈ H0 (U) a weak solution of the boundary value problem ( Lu = f in U (∗) u = 0 on ∂U

iﬀ

1 B[u, v] = hf, viL2(U) v ∈ H0 (U) (∆)

Recall that f ∈ L2(U).

Remark. (i) Note that (∆) is just (∗∗); in the motivation we argued that if u is a (classical) solution to (∗) then (∗∗) holds for u.

(ii) This deals with Dirichlet boundary conditions (i.e. u = 0 on ∂U). Later, we shall treat more general boundary conditions.

27 2.3 Existence via Lax-Milgram We aim to prove that weak solutions (as deﬁned in 2.2) exist (under certain conditions...) and that they are unique – via Functional Analysis.

Let’s deﬁne the setup for now. Let H be a (real!) Hilbert space, with inner product (·, ·), and norm k · k. Let h·, ·i denote the pairing of H with its (Banach) dual space: For 0 u ∈ H (dual), u(ϕ) = hu, ϕi, ϕ ∈ H (u: H −→ R, u linear and bounded (i.e. continuous), u ∈ H0). I.e. we do not identify H with its dual. For Hilbert spaces: See Evans Appendix D, or favourite FA-book, skript – or ask.

Theorem 2.3 (Lax-Milgram-Theorem). Assume that B : H × H −→ R, (u, v) 7−→ B[u, v] is a bilinear map, for which there exist constants α, β > 0 such that

(i) |B[u, v]| 6 αkuk · kvk for all u, v ∈ H (“bounded”). 2 (ii) βkuk 6 B[u, u] for all u ∈ H (“coercive”) 0 Let f : H −→ R be a bounded linear functional (i.e. f ∈ H ). Then there exists a unique u ∈ H such that B[u, v] = hf, vi ∀v ∈ H

Remark. Given a bilinear, bounded and coercive B: Then for every f ∈ H0 there is a unique u ∈ H such that B[u, v] = hf, vi for all v ∈ H.

Proof. For all u ∈ H the map v 7−→ B[u, v] is linear and (by (i)) bounded on H. By Riesz’ Representation Theorem (Riesz-Fischer) there exists a unique w ∈ H such that

B[u, v] = (w, v) ∀v ∈ H (∇)

Denote Au := w (the map that assigns w to u). Then (∇) becomes B[u, v] = (Au, v) for all v ∈ H. We claim that A: H −→ H is a bounded linear operator. Let λ1, λ2 ∈ R, u1, u2 ∈ H. For each (ﬁxed) v ∈ H we have

(A(λ1u1 + λ2u2), v) = B[λ1u1 + λ2u2, v] = λ1B[u1, v] + λ2B[u2, v]

= λ1(Au1, v) + λ2(Au2, v) = (λ1Au1 + λ2Au2, v)

Since this holds for all v ∈ H, it follows that A(λ1u1 + λ2u2) = λ1Au1 + λ2Au2 hence, 2 A is linear. Also kAuk = (Au, Au) = B[u, Au] 6 αkuk · kAuk. Hence kAuk 6 αkuk for all u ∈ H, so A is bounded (i.e. A ∈ B(H)). Next we claim A is injective and R(A) is 2 closed in H. Note that βkuk 6 B[u, u] = (Au, u) 6 kAuk · kuk so βkuk 6 kAuk (β > 0). From this, both claims follow. We claim R(A) = H (i.e. A is also surjective/onto). Assume, for contradiction, that R(A) = R(A) 6= H. Then there is w ∈ H, w 6= 0 ⊥ with w ∈ R(A) = R(A)⊥ (Hilbert space theory: H = R(A) ⊕ R(A)⊥ since R(A) is 2 0 closed). Then, however βkwk 6 B[w, w] = (Aw, w) = 0 . Now, since f ∈ H , Riesz’ Representation Theorem gives the existence of w˜ ∈ H such that hf, vi = (w, ˜ v) for all v ∈ H. Since R(A) = H, there is a (unique) u ∈ H such that Au =w ˜. Hence, for all v ∈ H, B[u, v] = (Au, v) = (w, ˜ v) = hf, vi. Finally we prove uniqueness: Assume u, u˜ ∈ H with B[u, v] = hf, vi = B[˜u, v] for all v ∈ H. Then, B[u − u,˜ v] = 0 for all v ∈ H. Let 2 v = u − u˜, then, since B is coercive βku − u˜k 6 B[u − u,˜ u − u˜] = 0. Hence u − u˜ = 0 since β > 0. Hence u =u ˜.

28 Note. The main interest of Lax-Milgram is that it works without B[·, ·] being symmetric (in fact: If B is symmetric, then ((u, v)) := B[u, v] is a new scalar/inner product on H. Using Riesz’ Representation Theorem directly for this scalar product gives the result, when B is symmetric (Exercise: When is B (the associated bilinear form to L) symmetric?). Now, we try to apply Lax-Milgram (abstract) to the existence of weak solutions (concrete!). Recall that the speciﬁc bilinear form B was given by

n n X X B[u, v] := aij(x)u v + bi(x)u v + cuv dx (∗) ˆ xi xj xi U i,j=1 i=1

1 Recall H0 (U) is a (real) Hilbert space. Theorem 2.4 (Energy estimate). Let B be given as in (∗) with aij, bi, c ∈ L∞(U). Then there exist constants α, β > 0 and γ > 0 such that

(i) |B[u, v]| αkuk 1 · kvk 1 6 H0 (U) H0 (U) 2 2 (ii) βkuk 1 6 B[u, u] + γkuk 2 (“not-quite-coercive”). H0 (U) L (U) 1 for all u, v ∈ H0 (U). Proof. We have

n n X X |B[u, v]| kaijk · |Du| · |Dv| dx + kbik |Du| · |v| dx + kck |u| · |v| dx 6 ∞ ˆ ∞ ˆ ∞ ˆ i,j=1 U i=1 U U

= αkuk 1 · kvk 1 H0 (U) H0 (U) for some appropriate α > 0. This proves (i). To prove (ii), recall the ellipticity condition: There is θ > 0 such that n X ij 2 hA(x)ξ, ξi = a (x)ξiξj > θ|ξ| = θhξ, ξi i,j=1

Hence (use: ξi := uxi ): n X θ |Du|2 dx aij(x)u u dx ˆ 6 ˆ xi xj U U i,j=1 n X = B[u, u] − biu u + cu2 dx ˆ xi U i=1 n X B[u, u] + kbik |Du| · |u| dx + kck |u|2 dx 6 ∞ ˆ ∞ ˆ i=1 U U

Recall “Cauchy’s inequality with ε > 0” (Peter-Paul): For a, b > 0, ε > 0: b2 ab εa2 + 6 4ε Then for all ε > 0

2 1 2 |Du| · |u| 6 ε |Du| dx + |u| dx Û Û 4ε Û

29 Pn i θ Choose ε > 0 so small that 0 < ε i=1 kb k∞ < 2 and insert to get (check!)

θ 2 2 |Du| dx 6 B[u, u] + c |u| dx 2 ˆU ˆU for some constant c > 0 (check! – what is c?!). Recall Poincaré’s inequality (Theorem 1.20, ˜ 1 p = 2) kukL2(U) 6 CkDukL2(U) for all u ∈ H0 (U). From (∆) follows then (exercise!) that

2 2 βkuk 1 B[u, u] + γkuk 2 H0 (U) 6 L (U)

(check! – what are β, γ?!) for some β > 0, γ > 0. Exercise: Do the proof without using Poincaré – and compare β, γ...! Remark. If γ > 0 (when is this the case...?) then B does not precisely satisfy condition (ii) in Lax-Milgram. This explains the presence of the “γ” in the next theorem: Theorem 2.5 (First existence Theorem for weak solutions). Let L be a second order (uniformly) elliptic operator in divergence form. Then there is a γ > 0 such that for all 2 1 µ > γ and all f ∈ L (U) there exists a unique weak solution u ∈ H0 (U) of the boundary value problem ( Lu + µu = f in U (∗ ∗ ∗) u = 0 on ∂U

Proof. Let γ be the γ from Theorem 2.4 and let µ > γ. Define Bµ[u, v] := B[u, v] + 1 µ(u, v)L2(U) for u, v ∈ H0 (U). Then this is the bilinear form corresponding (in the sence of Definition 2.2) to the uniformly elliptic operator in divergence form: Lµu := Lu + µu (check!). Then by Theorem 2.4, Bµ[·, ·] satisfies the hypothesis of Theorem 2.3 (Lax- 2 Milgram). Fix f ∈ L (U) and let hf, vi := (f, v)L2(U). Then v 7−→ hf, vi is a bounded 2 1 linear functional on L (U), hence (!) it is a bounded linear functional on H0 (U). 1 2 2 0 1 0 (H0 (U) ,→ L (U) so L (U) ,→ H0 (U) ). We apply Lax-Milgram to get a unique u ∈ 1 1 H0 (U) such that Bµ[u, v] = hf, vi for all v ∈ H0 (U) – that is, u is a weak solution to (∗∗∗) (see Definition 2.2 (ii)).

Example 2.1. Let Lu = −∆u. Then B[u, v] = U Du·Dv dx. Using Poincaré’s inequality, Theorem 2.4 (“Energy estimate”) holds with γ ´= 0. Hence for all f ∈ L2(U) there is a 1 unique u ∈ H0 (U) such that −∆u = f in U and u = 0 on ∂U in the weak sense. Similarly Pn ij for Lu = i,j=1(a uxi )xj + cu if c > 0 in U (for example H = −∆ + V in U with V > 0). Next, we use Fredholm theory for compact operators (Fredholm Alternative) to get more information (in particular, to “eliminate” µ in (∗ ∗ ∗) in Theorem 2.5). Deﬁnition 2.6. Let L be a second order partial diﬀerential operator. (i) The operator L∗, called the formal adjoint of L, is n n n ∗ X ij X i X i L v = − (a vxi )xj − b vxi + c − bxi )v i,j=1 i=1 i=1

i 1 provided b ∈ C (U), 1 6 i 6 n. ∗ 1 1 ∗ (ii) The adjoint bilinear form B : H0 (U) × H0 (U) −→ R is deﬁned by B [u, v] = B[v, u] 1 for all u, v ∈ H0 (U).

30 1 (iii) We say that v ∈ H0 (U) is a weak solution of the adjoint problem ( L∗v = f in U v = 0 on ∂U

∗ 1 iﬀ B [u, v] = (f, u)L2(U) for all u ∈ H0 (U). Remark. (1) For (i), we need (!?!) bi ∈ C1(U) (say). However, for (ii) this is not needed! Hence, it is not needed for (iii) either (the sense of L∗ is thus very formal!).

1 (2) Note that (iii) says B[u, v] = (f, v)L2 for all u ∈ H0 (U). Compare carefully (!!) with (ii) in Deﬁnition 2.3.

(3) If bi ∈ C1(U), then B∗ is the bilinear form associated to the operator L∗ (check!).

Recall the theory of compact operators in Functional Analysis. Let X,Y be Banach spaces.

Deﬁnition 2.7. Let K ∈ B(X,Y ). K : X −→ Y is called compact if for every bounded sequence (uk)k∈N ⊆ X, the sequence (Kuk)k∈N ⊆ Y has a convergent subsequence. Example 2.2. Recall a “compact embedding”!

Let H be a (real) Hilbert space.

Theorem 2.8 (Compactness of adjoints). If K : H −→ H is compact, then so is K∗ : H −→ H (the adjoint of K).

Notation. For T : X −→ Y we denote (T linear, bounded) N(T ) = ker(T ) (kernel), R(T ) = im T (image/range).

Deﬁnition 2.9. Let A ∈ B(H). Then

(i) The resolvent set of A is ρ(A) = {η ∈ R | (A − µI) is bijective}.

(ii) The spectrum of A is σ(A) = R \ ρ(A). (iii) η ∈ σ(A) is an eigenvalue of A iﬀ N(A − µI) 6= {0}. The set of eigenvalues is denoted σp(A) (the point spectrum). (iv) If η is an eigenvalue and w 6= 0 satisﬁes Aw = ηw, then w is an associated eigenvector.

Theorem 2.10 (Spectral Theorem for compact operators). Assume dim H = ∞ and K : H −→ H is compact. Then

(i) 0 ∈ σ(A)

(ii) σ(K) \{0} = σp(K) \{0} (i.e. σ(K) \{0} consists entirely of eigenvalues – apart from 0).

(iii) Either σ(K) \{0} is a ﬁnite set or σ(K) \{0} is a sequence tending to 0

Theorem 2.11 (Eigenvectors of a compact symmetric operator). Let H be a separable Hilbert space and S : H −→ H compact and symmetric (S∗ = S). Then there exists a countable ONB {uj}j∈N ⊆ H consisting of eigenvectors of S. Theorem 2.12 (Fredholm Alternative). Let K : H −→ H be compact. Then

31 (i) N(I − K) is ﬁnite dimensional.

(ii) R(I − K) is closed.

(iii) R(I − K) = N(I − K∗)⊥.

(iv) N(I − K) = {0} ⇐⇒ R(I − K) = H

(v) dim N(I − K) = dim(I − K∗). In particular: Either for every f ∈ H, the equation u − Ku = f has a unique solution or else the homogeneous equation u − Ku = 0 has non-trivial solutions u. We will use the Fredholm Alternative to get more/better information/results then that of Theorem 2.5.

n Theorem 2.13 (Second Existence Theorem for weak solutions). Let U ⊆ R be open and bounded, ∂U ∈ C1. Let L be a 2nd order uniformly elliptic partial diﬀerential operator in divergence form, and L∗ its formal adjoint. Then (i) Precisely one of the following two statements holds:

(1) Either, for all f ∈ L2(U) there exists a solution u of the boundary value problem ( Lu = f in U (∗) v = 0 on ∂U

(2) Or, There exists a weak solution u 6= 0 of the homogeneous bounday value problem ( Lu = 0 in U (∗∗) u = 0 on ∂U

1 (ii) Furthermore, if (2) holds, then the dimension of the nullspace N ⊆ H0 (U) of weak ∗ 1 solutions of (∗∗) is ﬁnite and it equals the dimension of the subspace N ⊆ H0 (U) of weak solutions of ( L∗v = 0 in U (∗ ∗ ∗) v = 0 on ∂U

2 (iii) The bounday value problem (∗) has a weak solution for a given f ∈ L (U) iﬀ (f, v)L2 = 0 for all v ∈ N ∗ (i.e. f ⊥ N ∗). Remark. The dichotomy (1) - (2) is called Fredholm Alternative.

Proof. Let µ := γ with γ as in Theorem 2.5 and deﬁne a bilinear form Bµ[u, v] := B[u, v]+ µ(u, v)L2(U). Note that this is the bilinear form associated to the operator Lγu := Lu + γu (which is uniformly elliptic and in divergence form. Then, by Theorem 2.5 for every 2 1 1 f ∈ L (U) there exists a unique u ∈ H0 (U) solving () Bµ[u, v] = (g, v) for all v ∈ H0 (U). 2 1 −1 −1 I.e. there is a map L 3 g 7−→ u ∈ H0 (U). We denote this map Lγ , i.e. Lγ g = u. Note −1 1 that () and uniqueness imply that Lγ is linear (check!). Now note that u ∈ H0 (U) is a weak solution of Lu = f in U and u = 0 on ∂U if and only if

1 Bµ[u, v] = (γu + f, v)L2(U) ∀v ∈ H0 (U) ()

−1 (just add/subtract γ(u, v)L2 from B[u, v] = hf, vi). That is, iﬀ u = Lγ (γu + f). Deﬁne −1 −1 2 2 Ku := γLγ u, h := Lγ f then u − Ku = (I − K)u = h. We claim K : L (U) −→ L (U)

32 is a bounded, linear, compact operator. Linearity we already argued (last term). By the choice of γ and Theorem 2.4 (“Energy Estimate”) follows: If () holds then

2 βkuk 1 Bγ[u, u] = (g, u) kgk2 · kuk2 kgk2 · kukH1(U) H0 (U) 6 6 6 0

−1 Hence, by deﬁnition of Kg = γLγ g = γu. This shows that

γ 2 kKgkH1(U) = γkukH1(U) 6 kgk2 ≡ Ckgk2 ∀g ∈ L (U) 0 0 β

2 1 Hence, K : L (U) −→ H0 (U) is bounded and linear (continuous). Now recall that Rellich- 1 2 1 2 Kondrachov (Theorem 1.29 the embedding H0 (U) ,→ L (U) is compact: H0 (U) b L (U). Hence, K : L2(U) −→ L2(U) is a composition of bounded, linear operators K : L2(U) −→ 1 1 2 H0 (U) and the compact operator id: H0 (U) −→ L (U). The composition of compact and bounded operators is always compact. Hence, K : L2(U) −→ L2(U) is compact1. Hence, we can apply the Fredholm Alternative for compact operators in Theorem 2.12: Either for each h ∈ L2(U) the equation u − Ku = f has a unique solution u ∈ L2(U) or else the equation u − Ku = 0 has non-zero solutions u in L2(U). If the former holds, there exists a weak solution of (∗). If the latter holds, then γ 6= 0 and (from Theorem 2.12). The dimension of the space N of solution of (2) is finite and equals the dimension N ∗ of solutions of the adjoint problem (∇∇) v −K∗v = 0. One sees (check!) that the latter holds iff u is a weak solution of (∗∗). Also, (∇∇) holds iff v is a weak solution of (∗ ∗ ∗). This proves (i) - (ii) in the theorem. Also (see Theorem 2.12), u − Ku = h has a non-trivial ∗ solution (for a given h) iff (h, v)2 = 0 for all v solving v −K v = 0. However, the definition −1 1 1 ∗ 1 of K (Ku = γLγ u) and h and (∇∇) gives: (h, v) = γ (Kf, v) = γ (f, K v) = γ (f, v). Hence (∗∗) has a weak solution u iff u − Ku = 0 has a non-trivial solution iff (h, v) = 0 for all v solving (∇∇) iff (f, v) = 0 for all weak solutions v of (∗ ∗ ∗) iff (f, v) = 0 for all v ∈ N ∗. This is (iii).

Theorem 2.14 (Third Existence Theorem for weak solutions). (i) There exists an at most countable set Σ ⊆ R such that the boundary value problem ( Lu = λu + f in U (∗) u = 0 on ∂U

has a unique solution for all (!) f ∈ L2(U) iﬀ λ∈ / Σ.

k→∞ (ii) If Σ is inﬁnite, then Σ = {λk}k∈N with λk ∈ R and λk −−−→∞. Deﬁnition 2.15. We call Σ the (real) spectrum of the operator L.

Remark 2.16. In particular: The boundary value problem Lu = λu in U and u = 0 on ∂U has a non-trivial solution iﬀ λ ∈ Σ – if so, we call λ an eigenvalue of L and u an associated/corresponding eigenfunction.

Recall that for L = −∆, Lu = λu turns into −∆u = λu is the Helmholtz’s equation (put λ ≡ k2, then −∆u = k2u).

Proof of 2.14. Let γ be the constant in Theorem 2.4 and assume λ > −γ. Assome wlog that γ > 0. By the Fredholm Alternative (Theorem 2.13). (∗) has a unique weak solution

1Important Step: One very important reason why compact Sobolev embeddings are so important!

33 for all f ∈ L2(U) iﬀ u ≡ 0 is the only weak solution to Lu = λu in U and u = 0 on ∂U iﬀ u ≡ 0 is the only solution to ( Lu + γu = (γ + λ)u in U u = 0 on ∂U

−1 γ+1 −1 But (∗∗) holds iff u = Lγ (γ +λ)u)) = γ Ku (Lγ = L+γ) with Ku := γLγ u) (as in the proof of Theorem 2.13). Recall that K : L2(U) −→ L2(U) is bounded, linear and compact. γ I.e. u ≡ 0 is the only solution to (∗∗) iff γ+λ is not an eigenvalue of K (recall γ+λ > 0). I.e. 2 γ (∗) has a unique weak solution for all f ∈ L (U) iff γ+λ is not in σ(K)\{0} = σp(K)\{0}, which is (at most) a countable set {µk}k∈N with µk → 0 (or, if is a finite set). Hence, by λ > −γ and (∗ ∗ ∗) the boundary value problem (∗) has a unique weak solution for all 2 f ∈ L (U) exept for a real sequence of {λk}k with λk → ∞ as k → ∞. γ Note that the values { }k∈ are real and tend to 0 as k → ∞. If we study instead K γ+λk N on complex functions this need not be all eigenvalues of K – just the real ones (compare linear algebra). Theorem 2.17 (Boundedness of the inverse (“Continuity in the data”)). If λ∈ / Σ, then there exists a constant C > 0 such that

kukL2(U) 6 CkfkL2(U) () 2 1 where f ∈ L (U) and u ∈ H0 (U) is the unique solution of ( Lu = λu + f in U u = 0 on U

The constant C depends on λ, U, and the coeﬃcients of L.

Remark 2.18. () could (!) also be read (using ()

kukL2(U) 6 Ck(L − λ)ukL2(U)

−1 or, with u = Lλ f, Lλ := L − λ: −1 kLλ fkL2(U) 6 CkfkL2(U) 2 for λ∈ / Σ. Note also () is “continuoity in the data” (i.e. f) in L -norm. Remark. The constant C = C(λ) will blow up (C → ∞) if λ approaches an eigenvalue λk in Σ. Proof. Assume for contradiction that such a constant does not exist. Then there would 2 1 exist sequences (fk)k∈N ⊆ L (U), (uk)k∈N ⊆ H0 (U) such that Luk = λuk + fk in U and uk = 0 on ∂U in the weak sense and kukkL2(U) > kkfkkL2(U). Wlog, we can assume 1 kukkL2(U) = 1 (divide everywhere with the number kukkL2(U). Hence k > kfkkL2(U), so 2 fk → 0, k → ∞ in L (U). Recall the Energy Estimate in Theorem 2.4

2 2 2 βkukk 1 B[uk, uk] + γkukk 2 = (λuk + fk, uk) + γkukk 2 H0 (U) 6 L (U) L (U) 2 6 (λ + γ)kukkL2(U) + kfkkL2(U)kukkL2(U) 1 γ + λ + γ + λ + 1 ∀k ∈ 6 k 6 N

34 1 1 1 Hence, (uk) is a bounded sequence in H0 (U) (i.e. in the H0 -norm), so (since H0 (U) is a Hilbert space – use Banach-Alaoglu) there exists a subsequence (ukj )j ⊆ (uk)k such 1 1 2 that ukj * u weakly in H0 (U). Also, since H0 (U) b L (U) (compact embedding) and 1 1 (ukj )j ⊆ H0 (U) is bounded (in H0 (U)), there is a subsequence (of this subsequence, denote j→∞ 2 it the same) such that ukj −−−→ u in L (U) (it converges; one checks that the limit has to be u). Now, uk is a weak solution of Luk = λuk + fk in U and u = 0 on ∂U iﬀ 1 B[uk, v] = (λuk + fk, v)L2(U) for all v ∈ H0 (U) () where

n n X X B[u, v] = aiju v + biu v + cuv dx ˆ xi xj xi U i,j=1 i=1 n n L2(U) n X X ij X i = uxi , a vxj + (uxi , b v)L2(U) + (u, cv)L2(U) () i=1 j=1 i=1

2 j→∞ 2 1 Since fk → 0 in L (U), and ukj −−−→ u in L (U) and ukj * u in H0 (U), it follows (from () and ()) that

j→∞ j→∞ B[u, v] ←−−− B[ukj , v] = (λukj + fkj , v)L2(U) −−−→ (λu, f)L2(U)

1 1 for all v ∈ H0 (U). Hence, B[u, v] = (λu, v) for all v ∈ H0 (U). So u is a weak solution to Lu = λu in U and u = 0 on ∂U. But since λ∈ / Σ, we have u ≡ 0 (by def of Σ). However, 2 kukj kL2(U) = 1 and ukj → u in L (U) implies kukL2(U) = 1 (compact embedding gave the L2-norm convergence!) – this is a contradiction.

Remark. The discussion of real solutions (u) above can be appropriately extended to complex solutions (u) – see brief (!) discussion in Evans.

2.4 Inhomogeneous bounday value problems So far, we have studied boundary value problems with homogeneous boundary conditions – speciﬁcally the Dirichlet boundary condition (u = 0 on ∂U). We now discuss how to apply the methods developed, to study inhomogeneous boundary value problems – as an 1 1 example, u = g on ∂U. More precisely, suppose ∂U is C and u ∈ H0 (U) is a weak solution to ( Lu = f in U u = g on ∂U

That is u = g on ∂U in the trace-sense (see Theorem 1.14) and

1 B[u, v] = (f, v)L2(U) ∀v ∈ H0 (U) (∗∗)

Note. When “deriving” (∗∗) from (∗) for Dirichlet boundary value problem (u = 0 on ∞ ∂U, for classical solutions), we multiplied by v ∈ Cc , then integrated by parts, and the ∞ 1 boundary term vanished – due to v ∈ Cc – we not need that u ∈ H0 (U (the “0” was for 1 “u = 0 on ∂U”) – hence also here we want v ∈ H0 (U). That u = g in the trace sense says that tr u = g (in L2(∂U)). I.e. g has, to start out with, to be the trace of something – i.e. g ∈ R(tr) ⊆ L2(∂U). So, assume g = tr w, w ∈ H1(U),

35 1 1 and let u solve (∗). Then the function u˜ := u − w ∈ H (U) saﬁsﬁes u˜ ∈ H0 (U) (since tr(˜u) = tr(u) − tr(w) = g − g = 0) and (wrong!)

Lu˜ = Lu − Lw = f − Lw i.e. u˜ is (! or not...) a weak solution to ( Lu˜ = f in U (∗ ∗ ∗) u˜ = 0 on ∂U with f˜ := f − Lw – our original problem! – except f − Lw∈ / L2(U), since Lw∈ / L2(U)). 1 Instead: Lw is (!) a continuous linear functional on H0 (U) – i.e. Lw belongs to the dual 1 of H0 (U) – which we need to study. Check that given g, f, if w, u˜ solve tr w = g and (∗∗∗) then u :=u ˜ + w solves (∗)). Hence, we return to Sobolev spaces!

2.5 The space H−1(U) 1 −1 Definition 2.19. We denote the dual space to the (Hilbert space) H0 (U) by H (U), i.e. −1 1 ∗ H (U) = (H0 (U)) . −1 1 Remark. I.e. f ∈ H (U) iff f is a bounded linear functional on H0 (U) (bounded in 1 k · k 1 -norm). We do not identify the space H (U) with its dual! They are of course H0 0 isomorphic (via Riesz’ Representation Theorem, but not equal (unlike for L2(U)). In fact, 1 2 −1 H0 (U) ⊆ L (U) ⊆ H (U). Notation. Recall that (·, · denotes the scalarproduct on L2(U). As when discussing general 1 Hilbert spaces – setup (before Theorem 2.3, we denote the pairing between H0 (U) and −1 1 0 −1 −1 H (U) (i.e. between H := H0 (U) and its dual H = H (U) by h·, ·i). For f ∈ H (U), 1 f : H0 (U) −→ R, u 7−→ f(u) is linear and bounded and we write hf, ui := f(u). Definition 2.20. For f ∈ H−1(U) we define the norm

1 kfk −1 := sup{|hf, ui| | u ∈ H (U), kuk 1 1} H (U) 0 H0 (U) 6 This is the usual norm from FA, it is also the operator norm of f.

Theorem 2.21 (Characterization of H−1). (i) Assume f ∈ H−1(U). Then there exist functions f 0, f 1, . . . , f n ∈ L2(U) such that n n 0 X i 0 X i 1 hf, vi = f v + f v dx = (f , v) 2 + (f , v ) 2 ∀v ∈ H (U) ˆ xi L (U) xi L (U) 0 U i=1 i=1 ()

(ii) Furthermore,

n 1/2 X i 2 0 n 2 kfkH1(U) = inf |f | dx | f satisﬁes ( ) for f , . . . , f ∈ L (U) 0 ˆ U i=0

1 2 −1 (iii) In particular, for all u ∈ H0 (U), v ∈ L (U) ⊆ H (U): (v, u)L2(U) = hv, ui Notation. When ( ) holse, we write “f = f 0 −Pn f i ”. Note that this is pure notation! i=1 xI i 2 i Is () after partial integraion – only: f ∈ L (U), so fxi not deﬁned (in this course...).

36 1 1 Proof. Given u, v ∈ H (U), we use the notation (·, ·) 1 for the scalar product on H (U) 0 H0 (U) 0

1 (u, v)H1(U) := (Du · Dv + uv) dx u, v ∈ H0 (U) 0 ˆU 1 1 Let f ∈ H (U). Apply Riesz’ Representation Theorem in H = H0 (U). There exists a 1 unique u ∈ H0 (U) such that 1 f(v) = hf, vi = (u, v) 1 ∀v ∈ H (U) (∆) H0 (U) 0 that is

1 hf, vi = (Du · Dv + uv) dx ∀v ∈ H0 (U) () ˆU 0 2 i 2 Setting f := u ∈ L (U), f := uxi ∈ L (U), i = 1, . . . , n. This gives that () holds for 0 1 this choice f , . . . , f1, . . . , fn. Furthermore, assume v ∈ H (U), kvk 1 1, then ( ) 0 H0 (U) 6 implies by Cauchy-Schwarz

n 1/2 X |hf, vi| |f i|2 dx 6 ˆ U i=0 Taking the inﬁmum on both sides yields

n 1/2 X i 2 kfk −1 |f | dx H (U) 6 ˆ U i=0 Now choosing v := u with the u ∈ H1(U) such that (∆) holse – and inserting in kuk 1 0 H0 (U) (∆) – shows that n 1/2 X i 2 kfk −1 = |f | dx H (U) ˆ U i=0 Here, f 0, . . . , f n are exactly the f’s chosen in (∆∆). Assume not that for some (other – no uniqueness) g0, . . . , gn ∈ L2(U) we have

n X hf, vi = g0v + giv dx ∀v ∈ H1(U) ˆ xi 0 U i=1

Now choose v := u in (), then n X (|Du|2 + u2) dx = hf, ui = g0u + giu dx ˆ ˆ xi U U i=1 n 1/2 n 1/2 X X |gi|2dx · u2 + (u )2 dx 6 ˆ ˆ xi U i=0 U i=1 n 1/2 1/2 X = |gi|2 dx · (|Du|2 + u2) dx ˆ ˆ U i=0 U Hence

1/2 n 1/2 X (|Du|2 + u2) dx |gi|2 dx ˆ 6 ˆ U U i=0

37 But by deﬁnition of the f i’s (see (∆∆)) this is

n 1/2 n 1/2 X X |f i|2 dx |gi|2 dx ˆ 6 ˆ U i=0 U i=0 for all choices g0, . . . , gn ∈ L2(U) such that (∆∆∆) holds. Hence, by

n 1/2 X i 2 kfk −1 = |f | dx (∗) H (U) ˆ U i=0 it follows that

n 1/2 X i 2 kfk −1 |g | dx H (U) 6 ˆ U i=0

0 2 for al choices g , . . . , gn ∈ L (U) such that

n X i hf, vi = (g0, v)L2(U) + (g , vxi )L2(U) (∆∆∆) i=1 Hence

n 1/2 X i 2 0 n 2 kfkH1(U) inf |f | dx | f satisﬁes ( ) for f , . . . , f ∈ L (U) 0 6 ˆ U i=0 Since f 0, . . . , f n ∈ L2(U) from (∆∆) is such a choice (∗) implies (ii). The identity in (iii) follows from (i).

Now, with L the usual uniformly elliptic 2nd order partial diﬀerential operator in divergence form with L∞-coeﬃcients and B the associated bilinear form, we consider the boundary value problem ( Lu = f 0 − Pn f i in U i=1 xi (∇) u = 0 on ∂U

0 Pn i −1 1 where, as above (i.e. last times) f − i=1 fxi ≡ f ∈ H (U), the dual of H0 (U). 1 Deﬁnition 2.22. We call u ∈ H0 (U) a weak solution of (∇) iﬀ B[u, v] = hf, vi for all 1 u ∈ H0 (U) with n X hf, vi = f 0v + f iv dx ˆ xi U i=1

−1 1 and h·, ·i the pairing of H (U) and H0 (U). Theorem 2.23. Let L be a uniformly elliptic 2nd order partial diﬀerntial operator in ∞ divergence form with L -coeﬃcients. There is a number γ > 0 such that for each µ > γ i 2 1 and all f ∈ L (U), i = 0, . . . , n there exists a unique weak solution u ∈ H0 (U) to the boundary value problem

( 0 Pn i Lu + µu = f − i=1 fxi in U u = 0 on ∂U

38 Proof. As for Theorem 2.5, noting that

n X hf, vi = f 0v + f iv dx ˆ xi U i=1

1 deﬁnes a bounded linear functional on H0 (U), so that Lax-Milgram can be applied. 1 −1 Remark 2.24. In particular: The mapping Lµ = L + µ: H0 (U) −→ H (U) (µ > γ) is an isomorphism.

2.6 Regularity of weak solutions Recall that under certain conditions, we have a (unique) weak solution to the boundary value problem ( Lu = f in U (∗) u = 0 on ∂U

Question: When do we have a classical solution (i.e. u ∈ C2(U) ∩ C(U))? Exercise 1 on ij 1 i 1 2 Problem Sheet 6, assume a ∈ C (U) and b , f, c ∈ C(U) and that u ∈ H0 (U) ∩ C (U) is a weak solution to (∗) – then u is a classical solution to (∗). Not surprisingly we need that aij, bi, c, f are continuous (aij ∈ C1(U) is needed because the equation is in divergence form). The claim is, it is suﬃcient (!) to prove the regularity u ∈ C2(U) of a weak solution to have a classical solution. Hence the study of regularity of weak solutions to (∗). Recall also PDG1 we proved: If u ∈ C2(U) and ∆u = 0 then u ∈ C∞(U). If u ∈ C2(U) and ∆u = 0 then u ∈ Cw(U) (real analytic). Not surprisingly: The better the regularity of the coeﬃcients in n X ij X i Lu = − (a uxi )xj + b uxi + cu = f i,j=1 i=1 (aij, bi, c, f) the better the regularity of u. One also needs more regularity of the boundary – i.e. “regular data =⇒ regular solution”. “All” of this is called “elliptic regularity”. This is a vast topic. We “only scratch the surface” here – remarks on more advanced & detailed results will be given at the end.

Heuristics & motivation In 1 dimension, if −u00 = f ∈ Ck then clearly (!) u ∈ Ck+2 (uniformly elliptic −a(x)u00 = k 00 f k k+2 f ∈ C , a(x) > a0 > 0 for all x then −u = a – so if a ∈ C , then u ∈ C ). However, n nd in U ⊆ R , n > 1, we do we have control of all derivatives of 2 order, if, for example, −∆u = f (∆u = tr(D2u); only sum of the diagonals)? One calculation that indicates the n n “reason”: Assume: −∆u = f in R and everything is “nice” (say, f ∈ Cc(R )). n 2 2 X f = (∆u) = uxixI uxj xj dx ˆ n ˆ n ˆ n R R i,j=1 R n n X X = − uxixixj uxj dx = uxixj uxixj dx ˆ ˆ n i,j=1 i,j=1 R Rn n X 2 2 2 = (uxixj ) = |D u| dx ˆ n ˆ n i,j=1 R R

39 ˜ ˜ Also, let u˜ := uxk , then diﬀerentiating the −∆u = f (!), we get −∆˜u = f with f = fxk , k = 1, . . . , n. I.e. by the same argument we get from the control of the 1st derivatives of f the control of the 3rd derivatives of u – generally: Control of the mth derivative of f (in L2 – in this case) gives control of (m + 2)nd derivative of u (again, in L2). This gives hope that for the poisson equation −∆u = f we could have: f ∈ Hm implies that u ∈ Hm+2 (and then hope (!) that we can generalise to all elliptic equations). In particular: If ∞ m m+2 f ∈ C (U) then f ∈ H (U) for all m ∈ N, so u ∈ H (U) for all m ∈ N – and by k ∞ Sobolev embedding, it follows u ∈ C (U) for all k ∈ N and hence u ∈ C (U) (as for −∆u = 0). However:

(1) The calculation above is not rigorous – we already assumed the existence of higher (3rd) derivatives of u (to control 2nd order derivative). The technique to show existence of derivatives involves diﬀerence quotients (just like for classical derivatives). We will need some material on that.

2 2 1 (2) Not even true yet (!) that u ∈ H (f ∈ L ). Recall that a weak solution is a u ∈ H0 (U) such that u solves Lu = f in U and u = 0 on ∂U in the weak sense. Hence, the ﬁrst step will be to prove that in fact u ∈ H2(U) (not only in H1). This is generally “always” the ﬁrst step: Prove a weak solution u ∈ H1 is in fact in H2 – also in the case of (some) nonlinear equations (semilinear, quasilinear).

Diﬀerence quotiens

1 Assume u ∈ U −→ R, u ∈ Lloc(U), V b U. Deﬁnition 2.25. (i) The ith diﬀerence quotient of size h is

u(x + he ) − u(x) (Dhu)(x) := i i = 1, . . . , n i h

for x ∈ V and h ∈ R, 0 < |h| < dist(V, ∂U). h h h h (ii) D u := (D1 u, D2 u, . . . , Dnu). h 1 1 Note. We still have that Di u ∈ Lloc(U) (since this is a linear combination of Lloc). Also, p h p if u ∈ L (U), then Di u ∈ L (V ) (again, since linear combination – watch h, U, V ) – even if u∈ / W 1,p (whole idea!)!

Theorem 2.26 (Diﬀerence quotients and weak derivatives). (i) Let p ∈ [1, ∞), u ∈ W 1,p(U). Then for each V b U we have

h kD ukLp(U) 6 CkDukLp(U)

1 for some C = C(V ) and all 0 < |h| < 2 dist(V, ∂U). (ii) For p ∈ (1, ∞), u ∈ Lp(V ), assume that there exists C > 0 such that

h kD ukLp(V ) 6 C ()

1 1,p for all 0 < |h| < 2 dist(V, ∂U). Then u ∈ W (V ) with

kDukLp(V ) 6 C () Remark. (a) (ii) is false for p = 1 (has to do with reﬂexitivity of Lp, p ∈ (1, ∞)).

40 (b) Both results are interesting, but in particular (ii): Allows to conclude the existence of weak derivatives in Lp by studying diﬀerence quotients: Prove uniform (in h) bound!

Proof. (i) p ∈ [1, ∞) and assume u is smooth (say, C∞). Let x ∈ V , i = 1, . . . , n, 1 0 < |h| < 2 dist(V, ∂U). Then

1 u(x + he ) − u(x) = h u (x + the ) dt i ˆ xi i 0 So

|Dhu|p dx C |Du|p dx ( ) ˆ 6 ˆ V U

1,p for u smooth. By approximation, we get () for all u ∈ W (U). ∞ (ii) Assume () holds: Choose i = 1, . . . , n and φ ∈ Cc (V ). Then, for |h| small enough (make change of variables for x + hei → x − hei):

φ(x + he ) − φ(x) u(x) − u(x − he ) u(x) i dx = − i φ(x) dx ˆ h ˆ h V V I.e. “integration by parts” for diﬀerence quotients

h −h u(Di φ) dx = ∓ (Di u)φ dx ˆV ˆV

The estimate () implies

−h sup kDi ukLp(V ) < ∞ 1 0<|h|< 2 dist(V,∂U)

p Hence, since p ∈ (1, ∞) (!), there exists a function vi ∈ L (V ) and a subsequence −kk p 2 hk → 0, k → ∞ such that Di u * vi weakly in L (V ) . Hence, by dominated 2(“every bounded sequence in Lp has a weakly convergent subsequence” – Banach-Alaoglu! – Needs p ∈ (1, ∞), so that Lp(V ) is reﬂexive.

41 convergence

uφxi dx = uϕxi dx ˆV ˆU

hk −hk = lim u(Di dx = − lim (Di u)φ dx hk→0 ˆ hk→ ˆ U V = − v φ dx = − v φ dx ˆ i ˆ i V U

q since φ ∈ L (q Hölder conjugate). Thus, vi = uxi in the weak sense for all i (i.e. u has a weak derivative) and Du ∈ Lp(V ) (!). We had u ∈ Lp(V ), so u ∈ W 1,p(V ). The estimate () follows from () and the weak convergence, since the norm is weak lower semicontinuous (Fatou!).

2 n 1 We are ready to prove Interior H -regularity. Let U ⊆ R open and bounded, u ∈ H0 (U) be a weak solution of Lu = f with L a uniformly elliptic 2nd order partial diﬀerential operator in divergence form with aij, bi, c ∈ L∞(U) – but, we will have to add assumptions on the coeﬃcients as need be.

2 ij 1 ∞ Theorem 2.27 (Interior H -regularity). Assume a ∈ C (U), bi, c ∈ L (U), i, j = 1, . . . , n and f ∈ L2(U). Assume u ∈ H1(U) is a weak solution of the elliptic partial 2 diﬀerential equation Lu = f in U. Then u ∈ Hloc(U). Furthermore, for each open subset ij i V b U, there exists a constant C = C(V, U, a , b , c) > 0:

kukH2(V ) 6 C(kfkL2(U) + kukL2(U))

Remark 2.28. (i) We have assumed additionally aij ∈ C1(U).

(ii) We have assumed u ∈ H1(U) solves Lu = f in U – nothing on the boundary condition 1 (not “u ∈ H0 (U)”), i.e. B[u, v] = (f, v)L2 . It follows (exercise!) that, in fact, since 2 u ∈ Hloc(U), (Lu)(x) = f(x) a.e. in U (the derivatives in L, on u, being weak derivatives). We say that u is a strong solution.

(iii) “Interior regularity” because it is for all V b U – no control “up to the boundary” (yet...). In particular, this does not say “u ∈ H2(U)”! This is a typical type of results (“interior regularity”) . Of course (!), C → ∞ if “V % U” (see proof...).

The proof will be somewhat long and somewhat (!) technical! But, it is an extremely important techniques (Nirenberg). Re-read it several times – and note the importance of ellipticity (hence “elliptic interior regularity”).

Proof. Fix an open set V b U and choose W open such that V b W b U. Then choose ∞ n n a cut-oﬀ function ζ ∈ Cc (R ), 0 6 ζ 6 1 and ζ = 1 on V , ζ = 0 on R \ W (i.e. supp ζ ⊆ W ). The purpuse of ζ is to restrict the following calculation to W . Needed since 1 we know nothing abount the behaviour of u near ∂U (not “u ∈ H0 (U)”). Recall that u is 1 a weak solution of Lu = f, i.e. B[u, v] = (f, v)L2 for all v ∈ H0 (U) that is

n X aiju v dx = fv˜ dx ∀v ∈ H1(U) ( ) ˆ xi xj ˆ 0 i,j=1 U

42 ˜ Pn i 2 where f = f − i=1 b uxi −cu ∈ L (U). Let know |h| > 0 be “small”, choose k ∈ {1, . . . , n} and let

−h 2 h v := −Dk (ζ Dk u) ()

h g(x+hek)−g(x) 1 where, recall, (Dk g)(x) = h , h ∈ R, h 6= 0. Note (exercise!) v ∈ H0 (U). 1 (1) This is the “reason” for the cut-off ζ to ensure v ∈ H0 (U) i.e. “allowed test-function”. (2) Many tricks and proofs in this business are abount choosing appropriate “test-functions”. Idea: We will plug this v into (), calculate and estimate! With v as in (), let n X A := aiju v dx B := v˜dx ( ) ˆ xi xj ˆ U i,j=1 U then () reads A = B. We first estimate A, i.e. we compute (since “partial integration for difference quotients” + Dh and D commute)

n X A = − aiju [D−h(ζ2Dhu)] dx ˆ xi k k xj U i,j=1 n X = Dh(aiju )(ζ2Dhu) dx ˆ k xi k xj U i,j=1 n X = aij,h(Dhu )(ζ2Dhu) + (Dhaij)u (ζ2Dhu) dx ˆ k xi k xj k xi k xj U i,j=1

h h h h h by “Leibniz’ rule”. Dk (vw) = v Dk w + wDk v with v (x) := v(x + hek) – check! This gives n X = aij,hDhu Dhu ζ2 dx ˆ ˆ k xi k xj U i,j=1 U n X + aij,h(Dhu Dhu) · (2ζζ ) + (Dhaij)u Dhu ζ2 ˆ k xi k xj k xi k xj i,j=1 U h ij h + ((Dk a )uxi Dk u) · (2ζζxj ) dx

=: A1 + A2

2 h by the real/true Leibniz’ rule to compute (ζ Dk u)xk – also note that (check/exercise) that k h Dh and (··· )xj commute (Hint: Keep an eye on terms with Dk Du ∼ “two derivatives on u”): By uniform ellipticity:

n X A = (aij,hDhu Dhu )ζ2 dx θ ζ2|DhDu|2 dx (∗) 1 ˆ k xi k xj > ˆ k i,j=1 U U

ij 1 h ij ij,h For A2, use a ∈ C (U) to get a uniform bound on both Dk a and a on W b U, also, ij ∞ kζxj kL 6 C and 0 6 ζ 6 1. All in all, for some C > 0 (dependent on V, W, U, a ), we have h h h h |A2| 6 C ζ|Dk Du| · |Dk u| + ζ|DK Du| · |Du| + ζ|Dk u| · |Du| dx ˆU

43 h (Hint: Keep your eyes on the Dk Du-terms! Want: A = A1 + A2 > A1 − |A2| and get h 2 2 b2 +|Dk Du| on the right side. Use: ab 6 εa + 4ε to get (exercise!)

2 h 2 C h 2 2 |A2| 6 ε ζ |Dk uDu| dx + |Dk u| + |Du| dx (∇) ˆU ε ˆW

We also used 0 6 ζ 6 1, supp ζ ⊆ W , the “C” has changed! From Theorem 2.26 (i): For 1 |h| > 0, small enough (0 < |h| < 2 dist(W, ∂U)) we have

h 2 2 |Dk u| dx 6 C |Du| dx (∇∇) ˆW ˆU

θ Choose now ε := 2 and use (∇∇) to get, from (∇):

θ 2 h 2 2 |A2| 6 ζ |Dk Du| dx + C |Du| dx (∗∗) 2 ˆU ˆU

(note: the constant C keeps changing). Now, from A = A1 + A2 and (∗) and (∗∗), follows

θ 2 h 2 2 A > A1 − |A2| > ζ |Dk Du| dx − C |Du| dx 2 ˆU ˆU ˜ ˜ Now, we estimate B, recall A = B (the equation). Recall B := U fv dx with f = Pn i −h 2 h 1 ´ f − i=1 b uxi − cu and v = −Dk (ζ Dk u) ∈ H0 (U). Hence:

|B| 6 c (|f| + |Du| + |u|)|v| dx (∇∇∇) ˆU

i ∞ −h 2 h since b , c ∈ L (U) – all that is needed. Use again Theorem 2.26 (i), but on Dk (ζ Dk u), h then on Dk u, by the Leibniz’ rule and supp ζ ⊆ W :

2 2 h 2 |v| dx 6 C |D(ζ Dk u)| dx Û Û = C |Dhu|2 + ζ2|DhDu|2 dx ˆ k k 2 2 h 2 6 C |Du| + ζ |Dk Du| dx Û

h 2 b2 (check!). Hint: Keep an eye on Dk Du! From this, (∇∇∇) and ab 6 εa + 4ε follows (!)

2 h 2 C 2 2 2 |B| 6 ε ζ |Dk Du| dx + (f + u + |Du| ) dx ˆU ε ˆU

θ Now, choose ε := 4 to get

θ 2 h 2 2 2 2 |B| 6 ζ |Dk Du| dx + C (f + u + |Du| ) dx (∗ ∗ ∗∗) 4 ˆU ˆU

Now A = B and so (∗ ∗ ∗) and (∗ ∗ ∗∗) and together give (recall 0 6 ζ 6 1, ζ = 1 on V !)

h 2 2 h 2 |Dk Du| dx 6 ζ |Dk Du| dx ˆV ˆW 2 2 2 6 C (f + u + |Du| ) dx = C(f, u, Du, . . .) ˆU

44 (check!). This holds for all k = {1, . . . , n} and all h suﬃciently small (|h| 6= 0). It follows 1 n 2 from Theorem 2.26 (ii) that Du ∈ Hloc(U; R ) – i.e. u ∈ Hloc(U) and we have

kukH2(V ) 6 C(kfkL2(U) + kukH1(U))

Note: If V b W b U, then the same type of argument gives the existence of C = C(V, W, . . .) > 0

kukH2(V ) 6 C(kfkL2(W ) + kukH1(W )) (∇)

2 Choose a new cut-oﬀ χ with χ ≡ 1 on W , 0 6 χ 6 1, supp χ ⊆ W , and let v := χ u ∈ 1 H0 (U)!). Plug this into n X aiju v = fv˜ dx ˆ xi xj ˆ i,j=1 U to get (exercise!)

2 2 2 2 2 |Du| dx 6 χ |Du| dx 6 C (f + u ) dx ˆW ˆU ˆU It follows that (exercise)

kukH1(W ) 6 C(kfkL2(U) + kukL2(U) (∇∇)

From (∇) and (∇∇) follows (exercise!) that

kukH2(V ) 6 C(kfkL2(U) + kukL2(U) Remarks. (1) Note that, since Lu = f a.e. in U (“strong solution”) (see Ex 3 Tut 9) the inequality can be read as

kukH2(V ) 6 C(kLukL2(U) + kukL2(U) One calls such an estimate an a priori estimate. There are other types of a priori estimates (given appropriate assumptions on the coeﬃcients) like

α kC2,α(V ) 6 C(kfkCα(U) + kukC (U))

“Schauder estimates”, or kukW 2,p(V ) 6 C(kfkLp(U) + kukLp(U)), 1 < p < ∞ (“interior W 2,p-estimates” —related to “Calerón-Zygmund estimates”)

1 (2) Recall Theorem 2.17: If 0 ∈/ Σ then kukL2(U) 6 CkfkL2(U) for u ∈ H0 (U) the unique weak solution to Lu = f in U and u = 0 on ∂U. Hence, combined we get (in this ˜ 2 case!) kukH2(V ) 6 CkfkL2(U) (“continuity in the data”) – in Hloc) for unique solutions 1 u ∈ H0 (U) to the problem above if 0 ∈/ Σ. Not iterate this argument (as described in the “motivation”) in the case both coeﬃcients aij, bi, c and f are more regular. Then the solutions get more regular.

ij i m+1 Theorem 2.29 (Higher interior regularity). Let m ∈ N0 and a , b , c ∈ C (U), i, j = 1, . . . , n and f ∈ Hm(U). Let u ∈ H1(U) be a weak solution of the uniformly elliptic PDE m+2 Lu = f in U. Then u ∈ Hloc (V ) () and for all V b U there is a C > 0 such that

kukHm+2(V ) 6 C(kfkHm(U) + kukL2(U))

45 Proof. Induction in m. m = 0 is (contained in) Theorem 2.27. Assume () and () hold ij i for some ﬁx m ∈ N0 and all open sets U, coeﬀs a , b , c, f etc. as in the theorem. Suppose aij, bi, c ∈ Cm+2(U) = C(m+1)+1(U) (∆) and f ∈ Hm+1(U) (∆∆) and u ∈ H1(U) is a m+2 weak solution to Lu = f. By induction hypothesis, u ∈ Hloc (U) (∆∆∆) and

kukHm+2(W ) 6 C(kfkHm(U) + kukL2(U) (∆∆∆∆) n for all W b U and appropriate C. Fix V b W b U. Let α ∈ N0 be a multiindex with ∞ |α| α ∞ 1 |α| = m + 1 and let v˜ ∈ Cc (W ). Deﬁne v := (−1) D v˜ ∈ Cc (W ) ⊆ H0 (U) and plug 1 this into (“the equation”) B[u, v] = (f, v)L2 for all v ∈ H0 (U). By integration by party (! Check – justiﬁed since u ∈ Hm+2(W )) we get ˜ B[˜u, v˜] = (f, v˜)L2 (∗)

α 1 m+2 for u˜ := D u ∈ H (W ) (as u ∈ Hloc (U), W b U) and n n X α X X f˜ := Dαf − − ((Dα−βaij)(Dβu )) + (Dα−βbi)Dβu ) + (Dα−βc)(Dβu) β xi xj xi β6α,β6=α i,j=1 i=1 (∇)

∞ 1 ˜ By density of Cc (W ) in H0 (W ) it follows from (∗) that u˜ is a weak solution of Lu˜ = f in ˜ 2 W . Using (∆) - (∆∆∆∆) and (∇) we see that f ∈ L (W ) with ˜ kfkL2(W ) 6 C(kfkHm+1(U) + kukL2(U)) (∇∇) 2 Using Theorem 2.27, this implies u˜ ∈ Hloc(W ) with estimate α ˜ kD ukH2(V ) = ku˜kH2(V ) 6 C(kfkL2(W ) + ku˜kL2(W ) 6 C(kfkHm+1(U) + kfkL2(U)) α here we used (∆∆∆∆), (∇∇) and u˜ = D u, |α| = m+1. THis holds for each multiindex α m+3 α with |α| = m+1 and u˜ = D u, hence u ∈ H (V ) with kukHm+3(V ) 6 C(kfkHm+1(U) + kukL2(U)). This proves the theorem by induction. By applying Theorem 2.29 repeatedly we can now prove: Theorem 2.30 (Inﬁnite diﬀerentiability in the interior). Assume aij, bi, c ∈ C∞(U), i, j = 1, . . . , n and f ∈ C∞(U). Assume u ∈ H1(U) is a weak solution of the uniformly elliptic PDE Lu = f in U. Then u ∈ C∞(U). Note. There is no assumption on the behaviour of u on ∂U – or on the regularity of the boundary. I.e. any possible singularity of u on ∂U does not “propagate” into the interior of U.

m Proof. By Theorem 2.29 we have u ∈ Hloc(U) for all m ∈ N. By Sobolev embedding k ∞ (Theorem 1.259 it follows that u ∈ C (U) for all k ∈ N, so u ∈ C (U). Note. If we do not have the C∞-situation, but sil enough regularity on f, aij, bi, c to get – for example – classical solutions u ∈ C2(U) then this method is not optimal. Bettey to stay in Hölder spaces Ck,α (instead of Ck,α Hm C`,β). See also the Remark after the proof of Theorem 2.27, on the Schauder estimate.; ; Next, we study the properties of solutions for more regular boundaries. Why did we “only” get inner regularity?? – because we wanted to bound (uniformly in h) on |DhDu|2 dx – we needed “space” to compute this. The idea for boundary regularity is: ´

46 (1) We still can move when close to the boundary – but only parallel to the boundary.

(2) The last (2nd order) derivative – orthogonal to ∂U: Control using the equation.

−h 2 h Recall that we used as “test-function” v = −Dk (ζ Dk u) – we needed a cut-oﬀ to ensure 1 1 1 v ∈ H0 (U), since we have u ∈ H (U) but (possibly) u∈ / H0 (U). Now, we will need 1 u ∈ H0 (U). Theorem 2.31 (Boundary H2-regularity). Assume aij ∈ C1(U), bi, c ∈ L∞(U), f ∈ 2 1 L (U). Let u ∈ H0 (U) be a weak solution of the (uniformly) elliptic boundary value problem ( Lu = f in U (∗) u = 0 on ∂U

Assume furthermore that ∂U is C2. Then u ∈ H2(U) and there is a constant C = C(U, aij, bi, c) > 0 such that

kukH2(U) 6 C(kfkL2(U) + kukL2(U) (∗∗)

Remark. (1) One can get similar results in the Ck,α and W k,p-cases (1 < p < ∞). This always speaks of “boundary regularity” – although it is the “regularity of solutions up to the boundary”)

1 (2) The result is for “u = 0 on ∂U” (i.e. u ∈ H0 (U)). Of course (!) it is the regularity of the boundary and the boundary data (“0”) that counts. I.e. also there also exist results for inhomogeneous boundary value problems.

(3) From Theorem 2.17 (“boundedness of the inverse”: kukL2(U) 6 CkfkL2(U) follows: If 1 u ∈ H0 (U) is the unique weak solution of (∗) (i.e. if (∗) has only one solution (0 ∈/ Σ), then the estimate (∗∗) becomes

kukH2(U) 6 CkfkL2(U) (“continuity in the data”, i.e. H2-norm up to the boundary)

The strategy of the proof will be as follows:

(1) We use a partition of unity argument: Cover ∂U with ﬁnitely many sets V1,...,Vn. Sn Add V0 that covers the innert, i.e. U ⊆ i=0 Vi

(2) For each Vi: “Straighten out the boundary” (we will not do this in all details – some in exercises...).

n 1 Proof. We do the case of a straight boundary. Assume U = B1(0) ∩ R+ and V = B(0, 2 ) ∩ n 1 R+. Choose a smooth cut-oﬀ ζ with 0 6 ζ 6 1 and ζ ≡ 1 on B(0, 2 ) and ζ ≡ 0 on n R \ B1(0) (i.e. in particular, ζ ≡ 1 on V and ζ ≡ 0 on the curved/upper part of ∂U). As 1 u is a weak solution of (∗) we have B[u, v] = (f, v)L2 for all v ∈ H0 (U), i.e.

n n X X aiju v dx = fv˜ dx f˜ := f − biu − cu (∇) ˆ xi xj ˆ xi i,j=1 U U i=1

47 Let |h| > 0 be small, and choose k ∈ {1, . . . , n − 1} (i.e. not k = n!) and let v := −h 2 h −Dk (ζ Dk u). Note that (compute!) 1 v(x) = − D−h ζ2(x)(u(x + he ) − u(x) h k k 1 = ζ2(x − he )[u(x) − u(x + he )] − ζ2(x)[u(x + he ) − u(x)] ( ) h2 k k k for x ∈ V (note: u(x ± hek) is ok for x ∈ V if k 6= n!). We claim (exercise – See also Tut 1 9, Ex 2(a)) since u ∈ H0 (U) (i.e. u = 0 along {xn = 0} in the trace sense) and ζ ≡ 0 1 near the curved/uppert part of ∂U it follows from () that v ∈ H0 (U) (i.e. “allowed test function”). Plug this v into (∇) and rewrite as A = B where

n X A := aiju v dx B := fv˜ dx (∇∇) ˆ xi xj ˆ i,j=1 U U Now estimate as in the proof of Theorem 2.27 to get

θ 2 h 2 2 A > ζ |Dk Du| dx − C |Du| dx (∇∇∇) 2 ˆU ˆU and

θ 2 h 2 2 2 2 |B| 6 ζ |Dk Du| dx + C (f + u + |Du| ) dx (∇∇∇∇) 4 ˆU ˆU Note, we here (see proof of Theorem 2.27) used part (i) of Theorem 2.26 (!) Exercise (since we do not have V bb U!) with U, V as here:

h p p Dk Du| dx 6 C |uxk | dx k = 1, . . . , n − 1, k 6= n ˆV ˆU As in the proof of Theorem 2.27, combining (∇∇), (∇∇∇) and (∇∇∇∇) we get

h 2 2 2 2 |Dk Du| dx 6 C (f + u + |Du| ) dx ˆV ˆU for all |h| small and for all k = 1, . . . , n − 1 (i.e. k 6= n). It follows from (ii) in Theorem 1 2.26 (Note as above!) that for k = 1, . . . , n − 1 (k 6= n) we have uxk ∈ H (V ) with the estimate X kuxkx` kL2(V ) 6 C kfkL2(U) + kukH1(U) (∗) k,`=1 k+`<2n

2 Note that we need an estimate of the L -norm (on V ) of uxnxn . The idea is to use the 2 equation and ellipticity. Recall (Remark 2.29 (i)) Since u ∈ Hloc, we have Lu = f, a.e. in U (“strong solution”), i.e.

n n X ij X ˜i − a uxixj + b uxi + cu = f () i,j=1 i=1

˜i i Pn ij with b := b − j=1 axj (i = 1, . . . , n. Hence (a.e. in U): n nn X ij X ˜i a uxnxn = − a uxixj + b uxi + cu − f () i,j=1 i=1 i+j<2n

48 Pn ij 2 n Recall the uniform ellipticity: i,j=1 a (x)ξiξj = hξ, A(x)ξi > θ|ζ| for all x ∈ U, ξ ∈ R . n nn Let ξ = en = (0,..., 0, 1) ∈ R , then this implies a (x) > θ > 0 for all x ∈ U. From ij i ∞ () and a ∈ U, b , c ∈ L (U) we get X |uxnxn 6 C |uxi,xj | + |Du| + |u| + |f| i,j=1 i+j<2n

Important! Notice how, for the only (double) derivative – uxnxn – for which we cannot takte the diﬀerence quotient – we can use the equation to relate it to other double derivatives 2 (plus “junk” – but only because L is elliptic). Using this and (∗) we get that u ∈ H (V ) and kukH2(V ) 6 C kfkL2(U) + kukL2(U)

Note that get rid of kukH1(U) on the right side, to get kukL2(U) like at the end of the proof of Theorem 2.27. Now we “straighten the boundary” (not in all details – ideas of 0 0 proof). We have U ∩ Br(x ) = {x ∈ Br(x ) | xn > γ(x1, . . . , xn−1} for some r > 0 2 n−1 and γ ∈ C , γ : R −→ R and y = Φ(x), x = Ψ(y) straighten the boundary. Let 0 U˜ = Bs(0) ∩ {yn > 0} ⊆ Φ(U ∩ Br(x )) and V˜ = B s (0) ∩ {yn > 0}. Let u˜(y) := u(Ψ(y)) 2 for y ∈ U˜. Claims:

1 ˜ 1 2 (1) u˜ ∈ H0 (U) (since u ∈ H (U) and Ψ ∈ C ). ˜ 1 (2) u˜ = 0 on ∂U ∩ {yn = 0} in the trace sence, since u ∈ H0 (U)). (3) u˜ is a weak solution of L˜u˜ = f˜ in U˜ with f˜(y) = f(Ψ(y)) and

n n ˜ X k` X ˜k Lu˜ = − (˜a u˜yk )y` + b u˜yk +c ˜u˜ k,`=1 k=1

with n k` X rs k a˜ (y) := a (Ψ(y))Ψxr (Ψ(y))Φxs (Ψ(y)) k, ` = 1, . . . , n r,s=1 n ˜k X r k b := b (Ψ(y))Φxr (Ψ(y)) k = 1, . . . , n k=1 c˜(y) := c(Ψ(y))

for y ∈ U˜ (proof: Long calculation!).

(4) The operator L˜ is uniformly-elliptic in U˜ (proof: a calculation, the fact that L is uniformly elliptic in U and that Ψ, Φ are diﬀeo’s).

(5) The coeﬃcients a˜k` ∈ C1 (proof: Since Φ ∈ C2 and ars ∈ C1).

One then applies the case of the ﬂat boundary to get u˜ ∈ H2(V ) and an estimate. Trans- lating back (via Φ and Ψ) to u, one gets the estimate for u on U (rather on Ψ(V˜ ). As mentioned at the beginning, one ﬁnishes the proof by a “partition of unity” argument.

As for inner regularity, we can prove higher regularity:

49 ij i m+1 Theorem 2.32 (Higher boundary regularity). Let m ∈ N0, assume a , b , c ∈ C (U) m 1 assume f ∈ H (U). Let u ∈ H0 (U) be a weak solution of the boundary value problem m+2 m+2 Lu = f in U, u = 0 on ∂U. Assume also that ∂U ∈ C . Then u ∈ H (U) () and kukHm+2(U) 6 C kfkHm(U) + kukL2(U) () for some C = C(m, U, aij, bi, c) > 0.

Remark. As in Remark ?? (3): If u is the unique solution of the boundary value problem then () becomes kukHm+2(U) 6 CkfkHm(U) (“continuity in the data”). Similar to the proof of Theorem 2.29, we prove the claim by induction in m. Similar to the proof of Theorem 2.31 we control the “parallel” derivatives by the induction hypothesis and the transversal derivatives by the equation and ellipticity. The plan is

(1) Partition of unity-argument (not done here).

(2) Straightening the boundary (not done here).

(3) Do the “ﬂat-boundary-case” (done in some detail).

n n Proof. Let U = Bs(0) ∩ R+, s > 0. Fix t ∈ (0, s) ⊆ R and let V := Bt(0) ∩ R+. We will prove by induction in m that () and () hold. For m = 0, Theorem 2.31 gives the claim (here we assume even more on bi, c...). Assume the statement of the theorem holds for ij i (k+1)+1 k+2 k+1 some k ∈ N0. Let a , b , c ∈ C (U) = C (U), f ∈ H (U), (i.e. assumption of the theorem for m = k + 1) and assume that u ∈ H1 is a weak solution of Lu = f in U n and u = 0 on ∂U in the trace sense. Fix r ∈ (t, s) and let W := Br(0) ∩ R+. By induction k+2 hypothesis we have u ∈ H (W ) with kukHk+2(W ) 6 C(kfkHk(U) + kukL2(U)). By inner k+3 n regularity (Theorem 2.29), u ∈ Hloc (U). Now let α ∈ N0 be a multiindex with |α| = k +1 α 1 and αn = 0 (i.e. only parallel derivatives along {xn = 0}). Let u˜ := D u, then u˜ ∈ H (U) k+2 α (as u ∈ H (U), |α| = k + 1) and (!) u˜ = 0 along {xn = 0}, and all derivatives in D are parallel to {xn = 0}. Also, u˜ is a weak solution of Lu˜ = f˜ in U with

n n X α X X f˜ := Dαf − −((Dα−βaij)(Dβu )) + (Dα−βbi)(Dβu ) + (Dα−βc)(Dβu) β xi xj xi β6α i,j=1 i=1 β6=α

(Proof: As in the proof of Theorem 2.29). Use now aij, bi, c ∈ Ck+2(U) (by assumption) k+1 and f ∈ H (U) (by assumption) and estimate (∗) (induction hypothesis!) to get that f˜ ∈ L2(W ) ˜ kfkL2(W ) 6 C(kfkHk+1(U) + kukL2(U)) (∇)

Hence, as in the proof ot Theorem 2.31, we get u˜ ∈ H2(V ) and ku˜kH2(V ) 6 C kfkHk+1(U) + kukL2(U) (∇)

α Recall that u˜ = D u, |α| = k + 1, αn = 0. From (∇) follows

β n kD ukL2(V ) 6 C kfkHk+1(U) + kukL2(U) ∀β ∈ N0 , |β| = k + 3, βn ∈ {0, 1, 2} (∇∇)

50 Missing is now the claim for (∇∇). We prove this by induction in βn (for all β). Assume (∇∇) holds for all β, |β| = k + 3, βn = 0, 1, . . . , j for some j ∈ {2, 3, . . . , k + 2}. We prove that the claim holds for βn = j + 1. Hence, assume |β| = k + 3, βn = j + 1. Let β = γ + δ, k+3 δ = (0,..., 0, 2), |γ| = k + 1. Note that u ∈ Hloc (U) and Lu = f in U (strongly). Hence (!) DγLu = Dγf a.e. in U. Now DγLu = annDβu + A, where A is a sum of terms involving at most j derivatives of u with respect to xn and at most k + 3 derivatives in total (by nn construction/choice of β, γ, δ). Again, by ellipticity, a (x) > θ > 0 for all x ∈ U, so, since (∇∇) holds, by induction hypothesis for βn = 0, 1, . . . , j, we get

β kD ukL2(V ) 6 C kfkHk+1(U) + kukL2(U) |β| = k + 3, βn = j + 1

This is (∇∇) for this β (βn = j+1). By induction in j we have kukHk+3(V ) 6 C(kfkHk+1(U)+ kukL2(U)). This is the estimate () for m = k + 1. So, by induction in m, () holds for all m ∈ N0. This ﬁnishes the proof of Theorem 2.32. As for inner regularity, we cann not use Sobolev embeddings to get:

Theorem 2.33 (Inﬁnity diﬀerentiability up to the boundary). Assume aij, bi, c ∈ C∞(U), ∞ 1 i, j = 1, . . . , n and f ∈ C (U). Let u ∈ H0 (U) be a weak solution to the (uniformly elliptic) boundary value problem ( Lu = f in U (∗) u = 0 on ∂U

Assume also ∂U is C∞. Then u ∈ C∞(U).

∞ m Proof. Since f ∈ C (U), we have f ∈ H (U) for all m ∈ N. By Theorem 2.32 we get m that u ∈ H (U) for all m ∈ N. Now, the Sobolev embedding, Theorem 1.25 (ii), gives k k,γ ∞ that u ∈ C (U) for all k ∈ N (in fact, u ∈ C (U)). Hence, u ∈ C (U). Remark 2.34. (1) In particular (Problem Sheet 6, Ex 1 (b)), in this case, any weak solution is a classical solution!

(2) Of course, we do not neet “C∞” everywhere (aij, bi, c, ∂U...) to “just” reach the conclusion that “any weak solution is a classical solution” (Exercise: How little regularity can we do with by now?!).

(3) The conclusion is: Assuming enough regularity on

(a) The coeﬃcients (aij, bi, c) (b) The inhomogenity f (c) The boundary condition (! – we did for “g = 0”). (d) The boundary ∂U.

one can prove

(i) Existence of weak solutions: u ∈ H1 (ii) Higher regularity of weak solutions: Hk, Cm,γ. (iii) The existence of classical solutions

51 (4) However, there is no 1:1 correspondence between weak and classical solutions: “Obvi- ously”: There are weak solutions which are not classical if there is not enough regularity of f, say. Also: There exist classical solutions, which are not weak solutions! This 1 2 is simple, because they are not in H (U)! Example U = B1(0) ⊆ R and −∆u = 0, u = g (∗∗). There is g ∈ C(∂U), u ∈ C2(U) ∩ C(U) such that u is a classical solution to (∗∗), but u∈ / H1(U)! Note that u ∈ C2(U), not u ∈ C2(U) (Hadamards example), using (real) Fourier series...), namely

∞ X 1 g(cos θ, sin θ) = cos(22nθ) 2n n=1 which is not in the image of the trace operator, hence belongs to a Fractional Sobolev space. The claim is that one can compute u (exists and has a formula via the Poisson 2 kernel – see PDE 1). One then has U |Du| dx = ∞. ´ For classical solutions to, say, Lu = f, u = 0 or Lu = 0, u = g, one can now go on to study properties of solutions – like we did for harmonic functions. Many of the (other) properties in PDE 1 for harmonic functions carry over if the equation is uniformly elliptic:

• Maximum principle

• Harnack inequalities

• regularity (already seen)

• a priori estimates (already seen)

• The mean value property is less easy to generalise!

• Unique continuation: If two solutions are equal on a small neighborhood, they are equal.

Example 2.3 (Maximum Principles, without proofs – see Pb. Sheet 11 & Evans p. 344-351). We will here assume that solutions are C2 and that L is elliptic of non-divergence form: n n X ij X i Lu = − a (x)uxixj + b uxi + cu i,j=1 i=1 with aij, bi, c ∈ C(U) and L uniformly elliptic.

Theorem 2.35 (Weak maximum principle; c ≡ 0). Assume u ∈ C2(U) ∩ C(U) and c ≡ i in U.

(i) If Lu 6 0 in U (a “sub-solution”), then maxU u = max∂U u.

(ii) If Lu > 0 in U (a “super-solution”) then minU u = min∂U u. 2 Theorem 2.36 (Weak max principle; c > 0). Assume u ∈ C (U) ∩ C(U) and c > 0 in U + (i) If Lu 6 0 in U (a “subsolution”), then maxU u 6 max∂U u − (ii) If Lu > 0 in U ( a “supersolution”), then minU > − max∂U u

(iii) Hence, if Lu = 0 in U (a solution) then maxU |u| = max∂U |u|.

52 + − ± Recall u = u − u , u > 0. Theorem 2.37 (Strong maximum principle). Assume u ∈ C2(U) ∩ C(U) and c ≡ 0 in U and U is open, bounded and connected.

(i) If Lu 6 0 in U, and if u attainsits max (over U) at an interior point, then u is constant in u.

(ii) If Lu > 0 and if u attains its min (over U) at an interior point, then u is constant in U.

2 Theorem 2.38 (Strong maximum principle with c > 0). Assume ∈ C (U) ∩ C(U) and c > 0 in U suppose u is open, bounded and connected

(i) If Lu 6 0 in U and if u attains a non-negative maximum over U at an interior point. Then u is constant in U.

(ii) If Lu > 0 in U and if u attains a non-positive minimum over U at an interior point, then u is constant in U.

Note. For the sign of c (c > 0), it is important that n n X ij X i Lu = − a uxixj + b uxi + cu i,j=1 i=1 n n X ij X i ˜ = − a uxixj + (−b )uxi + (−c)u ≡ −Lu i,j=1 i=1 ˜ Hence, Lu > 0 ⇐⇒ Lu 6 0. The sign of the operator is thus important.

2.6.1 Eigenvalues and eigenvectors of the Dirichtlet boundary value problem Recall the:

Theorem (Theorem 2.14). (i) There exists an at most countable set Σ ⊆ R such that the boundary value problem ( Lu = λu + f in U (∗) u = 0 on ∂U

has a unique solution for all (!) f ∈ L2(U) iﬀ λ∈ / Σ.

k→∞ (ii) If Σ is inﬁnite, then Σ = {λk}k∈N with λk ∈ R and λk −−−→∞. We study the special ase, where L is symmetric and given by

n X ij Lu = − (a uxi )xj i,j=1

Then, formally, L is symmetric (“self-adjoint”...) and the bilinear form B[·, ·] associated 1 to L satisﬁes B[u, v] = B[v, u] for all u, v ∈ H0 (U). We assume U open, bounded and connected.

Theorem 2.39 (Eigenvalues of symmetric elliptic operators). (i) Each eigenvalue is real.

53 (ii) Furthermore, if we repeat each eigenvalue according to its (ﬁnite) multiplicity, we

have Σ = {λk}k∈N, where 0 < λ1 6 λ2 6 λ3 6 ··· with λk → ∞, k → ∞. 2 1 (iii) There exists an ONB {wk}k∈N of L (U) where wk ∈ H0 (U) and wk are eigenfunctions corresponding to λk: ( Lw = λ w in U k k k (∗) wk = 0 on ∂U

for k = 1, 2,....

∞ Remark 2.40. (i) By regularity theory (Theorem 2.30 & 2.32) we have wk ∈ C (U) ∞ ij ∞ and if ∂U is smooth, wk ∈ C (U) – if a ∈ C .

ij (ii) The λk’s are called the “Dirichlet-eigenvalues” for U in the case a ≡ δij (i.e. L = −∆).

n (iii) An interesting (??) question is how “λk → ∞ as k → ∞”. For U ⊆ R open, bounded, connected, smooth and L = −∆ (and zero boundary condition: “Dirichlet Problem”). Fact: √ n/2 n (λk) (2π) 2π 2/n 2/n lim = λk ∼ |U| · k k→∞ k |U| · α(n) α(n)2/n

Here, |U| is the volume of U and α(n) = Vol(B1(0)). This is called Weyl’s Lax (or “Weyl asymptotics”). This type of questions is related to “spectral asymptotics”, “semi-classical analysis”, and “Lieb-Thirring inequalities”.

Proof of Theorem 2.39. As in the proof of Theorem 2.13 let S := L−1 (the solution operator). S is a bounded, linear, compact operator mapping L2(U) into L2(U). Claim: S 2 1 is symmetric (i.e. self-adjoint). Let f, g ∈ L (U), then Sf = u iﬀ u ∈ H0 (U) is a weak solution of Lu = f in U and u = 0 on ∂U – the same for Sf = v. We get

(Sf, g) = (u, g) = B[v, u] = B[u, v] = (f, v) = (f, Sg) since B[u, v] = B[v, u] (L, hence B symmetric). So (Sf, g)L2 = (f, Sg)L2 for all f, g ∈ 2 2 L (U), so S is symmetric. Also (Sf, f) = (u, f) = B[u, u] > 0 for all f ∈ L (U) by uniform ellipticity. Hence, S is a compact, symmetric operator on the Hilbert space L2(U). By 2 Theorem 2.11 there exists an ONB {wk}k∈N of L (U) of eigenfunctions of S – and the corresponding eigenvalues {ηk}k∈N are all real (S self-adjoint), positive (by the above and uniform ellipticity) and only can accumulate at 0; since inﬁnitely many (multiplicitiy of each ηk is ﬁnite) and S bounded, they have to accumulate (see also Theorem 2.10). (Re)- numbering the ηk’s in decreasing order, we have ηk → 0 as k → ∞. Finally note that 1 1 Sw = ηw (w 6= 0) if and only if Lw = λw for λ = (in the weak sense). Let λk := . η η k This proves the theorem.

Pn ij i ij ∞ ∞ Recall that Lu = − i,j=1(a uxi )xj , b ≡ c ≡ 0, a ∈ C (U) ∩ L (U).

Deﬁnition 2.41. We call λ1 > 0 the principal eigenvalue of L (or, “the ground state energy”, or “the ﬁrst Dirichlet eigenvalue”).

54 Theorem 2.42 (Variational principle for the principal eigenvalue). (i) We have

1 λ1 = min{B[u, u] | u ∈ H0 (U), kukL2(U) = 1} (∗) B[u, u] = min 1 2 u∈H0 (U) kukL2(U) u6=0

(“Rayleigh’s Formula”).

(ii) The minimum in (∗) is attained for a function w1, with w1(x) > 0 for all x ∈ U and solving (weakly) ( Lw1 = λ1w1 in U

w1 = 0 on ∂U

1 (iii) If u ∈ H0 (U) is any weak solution of ( Lu = λ1u in U u = 0 on ∂U

then u is a (real) multiple of w1.

Remark. (a) We call w1 the positive groundstate of L.

(b) Part (iii) says λ1 is a non-degenerate eigenvalue – i.e. λ1 is simple: 0 < λ1 < λ2 6 λ3 6 ··· B[u,u] (c) The quotient kuk2 is also called the Rayleigh-Ritz quotient. L2(U)

(d) The formula (∗) says that λ1 is given by constrained minimization – hence the Lagrange multiplier λ1 in “Lw1 = λ1w1” (Compare also Problem Sheet 6 Ex. 2(a) and Ex. 4, Problem Sheet 10, Ex 4).

2 2 1 2 Proof. Recall that {wk}k∈N ⊆ L (U) is an ONB of L (U). Let u ∈ H0 (U) ⊆ L (U) with P∞ 2 kukL2(U) = 1. Then with dk := (u, wk)L2(U), we have u = k=1 dkwk converging in L (U). Also, Parseval’s Theorem (Pythagoras...) tells that

∞ ∞ X 2 X 2 dk = |(u, wk)L2(U)| = kukL2(U) = 1 k=1 k=1

From Lwk = λkwk in U, wk = 0 on ∂U (weakly) for k ∈ N we get

2 B[wk, wk] = λkkwkkL2(U) = λk () B[wk, w`] = λk(wk, w`) = 0 k 6= `

1 Hence, consier H0 (U) with the new inner product B[·, ·]:

1 1 B[·, ·]: H0 (U) × H0 (U) −→ R (u, v) 7−→ B[u, v] is symmetric (!) and positive deﬁnite, bilinear – so, a scalar product (by uniform ellipticity). wk 1 Then ( ) shows that { √ }k∈ is an orthonormal subset of (H (U),B[·, ·]). Claim: In λk N 0

55 wk 1 fact, { √ }k∈ is an orthonormal basis for (H (U),B[·, ·]) ( ). It suﬃces to prove that λk N 0 1 if B[wk, u] = 0 for all k for some u ∈ H0 (U) then u ≡ 0. But:

0 = B[wk, u] = λk(wk, u)L2(U) ∀k 2 Then (wk, u)L2(U) = 0 for all k (since λk 6= 0 for all k). So u ≡ 0 (in L (U)) since {wk}k∈N 2 wk is an ONB for L (U). Hence ( ) holds. That is, with µk := B[u, √ ] we have λk ∞ ∞ X wk X µk u := µ √ = √ w k λ λ k k=1 k k=1 k 1 with convergence in H0 (U). Note that the convergence is in B[·, ·] – but this gives norm 1 convergence in k · k 1 since B[·, ·] is bounded, bilinear on H (U) and coercive – so, H0 (U) 0 2 deﬁnes equivalent norm. But then it also converges in L (U) since kuk 2 kuk 1 . L (U) 6 H0 (U) P∞ 2 However, recall that u = k=1 dkwk, dk = (u, wk)L2(U) converginv in L (U). Hence, √ 1 µk /2 P∞ dk = √ ir µk = dk λk = dkλ and it follows that u = dkwk with convergence in λk k k=1 1 2 H0 (U) (not only in L (U)). From this, and B[wk, w`] = λkδk` follows (exercise!) ∞ ∞ ∞ X X (!) X B[u, u] = B dkwk, d`w` = dkd`B[wk, w`] k=1 `=1 k,`=1 ∞ ∞ ∞ X X 2 X 2 = dkd`λkδk` = dkλk > λ1 dk = λ1 k,`=1 k=1 k=1

1 since λk > λ1 for all k. Hence, for all u ∈ H0 (U), kukL2(U) = 1, we have B[u, u] > λ1. On the other hand, B[w1, w1] = λ1 and kw1kL2(U) = 1. Hence

1 λ1 = min{B[u, u] | u ∈ H0 (U), kukL2(U) = 1} 1 Rayleigh’s Formula now follows from scaling (exercise!). Claim: Let u ∈ H0 (U), kukL2(U) = 1. Then u is a weak solution of ( Lu = λ1u in U u = 0 on ∂U iff B[u, u] = λ1. (⇒) is clear and for (⇐) with dk = (u, wk)L2(U) we have ∞ ∞ X 2 X 2 2 dkλ1 = λ1 dk = λ1kukL2(U) = λ1 k=1 k=1 ∞ X 2 = B[u, u] = dkλk k=1 Pn 2 So k=1 ∞(λk − λ1)dk = 0, hence dk = (u, wk)L2 = 0 if λk > λ1. Recall that λ1 has finite multiplicity, i.e. there exist only finitely many λ2, . . . , λm such that λk = λ1, k = 1, . . . , m. Pm Hence u = k=1(u, wk)L2(U)wk for some m ∈ N. Hence (in the weak sense!) m m m X X X Lu = L (u, wk)L2(U)wk = (u, wk)L2(U)Lwk = (u, wk)L2(U)λkwk k=1 k=1 k=1 m X = λ1 (u, wk)L2(U)wk = λ1u k=1

56 1 Now, we show that if u ∈ H0 (U), u 6≡ 0, is a weak solution of Lu = λ1u in U, u = 0 on ∂U. Then either u > 0 in U (∗∇) or u < 0 in U (∗). Wlog kukL2(U) = 1 and let

α := (u+)2 dx β := (u−)2 dx ˆU ˆU

+ − 1 Recall (Problem Sheet, Q3), u , u ∈ H0 (U) with ( ( Du a.e. on {u 0} 0 a.e. in {u 0} Du+ = > Du− = > 0 a.e. on {u 6 0} −Du a.e. on {u 6 0}

Hence, (compute! – exercise), B[u+, u−] = 0. Therefore, with B[u+, u−] = 0

+ + − − λ1 = B[u, u] = B[u , u + B[u , u ] + 2 − 2 > λ1ku kL2(U) + λ1ku kL2(U)

= (α + β)λ1 = λ1

Hence, we have = everywhere. So

+ + + 2 − − − 2 B[u , u ] = λ1ku kL2(U B[u , u ] = λ1ku kL2(U) But then (claim proved earlier, normalise...)

( ± ± Lu = λ1u in U u± = 0 on ∂U

From the fact (!! assumption !!) aij ∈ C∞(U) we can conclude that u+ ∈ C∞(U) (Theorem + + + + 2.30) and so Lu = λ1u pointwise in U (classical solution). Hence (Lu )(x) = λ1u (x) > 0 in U. Hence, u+ is a supersolution, u+ = 0 on ∂U. The operator L, however, satisﬁes the strong maximum principle (Theorem 2.37, c ≡ 0), i.e. either u+ > 0 in U or u+ ≡ 0 in − U. Similarly for u , hence either (∗) holds or (∗) holds (recall kukL2(U) = 1 so u 6≡ 0). This ﬁnishes the proof of (ii). We still need λ1 is simple.

57 Appendix

A Tutorials

A.1 Tutorial 1: Review of Integration Definition A.1. Let (X, M, µ) be a measure space and f : X −→ [−∞, ∞]. f ist measurable iff f −1(U) ∈ M for all U ⊆ [−∞, ∞] open. n n For this course X = R or X = U ⊆ R open, M = B the Borel-σ-algebra and µ is the Lebesgue measure – sometimes surface-/n − 1-dimensional Hausdorff measure S/Hn−1. Pn Definition A.2. S : X −→ [0, ∞) is a simple function iff S(x) = i=1 aiχAi (x) for all x ∈ X, where ai ∈ [0, ∞), Ai ∈ M. Then we define the integral of S as n X S dµ := a µ(A ) ˆ k k X i=1

Proposition A.3. f : X −→ [0, ∞] is µ-measurable iﬀ there exists a sequence (sj)j of n→∞ simple functions such that sj 6 sj+1 and sj(x) −−−→ f(x) for all x ∈ X. For such f, deﬁne

f(x) dµ(x) := lim sj dµ ˆX j→∞ ˆX If this is ﬁnite, f is said to be summable/integrable.

Definition A.4. Let f : X −→ [−∞, ∞]. Let f± := max{±f, 0}. If one of X f± dµ is finite, then define ´

f dµ := f+ dµ − f− dµ ∈ [−∞, ∞] ˆX ˆX ˆX The space of integrable functions is calles L1(µ). This is a normed vector space unter the 1 n 1 equivalence relation “equal µ-a.e.”. We shall usually talk about L (R ), L (U).

A.1.1 Convergence Theorems

Theorem A.5 (Monotone Convergence Theorem). Let fj : X −→ [0, ∞] be a increasing sequence of measurable functions on X, i.e. fj(x) 6 fj+1(x) for every x ∈ X. Then f(x) := limj→∞ fj(x) exists for all x ∈ X and

lim fj dµ = f dµ j→∞ ˆX ˆX

Lemma A.6 (Fatou’s Lemma). Let fj : X −→ [0, ∞] be measurable. Then

lim inf fj(x) dµ 6 lim inf fj(x) dµ ˆX j→∞ j→∞ ˆX

Theorem A.7 (Dominated Convergence Theorem). Let fj : X −→ [−∞, ∞] be measur- n→∞ 1 able, with fj(x) −−−→ f(x) µ-a.e. Suppose there is some g ∈ L (µ) such that |fj(x)| 6 g(x) 1 µ-a.e. for all j. Then f, fj ∈ L (µ) and

n→∞ |f − fj|dµ −−−→ 0 ˆX 1 (“fj converges to f in L (µ)”).

58 A.1.2 Ex 2

2 (b) ⇒ (a) Let ε > 0. Then there is J ∈ N such that X |fj − f| dµ < ε for every j > J. Then ´ εµ({|f − f| > ε} |f − f| dµ ε2 j 6 ˆ j 6 {|fj −f|>ε} (this always holds).

(a) ⇒ (b) Let ε > 0

|fj − f| dµ = |fj − f| dµ + |fj − f| dµ 6 2µ({|fj − f| > ε}) + εµ(X) ˆX ˆ | {z } ˆ {|fj −f|>ε} 62 {|fj −f|6ε}

1 Let µ(X) = ∞, f, fj : X −→ [−1, 1], then χ[j,j+1] shows (c);(b), (c);(a) and j χ[0,j] shows (a);(b), (c);(b). In the case µ(X) = ∞ and f, fj : X −→ R we have (b) ⇒ (a) only. COnsider µ(X) < ∞, f, fj : X −→ R, jχ 1 in (0, 1) shows (a);(b), (c);(b). (0, j )

Suppose we do not have (a). For some subsequence (fj)j there is ε > 0 such that µ(Ej) > ε for Ej := {|fj − f| > ε}. S Proposition A.8. If (Ej) ⊆ M, µ( Ej) < ∞, then lim sup µ(Ej) 6 µ(lim sup Ej) where T∞ S∞ lim sup Ej = k=1 j=k Ej

Since µ(X) < ∞, we have µ(lim sup Ej) > lim sup µ(Ej) > ε. If x ∈ lim sup Ej = T∞ S∞ k=1 j=k Ej > 0, this means ∀k ∈ N ∃j > k such that |fj(x) − f(x)| > ε i.e. fj does not converge to f, so not (c).

Hence for fj, f unbounded, µ(X) < ∞ we have • L1 convergence =⇒ convergence in measure (also if µ(X) = ∞).

• µ-a.e. convergence =⇒ convergence in measure DCT µ-a..e convergence + majorand implies L1 convergence. Convergence in measure + equiintegrability ⇐⇒ L1 convergence (Vitali Convergence Theorem).

A.2 Tutorial 2 n ∞ Proposition A.9. Let U ⊆ R open, V b U. Then there is ζ ∈ C (U) with ζ = 1 on V , ζ = 0 in a neighborhood of ∂U.

Proof. Take W open with V b W b U. Let ηε be a molliﬁer with ε < min{dist(V, ∂W ), dist(W, ∂U)} then ζ := ηε ∗ χW does the job.

A.2.1 Exercise 3 Let  1 − x1 x1 > 0, |x2| < x1 (A1)  1 + x x < 0, |x | < −x (A ) u(x) := 1 1 2 1 2 1 − x x , |x | < x (A )  2 2 1 2 3  1 + x2 x2 < 0, |x1| < −x2 (A4)

59 1 Then u ∈ C (Ai) for each i and u ∈ C(U) (no discontinuities along boundaries and  !  −1  A1  0   !  1  A2  0 ∇u(x1, x2) = ! 0   A3  −1  !  0   A4  1

∞ Let ϕ ∈ Cc (U). Then

4 4 ∂ϕ X ∂ϕ X ∂u (1) u(x) dx1 = u(x) dx = − ϕdx + uϕν dS ˆ ∂x ˆ ∂x ˆ ∂x ˆ Ai U 1 i=1 Ai 1 i=1 Ai 1 ∂Ai ∂u = − ˆU ∂xi

A.2.2 Lp Spaces 1 1 Lt 1 < p, q < ∞ with p + q = 1. Then we have Young’s inequality: For a, b > 0 we have ap bq qp bq ab 6 p + q (Consider f(b) = p + q − ab). Also we have Young’s Inequality with ε: q p q − −1 ab 6 εa + C(ε)b where C(ε) = (εp) p q . This is used to prove Hölder’s Inequality: 1 1 Theorem A.10 (Hölder’s Inequality). For 1 6 p, q 6 ∞, p + q = 1 (p = 1, q = ∞) let u ∈ Lp(U), v ∈ Lq(U). Then uv ∈ L1(U)

|uv|dx 6 kukpkvkq ˆU p Theorem A.11 (Minkowski’s Inequality). If u, v ∈ L (U) for 1 6 p 6 ∞, then u + v ∈ p L (U) and ku + vkp 6 kukp + kvkp. Proof. If p = 1, p = ∞ this is easy. By the convexity of t 7−→ tp we have u + v ∈ Lp. p p−1 p−1 q 1 1 Hence |u + v| ∈ L (U) = L (U) where p + q = 1. Further„

p−1 p−1 |u||u + v| dx 6 kukpk|u + v| kq ˆU p−1 p−1 |v||u + v| dx 6 kvkpk|u + v| kq ˆU But

1/q p−1 (p−1)q p−1 k|u + v| kq = |u + v| dx = ku + vkp ˆU Hence

ku + vkp = |u + v|p (|u| + |v|)|u + v|p−1 (kuk + kvk )ku + vkp−1 p ˆ 6 ˆ 6 p p p

60 Show that

pq pq p q p+q • If 1 6 p, q < ∞ with p+q > 1 and u ∈ L (U), v ∈ L (U), then uv ∈ L (U) and kuvk pq kukpkvkq p+q 6

p q p+q p+q 1 1 Proof. We have p+q + p+q = 1, set α = q > 1 and β = q so that α + β = 1. pq pq Then |u| p+q ∈ Lβ(U) and |v| p+q ∈ Lα(U). By Hölder

pq pq pq pq pq |uv| p+q ku p+q k kv p+q k = kuk p+q kvk p+q ˆ 6 β α p q

n ∞ p p 1 • If λ (U) < ∞ and 1 6 p, q 6 ∞ then L (U) ⊆ L (U) ⊆ L (U) ⊆ L (U).

Proof. L∞(U) ⊆ Lp(U) is easy:

p p n |u| 6 (kuk∞ + 1) λ (U) ˆU

q If 1 6 p < q < ∞ and u ∈ L (U)

q−p p p p p n q |u| = |u| · 1 6 ku k q k1k q = kukqλ (U) ˆ ˆ p q−p

q−p n Hence kukp 6 λ (U) pq kukq.

Hence Lq is continuously embedded in Lp.

For the Young’s Inequality with ε replace a with ra then

(ra)p bq (ra)b + 6 p q

rp−1ap r−1 q rp−1 which means ab 6 p + q b and set ε := p .

A.3 Tutorial 3: Norms Let u ∈ W k,p(U). There are many possible (equivalent) Norms on W k,p(U). For example

1/p X α X α p kukk,p := kD ukp kukk,p = |D u| dx ˆU |α|6k |α|6k Why are they equivalent? Because the norms aon ﬁnite dimensional spaces are equivalent and W k,p(U) is isomorphic to some closed linear subspace of [Lp(U)]M(k,n) (k = 1, M = p M p M n + 1) and we can make many norms on [L (U)] , e.g. for v = (v1, . . . , vM ) ∈ [L (U)] M let w = (kv1kp,..., kvmkp) ∈ R for instance kvk = kwk`q is a norm for any 1 6 q 6 ∞.

A.3.1 Exercise 2 (b) 1 Take any W b U open. Choose k ∈ N so that dist(W, ∂U) > k . So Uk b U and 6= ∅! Set V0 := Uk+1.

61 A.3.2 Density of smooth functions in W 1,p(U)

For 1 6 p < ∞ we have ∞ 1,p • C (U) is dense in W (Uε) (Molliﬁers, easy) • C∞(U) is dense in W k,p(U) (Partition of unity)

• If ∂U is C0, C∞(U) is dense in W k,p(U).

A.3.3 Exercise 3 PN (a) Note that for x ∈ U, u(x) = i=1 ζi(x)u(x). Hence

A.4 Tutorial 4 A.4.1 Gagliardo-Nirenberg-Sobolev Inequality

n 1 ∗ np Let U ⊆ R be open, bounded with ∂U C . Let 1 6 p < n and p = n−p . Then there is C = C(U, p, n) > 0 such that

1,p kukW 1,p(U) 6 kukLp∗ (U) ∀u ∈ W (U)

1,p p∗ n so W (U) ,→ L (U). The proof in the lectures extended u to u on R so that u ∈ 1,p n W (R ), u has compact support and kukW 1,p(Rn) 6 CkukW 1,p(U). Now take a sequence ∞ n 1,p n (u ) ⊆ C ( ) with u → u in W ( ), then show that kuk p∗ n Ckuk 1,p n . j j c R j R L (R ) 6 W (R ) n Taking a subsequence if necessary, we have uj(x) → u(x) for a.a. x ∈ R . Also kujkW 1,p(Rn) → kukW 1,p(Rn). Hence by Fatou’s Lemma

p∗ p∗ p∗ |u(x)| dx = lim inf |un(x)| lim inf |uj(x)| dx n→∞ 6 n→∞ ˆRn ˆRn ˆRn p∗ p p∗ p∗ lim inf C kDunk 1,p n = C kDuk 1,p n 6 n→∞ W (R ) W (R )

∗ ˜ So kukLp(U) 6 ukLp 6 CkukW 1,p(Rn) 6 CkukW 1,p(U).

62 n 1,p Let U ⊆ R be open and bounded, W open U ⊆ W and u ∈ W0 (U), 1 6 p < ∞. Now deﬁne ( u(x) x ∈ U w(x) = 0 x ∈ W \ U

∞ 1,p There is uj ⊆ Cc (U) with uj → u in W (U). Extend uj by zero to W (No problem ∞ since uj all have support b U!). So (uj) ⊆ Cc (W ). Need to check: p (1) uj → w in L (W ) (clear). (2) Weak derivatives ∂w of w exist in W and they equal ∂u on U and 0 outside U. Let ∂xj ∂xi ∞ ϕ ∈ Cc (W ) (can have ϕ 6= 0 on ∂U ∂ϕ ∂ϕ w dx = lim uj dx ˆW ∂xi j→∞ Û ∂xi ∂uj = − lim ϕ dx + ujϕ dx j→∞ Û ∂xi ˆ∂U ∂u = − lim j ϕ dx j→∞ Û ∂xi ∂u ∂u = − ϕ dx = − χU ϕ dx Û ∂xi ˆW ∂xi

(3) ∂u → ∂w in Lp(W ). ∂xi ∂xi Next we prove Morrey’s inequality (Theorem 1.21).

Proof. First, we claim that there is C, depending only on n, such that

|Du(y)| |u(y) − u(x)| dy 6 C n−1 dy (4) Br(x) ˆBr(x) |y − x|

1 |u(x + sw) − u(x)|dS(w) = n−1 |u(z) − u(x)|dS(z) ˆ∂B1(0) s ˆ∂Bs(x)

63 Hence,

n−1 |Du(y) |u(z) − u(x)| dS(z) 6 s n−1 dy ˆ∂Bs(x) ˆBr(x) |x − y| integrating the left hand side over (0, r): r n−1 |Du(y)| |u(y) − u(x)| dy 6 s n−1 dy ds ˆBr(x) ˆ0 ˆBr(x) |x − y| rn |Du(y)| = n−1 dy n ˆBr(x) |x − y| ∞ n n Next, we show that u ∈ L (R ). Fix x ∈ R and apply (4) as follows:

|u(x)| 6 |u(x) − u(y)| dy + |u(y)| dy B1(x) B1(x) |Du(y)| = C dy + Ckuk p n−1 L (B1(x) ˆB1(x) |x − y| p−1 1/p p p p (1−n) p−1 p 6 C |Du| |x − y| +CkukL (B1(x) ˆB1(x) ˆB1(x) | {z } =:I

6 CkukW 1,p(Rn) (1−n)p I < ∞ since p > n and we need p−1 > −n. So supx∈Rn |u(x)| 6 CkukW 1,p(Rn). Hence u n is bounded. Last we show that u is Hölder continuous. Let x, y ∈ R . Write r = |x − y|. Let W = Br(x) ∩ Br(y).. Then

|u(x) − u(z)|dz 6 C |u(x) − u(z)| dz W Br(x) p−1 1 p / (1−n)p p p 6 C |Du| dz |x − z| p−1 dz ˆBr(x) ˆBr(x) p−1 p−1 n−(n−1) p p 6 CkDukLp(Rn · (r ) Similarly

n 1− p |u(y) − u(z)| dz 6 Cr kDukLp(Rn W Substitute this into (5) to get

1−n/p 1−n/p |u(x) − u(y)| 6 Cr kDukLp(Rn) = CkDukLp(Rn)|x − y| So, for γ = 1 − n/p we have that

[u]0,γ 6 CkDukLp(Rn) Corollary A.12. Morrey’s inequality for u ∈ W 1,p(U) where U is open and bounded, ∂U ∈ C1 implies that u has a representative that is Hölder continuous with exponent n γ = 1 − /p and kukC0,γ (U) 6 CkukW 1,p(U). n 1 n Proof. Extend to u on R approximate u with (uj) ⊆ C (R ) etc. u has compact support.

64 A.5 Tutorial 6 (Sheet 7) A.5.1 Exercise 1

0 To make Cε (or γ) smaller, we need larger ε. So let r ∈ (0, 1) be close to 1 and choose ε > 0 Pn 0 Pn 2 ε i=1 kbik∞ 2 as in “0 < ε i=1 kbik∞ < rθ”. Then (1 − r)θkDukL2 < 4ε0 + kck∞ kukL2 + θ 2 B[u, u] where the coeﬃcient is still 1. Without Poincaré, we just add 2 kuk2 on both sides θ 2 2 of 2 kDuk2 6 B[u, u] + Cεkuk2 to get θ θ (kuk2 + kDuk2) B[u, u] + (C + )kuk2 2 2 2 6 ε 2 L2

To solve (e), assume c(x) > −µ for some µ > 0. Then n X B[u, u] = a u u dx + c|u|2 θ |Du|2 − µkuk2 ˆ ij xi xj ˆ > ˆ 2 i,j=1 U U U

θ 2 2 θ 2 2 θ 2 > kuk1,2 − µkuk1,2 > kuk1,2 − µkuk1,2 = ( − µ)kuk1,2 Cp Cp Cp

A.5.2 Exercise 4

2 2 2 Claim kuk∗ := U |∆u| dx is an equivalent norm to k · kW 2,2 on H0 (U). For, by Poincaré, 2 2 2,2 2 ∞ |D u| is equivalent´ to the usual W (U) norm on H0 (U). Suppose u ∈ Cc (U). Then ´

uxixI uxj xj = − uxi xxj xj xi = − uxi uxixj xi = uxixj uxixj Û Û Û ˆ = |u |2 ˆ xixj Sum over i and j to get

2 X 2 2 2 |∆u| = |uxixj | = |D u| ˆU ˆ ˆU Hence

2 2 2 2 B[u, u] = |∆u| = |D u| > Ckuk2,2 ˆU ˆU

A.6 Tutorial 7 If v is a weak solution of ( L∗v = 0 in U (2.0) v = 0 on ∂U then v − K∗v = 0. v being a weak solution to (2.0) means B∗[v, z] = B[z, v] = 0 for all 1 z ∈ H0 (U). Hence

1 Bγ[z, z] = γ(z, v) ∀z ∈ H0 (U) (2.1)

2 1 The previous theorem says: For all g ∈ L (U), there is a unique ug ∈ H0 (U) such that

1 Bγ[u, w] = (g, w) ∀w ∈ H0 (U)

65 −1 −1 2 1 −1 1 Write ug := Lγ G, whence Lγ : L −→ H0 . Set K := γLγ . Fix u ∈ H0 (U) and let u˜ solve (2.2) with g = γu, i.e.

1 Bγ[˜u, w] = (γu, w) = γ(u, w) ∀w ∈ H0 (U)

−1 −1 So u˜ = Lγ (γu) = γLγ (u) = Ku. Since

1 1 Bγ[Ku, w] = γ(u, w) ∀w ∈ H0 (U)∀u ∈ H0 (2.4)

Take w = v. Then (2.4) is Bγ[Ku, v] = γ(u, v). By (2.1) with z = Ku, Bγ[Ku, v] = ∗ ∗ 1 γ(Ku, v) = γu, K v). So since γ 6= 0 we have (u, K v) = (u, v) for all u ∈ H0 (U). By L2 1 2 ∗ H0 (U) = L (U) we have K v = v.

A.6.1 Exercise 1

Take a subsequence (ujk ) of (uj) such that

lim f(ujk ) dx = lim inf f(uj) =: ` k ˆU j ˆU

Since u → u in Lp(U; N ), there exists a further subsequence u such that u (x) → jk R jkm jkm u(x) for almost every x ∈ U. Now apply Fatou’s Lemma (f > 0) so that

` = lim f(u ) = lim inf f(u (x)) dx lim inf f(u (x)) dx = f(u(x)) dx jkm jkm > jkm m Û m Û Û m Û

A.6.2 Exercise 2

Use the dominated convergence theorem: uj(x) → u(x) a.e. So f(uj(x)) → f(u(x)) a.e. since f is continuous. Let M = supj kujk∞ < ∞. So for a.a. x ∈ U, uj(x), u(x) ∈ [−M,M]N . So

|f(uj(x))| 6 sup |f(r)| =: C < ∞ v∈[−M,M]N

N 1 (f continuous, [−M,M] compact). U bounded, so χU ∈ L (U). Hence by DCT f · uj → f ◦ u in L1(U). Show f ◦ f ◦ u → f ◦ u in L∞(U). Let ε > 0, f is uniformly continuous on N N [−M,M] . so there is a δ > 0 such that |f(v1) − f(v2)| < ε for all vi ∈ [−M,M] with ∞ |v1 − v2| < δ. Since uj → u in L , there is a τ ∈ N such that |uj(x) − u(x)| < δ a.e. in U for all j > τ. Then |(f ◦ uj)(x) − (f ◦ u)(x)| < ε for all j > τ for almost all x ∈ U.

A.6.3 Exercise 3

p N p Suppose uj, u ∈ L (U; R ), uj → u in L . Now apply Vitali’s Convergence Theorem:

For some subsequence (ujk ), ujk (x) → u(x) a.e.. f is continuous so f(ujk (x) → f(u(x)) n p p a.e. λ (U) < ∞, so f ◦ ujk → f ◦ u in measure. ujk → u in L , so by Vitali, |ujk | is p equiintegrable. Hence |f(ujk )| 6 C + C|ujk | is equi-integrable. Hence f(ujk ) → f(u) in 1 L (U) by Vitali. “True for any subsequence. So true for full sequence”. If f(uj) 6→ f(u) 1 in L then there is a subsequence and ε > 0 such that U |f(ujk (x)) − f(u(x))| dx > ε for all k. But then we apply the above argument to show the´ existence of a subsequence u jkm with f(u ) → f(u) in L1(U) . jkm

66 A.6.4 Weiter im Skript Now, if

F (u) = f(∇u(x)) dx (∗) ˆU

1,p N N×n u ∈ W (U; R ), f : R −→ R continuous we have 1,p N (1) For f > 0, F (·) strongly lowersemicontinuous in W (U; R ) 1,∞ N (2) F (·) is strongly continuous in W (U, R ) p N×n (3) If f satisﬁes |f(ξ)| 6 C(1 + |ξ| ) for all ξ ∈ R then F (·) is strongly continuous on 1,p N W (U; R ).

A.6.5 Calculus of Variations Consider f as in (∗). Interesed in the minimisation problem 1 M = inf f(∇u(x)) dx u ∈ Hg (U) (∗∗) ˆU

th 1 Hilbert’s 20 Problem: Do there exist minimisers for (∗∗) i.e. dows there exist u ∈ Hg (U) with F (u) = m?

Direct Method: Show that the set in (∗∗) is bounded below and non-empty. Then take (uj) ⊆ A (A the space of admissible functions) with F (uj) → m. Then use properties of X, f to show (for a subsequence perhaps) that “uj → u ∈ A”. Then show F is lower semi-continuous wrt “−→” i.e. lim inf F (uj) > F (u). Then u is a minimiser of F in A. 1,p N p We may, for example, take A = Wg (U; R ). Suppose f(ξ) > c|ξ| for 1 < p < ∞. Then 1,p N minimising sequence (uj). Then (uj) bounded in W (U; R ) – reﬂexive 1 < p < ∞. Then 1,p there exists a weakly convergent subsequence ujk * u and u ∈ Wg . Hence sequential weak lower semicontinuity in W 1,p is a key property.

p N Theorem A.13. F := U f(u(x)), u ∈ L (U; R ), f continuous, f > 0. F is sequentially weakly lower semicontinuous´ iﬀ f is convex.

A.7 Tutorial 9 2 h (a) We have ζ > 0, ζ 6 ζ and ζ|Dk aij| 6 max kDaijkL∞(W ). This suﬃces...

A.8 Tutorium 11 A.8.1 Exercise 1 Recall that 1 v(x) = ζ2(x + he )[u(x) − u(x − he )] − ζ2(x)[u(x + he ) − u(x)] h2 k k k

Write ∂U = S1 ∪ S2, S1 = ∂U \{xn = 0} – curved part, S2 = ∂U ∩ {xn ≡ 0} – ﬂat part. 2 2 v = 0 on S1 because of cut-oﬀ: if x ∈ S1, ζ (x − hek) = 0, ζ (x) = 0. For x ∈ S2 such that x + hek ∈ S2, xn = 0, k 6= n, so u(x ± hek) = 0 in a trace sence, so v(x) = 0. If x ∈ S2, x ± hek ∈ S2, x is close to S1, so v(x) = 0.

67 A.8.2 Weiter im Skript

Now suppose (going back to original case) x0 ∈ ∂U. Flatten the boundary: There is an r > 0 sich that in a suitable coordinate system

n B(x0, r) ∩ U = B(x0, r) ∩ {x ∈ R | xn > γ(x1, . . . , xn−1)} 2 n−1 n n 2 where γ ∈ C (R ). There is Φ: R −→ R which is C with | det DΦ(x)| = 1 for all x given by

n Φ(B(x0, r) ∩ U) = B(x0, r) ∩ R+ i Φ (x) = xi n Φ (x) = xn − γ(x1, . . . , xn−1)

−1 ˜ n ˜ s n Also let Ψ = Φ and wlog x0 = 0. Take U = B(0, s) ∩ R+, 0 < s < r, V = B(0, 2 ) ∩ R+. Define u˜(y) := u(Ψ(y)) u ∈ U˜ Aim: Show that u˜ satisfies the requirements of the previous step, i.e. u˜ ∈ H1(U˜), u˜ = 0 2 on ∂U˜ ∩ {yn = 0} and L˜u˜ = f˜ for some uniformly elliptic operator L˜, f˜ ∈ L (U˜). Hence u˜ ∈ H1(V˜ ), ˜ ku˜kH1(V˜ ) 6 C(kfkL2(U˜) + ku˜kL2(U)) 2 then go back to the non-flattened case to get H -estimate for u in a neighborhood of x0

A.8.3 Exercise 3

1 ∞ 1 For (a) and (b): u ∈ H0 (U), so there is a sequence (uj) ⊆ Cc (U) with uj → u in H (U). 1 uj(x) = 0 on ∂U. Also uj ◦ ψ → u ◦ ψ in H (U˜). True because

2 2 |uj(ψ(y)) − u(ψ(y))| dy 6 |uj(x) − u(x)| dx → 0 ˆU˜ ˆU∩B(x0,r)

Further D((uj ◦ ψ)(y)) = Dψ(y)Duj(ψ(y)) and

2 2 |D(uj ◦ ψ)(y) − D(u ◦ ψ)(y)| 6 kDψk∞ |Duj(x) − Du(x)| dx → 0 ˆU˜ ˆB(x0,r)∩U

We have tr(uj ◦ ψ) = 0 on ∂U˜ ∩ {yn = 0} and the trace is continuous with respect 1 strong (even weakly) H convergence, so tr(u ◦ ψ) = 0 on ∂U˜ ∩ {yn = 0}. Also u˜ is the limit of H1(U˜) functions, hence in H1(U˜). For (c), let B˜[·, ·] denote the bilinear form ˜ 1 ˜ ˜ ˜ associated with L. Let v˜ ∈ H0 (U) (need B[˜u, v˜] = (f, v˜)L2(U). Take v(x) :=v ˜(Φ(x)), 1 ˜ x ∈ U ∩ B(x0, s).So v ∈ H0 (U). (exent v˜ by 0 outside U). We have n n X X ˜ B[˜u, v˜] = ak`u˜yk v˜y` + bku˜yk v˜ + cu˜v˜ dy ˆ ˜ U k,`=1 k=1 Now n ∂ X i u˜y = u(ψ(y)) = ψ (y)ux (Ψ(y)) k ∂y yk i k i=1 n X v˜ = ψi v (ψ(y)) y` y` xi i=1

68 So (linear algebra)

n n n X X X a˜ u˜ v˜ = a˜ ψi Ψj u (x)v (x) k` yk y` k` yk y` xi xj k,`=1 i,j=1 k,`=1 x = ψ(y) and

n n n X X X a˜ (y)Ψi (y)Ψj (y) = a (x)Φk (x)Φ` (x)Ψi (y)Ψj (y) k` yk y` rs xr xs yk y` k,`=1 k,`=1 r,s=1

= aij(x) since DΦ(x)DΨ(y) = 1. Similarly

n n n n X X X X b u˜ v˜0 b (x)Ψi u (x) ··· = b u (x)v(x) k xk r yk xi i xi k=1 k=1 r=1 i=1 Hence (since | det DΦ| = 1) ˜ ˜ B[˜u, v˜] = B[u, v] = (f, v)L2(U) = (f, v˜)L2(U)

A.8.4 Covering argument We obtain

˜ ku˜kH2(V˜ ) 6 C(kfkL2(U˜) + ku˜kL2(U˜)) so

2 2 2 kukH (V ) 6 C(kfkL (U∩B(x0,r)) + kukL (U∩B(x0,r))) (∗) for all x ∈ ∂U there is Vx such that (∗) holds. Hence we can replace the right hand 2 side with L (U) norms. {Vx}x∈U cover ∂U. Take a ﬁnite subcover V1,...,VN . Let V0 = SN U \ i=1 Vi b U. So

kukH2(U) 6 C + NC(kfkL2(U) + kukL2(U))

69 B Solutions of the problem sheets

B.1 Solution to Sheet 1 B.1.1 Exercise 1 (a) Let ( 1 0 < x 1 w(x) := 6 0 1 < x < 2

∞ and we claim that w is the weak derivative of u. Let ϕ ∈ Cc (0, 2). Then

2 1 2 u(x)ϕ0(x) = xϕ0(x) dx + ϕ0(x) dx ˆ ˆ ˆ 0 0 1 1 = [xϕ(x)]1 − ϕ(x) dx + (ϕ(2) − ϕ(1)) 0 ˆ 0 1 2 = ϕ(1) − ϕ(x) dx − ϕ(1) = − w(x)ϕ(x) dx ˆ ˆ 0 0 So v is the weak derivative of (0, 2). Let now ( x 0 < x 1 v(x) = 6 2 1 < x < 2

1 ∞ Suppose z ∈ Lloc(0, 2) is the weak derivative of v on (0, 2). Then for all ϕ ∈ Cc (0, 2)

2 2 1 2 − z(x)ϕ(x) dx = v(x)ϕ0(x) dx = xϕ0(x) dx + 2 ϕ0(x) dx ˆ ˆ ˆ ˆ 0 0 0 1 1 1 = ϕ(1) − ϕ(x) dx + 2(ϕ(2) − ϕ(1) = −ϕ(1) − ϕ(x) dx ˆ ˆ 0 0 So

2 1 z(x)ϕ(x) dx = ϕ(1) + ϕ(x) dx ∀ϕ ∈ C∞(0, 2) ˆ ˆ c 0 0

∞ Take ϕj ∈ Cc (0, 2) such that ϕj(1) = 1 for all j but kϕjk1 → 0 (e.g. ϕj := ηεj ∗ 1 2 1 χ(1−εj ,1+εj ), εj := j ). Then 0 zϕj dx → 0 but ϕ(1) + 0 ϕj → 1. So v is not weakly diﬀerentiable. ´ ´

(b) Let n −α/2 1 X u(x) = = x2 |x|α i i=1 We show that u ∈ W k,p(U) for n > (α + 1)p.

70 B.1.2 Exercise 3 (Fundamental Lemma of Calculus of Variations)

Let E b U be open and let ηε be a molliﬁer with ε < dist(E, ∂U). Then, for each x ∈ E,

(η ∗ u)(x) = η (x − y)u(y) dy = 0 ε ˆ ε U ∞ by assumption since ηε(x − y) ∈ Cc (U). By 4 (c),

ε→0 |u| dx = |u − (ηε ∗ u)| dx −−−→ 0 ˆE ˆE So u = 0 a.e. in E. True for any E b U. So u = 0 a.e. in U.

B.1.3 Exercise 4 (a)

(b) Let u ∈ Cc(U) and x ∈ U. Then

|(ηε ∗ u)(x) − u(x)| = ηε(x − y)(u(y) − u(x)) dy ˆB(x,ε)

6 sup |u(y) − u(x)| · ηε(x − y) dy y∈B(x,ε) ˆB(x,ε)

n since u is uniformly continuous on R , for given γ > 0 there is ε0 > 0 such that |u(x) − u(y)| < γ whenever |x − y| < ε0. So

sup |(ηε ∗ u)(x) − u(x)| < γ ∀ε ∈ (0, ε0] x∈U

(c) Let γ > 0. Then there is v ∈ Cc(U) such that ku − vkp < γ. Then (by part (b)) there ε0 > 0 such that k(ηε ∗ v) − vkp < γ for all 0 < ε 6 ε0. Then

ku − (ηε ∗ u)kp = ku − (v − (ηε ∗ v)) + (v − (ηε ∗ v)) − (ηε ∗ u)kp 6 ku − vkp + k(ηε ∗ v) − vkp + k(ηε ∗ v) − (ηε ∗ u)kp 6 2ku − vkp + k(ηε ∗ v) − vkp < 3γ

B.2 Solution to Sheet 2 B.2.1 Exercise 1 (a)

Remark B.1. Check the scaling: If v(x) := µu(λx) (U a cone), λ, µ > 0. Then kvkr = nθ n −n/r θ θ − 1−θ 1−θ − (1−θ) µλ kukr, kvks = µ λ s kuks, kvkt = µ λ t kukt. Hence 1 = θ + (1 − θ) n n n X, − r = − s θ − (1 − θ) t X. We have by Hölders Inequality

r−s 1 p 1 q t−r / / s t t−s r s t−s t s t p q |u| dx = |u| |u| dx 6 |u| |u| = kuks kvkt Û Û Û Û

s t 1 t−r 1 r−s pr qr s t where p = r−s , q = t−s . So kukr 6 kuks kukt . Then θ = pr , 1 − θ = qr . In fact |u|s ∈ Lp, |u|t ∈ Lq.

71 |u(x)−u(y)| (b) Recall that kuk0,γ = kuk∞ + [u]0,γ where [u]0,γ = supx,y∈U |x−y|γ . For x, y ∈ U,

1−γ γ−β |u(x) − u(y)| |u(x) − u(y)| 1−β |u(x) − u(y)| 1−β = · |x − y|γ |x − y|β |x − y|

1−γ γ−β β−βγ+γ−β γ(1−β) Then β 1−β + 1−β = 1−β = 1−β = γ. Take the supremum to get

1−γ γ−β 1−β 1−β [u]0,γ 6 [u]0,β · [u]0,1

1−γ γ−β Set θ := 1−β , 1 − θ = 1−β . Hence

θ 1−θ kuk0,γ = kuk∞ + [u]0,γ 6 kuk∞ + [u]0,β · [u]0,1 | {z } | {z } =:a =:b

θ 1−θ θ 1−θ Wlog assume kuk∞ = 1. We need to show 1 + a b 6 (1 + a) (1 + b) . First write

θ 1 θ 1 θ θ (1 + a) = (1 θ + (a ) θ ) = k(1, a )k 1 θ 1−θ 1−θ (1 + b) = k(1, b )k 1 1−θ

θ 1−θ θ 1−θ 2 Hence 1 + a b = h(1, a ), (1, b )i (in R ) and so

θ 1−θ θ 1−θ 1 + a b k(1, a )k 1 · k(1, b )k 1 6 θ 1−θ = (1 + a)θ(1 + b)1−θ

B.2.2 Exercise 2 This is true if u ∈ C∞(U), i.e. by Taylor’s Theorem we have

: 0 1 Dαu(x ) k(x − x )α X 0 α X 0 k−1 α u(x) := (x − x0) + (1−t) D u(x0 + t(x − x0)) dt α! α! ˆ |α|6k |α|=k 0 1 Take V b U. Then u ∈ L (V ). Take ε0 < dist(V, ∂U) and for ε ∈ (0, ε0), let ηε be a ∞ molliﬁer and let u˜ := uχV . Then let u˜ε := (ηε ∗ u)(x), whence u˜ε ∈ C (U). Also, for α α x ∈ V (not U!), D u˜ε(x) = (ηε ∗ D u)(x) = 0 if |α| = k. By step 1, u˜ε = pε on V where pε ε→0 L1 is a polynomial of degree at most k − 1. But ku˜ε − u˜kL1(V ) −−−→ 0, hence u˜ ∈ Pk−1(V ) . Since Pk−1 is ﬁnite dimensional, hence closed, we have u˜ ∈ Pk−1(V ). Let (Vi)i be open S∞ 1 sets with Vi b U, Vi b Vi+1, U := i=1 Vi (e.g. Vi := {x ∈ U | dist(x, ∂U) > i } ∩ Bi(0)). By the above, for every i, u = pi ∈ Pk−1(Vi) and hence pi+1 is an extension of pi to Vi+1 hence (Identity Theorem) pi+1 = pi. Further, for every x ∈ U there is i ∈ N such that x ∈ Vi, So u(x) = p(x) for some p ∈ Pk−1(U).

B.2.3 Exercise 3

Definition B.2. u:[a, b] −→ R is absolutely continuous iff for all ε > 0 there is δ > 0 such that whenever {(ai, bi)}16i6k is a finite collection of disjoint subintervals of [a, b] and Pk Pk i=1(bi − ai) < δ then i=1 |u(bi) − u(ai)| < ε.

72 du 1 ∞ (a)( ⇒) By (i), v = dx exists a.e. and v ∈ L ([a, b]). Let ϕ ∈ Cc (a, b). Then by (i) ϕu is d(ϕu) dϕ absolutely continuous with dx = vϕ + u dx a.e. Hence b b b 0 = (ϕu)0 = vϕ + uϕ0 â â â So v is the weak derivative of u. (⇐) Suppose u is weakly differentiable. Write v for its weak derivative (by assumption 1 x v ∈ L [a, b]). Write w(x) := a v(y) dy. By (i), w is absolutely continuous. By (⇒), there is a weak derivative´ of w that equals the classical derivative a.e. So w0(x) = v(x) = u0(x) a.e. So (u − w)0 = 0. So u − w is constant by Question 2 and u = w + const. so u is absolutely continuous.

B.2.4 Exercise 4

For all x ∈ U there is Vi with x ∈ Vi. Hence there is εx > 0 with Bεx (x) ⊆ Vi.

Hence {Bεx (x)}x∈U is a cover of U. Since U is compact, there exists a ﬁnite subcover m Sm {Bεj (xj)}j=1. Hence U ⊆ j=1 Bεj (xj) and Bεj (xj) ⊆ Vi for some i. Not let Wi := S SN Bεj (xj). Then Wi Vi, 1 i N and U ⊆ Wi. Then (see Tutorial 2) Bεj (xj )⊆Vi b 6 6 i=1 ∞ ϕi(x) we can take ϕi ∈ C with supp(ϕi) ⊆ Wi and ϕi > 0 on Wi. Now let ζi(x) := PN . i=1 ϕi(x) This is welldeﬁned since the denominator is never 0.

B.3 Solution to Problem Sheet 3 B.3.1 Exercise 1

p p p n (a) Clearly, n |τhu| = n |u| < ∞. so τhu ∈ L (R ) and kτhukp = kukp for every p nR R u ∈ L (R´ ). τh is linear´ (trivial!). So τh is an isometry. For p = ∞ it is easy to see kuk∞ = kτhuk∞.

n j→∞ j→∞ p n (b) Take (hj)j ⊆ R , hj −−−→ 0 suﬃces to show τhj u −−−→ u in L (R ). Write uj = τhj u p Theorem B.3 (Vitali’s Convergence Theorem). Let (uj)n ⊆ L (U), 1 6 p < ∞, then p uj → u in L (U) if and only if

n ε→0 (1) uj → u in measure, i.e. for every ε > 0: λ ({x ∈ U | |uj(x) − u(x)| > ε} −−−→ 0. p (2) (uj)j is p-equiintegrable, i.e. for every ε > 0 there is δ > 0 such that supj E |uj| dx < ε whenever λn(E) < δ. ´

n j→∞ n First take B ⊆ R bounded. Then uj(x) −−−→ u(x) for every x ∈ B if u ∈ C(R ) (!). This implies convergence in measure (on B) u is p-equiintegrable (it’s one function) p n hence given ε > 0 there is δ > 0 such that E |u| < ε whenever λ (E) < δ. But p p n n |uj| = |u| < ε since λ (E + hj) =´ λ (E). So (uj) p-equi-integrable. So E E+hj p u´ j → u in L´ (B). Let ε > 0. Then ∃B such that

|u|p < ε ˆRn\B p and also n |uj| < ε Then R \B ´ p p |u − uj| 6 |u − uj| + 2ε < 3ε ˆRn ˆB

73 n p n if j large enough. Then use the density of C(R ) in L (R ). For p = ∞ this is not true. Take n = 1 and u = χ(0,1). Then kτhu − uk∞ = 1 for all h 6= 0.

(c) This is false for all 1 6 p 6 ∞. For p = ∞ this can’t be true due to (b). For 1 1 6 p < ∞. Suppose this holds. Let ε = 2 (suppose n = 1). Then there is δ > 0 1 such that kτhu − ukp < 2 whenever |h| < δ. Fix some h ∈ (0, δ). Now take γ > 0 such that (0, γ), (h, h + γ) are disjoint. Now let u(x) = 1 χ . Then kuk = 1 but γ1/p (0,γ) p |τ u − u|p = γ |u|p + h+δ |u|p = 2. R h 0 h ´ ´ ´ B.3.2 Exercise 2

1,p 1 0 ∞ Let 1 6 p < ∞, u ∈ W (U), F ∈ C (R) such that F ∈ L (R). Note that for t ∈ R we have t F (t) = F (0) + F 0(s) ds ˆ 0 so u(x) v(x)|p = |F (u(x)|p = |F (0) + F 0(s)ds|p C(|F (0)|p + kF 0kp |u(x)|p) ˆ 6 ∞ 0 p p n 0 p p 0 ∂u So |v| |F (0)| λ (U) + kF k∞ |u| < ∞. Now show w(x) = F (u(x)) (x) ∈ U 6 ∂xj Lp(U´ ). This is by ´

p 0 ∂u p |w| 6 kF k∞ | | < ∞ ˆU ˆU ∂xi

th ∞ ∞ Now show w is the i weak derivative of v. Let ϕ ∈ Cc (U) and take (uj)j ⊆ C (U) with kuj − ukW 1,p(U) → 0 as j → ∞. By the Chain rule for classical derivatives, Gauss-Green, we have writing v = f ◦ u , ∂vj = F 0(u ) ∂u in U, so j j ∂xj j ∂xj ∂ϕ F (uj(x)) (x)ϕ(x)dx ˆU ∂xj

Taking subsequences if necessary, we can assume uj(x) → u(x) ∂iuj(x) → ∂iu(x) a.e. in U. Then the right hand side is okay by dominated convergence. For we have

0 0 0 0 F (uj(x))Diuj(x)ϕ(x) − F (u(x))Diu(x)ϕ(x) = kϕk∞ |DiujF (uj) − DiuF (u)| ˆU ˆU ˆ 0 0 6 Ckϕk∞kDiujF (uj) − DiuF (u)kp Now

0 0 0 0 0 0 kDiujF (uj) − DiuF (u)kp 6 kDiujF (uj) − DiuF (uj)kp + kDiuF (uj) − DiuF (u)kp 0 0 0 6 kF k∞kDiuj − Diukp + |Diu| · kF (uj) − F (u)kp | {z } gj

74 p 0 p p 1 Now gj(x) → 0 a.e. |gj(x)| 6 2kF k∞|Diu(x)| ∈ L (U). So kgjkp → 0 as j → ∞. The left hand side is

F (uj)Diϕ − F (u)Diϕ = (F (uj) − F (u))Diϕ Û Û Û

6 kDϕk∞ |F (uj(x)) − F (u(x))|dx ˆU 0 j→∞ 6 kDϕk∞kF k∞ |uj − u| dx −−−→ 0 ˆU

0 1 since F is bounded, so F is Lipschitz and U is bounded, so uj → u in L . Hence, taking j → ∞ we get the result.

B.3.3 Exercise 3 Let ( 2 2 1/2 (z + ε ) − ε z > 0 fε(z) = 0 z < 0 We have

1 0 (i) fε ∈ C (R) and fε ∈ [0, 1] + (ii) 0 6 fε(z) 6 z for all z ∈ R + (iii) limε&0,ε>0 fε() = z This is easy to show. In particular ( (z2 + ε2)1/2 − ε z 0 f 0(z) = > ε 0 z < 0

∞ By Ex 2 for alle ϕ ∈ Cc (U)

fε(u)Diϕ dx = − Di(Fε(u)ϕ(x) dx Û Û 0 = − fε(u)Diuϕ dx Û uD u = − √ i ϕdx 2 2 Û u − ε

+ Let ε & 0, then fε((u(x))Diϕ(x) → u (x)ϕ(x) a.e.

0 fε(u(x))Diu(x)ϕ(x) → χ{u>0}(x)Diu(x)ϕ(x)

1 0 1 − Apply DCT, |fε(u)Diϕ| 6 |u| · kDiϕk∞ ∈ L and |fε(u)Diuϕ| 6 kϕk∞|Diu| ∈ L . For u we apply this result to −u. Part (b) is similar and for (c) we have by part (a): D(u+) = 0 − + − a.e. on {u 6 0} and D(u ) = 0 a.e. on {u > 0}.. So Du = D(u ) + D(u ) = 0 a.e. on {u = 0}.

75 B.4 Solution to Problem Sheet 4 B.4.1 Exercise 1

If there is such a T , there also is a C > 0 such that kT ukLp(∂U) 6 CkukLp(U) and (T u)(x) = u(x) for all x ∈ ∂U whenever u ∈ C(U) ∩ Lp(U). First suppose n = 1, U = (0, 1). Then, if the trace operator exists, we have u(0) = (T u)(0) and u(1) = (T u)(1) for all u ∈ C[0, 1]. For j ∈ let N  1 1 − jx j ∈ (0, j ) uj(x) = 1 ∈ C[0, 1] 0 x ∈ [ j , 1)

Then uj(0) = 1, uj(1) = 0 for all j and

p 0 |T uj| dλ = 1 ˆ{0,1} and

1 1 j 1 |u |pdλ1 dλ1 = ∀j ˆ j 6 ˆ j 0 0

C n→∞ So 1 = kT ukLp(∂U) 6 CkukLp(U) 6 j −−−→ 0 . For general n, U, adapt this idea.

B.4.2 Exercise 2

1 n n Let U bounded with ∂U ∈ C . Take a smooth vector ﬁeld α: R −→ R with hα, νi > 1. Let 1 6 p < ∞. Then

hα, D(|u|p)idx = − div α · |u|pdx + hα, νi ·|u|pdS ˆU ˆU ˆ∂U | {z } >1 So

|u|pdS hα, D(|u|p)dx + div α · |u|pdx ˆ 6 ˆ ˆ ∂U U U kαk · p|u|p−1 sgn(u)|Du|dx + k div αk |u|pdx 6 L∞(U) ˆ L∞(U) ˆ U A.2.2 (p−1)q p p|u| |Du| p 6 kαk∞ + dx + k div αk∞ |u| dx ˆU p p ˆ U C(α) |u|p + |Du|p dx 6 ˆ U

76 B.4.3 Exercise 3 (a) We have

|Du|pdx = hDu, Dui|Du|p−2dx = − u div(Du|Du|p−2dx Û Û Û = − u div(Du|Du|p−2)dx Û = − u∆u|Du|p−2 − uDu · D(|Du|p−2) Û Û Du = − u∆u |Du|p−2 dx − uDu · ((p − 2)|Du|p−3D2u Û |{z} Û |Du| p | {zp } 2 ∈L ∈L p−2 2 p−2 p p p 2 p p 6 C |u| 2 |D u| 2 · |Du| Û Û 1 1 2 p−2 2 2 p p C |u|p |D2u|p · |Du|p 6 ˆ ˆ ˆ

p 2 p−2 So kDukp 6 CkukpkD ukpkDukp .

B.4.4 Exercise 4 We need to check:

• For u ∈ C2(U), ∂U ﬂat (locally), we have u in W 2,p (it is not C2).

• Φ is a C2-diﬀeomorphism (so Φ ◦ u is C2, and Φ−1 ◦ (Φ ◦ u) is W 2,p)

• We get kukW 2,p(B) 6 CkukW 2,p(B+). 0 0 2 Fix x ∈ ∂U. Then there is r > 0 such that Br(x ) ∩ ∂U is locally the graph ot a C - 0 0 0 function. Assume ∂U ∩ Br(x ) is ﬂat, i.e. U ∩ Br(x ) = {x ∈ Br(x ) | xn > 0}. We have u ∈ C2(B+). It was proved in the lectures that u ∈ C1(B). But u∈ / C2(B) since

( + uxnxn (x) x ∈ B ux x = n n 0 0 xn − 3uxnxn (x , −xn) − uxnxn (x , − 2 ) x ∈ B

0 nd writing x = (x , xn) ∈ B. This is a 2 -order weak derivative. The rest is clear.

B.5 Solution to Exercise Sheet 5 If K is a compact Hausdorff space. A subset S of C(K) is equicontinuous at x ∈ K iff for all ε > 0 there is V open such that x ∈ V and |f(y) − f(x) < ε for all f ∈ S (V is independent of f ∈ S). S is equicontinuous (on K) iff S is equicontinuous at each point x ∈ K. S is uniformly bounded iff sup{kfk∞ | f ∈ S} < ∞. Theorem B.4 (Arzelá-Ascoli). A subset S of C(K) is relatively compact (or totally bounded) if and only if S is equicontinuous.

77 B.5.1 Exercise 1 (a) First we show that C0,β(U) ,→ C0,α(U). Let u ∈ C0,β(U).

|u(x) − u(y)| |u(x) − u(y)| |u(x) − u(y)| [u] = sup sup + sup 0,α |x − y|α 6 |x − y|α |x − y|α x,y∈U,x6=y x,y∈U,|x−y|>1 x,y∈U,|x−y|61

leq2kuk∞ + [u]0,β

So kukC0,α(U) 6 3kukC0,β (U). To show that the embedding is compact, it suﬃces to 0,β show that if (uj) ⊆ C (U) is bounded – kunkC0,β 6 M – then there is a subsequence 0,α (ujk ) and u ∈ C (U) such that kujk − ukC0,α(U) −→ 0. First show (uj) has a subsequence converging strongly in (C(U), k·k∞). Use Arzelá-Ascoli: (uj) is uniformly bounded by assumption. We show that (uj)j is equicontinuous. Let x ∈ U and ε > 0. Then |u (x) − u (y)| |u (x) − u (y)| = j j · |x − y|β j j |x − y|β β β 6 [uj]0,β|x − y| 6 M|x − y|

ε 1/β Let δ < ( /M) . This does the job. So there is a subsequence (ujk ) and u ∈ C(U)

with kujk − uk∞ → 0. Taking the limit of |ujk (x) − ujk (y)| and using the above, we β 0,β 0,α have |u(x) − u(y)| 6 M|x − y| so u ∈ C (U) ⊆ C (U). Show now that ujk → u in 0,α C (U). We need [ujk − u]0,α → 0. We have |u(x) − u (x)) − (u(y) − u (y)| |u(x) − u (x)) − (u(y) − u (y)| [u − u] sup jk jk + sup jk jk |x − y|β−α jk 0,α 6 |x − y|α |x − y|β |x−y|>δ |x−y|6δ −α β−α 6 2δ ku − ujk k∞ + δ [u − ujk ]0,β −α β−α 6 2δ ku − ujk k∞ + 2δ M

β−α ε For ε > 0 given, let δ > 0 be such that 2δ M < 2 . Then for this δ let k0 be large −α ε 2 so that 2δ ku − ujk k∞ < / for k > k0. 1,p q ∗ (b) Let 1 6 p < n. Rellich-Kondrachov gives W (U) b L (U) for p < q < p . Let p = n ∗ 1,p 1,r r∗ and r ∈ [1, n) be such that r > p = n. Then W (U) ,→ W (U) b L (U) ,→ n L (U). Let n < p 6 ∞ and let 0 < α < 1 − n/p then by Morrey’s inequality

1,p 0,1−n/p 0,α p W (U) ,→ C (U) b C (U) ,→ L (U)

B.5.2 Exercise 2

n 1 (a) Ist U ⊆ R open, connected and bounded with ∂U ∈ C . Then there is C > 0 such that p p 1,p u − u dx 6 C |Du| dx ∀u ∈ W (U) (∗) ˆU U ˆU

1,p Proof. Suppose (∗) holds for all u ∈ W (U) with U u = 0. Then for general u ∈ 1,p W (U), v(x) = u(x) − U u statisfies U v = 0, Dv =ffl Du. So (∗) holds for u, too. So 1,p it suffices to prove the statementffl for functionsffl u ∈ W (U) with U u = 0. Suppose 1,p (∗) does not hold for any constant C. Then for all j ∈ N there is ufflj ∈ W (U) with

78 p p uj 1,p u = 0 and |uj| j |Duj| . Now let vj = . Then vj ∈ W (U) and U U > U kuj kp ffl p ´ ´ U |vj| = 1 for all j. Also ´ p 1 p 1 j→∞ |Dvj| = p |Duj| dx < −−−→ 0 Û kujkp Û j

p n So Dvj → 0 in l (U; R ). By Rellich Kondrachov (or 1 (b) there is a subsequence vjk p p and v ∈ L (U) with vjk → v in L (U). We have kvjk kp = 1 for all k so kvkp = 1, too. p n 1,p p Since Dvjk → 0 so vjk is Cauchy in L (U, R ). So v ∈ W (U) and U |Dv| = 0. So Dv = 0 a.e. so v = const. a.e. Also U v = limk→∞ U vjk = 0. So´ v = 0 a.e. But kvkp = 1 . ﬄ ﬄ

(b) By (a), there is a constant such that

p p 1,p |u(y) − (u)B1(0)| dy 6 C |Du(y)| dy u ∈ W (B1(0)) (∗∗) ˆB1(0) ˆB1(0)

1,p 1,p Let u ∈ W (Br(x)). Then v(y) := u(x+ry) ∈ W (B1(0)) and Dv(y) = rDu(x+ry) and

u(y) dy = rn v(y) dy ˆBr(x) ˆBr(x)

So by (∗∗)

1 |u(y) − (u) |p = |v(y) − (v) |p dy n Br(x) B1(0) r ˆBr(x) ˆB1(0) p p Cr p 6 C |Dv(y)| dy = n |Du| dy ˆB1(0) r ˆBr(x)

An application are pointwise properties of Sobolev maps. Recall Lebesgue’s Diﬀerentiation THeorem. If u ∈ L1(U). Then for a.a. x ∈ U

r&0 u(y) dy −−−→ u(x) Br(x)

Corollary B.5. For u ∈ W 1,p(U)

p u(y) − u(x) − (Du)Br(x)(y − x) r→0 dy −−−→ 0 Br(x) r

So u admits an “aﬃne approximation” (in the integral sense) as r → 0, analogously to diﬀerentiable functions

B.5.3 Exercise 3

−1 x→0 ∞ (a) Let u(x) = ln ln(1 + |x| ) −−−→∞ (slowly). So u∈ / L (B1(0)). For x 6= 0 the strong derivative of u exists and ∂u 1 1 xi = −1 −1 − 3 ∂xi ln(1 + |x| ) 1 + |x| |x| We show

79 n (1) u ∈ L (B1(0)) (2) ∂u ∈ Ln(B (0)) ∂xi 1 (3) ∂u is the weak derivative of u on all of B (0). ∂xi 1

(1) We have ln(1 + s) 6 s for s > 0 so 1 n n−1 −1 n |u(x)| dx = ωn r (ln ln(1 + r )) dr ˆB1(0) ˆ 0 1 ∞ ω rn−1(ln r)ndr = ω e−(n−1)te−ttndt 6 n ˆ n ˆ 0 0 ∞ = ω e−nttndt < ∞ n ˆ 0 −t n using the change of variables r = e . So u ∈ L (B1(0)). (2) We have

1 n n ∂u n−1 1 1 1 dx 6 r −1 −1 · 2 dr ˆ ∂xi ˆ ln(1 + r ) 1 + r r B1(0) 0 1 1 1 n ω rn−1 · dr 6 n ˆ ln(1 + r−1 r2 + r 0 | {z1 } 6 r2 1 ω rn−1(ln(1 + r−1)−nr−n dr 6 n ˆ 0 1 −1 −1 −n = ωn r (ln(1 + r ) dr ˆ0 ∞ t −t t −n = ωn e · e (ln(1 + e )) dr ˆ | {z } 0 >et ∞ ω t−n dt < ∞ 6 n ˆ 0 for n > 1. ∞ (3) Let ϕ ∈ Cc (B1(0)). Then, for 0 < ε < 1: ∂ϕ(x) ∂ϕ ∂u u(x) dx = u − ϕ + uϕ dS ˆ ∂xi ˆ ∂xi ˆ ∂xi ˆ B1(0) Bε(0) B1(0)\Bε(0) ∂Bε(0) | {z } | {z } | {z } =:I1 =:I2 =:I3

We have I −−−→ε→0 0 and dominated convergence gives further I → − ϕ ∂u dx. 1 2 B1(0) ∂xi Eventually, we have ´

n−1 −1 ε→0 I3 6 kϕk∞ωnε ln ln(1 + ε ) −−−→ 0

80 1,n n n 1,n 1,1 (b) Let u ∈ W (Rbb ) and Br(x) ⊆ R . Then by 2 (b) u ∈ W (Br(x)) ,→ W (Br(x)) so

u − dy 6 Cr |Du| ˆBr(x) Br(x) ˆBr(x)

n Since |Du| ∈ Ln, 1 ∈ L n−1 , we have by Hölder

1 1 n n n n n−1 n n Cr Cr |Du| λ (B (x)) n = Cr |Du| ˆ 6 ˆ r ˆ Br(x) Br(x) | {z } Br(x) ≈rn−1 n 6 Cr kDukLn(Rn)

Hence [u]BMO 6 CkDukLn(Rn). (c) Easy/typos...

B.6 Solution to Exercise Sheet 7 B.6.1 Exercise 1 1,p (a) For 1 6 p < ∞, kukLp(U) 6 Cpk∂νukLp(U) for all u ∈ W0 (U), Cp = C(p, U) > 0. Use p Pn ∂u p p the norm |Du(x)|p = | | – ` -norm. i=1 ∂xi

∞ Choose coordinates such that ν = en. First suppose u ∈ Cc (U). Then if x ∈ U, 0 1 1 x = (x1, . . . , xn−1) and p + q = 1.

x x 1 n d 1 n ∂ p p 0 q |u(x)| = u(x , t) dt 6 (xn + d) u(x, t) dt ˆ−d dt ˆ−d ∂xn Hence

d p ∂ p p q |u(x)| dx 6 (xn + d) u(x, t) dt dx ˆU ˆU ˆ−d ∂xn

d p d q p 0 6 (xn + d) dxn |Du(x, t)|p dt dx ˆRn−1 ˆ−d ˆ−d p (2d) p = |Du(x)|p dx p ˆU

p (2d)p p 2 2 i.e. kuk k|Du| k . For p = 2 we get k|Du| k = kDuk and V ertuk 2 √ Lp 6 p p Lp 2 L2 L2 L 6 1,p ∞ 2dkDukL2 . For u ∈ W0 (U) use density of Cc (U) (easy). Equivalence of norms 1,p kDukLp and kukW 1,p on W0 is also easy. (b) Let

n X B[u, v] = a u v + cuv ˆ ij xi xj U i,j=1

1 ∞ P 2 where u, v ∈ H0 (U), aij, c ∈ L (U), c > 0 a.e., aij = aji and i,j aij(x)ξiξj > θ|ξ| n for all ξ ∈ R a.e. in U. Then

(a) B[·, ·] is linear X

81 (b) It is bounded X (c) It is coercive since

X 2 B[u, u] = aijuxi uxi + x|u| dx Û 2 2 > θ |Du| + c|u| dx Û Û 2 2 If (i) holds then this is bigger or equal to θkDukL2 > θCpkukW 1,2 (by (a)). If (ii) 2 holds then this is bigger or equal to min{θ, c0}kukW 1,2 . So B is coercive.

B.6.2 Exercise 2

1 For u, v ∈ H (U), deﬁne a(u, v) = U Du·Dv +uv dx. a(·, ·) is clearly symmetric, bilinear, and ´ 2 2 2 a(u, u) = |Du| + |u| dx = kukH1 ˆU Hence, a(·, ·) is an inner product on H1(U) and H1(U) is complete with respect to the norm induced by it. Since V is closed, (V, a) is a Hilbert space, hence by Riesz’ Representation ∗ Theorem, since v 7−→ fv ∈ V , there is a unique u ∈ V such that a(u, v) = U fv for all v ∈ V . ´ ´

B.6.3 Exercise 3

1 u solved (3) if and only if w := u − g ∈ H0 (U) solves 1 0 B[w, v] = hf, vi − B[g, v] ∀v ∈ H0 (U) (3 ) Note that v 7−→ hf, vi−B[g, v] ∈ H−1(U). By Lax-Milgram and Ex. 1(b) there is a unique 1 0 element w ∈ H0 (U) that solves (3 ). Then u := w + g solves (3) uniquely.

B.6.4 Exercise 4 (a) Assume also that U is connected,

(⇒) Let u ∈ H1(U) be a weak solution to Neumann’s Problem, take v = 1 ∈ H1(U). Then

0 = DuDv dx = f ˆU ˆU

(⇐) Suppose U f = 0. Now consider ´ 1 V := u ∈ H (U) u dx = 0 ˆU 1 V is a closed subspace of H (U). Claim: (u, v)1 := U Du · Dv dx is an inner product on V (with norm equivalent to k · kH1 . ´ – (·, ·)1 symmetric, bilinear X – Let (u, u)1 = 0. Then

2 1 2 0 = (u, u)1 = |Du| dx > |u| dx ˆU c ˆU Hence u = 0.

82 And

2 2 2 2 kukH1 = |u| + |Du| 6 (1 + c) |Du| = (1 + c)(u, u)1 ˆ ˆ ˆU 2 2 = (1 + c) |Du| dx 6 (1 + c)kukH1 ˆU

Hence, by Riesz, there is a unique u ∈ V , such that (u, v)1 = U fv for all v ∈ V . 1 Let v ∈ H (U). Then v − U v ∈ V . So ´ ﬄ (u, v) = (u, v − v) + (u, v) 1 1 1

= f(v − v) = fv dx − v fU ˆU ˆU U ˆ

(b) How can we define weak solutions? Suppose u ∈ C2(U) solves (4). Then for v ∈ C∞(U) we have ∂u − ∆uv dx = Du · Dv − v dS = Du · Dv + uv dS = fv dx Û Û ˆ∂U ∂ν Û ˆ∂U Û

Hence the weak formulation is: u is a weak solution to Robin’s Problem iﬀ

Du · Dv dx + (T u)(T v) dS = fv dx ∀v ∈ H1(U) (4.1) ˆU ˆ∂U ˆU For given f ∈ L2(U), take the bilinear form

B[u, v] = Du · Dv dx + (T u)(T v) dS u, v ∈ H1(U) ˆU ˆ∂U For existence and uniqueness via Lax-Milgram, we need to show

• B[·, ·] is bilinear X • B[·, ·] X 1 • B[·, ·] coercive: Assume not. Then for all j ∈ N there is uj ∈ H (U) such that 1 B[u, u] < ku k2 j j H1

uj 1 Take vj := . Then (vj) ⊆ H (U) with kujkH1 = 1 for all j. Then kuj kH1 1 1 2 1 B[vj, vj] < j . By Rellich-Kondrachov (H ⊂⊂ L , when ∂U ∈ C ), there is 2 2 a subsequence (vjk ) and v ∈ L with vjk → v in L (U). Hence

2 2 1 B[vjk , vjk ] = |Dvjk | + |T vjk | dS < ˆU ˆ∂U jk

2 n 1 So Dvjk → 0 in L (U, R ). HEnce v ∈ H (U) with Dv = 0. So v is constant, has 1 norm 1 but T v = 0 so v ∈ H0 (U) so v = 0 . So B[·, ·] is coercive and a weak solution exists.

83 B.7 Solution to problem sheet 8 B.7.1 Exercise 1 1 Take U = B(0, 4 ) and n = 2 and let p u(x, y) = (x2 − y2 log | log x2 + y2| Clearly u ∈ C∞(U \{0}). In polar coordinates x = r cos θ, y = r sin θ we have u(r, θ) = r2(cos2 θ − sin2 θ) ln | ln r| = r2(1 − 2 sin2 θ) ln | ln r|

1 (1) Show u ∈ C(U). We need to show that u is bounded as r → 0. For 0 < r < 1, r , 1 1 r 1 1 1 1 r 6 e . So − ln r 6 r , i.e. | ln r| 6 r . So ln | ln r| 6 ln r = − ln r 6 r . Hence 1 |u(r, θ)| 3r2 · −−→t→0 0 6 r So u ∈ C(U). (2) ∆u ∈ C(U). In polar coordinates we have 1 1 ∆u = u + u = u + u + u xx yy rr r r r2 θθ Also we have 1 u = (1 − 2 sin2 θ)r(2 ln | ln r| + ) r ln r 3 1 u = (1 − 2 sin2 θ)(2 ln | ln r| + − rr ln r (ln r)2 2 uθ = −4 sin θ cos θr ln | ln r| 2 2 uθθ = −4(1 − 2 sin θ)r ln | ln r| Hence 4 1 ∆u = ··· = (1 − 2 sin2 θ) − −−−→r→0 0 ln r (ln r)2 bad terms ln | ln r| cancel...

2 (3) u∈ / C (U). If we just consider uxx (say), ln | ln r| the bad terms don’t cancel, so u(r, θ) → ±∞ as r → 0. (4) u is a weak solution of Poissons equation on U. Deﬁne f as −∆u, as calculated in step 2. We have shown that u is a strong solution to −∆u = f on U \{0}. Now we show that it is a weak solution on all of U. We need

Du · Dϕ dx = fϕ dx Û Û 1 ∞ ∞ ϕ ∈ H0 (U). By density, it suffices to have it for Cc (U). We have u ∈ C (U \ Bε) for any ε > 0, so

Du · Dϕdx = Du · Dϕ dx + Du · Dϕ dx Û Û\Bε ˆBε ∂u = − ∆uϕ dx + ϕ dS + Du · Dϕ dx (∗) Û\Bε ˆ∂Bε ∂r ˆBε | {z } | {z } | {z } (1) (2) (3)

84 Now for (1) we have

∆u(x)ϕ(x)χU\Bε (x) = f(x)ϕ(x)χU\Bε (x) → f(x)ϕ(x)

for all x ∈ U \{0}. So by DCT (1) → U f(x)ϕ(x) dx. Further ´ ∂u 1 (2) 6 kϕk∞ | |dS 6 ωnkϕk∞ε · 3ε(2 ln | ln ε| + ) → 0 ε → 0 ˆ∂Bε ∂r ln ε Similarly

(3) 6 kDϕk∞ |ux| + |uy|d(x, y) 6 C |ur| + r|uθ| dx ˆBε ˆBε ε 1 = Cε r(2 ln | ln r| + ) + r3 ln | ln r| dr → 0 ε → 0 ˆ0 ln r So, letting ε → 0 in (∗), we are done.

∞ (5) u, f ∈ Cc(U): easy, Multiply u by a cut-oﬀ-function ψ ∈ Cc (U) such that ψ ≡ 1 on 1 B(0, 8 ) Then (recall only problem is at 0) we apply the above argument to u˜ = ψ · u, so

( 1 ˜ ∆u on B(0, 8 ) f = 1 ∆(uψ) on U \ B(0, 8 ) where ∆(uψ) is smooth away from 0.

B.7.2 Exercise 2 (a) Let A: H −→ H be a map (not necessarily linear!) satisfying

(i) kA(x) − A(y)k 6 γkx − yk (A continuous) 2 (ii) (x − y, A(x) − A(y)) > αkx − yk (A strongly monotone).

for all x, y ∈ H, some α, γ > 0. Then for all f ∈ H there is a unique uf ∈ H such that A(uf ) = f (A surjective). Fix f ∈ H and let R(v) := v − λA(v) + λf, where v ∈ H, λ > 0 to be ﬁxed later. We aim to show R is strongly contractive, and hence has a unique ﬁxed point. Let v, w ∈ H, then

kR(v) − R(w)k2 = (v − λA(v) + λf − w + λA(w) − λf, v − λA(v) + λf − w + λA(w) − λf) = (v − w − λ(A(v) − A(w)), v − w − λ(A(v) − A(w))) = kv − wk2 − 2λ(v − w, A(v) − A(w)) + λ2kA(v) − A(w)k2 2 2 2 2 6 kv − wk − 2λαkv − wk + λ γkv − wk = (1 − 2αλ + γλ2)kv − wk2

choose λ such that this is in (0, 1). Hence, by Banach’s Fixed Point Theorem, there is a unique u ∈ H such that R(u) = u. So u − λA(u) + λf = u So A(u) = f.

∗ (b) Suppose B : H × H −→ R satisﬁes:for all v ∈ H, w 7−→ B[v, w] ∈ H (2.5) (bounded linear functional) and

(1) |B[u, v] − B[u2, v]| 6 βku1 − u2k · kvk

85 2 (2) B[u1, u2 − u2] − B[u2, u1 − u2] > Cku1 − u2k ∗ Then for all f ∈ H there is a unique uf ∈ H such that B[u, w] = hf, wi for all w ∈ H (similar to Lax-Milgram, but B nonlinear in the 1st variable).

By (2.5), for all v, by Riesz’ there exists a unique v0 ∈ H such that B[v, w] = (v0, w). For all w ∈ H. Write A(v) = v0. Now show that A satisﬁes (1) and (2).

(a) Write (since X Banach space, kxk = supϕ∈X∗,kϕk=1 |ϕ(x)|): kA(x) − A(y)k = sup |(A(x) − A(y), w)| kwk=1 = sup B[x, w] − B[y, w]| kwk=1 6 βkx − yk (b) Here, we have (x − y, A(x) − A(y)) = (A(x), x − y) − (A(y), x − y) 2 = B[x, x − y] − B[y, x − y] > Ckx − yk Hence, by (a), A is onto. For f ∈ H∗, by Riesz there is a unique g ∈ H such that hf, wi = (g, w) for all w ∈ H. Since A is onto, there is a ug ∈ H such that hf, wi = (g, w) = (A(ug), w) = B[ug, w] for all w ∈ H.

B.7.3 Exercise 3

1 The weak formulation is B[u, v] = U fv dx for all v ∈ H0 (U) where ´ n X B[u, v] = a u v + g(u)v ˆ ij xi xj U i,j=1 This is nonlinear in u! We shall use 2 (b). We need to show (2.5), (3) and (4). 1 • For (2.5), clearly B[·, ·] linear in the second argument. Also for ﬁxed u ∈ H0 (U): n X |B[u, v]| 6 kaijkkukH1 kvkH1 + |g(u)v| dx ˆU i,j=1 | {z } =:I

Since g is Lipschitz, |g(t) − g(0)| 6 CL|t|, so g(t)| 6 C(1 + |t|) for all t ∈ R, g has linear growth (at inﬁnity). So, since U is bounded

I 6 C |v| + |u| · |v| 6 CkvkL2 + kukL2 kvkL2 ) Û 1 ∗ So |B[u, v]| 6 C(1 + kukH1 )kvkH1 . So B[u, ·] ∈ H (U) . • For (3) we have X |B[u, v] − B[u2, v] = aij(u1 − u2)xi vxj + (g(u1) − g(u2))v Û Û

6 Cku1 − u2kH1 kvkH1 + CL |u1 − u2| · |v| ˆU | {z } 6CLku1−u2kL2 kvkL2

So |B[u1, v] − B[u2, v]| 6 Cku1 − u2kH1 kvkH1 .

86 • For (4) we have since g is increasing

X B[u1, u1 − u2] − B[u2, u1 − u2] = aij(u1 − u2)xi (u1 − u2)xj dx + (g(u1) − g(u2))(u1 − u2) dx Û Û | {z } >0 2 2 > θ |D(u1 − u2)| dx > Cpθku1 − u2kH1 Û by Poincaré.

−1 Hence, the Theorem applies provided v 7→ U fv is in H (U). Lastly, we show that for m 1 ∗ 1 1,2 p f ∈ L (U), v 7−→ U fv dx ∈ (H0 (U)) , for´ v ∈ H0 (U). Note W (U) ,→ L (U) for ∗ 2n p 2n p ∗ ∼ q 1 6 p 6 2 = n−2 .´ So v ∈ L (U) for all 1 6 p 6 n−2 . Remember (L ) = L for 1 1 q 2n 1 < p 6 ∞ whenever p + q = 1. So we need f ∈ L (U). The lowest q is when p = n−2 so 1 n−2 2n q = 1 − 2n , q = n+2 .

B.8 Solution to Exercise Sheet 9 B.8.1 Exercise 1

1 n Theorem B.6. Suppose u ∈ H (R ) has compact support and is a weak solution to n 2 n 0 −∆u + g(u) = f in R (1) where f ∈ L (R ), g : R −→ R smooth, g(0) = 0, g > 0. 0 2 n Assume additionally g is bounded. Then u ∈ H (R ).

Note: The question as is statet in Problem Sheet/Evans PDE is wrong.

Proof. u is a weak solution fo (1) means

1 n 1 Du · Dv dx = fv − g(u)v dx ∀v ∈ H0 (R ) = H (R) (2) ˆRn ˆRn ˆRn

−h h Take v = −Dk (Dk u. Then (2) becomes

−h h −h h −h h − DuD(Dk (Dk u)) dx = − fDk (Dk u) + g(u)Dk (Dk u) ˆRn ˆRn ˆRn | {z } | {z } | {z } =:A =B1 =:B2 We now have

A = − Du(D−h(Dh(Du))) dx ˆ k k h h h 2 = Dk (Du)Dk (Du) dx = |Dk (Du)| dx ˆRn ˆRn Further

−h h |B1| 6 |f| · |Dk (Dk u)| dx ˆRn −h h 2 C 2 6 ε |Dk (Dk u)| dx + |f| dx ˆRn ε ˆRn h 2 C 2 6 C1ε |Dk (Du)| dx + |f| dx ˆRn ε ˆRn

87 For B2 note that u(x) 0 0 |g(u(x))| = g (t) dt 6 |u(x)|kg k∞ ˆ0 2 0 2 2 n So |g(u(x))| kg k kuk 2 . So g(u) ∈ L ( ). Note that if g(0) 6= 0 then g(u(x)) = Rn 6 ∞ L R ∞,´ since u has compact support. So ´

−h h h 2 C 0 2 2 |B2| 6 |g(u)| · |Dk (Dk u)| 6 C2ε |Dk (Du)| + kg k∞ · kukL2 ˆRn ˆRn ε 1 Take ε small enough so that (C1 + C2)ε = 2 . Then

1 h 2 C 2 0 2 2 |Dk (Du)| 6 kfkL2 + kg k∞ · kukL2 2 ˆRn ε 1 n 2 n So Du ∈ H (R ), i.e. u ∈ H (R ). By (2), g(u)v = fv − Du · Dv for all v ∈ H1( n). So T ∈ H−1( n), where Rn Rn R R hT, vi :=´ g(u)v. Then´ ´ ´ kT kH−1 = sup |hT, vi| = sup |(f, v) − Du · Dv| 6 kfkL2 + kDukL2 1 n 1 n ˆ v∈H (R ) v∈H (R ) kvkH1 =1 kvkH1 =1 and

−h h |B2| 6 kDukL2 + kfkL2 kDk (Dk u)kH1 not L2

B.8.2 Exercise 2

(a) u measurable means there is a sequence of simple functions (sj)j such that sj(x) → u(x) a. e. and if (uj) ⊆ M(U) and uj(x) → u(x) a.e. then u ∈ M(U).

Suppose u = χA and A ∈ B(U). Then if B ∈ B(R): (Nu)−1B) = {x ∈ U | f(x, u(x)) ∈ B} = ({x ∈ U | f(x, 0) ∈ B} ∩ Ac) ∪ ({x ∈ U | f(x, 1) ∈ B} ∩ A) ∈ B(U)

Pk Now suppose u is simple, i.e. u(x) = i=1 biχAi (x), bi ∈ R, Ai ∈ B(U), pairwise disjoint. Then similarly

k k −1 [ [ c (Nu) (B) = ({x ∈ U | f(x, bi) ∈ B} ∩ Ai) ∪ {x ∈ U | f(x, 0) ∈ B} ∩ ( Ai) ) i=1 i=1

So u is measurable if simple. For general u, there is a sequence (sj) of simple functions, such that sj(x) → u(x) a.e. Then since f(x, ·) is continuous

N(sj)(x) = f(x, sj(x)) → f(x, u(x)) = (Nu)(x) a.e. So Nu is measurable.

α β β β β (b) If u ∈ L (U), n |Nu| 6 n |g(x)| + |u(x)| < ∞. So Nu ∈ L (U). To show R R α α that N is continuous.´ Suppose´ (uj), u ⊆ L (U), uj → u in L (U). Need to show β Nuj → Nu in L (U).

88 Theorem B.7 (Generalised Dominated Convergence Theorem). If uj(x) → u(x) a.e. 1 1 and gj, g ∈ L (U) such that |uj(x)| 6 gj(x) a.e. and gj → g ∈ L (U). Then uj → u in L1(U).

α Since uj → u in L there is a subsequence ujk such that ujk → u a.e. Then

β β |Nujk − Nu| = |f(x, ujk (x)) − f(x, u(x))| dx =: hjk (x) dx ˆU ˆU ˆ

Then hjk → 0 a.e. since f(x, ·) is continuous. Further

β α α |hjk (x)| 6 C |g(x)| + |ujk (x)| + |u(x)|

| {z 1 } =:Gjk (x)∈L (U)

1 β α and Gjk → G in L , where G(x) = C(|g(x)| + 2|u(x)| ). Hence, by GDCT, Nujk → β 1 Nu in L (U) since hjk → 0 in L . This is true for any subsequence, hence the full β sequence Nuj → Nu in L (U). (c) We have

u(x) |F (x, u(x))| dx = | f(x, s) ds|dx Û Û ˆ0 u(x) u(x) p 6 |f(x, s)|ds dx 6 (g(x) + c|s| ) ds dx Û ˆ0 Û ˆ0 u(x) = u(x)g(x9 dx + c |s|p ds dx Û Û ˆ0 sp+1 u(x) = u(x)g(x) dx + c dx Û Û p 0 c p+1 6 u(x)g(x) dx + |u(x)| dx Û p + 1 Û | {z } | {z } I1 =:I2

1 n 2∗ ∗ 2n If u ∈ H0 (R ) then by Sobolev embedding, u ∈ L (U), where 2 = n−2 . For I1 < ∞, 2∗ ∗ 2n ∗ we need g ∈ (L (U)) = L n−2 . For I2 < ∞, need p + 1 6 2 . Hence Φ(u) is well-deﬁned (into R).

B.8.3 Exercise 4 Want m large enough so that Hm+2(U) ,→ C2(U). We want k large enough so that n n n k − b 2 c − 1 > 2 and k > 2 + b 2 c in Morrey’s Theorem. So m = 1 + b 2 c – integer part. m+1 m 2 aij, bi, c ∈ C (U), f ∈ H (U) and ∂U is C gives a weak solution u is a classical solution.

B.8.4 Solution to Problem Sheet 10 B.8.5 Exercise 1

1 Take a cut-oﬀ function η ∈ Cc (BR) such that η = 1 on Br and η = 0 oﬀ BR. Wlog 2 2 1 2 |∇η| 6 R−r . Then v := η u ∈ H0 (BR) and Dv = η Du + 2ηuDη. Since u is a weak

89 solution to −∆u = f, we have

1 Du · Dv = fv ∀v ∈ H0 (U) ˆBR ˆBR i.e.

η2|Du|2 dx + 2 ηuDη · Dudx = fη2udx ˆBR ˆBR ˆBR | {z } | {z } | {z } =:A1 =:A2 =:B Then

|A | 2 |η| · |Du| · |u| · |Dη|dx 2 6 ˆ BR 2 2 1 2 2 6 ε η |Du| + |u| |Dη| ˆBR ε ˆBR 1 Take ε = 2 and we have

1 2 2 4 2 |A2| 6 η |Du| + 2 |u| 2 ˆBR (R − r) ˆBR Further ε 1 |B| |f| · |η| · |η| · |u| dx η2f 2 + η2u2 6 ˆ 6 ˆ ˆ BR 2 BR 2ε BR Take ε = (R − r)2 then

2 (R − r) 2 2 1 2 2 |B| 6 η f + 2 η u 2 ˆBR 2(R − r) ˆBR

We have A1 + A2 = B, so |A1| 6 |B| + |A2| i.e. 2 2 2 1 2 2 4 2 (R − r) 2 2 1 2 2 η |Du| 6 η |Du| + 2 u + η f + 2 η u ˆBR 2 ˆBR (R − r) ˆBR 2 ˆBR 2(R − r) ˆBR So 2 2 2 2 2 4 1 2 |Du| 6 η |Du| 6 (R − r) f + 2 2 + 2 u ˆBr ˆBR ˆBR (R − r) 2(R − r) ˆBR Hence

2 2 2 9 2 kDuk 2 (R − r) kfk 2 + kuk 2 L (Br) 6 L (BR) (R − r)2 L (BR)

B.8.6 Exercise 2

Theorem B.8 (Conpanato’s Characterisation of Hölder continuity). Let P > 1, n < λ < n n n n+p. Let U ⊆ R be open, bounded and such that there is δ > 0 with λ (B(x, r)∩U) > δr p 0,γ λ−n for all x ∈ U, r > 0 (2.1) (this forbids cusps...). Then u ∈ L (U) is in C (U) for γ = p iﬀ there is K < ∞ such that p p u(x) − u dx 6 K rλ ∀x0 ∈ U, r > 0 ˆB(x0,r)∩U B(x0,r)

90 0,γ Proof. (⇒) Suppose u ∈ C (U). Then ﬁx x0 ∈ U, r > 0. For x ∈ B(x0, r) we have

u(x) − u(y) dy = u(y) − u(x) dy 6 |u(y) − u(x)| dy B(x0,r) B(x0,r) B(x0,r) γ γ 6 [u]0,γ |x − y| dx 6 (2r) kuk0,γ B(x0,r)

So p γp p n p γp+n u(x) − u dx 6 (2r) kuk0,γλ (B(x0, r) ∩ U) 6 Ckuk0,γr ˆB(x0,r)∩U B(x0,r) We haven’t used (2.1) here...

(⇐) Suppose (2.2) |u(x) − u |p dx Kprλ. We need to show u ∈ C0,γ(U). B(x0,r) B(x0,r) 6 Take x0 ∈ U, 0´< r < R. Then

p p−1 p p |uB(x0,R) − uB(x0,r)| 6 2 (|u(x) − uB(x0,R)| + |u(x) − uB(x0,r)| )

p n |uB(x0,R) − uB(x0,r)| λ (B(x0, r) ∩ U) | {z } >δrn Hence C |u − u |p (KpRλ + Kprλ) C(δ, p, K)Rλr−n B(x0,R) B(x0,r) 6 δrn >

Rλ/p R So |u − u | C . Take Rj = j → 0. Then B(x0,R) B(x0,r) 6 rλ/p R

n−λ λ−n j· p p |uB(x0,Rj ) − uB(x0,Rj+1)| 6 C2 R So for j < k:

λ−n p |uB(x0,Rj ) − uB(x0,Rk)| 6 CRj

−j So (uB(x0,Rj ))j is Cauchy, Also, taking rj = r2 we have

Rλ/p λ−n |u − u | C r p → 0 B(x0,Rj ) B(x0,rj ) 6 r j

By Lebesgues Diﬀerentiation Theorem, choosing a good representative of u uB(x0,Rj ) → u(x0) for all x0 and

91 Hence vR(x) := uB(x,R) converges uniformly to u as R → 0. Since vR is continuous, so is u.

We show that u ∈ C0,γ. Let x, y ∈ U, R = |x − y|. We have

C λ−n +n |u − u R p B(x,2R) B(y,2R) 6 λn(U ∩ B(x, 2R) ∩ B(y, 2R))

n n Since R = |x−y| we have B(x, R) ⊆ B(y, 2R). So λ (U∩B(x, 2R)∩B(y, 2R)) > δR . Take limit R → 0 in (2.5)

λ−n +n 1 |u(x) − u(y)| CR p R−n = C|x − y|γ 6 δ

B.8.7 Exercise 4 Via the Direct Method of the Calculus of Variations. We do the proof in 3 Steps.

1 (1) Show {I[u] | u ∈ H0 (U)} is bounded from below 1 (2) Take minimising sequence (uj)J ⊆ H0 (U) such that I[uj] → m := inf{I[u] | u ∈ 1 1 H0 (U)} and show that (uj) is bounded in H0 (U). Hence, by Banach Alaoglu, there is 1 1 a subsequence ujk , u ∈ H0 (U) such that ujk * u in H0 (U). 1 1 (3) Show that I is sequentially weakly lower semicontinuous in H0 (U), i.e. if vj ∈ H0 (U), 1 vj * v ∈ H0 (U), then

lim inf I[vj] I[u] j→∞ >

Then we have

m = lim inf I[uj ] I[u0] m k k > >

So u0 is a minimiser, by the Direct Method

Proof.

2 2 2 1 (1) By Poincaré there is Cp > 0 such that U |Du| > Cp U |u| +|Du| for all u ∈ H0 (U). Hence ´ ´

Cp 2 2 I[u] > |u| + |Du| − fudx 2 Û Û Cp 2 2 2 C 2 > |u| + |Du| − ε |u| − f 2 Û Û ε Û

92 Cp Take now ε = 4 to get

Cp 2 2 2 I[u] > |u| + |Du| − C f dx (∗) 4 ˆ ˆU 2 > −CkfkL2(U) > −∞

So the set is bounded from below. Let m := inf 1 I[u] > −∞. So take (uj)j ⊆ u∈H0 (U) 1 H0 (U), such that I[uj] → m. 2 (2) By (∗) we have kujkH1(U) 6 CI[uj]+CkfkL2(U) 6 K < ∞. Hence, there is a sequence 1 1 −1 ujk and H0 ∈ H0 (U) such that ujk * u in H0 (U) (i.e. for all T ∈ H (U), we have hT, uji → hT, ui.

1 (3) Show swlsc in H0 (U). Suppose vj → v.

Proposition B.9. Let X Banach and xj, x ∈ X, xj * x in X. Then lim inf kxjk > kxk.

Proof. By Hahn-Banach, there is a ϕ ∈ X∗ such that kϕk = 1 with hϕ, xi = kxk. Then lim inf kxjk > lim inf |hϕ, xji| = hϕ, i = kxk

2 1/2 1 kvk( |Du| ) is an equivalent norm in H0 (U) by Poincaré, so ´ 1 2 1 2 lim inf Dvj| > |Dv| ˆU 2 ˆU 2

Also fvj → fv, sinsce u 7−→ fv is a dual element. Here ´ ´ ´ 1 2 lim inf I[vj] = lim inf |Dvj| − fvj Û 2 Û 1 2 > lim inf |Dvj| − lim sup fvj 2 Û ˆ | {z } fv ´ 1 2 > |Du| − fv = I[v] 2 Û Û

1 Hence conclude that u0 is a minimiser of I in H0 (U).