LECTURENOTE FOR THE COURSE PARTIAL DIFFERENTIAL EQUATIONS 2, MATS340, 9 POINTS

MIKKO PARVIAINEN UNIVERSITY OF JYVASKYL¨ A¨

Contents 1. Introduction3 2. Sobolev spaces3 2.1. Notations3 2.2. Reminders (from the Measure and Integration)4 2.3. Weak derivatives6 2.4. Approximations 17 2.5. Global approximation in Sobolev space 22 k,p 2.6. Sobolev spaces with zero boundary values: W0 (Ω) 24 2.7. Properties of W 1,p(Ω), 1 ≤ p < ∞ 26 2.8. Difference quotient characterization of Sobolev spaces 29 2.9. Sobolev type inequalities 33 2.9.1. Gagliardo-Nirenberg-Sobolev inequality, 1 ≤ p < n 33 2.9.2. Poincare’s inequalities 39 2.9.3. Morrey’s inequality, p > n 43 2.9.4. Lipschitz functions and W 1,∞ 45 2.10. Compactness theorem 46 3. Elliptic linear PDEs 47 3.1. Weak solutions 49 3.2. Existence 53 3.3. Variational method 56 3.4. Uniqueness 60 3.5. Regularity 62 3.5.1. Local L2-regularity 63 3.5.2. Global L2-regularity 70 3.5.3. Local Lp-regularity 72 3.5.4. Cα regularity using De Giorgi’s method 73 3.6. Comparison and max principles 83 3.6.1. Strong maximum principle 85

Date: March 13, 2017. 1 2 PDE 2

4. Linear parabolic equations 87 4.1. Existence: Galerkin method 93 4.2. Standard time mollification 98 4.3. Steklov averages 99 4.4. Uniqueness 100 4.5. Cα regularity using Moser’s method 102 4.6. H¨oldercontinuity 112 4.7. Remarks 114 5. Schauder estimates 115 6. Notes and comments 126 PDE 2 3

1. Introduction This lecture note contains a sketch of the lectures. More illustrations and examples are presented during the lectures. Partial differential equations (PDEs) have a great variety of applica- tions to mechanics, electrostatics, quantum mechanics and many other fields of physics as well as to finance. In addition, PDEs have a rich mathematical theory. In the ICM at 1900, a German mathematician David Hilbert published part of a nowadays legendary list of 23 mathematical problems that have been very influential for 20th century mathematics. We are interested in particular with the problems: (1) 20th problem: Has not every regular variational problem a solu- tion provided certain assumptions regarding the given boundary conditions, and provided that, if needed, the notion of solutions shall be suitably extended? (2) 19 th problem: Are the solutions of regular problems in the calculus of variations always necessarily analytic? Comments: • Variational problems and PDEs have a tight connection. We will return to this later. • As Hilbert suggested, in most of the cases we will have to relax the definition of the solution to PDEs to obtain existence of a solution. Still we would like to preserve the uniqueness and to some extend regularity and stability. These are the question swe will deal with in this course.

2. Sobolev spaces 2.1. Notations. DOM = Lebesgue’s dominated convergence theorem, n Ω ⊂ R open set, bounded unless otherwise stated q 2 2 n |x| = x1 + ... + xn for x ∈ R , m(E) = |E| = a Lebesgue measure of a set E Z 1 Z . . . dy = . . . dy B(0,ε) |B(0, ε)| B(0,ε) f :Ω → R a function spt f = {x ∈ Ω: f(x) 6= 0} = the support of f C(Ω) = {f : f continuous in Ω} 4 PDE 2

C0(Ω) = {f ∈ C(Ω) : spt f is compact subset of Ω} Ck(Ω) = {f ∈ C(Ω) : f is k times continuously differentiable} k k C0 (Ω) = C (Ω) ∩ C0(Ω) ∞ ∞ k C (Ω) = ∩k=1C (Ω) = smooth functions

∞ ∞ C0 (Ω) = C (Ω) ∩ C0(Ω) = compactly supported smooth functions Remark 2.1. Recall that u ∈ Ck(Ω) ⇐⇒ Dαu ∈ C(Ω)

n for multi-index α = (α1, . . . , αn) ∈ N and |α| := α1 + ... + αn ≤ k, where ∂α1 ∂αn Dαu := ... . α1 αn ∂x1 ∂xn Example 2.2 (Warning). It is not always the case that spt f ⊂ Ω. Example 2.3. (1) ( x2, x ≥ 0 f : R → R, f(x) = −x2, x < 0 f ∈ C1(Ω) \ C2(Ω) (2)

( 1/(|x|2−1) n e , |x| < 1 ϕ : R → R, ϕ(x) = 0, |x| ≥ 1. ∞ ϕ ∈ C0 (Ω), spt ϕ ⊂ B(0, 1) Exercise. 2.2. Reminders (from the Measure and Integration). Let E be Lebesgue measurable, 1 ≤ p ≤ ∞, and f : E → [−∞, ∞] a Lebesgue measurable function. Then we define ( R p
1/p E |f| dx , p < ∞ ||f||Lp(E) = ess supE |f|, p = ∞. where ess sup |f| := inf{M : |f| ≤ M a.e. in E}. E PDE 2 5

Then we define Lp(E) to be a a linear space of all Lebesgue measur- able functions f : E → [−∞, ∞] for which

||f||Lp(E) < ∞. If we identify functions that coincide a.e., then this will be a Banach space with the norm defined above. We also recall

p p Lloc(E) := {f : E → [−∞, ∞]: f ∈ L (F ) for each F b E}, where b means that F is a compact subset of E. Remark 2.4. There is usually no inclusions between Lp spaces:

p q q p L * L L * L . This can be seen by recalling that xα ∈ L1((0, 1)) ⇐⇒ α > −1 xα ∈ L1((1, ∞)) ⇐⇒ α < −1. Thus if we let 1 ≤ p < q ≤ ∞ and choose β > 0 such that 1 1 − > −β > − q p we have x−β ∈ Lp((0, 1)), but x−β ∈/ Lq((0, 1)) x−β ∈/ Lp((1, ∞)), but x−β ∈ Lq((1, ∞)). Nonetheless, H¨older’sinequality is often a useful tool:

||fg||L1 ≤ ||f||Lp ||g||Lq , that is Z Z Z |fg| dx ≤ |f|p dx
1/p |g|q dx
1/q, where 1 ≤ p, q ≤ ∞ are H¨older-conjugates that is 1 1 + = 1. q p This implies, in particular, for 1 ≤ p0 < q0 ≤ ∞ and for a set |E| < ∞ that f ∈ Lq0 (E) ⇒ f ∈ Lp0 (E) 6 PDE 2 because (1 − p0/q0 = (q0 − p0)/q0)

Z 0 Z 0 0 0 Z 0 0 0 |f|p dx ≤ 1q0/(q0−p0) dx
(q −p )/q |f|q dx
p /q E E E (q0−p0)/q0 p0 ≤ |E| ||f|| 0 . Lq (E) Also the following inequalities are worth recalling. Young’s inequality: for each ε > 0, 1 < p, q < ∞, 1/p + 1/q = 1 and a, b ≥ 0 it holds ab ≤ εap + Cbq, where C = C(ε, p, q) (meaning that C depends on the quantities in the parenthesis). Minkowski’s inequality: for 1 ≤ p ≤ ∞ and f, g ∈ Lp(E) it holds that

||f + g||Lp(E) ≤ ||f||Lp(E) + ||g||Lp(E) . 1 ∞ 2.3. Weak derivatives. Let u ∈ C (Ω) and ϕ ∈ C0 (Ω). Then by integrating by parts Z ∂ϕ Z ∂u u dx = − ϕ dx, for i = 1, . . . , n. Ω ∂xi Ω ∂xi Observe that ϕ vanishes at the boundary and thus there is no boundary term above. More generally for multi-index α, |α| ≤ k, and u ∈ Ck(Ω), we have Z Z uDαϕ dx = (−1)|α| Dαuϕ dx. Ω Ω Remark 2.5. Observe that the left hand side does not require u to be continuously differentiable. This will be our starting point for defining weak derivatives for functions that are not continuous differentiable. 1 Definition 2.6. Let u, v ∈ Lloc(Ω) and α a multi-index. Then v is αth weak partial derivative of u if Z Z uDαϕ dx = (−1)|α| vϕ dx, Ω Ω ∞ for every test function ϕ ∈ C0 (Ω). We denote Dαu := v. We denote weak partial derivatives with the familiar notation ∂u . ∂xi We also use ∂u ∂u Du = ( ,..., ) ∂x1 ∂xn PDE 2 7 for the weak gradient. Example 2.7. ( x, 0 < x ≤ 1 u : (0, 2) → R, u(x) = 1, 1 < x < 2. We claim that the weak derivative is ( 1, 0 < x ≤ 1 u0 = v = 0, 1 < x < 2.

1 This is in Lloc so by definition, the task is to show that Z Z vϕ dx = −1 uϕ0 dx. (0,2) (0,2) To see this, we calculate Z Z Z uϕ0 dx = uϕ0 dx + uϕ0 dx (0,2) (0,1) (1,2) = u(1)ϕ(1) − u(0)ϕ(0) + u(2)ϕ(2) −u(1)ϕ(1) | {z } | {z } 0 0 Z Z − u0 ϕ dx − u0 ϕ dx |{z} |{z} (0,1) 1 (1,2) 0 Z = − ϕ dx (0,1) Z = − vϕ dx. (0,2) Note that above u∈ / C1((0, 2)) and u0 ∈/ C((0, 2)). Also observe that weak derivatives are only defined a.e. and thus it is irrelevant what is the point value for example at 1. We found one weak derivative but could there be several? Answer: No, weak derivatives are unique up to a set of measure zero. Theorem 2.8. A weak αth derivate of u is uniquely defined up to a set of measure zero.

1 Proof. Suppose that v, v ∈ Lloc(Ω) satisfy Z Z uDαϕ dx = (−1)|α| vϕ dx Ω Ω Z = (−1)|α| vϕ dx Ω 8 PDE 2

∞ for all ϕ ∈ C0 (Ω). It follows that Z (v − v)ϕ dx = 0 Ω ∞ for every ϕ ∈ C0 (Ω). This implies that v = v a.e. by the following reason: 0 ∞ 0 1 0 Let Ω b Ω and observe that C0 (Ω ) is dense in L (Ω ). Indeed, then there exists ∞ 0 ϕi ∈ C0 (Ω ), |ϕi| ≤ 2 such that 0 ϕi → sign(v − v) a.e. in Ω , (more about approximations later) where  1 x > 0  sign(x) = 0 x = 0 −1 x < 0. Then Z 0 = lim (v − v)ϕi dx i Ω0 DOM, below Z = lim(v − v)ϕi dx Ω0 i Z = (v − v) sign(v − v) dx Ω0 Z = |v − v| dx, Ω0 1 0 where the use of DOM is based on |(v − v)ϕi| ≤ 2(|v| + |v|) ∈ L (Ω ). 0 0 This implies that v = v a.e. in Ω , for any Ω b Ω, and thus a.e. in Ω.  The above proof also yields a useful result. 1 Lemma 2.9 (Fundamental lemma in calc var). If f ∈ Lloc(Ω), and Z fϕ dx = 0 Ω ∞ for every ϕ ∈ C0 (Ω), then f = 0 a.e. Example 2.10. ( x 0 < x ≤ 1 u : (0, 2) → R, u(x) = 2 1 < x < 2. PDE 2 9

This time u0 does not exist even in the weak sense. 1 Counterproposition: Suppose that there is v ∈ Lloc(Ω) such that Z Z uϕ0 dx = −1 vϕ dx, (0,2) (0,2)

∞ for every test function ϕ ∈ C0 (Ω). Then Z Z vϕ dx = − uϕ0 dx (0,2) (0,2) Z Z = − uϕ0 dx − uϕ0 dx (0,1) (1,2) Z Z = − xϕ0 dx − 2ϕ0 dx (0,1) (1,2) Z = −ϕ(1) + 2ϕ(1) + ϕ dx (0,1) Z = ϕ(1) + ϕ dx. (0,1)

∞ Then we can choose a sequence ϕi ∈ C0 (Ω), |ϕi| ≤ 2 such that ϕi(1) = 1 and ϕi(x) → 0 if x 6= 1. We obtain the desired contradiction by calculating Z Z
0 = lim vϕi dx − ϕi dx − ϕi(1) i (0,2) (0,1) Z Z DOM
= v lim ϕi dx − lim ϕi dx − ϕi(1) (0,2) i (0,1) i

= 0 − 0 − ϕi(1) = −1. The Sobolev spaces are named after a Soviet mathematician S.L. Sobolev for his significant contributions to the theory starting 1930’s.

Definition 2.11 (Sobolev space). Let 1 ≤ p ≤ ∞ and k ∈ N.A function u :Ω → [−∞, ∞] belongs to a Sobolev space W k,p(Ω) if u ∈ Lp(Ω) and its weak derivatives Dαu, |α| ≤ k exist and belong to Lp(Ω). k,p k,p 0 The function u belongs to the local Sobolev space Wloc , if u ∈ W (Ω ) 0 for each Ω b Ω. Remark 2.12. (1) Sobolev functions are only defined up to a mea- sure zero similarly as Lp functions. (2) Notation Hk := W k,2 as well as some further variants are en- countered in the literature 10 PDE 2

Example 2.13. For the function in Example 2.7, it holds

u ∈ W 1,p((0, 2)) for every p ≥ 1 and

u∈ / W k,p((0, 2)) for any k ≥ 2.

Example 2.14.

−β n u : B(0, 1) → [0, ∞], u(x) = |x| , x ∈ R , β > 0, n ≥ 2 will be in a Sobolev space for a suitable β. When x 6= 0

∂u −β−1 xi xi (x) = −β|x| = −β β+2 ∂xi |x| |x| as well as x Du(x) = −β . |x|β+2

We aim at showing that this function satisfies the definition of the weak derivative but we will have to be careful with the singularity. Therefore ∞ let ϕ ∈ C0 (B(0, 1)) and use Gauss’ theorem Z ∂(uϕ) Z dx =
uϕνi dS B(0,1)\B(0,ε) ∂xi ∂ (B(0,1)\B(0,ε) where ν = (ν1, . . . , νn) is the outer unit normal vector of the boundary. Recalling that ϕ = 0 on ∂B(0, 1) we get

Z ∂u Z ∂ϕ Z ϕ dx = − u dx + uϕνi dS B(0,1)\B(0,ε) ∂xi B(0,1)\B(0,ε) ∂xi ∂B(0,ε) (2.1) R If we can pass to the limit ε → 0 and to show that ∂B(0,ε) uϕνi dS → 0, we are done. To establish this we estimate Z Z −β uϕνi dS ≤ ||ϕ||L∞(B(0,1)) ε dS ∂B(0,ε) ∂B(0,ε) n−1−β ≤ ||ϕ||L∞(B(0,1)) ωn−1ε → 0 PDE 2 11 as ε → 0, if n − 1 − β > 0. Next we calculate Z Z ∂u |xi| dx = β β+2 dx B(0,1) ∂xi B(0,1) |x| Z |x| ≤ β β+2 dx B(0,1) |x| Z 1 = β β+1 dx B(0,1) |x| (2.2) Z 1 Z 1 = β β+1 dS dρ 0 ∂B(0,ρ) ρ Z 1 n−2−β = β ωn−1ρ dρ 0 .1 ρn−1−β = βωn−1 < ∞, 0 n − 1 − β whenever n − 1 − β > 0. Thus, we have integrable upper bound for χ ∂u and we have B(0,1)\B(0,ε) ∂xi

Z ∂u DOM Z ∂u lim ϕ dx = lim χB(0,1)\B(0,ε) ϕ dx ε→0 B(0,1)\B(0,ε) ∂xi B(0,1) ε→0 ∂xi Z ∂u = ϕ dx B(0,1) ∂xi Similarly as in (2.2), we see that Z Z 1 n−1−β |u| dx = ωn−1ρ dρ B(0,1) 0 .1 ρn−β = ωn−1 < ∞, 0 n − β whenever n − β > 0. Thus we can again pass to the limit Z ∂ϕ Z ∂ϕ lim u dx DOM= u dx. ε→0 B(0,1)\B(0,ε) ∂xi B(0,1) ∂xi Recalling (2.1), passing to the limit ε → 0 and combining the above estimates, we deduce Z ∂u Z ∂ϕ ϕ dx = − u dx + 0 B(0,1) ∂xi B(0,1) ∂xi

∞ for all ϕ ∈ C0 (Ω). 12 PDE 2

By modifying calculation (2.2), we have ∂u n − p ∈ Lp ⇐⇒ n − p(β + 1) > 0 ⇐⇒ β < ∂xi p and n u ∈ Lp(Ω) ⇐⇒ n − pβ > 0 ⇐⇒ > β. p As a conclusion n − p u ∈ W 1,p(Ω) ⇐⇒ β < p Observe: If p ≥ n, then u∈ / W 1,p(Ω) for all β > 0. Actually, we will later see that when p > n, Sobolev functions have a H¨oldercontinuous representative. Example 2.15. A Sobolev function can be rather singular! Indeed, let n qi be a set of points with rational coordinates in B(0, 1) ⊂ R . Then for ∞ X 1 −β u : B(0, 1) → [0, ∞], u(x) = |x − q | 2i i i=1 holds n − p u ∈ W 1,p(B(0, 1)) ⇐⇒ β < . p Observe: u explodes at every rational point! Example 2.16. Without a proof, we state that Cantor function is not in W 1,1(0, 1). Theorem 2.17 (Calculation rules). Let u, v ∈ W k,p(Ω) and |α| ≤ k. Then (1) Dαu ∈ W k−|α|,p(Ω). (2) Dα(Dβu) = Dβ(Dαu) for all multi-indexes with |α| + |β| ≤ k. (3) Let λ, µ ∈ R. Then λu + µv ∈ W k,p(Ω) and Dα(λu + µv) = λDαu + µDαv.

∞ k,p (4) If ξ ∈ C0 (Ω), then ξu ∈ W (Ω) and X α Dα(ξu) = Dβξ Dα−βu β β≤α where α α! = , α! = α ! · ... · α ! β β!(α − β)! 1 n PDE 2 13

and β ≤ α means βi ≤ αi for each i = 1, . . . , n. Proof. (1) Clear. ∞ (2) Let ϕ ∈ C0 (Ω). By the first statement, the weak derivatives exist and Z Z (−1)|β| DβDαuϕ dx ϕ smooth= (−1)|α| uDβDαϕ dx Ω Ω Z def= (−1)|α|(−1)|α+β| ϕDαDβu dx Ω Z = (−1)|β| ϕDαDβu dx. Ω (3) Clear. (4) When |α| = 1, then (4) says Dα(ξu) = uDαξ + ξDαu

which follows from the definition by observing Z Z ξuDαϕ dx = uDα(ξϕ) − uϕDαξ dx Ω Ω Z Z = − ξDαuϕ dx − u(Dαξ)ϕ dx Ω Ω Z = − (ξDαu + uDαξ)ϕ dx. Ω The rest follows by induction, but details are omitted.  Remark 2.18 (Reminder). Vector space with the norm satisfying (1) 0 ≤ ||u|| < ∞ (2) ||u|| = 0 ⇐⇒ u = 0 (3) ||cu|| = |c| ||u|| for each c ∈ R (4) ||u + v|| ≤ ||u|| + ||v|| is a normed vector space. If, in addition, the space is complete, it is called Banach space. Completeness means that all of its Cauchy sequences converge. Definition 2.19 (Sobo norm). If u ∈ W k,p(Ω), we define its norm to be  1/p  P R α p |α|≤k Ω |D u| dx 1 ≤ p < ∞ ||u||W k,p(Ω) = P α  |α|≤k ess supΩ |D u| p = ∞. 14 PDE 2

Remark 2.20. The norm ||u||W k,p(Ω) is equivalent with the norm

Z 1/p X  p  |Dαu| dx if 1 ≤ p ≤ ∞. |α|≤k Ω

This further gives that in the case p = ∞ the norm ||u||W k,∞(Ω) is equivalent with α α max ess sup |D u| = max ||D u||L∞(Ω) . |α|≤k Ω |α|≤k

k,p Definition 2.21. Let ui, u ∈ W (Ω). We say that ui converges to u in W k,p(Ω) denoted by

k,p ui → u in W (Ω), if

lim ||u − ui|| k,p = 0. i→∞ W (Ω)

k,p k,p Let ui, u ∈ Wloc (Ω). We say that ui converges to u locally in W (Ω) denoted by

k,p ui → u in Wloc (Ω), if

lim ||u − ui|| k,p 0 = 0 i→∞ W (Ω ) 0 for every Ω b Ω. The space C1(Ω) is not complete with respect to the Sobolev norm: to see this approximate in Example 2.7 the weak derivative by a smooth p 1 function vi in L . Then by integrating vi, we obtain ui ∈ C ((0, 2)) so that

1,p ui → u in W ((0, 2)), but clearly u∈ / C1((0, 2)). However, the Sobolev space ’fixes’ this issue. Theorem 2.22. The Sobolev space W k,p(Ω) is a Banach space.

Proof. First we check that ||u||W k,p(Ω) is a norm.

(1) ||u||W k,p(Ω) = 0 ⇐⇒ u = 0 a.e. in Ω ”⇒”

||u||W k,p(Ω) = 0 implies that ||u||Lp(Ω) = 0 and this implies by Chebysev’s inequality (see Measure and integration 1) that u = 0 a.e. in Ω. PDE 2 15

”⇐” Suppose that u = 0 a.e. in Ω. Then Z Z 0 = uDαϕ dx = (−1)|α| 0ϕ dx Ω Ω ∞ α for every ϕ ∈ C0 (Ω), i.e. D u = 0. (2) ||λu||W k,p(Ω) = |λ| ||u||W k,p(Ω) is clear. (3) Let (1 ≤ p < ∞, if p = ∞ a similar proof applies). Then 1/p  X α α p  ||u + v||W k,p(Ω) ≤ ||D u + D v||Lp(Ω) |α|≤k Minkowski 1/p  X α α
p  ≤ ||D u||Lp(Ω) + ||D v|| Lp(Ω) |α|≤k

Minkowski for |·|p X α p
1/p X α p
1/p ≤ ||D u||Lp(Ω) + ||D v||Lp(Ω) . |α|≤k |α|≤k

k,p Next we show that if ui is a Cauchy sequence in W (Ω), then it k,p k,p converges in W (Ω) i.e. W (Ω) is complete. To this end, let ui be a Cauchy sequence in W k,p(Ω). α p Claim: D ui is a Cauchy sequence in L (Ω) for each α, |α| ≤ k.

Proof: This follows by fixing ε > 0 and observing that α α ||D ui − D uj||Lp(Ω) ≤ ||ui − uj||W k,p(Ω) < ε k,p whenever i, j are large enough, since ui is a Cauchy sequence in W (Ω)./// p p The space L is complete and thus there exists uα ∈ L (Ω) such that α p D ui → gα in L (Ω). In particular for α = 0 p ui → u in L (Ω).

α Claim: D u = gα in the weak sense. ∞ Proof: Let ϕ ∈ C0 (Ω) 1 1 + = 1, p, q ≥ 1 p q and observe that Z H¨older Z Z α p
1/p α q
1/q (u − ui)D ϕ dx ≤ |u − ui| dx |D ϕ| dx → 0 Ω Ω Ω (2.3) 16 PDE 2 by Lp convergence. Thus Z Z α (2.3) α uD ϕ dx = lim uiD ϕ dx Ω i Ω Z |α| α = lim(−1) D uiϕ dx i Ω Z sim to (2.3) |α| = (−1) gαϕ dx. Ω This completes the proof of the auxiliary claim./// α p We have shown that D u := gα ∈ L (Ω) exists and α α p D ui → gα = D u in L (Ω) as desired.  Remark 2.23 (Warning). The Sobolev space W k,p(Ω) is not compact in the sense that from

||ui||W k,p(Ω) ≤ C < ∞ (2.4) it does not follow that there would be u ∈ W k,p(Ω) and a subsequence such that

k,p ui → u in W (Ω). If this were true some existence results would be much easier. For example, the functions  0 0 < x ≤ 1  ui : (0, 2) → R, ui(x) = (x − 1)i 1 ≤ x ≤ 1 + 1/i (2.5) 1 1 + 1/i < x < 2 are in W 1,1((0, 2)) and furthermore

||ui||W 1,1((0,2)) ≤ 2. However, there is no in W 1,1((0, 2)) convergent subsequence. If there was a limit, it should be (to have even L1 convergence) ( 0 0 < x ≤ 1 u(x) = 1 1 < x < 2 but this is not in W 1,1((0, 2)). When p > 1, W k,p(Ω) is a reflexive Banach space and thus from (2.4) it follows that there is weakly convergent subsequence ui (consequence PDE 2 17 of Banach-Alaoglu’s theorem). Especially, there is the weak limit u ∈ W k,p(Ω) such that

||u|| k,p ≤ lim inf ||ui|| k,p . W i W We omit the details here but observe that (2.5) shows that this fails in the case p = 1. By modifying the example to be  0 0 < x ≤ 1  √ u (x) = (x − 1) i 1 ≤ x ≤ 1 + 1/i i √ 1/ i 1 + 1/i < x < 2,

1,2 we have ui ∈ W ((0, 2)), ||ui||W 1,2((0,2)) ≤ C and 1,2 ui → u weakly in W (Ω), where u = 0. It clearly holds that

0 = ||u|| 1,2 ≤ lim inf ||ui|| 1,2 . W ((0,2)) i W ((0,2)) Observe carefully that strong convergence does not hold

1,2 ui 9 u in W ((0, 2)). 2.4. Approximations. Below we denote

Ωε = {x ∈ Ω : dist(x, ∂Ω) > ε} which is an open set by continuity of dist(x, ∂Ω). Definition 2.24 (Standard mollifier). Let

( 1/(|x|2−1) n Ce |x| < 1 η : R → R, η(x) = 0 |x| ≥ 1 where C is chosen so that Z η dx = 1. Rn Then we set for ε > 0 1 x η (x) := η( ) ε εn ε which is called a standard mollifier. Remark 2.25. Observe that

∞ n ηε ∈ C0 (R ), spt ηε ⊂ B(0, ε) 18 PDE 2 and Z 1 Z x ηε(x) dx = n η( ) dx Rn ε Rn ε y=x/ε,εn dy= dx Z = η(y) dy = 1. Rn Definition 2.26 (Standard mollification). Let

1 f :Ω → [−∞, ∞], f ∈ Lloc(Ω). Then we define the standard mollification for f by

fε :Ωε → R, fε := ηε ∗ f, R where ηε ∗ f = Ω ηε(x − y)f(y) dy denotes the convolution for x ∈ Ωε. Theorem 2.27. The standard mollification has the following properties 1 (f ∈ Lloc(Ω) unless otherwise specified) (1) α α D fε = f ∗ D ηε in Ωε and ∞ fε ∈ C (Ωε). (2) Let f ∈ Lp(Ω). Then

fε → f a.e. in Ω. (3) If f ∈ C(Ω), then

fε → f, uniformly in compact subsets of Ω.

p 0 00 (4) If f ∈ Lloc(Ω) for 1 ≤ p ≤ ∞, then for Ω b Ω b Ω

||fε||Lp(Ω0) ≤ ||f||Lp(Ω00) for small enough ε > 0, and for 1 ≤ p < ∞ p fε → f in Lloc(Ω). Warning: The convergence does not hold for p = ∞. k,p (5) If f ∈ Wloc (Ω) for 1 ≤ p ≤ ∞, k ∈ N, then α α D fε = ηε ∗ D f in Ωε.

k,p (6) If f ∈ Wloc (Ω), for 1 ≤ p < ∞, k ∈ N, then k,p fε → f in Wloc (Ω). PDE 2 19

Proof. (1) Let

x ∈ Ωε, ei = (0,..., 1 ,..., 0) |{z} ith

and h > 0 such that x + hei ∈ Ωε. Intuitive idea is ∂f Z ∂η (x − y) ε (x) = ε f(y) dy. ∂xi Ω ∂xi To make this rigorous we would like to deduce ∂f f (x + he ) − f (x) ε (x) = lim ε i ε ∂xi h→0 h ! 1 Z Z = lim ηε(x + hei − y)f(y) dy − ηε(x − y)f(y) dy h→0 h Ω0 Ω0 Z DOM,below 1 1  x + hei − y x − y  = n lim η( ) − η( ) f(y) dy ε Ω0 h→0 h ε ε Z ∂η (x − y) = ε f(y) dy Ω0 ∂xi ∂η = ε ∗ f ∂xi (2.6)

0 where B(x + hei, ε) ∪ B(x, ε) ⊂ Ω b Ω. For this we need to calculate the limit inside the integral and to look for an integrable upper bound to be able to use DOM: Claim 1: 1  x + he − y x − y  1 ∂η x − y η( i ) − η( ) → ( ). h ε ε ε ∂xi ε

Proof: This can be seen to hold by setting x − y ψ(x) = η( ) ε and the limit is ∂ψ 1 ∂η x − y (x) = ( ). /// ∂xi ε ∂xi ε

1 x+hei−y x−y Claim 2: h η( ε ) − η( ε ))f(y) has an integrable upper bound in Ω0. Proof: 20 PDE 2

Z h ∂ ψ(x + hei) − ψ(x) = ψ(x + tei) dt 0 ∂t Z h = Dψ(x + tei) · ei dt 0 Thus

|ψ(x + hei) − ψ(x)| ≤ h ||Dψ||L∞(Ω) and

1 x + hei − y x − y 1 0 η( ) − η( ))f(y) ≤ ||Dψ|| ∞ |f(y)| ∈ L (Ω ). /// h ε ε L (Ω) Thus the use of DOM in (2.6) was correct and the proof is complete. A similar argument shows that for every multi-index α α, D fε exists and α α D fε = D ηε ∗ f. Moreover, the convolution on the RHS is continuous (ex). Now, ∞ repeating the argument for the higher derivatives of fε ∈ C (Ωε) implies the result.

(2) Let x ∈ Ω0 Ω so that the convolution below is well defined b R for a small enough ε, recall n ηε dy = 1, and estimate R Z

|fε(x) − f(x)| = ηε(x − y)f(y) dy − f(x) Ω Z

= ηε(x − y)(f(y) − f(x)) dy

Ω (2.7) 1 Z ≤ ||η|| |f(y) − f(x)| dy L∞(Rn) n ε B(x,ε) Z ≤ C||η|| |f(y) − f(x)| dy →* 0 L∞(Rn) B(0,ε) a.e. in Ω, where at * we used Lebesgue’s differentiation theo- R 1 R rem. Above B(0,ε) . . . dy := |B(0,ε)| B(0,ε) . . . dy.

0 00 (3) Let Ω b Ω b Ω. Then f is uniformly continuous on a compact 00 subset Ω . Let ε > 0 be small enough so that for x ∈ Ω0 we have B(x, ε) ⊂ Ω00. By uniform continuity, for any δ > 0, there exists ε > 0 such that |x − y| < ε ⇒ |f(x) − f(y)| < δ PDE 2 21

00 for any x, y ∈ Ω . Then by this and (2.7), we have Z |f (x) − f(x)| ≤ C||η|| |f(y) − f(x)| dy ε L∞(Rn) B(x,ε) Z ≤ C||η|| δ dy L∞(Rn) B(x,ε) ≤ C ||η|| δ L∞(Rn) independent of x ∈ Ω0 for all small enough ε.

(4) Let 1 ≤ p < ∞ (bound in the case p = ∞ is straightforward) 0 00 and x ∈ Ω b Ω b Ω. Then Z

|fε(x)| = ηε(x − y)f(y) dy B(x,ε) Z 1−1/p 1/p ≤ ηε(x − y) ηε(x − y) |f(y)| dy B(x,ε) H¨older Z (p−1)/p Z 1/p    p  ≤ ηε(x − y) dy ηε(x − y)|f(y)| dy . B(x,ε) B(x,ε) | {z } 1 We apply this estimate together with Fubini’s/Tonelli’s theorem (R2n measurability ok). Thus, whenever ε > 0 is small enough, Z Z Z p p |fε(x)| dx ≤ ηε(x − y)|f(y)| dy dx Ω0 Ω0 B(x,ε) Z Z p = ηε(x − y)|f(y)| dy dx Ω0 Ω00 Z Z Fubini p = ηε(x − y)|f(y)| dx dy Ω00 Ω0 Z Z p = |f(y)| ηε(x − y) dx dy Ω00 Ω0 Z Z p ≤ |f(y)| ηε(x − y) dx dy Ω00 Rn | {z } 1 Z = |f(y)|p dy. Ω00 p It remains to show that fε → f in Lloc(Ω). Recall (not proven here) that C(Ω00) is dense in Lp(Ω00) ie. for any f ∈ 22 PDE 2

Lp(Ω00) and δ > 0, there exists g ∈ C(Ω00) such that Z |f − g|p dy
1/p < δ/3. Ω00 From this and the beginning of the proof, we deduce Z p
1/p |f − fε| dx Ω0 Minkowski Z Z Z p
1/p p
1/p p
1/p ≤ |f − g| dx + |g − gε| dx + |gε − fε| dx Ω0 Ω0 Ω0 Z Z p
1/p p
1/p ≤ δ/3 + |g − gε| dx + |g − f| dx Ω0 Ω00 Z p
1/p ≤ δ/3 + |g − gε| dx + δ/3 Ω0 ≤ δ/3 + δ/3 + δ/3, where the last inequality follows from fact we proved earlier: for continuous functions the convergence is uniform and thus Z p
1/p 0 1/p |g − gε| dx ≤ sup |g − gε||Ω | ≤ δ/3 Ω0 x∈Ω0 for small enough ε.

(5) Exercise. (6) Exercise.  2.5. Global approximation in Sobolev space. We already stated in Theorem 2.27(6) that Sobolev functions can be estimated locally by mollifying. At the vicinity of the boundary this does not hold as such since we need some space to mollify. To establish a global approx- imation the idea is to take smaller and smaller ε when approaching the boundary so that B(x, ε(x)) ⊂ Ω always holds. Theorem 2.28. Let u ∈ W k,p(Ω) for some 1 ≤ p < ∞. Then there is ∞ k,p a sequence ui ∈ C (Ω) ∩ W (Ω) of functions such that

k,p ui → u in W (Ω). Proof. We define

Ω0 = ∅

Ωi = {x ∈ Ω : dist(x, ∂Ω) > 1/i} ∩ B(0, i) PDE 2 23 and observe that Ωi are bounded sets such that Ω0 b Ω1 b ... b Ω and

∞ [ Ω = Ωi. i=1

∞ Claim: There are ξi ∈ C0 (Ωi+2 \ Ωi−1) such that

∞ X 0 ≤ ξi ≤ 1, ξi = 1 in Ω. i=1 This is called partition of unity. ˜ ∞ Proof: Clearly we can choose functions ξi ∈ C0 (Ωi+2 \Ωi−1) such that ˜ ˜ 0 ≤ ξi ≤ 1, and ξi = 1 in Ωi+1 \ Ωi.

We set ξ˜(x) ξ (x) = i , i = 1,... i P∞ ˜ j=1 ξj(x) Observe that for any fixed x ∈ Ω, only three terms in the sum will be nonzero. Similarly ξi is nonzero at the most for three indices. Then by ˜ P∞ P∞ ξi(x) i=1 ξi(x) = i=1 P∞ ˜ = 1 the claim follows./// j=1 ξj (x) We continue with the original proof. By Theorem 2.17(4) ξiu ∈ W k,p(Ω) and

spt(ξiu) ⊂ Ωi+2 \ Ωi−1.

Hence for small enough εi

∞ ηεi ∗ (ξiu) ∈ C0 (Ωi+2 \ Ωi−1) and δ ||η ∗ (ξ u) − ξ u|| ≤ . εi i i W k,p(Ω) 2i We define

∞ X v = ηεi ∗ (ξiu). i=1 24 PDE 2

Then it holds that v ∈ C∞(Ω) because at each point x ∈ Ω there are at the most three smooth functions that are nonzero in the sum. Then ∞ ∞ P ξi = 1 X X ||v − u|| = η ∗ (ξ u) − ξ u W k,p(Ω) εi i i i=1 i=1 W k,p(Ω) ∞ X ≤ ||ηεi ∗ (ξiu) − ξiu||W k,p(Ω) i=1 ∞ X δ ≤ ≤ δ. 2i  i=1 Corollary 2.29 (Approximation characterization of the Sobolev space).

u ∈ W k,p(Ω) ∞ if and only if there exists a sequence ui ∈ C (Ω) such that k,p ui → u in W (Ω). Proof. ”⇒”: This follows from the previous theorem. k,p ”⇐”: ui is a Cauchy sequence, and since W (Ω) is a Banach space k,p by Theorem 2.22, it follows that u ∈ W (Ω).  In other words: W k,p(Ω) can be characterized as a completion of ∞ ∞
C (Ω) (or C (Ω), ||·||W k,p(Ω) to be more precise).

k,p 2.6. Sobolev spaces with zero boundary values: W0 (Ω). Above, we showed that W k,p(Ω) can be characterized as a completion of C∞(Ω). By following this idea, we define Sobolev spaces with zero boundary ∞ values as a completion of C0 (Ω). k,p ∞ Definition 2.30. u ∈ W0 (Ω) if there exists a sequence ui ∈ C0 (Ω) such that

k,p ui → u in W (Ω).

k,p Remark 2.31 (Purpose). u ∈ W0 (Ω) has ”zero boundary values in the Sobolev sense”. Later, we want to set boundary values for weak solutions of PDEs: given v ∈ W 1,p(Ω), we say that u takes boundary values v in the ”Sobolev sense” if 1,p u − v ∈ W0 (Ω). Remark 2.32 (Warning). The regularity of Ω affect the outcome, 1,p and W0 (Ω) functions do not always look what one might intuitively PDE 2 25 expect by thinking smooth functions with zero boundary values. Set Ω = B(0, 1) \{0}. Then for

u :Ω → R, u(x) = dist(x, ∂B(0, 1)) = 1 − |x|

1,p it holds that u ∈ W0 (Ω) whenever p < n. Reason (with omitting some details): Choose a cut-off function ξε ∈ ∞ C0 (B(0, 1)), 0 ≤ ξi ≤ 1 such that ξε(x) = 1 in B(0, ε) and in B(0, 1)\ B(0, 1 − ε), ξε = 0 in B(0, 1 − 2ε) \ B(0, 2ε) and |Dξε| ≤ C/ε. Then ∞ (1 − ξε)u ∈ C0 (Ω) and 1,p (1 − ξε)u → u in W (Ω) as ε → 0, whenever p < n. Indeed, by MON (=Lebesgue’s monotone p convergence thm) (1 − ξε)u → u in L (Ω) and we may concentrate on ∂ ∂u p showing that ((1 − ξε)u) → in L (Ω). To see this, we calculate ∂xi ∂xi using Theorem 2.17 Z p ∂ ∂u ((1 − ξε)u) − dx Ω ∂xi ∂xi Z p ∂ξε ∂u ∂u = − u + (1 − ξε) − dx Ω ∂xi ∂xi ∂xi Z p Z p ∂ξε ∂ξε p ≤ C dx + C (1 − |x|) dx B(0,2ε) ∂xi B(0,1)\B(0,1−ε) ∂xi Z p ∂u + C ξε dx Ω ∂xi Z Z p 1−p+p p ≤ C |Dξε| dx + Cε + C ||Du||L∞(Ω) |ξε| dx B(0,2ε) | {z } Ω =1 ≤ Cεn/εp + Cε + C(2ε + (2ε)n) → 0, when ε → 0 and p < n. The problem in this example is that {0} is too small to be ”seen” by W 1,p(Ω) function when p < n. Let us also remark that Lebesgue measure is not the most accurate gauge to measure smallness of sets in the Sobolev theory. In a sense right gauge is so called p-capacity. The following lemma shows that when considering Sobolev spaces n 1,p n 1,p n over the whole R , W0 (R ) coincides with W (R ).

1,p n 1,p n Lemma 2.33. W0 (R ) = W (R ). Proof. Exercise.  26 PDE 2

2.7. Properties of W 1,p(Ω), 1 ≤ p < ∞. Lemma 2.34 (Chain rule). Let f ∈ C1( ), ||f 0|| < ∞, and R L∞(R) u ∈ W 1,p(Ω). Then ∂f(u) ∂u = f 0(u) , j = 1, . . . , n ∂xj ∂xj a.e. in Ω, and where ∂u , ∂f(u) denotes the weak derivative. ∂xj ∂xj ∞ 1,p Proof. We have proven that we can choose ui ∈ C (Ω) ∩ W (Ω) such that

1,p ui → u in W (Ω).

∞ Claim: For any ϕ ∈ C0 (Ω) Z ∂ϕ Z ∂ϕ f(u) dx = lim f(ui) dx. Ω ∂xj i→∞ Ω ∂xj

Proof: Let 1 < p < ∞ (the case p = 1 is similar). Then since 1/p + (p − 1)/p = 1, we have Z Z ∂ϕ ∂ϕ f(u) dx − f(ui) dx Ω ∂xj Ω ∂xj Z ≤ |f(u) − f(ui)||Dϕ| dx Ω H¨older Z Z p
1/p p/(p−1)
(p−1)/p ≤ |f(u) − f(ui)| dx |Dϕ| dx Ω Ω * Z Z ≤ ||f 0|| |u − u |p dx
1/p |Dϕ|p/(p−1) dx
(p−1)/p → 0, L∞(R) i Ω Ω

R u 0 0 where * follows from |f(u) − f(ui)| = f (t) dt ≤ ||f || ∞ |ui − u|./// ui L (R)

Z ∂ϕ Z ∂ϕ f(u) dx = lim f(ui) dx Ω ∂xj i→∞ Ω ∂xj Z calc for smooth functions 0 ∂ui = − lim f (ui) ϕ dx i→∞ Ω ∂xj Z * 0 ∂ui = − lim f (ui) ϕ dx Ω i→∞ ∂xj Z ∂u = − f 0(u) ϕ dx. Ω ∂xj PDE 2 27

Since the LHS above is as in the definition of the weak derivative of ∂f(u) , the proof is complete. At * we used ∂xj Z 0 ∂ui 0 ∂u (f (ui) − f (u) )ϕ dx Ω ∂xj ∂xj Z 0 ∂ui 0 ∂u 0 ∂u 0 ∂u = (f (ui) − f (ui) + f (ui) − f (u) )ϕ dx Ω ∂xj ∂xj ∂xj ∂xj Z 0 ∂ui ∂u 0 0 ∂u = f (ui)( − )ϕ + (f (ui) − f (u)) ϕ dx → 0. Ω ∂xj ∂xj ∂xj The first term converges because of H¨older’s inequality and the second p by the fact that since ui → u in L we can choose a.e. converging 0 0 0 subsequence to u. Moreover, as f is continuous, also f (ui) → f (u) a.e., and the conditions of DOM are satisfied.  1,p Theorem 2.35. If u ∈ W (Ω), then recalling u+ = max(u, 0) and 1,p u− = − min(u, 0), we have u+, u−, |u| ∈ W (Ω) and ( Du a.e. in {x ∈ Ω: u(x) > 0} Du = + 0 a.e. in {x ∈ Ω: u(x) ≤ 0} ( −Du a.e. in {x ∈ Ω: u(x) < 0} Du = − 0 a.e. in {x ∈ Ω: u(x) ≥ 0} and  Du a.e. in {x ∈ Ω: u(x) > 0}  D|u| = 0 a.e. in {x ∈ Ω: u(x) = 0} −Du a.e. in {x ∈ Ω: u(x) < 0}. Proof. We aim at using the previous theorem for a suitable f. Let (√ s2 + ε2 − ε s ≥ 0 f (s) = ε 0 s < 0.

1 It holds that fε ∈ C (R) and limε→0 fε(s) = f(s), where ( s s ≥ 0 f(s) = 0 s < 0. Also observe that ||f 0|| < ∞. ε L∞(R) Thus by Lemma 2.34 Z Z ∂ϕ 0 ∂u fε(u) dx = − fε(u) ϕ dx Ω ∂xj Ω ∂xj 28 PDE 2

∞ for every ϕ ∈ C0 (Ω). Observe that

lim fε(u) = u+ in Ω ε→0 and ( 0 1 in {x ∈ Ω: u(x) > 0} lim fε(u(x)) = ε→0 0 in {x ∈ Ω: u(x) ≤ 0}.

By DOM

Z ∂ϕ Z ∂ϕ u+ dx = lim fε(u) dx Ω ∂xj Ω ε→0 ∂xj Z ∂ϕ = lim fε(u) dx ε→0 Ω ∂xj Z prev. lemma 0 ∂u = lim − fε(u) ϕ dx ε→0 Ω ∂xj 0 Z DOM,||fε||L∞ 0} ∂xj

This proves the first part of the claim. The second and the third follow by observing

u− = (−u)+ and |u| = u+ + u−. 

Corollary 2.36. Let u, v ∈ W 1,p(Ω) and λ ∈ R. Then

min(u, v), max(u, v) ∈ W 1,p(Ω), and if Ω bounded

min(u, λ) ∈ W 1,p(Ω) and ( Du a.e. in {x ∈ Ω: u(x) < λ} D min(u, λ) = 0 a.e. in {x ∈ Ω: u(x) ≥ λ}. PDE 2 29

Proof. ( u, {x ∈ Ω: u(x) ≥ v(x)} max(u, v) = v, {x ∈ Ω: u(x) < v(x)}

( 1 2 (u + v + (u − v)), {x ∈ Ω: u(x) ≥ v(x)} = 1 2 (u + v − (u − v)), {x ∈ Ω: u(x) < v(x)} 1 = (u + v + |u − v|) 2 and 1 min(u, v) = (u + v − |u − v|). 2  Corollary 2.37. Let u ∈ W 1,p(Ω) and λ > 0. Then for  λ {x ∈ Ω: u(x) ≥ λ}  uλ := min(max(u, −λ), λ)) = u {x ∈ Ω: λ < u(x) < λ} −λ {x ∈ Ω: u(x) ≤ λ} we have

1,p uλ → u in W (Ω) when λ → ∞.

Proof. Exercise.  Theorem 2.38. If u, v ∈ W 1,p(Ω) ∩ L∞(Ω), then uv ∈ W 1,p(Ω) ∩ L∞(Ω), and ∂(uv) ∂u ∂v = v + u ∂xj ∂xj ∂xj almost everywhere in Ω. Proof. Exercise: The derivatives in the statement denote weak deriva- tives, so start from the integral definition and use similar techniques as in Lemma 2.34.  2.8. Difference quotient characterization of Sobolev spaces.

1 0 Definition 2.39. Let u ∈ L (Ω) and Ω ⊂ Ω and ei = (0,..., 1 , 0,..., 0). loc |{z} ith Then difference quotient of u to direction ei is u(x + he ) − u(x) Dhu(x) := i i h 30 PDE 2 for x ∈ Ω0 and |h| < dist(Ω0, ∂Ω). Further, we denote

h h h D u = (D1 u, . . . , Dnu).

Theorem 2.40. Let u ∈ W 1,p(Ω) for 1 ≤ p < ∞. Then there exists C = C(n, p) > 0 such that

h D u Lp(Ω0) ≤ C ||Du||Lp(Ω)

0 0 h h for every Ω b Ω and |h| < dist(Ω , ∂Ω). Here D u Lp(Ω0) := D u Lp(Ω0).

Proof. Let first u ∈ C∞(Ω) ∩ W 1,p(Ω). Then

Z h ∂ |u(x + hei) − u(x)| = u(x + tei) dt 0 ∂t Z h

= Du(x + tei) · ei dt 0 Z h ∂u(x + tei) = dt 0 ∂xi Z h ∂u(x + tei) ≤ dt. 0 ∂xi

Thus

h u(x + hei) − u(x) D u(x) = i h Z |h| 1 ∂u(x + tei) ≤ dt |h| 0 ∂xi Z |h| p 1/p H¨older  1 ∂u(x + tei)  ≤ dt |h| 0 ∂xi i.e.

Z |h| p h p 1 ∂u(x + tei) Di u(x) ≤ dt. |h| 0 ∂xi PDE 2 31

Using this Z Z Z |h| p h p 1 ∂u(x + tei) Di u(x) dx ≤ dt dx Ω0 |h| Ω0 0 ∂xi Z Z 1 p t = s|h| ∂u(x + s|h|ei) = ds dx Ω0 0 ∂xi Z 1 Z p Fubini ∂u(x + s|h|ei) = dx ds 0 Ω0 ∂xi Z p ∂u(x + s|h|ei) ≤ sup dx s∈[0,1] Ω0 ∂xi Z p ∂u(x) ≤ dx. Ω ∂xi Then we deduce the result for the full gradient n Z Z p/2 h p  X h 2 D u(x) dx = Di u(x) dx 0 0 Ω Ω i=1 Z n X h p ≤ C Di u(x) dx 0 Ω i=1 n Z X h p = C Di u(x) dx 0 i=1 Ω n Z p previous X ∂u(x) ≤ C dx ∂x i=1 Ω i Z n 2  X ∂u(x) p/2 ≤ C dx ∂x Ω i=1 i Z = C |Du(x)|p dx. Ω We assumed u ∈ W 1,p(Ω) ∩ C∞(Ω), but we can extend the result for 1,p W (Ω) by approximation.  p Theorem 2.41. Let Ω b Ω. If u ∈ L (Ω), 1 < p < ∞ and if there exists a uniform constant

h D u Lp(Ω0) ≤ C (2.8) for all |h| < dist(Ω0, ∂Ω), then u ∈ W 1,p(Ω0) and

||Du||Lp(Ω0) ≤ C 32 PDE 2 for the same constant C.

∞ 0 Proof. Let ϕ ∈ C0 (Ω ). Then Z ϕ(x + he ) − ϕ(x) u(x) i dx Ω0 h 1 Z 1 Z = u(x)ϕ(x + hei) dx − u(x)ϕ(x) dx h Ω0 h Ω0 Z Z y = x + hei 1 1 = u(y − hei)ϕ(y) dy − u(x)ϕ(x) dx h Ω0 h Ω0 Z u(x) − u(x − he ) = − i ϕ(x) dx Ω0 h Z u(x − he ) − u(x) = − i ϕ(x) dx Ω0 −h 0 for |h| so small that spt ϕ(· + hei) ⊂ Ω . Then Z Z h −h uDi ϕ dx = − (Di u)ϕ dx, (2.9) Ω0 Ω0 ”integration by parts for difference quotients”. From the assumption (2.8) it follows that −h sup Di u Lp(Ω0) < ∞, 0<|h| 1 is reflexive, there exist vi ∈ L (Ω ) and a subsequence hj → 0 such that (see Remark 2.42)

−hj p 0 Di u → vi weakly in L (Ω ). Next we check that this weak limit is a weak derivative. Recalling (2.9), it follows that Z ∂ϕ Z ∂ϕ u dx = u dx Ω0 ∂xi Ω ∂xi Z hj = lim Di ϕu dx Ω hj →0 Z DOM hj = lim Di ϕu dx hj →0 Ω Z (2.9) −hj = − lim ϕDi u dx hj →0 Ω weak convergence Z = − ϕvi dx. Ω PDE 2 33

∂u 1,p 0 As a conclusion vi = in a weak sense, and thus u ∈ W (Ω ). ∂xi Moreover, for weakly convergent sequence, we have ∂u −hj ||vi||Lp(Ω0) = ≤ lim inf Di u ≤ C. h →0 p 0 ∂xi Lp(Ω0) j L (Ω )  Remark 2.42 (Reminder).

p 0 fj → f weakly in L (Ω ) if Z Z fjg dx = fg dx Ω0 Ω0 for every g ∈ Lp0 (Ω0), where 1/p + 1/p0 = 1, 1 < p < ∞. If space is reflexive, it is weakly sequentially compact: every bounded (in the norm of the space) sequence has a weakly convergent subsequence. Moreover for this sequence

||f|| p 0 ≤ lim inf ||fj|| p 0 . L (Ω ) j→∞ L (Ω ) 2.9. Sobolev type inequalities. Study of Sobolev type inequalities is divided in three intervals of exponents: (1)1 ≤ p < n, Gagliardo-Nirenberg-Sobolev inequality (2) p = n (3) n < p ≤ ∞, Morrey’s inequality 1 R R Also recall the notation |B(x,r)| B(x,r) . . . dy = B(x,r) . . . dy.

2.9.1. Gagliardo-Nirenberg-Sobolev inequality, 1 ≤ p < n. We define a Sobolev conjugate pn p∗ = > p, n − p or in other words 1 1 1 − = . p n p∗

Motivation for this form of the Sobolev conjugate is as follows: We want to prove that an inequality of the form Z Z |u|q dx
1/q ≤ C |Du|p dx
1/p, Rn Rn 34 PDE 2

∞ n for every u ∈ C0 (R ) and constant independent of u. Then it should also hold for

∞ n uλ(x) = u(λx) ∈ C0 (R ), λ > 0. For this function Z Z q y = λx 1 q |u(λx)| dx = n |u(y)| dy Rn λ Rn and Z Z p p |Duλ(x)| dx = |λDu(λx)| dx Rn Rn Z y = λx 1 p = n−p |Du(y)| dy. λ Rn Thus we would have Z 1/q Z 1/p  1 q   1 p  n |u(y)| dy ≤ n−p |Du(y)| dy λ Rn λ Rn and constant would be independent of λ only if λn/q+1−n/p = λ0 that is 1 1 1 − = , p n q i.e. q = p∗. Next theorem shows that any function in W 1,p(Rn) can be controlled 1,p by its gradient. Later we will see that this holds in general for W0 (Ω)- 1,p n 1,p n functions (recall that W0 (R ) = W (R )). Also observe that the constant below does not depend on the function u itself.

Theorem 2.43 (Sobolev’s inequality, 1 ≤ p < n, Rn). Let 1 ≤ p < n. Then there exists C = C(n, p) such that

Z ∗ ∗ Z |u|p dx
1/p ≤ C |Du|p dx
1/p. Rn Rn for any u ∈ W 1,p(Rn). Proof. By approximation argument, as shown at the end of the proof, ∞ n we may again assume that u ∈ C0 (R ). Then Z xj ∂u u(x1, . . . , xj, . . . , xn) = (x1, . . . , tj, . . . , xn) dtj −∞ ∂xj PDE 2 35 implying Z |u(x)| ≤ |Du(x1, . . . , xj, . . . , xn)| dxj. R Multiplying we obtain n Z 1/(n−1) n/(n−1) Y   |u(x)| ≤ |Du(x1, . . . , xj, . . . , xn)| dxj j=1 R and further n Z Z 1/(n−1) Z Z 1/(n−1) n/(n−1)   Y   |u(x)| dx1 ≤ |Du| dx1 |Du| dxj dx1 R R R j=2 R n *  Z 1/(n−1) Y  Z Z 1/(n−1) ≤ |Du| dx1 |Du| dxj dx1 , R j=2 R R in * we used generalized H¨older’sinequality, Lemma 2.45, with powers Pn−1 1 i=1 n−1 = 1. We repeat the argument for x2: Z Z n/(n−1) |u(x)| dx1 dx2 R R n Z  Z 1/(n−1) Y  Z Z 1/(n−1) ≤ |Du| dx1 |Du| dxj dx1 dx2 R R j=2 R R  Z Z 1/(n−1) ≤ |Du| dx2 dx1 R R n Z  Z 1/(n−1) Y  Z Z 1/(n−1) · |Du| dx1 |Du| dxj dx1 dx2 R R j=3 R R gen H¨older  Z Z 1/(n−1) Z Z 1/(n−1) ≤ |Du| dx1 dx2 |Du| dx1 dx2 R R R R n Y  Z Z Z 1/(n−1) · |Du| dxj dx1 dx2 j=3 R R R Repeating the argument n times, we finally obtain Z Z Z Z n/(n−1) n/(n−1)   ... |u(x)| dx1 dx2 . . . dxn ≤ ... |Du| dx1 dx2 . . . dxn . R R R R This is the claim for p = 1. When 1 < p < n, we apply the estimate for v = |u|γ 36 PDE 2 where γ is to be selected. The above result yields  Z (n−1)/n |u|nγ/(n−1) dx Rn Z ≤ |D|u|γ| dx Rn Z = γ|u|γ−1|Du| dx Rn H¨older  Z (p−1)/p Z 1/p ≤ γ |u|(γ−1)p/(p−1) dx |Du|p dx Rn Rn Solving for γ so that on both sided u has a same power i.e. nγ/(n − 1) = (γ − 1)p/(p − 1) ⇐⇒ nγ(p − 1) = (γ − 1)(n − 1)p ⇐⇒ γ(pn − n − np + p) = −(n − 1)p p(n − 1) ⇐⇒ γ = . n − p Using this γ we have

 Z ∗ (n−1)/n p(n − 1) Z ∗ (p−1)/p Z 1/p |u|p dx ≤ |u|p dx |Du|p dx Rn n − p Rn Rn and since n − 1 p − 1 n − p − = n p np ∞ n we are done for C0 (R ). We complete the proof by justifying the smoothness assumption. Let 1,p n u ∈ W (R ) and ui a smooth sequence such that 1,p n ui → u in W (R ). We can also (not proven here) take a further subsequence so that

ui → u a.e.

p∗ n This ui is a Cauchy sequence in L (R ), since for any ε > 0 ∞ n ui − uj ∈ C0 (R ) ||u − u || ∗ ≤ ||D(u − u )|| ≤ ε, i j Lp (Rn) i j Lp(Rn) for all large enough i, j. Lp∗ ( n) is complete and thus there exists ∗ R u ∈ Lp (Rn) (more details at the end of the proof) such that p∗ n ui → u in L (R ). (2.10) PDE 2 37

Thus

||u|| ∗ Lp (Rn) Minkowski ≤ ||u − u|| ∗ + ||u || ∗ i Lp (Rn) i Lp (Rn) ≤ ||u − u|| ∗ + C ||Du || i Lp (Rn) i Lp(Rn) ≤ ||u − u|| ∗ + C ||Du − Du|| + C ||Du|| i Lp (Rn) i Lp(Rn) Lp(Rn) → 0 + 0 + C ||Du|| , Lp(Rn) which completes the proof, in case, we can show the following: We omitted one point above; why should Lp∗ -limit also be u? Claim: Lp∗ limit in (2.10) must be u. Reason: Assume the contrary:

p∗ n ui → g in L (R ). Choose a further subsequence

ui → g pointwise a.e. and by our earlier choices

ui → u a.e., a contradiction.  Corollary 2.44.

1,p n p n p∗ n u ∈ W (R ) ⇒ u ∈ L (R ) ∩ L (R ). Lemma 2.45 (Generalized H¨older). Let 1 1 + ... + = 1 p1 pm

p1 pm and suppose that u1 ∈ L (Ω), . . . , um ∈ L (Ω). Then m Z Z 1/pi Y  pi  |u1 · ... · um| dx ≤ |ui| dx . Ω i=1 Ω Theorem 2.46 (Sobolev’s inequality, 1 ≤ p < n, Ω). Let 1 ≤ p < n. Then there exists C = C(n, p) such that

Z ∗ ∗ Z |u|p dx
1/p ≤ C |Du|p dx
1/p Ω Ω 1,p for any u ∈ W0 (Ω). 38 PDE 2

∞ Proof. Idea. Similarly as before, we can concentrate on u ∈ C0 (Ω) and then obtain the general case by approximation. Now, u can be ∞ n extended by zero to have u ∈ C0 (R ). Then we can apply Theorem 2.43 to obtain the result.  Remark 2.47 (Warning). The above theorem does not hold without 1,p assumption u ∈ W0 (Ω) on zero boundary values: consider a constant function. Corollary 2.48. Let 1 ≤ p < n. Then there exists C = C(n, p) such that

Z ∗ ∗ Z |u|p dy
1/p ≤ Cr |Du|p dy
1/p. B(x,r) B(x,r) 1,p for any u ∈ W0 (B(x, r)). Theorem 2.49 (Sobolev’s inequality, n < p < ∞, Ω). Let n < p < ∞ and |Ω| < ∞. Then there exists C = C(n, p) such that  Z 1/p ess sup |u| ≤ C|Ω|(p−n)/pn |Du|p dx Ω Ω 1,p for any u ∈ W0 (Ω). Proof. This result is proven later in the section of Morrey’s inequality.  Corollary 2.50. Let n < p < ∞. Then there exists C = C(n, p) such that  Z 1/p ess sup |u| ≤ Cr(p−n)/p |Du|p dy B(x,r) B(x,r)  Z 1/p ≤ Cr |Du|p dy , B(x,r) 1,p for any u ∈ W0 (B(x, r)). Corollary 2.51. Let n < p < ∞. Then there exists C = C(n, p) such that  Z 1/q  Z 1/p |u|q dy ≤ Cr1−n/p+n/q |Du|p dy B(x,r) B(x,r) i.e.  Z 1/q  Z 1/p |u|q dy ≤ Cr |Du|p dy B(x,r) B(x,r) 1,p for any q ∈ (0, ∞] and u ∈ W0 (B(x, r)). PDE 2 39

The previous three results also extend to the case p = ∞. Theorem 2.52. Let p = n > 1. Then for any q ∈ (0, ∞) there exists C = C(n, q) such that  Z 1/q  Z 1/p |u|q dy ≤ Crp/q |Du|p dy B(x,r) B(x,r) i.e.  Z 1/q  Z 1/p |u|q dy ≤ Cr |Du|p dy , B(x,r) B(x,r) 1,p for u ∈ W0 (B(x, r)). Proof. Exercise.  R 2.9.2. Poincare’s inequalities. We denote uB(x,r) = B(x,r) u dy. Observe in particular that the constant in the next estimate is inde- pendent of p.

Theorem 2.53. Let Ω ⊂ Rn be an open bounded set, and 1 ≤ p < ∞. Then there is a constant C = C(n) such that Z Z |u|p dx ≤ Cp diam(Ω)p |Du|p dx, Ω Ω 1,p for every u ∈ W0 (Ω). ∞ Proof. By approximation, we may assume u ∈ C0 (Ω). Set/choose r = diam(Ω)

y = (y1, . . . , yn) ∈ Ω, n Y Ω ⊂ [yj − r, yj + r] j=1 Similarly as in the proof of Theorem 2.43

Z y1+r |u(x)| ≤ |Du(t1, x2, . . . , xn)| dt1 y1−r H¨older Z y1+r 1/p (p−1)/p p  ≤ (2r) |Du(x1, . . . , xn)| dx1 , y1−r so that Z y1+r p (p−1) p |u(x)| ≤ (2r) |Du(x1, . . . , xn)| dx1. y1−r 40 PDE 2

Using this

Z Z yn+r Z y1+r p p |u| dx ≤ ... |u| dx1 . . . dxn Ω yn−r y1−r Z yn+r Z y1+r p p ≤ (2r) ... |Du| dx1 . . . dxn yn−r y1−r Z ≤ (2r)p |Du|p dx. Ω 1,p The case u ∈ W0 (Ω) again by approximation.  For simplicity, we next work in cubes: n Q = [a1, b1] × ... × [an, bn] ⊂ R , (b1 − a1) = ... = (bn − an),

l(Q) = (b1 − a1) = side length of the cube, and l Q(x, l) = {y ∈ n : |y − x | ≤ , i = 1, . . . , n}. R i i 2 √ Observe that |Q| = ln and diam(Q) = n l.

Theorem 2.54 (1 ≤ p < ∞). Let 1 ≤ p < ∞ , Q ⊂ Rn and u ∈ W 1,p(Q). Then Z Z p p p p |u − uQ| dx ≤ l n |Du| dx. Q Q Proof. By approximation argument, we may again concentrate on u ∈ C∞(Rn). Let x, y ∈ Q and approximate

|u(x) − u(y)| ≤ |u(x) − u(x1, . . . , xn−1, yn)| + ... + |u(x1, y2, . . . , yn) − u(y)| n X Z bi ≤ |Du(x1, . . . , xi−1, t, yi+1, . . . , yn| dt. i=1 ai Thus |u(x) − u(y)|p n  X Z bi p ≤ |Du(x1, . . . , xi−1, t, yi+1, . . . , yn| dt i=1 ai n H¨older Z bi 1/p p  X (p−1)/p p   ≤ (bi − ai) |Du(x1, . . . , xi−1, t, yi+1, . . . , yn)| dt i=1 ai n * Z bi X p ≤ np−1lp−1 |Du(...)| dt i=1 ai PDE 2 41

convexity p n n p where at ∗ we used (c1 + ... + cn) = ( n c1 + ... + n cn) ≤ Pn p i=1(nci) /n. Now Z Z Z p p |u − uQ| dx = u(x) − u(y) dy dx Q Q Q Z Z p

≤ u(x) − u(y) dy dx Q Q H¨older Z Z ≤ |u(x) − u(y)|p dy dx Q Q n p−1 p−1 Z Z Z bi n l X p ≤ |Du(...)| dt dy dx |Q| Q Q i=1 ai p−1 p−1 n Z Fub+recall (...) n l X p = ln+1 |Du(z)| dz |Q| i=1 Q Z ≤ nplp |Du(z)|p dz. Q 1,p n The general case u ∈ W (R ) again follows by approximation.  Theorem 2.55 (1 ≤ p < n). Let 1 ≤ p < n and u ∈ W 1,p(B(x, r)). Then there exists a constant C = C(n, p) > 0 such that

Z 1/p∗ Z 1/p  p∗   p  u − uB(x,r) dy ≤ C |Du| dy B(x,r) B(x,r) i.e. Z 1/p∗ Z 1/p  p∗   p  u − uB(x,r) dy ≤ Cr |Du| dy . B(x,r) B(x,r) We do not prove the result in this form, but prove a weaker result in cubes with a bigger cube on the right hand side:

Theorem 2.56 (1 ≤ p < n). Let u ∈ W 1,p(2Q), Q := Q(z, l) ⊂ Rn and 2Q := Q(z, 2l). Then there exists a constant C = C(n, p) > 0 such that Z 1/p∗ Z 1/p  p∗   p  |u − uQ| dy ≤ C |Du| dy Q 2Q i.e. Z 1/p∗ Z 1/p  p∗   p  |u − uQ| dy ≤ Cl |Du| dy . Q 2Q 42 PDE 2

∞ n Proof. Let η ∈ C0 (R ) be a cut-off function such that C 0 ≤ η ≤ 1, |Dη| ≤ , l and ( 1 x ∈ Q η(x) = 0 x ∈ Rn \ 2Q. 1,p Then (u − uQ)η ∈ W0 (2Q) and Z 1/p∗  p∗  |u − uQ| dx Q spt η ⊂ 2Q Z 1/p∗  p∗  ≤ |(u − uQ)η| dx 2Q Sobo ineq Z 1/p  p  ≤ |D((u − uQ)η)| dx 2Q Z 1/p Z 1/p  p p   p p  ≤ C η |Du| dx + C |Dη| |u − uQ| dx 2Q 2Q Z 1/p Z 1/p  p  C  p  ≤ C |Du| dx + |u − uQ| dx . 2Q l 2Q

Further we may change uQ to u2Q as Z 1/p  p  |u − uQ| dx 2Q Z 1/p  p  = |u − uQ + u2Q − u2Q| dx 2Q Z 1/p Z 1/p  p   p  ≤ C |u − u2Q| dx + C |u2Q − uQ| dx 2Q 2Q Z 1/p Z Z p 1/p  p    ≤ C |u − u2Q| dx + C u2Q − u dy dx 2Q 2Q Q Poincar´e  Z 1/p ≤ Cl |Du|p dx + ... 2Q H¨older Z 1/p Z Z 1/p  p   p  ≤ Cl |Du| dx + C |u − u2Q| dy dx 2Q 2Q 2Q Poincar´e  Z 1/p ≤ Cl |Du|p dx , 2Q R R R where we used the facts that Q ≤ C 2Q and 2Q 1 dx = |2Q|. Combin- ing the above estimates, l will cancel out, and we obtain the claim.  PDE 2 43

Remark 2.57 (Warning). Global version

Z Z p p |u − uΩ| dy ≤ C |Du| dy. Ω Ω does not (in contrast with Sobolev’s inequality) hold without regularity assumptions on Ω. Exercise.

2.9.3. Morrey’s inequality, p > n.

Theorem 2.58. Let u ∈ W 1,p(Rn), p > n. Then there exists C = C(n, p) such that

|u(x) − u(y)| ≤ C|x − y|1−n/p ||Du|| Lp(Rn) for almost every x, y ∈ Rn.

∞ n 1,p n Proof. Let u ∈ C (R ) ∩ W (R ) and x, y ∈ Q := Q(x0, l). Again

Z 1 ∂u |u(x) − u(y)| = (y + t(x − y)) dt 0 ∂t Z 1

≤ Du((1 − t)y + tx) · (x − y) dt . 0

By using this

Z

|u(y) − uQ| = u(y) − u dx Q Z ≤ |u(y) − u(x)| dx Q Z Z 1

≤ Du((1 − t)y + tx) · (x − y) dt dx Q 0 n Z Z 1 def of grad 1 X ∂u ≤ ((1 − t)y + tx) |(x − y)i| dt dx |Q| ∂z i=1 Q 0 i | {z } ≤l n Z 1 Z Fub 1 X ∂u ≤ ((1 − t)y + tx) dx dt. ln−1 ∂z i=1 0 Q(x0,l) i 44 PDE 2

Then we change variables z = (1 − t)y + tx i.e. z0 = (1 − t)y + tx0 and dz = tn dx n Z 1 Z 1 X ∂u ((1 − t)y + tx) dx dt ln−1 ∂z i=1 0 Q(x0,l) i n Z 1 Z 1 X 1 ∂u ≤ (z) dz dt ln−1 tn ∂z i=1 0 Q(z0,tl) i n p H¨older Z 1 Z 1/p 1 X 1  ∂u  (p−1)/p ≤ (z) dz |Q(z0, tl)| dt ln−1 tn ∂z i=1 0 Q(z0,tl) i Q((1 − t)y + tx , tl) ⊂ Q(x , l) Z 1 0 0 n 1 (p−1)/p ≤ ||Du|| p |Q(x , l)| dt n−1 L (Q(x0,l) n 0 l 0 t Z 1 n 1 n(p−1)/p ≤ ||Du|| p (tl) dt n−1 L (Q(x0,l)) n l 0 t Z 1 (p−n)/p −n/p ≤ nl ||Du|| p t dt, L (Q(x0,l)) 0 where we also used n(p−1)/p−n+1 = (np−n−np+p)/p = (p−n)/p and n(p − 1)/p − n = (np − n − np)/p = −n/p. Since, and here we use the fact n < p, Z 1 n t−n/p dt = (1 − )−1 = p/(p − n), 0 p we get by combining the estimates that

np (p−n)/p |u(y) − uQ| ≤ l ||Du|| p p − n L (Q(x0,l)) np 1−n/p ≤ l ||Du|| p . p − n L (Q(x0,l)) To establish the final estimate, we write

|u(x) − u(y)| ≤ |u(x) − uQ| + |uQ − u(y)|

2np 1−n/p ≤ l ||Du|| p . p − n L (Q(x0,l))

n for every x, y ∈ Q. Hence, as for every x, y ∈ R there is Q(x0, l) such that l = 2|x − y| and x, y ∈ Q(x0, l), we finally have

|u(x) − u(y)| ≤ C|x − y|1−n/p ||Du|| , Lp(Rn) for u ∈ C∞(Rn) ∩ W 1,p(Rn). PDE 2 45

1,p n We extend this result to u ∈ W (R ) by approximation: Let uε be a standard mollification of u. Then by the above |u (x) − u (y)| ≤ C|x − y|1−n/p ||Du || . ε ε ε Lp(Rn) By passing to the limit ε → 0 and using the results, proved for approx- imations, we get for almost every x, y ∈ Rn (at Lebesgue points of u to be more precise) |u(x) − u(y)| ≤ C|x − y|1−n/p ||Du|| . Lp(Rn)  Remark 2.59. By Morrey’s inequality every u ∈ W 1,p(Rn) can be redefined in a set of measure zero to be H¨older-continuous. Remark 2.60. Let p > n. In the open set Ω the above only holds locally in the sense that

1,p 0,1−n/p u ∈ W (Ω) ⇒ u ∈ Cloc (Ω). Ex: Find an example showing that global implication is false. 2.9.4. Lipschitz functions and W 1,∞.

Theorem 2.61. A function u : Rn → R has a Lipschitz continuous representative if and only if u ∈ W 1,∞(Rn) . Proof.” ⇐”: Let u ∈ W 1,∞(Rn) and spt u is compact (if not, we may multiply by a cut-off function). By our results for approximations ∞ n uε ∈ C0 (R ) n uε → u a.e. R ||u || ≤ ||u|| , ε L∞(Rn) L∞(Rn) (the third one immediately follows from the def of mollification.) Thus we may estimate Z 1

|uε(x) − uε(y)| = Duε(y + t(x − y)) · (x − y) dt 0 ≤ ||Du || |x − y| ε L∞(Rn) ≤ ||Du|| |x − y|. L∞(Rn) Then, we pass to the limit ε → 0 and, since the left hand side converges almost everywhere we obtain that |u(x) − u(y)| ≤ ||Du|| |x − y|. L∞(Rn) ”⇒”: Suppose that u is Lipschitz continuous i.e. |u(x) − u(y)| ≤ L|x − y| 46 PDE 2 for all x, y ∈ Rn. We utilize the difference quotiens and estimate

−h u(x − hej) − u(x) D u(x) = ≤ L j h

−h 1 and thus D u(x) ≤ L|Ω| 2 for a bounded Ω . Since L2 is j L2(Ω) ∞ reflexive there exists a subsequence hi → 0 and functions vj ∈ L (Ω) such that

−hi 2 Dj u → vj weakly in L (Ω). Thus Z Z ∂ϕ def hi u dx = ( lim Dj ϕ)u dx Ω ∂xj Ω hi→0 Z DOM hi = lim (Dj ϕ)u dx hi→0 Ω Z −hi = lim ϕDj u dx hi→0 Ω Z = vjϕ dx Ω ∞ for every ϕ ∈ C0 (Ω). Thus ∂u = −vj in the weak sense.  ∂xj

2.10. Compactness theorem. Recall Remark 2.23 showing that Sobolev space is not compact. However, Sobolev space embeds compactly to suitable Lp spaces. This is sometimes useful for example in the exis- tence proofs. Theorem 2.62 (Rellich-Kontrachov compactness thm). Let B be a 1,p ball, ui ∈ W (B), 1 ≤ p < n and ||ui||W 1,p(B) < C < ∞ for each i = 1, 2,.... Then for each 1 ≤ q < p∗ there exists a subsequence and a limit u ∈ W 1,p(B) such that

q ui → u in L (B).

We don’t work out a detailed proof, but remark that the proof is based on the following steps: • By approximation, it holds that

q (ui)ε → ui in L (B) as ε → 0, uniformly in i. PDE 2 47

• Thus it suffices to prove the result for mollified functions. We show for mollified functions that C C |(u ) | ≤ , |D(u ) | ≤ . i ε εn i ε εn+1 • Arzela-Ascoli’s compactness result completes the proof. Remark 2.63. The case p > n is easier. Why?

3. Elliptic linear PDEs We consider the second order elliptic equations in the divergence form. Recall from PDE1 that divergence form equations have a natural physical interpretation as an equilibrium for diffusion. We consider the boundary value problem ( Lu = f in Ω u = g on ∂Ω, where Ω is a bounded open set, u :Ω → R is the (a priori unknown) solution to the problem, and g : Ω → R and f :Ω → R. Finally, L denotes a second order partial differential equation of the form n n X X Lu(x) = − Di(aij(x)Dju(x)) + bi(x)Diu(x) + c(x)u(x) i,j=1 i=1 for given coefficients aij, bi and c.

Example 3.1. Let bi = 0 and c = 0 and A(x) = [aij]i,j=1,2,...,n . Then n n n X X X Lu = − Di(aij(x)Dju(x)) = − Di( aij(x)Dju(x)) i,j=1 i=1 j=1 = − div(A(x)Du). This explains, why we say that the equation is in the divergence form.

We assume that A is a symmetric matrix ie. aij = aji. Definition 3.2 (uniformly elliptic). PDE is uniformly elliptic if there exists constants 0 < λ ≤ Λ < ∞ such that n 2 X 2 λ|ξ| ≤ aij(x)ξiξj ≤ Λ|ξ| i,j=1 for a.e. x ∈ Ω and ξ ∈ Rn. 48 PDE 2

Our standing assumptions are (unless otherwise stated) ∞ aij, ci, bi ∈ L (Ω), uniform ellipticity (3.11) A symmetric, Ω bounded. Intuitively, uniform ellipticity tells us how degenerate the diffusion determined by the diffusion coefficients to each direction can be: dif- fusion does not extinct or blow up. This helps in existence, regularity etc. Uniform ellipticity tells that real (due to symmetry) eigenvalues λi(x) of A satisfy λ ≤ λi(x) ≤ Λ. Example 3.3. x ∈ (0, 2) = Ω, b = 0 = c, a = 1 and ( 1 x ∈ (0, 1] f(x) = 2 x ∈ (1, 2). Consider the problem ( Lu = f, x ∈ Ω u(0) = 0 = u(2). Then solving formally in (0, 1] and (1, 2) as well as requiring that the solution is in C1, from the equation ( 1, x ∈ (0, 1] Lu = −u00 = 2, x ∈ (1, 2) we obtain ( 2 − x + 1.25x x ∈ (0, 1] u(x) = 2 −x2 + 2.25x − 0.5, x ∈ (1, 2). Clearly, this is not in C2. Is this a unique solution in some sense? Even more irregular examples are possible, see Example 3.12. Example 3.4. Let A = I. Then λ = Λ and n X − div(A(x)Du) = − div(Du) = − Diiu = ∆u i=1 ie. we obtain Laplacian. Example 3.5. Let α ∈ (0, 1), and

 x2+α2x2  1 2 (1 − α2) x1x2 |x|2 |x|2 A =  α2x2+x2  . (1 − α2) x1x2 1 2 |x|2 |x|2 PDE 2 49

Then 2 2 2 X 2 α |ξ| ≤ aij(x)ξiξj ≤ |ξ| , i,j=1 and α−1 u : B(0, 1) → R, u(x) = |x| x1 with x = (x1, x2) is a solution to − div(A(x)Du(x)) = 0 in a sense studied below . (Exercise) Remark 3.6. Observe that here A does not depend on u or Du. If it did, for example A = |Du|p−2I, yielding the so called p-Laplacian p−2 ∆pu := div(|Du| Du) = 0, the equation could be nonlinear. Our operator L instead is linear, a,˜ ˜b ∈ R L(˜au + ˜bv) =aLu ˜ + ˜bLv. 3.1. Weak solutions. In the spirit of Hilbert’s 20th problem, to guar- antee the existence of solutions, we can extend the class of functions to be studied. These less than C2 regular solutions are called weak solutions (in contrast with classical solutions that are C2 and satisfy the equation pointwise). We work in the spirit of Sobolev spaces, test the equation with smooth test functions and integrate by part to get rid of the second derivatives, so that only u ∈ W 1,2(Ω) is needed in the weak definition. 2 1 ∞ Let u ∈ C (Ω), aij ∈ C (Ω), f ∈ C(Ω) and ϕ ∈ C0 (Ω). Then starting from Lu = f we can calculate Z Z n n X X
fϕ dx = − Di(aijDju(x)) + bi(x)Diu + cu ϕ dx Ω Ω i,j=1 i=1 Z n n int by parts X X
= aijDjuDiϕ + bi(x)Diuϕ + cuϕ dx. Ω i,j=1 i=1 (3.12) On the other hand, if Z n n X X
0 = aijDjuDiϕ + biDiuϕ + cuϕ − fϕ dx Ω i,j=1 i=1 Z n n int by parts X X
= − Di(aijDju) + biDiu + cuϕ − f ϕ dx Ω i,j=1 i=1 50 PDE 2

∞ for every ϕ ∈ C0 (Ω), then by fundamental lemma in calc var Lemma 2.9, it holds for a.e. x ∈ Ω that n n X X − Di(aijDju) + biDiu + cuϕ − f = 0. i,j=1 i=1 Observe that the right hand side of (3.12) makes sense even with weaker assumptions, for example,

∞ 2 aij, bi, c ∈ L (Ω) and f ∈ L (Ω) and

1,2 u ∈ Wloc (Ω).

1,2 Definition 3.7 (Weak solution, local). The function u ∈ Wloc (Ω) is a weak solution to Lu = f if Z n n Z X X
aijDjuDiϕ + biDiuϕ + cuϕ dx = fϕ dx Ω i,j=1 i=1 Ω

∞ for every ϕ ∈ C0 (Ω). Remark 3.8 (Warning). This definition is useful when studying local properties such as local regularity of solutions. However, the solutions are not uniquely identified without fixing boundary values. Definition 3.9 (Weak solution to the boundary value problem). Let g ∈ W 1,2(Ω). The function u ∈ W 1,2(Ω) is a weak solution to ( Lu = f in Ω u = g on ∂Ω

1,2 if u − g ∈ W0 (Ω) and Z n n Z X X
aijDjuDiϕ + biDiuϕ + cuϕ dx = fϕ dx Ω i,j=1 i=1 Ω

∞ for every ϕ ∈ C0 (Ω). Remark 3.10. In the literature, the sums are sometimes dropped for brevity Z Z (aijDjuDiϕ + biDiuϕ + cuϕ) dx = fϕ dx. Ω Ω PDE 2 51

Example 3.11. Let us check that ( 2 − x + 1.25x, x ∈ (0, 1] u(x) = 2 −x2 + 2.25x − 0.5, x ∈ (1, 2). is a weak solution to Example 3.3. First task is to show that u ∈ 1,2 1,2 W (Ω) and u ∈ W0 (Ω), which is left as an exercise. Then we choose a symmetric cut-off function (−x + (1 + 2r))/r, x ∈ (1 + r, 1 + 2r)  (x − (1 − 2r))/r, x ∈ (1 − 2r, 1 − r) η(x) = 1, in B(1, r)  0, otherwise ∞ If ϕ ∈ C0 (Ω), then φ = (1 − (ηr)ε)ϕ is an admissible test function. Since all the ingredients are smooth, we see by integration by parts Z Z fφ dx classical= sol in spt φ au0φ0 dx Ω Ω Z 0 0 0 = u (−((ηr)ε) ϕ + (1 − (ηr)ε)ϕ ) dx. Ω By approximation results and DOM Z Z 0 0 0 0 u (1 − (ηr)ε)ϕ ) dx → u ϕ dx Ω Ω 0 1 as first ε → 0 and then r → 0. On the other hand, since |ηr| = r is symmetric and u0 ∈ C(Ω), we deduce Z 0 0 − u ((ηr)ε) ϕ dx → 0 Ω as ε → 0 and then r → 0. A more brief way to obtain the same conclusion is for ε > 0, ϕ ∈ ∞ C0 ((0, 2))) to write Z Z Z fϕ dx DOM← fϕ dx class.= sol x6=1 − (au0)0ϕ dx (0,2) (0,1−ε)∪(1+ε,2) (0,1−ε)∪(1+ε,2) Z = (au0)ϕ0 dx + a(1 − ε)u(1 − ε) − 0 + 0 − a(1 + ε)u(1 + ε) (0,1−ε)∪(1+ε,2) DOM, cancellation Z → (au0)ϕ0 dx (0,2) as ε → 0. Above the use of DOM can be justified, and at the last step cancellation a(1 − ε)u(1 − ε) − a(1 + ε)u(1 + ε) → 0 52 PDE 2 is important. Since here u ∈ C1, a ∈ C no confusion arises even if we immediately write Z Z fϕ dx = − (au0)0ϕ dx (0,2) (0,1)∪(1,2) Z = au0ϕ0 dx (0,2) + a(1)u0(1)ϕ(1) − a(0)u0(0)ϕ(0) + a(2)u0(2)ϕ(2) − a(1)u0(1)ϕ(1) Z = au0ϕ0 dx, (0,2) as we did with the weak derivatives. Look next at the example Ω = B(0, 1) ⊂ Rn, n > 1. The first task, 1,2 if no boundary conditions are considered, is to show that u ∈ Wloc (Ω). Suppose then that there would only be a singularity at the origin and everything is smooth elsewhere. We get by Gauss’ theorem that Z Z fϕ dx class.,=x6=0 − div(ADu)ϕ dx B(0,1)\B(0,ε) B(0,1)\B(0,ε) Z Z =
ADu · ν dS − ADu · Dϕ dx ∂ (B(0,1)\B(0,ε) B(0,1)\B(0,ε) Z Z = ADu · ν dS − ADu · Dϕ dx, ∂B(0,ε) B(0,1)\B(0,ε)

∞ where ϕ ∈ C0 (B(0, 1)). Then it remains to verify the following con- vergences Z ADu · ν dS → 0, ∂B(0,ε) Z Z ADu · Dϕ dx → ADu · Dϕ dx B(0,1)\B(0,ε) B(0,1) Z Z fϕ dx → fϕ dx B(0,1)\B(0,ε) B(0,1) in order to show that u is a weak solution. Example 3.12. x ∈ (0, 2) = Ω, f = 1, b = 0 = c, ( 1, x ∈ (0, 1] a(x) = 2, x ∈ (1, 2) PDE 2 53

Consider the problem ( Lu = f, x ∈ Ω u(0) = 0 = u(2). Then solving formally in (0, 1] and (1, 2) as well as requiring suitable conditions in the middle , we obtain ( x2 5 − 2 + 6 x x ∈ (0, 1] u(x) = x2 5 1 − 4 + 12 x + 6 , x ∈ (1, 2). This is not in C2 or even C1. Ex: Show that this is a weak solution to the above problem.

3.2. Existence. For simplicity, let bi = 0 and that we look for solu- tions with zero boundary values i.e. 1,2 u ∈ W0 (Ω). The Riesz representation theorem can be used to prove existence for weak solutions to n X − Di(aijDju) + cuϕ = f i,j=1 1,2 To this end, we define an inner product in W0 (Ω) by Z n X
hu, vi = aijDjuDiv + cuv dx. Ω i,j=1

Lemma 3.13. There is c0 such that if c ≥ c0, then h·, ·i is an inner 1,2 product in W0 (Ω). Proof. We intend to show that hu, ui = 0 implies u = 0 a.e. The other properties of inner product are easier. If c ≥ c0 ≥ 0, then the proof is immediate, but we can improve the bound for c0. Z n X 2
hu, ui = aijDjuDiu + cu dx Ω i,j=1 ellipticity Z 2 2 ≥ λ|Du| + c0u dx Ω Sob.-Poincar´e,Thm 2.53 Z λ 2 λ
2 ≥ |Du| + + c0 u dx Ω 2 µ2 2 ≥ α ||u||W 1,2(Ω) , 54 PDE 2

R 2 where α = min{λ/2, (c0 + λ/(2µ))}, and µ originates from Ω u dx ≤ R 2 2 µ Ω |Du| dx, µ = c diam(Ω) . Furthermore, we require c0+λ/(2µ) > 0 which gives the condition for c0.  Remark 3.14. If we set |||u||| := phu, ui, then by the above proof |||u||| ≥ c ||u|| 1,2 . On the other hand W0 (Ω) Z n 2 X 2
|||u||| = aijDjuDiu + cu dx Ω i,j=1 elliptic Z Z 2 2 ≤ Λ |Du| dx + ||c||L∞(Ω) u dx Ω Ω 2 ≤ C ||u|| 1,2 . W0 (Ω)

Thus the new norm |||·||| is equivalent to ||·|| 1,2 . W0 (Ω) ˆ 1,2 1,2 Lemma 3.15. Let W0 (Ω) be W0 (Ω) with the new inner product h·, ·i. Then Z F (v) = fv dx Ω ˆ 1,2 is a bounded linear functional in W0 (Ω). Proof. Z

|F (v)| = fv dx Ω H¨older  Z 1/2 Z 1/2 ≤ f 2 dx v2 dx Ω Ω

≤ ||f|| 2 ||v|| 1,2 L (Ω) W0 (Ω)

≤ C ||f||L2(Ω) |||v|||, where at the last step, we used the equivalence of the norms. 

Theorem 3.16. There is a constant c0 such that if c0 ≥ c then Lu = f 1,2 2 has a weak solution u ∈ W0 (Ω) for every f ∈ L (Ω). Proof. By the previous lemma Z F (v) = fv dx Ω ˆ 1,2 ˆ 1,2 is a bounded linear functional in W0 (Ω). Moreover, W (Ω) is a Banach space since the norms ||·||W 1,2(Ω) and |||·||| are equivalent. By PDE 2 55

Riesz representation theorem for Hilbert spaces, there exists a unique ˆ 1,2 u ∈ W0 (Ω) such that F (v) = hu, vi n Z X = aijDjuDiv + cuv dx Ω i,j=1 ˆ 1,2 for every v ∈ W0 (Ω). By the equivalence of norms we have shown 1,2 that there is a unique u ∈ W0 (Ω) such that n Z Z X fϕ dx = aijDjuDiϕ + cuϕ dx Ω Ω i,j=1 ∞ for every ϕ ∈ C0 (Ω).  Example 3.17. Consider Ω = (0, 2), c = 0 = b, f = 1 and ( x x ∈ (0, 1] a(x) = 1 x ∈ (1, 2) and a problem ( Lu = f, x ∈ Ω u(0) = 0 = u(2). Observe that this is not uniformly elliptic. Then by solving in (0, 1) and (1, 2) respectively the equation 1 = f = Lu = −(a(x)u0(x))0 we obtain ( −x + c1 ln(x) + c2, x ∈ (0, 1] u(x) = 1 2 − 2 x + c3x + c4, x ∈ (1, 2) Ex: Is there a weak solution? Remark 3.18. • Let g ∈ W 1,2(Ω) and consider the problem ( Lu = f in Ω u = g on ∂Ω. Then the problem ( Lv = f − Lg in Ω v = 0 on ∂Ω. 56 PDE 2

has a solution (Lg defines a bounded linear functional in the Sobolev space, and our proof extends to this setting as such), and u = v + g is a solution to the first problem. • Also observe that no regularity assumptions on ∂Ω is needed. Pn • If we included + i=1 biDiu to our operator, then L would not define an inner product. In this case, finding the element u as above is still based on Riesz representation theorem but requires more work. This is called Lax-Milgram theorem. Example 3.19. Let f ∈ L2(Ω). Then the Poisson problem ( −∆u = f in Ω u = 0 on ∂Ω has a unique weak solution. ”Research problem”: Show that there exists a solution (in a suitable p−2 sense) to −∆u = δ0, − div(A(x)Du) = δ0 and to − div(|Du| Du) = δ0. Observe that δ0 is not in the correct dual, so this requires more work. 3.3. Variational method. The existence can be shown by studying the corresponding variational integral. The variational integral related to PDE n X − Di(aijDju) + cu = f i,j=1 is n 1 Z X Z I(v) = ( a D vD v + cv2) dx − fv dx 2 ij j i Ω i,j=1 Ω Pn The PDE − i,j=1 Di(aijDju) + cu = f is called the Euler-Lagrange equation of this variational integral. Example 3.20. The variational integral corresponding to the Poisson equation −∆u = f is 1 Z Z |Dv|2 dx − fv dx. 2 Ω Ω 1,2 Definition 3.21. A function u ∈ W0 (Ω) is a minimizer to the vari- ational integral if I(u) ≤ I(v)

1,2 for every v ∈ W0 (Ω). PDE 2 57

Definition 3.22. Let g ∈ W 1,2(Ω). A function u ∈ W 1,2(Ω) with 1,2 u−g ∈ W0 (Ω) is a minimizer to the variational integral with boundary values if I(u) ≤ I(v)

1,2 1,2 for every v ∈ W (Ω) such that v − g ∈ W0 (Ω). 1,2 Theorem 3.23 (Dirichlet principle). If u ∈ W0 (Ω) is a minimizer to the variational integral I(u), then it is a weak solution to the corre- sponding Euler-Lagrange equation.

∞ Proof. Let ϕ ∈ C0 (Ω) and ε > 0. Now 1,2 u + εϕ ∈ W0 (Ω) I(u) ≤ I(u + εϕ) n 1 Z X Z = a D (u + εϕ)D (u + εϕ) + c(u + εϕ)2 dx − f(u + εϕ) dx 2 ij j i Ω i,j=1 Ω =: i(ε). We utilize the fact that if u is a minimizer, then i(ε) has a minimum at ε = 0 so that i0(0) = 0. Then n 1 Z X i(ε) = a (D uD u + εD uD ϕ + εD ϕD u + ε2D ϕD ϕ) 2 ij j i j i j i j i Ω i,j=1 1 Z Z + c(u2 + 2εuϕ + ε2ϕ2) dx − f(u + εϕ) dx. 2 Ω Ω and n 1 Z X i0(ε) = a (D uD ϕ + D ϕD u + 2εD ϕD ϕ) + c(2uϕ + 2εϕ2) dx 2 ij j i j i j i Ω i,j=1 Z − fϕ dx. Ω From this n 1 Z X Z i0(0) = a (D uD ϕ + D ϕD u) + c2uϕ dx − fϕ dx 2 ij j i j i Ω i,j=1 Ω Z n Z aij = aji X = aijDjuDiϕ + cuϕ dx − fϕ dx = 0.  Ω i,j=1 Ω 58 PDE 2

2 Lemma 3.24. Let f ∈ L (Ω). There is a constant c0 such that the 1,2 variational integral I(v) is bounded from below in W0 (Ω) if c ≥ c0. Further, we have the estimate Z Z 2 2 |Dv| dx + v dx ≤ c1 + c2I(v), Ω Ω where c1, c2 > 0 are independent of v.

R √ √ ε R 2 1 R 2 Proof. By Young’s inequality Ω | εfv/ ε| dx ≤ 2 Ω v dx+ 2ε Ω f dx, and thus ell Z λ c Z I(v) ≥ |Dv|2 + 0 v2 dx − |f||v| dx Ω 2 2 Ω Young Z λ c ε Z 1 Z ≥ |Dv|2 + 0 v2 dx − v2 dx − f 2 dx Ω 2 2 2 Ω 2ε Ω Poincar´e,Thm 2.53 Z Z Z λ 2 1 λ
2 1 2 ≥ |Dv| dx + + c0 − ε v dx − f dx 4 Ω 2 2µ Ω 2ε Ω λ where we choose c0 > −λ/(2µ) and ε such that 2µ + c0 − ε ≥ 0, so that 1,2 inequality holds for every v ∈ W0 (Ω). Recall that µ is the constant in Poincar´e’sinequality. The estimate in the claim is also build in the above proof.  Next we show existence of a minimizer. As shown above, minimizer is also a solution to the Euler-Lagrange equation. The following proof does not use Hilbert space structure (unlike the first proof) and works in the context of nonlinear equations as well.

Theorem 3.25. There is a constant c0 such that if c ≥ c0, then for any 2 1,2 f ∈ L (Ω), the variational integral I(v) has a minimizer u ∈ W0 (Ω). Proof. By the previous lemma I(v) is bounded from below and thus inf I(v) 1,2 v∈W0 (Ω) is a finite number. By the definition of inf there exists a minimizing 1,2 sequence uk ∈ W0 (Ω) such that

I(uk) → inf I(v) 1,2 v∈W0 (Ω) as k → ∞. Since the finite limit exists, we also have

I(uk) ≤ M PDE 2 59 for some M < ∞. By this and the estimate in the previous lemma, we have

Z Z 2 2 |uk| dx + |Duk| dx ≤ c1 + c2M. Ω Ω

2 Since uk and Duk are bounded in L (Ω), there is a subsequence, still denoted by uk such that

2 uk → u weakly in L (Ω), 2 n Duk → Du weakly in L (Ω) .

1,2 Since the space W0 (Ω) is closed under weak convergence so that u ∈ 1,2 W0 (Ω). Next we show

I(u) ≤ lim inf I(uk). k

To establish this, observe that a similar argument as in Lemma 3.24 implies

Z n X 2 aijDj(uk − u)Di(uk − u) + c(uk − u) dx Ω i,j=1 ell Z 2 2 ≥ λ|D(uk − u)| + c(uk − u) dx ≥ 0. Ω from which it follows that

Z n X 2 aijDjukDiuk + cuk dx Ω i,j=1 n Z X ≥ 2 aijDjukDiu + cuku dx Ω i,j=1 Z n X 2 − aijDjuDiu + cu dx. Ω i,j=1 60 PDE 2

Using this, we get Z n X 2 lim inf aijDjukDiuk + c(uk) dx k Ω i,j=1 n Z X ≥ 2 lim inf aijDjukDiu + cuku dx k Ω i,j=1 Z n X 2 − aijDjuDiu + c(u) dx Ω i,j=1 Z n X 2 = aijDjuDiu + c(u) dx, Ω i,j=1

2 since Djuk → Dju weakly in L (Ω). Combining this to the fact that weak convergence implies Z Z lim fuk dx = fu dx. k Ω Ω we obtain I(u) ≤ lim infk I(uk). Since we originally chose uk so that limk I(uk) = inf 1,2 I(v), we v∈W0 (Ω) finally obtain

I(u) ≤ lim inf I(uk) k

= lim I(uk) k = inf I(v). 1,2 v∈W0 (Ω) 1,2 Thus u ∈ W0 (Ω) is a minimizer to the variational integral.  3.4. Uniqueness. We start by showing that we can extend the class ∞ 1,2 C0 (Ω) of test functions to W0 (Ω). 1,2 Lemma 3.26. If u ∈ W0 (Ω) is a weak solution to Lu = f, then n Z X Z aijDjuDiv + cuv dx = fv dx Ω i,j=1 Ω

1,2 for every v ∈ W0 (Ω). 1,2 1,2 Proof. Let v ∈ W0 (Ω). By definition of W0 (Ω), we may take a ∞ sequence ϕk ∈ C0 (Ω) such that 1,2 ϕk → v in W (Ω). PDE 2 61

By using this, (3.11), and H¨older’sinequality, we obtain Z n X
aijDjuDiv + cuv − fv dx Ω i,j=1 Z n X
= aijDjuDi(v − ϕk) + cu(v − ϕk) − f(v − ϕk) dx Ω i,j=1 Z n X + aijDjuDiϕk + cuϕk − fϕk dx Ω i,j=1 n X Z ≤ ||aij||L∞(Ω) |DjuDi(v − ϕk)| dx i,j=1 Ω Z + |cu(v − ϕk)| + |f(v − ϕk)| dx + 0 Ω n Z 1/2 Z 1/2 X  2   2  ≤ ||aij||L∞(Ω) |Dju| dx |Di(v − ϕk)| dx i,j=1 Ω Ω Z 1/2 Z 1/2  2   2  + ||c||L∞(Ω) u dx |v − ϕk| dx Ω Ω Z 1/2 Z 1/2  2   2  + f dx |v − ϕk| dx Ω Ω → 0 as k → ∞.  1,2 Theorem 3.27 (Uniqueness). Let u1, u2 ∈ W0 (Ω) be two weak so- lutions. There is c0 such that if c ≥ c0 it holds almost everywhere that

u1 = u2. Proof. By the previous lemma, n Z X Z aijDju1Div + cu1v dx = fv dx Ω i,j=1 Ω n Z X Z aijDju2Div + cu2v dx = fv dx Ω i,j=1 Ω 1,2 for every v ∈ W0 (Ω). By subtracting the equations n Z X aijDj(u1 − u2)Div + c(u1 − u2)v dx = 0. Ω i,j=1 62 PDE 2

1,2 Now we choose v = (u1 − u2) ∈ W0 (Ω) and estimate Z n X 2 0 = aijDj(u1 − u2)Di(u1 − u2) + c(u1 − u2) dx Ω i,j=1 Z 2 2 ≥ λ|Di(u1 − u2)| + c(u1 − u2) dx. Ω Then Z Z 2 λ 2 c|u1 − u2| dx ≥ − |u1 − u2| dx Ω 2µ Ω with the choice c ≥ −λ/(2µ). Combining the facts and recalling R 2 R 2 Poincar´e’sinequality Ω v dx ≤ µ Ω |Dv| dx we have Z λ 2  λ λ  2 0 ≥ |Di(u1 − u2)| + − (u1 − u2) dx Ω 2 2µ 2µ Z λ 2 = |Di(u1 − u2)| dx. 2 Ω

Using Poincar´e’sinequality, we see that u1 = u2 a.e.  Example 3.28. The uniform ellipticity was utilized again: Choose Ω = (0, 2), b = 0 = c, ( 1, x ∈ (0, 1] f(x) = a(x) = 0, x ∈ [1, 2), and consider the problem ( Lu = f, in (0, 2), u(0) = 0 = u(2). Then ( −0.5x2 + x, x ∈ (0, 1] u (x) = 1 −x2 + 2.5x − 1, x ∈ [1, 2) and ( −0.5x2 + x, x ∈ (0, 1] u (x) = 2 1.5 − x, x ∈ [1, 2) are weak solutions to Lu = f. (Ex) 3.5. Regularity. PDE 2 63

3.5.1. Local L2-regularity. In the previous sections, we relaxed the con- cept of a solution and observed that weak solutions are not necessarily C2. Next we study what is the natural regularity class and which conditions are needed to have a better regularity. First we motivate our approach by a formal calculation. Let f ∈ L2(Ω) and u be a solution with zero bdr values to a Poisson equation

−∆u = f in Ω. Then Z Z f 2 dx = (∆u)2 dx Ω Ω n n Z X ∂2u X ∂2u = dx ∂x2 ∂x2 Ω i=1 i j=1 j n X Z ∂2u ∂2u = dx ∂x2 ∂x2 i,j=1 Ω i j n Z 3 int by parts X ∂ u ∂u = − dx ∂x2∂x ∂x i,j=1 Ω i j j n Z 2 2 int by parts X ∂ u ∂ u = dx ∂x ∂x ∂x ∂x i,j=1 Ω i j i j Z 2 2 = D u dx, Ω where we denoted

 ∂2u ∂2u  2 ... ∂x1 ∂x1∂xn  ∂2u ... ∂2u  2  ∂x2∂x1 ∂x2∂xn  D u =  . . .   . .. .   . .  ∂2u ∂2u ... 2 ∂xn∂x1 ∂xn

 2 2 and |D2u|2 = Pn ∂ u . i,j=1 ∂xi∂xj 1. guess: The L2 norm of second derivatives is estimated in terms of L2-norm of f. Then let us differentiate the Poisson equation ∂f ∂ ∂u = − ∆u = −∆ . ∂xk ∂xk ∂xk 64 PDE 2

Denote f := ∂f and u := ∂u ie. ∂xk ∂xk

−∆u = f.

Now we may apply the previous calculation to have 2. guess: L2-norm of the third derivatives of u can be estimated in terms of L2-norm of the first derivatives of f. 3. guess: A solution u has two more derivatives than f and L2-norm of the kth derivatives of u can be estimated in terms of L2-norm of the k − 2 derivatives of f. Next we make these formal calculations rigorous for

n n X X Lu = − Di(aijDju) + biDiu + cu. i,j=1 i=1 with the uniform ellipticity condition, and open, bounded Ω. Idea: We establish this by roughly speaking replacing derivatives of the formal calculation by difference quotients, and carefully deriving estimates for these.

Theorem 3.29 (local L2-regularity). Let

1 ∞ ∞ aij ∈ C (Ω), bi ∈ L (Ω), c ∈ L (Ω) and

f ∈ L2(Ω).

Further, let u ∈ W 1,2(Ω) be a weak solution to Lu = f. Then

2,2 u ∈ Wloc (Ω)

0 and for any Ω b Ω
||u||W 2,2(Ω0) ≤ C ||f||L2(Ω) + ||Du||L2(Ω) + ||u||L2(Ω) ,

0 where C may depend on Ω , Ω and aij, bi, c, but not on u. Remark 3.30. • Observe that c is uniform over all the boundary values, since we didn’t assume zero bdr values this time. • It follows from the theorem that

Lu = fa.e., PDE 2 65

2,2 because if u ∈ Wloc (Ω) then Z n n X X
aijDjuDiϕ + biDiuϕ + cuϕ − fϕ dx Ω i,j=1 i=1 Z n n weak deriv. X X
= − Di(aijDju) + biDiu + cu − f ϕ dx Ω i,j=1 i=1 ∞ holds for every ϕ ∈ C0 (Ω). Then the fundamental lemma in the calculus of variations, Lemma 2.9, implies the claim. Such solutions are sometimes called strong solutions. • Our earlier examples show that some regularity assumption on aij is needed. 0 00 Proof of Theorem 3.35. Let Ω b Ω b Ω and choose a test function ∞ η ∈ C0 (Ω), 0 ≤ η ≤ 1 such that ( 1 x ∈ Ω0 η(x) = 0 x ∈ Ω \ Ω00

1,2 Since u is a weak solution, for every v ∈ W0 (Ω) n Z X Z aijDjuDiv dx = fv dx Ω i,j=1 Ω Pn where f = f − i=1 biDiu − cu. We choose a test function, for h > 0 small enough

−h 2 h v = −Dk (η Dk u) where u(x + he ) − u(x) Dhu(x) = k k h is the difference quotient introduced in Section 2.8. Let Z n X −h 2 h
A := − aijDjuDi Dk (η Dk u dx Ω i,j=1 and Z −h 2 h B := − fDk (η Dk u) dx. Ω −h 2 h −h 2 h We first estimate A using DiDk (η Dk u) = Dk Di(η Dk u) at the first step, as well as the standard rules of calculus 66 PDE 2

Z n X −h 2 h A = − aijDjuDiDk (η Dk u) dx Ω i,j=1 n int by parts for Dh Z k X h
2 h = Dk aijDju Di(η Dk u) dx Ω i,j=1 Z n X h h
h 2 h = Dk aijDju + aijDk Dju 2ηDiηDk u + η Dk Diu) dx Ω i,j=1 Z n X h h h 2 h h h
= Dk aijDju(2ηDiηDk u) + Dk aijDju(η Dk Diu) + aijDk Dju(2ηDiηDk u) dx Ω i,j=1 Z n X h 2 h
+ aijDk Dju(η Dk Diu) dx Ω i,j=1

= A1 + A2.

h 2 Then since |Dη|, |aij|, Dk aij ≤ C and η ≤ Cη , we have

Z  h h h h  |A1| ≤ C η |Du| Dk u + |Du| Dk Du + Dk Du Dk u dx Ω Young Z Z 2 h 2 2 h 2
≤ ε η Dk Du dx + C(ε) |Du| + Dk u dx Ω Ω00 Z Z 2 h 2 2 ≤ ε η Dk Du dx + C(ε) |Du| dx Ω Ω

R h 2 R 2 where at the last step we used Theorem 2.40: Ω00 Dk u dx ≤ Ω |Du| dx. By uniform ellipticity

Z n X h 2 h
A2 = aijDk Dju(η Dk Diu) dx Ω i,j=1 Z 2 h 2 ≥ λ η Dk Du dx. Ω PDE 2 67

It remains to estimate B. We calculate Z

|B| = fv dx Ω Z −h 2 h = fDk (η Dk u) dx Ω n Z X = (f − b D u − cu)D−h(η2Dhu) dx i i k k Ω i=1 Young Z ≤ C(ε) |f|2 + |Du|2 + |u|2
dx Ω Z −h 2 h 2 + ε Dk (η Dk u) dx. Ω Next we estimate the last integral again by Theorem 2.40 Z Z −h 2 h
2 2 h
2 Dk η Dk u dx ≤ D η Dk u dx Ω Ω Z h 2 h 2 ≤ 2ηDηDk u + η DDk u dx Ω Z h 2 h 2 ≤ 2ηDηDk u + η Dk Du dx Ω |η||Dη| ≤ C, η4 ≤ Cη2 Z Z 2 h 2 2 h 2 ≤ C η Dk u dx + C η Dk Du dx Ω Ω Thm 2.40 Z Z 2 2 h 2 ≤ C |Du| dx + C η Dk Du dx. Ω Ω Thus Z Z 2 2 2
2 h 2 |B| ≤ C(ε) |f| + |Du| + |u| dx + ε η Dk Du dx. Ω Ω Combining the estimates with the fact

A2 − |A1| ≤ |A| = |B| we have Z Z Z 2 h 2 2 h 2 2 λ η Dk Du dx − ε η Dk Du dx − C(ε) |Du| dx Ω Ω Ω Z Z 2 2 2
2 h 2 ≤ C(ε) |f| + |Du| + |u| dx + ε η Dk Du dx Ω Ω 68 PDE 2 ie. Z Z 2 h 2 2 h 2 λ η Dk Du dx − 2ε η Dk Du dx Ω Ω Z Z ≤ C(ε) |f|2 + |Du|2 + |u|2
dx + C(ε) |Du|2 dx. Ω Ω Choosing ε = λ/4 and recalling η = 1 in Ω0, we have Z Z λ 2 h 2 2 2 2
η Dk Du dx ≤ C |f| + |Du| + |u| dx. 2 Ω0 Ω 1,2 This implies by Theorem 2.41 that Diu ∈ Wloc (Ω) and thus u ∈ 2,2 Wloc (Ω).  R 2 R 2 We can also obtain Ω |u| dx instead of Ω |Du| dx on the right hand side of the estimate in the previous theorem. Observe that u does not have zero bdr values so that we cannot use Sobolev-Poincare type inequalities directly. Nonetheless, a suitable estimate holds for solutions. This is a Caccioppoli type estimate.

Lemma 3.31 (Caccioppoli’s ie). Let u, aij, bi, c and f be as in the previous theorem. Then Z Z |Du|2 dx ≤ C |u|2 + f 2) dx Ω0 Ω 0 for Ω b Ω. Proof. Choose a test function v = η2u, where η is the same cut-off function as in the proof of the previous theorem so that Z n n X 2 X 2 2 aijDjuDi(η u) + biDiu(η u) + cu(η u) dx Ω i,j=1 i=1 Z n n X 2 X 2 2 = aijDju(2ηDiηu + η Diu) + biDiu(η u) + cu(η u) dx Ω i,j=1 i=1 Z n X 2 = η aijDjuDiu dx Ω i,j=1 Z n n X X 2 2 + 2aijuηDjuDiη + biDiu(η u) + cu(η u) dx Ω i,j=1 i=1

= A1 + A2. By the uniform ellipticity Z 2 2 A1 ≥ λ η |Du| dx Ω PDE 2 69

Recalling that |aij|, η, |Dη| ≤ C and using Young’s inequality yields Z Z 2 2 2 |A2| ≤ ε η |Du| dx + C u dx. Ω Ω Finally, again by Young’s inequality Z Z Z 2 2 2 fη u dx ≤ C f dx + C u dx. Ω Ω Ω Combining the above estimates with the PDE itself we have Z Z Z λ η2|Du|2 dx ≤ ε η2|Du|2 dx + C u2 + f 2 dx. Ω Ω Ω By choosing ε = λ/2 we can ”absorb” the first integral on the RHS into the left, and the proof is complete.  By adjusting the proof of Theorem 3.35 slightly to obtain some do- ˜ 00 ˜ main Ω, Ω b Ω b Ω on the right in the estimate, we could com- bine Theorem 3.35 with Caccioppoli’s inequality and have the following corollary. Corollary 3.32. Let 1 ∞ ∞ aij ∈ C (Ω), bi ∈ L (Ω), c ∈ L (Ω) and f ∈ L2(Ω). Further, let u ∈ W 1,2(Ω) be a weak solution to Lu = f. Then 2,2 u ∈ Wloc (Ω) 0 and for any Ω b Ω
||u||W 2,2(Ω0) ≤ C ||f||L2(Ω) + ||u||L2(Ω) , 0 where C may depend on Ω , Ω and aij, bi, c, but not on u. By a similar argument as above combined with the induction, we could prove the following higher regularity result if the coefficient and data are smooth enough. For details, see Evans: PDE p. 316. Consider ∆u = f, and suppose then that f ∈ W 1,2(Ω). By the above u ∈ 2,2 Wloc (Ω), and thus by the weak definition Z ∂ϕ Z ∂ϕ Du · D dx = f dx. ∂xi ∂xi so that Z ∂u Z ∂f − D · Dϕ dx = − ϕ dx. ∂xi ∂xi 70 PDE 2

Thus ∂u is a weak solution with the RHS ∂f ∈ L2. Thus u ∈ W 3,2(Ω). ∂xi ∂xi loc Iterating further, and using a generalized Sobolev imbedding gives that u is smooth. Theorem 3.33 (Local smoothness). Let ∞ aij, bi, c ∈ C (Ω) and f ∈ C∞(Ω). Further, let u ∈ W 1,2(Ω) be a weak solution to Lu = f. Then u ∈ C∞(Ω).

3.5.2. Global L2-regularity. Also a global regularity result holds. k Definition 3.34. We denote ∂Ω ∈ C (Ω), if for each point x0 ∈ ∂Ω there is a r > 0 and a Ck-function n−1 γ : R → R such that upon relabeling and reorienting the coordinate axes if neces- sary it holds that

Ω ∩ B(x0, r) = {x ∈ B(x0; r): xn > γ(x1, . . . , xn−1)}. We also denote for bounded Ω C1(Ω) = {u ∈ C1(Ω) : Dαu, |α| ≤ 1, is uniformly continuous in Ω}. Theorem 3.35 (Global regularity). Let 1 ∞ ∞ aij ∈ C (Ω), bi ∈ L (Ω), c ∈ L (Ω) and f ∈ L2(Ω). Further, assume that ∂Ω ∈ C2(Ω), let g ∈ W 2,2(Ω) and u ∈ W 1,2(Ω) be a weak solution to ( Lu = f in Ω, 1,2 u − g ∈ W0 (Ω). Then u ∈ W 2,2(Ω) and
||u||W 2,2(Ω) ≤ C ||f||L2(Ω) + ||u||L2(Ω) + ||g||W 2,2(Ω) , PDE 2 71 where C may depend on Ω and aij, bi, c, but not on u. Remark 3.36 (Warning). One might be tempted to think that all kind of properties of the boundary value function are inherited by the solution as long as the boundary and the coefficient are regular enough. This is false however! 1 Let α ∈ ( 2 , 1) and π π Ω = {z ∈ : |z| > 0, Arg z ∈ (− , )}. C 2α 2α Denote z = x + iy = reiθ with θ ∈ (−π, π]. Since

log z = iθ + log r

zα := eα log(z) = eα log reiαθ = rα(cos(αθ) + i sin(αθ)).

We take for granted that zα is an analytic function in Ω, and thus its real part

u(x, y) := Re zα = rα cos(αθ) is a harmonic function. Then for x > 0 it holds that

u(x, 0) = |x|α even if

u ≡ 0 on ∂Ω.

A harmonic function is actually locally but not necessarily globally Lip- schitz. The similar phenomenon happens even if the boundary is smooth. Indeed, consider the upper half plane and ( ∆u = 0 (x, y) ∈ 2 R+ (3.13) u(x, 0) = g(x), where g(x) = |x| close to 0 and continued in a suitable bounded and smooth fashion to the whole of R. Then y 7→ u(0, y) is only H¨older continuous close to y = 0. Similarly, if g ∈ C2(R) in (3.13) it does not always follow that u ∈ 2 2 C (R+). 72 PDE 2

3.5.3. Local Lp-regularity. Example 3.37 (Calder´on-Zygmund type inequality). First consider a classical approach to the problem ∆u = f where f ∈ Lp, 2 ≤ p < ∞. A solution u is of the form (we say nothing about a domain or uniqueness as this is just the idea on general level) Z f(y) u(x) = C dy. |x − y|n−2 One of the questions in the regularity theory of PDEs is, does u have the second derivatives in Lp i.e. ∂2u ∈ Lp? ∂xi∂xj If we formally differentiate u, we get ∂2u Z ∂2 1 = C f(y) n−2 dy. ∂xi∂xj Rn ∂xi∂xj |x − y| | {z } | · |≤C/|x−y|n

R ∂2 1 It follows that f(y) n−2 dy defines (the precise definitions are ∂xi∂xj |x−y| beyond our scope here) a singular integral T f(x). A typical theorem in the theory of singular integrals says

||T f||p ≤ C ||f||p and thus we can deduce that ∂2u ∈ Lp. Working further, we get ∂xi∂xj that u ∈ W 2,p. This was established by Calder´onand Zygmund (1952, Acta Math.) and thus the above inequality is often called the Calder´on- Zygmund inequality. Theorem 3.38 (Local Lp-regularity). Let 1 < p < ∞ and f ∈ Lp(Ω). Further, let u ∈ W 1,2(Ω) be a weak solution to ∆u = f. Then 2,p u ∈ Wloc (Ω) 0 and for any Ω b Ω
||u||W 2,p(Ω0) ≤ C ||f||Lp(Ω) + ||u||Lp(Ω) , where C may depend on p, n, Ω0, and Ω but not on u. PDE 2 73

3.5.4. Cα regularity using De Giorgi’s method. For expository reasons we only consider the Laplacian. Nonetheless, the method also applies to the equation with bounded measurable coefficients. 1,2 Lemma 3.39 (Caccioppoli’s inequality). Let u ∈ Wloc (Ω) be a weak solution to ∆u = 0 in Ω. Then there exists a constant c = c(n) such that Z 2 Z 2 2  C  2 η |D(u − k)+| dx ≤ |(u − k)+| dx, B(x0,R) R − r B(x0,R) where k ∈ R, 0 < r < R < ∞ s.t. B(x0,R) ⊂ Ω and u+ = max(u, 0). ∞ Proof. Let η ∈ C0 (B(x0,R)) cut-off function s.t. C 0 ≤ η ≤ 1, η = 1 in B(x , r), |Dη| ≤ 0 R − r and a test function 2 1,2 ϕ = (u − k)+η ∈ W0 (B(x0,R)). 2 Since Dϕ = D(u − k)+η + (u − k)+2ηDη we obtain using weak for- mulation Z Z 2 Du · D(u − k)+η dx = −2 Du · (u − k)+ηDη dx. (3.14) Ω Ω Recall ( Du a.e. {u > k} D(u − k) = (3.15) + 0 a.e. {u ≤ k}. Thus a.e. 2 |D(u − k)+| = Du · D(u − k)+ and combining this with (3.14), we get Z 2 2 |D(u − k)+| η dx B(x0,R) Z 2 = Du · D(u − k)+η dx B(x0,R) Z ≤ 2 |Du|(u − k)+η|Dη| dx B(x0,R) (3.15) Z = 2 |D(u − k)+|(u − k)+η|Dη| dx B(x0,R) Young Z Z 2 2 ≤ ε |D(u − k)+| η dx + C(ε) (u − k)+|Dη| dx. B(x0,R) B(x0,R) 74 PDE 2

From this the result follows by absorbing the first term on the RHS into the left, and recalling the definition of η.  1,2 Theorem 3.40 (ess sup-estimate). Let u ∈ Wloc (Ω) be a weak solution to ∆u = 0 in Ω. Then there exists c = c(n) such that

Z 1/2  2  ess sup u ≤ k0 + C |(u − k0)+| dx r B(x0, 2 ) B(x0,r) where k0 ∈ R and B(x0, r) ⊂ Ω. ∞ Proof. Let 0 < r/2 < ρ < r and η ∈ C0 (B(x0, r)) C 0 ≤ η ≤ 1, η = 1 in B(x , ρ), |Dη| ≤ , 0 r − ρ and use test function v = (u−k)+η. The proof will be based on the use of Sobolev-Poincare, Caccioppoli and iteration. To be more precise, Z Z 2 2 |Dv| dx ≤ |D((u − k)+η)| dx B(x0,r) B(x0,r) Z 2 ≤ |D(u − k)+η + (u − k)+Dη| dx B(x0,r) Z (3.16) 2 2 ≤ c |D(u − k)+η| + |(u − k)+Dη| dx B(x0,r) Cacc, def. of η Z C 2 ≤ 2 |(u − k)+| dx. (r − ρ) B(x0,r) Further,

Z 2/2∗ Z 2/2∗  2∗   2∗  |(u − k)+| dx ≤ C v dx B(x0,ρ) B(x0,r) 1,2 (3.17) v ∈ W0 (B(x0, r)), Sobo ie Z ≤ Cr2 |Dv|2 dx. B(x0,r) Combining (3.17) and (3.16), we get

Z 2/2∗ 2 Z  2∗  Cr 2 |(u − k)+| dx ≤ 2 |(u − k)+| dx. B(x0,ρ) (r − ρ) B(x0,r) (3.18) Define

A(k, ρ) = B(x0, ρ) ∩ {x ∈ Ω: u(x) > k} PDE 2 75 and observe Z 2 |(u − k)+| dx B(x0,ρ) Z 1 2 = |(u − k)+| dx |B(x0, ρ)| A(k,ρ) ∗ H¨older Z ∗ 2/2 2 1  2  1− ∗ ≤ |(u − k)+| dx |A(k, ρ)| 2 |B(x0, ρ)| A(k,ρ) Z 2/2∗ 1− 2 (3.19)  1 2∗   |A(k, ρ)|  2∗ = |(u − k)+| dx |B(x0, ρ)| A(k,ρ) |B(x0, ρ)| Z 2/2∗ 1− 2  2∗   |A(k, ρ)|  2∗ ≤ |(u − k)+| dx B(x0,ρ) |B(x0, ρ)| (3.18) 2 Z 1− 2 Cr 2  |A(k, ρ)|  2∗ ≤ 2 |(u − k)+| dx . (r − ρ) B(x0,r) |B(x0, ρ)| If h < k, then Z (k − h)2|A(k, ρ)| = (k − h)2 dx A(k,ρ) u > k in A(k, ρ) Z ≤ (u − h)2 dx A(k,ρ) h < k Z ≤ (u − h)2 dx A(h,ρ) Z ≤ (u − h)2 dx. B(x0,ρ) By this, denoting

Z 1/2  2  u(h, ρ) := |(u − h)+| dx , (3.20) B(x0,ρ) we get Z 1 2 |B(x0, ρ)| 2 |A(k, ρ)| ≤ 2 (u − h) dx = 2 u(h, ρ) . (3.21) (k − h) B(x0,ρ) (k − h) Using this with (3.19), we get

(3.19) 2 1− 2  r   |A(k, ρ)|  2∗ u(k, ρ)2 ≤ C u(k, r)2 r − ρ |B(x , ρ)| 0 (3.22) (3.21) 2 2(1− 2 )  r  u(h, ρ) 2∗ ≤ C u(k, r)2 . r − ρ k − h 76 PDE 2

This implies

u(k, r) ≤ u(h, r) r 1+1− 2 −(1− 2 ) u(k, ρ) ≤ C u(h, r) 2∗ (k − h) 2∗ r − ρ (3.23) r = C u(h, r)1+θ(k − h)−θ, r − ρ

2 where θ := 1 − 2∗ . Auxiliary claim: For k ∈ R it holds that

u(k0 + d, r/2) = 0

θ (1+θ)2/θ+1 θ where d := C2 u(k0, r) and c, θ are as above. Proof of the auxiliary claim: Let

−j kj = k0 + d(1 − 2 ) −j−1 ρj = r/2 + 2 r, j = 0, 1, 2,...

so that ρ0 = r, ρj & r/2 and kj % k + d as j → ∞. Then we show by induction that

−µj u(kj, ρj) ≤ 2 u(k0, r), j = 0, 1, 2,... (3.24) where µ = (1 + θ)/θ. Indeed j = 0 immediately follows since ρ0 = r. Assume then that the claim holds for some j, and observe that

−j−1 −j−2 −j−2 ρj − ρj+1 = (2 − 2 )r = 2 r, j−1 −j j−1 kj+1 − kj = (−2 + 2 )d = 2 d,

ρj ≤ r.

Using these with (3.23), we have

Cρj −θ 1+θ u(kj+1, ρj+1) ≤ (kj+1 − kj) u(kj, ρj+1) ρj − ρj+1 Cr ≤ (2−j−1d)−θ2−µj(1+θ)u(k , r)1+θ, 2−j−2r 0 PDE 2 77 where we used induction assumption to estimate u(kj, ρj+1). Then recalling the shorthand notations , we get (θ + 1)2 j+2+θ(j+1)− j θ (θ+1)2 | µ(1+{zθ) } +1 θ −1 1+θ u(kj+1, ρj+1) ≤ C2 (C2 θ u(k0, r) ) u(k0, r) | {z } d−θ 2 2 1+(1+θ)(j+1)− (θ+1) j− (θ+1) −1 = 2 θ θ u(k0, r) θ + 1 (θ+1)(j+1) (1 − ) θ | {z } − 1 = 2 θ u(k0, r) −µ(j+1) = 2 u(k0, r), and thus the induction is complete. Further, this implies

lim u(kj, ρj) = 0 j→∞ and Z 1 r kj ≤ k0 + d  2  2 0 ≤ u(k0 + d, ) = |(u − kj)+| dx 2 r B(x0, 2 ) r 1 ≤ ρj Z 2 B(x , ρ )  2 ≤ 0 j |(u − k ) |2 dx ≤ Cu(k , ρ ) → 0. B(x , r ) j + j j 0 2 B(x0,ρj ) It follows that r u(k + d, ) = 0, 0 2 and this ends the proof of the auxiliary claim. By using the auxiliary claim, we now finish the proof of the essup- estimate. Indeed, Z 1 r  2  2 0 = u(k0 + d, ) = |(u − (k0 + d))+| dx , 2 r B(x0, 2 ) θ (1+θ)2/θ+1 θ r where d = C2 u(k0, r) . Thus a.e. in B(x0, 2 ) it holds that Z 1  2  2 u ≤ k0 + d = k0 + C |(u − k0)+| dx B(x0,r) from which the claim Z 1  2  2 ess sup u ≤ k0 + C |(u − k0)+| dx r B(x0, 2 ) B(x0,r) follows.  78 PDE 2

Corollary 3.41. Let u be a weak solution to ∆u = 0 in Ω. Then there exists C = C(n) such that

1  Z  2 ess sup |u| ≤ c |u|2 dx r B(x0, 2 ) B(x0,r) r for all B(x0, 2 ) ⊂ Ω.

Proof. Choose k0 = 0 in the previous result and observe that Z 1  2  2 ess sup u ≤ C |u+| dx . r B(x0, 2 ) B(x0,r) Since −u is also a solution, we obtain

1  Z  2 − ess inf u = ess sup(−u) ≤ C |(−u) |2 dx . r + B(x0, ) r 2 B(x0, 2 ) B(x0,r) Combining the estimates

1  Z  2 ess sup |u| ≤ max{ess sup u, − ess inf u} ≤ C |u|2 dx . r  r r B(x0, ) B(x0, 2 ) B(x0, 2 ) 2 B(x0,r) The above result implies that (unlike Sobolev functions in general) are locally bounded. Next lemma is needed in order to prove H¨oldercontinuity for weak solutions. Lemma 3.42. [Measure decay] Let u be a weak solution to ∆u = 0 in Ω, B(x0, 2r) ⊂ Ω, m(2r) = ess inf u, M(2r) = ess sup u, B(x ,2r) 0 B(x0,2r) and

|A(k0, r)| ≤ γ|B(x0, r)|, 0 < γ < 1, where A(k0, r) = B(x0, r) ∩ {x ∈ Ω: u(x) > k0} and k0 = (m(2r) + M(2r))/2. Then lim |A(k, r)| = 0. k%M(2r) We postpone the proof and go to the proof of the H¨oldercontinuity immediately. Theorem 3.43 (H¨older-continuity). If u be a weak solution to ∆u = 0 in Ω, then u is locally H¨older-continuous (or has such a representative to be more precise). PDE 2 79

Proof. Let m(2r) + M(2r) k = 0 2 similarly to the previous lemma. Without loss of generality we may assume that 1 |B(x , r) ∩ {x ∈ Ω: u(x) > k }| = |A(k , r)| ≤ |B(x , r)|, (3.25) 0 0 0 2 0 1 since otherwise if |A(k0, r)| > 2 |B(x0, r)| it holds that 1 |{x ∈ B(x , r): −u(x) ≤ −k }| > |B(x , r)| 0 0 2 0 1 |{x ∈ B(x , r): −u(x) > −k }| < |B(x , r)| 0 0 2 0 and the argument below works for −u and −k0 instead. Here we need that both u and −u are solutions. Using the esssup-estimate with −(l+1) kl = M(2r) − 2 (M(2r) − m(2r)) we have Z 1 r  2  2 M( ) ≤ kl + C |(u − kl)+| dx 2 B(x0,r) | {z 2} ≤(M(2r)−kl) 1  |A(kl, r)|  2 ≤ kl + C(M(2r) − kl) |B(x0, r)| since the integrand can only be nonzero in the set A(kl, r). By Lemma 3.42, we may choose l large enough so that 1  |A(k , r)|  2 1 C l < . |B(x0, r)| 2 This fixes l. Combining the previous two estimates we obtain r 1 M( ) ≤ k + (M(2r) − k ) 2 l 2 l ≤ M(2r) − 2−(l+1)(M(2r) − m(2r)) 1 + M(2r) − (M(2r) − 2−(l+1)(M(2r) − m(2r)))
2 ≤ M(2r) − 2−(l+2)(2 − 1)(M(2r) − m(2r)). From this we get r r r M( ) − m( ) ≤ M( ) − m(2r) 2 2 2 ≤ (1 − 2−(l+2))(M(2r) − m(2r)). 80 PDE 2

Using the notation oscB(x0,2r) u := M(2r) − m(2r) = ess supB(x0,2r) u − −(l+2) ess infB(x0,2r) u and λ := (1 − 2 ) < 1, the above reads as

osc r u ≤ λ osc u. B(x0, 2 ) B(x0,2r) The H¨oldercontinuity follows from this by a standard iteration. To be more precise, choose j ∈ N such that R 4j−1 ≤ < 4j. r Then j−1 osc r u ≤ λ osc j−1 u B(x0, 2 ) B(x0,4 r) 4j−1r≤R j−1 ≤ λ oscB(x ,R) u 0 (3.26) j−1 ≤ λ oscB(x0,R) u  r α ≤ C osc u, R B(x0,R) where we denoted α = − log λ/ log 4 ∈ (0, 1) and observed R R−α  r α < 4j ⇒ 4−jα ≤ = r r R j−1  r α λj−1 = 4log4(λ ) = 4(j−1) log(λ)/ log(4) = 4(1−j)α ≤ 4α . R 1 Let y ∈ Ω s.t. |x0 − y| ≤ 8 dist(x0, ∂Ω), R = dist(x0, ∂Ω), r = |x0 − y| (actually only for a.e. point but then the below deduction can be used to define the H¨oldercontinuous representative). Then by the estimate (3.26)

|u(x0) − u(y)| ≤ oscB(x0,2r) u  2r α ≤ C osc 3 u 3 B(x0, 8 R) 8 R α 2|x0 − y| ≤ C osc 3 u 3 B(x0, 8 R) 8 R α −α ≤ C|x0 − y| R 2 ess sup |u| 3 B(x0, 8 R) esssup-est Z 1 α −α 2  2 ≤ C|x0 − y| R |u| dx 6 B(0, 8 R) α ≤ C|x0 − y| .  It remains to prove the lemma used above. PDE 2 81

Proof of Lemma 3.42. The proof is based on deriving an estimate con- taining |A(h, r)| − |A(k, r)| by using Poincare’s and Caccioppoli’s in- equalities. To this end, we let k > h > k0 and define an auxiliary function  k − h, u(x) ≥ k,  v(x) = u(x) − h, h < u(x) < k, 0, u(x) ≤ h.

It immediately follows

{x ∈ B(x0, r): v(x) = 0} = |{x ∈ B(x0, r): u(x) ≤ h}|

= |B(x0, r)| − |{x ∈ B(x0, r): u(x) > h}|

= |B(x0, r)| − |A(h, r)|

k0 < h ≥ |B(x0, r)| − |A(k0, r)| assump. ≥ (1 − γ)|B(x0, r)|. R From this denoting vB(x ,r) := v dx, we get 0 B(x0,r) Z

k − h − vB(x0,r) = (k − h − v) dx B(x0,r) Z ≥ (1 − γ) (k − h − v) dx {x∈B(x0,r): v(x)=0} = (1 − γ)(k − h).

Integrating this over A(k, r) and using Poincare’s inequality, we obtain

1 Z (k − h)|A(k, r)| ≤ (k − h − vB(x0,r)) dx 1 − γ A(k,r) v = k − h in A(k, r) 1 Z = (v − vB(x0,r)) dx 1 − γ A(k,r) 1 Z ≤ (v − vB(x0,r)) dx 1 − γ B(x0,r) n−1 Z n H¨older 1   n n−1 ≤ |B(x0, r)| v − vB(x0,r) dx 1 − γ B(x0,r) Poinc., p=1 1 Z ≤ |B(x0, r)|cr |Dv| dx. 1 − γ B(x0,r) 82 PDE 2

Then using the above estimate and by H¨older’sas well as Caccioppoli’s inequalities 1 Z (k − h)|A(k, r)| ≤ Cr |Dv| dx 1 − γ B(x0,r) def. v 1 Z ≤ Cr |Du| dx 1 − γ A(h,r)\A(k,r) (3.27) H¨older Z 1/2 1  2  1 ≤ Cr |Du| dx (|A(h, r)| − |A(k, r)|) 2 1 − γ A(h,r) Cacc. Z 1/2  2  1 ≤ C |u − h| dx (|A(h, r)| − |A(k, r)|) 2 , A(h,2r) where at the last step we observed that r/(2r − r) ≤ C. Next replace k and h in the above inequality by kj and kj−1 where

−(j+1) kj = M(2r) − 2 (M(2r) − m(2r)).

Then since

−(j+1) kj − kj−1 = (M(2r)−2 (M(2r) − m(2r))) − M(2r) − 2−j(M(2r) − m(2r))
= 2−(j+1)(M(2r) − m(2r)), we get by (3.27) that

−(j+1) 2 (M(2r) − m(2r))|A(kj, r)| = (kj − kj−1)|A(kj, r)| Z 1/2  2  1 ≤ C |u − kj−1| dx (|A(kj−1, r)| − |A(kj, r)|) 2 . A(kj−1,2r)

−j Then observing u−kj−1 ≤ M(2r)−kj−1 = M(2r)− M(2r)−2 (M(2r)− m(2r))
= 2−j(M(2r)−m(2r)) and using the above estimate, we obtain

−(j+1) 2 (M(2r) − m(2r))|A(kj, r)| 1 1 ≤ C(M(2r) − kj−1)|A(kj−1, 2r)| 2 (|A(kj−1, r)| − |A(kj, r)|) 2 −j 1 1 ≤ C2 (M(2r) − m(2r))|A(kj−1, 2r)| 2 (|A(kj−1, r)| − |A(kj, r)|) 2 .

Cancelling 2−j(M(2r) − m(2r)) on both sides and choosing l, l ≥ j, we end up with

1 1 |A(kl, r)| ≤ |A(kj, r)| ≤ C|B(x0, 2r)| 2 (|A(kj−1, r)| − |A(kj, r)|) 2 . PDE 2 83

Taking squares and summing over j, this gives by telescoping

l 2 X 2 l|A(kl, r)| = |A(kl, r)| j=1 l X ≤ C|B(x0, 2r)| (|A(kj−1, r)| − |A(kj, r)|) j=1

≤ C|B(x0, 2r)|(|A(k0, r)| − |A(kl, r)|) ≤ cC|B(x0, 2r)||B(x0, r)|. Dividing by l, we finally get

lim |A(kl, r)| = 0. l→∞ 

3.6. Comparison and max principles. In this section we consider n X Lu = − Di(aij(x)Dju(x)) + c(x)u(x) = f. i,j=1 Theorem 3.44 (Comparison principle). Let u, w ∈ W 1,2(Ω) be weak 1,2 solutions and (u − w)+ ∈ W0 (Ω). Then there is c0 such that if c ≥ c0 it holds that u ≤ w in Ω.

Proof. The idea is the same as in the proof of the uniqueness. First n Z X Z aijDjuDiv + cuv dx = fv dx Ω i,j=1 Ω n Z X Z aijDjwDiv + cwv dx = fv dx Ω i,j=1 Ω

1,2 for every v ∈ W0 (Ω). By subtracting the equations n Z X aijDj(u − w)Div + c(u − w)v dx = 0. Ω i,j=1

1,2 Now we choose v = (u − w)+ ∈ W0 (Ω) and estimate Z n X 2 0 = aijDj(u − w)Di(u − w)+ + c(u − w)+ dx Ω i,j=1 Z 2 2 ≥ λ|Di(u − w)+| + c(u − w)+ dx. Ω 84 PDE 2

Since Z Z 2 λ 2 c(u − w)+ dx ≥ − (u − w)+ dx Ω 2µ Ω with choice c ≥ −λ/(2µ). Combining the facts and recalling Poincar´e’s R 2 R 2 inequality Ω v dx ≤ µ Ω |Dv| dx we have Z λ 2  λ λ  2 0 ≥ |Di(u − w)+| + − (u − w)+ dx Ω 2 2µ 2µ Z λ 2 = |Di(u − w)+| dx. 2 Ω Again using Poincar´e’sinequality (recall exercise set 1), we see that (u − w)+ = 0 a.e., that is u ≤ w a.e.  Remark 3.45. By analyzing the above proof, we see that also the fol- lowing holds: Let u, w ∈ W 1,2(Ω) and u and w be sub- and supersolu- tions respectively ie. n Z X Z aijDjuDiv + cuv dx ≤ fv dx Ω i,j=1 Ω n Z X Z aijDjwDiv + cwv dx ≥ fv dx Ω i,j=1 Ω 1,2 1,2 for every v ≥ 0, v ∈ W0 (Ω), and (u − w)+ ∈ W0 (Ω). Then u ≤ w in Ω.

For the next theorem, we define 1,2 sup u := inf{l ∈ R :(u − l)+ ∈ W0 (Ω)}. ∂Ω Theorem 3.46 (Weak max principle). Let u ∈ W 1,2(Ω) be a weak Pn solution to − i,j=1 Di(aij(x)Dju(x)) + c(x)u(x) = 0, with c ≥ 0. Then

ess sup u ≤ sup u+. Ω ∂Ω 1,2 Proof. Set M := sup∂Ω u+ ≥ 0. It holds that (u − M)+ ∈ W0 (Ω). To see this, choose decreasing sequence li → M so that (u − li)+ = 1,2 (u+ −li)+ ∈ W0 (Ω). Then since Ω is bounded, it follows that u−li → u − M in W 1,2(Ω). By it holds , 1,2 (u − li)+ → (u − M)+ in W (Ω) 1,2 and thus the claim (u − M)+ ∈ W0 (Ω) follows. PDE 2 85

We may use v = (u − M)+ as a test function in n Z X aijDjuDiv + cuv dx = 0 Ω i,j=1 n Z X aijDjMDiv + cMv dx ≥ 0, Ω i,j=1 where M, c, v ≥ 0 was used. We subtract these to have Z 2 2 λ |D(u − M)+| + c(u − M)+ dx ≤ 0. Ω From this it follows that u ≤ M a.e.  3.6.1. Strong maximum principle. Strong comparison principle for weak solutions follows Harnack type arguments that we have not proven yet. Nonetheless, we show that due to the classical theory this is something to be expected anyway. Recall C1(Ω) = {u ∈ C1(Ω) : Dαu is uniformly continuous for all |α| ≤ 1}. The argument does not rely on divergence form. For simplicity of notation we consider n X Lu = − aijDiDju = 0. i,j=1

By interior ball condition for Ω at x0 ∈ ∂Ω, we mean that there is a ball B ⊂ Ω such that x0 ∈ ∂B. Lemma 3.47 (Hopf). Let u ∈ C2(Ω) ∩ C1(Ω) satisfy Lu ≤ 0, and suppose that there is x0 satisfying interior ball condition for B and

u(x0) > u(x) for all x ∈ Ω. Then ∂u (x ) > 0 ∂ν 0 where ν is exterior unit normal for B at x0.

Proof. We may assume that B = B(0, r) and u(x0) ≥ 0. Set for γ > 0 2 v(x) = e−γ|x| − e−γr2 , x ∈ B(0, r). Then

−γ|x|2 Djv = −2xjγe 86 PDE 2 and

2 −γ|x|2 DiDjv = (−2δijγ + 4γ xixj)e . Thus n X Lv = − aijDiDjv i,j=1 n X 2 −γ|x|2 = − aij(−2δijγ + 4γ xixj)e i,j=1 n ell X 2 2 −γ|x|2 ≤ (2γ aii − 4γ λ|x| )e . i=1 Thus for large enough γ, we have n X 2 2 r Lv ≤ (2γ a − 4γ2λ|x| )e−γ|x| ≤ 0, x ∈ B(0, r) \ B(0, ). ii 2 i=1

By the assumption u(x0) > u(x) for all x ∈ Ω, for small enough ε > 0, it holds that

u(x0) ≥ u(x) + εv(x) on ∂B(0, r/2) ⊂ Ω. The same holds on ∂B(0, r) since there v = 0. We have

L(u + εv − u(x0)) = Lu + εLv ≤ 0, and therefore the weak maximum principle for classical solutions (ex.) implies

u + εv − u(x0) ≤ 0 in B(0, r) \ B(0, r/2). But

u(x0) + εv(x0) − u(x0) = 0 so that ∂(u + εv − u(x )) 0 (x ) ≥ 0. ∂ν 0 This yields

∂u ∂v x0 x0 2 (x ) ≥ −ε (x ) = −ε Dv(x ) = −ε (−2x γe−γ|x0| ) > 0. ∂ν 0 ∂ν 0 r 0 r 0  Remark 3.48. The nontrivial point on Hopf’s lemma is that the in- ∂u equality ∂ν (x0) > 0 is strict! PDE 2 87

Theorem 3.49 (Strong max principle). Let u ∈ C2(Ω) ∩ C(Ω) satisfy Lu ≤ 0, and let Ω be a bounded, open and connected set. Then if u attains its max at the interior of Ω, it follows that u ≡ sup u. Ω

Proof. Let M := maxΩ u and C = {x ∈ Ω: u(x) = M}, V = {x ∈ Ω: u(x) < M}. Let us make a counter proposition that V is not empty. Take a point y ∈ V with dist(y, C) < dist(y, ∂Ω), which exist since dist(C,V ) = 0 by continuity of u. Let B = B(y, r) ⊂ V be a largest possible ball in V centered at y. Then B touches C at some point x0, and thus V satisfies interior ball condition at this point. By Hopf’s lemma, ∂u (x ) > 0 ∂ν 0 but this is a contradiction since x0 is a max point for u implying Du(x0) = 0.  4. Linear parabolic equations Next we study generalizations of the heat equation. We denote

ΩT = Ω × (0,T ) and

∂pΩT = (Ω × {0}) ∪ (∂Ω × (0,T )). Definition 4.1 (parabolic Sobolev space). The Sobolev space L2(0,T ; W 1,2(Ω)), consists of all measurable(in ΩT ) functions u(x, t) such that u(x, t) be- longs to W 1,2(Ω) for almost every 0 < t < T ,(u(x, t) is measurable as a mapping from (0,T ) to W 1,2(Ω), and the norm ZZ 1/2 (|u(x, t)|2 + |Du(x, t)|2) dx dt ΩT 2 1,2 is finite. The definition of the space L (0,T ; W0 (Ω)) is analogous. 88 PDE 2

The notation above refers to Banach valued functions (0,T ) 7→ W 1,2(Ω) and thus refers to Bochner integration theory. However, we do not pursue this analysis here. Definition 4.2. The space C(0,T ; L2(Ω)) consists of all measurable functions u :ΩT → R such that Z 1/2  2  ||u||C(0,T ;L2(Ω)) := max |u(x, t)| dx < ∞ t∈[0,T ] Ω and for any ε > 0 and t1 ∈ [0,T ] there is δ > 0 such that if |t1 − t2| ≤ δ, where t2 ∈ [0,T ], then Z 1/2  2  |u(x, t1) − u(x, t2)| dx ≤ ε. Ω ∞ 2 1,2 Theorem 4.3. The space C (ΩT ) is dense in L (0,T ; W (Ω)). Proof. The space W 1,2(Ω) is separable (not proven here). The proof consists of three steps. First, by separability, we can approximate any function u ∈ L2(0,T ; W 1,2(Ω)), denoted by u(t) = u(x, t), with simple functions. By modifying the simple functions in the set where the norm is large compared to the norm of the original function, and using Lebesgue’s dominated convergence theorem, we obtain a L2-convergent sequence. Finally, we mollify the simple function. Next we work out the details. Utilizing the separability of W 1,2(Ω), we can choose a countable dense set ∞ {ak}k=1 ⊂ u(0,T ). We define for k = 1, . . . , n n 1,2 Fk = {f ∈ W (Ω) : ||f − ak|| 1,2 = min ||f − ai|| 1,2 } W (Ω) 1≤i≤n W (Ω) and n −1 n n n n n k−1 n Bk = u (Fk ),D1 = B1 ,Dk = Bk \ (∪i=1 Bi ) for k = 2, 3,.... n It follows from the measurability of u(t) that the sets Dk are measur- able, and thus n X u (t) = a χ n (t) n k Dk k=1 ∞ is a simple function. Because {ak}k=1 is a dense set, it follows that a.e. 1,2 un(t) → u(t) in W (Ω) as n → ∞. PDE 2 89

In order to use Lebesgue’s dominated convergence theorem, we mod- ify un whenever ||un(t)||W 1,2(Ω) is large compared to ||u(t)||W 1,2(Ω), and define ( un(t), if ||un(t)||W 1,2(Ω) ≤ 2 ||u(t)||W 1,2(Ω) , vn(t) = 0, if ||un(t)||W 1,2(Ω) > 2 ||u(t)||W 1,2(Ω) .

If ||u(t)||W 1,2(Ω) = 0, then vn(t) = 0 and if ||u(t)||W 1,2(Ω) > 0, then vn(t) = un(t) for n large enough. We deduce 1,2 vn(t) → u(t) in W (Ω), and ||vn(t)||W 1,2(Ω) ≤ 2 ||u(t)||W 1,2(Ω) . Thus Lebesgue’s dominated convergence theorem implies Z T 2 ||vn(t) − u(t)||W 1,2(Ω) dt → 0, as n → ∞. 0 Next we denote n o ˆ n n Dk = Dk \ t ∈ (0,T ): ||un(t)||W 1,2(Ω) > 2 ||u(t)||W 1,2(Ω) and get n X vn(t) = akχ ˆ n (t). Dk k=1 We have shown earlier, using approximations that C∞(Ω) is dense in 1,2 ∞ W (Ω), and hence we can choose ϕk ∈ C (Ω) such that ε ||ϕ − a ||2 < . k k W 1,2(Ω) T This implies n n 2 Z T X X akχ ˆ n (t) − ϕkχ ˆ n (t) dt < ε. Dk Dk 0 k=1 k=1 W 1,2(Ω)

Finally, we may mollify in t with a mollification parameter δn (this follows from the approximation results applied in 1D) such that for each k = 1, . . . , n Z T 2 ε

χ ˆ n (t) − (χ ˆ n )δn (t) dt < . Dk Dk 0 n ||ϕk||W 1,2(Ω) Accomplishing this approximation for each k = 1, 2, . . . , n, we obtain the desired smooth function n X ϕk(χ ˆ n )δn (t), (4.28) Dk k=1 which completes the proof.  90 PDE 2

2 Lemma 4.4. Let u ∈ L (ΩT ), extend u as zero to (−∞, 0) and (T, ∞) and set Z uε(x, t) = u(x, t − s)ηε(s) ds R where ηε is a standard mollifier. Then

2 uε → u in L (ΩT )

Proof. By repeating the argument in the previous proof (cf. (4.28)), we can produce a smooth approximation g such that Z |u − g|2 dy
1/2 < δ/3. ΩT

We extend u by zero to (−∞, 0) and (T, ∞), and denote by uε a stan- dard mollification in the time direction. Similarly as for space mollifi- cations Z t+ε

|uε(x, t)| = ηε(t − s)u(x, s) ds t−ε Z t+ε 1/2 1/2 ≤ ηε(t − s) ηε(t − s) |u(x, s)| ds t−ε H¨older Z t+ε 1/2 Z t+ε 1/2    2  ≤ ηε(t − s) ds ηε(t − s)|u(x, s)| ds . t−ε t−ε | {z } 1 and Z Z Z t+ε 2 2 |uε(x, t)| dx dt ≤ ηε(t − s)|u(x, s)| ds dx dt ΩT ΩT t−ε Z T Z Z 2 = ηε(t − s)|u(x, s)| ds dx dt 0 Ω R Z Z T Z Fubini 2 = ηε(t − s)|u(x, s)| dx dt ds R 0 Ω Z Z T Z 2 = ηε(t − s) dt |u(x, s)| dx ds R 0 Ω Z Z ≤ |u(x, s)|2 dx ds R Ω Z = |u(x, s)|2 dx ds. ΩT PDE 2 91

We deduce Z 2
1/2 |u − uε| dx dt ΩT Minkowski Z Z Z 2
1/2 2
1/2 2
1/2 ≤ |u − g| dx dt + |g − gε| dx dt + |gε − uε| dx dt ΩT ΩT ΩT Z Z 2
1/2 2
1/2 ≤ δ/3 + |g − gε| dx dt + |g − u| dx dt ΩT ΩT Z 2
1/p ≤ δ/3 + |g − gε| dx dt + δ/3. ΩT By adjusting the argument we used in with x-approximations, we see 2 that gε → g pointwise in Ω×(0,T ). Moreover, |g − gε| ≤ 4 maxΩT |g| ∈ 1 L (ΩT ) and thus by DOM, for all small enough ε Z 2
1/2 |g − gε| dx dt ≤ δ/3.  ΩT 2 ∂u 2 Theorem 4.5. Let u ∈ L (ΩT ) and ∂t ∈ L (ΩT ). Then there is such a representative that u ∈ C(0,T ; L2(Ω)). Proof. By the previous lemma ( u → u, in L2(Ω ) ε T (4.29) ∂uε ∂u 2 ∂t → ∂t , in L (Ω × (h, T − h)), where ε < h and the proof of the second statement again follows the guidelines of the space approximations. By Fubini’s theorem for a.e. x 2 1 the function t 7→ u(x, t) in L (0,T ) ⊂ L (0,T ). Thus for a.e. x uε(x, t) is a smooth function so that Z t2 ∂uε uε(x, t1) − uε(x, t2) = dt t1 ∂t and 2 Z t2 2 ∂uε ||uε(x, t1) − uε(x, t2)||L2(Ω) = dt . t1 ∂t L2(Ω) We apply (4.29) to the RHS together with Fubini’s thm and state without a proof (cf. approx section) that LHS converges for a.e. t1, t2. Thus Z t2 Z 2 2 ∂u ||u(x, t1) − u(x, t2)||L2(Ω) ≤ C(t1 − t2) dx dt. t1 Ω ∂t This also implies the continuity on the whole interval [0,T ].  92 PDE 2

We study initial-boundary value problem for given g : ΩT → R, f :ΩT → R ( ut + Lu = f, x ∈ ΩT

u = g, x ∈ ∂pΩT . Here n n X X Lu(x, t) = − Di(aij(x, t)Dju(x, t)) + bi(x, t)Diu(x, t) i,j=1 i=1 + c(x, t)u(x, t). Definition 4.6 (uniformly parabolic). The operator is uniformly par- abolic if there are 0 < λ ≤ Λ < ∞ such that n 2 X 2 λ|ξ| ≤ aijξiξj ≤ Λ|ξ| . i,j=1

2 Definition 4.7 (local weak solution). A functions u ∈ Cloc(0,T ; Lloc(Ω))∩ 2 1,2 Lloc(0,T ; Wloc (Ω)) is a weak solution to the above PDE if n n Z Z X X − uϕt dx dt + aijDjuDiϕ + biDiuϕ + cuϕ dx dt ΩT ΩT i,j=1 i=1 Z = fϕ dx dt ΩT ∞ for every ϕ ∈ C0 (ΩT ). Definition 4.8 (global weak solution). Let g ∈ C(0,T ; L2(Ω)) ∩ L2(0,T ; W 1,2(Ω)). A functions u ∈ C(0,T ; L2(Ω)) ∩ L2(0,T ; W 1,2(Ω)) is a weak solution with boundary values g to the above PDE if n n Z Z X X − uϕt dx dt + aijDjuDiϕ + biDiuϕ + cuϕ dx dt ΩT ΩT i,j=1 i=1 Z = fϕ dx dt ΩT ∞ for every ϕ ∈ C0 (ΩT ), and 2 1,2 u − g ∈ L (0,T ; W0 (Ω)) as well as Z Z ∞ u(x, 0)φ(x) dx = g(x, 0)φ(x) dx for every φ ∈ C0 (Ω). Ω Ω PDE 2 93

2 4.1. Existence: Galerkin method. Let f ∈ L (ΩT ). For simplicity we only consider the problem  u = ∆u + f, in Ω  t T u = 0, on ∂Ω × [0,T ] u(x, 0) = g(x), on Ω,

1,2 where g ∈ W0 (Ω), but intend to use methods that also work in greater generality. In the weak form, Z ∂ϕ Z Z − u dx dt + Du · Dϕ dx dt = fϕ dx dx (4.30) ΩT ∂t ΩT Ω ∞ for every ϕ ∈ C0 (ΩT ). 2 Idea in Galerkin’s method is to take a basis ωi i = 1, 2,... in L and 1,2 W0 (Ω) and approximate solution as m X m um(x, t) = ci (t)ωi(x). i=1 Choosing the coefficients properly, we can show that this approximation converges to a weak solution. Galerkin’s method has turned out to be useful in numerical approximations to solutions of PDEs as well. Step 1(basis): Let

ωi, i = 1, 2,... 1,2 be orthogonal basis in W0 (Ω) (wrt the standard inner product in 1,2 2 W0 (Ω)), and orthonormal in L (Ω) (with respect to inner prod of L2). Step 2 (approx solutions): Construct approximating solutions by m X m um(x, t) = ci (t)ωi(x). i=1 where the coefficients satisfy Z Z Z ∂um ωk dx = − Dum · Dωk dx + fωk dx (4.31) Ω ∂t Ω Ω for k = 1, 2, . . . , m. Then for LHS Z Z m m ∂um X ∂c ω dx = i ω ω dx ∂t k ∂t i k Ω Ω i ∂cm orthonormality= k ∂t 94 PDE 2 and Z Z m m − Dum · Dωk dx = − ck (t)Dωk · Dωk dx = −ck (t)/λk. Ω Ω Altogether, we obtain ODE ∂cm(t) k = −cm(t)/λ + f (t), ∂t k k k R where fk(t) = Ω f(x, t)ωk(x) dx. It follows that  Z t  m −t/λk τ/λk ck (t) = e ck + e fk(τ) dτ 0 where ck are chosen so that ∞ X g(x) = ckωk(x) k=1 1,2 which is possible since ωi, i = 1, 2,... forms a basis for W0 (Ω). Step 3 (uniform estimates for solutions): Multiplying (4.31) by the coefficients and summing, we obtain Z Z Z ∂um um dx = − Dum · Dum dx + fum dx Ω ∂t Ω Ω i.e. Z Z Z 1 ∂ 2 2 um dx = − |Dum| dx + fum dx. 2 ∂t Ω Ω Ω Further, by integrating over (0, τ), we obtain Z Z 1 2 1 2 um(x, τ) dx − um(x, 0) dx 2 Ω 2 Ω Z Z 2 = − |Dum| dx dt + fum dx dt. Ωτ Ωτ We further estimate by using Young’s and Sobolev-Poincar´e’sinequal- ities Z Z Z T Z 2 2 2 f(x, t)um dx dt ≤ C f dx dt + ε/µ um dx dt Ωτ ΩT 0 Ω Z Z T Z 2 2 ≤ C f dx dt + ε |Dum| dx dt ΩT 0 Ω where µ is again the constant in Sobolev-Poincar´e’sinequality. By choosing ε > 0 small enough, we can absorb the gradient term and PDE 2 95 obtain an important energy estimate Z Z 1 2 2 sup um(x, t) dx+ |Dum| dx dt t∈[0,T ] 2 Ω ΩT Z Z (4.32) 2 2 ≤ C um(x, 0) dx + C f dx dt. Ω ΩT ∂ m Multiplying (4.31) by ∂t ck (t) and summing over k, we obtain Z Z Z ∂um ∂um ∂um ∂um dx = − Dum · D dx + f(x, t) dx Ω ∂t ∂t Ω ∂t Ω ∂t and again integrating over (0,T ) and using Fubini, we have Z 2 Z Z T Z ∂um 1 ∂ 2 ∂um dx dt = − |Dum| dt dx + f dx dt. ΩT ∂t 2 Ω 0 ∂t ΩT ∂t Again by using Young’s inequality Z 2 Z Z ∂um 1 2 2 dx dt+ ( |Dum(x, T )| − |Dum(x, 0)| dx) ∂t 2 ΩT Ω Ω (4.33) Z 2 Z ∂um 2 ≤ ε dx dt + C f dx dt. ΩT ∂t ΩT Combining (4.32) and (4.33), we have Z 2 2 ∂um 2 |Dum| + + |um| dx dt ΩT ∂t Z Z 2 2 (4.34) ≤ C |Dum(x, 0)| dx +C f dx dt, Ω ΩT | {z } →R |Dg(x)|2 dx, as m→∞ where the right hand side is independent of m . Altogether, we have Z 2 2 ∂um 2 |Dum| + + |um| dx dt ≤ C (4.35) ΩT ∂t where C is independent of m. Step 4 (taking limits): Since the estimate (4.35) is uniform in m, 2 1,2 ∂um the sequence um is uniformly bounded in L (0,T ; W (Ω)) and ∂t in 2 L (ΩT ). Thus, there exists a weak limit such that ∂u u ∈ L2(0,T ; W 1,2(Ω)), ∈ L2(Ω ). 0 ∂t T Further, by Thm 4.5, u ∈ C(0,T ; L2(Ω)). 96 PDE 2

Step 5 (weak solution): A priori, um satisfies the weak formula- tion for basis functions, so it remains first to check that u is a weak solution. To this end, let

∞ ∞ h ∈ C0 (Ω) and ψ ∈ C0 ([0,T ]), and choose a sequence

j X 1,2 hj(x) := αkjωk(x) → h in W (Ω) as j → ∞. k=1 We multiply (4.31) by ψ(t), integrate over (0,T ), and pass to the limit m → ∞ to have Z T Z ∂u Z T Z ωkψ dx dt = − Du · Dωkψ dx 0 Ω ∂t 0 Ω Z T Z + fωkψ dx dt. 0 Ω

Then, we multiply this by αkj, and sum up to have

j j Z T Z ∂u X Z T Z X α ω ψ dx dt = − Du · D α ω ψ dx dt ∂t kj k kj k 0 Ω k=1 0 Ω k=1 j Z T Z X + f(x, t) αkjωkψ dx dt. 0 Ω k=1 Then passing to the limit with j, we end up with Z T Z ∂u(x, t) Z T Z h(x)ψ(t) dx dt = − Du(x, t) · Dh(x)ψ(t) dx dt 0 Ω ∂t 0 Ω Z T Z + f(x, t)h(x)ψ(t) dx dt. 0 Ω By modifying the proof of Thm 4.3, see in particular (4.28), we see that by summing up the functions of the type h(x)ψ(t) we may approximate 2 1,2 functions in L (0,T ; W0 (Ω)). Thus, in particular, Z T Z ∂u Z T Z Z T Z ϕ dx dt = − Du · Dϕ dx dt + fϕ dx dt 0 Ω ∂t 0 Ω 0 Ω ∞ for all ϕ ∈ C0 (ΩT ). Step 5 (initial condition): It remains to check that the ini- tial condition is satisfied. Similarly as above, denoting vj(x, t) := PDE 2 97

Pj k=1 βkj(t)ωk(x), j ≤ m, and for which vj(x, T ) = 0 we obtain Z T Z ∂u Z T Z m v dx dt = − Du · Dv (x) dx dt ∂t j m j 0 Ω 0 Ω (4.36) Z T Z + f(x, t)vj(x) dx dt. 0 Ω Integrating by parts wrt t, Z Z T Z ∂vj − um(x, 0)vj(x, 0) dx − um dx dt Ω 0 Ω ∂t Z T Z Z T Z = − Dum · Dvj dx dt + fvj dx dt. 0 Ω 0 Ω Then we pass to the limit m → ∞, and then with j → ∞, where ∞ 2 we may choose βkj so that vj(x, 0) → φ(x) ∈ C0 (Ω) in L (Ω) and vj converges to a suitable test function v. This produces Z Z T Z ∂v g(x)φ(x) dx − u dx dt ∂t Ω 0 Ω (4.37) Z T Z Z T Z = − Du · Dv dx dt + fv dx dt. 0 Ω 0 Ω On the other hand, passing first to the limit m → ∞ and then j → ∞ in (4.36), as well as integrating by parts wrt t after that, we get Z Z T Z ∂v − u(x, 0)φ(x) dx − u dx dt ∂t Ω 0 Ω (4.38) Z T Z Z T Z = − Du · Dv dx dt + fv dx dt. 0 Ω 0 Ω Comparing (4.37) and (4.38), we see that u satisfies the initial condi- tion. We have proven the following.

1,2 2 Theorem 4.9. Let g ∈ W0 (Ω) and f ∈ L (ΩT ). There exists a weak solution to the problem  u = ∆u + f, in Ω  t T u = 0, on ∂Ω × [0,T ] u(x, 0) = g(x), on Ω.

1,2 Remark 4.10. The condition g ∈ W0 (Ω) can be relaxed as well as the operator with ∞ 2 aij, bi, c ∈ L (ΩT ), f ∈ L (ΩT ) 98 PDE 2 is ok, see Evans p. 356. The method remains essentially the same. Method also generalizes for more general bdr conditions g ∈ C(0,T ; L2(Ω))∩ L2(0,T ; W 1,2(Ω)).

∂u 4.2. Standard time mollification. Now ∂t exists but in more gen- eral situation (for example ut = div(A(x, t, Du)) for a suitable nonlin- ear operator), u does not necessarily have time derivative. Nonetheless, it is often useful to have u in the test function, and thus we would have ∂u ∂t in the weak formulation, which does not necessarily exist as a func- tion. This problem is treated by time mollification. ∞ Let φ ∈ C0 (ΩT ). Our goal is to show Z T Z ∂φ Z T Z − uε dz + (Du)ε · Dφ dz = 0, (4.39) 0 Ω ∂t 0 Ω where ε in uε and (Du)ε denote the mollification with respect to t. Let spt φ(x, ·) ⊂ (ε, T − ε). We can use Lebesgue’s dominated con- vergence theorem to see that D R = R D in this case. Further, by Fubini’s theorem and by taking into account the support of φ(x, ·), we see that Z T Z Du(x, t) · Dφε dz 0 Ω Z T Z Z = Du(x, t) · D φ(x, s)ηε(t − s) ds dz 0 Ω R Z Z Z T = Du(x, t) · Dφ(x, s)ηε(t − s) dt dx ds R Ω 0 (4.40) Z Z Z T = Du(x, t) ηε(t − s) dt · Dφ(x, s) dx ds. R Ω 0

Since ηε is an even function, we have Z T Z T Du(x, t) ηε(t − s) dt = Du(x, t) ηε(s − t) dt = (Du(x, s))ε, 0 0 (4.41) where we can restrict ε ≤ s ≤ T − ε. This is due to assumption spt(φ(x, ·)) ⊂ (ε, T − ε). By subtracting (4.41) into (4.40), we obtain Z T Z Du(x, t) · Dφε(x, t) dz 0 Ω (4.42) Z T −ε Z = (Du(x, s))ε · Dφ(x, s) dx ds. ε Ω PDE 2 99

Similarly

Z T Z ∂φ u(x, t) ε dx dt 0 Ω ∂t Z T Z Z ∂ = u(x, t) φ(x, s)η (t − s) ds dx dt ∂s ε 0 Ω R (4.43) Z T −ε Z Z T ∂ = u(x, t)ηε(t − s) dt φ(x, s) dx ds ε Ω 0 ∂s Z T −ε Z ∂ = uε(x, s) φ(x, s) dx ds. ε Ω ∂s

The definition of a weak solution combined with (4.42) and (4.43) imply (4.39).

4.3. Steklov averages. Another alternative is to use Steklov aver- 1 ages. Let u ∈ L (ΩT ). Then the Steklov average is defined as

1 Z t+h uh = u(x, τ) dτ, t ∈ (0,T − h). h t

Weak formulation can also be written (ex) for 0 < t1 < t2 < T as Z Z u(x, t2)ϕ(x, t2) dx − u(x, t1)ϕ(x, t1) dx Ω Ω Z Z − uϕt dx dt + Du · Dϕ dx dt = 0. Ω×(t1,t2) Ω×(t1,t2)

∞ Then choose ϕ(s, x) ∈ C0 (ΩT ) independent of t (this is not compactly supported in t as it is constant in t, but it does not matter). Since ϕ is compactly supported in s, we can choose t1 = s, t2 = s + h for small enough h. Then divide by h, and observe that ϕt = 0 so that

1 Z 0 = (u(x, s + h) − u(x, s))ϕ(x, s) dx h Ω 1 Z Z s+h + Du(x, t) dt · Dϕ(x, s) dx h Ω s Z Z ∂uh(x, s) = ϕ(x, s) dx + Du)h(x, s) · Dϕ(x, s) dx. Ω ∂s Ω 100 PDE 2

Integrate wrt s over (0,T ) to obtain Z Z ∂uh(x, s) 0 = ϕ(x, s) dx ds + (Du)h(x, s) · Dϕ(x, s) dx ds ΩT ∂s ΩT Z ∂ϕ Z = − uh dx ds + (Du)h · Dϕ dx ds, ΩT ∂s ΩT (4.44)

∞ for every ϕ ∈ C0 (ΩT ).

4.4. Uniqueness. In this section, similarly in the elliptic case, we sim- plify the treatment considering

n X Lu = − Di(aijDju) + cu i,j=1 with c ≥ c0, c0 ∈ R. In the proof below, we want avoid using the time derivative of a solution and therefore use mollifications.

Theorem 4.11. There is c0 such that if c ≥ c0, then the weak solution to ( ut + Lu = 0, in ΩT

u = g on ∂pΩT . with g ∈ C(0,T ; L2(Ω)) ∩ L2(0,T ; W 1,2(Ω)) is unique.

Proof. Let u and w be two weak solutions. Then similarly as in (4.39)

n Z ∂vε Z X − u dx dt + (a D u) D vε + cu(vε) dx dt = 0 ε ∂t ij j ε i ε ΩT ΩT i,j=1 where spt v ⊂ ΩT , and ε is small enough, and a similar equation for w. Then by subtracting the equations, we have

n Z ∂vε Z X − (u − w) dx dt + (a (D (u − w)) D (vε) ) ε ∂t ij j ε i ε ΩT ΩT i,j=1 (4.45) ε + c(u − w)(v )ε dx dt = 0. We choose

ε v (x, t) = (χh(t))ε(u − w)ε PDE 2 101 with

0 t ≤ h,   (t − h)/h h < t ≤ 2h, h  χ0,T = 1, 2h < t ≤ T − 2h,  (−t + T − h)/h, T − 2h < t ≤ T − h,  0,T − h < t.

Moreover, by density we can extend the class of test functions so that ε (v )ε is admissible (ex). We estimate

Z ∂vε (u − w)ε dx dt ΩT ∂t Z ∂(χh)ε(u − w)ε = (u − w)ε dx dt ΩT ∂t Z ∂(χh)ε ∂(u − w)ε = (u − w)ε (u − w)ε + (χh)ε dx dt ΩT ∂t ∂t Z Z 2 2 ∂(χh)ε 1 ∂(u − w)ε = (u − w)ε dx dt + (χh)ε dx dt ΩT ∂t ΩT 2 ∂t

Then we integrate by parts and pass to the limit

Z Z int by parts 2 ∂(χh)ε 1 ∂(χh)ε 2 = (u − w)ε dx dt − (u − w)ε dx dt ΩT ∂t 2 ΩT ∂t Z 1 ∂(χh)ε 2 = (u − w)ε dx dt 2 ΩT ∂t 1 Z ∂χ ε =→ 0 h (u − w)2 dx dt 2 ΩT ∂t 1 Z 2h Z 1 Z T −h Z = (u − w)2 dx dt − (u − w)2 dx dt 2h h Ω 2h T −2h Ω 1 Z h =→ 0 0 − (u(x, T ) − w(x, T ))2 dx, 2 Ω where at the last step we used continuity and the initial condition. The other terms in (4.45) converge by similar approximation argu- ments as before, and combining the above calculation together with 102 PDE 2

(4.45), we obtain by first letting ε → 0 and then h → 0 1 Z 0 = (u(x, T ) − w(x, T ))2 dx 2 Ω n Z X + aijDj(u − w)Di(u − w) + c(u − w)(u − v) dx dt ΩT i,j=1 parab 1 Z ≥ (u(x, T ) − w(x, T ))2 dx 2 Ω Z 2 2 + λ|Dj(u − w)| + c(u − w) dx dt ΩT Z Z 1 2 λ  2 ≥ (u(x, T ) − w(x, T )) dx + + c0 (u − w) dx dt 2 Ω ΩT µ where we used Sobolev-Poincar´e’sinequality with a constant µ. If 1 −γ := 2 (λ/µ + c0) > 0 then the result is immediate. Otherwise, let R 2 us denote η(T ) := Ω(u(x, T ) − w(x, T )) dx. Then the above estimate reads as Z T γ η(t) dt ≥ η(T ). 0 We can repeat the argument for a.e. t ∈ (0,T ) instead of T , and have R t γ 0 η(s) ds ≥ η(t). But this is as in well-known Gr¨onwall’s inequality (proof is ex.) which now says η(t) = 0 a.e. completing the proof.  4.5. Cα regularity using Moser’s method. For simplicity we con- ∂u ∞ centrate on ∂t = ∆u+f, f ∈ L (ΩT ) but method immediately extends to more general linear PDEs.

2 1,2 Definition 4.12 (supersolution). A function u ∈ Lloc(0,T ; Wloc (Ω)) ∂u is a weak supersolution to ∂t = ∆u + f, if Z ∂ϕ Z Z − u dx dt + Du · Dϕ dx dt ≥ fϕ dx dt ΩT ∂t ΩT ΩT ∞ for every ϕ ∈ C0 (ΩT ), ϕ ≥ 0.

2 1,2 Definition 4.13 (subsolution). A function u ∈ Lloc(0,T ; Wloc (Ω)) is ∂u a weak subsolution to ∂t = ∆u + f, if Z ∂ϕ Z Z − u dx dt + Du · Dϕ dx dt ≤ fϕ dx dt ΩT ∂t ΩT ΩT ∞ for every ϕ ∈ C0 (ΩT ), ϕ ≥ 0. PDE 2 103

∂u Formally we can write for example for subsolution ∂t − ∆u ≤ f. Rough plan:

We will describe details, notation etc. later, but look at rough ideas to begin with. We look at parabolic Harnack’s inequality. The elliptic Harnack’s inequality for positive harmonic function in Ω reads as ess sup u ≤ C ess inf u B B where 2B ⊂ Ω (local estimate). In contrast with this, in parabolic Harnack’s inequality the sets on RHS/LHS are not the same. Instead, they take into account the flow of information from the past to the future. Indeed, parabolic Harnack’s inequality for a positive solution to the heat equation can be stated as ess sup u ≤ C ess inf u Q− Q+ where Q− lies in the past compared to Q+, where the cylinder lie well within the domain (again a local estimate) . There are counterexamples (ex) showing that this so called waiting time is indispensable. (1) (Easier part ) We intend to show that a positive subsolution is bounded from above with explicit estimate Z ess sup u ≤ C u dx dt Q Q˜ where Q, Q˜ are parabolic cylinders . (2) (Harder part ) We partly show a lower bound for a positive weak supersolution in a form Z u dx dt ≤ C ess inf u. Q− Q˜+ In this estimate, direction of time plays a crucial role. Lemma 4.14 (Energy estimate). Let u ≥ 0 be a weak subsolution to ∂u ∂t − ∆u ≤ 0 in ΩT and γ ≥ 1. Then there exists C = C(γ) such that Z Z |Du|2uγ−1η2 dx dt + ess sup u1+γη2 dx dt ΩT t∈(0,T ) Ω Z Z 1+γ 2 1+γ ∂η ≤ C u |Dη| dx dt + C u η dx dt ΩT ΩT ∂t ∞ for every η ∈ C0 (ΩT ), η ≥ 0. 104 PDE 2

2 h γ Proof. Use (formal, details are ex.) test function ϕ = η χ0,tu (now γ is a power) in Z ∂ϕ Z − u dx ds + Du · Dϕ dx ds ≤ 0. ΩT ∂t ΩT First term can be estimated by integration by parts as Z ∂ϕ Z ∂(η2χh uγ) u dx ds = u 0,t dx ds ΩT ∂t ΩT ∂t Z 2 h ∂η χ0,t γ 2 h γ−1 ∂u = u( u + η χ0,tγu ) dx ds ΩT ∂t ∂t 1 Z ∂η2χh = 0,t uγ+1 dx ds. 1 + γ ΩT ∂t For the second term Z Du · Dϕ dx ds ΩT Z 2 h γ = Du · D(η χ0,tu ) dx ds ΩT Z 2 h 2 γ−1 h 2 γ = η χ0,tγ|Du| u + Du · χ0,tDη u dx ds ΩT Z 2 h 2 γ−1 (γ−1)/2 h 2 (γ+1)/2 = γη χ0,t|Du| u + u Du · χ0,t(Dη )u dx ds. ΩT Then use Young’s inequality to estimate the second term. Finally combine the estimates and absorb the gradient term into the left, choose t suitably so that one of the terms is close to ess sup-term (detailed calculation was presented during the lecture).  Choosing γ = 2qk − 1, k = 0, 1, 2, . . . , q ≥ 1 gives the following corollary. Also observe that as γ increases, the constants in the previous lemma remain bounded. Thus we can choose the constant independent of k below.

Corollary 4.15. Let u ≥ 0 be a subsolution in ΩT and q ≥ 1. Then there exists C = C(q) such that

Z k 2 Z k q 2 2q 2 Du η dx dt + ess sup u η dx dt ΩT t∈(0,T ) Ω Z Z 2qk 2 2qk ∂η ≤ C u |Dη| dx dt + C u η dx dt ΩT ΩT ∂t ∞ for every η ∈ C0 (ΩT ), η ≥ 0. PDE 2 105

2 1,2 Lemma 4.16 (Parabolic Sobolev’s inequality). Let u ∈ L (0,T ; W0 (Ω)) and q = 1 + 2/n. Then there exists C = C(n) such that

Z  Z 2/n Z |u|2q dx dt ≤ C ess sup |u|2 dx |Du|2 dx dt ΩT t∈(0,T ) Ω ΩT Proof. By H¨older’sinequality for a.e. t ∈ (0,T ) Z Z |u|2q dx ≤ |u|1−q+2q+(q−1) dx Ω Ω  Z (n−1)/n Z 1/n ≤ |u|(1+q)n/(n−1) dx |u|(q−1)n dx Ω Ω  Z (n−1)/n Z 1/n ≤ |u|(1+q)n/(n−1) dx |u|2 dx Ω Ω Then using Sobolev’s inequality with 1∗ = n/(n − 1) and 1, we have

 Z (n−1)/n Z 1 (1+q)n/(n−1) (1+q) |u| dx ≤ C D(|u| ) dx Ω Ω Z = C |u|q|Du| dx Ω H¨older  Z 1/2 Z 1/2 ≤ C |u|2q dx |Du|2 dx . Ω Ω Then we combine the estimates, integrate over (0,T ) and estimate by ess sup to obtain the result.  For notational convenience we consider the domain around the origin. This can be done without loss of generality.

2 2 QR = B(0,R) × (−R ,R ),

We assume that 2 < n and R ≤ 1.

∂u Lemma 4.17. Let u ≥ 0 be a subsolution to ∂t − ∆u ≤ f in Q2R . Then there are C = C(n) such that

Z 1/2  2  ess sup u ≤ C u dx dt + C ||f|| ∞ . L (QR) QR/2 QR Proof. The proof consists of several steps:

Step 1 (simplification): Set w = u + (t − t) ||f|| ∞ , where max L (QR) 106 PDE 2 tmax is a suitable constant so that tmax − t ≥ 0, and observe that Z ∂ϕ Z − w dx dt + Dw · Dϕ dx dt QR ∂t QR Z ∂ϕ Z Z = − u dx dt + Du · Dϕ dx dt − ||f|| ∞ ϕ dx dt L (QR) QR ∂t QR QR Z ≤ (f − ||f|| ∞ )ϕ dx dt ≤ 0. L (QR) QR

∂w Thus we may concentrate on the homogenous equation ∂t − ∆w ≤ 0. If the results holds for w  Z 1/2 ess sup w ≤ C w2 dx dt QR/2 QR this then implies

Z 1/2  2  ess sup u ≤ C u dx dt + C ||f|| ∞ . L (QR) QR/2 QR

Step 2 (reverse H¨older): By step 1, let u ≥ 0 be a subsolution to ∂u ∂t − ∆u ≤ 0. Let ρ, σ be such that R ≤ ρ < σ ≤ R 2

∞ and choose a cut-off function η ∈ C0 (Qσ), 0 ≤ η ≤ 1 such that η = 1 in Qρ and

1 ∂η 2 C |Dη| + ≤ ∂t σ − ρ

By Corollary 4.15 (the same proofs give the estimates in Qσ instead of ΩT ) choosing k = 0, we have Z Z |Du|2η2 dx dt + ess sup u2η2 dx dt 2 2 Qσ t∈(−σ ,σ ) B(0,σ) Z Z 2 2 2 ∂η ≤ C u |Dη| dx dt + C u η dx dt Qσ Qσ ∂t Z C 2 ≤ 2 u dx dt. (ρ − σ) Qσ PDE 2 107

By using parabolic Sobolev’s inequality in Lemma 4.16 and then the previous estimate, we deduce  Z 1/q u2q dx dt Qρ  Z 2/(nq) Z 1/q ≤ C1/q ess sup (ηu)2 dx |D(ηu)|2 dx dt t σB Qσ (2/n + 1)/q Z Z 1/q 2 2 2 | {z } ≤ C ess sup η u dx + |D(ηu)| dx dt 1 t σB Qσ  Z Z  ≤ C1/q ess sup η2u2 dx + |Dη|2u2 + η2|Du|2 dx dt t σB Qσ 1/q Z C 2 ≤ 2 u dx dt. (σ − ρ) Qσ Similarly

2  Z 2 1/q u2q dx dt Qρ 2 2 parab Sobo 2  Z 2/(nq ) Z 1/q ≤ C1/q ess sup (ηuq)2 dx |D(ηuq)|2 dx dt t B(0,σ) Qσ (2/n + 1)/q2 Z Z 2  | {z } ≤ C1/q ess sup η2u2q dx + |D(ηuq)|2 dx dt 1/q t B(0,σ) Qσ

2  Z Z 1/q ≤ C1/q ess sup η2u2q dx + |Dη|2u2q + η2|Duq|2 dx dt t B(0,σ) Qσ Cor 4.15, k = 1 1/q Z 1/q  C 2q  ≤ 2 u dx dt . (σ − ρ) Qσ This argument in general yields

Z 1/(2qk+1)  1 2qk+1  n+2 u dx dt R Q ρ (4.46) 1/q Z 1/(2qk)  C 2qk  ≤ n 2 u dx dt . R (ρ − σ) Qσ

Step 3 (Moser’s iteration): Replace in (4.46) ρ by ρk+1 and σ by ρk where R ρ = (1 + 2−k), k = 0, 1,... k 2 108 PDE 2

R −k −k−2 so that ρk − ρk+1 = 2 2 (1 − 1/2) = R2 . Thus

k+1 1/q 2(k+2) k  1 Z k+1 1/(2q ) C 2 Z k 1/(2q ) u2q dx dt ≤ u2q dx dt . Rn+2 Rn+2 Qρk+1 Qρk We iterate this

k+1  1 Z k+1 1/(2q ) u2q dx dt Rn+2 Qρk+1 k k+1 k 1 Z k 1/(2q ) ≤ C1/(2q )2(k+2)/q ( u2q dx dt Rn+2 Qρk k−1 k+1 k k k−1  1 Z k−1 1/(2q ) ≤ C1/(2q )2(k+2)/q C1/(2q )2((k−1)+2)/q u2q dx dt Rn+2 Qρk−1 k−1 k+1 k k k−1  1 Z k−1 1/(2q ) ≤ C1/(2q )2(k+2)/q C1/(2q )2((k−1)+2)/q u2q dx dt Rn+2 Qρk−1 ...

0 ∗  Z 1/2 ≤ Cγ 2γ u2 dx dt ,

Qρ0 where ∞ ∞ X 1 X 2(i + 2) γ0 = , γ∗ = . 2qi qi i=1 i=0 Then let k → ∞ and observe that the LHS in the above estimate converges (see Measure and integration 1) to ess sup u. QR/2  Lemma 4.18 (Iteration lemma). Let G(s) be a bounded and nonneg- ative function for s ∈ [R/2,R] C G(ρ) ≤ θG(σ) + 0 (σ − ρ)α where θ < 1 and R/2 ≤ ρ < σ ≤ R. Then there is C = C(α, θ) such that  C  G(ρ0) ≤ C 0 , (σ0 − ρ0)α where R/2 ≤ ρ0 < σ0 ≤ R.

Proof. Ex.  PDE 2 109

∂u Corollary 4.19. Let u ≥ 0 be a subsolution to ∂t − ∆u ≤ f in Q2R. Then there is C = C(n, s) such that

Z 1/s  s  ess sup u ≤ C u dx dt + C ||f|| ∞ . L (QR) QR/2 QR for s > 0. Proof. First, we may again without loss of generality restrict ourselves to the homogenous case. If s ≥ 2, then the result follows directly from the previous lemma and H¨older’sinequality. Let then 0 < s < 2. Using the result of previous −i lemma with σ, ρ instead of R/2,R, and ρi = ρ + 2 (σ − ρ) as well as inspecting carefully the proof, we get

k+1  1 Z k+1 1/(2q ) u1/(2q ) dx dt σn Qρk+1 k  C Z k 1/(2q ) ≤ u2q dx dt σn(σ − ρ)2 Qρk k  1/q k σ  C Z k 1/(2q ) ≤ u2q dx dt . σ − ρ σn+2 Qρk P∞ i Iterating this, observing that i=0 1/q = 1/(1−1/q) = 1/((q−1)/q) = q/(q − 1) = (1 + 2/n)/(2/n) = (n + 2)/2, and then using Young’s inequality to the resulting inequality we have

P∞ i   i=0 1/q Z 1/2 γ0 γ∗ σ  1 2  ess sup u ≤ C 2 n+2 u dx dt Qρ σ − ρ σ Qσ  (n+2)/2 Z 1/2 σ  1 2  ≤ C n+2 u dx dt σ − ρ σ Qσ  1 Z 1/2 ≤ C (ess sup u)2−sus dx dt (σ − ρ)n+2 Qσ Qρk Young Z 1/s 1  1 s  ≤ ess sup u + C n+2 u dx dt , 2 Qσ (σ − ρ) Qσ since (2−s)/2+s/2 = 1. Then use iteration Lemma 4.18 with ρ0 = R/2 and σ0 = R , we get Z 1/s C  s  ess sup u ≤ (n+2)/s u dx dt .  QR/2 (R − R/2) QR Next we consider the second part. 110 PDE 2

∂u Lemma 4.20. Let u ≥ δ > 0 be a weak supersolution to ∂t − ∆u ≥ 0. Then w = u−1 is a weak subsolution. Proof. First observe that u−1 ≤ δ−1 and |Du−1| = |u−2Du| ≤ δ−2|Du| so that u−1 is in the right parabolic Sobolev space. We choose (for- −2 ∞ mally) a test function ϕ = ηu with η ∈ C0 (ΩT ), η ≥ 0, and calculate Z ∂ϕ 0 ≤ −u + Du · Dϕ dx dt ΩT ∂t Z ∂(ηu−2) = −u + Du · D(ηu−2) dx dt ΩT ∂t Z ∂η ∂u = −u( u−2 − 2ηu−3 ) + Du · (u−2Dη − 2ηu−3Du) dx dt ΩT ∂t ∂t Z ∂η ∂u−1 = − u−1 − 2η − Du−1 · Dη − 2ηu−3|Du|2 dx dt ΩT ∂t ∂t Z ∂η ≤ u−1 − Du−1 · Dη dx dt ΩT ∂t where at the last step we integrated by parts and dropped the negative term. Thus Z ∂η −1 −1 0 ≥ − u + Du · Dη dx dt.  ΩT ∂t

∂u Lemma 4.21. Let u ≥ δ > 0 be a weak supersolution to ∂t − ∆u ≥ 0. Then there is C = C(n, s) such that

 Z −1/s u−s dx dt ≤ C ess inf u. Q QR R/2 for any s > 0. Proof. By the previous lemma, u−1 is a subsolution. Then by Corollary 4.19, we have  Z 1/s ess sup u−1 ≤ C (u)−s dx dt QR/2 QR so that  Z −1/s u−s dx dt ≤ ess inf u. Q QR R/2  PDE 2 111

We denote Q˜ = B(0,R) × (−3R2, 3R2), Q˜+ = B(0,R) × (R2, 3R2), Q˜− = B(0,R) × (−3R2, −R2), (4.47) Q+ = B(0,R/2) × (2R2 − (R/2)2, 2R2 + (R/2)2), Q− = B(0,R/2) × (−2R2 − (R/2)2, −2R2 + (R/2)2). The proof of the following deep theorem can be found in Fabes and Garofalo: Parabolic B.M.O and Harnack’s inequality.

∂u Theorem 4.22. Let u ≥ δ > 0 be weak supersolution to ∂t − ∆u ≥ 0 in Q2R. Then there is s > 0 and C = C(n) such that  Z 1/s  Z −1/s us dx dt ≤ C u−s dx dt . Q˜− Q˜+ Combining the previous two results i.e. Lemma 4.21 and Theorem 4.22, we immediately obtain weak Harnack’s inequality. One could show (not done here) that this holds for 0 < s < (n + 2)/2 with C = C(n, s) and in particular with s = 1.

∂u Theorem 4.23. Let u ≥ δ > 0 be a weak supersolution to ∂t −∆u ≥ 0 in Q2R. Then there is s > 0 and C = C(n) such that  Z 1/s us dx dt ≤ C ess inf u. + Q˜− Q

∂u Corollary 4.24. Let u ≥ δ > 0 be a weak supersolution to ∂t −∆u ≥ f in Q2R. Then there is s > 0 and C = C(n) such that

Z 1/s  s  u dx dt ≤ C ess inf u + C ||f|| ∞ ˜ . + L (Q) Q˜− Q

Proof. Observe that u+(t−tmin) ||f||L∞(Q˜) ≥ δ, where tmin is a constant ∂u such that t − tmin ≥ 0, is a weak supersolution to ∂t − ∆u ≥ 0 and thus by the previous theorem

Z 1/s  s  (u + (t − tmin) ||f||L∞(Q˜)) dx dt Q˜−

≤ C ess inf(u + (t − tmin) ||f||L∞(Q˜)). Q+

This implies the result.  112 PDE 2

Then by weak Harnack’s inequality (Corollary 4.24 ) and local bound- ∂u edness estimate (Corollary 4.19), we get for a weak solution of ∂t − ∆u = f that

Z 1/s  s  ess sup u ≤ C u dx dt + C ||f||L∞(Q˜) Q− Q˜− 2 ≤ C ess inf u + C ||f||L∞(Q˜) . Q+ This finally gives us parabolic Harnack’s inequality.

∂u Theorem 4.25 (Harnack). Let u ≥ δ > 0 be a weak solution to ∂t − ∆u = f in Q˜. Then there is C = C(n) such that

ess sup u ≤ C ess inf u + C ||f||L∞(Q˜) . Q− Q+ Remark 4.26. The assumption u ≥ δ > 0 is only technical: if u ≥ 0, we may consider u + δ and since the constant in Harnack’s inequality is independent of δ, we may let δ → 0.

Example 4.27. ”Elliptic” Harnack’s ie., where we have same cylinder ∂u on both sides, does not hold in the parabolic case: the equation ∂t − 2 2 uxx = 0 has a nonnegative solution in (−R,R) × (−R ,R ) (translated fundamental solution)

2 1 − (x+ξ) u(x, t) = √ e 4(t+2R2) t + 2R2 where ξ is a constant. Let x ∈ (−R/2,R/2), x 6= 0 and t ∈ (−R2,R2). Then

2 2 2 2 u(0, t) − ξ −(x+ξ) − −x −2xξ x +2xξ = e 4(t+2R2) = e 4(t+2R2) = e 4(t+2R2) → 0 u(x, t) as ξ sign x → −∞.

4.6. H¨oldercontinuity. By iterating (weak) Harnack’s inequality we may prove the local H¨oldercontinuity of weak solutions.

∂u Theorem 4.28. Let u be a positive weak solution to ∂t − ∆u = 0. Then there exists γ ∈ (0, 1) and a representative such that

|u(x, t) − u(y, s)| ≤ C(|x − y| + |t − s|1/2)γ locally. PDE 2 113

Proof. We take the weak Harnack for s = 1 for 1granted and using that for a weak (super)solutions u − ess infQ˜ u and ess supQ˜ u − u, we have Z   u dx − ess inf u ≤ C ess inf u − ess inf u + Q˜ Q˜ Q Q˜ Z   ess sup u − u dx ≤ C ess sup u − ess sup u . Q˜ Q˜ Q˜ Q+ Summing up yields   oscQ˜ u ≤ C oscQ˜ u − oscQ+ u , where we denoted

oscQ˜ u := ess sup u − ess inf u. ˜ Q˜ Q Rearranging the terms, we have  1  osc + u ≤ 1 − osc u. Q C Q˜ Thus by setting θ := 1 − 1/C ∈ (0, 1) we obtain

oscQ+ ≤ θ oscQ˜ u. (4.48) The proof of (weak) Harnack would also work in the geometry Q˜ := B(0,R) × (−R2,R2) Q+ := B(0,R/2) × (R2/2 − (R/2)2,R2/2 + (R/2)2). Using this and denoting, with a slight abuse of notation,

k  k 2 k 2 Qk := B(0,R/2 ) × tk − (R/2 ) , tk + (R/2 ) for a suitable tk we obtain oscQ1 u ≤ θ oscQ0 u. Repeating the argument, we deduce k oscQk u ≤ θ oscQ0 u. Then fix ρ < R and k such that 2k < R/ρ ≤ 2k+1, k = 0, 1, 2,... so that 2−kR > ρ ≥ 2−(k+1)R and k ≤ log(R/ρ)/ log(2) ≤ k + 1 i.e. log(R/(2ρ))/ log(2) ≤ k

1 R Or use (strong) Harnack to have Q− u dx ≤ C ess infQ+ u. Then applying this to u − ess infQ˜ u and ess supQ˜ u − u gives the same oscillation estimate. 114 PDE 2

Thus

log(R/(2ρ))/ log(2) oscQρ u ≤ θ oscQ0 u log(R/(2ρ)) log(θ)/(log(θ) log(2)) = θ oscQ0 u  ρ log(θ)/ log(2) = osc u 2R Q0 − log(θ)/ log(2)  ρ | {z } = C =:γ osc u. R Q0

Since ρ can be chosen arbitrarily small and u is locally bounded, this implies H¨older-continuity. 

4.7. Remarks. Also a similar regularity theory that we established ∂u for the elliptic equations can be developed for ∂t + Lu = f if the coefficients are smooth enough. Intuition: We consider formally the heat equation

( ∂u n ∂t − ∆u = f in R × (0,T ] u = g on Rn × {0} and u decays fast enough at infinity. Then integration by parts gives

Z Z ∂u f 2 dx = ( − ∆u)2 dx Rn Rn ∂t Z ∂u 2 ∂u = − 2 ∆u + ∆u2 dx Rn ∂t ∂t Z ∂u 2 ∂Du = + 2 · Du + ∆u2 dx Rn ∂t ∂t

Then we calculate

Z t Z ∂Du Z t Z ∂|Du|2 · Du dx ds = dx ds 0 Rn ∂t 0 Rn ∂t Z Z init= cond |Du(x, t)|2 dx − |Dg|2 dx. Rn Rn PDE 2 115

Moreover, similarly as with the elliptic equations

Z Z n 2 n 2 2 X ∂ u X ∂ u (∆u) dx = 2 2 dx n n ∂x ∂x R R i=1 i j=1 j n X Z ∂2u ∂2u = 2 2 dx n ∂x ∂x i,j=1 R i j n Z 3 int by parts X ∂ u ∂u = − 2 dx n ∂x ∂x ∂x i,j=1 R i j j n Z 2 2 int by parts X ∂ u ∂ u = dx n ∂x ∂x ∂x ∂x i,j=1 R i j i j Z 2 2 = D u dx. Rn Choosing t so that R |Du|2(x, t) dx ≥ 1 sup R |Du(x, t)|2 dx Rn 2 t∈(0,T ) Rn and combining the estimates, we end up with

Z T Z 2 Z ∂u 2 2 2 + D u dx dt + sup |Du(x, t)| dx 0 Rn ∂t t∈(0,T ) Rn Z T Z Z ≤ C |f|2 dx dt + C |Dg|2 dx. 0 Rn Rn Continuing in this way (cf. elliptic), we would obtain higher regularity estimates as well. The solution has two more space derivatives than f etc. To make above conclusions rigorous, we could again utilize difference quotients both in space and time.

5. Schauder estimates We finish the course by briefly returning to the elliptic theory, and sketching the Schauder theory because this is needed to finish the story with Hilbert’s 19th problem. Recall H¨oldercontinuity

Definition 5.1. Let u :Ω → R. For α ∈ (0, α), we denote the semi- norm |f(x) − f(y)| |u|Cα(Ω) = sup α , x,y∈Ω,x6=y |x − y| 116 PDE 2

α and a set of all functions satisfying |u|Cα(Ω) < ∞ by C (Ω). This space can be equipped with the norm |f(x) − f(y)| ||f||Cα(Ω) = ||f||L∞(Ω) + sup α . x,y∈Ω,x6=y |x − y| Similarly Ck,α(Ω) = {u : Dβu ∈ Cα(Ω) for |β| ≤ k}, where β is a multi-index. The main result of this section is Theorem 5.2. Let u be a weak solution to −∆u = f in B(0, 2R) with f ∈ Cα(B(0, 2R)). Then u ∈ C2,α(B(0,R/4)) with an explicit estimate. Remark 5.3. • Theorem 5.2 actually comes with estimate, see Theorem 5.13. • The result can be extended to Lu = f with

||aij||Cα(B(0,2R)) , ||bi||Cα(B(0,2R) , ||c||Cα(B(0,2R) ≤ M,

uniform ellipticity, and aij = aji by the freezing of coefficients technique. • In regular domains, there is also a corresponding global result. First, we look at the important step i.e. how to pass from integral estimates (natural from the point of view what we have done so far) to the pointwise H¨oldernorms. For this, we use a theory of Campanato spaces. Denote 1 Z ux,ρ = u(y) dy. |Ω ∩ B(x, ρ)| Ω∩B(x,ρ) where Ω is a regular domain, for example Ω = B(0,R). Definition 5.4 (Campanato space). Let µ ≥ 0 and u ∈ L2(Ω). Then the functions satisfying Z 1/2  1 2  |u|L2,µ(Ω) = sup µ |u(y) − ux,ρ| dy < ∞ x∈Ω,0<ρ

||u||L2,µ(Ω) = |u|L2,µ(Ω) + ||u||L2(Ω) . Lemma 5.5 (Mean value lemma). Let u ∈ L2,µ(Ω), x ∈ Ω and 0 < ρ < R < diam(Ω). Then −n/2 µ/2 |ux,R − ux,ρ| ≤ Cρ R |u|L2,µ(Ω). PDE 2 117

Proof. Let y ∈ B(x, ρ) ∩ Ω and write

2 2 2 |ux,R − ux,ρ| ≤ C(|ux,R − u(y)| + |u(y) − ux,ρ| , integrate over B(x, ρ) ∩ Ω ⊂ B(x, R) ∩ Ω to have Z 2 2 |ux,R − ux,ρ| = |ux,R − ux,ρ| dy B(x,ρ)∩Ω Z 2 2 ≤ C |ux,R − u(y)| + |u(y) − ux,ρ| dy B(x,ρ)∩Ω Z 2 2 ≤ C |ux,R − u(y)| + |u(y) − ux,ρ| dy B(x,ρ)∩Ω Rµ ρµ  ≤ C + |u|2 ρn ρn L2,µ(Ω) Rµ ≤ 2C |u|2 . ρn L2,µ(Ω) 

In the proof of the key result, we need integral characterization of H¨oldercontinuous functions i.e. Campanato estimate. Lemma 5.6 (Integral characterization of H¨oldercontinuous functions). Let n < µ ≤ n + 2. Then L2,µ(Ω) = Cα(Ω) and

−1 2 C |u|Cα(Ω) ≤ |u|L2,µ(Ω) ≤ C |u|Cα(Ω) with α = (µ − n)/2 and C = C(n, µ).

Interpretation: Cα(Ω) ⊂ L2,µ(Ω) and each u ∈ L2,µ(Ω) has a presen- tativeu ˜ in Cα(Ω).

Proof. The second inequality: Let u ∈ Cα(Ω), x ∈ Ω and 0 < ρ < diam(Ω) and y ∈ Ω ∩ B(x, ρ). We have Z

|u(y) − ux,ρ| = u(y) − u(z) dz B(x,ρ)∩Ω Z ≤ |u(y) − u(z)| dz B(x,ρ)∩Ω Z α ≤ |u|Cα(Ω) |y − z| dz B(x,ρ)∩Ω Z C|u|Cα(Ω) α ≤ n |y − z| dz = ∗ ρ B(x,ρ)∩Ω 118 PDE 2 since |Ω ∩ B(x, ρ)|C ≥ ρn . Moreover, since y − x ∈ B(0, ρ) it follows that B(y −x, ρ)∩Ω ⊂ B(0, 2ρ)∩Ω and by the change of variables that Z C|u|Cα(Ω) α ∗ ≤ n |z| dz ρ B(0,2ρ)∩Ω Z C|u|Cα(Ω) α ≤ n |z| dz ρ B(0,2ρ) (5.49) Z 2ρ C|u|Cα(Ω) n−1+α ≤ n r dz ρ 0 α ≤ C|u|Cα(Ω)ρ . Hence Z 1 2 2 2α−µ µ |u(y) − ux,ρ| dy ≤ C|u|Cα(Ω)ρ |B(x, ρ) ∩ Ω| ρ B(x,ρ)∩Ω 2α − µ + n

2 | {z0 } ≤ C|u|Cα(Ω)ρ and thus the second inequality follows. The proof of the first inequality is in three steps: construction of the representativeu ˜, showing thatu ˜ = u a.e., and showing thatu ˜ ∈ Cα(Ω) Step1(construction of the representative u˜): Let x ∈ Ω, 0 < R < −i diam(Ω) and Ri = 2 R, i = 0, 1,.... Then by Lemma 5.5 −n/2 µ/2 ux,Rj − ux,Rj+1 ≤ CRj+1 Rj |u|L2,µ(Ω) j(n−µ)/2+n/2 (µ−n)/2 = C2 R |u|L2,µ(Ω). Thus for 0 ≤ j < i

ux,Rj − ux,Rj+1 + ux,Rj+1 − ... + ux,Ri−1 − ux,Ri i−1 (µ−n)/2 X k(n−µ)/2+n/2 ≤ CR |u|L2,µ(Ω) 2 k=j i−j−1 (µ−n)/2 j(n−µ)/2+n/2 X k(n−µ)/2+n/2 = CR |u|L2,µ(Ω)2 2 k=0 1 − 2(i−j)(n−µ)/2+n/2 = CR(µ−n)/2|u| 2j(n−µ)/2+n/2 L2,µ(Ω) 1 − 2(n−µ)/2+n/2 (µ−n)/2 = CRj |u|L2,µ(Ω), where C = C(n, µ). We have derived the estimate (µ−n)/2 ux,Rj − ux,Ri ≤ CRj |u|L2,µ(Ω), (5.50) PDE 2 119

It follows that ux,Ri , i = 0, 1, 2,... is a Cauchy sequence. Hence we may define

u˜(x) = lim ux,R , x ∈ Ω. i→∞ i It also holds that the limit does not depend on the particular choice −i of R. To see this, take 0 < r < R and let ri = 2 r, i = 0, 1,.... Then again by Lemma 5.5 −n/2 µ/2 |ux,Ri − ux,ri | ≤ Cri Ri |u|L2,µ(Ω)  n/2 Ri (µ−n)/2 ≤ C Ri |u|L2,µ(Ω) ri Rn/2 ≤ C R(µ−n)/2|u| → 0 r i L2,µ(Ω) as i → ∞, since µ > n. Thusu ˜R(x) =u ˜r(x). Moreover, by (5.50) setting j = 0 and letting i → ∞

(µ−n)/2 |ux,R − u˜(x)| ≤ CR |u|L2,µ(Ω) (5.51) so thatu ˜(x) = limR→0 ux,R. Step2 (u˜ = u a.e): By Lebesgue’s theorem

Leb. u˜(x) = lim ux,R = u(x) a.e. in Ω. R→0 Step3 (u˜ ∈ Cα(Ω)): Let x, y ∈ Ω, x 6= y and set R := |x − y|. By (5.51)

|u˜(x) − u˜(y)| ≤ |u˜(x) − ux,R| + |ux,R − uy,R| + |uy,R − u˜(y)| (µ−n)/2 ≤ CR |u|L2,µ(Ω) + |ux,R − uy,R|. Set G = Ω ∩ B(x, 2R) ∩ B(y, 2R). Observe that G ⊂ Ω ∩ B(x, 2R) and G ⊂ Ω ∩ B(y, 2R)), and C|G| ≥ Rn because Ω is smooth. Estimate the second term on the RHS as

|ux,R − uy,R| Z = |ux,R − uy,R| dz G 1−1/2 Z 1/2 |Ω ∩ B(x, 2R)|  2  ≤ |ux,R − u(z)| dz |G| Ω∩B(x,2R) 1−1/2 Z 1/2 |Ω ∩ B(y, 2R)|  2  + |u(z) − uy,R| dz |G| Ω∩B(y,2R) (µ−n)/2 ≤ CR |u|L2,µ(Ω). 120 PDE 2

Combining the estimates, we have

(µ−n)/2 |u˜(x) − u˜(y)| ≤ CR |u|L2,µ(Ω) so thatu ˜ ∈ Cα(Ω) with α = (µ − n)/2, and

|u|Cα(Ω) ≤ C|u|L2,µ(Ω).  Calculation (5.49) gives us a useful estimate, that we state separately as a lemma Lemma 5.7. Let u ∈ Cα(Ω), x ∈ Ω. Then Z 2 2 n+2α |u(y) − ux,ρ| dy ≤ C|u|Cα(Ω)ρ . B(x,ρ)∩Ω Without loss of generality, as long as we have the uniqueness, we may derive the Schauder estimates assuming smoothness, by using smooth approximations, and passing to the limit at the end. Lemma 5.8. Let u be a weak solution ∆u = 0 in B(0, 2R). Then for any 0 < ρ ≤ R, it holds Z  ρ n Z u2 dx ≤ C u2 dx B(0,ρ) R B(0,R) Z n+2 Z 2  ρ  2 (u − uρ) dx ≤ C (u − uR) dx, B(0,ρ) R B(0,R) with C = C(n). Proof. The first estimate: By the elliptic counterpart of the (ess)sup- estimate (cf. Lemma 4.17, and ex 13 in set 3 ), we have for 0 < ρ < R/2 Z  ρ n Z u2 dx ≤ Cρn sup u2 ≤ C u2 dx. B(0,ρ) B(0,ρ) R B(0,R) For R/2 ≤ ρ ≤ R the result immediately follows Z  ρ n Z u2 dx ≤ C u2 dx. R B(0,ρ) |{z} B(0,R) ≥C The second estimate: The second follows from the first one by observing that w = Diu is also a solution to the Laplace equation, and thus by the first estimate Z n Z 2  ρ  2 (Diu) dx ≤ C (Diu) dx (5.52) B(0,ρ) R B(0,R) PDE 2 121

Summing over i, assuming 0 < ρ < R/2, and using Poincar´e’sinequal- ity Z Poinc. Z 2 2 2 (u − uρ) dx ≤ Cρ |Du| dx B(0,ρ) B(0,ρ)  ρ n Z ≤ Cρ2 |Du|2 dx R B(0,R/2) By Caccioppoli’s inequality (ex) Z Z 2 C 2 |Du| dx ≤ 2 (u − uR) dx. B(0,R/2) R B(0,R) Combining the previous two estimates, we have Z n+2 Z 2  ρ  2 (u − uρ) dx ≤ C (u − uR) dx. B(0,ρ) R B(0,R) The case R/2 ≤ ρ ≤ R is again easier: Z 2 (u − uρ) dx B(0,ρ) Z 2 ≤ (u − uR + uR − uρ) dx B(0,ρ) Z Z Z 2 2 ≤ C (u − uR) dx + C (uR − u) dx dx B(0,ρ) B(0,ρ) B(0,ρ) (5.53) Z 2 ≤ C (u − uR) dx B(0,R)  ρ n+2 Z ≤ C (u − u )2 dx. R R |{z} B(0,R) ≥C  Lemma 5.9. Let u be a solution to ∆u = f in B(0, 2R), and let α w = Diu, f ∈ C (B(0, 2R)). Then for 0 < ρ ≤ R Z 1 2 n+2α |Dw − (Dw)ρ| dx ρ B(0,ρ) Z C 2 2 ≤ n+2α |Dw − (Dw)R| dx + C|f|Cα(B(0,R)). R B(0,R)

Proof. Decompose w = w1 + w2, where ( −∆w1 = 0 in B(0,R)

w1 = w on ∂B(0,R) 122 PDE 2 and ( −∆w2 = Dif = Di(f − fR) in B(0,R)

w2 = 0 on ∂B(0,R) in the weak sense. Then use Lemma 5.8 for Diw1 (this is also a solution to Laplace eq) to have

Z n+2 Z 2  ρ  2 (Diw1 − (Diw1)ρ) dx ≤ C (Diw1 − (Diw1)R) dx B(0,ρ) R B(0,R) Summing over i Z n+2 Z 2  ρ  2 (Dw1 − (Dw1)ρ) dx ≤ C (Dw1 − (Dw1)R) dx B(0,ρ) R B(0,R) and further (change of radius as before in (5.53)) Z 2 (Dw − (Dw)ρ) dx B(0,ρ) Z Z 2 2 ≤ C (Dw1 − (Dw1)ρ) dx + C (Dw2 − (Dw2)ρ) dx B(0,ρ) B(0,ρ) n+2 Z Z  ρ  2 2 ≤ C (Dw1 − (Dw1)R) dx + C (Dw2 − (Dw2)R) dx R B(0,R) B(0,R) n+2 Z Z  ρ  2 2 ≤ C (Dw − (Dw)R) dx + C (Dw2 − (Dw2)R) dx R B(0,R) B(0,R) n+2 Z Z  ρ  2 2 ≤ C (Dw − (Dw)R) dx + C |Dw2| dx, R B(0,R) B(0,R) where we wrote Dw1 = D(w1 + w2) − Dw2 etc. R R By using ϕ = w2 as a test function in Dw2 · Dϕ dx = − fDiϕ dx we get (recall zero bdr values) Z Z 2 |Dw2| dx = − (f − fR)Diw2 dx B(0,R) B(0,R) Z Z 1 2 1 2 ≤ (f − fR) dx + |Dw2| dx. 2 B(0,R) 2 B(0,R) Thus Z Z 2 2 |Dw2| dx ≤ C (f − fR) dx B(0,R) B(0,R) n+2α 2 ≤ CR |f|Cα(B(0,R)) where we also used estimate similar to Lemma 5.7. PDE 2 123

Combining the estimates Z 2 (Dw − (Dw)ρ) dx B(0,ρ) n+2 Z  ρ  2 n+2α 2 ≤ C (Dw − (Dw)R) dx + CR |f|Cα(B(0,R)). R B(0,R) Then by Lemma 5.10 Z 2 (Dw − (Dw)ρ) dx B(0,ρ) n+2α Z  ρ   2 n+2α 2  ≤ C (Dw − (Dw)R) dx + R |f|Cα(B(0,R)) .  R B(0,R) In the previous proof, we used the following iteration lemma for Z 2 G(r) = (Dw − (Dw)r) dx B(0,r) where r ∈ [0,R]. Lemma 5.10 (another iteration lemma). If  ρ γ G(ρ) ≤ A G(R) + BRβ, 0 < ρ < R ≤ R R 0 where 0 < β < γ, then there is C = C(A, γ, β) such that  ρ β G(ρ) ≤ C (G(R) + BRβ), 0 < ρ < R ≤ R . R 0 Proof. Ex.  We also need a Caccioppoli type estimate. Lemma 5.11. Let u be a solution to −∆u = f in B(0, 2R), f ∈ Cα(B(0, 2R)). Then there is C = C(n) such that Z 2 2 D u dx B(0,R/2) Z  1 2 n 2 n+2α 2  ≤ C 4 u dx + R ||f||L∞(Ω) + R |f|Cα(B(0,R)) . R B(0,R) 2,2 Proof. Since u ∈ Wloc (B(0, 2R)) by our earlier regulatity results, we may test with ϕ = Diφ with a smooth function φ. Integrating by parts Z Z fDiφ dx = Du · DDiφ dx

int by parts Z = − DDiu · Dφ dx. 124 PDE 2

Thus in the weak sense w = Diu

−∆w = Dif = Di(f − fR).

2 ∞ Testing this with ϕ = η w, where η ∈ C0 (B(0,R)), 0 ≤ η ≤ 1, η = 1 in B(0,R/2) and |Dη|2 ≤ C/R2, we have (ex) Z Z Z 2 2 2 2 2 2 |Dw| η dx ≤ C w |Dη| dx + C η |f − fR| dx B(0,R) B(0,R) B(0,R) Z C 2 n+2α ≤ 2 w dx + CR |f|Cα(B(0,R)), R B(0,R) where at the last step we also used Lemma 5.7. Further testing the weak 2 ∞ formulation of −∆u = f by ϕ = η u where η ∈ C0 (B(0, 3R/2)), 0 ≤ η ≤ 1, η = 1 in B(0,R) and |Dη|2 ≤ C/R2, we have Z Z w2 dx ≤ |Du|2η2 dx B(0,R) B(0,3R/2) Z Z ≤ C u2|Dη|2 dx + CR2 η2|f|2 dx B(0,3R/2) B(0,3R/2) Z C 2 n+2 2 ≤ 2 u dx + CR ||f||L∞(B(0,3R/2)) , R B(0,3R/2)

R R 2 2 R 2 2 where we estimated for example B(0,3R/2) R η uf dx ≤ CR B(0,3R/2) η |f| dx+ C R 2 2 R2 B(0,3R/2) η u dx. Combining the previous two estimates we have Z 2 2 D u dx B(0,R/2) Z C 2 n 2 n+2α ≤ 4 u dx + CR ||f||L∞(B(0,3R/2)) + CR |f|Cα(B(0,R)). R B(0,3R/2)  Lemma 5.12. Let u be a solution to −∆u = f in B(0, 2R), and let α w = Diu, f ∈ C (B(0, 2R)). Then for 0 < ρ ≤ R/2 there is C = C(n) st Z 2 n+2α |Dw − (Dw)ρ| dx ≤ ρ MRC B(0,ρ) where 1 1 M = ||u||2 + ||f||2 + |f|2 . R R4+2α L∞(B(0,R)) R2α L∞(B(0,R)) Cα(B(0,R)) PDE 2 125

Proof. By Lemma 5.9 Z 2 |Dw − (Dw)ρ| dx B(0,ρ) Z n+2α 1 2 2  ≤ Cρ n+2α |Dw − (Dw)R| dx + |f|Cα(B(0,R/2)) R B(0,R/2) Z n+2α 1 2 2  ≤ Cρ n+2α |Dw| dx + |f|Cα(B(0,R)) . R B(0,R/2) Then by the Caccioppoli type estimate Lemma 5.11 we have Z 2 |Dw − (Dw)ρ| dx B(0,ρ) Z n+2α 1 2 2  ≤ Cρ n+2α |Dw| dx + |f|Cα(B(0,R)) R B(0,R/2) Z n+2α 1  1 2 ≤ Cρ n+2α 4 u dx R R B(0,R) ! n 2  2 + R ||f||L∞(B(0,R)) + |f|Cα(B(0,R)) .

 Theorem 5.13. Let u be a solution to −∆u = f in B(0, 2R) with f ∈ Cα(B(0, 2R)). Then 2 D u Cα(B(0,R/4)) 1 1 ≤ C( ||u|| + ||f|| + |f| ). R2+α L∞(B(0,R)) Rα L∞(B(0,R)) Cα(B(0,R)) Proof. What we have in Lemma 5.12 looks very much like the Cam- panato seminorm. Indeed, this is exactly the idea of the proof. To be more precise, by Lemma 5.6 it suffices to bound the Campanato seminorm. Z 1/2  1 2  |u|L2,µ(B(0,R/4)) = sup µ |u(y) − ux,ρ| dx . x∈Ω,0<ρ

Then by Lemma 5.12, Z 2 2 2 D u(y) − (D u)B(x,ρ) dx B(x,ρ) ! 1 1 ≤ Cρn+2α ||u||2 + ||f||2 + |f|2 . R4+2α L∞(B(0,R))) R2α L∞(B(0,R)) Cα(B(0,R)) We combine the estimates, divide on both sides by ρn+2α, take 1 supx∈Ω,0<ρ

6. Notes and comments I would like to thank Juha Kinnunen for providing his lecture notes at my disposal when designing this course. Other material includes ”Elliptic & Parabolic Equations” (Wu, Yin and Wang, 2006, World Scientific), ”Partial Differential Equations” (Evans, 1998, American Mathematical Society), ”Elliptic Partial Differetential Equations of Second Order” (Gilbarg, Trudinger, 1977, Springer), ”Second Order Elliptic Equations and Elliptic Systems” (Chen, Wu, 1998, American Mathematical Society), ”Direct Methods in the Calculus of Variations” (Giusti, 2003, World Scientific), ”Partial Differential Equations” (Jost, 2002, Springer), and ”Partial Differential Equations” (DiBenedetto, 2010, Birkh¨auser).

Department of Mathematics and Statistics, University of Jyvaskyl¨ a,¨ PO Box 35, FI-40014 Jyvaskyl¨ a,¨ Finland E-mail address: Please email typos/errors to:[email protected]