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1. Harmonic Functions 2. Perron's Method 3. Potential Theor

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1. Harmonic Functions 2. Perron's Method 3. Potential Theor Elliptic and Parabolic Equations by J. Hulshof Elliptic equations: 1. Harmonic functions 2. Perron’s method 3. Potential theory 4. Existence results; the method of sub- and supersolutions 5. Classical maximum principles for elliptic equations 6. More regularity, Schauder’s theory for general elliptic operators 7. The weak solution approach in one space dimension 8. Eigenfunctions for the Sturm-Liouville problem 9. Generalization to more dimensions Parabolic equations: 10. Maximum principles for parabolic equations 11. Potential theory and existence results 12. Asymptotic behaviour of solutions to the semilinear heat equation Functional Analysis: A. Banach spaces B. Hilbert spaces C. Continous semigroups and Liapounov functionals 1 1. Harmonic functions Throughout this section, Ω ⊂ Rn is a bounded domain. 1.1 Definition A function u ∈ C2(Ω) is called subharmonic if ∆u ≥ 0 in Ω, harmonic if ∆u ≡ 0 in Ω, and superharmonic if ∆u ≤ 0 in Ω. 1.2 Notation The measure of the unit ball in Rn is Z n/2 n 2 2 2π ωn = |B1| = |{x ∈ R : x1 + ... + xn ≤ 1}| = dx = . B1 nΓ(n/2) The (n − 1)-dimensional measure of the boundary ∂B1 of B1 is equal to nωn. 1.3 Mean Value Theorem Let u ∈ C2(Ω) be subharmonic, and n BR(y) = {x ∈ R : |x − y| ≤ R} ⊂ Ω. Then 1 Z u(y) ≤ n−1 u(x)dS(x), nωnR ∂BR(y) where dS is the (n − 1)-dimensional surface element on ∂BR(y). Also 1 Z u(y) ≤ n u(x)dx. ωnR BR(y) Equalities hold if u is harmonic. Proof We may assume y = 0. Let ρ ∈ (0,R). Then Z Z ∂u Z ∂u 0 ≤ ∆u(x)dx = (x)dS(x) = (x)dS(x) = Bρ ∂Bρ ∂ν ∂Bρ ∂r (substituting x = ρω) Z ∂u Z ∂ d Z (ρω)ρn−1dS(ω) = ρn−1 ( u(ρω))dS(ω) = ρn−1 u(ρω)dS(ω) ∂B1 ∂r ∂B1 ∂ρ dρ ∂B1 (substituting ω = x/ρ) Z n−1 d 1 = ρ n−1 u(x)dS(x), dρ ρ ∂Bρ 2 which implies, writing 1 Z f(ρ) = n−1 u(x)dS(x), nωnρ ∂Bρ that f 0(ρ) ≥ 0. Hence 1 Z u(0) = lim f(ρ) ≤ f(R) = u(x)dS(x), ρ↓0 n−1 nωnR ∂BR which proves the first inequality. The second one follows from Z Z R Z Z R n−1 n u(x)dx = u(x)dS(x) dρ ≥ nωnρ u(0)dρ = ωnR u(0). BR 0 ∂Bρ 0 This completes the proof. 1.4 Corollary (Strong maximum principle for subharmonic functions) Let u ∈ 2 C (Ω) be bounded and subharmonic. If for some y ∈ Ω, u(y) = supΩ u, then u ≡ u(y). Proof Exercise (hint: apply the mean value theorem to the functionu ˜(x) = u(x) − u(y), and show that the set {x ∈ Ω :u ˜(x) = 0} is open). 1.5 Corollary (weak maximum principle for subharmonic functions) Suppose u ∈ C2(Ω) ∩ C(Ω) is subharmonic. Then sup u = max u = max u. Ω Ω ∂Ω Proof Exercise. 1.6 Corollary Let u ∈ C2(Ω)∩C(Ω) be subharmonic. If u ≡ 0 on ∂Ω, then u < 0 on Ω, unless u ≡ 0 on Ω. Proof Exercise. 1.7 Corollary Let ϕ ∈ C(∂Ω). Then there exists at most one function u ∈ C2(Ω) ∩ C(Ω) such that ∆u = 0 in Ω and u = ϕ on ∂Ω. Proof Exercise. 3 1.8 Theorem (Harnack inequality) Let Ω0 ⊂⊂ Ω (i.e. Ω0 ⊂ Ω0 ⊂ Ω) be a subdomain. Then there exists a constant C which only depends on Ω0 and Ω, such that for all harmonic nonnegative functions u ∈ C2(Ω), sup u ≤ C inf u. Ω0 Ω0 Proof Suppose that B4R(y) ⊂ Ω. Then for any x1, x2 ∈ BR(y) we have BR(x1) ⊂ B3R(x2) ⊂ B4R(y) ⊂ Ω, so that by the mean value theorem, Z n Z 1 3 n u(x1) = n u(x)dx ≤ n u(x)dx = 3 u(x2). ωnR BR(x1) ωn(3R) B3R(x2) Hence, x1, x2 being arbitrary, we conclude that sup u ≤ 3n inf u. B (y) BR(y) R 0 Thus we have shown that for Ω = BR(y), with B4R(y) ⊂ Ω, the constant in the inequality can be taken to be 3n. Since any Ω0 ⊂⊂ Ω can be covered with finitely many of such balls, say 0 Ω ⊂ BR1 (y1) ∪ BR2 (y2) ∪ ... ∪ BRN (yN ), we obtain for Ω0 that C = 3nN . Next we turn our attention to radially symmetric harmonic functions. Let u(x) be a function of r = |x| alone, i.e. u(x) = U(r). Then u is harmonic if and only if n n n X ∂ X ∂ ∂r X ∂ xi 0 = ∆u(x) = ( )2u = (U 0(r) ) = ( U 0(r)) ∂x ∂x ∂x ∂x r i=1 i i=1 i i i=1 i n X 1 xi ∂r ∂ 1 = U 0(r) + U 00(r) − x U 0(r) ( ) r r ∂x i ∂x r i=i i i n 2 n n X x X xi = U 0(r) + i U 00(r) − x U 0(r) = r r i r3 i=1 i=1 n − 1 1 U 00(r) + U 0(r) = (rn−1U 0(r))0, r rn−1 4 implying n−1 0 r U (r) = C1, so that C1r + C2 n = 1; C log r + C n = 2; U(r) = 1 2 (1.1) C1 1 + C n > 2. 2 − n rn−2 2 We define the fundamental solution by 1 |x| n = 1 2 1 Γ(x) = log |x| n = 2 (1.2) 2π 1 1 n−2 n > 2, nωn(2 − n) |x| i.e. C1 = 1/nωn and C2 = 0 in (1.1). Whenever convenient we write Γ(x) = Γ(|x|) = Γ(r). 1.9 Theorem The fundamental solution Γ is a solution of the equation ∆Γ = δ in the sense of distributions, i.e. Z n Γ(x)∆ψ(x)dx = ψ(0) ∀ψ ∈ D(R ). n R ∞ P Proof First observe that for all R > 0, we have Γ ∈ L (BR) if n = 1, Γ ∈ L (BR) P n for all 1 ≤ p < ∞ if n = 2, and Γ ∈ L (BR) for all 1 ≤ p < n−2 if n > 2, so for all ψ in D(Rn), choosing R large enough, we can compute Z Z Z Γ(x)∆ψ(x)dx = Γ(x)∆ψ(x)dx = lim Γ(x)∆ψ(x)dx = n ρ↓0 R BR AR,ρ (here AR,ρ = {x ∈ BR : |x| > ρ}) Z ∂ψ Z Z ∂ψ ∂Γ Z lim Γ − ∇Γ∇ψ = lim (Γ − ψ) + ψ∆Γ = ρ↓0 ρ↓0 ∂AR,ρ ∂ν AR,ρ ∂AR,ρ ∂ν ∂ν AR,ρ Z −∂ψ/∂ν ψ lim + = ψ(0). ρ↓0 n−2 n−1 ∂Bρ nωn(2 − n)ρ nωnρ For n = 1, 2 the proof is similar. Closely related to this theorem we have 5 1.10 Theorem (Green’s representation formula) Let u ∈ C2(Ω) and suppose ∂Ω ∈ C1. Then, if ν is the outward normal on ∂Ω, we have Z ∂ ∂u Z u(y) = {u(x) Γ(x − y) − Γ(x − y) (x)}dS(x) + Γ(x − y)∆u(x)dx. ∂Ω ∂ν ∂ν Ω Here the derivatives are taken with respect to the x-variable. Proof Exercise (Hint: take y = 0, let Ωρ = {x ∈ Ω: |x| > ρ}, and imitate the previous proof). If we want to solve ∆u = f on Ω for a given function f, this representation formula strongly suggests to consider the convolution Z Γ(x − y)f(x)dx Ω as a function of y, or equivalently, Z (Γ ∗ f)(x) = Γ(x − y)f(y)dy (1.3) Ω as a function of x. This convolution is called the Newton potential of f. For any harmonic function h ∈ C2(Ω) we have Z Z ∂u ∂h h∆u = (h − u ), Ω ∂Ω ∂ν ∂ν so that, combining with Green’s representation formula, Z ∂G ∂u Z u(y) = {u − G } + G∆u, (1.4) ∂Ω ∂ν ∂ν Ω where G = Γ(x − y) + h(x). The trick is now to take instead of a function h(x) a function h(x, y) of two variables x, y ∈ Ω, such that h is harmonic in x, and for every y ∈ Ω, G(x, y) = Γ(x − y) + h(x, y) = 0 ∀x ∈ ∂Ω. This will then give us the solution formula Z ∂G Z u(y) = u + G∆u. ∂Ω ∂ν Ω In particular, if u ∈ C2(Ω) is a solution of ( ∆u = f in Ω; (D) u = ϕ on ∂Ω, 6 then Z ∂G(x, y) Z u(y) = ϕ(x) + G(x, y)f(x)dx. (1.5) ∂Ω ∂ν Ω The function G(x, y) = Γ(x − y) + h(x, y) is called the Green’s function for the Dirichletproblem. Of course h(x, y) is by no means trivial to find. The function h is called the regular part of the Green’s function. If we want to solve ( ∆u = 0 in Ω; u = ϕ on ∂Ω, (1.5) reduces to Z ∂G(x, y) u(y) = ϕ(x) dS(x). (1.6) ∂Ω ∂ν n We shall evaluate (1.6) in the case that Ω = B = B1(0) = {x ∈ R : |x| < 1}. Define the reflection in ∂B by y S(y) = if y 6= 0; S(0) = ∞; S(∞) = 0. (1.7) |y|2 Here ∞ is the point that has to be added to Rn in order to construct the one-point compactification of Rn. If 0 6= y ∈ B, then y = S(y) is uniquely determined by asking that 0, y and y lie (in that order) on one line l, and that the boundary ∂B of B is tangent to the cone C with top y spanned by the circle obtained from intersecting the ball B with the plane perpundicular to l going through y (you’d better draw a picture here).
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