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Harmonic Functions, Second Order Elliptic Equations I Harmonic functions, Second Order Elliptic Equations I Introduction to PDE 1 Green's Identities, Fundamental Solution Let Ω be a bounded open set in Rn, n ≥ 2, with smooth boundary @Ω. The fact that the boundary is smooth means that at each point x 2 @Ω the external unit normal vector ν(x) is a smooth function of x. (If the boundary is Ck then this function is Ck−1. Locally, @Ω is an embedded hypersurface - a manifold of codimension 1). Green's identities are Z Z (v∆u + ru · rv)dx = (v@νu)dS (1) Ω @Ω and Z Z (v∆u − u∆v)dx = (v@νu − u@νv)dS: (2) Ω @Ω In the two identities above, u and v are real valued functions twice contin- iously differentiable in Ω, (u; v 2 C2(Ω)), with first derivatives that have continuous extensions to Ω. The Laplacian ∆ is n n X @2u X ∆u = = @2u; @2x j j=1 j j=1 the gradient is ru = (@1u; : : : ; @nu); partial derivatives are @ @i = ; @xi 1 the normal derivative is @νu = ν · ru Pn where x · y = j=1 xjyj. Finally, dS is surface measure. Exercise 1 Prove that (2) follows from (1) and that (1) follows from the familiar Green's identity Z Z (r · Φ) dx = (Φ · ν)dS (3) Ω @Ω where Φ is a C1(Ω) vector field (n functions) that has a continuous extension to Ω. Exercise 2 Prove that the Laplacian commutes with rotations: ∆(f(Rx)) = (∆f)(Rx), where R is a O(n) matrix, i.e. RRt = RtR = I with Rt the transpose. Consequently, rotation invariant functions (i.e., radial functions, functions that depend only on r = jxj) are mapped to rotation invariant functions. Show that, for radial functions, n − 1 ∆f = f + f rr r r x where r = jxj, fr = r · rf. The equation ∆f = 0 in the whole plane has affine solutions. If we seek solutions that are radial, that is n − 1 f + f = 0; for r > 0; rr r r n−1 n−1 then, multiplying by r , we see that r fr should be constant. There are only two possibilities: this constant is zero, and then f itself must be a constant, or this constant is not zero, and then f is a multiple of r2−n (plus a constant) if n > 2 or a multiple of log r (plus a constant). The fundamental solution is 1 log jxj; if n = 2; N(x) = 2π (4) 1 jxj2−n; if n ≥ 3: (2−n)!n n Here !n is the area of the unit sphere in R . Note that N is radial, and it is singular at x = 0. 2 2 n Proposition 1 Let f 2 C0 (R ), and let Z u(x) = N(x − y)f(y)dy: (5) Rn Then u 2 C2(Rn) and ∆u = f 1 n 1 n Lemma 1 Let N 2 Lloc(R ) and let φ 2 C0 (R ). Then u = N ∗ φ is in C1(Rn) and Z ru(x) = N(x − y)rφ(y)dy p Idea of proof of the lemma. First of all, the notation: Lloc is the space of functions that are locally in Lp and that means that their restrictions to p 1 compacts are in L . The space C0 is the space of functions with continu- ous derivatives of first order, and having compact support. Because φ has compact support Z Z u(x) = N(x − y)φ(y)dy = N(y)φ(x − y)dy is well defined. Fix x 2 Rn and take h 2 Rn with jhj ≤ 1. Note that 1 R jhj u(x + h) − u(x) − h · N(x − y)rφ(y)dy = R 1 N(y) jhj (φ(x + h − y) − φ(x − y) − h · rφ(x − y)) dy 1 The functions y 7! jhj (φ(x + h − y) − φ(x − y) − h · rφ(x − y)) for fixed x and jhj ≤ 1 are supported all in the same compact, are uniformly bounded, and converge to zero as h ! 0. Because of Lebesgue dominated, it follows that u is differentiable at x and that the derivative is given by the desired expression. Because Z ru(x) = N(y)rφ(x − y)dy and rφ is continuous, it follows that ru is continuous. This finishes the proof of the lemma. Idea of proof of the Proposition 1. By the Lemma, u is C2 and Z ∆u(x) = N(x − y)∆f(y)dy 3 Fix x. The function y 7! N(x − y)∆f(y) is in L1(Rn) and compactly sup- ported in jx − yj < R for a large enough R that we'll keep fixed. Therefore Z ∆u(x) = lim N(x − y)∆f(y)dy !0 fy;<jx−yj<Rg R We will use Green's identities for the domains Ω = fy; < jx − yj < Rg. This is legitimate because the function y 7! N(x−y) is C2 in a neighborhood R of Ω . Note that, because of our choice of R, f(y) vanishes identically near R the outer boundary jx − yj = R. Note also that ∆yN(x − y) = 0 for y 2 Ω . From (2) we have R fy;<jx−yj<Rg N(x − y)∆ f(y)dy = R R jx−yj= N(x − y)@νf(y)dS − jx−yj= f(y)@νN(x − y)dS The external unit normal at the boundary is ν = −(x − y)=jx − yj. The first integral vanishes in the limit because jrfj is bounded, N(x−y) diverges like 2−n (or log ) and the area of boundary vanishes like n−1: Z N(x − y)@νf(y)dS ≤ Ckrfk1 jx−yj= in n > 2, and the same thing replacing by log −1 in n = 2. The second integral is more amusing, and it is here that it will become clear why the constants are chosen as they are in (4). We start by noting carefully that 1 1−n −@νN(x − y) = jx − yj : !n Therefore, in view of the fact that jx − yj = on the boundary we have Z 1 Z − f(y)@νN(x − y)dS = n−1 f(y)dS jx−yj= !n jx−yj= Passing to polar coordinates centered at x we see that Z 1 Z − f(y)@νN(x − y)dS = f(x + z)dS jx−yj= !n jzj=1 and we do have 1 Z lim f(x + z)dS = f(x) !0 !n j zj=1 because f is continuous. 4 Remark 1 The fundamental solution solves ∆N = δ: This is rigorously true in the sense of distributions, but formally it can be appreciated without knowledge of distributions by using without justification the fact that δ is the identity for convolution δ ∗ f = f, the rule ∆(N ∗ f) = ∆N ∗ f and the Proposition above. 2 Harmonic functions Definition 1 We say that a function u is harmonic in the open set Ω ⊂ Rn if u 2 C2(Ω) and if ∆u = 0 holds in Ω. We denote by B(x; r) the ball centered at x of radius r, by Af (x; r) the surface average 1 Z Af (x; r) = n−1 f(y)dS !nr @B(x;r) and by Vf (x; r) the volume average n Z Vf (x; r) = n f(y)dy !nr B(x;r) Proposition 2 Let Ω be an open set, let u be harmonic in Ω and let B(x; r) ⊂ Ω. Then u(x) = Au(x; r) = Vu(x; r) holds. Idea of proof. Let ρ ≤ r and apply (2) with domain B(x; ρ) and functions v = 1 and u. We obtain Z @νu dS = 0 @B(x,ρ) On the other hand, because 1 Z Au(x; ρ) = u(x + ρz)dS !n jzj=1 5 it follows that d 1 Z 1 Z Au(x; ρ) = z · ru(x + ρz) dS = n−1 @νu dS = 0 dρ !n jzj=1 ρ !n @B(x,ρ) So Au(x; ρ) does not depend on ρ for 0 < ρ ≤ r. But, because the function u is continuous, limρ!0 Au(x; ρ) = u(x), so that proves u(x) = Au(x; r): The relation Z r −n n−1 Vf (x; r) = nr ρ Af (x; ρ)dρ, 0 valid for any integrable function, implies the second equality, in view of the fact that Af (x; ρ) = u(x) does not depend on ρ. Remark 2 The converse of the mean value theorem holds. If u is C2 and u(x) = Au(x; r) for each x and r sufficiently small, then u is harmonic. R Indeed, in view of the above, the integral B(x;r) ∆u(y)dy must vanish for each x and r small enough, and that implies that there cannot exist a point x where ∆u(x) does not vanish. Theorem 1 (Weak maximum principle.) Let Ω be open, bounded. Let u 2 C2(Ω) \ C0(Ω) satisfy ∆u(x) ≥ 0; 8x 2 Ω: Then max u(x) = max u(x): x2Ω x2@Ω Idea of proof. If ∆u > 0 in Ω then clearly u cannot have an interior maximum, so its maximum must be achieved on the boundary. If ∆u ≥ 0, then a useful trick is to add jxj2. The function u(x) + jxj2 achieves its maximum on the boundary, for any positive . Therefore, max u(x) + min jxj2 ≤ max u(x) + max jxj2 x2Ω x2Ω x2@Ω x2@Ω and the result follws by taking the limit ! 0. Remark 3 If u is harmonic, then by applying the previous result to both u and −u we deduce that max ju(x)j = max ju(x)j x2Ω x2@Ω 6 Definition 2 A continuous function u 2 C0(Ω) is subharmonic in Ω if 8x 2 Ω, 9r > 0 so that u(x) ≤ Au(x; ρ) holds 8ρ, 0 < ρ ≤ r.
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