Math The Laplacian. 1 Green s Identities, Fundamental Solution

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1 Math. 209 The Laplacian Green s Identities, Fundamental Solution Let be a bounded open set in R n, n 2, with smooth boundary. The fact that the boundary is smooth means that at each point x the external unit normal vector ν(x) is a smooth function of x. (If the boundary is C k then this function is C k. Locally, is an embedded hypersurface - a manifold of codimension ). Green s identities are (v u + u v)dx = (v ν u)ds () and (v u u v)dx = (v ν u u ν v)ds. (2) In the two identities above, u and v are real valued functions twice continiously differentiable in, (u, v C 2 ()), with first derivatives that have continuous extensions to. The Laplacian is u = n j= 2 u x 2 j = n j 2 u, j= the gradient is the normal derivative is u = ( u,..., n u), ν u = ν u where x y = n j= x jy j. The surface measure is ds = dh n. Recall that (2) follows from () and that () follows from the divergence theorem ( Φ) dx = (Φ ν)ds (3)

2 where Φ is a C () vector field (n functions) that has a continuous extension to. Recall also that the Laplacian commutes with rotations: (f(ox)) = ( f)(ox), where O is a O(n) matrix, i.e. OO = O O = I with O the transpose. Consequently, rotation invariant functions (i.e., radial functions, functions that depend only on r = x ) are mapped to rotation invariant functions. For radial functions, f = f rr + n f r r where r = x, f r = x r f. The equation f = 0 in the whole plane has affine solutions. If we seek solutions that are radial, that is f rr + n f r = 0, for r > 0, r then, multiplying by r n, we see that r n f r 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 r 2 n (plus a constant) if n > 2 or a multiple of log r (plus a constant). The fundamental solution is N(x) = { 2π log x, if n = 2, (2 n)ω n x 2 n, if n 3. Here ω n is the area of the unit sphere in R n. Note that N is radial, and it is singular at x = 0. Proposition Let f C0(R 2 n ), and let u(x) = N(x y)f(y)dy. (5) R n Then u C 2 (R n ) and u = f Lemma Let N L loc (Rn ) and let φ C0(R n ). Then u = N φ is in C (R n ) and u(x) = N(x y) φ(y)dy (4) 2

3 Idea of proof of the lemma. First of all, the notation: L p loc is the space of functions that are locally in L p and that means that their restrictions to compacts are in L p. The space C0 is the space of functions with continuous derivatives of first order, and having compact support. Because φ has compact support u(x) = N(x y)φ(y)dy = N(y)φ(x y)dy is well defined. Fix x R n and take h R n with h. Note that ( ) h u(x + h) u(x) h N(x y) φ(y)dy = N(y) (φ(x + h y) φ(x y) h φ(x y)) dy h The functions y (φ(x + h y) φ(x y) h φ(x y)) for fixed x h and h 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 u(x) = N(y) φ(x y)dy and φ is continuous, it follows that u is continuous. This finishes the proof of the lemma. Idea of proof of the Proposition. By the Lemma, u is C 2 and u(x) = N(x y) f(y)dy Fix x. The function y N(x y) f(y) is in L (R n ) and compactly supported in x y < R for a large enough R that we ll keep fixed. Therefore u(x) = lim N(x y) f(y)dy ɛ 0 {y;ɛ< x y <R} We will use Green s identities for the domains R ɛ = {y; ɛ < x y < R}. This is legitimate because the function y N(x y) is C 2 in a neighborhood of R ɛ. Note that, because of our choice of R, f(y) vanishes identically near 3

4 the outer boundary x y = R. Note also that y N(x y) = 0 for y R ɛ. From (2) we have N(x y) f(y)dy = {y;ɛ< x y <R} N(x y) x y =ɛ νf(y)ds f(y) x y =ɛ νn(x y)ds The external unit normal at the boundary is ν = (x y)/ x y. The first integral vanishes in the limit because f is bounded, N(x y) diverges like ɛ 2 n (or log ɛ) and the area of boundary vanishes like ɛ n : N(x y) ν f(y)ds Cɛ f x y =ɛ in n > 2, and the same thing replacing ɛ by ɛ log ɛ 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 two minuses make a plus, and ν N(x y) = ω n x y n. Therefore, in view of the fact that x y = ɛ on the boundary we have f(y) ν N(x y)ds = f(y)ds ɛ n ω n x y =ɛ x y =ɛ Passing to polar coordinates centered at x we see that f(y) ν N(x y)ds = f(x + ɛz)ds ω n x y =ɛ z = and we do have lim f(x + ɛz)ds = f(x) ɛ 0 ω n z = because f is continuous. Remark 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. 4

5 2 Harmonic functions Definition We say that a function u is harmonic in the open set R n if u C 2 () and if u = 0 holds in. We denote by B(x, r) the ball centered at x of radius r, by A f (x, r) the surface average A f (x, r) = f(y)ds ω n r n B(x,r) and by V f (x, r) the volume average V f (x, r) = n ω n r n B(x,r) f(y)dy Proposition 2 Let be an open set, let u be harmonic in and let B(x, r). Then holds. u(x) = A u (x, r) = V u (x, r) Idea of proof. Let ρ r and apply (2) with domain B(x, ρ) and functions v = and u. We obtain ν u ds = 0 On the other hand, because B(x,ρ) A u (x, ρ) = ω n z = u(x + ρz)ds it follows that d dρ A u(x, ρ) = ω n z = z u(x + ρz) ds = ρ n ω n B(x,ρ) ν u ds = 0 So A u (x, ρ) does not depend on ρ for 0 < ρ r. But, because the fiunction u is continuous, lim ρ 0 A u (x, ρ) = u(x), so that proves u(x) = A u (x, r). 5

6 The relation r V f (x, r) = n ρ n A f (x, ρ)dρ, 0 valid for any integrable function, implies the second equality, because A u (x.ρ) = u(x) does not depend on ρ. Remark 2 The converse of the mean value theorem holds. If u is C 2 and u(x) = A u (x, r) for each x and r sufficiently small, then u is harmonic. Indeed, in view of the above, the integral u(y)dy must vanish for B(x,r) each x and r small enough, and that implies that there cannot exist a point x where u(x) does not vanish. Theorem (Weak maximum principle.) Let be open, bounded. Let u C 2 () C 0 () satisfy u(x) 0, x. Then max u(x) = max u(x). x x Idea of proof. If u > 0 in 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 ɛ x 2. The function u(x) + ɛ x 2 achieves its maximum on the boundary, for any positive ɛ. Therefore, max x u(x) + ɛ min x x 2 max u(x) + ɛ max x x x 2 and the result follws by taking the limit ɛ 0. Remark 3 If u is harmonic, then by applying the previous result to both u and u we deduce that max u(x) = max u(x) x x Definition 2 A continuous function u C 0 () is subharmonic in if x, r > 0 so that u(x) A u (x, ρ) holds ρ, 0 < ρ r. 6

7 Exercise Let u C 2 (), and assume that u(x) 0 holds for any x. Prove that u is subharmonic. Theorem 2 (Strong maximum principle) Let be open, bounded and connected. Let u C 0 () be subharmonic. Then, either u is constant, or holds. u(x) < sup u(x) x Idea of proof. Let M = sup x u(x). Consider the sets S = {x ; u(x) < M} and S 2 = {x ; u(x) = M}. The two sets are disjoint, and S is open. We show that S 2 is open as well. Indeed, take x S. Then 0 A u (x, ρ) M = (u(y) M)dS 0. ρ n ω n B(x,ρ) Because the integrand is non-positive we deduce that u(y) = M for all y so that y x = ρ, with 0 < ρ < r. This means that S 2 is open. Remark 4 The result implies that, if u C 2 () C 0 () satisfies u 0 in the bounded open connected domain, then either u is constant, or holds. u(x) < max x u(x) 3 Green s functions, Poisson kernel Exercise 2 Let be open, bounded, with smooth boundary. Let u C 2 () with first derivatives that are continuous up to the boundary u C(). Let x Then u(x) = (u ν N N ν u) ds + N(x y) u(y)dy (6) is true. The function N in the boundary integral is computed at x y, with y ; the normal derivative refers to the external normal at y. 7

8 Hint: Take a small ball B(x, r) and write (2) in \ B(x, r), with functions u and N(x y). Then use the calculation from the proof of the Proposition. Let us consider now the inhomogeneous Dirichlet problem { u = f, u = g where f is some given function in and g is a continuous function on. Suppose that we can find, for each x, a harmonic function n (x) (y) such that { y n (x) (y) = 0 n (x) (y) = N(x y), for y. Then, applying (2) we have 0 = n (x) (y) u(y)dy + Adding to (6) we deduce that G(x, y) = N(x y) n (x) (y) (N ν u u ν n (x) )ds. provides the solution to the Dirichlet problem, u(x) = G(x, y)f(y)dy + g(y) ν G(x, y)ds (7) Note that the Green s function solves { y G(x, y) = δ(x y) G(x, y) = 0 for y with δ(x y) the δ function concentrated at x. When f = 0 we obtain the representation of harmonic functions u(x) for x that satisfy u(y) = g(y) for y : u(x) = P (x, y)g(y)ds (8) with P (x, y) = ν G(x, y) (9) the Poisson kernel. The representations (7) and (8) are sublime, but the Green s function and the Poisson kernel for a general domain are hard to 8

9 obtain explicitly. Two important examples that can be computed are the half plane and the ball. The main idea, in both cases, is to reflect the singularity away. The reflection is rather straightforward for the half-plane, less so for the ball. Let = {x R n, x n > 0}. Set, G(x, y) = N(x y) N(x y) where x = (x,..., x n, x n ). This is a Green s function for. It is convenient to write a point in as x = (x, x n ) with x R n and x n > 0. Then x = (x, x n ), the Poisson kernel is a function of x y and x n, where y = (y, 0) represents a point on the boundary, P (x y, x n ) = 2x n ω n ( ) x y 2 + x 2 n 2 n We ought to cast a suspicious eye on the calculations leading to the Poisson kernel for the half-space, and prove rigorously that the result works: Proposition 3 Let g be a continuous bounded function of n variables, and let n >. The function u(x) = P (x y, x n )g(y )dy R n is harmonic in x n > 0 and the limit lim xn 0 u(x, x n ) = g(x ) holds. The calculations for a ball follow. Let = B(0, R). Let x and set x = R2 x (yes, trouble when x = 0, ignore for a moment). Note that if x 2 y then x y x y = R x does not depend on y. For n > 2 take the fundamental solutions N(x y) and N(x y) and write ( ) 2 n x G(x, y) = N(x y) N(x y) R Clearly, for y, G(x, y) vanishes. The second term is not singular in y, so it is harmonic in y. The Poisson kernel is This works even in n = 2. P (x, y) = Rω n R 2 x 2 x y n (0) 9

10 Proposition 4 Let g be a continuous function on B(0, R) in R n, n 2. The function { P (x, y)g(y)ds, for x < R B(0,R) u(x) = u(x) = g(x), for x = R given by the Poisson kernel (0) is harmonic in x < R, and continuous in x R. The proof uses the following properties of the Poisson kernel: P (x, y) C, P (x, y) > 0 for x < R, y = R, x P (x, y) = 0, for x < R, y = R, P (x, y)ds = x, x < R, B(0,R) lim x z, x <R P (x, y) = 0, uniformly for y = z = R, z y δ > 0. 4 Dirichlet principle, variational solutions Let be a bounded open set in R n. Let f C(), g C( ) and let A = {w C 2 (); w = g}. Let I[w] = ( ) 2 w 2 + wf dx. () Proposition 5 u C 2 () solves { u = f in, u = g on (2) if, and only if u = arg min w A I[w]. Note carefully that this is not an existence theorem, rather, it states the equivalence of two possible existence theorems. One theorem asserts that the Poisson problem with data f and g has a solution u with the desired smoothness. The other theorem asserts that one can minimize the integral I[w] and find a true minimum, in the class of admissible functin s A. The 0

11 variational method considers the minimization program. The program consists in two steps. The first step is to establish the existence of a minimum. Unfortunately, the natural function spaces for I[w] are not spaces of continuous functions, but rather Sobolev spaces based on L 2. This presents an opportunity to generalize: the right-hand side f will be allowed to be in L 2, because the method does not require more. This comes at a price: the minimum thus obtained is not smooth. The second step is to show that if the function f is smooth then the solution is smooth. Here, the natural smoothness requirement for f is more stringent than C(), it is the Hölder space C 0,α (). If is smooth, then the solution obtained is smoother than just C 2 (), it is C 2,α (). So, the innocent looking proposition above is both a begining and an end: it is the begining of a search for more general solutions, and it is the end of the classical problem in C H 0() Definition 3 Let R n be an open bounded set with smooth boundary. Th e space H () is the completion of C (R n ) with norm u 2 H () = ( u 2 + u 2 )dx. The space H 0() is the closure of C 0 () in H (). Similar definitions are given for the whole space and the torus. In those cases H 0 = H. However, in bounded domains H 0 is strictly smaller: it repres ents functions that vanish at the boundary, in a weak sense. Exercise 3 Show that H 0((0, )) H ((0, )). Lemma 2 (Poincaré Inequality) There exists a constant C depending on the bounded domain such that u 2 C u 2 dx holds for all u H 0(). Proof. Because is bounded in the direction x, we know that there exists an interval [a, b] such that t [a, b] holds for all t such that there exists

12 x 2,... x n so that x = (t, x 2,..., x n ). Because both sides of the inequality are continuous in H and because of the definition of H0, we may assume, WLOG that u C0 (). Then By Schwartz: u(x,... x n ) 2 = 2 x a u(t, x 2,..., x n )( u)(t, x 2,..., x n )dt { b } { u(x,... x n ) 2 2 u 2 2 b } (t, x 2,..., x n )dt u(t, x 2,..., x n ) 2 2 dt a a We keep x fixed, integrate dx 2... dx n and use Schwartz again: u 2 (x, x 2,... x n )dx 2... dx n 2 u L 2 u L 2. We integrate dx on [a, b], noting that the RHS is independent of x : u 2 L 2 2(b a) u L 2 u L 2. Dividing by u L 2 we obtain the inequality with C = 4(b a) 2. It is clear from this proof that we do not need to use the boundedness of, only the existence of some direction in which is bounded. Exercise 4 Show that the Poincaré inequality fails in H () if is bounded. Show that the Poincaré inequality fails in some unbounded domains: for instance in H 0(R). The Poincaré inequality implies that the scalar product (u, v) = u vdx is equivalent (gives the same topology) with the scalar product in H0(): < u, v >= (uv + u v)dx 2

13 Theorem 3 Let be a bounded open set with smooth boundary. Let f L 2 (). Then, there exists a unique u H 0() that solves where I[w] = I[u] = min I[w] w H0 () ( ) 2 w 2 + wf dx. The function u satisfies the variational formulation of the problem (2) with g = 0: (u, v) + vfdx = 0 v H0() (3) Proof. The function I[w] is bounded below. Indeed, using Schwartz we have I[w] w 2 dx f L 2 w L 2 2 and using the Poincaré inequality I[w] 2C w 2 L 2 f L 2 w L 2 and that is bounded below. Thus, the infimum exists: m = inf I[w]. w H0 () Let w k be a minimizing sequence, I[w k ] m. The sequence is bounded in H 0() because (w k, w k ) are bounded. Therefore there exists a subsequence (denoted again w k ) and an element u in H 0() such that w k converges weakly to u, lim k (w k, v) = (u, v) v H 0(). Then, because v vfdx is continouous in H 0, hence weakly continuous, it follows that ufdx = lim w k fdx k Exercise 5 In a Hilbert space, the square of the norm is weakly lower semicontinuous. 3

14 By the above exercise, 2 u 2 dx 2 lim inf k This shows that u achieves the minimum. I(u) = m. w k 2 dx The variational formulation (3) follows by looking at the function q(t) = I(u + tv) for fixed, but arbitrary v H0() and t real. q(t) is a quadratic polynomial in t, { } q(t) = I(u) + t (u, v) + vfdx + t2 (v, v) 2 that has a minimum at t = 0. The variational formulation is equivalent to the fact that q (0) = 0. The uniqueness of u follows immediately from variational formulation with v = u, the inequality thus u 2 C(u, u) 2C f L 2 u L 2, u L 2 2C f L 2, and because the variational formulation is linear. 4