Theory of PDE Homework 2

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1 Theory of PDE Homework 2 Adrienne Sands April 18, 2017 In the following exercises we assume the coefficients of the various PDE are smooth and satisfy the uniform ellipticity condition. R n is always an open, bounded set, with smooth boundary. 1. (Evans 6.6/3 A function u H0 2 ( is a weak solution of this boundary-value problem for the biharmonic equation 2 u = f in (1 provided u = u ν = 0 u v dx = on fv dx for all v H 2 0 (. Given f L2 (, prove that there exists a unique weak solution of (1. Proof. Define a bilinear form B : H0 2( H2 0 ( R as follows: B[u, v] := u v dx = u xi x i v xj x j dx, u, v H0 2 ( We claim B[u, v] is bounded and coercive. Crude estimates suffice for boundedness: B[u, v] D 2 u L 2 ( D 2 v L 2 ( n2 u H 2 0 ( v H0 2( For coercivity, we use the Gagliardo-Nirenberg-Sobolev inequality for u H0 2( H1 0 (: u 2 L 2 ( C Du 2 L 2 (. By integration by parts and Cauchy s inequality with ε, Du 2 L 2 ( = Du Du dx = u u dx ε u 2 L 2 ( + 1 4ε u 2 L 2 ( Taking Cε = 1 2 Cε Du 2 L 2 ( + 1 4ε u 2 L 2 ( = Cε Du 2 L 2 ( + 1 B[u, u] 4ε and rearranging, 1 2 Du 2 L 2 ( C 2 B[u, u] = u 2 L 2 ( C2 B[u, u] such that by Theorem 4 (Boundary H 2 -regularity, u 2 H 2 ( C( f 2 L 2 ( + u 2 L 2 ( = C(B[u, u] + u 2 L 2 ( C B[u, u] for u H0 2( a weak solution to (1. Since H2 0 ( and B[, ] satisfy the hypotheses for Lax-Milgram, we have our result. 1

2 2. (Evans 6.6/4 Assume is connected. A function u H 1 ( is a weak solution of Neumann s problem u = f in (2 u ν = 0 on if Du Dv dx = fv dx for all v H 1 (. Suppose f L 2 (. Prove (2 has a weak solution if and only if f dx = 0 Proof. First assume u is a weak solution and let v 0 constant. Since Dv = 0, our result follows immediately. Now assume f dx = 0. Define B : H1 ( H 1 ( R and S H 1 ( as follows: B[u, v] := Du Dv dx, Su = u dx u, v H 1 ( To accommodate the boundary condition, consider the subspace S 1 ( := H 1 ( ker S. Since S is a continuous linear map on H 1 (, S 1 ( is a Hilbert space (as a closed subspace of a Hilbert space with induced norm u S 1 ( = u H 1 (. We claim B[u, v] satisfies the hypotheses of Lax-Milgram on S 1 (. Crude estimates suffice for boundedness: B[u, v] Du L 2 ( Dv L 2 ( u H 1 ( v H 1 ( = u S 1 ( v S 1 ( Since u u Su for u S 1 (, Poincare s inequality suffices for coercivity: u 2 S 1 ( = u Su 2 L 2 ( + Du 2 L 2 ( (C + 1 Du 2 L 2 ( = (C + 1B[u, u] By Lax-Milgram, there is a unique element ũ S 1 ( such that B[ũ, ṽ] = fṽ dx, ṽ S 1 ( We claim ũ is also a weak solution for (2. Let v H 1 ( arbitrary and ṽ := v Sv S 1 (. By assumption and linearity of the L 2 inner product, B[ũ, v] = B[ũ, ṽ] + B[ũ, Sv] = fṽ dx + Dũ D(Sv dx = fṽ dx + 0 = fṽ dx + Sv f dx = fv dx as desired. 3. (Evans 6.6/5 Explain how to define u H 1 ( to be a weak solution of Poisson s equation with Robin boundary conditions: u = f in (3 u + u ν = 0 on Discuss the existence and uniqueness of a weak solution for a given f L 2 (. 2

3 Proof. To define a reasonable weak solution, multiply Poisson s equation by v H 1 ( and integrate by parts, using the given conditions when appropriate: du f, v L 2 ( = u v dx = Du Dv dx dν v ds = Du Dv dx + u v ds where u, v are defined in a trace sense. In other words, u v ds := T u T v ds for T : H 1 ( L 2 (, a bounded and linear trace operator. Thus, define B : H 1 ( H 1 ( R and S H 1 ( as follows: B[u, v] := Du Dv + T u T v ds, Su = u dx u, v H 1 ( Thus, any reasonable weak solution would satisfy B[u, v] = f, v, v H 1 ( We claim B[u, v] satisfies the hypotheses of Lax-Milgram on H 1 (. Since T is bounded, B[u, v] is bounded: B[u, v] u H 1 ( v H 1 ( + T u L 2 ( T v L 2 ( (C + 1 u H 1 ( v H 1 ( If B[u, v] is not coercive, there are u k H 1 ( such that Du k 2 dx + u 2 k ds < 1 k u k 2 H 1 (, k = 1, 2,... and u k H 1 ( = 1 without loss of generality. Passing to a subsequence, u k v u k v weakly in H 1 ( strongly in L 2 ( For any ϕ C0 (, we have the following by strong convergence of u k in L 2 ( and integration by parts: Dv ϕ = v Dϕ = lim u k Dϕ = lim Du k ϕ k k Furthermore, lim k Du k 2 1 dx lim k k = 0 = Du k 0 strongly in L 2 ( Thus, Dv ϕ = 0 for all ϕ C 0 (, and since v H1 (, v is constant. By linearity, we also have D(u k v = Du k 0 strongly in L 2 (, so u k v strongly in H 1 (. Finally, the trace theorem implies u k v in L 2 ( with v 2 L 2 ( = lim k u 2 k 1 ds lim k k = 0 Thus, v H 1 ( is the unique trace zero constant ( 0. But, u k v strongly in H 1 ( implies v H 1 ( = u k H 1 ( = 1 0. Thus B[u, v] is coercive, and (3 has a unique weak solution in H 1 ( by Lax-Milgram. 3

4 4. (Evans 6.6/8 Let u be a smooth solution of the uniformly elliptic equation Lu = a ij (xu xi x j = 0 in Assume that the coefficients have bounded derivatives. Set v := Du 2 + λu 2 and show that Lv 0 in if λ is large enough. Deduce Du L ( ( Du C L ( + u L (. Proof. For brevity, denote u i := u xi and u ij = u xi x j. By direct computation, and by assumption, such that Du 2 ij = (Du Du ij = (2Du Du i j = 2(Du j Du i + Du Du ij (u 2 ij = (2uu i j = 2(u i u j + uu ij Lu = a ij u ij = D( Lu = Lv 2 = = ( Da ij u ij + a ij Du ij = 0 ( a ij Du ij Du + a ij Du i Du j + λa ij u i u j + λa ij uu ij ( Da ij u ij Du + a ij Du i Du j + λa ij u i u j + 0 Da ij L ( D2 u Du θ D 2 u 2 λθ Du 2 Da ij L ( ( D2 u 2 + Du 2 θ D 2 u 2 λθ Du 2 = (C θ D 2 u 2 + (C λθ Du 2 0 for sufficiently large λ > 1. By the weak maximum principle, v achieves its maximum value on such that Du 2 L ( Du 2 + λu 2 L ( Du 2 + λu 2 L ( Du 2 L ( + λ u 2 L ( λ ( Du 2 L ( + u 2 L ( 5. (Evans 6.6/10 Assume is connected. se (a energy methods and (b the maximum principle to show that the only smooth solutions of the Neumann boundary-value problem u = 0 in are u C, for some constant C. u ν = 0 4 on

5 Proof. (a For any test function v on, du 0 = uv dx = dν v dx + Du Dv dx = Du Dv dx Take v = Du. Since Du 2 L 2 ( = 0, Du 0. Thus, smooth u C, for some constant C. (b If u attains an interior maximum, we have our result by the strong maximum principle. Otherwise, there is some x 0 such that u(x 0 > u(x for all x. By Hopf s lemma, u ν (x 0 > 0, contradicting our assumption that u ν = 0 on. Thus, u attains an interior maximum, and u C by the strong maximum principle. 6. (Evans 6.6/12 We say that the uniformly elliptic operator Lu = a ij u xi x j + b i u xi + cu satisfies the weak maximum principle if for all u C 2 ( C(Ū Lu 0 in u 0 i on implies that u 0 in. Suppose that there exists a function v C 2 ( C(Ū such that Lv 0 in and v > 0 on Ū. Show that L satisfies the weak maximum principle. Hint: Find an elliptic operator M with no zeroth-order term such that w := u/v satisfies Mw 0 in the region u > 0}. To do this, first compute (v 2 w xi xj. Proof. sing the hint, we can discover a suitable elliptic operator M. By direct computation, w xi = u x i v uv xi v 2, (v 2 w xi xj = u xi x j v + u xi v xj u xj v xi uv xi x j By ellipticity of L and symmetry of (a ij, a ij (v 2 w xi xj = = v = v a ij ( u xi x j v + u xi v xj u xj v xi uv xi x j ( a ij u xi x j + u Lu a ij v xi x j ( b i u xi cu u Lv = vlu ulv b i (vu xi uv xi Thus, it is reasonable to consider the following elliptic operator: a ij (v 2 w xi xj Mw := v 2 + b i vlu ulv w xi = v 2 b i v xi cv 5

6 By assumption, Lu 0, v > 0, and Lv 0 in. Thus, Mw has no zeroth order term and satisfies Mw 0 on u > 0}. By the weak maximum principle (multiplying by v > 0, max Ū u>0} vw = max vw u>0} Since vw = u 0 on by assumption, necessarily u 0 on Ū. 6