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EXAM PDE 18.02.13 1. Exercise 3 Let Ω ⊂ R be a bounded connected open set with ∂Ω ∈ C 4 , and for coefficients aij ∈ C 3 (Ω) PN 2 satisfying aij (x) = aji (x) consider the elliptic operator Lu = − i,j=1 aij (x) ∂x∂i ∂xj u. Let g ∈ H 2 (Ω) and ϕ ∈ C ∞ (Ω) . (a) Justify the following statement: the Dirichlet problem ( Lu = g in Ω u = ϕ on ∂Ω has a unique solution u ∈ C 2 (Ω) ∩ C 1 (Ω) . (b) Justify the following statement: Ω satisfies the interior ball condition. (c) Denote with ν(x) the unit outer normal to ∂Ω and let a C 2 vector field η(x) be given such that η(x) · ν(x) > 0 for every x ∈ ∂Ω . Consider the following problem ( Lu = g in Ω (1.1) ∂u ∂η = ϕ on ∂Ω and assume it has a solution u ∈ C 2 (Ω) ∩ C 1 (Ω) . Prove that v ∈ C 2 (Ω) ∩ C 1 (Ω) solves (1.1) if and only if v − u is constant in Ω . Solution: (a) It is clear that u solves the problem if and only if u = v + ϕ where v solves ( Lv = g − Lϕ in Ω v = 0 on ∂Ω . It is clear (Lax-Milgram’s Theorem) that such a problem has a unique weak solution v . In view of the regularity of ϕ, it suffices to prove that v ∈ C 2 (Ω) ∩ C 1 (Ω) . Since g − Lϕ ∈ H 2 (Ω) and aij ∈ C 3 (Ω) , by the smoothness of the boundary and the regularity Theorem (Evans, Thm. 5, Section 6.3.2) we have v ∈ H 4 (Ω) . Since we are in dimension N = 3 , Morrey’s imbedding Theorem gives that H 4 (Ω) ⊂ C 2 (Ω) , as required. (b) It has already been observed, before proving Hopf’s Lemma, that the interior ball condition is automatically satisfied by a C 2 boundary, a fortiori it is in the situation in object. (c) One implication is clear. To prove the other one, let w = v − u. Then w solves ( Lw = 0 in Ω ∂w ∂η = 0 on ∂Ω . We are in setting of Hopf’s Lemma and of the strong maximum principle (Evans, Section 6.4), therefore either w is constant or there exists a maximum point x0 ∈ ∂Ω such that w(x0 ) > w(x) for every x ∈ Ω . Since x0 is a maximum point, for every tangent vector τ to ∂Ω at x0 , one has ∂w (x0 ) = 0 ∂τ while Hopf’s Lemma states that ∂w (x0 ) > 0 . ∂ν 1 2 EXAM Considering an orthonormal basis of R3 of the form {ν(x0 ), τ1 , τ2 } where τ1 and τ2 are tangent vectors to ∂Ω at x0 , one has then (η(x0 ) · ν(x0 )) ∂w ∂ν (x0 ) + (η(x0 ) · ∂w ∂η (x0 ) = ∂w (x0 ) τ1 ) ∂τ 1 ∇w(x0 ) · η(x0 ) = ∂w + (η(x0 ) · τ2 ) ∂τ (x0 ) = (η(x0 ) · ν(x0 )) ∂w ∂ν (x0 ) > 0 , 2 a contradiction. Therefore w must be constant as required. 2. Exercise N Let Ω ⊂ R , N ≥ 2 , be a bounded connected open set with ∂Ω ∈ C 1 and 1 ≤ p < N . For any q ≥ p0 the following statement is satisfied: for every f ∈ W 1,p (Ω) and g ∈ W 1,q (Ω) the product f g ∈ W 1,1 (Ω) . Compute also the weak gradient of the product. Solution: (we will compute here the optimal exponent q . Actually it is an exponent less than p0 , so the bound is even larger. But the arguments leading to a proof are essentially the same) For the p , let q = (p∗ )0 , that is, by a direct computation Sobolev exponent p∗ = NN−p q= Np . N (p − 1) + p It is easy to check that q < p0 . Furthermore, again by directly computing q ∗ = p0 . Now, given f ∈ W 1,p (Ω) and g ∈ W 1,q (Ω) we can find a sequence fn ∈ C ∞ (Ω) and gn ∈ C ∞ (Ω) such that fn → f in W 1,p (Ω) and gn → g in W 1,q (Ω) . In particular, by Sobolev’s embedding ∗ 0 Theorem, gn → g in Lq (Ω) = Lp (Ω) , where the last equality follows by our choice of q . It then easily follows by Hölder’s inequality that fn gn → f g in L1 (Ω) . On the other hand D(fn gn ) = fn Dgn + gn Dfn . Again by Sobolev’s embedding Theorem and our choice of q we have now ∗ fn → f in Lp (Ω) q∗ and Dgn → Dg in Lq (Ω) = L(p p0 gn → g in L (Ω) = L (Ω) ∗ 0 ) (Ω) p and Dfn → Df in L (Ω) so that another application of the Hölder’s inequality assures that D(fn gn ) → f Dg + gDf in L1 (Ω) . It follows that f g ∈ W 1,1 (Ω) and that D(f g) = f Dg + gDf . Since q < p0 , the statement holds true for any larger exponent, as required. 3. Exercise N N (a) Let T : R → R be an affine map of the form T x := Ax + b with A ∈ O(N ) , the group of orthogonal matrices, and b ∈ RN . Let Ω ⊂ RN be an open set. Show that for every u ∈ C 2 (Ω) one has ∆(u ◦ T ) = (∆u) ◦ T . (b) Consider in R2 the open set Ω := B(0, 1) \ {0} with B(0, 1) the unit ball. Let the Dirichlet problem ( −∆u = −1 in Ω (3.1) u = 0 on ∂Ω be given. Show that such a problem has no solution u ∈ C 2 (Ω) ∩ C(Ω) . (Hint: deduce from the previous step that a solution must be radially symmetric. This reconducts the calculation of a solution to an ordinary differential equation). (c) Use the calculations of the previous step to construct a weak solution u ∈ H01 (Ω) . Is it unique? What is the limit of u(x) as x tends to 0 ? EXAM 3 Solution: (a) We convene that a vector field f : RN → RN is represented by a row vector, that is a 1 × N matrix. According to this convention, for a N × N matrix Q, one has by a direct computation D(f Q) = QT (Df ) . Since one has, by the differentation theorem for composite functions ∇(u ◦ T ) = ((∇u) ◦ T )A we get to the well-known change of variable formula for the Hessian Matrix under rigid transformations D2 (u ◦ T ) = AT [(D2 u) ◦ T ] A (NB: if someone remembers the formula, it can be directly written without proof). Now, for any N × N matrix B one has tr(AT BA) = tr(B) since AT A = I ; being the Laplacian the trace of the Hessian, the claim is proved. (b) Assume that u ∈ C 2 (Ω) ∩ C(Ω) is a solution of (3.1). It is easily seen by means of part (a) that given a rotation matrix A (which maps bijectively both Ω onto Ω and ∂Ω onto ∂Ω ), u(Ax) is again a solution. If a solution exists, it is unique by the maximum principle, therefore u must be invarian under rotations. This implies u(x) = v(|x|) with v ∈ C 2 ((0, 1) ∩ C([0, 1]) . The Laplacian of a radial function has already been calculated to derive fundamental solutions; it is well-known that imposing that u solves (3.1) we get to the following ODE for v (taking also into account N = 2 ): 1 v 00 (r) + v 0 (r) = 1 r whose solutions are of the form 1 v(r) = r2 + c1 log r + c2 . 4 Since v ∈ C([0, 1]) it must be c1 = 0 ; it must be also v(1) = v(0) = 0 to fulfill the boundary conditions! But v(1) = 0 leads to c2 = − 14 , which is incompatible with v(0) = 0 ! (c) The function 1 1 u(x) = |x|2 − 4 4 satisfies by construction −∆u = −1 in Ω . Furthermore u ∈ H01 (B(0, 1)) , since u(x) = 0 when |x| = 1 . It has already been proved (Exercise 2, 14.2.13) that H01 (B(0, 1)) = H01 (B(0, 1) \ {0}) therefore u is easily seen to be a weak solution of (3.1). It is also the unique one, by Lax-Milgram’s theorem. Maybe surprisingly, one has u(x) → − 41 when x → 0 !