UNIVERSITETET I OSLO

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1 NIVERSITETET I OSLO Det matematisk-naturvitenskapelige fakultet Examination in: Trial exam Partial differential equations and Sobolev spaces I. Day of examination: November Examination hours: This examination set consists of 7 pages. Appendices: Permitted aids: None. Approved calculator. Make sure that your copy of the examination set is complete before you start solving the problems. Problem 1. a. If is a domain, and u C and v H 2, show that uv xj x i = u xi x j v + u xi v xj + u xj v xi + uv xi x j. Answer: Let ϕ be a test function, we compute ϕ u xi x j v + u xi v xj + u xj v xi + uv xi x j dx = ϕu xi x j v u xi ϕ xj v u xj ϕ x i v + uϕ xi x j v dx = v u xi x j ϕ u xi x j ϕ u xi ϕ xj u xi x j ϕ u xj ϕ xi + u xi x j ϕ = + u xi ϕ xj + u xj ϕ xi + uϕ xi x j dx vuϕ xi x J dx = uv xi x j ϕ dx Continued on page 2.

2 Examination in Trial exam, November Page 2 For the remainder of this problem, let be a bounded open subset of R 3, with a C 1 boundary. You may find the theorem in Figure 1 useful. Figure 1: Theorem from Evans: Partial differential equations. b. Show that for u and v in H 2 uxi x j v C v L 2 H 2 where C is some constant depending on only. uxi L x j 2 Answer: We have R 3 and u, v W 2,2. Thus in the terminology of the theorem; p = 2, k = 2 and n = 3, i.e., we are in case ii since 2 > 3/2. Hence u and v are in C 0,1/2, and we have the estimate u L u C 0,1/2 C u H 2. Thus uv xi x 2 j dx u 2 L v xi x 2 j dx. From these two estimates follows that uv L xi x u j 2 L vxi L x C u j 2 H 2 vxi L x. j 2 The other estimate is proved in the same way. Continued on page 3.

3 Examination in Trial exam, November Page 3 c. Show that if u is in H 2 then u xi L 6 for i = 1, 2, 3, and that u xi L p C u H 2, for 1 p 6. Answer: We have that u xi H 1 = W 1,2, so in the terminology of the theorem; p = 2, k = 1 and n = 3. Thus we are in case i, since 1 < 3/2. Thus u xi is in L q, with 1 q = = 1 6. By the theorem we have that u xi L 6 C u H 2, but since is bounded, u xi L p C u xi L 6 for 1 p 6 since is bounded. d. Show that if u and v are in H 2, then u xi v xj is in L 2, with the estimate uxi v xj C u L 2 H 2 v H 2. Answer: sing Hölder s inequality with p = 3, q = 3/2 uxi v xj 2 = L 2 uxi v xj 2 dx 1/3 u xi 2 3 dx vxj 2 3/2 2/3 dx = uxj 2 L 6 v x i 2 L 3 C uxj 2 v L 2 x i 2 L 2, since is bounded, C u 2 H 2 v 2 H 2. e. Show that if u and v are in H 2, then so is the product uv, and we have the estimate uv H 2 C u H 2 v H 2. Hint: use approximation of u by smooth functions in H 2 C. We call H 2 an algebra due to this property. Answer: Let {u n } H 2 C such that u n u in H 2. First we note that u n v 2 L 2 un 2 L v 2 H 2 C un 2 H 2 v 2 H 2. Continued on page 4.

4 Examination in Trial exam, November Page 4 By b, Hence u n v xi = u n x i v + u n v xi u n v xi x j = u n x i x j v + u n x i v xi + u n x j v xi + u n v xi x j. u n v xi L 2 u n x i v L 2 + un v xi L 2 u n xi L 2 v L + un L v L 2 C u n H 2 v H 2, and u n L v xi x j 2 u n x i x j v + u n xi v L u + n L xi L 2 2 xj v xi + u n v L xi x j 2 2 u n L x i x j v L 2 + C un H 2 v H 2 + un L vxi L x j 2 C u n H 2 v H 2. Hence u n v H 2 with the estimate Thus {u n v} H 2 satisfies u n v H 2 C un H 2 v H 2. u n v u m v H 2 C un u m H 2 v H 2 0, as m, n. Therefore {u n v} is a Cauchy sequence and u n v w H 2. We also have that the product uv L 2, u n v uv L 2 v L un u L 2 0, so u n v uv in L 2. So for any multiindex α and test function ϕ, D α wϕ dx = 1 α wd α ϕ dx and thus w = uv in H 2. = lim 1 α u n n vd α ϕ dx = 1 α uvd α ϕ dx = D α uvϕ dx, Continued on page 5.

5 Examination in Trial exam, November Page 5 Problem 2. a. For functions in L 2, the Fourier transform is given by 1 ûy = e ix y ux dx. 2π n/2 Show that if u L 2, then uy = y 2 ûy. Answer: We have that û xi x i y = y 2 i ûy. The answer follows by summing over i. b. On the space H m, we use the norm u H m = 1 + y m û L 2. Show that if u H m+1 and u H m then u H m+2 C u H m + u L 2 for some constant C. Hint: The following facts may be useful: i a + b 2 2a 2 + b 2, ii 1 + y m y m y 2., Answer: We have that 1 + y m y m y 2. sing a + b 2 2a 2 + b 2 we find that 1 + y m y m 2 y 2 2. Therefore u 2 H m+2 R = 1 + y m+2 2 ûy 2 dy n 2 ûy 2 dy y m 2 y 2 ûy 2 dy R n = 2 u 2 L 2 + u 2 H m 2 2 u L 2 + u H m R. n Continued on page 6.

6 Examination in Trial exam, November Page 6 c. Let cx be a bounded positive function such that 0 < c 1 cx c 2 < for all x. Here c 1 and c 2 are constants. Define a weak solution to the differential equation u + cxu = fx, x, 1 where f L 2. se the Riesz representation theorem to show that there exists a unique weak solution to 1 in H 1. Answer: The weak formulation is B[u, v] := u xi v xi + cxuv dx = fv dx = f, v L 2 u is called a weak solution if this holds for any v H 1. We have that B[u, v] max {1, c 2 } u xi v xi + uv dx C u H 1 v H 1, B[u, u] min {1, c 1 } u 2 H 1. Hence B is continuous and coersive. Furthermore, B[u, v] = B[v, u], and therefore B defines an alternative inner product on H 1. Consider the linear functional v f, v L 2. This is in H 1, and Riesz representation theorem says that there exists a unique u H 1 such that B[u, v] = f, v L 2. i d. Show that if u is a weak solution in H 1 of 1 then u H 2 and that u H 2 C f L 2, for some constant C. Answer: Let u be the weak solution, and consider the linear functional on L 2 given by v fv cuv dx. By Riesz representation theorem, there is a unique h L 2 such that vh dx = fv cuv dx, Continued on page 7.

7 Examination in Trial exam, November Page 7 for all v L 2. Furthermore h 2 L 2 fh + cuh dx h L 2 f L 2 + c 2 u L 2, or h L 2 C f L 2 + u L 2 Also for every v Cc L 2, vh dx = fv cuv dx = Du Dv dx = u v dx. This means that u = h in the weak sense. Next, use b with m = 0, this gives u H 2 C f L u L 2 since u L 2 C f L 2 + u L 2. However,, u 2 L 2 CB[u, u] C f L 2 u L 2, so u L 2 C f L 2. END