Solution Sheet 3. Solution Consider. with the metric. We also define a subset. and thus for any x, y X 0

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1 Solution Sheet Throughout this sheet denotes a domain of R n with sufficiently smooth boundary. 1. Let 1 p <. Show that Sobolev space W 1,p () is separable. [Hint: Reduction to Remark 2.10 ] Solution Consider X (L p ()) n+1 {x (u 1, u 2,..., u n+1 ) : u i L p (), i 1, 2,..., n + 1} with the metric ϱ(x, y) u 1 v 1 p, + u 2 v 2 p, u n+1 v n+1 p,. We also define a subset X 0 {x (u 1, u 2,..., u n+1 ) : u 1 u W 1,p (), u i D i 1 u, i 2,..., n+1} and thus for any x, y X 0 ϱ(x, y) u v p, + D 1 u D 1 v p, D n+1 u D n+1 v p, u v p,1,. The result follows from Remark 4.10 and separability of L p () for 1 p <. 2. Prove Ladyzhenskaya s inequality: u 2 4,R 2 2 u 2,R 2 u 2,R 2 u C 1 0(R 2 ). [Hint: Lemma 5.1 applied to u 2 ]. Solution From Lemma 5.1 we have u 2 2,R 2 u 2 4,R 2 D 1u ,R 2 D 2 u ,R 2. On the other hand by Hölder inequality we have and thus D i u 2 1,R 2 2 ud 2 u 1,R 2 2 u 2,R 2 D i u 2,R 2 u 2 4,R 2 2 u 2,R 2 D 1u 1 2 2,R 2 D 2 u 1 2 2,R 2 2 u 2,R 2( D 1 u 2 2,R 2+ D 2u 2 2,R 2)

2 . Prove a multiplicative inequality u c u 4,R 2,R u 2 2,R u C1 0(R 2 ). [Hint: use interpolation between L 2 and L 6, see Qn.9, Sheet 1, then apply Theorem 5.2 to L 6 -norm]. Solution Interpolation: 10 2λ + 6(1 λ). So, λ 2. So, ( ) u u dx u ( dx u 2 dx,r R R R According to Theorem 5.2, we have u 6,R 2( 1) 2 u 2,R. From the latter inequalities, we get our statement with c 4 2. R u 6 dx 4. Suppose that is a star-shaped domain with respect to the origin x 0 0. Let f L p () with 1 p < be extended by zero to the whole R n. Show that ( ) 1 f( /λ) f( ) p, : f(x/λ) f(x) p p dx 0 as λ 1. [Hint: similar to the proof of integral continuity, see Theorem 2.11 ] Solution Fix a big cube Q centered at the origin so that λ Q for all 1/2 < λ < 2. We know that f L p (2Q) and given ε > 0 there exists a function g C(2Q) such that f g p,2q < ε. Letting g λ (x) g(x/λ), we see that ( f λ g λ p,q λ n p Q ( Q/λ ( x ( x p ) 1 p f g dx λ) λ) ) 1 f(y) g(y) p p dy 2 λ n p ε. { } y x/λ ) 1.

3 Since g C(2Q), it is uniformly continuous there and thus ( x g g(x) < ε λ) for any x Q and λ 1 < δ(ε). So, f λ f p, f λ f p,q f g p,q + g λ g p,q + f λ g λ p,q ε(1 + λ n p ) + ε Q 1 p ε(1 + 2 n p ) + ε Q 1 p. 5. Let {x (x 1, x 2 ) R 2 : x < 1, x 2 > 0} and u W 1,p ( ) with 1 p <. Denote by ũ an even extension of u to the whole disk B { x < 1}, i.e., ũ(x) u(x) if x and ũ(x) ũ(x 1, x 2 ) u(x 1, x 2 ) if x { x < 1 : x 2 < 0}. Show that ũ W 1,p (B) and find the norm of ũ in W 1,p (B) in terms of the norm of u in W 1,p ( ). [Hint: Note that is a star-shaped domain, approximate u by smooth functions, extend smooth functions in the even way and show that extended functions is a Cauchy sequence in W 1,p (B).] Solution Let u (n) C ( ) u in W 1,p ( ). Let ũ (n) be the even extension of u (n) into, i.e., ũ (n) (x 1, x 2 ) u (n) (x 1, x 2 ) if x 2 < 0. Obviously, ũ (n) C(B). Let us show ũ (n) has weak derivatives of order one and u(n), u(n) in, Indeed, B (x 1, x 2 ) u(n) (x 1, x 2 ) if x 2 < 0, (x 1, x 2 ) u(n) (x 1, x 2 ) if x 2 < 0. (n) ϕ ũ dx dx dx

4 and 1 B ϕdx + (n) ϕ ũ dx u (n) (x 1, +0)ϕ(x 1, +0)dx 1 (x 1, x 2 )ϕdx dx 1 dx u (n) (x 1, 0)ϕ(x 1, 0)dx 1 + and 1 + Simple calculations give ϕdx ϕdx 1 (x 1, x 2 )ϕdx (x 1, x 2 )ϕdx. ũ (n) p p,b 2 u(n) p p,, Dũ (n) p p,b 2 Du(n) p p,, ũ (n) ũ (m) p p,b 2 1 p u (n) u (m) p p,. Therefore, ũ (n) ũ in W 1,p (B) and ũ is just an even extension of u. Moreover, ũ p p,b 2 1 p u p p,. 6. Assume that is a bounded Lipschitz domain. Let 1 < p < and let a sequence u j is bounded in W 1,p (). Show that there exist a subsequence u jk and a function u W 1,p () such that u jk u in L p () and D i u jk D i u in L p (), i 1, 2,, n, as k. Solution Since norms u j p, and Du j p, are bounded there exists a subsequence u jk such that u jk u and D i u jk v i, i 1, 2,..., n, in L p (). By the Reillich-Kondrashov theorem u jk u in L p (). It remains to show that v i D i u. Indeed, we have by the definition D i u jk ϕdx u jk D i ϕdx. 4

5 Passing to the limit as k, we find from the latter identity v i ϕdx ud i ϕdx for all ϕ C 0 () that implies v i D i u. 7. Let be a bounded domain of class C 1 and 1 p <. Suppose that u W 1,p () with u 0. Let 0 and ũ is an extension of u by zero from to 0. Show that ũ W 1,p 0 ( 0 ). [Hint: use the formula of integration by parts ] Solution Let ϕ C0 ( 0 ). Then by formula integration by parts, see Lemma 6.4, ũd i ϕdx ud i ϕdx uϕν i ds + ϕd i udx ϕd i udx 0 and thus D i ũ D i u in, D i ũ 0 in 0 \. This means that ũ W 1,p ( 0 ). Obviously, ũ ϱ C0 ( 0 ) for sufficiently small ϱ > 0 and ũ ϱ ũ in W 1,p ( 0 ) as ϱ 0. So, ũ W 1,p 0 ( 0 ). 8. Suppose that be a bounded domain of class C 1. Consider the following Dirichlet problem u + b u div g in u 0 on with g L 2 () and b L () such that div b 0 in the sense of distributions. Show that the above Dirichlet problem has a unique weak solution. Solution We let L(u, v) ( u v b uv)dx. Then L(u, u) u 2 b i u,i udx. 5

6 We state L(u, u) u 2 dx for all u C0 (). To prove that, we first notice that if u C0 (), then u 2 C0 () as well. And then b i u,i udx 1 b u 2 dx T b( u 2 ) 1 2 divt b( u 2 ) 0. By Poincare s inequality, L(u, u) c u 2 2, for all u C 0 (). The statement now follows from Theorem 7.4 and Corollary Consider the following multiple integrals I(v) : v 2 dx + v dx + 2 fvdx. Suppose that R and f L 2 (). Show that: (i) I(v) < for each v H 1 (); (ii) C : inf I(v) >. v H 1 () (iii) there exists u H 1 () such that I(u) C. Solution (i) By imbedding theorem, we have v, c 1 v 2,1, and Hence, v 2, v 2,1,. I(v) v 2 2, + (c 1 v 2,1, ) + 2 f 2, c 2 v 2,1, < 6

7 for each v H 1 (). (ii) From the above inequalities, it follows I(v) v 2 2, + v, 2 f 2, v, v 2 2, v, c f 2 2,. Since v 2 v + c and f 2, c() f 2 2,, we have I(v) 1 2 v 2 2,1, c(, f 2, ). (iii) Let v (i) be a minimizing sequence, i.e., I(v (i) ) C. By the previous estimate, the sequence v (i) is bounded in H 1 () and thus WLOG we may assume that it is weakly converging to u in H 1 (). We know that the function I 0 (h) : h 2 dx is continuous in L 2 () and convex. Therefore, it is sequentially weakly lower semi-continuous in L 2 (). So, we have lim inf v (i) 2 dx u 2 dx. i On the other hand, by compactness of embedding H 1 () into L (), we may assume without loss of generality that v (i) u in L () and thus lim ( v (i) + 2fv (i) )dx ( u + 2fu)dx. i Hence, since lim inf i I(u) C. I(v(i) ) I(u), we find that I(u) C and therefore 7