Theory of PDE Homework 5

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1 Theory of PDE Homework 5 Adrienne Sands March 22, 27. (Evans 5./2) Show by eamle that if we have D h u L (V ) C for all < h < 2 dist(v, ), it does not necessarily follow that u W, (V ). Proof. Let u() χ (,) (), V (, ), ( 2, 2), and < h < 2 dist(, V ) 2. Then h D h u() h h < < < h h h D h u D h u d L (V ) h h d + h h d 2 Suose u W, (V ) with u v in the weak sense for some v L (V ). ϕ Cc (V ), vϕ uϕ ϕ ϕ() ϕ() ϕ() Then for all by the fundamental theorem of calculus and comact suort of ϕ. Choose a sequence ϕ m } of smooth functions on V satisfying ϕ m ϕ m () 2 ϕ m () for all Then by construction, a contradiction. 2 lim ϕ m() lim vϕ, m m 2. (Evans 5./3) Give an eamle of an oen set R n and a function u W, (), such that u is not Lischitz continuous on. (Hint: Take to be the oen unit disk in R 2, with a slit removed.) Proof. Let (, 2 ) R 2 : 4 < <, > or y }, an oen annulus with the negative -ais removed and u(, y) 2 arctan ( 2 + ). Clearly u L () π, and u is

2 classically differentiable on with u () ( ) ( ) ( + ) 2 + ( ) u 2 () ( ) ( + ) such that Du L () 4. Consider ε(r, θ) ( 2, π ε) and y ε (r, θ) ( 2, ε π) for small ε >. Observe u() u(y) (π ε) (ε π) 2π 2ε su lim lim y y ε + ε y ε ε + sin(π ε) Thus, u W, () is not Lischitz continuous. 3. (Evans 5./6, Variant of Hardy s inequality) Show that for each n 3 there eists a constant C so that u 2 R n 2 d C Du 2 d R n for all u H (R n ). (Hint: Du + λ 2 u 2 for each λ R.) Proof. By the density of test functions in H (R n ), it suffices to rove the inequality for u C c (R n ). For all λ R, Du + λ 2 u 2 Du 2 + 2λu u2 Du + λ2 2 2 Mimicking Evans roof of Hardy s inequality, we have Rn u 2 2 u 2 R n 3 u 2 D R ( n R n D u 2 ) R n Rn 2uDu R n 2 + (n ) u2 2 such that Thus, (2 n) Rn u 2 ( ) ( 2uDu 2 2 u R n 2 Du Du 2 + 2λ u R n R n 2 Du + Rn Du 2 + ((2 n)λ + λ 2 u 2 ) R n u2 Rn λ 2 u2 2 ( )) n

3 or equivalently Rn u 2 2 ((n 2)λ λ 2 ) R n Du 2 4. (Evans 5./7, Chain rule) Assume F : R R is C with F bounded. Suose is bounded and u W, () for some. Show v : F (u) W, () and v i F (u)u i (i,..., n) Remark: Per the instructions for Evans Section 5., we also assume is smooth. Proof. Let <. By the fundamental theorem of calculus and boundedness of F, u() v L () F (u()) F u() F L (R) L (R) u L < By global aroimation, there are u k W, () C () such that u k u in W, () and ointwise a.e. Let v k : F (u k ) C () L () with Dv k F (u k )Du k C() L (). Then, u() v v k L () F (u()) F (u k ()) u k () F u() u L (R) k () : I Dv k F (u)du L () F (u k )Du k F (u)du F (u k )Du k F (u k )Du + F (u k )Du F (u)du F Du L (R) k Du + u W, () F (u k ) F (u) : J + K We have I, J by convergence of u k } in W, (). Since u k u ointwise and F continuous, F (u k ) F (u) ointwise; thus, K by Lebesgue Dominated Convergence with majorant 2 u W, () F L (R), yielding our result for <. Now let. Since is oen and bounded with of class C, u W, () is Lischitz such that for all, y, u() F (u()) F (u(y)) F L () u() u(y) F L M y () u(y) Thus, v W, () with Dv F (u)du a.e. by Rademacher s Theorem. 5. (Evans 5./8) Assume and is bounded. Remark: Per the instructions for Evans Section 5., we also assume is smooth. 3

4 (a) Prove that if u W, (), then u W, (). Proof. As in Evans, we use the following convention u + ma(u, ) u min(u, ) Since u u + + u, this follows immediately from art (b) and linearity of the weak gradient. (b) Prove u W, () imlies u +, u W, (), and Du + Du a.e. on u > } a.e on u } Du a.e. on u } Du a.e. on u < } Hint: u + lim ε F ε (u), for F ε (z) : (z 2 + ε 2 ) /2 ε if z if z < Proof. For fied ε >, F ε (z) C (R) with F ε(z) z z z > 2 +ε 2 z and F ε L (R). By the revious roblem, F ε(u) W, () and DF ε (u) F ε(u)du such that for all ϕ Cc (), F ε (u)dϕ ϕf ε(u)du ϕ udu u> u 2 + ε 2 Taking the limit as ε +, we have u + Dϕ u> ϕdu yielding our result for Du +. Since u ( u) +, our result for Du follows immediately by linearity of the weak gradient. (c) Prove that if u W, (), then Du a.e. on the set u } Proof. Since u u + u, this follows immediately from art (b) and linearity of the weak gradient. 6. (Evans 5./2) se the Fourier transform to rove that if u H s (R n ) for s > n/2, then u L (R n ), with the bound u L (R n ) C u H s (R n ) for a constant C deending only on s and n. 4

5 Proof. For convenience, let H s H s (R n ) and L L (R n ). Since u H s L 2, we have Fourier inversion such that û u( ); futhermore, it suffices to show û L since Fourier transform mas L to L and û u L L. First note ( + ξ s ) L 2 < for s > n/2: ( + ξ s ) 2 ( + ξ s ) 2 L 2 R n ( + r s ) 2 r n dr r n 2s dr < By Cauchy-Schwarz-Bunyakovsky and the characterization of H s by Fourier transform, û(ξ) + ( ) ξ s /2 ( ) /2 û R n R n + ξ s (û( + ξ s )) 2 ( + ξ s ) 2 R n R n ( + ξ s )û L 2 ( + ξ s ) L 2 C u H s ( + ξ s ) L 2 C u H s Since û L, û L with u L û su L (2π) n/2 e i ξ û(ξ) R n (2π) n/2 û(ξ) C u H s R n 5