Solutions to Partial Differential Equations by Lawrence Evans

1 Solutions to Partial DifferentialEquations by Lawrence EvansMatthew KehoeMay 22, are my Solutions to selected problems from chapters 5 9 ofPartial Differential Equations by Lawrence Evans . Any mistakes in thesesolutions are my own. I plan to write more Solutions in the future. If youwould like to speak with me about these Solutions (or about anything relatedto PDEs) then I can be contacted at 5 .. 14 Exercise (Hardy s Inequality onR+).. 161 Evans Chapters 5 - 9 Chapter 6 .. 27 Elasticity Exercise.. 28 Chapter 7 .. 44 Strong Maximum Principle Exercise.. 45 Chapter 8 .. 522 Evans Chapters 5 - 9 Chapter 9 .. 62 Chapter 5 1 andu W1,p(0,1) for some 1 p < .(a) Show thatuis equal to an absolutely continuous functionu (whichexists ) belongs toLp(0,1).

2 (b) Prove that if 1< p < , then|u(x) u(y)| |x y|1 1p( 10|u |pdt)1/pfor ,y [0,1].We first state a lemma summarizing the relationship between absolutelycontinuous functions and the fundamental theorem of functionUon[a,b]is absolutely continuous if and only ifU(x) =U(a) + xau(t)dtfor some integrable functionuon[a,b]. sufficiency part of the lemma follows directly from the fundamentaltheorem of calculus. That is, ifuis integrable on [a,b], and ifUis defined byU(x) := xau(t)dt, a x b,thenU (x) =u(x) for almost everyxin [a,b]. To prove the necessity part, weletUbe an absolutely continuous function on [a,b]. ThenUis differentiablealmost everywhere andU is integrable on [a,b]. LetG(x) :=U(a) + xaU (t)dt, x [a,b].3 Evans Chapters 5 - 9By the fundamental theorem of calculus,G (x) =U (x) for almost everyx [a,b].

3 It then follows that (U G) (x) = 0 for almost everyx [a,b].Therefore,U Gis a constant. ButU(a) =G(a). Therefore,U(x) =G(x) foralmost everyx [a,b].For (a), let [a,b] = [0,1] andv(x) = x0u (s)ds. Thenvis an absolutelycontinuous function by Lemma 1 (also see Rudin [7] for a proof). Therefore forany test function C c(0,1) : 10(v(x) u(x)) (x)dx= 10 x0u (s)ds (x)dx 10u(x) (x)dx= 10 1s (x)dxu (s)ds 10u(x) (x)dx= 10 (s)u (s)ds+ 10u (x) (x)dx= 10 (x)u (x)dx+ 10u (x) (x)dx= was chosen arbitrarily, we see thatuis equal to an absolutelycontinuous functionu as required. For (b), sinceu is inLp(0,1) we applyHolder s inequality withq=p/(p 1):|u(x) u(y)| xy|u (t)|dt ( xydt)(p 1)/p( xy|u |pdt)1/p |x y|1 1/p( 10|u |pdt)1 bounded andU Ni=1Vi. Show there existC functions i(i= 1,2.)

4 ,N) such that{0 i 1,spt i Vi(i= 1,2,..,N) Ni=1 i= 1 functions{ i}Ni=1form apartition of first complete problem LetUandVbe open sets withV U. We need to show that there exists a smooth function such that 1 onVand = 0 near Chapters 5 - 9As suggested in the hint by Evans , takeV W Uand let >0 be thedistance betweenVand U. Then defineW:={x U:d(x,V)< 2}.By making this distance small enough, we have constructed an open setWwhich is contained betweenUandV. Let = /8. Then (x) =1 n (x )is the required mollifier as suggested in Appendix C of Evans . Define (x) := W(x),where spt( ) spt( ) + spt( W) Uand is therefore a smooth (x) = Rn (x y) W(y)dy= B(x, ) W (x y) W(y)dy= B(x, ) W (x y) ifB(x, ) W, we see Rn (y)dy= B(0, ) (y)dy= 1,which implies that the support is inW B(0, ).}

5 As Wis compact, we willcover it and its boundary by open balls. Let W N i=1 WiwhereWidenotes an open ball which covers a portion ofWand possibly theboundary. Then we may observe that we can use the mollifier ifor everyopen ballWiwhere i(x) := i W(x).By defining (x) :=N i=1 i(x) Ni=1 i(x),we observe that for any fixedx U, only three terms in the sum will benonzero. AsV W U, it is clear that 1 onVand = 0 near Chapters 5 - 9To complete , we assumeUis bounded andU Ni=1Wi Ni= ,Uhas a finite cover{V1,..,Vn}where for everyVi, we have a iasconstructed above. The support of iis contained entirely inVi, i 1 onWi, and every iis smooth by definition. Therefore, we define i(x) := i(x) Ni=1 i(x)where i i 1 by construction and for allx U, the support of iiscontained inVi.

6 Also, iis smooth because the iare smooth. Thus, thecollection{ i}Ni=1fulfills all of the requirements and is a partition of unitysubordinate to the cover{V1,..,Vn}. thatUis bounded and there exists a smooth vector field such that 1 along U, where as usual denotes the outward unitnormal. Assume 1 p < .Apply the Gauss-Green Theorem to U|u|p dS, to derive a new proof ofthe trace inequality U|u|pdS C U|Du|p+|u|pdxfor allu C1( U). 1 along U, we have|u|p |u|p . Then U|u|pdS U|u|p dS= U (|u|p )dx(Gauss Green)= U|u|p( ) + |u|pdx C U|u|p+| |u|p| , since |u|p=p|u|p 1(sgnu)Du,we have forp= 1 U|u|dS C U|u|+|Du| the other hand, ifp >1 then we apply Young s inequality witha=|Du|, b=|u|p 1, q=p/(p 1) to form U| |u|p|dx C Up|u|p 1|Du|dx C U|Du|p+ (p 1)|u| Chapters 5 - 9 The constants above are different at every inequality.

7 We may now observethat U|u|pdS C U|u|p+|Du|p+ (p 1)|u|pdx C U|u|p+|Du|pdx,as bounded, with aC1boundary. Show that a typical functionu Lp(U) (1 p < ) does not have a trace on U. More precisely,prove that there does not exist a bounded linear operatorT:Lp(U) Lp( U)such thatTu=u| Uwheneveru C( U) Lp(U). will construct a counterexample inL2(U). We need to show thatthere does not exist a constantC >0 such that Tu L2( U) C u L2(U)foreveryu L2(U). Let s consider the following sequence of continuous functions:un(x) =11 +nd(x, U), x un(x) 1,un(x) = 1, x , for everyx U, we see thatun(x) 0 pointwise, so by thedominated convergence theorem un 2L2(U) , for everynwe have Tun 2L2( U) C2 un 2L2(U) 0,which implies that the area of the boundary is equal to zero.

8 AsTun= 1 foreverynwe see that we have arrived at a contradiction. The same analysisworks forLp(U) when 1 p < .7 Evans Chapters 5 - by parts to prove the interpolation inequality: Du L2 C u 1/2L2 D2u 1/2L2for allu C c(U). AssumeUis bounded, Uis smooth, and prove thisinequality ifu H2(U) H10(U).(Hint: Take the sequences{vk} k=1 C c(U) converging touinH10(U) and{wk} k=1 C ( U) converging touinH2(U).) C c(U). Integrating by parts and applying Cauchy Schwarz inthe last inequality forms UDu Dudx C U|u||D2u|dx C( U|u|2dx)1/2( U|D2u|2dx)1/2,where the boundary term disappears sinceuhas compact support. Taking thesquare root yields Du L2(U) C u 1/2L2(U) D2u 1/2L2(U).Following the hint provided by Evans , sinceH10(U) is the closure ofC 0(U)with the norm ofH1(U), we can find a sequence{vk} k=1inH1(U) C c(U)converging touinH10(U).

9 Also, since Uis smooth, we can extendUto a setVsuch thatU V. Then, by the density ofC c(V), we can find a sequence{wk} k=1inC ( U) converging touinH2(U). So, we apply the Gauss GreenTheorem and evaluateDvk Dwkwhich after one application ofCauchy Schwarz yields UDvk Dwkdx C U|vk||D2wk|dx C( U|vk|2dx)1/2( U|D2wk|2dx)1/2,where the boundary term once again vanishes becausevkhas compactsupport. Ask ,C( U|vk|2dx)1/2( U|D2wk|2dx)1/2 C( U|u|2dx)1/2( U|D2u|2dx)1/2,which is equivalent toC u L2(U) D2u L2(U). To show that the left-hand sideconverges to Du 2L2(U), we evaluate the difference and once again applyCauchy Schwarz U(Dvk Dwk Du Du)dx= U(Dvk (Dwk Du) +Du (Dvk Du))dx U|Dvk| |Dwk Du|+|Du| |Dvk Du|dx Dvk L2(U) Dwk Du L2(U)+ Du L2(U) Dvk Du L2(U),8 Evans Chapters 5 - 9where ask , the right-hand side goes to zero since bothvk,wk uwhereu H2(U) H10(U).

10 This implies that UDvk Dwkdx U|Du|2dx,and we may conclude Du 2L2(U) C u L2(U) D2u L2(U)which after taking the square root forms Du L2(U) C u 1/2L2(U) D2u 1/2L2(U). connected andu W1,p(U) satisfiesDu= 0 constant Solution:Consider the domainU ={x U: dist(x, U)> }.Forx U consider the functionu (x) = (x y)u(y)dywhere is astandard mollifier. Thenu C (U ) andDu (x) = (x y)Du(y)dyby the definition of the weak derivative. SinceDu= 0 , we have thatDu (x) = 0 for allx U and henceu is constant inU . Since u u Lp(U ) 0 as 0, we have thatuis constant Solution:Let >0. Then consider the smooth functionsu = u C (U ),whereU ={x U:d(x, U)> }. By Theorem in Evans , we haveDxi(u ) = , by assumption,Dxiu = 0 inU . Sou is constant on eachconnected subset ofU.