evans-solutions

Authors:JoeBenson,DenisBashkirov,MinsuKim,HelenLi,AlexCsarEvansPDESolutions,Chapter2Joe:1,2,11;Denis:4,6,14,18;Minsu:2,3,15;Helen:5,8,13,17.Alex:10,16Problem1.Writedownanexplicitformulaforafunctionusolvingtheinitial-valueproblem(ut+b
Du+cu=0onRn(0;1)u=gonRnft=0gHerec2Randb2Rnareconstants.Sol:Fixxandt,andconsiderz(s):=u(x+bs;t+s)Then˙z(s)=b
Du+ut=cu(x+bs;t+s)=cz(s)Therefore,z(s)=Decs,forsomeconstantD.WecansolveforDbylettings=t.Then,z(t)=u(xbt;0)=g(xbt)=Decti.e.D=g(xbt)ectThus,u(x+bs;t+s)=g(xbt)ec(t+s)andsowhens=0,wegetu(x;t)=g(xbt)ect.Problem2.ProvethatLaplace’sequation
u=0isrotationinvariant;thatis,ifOisanorthogonalnnmatrixandwedefinev(x):=u(Ox)(x2R)then
v=0.Solution:Lety:=Ox,andwriteO=(aij).Thus,v(x)=u(Ox)=u(y)whereyj=Pni=1ajixi.Thisthengivesthat@v@xi=nXj=1@u@yj@yj@xi=nXj=1@u@yjaji12Thus,266666666664@v@x1:::@v@xn377777777775=26666666664a11:::an1::::::a1n:::ann37777777775266666666664@u@y1:::@u@yn377777777775=OT266666666664@u@y1:::@u@yn377777777775Dx
v=OTDy
uNow,
v=Dxv
Dxv=(OTDyu)
(OTDyu)=(OTDyu)TOTDyu=(Dyu)T(OT)TOTDyu=(Dyu)TOOTDyu=(Dyu)TDyubecauseOisorthogonal=(Dyu)
(Dyu)=
u(y)=0Problem3.Modifytheproofofthemeanvalueformulastoshowforn3thatu(0)=1n(n)rn1Z@B(0;r)gdS+1n(n2)(n)ZB(0;r)1jxjn21rn2fdx;provided8:
u=finB0(0;r)u=gon@B(0;r).Solution:Set(t)=1n(n)tn1Z@B(0;t)u(y)dS(y);0tr;and(r)=1n(n)rn1Z@B(0;r)u(y)dS(y)=1n(n)rn1Z@B(0;r)gdS:Then,0(t)=tn1(n)tnZB(0;t)
u(y)dy=tn1(n)tnZB(0;t)fdy=1(n)tn1ZB(0;t)fdy:(SeetheproofofThm2)3Let0begiven.(1)()=(r)Zr0(t)dt=1n(n)rn1Z@B(0;r)gdSZr0(t)dt:Usingintegrationbyparts,wecomputeZr0(t)dt=Zr1n(n)tn1ZB(0;t)fdydt=1n(n)Zr1tn1ZB(0;t)fdydt=1n(n)12n1tn2ZB(0;t)fdyrZr12n1tn2Z@B(0;t)fdSdt=1n(n2)(n)Zr1tn2Z@B(0;t)fdSdt1rn2ZB(0;r)fdy+1n2ZB(0;)fdy=:1n(n2)(n)I1rn2ZB(0;r)fdy+J:ObservethatJ:1n2ZB(0;)fdyC
2;forsomeconstantC0andZB(0;)1jxjn2f(x)dx=Zr0dtZ@B(0;t)1tn2fdS:As!0,I+J!RB(0;)1jxjn2f(x)dx:Thus,lim!0Zr0(t)dt=1n(n2)(n)ZB(0;r)1jxjn2f(x)dx1rn2ZB(0;r)fdy=1n(n2)(n)ZB(0;r)1jxjn21rn2fdx:Therefore,letting!0,wehavefrom(1)u(0)=(0)=1n(n)rn1Z@B(0;r)gdS+1n(n2)(n)ZB(0;r)1jxjn21rn2fdx:Problem4.Wesayv2C2(¯U)issubharmonicif
v0inU:(a)Proveforsubharmonicvthatv(x)?B(x;r)vdyforallB(x;r)U:(b)Provethatthereforemax¯Uv=max@Uv.(c)Let:R!Rbesmoothandconvex.Assumeuisharmonicandv:=(u).Provevissubharmonic.4(d)Provev:=jDuj2issubharmonic,wheneveruisharmonic.Solution.(a)AsintheproofofTheorem2,set(r):=@B(x;r)vdS(y)andobtain0(r)=rn?B(x;r)
v(y)dy0:For0r,Zr0(s)ds=(r)()0:Hence,(r)lim!0()=v(x).Therefore,?B(x;r)vdy=1(n)rnZB(x;r)vdy=1(n)rnZr0Z@B(x;s)v(z)dS(z)!ds=1(n)rnZr0n(n)sn1(s)ds1rnZr0nsn1v(x)ds=v(x)(b)WeassumethatURnisopenandbounded.Foramoment,weassumealsothatUisconnected.Supposethatx02Uissuchapointthatv(x0)=M:=max¯Uv.Thenfor0rdist(x0;@U),M=v(x0)?B(x0;r)vdyM:Duetocontinuityofv,anequalityholdsonlyifvMwithinB(x0;r).Therefore,thesetu1(fMg)\U=fx2Uju(x)=MgisbothopenandrelativelyclosedinU.Bytheconnect-ednessofU,visconstantwithinthesetU.Hence,itisconstantwithin¯Uandweconcludethatmax¯Uv=[e-mail]2Uandfindj2Isuchthatx2Uj.Weobtainv(x)max¯Ujv=max@Ujvmax@Uvandconcludethatmax¯Uv=max@Uv.(c)Forx=(x1;:::;xn)2Uand1i;jn,@2v@xi@xj(x)=@2@xi@xj(u(x))=00(u(x))
@u@xi(x)
@u@xj(x)+0(u(x))
@2u@xi@xj(x):Sinceisconvex,then00(x)0foranyx2R.Recallthatuisharmonicandobtain
v=00(u)
nXi=1@u@xi!2+
u=00(u)
nXi=1@u@xi!20:(d)Wesetv:=jDuj2=nPk=1@u@xk2.Forx=(x1;:::;xn)2Uand1i;jn,@2v@xi@xj(x)=2nXk=1@2u@xi@xk(x)
@2u@xi@xj(x)+@u@xk(x)
@3u@xi@xj@xk(x)#:5Therefore,@2v@xi2=2nXk=12666664@2u@xi@xk!2+@u@xk
@@xk@2u@xi2!3777775;
v=2X1i;kn@2u@xi@xk!2+nXk=1@u@xk
@@xk
u=2X1i;kn@2u@xi@xk!20:Problem5:ProvethatthereexistsaconstantC,dependingonlyonn,suchthatmaxB(0;1)jujCmax@B(0;1)jgj+maxB(0;1)jfj!wheneveruisasmoothsolutionof8:4u=finB0(0;1)u=gon@B(0;1):Proof:LetM:=maxB(0;1)jfj,thenwedefinev(x)=u(x)+M2njxj2andw(x)=u(x)+M2njxj2.Wefirstconsiderv(x).Notethat4v=4uM=fM0:So,v(x)isasubharmonicfuncion.FromProblem4(b),wehavemaxB(0;1)v(x)=max@B(0;1)v(x)max@B(0;1)jgj+M2n:ThatismaxB(0;1)u(x)maxB(0;1)v(x)max@B(0;1)jgj+12nmaxB(0;1)jfj:Then,forw(x),wehave4w=4uM=fM0:Again,wecangetmaxB(0;1)w(x)=max@B(0;1)w(x)max@B(0;1)jgj+M2n:i.e.maxB(0;1)u(x)maxB(0;1)w(x)max@B(0;1)jgj+12nmaxB(0;1)jfj:Combiningthesetwotogether,wefinallyprovedtheproblem.Problem6.UsePoisson’sformulafortheballtoprovern2rjxj(r+jxj)n1u(0)u(x)rn2r+jxj(rjxj)n1u(0)wheneveruispositiveandharmonicinB0(0;r).ThisisanexplicitformofHarnack’sinequality.6Solution.Sincey2@B(0;r),thenjxyjjxj+r.Therefore,u(x)=r2jxj2n(n)rZ@B(0;r)g(y)jxyjndS(y)r2jxj2n(n)rZ@B(0;r)g(y)(r+jxj)ndS(y)=rn2rjxj(r+jxj)n1
1n(n)rn1Z@B(0;r)g(y)dS(y)=rn2rjxj(r+jxj)n1?@B(0;r)g(y)dS(y)=rn2rjxj(r+jxj)n1u(0)Theinequalityu(x)rn2r+jxj(rjxj)n1u(0)canbeproveninasimilarway.Problem7.ProvePoisson’sformulaforaball:Assumeg2C(@B(0;r))andletu(x)=r2x2n(n)rZ@B(0;r)g(y)jxyjndS(y)forx2B0(0;r):ShowthatPr