3 Laplace’s Equation - Stanford University

1 3 Laplace s EquationWe now turn to studyingLaplace s Equation u= 0and its inhomogeneous version,Poisson s Equation , u= say a functionusatisfying Laplace s Equation is aharmonic The Fundamental SolutionConsider Laplace s Equation inRn, u= 0x , there are a lot of functionsuwhich satisfy this Equation . In particular, anyconstant function is harmonic. In addition, any function of the formu(x) =a1x1+..+anxnfor constantsaiis also a solution. Of course, we can list a number of others. Here, however,we are interested in finding a particular solution of Laplace s Equation which will allow usto solve Poisson s the symmetric nature of Laplace s Equation , we look for aradialsolution. Thatis, we look for a harmonic functionuonRnsuch thatu(x) =v(|x|). In addition, to beinga natural choice due to the symmetry of Laplace s Equation , radial solutions are natural tolook for because they reduce a PDE to an ODE, which is generally easier to solve. Therefore,we look for a radial (x) =v(|x|), thenuxi=xi|x|v (|x|)|x|6= 0,which impliesuxixi=1|x|v (|x|) x2i|x|3v (|x|) +x2i|x|2v (|x|)|x|6= , u=n 1|x|v (|x|) +v (|x|).

2 Lettingr=|x|, we see thatu(x) =v(|x|) is a radial solution of Laplace s Equation impliesvsatisfiesn 1rv (r) +v (r) = ,v =1 nrv = v v =1 nr= lnv = (1 n) lnr+C= v (r) =Crn 1,1which impliesv(r) ={c1lnr+c2n= 2c1(2 n)rn 2+c2n these calculations, we see that for any constantsc1, c2, the functionu(x) {c1ln|x|+c2n= 2c1(2 n)|x|n 2+c2n 3.( )forx Rn,|x| 6= 0 is a solution of Laplace s Equation inRn {0}. We notice that thefunctionudefined in ( ) satisfies u(x) = 0 forx6= 0, but atx= 0, u(0) is claim that we can choose constantsc1andc2appropriately so that xu= 0in the sense of distributions. Recall that 0is the distribution which is defined as all D,( 0, ) = (0).Below, we will prove this claim. For now, though, assume we can prove this. That is, assumewe can find constantsc1, c2such thatudefined in ( ) satisfies xu= 0.( )Let denote the solution of ( ). Then, definev(x) = Rn (x y)f(y) , we compute the Laplacian ofvas follows, xv= Rn x (x y)f(y)dy= Rn y (x y)f(y)dy= Rn xf(y)dy=f(x).}}

3 That is,vis a solution of Poisson s Equation ! Of course, this set of equalities above is entirelyformal. We have not proven anything yet. However, we have motivated a solution formulafor Poisson s Equation from a solution to ( ). We now return to using the radial solution( ) to find a solution of ( ).Define the function as follows. For|x|6= 0, let (x) ={ 12 ln|x|n= 21n(n 2) (n)1|x|n 2n 3,( )where (n) is the volume of the unit ball inRn. We see that satisfies Laplace s equationonRn {0}. As we will show in the following claim, satisfies x = 0. For this reason,we call thefundamental solutionof Laplace s defined in( ), satisfies x = 0in the sense of distributions. That is, for allg D, Rn (x) xg(x)dx=g(0). be the distribution associated with the fundamental solution . That is, letF :D Rbe defined such that(F , g) = Rn (x)g(x)dxfor allg D. Recall that the derivative of a distributionFis defined as the distributionGsuch that(G, g) = (F, g )for allg D.}

4 Therefore, thedistributional Laplacianof is defined as the distributionF such that(F , g) = (F , g)for allg D. We will show that(F , g) = ( 0, g) = g(0),and, therefore,(F , g) = g(0),which means x = 0in the sense of definition,(F , g) = Rn (x) g(x) , we would like to apply the divergence theorem, but has a singularity atx= 0. Weget around this, by breaking up the integral into two pieces: one piece consisting of the ballof radius about the origin,B(0, ) and the other piece consisting of the complement of thisball inRn. Therefore, we have(F , g) = Rn (x) g(x)dx= B(0, ) (x) g(x)dx+ Rn B(0, ) (x) g(x)dx=I+ look first at termI. Forn= 2, termIis bounded as follows, B(0, )12 ln|x| g(x)dx C| g|L B(0, )ln|x|dx C 2 0 0ln|r|r dr d C 0ln|r|r dr Cln| | 3, termIis bounded as follows, B(0, )1n(n 2) (n)1|x|n 2 g(x)dx C| g|L B(0, )1|x|n 2dx C 0( B(0,r)1|y|n 2dS(y))dr= 01rn 2( B(0,r)dS(y))dr= 01rn 2n (n)rn 1dr=n (n) 0r dr=n (n)2 , as 0+,|I| , we look at termJ.

5 Applying the divergence theorem, we have Rn B(0, ) (x) xg(x)dx= Rn B(0, ) x (x)g(x)dx (Rn B(0, )) g(x)dS(x)+ (Rn B(0, )) (x) g dS(x)= (Rn B(0, )) g(x)dS(x) + (Rn B(0, )) (x) g dS(x) J1 + the fact that x (x) = 0 forx Rn B(0, ).We first look at termJ1. Now, by assumption,g D, and, therefore,gvanishes at . Consequently, we only need to calculate the integral over B(0, ) where the normalderivative is the outer normal toRn B(0, ). By a straightforward calculation, we seethat x (x) = xn (n)|x| outer unit normal toRn B(0, ) onB(0, ) is given by = x|x|.4 Therefore, the normal derivative of onB(0, ) is given by =( xn (n)|x|n) ( x|x|)=1n (n)|x|n ,J1 can be written as B(0, )1n (n)|x|n 1g(x)dS(x) = 1n (n) n 1 B(0, )g(x)dS(x) = B(0, )g(x)dS(x).Now ifgis a continuous function, then g(x)dS(x) g(0) as , we look at termJ2. Now using the fact thatgvanishes as|x| + , we only needto integrate over B(0, ). Using the fact thatg D, and, therefore, infinitely differentiable,we have B(0, ) (x) g dS(x) g L ( B(0, )) B(0, )| (x)|dS(x) C B(0, )| (x)|dS(x).

6 Now first, forn= 2, B(0, )| (x)|dS(x) =C B(0, )|ln|x||dS(x) C|ln| || B(0, )dS(x)=C|ln| ||(2 ) C |ln| ||.Next, forn 3, B(0, )| (x)|dS(x) =C B(0, )1|x|n 2dS(x) C n 2 B(0, )dS(x)=C n 2n (n) n 1 C .Therefore, we conclude that termJ2 is bounded in absolute value byC |ln |n= 2C n ,|J2| 0 as 0+.5 Combining these estimates, we see that Rn (x) xg(x)dx= lim 0+I+J1 +J2 = g(0).Therefore, our claim is Poisson s now return to solving Poisson s Equation u=fx our discussion before the above claim, weexpectthe functionv(x) Rn (x y)f(y)dyto give us a solution of Poisson s Equation . We now prove that this is in fact true. First, wemake a we hope that the functionvdefined above solves Poisson s Equation , we mustfirst verify that this integral actually converges. If we assumefhas compact support onsome bounded setKinRn, then we see that Rn (x y)f(y)dy |f|L K| (x y)| we additionally assume thatfis bounded, then|f|L C. It is left as an exercise toverify that K| (x y)|dy <+ on any compact C2(Rn)and has compact support.

7 Letu(x) Rn (x y)f(y)dywhere is the fundamental solution of Laplace s Equation ( ). C2(Rn)2. u= : Evans, p. a change of variables, we writeu(x) = Rn (x y)f(y)dy= Rn (y)f(x y) (.. ,0,1,0, ..)be the unit vector inRnwith a 1 in theithslot. Thenu(x+hei) u(x)h= Rn (y)[f(x+hei y) f(x y)h] C2impliesf(x+hei y) f(x y)h f xi(x y) ash 0uniformly onRn. Therefore, u xi(x) = Rn (y) f xi(x y) , 2u xixj(x) = Rn (y) 2f xixj(x y) function is continuous because the right-hand side is the above calculations and Claim 1, we see that xu(x) = Rn (y) xf(x y)dy= Rn (y) yf(x y)dy= f(x). Properties of Harmonic Mean Value PropertyIn this section, we prove a mean value property which all harmonic functions satisfy. First,we give some definitions. LetB(x, r) = ball of radiusraboutxinRn B(x, r) = boundary of ball of radiusraboutxinRn (n) = volume of unit ball inRnn (n) = surface area of unit ball a functionudefined onB(x, r), theaverage ofuonB(x, r) is given by B(x,r)u(y)dy=1 (n)rn B(x,r)u(y) a functionudefined on B(x, r), theaverage ofuon B(x, r) is given by B(x,r)u(y)dS(y) =1n (n)rn 1 B(x,r)u(y)dS(y).

8 Theorem 3.(Mean-Value Formulas)Let Rn. Ifu C2( )is harmonic, thenu(x) = B(x,r)u(y)dS(y) = B(x,r)u(y)dyfor every ballB(x, r) . C2( ) is harmonic. Forr >0, define (r) = B(x,r)u(y)dS(y).Forr= 0, define (r) =u(x). Notice that ifuis a smooth function, then limr 0+ (r) =u(x), and, therefore, is a continuous function. Therefore, if we can show that (r) = 0,then we can conclude that is a constant function, and, therefore,u(x) = B(x,r)u(y)dS(y).We prove (r) = 0 as follows. First, making a change of variables, we have (r) = B(x,r)u(y)dS(y)= B(0,1)u(x+rz)dS(z).Therefore, (r) = B(0,1) u(x+rz) z dS(z)= B(x,r) u(y) y xrdS(y)= B(x,r) u (y)dS(y)=1n (n)rn 1 B(x,r) u (y)dS(y)=1n (n)rn 1 B(x,r) ( u)dy(by the Divergence Theorem)=1n (n)rn 1 B(x,r) u(y)dy= 0,8using the fact thatuis harmonic. Therefore, we have proven the first part of the remains to prove thatu(x) = B(x,r)u(y) do so as follows, using the first result, B(x,r)u(y)dy= r0( B(x,s)u(y)dS(y))ds= r0(n (n)sn 1 B(x,s)u(y)dS(y))ds= r0n (n)sn 1u(x)ds=n (n)u(x) r0sn 1ds= (n)u(x)sn|s=rs=0= (n)u(x) , B(x,r)u(y)dy= (n)rnu(x),which impliesu(x) =1 (n)rn B(x,r)u(y)dy= B(x,r)u(y)dy,as Converse to Mean Value PropertyIn this section, we prove thatifa smooth functionusatisfies the mean value propertydescribed above, thenumust be C2( )satisfiesu(x) = B(x,r)u(y)dS(y)for allB(x, r) , thenuis (r) = B(x,r)u(y)dS(y).

9 Ifu(x) = B(x,r)u(y)dS(y)9for allB(x, r) , then (r) = 0. As described in the previous theorem, (r) =rn B(x,r) u(y) not harmonic. Then there exists some ballB(x, r) such that u >0 or u <0. Without loss of generality, we assume there is some ballB(x, r) such that u > , (r) =rn B(x,r) u(y)dy >0,which contradicts the fact that (r) = 0. Therefore,umust be Maximum PrincipleIn this section, we prove that ifuis a harmonic function on a bounded domain inRn,thenuattains its maximum value on the boundary of .Theorem Rnis open and bounded. Supposeu C2( ) C( )is (Maximum principle)max u(x) = max u(x).2.(Strong maximum principle)If is connected and there exists a pointx0 suchthatu(x0) = max u(x),thenuis constant within . prove the second assertion. The first follows from the second. Suppose thereexists a pointx0in such thatu(x0) =M= max u(x).Then for 0< r <dist(x0, ), the mean value property saysM=u(x0) = B(x0,r)u(y)dy , therefore, B(x0,r)u(y)dy=M,andM= max u(x).

10 Therefore,u(y) Mfory B(x0, r). To proveu Mthroughout , you continue with this argument, filling with replacinguby uabove, we can prove the Minimum , we use the maximum principle to prove uniqueness of solutions to Poisson s equa-tion on bounded domains 6.(Uniqueness)There exists at most one solutionu C2( ) C( )of theboundary-value problem,{ u=f x u=gx . there are two solutionsuandv. Letw=u vand let w=v u. Thenwand wsatisfy{ w= 0x w= 0x .Therefore, using the maximum principle, we concludemax |u v|= max |u v|= Smoothness of Harmonic FunctionsIn this section, we prove that harmonic functions areC .Theorem be an open, bounded subset ofRn. Ifu C( )andusatisfies the meanvalue property,u(x) = B(x,r)u(y)dS(y)for every ballB(x, r) , thenu C ( ). proven earlier, ifu C2( ) C( ) anduis harmonic, thenusatisfies the meanvalue property, and, therefore,u C ( ). fact, ifusatisfies the hypothesis of the above theorem, thenuis analytic, but wewill not prove that here.}}