4 Green’s Functions - Stanford University

1 4 green s FunctionsIn this section, we are interested in solving the following problem. Let be an open, boundedsubset ofRn. Consider{ u=f x Rnu=gx .( ) Motivation for green s FunctionsSuppose we can solve the problem,{ yG(x, y) = xy G(x, y) = 0y ( )for eachx . Then,formally, we can say that forua solution of ( ),u(x) = xu(y)dy= yG(x, y)u(y)dy= yG(x, y) yu(y)dy G (x, y)u(y)dS(y)= G(x, y) yu(y)dy+ G(x, y) u (y)dS(y) G (x, y)u(y)dS(y)= G(x, y)f(y)dy G (y)g(y)dS(y).Now, we do know that the fundamental solution of Laplace s equation (y) satisfies y (y) = 0and, moreover, y (x y) = course, (x y) does not satisfy our boundary conditions, but we will use that as astarting ground to try and construct a solution of ( ), and, ultimately ( ). Below, wewill also make the formal argument given above more the definition of distributional derivative, we will start by looking at (x y) yu(y) would like to integrate this term by parts.}}

2 However, we know that (x y) has asingularity aty=x. Therefore, in order to integrate by parts, we must proceed as and >0 such that dist(x, )< and therefore,B(x, ) . LetV B(x, ).1x V Let be the fundamental solution of Laplace s equation. That is, (x) ={ 12 ln|x|n= 21n(n 2) (n)1|x|n 2n C2( ). By the Divergence Theorem, we have V (y x) u(y)dy= V y (y x) yu(y)dy+ V (y x) u dS(y)= V y (y x)u(y)dy V (y x)u(y)dS(y)+ V (y x) u dS(y).where u denotes the derivative ofuin the outer normal direction toV . Now onV , y (y x) = 0. Therefore, V (y x) u(y)dy= V (y x)u(y)dS(y) + V (y x) u dS(y).Now, we note thatlim 0+ V (y x) u(y)dy= (y x) u(y) make the following claims about the limits of the other two terms as 0+.Claim 0+[ V (y x)u(y)dS(y)]= (y x)u(y)dS(y) u(x).Claim 0+ V (y x) u (y)dS(y) = (y x) u dS(y).Assuming these claims for a moment, we conclude that for anyu C2( ),u(x) = [ (y x) u (y) (y x)u(y)]dS(y) (y x) u(y)dy.}

3 ( )2We would now like to use the representation formula ( ) to solve ( ). If we knew uon anduon and u on , then we could solve foru. But, we don t know all thisinformation. We know uon anduon .We proceed as follows. For eachx , we introduce acorrector functionhx(y) whichsatisfies the following boundary-value problem,{ yhx(y) = 0y hx(y) = (y x)y .( )Now suppose we can find such a (smooth) functionhxwhich satisfies ( ). Then usingthe same analysis as above, we have hx(y) u(y)dy= yhx(y) yu(y)dy+ hx(y) u (y)dS(y)= yhx(y)u(y)dy hx (y)u(y)dS(y)+ hx(y) u (y)dS(y).Now using the fact thathxis a solution of ( ), we conclude that0 = [ (y x) u (y) hx (y)u(y)]dS(y) hx(y) u(y)dy.( )Now subtracting ( ) from ( ), we conclude thatu(x) = [ (y x) hx (y)]u(y)dS(y) [ (y x) hx(y)] u(y) (x, y) = (y x) hx(y).Then,ucan be written asu(x) = G (x, y)u(y)dS(y) G(x, y) u(y)dy.}

4 ( )We define this functionGas theGreen s functionfor . That is, the green s function fora domain Rnis the function defined asG(x, y) = (y x) hx(y)x, y , x6=y,where is the fundamental solution of Laplace s equation and for eachx ,hxis asolution of ( ). We leave it as an exercise to verify thatG(x, y) satisfies ( ) in the senseof :Ifuis a (smooth) solution of ( ) andG(x, y) is the green s function for ,thenu(x) = G (x, y)g(y)dS(y) + G(x, y)f(y)dy.( )3We will show below thatconverselya function of the form ( ) will give us a solution of( ). First, however, we prove the two claims given of Claim 1. V (y x)u(y)dS(y) = (y x)u(y)dS(y) + B(x, ) (y x)u(y)dS(y).Now y (y) = 1n (n)y|y|nand the outward normal onB(x, ) is =y x|y x|.Therefore, (y x) = y (y x) = 1n (n)y x|y x|n y x|y x|= 1n (n) 1|y x|n , B(x, ) (y x)u(y)dS(y) = 1n (n) B(x, )1|y x|n 1u(y)dS(y)= 1n (n) n 1 B(x, )u(y)dS(y)= B(x, )u(y)dS(y).

5 As 0+, B(x, )u(y)dS(y) u(x).Therefore, we havelim 0+ B(x, ) (y x)u(y)dS(y) = u(x),and the claim follows. Proof of Claim we know V (y x) u (y)dS(y) = (y x) u (y)dS(y) B(x, ) (y x) u (y)dS(y).4We just need to show that B(x, ) (y x) u (y)dS(y) 0as 0+.Substituting in the explicit formula for forn 3 (the casen= 2 can be handled similarly),we see that B(x, ) (y x) u (y)dS(y) 1n (n)(n 2) B(x, )1|y x|n 2 u (y) dS(y) u L (B(x, ))1n (n) n 2 B(x, )dS(y) C B(x, )dS(y)=C .Therefore, as 0+, B(x, ) (y x) u (y)dS(y) 0as claimed. Therefore, the claim follows. Above we have proven the following C2( )is a solution of{ u=f x Rnu=gx ,wherefandgare continuous, thenu(x) = g(y) G (x, y)dS(y) + f(y)G(x, y)dy( )forx , whereG(x, y)is the green s function for .Corollary harmonic in andu=gon , thenu(x) = g(y) G (x, y)dS(y). Finding green s FunctionsFinding a green s function is difficult.}

6 However, for certain domains with special geome-tries, it is possible to find green s Functions . We show some examples +be the upper half-plane inR2. That is, letR2+ {(x1, x2) R2:x2>0}.5We will look for the green s function forR2+. In particular, we need to find a correctorfunctionhxfor eachx R2+, such that{ yhx(y) = 0y R2+hx(y) = (y x)y R2+.Fixx R2+. We know y (y x) = 0 for ally6=x. Therefore, if we choosez / , then y (y z) = 0 for ally . Now, if we choosez=z(x) appropriately,z / , such that (y z) = (y x) fory , then lettinghx(y) = (y z(x)), we will have found acorrector function. Recall that forn= 2, (y z) = 12 ln|y z|.Therefore, (y z) is a function of|y z|. Consequently, forx= (x1, x2) R2+, we see thatfor ally R2+,|y x|=|(y1,0) (x1, x2)|=|(y1,0) (x1, x2)|=|y x|where x (x1, x2)is thereflection of xin the ~R2+Therefore, lettinghx(y) = (y x), we have found a corrector function forR2+.}

7 Therefore,a green s function for the upper half-plane is given byG(x, y) = (y x) (y x)= 12 [ln|y x| ln|y x|]. Example generally, for the upper half-space inRn,Rn+ {(x1, .. , xn) Rn:xn>0},the corrector functionhx(y) is given byhx(y) = (y x)6where x (x1, .. , xn 1, xn)is thereflection of xin the plane. Therefore, a green s function for the upper half-spaceRn+is given byG(x, y) = (y x) (y x). Example (0,1) be the unit ball inR2. That is, letB2(0,1) {(x1, x2) R2:x21+x22<1}.Fixx B2(0,1). We need to find a corrector functionhxforB2(0,1). Again, (y x) = 12 ln|y x|.Therefore, (y x) is a function of|y x|. We needhx(y) = (y x) for ally B2(0,1),that is, allysuch that|y|= 1. Now fory B2(0,1),|y x|2= (y x) (y x)=|y|2 2y x+|x|2=|x|2 2x y+ 1=|x|2|y|2 2x y+ 1=|x|2(|y|2 2x y|x|2+1|x|2)=|x|2(|y|2 2y x|x|2+|x|2|x|4)=|x|2|y x |2,wherex =x|x|2is called thepoint dual to B2(0,1)7 Notice that forx B2(0,1),x isnotinB2(0,1).

8 Consequently, we can conclude that (|x|(y x )) is harmonic for allyin . In addition, (|x|(y x )) = (y x) for ally B2(0,1). Therefore, lettinghx(y) = (|x|(y x )),we see thathxis a corrector function for the unit ballB2(0,1).Consequently, the green s function forB2(0,1) is given byG(x, y) = (y x) (|x|(y x ))= 12 [ln|y x| ln[|x||y x |]]. Example the unit ball inRn,Bn(0,1) {(x1, .. , xn) :x21+..+x2n= 1},a corrector functionhxis given byhx(y) = (|x|(y x ))wherex =x|x|2is thepoint dual to x. Therefore, a green s function forBn(0,1) is given byG(x, y) = (y x) (|x|(y x )). Using green s Functions to Solve Poisson s EquationWe have shown above thatifuis a smooth solution of the Dirichlet problem{ u=f x u=gx ,thenucan be represented in terms of the green s function for by ( ). It remains toshow the converse. That is, it remains to show that for continuous functionsf,gand a givendomain Rn, the representation formula ( ) does give us a solution of the Dirichletproblem.}

9 First, however, we will use the representation formula ( ) to write the proposedformula for the solution in the cases above, where we can explicitly calculate the green sfunction for the domain .8 Example +be the upper half-space inRn,Rn+={(x1, .. , xn) R2:xn>0}.From above, we calculated thatG(x, y) = (y x) (y x)is a green s function forRn+, where x= (x1, .. , xn 1, xn) and is the fundamental solutionof Laplace s equation inRn. Now, from ( ), our proposed solution has the formu(x) = Rn+g(y) G (x, y)dS(y) + Rn+f(y)G(x, y)dy.(Note: The analysis in Section was carried out under the assumption that was abounded domain. However, for now we will assume we can use the same representationformula to derive a solution for the unbounded half-space. Later, we will need to prove thatthis representation formula actually gives us a solution.)Now, we calculate G on{yn= 0}to find an explicit formula for solutions to{ u= 0x u=gx.}

10 Now yn(y) = ynn (n)|y| , the normal derivative ofGon{yn= 0}is given by G (x, y) = yn(y x) yn(y x)=yn xnn (n)|y x|n yn xnn (n)|y x|n= 2xnn (n)|y x| ,ifuis the solution of Laplace s equation on the upper half-space with Dirichletboundary conditions, then we suspect thatuwill have the formu(x) =2xnn (n) Rn+g(y)|y x|ndy.( )This is calledPoisson s formulafor the half-spaceRn+. The functionK(x, y) =2xnn (n)1|x y|nis calledPoisson s kernelfor the half-spaceRn+. 9 Example (0,1) be the unit ball inRn. We look for a formula for the solution ofLaplace s equation inBn(0,1) with Dirichlet boundary conditions,{ u= 0x Bn(0,1)u=gx Bn(0,1).( )By ( ), ifuis a solution of ( ), thenuwill have the formu(x) = Bn(0,1)g(y) G (x, y)dS(y).Now we just need to calculate G on Bn(0,1) whereGis a green s function forBn(0,1).As shown above,G(x, y) = (y x) (|x|(y x ))is a green s function for the unit ball inRnwherex =x|x|2is the point dual tox.}