Section 3.10: THE CONSTRUCTION OF GREEN'S FUNCTIONS FOR NEUMANN PROBLEMS

We now consider the Neumann problem

--2u = f on D

[[partialdiff]]u/[[partialdiff]]h = g on [[partialdiff]]D.

As for the Dirchlet Problem, we use this fundamental identity: For all u and v in the domain of --2,

òòD[u --2v - --2u v] dA = òòD -- [u --v - --u v]

= ò[[partialdiff]]D [u [[partialdiff]]v/[[partialdiff]][[eta]] - [[partialdiff]]u/[[partialdiff]][[eta]] v ] ds.

Choose u to satisfy the Neumann Problem and v to be 1. Then

- òòf = - òg.

D [[partialdiff]]D

Thus, if one poses a Neumann problem and this relationship does not hold, the problem cannot have a solution.

In particular, if g = 0 then òòD f = 0. Physically, this is reasonable for if nothing crosses the boundary, then the total input from the forcing function must sum to zero.

More important in the context of this course, one could have predicted this from the Fredholm Alternative theorems. Recall that Ly = f has a solution if and only if < f , z > = 0 for all solutions z of L*z = 0. Then realize that L*z = --2z = 0 on M* = {z : [[partialdiff]]z/[[partialdiff]][[eta]] = 0 } has a solution z = 1. Hence, the Fredholm Alternative theorem requires that òò D 1 f = 0.

The problem before us is to construct a Green function for this Neumann problem. In constructing the Green function for a Dirchlet problem, G was made to satisfy

--2G = d on D

G(P,Q) = 0 for P in [[partialdiff]]D.

Thus, one might guess to make the Green function for the Neumann problem to satisfy

--2N = d on D

[[partialdiff]]N/[[partialdiff]][[eta]] = < --PN , [[eta]] > = 0 on [[partialdiff]]D.

But, this would be a mistake! For, it would imply that

0 = ò [[partialdiff]]N/[[partialdiff]][[eta]] = òò--2N = òò d = 1.

as we saw from the above work. This can be repaired by asking that [[partialdiff]]N/[[partialdiff]][[eta]] should be k(p) where k satisfies

ò[[partialdiff]]D k(p) dp s = 1.

Here is the important result from conformal mapping. Suppose that f is a conformal mapping from D onto the unit disk. Then, the Green function for the Neumann problem is

N(z,[[alpha]]) = [ln( |f(z) - f([[alpha]]) | ) + ln(|1 - f(z) f([[alpha]])*|)].F(1,2[[pi]])

Example. Give N for the unit disk and verify that ò[[partialdiff]]N/[[partialdiff]][[eta]] ds = 1.

To do this example, we take f(z) = z. Then

N(z,[[alpha]]) = [ ln( |z-[[alpha]]|) + ln(|1-z[[alpha]]*|)/ 2[[pi]]

= [ln([[rho]]2 +r2 - 2[[rho]]r cos([[theta]]-[[phi]]))+ ln( 1+[[rho]]2r2-2[[rho]]rcos([[theta]]-[[phi]]))]/4[[pi]].

Here, again, z = rei[[theta]] and [[alpha]] = [[rho]]ei[[phi]]. The normal derivative is found as follows:

[[partialdiff]]N/[[partialdiff]][[eta]] = [[partialdiff]]n/[[partialdiff]]r|r=1

= F(1,4[[Pi]]) ( [2r-2[[rho]]cos([[theta]]-[[phi]])/ ([[rho]]2 +r2 -2[[rho]]rcos([[theta]]-[[phi]])]|r=1

+[2r[[rho]]2 - 2[[rho]]cos([[theta]]-[[phi]])]/[1+[[rho]]2r2-2[[rho]]rcos[[theta]]-[[phi]])]|r=1 )

= F(1,4[[Pi]]) ( [2-2[[rho]]cos([[theta]]-[[phi]])]/[1+[[rho]]2-2[[rho]]cos([[theta]]-[phi]])]

+[2[[rho]]2-2[[rho]]cos([[theta]]-[[phi]])]/[1+[[rho]]2-2[rho]]cos([[theta]]-[[phi]])] )

= 1/2[[pi]]. Thus, ò[[partialdiff]]n/[[partialdiff]][[eta]] d[[sigma]] = I(0,2[[pi]], ) d[[theta]] / 2[[pi]] = 1.

EXAMPLE. Give N for the upper half plane. Compute [[partialdiff]]N/[[partialdiff]][[eta]]|[[partialdiff]]D.

For this example, N(z,[[alpha]]) = [ln|f(z) - f([[alpha]])| + ln(| 1-f(z)f([[alpha]])*|)]/2[[pi]]

where f(z) = (z-i)/(z+i).

Then f(z) - f([[alpha]]) = ((z-i)/(z+i) - ([[alpha]]-i)/([[alpha]]+i) = 2i(z-[[alpha]])/(z+i)([[alpha]]+i)

and 1-f(z) f([[alpha]])* = -2i(z-[[alpha]]*)/(z-i)([[alpha]]*-i).

So, N(z,a) = ln(|(z - [[alpha]]) (z - [[alpha]]*) 4/(z + i)2 ([[alpha]] + i)([[alpha]]*-i)|)

= [ln(x-a)2+(y+b)2) + ln((x-a)2+(y-b)2))-2ln(x2+(y+1)2) +2ln(4)-2ln(a2+(b+1)2)] / 4[[pi]]

Finally, we compute [[partialdiff]]N/[[partialdiff]][[eta]] for the upper half plane.

[[partialdiff]]N/[[partialdiff]][[eta]] = - [[partialdiff]]N/[[partialdiff]]y|y=0 =

2b/(4[[pi]][(x-a)2 + b2]) -2b/4[[pi]][(x-a)2+b2] - 1/[[pi]](x2+1) = 1/[[pi]](x2+1).

Thus,

I(-*,*, ) [[partialdiff]]N/[[partialdiff]][[eta]](x,0) dx = 1/[[pi]] I(-*,*, ) dx/(x2+1) = 1/[[pi]] arctan(x) ]O(-*, *) = 1.

We can also check that --2N = d: already --2 1/2[[pi]] ln(| z-[[alpha]]|) = d(z,[[alpha]]).

EXERCISE: Get the Green function for the right half plane. Verify that

[[partialdiff]]N/[[partialdiff]][[eta]]({x,y},{a,b}) is independent of {a,b}.

SUMMARY. FORMULAS FOR U AND V WHICH SOLVE

--2u = f on D and --2v = f on D

u = g on [[partialdiff]]D [[partialdiff]]v/[[partialdiff]][[eta]] = h on [[partialdiff]]D.

Recall the fundamental identity

òòU --2V - òò--2U V = òò--[U--V - --U V} = ò[U [[partialdiff]]V/[[partialdiff]][[eta]] - [[partialdiff]]U/[[partialdiff]][[eta]] V].

Thus, if U = u and V = G, then

òòu(P) d(P,Q) dPA - òòf(P) G(P,Q) dpA = ò g(P) [[partialdiff]]pG(P,Q)/[[partialdiff]][[eta]] dps,

so that

u(Q) = òò f(P) G(P,Q) dpA + ò g(P) [[partialdiff]]pG(P,Q)/[[partialdiff]][[eta]] dps.

On the other hand, if V = v and u = N then

òòf(P) [[Nu]](P,Q) dPA - òòv(P) d(P,Q) dpA

= -ò V(P) [[partialdiff]]pN(P,Q)/[[partialdiff]][[eta]] dps+òh(P)N(P,Q)dPs

so that

V(Q) = òòf(P)N(P,Q)dPA - òh(P)N(P,Q)dPs +ò[[partialdiff]]N/[[partialdiff]][[eta]]V(P)dPs.

ò [[partialdiff]]N/[[partialdiff]][[eta]](P,Q) V(P) dPs is independent of Q. It follows that

V(Q) = òòf(P)N(P,Q)dPA - òh(P) N(P,Q)dPs + constant.

A COMPENDIUM OF PROBLEMS

Consider L(U) = [[partialdiff]]2U/[[partialdiff]]x2 + [[partialdiff]]2U/[[partialdiff]]y2 - [[partialdiff]]2U/[[partialdiff]]z2 - [[partialdiff]]U/[[partialdiff]]x restricted to

M= { U(x,y,z): U(0,y,z) =0, U(1,y,z) = 0, [[partialdiff]]U/[[partialdiff]]y (x,0,z) = 3 U(x,0,z),

[[partialdiff]]U/[[partialdiff]]y (x,1,z) = 5 U(x,1,z), U(x,y,0) = U(x,y,1), [[partialdiff]]U/[[partialdiff]]z (x,y,0) = [[partialdiff]]U/[[partialdiff]]z (x,y,1)}

(1) L is a (parabolic, hyperbolic, or elliptic) operator.

(2) The formal adjoint, L* , of L is given by......

(3) The divergence theorem implies that if D is the box described by D = { (x,y,z): 0 < x < 1, 0 < y < 1, 0 < z < 1 }

and F has continuous partials then

òòòD--*F dV = .........

(4) We will use the following identity: V L(U) - L*(V) U =

--*[..........].

(5) This identity is established as follows:..........

(6) Considering (3) and (4) we guess that M* = { U(x,y,z): .....}.

(7) That this is M* may be established as follows: ......

(8). Explain why this problem is "really" self-adjoint, or not:

3 F([[partialdiff]]2U,[[partialdiff]]x2) - 5 F([[partialdiff]]2U,[[partialdiff]]y2) = 0 on 0 < x < a, 0 < y < b,

U(x,y) = 0 for {x,y} on the boundary of the rectangle.

(9) Suppose that D is a region in the plane and that h is a one-to-one analytic function from D onto the unit disk. Explain, in some detail, how to make up the Green function and the solution for the problem --2U = f on D, with U = g on [[partialdiff]]D by using this f, g, and h.

(10) Let L(U) = F([[partialdiff]]2U,[[partialdiff]]x2) + F([[partialdiff]]2U,[[partialdiff]]y2) - F([[partialdiff]]2U,[[partialdiff]]z2) - F([[partialdiff]]U,[[partialdiff]]x) + U. Find F such that

--.F = V L(U) - L*(V) U.

(11) Let L(U) = --2U on the rectangle [0,1] x [0,1] and

M = {U: U(x,0) = 0, U(1,y) - 3 F([[partialdiff]]U,[[partialdiff]]x) (1,y) = 0, U(x,1) = 0, U(0,y) + 5 F([[partialdiff]]U,[[partialdiff]]x) (0,y) = 0}. Give L* and M*.

(12) Find F(x,y) in terms of f so that the solution of --2W + W = F provides the solution for 5F([[partialdiff]]2u,[[partialdiff]]x2) + 4 F([[partialdiff]]2u,[[partialdiff]]y2) + 3F([[partialdiff]]u,[[partialdiff]]x) + 2F([[partialdiff]]u,[[partialdiff]]y) + u = f.