# Thread: Laplacian and Fredholm Alternative

1. ## Laplacian and Fredholm Alternative

Here is the question:
"Consider the boundary value problem in lR^2:

Laplacian(u)=(1/r)*d/dr(r*du/dr)=c .for r^2=x^2+y^2<1
du/dr=2 .on r=1

Show there is no solution unless c=4, and find solutions in this case."

I think I am supposed to use the Fredholm alternative:
L(u) = f has a solution if and only if <f,v>=0 for all v in ker(L*) where L* is the adjoint.

My problem is basically finding the boundary conditions for L*. Since <u,v> involves the integral between 0 and 1, and so the 1/r is infinity when r=0.

I get my adjoint problem to be:
v''-d/dr(v/r)=0
subject to
[v(r)u'(r)-u(r)v'(r)+u(r)v(r)/r](1,0)=0
where [f(r)](1,0)=f(1)-f(0)

You can see the third term doesn't really work when I stick in r=0.

Any help would be much appreciated.

3. I'm not confident that I know what I'm talking about here, but it seems to me that if you're looking at a Laplacian in polar coordinates then you should be using the polar version of areal measure, in other words rdrdθ:

$\displaystyle \int_0^{2\pi}\int_0^1L(u)v\,r\,dr\,d\theta = \int_0^{2\pi}\int_0^1uL^*(v)\,r\,dr\,d\theta$.

That has the big advantage of killing off the embarrassing factor 1/r, because it gets multiplied by r.

Since θ does not figure explicitly in the calculation, we can ignore it and just do the r integration by parts. I get the answer to be that L* is the same differential operator as L, with boundary condition 2v(1) – u(1)v'(1) = 0.

4. Cheers for looking at that Opalg...

think you have spotted my mistake... didn't bother actually checking my adjoint worked, but rather just used a formula for what it should be.

Obviously the formula only works when it is d/dx.

I guess this is why you should understand how formulas work before you use them!