Seperable solutions of Laplace equation in circular polar coords

[francis.lane@]

Homework Statement

I'm considering Laplace's equation in 2D, written in circular polar coordinates (so that's u_rr + 1/r*u_r + (1/r^2)*u_theta,theta). I've worked out what all the seperable solutions are.

My question is: is this set of seperable solutions complete.

(That is, can all solutions of Laplace's equation be expressed as a sum of these solutions.)

No speculation please.

Homework Equations

The Attempt at a Solution

 

[hunt_mat]

No. There are other ways of solving the Laplace equation which don't have anything to do with seperable solutions. The boundary conditions are also an important factor.

 

[francis.lane@]

OK, the full question is this:

I'm given the 2D Laplace equation in circular polars. The problem is defined on the annulus
1<= r <= 2.
The boundary conditions are

a*u + diff(u,r) = 0

on r=1 and r=2.

The question says: show there are nonzero solutions if a = 0 or a = 1/(2*ln(2)).
This part is easy; constants will work for a = 0, and you can find infinitely many solutions of the form c + d*ln(r) in case a=1/(2*ln(2)).

But then the question says:

Are there nonzero solutions for any other values of a?

I can't do this part of the question.

 

[hunt_mat]

So you have the Laplace equation:

$$\frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial\varphi}{\partial r}\right) +\frac{1}{r^{2}}\frac{\partial^{2}\varphi}{\partial \theta^{2}}=0$$

along with the boundary conditions:

$$a\varphi +\frac{\partial\varphi}{\partial r}=0\quad r=1,2$$

I am assuming that the solution you seek in radially symmetric solution? My suggestion would be solve the equation and see what you get. You wrote the solution down, so write out the BC for the general case and you should have a system for c and d in terms of a.

 

[francis.lane@]

No. The question is not whether there are radially symmetric solutions for other values of a, but whether there are any solutions. The solution could be theta-dependent (unless the fact that the BC's are radially symmetric implies that the solution is, but I don't think this is true).

In that case, you can't write down a general solution for the equation, and so you can't obtain a simple set of linear equations for a. (If I could, then, as you say, I would be able to establish conditions for the existence of solutions in terms of a). I could write down a general series solution consisting of an infinite sum of seperable solutions, and then proceed as you suggest. However, this would not be a proof, as the person above says that the seperable solutions do not form a complete set - that is, there may be solutions to the problem that cannot be written as an infinite sum of seperable solutions.

 

[hunt_mat]

My initial thoughts are that you can, as you have only been given BC for r and not $$\theta$$, so this should allow enough freedom for allow values for other a's.

What I meant by my previous comment was that the general solution (for a seperable solution) is

$$\varphi \alpha +\beta\log r$$

Evaluate this at r=1,2 to obtain:

$$a\alpha +\beta =0\quad a\alpha +\beta\log 2+\frac{\beta}{2}=0$$

Can you find an $$\alpha ,\beta$$ that will satisfy the above equations?

 

[francis.lane@]

Yes you can, and it turns out that there are only nonzero solutions in case a = 0 or a = 1/(2*ln(2)), and in those cases there are infinitely many solutions.

However, it doesn't strike me as obvious that just because the BC's are radially symmetric, it must follow that the entire solution is. This really is the crux of the matter - if I can prove this, then the above analysis will complete the proof.

 

[hunt_mat]

Hmm, Not done any formal courses on PDEs, I picked up what I know from books or on the job really.