# Eigenvalue problem.

1. Apr 12, 2012

### hawaiifiver

1. The problem statement, all variables and given/known data

I have a problem

u'' + lambda u = 0

with BCs: u'(0) = b*u'(pi), u(0) = u(pi).

where b is a constant.

I have to find b which makes the BCs and problem self-adjoint.

2. Relevant equations

see below

3. The attempt at a solution

I see in my notes that it says that when there are periodic BCs, then the problem is self-adjoint. I think b = 1, and not any other number. Won't the value of the derivative of the solution change if b does not equal one? And therefore the BCs will not be periodic any more? Thoughts? Thanks.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Apr 12, 2012

### dikmikkel

For a Sturm-Liouville problem(Self-adjoint) on the form:
$\dfrac{d}{dx}\left(p(x)\dfrac{dU}{dx}\right) +q(x)U +\lambda w(x)U = 0$
the following integrated value should vanish, then the problem is self-adjoint:
$\left[p\left(u^*\dfrac{dv}{dx}-\dfrac{du^*}{dx}v\right)\right]_{0}^{\pi}$
Where u and v are solutions to the problem. so in your case:
$p(x)=1, q(x)=0, w(x)=1\\$
With periodic boundary condiitions:
$u\dfrac{du}{dx} |_0 = u\dfrac{du}{dx}|_\pi = 0$
Only holds true for b=1 in my opinion.
But maybe diriclet or legendre or neumann or mixed BCS can lead to another value of b?

3. Apr 12, 2012

### dikmikkel

actually i think that mixed BCS are the only other option in your case.

4. Apr 12, 2012

### hawaiifiver

Ah yes, now I comprehend the method. Thank you.