Solving Eigenvalue Problem with Periodic BCs: Find b for Self-Adjointness

[hawaiifiver]

Homework Statement

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.

Homework Equations

see below

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.

 

[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?

 

[dikmikkel]

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

 

[hawaiifiver]

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