# Finding eigenvalues and eigenfunctions

1. May 11, 2009

### jaejoon89

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

Given
X''(x) + lambda*X(x) = 0
X(0) = X'(0), X(pi) = X'(pi)

Find all eigenvalues and eigenfunctions.

2. Relevant equations

Case lambda = 0
Case lambda > 0
Case lambda < 0

3. The attempt at a solution

First case, X(x) = Ax + B but the function doesn't satisfy the boundary condition at pi.
Second case, lambda = k^2
X(x) = Acos(kx) + Bsin(kx)
X'(x) = Bkcos(kx) - Aksin(kx)
X(pi) - X'(pi) = Bsin(kpi) + Bk^2 sin(kpi)
=> 1+ k^2 = 0 so lambda = -1

a) I was told the eigenfunction for that is X(x) = Ce^x. How?

b) Also, is that just the lambda_0 or does it satisfy all possibly eigenfunctions (i.e. eigenvalue lambda_n)?

2. May 11, 2009

### Cyosis

This is an eigenvalue problem of the form $D^2 f=-\lambda f$, $-\lambda$ the eigenvalue. With $D^2$ being the operator that takes the derivative of f(x) with respect to x twice.

Now if Ce^x is an eigenfunction it needs to satisfy the eigenvalue problem. Plugging it in $D^2 f=-\lambda f$ yields $Ce^x=-\lambda C e^x$. For which $\lambda$ is this equation satisfied? This should also answer question b.

Your argument to why $\lambda=0$ isn't an eigenvalue isn't entirely water tight (although $\lambda=0$ is indeed not an eigenvalue). For example for A=B=0, $f(x)=Ax+B=0$ satisfies the requirements just fine. Why is f(x)=0 not an eigenfunction?

For the second case you've written down the correct eigenfunction, $f(x)=A \cos (\sqrt{\lambda}x)+B \sin (\sqrt{\lambda}x)$. Emphasis on $\lambda>0$. You solve this equation for a $\lambda<0$. Now you get a cosine and a sine with an imaginary argument, which can be written as real exponents, which is the eigenfunction for $\lambda<0$. Try to find eigenvalues for $\lambda>0$ with regards to the second case.

Last edited: May 11, 2009