# Periodic solution to DE

1. Sep 3, 2005

### Zaare

I don't know how to use the extra condition given in the this problem:

$$f^{\prime\prime}+\lambda f = 0$$, $$f=f\left(r\right)$$
$$f\left(r\right) = f\left(r+\pi\right)$$

For $$\lambda = 0$$, the solution is some constant.
For other $$\lambda$$, I put $$\lambda=k^2$$, and get

$$f^{\prime\prime}+k^2 f = 0$$,

which has the solution

$$f\left(r\right)=A\cos\left(kr\right)+B\sin\left(kr\right).$$

The condition now gives

$$A\cos\left(kr\right)+B\sin\left(kr\right)= A\cos\left(kr+k\pi\right)+B\sin\left(kr+k\pi\right).$$

How can I use this to further specify the solution?

2. Sep 3, 2005

### LeonhardEuler

Well, you know that $$\cos{x} + \sin{x} = \cos{(x+2n\pi)} + \sin{(x+2n\pi)}$$ for all x only when n is an integer. What does that tell you about k, and therefore $\lambda$?

3. Sep 3, 2005

### Zaare

Oh, right, so k must be an even integer.
Thank you.