Solve the eigenvalue problem

1. Jun 30, 2008

mathwizarddud

Solve the eigenvalue problem

$$\frac{d^2 \phi}{dx^2} = -\lambda \phi$$

subject to

$$\phi(0) = \phi(2\pi)$$

and

$$\frac{d \phi}{dx} (0) = \frac{d \phi}{dx} (2 \pi).$$

I had the solution already, but am looking for a much simpler way, if any.

EDIT:

Sorry that I accidentally posted this twice; I meant to edit the post, but not sure why the edited post becomes a new one...

2. Jul 1, 2008

malawi_glenn

How do we know that our suggested solution is simpler than yours if you dont demonstrate your attempt?

3. Jul 4, 2008

mathwizarddud

after solving the ODE, we have the general solution

$$\phi = C_1 \sin(\sqrt{\lambda}x) + C_2 \cos(\sqrt{\lambda}x)$$

applying the conditions we have the system

$$C_2 = C_1 \sin(\sqrt{\lambda}2\pi) + C_2 \cos(\sqrt{\lambda}2\pi)$$

$$C_1 \sqrt{\lambda} = C_1 \sqrt{\lambda} \cos(\sqrt{\lambda}2\pi) - C_2 \sqrt{\lambda} \sin(\sqrt{\lambda}2\pi)$$

then using a little knowledge of linear algebra and determinant, we get, in order not to have trivial solutions,

$$C_1 C_2 u^2 + C_1 C_2 (v-1)^2 = 0$$

where $$u = \sin(\sqrt{\lambda}2\pi)$$ and
$$v = \cos(\sqrt{\lambda}2\pi)$$

or simply

$$u^2 + (v-1)^2 = 0$$

So $$u^2 = (v-1)^2 = 0$$ or
$$u = 0; v = 1$$

$$\sqrt{\lambda_n}2\pi = 2n\pi$$
$$\lambda_n = n^2$$

So the eigenfunction is

$$\phi_n = C_1 \sin(nx) + C_2 \cos(nx)$$

I don't think this is complete because we haven't determined $$C_1$$ and $$C_2$$ yet.

4. Jul 5, 2008

dirk_mec1

Use the BC's to get your constants.

5. Jul 5, 2008

mathwizarddud

I thought that I've already used them in first determining the eigenvalue.

6. Sep 28, 2008

anyone?