Eigenvalue Problem Simplified: A Simple Solution to the Eigenvalue Problem

[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...

 

[malawi_glenn]

How do we know that our suggested solution is simpler than yours if you don't demonstrate your attempt?

 

[mathwizarddud]

Here's what I had:

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.

 

[dirk_mec1]

Use the BC's to get your constants.

 

[mathwizarddud]

Use the BC's to get your constants.

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

 

[mathwizarddud]

anyone?