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?