Eigenvalue Problem Simplified: A Simple Solution to the Eigenvalue Problem

[mathwizarddud]

Solve the eigenvalue problem

[tex]\frac{d^2 \phi}{dx^2} = -\lambda \phi[/tex]

subject to

[tex]\phi(0) = \phi(2\pi)[/tex]

and

[tex] \frac{d \phi}{dx} (0) = \frac{d \phi}{dx} (2 \pi).[/tex]

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

[tex]\phi = C_1 \sin(\sqrt{\lambda}x) + C_2 \cos(\sqrt{\lambda}x)[/tex]

applying the conditions we have the system

[tex]C_2 = C_1 \sin(\sqrt{\lambda}2\pi) + C_2 \cos(\sqrt{\lambda}2\pi)[/tex]

[tex] C_1 \sqrt{\lambda} = C_1 \sqrt{\lambda} \cos(\sqrt{\lambda}2\pi) - C_2 \sqrt{\lambda} \sin(\sqrt{\lambda}2\pi)[/tex]

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

[tex]C_1 C_2 u^2 + C_1 C_2 (v-1)^2 = 0[/tex]

where [tex]u = \sin(\sqrt{\lambda}2\pi)[/tex] and
[tex]v = \cos(\sqrt{\lambda}2\pi)[/tex]

or simply

[tex] u^2 + (v-1)^2 = 0[/tex]

So [tex]u^2 = (v-1)^2 = 0[/tex] or
[tex] u = 0; v = 1[/tex]

[tex] \sqrt{\lambda_n}2\pi = 2n\pi[/tex]
[tex]\lambda_n = n^2[/tex]

So the eigenfunction is

[tex]\phi_n = C_1 \sin(nx) + C_2 \cos(nx)[/tex]

I don't think this is complete because we haven't determined [tex]C_1 [/tex] and [tex] C_2[/tex] 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?