# Lost on eigenvalues

1. Nov 20, 2004

### Chris_K

I'm having trouble getting started on this problem... I just really don't understand what to do.

Solve
$$X'+2X'+(\lambda-\alpha)X=0, 0<x<1$$
$$X(0)=0$$
$$X'(1)=0$$

a. Is $$\lambda=1+\alpha$$ an eigenvalue? What is the corresponding eigenfunction?
b. Find the equation that the other eigenvalues satisfy.

I appreciate any help you can give me!

Thanks,
Chris

Last edited: Nov 20, 2004
2. Nov 20, 2004

### ReyChiquito

are u sure that the edo is correct?

you have there a linear second order d.o with constant coeficients and initial values, so your solution will be some linear comb. of exp[rt] (you have to calculate r of course).

i dont really understand why would you end with an eigenvalue problem in this way but maybe im not well informed.

3. Nov 21, 2004

### Chris_K

Yeah, everything on there is correct. I'm not sure what you mean though...

nm... figured it out.

Last edited: Nov 22, 2004
4. Nov 22, 2004

### ReyChiquito

$$X''+2X'+(\lambda-\alpha)X=0$$

let $X=e^{rt}$ implies

$$r^2+2r+(\lambda-\alpha)=0$$

so

$$r=-1\pm\sqrt{1-(\lambda-\alpha)}$$

$$X(t)=Ae^{(-1+\sqrt{1-(\lambda-\alpha)})t}+Be^{(-1-\sqrt{1-(\lambda-\alpha)})t}$$

$$X(0)=A+B$$

so $A=-B[/tex] $$X'(1)=A[r_{+}e^{r+}-r_{-}e^{r_{-}}]=0$$ [itex]A=0$ would lead to the trivial solution, so

$$r_{+}e^{\sqrt{1-(\lambda-\alpha)}}=r_{-}e^{-\sqrt{1-(\lambda-\alpha)}}$$

$\lambda=1+\alpha$ is clearly an eigenvalue

and $\lambda_{m}$ satisfy the equation

$$\frac{-1+\sqrt{1-(\lambda_{m}-\alpha)}}{-1-\sqrt{1-(\lambda_{m}-\alpha)}}e^{2\sqrt{1-(\lambda_{m}-\alpha)}}=1$$

Last edited: Nov 22, 2004