View Full Version : Lost on eigenvalues
Chris_K
Nov20-04, 04:02 PM
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
ReyChiquito
Nov20-04, 05:36 PM
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.
Chris_K
Nov21-04, 03:54 PM
Yeah, everything on there is correct. I'm not sure what you mean though...
nm... figured it out.
ReyChiquito
Nov22-04, 12:15 AM
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
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