# Boundary value problem involving eigenvalues

1. Apr 26, 2006

### dak246

I need a bit of help with these boundary value problems. I'm trying to find their eigenvalues and eigenfunctions and although I pretty much know how to do it, I want to exactly WHY I'm doing each step. I attached part of my work, and on it I have a little question next to the steps I need explained. The biggest question I have is how do you know what form to choose for the eigenfunctions. Thanks a lot for taking the time to help me out.

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2. Apr 26, 2006

### HallsofIvy

Staff Emeritus
I HATE trying to read those Jpg things!

In the first problem, you have
$$y"+ \lambda y= 0$$
with boundary conditions y'(0)= 0, y(0)= 0.
You first try $\lambda< 0$ and determine that there are no non-trivial functions satisfying that so no negative $\lambda$ is an eigenvalue.

You then try $\lambda= 0$ so your equation is y"= 0 which has general solution y(x)= C1x+ C2. y'= C1 so the boundary conditions give C1= 0. But C2 can be anything so there exist non-trivial solutions of the form y= C. That is,
0 is in fact an eigenvalue. Since the set of all such solutions is an "subspace" of the space of all functions, you can find a basis for it. The simplest is y= 1 so that y= C= C(1). Actually, any number (except 0) would have worked.

If $\lamba> 0$, then you get general solution
$$y= C_1 cos(\sqrt{\lambda}x)+ C_2 sin(\sqrt{\lamba x})[/itex] [tex]y'= -C_1/sqrtr{\lambda}sin(\sqrt{\lambda}x)+ C_2\sqrt{\lamba}cos{\lamba x})$$
The boundary conditions give
$$y'(0)= C_2= 0$$
and then
$$y'(L)= -C_1\sqrt{\lamba}sin(\sqrt{\lambda}L)= 0$$
which says that $\sqrt{\lamba}L) must be a multiple of $\pi$ and so we must have $$\sqrt{\lamba}= \frac{n\pi}{L}$$ so the eigenvalues are of the form $$\frac{n^2\pi^2}{L^2}= $$\frac{n\pi}{L}$$^2$$ for n an integer (that, by the way, includes 0 which you had already found). That means that the "eigenvectors" or solutions are $$C_1 cos(\frac{n\pix}{L})$$ Again, any multiple of a single solution is also a solution so we might as well take C1= 1. The way you've written it, you have $\lamba_0= 0, y_0(x)= 1$ separate but you could as easily take [tex]\lamba_n= $$\frac{n\pi}{L}$$^2 y_n(x)= cos(\frac{n\pix}{L})$
with n= 0, 1, 2,... (since cosine is an even function, we don't get anything new taking negative values for n).

I think your basic question in both problems is why, after having gotten a non-trivial solution, involving an undetermined constant C1, do they then drop the constant. The answer is that the set of all vectors satisfying the equation $Ay= \lamba y$ is a subspace of the set of all vectors and so there exists a basis for it. In particular, in these problems the "eigenspaces" are 1 dimensiona. Since we want a solution yn such that any solution can be written y= Cyn for some number C, and we already have y= C1f(x) we might as well take C1= 1 and use that. In fact, we could have as easily let C1 be any non-zero number.

Last edited: May 14, 2011
3. Apr 26, 2006

### dak246

That explained it perfectly! Thanks sooo much for taking the time help me out with that...and sorry about the .jpg

4. May 13, 2011

### DonRico

I also face a big difficulty in this question! Could anyone help me though, please?

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5. May 13, 2011

### LCKurtz

Hello DonRico. Welcome to PF. I have quoted a few lines that you will see if you click on the Rules tab above: