Boundary value problem involving eigenvalues (1 Viewer)

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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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I HATE trying to read those Jpg things!

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

You then try [itex]\lambda= 0[/itex] 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 [itex]\lamba> 0[/itex], then you get general solution
[tex]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})[/tex]
The boundary conditions give
[tex]y'(0)= C_2= 0[/tex]
and then
[tex]y'(L)= -C_1\sqrt{\lamba}sin(\sqrt{\lambda}L)= 0[/tex]
which says that [itex]\sqrt{\lamba}L) must be a multiple of [itex]\pi[/itex] and so we must have
[tex]\sqrt{\lamba}= \frac{n\pi}{L}[/tex]
so the eigenvalues are of the form
[tex]\frac{n^2\pi^2}{L^2}= \(\frac{n\pi}{L}\)^2[/tex]
for n an integer (that, by the way, includes 0 which you had already found). That means that the "eigenvectors" or solutions are
[tex]C_1 cos(\frac{n\pix}{L})[/tex]
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 [itex]\lamba_0= 0, y_0(x)= 1[/itex] separate but you could as easily take
[tex]\lamba_n= \(\frac{n\pi}{L}\)^2 y_n(x)= cos(\frac{n\pix}{L})[/itex]
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 [itex]Ay= \lamba y[/itex] 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.
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That explained it perfectly! Thanks sooo much for taking the time help me out with that...and sorry about the .jpg
I also face a big difficulty in this question! Could anyone help me though, please?



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

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=) Sorry for breaking rules ! I just read HallsofIvy's solution again wisely ! and I found a way though =) thank you all

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