What ideas have you had so far?
Okay, I'm a little confused with what to do with the second BC (i.e. , but here is what I have - I think I'm on the right path with the first BC).
For , set . This gives the simple harmonic equation
for which the general solution is
Applying the BC we get
and recalling that and we have
which implies
(just a question: Why can't and be some other value)
Applying the BC we have (note that this is where I'm not sure of what to do)
but
so
For non-trivial solutions we require that so we have
and then I am supposed to be able to get the eigenvalues and functions for these.
I do a similar thing for . Set which gives the equation
which has general solution
Applying the first BC, we have
so
(again Im not sure what to do with the 2nd BC).
For , the equation reduces to
which has general solution
The first BC is satisfied by .
This is as far as I can go...... I still have no idea how to :
Using the explicit form of show that and
Good work there! For the case, I'd say that you can't have , because you already assumed that and that's a strict inequality. Now you're down to the equation Since by assumption, you must have For what values of does this equation hold?
For the case, I'd clean up your notation there a bit. You're confusing the notation with the case. You end up with the solution
But you've shown that , hence
You must now apply the other boundary condition: , which implies
. But now and you can't have or else you don't have eigenvectors (eigenvectors must, by definition, be nonzero). Hence, you must have But , and . Can you ever satisfy this equation? What does this tell you?
For the case, I'm not sure you took the derivative correctly. Think through that case a bit more carefully.
LOL. How obvious.....
but so , ectNow you're down to the equation Since by assumption, you must have For what values of does this equation hold?
or alternatively
I am going to guess that therefore
for
and the corresponding function
I don't know how to determine from the values of , but I took a guess based upon a worked example. Even is my guess is correct, how is determined here?
I think this just says that there are no negative eigenvalues.For the case, I'd clean up your notation there a bit. You're confusing the notation with the case. You end up with the solution
But you've shown that , hence
You must now apply the other boundary condition: , which implies
. But now and you can't have or else you don't have eigenvectors (eigenvectors must, by definition, be nonzero). Hence, you must have But , and . Can you ever satisfy this equation? What does this tell you?
For the equation reduces toFor the case, I'm not sure you took the derivative correctly. Think through that case a bit more carefully.
which has general solution
(which is what I originally meant to write)
The first BC is satisfied by , so A is arbitrary, and is an eigenvalue, with an eigenfunction of
I am also really confused and stuck on the second question.
Hit reply too soon.
I agree that I don't think there are any negative eigenvalues. You do have the zero eigenvalue with the eigenvector you described.
To do the second problem, I'd recommend collecting all your results into one place. Then start performing the integrals mentioned and see where that leads.
I'm tossing and turning about the next part to this question.
I thought about trying
which gives
but that doesn't make sense to integrate between 0 and L.
So I was thinking about or something to that affect. Really at a loss....
I would clean up your parentheses a bit, but your integration works fine for the m not equal to n case. Why can't you integrate between 0 and L? Can't you just plug in the limits, as per the Fundamental Theorem of the Calculus?
Incidentally, I would to the m = n case separately, since your antiderivative doesn't look so nice in this case. That is, do the integral from the beginning, using a simplification.
Does this shine a little light on your path?