# Thread: Eigenvalues and Eigenfunctions of BVP

1. ## Eigenvalues and Eigenfunctions of BVP

Find the eigenvalues $\lambda_n$ and eigenfunctions $X_n$ of the boundary value problem

$X'' + \lambda X = 0 \ \ \ X'(0) = 0 \ \ \ X'(L) = 0$

Using the explicit form of $X_n(x)$ show that $\int_0^L X_nX_m \ dx = 0 \ (m \neq n)$ and $\int_0^L X_n X_m \ dx = \frac{L}{2} \ (m = n)$

2. What ideas have you had so far?

3. Okay, I'm a little confused with what to do with the second BC (i.e. $X'(L) = 0$, but here is what I have - I think I'm on the right path with the first BC).

For $\lambda > 0$, set $\lambda = \omega^2$. This gives the simple harmonic equation

$X'' + \omega^2X = 0$

for which the general solution is

$X(x) = A \cos(\omega x) + B \sin(\omega x)$
$X'(x) = -\omega A \sin(\omega x) + \omega B \cos(\omega x)$

Applying the BC $X'(0) = 0$ we get

$-\omega A \sin(0) + \omega B \cos(0)$

and recalling that $\sin(0) = 0$ and $\cos(0) = 1$ we have

$\omega B = 0$ which implies $B = 0$

(just a question: Why can't $\omega = 0$ and $B$ be some other value)

Applying the BC $X'(L) = 0$ we have (note that this is where I'm not sure of what to do)

$X'(L) = -\omega A \sin(\omega L) + \omega B \cos(omega L)$

but $B = 0$

so $-\omega A \sin(\omega L) = 0$
For non-trivial solutions we require that $A \neq 0$ so we have $- \omega \sin(\omega L) = 0$

and then I am supposed to be able to get the eigenvalues and functions for these.

I do a similar thing for $\lambda < 0$. Set $\lambda = -p^2$ which gives the equation

$X'' - p^2 x = 0$

which has general solution

$X(x) = A \cosh(\omega x) + B \sinh(\omega x)$
$X'(x) = p A \sinh(\omega x) + p B \cosh(\omega x)$

Applying the first BC, we have

$p A \sinh(0) + p B \cosh(0)$

so $pB = 0$

(again Im not sure what to do with the 2nd BC).

For $\lambda = 0$, the equation reduces to

$X'' = 0$

which has general solution

$X(x) = Ax + B$
$X'(x) = B$

The first BC is satisfied by $B = 0$.

This is as far as I can go...... I still have no idea how to :

Using the explicit form of $X_n(x)$ show that $\int_0^L X_nX_m \ dx = 0 \ (m \neq n)$ and $\int_0^L X_n X_m \ dx = \frac{L}{2} \ (m = n)$

4. Good work there! For the $\lambda>0$ case, I'd say that you can't have $\omega=0$, because you already assumed that $\lambda=\omega^{2}>0,$ and that's a strict inequality. Now you're down to the equation $- \omega \sin(\omega L) = 0.$ Since $\omega\not=0$ by assumption, you must have $\sin(\omega L) = 0.$ For what values of $\omega$ does this equation hold?

For the $\lambda<0$ case, I'd clean up your notation there a bit. You're confusing the notation with the $\lambda>0$ case. You end up with the solution

$X'(x) = p A \sinh(p x) + p B \cosh(p x).$

But you've shown that $B=0$, hence

$X'(x) = p A \sinh(p x).$ You must now apply the other boundary condition: $X'(L)=0$, which implies

$p A \sinh(p L)=0$. But now $p\not=0,$ and you can't have $A=0,$ or else you don't have eigenvectors (eigenvectors must, by definition, be nonzero). Hence, you must have $\sinh(pL)=0.$ But $p\not=0$, and $L\not=0$. Can you ever satisfy this equation? What does this tell you?

For the $\lambda=0$ case, I'm not sure you took the derivative correctly. Think through that case a bit more carefully.

5. Originally Posted by Ackbeet
Good work there! For the $\lambda>0$ case, I'd say that you can't have $\omega=0$, because you already assumed that $\lambda=\omega^{2}>0,$ and that's a strict inequality.
LOL. How obvious.....

Now you're down to the equation $- \omega \sin(\omega L) = 0.$ Since $\omega\not=0$ by assumption, you must have $\sin(\omega L) = 0.$ For what values of $\omega$ does this equation hold?
$0$ but $\omega \neq 0$ so $\omega = \frac{\pi}{L} , \frac{2 \pi}{L}, \frac{3 \pi}{L}$, ect

or alternatively $\omega = \frac{n \pi}{L} \ \text{for} \ n = 1, 2, ...$

I am going to guess that therefore

$\lambda = \frac{n^2 \pi^2}{L}$ for $n = 1,2,....$

and the corresponding function $X_n (x) = \cos(\frac{n \pi x}{L})$

I don't know how to determine $\lambda$ from the values of $\omega$, but I took a guess based upon a worked example. Even is my guess is correct, how is $\lambda$ determined here?

For the $\lambda<0$ case, I'd clean up your notation there a bit. You're confusing the notation with the $\lambda>0$ case. You end up with the solution

$X'(x) = p A \sinh(p x) + p B \cosh(p x).$

But you've shown that $B=0$, hence

$X'(x) = p A \sinh(p x).$ You must now apply the other boundary condition: $X'(L)=0$, which implies

$p A \sinh(p L)=0$. But now $p\not=0,$ and you can't have $A=0,$ or else you don't have eigenvectors (eigenvectors must, by definition, be nonzero). Hence, you must have $\sinh(pL)=0.$ But $p\not=0$, and $L\not=0$. Can you ever satisfy this equation? What does this tell you?
I think this just says that there are no negative eigenvalues.

For the $\lambda=0$ case, I'm not sure you took the derivative correctly. Think through that case a bit more carefully.
For $\lambda = 0$ the equation reduces to

$X'' = 0$

which has general solution

$X(x) = A + Bx$ (which is what I originally meant to write)
$X'(x) = B$

The first BC is satisfied by $B = 0$, so A is arbitrary, and $\lambda = 0$ is an eigenvalue, with an eigenfunction of

$X_0 (x) = 1$

I am also really confused and stuck on the second question.

6. For the $\lambda>0$ case: don't forget that $\omega$ can be negative. I agree that it can't be zero, since then $\lambda=0$. However, you can have

$\displaystyle{\omega = \pm\frac{n \pi}{L} \ \text{for} \ n = 1, 2, ...}$

I don't think you have to guess what $\lambda$ is if you have $\omega$. How did you define $\omega$ again?

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.

8. I'm tossing and turning about the next part to this question.

I thought about trying $\int(\cos(\frac{n \pi x}{L}) \ \cos(\frac{m \pi x}{L})) \ dx$

which gives $1/2\,L\sin \left( {\frac {\pi \, \left( -n+m \right) x}{L}} \right) {
\pi }^{-1} \left( -n+m \right) ^{-1}+1/2\,L\sin \left( {\frac {\pi \,
\left( n+m \right) x}{L}} \right) {\pi }^{-1} \left( n+m \right) ^{-1
}
$

but that doesn't make sense to integrate between 0 and L.

So I was thinking about $\int X_n X_m dx = X_n^'$ $X_m^'$ or something to that affect. Really at a loss....

9. 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?

10. Originally Posted by Maccaman
$1/2\,L\sin \left( {\frac {\pi \, \left( -n+m \right) x}{L}} \right) {
\pi }^{-1} \left( -n+m \right) ^{-1}+1/2\,L\sin \left( {\frac {\pi \,
\left( n+m \right) x}{L}} \right) {\pi }^{-1} \left( n+m \right) ^{-1
}
$
When trying to integrate between 0 and L I get

$
1/2\,{\frac {L\sin \left( \pi\, \left( -n+m \right) \right) }{\pi\,
\left( -n+m \right) }}+1/2\,{\frac {L\sin \left( \pi\, \left( n+m
\right) \right) }{\pi\, \left( n+m \right) }}

$

but we can't have $n = m$ in this case because we'll be dividing by 0, and I can't see how to simplify this if $n \neq m$

I'm Lost, Really Lost

11. Right. You can't have $n = m$ in this case. You have to go back to the original integral, plug in $n = m$, and go from there. It's basically a different integral.

For $n \not= m$, you can simplify quite a bit. What is $\sin(\pi(n\pm m))$ for $n,m\in\mathbb{Z}?$

12. Thanks Ackbeet. I got out the one for $m \neq n$.

For $n = m$ I get

$\int(\cos(\frac{n \pi x}{L}) \ \cos(\frac{n \pi x}{L})) \ dx = \frac{\left( \cos \left( {\frac {n\pi\,x}{L}} \right) \sin \left( {
\frac {n\pi\,x}{L}} \right) L+n\pi\,x \right)}{2n \pi}
$

but again I'm having trouble getting the required solution \frac{L}{2}

13. I agree with your antiderivative. Isn't the integral a definite integral, though?