1. ## boundary value problem

Find the eigenvalues , and normalised eigenfunctions of the problem:

I'm fine of the case but unsure about the other two.

Let
DE transforms to
General solution:

Hence

Hence eigenvalues are

Let
DE transforms to
General solution:

Hence

Which only has solution w = 0 hence trivial solution and we reject.

Could some please check this is OK?

2. Originally Posted by hunkydory19
Find the eigenvalues , and normalised eigenfunctions of the problem:

I'm fine of the case but unsure about the other two.

Let
DE transforms to
General solution:

Hence

Hence eigenvalues are

Let
DE transforms to
General solution:

Hence

Which only has solution w = 0 hence trivial solution and we reject.

Could some please check this is OK?

Yeah, that looks okay. One question though. I agree with $\lambda_n=\left(n-\frac{1}{2}\right)^2\pi^2$, but where are you getting $\lambda_n=mu^2$? What are $u$ and $m$?

3. Originally Posted by redsoxfan325
Yeah, that looks okay. One question though. I agree with $\lambda_n=\left(n-\frac{1}{2}\right)^2\pi^2$, but where are you getting $\lambda_n=mu^2$? What are $u$ and $m$?
It's mu ( $\mu$)

$\lambda_n = \mu^2 = \left(n-\frac{1}{2}\right)^2\pi^2$

4. Thank you so much for the reply.

I'm also confused on how to get the eigenfunctions. The answer given is $\sqrt 2 sin ((n - \frac{1}{2}) \pi x)$ but I don't see where this comes from...could you possibly explain?

5. Originally Posted by hunkydory19
Thank you so much for the reply.

I'm also confused on how to get the eigenfunctions. The answer given is $\sqrt 2 sin ((n - \frac{1}{2}) \pi x)$ but I don't see where this comes from...could you possibly explain?
As you said earlier, in order to satisfy the BC's you need $B = 0,\; \text{and}\; \mu = \left(n-\frac{1}{2}\right) \pi$ . Now from your solution

$
y = A \sin \mu x = A\sin \left(n-\frac{1}{2}\right) \pi x
$

This satisfies both the equation and BCs (for $\lambda = \mu^2$). Not sure where the $\sqrt{2}$ came from unless there's another BC floating around.

6. Yes it was the root 2 bit that I was confused about, there are no other boundary conditions so I don't see how they have got this value for A but thank you again for your help!

7. Originally Posted by danny arrigo
It's mu ( $\mu$)

$\lambda_n = \mu^2 = \left(n-\frac{1}{2}\right)^2\pi^2$
Yikes. I should've realized that. I'm starting to think like Maple. In my defense, the actual symbol $\mu$ was used throughout the rest of the post, so when it got suddenly switched up like that it confused my poor brain.

8. There is one other possiblity you left out.

You considered $\lambda> 0$ and $\lambda< 0$. What if $\lambda= 0$?

9. He said he was fine in the case of $\lambda=0$.

10. The $\sqrt{2}$ comes from the fact that the question states normalised eigenfunctions hence the square should integrate to 1 over the relevant domain.

11. Also just noticed you made a mistake with you $\lambda < 0$ case. You should have found that

$A = 0$

so that

$0 = B \omega \cosh \omega$

which (since the minimum value o cosh function is 1) means that

$B = 0$

so there is no solution for these BCs with $\lambda < 0$ .

12. Originally Posted by hunkydory19
Find the eigenvalues , and normalised eigenfunctions of the problem:

Yep - missed that part.