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Question 9 from 4.2P

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Tristan Fraser:
The problem states:

--- Quote ---Consider wave equation with the Neumann boundary condition on the left and weird b.c. on the right:
$u_{tt}−c^2 u_{xx}=0 , \ \ u_x(0,t)=0, \ (u_x+i\alpha u_t)(l,t)=0, \ \ \ 0<x<l$
with $\alpha$∈ℝ.

Separate variables;
Find weird eigenvalue problem for ODE;
Solve this problem;
Find simple solution u(x,t)=X(x)T(t).
Hint. You may assume that all eigenvalues are real (which is the case).
--- End quote ---

I used the usual separation of variables: $u = X(x)T(t)$ and made

$\frac{T''}{c^2 T} = \frac{X''}{X} = - \lambda$, *

to then get
$$X(x) = A\cos(\sqrt{\lambda}x) + B\sin(\sqrt{\lambda}x).$$
Resolving the "easy" boundary condition gives us

$X(x) = A\cos(\sqrt{\lambda}x)$ OK

While for T:

T  = $C\cos(\frac{\sqrt{\lambda}}{c}t) + D\sin(\frac{\sqrt{\lambda}}{c}t)$

And then taking the "weird" boundary condition gives me something like: $\sqrt{\lambda}\tan(\sqrt{\lambda}l) = i\alpha\frac{T'}{T}$

But I wasn't quite sure where to go from there in terms of solving the problem. One approach that I think brings me closer (hopefully) to the solution is:

Note that $(\frac{1}{c^2}\frac{T'}{T})'$ = $\frac{1}{c^2} (\frac{T''}{T} - (\frac{T'}{T})^2)$

Then knowing that the first term on the RHS is constant ($-\lambda$), we can rewrite the expression wrt to $\frac{T'}{T}$ as follows:

$\frac{T'}{T} = c\sqrt{\frac{T''}{c^2 T} - \frac{1}{c^2}(\frac{T'}{T})'} = c \sqrt{-\lambda - (\frac{T'}{c^2T})'}$

which can be incorporated into the weird BC above :

$\frac{X'}{X}(l) = \alpha c \sqrt{\lambda + (\frac{T'}{c^2T})'}$

and then:

$\sqrt{\lambda} \tan(\sqrt{\lambda l}) = \alpha c \sqrt{\lambda + (\frac{T'}{c^2T})'}$

That being said, I'm worried that I've gone off an a tangent and have failed to simplify the problem. Any hints would be appreciated.

Victor Ivrii:
Look at place I marked OK. Plug it into boundary condition as $x=l$ and find $T(t)$ from here, and plug it into the second order equation *

Jingxuan Zhang:
Tristan,

Two things: first, shouldn't it rather be $C\cos c\sqrt{\lambda}t + D\sin c\sqrt{\lambda}t$? second, how do you thereof derive the $\tan$ equation?

Edit: never mind. I think I understand you now. I think your problem is that your supposed form of $T$ and the condition for $T$ doesn't match. Indeed if $\frac{T'}{T}$ is some constant then $T$ should be rather of exponential form.

From your tangent equatoin after some calculation I found, if $u\neq\text{const}$,
$$T=-\frac{\tan\omega l}{c\alpha}e^{ic\omega t},\frac{\omega l}{\pi}\notin\mathbb{Z}.$$

EDIT: Above are wrong. But not very wrong: it is right to consider $T$ as complex exponential.

Tristan Fraser:
So we basically take:

$X'(l) = -A \sqrt{ \lambda}\sin \sqrt{ \lambda} l$
$T'(t) = c \sqrt{ \lambda} [D \cos(c \sqrt{ \lambda}t) -C \sin(c \sqrt{ \lambda}t)$

Applying the tricky boundary condition:

$A \sqrt{ \lambda} \sin \sqrt{\lambda}l = i \alpha (c \sqrt{ \lambda} [D\cos(c \sqrt{ \lambda}t) -C\sin(c \sqrt{ \lambda}t))$

Now here's where I might have done something very foolish;

$A \sqrt{\lambda}\sin\sqrt{\lambda}l = i\alpha (T'(t))$ and integrated both sides:

$A \sqrt{\lambda}\sin(\sqrt{\lambda}l) t + F= i\alpha T(t)$

Then I plugged it into *

$\frac{T''}{c^2{\frac{A\sqrt{\lambda}\sin(\sqrt{\lambda}l) t + F}{i\alpha}}} = \frac{X''}{X} = - \lambda$

I'm afraid I'm a little lost. Perhaps another hint might help?

EDIT: I'm not sure why the LaTeX is producing this output, it looked fine in the editor I was working in (sharelatex)...

For clarity, refer to the screenshot:

Victor Ivrii:
MathJax produced this output because of unbalanced { . I fixed them.

You are trying your best to make things as complicated as possible. We considered it during Review lecture. After we found $X'(l(/X(l)$ we can find $T(t)$ from the boundary condition, then plug it into equation $T''=-\lambda T$ and get extremely simple equation to $\omega =\sqrt{\lambda}$.