### Author Topic: Problem 3  (Read 12587 times)

#### James McVittie

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• Karma: 1 ##### Problem 3
« on: October 20, 2012, 10:01:33 PM »
What is implied by the word "weird", is it just something unexpected that comes up or something that hasn't been discussed in the course? Thanks!

#### Victor Ivrii ##### Re: Problem 3
« Reply #1 on: October 21, 2012, 03:18:11 AM »
What is implied by the word "weird", is it just something unexpected that comes up or something that hasn't been discussed in the course? Thanks!

Why you don't try to solve and see by yourself?
« Last Edit: October 21, 2012, 04:19:25 AM by Victor Ivrii »

#### James McVittie

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• Karma: 1 ##### Re: Problem 3
« Reply #2 on: October 22, 2012, 11:58:32 AM »
Are we allowed to assume that solutions are real or must we always assume in greatest generality that the solution could be complex?
Thanks

#### Victor Ivrii ##### Re: Problem 3
« Reply #3 on: October 22, 2012, 12:13:10 PM »
Are we allowed to assume that solutions are real or must we always assume in greatest generality that the solution could be complex?
Thanks

Have you read the preamble of this HA? It says that you may assume that eigenvalues are real (but nothing about eigenfunctions).

#### Shu Wang

• Jr. Member
•  • Posts: 11
• Karma: 0 ##### Re: Problem 3
« Reply #4 on: October 23, 2012, 05:20:03 AM »
uh, say we assume some form of solution for X(x) and T(t), so there will be 3 coefficients. When we setup the matrix for A,B,C it becomes 2x3 matrix since we're given 2 B.C. I was wondering if we were to solve for eigenvalue for the matrix, do we need to take into account for all combinations of 2x2 matrices? (In other words, break the matrix into A&B, B&C, A&C). Or am I just completely off the question? o.O

#### Victor Ivrii ##### Re: Problem 3
« Reply #5 on: October 23, 2012, 08:05:21 AM »
uh, say we assume some form of solution for X(x) and T(t), so there will be 3 coefficients.

Where? and don't mix coefficients for $X$ and $T$!

#### Bowei Xiao

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• Karma: 2 ##### Re: Problem 3
« Reply #6 on: October 23, 2012, 12:42:40 PM »
Is there supposed to be condition at t=0? Because I feel there's no restriction for interior and that can go wildly I guess?

#### Victor Ivrii ##### Re: Problem 3
« Reply #7 on: October 23, 2012, 04:31:06 PM »
We are looking at $u(x,t)=X(x)T(t)$.

Plug it in the equation and separating variables find equations to $X(x)$ and $T(t)$ in the standard way.

Plug into both boundary conditions. $u(0,x)=0$ implies what? (Standard)

Another b.c. implies ODE to $T(t)$. Solve it and use in conjugation with everything else. This is the only non-standard (albeit rather easy) part.

#### Peishan Wang

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• Karma: 6 ##### Re: Problem 3
« Reply #8 on: October 24, 2012, 11:20:15 PM »
Please let me know if there's anything wrong with the answer attached. Thanks
« Last Edit: October 24, 2012, 11:30:22 PM by Peishan Wang »

#### Peishan Wang

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• Karma: 6 ##### Re: Problem 3
« Reply #9 on: October 24, 2012, 11:21:14 PM »
Part 3 (to check if there's 0 or negative eigenvalues).

OK. -- V.I.
« Last Edit: October 25, 2012, 12:19:12 PM by Victor Ivrii »

#### Victor Ivrii ##### Re: Problem 3
« Reply #10 on: October 25, 2012, 03:33:06 PM »
So, as Peishan did: plugging $u=X(x)T(t)$ into equation and boundary conditions we get after separation of variables
\begin{align}
& \frac{T''}{T}=c^2\frac{X''}{X}=-c^2\lambda, \label{eq-1}\\
& X(0)=0,\label{eq-2}\\
& \frac{T'}{T}=-i\alpha \frac{X'(l)}{X(l)}=-i\beta\label{eq-3}
\end{align}
and from(\ref{eq-3}) we conclude that $\beta$ is a constant and $T=e^{-i\beta t}$ (do not care about constant factor) and then $\lambda=c^{-2}\beta^2$
\begin{align}
&X''+c^{-2}\beta^2 X=0 \label{eq-4}\\
& X(0)=0,\label{eq-5}\\
& X'(l)=\alpha^{-1}\beta X(l)\label{eq-6}
\end{align}
and the weirdness of this problem is that spectral parameter $\beta$ is present in both equation and the boundary condition. Then (\ref{eq-4})--(\ref{eq-5}) imply that $X= \sin (c ^{-1}\beta x)$ and (\ref{eq-6}) that $c^{-1}\beta \cos (c^{-1}\beta l)=\alpha{-1}\beta \sin (c^{-1}\beta l)$ which is equivalent to $\tan (c^{-1}\beta l)= c^{-1}\alpha$ i.e. $\beta = cl^{-1}\arctan c^{-1}\alpha +cl^{-1} n$ with $n\in \mathbb{Z}$.

Then exactly like in Peisan HA.

#### Chiara Moraglia

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• Karma: 1 ##### Re: Problem 3
« Reply #11 on: November 09, 2012, 05:55:45 PM »
Hi,
I have a question regarding (3) in your above response Professor. Where is beta coming from? Is it just equal to alpha*Xâ€²(l)X(l) and this is why beta is constant? I am also unclear as to how you obtain Î».

#### Victor Ivrii ##### Re: Problem 3
« Reply #12 on: November 09, 2012, 07:44:11 PM »
Hi,
I have a question regarding (3) in your above response Professor. Where is beta coming from? Is it just equal to alpha*Xâ€²(l)X(l) and this is why beta is constant? I am also unclear as to how you obtain Î».

Yes