Toronto Math Forum
APM346-2012 => APM346 Math => Home Assignment 4 => Topic started by: James McVittie on October 20, 2012, 10:01:33 PM
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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!
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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?
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Are we allowed to assume that solutions are real or must we always assume in greatest generality that the solution could be complex?
Thanks
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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).
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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
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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 $T$!
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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?
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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.
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Please let me know if there's anything wrong with the answer attached. Thanks
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Part 3 (to check if there's 0 or negative eigenvalues).
OK. -- V.I.
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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.
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Hi,
I have a question regarding (3) in your above response Professor. Where is beta coming from? Is it just equal to alpha*X′(l)X(l) and this is why beta is constant? I am also unclear as to how you obtain λ.
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Hi,
I have a question regarding (3) in your above response Professor. Where is beta coming from? Is it just equal to alpha*X′(l)X(l) and this is why beta is constant? I am also unclear as to how you obtain λ.
Yes