APM346-2018S > Term Test 2

TT2--P1

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Victor Ivrii:
Solve by Fourier method
\begin{align}
& u_{tt}-u_{xx}=0\qquad 0<x<\pi,\label{1-1}\\
& u|_{x=0}= 0,\qquad (u_x+\alpha u)|_{x=\pi}=0\label{1-2}\\
&u| _{t=0}=\sin (x),\qquad u_t|_{t=0}=0\label{1-3}
\end{align}
with $\alpha\in \mathbb{R}$.

Hint: We know that $\lambda_n$ are real but since we do not know the sign of $\alpha$ we do not know if it all $\lambda_n\ge 0$; so you must consider the case of some of $\lambda_n<0$.

Note: Only find equations for eigenvalues.

Jingxuan Zhang:
The associated eigenproblem is
\begin{equation}\left\{\begin{split}&X''=\lambda X,\\&X|_{x=0}=(X'-\alpha X)|_{x=0}=0.\end{split}\right.\label{1-4}\end{equation}
If $\alpha=0$ the we know the solution are half-integer $\sin$'s
\begin{equation}\label{error}\lambda_n=-\Bigl(n+\frac{1}{2}\Bigr)^2, X_n(x)=\sin \Bigl(n+\frac{1}{2}\Bigr)x,n=0,1,....\end{equation}
If $\alpha\neq 0,\lambda>0$ then the general solution for the DE in \eqref{1-4} is
$$X(x)=A\cosh \gamma x + B\sinh \gamma x, \gamma>0.$$
Plugging in boundary condition we find $A=0$ and
$$\gamma B\cosh \gamma\pi+\alpha B\sinh \gamma\pi=0.$$
Hence the various eigenvalues are given by $\lambda_n=\gamma_n^2$ where $\gamma_n$ is a nonzero root of
$$\gamma=-\alpha\tanh \gamma\pi.$$
If $\alpha\neq 0,\lambda<0$ then the general solution for the DE in \eqref{1-4} is
$$X(x)=A\cos \omega x + B\sin \omega x, \omega>0.$$
Plugging in boundary condition we find $A=0$ and
$$\omega B\cos \omega\pi+\alpha B\sin \omega\pi=0.$$
Hence the various eigenvalues are given by $\lambda_n=-\omega_n^2$ where $\omega_n$ is a nonzero root of
$$\omega=-\alpha\tan \omega\pi.$$
If $\lambda=0$ then we have only trivial solution.

Victor Ivrii:
Not
$$
 X|_x=0=(X'-\alpha X)|_x=0=0$$
 but
$$X|_{x=0}=(X'-\alpha X)|_{x=0}=0$$.

and also correct signs at $\lambda_n$

I attach pictures for $\lambda<0$ and $\lambda >0$. On the first, brown line for $\alpha >0$, red line for $\alpha<0$.  On the second brown line for $\alpha> -1/\pi$, red line for $\alpha<-1/\pi$

Jingxuan Zhang:
That subscript is really awkward but I don't see sign problem at $\lambda_n$?

Victor Ivrii:
If $X''=n^2$ (etc) then $X''=-\lambda_n X$

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