Toronto Math Forum
APM3462018S => APM346Tests => Term Test 2 => Topic started by: Victor Ivrii on March 23, 2018, 06:08:24 AM

Solve by Fourier method
\begin{align}
& u_{tt}u_{xx}=0\qquad 0<x<\pi,\label{11}\\
& u_{x=0}= 0,\qquad (u_x+\alpha u)_{x=\pi}=0\label{12}\\
&u _{t=0}=\sin (x),\qquad u_t_{t=0}=0\label{13}
\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.

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{14}\end{equation}
If $\alpha=0$ the we know the solution are halfinteger $\sin$'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{14} 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{14} 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.

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 $\alpha<1/\pi$

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

If $X''=n^2$ (etc) then $X''=\lambda_n X$

Done. To the posterity: my mistake was on \eqref{error}.