### Author Topic: TT2--P1N  (Read 2892 times)

#### Victor Ivrii ##### TT2--P1N
« on: March 23, 2018, 06:14:32 AM »
Solve by Fourier method
\begin{align}
\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

• Elder Member
•     • Posts: 106
• Karma: 20 ##### Re: TT2--P1N
« Reply #1 on: March 24, 2018, 08:46:48 AM »
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 integer $\cos$'s
\begin{equation}\label{error}\lambda_n=-n^2, X_n(x)=\cos nx, 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 $B=0$ and
$$\gamma A\sinh \gamma\pi+\alpha A\cosh \gamma\pi=0.$$
Hence the various eigenvalues are given by $\lambda_n=\gamma_n^2$ where $\gamma_n$ is a nonzero root of
$$\gamma\tanh \gamma\pi+\alpha=0.$$
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 $B=0$ and
$$-\omega A\sin\omega\pi+\alpha A\cos \omega\pi=0.$$
Hence the various eigenvalues are given by $\lambda_n=-\omega_n^2$ where $\omega_n$ is a nonzero root of
$$\omega\tan \omega\pi-\alpha=0.$$
If $\lambda=0$ then we have only trivial solution.
« Last Edit: March 25, 2018, 10:00:56 AM by Jingxuan Zhang »

#### Victor Ivrii ##### Re: TT2--P1N
« Reply #2 on: March 25, 2018, 04:37:29 AM »
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 $\alpha<-1/\pi$

Also correct sign at $\lambda_n$

#### Jingxuan Zhang

• Elder Member
•     • Posts: 106
• Karma: 20 ##### Re: TT2--P1N
« Reply #3 on: March 25, 2018, 07:36:09 AM »
Again I don't quite see the issue of sign, please inform me the mistake.
Found it. To the posterity: my mistake was on \eqref{error}.

By the way, how can I draw with tikz a picture like this? or did you use latex at all?
« Last Edit: March 25, 2018, 10:00:35 AM by Jingxuan Zhang »

#### Victor Ivrii ##### Re: TT2--P1N
« Reply #4 on: March 25, 2018, 10:02:32 AM »