# Eigenvalue problem for difference equations

• May 21st 2009, 09:44 AM
bkarpuz
Eigenvalue problem for difference equations
Let $m\in\mathbb{N}$.
Consider the following boundary value problem
$\begin{cases}L\psi(n)+\lambda\psi(n)=0,&n\in\{1,2, \ldots,m\}\\ \psi(n)=0,&n\in\{0,m+1\},\end{cases}\rule{4cm}{0cm }(\star)$
where the operator $L$ is defined to be
$L\psi(n):=\Delta^{2}\psi(n-1)$ and $\Delta$ is the usual forward difference operator.

When I try to find a nontirivial real sequence satisfying $(\star)$ with $\lambda>0$ and $m=2$ or $m=3$, I always fail.
I need help to find a suitable $\lambda>0$ and $m\in\mathbb{N}$ for which $(\star)$ admits a nontrivial solution $\psi$.

Thanks.
• May 21st 2009, 03:22 PM
the_doc
I believe it to be impossible to find any non-trivial solution for this problem. The only solutions that I can see are with $\lambda = 0$ and $\lambda \in \mathbb{C}$ .
• May 21st 2009, 10:37 PM
bkarpuz
Consider the following difference equation

$\begin{cases}
L\psi (m)+\alpha \psi (m)=0,&m\in \left\{ 1,2,3\right\}\\
\psi (0)=0,\ \psi (4)=0.
\end{cases}
$

Clearly, the set of eigenvalues of the equation is $\left\{ 2-\sqrt{2},2,2+\sqrt{2}\right\}$. Thus the least eigenvalue is $\alpha _{0}=2-\sqrt{2}$.
Moreover one can check that
$
\psi (m):=\left( \frac{1-\mathrm{i}}{\sqrt{2}}\right) ^{m}-\left( \frac{1+
$

is a nontrivial solution of
$
\begin{cases}
L\psi (m)+(2-\sqrt{2}) \psi (m)=0,&m\in \left\{ 1,2,3\right\}\\
\psi (0)=0,\ \psi (4)=0.
\end{cases}
$

But this is not a real sequence. (Doh)
• May 22nd 2009, 08:35 AM
the_doc
Oops! Sorry, it was quite late here (at night) when I responded and I didn't expand the L operator out correctly in my head.

I haven't checked your eigenvalues so I'll assume they're correct in which case I think you'll find multiplying your solution by $\color[rgb]{0,0,1} i$ will render it to a sequence of reals.

This is perfectly valid as your equation is homogeneous so you can scale up by whatever constant you like!

Hope that's of more help to you!
• May 22nd 2009, 01:11 PM
bkarpuz
Quote:

Originally Posted by the_doc
Oops! Sorry, it was quite late here (at night) when I responded and I didn't expand the L operator out correctly in my head.

I haven't checked your eigenvalues so I'll assume they're correct in which case I think you'll find multiplying your solution by $\color[rgb]{0,0,1} i$ will render it to a sequence of reals.

This is perfectly valid as your equation is homogeneous so you can scale up by whatever constant you like!

Hope that's of more help to you!

You are telling what was in my mind, but I am not sure whether it will work, because multiplying by i will make the real parts imaginary this time.
May be I should try to obtain some linear combinations of such solutions to make them real, but i dont think it is possible...

ty anyways.
• May 22nd 2009, 04:33 PM
the_doc
No! The point is that there are no real parts in the solution that you just described since

$\psi (m)=\left( \frac{1-\mathrm{i}}{\sqrt{2}}\right) ^{m}-\left( \frac{1+
\mathrm{i}}{\sqrt{2}}\right) ^{m}
$

$\Leftrightarrow \psi (m)= \frac{1}{(\sqrt{2})^m} \sum_{k=0}^{k=m} \begin{pmatrix} m \\
k
\end{pmatrix} [(-1)^k - 1] i^{k}
$
,

so the real terms correspond to
$k$ even but those terms are clearly 0. So like I said multiply by $i$.
• May 26th 2009, 11:30 AM
bkarpuz
Quote:

Originally Posted by the_doc
No! The point is that there are no real parts in the solution that you just described since

$\psi (m)=\left( \frac{1-\mathrm{i}}{\sqrt{2}}\right) ^{m}-\left( \frac{1+
\mathrm{i}}{\sqrt{2}}\right) ^{m}
$

$\Leftrightarrow \psi (m)= \frac{1}{(\sqrt{2})^m} \sum_{k=0}^{k=m} \begin{pmatrix} m \\
k
\end{pmatrix} [(-1)^k - 1] i^{k}
$
,

so the real terms correspond to
$k$ even but those terms are clearly 0. So like I said multiply by $i$.

ty the_doc, multiplying the solution by $\mathrm{i}$ gave the following solution
$\psi^{\dagger}=[0, \sqrt{2}, 2, \sqrt{2}, 0]$.
I was just thinking that there would be $\mathrm{i}\sqrt{2}$ instead of $\sqrt{2}$, but I missed that all the terms were purely complex before multiplying by $\mathrm{i}$.
• May 27th 2009, 08:05 AM
the_doc
Yes, I understand completely - it wasn't necessarily obvious. :)

Also I should add that the general solution, for any $m$, for the eigenvalue $\lambda$ is:
$\lambda_k = 4 \sin^2 \left(\frac{k}{m+1} \, \frac{\pi}{2} \right)$ with $k = 1,2, \ldots , (m-1) , m$ .
The corresponding $\psi$ are:
$\psi_n = A \sin \left( \frac{k}{m+1} \, n \pi \right)$
where $A \in \Re$ .