Results 1 to 9 of 9

Math Help - Eigenvalue problem for difference equations

  1. #1
    Senior Member bkarpuz's Avatar
    Joined
    Sep 2008
    From
    R
    Posts
    481
    Thanks
    2

    Exclamation 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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    May 2009
    Posts
    127
    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} .
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member bkarpuz's Avatar
    Joined
    Sep 2008
    From
    R
    Posts
    481
    Thanks
    2

    Angry

    Consider the following difference equation

    \begin{cases}<br />
L\psi (m)+\alpha \psi (m)=0,&m\in \left\{ 1,2,3\right\}\\<br />
\psi (0)=0,\ \psi (4)=0.<br />
\end{cases}<br />
    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
    <br />
\psi (m):=\left( \frac{1-\mathrm{i}}{\sqrt{2}}\right) ^{m}-\left( \frac{1+<br />
\mathrm{i}}{\sqrt{2}}\right) ^{m},\quad\text{for}\ m=0,1,2,3,4<br />
    is a nontrivial solution of
    <br />
\begin{cases}<br />
L\psi (m)+(2-\sqrt{2}) \psi (m)=0,&m\in \left\{ 1,2,3\right\}\\<br />
\psi (0)=0,\ \psi (4)=0.<br />
\end{cases}<br />
    But this is not a real sequence.
    Last edited by bkarpuz; February 25th 2010 at 10:30 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    May 2009
    Posts
    127
    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!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member bkarpuz's Avatar
    Joined
    Sep 2008
    From
    R
    Posts
    481
    Thanks
    2

    Exclamation

    Quote Originally Posted by the_doc View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    May 2009
    Posts
    127
    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+<br />
\mathrm{i}}{\sqrt{2}}\right) ^{m}<br />

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

    so the real terms correspond to
    k even but those terms are clearly 0. So like I said multiply by i.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member bkarpuz's Avatar
    Joined
    Sep 2008
    From
    R
    Posts
    481
    Thanks
    2

    Red face

    Quote Originally Posted by the_doc View Post
    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+<br />
\mathrm{i}}{\sqrt{2}}\right) ^{m}<br />

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

    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}.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    May 2009
    Posts
    127
    Yes, I understand completely - it wasn't necessarily obvious.

    Your welcome Bkarpuz!
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    May 2009
    Posts
    127
    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 .
    Last edited by the_doc; May 27th 2009 at 09:09 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Eigenvalue Problem
    Posted in the Advanced Algebra Forum
    Replies: 10
    Last Post: July 5th 2011, 01:35 AM
  2. Coupled difference equations - Change of coordinates problem
    Posted in the Differential Equations Forum
    Replies: 14
    Last Post: April 9th 2010, 04:28 AM
  3. Eigenvalue problem
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: April 4th 2010, 07:52 PM
  4. Replies: 0
    Last Post: October 18th 2009, 11:21 AM
  5. eigenvalue problem
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: March 10th 2007, 06:29 AM

Search Tags


/mathhelpforum @mathhelpforum