1
$\begingroup$

Consider the following Sturm-Liouville problem: $$X''+\lambda X=0,\quad X'(0) = 0,\, X(\pi) = 0,$$ where $X = X(x)$.

  1. Find all positive eigenvalues and corresponding eigenfunctions of the problem.
  2. Is $\lambda = 0$ an eigenvalue for this problem? If yes, find its eigenfunction. If no, explain why it is not.
$\endgroup$
  • $\begingroup$ What have your tried? $\endgroup$ – Artem Nov 30 '12 at 2:34
  • $\begingroup$ Have you seen the case $X(0)=0, X(\pi)=0$? Just run the same analysis. Break it into three cases $\lambda>0$, $\lambda=0$ and $\lambda<0$ and see if each are possible. $\endgroup$ – Matt Nov 30 '12 at 2:50
2
$\begingroup$

Related problems: (I), (II). First solve the differential equation

$$ X(x)=c_1\,\sin \left( \sqrt {\lambda}x \right) + c_2\,\cos\left( \sqrt {\lambda}x \right) .$$ Applying the boundary conditions to the solution results in the two equations

$$ { c_1}\,\sin \left( \sqrt {\lambda}\pi \right) +{c_2}\,\cos\left( \sqrt {\lambda}\pi \right) =0 \rightarrow (1) $$

$$ {c_1}\,\sqrt{\lambda} = 0 \rightarrow (2), $$

where $c_1$ and $c_2$ are arbitrary constants. From (2), we assume $\lambda\neq 0$, then we will have $c_1=0.$ Substituting $c_1=0$ in (1) gives

$$ {c_2}\,\cos\left( \sqrt {\lambda}\pi \right) = 0 \implies \cos\left( \sqrt {\lambda}\pi \right)=0 \implies \sqrt{\lambda} = \frac{2n+1}{2} $$

$$ \implies \lambda = \frac{(2n+1)^2}{4},\quad n=0,1,2,3\dots $$

I will leave it here for you to finish the task. Note that, $\lambda = 0 $ is a special case. Subs $\lambda=0$ in the diff. eq. and follow the above technique and see what you get.

$\endgroup$
  • $\begingroup$ And the final part? $\endgroup$ – Olinda Fernandes Jun 5 '13 at 16:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.