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.
  • $\begingroup$ 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

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.

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

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