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I have the Sturm-Liouville problem $$y''(t) + \lambda\ y(t) = 0,\hspace{1cm} y(0) = y(\pi) = 0.$$

When I reach the case where $$\Delta < 0\ \implies \lambda > 0$$ I find $$y(t) = C_1 \cos( \lambda^{\frac{1}{2}} t) + C_2 \sin( \lambda^{\frac{1}{2}} t),$$ and both coefficients $C_1$ and $C_2$ are equal to zero.

How to solve this problem, meaning that I need to find the eigenvalues and eigenfunctions?

BugShotGG
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2 Answers2

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Your general solution is correct and the given boundary conditions imply $y(t) = 0$. Overall, you have not made any mistakes as you have arrived at the correct solution. However, to solve for the eigenvalues and the eigenfunctions use the fact that $y(\pi) = 0$ and $\sin x = 0 \Longleftrightarrow x = n\pi$ whenever $n \in \mathbb Z$. Do not forget to apply the Superposition Principle. I hope this helps.

glebovg
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    @GeoPapas No mistakes except for the "there is no solution" part. There is a solution, it's $y(t)=0$ :) – rurouniwallace Jul 27 '13 at 14:35
  • @GeoPapas Precisely. – glebovg Jul 27 '13 at 14:36
  • Those are not inital conditions, but boundary conditions, and these leave possibilities for certain values of $\lambda$. See my answer. – ccorn Jul 27 '13 at 14:45
  • @ccorn Thanks. By the way, although the general solution will change depending on the sign of $\lambda$, given boundary conditions lead to $y(t) = 0$. – glebovg Jul 27 '13 at 15:15
  • @glebovg: You essentially state that there are no oscillatory modes for a linearized guitar string. Think again. – ccorn Jul 27 '13 at 15:15
  • @ccorn No. I am simply pointing out that there is a difference between solving for $y(t)$ while assuming $\lambda$ is some constant, and solving for the eigenvalues and eigenfunctions. The OP did not understand the distinction and simply solved for $y(t)$. – glebovg Jul 27 '13 at 15:41
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Since $C_1 = 0$ follows from the first boundary condition, we have $$y(\pi) = 0 = C_2 \sin(\sqrt{\lambda}\pi) \implies \sqrt{\lambda}\pi = n\pi \implies \lambda = n^2 \quad (\text{for } n = 1, 2, \ldots).$$ Now you can find the eigenfunctions because $$y(t) = C_2 \sin(\sqrt{\lambda}t) = C_2 \sin(nt).$$ Note that the case $n = 0$ needs to be studied separately. Just plug in $\lambda = 0$ in the ODE and see what you get. Also, see this related question.

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Mhenni Benghorbal
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