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For the boundary value problem,
$y''+\lambda y=0$
$y(-π)=y(π)$ , $y’(-π)=y’(π)$
to each eigenvalue $\lambda$, there corresponds

  1. Only one eigenfunction
  2. Two eigenfunctions
  3. Two linearly independent eigenfunctions
  4. Two orthogonal eigenfunctions

I have tried to solve the problem but could not get my calculations right. Can somebody help?

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Surely you know the general solution of the equation $y''+\lambda y=0$. Once this is written down, you might want to examine points 1. to 4. –  Did Dec 19 '12 at 15:39

2 Answers 2

This is elementary. Find for each $\lambda$ general form of a solution of the equation without boundary conditions, and try to apply the boundry condition. You won't get stuck in the calcualtions again, as there is almost nothing to calculate.

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Here is a related problem. First solve the differential equation

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

$$ y(\pi)=y(-\pi) \implies 2{ c_1}\,\sin \left( \sqrt {\lambda}\pi \right)= 0 \rightarrow (1) $$

$$y'(\pi)=y'(-\pi)\implies 2{c_2}\sqrt{\lambda}\sin \left( \sqrt {\lambda}\pi \right) = 0 \rightarrow (2), $$

where $c_1$ and $c_2$ are arbitrary constants. From $(1)$, since $c_1$ is an arbitrary constant, then we have

$$ \sin\left( \sqrt {\lambda}\pi \right) = 0 \implies \sqrt {\lambda}\pi = n\pi \implies \lambda = n^2. $$

Eq. $(2)$ gives the same eigenvalues with the other eigenvalue $\lambda=0$, since $c_2$ is an arbitrary constant. So, can you find the eigenvectors now and answer the question?

Note: This problem is known as "The Sturm-Liouville Eigenvalue Problems". One of the properties of this kind of problems is that "for each eigenvalue $\lambda_n$ there exists an eigenfunction $\phi_n$.

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The explanation since $c_1$ is an arbitrary constant and the line after it are misleading. One should first assume that $\sin(\sqrt{\lambda}\pi)\ne0$, show that this leads to a contradiction, then proceed with the case $\sin(\sqrt{\lambda}\pi)=0$. Similar remark about since $c_2$ is an arbitrary constant. –  Did Dec 19 '12 at 16:38

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