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Given the boundary problem

$$X''=\mu X,\;X(0)=0,\;X'(L)=0$$

We need to find the non trivial solution $X(x)$ that satisfies the above equations. Let $\mu<0$, $\mu=-k^2$ for some $k>0$. The general solution for $\mu<0$ can be described as

$$X(x)=A\sin(k x)+B\cos(k x)$$

For some constants $A$, $B$. How do I formulate the solution $X(x)$ that satisfies $X(0)=0$ and $X(L)=0$

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Hint.

From the boundary conditions

$$ A\sin(0)+B\cos(0) = 0\\ A k\cos(k L)-B k\sin(k L) = 0 $$

or

$$ \left( \begin{array}{cc} 0& 1\\ k\cos(kL) & -k\sin(kL) \end{array} \right) \left( \begin{array}{c} A\\ B \end{array} \right)=\left( \begin{array}{c} 0\\ 0 \end{array} \right) $$

This linear system has a non trivial solution for

$$ \det\left( \begin{array}{cc} 0& 1\\ k\cos(kL) & -k\sin(kL) \end{array} \right) = -k\cos(k L) = 0 $$

or for $k L = \frac{\pi}{2}+\nu \pi$

hence

$$ k = \frac 1L\left(\frac{\pi}{2}+\nu \pi\right)\ \ \ \nu = 1,2,3,\cdots $$

now once we have the set of eigenfunctions we proceed with the determination of $A, B$

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Any non-trivial solution where $X(0)=0$ can be normalized so that $X'(0)=1$, which gives a unique normalized solution $$ X_{\mu}(x)=\frac{\sin(\sqrt{-\mu}x)}{\sqrt{-\mu}}. $$ This works in the limiting case as $\mu\rightarrow 0$ as well, where $X_0(x)=x$. Normalization is a handy way to eliminate special cases for general problems of this nature.

The set of $\mu$ for which there is a non-trivial solution of your problem is the set of $\mu$ for which $X'(L)=0$, or

$$ \cos(\sqrt{-\mu}L)=0 \\ \implies \sqrt{-\mu}L=\frac{\pi}{2}+n\pi,\;\; n=0,\pm 1,\pm 2,\cdots, \\ \mu = -\frac{1}{L^2}\left(\frac{\pi}{2}+n\pi\right)^2. $$ So $X$ can be indexed by $n$: $$ X_n(x)=\sin((n+1/2)\pi x/L),\;\;\; n=0,1,2,3,\cdots . $$ The cases for $n=-1,-2,-3,\cdots$ do not lead to additional solutions.

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