# non-trivial solution boundary problems

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 $$X(L)=0$$

## 2 Answers

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 $$A, B$$

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.