Question about vibrating string problem,Sturm-Liouville problem

Question

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I'm stuck on how to simplify the BC condition. I get '(0)=(0); '(1)=(1). How can I connect this to question? Bty, what's Sturm-Liouville problem for (x)?

Answer 1

The eigenvalue problem for your equation in $x$ is $$ X_{\lambda}''(x) = -\lambda X_{\lambda}(x) \\ X_{\lambda}'(0)=X_{\lambda}(0),\\ X_{\lambda}'(1)=X_{\lambda}(1). $$ Part of the problem in solving such an equation is that $X_{\lambda}$ is unique only up to a multiplicative constant, which is something remedied by choosing a particular normalization. To this end, replace the condition at $x=0$ with $$ X_{\lambda}'(0)=1=X_{\lambda}(0). $$ Without trying to satisfy the right endpoint condition, the unique $X_{\lambda}$ has the form $$ X_{\lambda}(x)=\cos(\sqrt{\lambda}x)+\frac{\sin(\sqrt{\lambda}x)}{\sqrt{\lambda}}. $$ At $\lambda=0$, the above takes on the correct limiting form, $$ X_{0}(x) = 1 + x. $$ In order for $\lambda$ to be an eigenvalue, it is necessary and sufficient that $$ X_{\lambda}'(1)=X_{\lambda}(1), $$ which leads to the eigenvalue equation $$ -\sqrt{\lambda}\sin(\sqrt{\lambda})+\cos(\sqrt{\lambda})=\cos(\sqrt{\lambda})+\frac{\sin(\sqrt{\lambda})}{\sqrt{\lambda}}. $$ $\lambda=0$ is not an eigenvalue. So, after excluding this case multiplying by $\sqrt{\lambda}$ give an equivalent equation: $$ (\sqrt{\lambda}+1/\sqrt{\lambda})\sin(\sqrt{\lambda})=0, \\ (\lambda+1)\sin(\sqrt{\lambda})=0. $$ The eigenvalues are $$ \lambda = -1,\pi^2,(2\pi)^2,\cdots. $$ The corresponding eigenfunctions are $$ X_{-1}(x) = \cosh(x)+\sinh(x)=e^{x}, \\ X_{n^2\pi^2}(x)= \cos(n\pi x)+\frac{\sin(n\pi x)}{n\pi},\;\; n=1,2,3,\cdots. $$