Consider the Sturm-Liouville problem:
where h > 0.
I need to determine the transcendental equation for the eigenvalues, as well as the "form" of the eigenfunctions. I tried finding the Fourier coefficients, as that has been the typical class process, but I don't know how that relates to the transcendental equation.
Any help/guidance would be appreciated. Thanks.
Best Answer
The non-zero solutions of $X''+\mu X = 0$ with $X(\pi)=0$ can be normalized so that $X'(\pi)=1$. The resoluting solution $X_{\mu}(x)$ has the form $$ X_{\mu}(x) = \frac{\sin(\sqrt{\mu}(x-\pi))}{\sqrt{\mu}}. $$ This is correct even for $\mu=0$ if you take the limiting case as $\mu\rightarrow 0$: $$ X_{0}(x)=\lim_{\mu\rightarrow 0}X_{\mu}(x)=\lim_{\mu\rightarrow 0}\frac{\sin(\sqrt{\mu}(x-\pi)}{\sqrt{\mu}(x-\pi)}(x-\pi)=x-\pi. $$ Using this form of $X_{\mu}(x)$ , the eigenvalue equation becomes \begin{align} 0 & =X_{\mu}(0)-hX_{\mu}'(0) \\ & =\frac{\sin(\sqrt{\mu}(-\pi))}{\sqrt{\mu}}-h\cos(\sqrt{\mu}(-\pi)) \\ & = -\frac{\sin(\sqrt{\mu}\pi)}{\sqrt{\mu}}-h\cos(\sqrt{\mu}\pi). \end{align} The solutions $\mu$ are the zeros of an entire function of $\mu$. Note that $\mu=0$ is a solution iff $h=-\pi$. For $\mu\ne 0$, the the above may be written as the transcendental equation $$ \tan(\sqrt{\mu}\pi)=-h\sqrt{\mu}. $$ If you plot both of these functions, you can graphically see the non-zero values of $\sqrt{\mu}$ that are solutions.