Spectrum via Green's Function

In order to evaluate the contour integral of $G_\lambda(x;\xi)$ one must know its singular points in the complex $\lambda$ -plane. It is clear that on this domain the Green's function has the form

$$G_\lambda(x;\xi)=\frac{g(\lambda)}{c(\lambda)} ~~,$$

where both $g(\lambda)$ and $c(\lambda)$ are analytic for all $\lambda$ , even though each one depends manifestly on the nonanalytic function $\sqrt{\lambda}$ . Thus the singular points of $G_\lambda$ are located at the zeroes of $c(\lambda)$ , the eigenvalues of the Sturm-Liouville system:

$$c(\lambda)\equiv \sqrt{\lambda}\sin \sqrt{\lambda}\ell=0~~\Rightarrow~~ \lambda_n=\left(\frac{n\pi}{\ell}\right)^2,~~n=0,1,2,\cdots~.$$

At these points $c'(\lambda_n)\ne 0$ . Consequently, $\lambda =\lambda_n$ is a simple pole in whose neighborhood the ratio $g/c$ has the expansion

$$G_\lambda=\frac{g(\lambda)}{c(\lambda)} =\frac{\alpha_0}{\lambda -\lambda_n} +\alpha_1+\alpha_2(\lambda -\lambda_n)+\cdots~~.$$

Here $\alpha_0$ is the residue of $G_\lambda$ , and one must find it. To do this, consider

$$\frac{g(\lambda)}{c(\lambda)}(\lambda -\lambda_n) =\alpha_0 +\alpha_1(\lambda -\lambda_n)+\alpha_2(\lambda -\lambda_n)^2+\cdots$$

and take the limit. Thus

$$\alpha_0 = \lim_{\lambda \to \lambda_n} \frac{g(\lambda)(\lambda -\lambda_n)}{c(\lambda)}$$

$$= \lim_{\lambda \to \lambda_n} \frac{g(\lambda)}{c'(\lambda)}$$

$$= \frac{g(\lambda_n)}{c'(\lambda_n)}$$

where the second step used L'Hospital's rule. The residue of $G_\lambda$ is therefore

$$\begin{array}[t]{c} \textrm{Res}\\ \lambda =\lambda _n \end{array}G_\lambda (x;\xi)=\frac{g(\lambda_n)}{c'(\lambda_n)}$$

Its evaluation is based on the following expressions

$$g(\lambda) = -\cos \sqrt{\lambda} x_< ~ \cos \sqrt{\lambda} (x_> -\ell)$$

$$c(\lambda) = \sqrt{\lambda}\sin \sqrt{\lambda}\ell$$

$$c(\lambda_n) = 0 ~~\rightarrow~~\sqrt{\lambda_n}=\frac{n\pi}{\ell},~~ n=0,1,2,\cdots$$

$$c'(\lambda_n) = \frac{1}{2\sqrt{\lambda}} \sin\sqrt{\lambda}\ell~+~ \left. \sqrt{... ...rt{\lambda}\ell) \frac{\ell}{2\sqrt{\lambda}} \right\vert _{\lambda =\lambda_n}$$

$$~ = \frac{\ell}{2}\cos n\pi~~\textrm{when}~~\lambda_n \ne 0$$

$$~ = \frac{\ell}{2}+ \frac{\ell}{2} ~~\textrm{when}~~\lambda_n = 0$$

It follows that $G_\lambda$ has a closed contour integral given by

$$\frac{1}{2\pi i} \oint_C G_\lambda (x;\xi)~d\lambda = -\frac{1}{\ell}-\sum_{n=1}^\infty \left. \frac{\cos \sqrt{\lambda... ...da} (x_> -\ell) } {c'(\lambda)} \right\vert _{\sqrt{\lambda}=\frac{n\pi}{\ell}}$$

$$= -\frac{1}{\ell}-\sum_{n=1}^\infty \frac{\cos \frac{n\pi}{\ell} x_< ~ \cos \frac{n\pi}{\ell} (x_> -\ell) } {\frac{\ell}{2}\cos n\pi }$$

$$= -\frac{1}{\ell}- \frac{2}{\ell}\sum_{n=1}^\infty \frac{\cos \frac{n\pi}{\ell} x_< ~ \cos \frac{n\pi}{\ell} x_> \cos n\pi } {\cos n\pi }$$

$$= -\frac{1}{\ell}- \frac{2}{\ell}\sum_{n=1}^\infty \cos \frac{n\pi}{\ell} x ~ \cos \frac{n\pi}{\ell} \xi$$ (432)

Compare this bilinear expression with the fundamental formula, Eq.(4.29) on page , and read out the complete set of orthonormalized eigenfunctions

$$\{u_m(x)\}=\left\{ \sqrt{\frac{1}{\ell}},\sqrt{\frac{2}{\ell}}\cos\frac{m\pi}{\ell}x~~ :~~~m=1,2,\cdots \right\}~~.$$