Spectral Resolution of the Green's Function

Instead of representing the Green's function $G_\lambda(x;\xi)$ of a system in terms of two solutions to the homogeneous differential equation, as by Eq.(4.19), we shall now represent it as a generalized Fourier series. The Green's function is a solution to the inhomogeneous boundary value problem

$$\left( \frac{1}{\rho(x)}~\left[ -\frac{d}{dx} p(x) \frac{d}{dx}+q(x)\right] ~-\lambda \right) u=f(x)$$

or

$$({\mathcal L}-\lambda)u=f$$ (424)

with

$$B_1(u) =$$ 0

$$B_2(u) =$$ 0

Here $\rho (x)$ is the weight function of the Sturm-Liouville differential equation.

We are led to the spectral representation of $G_\lambda$ by the following three-step line of reasoning:

(1) The non-trivial solutions to the homogeneous boundary value problem are the eigenfunctions, $u_n(x)~~~n=1,2,\cdots,$ each of which satisfies

$$\left. \begin{array}{rcl} {\mathcal L}u_n&=&\lambda u_n \\ B_1(u_n)&=&0\\ B_2(u_n)&=&0 \end{array}\right\} ~~~~n=1,2,\cdots$$

These eigenfunction are used to represent the solution $u(x)$ as a generalized Fourier series,

$$u=\sum _{m=1}^\infty c_m u_m ~~,$$

whose coefficients $c_n$ are to be determined. To obtain them, take the inner product of Eq.(4.24) with these eigenfunctions $u_n$ ,

$$\langle u_n ,{\mathcal L}u \rangle -\lambda \langle u_n, u \rangle= \langle u_n,f \rangle$$ (425)

(2) For illustrative purposes consider the case where ${\mathcal L}$ is self-adjoint with respect to the inner product

$$\langle u_n,u\rangle \equiv \int_a^b \overline{u_n}(x)\, u(x)\, \rho(x)\, dx~$$

Consequently, the first term of Eq.(4.25) is

$$\langle u_n ,{\mathcal L}u \rangle = \langle {\mathcal L}u_n ,u \rangle$$

$$~ = \lambda _n \langle u_n ,u \rangle$$

$$~ = \lambda _n c_n~~.$$

The equation itself becomes

$$(\lambda _n -\lambda ) c_n=\langle u_n,f \rangle$$

and thus determines the coefficient $c_n$ . Thus the Fourier series representation of the solution is

$$\boxed{ u (x) =\sum_{n=1}^\infty u_n\frac{\langle u_n,f \rangle}{\lambda _n -\lambda} }$$

(3) This representation can be applied to the Green's function. Letting $f(x)=\delta(x-\xi)/\rho(x)$ , one obtains

$$\langle u_n,\frac{\delta(x-\xi)}{\rho(x)}\rangle = \overline u_n(\xi) \rangle~~.$$

Thus the Fourier series becomes

$$\boxed{ G_\lambda (x;\xi) =\sum_{n=1}^\infty \frac{u_n(x)\overline {u}_n (\xi )}{\lambda _n -\lambda } }$$ (426)

This is the spectral representation of the Green's function.

For the purpose of comparison consider the spectral representation of the identity. It is obtained from the Fourier representation of a generic function satisfying the given boundary conditions,

$$u(x) = \sum_{n=1}^\infty u_n(x) c_n$$

$$~~ = \sum_{n=1}^\infty u_n(x) \int_a^b \overline u_n(\xi) \rho u(\xi)~d\xi$$

$$~~ = \int_a^b \sum_{n=1}^\infty u_n(x) \rho \overline u_n(\xi) u(\xi)~d\xi~~.$$

This holds for any function $u$ . Consequently, the expression in front of $u(\xi)$ is the Dirac delta,

$$\sum_{n=1}^\infty u_n(x) \rho \overline u_n(\xi) =\delta(x-\xi)$$

or

$$\boxed{ \frac{\delta (x-\xi)}{\rho (\xi)} = \sum^\infty_{n=1}u_n(x)\overline{u}_n(\xi) }$$

This is the spectral representation of the identity operator in the Hilbert space, the same as Eq.(1.14) on page , which was obtained by essentially the same line of reasoning. Note the perspicuous similarity with the Green's function $G_\lambda(x;\xi)$ , the resolvent of the Sturm-Liouville operator in this Hilbert space.