Infinite String

Even though in nature one never observes an infinite string, such a string is a concept with properties which are directly observable and which lend themselves to easy mathematical analysis. This means that the infinite string is a natural way by which to grasp the properties and behavior of any system which exhibits these attributes. All the essential properties of a string are contained in the solution to the following

Problem:  Construct the Green's function for the system

$$\frac{d^2 G_\lambda}{dx^2} +\lambda G_\lambda =-\delta (x-\xi )~~\qquad~~ 0<x<\infty$$

subject to

1. $G_\lambda (0;\xi )=0$
2. $G_\lambda(x;\xi)$ expresses an ``outgoing'' wave for very large $x$ , i.e., $G_\lambda\sim e^{i\lambda^{1/2}x}$ .
3. $G_\lambda(x;\xi)$ is square-integrable.

Comment

Such a boundary value problem arises in the solution to a vibrating semi-infinite string which is imbedded in an elastic medium, and which responds to a harmonically varying force:

$$\frac{\partial^2\psi}{\partial x^2} - \frac{\partial^2\psi}{\partial t^2} -k^2\psi = -f(x)e^{-i\omega t}\,.$$

The steady state solution to this system is

$$\psi (x,t) = u(x)e^{-i\omega t}$$

where

$$\frac{d^2 u}{dx^2}+\lambda u(x) = -f(x)$$

with $\lambda =\omega^2 -k^2$ . If the harmonic driving force is localized to a point, then the solution is

$$\psi (x,t) = G_\lambda (x;\xi )e^{-i\omega t}\,.$$

Being square integrable, for large ($\xi<x$ ) the solution is

$$\psi\sim e^{\pm i\sqrt\lambda x} e^{-i\omega t}\,,$$

where

$$\sqrt \lambda = \pm \vert a\vert+i\epsilon \quad \textrm{with}\quad 0<\epsilon \lll 1~.$$

It is evident that the upper sign expresses an outgoing wave and the lower sign an incoming wave. This is because the locus of the constant amplitude

$$\pm\vert a\vert x-\omega t=\textrm{const}$$

is a point with phase velocity

$$\frac{dx}{dt} =\pm \frac{\omega}{\vert a\vert}\,.$$

The upper sign refers to a wave moving towards larger $x$ . It is an outgoing wave. The difference between an outgoing and an ingoing wave is the difference between the driving force emitting and absorbing wave energy.

The Green's function is constructed in the usual way:

$$G_\lambda (x;\xi )=\frac{-1}{c} u_1 (x_<)u_2(x_>)$$

where

$$u_1(x) = \sin\lambda^{1/2}x~~\qquad~~(u_1(0)=0)$$

$$u_2(x) = \exp i\lambda^{1/2}x~~\quad~~(\textrm{outgoing~b.c.}:\lambda^{1/2}=\vert a\vert+i\epsilon )$$

$$c = u_1u'_2-u'_1u_2$$

$$= \lambda^{1/2}(i\sin\lambda^{1/2}x-\lambda^{1/2} \cos\lambda^{1/2}x)e^{i\lambda^{1/2}x}$$

$$= -\lambda^{1/2}\,.$$

Thus the Green function is

$$G_\lambda (x;\xi )=\frac{\sin\lambda^{1/2} x_<}{\lambda^{1/2}} e^{i\lambda^{1/2}x_>}~.$$

Here

$$\lambda^{1/2}=\left\{ \begin{array}{r} \vert\lambda\vert^{1/2}+i\... ...t^{1/2}+i\epsilon\textrm{ for \lq\lq incoming'' wave at large }x \end{array}\right.$$

From the perspective of physics, a non-zero but neglegible $\epsilon$ expresses the presence of damping. The exponential damping factor $e^{-\epsilon x}$ gurantees that $G_\lambda$ is square integrable on $[0,\infty)$ , but it has no effect on the shape of the Green's function because $\epsilon \to 0$ . It can thus be written as

$$G^{\textrm{outgoing}}_\lambda (x;\xi )=\frac{\sin\lambda^{1/2}x_<}{\lambda^{1/2}} e^{(i\sqrt{\lambda}-0^+) x_>}\,,$$

while

$$G^{\textrm{incoming}}_\lambda (x;\xi )=\frac{\sin\lambda^{1/2}x_<}{\lambda^{1/2}} e^{(-i\sqrt{\lambda}-0^+) x_>}\,.$$

The branch cut of $\lambda ^{1/2}$ has a significant effect on the exponential. However, the sinc function remains uneffected because it is an analytic function of $\lambda$ .