Green's Function via Dimensional Reduction

To find the Green's function whose domain dimension is two or higher, introduce a technique whose virtue is that it reduces the problem to a Green's function problem in just one dimension. The potency of this technique is a consequence of the fact that it leads to success whenever the Helmholtz equation is separable relative to the curvilinear coordinate system⁵⁴ induced by the boundarie(s) of the given domain.

Polar coordinates is a case in point. It is illustrated by the following

Problem (Green's Function for Radiation in the Euclidean Plane) The equation for the Green's function relative to polar coordinates is

$$\left[ \frac{1}{r} \frac{\partial}{\partial r} r \frac{\partial}{... ...\right] G_k(\vec x;\vec x_0)= -\frac{\delta(r-r_0)}{r}\delta(\theta-\theta_0)~.$$ (564)

Let the homogeneous boundary conditions for $G$ be

(i)

Sommerfeld's outgoing radiation condition

$$0= \left.\sqrt{r}\left( a(\vec x)G_k(\vec x;\vec x_0)+ \vec n\cdo... ...\lim_{r\to\infty} \sqrt{r} \left( (-)ik G+\frac{\partial}{\partial r}G \right)$$

(ii)

$G_k(\vec x;\vec x_0)$ is finite at $r=0$ , where $\theta$ is undefined.

Solution

This problem is solved by expanding the Green's function as a Fourier series on $[0,2\pi ]$ :

$$G_k(\vec x;\vec x_0)=\sum_{-\infty}^\infty g_m(r;r_0) \frac{e^{im(\theta-\theta_0)}}{2\pi}$$ (565)

with to-be-determined Fourier coefficients. The method of dimensional reduction consists of establishing that each of them satisfies a 1-dimensional Green's function problem. Next one constructs its solution using formula Eq.(4.19) on page . Finally one introduces this solution into the Fourier series expansion. This yields the desired 2-dimensional Green's function. As an additional benefit one finds that this expansion can be summed into a closed form expression given by a familiar function.

The details of this four step procedure are as follows: Introduce the Fourier expansion into the 2-dimensional Green's function equation and obtain

$$\sum_{-\infty}^\infty \frac{e^{im(\theta-\theta_0)}}{2\pi} \left[... ...2}\right)\right] g_m(r;r_0) = -\frac{\delta(r-r_0)}{r}\delta(\theta-\theta_0)~.$$ (566)

To isolate the equation obeyed by the each of the coefficient functions $g_m(r;r_0)$ introduce the Fourier representation of the Dirac delta function restricted to $[0,2\pi ]$ :

$$\delta(\theta-\theta_0)= \sum_{-\infty}^\infty \frac{e^{im(\theta-\theta_0)}}{2\pi}~,$$

and make use of the linear independence of the functions $e^{im(\theta-\theta_0)}$ . Alternatively, multiply both sides of Eq.(5.66) by $e^{-im'(\theta-\theta_0)}$ , integrate $\int e^{-im'(\theta-\theta_0)}(\cdots)d\theta$ , make use of orthogonality, and finally drop the prime to obtain

$$\left[ \frac{d}{d r} r \frac{d}{d r} +\left(r k^2 -\frac{m^2}{r}\right)\right] g_m(r;r_0)~, = -\delta(r-r_0)$$

the equation for the 1-dimensional Green's function. The boundary conditions for $G$ imply that the solution

$$g_m(r;r_0)=\frac{-1}{c} \left\{ \begin{array}{ccl} u_1(r) u_... ... u_1(r_0) u_2(r) & \textrm{for} & r_0 < r \end{array} \right.$$

satisfies

$$g_m(r=0;r_0) = finite$$

$$g_m(r;r_0) \sim \frac{e^{ikr}}{\sqrt{r}}~\textrm{for very large }r$$

This is a set of 1-dimensional Green's function problems whose solutions yield the 2-d Green's funtion, Eq.(5.65). The two functions which satisfy the homogeneous differential equation and the respective boundary conditions are

$$u_1 =J_m(kr)$$
and

$$u_2 =H_m^{(1)}(kr)~,$$

while their Wronskian is

$$u_1u_2'-u_1'u_2=\frac{2i}{\pi r}~.$$

Consequently, the 1-dimensional Green's function is

$$g_m(r;r_0)=\left\{\begin{array}{c} \displaystyle \frac{\pi i}{2} ... ...laystyle \frac{\pi i}{2} J_m(kr_0)H_m^{(1)}(kr)~~~r_0\le r \end{array} \right.$$

The 2-dimensional Green's function, Eq.(5.65), for outgoing radiation in the Euclidean plane is therefore

$$G_k(\vec x;\vec x_0)=\sum_{-\infty}^\infty \frac{i}{4} e^{im(\theta-\theta_0)} J_m(kr_<)H_m^{(1)}(kr_>)~.$$

This expression can be simplified by means of the displacement formula for cylinder modes, Property 19, on page ,

$$H_\nu(k\vert\vec x-\vec x_0\vert)e^{i\nu(\Omega-\theta_0)}= \sum_{m=-\infty}^\infty e^{i(\nu+m)(\theta-\theta_0)} J_m(kr_0) H_{\nu+m}(kr)$$

Set $\nu=0$ , compare the Green's function with the right hand side of the displacement formala, and conclude that

$$G_k(\vec x;\vec x_0)=\frac{i}{4} H_0(k\vert\vec x-\vec x_0\vert)~;$$ (567)

in other words,

$$\boxed{ \left( \nabla^2 +k^2\right) \frac{i}{4} H_0(k\vert\vec x-\vec x_0\vert) =-\delta^2(\vec x-\vec x_0) }$$ (568)

Thus one has obtained an expression for the 2-dimensional Green's function which exhibits the rotational and translational symmetry of the linear system. It represents an asymtotically (large $\vert\vec x\vert$ !) outgoing wave whose source is located at $\vert\vec x_0\vert$ . This is the amplitude profile of a wave that you make when you stick your wiggling finger at $\vert\vec x_0\vert$ into an otherise motionless pond.

Footnotes

... system⁵⁴

In three dimensions the Helmholtz equation is separable in eleven coordinate systems. They are listed and depicted at the end of chapter five of reference [#!Morse_SPMamp_Feshbach5!#]