Green's Function: 2-D Laplace vs. (Limit of) 2-D Helmholtz

It is instructive to attempt to solve Eq.(5.63) and its boundary conditions by simply computing the limit of the solution, Eq.(5.66),

$$\lim_{k\to 0}\frac{i}{4} H_0(kR);~~~~~~R=\vert\vec x-\vec x_0\vert~,$$

and compare this limit with the result obtained by directly solving the boundary value

Problem (Green's Function for the Potential of an Isolated Source)

Setting $k=0$ in the previous problem, find the Green's function which satisfies

$$\left[ \frac{1}{r} \frac{\partial}{\partial r} r \frac{\partial}{... ...\right] G_0(\vec x;\vec x_0)= -\frac{\delta(r-r_0)}{r}\delta(\theta-\theta_0)~.$$ (568)

together with

(i)

$$0= \left.\sqrt{r}\vec n\cdot \vec\nabla G_0 (\vec x;\vec x_0)\rig... ..._{\partial R} \equiv \lim_{r\to\infty} \sqrt{r} \frac{\partial}{\partial r}G_0$$

and

(ii)

$G_0(\vec x;\vec x_0)$ is finite at $r=0$ , where $\theta$ is undefined.

Solution: Using the method of dimensional reduction, one again starts with

$$G_0(\vec x;\vec x_0)=\sum_{-\infty}^\infty g_m(r;r_0) \frac{e^{im(\theta-\theta_0)}}{2\pi}$$ (569)

and the implied boundary conditions

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

$$\lim_{r\to \infty} = \sqrt{r}\frac{\partial g_m}{\partial r}=0~.$$

The solution can be summed into the closed form

$$G_0(\vec x;\vec x_0)=-\frac{1}{2\pi}\ln \vert\vec x-\vec x_0\vert \textrm{ whenever } r\ne r_0$$ (570)

How does this Green's function compare with the asymptotic limit of the Helholtz Green's function, Eq.(5.66),

$$\lim_{k\to 0} G_k(\vec x;\vec x_0) ~?$$

The answer is contained in a comparison with the following

Problem (Potential Green's Function as an Asymtotic Limit)

Exhibit the limiting form of Eq.(5.66) as $k\to 0$ .

Solution

Using Property 14 on page ,

$$H^{(1)}_\nu (kR ) = e^{-i\nu\pi/2}~ \frac{e^{-i\nu\pi/2} J_\nu (kR ) - e^{i\nu\pi/2}J_{-\nu}(kR)} {-i\sin\pi\nu}~,$$

one needs to calculate

$$\lim_{k\to 0} H^{(1)}_0(kR )= \lim_{k\to 0}\lim_{\nu\to 0}H^{(1)}_\nu (kR )$$

This is a double limit which one obtains by evaluating

$$\lim_{\nu\to 0}H^{(1)}_\nu (kR )$$

for $kR\ll 1$ , and then by examining the behaviour of the resulting expression as $k\to 0$ .

The evaluation yields

$$\lim_{\nu\to 0}H^{(1)}_\nu (kR )=\frac{0}{0}~.$$

Consequently, l'Hospital's rule

$$\lim_{\nu\to 0}H^{(1)}_\nu (kR ) = \lim_{\nu\to 0} \frac{ \displaystyle \frac{d}{d\nu}\left(e^{-i\nu... ...}J_{-\nu}(kR)\right)} {\displaystyle \frac{d}{d\nu}\left( -i\sin\pi\nu \right)}$$ (571)

$$= \lim_{\nu\to 0} \frac{ e^{-i\nu\pi/2}\left[ \frac{-i\pi}{2} J_\nu... ...} J_{-\nu} (kR) +\displaystyle\frac{dJ_{-\nu}}{d\nu}\right] }{-i\pi\cos \pi\nu}$$

must be used. Taking advantage of the asymptotic small-$k$ expansion of $J_\nu (kR)$ , Eq.(5.11),

$$J_\nu (kR) \approx\left( \frac{kR}{2} \right)^\nu \frac{1}{\Gamma(1+\nu)}~,$$

$$\frac{d}{d\nu}\left( \frac{kR}{2} \right)^\nu =\left( \frac{kR}{2} \right) \ln\left( \frac{kR}{2} \right)~,$$
and

$$\lim_{\nu\to 0}\frac{d}{d\nu}\left(\frac{1}{\Gamma(1+\nu)}\right) = \lim_{\nu\to 0}\frac{-1}{\Gamma(1+\nu)}\times \frac{\Gamma'(1+\nu)}{\Gamma(1+\nu)}$$

$$=-1\times(-)C~;~~~~C=.5772~(\textrm{Euler-Mascheroni constant})$$

$$\equiv \ln\gamma~; ~~~~\gamma=1.781$$
one finds that

$$\frac{dJ_\nu(kR)}{d\nu} =\left( \frac{kR}{2} \right)^\nu\ln\left( \frac{kR}{2} \right) +\ln\gamma$$

$$= \left( \frac{kR}{2} \right)^\nu\ln\left( \frac{\gamma kR}{2} \right)$$

Consequently, l'Hospital's rule tells us that

$$\lim_{\nu\to 0}H^{(1)}_\nu (kR ) = \frac{\left[ \displaystyle\frac{-i\pi}{2}+\ln\displaystyle\frac{\gamma kR}{2} \right]\times 2}{-i\pi}$$

$$= 1+\frac{2i}{\pi}\ln\frac{\gamma kR}{2}$$

Thus the small-$k$ form of the 2-D helmholtz Green's function, Eq.(5.66), is

$$G_k(\vec x;\vec x_0) = \frac{i}{4} H_0(k\vert\vec x-\vec x_0\vert)$$ (572)

$$= -\frac{1}{2\pi}\ln \vert\vec x-\vec x_0\vert+\left( \frac{i}{4} -\frac{1}{2\pi}\ln \frac{\gamma kR}{2} \right) \parbox{1.2in}{whenever $k\ll 1$.}$$

This is the solution to the problem and it expresses the amplitude profile of a membrane responding to a unit force applied at $\vec x_0$ . This membrane is imbedded in an elastic medium whose local force of restitution (per unit area) is proportional to $k^2$ :

$$G_k \times k^2 \Delta (\textrm{area})=\Delta(\textrm{force of restitution})$$

Thus $k^2$ is the Young's modulus of the elastic medium in which the membrane is imbedded. As $k^2\to 0$ there is nothing to push the membrane back towards its zero-amplitude equilibrium. Consequently, the smaller that Young's modulus is, the further the membrane gets pushed away from this equilibrium by the Dirac delta function force density. This is why $G_k(\vec x;\vec x_0)\to \infty$ as $k^2\to 0$ . Equation (5.72) expresses this fact quantitatively.

By contrast $G_0(\vec x;\vec x_0)$ as given by Eq.(5.70) does not presume any elastic medium. The asymptotic Neumann boundary condition that went into $G_0$ forbids it being interpreted as the amplitude of any membrane. Instead $G_0$ expresses the potential due to an electrostatic charge.