Wanted: Rotation Invariant Solutions to the Helmholtz Equation

A plane wave solution $e^{i\vec k\cdot\vec x}$ is also an eigenfunction of the translation operator:

$$X_{a\ast}e^{i\vec k\cdot\vec x} = e^{-ik_xa}e^{i\vec k\cdot\vec x}$$

$$Y_{b\ast}e^{i\vec k\cdot\vec x} = e^{-ik_yb}e^{i\vec k\cdot\vec x}$$
but

$$R_{\gamma\ast}e^{ikr\cos (\alpha -\theta )} \equiv e^{ikr\cos (\alpha - \theta -\gamma )}\not= \lambda e^{ikr\cos (\alpha -\theta )}$$

for any $\lambda !$ . In other words, a plane wave solution is not an eigenfunction of the rotation operator! Nevertheless, we know that $R_\gamma$ is a transformation which takes eigenfunctions of $\nabla^2$ into eigenfucntions belonging to the same eigenvalue. This leads to the following question:

Which linear combination of plane waves (having the same $k^2$ ) is an eigenfunction of $R_\gamma$ ?

We need a solution to the Helmholtz equation of the form

$$\psi =Z(kr)e^{i\nu \theta}~~\qquad~~\qquad~~(\nu =~\textrm{complex constant})$$
so that

$$R_{\gamma\ast}\psi = e^{-i\nu\gamma}\psi~~\qquad~~\textrm{(\lq\lq Rotation eigenfunction~} \psi\textrm{'')}\,.$$

If we can find $Z(kr)$ such that

$$(\nabla^2 +k^2)Z(kr)e^{i\nu\theta} = 0\,,$$

then we shall have what we are looking for, namely a solution which is also an eigenfunction of the rotation operator.

Using the polar representation of $\nabla^2$ , and cancelling out the factor $e^{i\nu\theta}$ , we have

$$\left\{ \frac{d^2}{dr^2} +\frac{1}{r}~\frac{d}{dr} +\left( k^2-\frac{\nu ^2}{r^2} \right)\right\} Z(kr)=0~~,$$

or with $\rho =kr$ ,

$$\left\{\frac{d^2}{d\rho^2}+\frac{1}{\rho}~\frac{d}{d\rho} + \left( 1- \frac{\nu ^2}{\rho^2}\right)\right\} Z(\rho )=0\,.$$

In other words, $Z(\rho )$ must satisfy Bessel's equation.

The first impulse is to solve this equation using infinite series. However, we shall take note of STOKE'S observation: ``series solutions have the advantage of being generally applicable, but are wholly devoid of elegance''. In our case ``elegance'' means ability to capture the geometric and physical properties of the Euclidean plane.

Lecture 40

Instead of a series solution, we shall take the question on the previous page seriously and construct an appropriate superposition of plane wave solutions, namely

$$\psi =\int^{\alpha_2}_{\alpha_1}e^{ikr\cos (\alpha -\theta )}e^{i\nu\alpha}d\alpha ~~.$$ (52)

Is this solution invariant (i.e. gets changed only by a constant phase factor)? To find out, let $\alpha=\overline{\alpha} + \theta$ so that

$$\psi=\underbrace{\int^{\overline{\alpha}_2 =\alpha_2-\theta}_{\ov... ...\overline{\alpha}}d\overline{\alpha}}_{\displaystyle Z(kr)} e^{i\nu\theta}\,.$$

This superposition has the desired form

$$Z(kr)e^{i\nu\theta}$$

provided the effect of the $\theta$ -dependence in the integration limits can be removed. In other words, expression (5.2), which is a solution of

$$0=\nabla^2+k^2=\left(\frac{1}{r}~\frac{\partial}{\partial r}r\fra... ...partial r}+\frac{1}{r^2}~\frac{\partial^2}{\partial\theta^2}+k^2\right)\psi~~,$$

is an eigenfunction of $L_\theta =\frac{1}{i}~\frac{\partial}{\partial\theta}$ if

$$Z\equiv\int^{\alpha_2-\theta}_{\alpha_1-\theta} e^{ikr\cos\overline{\alpha}} e^{i\nu\overline{\alpha}}d\overline{\alpha}$$

can be shown to be independent of $\theta$ . In that case $Z=Z(kr)$ , and it necessarily satisfies

$$\left[\frac{1}{r}~\frac{d}{dr}r\frac{d}{dr}+k^2-\frac{\nu^2}{r^2}\right] Z(kr) =0\,,$$

which is Bessel's equation, with $\nu$ equal to any complex constant.

Let us, therefore, consider more closely the complex line integral

$$Z_\nu (\rho )=\int_C e^{i\rho\cos\alpha +i\nu\alpha}d\alpha\,$$

Here we assume, for the time being, that $\rho =\vert\rho\vert$ because

$$\rho =kr\,,$$

a product of two positive numbers. The integration contour $C$ is a curve in the complex $\alpha$ -plane, whose points are

$$\alpha = p+iq~~\qquad~~p,q~\textrm{real}\,.$$

We shall find that the chosen integration contour will start far away from the origin at a point with large positive or negative imaginary part, $q=\pm\infty$ , and terminate at another such point, again with $q=\infty$ or $q=-\infty$ . This choice has a dual purpose. (i) It guarantees, as we shall see, that the contour integral will be independent of the real angle $\theta$ , which is the amount by which the two end points get shifted horizontally in the complex $\alpha$ -plane, and (ii) it guarantees, as we shall see, that the integral converges. The value of the integral itself is independent of the integration path because the integrand is analytic in the whole complex $\alpha$ -plane.

Where shall the starting and termination points of the contour integral be located? This question is answered by the asymptotic behaviour of the dominant terms in the exponent of the integrand,

$$i\cos\alpha = i\cos (p+iq)$$

$$= i\cos p\cosh q+\sin p\sinh q\,.$$

When the real part of this expression becomes large and negative $(\sin p \sinh q\to -\infty )$ , then the convergence of the integral will be guaranteed because in that case the term $\sin p \sinh q$ dominates over all other contributions to the exponent of the integrand. This is true for all complex numbers $\nu$ .

Figure 5.3: Contour integration paths $C_1$ and $C_2$ for the two Hankel functions $H^{(1)}_\nu$ and $H^{(2)}_\nu$ . The shaded regions are the regions of convergence as $q\to \pm \infty$ .

The integration contour we choose has endpoints which lie far in the upper $\alpha$ -plane or in the lower plane ( $q\to\infty$ or $q\to-\infty$ ).

To obtain an integral which converges, one must have $\sin p\sinh q\to -\infty$ at both endpoints. This implies that if $q\to\infty$ , then the value of $p$ must satisfy

$$\sin p<0\,,~~\quad~~\textrm{i.e.,}~~-\pi <p<0~~\mod 2\pi\,.$$

On the other hand, if $q\to-\infty$ , then the value of $p$ must satisfy

$$0<\sin p\,,~~\qquad~~\textrm{i.e.,}~~0<p<\pi~~\mod 2\pi\,.$$

Thus the integration contour can start and terminate only in one of the shaded regions in the complex $\alpha$ -plane of Figure 5.3.

There are only two basic contour integrals that one needs to consider, and they give rise to the two kinds of fundamental functions. They are $H^{(1)}_\nu (\rho )$ , the Hankel function of the first kind , and $H^{(2)}_\nu (\rho )$ , the Hankel function of the second kind. All other integration contours give rise to contour integrals which merely are linear combinations of these two fundamental functions.

Moving forward, we shall use in the next subsection these two functions to deduce $23$ of their mathematical wave mechanical properties and applications.

Exercise 51.1 (DIFFERENT INTEGRATION CONTOUR)
Evaluate the integral

$$\int\limits_Ce^{i\rho\cos\alpha+i\nu\alpha}d\alpha$$

along the curve $C$ (in the complex $\alpha$ -plane below) in terms of the two kinds of Hankel functions $H^{(1)}_{\nu}(\rho)$ and $H^{(2)}_{\nu}(\rho)$

Exercise 51.2 (STRIPS OF CONVERGENCE)
In the complex $\beta$ -plane determine those semi-infinite strip regions where the line integral

$$\int\limits_Ce^{i\rho\cosh\beta-i\omega\beta}d\beta$$

converges if the integration limits of the integration path $C$ are extended to infinity in each of a pair of such strips.

Exercise 51.3 (HANKEL FUNCTION AS A DEFINITE INTEGRAL)
By slightly deforming the integration path prove or disprove that the integral

$$\int\limits_{-\infty}^{\infty}e^{i\rho\cosh\beta-i\omega\beta}d\beta$$

can be expressed in terms of a Hankel function. Which kind and which order?

Exercise 51.4 (WAVE EQUATION IN PSEUDOPOLAR COORDINATES)
Instead of applying

$$x = r\cos\theta$$

$$y = r\sin\theta$$

to the Helmholtz equation

$${\partial^2\psi\over{\partial x^2}}+{\partial^2\psi\over{\partial y^2}}+k^2\psi=0$$

to obtain

$$\frac{1}{r} \frac{\partial}{\partial r}r {\partial \psi \over \partial r} +{1\over r^2} {\partial^2 \psi \over \partial \theta^2} +k^2\psi=0~~~,$$

apply

$$t = \xi\cosh\tau \qquad~~~~~~ 0<\xi<\infty$$

$$z = \xi\sinh\tau \qquad -\infty<\tau<\infty$$

to the wave equation

$$-{\partial^2\psi\over\partial t^2}+{\partial^2\psi\over\partial z^2} -k^2\psi=0$$ (53)

in order to obtain the wave equation relative to the coordinates $\xi$ and $\tau$ . To do this, take advantage of the fact that letting

$$r = \xi$$

$$\theta = i\tau$$

and

$$x = t$$

$$y = iz$$

yields the hyperbolic transformation and the wave equation (5.3).

a)

Write down the wave equation in terms of the (``pseudo'') polar coordinates $\xi$ and $\tau$ .

b)

Consider a solution which is a (``pseudo'') rotation eigenfunction $\psi_{\omega}$ :

$${\partial\psi_{\omega}\over{\partial\tau}}=-i\omega\psi_{\omega}$$

and determine the differential equation

$$[\alpha(\xi){d^2\over{d\xi^2}}+\beta(\xi){d\over{d\xi}}+\gamma(\xi)]\psi_{\omega}=0$$

it satisfies.

c)

Verify that the translation (in the $t,z$ -plane) eigenfunction

$$\psi=e^{-i(k_0t-k_zz)}$$

is a solution to the wave Eq.(5.3) whenever the two constants $k_0$ (``frequency'') and $k_z$ (``wave number'') satisfy the dispersion relation

$$k_0^2 -k_z^2=k^2 ~~.$$

Then, using $k_0=k\cosh \alpha$ , $k_z=k\sinh\alpha$ (with $k>0$ ) and $t=\xi\cosh\tau,~~z=\xi\sinh\tau$ , and the hyperbolic angle addition formula, rewrite the phase and hence the wave function $\psi$ in terms of $\xi$ and $\tau$ .

d)

Construct a superposition (as an integral over $\alpha$ ) of waves $\psi$ which is a (``pseudo'') rotation eigenfunction, i.e. satisfies

$${\partial\psi_{\omega}\over{\partial\tau}}=-i\omega\psi_{\omega},$$

where $\psi_{\omega}$ is that superposition.

e)

Exhibit two independent solutions $\psi_\omega$ to Eq.(5.3) corresponding to two different integration contours. What are they? If your solutions are proportional to Hankel functions, specify what kind, and identify their order.