More Properties of Hankel and Bessel Functions

Plane waves, i.e. disturbances with planar wave fronts, can be subjected to translations in the Euclidean plane. They can also be used as basis functions for the two-dimensional Fourier transform. Both of these features extend to cylinder harmonics. The first one is captured by Property 19, the second one by Eq.(5.43) of Property 21. An example of a problem which uses the translation property for cylinder harmonics is a scattering problem similar to the one on page :

Consider a cylindrical source of waves and some distance away from it there is a scatterer also cylindrical in shape. Given the distance between these two cylinders, find the scattered wave field.

Figure 5.15: Scattering of a cylindrical disturbance by a cylinder. A cylindrical wave $\psi _{incident}$ emanating from a source on the left gives rise in the presence of a cylindrical boundary to a circular scattered wave $\psi _{scattered}$ on the right.

Property 19 (Addition theorem for cylinder harmonics)
A displaced cylinder harmonic is a linear superposition of the undisplaced cylinder harmonics. Mathematically one states this fact by the equation

$$H_\nu(kR)e^{i\nu(\Omega-\theta_0)}=\sum_{m=-\infty}^\infty J_m(kr_0) H_{\nu+m}(kr)e^{i(\nu+m)(\theta-\theta_0)}~~.$$ (534)

This equation is also known as the ``addition theorem'' for cylinder harmonics, be they singular or non-singular at the origin $R=0$ . The geometrical meaning of this theorem is as follows: Consider a displacement in the Euclidean plane by the vectorial amount $\vec x_0$ and express this displacement in terms of polar coordinates:

$$\vec x_0: ~ x_0+iy_0=r_0e^{i\theta_0}~~.$$

Next, consider a point of observation, also expressed in terms of polar coordinates,

$$\vec x: ~ x+iy=re^{i\theta}~~.$$

Finally, consider this same point of observation, but relative to the displaced origin at $\vec x_0$ . In terms of polar coordinates one has

$$\vec X\equiv \vec x -\vec x_0:~ X+iY = (x-x_0) +i(y-y_0)$$

$$Re^{i\Omega} = re^{i\theta}-r_0e^{i\theta_0} =\vert \vec x -\vec x_0\vert e^{i\Omega}~~,$$ (535)

where

$$R\cos \Omega = r\cos \theta -r_0\cos \theta_0\equiv x -x_0$$

$$R\sin \Omega = r\sin \theta -r_0\sin \theta_0\equiv y -y_0$$

$$R^2 = r^2+r_0^2 -2rr_0\cos (\theta-\theta_0)$$

$$e^{2i\Omega}=\frac{Re^{i\Omega}}{Re^{-i\Omega}} = \frac{re^{i\theta}-r_0e^{i\theta_0}}{re^{-i\theta}-r_0e^{-i\theta_0}}$$

are the observation coordinates relative to the displaced origin.

Figure 5.16: Displaced cylinder harmonic and its displaced coordinate system. The observation point is labelled (``coordinatized'') in two different ways; by $(r,\theta )$ and by $(R,\Omega )$ . The dotted vector is the displacement vector $\vec x_0:~ r_0e^{i\theta_0}$ .

The problem is this: express a typical displaced cylinder harmonic,

$$H_\nu(kR)e^{i\nu\Omega}= H_\nu(k\vert \vec{x}-\vec{x}_0\vert)e^{i\nu\Omega}$$

a solution to the Helmholtz equation, in terms of the undisplaced cylinder harmonics,

$$H_\nu(kr)e^{i\nu\theta}=H_\nu(k\vert \vec{x}\vert)e^{i\nu\theta},~~ \vert \vec{x}\vert =\sqrt{x^2+y^2}~~,$$ (536)

which are also solutions to the same Helmholtz equation.

The solution to this problem is given by the ``addition theorem'', Eq.(5.34).

It is interesting to note that both $R$ and $\Omega$ , and hence $H_\nu(kR)e^{i\nu\Omega}$ are periodic functions of $\theta$ . Indeed, one notices that

$$\vec X=\vec x-\vec x_0$$

or, equivalently, that

$$Re^{i\Omega} =re^{i\theta}-r_0e^{i\theta_0}~~.$$
As a consequence, the old and the new polar coordinates are related by

$$R^2 =r^2+r_0^2-2rr_0\cos(\theta-\theta_0)$$
and

$$e^{2i\Omega} =\frac{Re^{i\Omega}}{Re^{-i\Omega}}= \frac{re^{i\theta}-r_0e^{i\theta_0}}{re^{-i\theta}-r_0e^{-i\theta_0}}\, .$$

Thus one is confronted with the problem of finding the Fourier series of the periodic function

$$H_\nu(kR)e^{i\nu(\Omega-\theta_0)}= H_\nu\left(k\sqrt{r^2+r_0^2 -... ...rac{r_0-re^{i(\theta-\theta_0)}}{r_0-re^{-i(\theta-\theta_0)}} \right)^{\nu/2}$$

The solution to this problem is given by the ``addition theorem'', Eq.(5.34). We shall refrain from validating this Fourier series by a frontal assault. Instead, we give a simple three-step geometrical argument. It accomplishes the task of expressing the displaced cylinder harmonics in terms of the undisplaced cylinder harmonics

(i)

Represent the displaced harmonic as a linear combination of plane waves in the usual way

$$H_\nu(kR)e^{i\nu\Omega}= \frac{e^{-i\nu\pi /2}}{\pi} \int e^{ikR\cos (\alpha -\Omega )} e^{i\nu\alpha}d\alpha ~~,$$ (537)

(ii)

take each of these plane waves and reexpress them relative to the undisplaced origin:

$$e^{ikR\cos (\alpha -\Omega )}\equiv e^{i\vec k \cdot \vec X}= e^{... ...cdot (\vec x -\vec x_0)}= e^{-i\vec k \cdot \vec x_0} e^{i\vec k \cdot \vec x}$$

The phase shift factor is a plane wave amplitude in its own right, which depends periodically on the angel $\theta_0$ , and is therefore, according to Property 18, a linear combination of Bessel harmonics

$$e^{-i\vec k\cdot\vec x_0} \equiv e^{-ikr_0\cos (\alpha -\theta_0)}$$

$$= \sum^\infty_{m=-\infty} J_m(kr_0) e^{im\alpha} e^{-im\left(\theta_0 +\frac{\pi}{2}\right)}\,.$$

(iii)

Reintroduce the translated plane wave

$$e^{i\vec k\cdot\vec x}=e^{ikr\cos(\alpha-\theta)}$$

and its concomitant phase shift factor $e^{-i\vec k\cdot\vec x_0}$ from step (ii) into the displaced cylinder harmonic. The result is a linear sum of phase shifted cylinder harmonics, Eq.(5.37),

$$H_\nu (kR)e^{i\nu\Omega} \equiv \frac{e^{-i\nu\pi /2}}{\pi} \int e^{i\vec k \cdot\vec x} e^{-i\vec k \cdot\vec x_0}e^{i\nu\alpha} d\alpha$$

$$= \frac{e^{-i\nu\pi /2}}{\pi} \sum^\infty_{m=-\infty} J_m(kr_0) \in... ...a )} e^{i(\nu +m)\alpha} ~d\alpha ~e^{-im\left( \theta_0+\frac{\pi}{2} \right)}$$

$$= \sum^\infty_{m=-\infty} J_m(kr_0) \underbrace{ \frac{e^{-i(\nu+m)... ... +m)\alpha} ~d\alpha }_{H_{\nu +m}(kr)e^{i(m+\nu )\theta}} ~e^{-im\theta_0}\, .$$

According to the definitions, Eqs.(5.4)-(5.5), the integral is a cylinder harmonic of order $\nu +m$ . Consequently, one obtains

$$H_\nu (kR)e^{i\nu\Omega} = \sum^\infty_{m=-\infty} J_m (kr_0)H_{\nu +m} (kr)e^{i(m+\nu )\theta} e^{-im\theta_0}\,.$$

Multiplying both sides by $e^{-i\nu\theta_0}$ yields the following geometrically perspicuous result:

$$H_\nu (kR)e^{i\nu (\Omega -\theta_0)} = \sum^\infty_{m=-\infty} J_m(kr_0) H_{\nu +m}(kr)e^{i(m+\nu )(\theta-\theta_0)}$$

Note that the left hand side is a displaced cylinder harmonic of order $\nu$ relative to the new $x$ -axis which point along the displacement vector $\vec x_0$ and whose origin lies along the tip of this vector. The angle $\Omega -\theta_0$ is the new angle of observation relative to the new tilted $x$ -axis and the new origin.

The sum on the right is composed of the cylinder harmonics of order $\nu +m$ undisplaced relative to the tilted $x$ -axis. The angle $\theta -\theta_0$ is the old angle of observation relative to the tilted $x$ -axis and the old origin.

The displacement formula can be summarized as follows

$$\left( \begin{array}{c} \textrm{displaced~wave}\\ \textrm{o... ...placed~wave}\\ \textrm{of~order~}\nu+m \end{array}\right) \,.$$

Property 20 (Translations represented by cylinder harmonics)
It is amusing to specialize to the case where $\nu=n$ is an integer and $H_\nu=J_n$ is a Bessel function of integral order $n$ . In that case the displacement formula becomes

$$J_n (kR)e^{in\Omega} = \sum^\infty_{m=-\infty} \left[ J_{n+m} (kr) e^{i(n+m)\theta} \right] J_{m} (kr_0)e^{-im\theta_0}$$

$$= \sum^\infty_{m=-\infty} \left[ J_{n-m} (kr) e^{i(n-m)\theta} \right] J_{m} (kr_0)e^{im(\theta_0+\pi)}~~,$$

or equivalently, after changing the summation index,

$$J_{n-\ell} (kR)e^{i(n-\ell)\Omega} = \sum^\infty_{m=-\infty} \left[ J_{n-m} (kr) e^{i(n-m)\theta} \right] \left[J_{m-\ell} (kr_0) e^{i(m-\ell)\overline{\theta_0}}\right]$$ (538)

where

$$\overline{\theta_0}=\theta_0+\pi~~,$$

while Eq.(5.35) for the vector triangle becomes

$$Re^{i\Omega}=re^{i\theta}+r_0e^{i\overline{\theta_0}}~~.$$ (539)

Compare Eq.(5.39) with Eq.(5.38). Observe that (i) for each translation in the Euclidean plane, say $re^{i\theta}$ , there is a corresponding infinite dimensional matrix

$$\{ J_{n-m} (kr) e^{i(n-m)\theta}~: ~~ n,m=0,\pm 1,\pm 2, \cdots \}$$

and (ii) the result of successive translations, such as Eq.(5.39), is represented by the product of the corresponding matrices, Eq.(5.38).

Exercise 54.1 (ADDITION FORMULA FOR BESSEL FUNCTIONS)
Express $J_n(x_1+x_2)$ as a sum of products of Bessel functions of $x_1$ and $x_2$ respectively.

Property 21 (Completeness)
The cylinder waves form a complete set. More precisely,

$$\frac{\delta (r-r_0)\delta (\theta -\theta_0)}{r} = \sum^\infty_{m=-\infty} \frac{1}{2\pi}\int^\infty_0 kdk J_m (kr)J_m(kr_0) e^{im(\theta -\theta_0)}~.~~~~~~$$ (540)

This relation is the cylindrical analogue of the familiar completeness relation for plane waves,

$$\delta (x-x_0)\delta (y-y_0) = \int^\infty_{-\infty} \int^\infty_{-\infty} dk_x dk_y\frac{e^{i(k_xx+k_y y)}}{2\pi}~\frac{e^{-i(k_xx_0 +k_y y_0)}}{2\pi}$$

$$= \int^\infty_0 \int^{2\pi}_0 kdkd\alpha \frac{e^{ikr\cos (\alpha -\theta )}} {2\pi}~\frac{e^{-ikr_0\cos (\alpha -\theta_0)}}{2\pi}~.~~~~~~$$ (541)

In fact, the one for plane waves is equivalent to the one for cylinder waves. The connecting link between the two is the plane wave expansion, Eq.(5.24),

$$e^{ikr\cos (\theta-\alpha)} = \sum i^m J_m(kr)e^{im(\theta-\alpha)}~~.$$

Introduce it into Eq.(5.41) and obtain

$${\delta (x-x_0)\delta (y-y_0) =}$$

$$\sum_m\int^\infty_0 kdk\int^{2\pi}_0 d\alpha J_m(kr)i^m\frac{e^{i... ...\pi} \sum_{m'} J_{m'}(kr_0) \frac{e^{-im'(\theta_0-\alpha )}}{2\pi}(-i)^{m'}\,.$$

Using the orthogonality property

$$\int^{2\pi}_0 d\alpha ~\frac{e^{i(m'-m)\alpha}}{2\pi} = \delta_{mm'}~~,$$ (542)

the definition

$$\delta (x-x_0)\delta (y-y_0)dxdy=\delta (r-r_0)\delta (\theta -\theta_0) drd\theta~~,$$

and

$$dxdy = rdrd\theta\,,$$

one obtains

$$\frac{\delta (r-r_0)\delta (\theta -\theta_0)}{r} = \sum^\infty_{m=-\infty} \frac{1}{2\pi}\int^\infty_0 kdk J_m (kr)J_m(kr_0) e^{im(\theta -\theta_0)}\,,$$ (543)

the completeness relation for the cylinder waves.

Property 22 (Fourier-Bessel transform)
The Bessel functions $\{ J_m(kr)\colon 0\le k<\infty\}$ of fixed integral order form a complete set

$$\frac{\delta (r-r_0)}{r} = \int^\infty_0 J_m(kr)J_m (kr_0)kdk\,.$$ (544)

This result is a direct consequence of Property 21. Indeed, multiply the cylinder wave completeness relation, Eq.(5.40) by $e^{-im'\theta}$ , integrate over $\theta$ from 0 to $2\pi$ , again use the orthogonality property, Eq. 5.42, and cancel out the factor common factor $e^{- im'\theta_0}$ from both sides. The result is Eq.(5.44), the completeness relation for the Bessel functions on the positive $r$ -axis.

Remark: By interchanging the roles of $k$ and $r$ one obtain from Eq.(5.44)

$$\frac{\delta (k-k_0)}{k}=\int^\infty_0 J_m(kr)J_m(k_0 r)rdr~~.$$

Remark: The completeness relation, Eq.(5.44), yields

$$f(r)=\int^\infty_0 F(k)J_m(kr)kdk$$

where

$$F(k) = \int^\infty_0 f(r) J_m(kr)rdr\,.$$

This is the Fourier-Bessel transform theorem.

It is interesting to note that the completeness relation, Eq.(5.44), is independent of the integral order of $J_m(kr)$ . One therefore wonders whether Eq.(5.44) also holds true if one uses $J_\nu(kr)$ , Bessel functions of any complex order $\nu$ . This is ideed the case.

Property 23 (Bessel transform)
The Bessel functions $\{ J_\nu(kr): 0<k,\infty~\}$ of complex order $\nu$ form a complete set

$$\frac{\delta (r-r_0)}{r} = \int^\infty_0 J_\nu(kr)J_\nu (kr_0)kdk\,.$$ (545)

This result gives rise to the transform pair

$$f(r) = \int^\infty_0 F(k)J_\nu(kr)kdk$$ (546)

$$F(k) = \int^\infty_0 f(r) J_\nu(kr)rdr\,.$$ (547)

and it is obvious that mathematically Property 22 is a special case of Property 23.