Properties of Hankel and Bessel Functions

Associated with the two kinds of Hankel functions are two solutions to the Helmholtz equation. They are the ``cylinder harmonics'' or order $\nu$ ,

$$\psi_1(kr,\theta ) = c_1\int^{\varepsilon -i\infty}_{-\varepsilon + i\infty } e^{i\rho\cos(\alpha-\theta) +i\nu\alpha}d\alpha$$

$$= c_1\int^{\varepsilon -i\infty -\theta}_{-\varepsilon + i\infty -\... ...cos\alpha +i\nu\alpha}d\alpha e^{i\nu\theta} \equiv H^{(1)}_\nu ~e^{i\nu\theta}$$ (54)

and

$$\psi_2(kr,\theta ) = c_2\int^{\pi+\varepsilon +i\infty }_{\pi-\varepsilon -i\infty} e^{i\rho\cos(\alpha-\theta) +i\nu\alpha}d\alpha$$

$$= c_2\int^{\pi+\varepsilon +i\infty -\theta}_{\pi - \varepsilon -i\... ...cos\alpha +i\nu\alpha}d\alpha e^{i\nu \theta}\equiv H^{(2)}_\nu ~e^{i\nu\theta}$$ (55)

Here

$$c_1=c_2=\frac{e^{-i\nu \pi /2}}{\pi}$$

are normalization constants whose values are derived below (see Property 11 below).

The name ``cylinder harmonic'' arises from the fact that these two functions emerge from those solutions of the Helmholtz equation whose level surfaces mold themselves naturally to the cylindrical geometry of its domain. These functions have the following properties:

Property 1
They are linear superpositions of plane waves.

Property 2
Their integration contours in the complex $\alpha$ -plane are as indicated in Figure 5.3.

Property 3 (No angular dependence)
The integral representatives

$$\left.\begin{array}{c} H^{(1)}_\nu\\ H^{(2)}_\nu\end{array}\right... ...\alpha_2 -\theta}_{\alpha_1-\theta} e^{i\rho\cos\alpha} e^{i\nu\alpha} d\alpha$$

of the two Hankel functions do not depend on any real changes in the integration limits.

This means that the $\theta$ -dependent shift in the limits of the integral has no effect on the value of the integral itself, whenever the integration limits $\alpha_1$ and $\alpha_2$ each lie near infinity in a strip of convergence of the integral.

Figure: Integration contour for a Hankel function. If the integration limits are such that $Im~\alpha_1=+\infty$ and $Im~\alpha_2=-\infty$ then the integral does not change under horizontal shifts.

Suppose the integration contour is taken to be the curve labelled $\hbox{${\bigcirc}\!\!\!\!\scriptstyle 1$}$  , where $\alpha_1$ is near the vertical line $p=0$ and $\alpha_2$ is near $p=\pi$ . Then for $0\le\theta <\pi$ we see that

$$\int^{\alpha_2}_{\alpha_1} e^{i\rho\cos\alpha+i\nu\alpha} d\alpha... ...^{\alpha_2-\theta}_{\alpha_1-\theta} e^{i\rho\cos\alpha +i\nu\alpha} d\alpha\,.$$ (56)

This equality is a result of two facts:

1. The dominant contribution comes from the path between $\alpha_1- \theta$ and $\alpha_2-\theta$ , and that path can be deformed into the original curve $\hbox{${\bigcirc}\!\!\!\!\scriptstyle 1$}$ and by the Cauchy-Goursat theorem the integral will remain unchanged.
2. The $\theta$ -dependent change due to the shift in the end points $\alpha_1- \theta$ and $\alpha_2-\theta$ is neglegible, because the integrand is already neglegibly small at these points.

When $-\pi <\theta \le0$ , then the path of integration can again be deformed into a standard one, but in that case one starts with $\alpha_1$ and $\alpha_2$ are near $p=-\pi$ and $p=0$ instead.

To summarize:

$$\int^{\alpha_2}_{\alpha_1} e^{i\rho\cos(\alpha-\theta)+i\nu\alpha} d\alpha$$

represents a continuous function of $\theta$ whenever $-\pi<\theta<\pi$ . When this inequality is fulfilled one has

$$\int^{\alpha_2}_{\alpha_1} e^{i\rho\cos(\alpha-\theta)+i\nu\alpha... ...t^{\alpha_2}_{\alpha_1} e^{i\rho\cos\alpha+i\nu\alpha} d\alpha \,e^{i\nu\theta}$$ (57)

If this inequality is not fulfilled, then the right hand side of this equation diverges and is therefore not valid. This is because in this case the contour of the integral on r.h.s. of Eq.(5.6) cannot be deformed into that of the l.h.s.: The integration limits would fall outside the shaded strips of convergence, the integral would diverge, and the r.h.s. of Eq.(5.7) would loose its meaning.

As usual, the circumstance $\theta=-\pi$ or $\theta=\pi$ are defined in terms of limits as $\theta \to \pm \pi$ from the inside of the interval..

Thus we conclude that $H^{(1)}_\nu =H^{(1)}_\nu (kr)$ and $H^{(2)}_\nu = H^{(2)}_\nu (kr)$ are independent of $\theta$ indeed. The result is that the two cylinder harmonics have the form

$$\psi (r,\theta )=Z_\nu (\rho )e^{i\nu\theta}\,,~~\qquad~~\rho =kr$$

a product of two functions, each one depending separately on its own variable, but independent of the other.

Property 4 (Eigenfunction of rotations in the Euclidean plane)
The cylinder harmonics are eigenfunctions of the rotation generator $L_\theta =\frac{1}{i}~\frac{\partial}{\partial\theta}$ ,

$$L_\theta \psi = \nu\psi\,;$$

that is to say, they are invariant (modulo $a$ constant multiplicative factor) under rotation around the origin

$$R_{\gamma\ast}\psi \equiv e^{-i\gamma L_\theta}\psi$$

$$= e^{-i\gamma\nu} Z_\nu (\rho )e^{i\nu\theta}\,.$$

Property 5 (Solution to the Helmholtz equation)
They satisfy the Helmholtz's equation, which in polar coordinates becomes Bessel's equation

$$= [\nabla^2+k^2]\psi$$ 0

$$= \left[\frac{1}{r}~\frac{\partial}{\partial r} r\frac{\partial}{\partial r} +k^2+\frac{1}{r^2}~\frac{\partial^2}{\partial\theta^2}\right]\psi$$

$$= \left[\frac{1}{r}~\frac{\partial}{\partial r}r\frac{\partial}{\partial r} +k^2-\frac{\nu^2}{r^2}\right] Z_\nu (kr)e^{i\nu\theta}$$

$$= k^2\left[\frac{d^2}{d\rho^2} +\frac{1}{\rho}~\frac{d}{d\rho}+... ...o ) \\ H^{(2)}_\nu (\rho)\end{array}\right\} e^{i\nu\theta}\,.$$

Even through the eigenvalue $k^2$ of the operator $-\nabla^2$ is infinitely degenerate, the eigenvalues of $L_\theta$ in the equation

$$\frac{1}{i}~\frac{\partial}{\partial\theta}\left\{ \begin{array}{... ...{(1)}_\nu (\rho ) \\ H^{(2)}_\nu (\rho)\end{array}\right\} e^{i\nu\theta} \,,$$

serve to distinguish the elements of the degenerate set.

Property 6 (Cylinder waves)
The domain of a cylinder harmonic is the $r$ and $\theta$ coordinatized transverse cross section of a cylinder. A cylinder harmonic itself is the $r$ and $\theta$ dependent part of a cylinder wave

Figure 5.5: The domain of a cylinder harmonic is the perpendicular cross section of a cylindrical configuration.

$$\psi=H_\nu (kr)e^{i\nu\theta} e^{ik_z z} e^{-i\omega t}~~,$$

which satisfies the wave equation

$$\left[\frac{1}{r}~\frac{\partial}{\partial r}r\frac{\partial}{\pa... ...} {\partial z^2} -\frac{1}{c^2}~\frac{\partial^2}{\partial t^2}\right] \psi =0$$

whenever the constants $k$ , $k_z$ and $\omega$ satisfy the dispersion relation

$$\frac{\omega^2}{c^2} = k^2+k^2_z\,.$$

Lecture 41

Property 7 (Two kinds of Hankel functions)
The $\rho =kr$ dependent factors of these cylinder harmonics,

$$H^{(1)}_\nu (\rho ) = c_1\int_{C_1} e^{i\rho\cos \alpha +i\nu\alpha} d\alpha$$

$$H^{(2)}_\nu (\rho ) = c_2 \int_{C_2} e^{i\rho\cos \alpha +i\nu\alpha} d\alpha~~,$$

are called Hankel functions of the first and second kind of (complex) order $\nu$ .

It is worthwhile to reemphasize that the integral representations of $H^{(1)}_\nu$ and $H^{(2)}_\nu$ converge and are well defined for any complex number $\nu$ .

Property 8 (Bessel function)
Having equal normalization constants,

$$c_1=c_2\equiv c(\nu )\,.$$

the two Hankel functions, Eqs. (5.4) and (5.5), determine the Bessel function of (complex) order $\nu$ ,

$$J_\nu (\rho )= \frac{1}{2} [H^{(1)}_\nu (\rho )+H^{(2)}_\nu (\rho )]\,.$$ (58)

One arrives at this definition by means of the union of the two paths $C_1$ and $C_2$ which define $H^{(1)}_\nu$ and $H^{(2)}_\nu$ . By the Cauchy-Goursat theorem these paths can be deformed into a single path as depicted in Figure 5.7

Figure 5.6: Contour integration paths for the Bessel function $J_\nu$ . The integration contour for the Neumann function $N_\nu$ is the union of the two indicated paths.

If $c_1$ were not equal to $c_2$ , then the contributions to the complex integral from the lower parts of $C_1$ and $C_2$ would not cancel.

Property 9 (Neumann function)
Their difference

$$N_\nu (\rho )\equiv Y_\nu (\rho )\equiv\frac{1}{2i} [H^{(1)}_\nu (\rho )- H^{(2)}_\nu (\rho )]$$ (59)

is the Neumann function of (complex) order $\nu$ .

Its integral representation requires the two integration contours depicted in Figure 5.6.

Property 10 (Analogue to trigonometric and exponential functions)
The Hankel functions are the analogues of the exponential functionsin trigonometry. In fact, as we shall see, one has the following scheme

$$\cong \begin{array}[b]{cccc} e^{ix}\,; &e^{-ix}\,; &\cos x\,; &\s... ...{2}{\pi\rho}}\cos[\cdots]\,;& \sqrt{\frac{2}{\pi\rho}}\sin [\cdots] \end{array}$$

for large $\rho$ , as we shall see later.

The next property asks and answers the the question: How do the Bessel and the Neuman functions depend on their complex order $\nu$ ? With the universally agreed-upon value for the normalization constant $c(\nu)$ , the answer could be no simpler: For real $\nu$ these functions are real and for complex $\nu$ these functions are their analytic extensions into the complex domain. More precisely, we have

Property 11 (Reflection Principle)

1. If the order $\nu$ is real then their sum (the ``Bessel function'')
   1. $J_\nu (\rho )=\frac{1}{2}[H^{(1)}_\nu (\rho )+H^{(2)}_\nu (\rho )]$ is real when $\nu$ is real and
   2. $J_0(0)=1$
   provided the normalization constant $c(\nu)$ is
   $$c_1=c_2\equiv c(\nu ) = \frac{e^{-i\nu\pi /2}}{\pi}\,.$$

2. If $\nu$ is complex, then, for fixed positive $\rho$ , both $J_\nu (\rho )$ and $N_\nu(\rho)$ are analytic functions of their order $\nu$ . Furthermore, they obey the reflection principle:
   $$\overline{J_\nu (\rho)} = J_{\overline{\nu}}(\rho )~\textrm{and}~ \overline{N_\nu (\rho)} = N_{\overline{\nu}}(\rho )$$

That the Bessel and the Neumann functions are analytic in their order $\nu$ follows from their defining integral representations and the form of the normalization constant $c(\nu)$ .

The reflection principle is a general property which analytic functions enjoy whenever their values are real on the real ($\nu$ ) axis. It is shown below that the form of the normalization constant $c(\nu)$ guarantees this. Indeed, for the Bessel function $J_\nu (\rho )$ the proof consists of three steps below. (We delay the application of the reflection principle to the Neumann function until after we have exhibited the complex conjugation property applied to the two Hankel functions on page .)

Step 1: Deform the integration path into straight lines. The result is

$$J_\nu (\rho )= \frac{c(\nu )}{2}\left[ \int^{-\frac{\pi}{2}}_{-\f... ...} +i\infty}_{\frac{3\pi}{2}}\right] e^{i\rho\cos\alpha +i\nu\alpha} d\alpha\,.$$

Figure 5.7: Three-part integration contour for the Bessel function.

Step 2: Symmetrize the integrals by shifting the integration limits to the left. This is achieved by introducing the new dummy variable

$$\beta = \alpha -\frac{\pi}{2}\,,~~\qquad~~\alpha =\beta +\frac{\pi}{2}~~.$$

The result is

$$J_\nu (\rho ) = \frac{c(\nu )}{2}\left[ \overbrace{\int^{-\pi}_{... ...tstyle 3$}} \right] e^{i\rho\sin\beta} e^{i\nu\beta} d\beta ~e^{i\nu\pi /2}\,.$$

Reminder:  We have not shifted the path of integration. Instead, we have only altered the coordinate labelling used to describe that path in the complex plane.

Step 3:

1. Fix the normalization constant $c(\nu)$ by requiring that $J_\nu$ be real when $\nu$ is real. This is achieved by setting $c(\nu )=\frac{1}{\pi} e^{-i\nu\pi /2}$ . This cancels out the last factor.
2. To bring this reality of $J_\nu$ to light, combine the first and third integral by introducing
   $$\left.\begin{array}{lcl} \beta = -\pi+i\gamma &~~~~~ &\textrm{for... ...xtrm{to~make~the}\\ \textrm{integration~limits}\\ \textrm{equal!} \end{array}$$
   The result, after dropping the bar, is
   

   $$J_\nu (\rho ) = \frac{1}{2\pi} \int^{\gamma =\infty}_0 e^{-\rho\sinh \gamma -\nu\... ...i}}^{\hbox{${\bigcirc}\!\!\!\!\scriptstyle 3$}} ]}_{\displaystyle -2\sin\nu\pi}$$

   $$~ +\underbrace{\frac{1}{2\pi} \int^\pi_{-\pi} e^{-i\rho\sin\beta +i... ...a}_{\displaystyle \frac{1}{\pi}\int^\pi_0\cos (\rho\sin\beta -\nu\beta )d\beta}$$

   
   

Conclusion: When $\nu$ is real, then

$$J_\nu (\rho)~~\textrm{is~real}$$

$$J_0(0)=1\,,$$

also, if $\nu$ is complex, then by inpection one finds

$$\overline{J_\nu (\rho )} = J_{\overline{\nu}}(\rho )~~,$$

which is what we set out to show.

Property 12 (Bessel function of integral order)
The Bessel functions of integral order $(\nu =m=0,1,2,\dots )$ is given by

$$J_m(\rho ) = \frac{1}{2\pi} \int^\pi_{-\pi} e^{-i\rho\sin\beta +im\beta} d\beta$$

$$= \frac{1}{2\pi}\int^\pi_{-\pi} e^{i\rho\sin\beta -im\beta}d\beta$$

$$= \frac{1}{\pi}\int^\pi_0 \cos (\rho\sin\beta -m\beta )d\beta\,.$$

Furthermore, the set $\{ J_m,J_{-m}\}$ forms a linearly dependent set. Indeed,

$$J_{-m}(\rho )= (-1)^mJ_m(\rho )~~\qquad~~ m=0,1,2\dots\,.$$ (510)

This equation is the result of changing the integration variable $\beta$ . Letting $\beta =\pi -\overline{\beta}$ , one obtains

$$J_m(\rho ) = \frac{1}{2\pi}\int^0_{2\pi}e^{-i\rho\sin\overline{\beta}- im\overline{\beta}}(-1)^m(-)d\overline{\beta}$$

$$= (-1)^m\frac{1}{2\pi}\int^\pi_{-\pi} e^{-i\rho\sin\overline{\beta}-im \overline{\beta}}d\overline{\beta}$$

$$= (-1)^m J_{-m} (\rho )\,.$$

Exercise 52.1 (HANKEL AND NEUMANN FUNCTIONS OF INTEGRAL ORDER)
Show that

$$H^{(1)}_{-n}(\rho) = (-1)^n H^{(1)}_n(\rho)$$

$$H^{(2)}_{-n}(\rho) = (-1)^n H^{(2)}_n(\rho)$$

$$N_{-n}(\rho) = (-1)^nN_n(\rho)$$

Lecture 42

Property 13 (Power series)
The Bessel function $J_\nu (\rho )$ of complex order and for real $\rho$ has the following Frobenious expansion around the origin

$$J_\nu (\rho ) = \left(\frac{\rho}{2}\right)^\nu\left[\frac{1}{\Gamma (\nu +1)}-\f... ...ght)^2 +\frac{1}{2!\Gamma (\nu +3)} \left(\frac{\rho}{2}\right)^4-\cdots\right]$$

$$= \left(\frac{\rho}{2}\right)^\nu \sum_{k=0}^\infty \frac{(-\rho/2)^{2k}}{\Gamma(1+\nu+k)k!}$$ (511)

Remark: Near $\rho =0$ the dominant behaviour of $J_\nu (\rho )$ is given by

$$\boxed{ J_\nu (\rho )=\frac{1}{\Gamma(1+\nu)}\left(\frac{ \rho}{2}\right)^\nu~~\qquad~~\qquad~~\rho\ll 1~. }$$

The power series, together with its normalization constant, follows from the integral representation

$$J_\nu (\rho ) =\frac{e^{-i\nu \pi/2}}{2}\int_C e^{i\rho\cos\alpha +i\nu\alpha} d\alpha\,,$$

where $C$ is the integration contour indicated in Figure 5.7. Indeed, introduce the new variable of integration

$$z=\frac{\rho}{2}e^{-i(\alpha -3\pi/2)},~~d\alpha=i\frac{dz}{z},~~ e^{i \alpha}=\frac{\rho}{2}\frac{1}{z}e^{i3\pi/2}~~.$$

Under this change, the new integration contour is the one depicted in Figure 5.8, which is based on the following scheme:

$$\begin{array}{cccc cccc cccc cccc} \alpha= & i\infty-\frac{\... ...rho}{2}\exp\frac{\pi i}{2} &\vert&\frac{\rho}{2} \end{array}~.$$

The integral becomes

$$J_\nu (\rho ) =-\frac{e^{i\nu \pi}}{2\pi i}\left( \frac{\rho}{2}\right)^\nu \int _{Z_0} e^{-z+\frac{\rho^2}{4z}} z^{-\nu-1}dz ~~.$$

Figure 5.8: The transformed integration contour in the complex $z$ -plane surrounds and touches the branch cut of the multiple-valued function $z^{-(\nu +1)}$ .

By expanding the exponential $e^{\frac{\rho^2}{4z}}$ in a Taylor series one obtains Eq.(5.11), provided one sets

$$\frac{1}{\Gamma (\nu +1)}=\frac{e^{i\pi(\nu+1)}}{2\pi i} \int _{Z_0} e^{-z} z^{-(\nu+1)}dz ~~,$$

which is one of the definitions of the gamma function. This contour integral is meaningless unless one specifies the branch of the multiple-valued function $z^{-(\nu +1)}$ . The branch is dictated by the requirement that

$$\frac{1}{\Gamma (m +1)}=\frac{1}{m!}~.$$

For this branch the domain is restricted to $0<\arg z<2\pi$ whenever the cut for this branch is the positive $x$ -axis, as in Figure 5.8.

Let us look at the solutions to Bessel's equation from the viewpoint of linear algebra. The solution space is two dimensional. There are two important spanning sets. The first one,

$$\left\{ J_\nu \simeq \frac{(\rho/2)^\nu}{\Gamma(1+\nu)},~ J_{-\nu} \simeq \frac{(\rho/2)^{-\nu}}{\Gamma(1-\nu)} \right\}$$

is simple whenever $\rho \ll 1$ . By contrast, the second one

$$\left\{H^{(1)}_\nu \simeq\sqrt{\frac{2}{\pi\rho}}e^{i[\rho-(\nu+\... ...q\sqrt{\frac{2}{\pi\rho}}e^{-i[\rho-(\nu+\frac{1}{2}) \frac{\pi}{2}]} \right\}$$

is simple whenever $1\ll\rho$ . However, we know that these two bases are related by a linear transformation.

$$\left[ \begin{array}{cc} H^{(1)}_\nu &H^{(2)}_\nu \end{ar... ...right] \left[ \begin{array}{cc} a&b\\ c&d \end{array} \right]$$

The question is: what is this linear transformation? The answer is provided by the following

Property 14 (Hankel-Bessel relation)
When the order of a Bessel function is not an integer ( $\nu\not= m$ ), then the set of Bessel functions $\{ J_\nu , J_{-\nu}\}$ form an independent set. Moreover, one has

$$H^{(1)}_\nu (\rho ) = \frac{e^{-i\nu\pi} J_\nu (\rho ) - J_{-\nu}(\rho )} {-i\sin\pi\nu}$$ (512)

$$H^{(2)}_\nu (\rho ) = \frac{e^{i\nu\pi} J_\nu (\rho )-J_{-\nu}(\rho )} {i\sin\pi\nu}\,.$$ (513)

Once one recalls the defining equation for the Bessel equation, Eq.(5.8), i.e. that $H^{(2)}_\nu=2J_\nu -H^{(1)}_\nu$ , one finds that Eq.(5.12) is a mere consequence of Eq.(5.13). Thus one's primary concern is the validity of that second equation. However, before validating it, let us identify some consequences of the fundamental identities, Eqs.(5.12) and (5.13).

Property 15 (Complex conjugation property)
If the Hankel functions are of real order $\nu$ (and $\rho =\vert\rho\vert$ ), then

$$\overline{H^{(2)}_\nu (\rho )} = H^{(1)}_\nu (\rho )\,,~~\qquad~~... ...overline{e^{-ix}} = e^{ix}~~\textrm{for}\\ \textrm{real}~x \end{array}\right)$$

i.e., they are complex conjugates of each other. This follows from equations 5.12 and 5.13 of Property 13.

Remark: There are three additional consequences:

First of all, it follows from Property 12 that if $\nu$ is complex, then

$$\overline{J_\nu (\rho)} = J_{\overline{\nu}}(\rho )\Rightarrow \overline{H^{(2)}_\nu (\rho )}=H^{(1)}_{\overline{\nu}}(\rho )\,.$$

Second, apply this complex conjugation property to the defining Eq.(5.9) and obtain the reflection principle applied to the Neumann functions

$$\overline{N_\nu (\rho)} = N_{\overline{\nu}}(\rho )~.$$

Third, if $\rho$ is complex also, then

$$\overline{H^{(2)}_\nu (\rho )} = H^{(1)}_{\overline{\nu}}(\overline{\rho}) ~.$$

Returning to the validation of the Hankel-Bessel identities, one finds that this process consists of four steps. They consist primarily of manipulating the intergration paths of the integral representations of $J_\nu$ and $J_{-\nu}$ .

Step 1. Recall the definition of $J_{-\nu}(\rho )$ :

$$2\pi e^{-i\nu\pi /2}J_{-\nu}(\rho ) = \int_{C_0} e^{i\rho\cos\alpha -i\nu \alpha}d\alpha$$

$$= -\int_{\overline{C}} e^{i\rho\cos\alpha+i\alpha} d\overline{\alpha}\qquad~~\overline{\alpha}=-\alpha~,\textrm{and~then~drop~the~\lq\lq bar''}$$

Here the $\overline {C}$ is the complex conjugate of the integration path $C_0$

Figure 5.9: Integration path $C_0$ and its complex conjugate $\overline {C}$ .

The rest of the proof consists of

Step 2. Subtract this from an analogous expression for $J_\nu$ .

Step 3. Deform the contour and reexpress the r.h.s. in terms of $H^{(2)}_\nu$ . This yields the desired equation.

Step 4. Use Property 8 to obtain the expression for $H^{(1)}_{\nu}$ .

The details of these remaining steps are

Step 2. Subtract the expression in Step 1 from the analogous expression for $J_\nu$ . After a slight deformation of the two respective integration path, obtain

$$2\pi (e^{i\nu\pi /2}J_\nu - e^{-i\nu\pi /2}J_{-\nu}) = \left[\int_{C_0} + \int_{\overline{C}}\right] e^{i\rho\cos\alpha +i\nu\alpha} d\alpha$$

$$= \left[ \int_{C_2}+\int_{C'_2}\right] e^{i\rho\cos\alpha+i\nu\alpha} d\alpha\,.$$

Here $C_2$ is the integration contour for $H^{(2)}_\nu$ and $C'_2$ is $-C_2$ shifted by $2\pi$ to the left.

Figure 5.10: The two paths in Figure 5.9 have been deformed into $C'_2$ and $C_2$ .

Step 3.

1. Recall from Property 7 that
   $$\int_{C_2} e^{i\rho\cos\alpha +i\nu\alpha}d\alpha = \pi e^{i\nu\pi /2}H^2_\nu (\rho )\,.$$

2. In addition, we have
   

   $$\int_{C'_2} e^{i\rho\cos\alpha +i\nu\alpha}d\alpha = \int^{-\pi -\varepsilon -i\infty}_{-\pi +\varepsilon +i\infty} e^{i\rho\cos\alpha +i\nu\alpha} d\alpha~~\qquad~~\overline{\alpha}=\alpha +2\pi$$

   $$= \int^{\pi -\varepsilon - i\infty}_{\pi -\varepsilon +i\infty} e^{... ...o \cos\overline{\alpha}+i\nu\overline{\alpha}}d\overline{\alpha} e^{-i\nu 2\pi}$$

   $$= (-)\pi e^{i\nu\pi /2}H^{(2)}_\nu (\rho )e^{-i\nu 2\pi}\,.$$

   
   
3. Introduce the results of 1. and 2. into the last expression in Step 2, and obtain
   $$2\pi (e^{i\nu\pi /2}J_\nu -e^{-i\nu\pi /2}J_{-\nu}) = \pi (e^{i\nu\pi /2} -e^{-i3\pi\nu /2})H^{(2)}_\nu\,.$$
   Using
   $$e^{\nu\pi /2}-e^{-3\pi \nu /2} = 2ie^{-i\pi \nu /2}\sin\nu\pi\,,$$
   and solving for $H^{(2)}_\nu$ one obtains
   $$H^{(2)}_\nu = \frac{e^{i\nu\pi} J_\nu -J_{-\nu}}{i\sin\nu\pi}\,.$$

Step 4. Use Property 8 to obtain

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

$$= \frac{-e^{-i\nu\pi}J_\nu +J_{-\nu}}{i\sin\nu\pi}\,.$$

These are the two expressions for the two kinds of Hankel functions in terms of Bessel functions of order $\nu$ and $-\nu$ .

Property 16 (Contiguity relations)
Let $Z_\nu = H^{(1)}_\nu$ or $H^{(2)}_\nu$ or $J_\nu$ or $N_\nu (\equiv Y_\nu )$ be any solution to Bessel's equation of complex order $\nu$ . Then

$$Z_{\nu +1}(\rho ) +Z_{\nu -1}(\rho ) = \frac{2\nu}{\rho} Z_\nu (\rho )$$ (514)

$$Z_{\nu +1}(\rho ) -Z_{\nu -1}(\rho ) = -2\frac{d}{d\rho} Z_\nu (\rho )\,.$$ (515)

Proof (in two steps):

Step 1. Apply the definition to the sum and difference

$$\left.\begin{array}{c} I_1\\ I_2\end{array}\right\} = \frac{\pi}{2} \{ H_{\nu +1}\pm H_{\nu -1}\}$$

$$= \int_C e^{i\rho\cos\alpha} e^{i\nu\left(\alpha -\frac{\pi}{2}\rig... ... -\frac{\pi}{2}\right)}\pm e^{-i\left(\alpha - \frac{\pi}{2}\right)}]d\alpha\,.$$

Step 2. Observe that

$$\frac{1}{2} [e^{i\left(\alpha -\frac{\pi}{2}\right)}\pm e^{-i\lef... ...a\\ i\sin\left(\alpha -\frac{\pi}{2}\right) = -i\cos\alpha \end{array}\right.$$

Consequently,

$$I_1 = \frac{-1}{i\rho}\int_C\frac{d}{d\alpha} (e^{i\rho\cos\alpha})e^{i\nu \left(\alpha -\frac{\pi}{2}\right)}d\alpha$$

$$= \frac{-1}{i\rho}\int_C (-) e^{i\rho\cos\alpha}\frac{d}{d\alpha} (e^{i\nu \left(\alpha -\frac{\pi}{2}\right)}) d\alpha = \pi \frac{\nu}{\rho} H_\nu$$

$$I_2 = -\pi\frac{d}{d\rho} H_\nu\,.$$

These are the two recursion relations (5.14) and (5.15).

These relations are quite useful. Note that by adding and subtracting the recursion relations one obtains

$$Z_{\nu+1}(\rho ) e^{i(\nu +1)\theta} = -e^{i\theta}\left(\frac{\partial} {\partial\rho} +\frac{i}{\rho}~\frac{\partial}{\partial\theta}\right) Z_\nu (\rho )e^{i\nu\theta}$$

$$Z_{\nu-1}(\rho )e^{i(\nu -1)\theta} = e^{-i\theta}\left(\frac{\partial} {\partial\rho} -\frac{i}{\rho}~\frac{\partial}{\partial\theta}\right) Z_\nu (\rho )e^{i\nu\theta}~~.$$

Let us call

$$e^{i\theta}\left(\frac{\partial}{\partial\rho}+\frac{i}{\rho}~\fr... ...{\partial y}\right)\equiv L_+~~\qquad~~\textrm{the}~ \textit{raising~operator}$$

and

$$-e^{-i\theta}\left(\frac{\partial}{\partial\rho}- \frac{i}{\rho}~... ...rtial y}\right)\equiv L_-~~\qquad~~\textrm{the}~ \textit{lowering~operator}\,.$$

These operators yield

Property 17 (Raising and lowering the order of $Z_\nu$ )

$$L_+ Z_\nu (\rho )e^{i\nu\theta} = -Z_{\nu+1} (\rho )e^{i(\nu+1)\theta}$$

$$L_- Z_\nu (\rho )e^{i\nu\theta} = -Z_{\nu-1} (\rho )e^{i(\nu-1)\theta}$$

and

$$L_+L_-=L_-L_+ = -\frac{1}{k^2}\left( \frac{\partial^2}{\partial x^2}+ \frac{\partial^2}{\partial y^2} \right)$$

$$= -\left[ \frac{\partial^2}{\partial\rho^2}+\frac{1}{\rho}~\frac{\p... ...2}~\frac{\partial^2}{\partial\theta^2} \right] \equiv -\frac{1}{k^2}\nabla^2\,,$$ (516)

which is the (rotationally and translation) invariant Laplacian operator.

Comment: (Factorization Method for Finding Cylinder Harmonics.)

It is difficult to exclude these raising and lowering operators as the fastest way for establishing relationships between normal modes in a cylindrical cavity. For example, suppose one knows explicitly the rotationally symmetric $(\nu=0)$ mode $J_0(\rho )e^{i0\cdot\theta}$ . Then all the other modes

$$J_m(\rho )e^{im\theta} = (-1)^m(L_+)^mJ_0(\rho )~~,$$

can be obtained by repeated application of the raising operator $L_+$ . The lowering operator undoes the work of the raising operators

$$L_-L_+Z_\nu e^{i\nu\theta} = Z_\nu e^{i\nu\theta}\,,$$

i.e.,

$$(\nabla^2_\rho +1)Z_\nu e^{i\nu\theta} = 0\,.$$

This feature also illustrates the fact that the $2$ -dimensional Laplacian $\nabla^2 = -L_-L_+$ (in $E^2$ ) can be factorized into a product of first order operators. This method is purely algebraic and its essentials are as follows:

Recalling the definition of the rotation generator $L_\theta=\frac{1}{i} \frac{\partial}{\partial \theta}$ in Section 5.1.4, notice that

$$[L_\theta,L_\pm ]\equiv L_\theta L_\pm -L_\pm L_\theta=\pm L_\pm~.$$

This commutation relation is fundamental for the following reason:

Suppose we have a solution to the Helmholtz equation

$$(\nabla^2 +k^2)\psi_m =0~,$$

and suppose that the solution is a rotation eigenfunction, i.e.

$$L_\theta \psi_m =m\psi_m ~.$$

Then the commutation relation implies

$$L_\theta L_+ \psi_m = (L_+ L_\theta +L_+)\psi_m$$

$$= (m+1)L_+ \psi_m ~.$$

In other words, $L_+ \psi_m$ is another rotation eigenfunction. Furthermore,

$$(\nabla^2 +k^2)L_+\psi_m = (-k^2 L_+ L_- +k^2)L_+\psi_m$$

$$= L_+(-k^2 L_- L_+ +k^2)\psi_m$$

$$= L_+(\nabla^2 +k^2)\psi_m$$

$$= 0 ~,$$ (517)

i.e. the new rotation eigenfunction

$$L_+\psi_m \equiv \psi_{m+1}$$

is again a solution to the Helmholtz equation. The analogous result holds for $L_-\psi_m$ .

To summarize: The algebraic method for solving the Helmholtz equation is a two step process: (i) Factor the Laplacian, Eq.(5.1) into two factors $L_+$ and $L_-$ , and (ii) for each eigenspace of $\nabla^2$ construct a basis using $L_+$ and $L_-$ , whose capability as raising and lowering operators is implied by the two commutation relations

$$[L_\theta,L_\pm ]=\pm L_\pm~.$$ (518)

These operators obviously commute,

$$[L_+,L_-]=0~.$$ (519)

This is evident from Eq.(5.16). Furthermore, as we have seen from Eq.(5.17), the fact that

$$[\nabla^2,L_\pm]=0$$ (520)

acts as a guarantee that all the basis elements obtained from these raising and lowering operators lie in the same subspace characterized by the degenerate eigenvalue $-k^2$ of the Laplacian.

For illustrative purposes we compute the first few cylinder harmonics. Starting with $Z_0(\rho)=Z_0(kr)$ , one obtains:

$$Z_1e^{i\theta} = -e^{i\theta}\left(\frac{\partial}{\partial\rho} + \frac{i}{\rho}~\frac{\partial}{\partial\theta}\right) Z_0~~\qquad~~$$

$$\Rightarrow Z_1 = -Z'_0$$ (521)

$$Z_2 e^{2i\theta} = -e^{i\theta}\left(\frac{\partial}{\partial\rho} + \frac{i}{\rho}~\frac{\partial}{\partial\theta}\right) (-)e^{i\theta} Z'_0$$

$$= e^{2i\theta}\left( Z''_0 - \frac{1}{\rho} Z'_0\right)~\quad~$$

$$\Rightarrow Z_2 = Z''_0 -\frac{1}{\rho} Z'_0$$

$$Z_3 e^{3i\theta} = -e^{i\theta}\left[\frac{\partial}{\partial\rho} + \frac{i}{\rho}~\frac{\partial}{\partial\theta}\right] e^{2i\theta} Z_2$$

$$= -e^{3i\theta}\left[\left( Z''_0-\frac{1}{\rho} Z'_0\right)'- \frac{2}{\rho}\left( Z''_0-\frac{1}{\rho} Z'_0\right)\right]$$

$$= -e^{3i\theta}\left[ Z'''_0 -\frac{3}{\rho} Z''_0+\frac{3}{\rho^2} Z'_0 \right]$$

$$\Rightarrow Z_3 = -Z'''_0+\frac{3}{\rho} Z''_0-\frac{3}{\rho^2} Z'_0$$

$$Z_4 e^{4i\theta} = -e^{i\theta}\left[\frac{\partial}{\partial\rho} + \frac{i}{\rho}~\frac{\partial}{\partial\theta}\right] e^{3i\theta} Z_3$$

$$= -e^{4i\theta}\left[\left( -Z'''_0+\frac{3}{\rho} Z''_0-\frac{3}{\... ...}{\rho}\left( -Z'''_0+\frac{3}{\rho} Z''_0-\frac{3} {\rho^2} Z'_0\right)\right]$$

$$= e^{4i\theta}\left[ Z^{iv}_0 - \frac{6}{\rho} Z'''_0 +\frac{12}{\rho^2} Z''_0-\frac{15}{\rho^3} Z'_0\right]$$

$$\Rightarrow Z_4 = Z^{iv}_0 - \frac{6}{\rho} Z'''_0 +\frac{12}{\rho^2} Z''_0 - \frac{15}{\rho^3} Z'_0\,.$$ (522)

Lecture 43

Property 18 (Plane wave as a combination of cylinder waves)
Recall that the cylinder harmonics of (complex) order $\nu$ where constructed as a linear superposition of plane wave solutions

$$\psi (r,\theta ) = H_\nu (kr)e^{i\nu\theta}$$

$$= \frac{e^{-i\nu\pi /2}}{\pi} \int_C e^{ikr\cos (\alpha -\theta )} e^{i\nu\alpha}d\alpha\,.$$

Let us now consider that harmonic which satisfies (i) the periodicity requirement

$$\psi (r,\theta +2\pi )= \psi (r,\theta )$$

and (ii) the requirement of being finite at the origin $r=0$ . These boundary conditions give rise to the cylinder waves

$$J_m(kr)e^{im\theta} = \frac{e^{-im\pi /2}}{2\pi} \int^{2\pi}_0 e^{ikr\cos \alpha}e^{im\alpha} d\alpha ~e^{im\theta}~\quad~ m=0,\pm 1,\dots$$

$$= \frac{1}{2\pi}\int^{2\pi}_0 e^{ikr\sin\overline{\alpha}} e^{-im \... ...\overline{\alpha} ~e^{im\theta}~\quad~ \overline{\alpha} =\frac{\pi} {2}-\alpha$$

We see that this is the $m$ th Fourier coefficient of $e^{ikr\sin\theta}$ in disguise:

$$e^{ikr\sin\theta} = \sum^\infty_{m=-\infty} J_m(kr)e^{im\theta}\,.$$ (523)

This represents a plane wave propagating along the $y$ axis, and it is easy to remember.

Figure: Isograms of $Re~\displaystyle e^{ikr\sin\theta}$ . The arrow points into the direction of increasing phase.

More generally, an arbitrary plane wave is also represented as a linear combination of cylinder waves.

Figure: Isograms of $$Re~e^{ikr\cos (\theta -\alpha )}$$ .

Indeed, if one replaces $\theta$ with $\frac{\pi}{2}-(\alpha-\theta)$ in Eq.(5.23), one obtains

$$e^{i(k_xx+k_yy)} = e^{ikr\cos (\theta -\alpha )}$$

$$= \sum^\infty_{m=-\infty} J_m(kr)i^m e^{im(\theta -\alpha)}~~.$$ (524)

This means that any plane wave in the Euclidean plane can be represented as a linear combination of cylinder harmonics of integral order. Remark: The plane waves $e^{ikr\sin\theta}$ and $e^{ikr\cos (\theta -\alpha )}$ are sometimes called generating functions of the Bessel functions of integral order. By considering appropriate derivatives and power series expansions one obtains various identities among these Bessel functions.