Return to computing page for the first course APMA0330
Return to computing page for the second course APMA0340
Return to computing page for the fourth course APMA0360
Return to Mathematica tutorial for the first course APMA0330
Return to Mathematica tutorial for the second course APMA0340
Return to Mathematica tutorial for the fourth course APMA0360
Return to the main page for the course APMA0330
Return to the main page for the course APMA0340
Return to the main page for the course APMA0360
Introduction to Linear Algebra with Mathematica
Glossary
Preface
Hankel transformations are integral transformations whose kernels are Bessel functions. They are sometimes referred to as Bessel or Fourier--Bessel transformations. These transformations were first developed by the German mathematician Hermann Hankel (1839--1873). When you are dealing with problems that show circular symmetry, Hankel transforms may be very useful. Laplace's partial differential equation in ctlindrical coordinates can be transformed into ordinary differential equation because its radial part is transferted into a constant by Hankel transform, being its spectral representaion.
Hankel Transformation
All functions we will consider will be assumed to define on half line [0, ∞) such that
\[
\int_0^{+\infty} | f(r)|\, r^{1/2} {\text d} r < \infty .
\]
The Hankel transform of order ν of a function f(r) is given by
For Hankel transformations, we have
\begin{equation} \label{EqHankel.1}
F_{\nu} (k) = {\cal H}_{\nu} \left[ f \right] = \int_0^{\infty} f(r)\,J_{\nu} (kr)\,r\,{\text d}r ,
\end{equation}
where Jν is the Bessel function of the first kind of order
ν with ν ≥ −1/2. The inverse Hankel transform of
Fν(k) is defined as
\begin{equation} \label{EqHankel.2}
f(r) = \int_0^{\infty} F_{\nu} (k)\,J_{\nu} (kr)\,k\,{\text d}k .
\end{equation}
\begin{equation} \label{EqHankel.3}
\int_0^{\infty} r\left( \frac{{\text d}^2 f}{{\text d} r^2} + \frac{1}{r} \,
\frac{{\text d} f}{{\text d} r} - \frac{\nu^2}{r^2} \, f \right) J_{\nu} (rk)\,{\text d}r = - k^2 F_{\nu} (k) = - k^2 \int_0^{\infty} f(r)\,J_{\nu} (kr)\,r\,{\text d}r .
\end{equation}
Besides spectral decomposition idenity \eqref{EqHankel.3}, we mention several other properties of the Hankel transformation.
-
Similarity:
\[ {\cal H}_{\nu} \left[ f(ar) \right] = \frac{1}{a^2}\,F_{\nu} \left( \frac{k}{\nu} \right) . \]
-
Derivative:
\[ {\cal H}_{\nu} \left[ f' (r) \right] = k \left[ \frac{\nu + 1}{2\nu}\, F_{\nu -1} (k) - \frac{\nu - 1}{2\nu}\, F_{\nu +1} (k) \right] . \]
-
Division by r:
\[ {\cal H}_{\nu} \left[ \frac{f(r)}{r} \right] = \frac{k}{2\nu} \left[ F_{\nu -1} (k) + F_{\nu +1} (k) \right] . \]
-
Parseval's theorem: If
\[ F_{\nu} (k) = {\cal H}_{\nu} \left[ f(r) \right] , \qquad G_{\nu} (k) = {\cal H}_{\nu} \left[ g(r) \right] , \]then\[ \int_0^{+\infty} F_{\nu}(k)\,G_{\nu} (k) \,k\,{\text d}k = \int_0^{+\infty} r\,f(r)\,g(r)\,{\text d} r . \]
-
\[ {\cal H}_{\nu} \left[ r^{\nu -1} \frac{\text d}{{\text d}r} \left( r^{1-\nu} f(r) \right) \right] = k{\cal H}_{\nu +1} \left[ f (r) \right] . \]
-
\[ {\cal H}_{\nu} \left[ r^{-\nu -1} \frac{\text d}{{\text d}r} \left( r^{1+\nu} f(r) \right) \right] = -k{\cal H}_{\nu -1} \left[ f (r) \right] . \]
Example 1: Let r² = x² + y². Then
\[
{\cal H}_{\nu} \left[ e^{-a r^2} \right] = \frac{1}{2a}\, e^{- k^2 /(4a)} .
\]
■ Example 2: We consider a piecewise continuous function:
\[
h_a (r) = \begin{cases}
1 , & \quad\mbox{if } |r| < a , \\
0, & \quad\mbox{otherwise}.
\end{cases}
\]
Its Hankel transform becomes
\[
{\cal H}_{\nu} \left[ h_a (r) \right] = \frac{a}{k}\, J_1 (ak) .
\]
■
End of Example 2
Example 3: From the formula
\[
\int_0^{\infty} r\,\delta (r-a)\, J_0 (kr)\,{\text d}r = \frac{1}{k}
\]
we find
\[
{\cal H}_{0} \left[ \frac{1}{r} \right] = \left[ \frac{1}{k} \right] .
\]
■ Example 4: From the identity
\[
{\cal H}_{0} \left[ \delta (r-1) \right] = a\,J_0 (ak) ,
\]
we get
\[
{\cal H}_{0} \left[ a\,J_0 (ar) \right] = \delta (k-a ) .
\]
■
End of Example 4
Example 5: According to the definition of Hankel transformation,
\[
{\cal H}_{\nu} \left[ r^{\nu} H(a- r) \right] = \int_0^a r^{\nu +1} J_{\nu} (kr)\,{\text d} r = \frac{1}{k^{\nu +2}} \int_0^{ak} x^{\nu +1} J_{\nu} (x)\,{\text d} x .
\]
Since the Heaviside function is a unit function, we have
\[
{\cal H}_{\nu} \left[ r^{\nu -1} H(a-r) \right] = \frac{(ak)^{\nu +1}}{k^{\nu +2}} \, J_{\nu +1} (ak) = \frac{a^{\nu +1}}{k} \, J_{\nu +1} (ak) .
\]
■
End of Example 5
Example 6: The Hankel transform of \( \displaystyle r^{\nu -1} e^{-ar} , \quad a > 0, \) is given by
\begin{align*}
{\cal H}_{\nu} \left[ r^{\nu -1} e^{-ar} \right] &= \int_0^{\infty} r^{\nu} e^{-ar} J_{\nu} (kr)\,{\text d} r = \frac{1}{k^{\nu +1}} \int_0^{\infty} t^{\nu} J_{\nu} (t)\,e^{-at/k} {\text d} t
\\
&= \frac{1}{k^{\nu +1}} {\cal L} \left[ t^{\nu} J_{\nu} (t)\ \lambda \to \frac{a}{k} \right] ,
\end{align*}
where we set y = rk and ℒ is the Laplace transform operator. Since
\[
t^{\nu} J_{\nu} (t) = \sum_{n\ge 0} \frac{(-1)^n}{n! \Gamma (n+\nu +1)\, 2^{2n + \nu}}\, t^{2n + 2\nu} ,
\]
we have
\begin{align*}
{\cal L}_{t \to \lambda} \left[ t^{\nu} J_{\nu} (t) \right] &= \sum_{n\ge 0} \frac{(-1)^n}{n! \Gamma (n+\nu +1)\, 2^{2n + \nu}}\, {\cal L}_{t \to \lambda} \left[ t^{2n + 2\nu} \right]
\\
&= \sum_{n\ge 0} \frac{(-1)^n}{n! \Gamma (n+\nu +1)\, 2^{2n + \nu}}\cdot \frac{2n + 2\nu +1}{\lambda^{2n + 2\nu +1}} .
\end{align*}
Using formula
\[
\frac{\Gamma (2n+ 2\nu +1)}{\Gamma (n+\nu +1)} = \frac{1}{\sqrt{\pi}} \,2^{2n+ 2\nu} \Gamma \left( n + \nu + \frac{1}{2} \right) ,
\]
we rewrite the Laplace transform as
\[
{\cal L}_{t \to \lambda} \left[ t^{\nu} J_{\nu} (t) \right] = \frac{2^{\nu}}{\sqrt{\pi}\,\lambda^{2\nu +1}} \sum_{n\ge 0} \frac{(-1)^n \Gamma \left( n+\nu + \frac{1}{2} \right)}{n!} \left( \frac{1}{\lambda^2} \right)^n
\]
Using the binomial formular
\[
(1+x)^{-c} = \sum_{n\ge 0} \binom{-c}{n} x^n = \sum_{n\ge 0} \frac{(-1)^n \Gamma (n+c)}{n! \Gamma (c)}\, x^n ,
\]
we rewrite the latter as
\[
{\cal L}_{t \to \lambda} \left[ t^{\nu} J_{\nu} (t) \right] = \frac{2^{\nu} \Gamma \left( \nu + \frac{1}{2} \right)}{\sqrt{\pi}\,\left( \lambda^2 + 1 \right)^{\nu + 1/2}}
\]
Since λ = 𝑎/k, we get
\[
{\cal H}_{\nu} \left[ r^{\nu -1} e^{-ar} \right] = \frac{k^{\nu} 2^{\nu} \Gamma \left( \nu + \frac{1}{2} \right)}{\sqrt{\pi} \left( a^2 \k^2 \right)^{\nu + 1/2}} .
\]
For ν = 0, we have
\[
{\cal H}_{0} \left[ r^{-1} e^{-ar} \right] = \int_0^{+\infty} e^{-ar} J_0 (kr)\,{\text d} r = \frac{1}{\sqrt{a^2 + k^2}} , \qquad a >' 0 .
\]
■
End of Example 6
Example 7: The Hankel transformation of the exponential function is
\begin{align*}
{\cal H}_{0} \left[ e^{-ar} \right] &= {\cal L}_{r \to a} \left[ r\,J_0 (kr) \right]
\\
&= - \frac{\text d}{{\text d}a} \left[ \frac{1}{\sqrt{k^2 + a^2}} \right]
\\
&= a \left( k^2 + a^2 \right)^{-3/2} , \qquad a > 0 .
\end{align*}
■
End of Example 7
Example 8: Let v be the electric potential due to flat circular electrified disc, with radius R = 1, the center of the disc being at the origin of the three-dimensional space and its axis alongthe z-axis. In polar coordinates, the potential satisfies Laplace's equation
\[
\nabla^2 v = \frac{\partial^2 v}{\partial r^2} + \frac{1}{r}\,\frac{\partial v}{\partial r} + \frac{\partial^2 v}{\partial z^2} = 0 .
\tag{8.1}
\]
The boundary conditions are
\[
v(r, 0) = v_0 , \qquad 0 \le r \le R,
\tag{8.2}
\]
\[
\frac{\partial v}{\partial z} \left( r, 0 \right) = 0 , \qquad 1=R < r .
\tag{8.3}
\]
Let
\[
V(k, z) = {\cal H}_0 \left[ v(r, z) \right] = \int_0^{\infty} r\, v(r, z)\, J_0 (kr)\,{\text d} r ,
\]
which leads to
\[
{\cal H}_0 \left[ \nabla^2 v(r, z) \right] = - k^2 V(k, z) + \frac{\partial^2 v}{\partial z^2} = 0 .
\]
The general solution of this ordinary differential equation is
\[
V(k, z) = A(k)\, e^{-kz} + B(k)\, e^{kz} ,
\]
where A and B are have to be determined to satisfy the boundary conditions. We have to set B ≡ 0 because its multiple grows up exponential. Hence,
\[
v(r, z) = \int_0^{+\infty} k\, A(k)\,e^{-kz} J_0 (kr)\,{\text d} k .
\]
From the boundary conditions, we get two equations
\[
v(r, 0) = \int_0^{+\infty} k\, A(k)\, J_0 (kr)\,{\text d} k = v_0 ,
\tag{8.4}
\]
and
\[
\frac{\partial v}{\partial z} \left( r, 0 \right) = \int_0^{\infty} k^2 A(k)\,J_0 (kr)\,{\text d} k = 0.
\tag{8.5}
\]
They are satisfied when
\[
v(r, z) = \frac{2\,v_0}{\pi} \int_0^{\infty} \frac{\sin k}{k}\, e^{-kz} J_0 (kr)\,{\text d} k .
\]
■
End of Example 8
Example 9: Let us find a solution to the boundary value problem
\[
\nabla^2 v = \frac{\partial^2 v}{\partial r^2} + \frac{1}{r}\,\frac{\partial v}{\partial r} + \frac{\partial^2 v}{\partial z^2} = 0 , \qquad 0 < r, z < \infty ,
\tag{9.1}
\]
\[
v(r, 0) = f(r) .
\tag{9.2}
\]
Taking the Hankel transformation of zero order yields
\[
- k^2 V(k, z) + \frac{\partial^2 V}{\partial z^2} = 0 ,
\tag{9.3}
\]
subject
\[
V(k, 0) = \int_0^{\infty} r\, f(r) \, J_0 (kr} \,{\text d} r .
\]
Its solution is known to be
\[
V(k, z) = e^{-kz} \int_0^{\infty} r\,f(r)\, J_0 (kr)\,{\text d} r.
\]
Then
\[
v(r, z) = \int_0^{\infty} k\, e^{-kz} J_0 (kr)\,{\text d}k \int_0^{\infty} p\,f(p)\, J_0 (kr)\,{\text d} p .
\]
■
End of Example 9
The finite Hankel transform
\begin{equation} \label{EqHankel.4}
F_{\nu} (k) = {\cal H}_{\nu} \left[ f; k \right] = \int_0^1 r\,f(r)\,J_{\nu} (kr)\,{\text d} r .
\end{equation}
Let &Deltaν be the Bessel operator
\[
\Delta_{\nu} = \frac{{\text d}^2}{{\text d} r^2} + \frac{1}{r}\,\frac{\text d}{{\text d}r} - \frac{\nu^2}{r^2} .
\]
Then finite Hankel transformation gives
\begin{equation} \label{EqHankel.5}
{\cal H}_{\nu} \left[ \Delta_{\nu} f; k \right] = -k^2 {\cal H}_{\nu} \left[ f; k \right] + J_{\nu} (k)\,f' (1) - k\,J'_{\nu} (k)\, f(1) .
\end{equation}
Example 10: Let v(r, t) be a temberature at them t of a long solid cylinder of unit radius. The nitial temperature in unity and radiation takes place at the surface into the surrounding medium maintained at zero temperature.
The mathematical model of this problem consists of heat differential equation
\[
\frac{\partial v}{\partial t} = \frac{\partial^2 v}{\partial r^2} + \frac{1}{r}\,\frac{\partial v}{\partial r} , \qquad 0 < r < 1, \quad t > 0,
\tag{10.1}
\]
with the initial condition
\[
v(r, 0) = 1 = g(r) , \qquad 0 < r < 1 .
\tag{10.2}
\]
The radial condition on the surface of the cylinder is specified by mixed boundary condition
\[
\frac{\partial v}{\partial r} \left( 1, t) + h\,v(1, t) = \varphi (t) = 0,
\tag{10.3}
\]
wherw h is a positive constant. Let
\[
V(r, t) = \int_0^1 r\, v(r, t)\, J_0 (kr)\, {\text d} r .
\]
■
End of Example 10