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[parent] Bessel's equation (Definition)

The linear differential equation

$\displaystyle x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}+(x^2-p^2)y = 0,$ (1)

in which $ p$ is a constant (non-negative if it is real), is called the Bessel's equation. We derive its general solution by trying the series form
$\displaystyle y = x^r\sum_{k=0}^\infty a_kx^k = \sum_{k=0}^\infty a_kx^{r+k},$ (2)

due to Frobenius. Since the parameter $ r$ is indefinite, we may regard $ a_0$ as distinct from 0.

We substitute (2) and the derivatives of the series in (1):

$\displaystyle x^2\sum_{k=0}^\infty(r+k)(r+k-1)a_kx^{r+k-2}+ x\sum_{k=0}^\infty(r+k)a_kx^{r+k-1}+ (x^2-p^2)\sum_{k=0}^\infty a_kx^{r+k} = 0. $
Thus the coefficients of the powers $ x^r$, $ x^{r+1}$, $ x^{r+2}$ and so on must vanish, and we get the system of equations
\begin{align*}\begin{cases}{[}r^2-p^2{]}a_0 = 0,\\ {[}(r+1)^2-p^2{]}a_1 = 0,\\ {... ...\qquad \qquad \ldots\\ {[}(r+k)^2-p^2{]}a_k+a_{k-2} = 0. \end{cases}\end{align*} (3)

The last of those can be written
$\displaystyle (r+k-p)(r+k+p)a_k+a_{k-2} = 0.$
Because $ a_0 \neq 0$, the first of those (the indicial equation) gives $ r^2-p^2 = 0$, i.e. we have the roots
$\displaystyle r_1 = p,\,\, r_2 = -p.$
Let's first look the the solution of (1) with $ r = p$; then $ k(2p+k)a_k+a_{k-2} = 0$, and thus
$\displaystyle a_k = -\frac{a_{k-2}}{k(2p+k).}$
From the system (3) we can solve one by one each of the coefficients $ a_1$, $ a_2$, $ \ldots$ and express them with $ a_0$ which remains arbitrary. Setting for $ k$ the integer values we get
\begin{align*}\begin{cases}a_1 = 0,\,\,a_3 = 0,\,\ldots,\, a_{2m-1} = 0;\\ a_2 =... ...^ma_0}{2\cdot4\cdot6\cdots(2m)(2p+2)(2p+4)\ldots(2p+2m)} \end{cases}\end{align*} (4)

(where $ m = 1,\,2,\,\ldots$). Putting the obtained coefficients to (2) we get the particular solution
$\displaystyle y_1 := a_0x^p \left[1\!\!\frac{x^2}{2(2p\!+\!2)}\! +\!\frac{x^4}{... ...frac{x^6}{2\!\cdot\!4\!\cdot\!6(2p\!+\!2)(2p\!+\!4)(2p\!+\!6)}\!+-\ldots\right]$ (5)

In order to get the coefficients $ a_k$ for the second root $ r_2 = -p$ we have to look after that

$\displaystyle (r_2+k)^2-p^2 \neq 0,$
or $ r_2+k \neq p = r_1$. Therefore
$\displaystyle r_1-r_2 = 2p \neq k$
where $ k$ is a positive integer. Thus, when $ p$ is not an integer and not an integer added by $ \frac{1}{2}$, we get the second particular solution, gotten of (5) by replacing $ p$ by $ -p$:
$\displaystyle y_2 := a_0x^{-p}\!\left[1 \!-\!\frac{x^2}{2(-2p\!+\!2)}\!+\!\frac... ...c{x^6}{2\!\cdot\!4\!\cdot\!6(-2p\!+\!2)(-2p\!+\!4)(-2p\!+\!6)}\!+-\ldots\right]$ (6)

The power series of (5) and (6) converge for all values of $ x$ and are linearly independent (the ratio $ y_1/y_2$ tends to 0 as $ x\to\infty$). With the appointed value

$\displaystyle a_0 = \frac{1}{2^p\,\Gamma(p+1)},$
the solution $ y_1$ is called the Bessel function of the first kind and of order $ p$ and denoted by $ J_p$. The similar definition is set for the first kind Bessel function of an arbitrary order $ p\in \mathbb{R}$ (and $ \mathbb{C}$). For $ p\notin \mathbb{Z}$ the general solution of the Bessel's differential equation is thus
$\displaystyle y := C_1J_p(x)+C_2J_{-p}(x),$
where $ J_{-p}(x) = y_2$ with $ a_0 = \frac{1}{2^{-p}\Gamma(-p+1)}$.

The explicit expressions for $ J_{\pm p}$ are

$\displaystyle J_{\pm p}(x) = \sum_{m=0}^\infty \frac{(-1)^m}{m!\,\Gamma(m\pm p+1)}\left(\frac{x}{2}\right)^{2m\pm p},$ (7)

which are obtained from (5) and (6) by using the last formula for gamma function.

E.g. when $ p = \frac{1}{2}$ the series in (5) gets the form

$\displaystyle y_1 = \frac{x^{\frac{1}{2}}}{\sqrt{2}\,\Gamma(\frac{3}{2})}\left[... ...\frac{2}{\pi x}}\left(x\!-\!\frac{x^3}{3!}\!+\!\frac{x^5}{5!}\!-+\ldots\right).$
Thus we get
$\displaystyle J_{\frac{1}{2}}(x) = \sqrt{\frac{2}{\pi x}}\sin{x};$
analogically (6) yields
$\displaystyle J_{-\frac{1}{2}}(x) = \sqrt{\frac{2}{\pi x}}\cos{x},$
and the general solution of the equation (1) for $ p = \frac{1}{2}$ is
$\displaystyle y := C_1J_{\frac{1}{2}}(x)+C_2J_{-\frac{1}{2}}(x).$

In the case that $ p$ is a non-negative integer $ n$, the “+” case of (7) gives the solution

$\displaystyle J_{n}(x) = \sum_{m=0}^\infty \frac{(-1)^m}{m!\,(m+n)!}\left(\frac{x}{2}\right)^{2m+n}, $
but for $ p = -n$ the expression of $ J_{-n}(x)$ is $ (-1)^nJ_n(x)$, i.e. linearly dependent of $ J_n(x)$. It can be shown that the other solution of (1) ought to be searched in the form $ y = K_n(x) = J_n(x)\ln{x}+x^{-n}\sum_{k=0}^\infty b_kx^k$. Then the general solution is $ y := C_1J_n(x)+C_2K_n(x)$.

Other formulae

The first kind Bessel functions of integer order have the generating function $ F$:

$\displaystyle F(z,\,t) = e^{\frac{z}{2}(t-\frac{1}{t})} = \sum_{n=-\infty}^\infty J_n(z)t^n$ (8)

This function has an essential singularity at $ t = 0$ but is analytic elsewhere in $ \mathbb{C}$; thus $ F$ has the Laurent expansion in that point. Let us prove (8) by using the general expression
$\displaystyle c_n = \frac{1}{2\pi i}\oint_{\gamma} \frac{f(t)}{(t-a)^{n+1}}\,dt$
of the coefficients of Laurent series. Setting to this $ a := 0$, $ f(t) := e^{\frac{z}{2}(t-\frac{1}{t})}$, $ \zeta := \frac{zt}{2}$ gives
$\displaystyle c_n = \frac{1}{2\pi i} \oint_\gamma\frac{e^{\frac{zt}{2}}e^{-\fra... ...{2}\right)^{2m+n}\! \frac{1}{2\pi i}\oint_\delta \zeta^{-m-n-1}e^\zeta\,d\zeta.$
The paths $ \gamma$ and $ \delta$ go once round the origin anticlockwise in the $ t$-plane and $ \zeta$-plane, respectively. Since the residue of $ \zeta^{-m-n-1}e^\zeta$ in the origin is $ \frac{1}{(m+n)!} = \frac{1}{\Gamma(m+n+1)}$, the residue theorem gives
$\displaystyle c_n = \sum_{m=0}^\infty \frac{(-1)^m}{m!\Gamma(m+n+1)}\left(\frac{z}{2}\right)^{2m+n} = J_n(z).$
This means that $ F$ has the Laurent expansion (8).

By using the generating function, one can easily derive other formulae, e.g. the integral representation of the Bessel functions of integer order:

$\displaystyle J_n(z) = \frac{1}{\pi}\int_0^\pi\cos(n\varphi-z\sin{\varphi})\,d\varphi$
Also one can obtain the addition formula
$\displaystyle J_n(x+y) = \sum_{\nu=-\infty}^{\infty}J_\nu(x)J_{n-\nu}(y)$
and the series representations of cosine and sine:
$\displaystyle \cos{z} = J_0(z)-2J_2(z)+2J_4(z)-+\ldots$
$\displaystyle \sin{z} = 2J_1(z)-2J_3(z)+2J_5(z)-+\ldots$

Bibliography

1
N. PISKUNOV: Diferentsiaal- ja integraalarvutus kõrgematele tehnilistele õppeasutustele. Kirjastus Valgus, Tallinn (1966).
2
K. KURKI-SUONIO: Matemaattiset apuneuvot. Limes r.y., Helsinki (1966).



"Bessel's equation" is owned by pahio.
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See Also: Laplace equation in cylindrical coordinates, Cauchy residue theorem, properties of entire functions, Frobenius method, table of Laplace transforms

Other names:  Bessel's differential equation, Bessel equation
Also defines:  Bessel's function, Bessel function

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Cross-references: sine, cosine, addition formula, residue, origin, paths, coefficients of Laurent series, point, Laurent expansion, analytic, essential singularity, function, generating function, linearly dependent, gamma function, expressions, similar, ratio, linearly independent, converge, power series, positive, particular solution, integer, solution, roots, indicial equation, equations, vanish, powers, coefficients, derivatives, indefinite, parameter, series, general solution, real, linear differential equation
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This is version 19 of Bessel's equation, born on 2007-01-16, modified 2008-02-21.
Object id is 8774, canonical name is BesselsEquation.
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Classification:
AMS MSC: 33C10 (Special functions :: Hypergeometric functions :: Bessel and Airy functions, cylinder functions, $_0F_1$)
 34A05 (Ordinary differential equations :: General theory :: Explicit solutions and reductions)
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Rendering problem in HTML mode by pahio on 2007-02-14 02:50:05
Hi, experts of TeX! Rendering the entry "Bessel's equation" does not succeed in HTML mode; it forms great black blocks and the formulas after them are in a mess. I have tried to alter the preamble, but not succeeded. I think the below portion of the code causes the problem, but I cannot fix the thing =o( Please help!

\begin{align}
\begin{cases}
[r^2-p^2]a_0 = 0,\\
[(r+1)^2-p^2]a_1 = 0,\\
[(r+2)^2-p^2]a_2+a_0 = 0,\\
\qquad \qquad \ldots\\
[(r+k)^2-p^2]a_k+a_{k-2} = 0.
\end{cases}
\end{align}

Jussi
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