# PlanetPhysics/Bessel Equation

The linear differential equation
   ${\begin{matrix}x^{2}{\frac {d^{2}y}{dx^{2}}}+x{\frac {dy}{dx}}+(x^{2}-p^{2})y=0,\end{matrix}}$
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
   ${\begin{matrix}y=x^{r}\sum _{k=0}^{\infty }a_{k}x^{k}=\sum _{k=0}^{\infty }a_{k}x^{r+k},\end{matrix}}$
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): $x^{2}\sum _{k=0}^{\infty }(r+k)(r+k-1)a_{k}x^{r+k-2}+x\sum _{k=0}^{\infty }(r+k)a_{k}x^{r+k-1}+(x^{2}-p^{2})\sum _{k=0}^{\infty }a_{k}x^{r+k}=0.$ 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
   ${\begin{matrix}{\begin{cases}{r^{2}-p^{2}}a_{0}=0,\\{(r+1)^{2}-p^{2}}a_{1}=0,\\{(r+2)^{2}-p^{2}}a_{2}+a_{0}=0,\\\qquad \qquad \ldots \\{(r+k)^{2}-p^{2}}a_{k}+a_{k-2}=0.\end{cases}}\end{matrix}}$
The last of those can be written $(r+k-p)(r+k+p)a_{k}+a_{k-2}=0.$ Because\, $a_{0}\neq 0$,\, the first of those (the indicial equation) gives\, $r^{2}-p^{2}=0$,\, i.e. we have the roots $r_{1}=p,\,\,r_{2}=-p.$ Let's first look the the solution of (1) with\, $r=p$;\, then\, $k(2p+k)a_{k}+a_{k-2}=0$,\, and thus\, $a_{k}=-{\frac {a_{k-2}}{k(2p+k).}}$ 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
   ${\begin{matrix}{\begin{cases}a_{1}=0,\,\,a_{3}=0,\,\ldots ,\,a_{2m-1}=0;\\a_{2}=-{\frac {a_{0}}{2(2p+2)}},\,\,a_{4}={\frac {a_{0}}{2\cdot 4(2p+2)(2p+4)}},\,\ldots ,\,\,a_{2m}={\frac {(-1)^{m}a_{0}}{2\cdot 4\cdot 6\cdots (2m)(2p+2)(2p+4)\ldots (2p+2m)}}\end{cases}}\end{matrix}}$
(where\, $m=1,\,2,\,\ldots$). Putting the obtained coefficients to (2) we get the particular solution
   ${\begin{matrix}y_{1}:=a_{0}x^{p}\left[1\!\!{\frac {x^{2}}{2(2p\!+\!2)}}\!+\!{\frac {x^{4}}{2\!\cdot \!4(2p\!+\!2)(2p\!+\!4)}}\!-\!{\frac {x^{6}}{2\!\cdot \!4\!\cdot \!6(2p\!+\!2)(2p\!+\!4)(2p\!+\!6)}}\!+-\ldots \right]\end{matrix}}$
In order to get the coefficients $a_{k}$ for the second root\, $r_{2}=-p$\, we have to look after that $(r_{2}+k)^{2}-p^{2}\neq 0,$ or\, $r_{2}+k\neq p=r_{1}$.\, Therefore $r_{1}-r_{2}=2p\neq k$ 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 $-p$:
   ${\begin{matrix}y_{2}:=a_{0}x^{-p}\!\left[1\!-\!{\frac {x^{2}}{2(-2p\!+\!2)}}\!+\!{\frac {x^{4}}{2\!\cdot \!4(-2p\!+\!2)(-2p\!+\!4)}}\!-\!{\frac {x^{6}}{2\!\cdot \!4\!\cdot \!6(-2p\!+\!2)(-2p\!+\!4)(-2p\!+\!6)}}\!+-\ldots \right]\end{matrix}}$
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 $a_{0}={\frac {1}{2^{p}\,\Gamma (p+1)}},$ the solution $y_{1}$ is called the \htmladdnormallink{Bessel function {http://planetphysics.us/encyclopedia/BesselEquation2.html} 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 $y:=C_{1}J_{p}(x)+C_{2}J_{-p}(x),$ where\, $J_{-p}(x)=y_{2}$\, with\, $a_{0}={\frac {1}{2^{-p}\Gamma (-p+1)}}$.
The explicit expressions for $J_{\pm p}$ are
   ${\begin{matrix}J_{\pm p}(x)=\sum _{m=0}^{\infty }{\frac {(-1)^{m}}{m!\,\Gamma (m\pm p+1)}}\left({\frac {x}{2}}\right)^{2m\pm p},\end{matrix}}$
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 $y_{1}={\frac {x^{\frac {1}{2}}}{{\sqrt {2}}\,\Gamma ({\frac {3}{2}})}}\left[1\!-\!{\frac {x^{2}}{2\!\cdot \!3}}\!+\!{\frac {x^{4}}{2\!\cdot \!4\!\cdot \!3\!\cdot \!5}}\!-\!{\frac {x^{6}}{2\!\cdot \!4\cdot \!6\!\cdot \!3\!\cdot \!5\!\cdot \!7}}\!+-\ldots \right]={\sqrt {\frac {2}{\pi x}}}\left(x\!-\!{\frac {x^{3}}{3!}}\!+\!{\frac {x^{5}}{5!}}\!-+\ldots \right).$ Thus we get $J_{\frac {1}{2}}(x)={\sqrt {\frac {2}{\pi x}}}\sin {x};$ analogically (6) yields $J_{-{\frac {1}{2}}}(x)={\sqrt {\frac {2}{\pi x}}}\cos {x},$ and the general solution of the equation (1) for\, $p={\frac {1}{2}}$\, is $y:=C_{1}J_{\frac {1}{2}}(x)+C_{2}J_{-{\frac {1}{2}}}(x).$
In the case that $p$ is a non-negative integer $n$, the "+" case of (7) gives the solution $J_{n}(x)=\sum _{m=0}^{\infty }{\frac {(-1)^{m}}{m!\,(m+n)!}}\left({\frac {x}{2}}\right)^{2m+n},$ but for\, $p=-n$\, the expression of $J_{-n}(x)$ is $(-1)^{n}J_{n}(x)$, 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\, $y=K_{n}(x)=J_{n}(x)\ln {x}+x^{-n}\sum _{k=0}^{\infty }b_{k}x^{k}$.\, Then the general solution is\, $y:=C_{1}J_{n}(x)+C_{2}K_{n}(x)$.\\
Other formulae
The first kind Bessel functions of integer order have the generating function $F$:
   ${\begin{matrix}F(z,\,t)=e^{{\frac {z}{2}}(t-{\frac {1}{t}})}=\sum _{n=-\infty }^{\infty }J_{n}(z)t^{n}\end{matrix}}$
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 $c_{n}={\frac {1}{2\pi i}}\oint _{\gamma }{\frac {f(t)}{(t-a)^{n+1}}}\,dt$ 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 $c_{n}={\frac {1}{2\pi i}}\oint _{\gamma }{\frac {e^{\frac {zt}{2}}e^{-{\frac {z}{2t}}}}{t^{n+1}}}\,dt={\frac {1}{2\pi i}}\left({\frac {z}{2}}\right)^{n}\!\oint _{\delta }{\frac {e^{\zeta }e^{-{\frac {z^{2}}{4\zeta }}}}{\zeta ^{n+1}}}\,d\zeta =\sum _{m=0}^{\infty }{\frac {(-1)^{m}}{m!}}\left({\frac {z}{2}}\right)^{2m+n}\!{\frac {1}{2\pi i}}\oint _{\delta }\zeta ^{-m-n-1}e^{\zeta }\,d\zeta .$ 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 $c_{n}=\sum _{m=0}^{\infty }{\frac {(-1)^{m}}{m!\Gamma (m+n+1)}}\left({\frac {z}{2}}\right)^{2m+n}=J_{n}(z).$ 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: $J_{n}(z)={\frac {1}{\pi }}\int _{0}^{\pi }\cos(n\varphi -z\sin {\varphi })\,d\varphi$ Also one can obtain the addition formula $J_{n}(x+y)=\sum _{\nu =-\infty }^{\infty }J_{\nu }(x)J_{n-\nu }(y)$ and the series representations of cosine and sine: $\cos {z}=J_{0}(z)-2J_{2}(z)+2J_{4}(z)-+\ldots$ $\sin {z}=2J_{1}(z)-2J_{3}(z)+2J_{5}(z)-+\ldots$