PlanetPhysics/Bessel Equation

The linear [[../DifferentialEquations/|differential equation]]

$\begin{array}{c}{x}^2\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}+(x^2-p^2)y=0,\end{array}$

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

$\begin{array}{c}y=x^r\sum _{k=0}^{\mathrm{\infty }}{a}_k{x}^k=\sum _{k=0}^{\mathrm{\infty }}{a}_k{x}^{r+k},\end{array}$

due to Frobenius.\, Since the [[../Parameter/|parameter]] $r$ is indefinite, we may regard $a_0$ as distinct from 0.

We substitute (2) and the derivatives of the series in (1): $x^2\sum _{k=0}^{\mathrm{\infty }}(r+k)(r+k-1)a_k{x}^{r+k-2}+x\sum _{k=0}^{\mathrm{\infty }}(r+k)a_k{x}^{r+k-1}+(x^2-p^2)\sum _{k=0}^{\mathrm{\infty }}{a}_k{x}^{r+k}=0.$ Thus the coefficients of the [[../Power/|powers]] $x^r$, $x^{r+1}$, $x^{r+2}$ and so on must vanish, and we get the [[../SimilarityAndAnalogousSystemsDynamicAdjointnessAndTopologicalEquivalence/|system]] of equations

$\begin{array}{c}\{\begin{array}{c}[r^2-p^2]a_0=0,\\ [(r+1{)}^2-p^2]a_1=0,\\ [(r+2{)}^2-p^2]a_2+a_0=0,\\ \dots \\ [(r+k{)}^2-p^2]a_k+a_{k-2}=0.\end{array}\end{array}$

The last of those can be written $(r+k-p)(r+k+p)a_k+a_{k-2}=0.$ Because\, $a_0\ne 0$,\, 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 $a_1$, $a_2$, $\dots$\, and express them with $a_0$ which remains arbitrary.\, Setting for $k$ the integer values we get

$\begin{array}{c}\{\begin{array}{c}{a}_1=0,a_3=0,\dots ,a_{2m-1}=0;\\ a_2=-\frac{a_0}{2(2p+2)},a_4=\frac{a_0}{2\cdot 4(2p+2)(2p+4)},\dots ,a_{2m}=\frac{(-1{)}^m{a}_0}{2\cdot 4\cdot 6\cdots (2m)(2p+2)(2p+4)\dots (2p+2m)}\end{array}\end{array}$

(where\, $m=1,2,\dots$). Putting the obtained coefficients to (2) we get the particular solution

$\begin{array}{c}{y}_1:=a_0{x}^p[1\frac{x^2}{2(2p+2)}+\frac{x^4}{2\cdot 4(2p+2)(2p+4)}-\frac{x^6}{2\cdot 4\cdot 6(2p+2)(2p+4)(2p+6)}+-\dots ]\end{array}$

In order to get the coefficients $a_k$ for the second root\, $r_2=-p$\, we have to look after that $(r_2+k{)}^2-p^2\ne 0,$ or\, $r_2+k\ne p=r_1$.\, Therefore $r_1-r_2=2p\ne k$ 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 $-p$:

$\begin{array}{c}{y}_2:=a_0{x}^{-p}[1-\frac{x^2}{2(-2p+2)}+\frac{x^4}{2\cdot 4(-2p+2)(-2p+4)}-\frac{x^6}{2\cdot 4\cdot 6(-2p+2)(-2p+4)(-2p+6)}+-\dots ]\end{array}$

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\, $x\to \mathrm{\infty }$).\, With the appointed value $a_0=\frac{1}{2^p\mathrm{\Gamma }(p+1)},$ 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\, $p\in ℝ$ (and $ℂ$). 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\, $a_0=\frac{1}{2^{-p}\mathrm{\Gamma }(-p+1)}$.

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

$\begin{array}{c}{J}_{\pm p}(x)=\sum _{m=0}^{\mathrm{\infty }}\frac{(-1{)}^m}{m!\mathrm{\Gamma }(m\pm p+1)}{\left(\frac{x}{2}\right)}^{2m\pm p},\end{array}$

which are obtained from (5) and (6) by using the last formula for [[../GammaFunction/|gamma function]].

E.g. when\, $p=\frac{1}{2}$\, the series in (5) gets the form $y_1=\frac{x^{\frac{1}{2}}}{\sqrt{2}\mathrm{\Gamma }(\frac{3}{2})}[1-\frac{x^2}{2\cdot 3}+\frac{x^4}{2\cdot 4\cdot 3\cdot 5}-\frac{x^6}{2\cdot 4\cdot 6\cdot 3\cdot 5\cdot 7}+-\dots ]=\sqrt{\frac{2}{\pi x}}(x-\frac{x^3}{3!}+\frac{x^5}{5!}-+\dots ).$ 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 $J_n(x)=\sum _{m=0}^{\mathrm{\infty }}\frac{(-1{)}^m}{m!(m+n)!}{\left(\frac{x}{2}\right)}^{2m+n},$ 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\, $y=K_n(x)=J_n(x)\ln x+x^{-n}\sum _{k=0}^{\mathrm{\infty }}{b}_k{x}^k$.\, 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 [[../Bijective/|function]] $F$:

$\begin{array}{c}F(z,t)=e^{\frac{z}{2}(t-\frac{1}{t})}=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{J}_n(z)t^n\end{array}$

This function has an essential singularity at\, $t=0$\, but is analytic elsewhere in $ℂ$; 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 $c_n=\frac{1}{2\pi i}{\oint }_{\gamma }\frac{e^{\frac{zt}{2}}{e}^{-\frac{z}{2t}}}{t^{n+1}}dt=\frac{1}{2\pi i}{\left(\frac{z}{2}\right)}^n{\oint }_{\delta }\frac{e^{\zeta }{e}^{-\frac{z^2}{4\zeta }}}{{\zeta }^{n+1}}d\zeta =\sum _{m=0}^{\mathrm{\infty }}\frac{(-1{)}^m}{m!}{\left(\frac{z}{2}\right)}^{2m+n}\frac{1}{2\pi i}{\oint }_{\delta }{\zeta }^{-m-n-1}{e}^{\zeta }d\zeta .$ 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\, $\frac{1}{(m+n)!}=\frac{1}{\mathrm{\Gamma }(m+n+1)}$,\, the residue theorem gives $c_n=\sum _{m=0}^{\mathrm{\infty }}\frac{(-1{)}^m}{m!\mathrm{\Gamma }(m+n+1)}{\left(\frac{z}{2}\right)}^{2m+n}=J_n(z).$ 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: $J_n(z)=\frac{1}{\pi }\underset{0}{\overset{\pi }{\int }}\cos (n\phi -z\sin \phi )d\phi$ Also one can obtain the addition [[../Formula/|formula]] $J_n(x+y)=\sum _{\nu =-\mathrm{\infty }}^{\mathrm{\infty }}{J}_{\nu }(x)J_{n-\nu }(y)$ and the series representations of cosine and sine: $\cos z=J_0(z)-2J_2(z)+2J_4(z)-+\dots$ $\sin z=2J_1(z)-2J_3(z)+2J_5(z)-+\dots$