Bessel’s equation

The linear differential equation

$x^2{\displaystyle \frac{d^{2}y}{d{x}^2}}+x{\displaystyle \frac{dy}{dx}}+(x^2-p^2)y=\mathrm{\hspace{0.33em}0},$(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

$y=x^r{\displaystyle \sum _{k=0}^{\mathrm{\infty }}}{a}_k{x}^k={\displaystyle \sum _{k=0}^{\mathrm{\infty }}}{a}_k{x}^{r+k},$(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):

$$x^2\sum _{k=0}^{\mathrm{\infty }}(r+k)(r+k-1)a_k{x}^{r+k-2}+x\sum _{k=0}^{\mathrm{\infty }}(r+k)a_k{x}^{r+k-1}+(x^2-p^2)\sum _{k=0}^{\mathrm{\infty }}{a}_k{x}^{r+k}=\mathrm{\hspace{0.33em}0}.$$

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

$\{\begin{array}{cc}[r^2-p^2]a_0=\mathrm{\hspace{0.33em}0},\hfill & \\ [{(r+1)}^2-p^2]a_1=\mathrm{\hspace{0.33em}0},\hfill & \\ [{(r+2)}^2-p^2]a_2+a_0=\mathrm{\hspace{0.33em}0},\hfill & \\ \mathrm{\dots }\hfill & \\ [{(r+k)}^2-p^2]a_k+a_{k-2}=\mathrm{\hspace{0.33em}0}.\hfill & \end{array}$(3)

The last of those can be written

$$(r+k-p)(r+k+p)a_k+a_{k-2}=\mathrm{\hspace{0.33em}0}.$$

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 $a_1$, $a_2$, $\mathrm{\dots }$  and express them with $a_0$ which remains arbitrary.  Setting for $k$ the integer values we get

$\{\begin{array}{cc}{a}_1=\mathrm{\hspace{0.33em}0},a_3=\mathrm{\hspace{0.33em}0},\mathrm{\dots },a_{2m-1}=\mathrm{\hspace{0.33em}0};\hfill & \\ a_2=-\frac{a_0}{2(2p+2)},a_4=\frac{a_0}{2\cdot 4(2p+2)(2p+4)},\mathrm{\dots },a_{2m}=\frac{{(-1)}^m{a}_0}{2\cdot 4\cdot 6\mathrm{\cdots }(2m)(2p+2)(2p+4)\mathrm{\dots }(2p+2m)}\hfill & \end{array}$(4)

(where  $m=1,\mathrm{\hspace{0.17em}2},\mathrm{\dots }$).Putting the obtained coefficients to (2) we get the particular solution

$y_1:=a_0{x}^p[{\displaystyle \frac{x^2}{2(2p+2)}}+{\displaystyle \frac{x^4}{2\cdot 4(2p+2)(2p+4)}}-{\displaystyle \frac{x^6}{2\cdot 4\cdot 6(2p+2)(2p+4)(2p+6)}}+-\mathrm{\dots }]$(5)

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

$${(r_2+k)}^2-p^2\ne \mathrm{\hspace{0.33em}0},$$

or  $r_2+k\ne p=r_1$.  Therefore

$$r_1-r_2=\mathrm{\hspace{0.33em}2}p\ne k$$

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 $-p$:

$y_2:=a_0{x}^{-p}[1-{\displaystyle \frac{x^2}{2(-2p+2)}}+{\displaystyle \frac{x^4}{2\cdot 4(-2p+2)(-2p+4)}}-{\displaystyle \frac{x^6}{2\cdot 4\cdot 6(-2p+2)(-2p+4)(-2p+6)}}+-\mathrm{\dots }]$(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  $x\to \mathrm{\infty }$).  With the appointed value

$$a_0=\frac{1}{2^p\mathrm{\Gamma }(p+1)},$$

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  $p\in ℝ$ (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

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

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

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

$$y_1=\frac{x^{\frac{1}{2}}}{\sqrt{2}\mathrm{\Gamma }(\frac{3}{2})}[1-\frac{x^2}{2\cdot 3}+\frac{x^4}{2\cdot 4\cdot 3\cdot 5}-\frac{x^6}{2\cdot 4\cdot 6\cdot 3\cdot 5\cdot 7}+-\mathrm{\dots }]=\sqrt{\frac{2}{\pi x}}(x-\frac{x^3}{3!}+\frac{x^5}{5!}-+\mathrm{\dots }).$$

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 on $J_n(x)$.  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 function $F$:

$F(z,t)=e^{\frac{z}{2}(t-\frac{1}{t})}={\displaystyle \sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}}{J}_n(z)t^n$(8)

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 (http://planetmath.org/CauchyResidueTheorem) 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 that $F$ has the Laurent expansion (8).

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

$$J_n(z)=\frac{1}{\pi }{\int }_0^{\pi }\cos (n\phi -z\sin \phi )𝑑\phi$$

Also one can obtain the addition formula

$$J_n(x+y)=\sum _{\nu =-\mathrm{\infty }}^{\mathrm{\infty }}{J}_{\nu }(x)J_{n-\nu }(y)$$

and the series of cosine and sine:

$$\cos z=J_0(z)-2J_2(z)+2J_4(z)-+\mathrm{\dots }$$

$$\sin z=\mathrm{\hspace{0.33em}2}{J}_1(z)-2J_3(z)+2J_5(z)-+\mathrm{\dots }$$