PlanetPhysics/Bessel Equation

The linear differential equation $$\begin{matrix} x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}+(x^2-p^2)y = 0, \end{matrix}$$ 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 $$\begin{matrix} y = x^r\sum_{k=0}^\infty a_kx^k = \sum_{k=0}^\infty a_kx^{r+k}, \end{matrix}$$ 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): $$ x^2\sum_{k=0}^\infty(r+k)(r+k-1)a_kx^{r+k-2}+ x\sum_{k=0}^\infty(r+k)a_kx^{r+k-1}+ (x^2-p^2)\sum_{k=0}^\infty a_kx^{r+k} = 0. $$ 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 $$\begin{matrix} \begin{cases} {[}r^2-p^2{]}a_0 = 0,\\ {[}(r+1)^2-p^2{]}a_1 = 0,\\ {[}(r+2)^2-p^2{]}a_2+a_0 = 0,\\ \qquad \qquad \ldots\\ {[}(r+k)^2-p^2{]}a_k+a_{k-2} = 0. \end{cases} \end{matrix}$$ 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 $$\begin{matrix} \begin{cases} a_1 = 0,\,\,a_3 = 0,\,\ldots,\, a_{2m-1} = 0;\\ a_2 = -\frac{a_0}{2(2p+2)},\,\,a_4 = \frac{a_0}{2\cdot4(2p+2)(2p+4)},\,\ldots,\,\, a_{2m} = \frac{(-1)^ma_0}{2\cdot4\cdot6\cdots(2m)(2p+2)(2p+4)\ldots(2p+2m)} \end{cases} \end{matrix}$$ (where\, $$m = 1,\,2,\,\ldots$$). Putting the obtained coefficients to (2) we get the particular solution $$\begin{matrix} y_1 := a_0x^p \left[1\!\!\frac{x^2}{2(2p\!+\!2)}\! +\!\frac{x^4}{2\!\cdot\!4(2p\!+\!2)(2p\!+\!4)} \!-\!\frac{x^6}{2\!\cdot\!4\!\cdot\!6(2p\!+\!2)(2p\!+\!4)(2p\!+\!6)}\!+-\ldots\right] \end{matrix}$$

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 $$-p$$: $$\begin{matrix} y_2 := a_0x^{-p}\!\left[1 \!-\!\frac{x^2}{2(-2p\!+\!2)}\!+\!\frac{x^4}{2\!\cdot\!4(-2p\!+\!2)(-2p\!+\!4)} \!-\!\frac{x^6}{2\!\cdot\!4\!\cdot\!6(-2p\!+\!2)(-2p\!+\!4)(-2p\!+\!6)}\!+-\ldots\right] \end{matrix}$$

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 $$y := C_1J_p(x)+C_2J_{-p}(x),$$ where\, $$J_{-p}(x) = y_2$$\, with\, $$a_0 = \frac{1}{2^{-p}\Gamma(-p+1)}$$.

The explicit expressions for $$J_{\pm p}$$ are $$\begin{matrix} J_{\pm p}(x) = \sum_{m=0}^\infty \frac{(-1)^m}{m!\,\Gamma(m\pm p+1)}\left(\frac{x}{2}\right)^{2m\pm p}, \end{matrix}$$ 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 $$y := C_1J_{\frac{1}{2}}(x)+C_2J_{-\frac{1}{2}}(x).$$

In the case that $$p$$ is a non-negative integer $$n$$, the "+" case of (7) gives the solution $$J_{n}(x) = \sum_{m=0}^\infty \frac{(-1)^m}{m!\,(m+n)!}\left(\frac{x}{2}\right)^{2m+n}, $$ 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 $$F$$: $$\begin{matrix} F(z,\,t) = e^{\frac{z}{2}(t-\frac{1}{t})} = \sum_{n=-\infty}^\infty J_n(z)t^n \end{matrix}$$ 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: $$\cos{z} = J_0(z)-2J_2(z)+2J_4(z)-+\ldots$$ $$\sin{z} = 2J_1(z)-2J_3(z)+2J_5(z)-+\ldots$$