\documentclass[a4paper,12pt]{article} \newcommand{\ds}{\displaystyle} \newcommand{\pl}{\partial} \newcommand{\al}{\alpha} \parindent=0pt \begin{document} \begin{center} {\bf Bessel Functions} \end{center} ${}$ {\bf Vibrations of a Membrane} ${}$ The governing equation for the displacement $w(x, y, t)$from the equilibrium position (the plane $z=0$) is $\ds \hspace{2in} \nabla_1^2w = \frac{1}{c^2}\frac{\pl ^2w}{\pl t^2} \hfill(1)$ $$\nabla_1^2 = \frac{\pl^2}{\pl x^2} + \frac{\pl^2}{\pl y^2}$$ [ The assumptions are:- \begin{itemize} \item[(i)] That the action across the element $Delta S$ is a force $T \Delta S$ perpendicular to $\Delta S$, where $T' \to T$ as $\Delta S \to 0$ \item[(ii)] the displacement $w$ of any point of the membrane is purely transverse. \item[(iii)] that $\ds \left[\left( \frac{\pl w}{\pl x}\right)^2 + \left( \frac{\pl w}{\pl y} \right)^2 \right]^\frac{1}{2} $ is small. \hfill] \end{itemize} It can be shown that the tension is isotropic at a point at a point (i.e. independent of the orientation of $\Delta S$), independently of (ii) and (iii), and for equilibrium or motion from (ii) it can be shown that $T$ is uniform over the membrane and $c = T$/ mass/unit area. We assume $c^2$=constant. The usual boundary condition for a finite membrane) is that $w=0$ on the boundary. ${}$ {\bf Simple Harmonic Vibrations} ${}$ $\ds \hspace{1.75in} w(x,y,t) = W(x,y) \cos (wt + \epsilon)$ \hfill(2) Then $\ds \hspace{1.5in} \nabla_1^2W + k^2W =0 \hspace{1in} k^2 = \frac{w^2}{c^2}$ \hfill(3) For a circular membrane (complete or annular) we use plane polar coordinates $r, \, \theta$ $$ \nabla_1^2 = \frac{1}{r} \frac{\pl}{\pl r} r \frac{\pl}{\pl r} + \frac{1}{r^2} \frac{\pl^2}{\pl \theta^2}$$ Therefore $$\frac{1}{r} \frac{\pl }{\pl r} r \frac{\pl W}{\pl r} + \frac{1}{r^2} \frac{\pl^2W}{\pl \theta^2} + k^2 W =0$$ This is separable. i.e. we can find solutions of the form $\ds W(r \theta) = F(r) G(\theta)$ by substitution we have $$ \frac{1}{F} \frac{1}{r} \frac{d}{dr} \left( r \frac{dF}{dr} \right) + k^2 + \frac{1}{r^2} \frac{1}{G} \frac{d^2G}{d\theta^2} =0$$ Therefore $\ds \hspace{2in} \frac{1}{G} \frac{d^2G}{d\theta^2}=$constant$=-n^2$\hfill(4) Therefore $$ G(\theta) = A \cos n \theta + B \sin n \theta$$ $\ds \hspace{1.5in} r \frac{d}{dr} \left( r \frac{dF}{dr} \right) + (k^2r^2 - n^2)F =0$ \hfill(5) ${}$ ${}$ [Note that $\frac{d}{dr}(r \frac{dF}{dr}) + (k^2r - \frac{n^2}{r})F =0$ is the self adjoint for ?????????] ${}$ In (5) write $kr = x$ (not the co-ordinate) $\ds \hspace{1.5in} x \frac{d}{dx} \left(x \frac{dF}{dx}\right) + (x^2 - n^2)F =0$ \hfill(6) This is {\bf Bessel's Equation of order ${\bf n}$}. ${}$ The solution of the original equation $\nabla^2 W+ k^2 W=0$ must be periodic in $\theta$ of period $2\pi$, for otherwise $W$ would not be a one valued function of position. Therefore $n$ is an integer. ${}$ {\bf Series solution for ${\bf F}$} ${}$ We assume $\ds F \sum_{m=0}^\infty a_mx^{m+c}$ where $c$ is to be found. $\ds \left[ \left( x \frac{d}{dx} \right)^2 - n^2 \right]x^{(m+c)} = [(m+c)^2 - n^2]x^{m+c}$ $\ds \left[ \left( x \frac{d}{dx} \right)^2 - n^2 \right] F = \sum_{m=0}^\infty a_m[(m+c)^2 - n^2]x^{m+c}$ $\ds x^2 F = \sum_{m=0}^\infty a_mx^{n+c+2} = \sum_{m=2}^\infty a_{m-2}x^{m+c}$ Hence we require $\ds \sum_{m=0}^\infty a_m [(m+c)^2 - n^2] x^{m+c} + \sum_{m=2}^\infty a_{m-2}x^{m+c} =0$ this is true if $\ds a_0(c^2 - n^2)=0$ Indicial equation $\ds a_1(\overline{c+1}^2 - n^2)=0$ $a_m(\overline{m+c}^2 - n^2) + a_{m-2} \hspace{.2in} m=2, 3, \ldots$ Since $a_0 \not=0, \, c = \pm n$ In the second, putting $c=n$ $$a_1(2n+1)=0$$ In the third, putting $c=n, \, m = 3, 5, \ldots$ $$a_3 3(2n+3) + a_1 = 0 \cdots$$ We suppose $n \not= -\frac{1}{2} , -\frac{3}{2}, \ldots$ Then $a_1 = a_3 = \ldots =0$ For $m = 2, 4, 6, \ldots$ in the third relation $\ds a_2 2(2 + 2n) + a_0 \Rightarrow a_2 = \frac{-a_0}{2^2 \cdot 1(n+1)}$ $\ds a_4 4(4 + 2n) + a_2 \Rightarrow a_4 = \frac{a_0}{2^4 \cdot 1\cdot 2(n+1)(n+2)}$ $\ds a_{2m} = \frac{(-1)^ma_0}{2^{2m} m!(n+1)\ldots(n+m)}$ Hence we have one solution (taking $c=n$) $\ds a_0 x^n \sum_{m=0}^\infty \frac{(-1)^m}{m!(n+1)\ldots(n+m)} \left( \frac{x}{2} \right)^{2m}$ The Bessel function $J_n(x)$ is defined by taking $\ds a_0 = \frac{1}{2^n \Gamma(n+1)}$ \begin{eqnarray*} J_n(x) & = & \left( \frac{x}{2} \right)^n \frac{1}{\Gamma(n+1)} \sum_{m=0}^\infty \frac{(-1)^m}{m!(n+1)\ldots(n+m)}\left( \frac{x}{2}\right)^{2m} \\ & = & \sum_{m=1}^\infty \frac{(-1)^m}{m!\Gamma(m+n+1)}\left( \frac{x}{2}\right)^{2m} \end{eqnarray*} The series converges for all $x$, by the ratio test. So $J_n(x)$ is an integral function of $x$. If $L_n$ is the operation in Bessels equation $L_{-n}(y) = L_n(y)$ since n appears as $n^2$ $L_n[J_n(x)] =0$ for all n Therefore $L_{-n} [j_{-n}(x)]=0$ i.e. $L_nJ_{-n}(x)]=0$ Hence $J_{-n} (x) $ is a solution. $$J_{-n}(x) = \sum_{m=0}^\infty \frac{(-1)^m}{m!\Gamma(m-n+1}\left( \frac{x}{2}\right)^{-n+2m}$$ When $n$ is a non-negative integer $\Gamma(m+n+1)$ is infinite from $ m = 0, 1, 2, \ldots , n-1$ since $\Gamma(z)$ has poles at $z = 0 , -1, -2, \ldots$ Therefore $$J_{-n}(x) = \sum_{m=n}^\infty \frac{(-1)^m}{m!\Gamma(m-n+1}\left( \frac{x}{2}\right)^{-n+2m}$$ Write $m = n+k$ $$J_{-n}(x) = \sum_{k=o}^\infty \frac{(-1)^k}{(n+k)!k!}\left( \frac{x}{2}\right)^{2k+n} = (-i)^n J_n(x)$$ ${}$ {\bf Wronskin of $J_n$ and $J_{-n}$} ${}$ $$W = \left| \begin{array}{cc} J_n & J_{-n} \\ J'_n & J'_{-n} \end{array} \right|$$ ${}$ we have $\ds \frac{d}{dx} \left( x \frac{d}{dx} j_{\pm n}\right) + \left(x - \frac{n^2}{x}\right) J_{\pm n}=0$ Therefore $\ds \frac{d}{dx} \left\{ x [J'_n J_{-n} - J_n J'_{-n}] \right\} = 0$ i.e. $\ds x[J'_n J_{-n} - J_n J'_{-n}] = C$ (constant). $\ds C = \lim_{x \to 0} x[J'_n J_{-n} - J_n J'_{-n}]$ \begin{eqnarray*} J_n & = & \frac{\left( \frac{x}{2} \right)^n}{r(n+1)} \left[ 1 - \frac{\left(\frac{x}{2}\right)^2}{1!(n+1)} + \cdots \right] \\ J_{-n} & = & \frac{\left( \frac{x}{2} \right)^{-n}}{r(-n+1)} \left[ 1 - \frac{\left(\frac{x}{2}\right)^2}{1!(-n+1)} + \cdots \right] \\ xJ'_n & = & \frac{n\left( \frac{x}{2} \right)^n}{r(n+1)} \left[ 1 + 0x^2 \right] \\ xJ'_{-n} & = & \frac{-n\left( \frac{x}{2} \right)^{-n}}{r(-n+1)} \left[ 1 + 0x^2 \right] \end{eqnarray*} Therefore $\ds J_{-n} \cdot x J'_n - xJ'_{-n} J_n = \frac{2n}{\Gamma(1+n)\Gamma(1-m)}[1+0x^2]$ ${}$ Therefore $\ds C = \frac{2n}{\Gamma(1+n)\Gamma(1-m)} = \frac{2}{\Gamma (1-n) \Gamma(n)} = \frac{2\sin n\pi}{\pi}$ ${}$ Therefore $\ds J_{-n} \cdot x J'_n - xJ'_{-n} J_n =\frac{2}{x} \frac{\sin n \pi}{\pi}$ Thus $J_n$ and $J_{-n}$ are linearly independent when $n$ is not an integer. and $J_n$ and $J_{-n}$ are linearly dependent when $n$ is an integer. ${}$ {\bf Definition of the second solution [Weber]} ${}$ $$Y_n(x) = \frac{\cos n \pi J_n - J_{-n}}{\sin n \pi}$$ when $n$ tends to an integer the numerator tends to 0 nd the denominator tens to 0. Hence when $m$ is an integer we define $\ds Y_m(x) = \lim_{n\to m} Y_n(x)$ Therefore \begin{eqnarray*} Y_m(x) = \left. \frac{\frac{\pl}{\pl n} [\cos n \pi J_n - J_{-n}]}{\frac{\pl}{\pl n} \sin n \pi} \right\}_{n=m} & = & \left.\frac{\pi\sin n \pi J_n + \cos n \pi \frac{\pl}{\pl n} - \frac{\pl}{\pl n} J_{-n}}{\pi \cos \pi n}\right\}_{n=m} \\ & = & \frac{1}{\pi} \left[\frac{\pl}{\pl n} (J_n) - (-1)^m \frac{\pl}{\pl n} (J_{-n})\right]_{n=m} \end{eqnarray*} In particular $\ds Y_0(x) = \frac{1}{\pi}\left[\left(\frac{\pl}{\pl n} J_n\right)_{n=0} - \left(-\frac{\pl}{\pl n} J_n\right)_{n=0} \right] = \frac{2}{\pi} \left[ \frac{\pl}{\pl n} J_n\right]_{n=0}$ The functions of the second kind of order $n$. They are unbounded at $x=0$, for fro the relation between $J_n$ and and other solution of Bessel's equations, say $Y_n(x)$, we have $\ds x[Y_nJ'_n - Y'_nJ_n] = C$ $\ds Y_nJ'_N - Y' _n J_n = \frac{c}{x}$ ${}$ If $c \not=0$ then $Y_n$ and $Y'_n$ can not exist at x=0. Hence the general solution of Bessels equations is either $A_1J_n(x) + B_1J_{-n}(x)$ (n not an integer) or $A_1J_n(x) + B_1Y_{n}(x)$ (all cases) A solution bounded at $x=0$ is necessarily $AJ_n(x)$ ${}$ {\bf Returning to the membrane problem} ${}$ \begin{enumerate} \item Complete membrane ($0 \leq r \leq a$). Since $W$ must be bounded at $r=0$, we have $F(r) = AJ_n(kr)$ and $W(r\theta) = AJ_n(kr)\cos(n\theta+\varepsilon)$ (n integer $\geq 0$) Since $W=0$ on $r=a$. $AJ_n(ka)=0$ $A=0$ is trivial therefore $J_n(ka)=0$. This is an equation for the eigenvalues $k_1^2, \, k_2^2, \ldots$. Hence for a given $n$ the values of $k$ are given by $ka = j_{n1},\, j_{n2}, \ldots$ where $j_{n1},\, j_{n2}, \ldots$ are the positive zeros of $J_n(x)$. The allowed frequencies (frequencies of normal modes) are $\ds \frac{w}{a2pi}, \, \, w = kc$ Therefore $$w = \frac{c}{a}(j_{n1},\, j_{n2}, \ldots) \hspace{.2in} n = 0, 1, \ldots$$ \item Annular Membrane $b \leq r \leq a$. In this case $r=0$ is not in the \lq \lq physical space \rq \rq. Therefore we must take $F(r) = AJ_n(kr) +BY_n(kr)$ $F(a) =0 \, \, \, \, F(b) = 0$ give:- \begin{eqnarray*} AJ_n(ka) +BY_n(ka) & = & 0 \\ AJ_n(kb) +BY_n(kb) & = & 0 \end{eqnarray*} Hence for non-trivial A, B $$\left| \begin{array}{cc} AJ_n(ka) & BY_n(ka) \\ AJ_n(kb) & BY_n(kb) \end{array} \right|=0$$ \end{enumerate} ${}$ \underline{ Sketch of $J_0\, J_1\, J_2$} ${}$ $\ds J_0 = 1 = \left( \frac{x}{2}\right)^2 \cdot \frac{1}{(1!)^2} + \left( \frac{x}{4}\right)^4 \frac{1}{(2!)^2} + \cdots$ $\ds J_1 = \left( \frac{x}{2} \right) -\left( \frac{x}{2}\right)^3 \frac{1}{1!2!} + \left( \frac{x}{2} \right)^5 \frac{1}{2!3!}$ Note also $J'_0(x) = -J_1(x)$ ${}$ PICTURE ${}$ The asymptotic formula for $J_n(x)$ is $\ds J_n(x) \sim \left( \frac{2}{\pi x} \right)^\frac{1}{2}\cos\left( x - \frac{\pi}{4} + n\frac{\pi}{2}\right)$ ${}$ ${}$ {\bf Orthogonal and normal Properties of ${\bf J_0\, \, (j_mx)}$} ${}$ where $j1, \, j_2\ldots$ are the zeros of $j_0(x)$. We show that \begin{eqnarray*} \int_0^1xJ_0(j_mx)J_0(j_px) & = & 0 \hspace{.7in} p \not=m \\ & = & \frac{1}{2} J_1^2(j_m) \hspace{.2in} p=m \end{eqnarray*} the functions are orthogonal to weight function $x$ over $[a,b]$. $\ds \frac{d}{dx} x \frac{d}{dx} J_0(\alpha x) + \alpha^2xJ_0(\alpha x) =0$ $\ds \frac{d}{dx} x \frac{d}{dx}J_0(\beta x) + \beta^2xJ_0(\beta x)=0$ $\ds \frac{d}{dx} x \left[J_0(\beta x)\frac{d}{dx}J_0(\alpha x) - J_0(\alpha x)\frac{d}{dx}J_0(\beta x) \right] = (\alpha^2 - \beta^2)x J_0(\alpha x) J_0(\beta x)$ Therefore $\ds \left.x\left[J_0(\beta x)\frac{d}{dx}J_0(\alpha x) - J_0(\alpha x)\frac{d}{dx}J_0(\beta x) \right]\right\}_0^1 $ $I(\alpha^2 - \beta^2) \int_0^1 J_0(\alpha x) J_0(\beta x) \, dx$ i.e. \begin{eqnarray*} (\alpha^2 - \beta^2) \int_0^1 x J_0(\alpha x) - J_0(\beta x) \, dx & = & J_0(\alpha)\beta J'_0(\beta) - J_0(\beta) \alpha J'_0(\alpha) \hspace{.4in} (1)\\ & = & J_0(\beta) J_1(\alpha) - J_0(\alpha) \beta J_1(\beta) \end{eqnarray*} If $\alpha = j_m \hspace{.2in} \beta = j_p \hspace{.2in} m \not=p$ $$ \int_0^1 x J_0(j_mx) J_0(j_px) dx =0$$ \begin{eqnarray*} \int_0^1xJ_0^2(\alpha x) & = & \lim_{\beta \to \alpha} \frac{-\alpha J_0(\beta) J'_0(\alpha) + \beta J_0(\alpha) J_0' \beta}{\alpha^2 - \beta^2} \\ & = & \left[ \frac{\frac{\pl}{\pl \beta} {\rm Numerator}}{\frac{\pl}{\pl \beta} {\rm Denominator}}\right]_{\beta=\alpha} \\ & = & \left[ \frac{-j'_0(\beta) \cdot \alpha J'_0(\alpha) \frac{\pl}{\pl \beta} (\beta J'_0\beta)}{-2\beta} \right]_{\beta = \alpha} \\ & = & -\frac{\alpha J_0^{'2} (\alpha) - \alpha J_0^2(\alpha)}{-2\alpha} \left[ \frac{\pl}{\pl\beta} (\beta J'_0(\beta)) + \beta J_0(\beta) =0\right] \end{eqnarray*} Therefore $\ds \int_0^1 x J_o^2(\alpha x) = \frac{1}{2} J_1^2(\alpha) + J_0^2(\alpha) \hspace{.3in} [J_0' = -J_1]$ \hfill (2) Therefore then $\alpha = j_m$ $\ds \int_0^1 x J_0(j_m x) J_0(j_p x) \, dx = \frac{1}{2} J_1^2(j_m) \delta _{mp}$ It also follows that if $f'_1, \, f'_2 \, , \ldots$ are the zeros of $J'_0(x)$ then $$ \int_0^1 x J_0(j'_mx)J_0(j'_px) \, dx = \frac{1}{2} J_0^2 (j_m) \delta _{mp}$$ \newpage {\bf Special Cases of ${\bf \ds \int_0^1 J_0(\alpha_n x)f(x) dx}$} ${}$ In (1) take $\alpha = \alpha\-n$ ($a$ zero) $\beta \not= \alpha_m \, \, \, m = 1,2, \ldots$ Then $\ds \int_0^1 xJ_0 (\alpha_n x)J_0(\beta x) \, dx = \frac{\alpha_n}{\alpha_n^2 - \beta ^2} J_1(\alpha_n)J_0(\beta)$\hfill (3) In this put $\beta =0$ and $J_0(0) =1$ and so $\ds \int_0^1xJ_0(\alpha_n x) \, dx = \frac{J_1(\alpha_n)}{\alpha_n}$\hfill (4) (4) can also be obtained as follows: $\ds \frac{d}{dx} x \frac{d}{dx}J_0(\alpha_n x) = -\alpha_n^2 xJ_0(\alpha_n x)$ Therefore \begin{eqnarray*}-\alpha_n^2 \int_0^1 x J_0(\alpha_n x) dx & = & \left[x \frac{d}{dx} J_0(\alpha_n x)\right]_0^1 \\ & = & a_n J'_0(\alpha_n) \\ & = & -\alpha_n J_1(\alpha_n) \end{eqnarray*} Next we consider $$I_k = \int_0^1 x J_0(\alpha x)x^k\,dx$$ We have $\ds \hspace{1in} \frac{d}{dx} x \frac{d}{dx} J_0(\alpha x) = -\alpha^2xJ_0(\alpha x)$\hfill(i) $\ds\hspace{1in} \frac{d}{dx} x \frac{d}{dx} x^k = k^2x^{k-1}$ \hfill(ii) Therefore $$ \frac{d}{dx} x \left\{ x^k \frac{d}{dx} J_0(\alpha x) - J_0(\alpha x) \frac{d}{dx} x^k\right\} = -J_0(\alpha x) \{ x \alpha^2 x^k + k^2x^{k-1}\}$$ Therefore by integration over [0,1] $$-\alpha^2I_k - k^2I_{k-2} = \alpha J'_0(\alpha) - k J_0(\alpha)$$ Therefore $I_0\, I_2, \ldots$ can be found in terms of $J_0(\alpha)$ and $J_1(\alpha)$. $I_1$ and $I_£$ can be found in terms of the \lq \lq Sturve function \rq \rq $$\rightarrow \int_0^1J_0(\alpha x) dx q J_0(\alpha) J_1(\alpha)$$ \newpage {\bf Formal Fourier - Bessel Explanations} ${}$ Assume that $f(x)$, defined in [0,1], possesses and expansion $$f(x) = \sum_{n=1}^\infty A_n J_0(\alpha_n x)$$ Then \begin{eqnarray*} \int_0^1 x J_0 (\alpha_m x) f(x) dx & = & \sum_{n=1}^\infty A_m \int_0^1 xJ_0(\alpha_n x) J_0(\alpha_m x) dx \\ & = & A_m \int_0^1 x J_0^2 (\alpha_m x) \, dx \\ & = & A_m \frac{J_1^2(\alpha _m)}{2} \\ A_m & = & \frac{2}{J_1^2(\alpha_m) } \int_0^1 x J_0 (\alpha _m) f(x) \, dx \end{eqnarray*} ${}$ {\bf Initial and Boundary Problem for the Vibrating Membrane} ${}$ We have, for the displacement $w(r,t)$ in radially symmetric vibrations $$\frac{1}{r} \frac{\pl}{\pl r} \left( r \frac{\pl w}{\pl r} \right) = \frac{1}{c^2} \frac{\pl^2w}{\pl t^2} \hspace{.5in} 0 \leq r \leq a$$ $w(0,t) $ exists and $w(a,t) =0$ Also $\ds w(r,0) = f(r) \hspace{.3in} 0 \leq r \leq a \hspace{.5in} \frac{\pl w} {\pl t} (r,0) =0 \hspace{.3in} 0 \leq r \leq a$ ${}$ Choose $a$ - unit of length and choose unit on time so that $c=1$. Also replace $r$ by $x$, giving $$\frac{1}{x} \frac{\pl}{\pl x} x\frac{ \pl w}{\pl x} = \frac{\pl^2w}{\pl t^2}$$ $w(0,t) $ exists and $w(1,t) =0$ Also $\ds w(x,0) = f(x) \hspace{.5in} \frac{\pl w} {\pl t} (x,0) =0$ ${}$ Assume $\ds w(x,t) = \sum_{n=1}^\infty J_0(\al_nx)\phi_n(t)$ This satisfies the boundary conditions $$\frac{1}{2} J_1^2 (\al _n) \phi_n(t) = \int_0^1 x w(x,t) J_0(\al_n x) dx$$ \begin{eqnarray*} \frac{1}{2} J_1^2(\al_n)\ddot\phi_n(t) & = & \int_0^1 xJ_0(\al _n x) \frac{\pl^2w}{\pl t^2}(xt) \, dx \\& = & \int_0^1 J_0(\al_n x) \frac{\pl}{\pl x} \left( x \frac{\pl w}{\pl x} \right) \, dx \\ & = & \left[J_0(\al_n x) x \frac{\pl w}{\pl x} \right]_0^1 - \int_0^1 x \frac{\pl w}{\pl x} \frac{d}{dx} J_0(\al _nx) dx \\ & = & 0 + \left[ -w x \frac{d}{dx} J_0(\al _n x) \right]_0^1 + \int_0^1 w \frac{d}{dx} x \frac{d}{dx} J_0(\al _n x) \, dx \\ & = & 0 + \int_0^1 w [-\al_n^2 x J_0(\al_n x)]\, dx \\ & = & -\al^2 \cdot \frac{1}{2} J_1^2(\al_n) \phi_n(t) \end{eqnarray*} Therefore $$\ddot \phi(t) + \al_n^2 \phi(t) =0$$ and $$\phi_n(t) = A_n \cos(\al_n(t) + B_n \sin(\al_n t))$$ $\begin{array}{ll} w(x,0) =f(x) \hspace{.2in} & 0 \leq x \leq 1 \\ \ds \frac{\pl}{\pl t} w(x.0) =0 & 0 \leq x \leq i\end{array}$ Therefore \begin{eqnarray*} f(x) & = & \sum_0^\infty J_0(\al_n x) \phi_n(0) \\ 0 & = & \sum_9^\infty J_0(\al_n x) \dot \phi_n(0) \\ \dot \phi_n = 0, \hspace{.75in} \phi_n(0) \cdot \frac{1}{2} J_1^2(\al_n) & = & \int_0^1 xJ_0(\al_n x) f(x) \, dx \\ {\rm i.e.\ }B_n=0 \hspace{1.5in} A_n & = & \frac{2}{J_1^2(\al _n)} \int_0^1 xJ_0(\al_n x) f(x) \, dx \end{eqnarray*} Hence we have the series solution for $w(x,t)$ $$w(x,t) = \sum_1^\infty J_0(\al _n x) \cos \al_nt \frac{2}{J_1^2(\al_n)} \int_0^1 y J_0(\al_n y) f(y) \, dy$$ \newpage {\bf Alternative procedure} ${}$ Solutions of the differential equation bounded at $x=0$ are $J_0(kx)[A\cos kt + \sin kt]$ This satisfies the boundary conditions $w(1,t) =0$ if $J_0(k) =0$ thus $k = \al_1, \, \al_2 , \ldots$ The solution also satisfies $\ds \frac{\pl w}{\pl t} (x0) =0$ if $B_n=0$. Formally the series $\ds \sum_{n=1}^\infty A_n J_0(\al_n x) \cos \al_n t$ satisfies both boundary conditions and the initial conditions on $\ds \frac{\pl w}{\pl t}$. Therefore as $w(x_0)=f(x)$ thus $f(x) = \sum A_n J_0(\al_n x)$ Therefore $$A_n \cdot \frac{1}{2} J_1^2(\al_n) = \int_0^1 xJ_0(\al_n x) f(x) \, dx.$$ ${}$ \underline{Example} $$f(x) = 1- \frac{J_0(kx)}{J_0(k)}$$ $k$ real and $J_0(k) \not= 0$ \begin{eqnarray*} \int_0^1 x J_0(\al_n x)J_0(kx) dx & = & \frac{J_0(k) \al_nJ_1(\al_n) - J_0(\al_n) kJ_1(k)}{\al_n^2 - k^2} \\ & = & J_0(k) \frac{\al_nJ_1(\al_n)}{\al_n^2 - k^2} \\ k=0 {\rm \ gives\ \ } \int_0^1 x J_0(\al_n x) dx & = & \frac{J_1(\al_n)}{\al_n} \\ \int_0^1 x J_0(\al_n x) f(x) dx & = & J_1(\al_n) \left\{ \frac{1}{\al_n} - \frac{\al_n}{\al_n^2 - k^2}\right\} \\ & = & \frac{k^2 J_1(\al_n)}{\al_n(k^2 - \al_n^2)} \end{eqnarray*} Hence in this case: $$ w(x,t) = \sum_{n=1}^\infty J_0(\al_n x) \cos \al_n t\cdot \frac{2k^2}{\al_n(k^2-\al_n^2)}\frac{1}{J_1(\al_n)}$$ Since $\al_n = O(n)$ for large $N$, and $\ds J_1(\al_n) = O\left( \frac{1}{\al_n^\frac{1}{2}}\right) = O\left( \frac{1}{n^\frac{1}{2}} \right)$ Then the coefficient of $J_0(\al_n x) \cos(\al_n t)$ is $\ds O\left(\frac{1}{n^\frac{5}{2}}\right)$ \newpage {\bf Solution of a linear differential equation by definite integral (or a contour integral)} ${}$ {\bf Preliminary Remarks} ${}$ Consider the differential equation $$x \phi (D) y + \psi (D) y=0$$ where \begin{eqnarray*} \phi(p) & = & a_0 p^n + a_1 p^{n-1} + \ldots +a_n \\ \psi(p) & = & b_0p^m + b_1p^{m-1} + \ldots + b_m \end{eqnarray*} Seek a solution $$y = \int_a^b e^{px} K(p) dp h\space{.2in}{\rm ot} \hspace{.2in} y = \int_C e^{px} K(p) dp$$ where $K(p)$ is to be found and $a, b$ (or $C$) are also to be found (and are independent of $x$.) Then \begin{eqnarray*} \phi(D) y & = & \int_a^b \phi(D) e^{px} K(p) dp \\ & = & \int_a^b \phi(p) e^{px} K(p) dp \\ \psi(D) y & = & \int_a^b \psi(p) e^{px} K(p) dp \\ {\rm then\ }\hspace{.75in} x \phi(D) y & = & \int_a^b x e^{px} \phi(p)K(p) dp \\ & = & \int_a^b \frac{d}{dp} (e^{px}) \phi(p) K(p) dp \\ & = & \left[ e^{px} \phi(p) K(p) \right]_a^b - \int_a^b e^{px} \frac{d}{dp} (\phi(p) K(p) ) dp \end{eqnarray*} Hence $$x \phi(D) y + \psi(D) y = \left[ e^{px} \phi(p) K(p)\right]_a^b + \int_a^b e^{px} \left\{ \psi(p) K(p) - \frac{d}{dp} \phi(p) K(p) dp\right\} dp$$ \begin{itemize} \item[(i)] Choose $K(p)$ so that the integrand is zero. i.e.$$\frac{d}{dp} \{ \phi(p) K(p) \} = \psi(p)K(p) = \frac{\psi(p)}{\phi(p)} \phi(p) K(p)$$ therefore $$\phi(p) k(p) = C \exp\left\{\int^p \frac{\psi(q)}{\phi(q)} dq\right\}$$ \underline{Note:} if all the zeros of $\phi$ are simple $\ds \frac{\psi}{\phi} = \sum_1^n \frac{a_r}{q-q_r} \Rightarrow \int^p\frac{\psi}{\phi} = \log(p - p_r)^{a_r}$ Therefore $\ds K_r = \prod_1^n (p-p_r)^{a_{r-1}}$ \item[(ii)] when $K(p)$ is known we choose $a$ and $b$ (or $C$) so that $\ds \left[ e^{px} K(p) \phi(p)\right]_a^b =0$ \end{itemize} ${}$ ${}$ {\bf Consider Bessel's equation on order n} ${}$ $$\left\{ \left( x \frac{d}{dx} \right)^2 + x^2 - n^2\right\} y =0$$ $$x^2 \left\{ \frac{d^2y}{dx^2} + y\right\} + x\frac{dy}{dx} - n^2y =0$$ Substitute $y = x^nz$ \begin{eqnarray*} x\frac{dy}{dx} & = & x^n \left\{ x \frac{d}{dx} + n \right\} \\ \left( x \frac{d}{dx} \right)^2 y & = & x^n \left\{ x \frac{d}{dx} +n \right]^2 z \\ & = & x^n \left\{ \left( x \frac{d}{dx}\right)^2 + 2n x \frac{d}{dx} + n^2 \right\}z \\ & =& x^n \left\{ x^2 \frac{d^2}{dx^2} + (2n+1) x \frac{d}{dx} + n^2 \right\} \\ \left\{ \left( x \frac{d}{dx}\right)^2 + x^2 - n^2\right\}y & = & x^n \left\{ x^2 \frac{d^2}{dx^2} + (2n+1) x \frac{d}{dx} + x^2\right\} z \\ & = & x^{n+1} \left\{ x \left( \frac{d^2}{dx^2} _1 \right) + (2n+1) \frac{d}{dx} \right\} z \end{eqnarray*} Hence $y$ satisfies Bessel's equation if $z$ satisfies $$\left\{ x \left( \frac{d^2}{dx^2} +1\right) + (2n+1) \frac{d}{dx} \right\} z =0$$ Consider a solution for $z$ of the form $$\int_a^b e^{itx} K(t) dt$$ Then $\ds \left\{ x \left( \frac{d^2}{dx^2} +1\right) + (2n+1) \frac{d}{dx} \right\} \int_a^b e^{itx} K(t) dt$ \begin{eqnarray*}\hspace{.1in} & = & \int_a^b \{ x(1-t^2) + (2n+1)it \} K(t) e^{itx} dt \\ & = & \left[ \frac{1-t^2}{i} e^{itx} K(t) \right]_a^b + \int_a^b e^{itx} \left[ -\frac{d}{dt} \left\{ \frac{1-t^2}{i}K(t) \right\} + it (2n+1)K(t) \right] dt \end{eqnarray*} Hence choose $K(t)$ so that $$\frac{d}{dt}(1-t^2)K(t) = -(2n-1) t K(t)$$ $$K(t) = c(1-t^2)^{n - \frac{1}{2}}$$ hence a solution of Bessel's equation is $$ y = x^n \int_a^b e^{itx} (1-t^2)^{n - \frac{1}{2}} dt$$ if $a$ and $b$ are chosen so that $$[(1-t^1)^{n+\frac{1}{2}}e^{itx}]_a^b =0$$ Suppose $n>-\frac{1}{2}$ and also $x$ is real and positive [There is no difficulty if $x=z$ is complex] ${}$ {\bf Admissible Pairs of limits} ${}$ (i) $(-1, +1)$ (ii) $(-1, -1+i\infty)$ (iii) $(+1, +1+i\infty)$ PICTURE ${}$ These integrals must be linearly dependent since that are solutions of a second order equation. With proper specification of $(1-t^2)^{n-\frac{1}{2}}$ the relation is 1 = (ii) - (iii) Consider $$\int e^{itx}(1 -t^2)^{n-\frac{1}{2}} dt$$ round the contour shown. PICTURE We choose that branch of $(1-t2)^{n-\frac{1}{2}}$ which is real and positive on $AA'$. i.e. $\arg(1-t)=0, \, \arg(1+t)=0$ on $AA'$. AS $t$ passes from $a$ to $b$ round $AB \arg(1-t)$ decreases by $\frac{\pi}{2}$; as $t$ passes from $A'$ to $B'$ round $A'B' \arg(1+t)$ increases by$\frac{\pi}{2}$. Since the integrand $e^{itx}(1-t^2)^{n-\frac{1}{2}}$ is now one-valued and regular on and inside the countour, by Cauchy's Theorem we have $$\int_{\{A'A\}} = \int_{\{A'A\}}+ \int_{\{B'C'\}}+ \int_{\{CB\}}+\int_{\{C'C\}}+\int_{\{A'B'\}}+\int_{\{BA\}}$$ We show \begin{itemize} \item[(a)] $\ds \lim_{h \to \infty} \int_{\{C'C\}} =0$ \item[(b)] $\ds \lim_{\epsilon \to \infty} \int_{\{A'B'\}} = \lim_{\epsilon\to \infty}\int_{\{BA\}}=0$ \end{itemize} We shall then have $$\int_{-1}^1 e^{itx}(1-t^2)^{n-\frac{1}{2}} dt = \int_{-1}^{-1_i\infty} e^{itx}(1-t^2)^{n-\frac{1}{2}} - \int_{1}^{1+i\infty} e^{itx}(1-t^2)^{n-\frac{1}{2}}$$ i.e. $$(i) = (ii) - (iii) $$ since the limits if the three integrals exist as $\epsilon\to\infty$ if $n> -\frac{1}{2}$ and as $h \to \infty.$ on $CC'$, $|1 - t^2| =PX \cdot PX' \leq C'X \cdot CX' = h^2 +4$ $|e^{itx}| = |e^{ix(u+ih)}| = e^{-x+h}$ Therefore $\left| \int_{CC'} \right| \leq e^{-xa} (h^2 +4)^{n-\frac{1}{2}} \cdot 2 \to 0$ as $h \to \infty$ on $AB$ we have $t-1 = \epsilon e^{t\theta}$ Therefore $|t-1| ^{n-\frac{1}{2}} = \epsilon^{n - \frac{1}{2}}$ $e^{itx} (t^2-1)^{n-\frac{1}{2}}$ is bounded in the neighbourhood of $t=1$ with bound $M$ say. Therefore $\ds |e^{itx} (t^2-1(^{n-\frac{1}{2}}| \leq M\epsilon^{n-\frac{1}{2}}$ Thus $\ds \left| \int_{AB}\right| \leq M\epsilon^{n-\frac{1}{2}} \cdot \frac{\pi}{2} \epsilon \frac{M \pi}{2} e^{n-\frac{1}{2}} \to 0 $ as $\epsilon\to 0$ Similarly $\ds \int_{B'A'} \to 0$ as $\epsilon \to 0$ ($n > -\frac{1}{2}$) \newpage Hence we have the following solutions of Bessel's equation \begin{itemize} \item[(i)] $\ds x^n \int_0^1 e^{itx}(1 - t^2)^{n-\frac{1}{2}} dt $ \item[(ii)] $\ds x^n \int_{-1}^{-1+i\infty}e^{itx}(1 - t^2)^{n-\frac{1}{2}} dt $ \item[(iii)] $\ds x^n \int_{+1}^{+1+i\infty}e^{itx}(1 - t^2)^{n-\frac{1}{2}} dt $ \end{itemize} ${}$ ${}$ {\bf Series Expansions of (i)} ${}$ $$(i) = x^n \int_{-1}^1 (i-t^2)^{n-\frac{1}{2}} \sum_0^\infty \frac{(itx)^m}{m!} dt$$ the series for all $x, \, t$ absolutely and uniformly. \begin{eqnarray*} (i) & = & x^n \sum_{m=0}^\infty \frac{(ix)^m}{m!} \int_{-1}^1 t^m(1 - t^2)^{n-\frac{1}{2}} dt \\ \int_{-1}^1 t^m(1 - t^2)^{n-\frac{1}{2}} dt & = & 0 {\rm if\ m\ is\ odd} \\ \int_{-1}^1 t^{2m}(1 - t^2)^{n-\frac{1}{2}} dt & = & 2 \int_{0}^1 t^{2m}(1 - t^2)^{n-\frac{1}{2}} dt \\ & = & 2 \int_{0}^1 u^{m}(1 - u)^{n-\frac{1}{2}}\cdot \frac{1}{2} u^\frac{1}{2} du \\ & = & \int_{0}^1 u^{m-\frac{1}{2}}(1 - u)^{n-\frac{1}{2}} du \\ & = & \frac{\Gamma (n+\frac{1}{2}) \Gamma (m+\frac{1}{2})}{\Gamma (m+n+1)} \\ \Rightarrow (i) & = &x^n \sum_{m=0}^\infty \frac{(ix)^{2m}}{(2m)!}\frac{\gamma (n+\frac{1}{2}) \Gamma (m+\frac{1}{2})}{\Gamma (m+n+1)} \\ \frac{\Gamma (m+\frac{1}{2})}{(2m)!} & = & \frac{\Gamma (\frac{1}{2} )\frac{1}{2} \frac{3}{2} \cdots m - \frac{1}{2}}{1 \cdot 2 \cdot 3 \cdots (2m-1)2m} \\ & = & \frac{\Gamma (\frac{1}{2})}{2^{2m} \cdot m!} \\ \Rightarrow (i) & = & \Gamma (n+\frac{1}{2}) \Gamma \left(\frac{1}{2}\right) x^n \sum_{m=0}^\infty \frac{(-1)^m \left(\frac{x}{2}\right)^{2m}}{m! \Gamma(m+n+1)} \\ & = & 2^n \Gamma (n+\frac{1}{2}) \Gamma \left(\frac{1}{2}\right) J_n(x) \Rightarrow J_n(x) \\ & = & \frac{1}{2^n \Gamma (n+\frac{1}{2}) \Gamma \left(\frac{1}{2}\right)} \times (i) \end{eqnarray*} ${}$ {\bf Hanbel Functions of order n (Bessel functions of the third kind)} ${}$ \underline{Definition} \begin{eqnarray*} H_n^{(1)} (x) & = & -2 \left( \frac{1}{2^n \Gamma (\frac{1}{2}) \Gamma (n + \frac{1}{2}) }\right) \cdot (iii) \\H_n^{(2)} (x) & = & 2 \left( \frac{1}{2^n \Gamma (\frac{1}{2}) \Gamma (n + \frac{1}{2}) }\right) \cdot (ii) \end{eqnarray*} Then $$J_n(x) = \frac{1}{2} (H_n^{(1)}(x) + H_n^{(2)}(x))$$ we also define $$Y_n(x) = \frac{1}{2i} (H_n^{(1)}(x) - H_n^{(2)}(x))$$ ${}$ {\bf Alternative integral representation of ${\bf H_n^{(1)}(x)}$ and $\bf{ H_n^{(1)}(x)}$} ${}$ In the integral representation for $H_n^{(1)}(x)$, $\arg (1-t) = -\frac{\pi}{2}$ Therefore we write $(1-t) = \eta e^{-\frac{\pi i}{2} (= -i\eta}$. Where $\eta$ goes from 0 to $\infty$ through real values as $t$ goes from 1 to $1+i\infty$. Thus $$ (1-t)^{n - \frac{1}{2}} e^{-\frac{\pi i}{2}( n - \frac{1}{2})}$$ Also \begin{eqnarray*} (1+t) & = & 2 - (1-t) = 2 - \eta e^{-\frac{\pi i}{2}} = 2(1 + \frac{i \eta}{2}) \\ \Rightarrow (1+t)^{n - \frac{1}{2}} & = & 2 ^{n - \frac{1}{2}}( 1 + \frac{in}{2}) ^{n - \frac{1}{2}} \\ \Rightarrow (1-t)^{n - \frac{1}{2}} & = & 2^{n - \frac{1}{2}} \eta ^{n - \frac{1}{2}} e^{-\frac{\pi i}{2}(n - \frac{1}{2})} \left( 1 + \frac{i\eta}{2} \right)^{n - \frac{1}{2}} \\ \\ e^{itx} & = & e^{ix ( 1+ i\eta)} = e^{ix} e^{-x\eta} \\ dt & = & -e^{-\frac{\pi i}{2} d \eta} \\ \Rightarrow e^{itx}(1 - t^2)^{n - \frac{1}{2}} dt & = & 2^{n - \frac{1}{2}} e^{i(x - \frac{n\pi}{2} + \frac{pi}{4})} e^{-x\eta} \eta^{n - \frac{1}{2}} \left( 1 + \frac{i\eta}{2}\right)^{n - \frac{1}{2}} d\eta \\ \Rightarrow H_n^{(1)} (x) & = & \frac{2^\frac{1}{2}x^ne^{i(x - \frac{n\pi}{2} - \frac{\pi}{4})}}{\Gamma (\frac{1}{2}) \Gamma (n + \frac{1}{2})} \int_0^\infty e^{-x\eta}\eta ^{n - \frac{1}{2}}\left( 1 + \frac{i\eta}{2}\right)^{n - \frac{1}{2}} d\eta \hspace{.2in}(I) \end{eqnarray*} In the integral for $H_n^{(2)} (x) \arg(t+1) = \frac{\pi}{2}$ Therefore $(t+1) = \eta e^\frac{\pi i}{2} = i\eta$ where $\eta$ goes from 0 to $\infty$ as $t$ goes from -1 to $+1+i\infty$ we get similarly $$H_n^{(2)} (x) = \frac{2^\frac{1}{2}x^ne^{-i(x - \frac{n\pi}{2} - \frac{\pi}{4})}}{\Gamma (\frac{1}{2}) \Gamma (n + \frac{1}{2})} \int_0^\infty e^{-x\eta}\eta ^{n - \frac{1}{2}}\left( 1 - \frac{i\eta}{2}\right)^{n - \frac{1}{2}} d\eta \hspace{.2in}(I) $$ [Note: when $x$ is real and positive, $H_n^{(1)} (x)$, $H_n^{(2)} (x)$ are complex conjugates. Therefore $\frac{1}{2}[H_n^{(1)} (x) + H_n^{(2)}(x)]$ is real, so is $\frac{1}{2i}[H_n^{(1)} (x) - H_n^{(2)}(x)]$.] ${}$ Finally substitute $\eta x = u$ in both integrals. For real and positive $u$ goes from 0 to $\infty$ as $\eta$ goes from 0 to $\infty$ $$H_n^{(1)} (x) = \left( \frac{2}{x}\right)^\frac{1}{2} \frac{1}{\Gamma(\frac{1}{2})} \frac{e^{i(x - \frac{n\pi}{2} - \frac{\pi}{4})}}{\Gamma (n+\frac{1}{2}) } \int_0^\infty e^{-u}u ^{n - \frac{1}{2}}\left( 1 + \frac{iu}{2x}\right)^{n - \frac{1}{2}} du $$ $$H_n^{(2)} (x) = \left( \frac{2}{x}\right)^\frac{1}{2} \frac{1}{\Gamma(\frac{1}{2})} \frac{e^{-i(x - \frac{n\pi}{2} - \frac{\pi}{4})}}{\Gamma (n+\frac{1}{2}) } \int_0^\infty e^{-u}u ^{n - \frac{1}{2}}\left( 1 - \frac{iu}{2x}\right)^{n - \frac{1}{2}} du $$ ${}$ ${}$ {\bf Asymptotic Expansions of ${\bf \ds \int_0^\infty e^{-u}u ^{n - \frac{1}{2}}\left( 1 \pm \frac{iu}{2x}\right)^{n - \frac{1}{2}} du }$} ${}$ We consider the case $n=0$ i.e.$$ \int_0^\infty e^{-u}u ^{ - \frac{1}{2}}\left( 1 \pm \frac{iu}{2x}\right)^{ - \frac{1}{2}} du$$ We apply Taylor's formula $$f(t) = \sum_{r=0}^{n-1} \frac{f^{(r)}(0) t^r}{r!} + \frac{1}{(n-1!)} \int_0^t (t-s)^{n-1} f^{(n)} (s) ds$$ to the function $\ds (1 -t)^{-\frac{1}{2}}$ this gives $$ (1 - t)^{-\frac{1}{2}} = \sum_{r=0}^{n-1} \frac{\frac{1}{2} \cdot \frac{3}{2} \cdots r - \frac{1}{2}}{r!}t^r +\frac{\frac{1}{2} \cdot \frac{3}{2} \cdots n - \frac{1}{2}}{(n-1)!} \int_0^t(t-s)^{n-1} (1-s)^{-n - \frac{1}{2}} ds $$ the last term is $$ \frac{\frac{1}{2} \cdot \frac{3}{2} \cdots n - \frac{1}{2}}{(n-1)!} t^n \int_0^1 (1-v)^{n-1}(1 - tv)^{-n - \frac{1}{2}} dv$$ writing $\ds t = \frac{u}{2ix}$ $$\left( 1 - \frac{u}{2ix}\right)^{-\frac{1}{2}} = \sum_{r=0}^{n-1} \frac{\frac{1}{2} \frac{3}{2} \cdots n - \frac{1}{2}}{r!} \frac{u^n}{(2ix)^n} + r_n \left( \frac{u}{x} \right) $$ $$ r_n\left( \frac{u}{x} \right) = \frac{\frac{1}{2} \frac{3}{2} \cdots n - \frac{1}{2}}{(n-1)!} \frac{u^r}{(2ix)^r} \int_0^1 (1 - v)^{n-1} \left( 1 - \frac{u}{2ix} v \right) ^{-n - \frac{1}{2}} dv $$ If $x$ is real and positive $$\left| 1 - \frac{u}{2ix}v \right| = \left( 1 + \frac{u^2v^2}{4x^2} \right)^\frac{1}{2} \geq 1$$ Therefore \begin{eqnarray*} \left| r_n \left( \frac{u}{x} \right) \right| & \leq & \frac{\frac{1}{2} \frac{3}{2} \cdots n - \frac{1}{2}}{n!} \frac{u^n}{(2x)^n} \int_0^1 (1 - v)^{n-1} \cdot 1 dv \\ & = &\frac{\frac{1}{2} \frac{3}{2} \cdots n - \frac{1}{2}}{n!} \frac{u^n}{(2x)^n} \end{eqnarray*} Hence $$ \int_0 ^\infty e^{-u} u ^{-\frac{1}{2}} \left( 1 - \frac{u}{2ix} \right)^{-\frac{1}{2}} du = \sum_{r=0}^{n-1} \frac{\frac{1}{2} \frac{3}{2} \cdots r - \frac{1}{2}}{r!} \frac{1}{(2ix)^r} \int_0^\infty e^{-u} y^{r - \frac{1}{2}} du + R_n(x)$$ $$R_n(x) = \int_0^\infty e^{-u} u^{-\frac{1}{2}} r_n\left( \frac{u}{x} \right) du $$ $$ \int_0^\infty e^{-u} u^{r-\frac{1}{2}}du = \Gamma (r+ \frac{1}{2} = \Gamma \left( \frac{1}{2} \right) \frac{1}{2} \cdot \frac{3}{2} \cdots r - \frac{1}{2}$$ Also \begin{eqnarray*} |R_n(x)| & \leq & \int_0^\infty e^{-u} u^{n - \frac{1}{2}} \left| r_n \frac{u}{x} \right| du \\ & \leq & \frac{\frac{1}{2} \cdot \frac{3}{2} \cdots n - \frac{1}{2}}{n!} \frac{1}{(2x)^n} \int_0^\infty e^{-u} u^{n-\frac{1}{2}} du \end{eqnarray*} Therefore $$ \int_0^\infty e^{-u} u^{-\frac[1}{2} \left( 1 - \frac{u}{2ix} \right)^{-\frac{1}{2}} du = \Gamma \left( \frac{1}{2} \right) \sum_{r=0} ^{n-1} \frac{[\frac{1}{2} \frac{3}{2} \cdots r - \frac{1}{2}]^2}{r!} \frac{1}{(2ix)^r} + \bar R_n \bar x$$ Where $\ds |\bar R_n| \leq \frac{[\frac{1}{2} \cdot \frac{3}{2} \cdots n - \frac{1}{2}]^2}{n!} \frac{1}{(2x)^n} \cdot \lim_{x \to \infty} x^{n-1} =0$ There the series is the asymptotic expansion of the left hand side. $\left({\rm In\ fact\ } R_n = 0\left( \frac{1}{x^n}\right)\right)$ \newpage {\bf Divergence of the Infinite series} ${}$ The D'Alernbert ratio is $$\left| \frac{(n+\frac{1}{2})^2}{n} \cdot \frac{1}{2ix}\right| = \frac{1}{2x} \frac{(n+\frac{1}{2})^2}{n} $$ which tends to infity for and $x$. ${}$ {\bf Asymptotic Expansion of $\bf H_0^{(1)}(x) \, H_0^{(2)}(x)$} $$H_0^{(1)}(x) \sim \left( \frac{2}{\pi x}\right)^\frac{1}{2}e^{i(x - \frac{\pi}{4})} \sum_{r=0}^\infty \frac{\left[ \frac{1}{2} \frac{3}{2} \cdots r - \frac{1}{2}\right]^2 }{r!} \frac{1}{(2ix)^r}$$ $$H_0^{(2)}(x) \sim \left( \frac{2}{\pi x}\right)^\frac{1}{2}e^{-i(x - \frac{\pi}{4})} \sum_{r=0}^\infty \frac{\left[ \frac{1}{2} \frac{3}{2} \cdots r - \frac{1}{2}\right]^2 }{r!} \frac{1}{(2ix)^r}$$ where the remainder after the term in $\ds \frac{1}{x^{n-1}}$ has modulas $\leq | {\rm term\ in\ } \frac{1}{x^n}|$ Write $\ds A(x) = \frac{1}{2} \frac{1}{\Gamma (\frac{1}{2})} \left[ \int_0^\infty e^{-u} u^{-\frac{1}{2}}\left(1 - \frac{u}{2ix}\right)^{-\frac{1}{2}}du+\int_0^\infty e^{-u} u^{-\frac{1}{2}} \left( 1+ \frac{u}{2ix}\right)^{-\frac{1}{2}}du\right] $ $\ds B(x) = \frac{1}{2} \frac{1}{\Gamma (\frac{1}{2})} \left[ \int_0^\infty e^{-u} u^{-\frac{1}{2}}\left(1 - \frac{u}{2ix}\right)^{-\frac{1}{2}}du-\int_0^\infty e^{-u} u^{-\frac{1}{2}} \left( 1+ \frac{u}{2ix}\right)^{-\frac{1}{2}}du\right] $ $\ds H_0^{(1)}(x) = \left(\frac{2}{\pi x}\right)^\frac{1}{2} e^{i(x - \frac{\pi}{4})} [A(x) + iB(x)]$ $\ds H_0^{(2)}(x) = \left(\frac{2}{\pi x}\right)^\frac{1}{2} e^{-i(x - \frac{\pi}{4})} [A(x) - iB(x)]$ \begin{eqnarray*} J_0(x) & = & \frac{1}{2}\left( H_0^{(1)}(x) + H_0^{(2)}(x)\right)\\ & = & \left( \frac{2}{\pi x}\right)^\frac{1}{2} \left[A(x) \cos \left( x - \frac{\pi}{4} \right) - B(x) \sin \left( x - \frac{\pi}{4} \right)\right]\end{eqnarray*} \begin{eqnarray*} Y_0(x) & = & \frac{1}{2i}\left( H_0^{(1)}(x) - H_0^{(2)}(x)\right] \\ & = & \left( \frac{2}{\pi x}\right)^\frac{1}{2} \left[A(x) \sin \left( x - \frac{\pi}{4} \right) - B(x) \cos \left( x - \frac{\pi}{4} \right)\right]\end{eqnarray*} The general Bessel function of zero order is $\ds A_1 J_0(x) + B_1Y_0(x) = C(\cos x J_0() + \sin x Y_0(x))$ From the definitions of $A(x)$ and $B(x)$ $\ds A(x) \sim \sum_{r=0}^\infty \frac{[\frac{1}{2} \frac{3}{2} \cdots 2r - \frac{1}{2}]^2}{(2r)!} \frac{(-1)^r}{(2x)^{2r}}$ ${}$ ${}$ $\ds B(x) \sim \sum_{r=0}^\infty \frac{[\frac{1}{2} \frac{3}{2} \cdots 2r + \frac{1}{2}]^2}{(2r+1)!} \frac{(-1)^{r+1}}{(2x)^{2r+1}}$ ${}$ ${}$ {\bf Zeros of a Bessel Function of zero order} ${}$ The zeros are given by $\ds \cot (x - \frac{\pi}{4} - \al) = \frac{B(x)}{A(x)}$ $$A(x) = 1+O\left( \frac{1}{x^2}\right) \hspace{.4in} B(x) = -\frac{1}{8x} + O\left(\frac{1}{x^3}\right)$$ Therefore $$\frac{A(x)}{B(x)} = -\frac{1}{8x} + O\left( \frac{1}{x^3}\right)$$ Therefore for large $x$, the zeros are approximately given by $\cot (x - \frac{\pi}{4} - \al) =0$ i.e. $\ds x - \frac{\pi}{4} -\al = \left( k + \frac{1}{2} \right) \pi$ k = large integer. $\ds x = \al + \left( k + \frac{3}{4}\right) \pi$ ${}$ ${}$ {\bf Asymptotic Expansions of ${\bf H_0^{(1)}(x) \, \, \& \, \, H_0^{(2)}(x)}$} $\ds H_0^{(1)}(x) \sim\left( \frac{2}{\pi x} \right)^\frac{1}{2}e^{i(x - \frac{n\pi}{2} - \frac{\pi}{4})} \sum_{m=0}^\infty \frac{(-1)^m(m,n)}{(2ix)^m}$ $\ds H_0^{(2)}(x) \sim\left( \frac{2}{\pi x} \right)^\frac{1}{2}e^{-i(x - \frac{n\pi}{2} - \frac{\pi}{4})} \sum_{m=0}^\infty \frac{(-1)^m(m,n)}{(2ix)^m}$ where $(0,n) =1.$ $\ds (m,n) = \frac{(4n^2 - 1^2)(4n^2 - 3^2) \cdots (4n^2 - (2m - 1)^2)}{2^{2m} m!}$ There expansion are only useful when $x >> n$ ${}$ {\bf Bessel Functions of order $\bf {(k+\frac{1}{2}) \, k = 0, 1, \cdots}$} We have $$ J_n(x) = \sum_{m=0}^\infty \frac{(-1)^m \left( \frac{x}{2} \right) ^{n+2m}}{m! \Gamma (n+m+1)}$$ $$ J_\frac{1}{2} (x) = \left(\frac{2}{\pi x}\right)^\frac{1}{2} \sin x \hspace{.3in} J_{-\frac{1}{2}} (x) = \left(\frac{2}{\pi x}\right)^\frac{1}{2} \cos x$$ $\ds J_\frac{1}{2} (x) = \sum_{m=0}^\infty\left(\frac{x}{2}\right)^\frac{1}{2}\frac{(-1)^m x^{2m}}{2^{2m}m! \Gamma (m+\frac{3}{2})}$ \begin{eqnarray*} 2^{2m}m! \Gamma (m+\frac{3}{2}) & = & 2^{2m }m! \Gamma (\frac{3}{2}) \frac{3}{2} \frac{5}{2} \cdots m + \frac{1}{2} \\ & = & \Gamma (\frac{3}{2}) 2 \cdot 4 \cdots 2m 3 \cdot 5 \cdots (2m+1) \\ & = & (2m+1)! \Gamma (\frac{3}{2}) \\ & = & (2m+1)! \frac{1}{2} \Gamma(\frac{1}{2}) \\ \Rightarrow J_\frac{1}{2} (x) & = & \left( \frac{x}{2} \right)^\frac{1}{2} \frac{1}{\frac{1}{2} \Gamma \frac{1}{2}} \sum_{m=0}^\infty\frac{(-1)^m x^{2m}}{(2m+1)!} \\ &= & \left( \frac{2x}{\pi} \right)^\frac{1}{2} \frac{\sin x}{x}\\ & = & \left( \frac{2}{\pi x} \right)^\frac{1}{2} \sin x \end{eqnarray*} Similarly $$J_{-\frac{1}{2}} (x) = \left( \frac{2}{\pi x} \right)^\frac{1}{2} \cos x$$ \begin{eqnarray*} H_\frac{1}{2}^{(1)} & = & -i e^{ix} \left( \frac{2}{\pi x}\right)^\frac{1}{2} \hspace{,2in} H_{-\frac{1}{2}}^{(2)} = i e^{-ix} \left( \frac{2}{\pi x}\right)^\frac{1}{2} \\ H_{k + \frac{1}{2}}^{(1)} (x) & = & -2 \left( \frac{x}{2} \right)^{k+ \frac{1}{2} } \frac{1}{\Gamma (\frac{1}{2}\Gamma(k+1)} \int_1^{1+i\infty} e^{itx} (1 - t^2)^k dt \\ & = &-2 \left( \frac{x}{2} \right)^{k+\frac{1}{2}} \frac{1}{ \Gamma (\frac{1}{2})k!} \left( 1+ \frac{d^2}{dx^2} \right)\int_1^{1+i\infty} e^{itx} dt \\ & =& 2 \left( \frac{x}{2} \right)^{k+\frac{1}{2}} \frac{1}{ \Gamma (\frac{1}{2})k!} \left( 1+ \frac{d^2}{dx^2} \right)\frac{e^{itx}}{x} \\ & = & \frac{e^x}{x^\frac{1}{2}} \{ {\rm Polynimial\ in\ } \frac{1}{x}, {\rm degree\ } k \} \end{eqnarray*} The functions $\ds H_{k + \frac{1}{2}}^{(1)} (x), \, H_{k + \frac{1}{2}}^{(2)} (x)$ are called spherical Bessel functions. They arise in solution of the wave equation in spherical Polar coordinates. \newpage {\bf Radially Progressive Waves in two dimensions} ${}$ We had for the membrane $$ \nabla_1^2 w = \frac{1}{c^2} \frac{\pl^2w}{\pl t^2} \hspace{.3in} \nabla_1^2 = \frac{\pl^2}{\pl x^2} +\frac{\pl^2}{\pl y^2}$$ and found solutions (in the case of radial symmetry) $$w = [AJ_0(kr) + BY_0(kr)]\cos (wt + \epsilon) \hspace{.2in} k = \frac{w}{c}$$ assuming the form $$w = f(r) e^{iwt}$$ (real parts to be taken eventually) we find similar form $$w = [A_1H_0^{(1)} (kr) + A_2H_0^{(2)}(kr)]e^{iwt}$$ since $$H_0^{(1)}(kr) \sim \left( \frac{2}{\pi k r}\right)^\frac{1}{2} e^{i(kr - \frac{\pi}{4})} {\rm \ \ as\ } r \to \infty$$ $$H_0^{(2)}(kr) \sim \left( \frac{2}{\pi k r}\right)^\frac{1}{2} e^{i(kr - \frac{\pi}{4})} {\rm \ \ as\ } r \to \infty$$ we get $$H_0^{(1)}(kr) e^{iwt} \sim \left( \frac{2}{\pi k r}\right)^\frac{1}{2} e^{i(w (t+ \frac{r}{c}) - \frac{\pi}{4})}$$ $$H_0^{(2)}(kr) e^{iwt} \sim \left( \frac{2}{\pi k r}\right)^\frac{1}{2} e^{i(w (t- \frac{r}{c}) - \frac{\pi}{4})}$$ The first repents a wave converging to the origin with velocity $c$, the second a wane diverging from the origin with velocity $c$. \end{document}