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inttrans

 hankel
 Hankel transform

 Calling Sequence hankel(expr, t, s, nu)

Parameters

 expr - expression, equation, or set of expressions and/or equations to be transformed t - variable expr is transformed with respect to t s - parameter of transform nu - order of the transform opt - option to run this under (optional)

Description

 • The hankel function computes the Hankel transform (F[nu](s)) of expr (f(t)), a linear transformation $C\left[0,\infty \right)\to C\left[0,\infty \right)$ defined by:

${F}_{\mathrm{\nu }}\left(s\right)={\int }_{0}^{\mathrm{\infty }}f\left(t\right)\sqrt{ts}\mathrm{BesselJ}\left(\mathrm{\nu },ts\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}ⅆt$

 • The function $F\left(s\right)$ returned is defined on the positive real axis only.
 • Expressions involving exponentials, products involving powers of t, trigonometrics (sin, cos) with linear arguments, and a variety of other functions can all be transformed.
 • The Hankel transform is self-inverting for nu >$-\frac{1}{2}$ .
 • The hankel function can transform Bessel's operator, and if nu$\ne 0$, derivatives of functions as well, and can be used in the solution of ODEs and PDEs.
 • The hankel function attempts to simplify an expression according to a set of heuristics and then match the result with a table of patterns.  Entries can be added to this table by addtable(hankel, f(t), F(s), t, s), where F(s) is the transform of f(t), which may have an arbitrary number of parameters.
 • hankel  recognizes the Dirac-delta (or unit-impulse) function as Dirac(t) and Heaviside's unit step function as Heaviside(t).
 • If the option opt is set to 'NO_INT', then the program will not resort to integration of the original problem if all other methods fail.  This will increase the speed at which the transform will run.
 • The command with(inttrans,hankel) allows the use of the abbreviated form of this command.

Examples

 > $\mathrm{with}\left(\mathrm{inttrans}\right):$
 > $\mathrm{assume}\left(0
 > $\mathrm{assume}\left(-\frac{1}{2}<\mathrm{\nu }\right):$$\mathrm{additionally}\left(\mathrm{\nu }\ne 0\right):$
 > $F≔\mathrm{hankel}\left(g\left(x\right),x,y,\mathrm{\nu }\right)$
 ${F}{≔}{\mathrm{hankel}}{}\left({g}{}\left({x}\right){,}{x}{,}{y}{,}{\mathrm{ν~}}\right)$ (1)
 > $\mathrm{hankel}\left(F,y,z,\mathrm{\nu }\right)$
 ${g}{}\left({z}\right)$ (2)

By default, derivatives of integral transforms are computed:

 > $\mathrm{setup}\left(\mathrm{computederivatives}\right)$
 ${\mathrm{computederivatives}}{=}{\mathrm{true}}$ (3)
 > $\mathrm{diff}\left(\mathrm{hankel}\left(f\left(t\right),t,s,1\right),s\right)$
 ${-}\frac{{-}{\mathrm{hankel}}{}\left({f}{}\left({t}\right){}{t}{,}{t}{,}{s}{,}{4}\right){}{s}{+}{6}{}{\mathrm{hankel}}{}\left({f}{}\left({t}\right){,}{t}{,}{s}{,}{3}\right)}{{s}}{+}\frac{{\mathrm{hankel}}{}\left({f}{}\left({t}\right){,}{t}{,}{s}{,}{1}\right)}{{s}}$ (4)

Therefore, the following, that can be expressed using derivatives of hankel, results in

 > $\mathrm{hankel}\left(tf\left(t\right),t,s,\mathrm{\nu }\right)$
 $\frac{{2}{}{\mathrm{hankel}}{}\left({f}{}\left({t}\right){,}{t}{,}{s}{,}{\mathrm{ν~}}{+}{1}\right){}{\mathrm{ν~}}{-}{\mathrm{hankel}}{}\left({f}{}\left({t}\right){}{t}{,}{t}{,}{s}{,}{\mathrm{ν~}}{+}{2}\right){}{s}{+}{2}{}{\mathrm{hankel}}{}\left({f}{}\left({t}\right){,}{t}{,}{s}{,}{\mathrm{ν~}}{+}{1}\right)}{{s}}$ (5)

The same computation but not computing the derivatives

 > $\mathrm{setup}\left(\mathrm{computederivatives}=\mathrm{false},\mathrm{alternativehankeldefinition}=\mathrm{false}\right)$
 ${\mathrm{alternativehankeldefinition}}{=}{\mathrm{false}}{,}{\mathrm{computederivatives}}{=}{\mathrm{false}}$ (6)
 > $\mathrm{hankel}\left(tf\left(t\right),t,s,\mathrm{\nu }\right)$
 $\frac{\left(\frac{{\partial }}{{\partial }{s}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{hankel}}{}\left({f}{}\left({t}\right){,}{t}{,}{s}{,}{\mathrm{ν~}}{+}{1}\right)\right){}{s}{+}{\mathrm{hankel}}{}\left({f}{}\left({t}\right){,}{t}{,}{s}{,}{\mathrm{ν~}}{+}{1}\right){}{\mathrm{ν~}}{+}{\mathrm{hankel}}{}\left({f}{}\left({t}\right){,}{t}{,}{s}{,}{\mathrm{ν~}}{+}{1}\right)}{{s}}$ (7)
 > $\mathrm{hankel}\left(\frac{f\left(t\right)}{t},t,s,\mathrm{\nu }\right)$
 $\frac{{s}{}\left({\mathrm{hankel}}{}\left({f}{}\left({t}\right){,}{t}{,}{s}{,}{\mathrm{ν~}}{-}{1}\right){+}{\mathrm{hankel}}{}\left({f}{}\left({t}\right){,}{t}{,}{s}{,}{\mathrm{ν~}}{+}{1}\right)\right)}{{2}{}{\mathrm{ν~}}}$ (8)
 > $\mathrm{hankel}\left(\mathrm{D}\left(f\right)\left(t\right),t,s,\mathrm{\nu }\right)$
 ${-}\frac{{s}{}\left({\mathrm{hankel}}{}\left({f}{}\left({t}\right){,}{t}{,}{s}{,}{\mathrm{ν~}}{-}{1}\right){}{\mathrm{ν~}}{-}{\mathrm{hankel}}{}\left({f}{}\left({t}\right){,}{t}{,}{s}{,}{\mathrm{ν~}}{+}{1}\right){}{\mathrm{ν~}}{+}{\mathrm{hankel}}{}\left({f}{}\left({t}\right){,}{t}{,}{s}{,}{\mathrm{ν~}}{-}{1}\right){+}{\mathrm{hankel}}{}\left({f}{}\left({t}\right){,}{t}{,}{s}{,}{\mathrm{ν~}}{+}{1}\right)\right)}{{2}{}{\mathrm{ν~}}}$ (9)

Bessel's differential equation

 > $\mathrm{pde}≔\mathrm{diff}\left(f\left(r,z\right),z,z\right)+\mathrm{diff}\left(f\left(r,z\right),r,r\right)+\frac{\mathrm{diff}\left(f\left(r,z\right),r\right)}{r}-{\left(\frac{\mathrm{\nu }}{r}\right)}^{2}f\left(r,z\right)=0$
 ${\mathrm{pde}}{≔}\frac{{{\partial }}^{{2}}}{{\partial }{{z}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{f}{}\left({r}{,}{z}\right){+}\frac{{{\partial }}^{{2}}}{{\partial }{{r}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{f}{}\left({r}{,}{z}\right){+}\frac{\frac{{\partial }}{{\partial }{r}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{f}{}\left({r}{,}{z}\right)}{{r}}{-}\frac{{{\mathrm{ν~}}}^{{2}}{}{f}{}\left({r}{,}{z}\right)}{{{r}}^{{2}}}{=}{0}$ (10)
 > $\mathrm{hankel}\left(\mathrm{pde},r,p,\mathrm{\nu }\right)$
 $\frac{{{\partial }}^{{2}}}{{\partial }{{z}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{hankel}}{}\left({f}{}\left({r}{,}{z}\right){,}{r}{,}{p}{,}{\mathrm{ν~}}\right){-}{{p}}^{{2}}{}{\mathrm{hankel}}{}\left({f}{}\left({r}{,}{z}\right){,}{r}{,}{p}{,}{\mathrm{ν~}}\right){=}{0}$ (11)
 > $\mathrm{setup}\left(\mathrm{alternativehankeldefinition}=\mathrm{true}\right)$
 ${\mathrm{alternativehankeldefinition}}{=}{\mathrm{true}}$ (12)
 > $\mathrm{hankel}\left({t}^{\frac{1}{2}},t,s-2,1\right)$
 $\frac{{1}}{{\left({s}{-}{2}\right)}^{{3}}{{2}}}}$ (13)
 > $\mathrm{hankel}\left(\mathrm{exp}\left(-at\right),t,s,\mathrm{\nu }\right)$
 $\frac{{\mathrm{\Gamma }}{}\left({\mathrm{ν~}}{+}\frac{{3}}{{2}}\right){}{\mathrm{hypergeom}}{}\left(\left[\frac{{\mathrm{ν~}}}{{2}}{+}\frac{{5}}{{4}}{,}\frac{{\mathrm{ν~}}}{{2}}{+}\frac{{3}}{{4}}\right]{,}\left[{\mathrm{ν~}}{+}{1}\right]{,}{-}\frac{{{s}}^{{2}}}{{{\mathrm{a~}}}^{{2}}}\right){}{{s}}^{{\mathrm{ν~}}{+}\frac{{1}}{{2}}}{}{{2}}^{{-}{\mathrm{ν~}}}{}{{\mathrm{a~}}}^{{-}{\mathrm{ν~}}{-}\frac{{3}}{{2}}}}{{\mathrm{\Gamma }}{}\left({\mathrm{ν~}}{+}{1}\right)}$ (14)

 > $\mathrm{addtable}\left(\mathrm{hankel},\mathrm{myfunc}\left(t\right),\mathrm{MyFunc}\left(s,\mathrm{\nu }\right),t,s,\mathrm{hankel}=\mathrm{\nu }::\mathrm{Range}\left(-\mathrm{\infty },\mathrm{\infty }\right)\right)$
 > $\mathrm{hankel}\left(\mathrm{myfunc}\left(x\right),x,y,\mathrm{\nu }\right)$
 ${\mathrm{MyFunc}}{}\left({y}{,}{\mathrm{ν~}}\right)$ (15)