gensys - Maple Help

DEtools

 gensys
 return the determining PDE system for either the coefficients of the symmetry generator or the integrating factor

 Calling Sequence gensys(ODE, [_xi = f(x, y, ..), _eta = g(x, y, ..)], y(x)) gensys(ODE, _mu = f(x, y, ..), y(x))

Parameters

 ODE - ODE (ordinary differential equation) [_xi=.., _eta=..] - (optional) list containing the infinitesimal; right hand sides may be functions or algebraic expressions _mu =..] - (optional) equation where the right hand side may be a function or an algebraic expression y(x) - (optional) required if the ODE has derivatives of more than one function

Description

 • Given an ODE, gensys returns the determining PDE system related to either its symmetries or its integrating factors, depending on the second argument. This command is typically used together with other commands to determine symmetries and integrating factors for ODEs. gensys first calls odepde to get the determining PDE and then splits it -- when possible -- by taking coefficients with respect to the derivatives of the dependent variable.
 • Concerning the optional second argument, gensys works like odepde; this second argument may be either an explicit form for the symmetry or the integrating factor to be taken into account at the time of calculating the determining PDE.
 • This function is part of the DEtools package, and so it can be used in the form gensys(..) only after executing the command with(DEtools). However, it can always be accessed through the long form of the command by using DEtools[gensys](..).

Examples

 > $\mathrm{with}\left(\mathrm{DEtools}\right):$
 > $\mathrm{with}\left(\mathrm{PDEtools}\right):$
 > $\mathrm{declare}\left(y\left(x\right),\mathrm{prime}=x\right)$
 ${y}{}\left({x}\right){}{\mathrm{will now be displayed as}}{}{y}$
 ${\mathrm{derivatives with respect to}}{}{x}{}{\mathrm{of functions of one variable will now be displayed with \text{'}}}$ (1)

1. Determine if an integrating factor of the form $\mathrm{\mu }\left(x,y\right)$ exists for the second order nonlinear ODE

 > $\mathrm{ODE}≔\mathrm{diff}\left(y\left(x\right),x,x\right)=-\frac{\mathrm{exp}\left(x\right)\mathrm{diff}\left(y\left(x\right),x\right)x{y\left(x\right)}^{2}+{y\left(x\right)}^{2}-\mathrm{diff}\left(y\left(x\right),x\right)x}{x{y\left(x\right)}^{2}\mathrm{exp}\left(x\right)}$
 ${\mathrm{ODE}}{≔}{\mathrm{y\text{'}\text{'}}}{=}{-}\frac{{{ⅇ}}^{{x}}{}{\mathrm{y\text{'}}}{}{x}{}{{y}}^{{2}}{+}{{y}}^{{2}}{-}{\mathrm{y\text{'}}}{}{x}}{{x}{}{{y}}^{{2}}{}{{ⅇ}}^{{x}}}$ (2)

and if so, determine the integrating factor itself.

Set up the determining PDE system for these integrating factors (gensys enters in this step), then try to simplify it (perhaps using casesplit), then try to solve it. The determining PDE system is set up as follows:

 > $\mathrm{gensys}\left(\mathrm{ODE},\mathrm{_μ}=\mathrm{\mu }\left(x,y\left(x\right)\right)\right)$
 ${{\mathrm{\mu }}}_{{y}}{}{{ⅇ}}^{{x}}{}{x}{,}{{\mathrm{\mu }}}_{{y}{,}{y}}{}{{ⅇ}}^{{x}}{}{x}{,}{{\mathrm{\mu }}}_{{x}{,}{y}}{}{{ⅇ}}^{{x}}{}{x}{,}{-}\frac{\left({{ⅇ}}^{{x}}{}{{y}}^{{2}}{-}{1}\right){}{{\mathrm{\mu }}}_{{x}}{}{x}}{{{y}}^{{2}}}{-}\frac{{\mathrm{\mu }}{}\left({x}{,}{y}\right){}{x}}{{{y}}^{{2}}}{+}{{\mathrm{\mu }}}_{{x}{,}{x}}{}{{ⅇ}}^{{x}}{}{x}{+}{{\mathrm{\mu }}}_{{y}}$ (3)

This system (each equation above is assumed to be equal to zero) has four equations,

 > $\mathrm{nops}\left(\left[\right]\right)$
 ${4}$ (4)

and by taking into account their integrability conditions it is simplified to

 > $\mathrm{casesplit}\left(\left[\right]\right)$
 $\left[{{\mathrm{\mu }}}_{{x}}{=}{\mathrm{\mu }}{}\left({x}{,}{y}\right){,}{{\mathrm{\mu }}}_{{y}}{=}{0}\right]\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{&where}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\left[\right]$ (5)

Thus, mu does not depend on $y$ and the remaining equation leads to

 > $\mathrm{\mu }=\mathrm{exp}\left(x\right)$
 ${\mathrm{\mu }}{=}{{ⅇ}}^{{x}}$ (6)

Note that this result can be tested using mutest

 > $\mathrm{mutest}\left(\mathrm{exp}\left(x\right),\mathrm{ODE}\right)$
 ${0}$ (7)

and can be used to obtain a first integral (that is, to reduce the order of ODE).

 > $\mathrm{firint}\left(\mathrm{exp}\left(x\right)\mathrm{ODE}\right)$
 $\frac{{1}}{{y}}{+}{{ⅇ}}^{{x}}{}{\mathrm{y\text{'}}}{+}{\mathrm{ln}}{}\left({x}\right){+}{\mathrm{_C1}}{=}{0}$ (8)

First integrals can be tested using firtest.

2. Determine if an integrating factor of the form $\mathrm{mu}\left(y,y'\right)$ exists for the same second order ODE.

 > $\mathrm{gensys}\left(\mathrm{ODE},\mathrm{_μ}=\mathrm{\mu }\left(y,\mathrm{_y1}\right)\right)$
 ${-}{{\mathrm{\mu }}}_{{\mathrm{_y1}}{,}{\mathrm{_y1}}}{}{{ⅇ}}^{{x}}{}{\mathrm{_y1}}{}{{x}}^{{2}}{}{{y}}^{{3}}{+}{{\mathrm{\mu }}}_{{\mathrm{_y1}}{,}{y}}{}{{ⅇ}}^{{x}}{}{\mathrm{_y1}}{}{{x}}^{{2}}{}{{y}}^{{3}}{-}{2}{}{{\mathrm{\mu }}}_{{\mathrm{_y1}}}{}{{ⅇ}}^{{x}}{}{{x}}^{{2}}{}{{y}}^{{3}}{+}{2}{}{{\mathrm{\mu }}}_{{y}}{}{{ⅇ}}^{{x}}{}{{x}}^{{2}}{}{{y}}^{{3}}{+}{{\mathrm{\mu }}}_{{\mathrm{_y1}}{,}{\mathrm{_y1}}}{}{\mathrm{_y1}}{}{{x}}^{{2}}{}{y}{-}{{\mathrm{\mu }}}_{{\mathrm{_y1}}{,}{\mathrm{_y1}}}{}{x}{}{{y}}^{{3}}{+}{2}{}{{\mathrm{\mu }}}_{{\mathrm{_y1}}}{}{{x}}^{{2}}{}{y}{,}{-}{{\mathrm{\mu }}}_{{\mathrm{_y1}}{,}{y}}{}{{ⅇ}}^{{x}}{}{{\mathrm{_y1}}}^{{2}}{}{{x}}^{{2}}{}{{y}}^{{3}}{+}{{\mathrm{_y1}}}^{{2}}{}{{\mathrm{\mu }}}_{{y}{,}{y}}{}{{ⅇ}}^{{x}}{}{{x}}^{{2}}{}{{y}}^{{3}}{+}{{\mathrm{\mu }}}_{{\mathrm{_y1}}{,}{y}}{}{{\mathrm{_y1}}}^{{2}}{}{{x}}^{{2}}{}{y}{-}{{\mathrm{\mu }}}_{{\mathrm{_y1}}{,}{y}}{}{\mathrm{_y1}}{}{x}{}{{y}}^{{3}}{-}{2}{}{{\mathrm{\mu }}}_{{\mathrm{_y1}}}{}{{\mathrm{_y1}}}^{{2}}{}{{x}}^{{2}}{-}{{\mathrm{\mu }}}_{{\mathrm{_y1}}}{}{\mathrm{_y1}}{}{{x}}^{{2}}{}{y}{+}{{\mathrm{\mu }}}_{{\mathrm{_y1}}}{}{x}{}{{y}}^{{3}}{+}{{\mathrm{\mu }}}_{{y}}{}{x}{}{{y}}^{{3}}{+}{{\mathrm{\mu }}}_{{\mathrm{_y1}}}{}{{y}}^{{3}}{-}{\mathrm{\mu }}{}\left({y}{,}{\mathrm{_y1}}\right){}{y}{}{{x}}^{{2}}$ (9)
 > $\mathrm{nops}\left(\left[\right]\right)$
 ${2}$ (10)
 > $\mathrm{casesplit}\left(\left[\right]\right)$
 $\left[{\mathrm{\mu }}{}\left({y}{,}{\mathrm{_y1}}\right){=}{0}\right]\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{&where}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\left[\right]$ (11)

So this ODE has no integrating factor of the form $\mathrm{mu}\left(y,y'\right)$.

3. Determine whether or not this ODE has point symmetries.

Set the determining system for the coefficients of the infinitesimal symmetry generator (so-called "infinitesimals") as follows:

 > $\mathrm{gensys}\left(\mathrm{ODE},\left[\mathrm{\xi }\left(x,y\right),\mathrm{\eta }\left(x,y\right)\right]\right)$
 ${{\mathrm{\xi }}}_{{y}{,}{y}}{}{{ⅇ}}^{{x}}{}{{x}}^{{2}}{}{{y}}^{{3}}{,}{2}{}{{ⅇ}}^{{x}}{}{{\mathrm{\xi }}}_{{y}}{}{{x}}^{{2}}{}{{y}}^{{3}}{+}{{\mathrm{\eta }}}_{{y}{,}{y}}{}{{ⅇ}}^{{x}}{}{{x}}^{{2}}{}{{y}}^{{3}}{-}{2}{}{{\mathrm{\xi }}}_{{x}{,}{y}}{}{{ⅇ}}^{{x}}{}{{x}}^{{2}}{}{{y}}^{{3}}{-}{2}{}{{\mathrm{\xi }}}_{{y}}{}{{x}}^{{2}}{}{y}{,}{{ⅇ}}^{{x}}{}{{\mathrm{\xi }}}_{{x}}{}{{x}}^{{2}}{}{{y}}^{{3}}{+}{2}{}{{\mathrm{\eta }}}_{{x}{,}{y}}{}{{ⅇ}}^{{x}}{}{{x}}^{{2}}{}{{y}}^{{3}}{-}{{\mathrm{\xi }}}_{{x}{,}{x}}{}{{ⅇ}}^{{x}}{}{{x}}^{{2}}{}{{y}}^{{3}}{+}{3}{}{{\mathrm{\xi }}}_{{y}}{}{x}{}{{y}}^{{3}}{-}{{\mathrm{\xi }}}_{{x}}{}{{x}}^{{2}}{}{y}{+}{\mathrm{\xi }}{}\left({x}{,}{y}\right){}{{x}}^{{2}}{}{y}{+}{2}{}{\mathrm{\eta }}{}\left({x}{,}{y}\right){}{{x}}^{{2}}{,}{{ⅇ}}^{{x}}{}{{\mathrm{\eta }}}_{{x}}{}{{x}}^{{2}}{}{{y}}^{{3}}{+}{{\mathrm{\eta }}}_{{x}{,}{x}}{}{{ⅇ}}^{{x}}{}{{x}}^{{2}}{}{{y}}^{{3}}{+}{2}{}{{\mathrm{\xi }}}_{{x}}{}{x}{}{{y}}^{{3}}{-}{{\mathrm{\eta }}}_{{y}}{}{x}{}{{y}}^{{3}}{-}{\mathrm{\xi }}{}\left({x}{,}{y}\right){}{x}{}{{y}}^{{3}}{-}{\mathrm{\xi }}{}\left({x}{,}{y}\right){}{{y}}^{{3}}{-}{{\mathrm{\eta }}}_{{x}}{}{{x}}^{{2}}{}{y}$ (12)

(in above, [_xi = xi(x,y), _eta = eta(x,y)] also works). This system consists of 4 equations,

 > $\mathrm{nops}\left(\left[\right]\right)$
 ${4}$ (13)

and in simplifying these equations by taking into account their integrability conditions one obtains

 > $\mathrm{casesplit}\left(\left[\right]\right)$
 $\left[{\mathrm{\eta }}{}\left({x}{,}{y}\right){=}{0}{,}{\mathrm{\xi }}{}\left({x}{,}{y}\right){=}{0}\right]\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{&where}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\left[\right]$ (14)

showing that this ODE has no point symmetries.