DEtools/IVPsol - Maple Help

DEtools

 IVPsol
 specializes ODE solutions to match given initial or boundary conditions

 Calling Sequence IVPsol(ics, sol)

Parameters

 ics - a set or list of initial or boundary conditions for an ODE problem. This value can include the ODE itself. sol - a solution of the form y(x) = ... where the lhs is the unknown of the ODE problem

Description

 • The IVPsol command specializes the integration constants of the form $\mathrm{_Cn}$ (where $n$ is an integer) by entering the given solution sol to match the given initial or boundary conditions ics.
 • This command is part of the DEtools package, and so it can be used in the form IVPsol(..) after executing the command with(DEtools). However, it can always be accessed through the long form of the command by using DEtools[IVPsol](..).

Examples

 > $\mathrm{with}\left(\mathrm{DEtools}\right):$
 > ${\mathrm{ode}}_{1}≔x\left(\frac{{ⅆ}^{2}}{ⅆ{x}^{2}}y\left(x\right)\right)+4{y\left(x\right)}^{2}=0$
 ${{\mathrm{ode}}}_{{1}}{≔}{x}{}\left(\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){+}{4}{}{{y}{}\left({x}\right)}^{{2}}{=}{0}$ (1)

A general solution for ode[1] is not known. Therefore, in principle, it would be not possible to solve an initial value problem that contains it, for example, by using the following initial conditions.

 > ${\mathrm{ic}}_{1}≔y\left(1\right)=-\frac{1}{2},\mathrm{D}\left(y\right)\left(1\right)=\frac{1}{2}$
 ${{\mathrm{ic}}}_{{1}}{≔}{y}{}\left({1}\right){=}{-}\frac{{1}}{{2}}{,}{\mathrm{D}}{}\left({y}\right){}\left({1}\right){=}\frac{{1}}{{2}}$ (2)

However, suppose that you find a solution for ode[1] that has been computed by another means, for example

 > ${\mathrm{sol}}_{1}≔y\left(x\right)=\frac{\mathrm{_C1}}{2x}$
 ${{\mathrm{sol}}}_{{1}}{≔}{y}{}\left({x}\right){=}\frac{{\mathrm{_C1}}}{{2}{}{x}}$ (3)

You can pass the initial conditions and the solution directly to the DEtools[IVPsol] command.

 > ${\mathrm{DEtools}}_{\mathrm{IVPsol}}\left(\left[{\mathrm{ic}}_{1}\right],{\mathrm{sol}}_{1}\right)$
 ${y}{}\left({x}\right){=}{-}\frac{{1}}{{2}{}{x}}$ (4)

For the following example, the general solution of the ODE involves the inverse of an elliptic integral, represented by the elliptic function JacobiSN.

 > ${\mathrm{ode}}_{2}≔\frac{{ⅆ}^{2}}{ⅆ{x}^{2}}y\left(x\right)-2{y\left(x\right)}^{3}=0$
 ${{\mathrm{ode}}}_{{2}}{≔}\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){-}{2}{}{{y}{}\left({x}\right)}^{{3}}{=}{0}$ (5)
 > ${\mathrm{gensol}}_{2}≔\mathrm{dsolve}\left({\mathrm{ode}}_{2}\right)$
 ${{\mathrm{gensol}}}_{{2}}{≔}{y}{}\left({x}\right){=}\mathrm{c__2}{}{\mathrm{JacobiSN}}{}\left(\left({I}{}{x}{+}\mathrm{c__1}\right){}\mathrm{c__2}{,}{I}\right)$ (6)

Consider the following initial conditions for ode[2]:

 > ${\mathrm{ic}}_{2}≔y\left(0\right)=1,\mathrm{D}\left(y\right)\left(0\right)=1$
 ${{\mathrm{ic}}}_{{2}}{≔}{y}{}\left({0}\right){=}{1}{,}{\mathrm{D}}{}\left({y}\right){}\left({0}\right){=}{1}$ (7)

When solving the complete ODE-IVP, instead of directly using the general solution gensol[2], dsolve first searches for symmetries from where a particular solution could be computed. The computed solution could be easier to specialize to satisfy the given conditions y(0) = 1, D(y)(0) = 1, resulting in the following ODE-IVP solution.

 > $\mathrm{dsolve}\left(\left[{\mathrm{ode}}_{2},{\mathrm{ic}}_{2}\right]\right)$
 ${y}{}\left({x}\right){=}{-}\frac{{1}}{{x}{-}{1}}$ (8)

If you use the following command to view the results of trying to specialize the general solution gensol[2], instead of using a short-hand form for the process using symmetries, the following results are returned. This is the solution returned in Maple 12.

 > ${\mathrm{DEtools}}_{\mathrm{IVPsol}}\left(\left[{\mathrm{ic}}_{2}\right],{\mathrm{gensol}}_{2}\right)$
 ${y}{}\left({x}\right){=}{\mathrm{RootOf}}{}\left({I}{}{{\mathrm{_Z}}}^{{2}}{}{\mathrm{JacobiCN}}{}\left({\mathrm{InverseJacobiSN}}{}\left(\frac{{1}}{{\mathrm{_Z}}}{,}{I}\right){,}{I}\right){}{\mathrm{JacobiDN}}{}\left({\mathrm{InverseJacobiSN}}{}\left(\frac{{1}}{{\mathrm{_Z}}}{,}{I}\right){,}{I}\right){-}{1}\right){}{\mathrm{JacobiSN}}{}\left(\left({I}{}{x}{+}\frac{{\mathrm{InverseJacobiSN}}{}\left(\frac{{1}}{{\mathrm{RootOf}}{}\left({I}{}{{\mathrm{_Z}}}^{{2}}{}{\mathrm{JacobiCN}}{}\left({\mathrm{InverseJacobiSN}}{}\left(\frac{{1}}{{\mathrm{_Z}}}{,}{I}\right){,}{I}\right){}{\mathrm{JacobiDN}}{}\left({\mathrm{InverseJacobiSN}}{}\left(\frac{{1}}{{\mathrm{_Z}}}{,}{I}\right){,}{I}\right){-}{1}\right)}{,}{I}\right)}{{\mathrm{RootOf}}{}\left({I}{}{{\mathrm{_Z}}}^{{2}}{}{\mathrm{JacobiCN}}{}\left({\mathrm{InverseJacobiSN}}{}\left(\frac{{1}}{{\mathrm{_Z}}}{,}{I}\right){,}{I}\right){}{\mathrm{JacobiDN}}{}\left({\mathrm{InverseJacobiSN}}{}\left(\frac{{1}}{{\mathrm{_Z}}}{,}{I}\right){,}{I}\right){-}{1}\right)}\right){}{\mathrm{RootOf}}{}\left({I}{}{{\mathrm{_Z}}}^{{2}}{}{\mathrm{JacobiCN}}{}\left({\mathrm{InverseJacobiSN}}{}\left(\frac{{1}}{{\mathrm{_Z}}}{,}{I}\right){,}{I}\right){}{\mathrm{JacobiDN}}{}\left({\mathrm{InverseJacobiSN}}{}\left(\frac{{1}}{{\mathrm{_Z}}}{,}{I}\right){,}{I}\right){-}{1}\right){,}{I}\right)$ (9)

Yet another general solution, involving elliptic integrals themselves, is obtained using Lie symmetry methods to compute it; this is the solution returned in Maple 11 and earlier. It can also be specialized.

 > ${\mathrm{gensol}}_{2.1}≔\mathrm{dsolve}\left({\mathrm{ode}}_{2},\mathrm{Lie},\mathrm{explicit}\right)$
 ${{\mathrm{gensol}}}_{{2.1}}{≔}{y}{}\left({x}\right){=}{\mathrm{RootOf}}{}\left({-}\left({{\int }}_{{}}^{{\mathrm{_Z}}}\frac{{1}}{\sqrt{{{\mathrm{_f}}}^{{4}}{+}\mathrm{c__1}}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{\mathrm{_f}}\right){+}{x}{+}\mathrm{c__2}\right){,}{y}{}\left({x}\right){=}{\mathrm{RootOf}}{}\left({{\int }}_{{}}^{{\mathrm{_Z}}}\frac{{1}}{\sqrt{{{\mathrm{_f}}}^{{4}}{+}\mathrm{c__1}}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{\mathrm{_f}}{+}{x}{+}\mathrm{c__2}\right)$ (10)
 > ${\mathrm{DEtools}}_{\mathrm{IVPsol}}\left(\left[{\mathrm{ic}}_{2}\right],{{\mathrm{gensol}}_{2.1}}_{1}\right)$
 ${y}{}\left({x}\right){=}{\mathrm{RootOf}}{}\left({{\int }}_{{\mathrm{_Z}}}^{{1}}\frac{{1}}{\sqrt{{{\mathrm{_f}}}^{{4}}}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{\mathrm{_f}}{+}{x}\right)$ (11)

As illustrated, the solution returned by dsolve without extra arguments (that is, dsolve([ode[2], ic[2]])) is the simplest one. This solution is obtained by first computing a particular solution from the symmetries (see DEtools,particularsol).

 > ${\mathrm{partsol}}_{2}≔{{\mathrm{DEtools}}_{\mathrm{particularsol}}\left({\mathrm{ode}}_{2}\right)}_{2}$
 ${{\mathrm{partsol}}}_{{2}}{≔}{y}{}\left({x}\right){=}\frac{{1}}{{x}{+}\mathrm{c__1}}$ (12)

This particular solution, which depends on only one arbitrary constant $\mathrm{c__1}$, can also be specialized.

 > ${\mathrm{DEtools}}_{\mathrm{IVPsol}}\left(\left[{\mathrm{ic}}_{2}\right],{\mathrm{partsol}}_{2}\right)$