Solving Abel's ODEs of the First Kind - Maple Programming Help

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Solving Abel's ODEs of the First Kind

Description

 • The general form of Abel's equation of the first kind is given by:
 > Abel_ode := diff(y(x),x)=f3(x)*y(x)^3+f2(x)*y(x)^2+f1(x)*y(x)+f0(x);
 ${\mathrm{Abel_ode}}{≔}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){=}{\mathrm{f3}}{}\left({x}\right){}{{y}{}\left({x}\right)}^{{3}}{+}{\mathrm{f2}}{}\left({x}\right){}{{y}{}\left({x}\right)}^{{2}}{+}{\mathrm{f1}}{}\left({x}\right){}{y}{}\left({x}\right){+}{\mathrm{f0}}{}\left({x}\right)$ (1)
 where f3(x), f2(x), f1(x) and f0(x) are arbitrary functions.
 See Differentialgleichungen, by E. Kamke, p. 24. There is as yet no general solution for this ODE. For Abel's equation of the second kind, see Abel2A and Abel2C.
 • The most general method available at the moment to solve Abel ODEs seems to be the method of "Abel's invariant", described in E. Kamke, p. 26, as sub-method (g) due to M. Chini. The invariant of an Abel equation with f2=0 is the following quantity:
 > Abel_invariant := -1/27/f3(x)^4*(-diff(f0(x),x)*f3(x)+f0(x)*diff(f3(x),x)+ 3*f0(x)*f3(x)*f1(x))^3/f0(x)^5;
 ${\mathrm{Abel_invariant}}{≔}{-}\frac{{\left({-}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{f0}}{}\left({x}\right)\right){}{\mathrm{f3}}{}\left({x}\right){+}{\mathrm{f0}}{}\left({x}\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{f3}}{}\left({x}\right)\right){+}{3}{}{\mathrm{f0}}{}\left({x}\right){}{\mathrm{f3}}{}\left({x}\right){}{\mathrm{f1}}{}\left({x}\right)\right)}^{{3}}}{{27}{}{{\mathrm{f3}}{}\left({x}\right)}^{{4}}{}{{\mathrm{f0}}{}\left({x}\right)}^{{5}}}$ (2)
 If the invariant does not depend on x, then the equation can be solved directly.
 For an Abel equation with f2<>0, the f2 term can be removed by using the following transformation:
 > y(x)=u(x)-f2(x)/3/f3(x);
 ${y}{}\left({x}\right){=}{u}{}\left({x}\right){-}\frac{{\mathrm{f2}}{}\left({x}\right)}{{3}{}{\mathrm{f3}}{}\left({x}\right)}$ (3)
 The invariant can then be calculated as in the previous case. Note that if an Abel ODE has a constant invariant, then any other Abel ODE obtained from it by a transformation of the form
 > {y(x)=G(t)*u(t)+H(t), x=F(t)};
 $\left\{{x}{=}{F}{}\left({t}\right){,}{y}{}\left({x}\right){=}{G}{}\left({t}\right){}{u}{}\left({t}\right){+}{H}{}\left({t}\right)\right\}$ (4)
 will also have a constant invariant (that is, is also solvable by this method).
 The method "Chini" (see ?odeadvisor,Chini), also due to Chini, generalizes this method of the constant invariant for Abel ODEs.

Examples

 > $\mathrm{Abel_ode}≔\frac{ⅆ}{ⅆx}y\left(x\right)=\mathrm{f3}\left(x\right){y\left(x\right)}^{3}+\mathrm{f2}\left(x\right){y\left(x\right)}^{2}+\mathrm{f1}\left(x\right)y\left(x\right)+\mathrm{f0}\left(x\right)$
 ${\mathrm{Abel_ode}}{≔}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){=}{\mathrm{f3}}{}\left({x}\right){}{{y}{}\left({x}\right)}^{{3}}{+}{\mathrm{f2}}{}\left({x}\right){}{{y}{}\left({x}\right)}^{{2}}{+}{\mathrm{f1}}{}\left({x}\right){}{y}{}\left({x}\right){+}{\mathrm{f0}}{}\left({x}\right)$ (5)
 > $\mathrm{Abel_invariant}≔-\frac{1{\left(-\left(\frac{ⅆ}{ⅆx}\mathrm{f0}\left(x\right)\right)\mathrm{f3}\left(x\right)+\mathrm{f0}\left(x\right)\left(\frac{ⅆ}{ⅆx}\mathrm{f3}\left(x\right)\right)+3\mathrm{f0}\left(x\right)\mathrm{f3}\left(x\right)\mathrm{f1}\left(x\right)\right)}^{3}}{27{\mathrm{f3}\left(x\right)}^{4}{\mathrm{f0}\left(x\right)}^{5}}$
 ${\mathrm{Abel_invariant}}{≔}{-}\frac{{\left({-}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{f0}}{}\left({x}\right)\right){}{\mathrm{f3}}{}\left({x}\right){+}{\mathrm{f0}}{}\left({x}\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{f3}}{}\left({x}\right)\right){+}{3}{}{\mathrm{f0}}{}\left({x}\right){}{\mathrm{f3}}{}\left({x}\right){}{\mathrm{f1}}{}\left({x}\right)\right)}^{{3}}}{{27}{}{{\mathrm{f3}}{}\left({x}\right)}^{{4}}{}{{\mathrm{f0}}{}\left({x}\right)}^{{5}}}$ (6)
 > $y\left(x\right)=u\left(x\right)-\frac{\mathrm{f2}\left(x\right)}{3\mathrm{f3}\left(x\right)}$
 ${y}{}\left({x}\right){=}{u}{}\left({x}\right){-}\frac{{\mathrm{f2}}{}\left({x}\right)}{{3}{}{\mathrm{f3}}{}\left({x}\right)}$ (7)
 > $\left\{y\left(x\right)=G\left(t\right)u\left(t\right)+H\left(t\right),x=F\left(t\right)\right\}$
 $\left\{{x}{=}{F}{}\left({t}\right){,}{y}{}\left({x}\right){=}{G}{}\left({t}\right){}{u}{}\left({t}\right){+}{H}{}\left({t}\right)\right\}$ (8)
 > $\mathrm{with}\left(\mathrm{DEtools},\mathrm{odeadvisor}\right)$
 $\left[{\mathrm{odeadvisor}}\right]$ (9)
 > $\mathrm{with}\left(\mathrm{PDEtools},\mathrm{dchange}\right)$
 $\left[{\mathrm{dchange}}\right]$ (10)
 > $\mathrm{odeadvisor}\left(\mathrm{Abel_ode}\right)$
 $\left[{\mathrm{_Abel}}\right]$ (11)

1) An example of an Abel ODE having a constant invariant solved using the related scheme:

 > $\mathrm{ODE}≔\frac{ⅆ}{ⅆx}y\left(x\right)=\frac{1\left(12x+27{x}^{3}+27{x}^{3}{y\left(x\right)}^{2}+18{x}^{2}y\left(x\right)+27{y\left(x\right)}^{3}{x}^{3}+27{x}^{2}{y\left(x\right)}^{2}+9xy\left(x\right)+1\right)}{27{x}^{3}}:$
 > $\mathrm{ans}≔\mathrm{dsolve}\left(\mathrm{ODE}\right)$
 ${\mathrm{ans}}{≔}{y}{}\left({x}\right){=}\frac{{29}{}{\mathrm{RootOf}}{}\left({-}{81}{}\left({{\int }}_{{}}^{{\mathrm{_Z}}}\frac{{1}}{{841}{}{{\mathrm{_a}}}^{{3}}{-}{27}{}{\mathrm{_a}}{+}{27}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{\mathrm{_a}}\right){+}{x}{+}{3}{}{\mathrm{_C1}}\right){}{x}{-}{3}{}{x}{-}{3}}{{9}{}{x}}$ (12)

Any "linear transformation" of ODE will also be solved by the same method. For example:

 > $\mathrm{TR_LIN}≔\left\{y\left(x\right)=G\left(t\right)u\left(t\right)+H\left(t\right),x=F\left(t\right)\right\}$
 ${\mathrm{TR_LIN}}{≔}\left\{{x}{=}{F}{}\left({t}\right){,}{y}{}\left({x}\right){=}{G}{}\left({t}\right){}{u}{}\left({t}\right){+}{H}{}\left({t}\right)\right\}$ (13)
 > $\mathrm{ODE_p}≔\mathrm{dchange}\left(\mathrm{TR_LIN},\mathrm{ODE},\left[u,t\right]\right):$
 > $\mathrm{ans_p}≔\mathrm{dsolve}\left(\mathrm{ODE_p},u\left(t\right)\right)$
 ${\mathrm{ans_p}}{≔}{u}{}\left({t}\right){=}{-}\frac{{9}{}{H}{}\left({t}\right){}{F}{}\left({t}\right){-}{29}{}{\mathrm{RootOf}}{}\left({-}{81}{}\left({{\int }}_{{}}^{{\mathrm{_Z}}}\frac{{1}}{{841}{}{{\mathrm{_a}}}^{{3}}{-}{27}{}{\mathrm{_a}}{+}{27}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{\mathrm{_a}}\right){+}{F}{}\left({t}\right){+}{3}{}{\mathrm{_C1}}\right){}{F}{}\left({t}\right){+}{3}{}{F}{}\left({t}\right){+}{3}}{{9}{}{G}{}\left({t}\right){}{F}{}\left({t}\right)}$ (14)

2) A case for which the solving method is known: f0(x) = f1(x) = 0, and diff(f3(x)/f2(x),x)=a*f2(x).

In this case, one can proceed as follows:

 > $\mathrm{ode}≔\mathrm{subs}\left(\left\{\mathrm{f1}\left(x\right)=0,\mathrm{f0}\left(x\right)=0\right\},\mathrm{Abel_ode}\right)$
 ${\mathrm{ode}}{≔}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){=}{\mathrm{f3}}{}\left({x}\right){}{{y}{}\left({x}\right)}^{{3}}{+}{\mathrm{f2}}{}\left({x}\right){}{{y}{}\left({x}\right)}^{{2}}$ (15)

First introduce r(t) and t as new variables using:

 > $\mathrm{ITR}≔\left\{y\left(x\right)=\frac{\mathrm{f2}\left(t\right)r\left(t\right)}{\mathrm{f3}\left(t\right)},x=t\right\}$
 ${\mathrm{ITR}}{≔}\left\{{x}{=}{t}{,}{y}{}\left({x}\right){=}\frac{{\mathrm{f2}}{}\left({t}\right){}{r}{}\left({t}\right)}{{\mathrm{f3}}{}\left({t}\right)}\right\}$ (16)
 > $\mathrm{new_ode}≔\mathrm{dchange}\left(\mathrm{ITR},\mathrm{ode},\left[r\left(t\right),t\right]\right):$

Now, introduce the condition on the derivative of f3(t)/f2(t):

 > $\mathrm{constraint}≔\frac{ⅆ}{ⅆt}\left(\frac{\mathrm{f3}\left(t\right)}{\mathrm{f2}\left(t\right)}\right)-a\mathrm{f2}\left(t\right)=0$
 ${\mathrm{constraint}}{≔}\frac{\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{f3}}{}\left({t}\right)}{{\mathrm{f2}}{}\left({t}\right)}{-}\frac{{\mathrm{f3}}{}\left({t}\right){}\left(\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{f2}}{}\left({t}\right)\right)}{{{\mathrm{f2}}{}\left({t}\right)}^{{2}}}{-}{a}{}{\mathrm{f2}}{}\left({t}\right){=}{0}$ (17)

and simplify new_ode with regard to this relation:

 > $\mathrm{new_ode2}≔\mathrm{simplify}\left(\mathrm{new_ode},\left\{\mathrm{constraint}\right\},\left\{\frac{ⅆ}{ⅆt}\mathrm{f3}\left(t\right)\right\}\right)$
 ${\mathrm{new_ode2}}{≔}\frac{{-}{r}{}\left({t}\right){}{{\mathrm{f2}}{}\left({t}\right)}^{{3}}{}{a}{+}{\mathrm{f2}}{}\left({t}\right){}\left(\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{r}{}\left({t}\right)\right){}{\mathrm{f3}}{}\left({t}\right)}{{{\mathrm{f3}}{}\left({t}\right)}^{{2}}}{=}\frac{{{\mathrm{f2}}{}\left({t}\right)}^{{3}}{}{{r}{}\left({t}\right)}^{{2}}{}\left({r}{}\left({t}\right){+}{1}\right)}{{{\mathrm{f3}}{}\left({t}\right)}^{{2}}}$ (18)

This ODE is separable.

 > $\mathrm{odeadvisor}\left(\mathrm{new_ode2}\right)$
 $\left[{\mathrm{_separable}}\right]$ (19)

3) Rewrite in "normal form" (no square term in the RHS) when: f3(x)=1/x, f2(x)=1/x, f1(x)=0, f0(x)=4

 > $\mathrm{ode}≔\mathrm{eval}\left(\mathrm{subs}\left(\left\{\mathrm{f3}\left(x\right)=\frac{1}{x},\mathrm{f2}\left(x\right)=\frac{1}{x},\mathrm{f1}\left(x\right)=0,\mathrm{f0}\left(x\right)=4\right\},\mathrm{Abel_ode}\right)\right)$
 ${\mathrm{ode}}{≔}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){=}\frac{{{y}{}\left({x}\right)}^{{3}}}{{x}}{+}\frac{{{y}{}\left({x}\right)}^{{2}}}{{x}}{+}{4}$ (20)

First of all, Abel's ODEs of the first kind can be rewritten in normal form (which is sometimes useful) by making the appropriate change of variables. The transformation is of a general type. After introducing

 > $w\left(x\right)≔{ⅇ}^{∫\left(\mathrm{f1}\left(x\right)-\frac{{\mathrm{f2}\left(x\right)}^{2}}{3\mathrm{f3}\left(x\right)}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}ⅆx}$
 ${w}{}\left({x}\right){≔}{{ⅇ}}^{{\int }\left({\mathrm{f1}}{}\left({x}\right){-}\frac{{{\mathrm{f2}}{}\left({x}\right)}^{{2}}}{{3}{}{\mathrm{f3}}{}\left({x}\right)}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}}$ (21)

the following transformations (where {x,y(x)} = old vars; {t,r(t)} = new vars) will yield the desired normal form:

 > $\mathrm{tr}≔\left\{t=∫\mathrm{f3}\left(x\right){'w'\left(x\right)}^{2}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}ⅆx,r\left(t\right)=\frac{1\left(3y\left(x\right)\mathrm{f3}\left(x\right)+\mathrm{f2}\left(x\right)\right)}{3{ⅇ}^{∫\left(\mathrm{f1}\left(x\right)-\frac{1{\mathrm{f2}\left(x\right)}^{2}}{3\mathrm{f3}\left(x\right)}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}ⅆx}\mathrm{f3}\left(x\right)}\right\}$
 ${\mathrm{tr}}{≔}\left\{{t}{=}{\int }{\mathrm{f3}}{}\left({x}\right){}{{w}{}\left({x}\right)}^{{2}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{,}{r}{}\left({t}\right){=}\frac{{3}{}{y}{}\left({x}\right){}{\mathrm{f3}}{}\left({x}\right){+}{\mathrm{f2}}{}\left({x}\right)}{{3}{}{{ⅇ}}^{{\int }\left({\mathrm{f1}}{}\left({x}\right){-}\frac{{{\mathrm{f2}}{}\left({x}\right)}^{{2}}}{{3}{}{\mathrm{f3}}{}\left({x}\right)}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}}{}{\mathrm{f3}}{}\left({x}\right)}\right\}$ (22)

The transformation equations required for this case are obtained from the general transformation tr (above) as follows:

 > $\mathrm{TR}≔\mathrm{eval}\left(\mathrm{subs}\left(\left\{\mathrm{f3}\left(x\right)=\frac{1}{x},\mathrm{f2}\left(x\right)=\frac{1}{x},\mathrm{f1}\left(x\right)=0,\mathrm{f0}\left(x\right)=4\right\},\mathrm{tr}\right)\right)$
 ${\mathrm{TR}}{≔}\left\{{t}{=}{-}\frac{{3}}{{2}{}{{x}}^{{2}}{{3}}}}{,}{r}{}\left({t}\right){=}\frac{{{x}}^{{4}}{{3}}}{}\left(\frac{{3}{}{y}{}\left({x}\right)}{{x}}{+}\frac{{1}}{{x}}\right)}{{3}}\right\}$ (23)
 > $\mathrm{ITR}≔\mathrm{solve}\left(\mathrm{TR},\left\{y\left(x\right),x\right\}\right)$
 ${\mathrm{ITR}}{≔}\left\{{x}{=}{-}\frac{{3}{}{\mathrm{RootOf}}{}\left({2}{}{t}{}{{\mathrm{_Z}}}^{{2}}{+}{3}\right)}{{2}{}{t}}{,}{y}{}\left({x}\right){=}{-}\frac{{\mathrm{RootOf}}{}\left({2}{}{t}{}{{\mathrm{_Z}}}^{{2}}{+}{3}\right){-}{3}{}{r}{}\left({t}\right)}{{3}{}{\mathrm{RootOf}}{}\left({2}{}{t}{}{{\mathrm{_Z}}}^{{2}}{+}{3}\right)}\right\}$ (24)

and the change of variables is implemented as follows:

 > $\mathrm{new_ode}≔\mathrm{dchange}\left(\mathrm{ITR},\mathrm{ode},\left[t,r\left(t\right)\right],'\mathrm{known}'=\mathrm{indets}\left(\mathrm{ode},'\mathrm{unknown}'\right)\right):$
 > $\mathrm{new_ode2}≔\mathrm{simplify}\left(\mathrm{op}\left(1,\mathrm{map}\left(\mathrm{allvalues},\left[\mathrm{solve}\left(\mathrm{new_ode},\left\{\frac{ⅆ}{ⅆt}r\left(t\right)\right\}\right)\right]\right)\right)\right)$
 ${\mathrm{new_ode2}}{≔}\left\{\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{r}{}\left({t}\right){=}\frac{{18}{}{{t}}^{{3}}{}{{r}{}\left({t}\right)}^{{3}}{-}{{t}}^{{2}}{}\sqrt{{6}}{}\sqrt{{-}\frac{{1}}{{t}}}{-}{243}}{{18}{}{{t}}^{{3}}}\right\}$ (25)

Finally, the normal form can be made explicit as follows:

 > $\mathrm{collect}\left(\mathrm{new_ode2},r\left(t\right),\mathrm{factor}\right)$
 $\left\{\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{r}{}\left({t}\right){=}{{r}{}\left({t}\right)}^{{3}}{-}\frac{{{t}}^{{2}}{}\sqrt{{6}}{}\sqrt{{-}\frac{{1}}{{t}}}{+}{243}}{{18}{}{{t}}^{{3}}}\right\}$ (26)