Integrable Cases of the Polynomial Liénard-type Equation with Resonance in the Linear Part

Abstract

The paper considers the possible relationship between the local integrability of an autonomous two-dimensional ODE system with polynomial right hand sides and its global integrability. A hypothesis is put forward that local integrability in a neighborhood of each point of some region of the phase space is necessary for the existence of the first integral in this region. By integrability in some domain of the phase space it is meant the existence there of a differentiable function which is constant along the orbit of the system. Using the example of a polynomial resonance case of the Liénard-type equation with parameters, we have written out the conditions for local integrability near stationary points and found restrictions on the parameters under which these conditions are satisfied. The resulting constraint is written as a system of algebraic equations for the ODE parameters. It is shown that for parameter values that are solutions of such an algebraic system, the ODE turns out to be integrable. In this way we have found several cases of integrability. We propose a heuristic method that allows one to a priori determine the cases of integrability of an autonomous ODE with a polynomial right-hand side. The paper has an experimental character.

Appendices

Appendix A

Below are the first integrals I(x(t), y(t)) for the found cases. In the formulas below, we exclude the dependence of x(t) and y(t) on time.

$$\begin{aligned}&(\alpha )\quad 6 b_1 x^2 + 4 b_2 x^3 + 3 b_3 x^4 - 6 y^2;\\&(\beta )\quad \left( a_1 y \left( b_1+b_3 x^2\right) +\left( b_1+b_3 x^2\right) ^2-2 b_3 y^2\right) \exp \left( -\frac{2 a_1 \textrm{Artanh}\left( \frac{a_1 y+2 b_1+2 b_3 x^2}{y \sqrt{a_1^2+8 b_3}}\right) }{\sqrt{a_1^2+8 b_3}}\right) ;\\&(\textrm{a})\quad (a_1 x^2 y+b_3 x^4-2 y^2) \exp \left\{ \frac{2 a_1 \textrm{Artanh}\left[ \frac{a_1 x^2-4 y}{ x^2 \sqrt{a_1^2+8 b_3}}\right] }{\sqrt{a_1^2+8 b_3}}\right\} ;\\&(\textrm{b})\quad \frac{({a_1} x-2 {a_0}) \sinh \left( \frac{1}{2} R(x,y)\right) +{a_0} R(x,y) \cosh \left( \frac{1}{2} R(x,y)\right) }{({a_1} x-2 {a_0}) \cosh \left( \frac{1}{2} R(x,y)\right) +{a_0} R(x,y) \sinh \left( \frac{1}{2} R(x,y)\right) },\\&\quad \hbox {where}\quad R(x,y)=\sqrt{\frac{{a_1} (x ({a_1} x-2 {a_0})-2 y)}{{a_0}^2}};\\&(\textrm{c})\quad \frac{4802 a_0^4-2058 a_0^3 a_1 x-294 a_0^2 a_1 (a_1 x^2-7 y)+98 a_0 a_1^2 x (a_1 x^2-3 y)+3 a_1^2 (2 a_1 x^2-7 y)^2}{21 a_1 (14 a_0 a_1 x+3 a_1^2 x^2-7 (7 a_0^2+2 a_1 y))\sqrt{\frac{7 a_0^2}{2 a_1}-a_0 x-\frac{3}{14} a_1 x^2+y}};\\&(\textrm{d})\quad \frac{(a_1 (3 a_0 x+a_1 x^2-3 y)-18 a_0^2)^3}{(6 a_0 x+a_1 x^2-3 y)^2 (3 a_0 x-a_1 x^2+3 y)};\\&(\textrm{e})\quad (2 a_0 x+a_1 x^2-y)^2 (a_0 x+\frac{1}{2} a_1 x^2+y);\\&(\textrm{f})\quad (2 a_0 x - y)^2 (a_0 x + y);\\&(\textrm{a}')\quad \hbox {the same as } a);\\&(\textrm{b}')\quad a_0 Q\left( \frac{4 a_1 (x (a_0-2 a_1 x)+2 y)}{a_0 (a_0-2 a_1 x)+4 a_1 y}\right) \\&\qquad \quad +\frac{8 \root 3 \of {2} a_1 y \sqrt{-\frac{(a_0-2 a_1 x)^2}{a_0 (a_0-2 a_1 x)+4 a_1 y}} \root 3 \of {\frac{a_1 (x (a_0-2 a_1 x)+2 y)}{a_0 (a_0-2 a_1 x)+4 a_1 y}} \root 6 \of {-\frac{(a_0-2 a_1 x)^2}{a_0 (a_0-2 a_1 x)+4 a_1 y}-2}}{a_0-2 a_1 x},\\&\quad \hbox {where}\quad Q(v)=\frac{\root 3 \of {v} (-2 (6-v)^{5/6} \root 6 \of {6-3 v} \, _2F_1[\frac{1}{3},\frac{5}{6};\frac{4}{3};\frac{2 v}{3 (v-2)}]+v^2-8 v+12)}{(v-6)^{5/6} \sqrt{v-2}},\\&_2F_1[\alpha ,\beta ;\gamma ;z]\hbox {---the hypergeometric function};\\&(\textrm{c}')\quad \frac{K_{\frac{1}{3}}[-\frac{2}{3} P(x,y)]\cdot (a_0-a_1 x)-a_0 P(x,y)\cdot K_{\frac{2}{3}}[-\frac{2}{3} P(x,y)]}{I_{\frac{1}{3}}[-\frac{2}{3} P(x,y)]\cdot (a_1 x-a_0)-a_0 P(x,y)\cdot I_{-\frac{2}{3}}[-\frac{2}{3} P(x,y)]},\\&\quad \hbox {where}\quad P(x,y)=\sqrt{\frac{a_1 (-a_0 x+a_1 x^2-2 y)}{a_0^2}},\\&I_\alpha [z], K_\alpha [z]\hbox {---the modified Bessel functions of the first and the second}\\&\hbox {kind respectively}; \end{aligned}$$

$$\begin{aligned} \begin{array}{l}(\textrm{d}')\quad (2 (7 a_0+a_1 x)^3 (235298 a_0^6-403368 a_0^5 a_1 x+7203 a_0^4 a_1 (13 a_1 x^2+56 y)\\ \qquad \qquad + 686 a_0^3 a_1^2 x (83 a_1 x^2-420 y)+147 a_0^2 a_1^2 (79 a_1^2 x^4-420 a_1 x^2 y+980 y^2)\\ \qquad \qquad +84 a_0 a_1^3 x (2 a_1 x^2-7 y) (11 a_1 x^2-28 y)+16 a_1^3 (2 a_1 x^2-7 y)^3)/F(x,y),\\ \hbox {where}\quad F(x,y)=3 (a_0 (7 a_0-2 a_1 x))^{3/2} (3 a_1 x (7 a_0+a_1 x)-14 a_1 y)^3\\ \qquad \qquad \qquad \times \left( \frac{(7 a_0+a_1 x)^2 (56 a_1 y (2 a_1 x-7 a_0) (7 a_0+a_1 x)+(8 a_1 x-7 a_0) (7 a_0-2 a_1 x) (7 a_0+a_1 x)^2-196 a_1^2 y^2)}{a_0 (7 a_0-2 a_1 x) (3 a_1 x (7 a_0+a_1 x)-14 a_1 y)^2}\right) ^{3/2};\\ (\textrm{e}')\quad (4 \sqrt{\frac{(11 a_0+2 a_1 x)^2}{33 a_0^2+6 a_0 a_1 x+4 a_1 y}} (220 a_1 y ((121 a_0^2-176 a_0 a_1 x+54 a_1^2 x^2) (11 a_0+2 a_1 x)^2\\ \qquad \qquad -66 a_1 y (6 a_1 x-11 a_0) (11 a_0+2 a_1 x)+968 a_1^2 y^2)+(11 a_0-10 a_1 x) (121 a_0^2\\ \qquad \qquad -176 a_0 a_1 x+54 a_1^2 x^2) (11 a_0+2 a_1 x)^3))/S(x,y),\\ \hbox {where}\quad S(x,y)=(5 (11 a_0+2 a_1 x) ((11 a_0+2 a_1 x) (11 a_0-10 a_1 x)+88 a_1 y)^2\\ \qquad \qquad \times \sqrt{\frac{5 (11 a_0+2 a_1 x)^2}{33 a_0^2+6 a_0 a_1 x+4 a_1 y}-22});\\ \textrm{f}')\quad \frac{(27 a_0^2-4 a_1 (3 a_0 x+a_1 x^2-3 y))^4}{(-3 a_0 x+2 a_1 x^2-6 y) (9 a_0 x+2 a_1 x^2-6 y)^3};\\ \textrm{g}')\quad (6 a_0 x + 3 a_1 x^2 - 4 y)^3 (2 a_0 x + a_1 x^2 + 4 y);\\ \textrm{h}')\quad \hbox {The same as } (\textrm{f}).\\ \end{array} \end{aligned}$$

All of the above integrals were verified by substituting the right-hand sides of the corresponding equation into the total derivative of the integral with respect to time. In all cases it turned out to be zero. However, it is difficult to verify case $(\textrm{b}')$ by such calculations because of the nested radicals. But the corresponding derivative vanishes when different numerical values of the parameters and variables are substituted.

Note, the cases $(\textrm{d}), (\textrm{f}')$ (and at the higher resonances) are the same. In a heuristic tradition we can suppose that $b_1$ does not depend on the resonance order, that is the parameter $b_1$ in (d) can be arbitrary

$$\begin{aligned} \ddot{x}=({a_0}+{a_1} x)\,\dot{x}+b_1 x-\frac{1}{3} {a_0} {a_1} x^2-\frac{1}{9} {a_1}^2 x^3. \end{aligned}$$

This case is exactly up to a multiplier the example Sects. 2.2.3-$-$2.4 from book [15] at $a_1 = 3 a, a_0 = b$. It should be integrable at arbitrary $b_1$. This is true. The corresponding first integral is

$$\begin{aligned} \begin{array}{l} \left[ \frac{x (-3 \sqrt{a_0^2+4 b_1}+3 a_0+2 a_1 x)-6 y}{3 (\sqrt{a_0^2+4 b_1}+a_0) x+2 a_1 x^2-6 y}\right] ^{a_0}\times \left[ \frac{(6 a_0 x-6 (\frac{3 b_1}{a_1}+y)+2 a_1 x^2)^2}{3 a_0 a_1 x^3-9 a_0 x y+a_1^2 x^4-3 x^2 (2 a_1 y+3 b_1)+9 y^2}\right] ^{\sqrt{a_0^2+4 b_1}}.\\ \end{array} \end{aligned}$$

Appendix B

The three equations of the condition of integrability A for $M=3$ are

$$\begin{aligned}&a_0^{24} (-11872 a_0 a_1 b_2^3 - 11968 b_2^4 + 24 a_0^3 a_1 b_2 (333 a_1^2 + 619 b_3) + 4 a_0^2 b_2^2 (1869 a_1^2 + 8216 b_3) \\&\quad \qquad + 63 a_0^4 (9 a_1^4 - 138 a_1^2 b_3 - 175 b_3^2)) = 0,\\&a_0^{12} (197068017664 a_0 a_1 b_2^7 + 58885623808 b_2^8 + 512 a_0^2 b_2^6 (378756343 a_1^2 - 731989864 b_3) \\&\quad - 256 a_0^3 a_1 b_2^5 (83771767 a_1^2 + 3821209124 b_3) - 16 a_0^4 b_2^4 (8660719683 a_1^4 + 41594560224 a_1^2 b_3 \\&\quad -46598666552 b_3^2) - 192 a_0^5 a_1 b_2^3 (338543814 a_1^4 - 906067049 a_1^2 b_3 - 7047641693 b_3^2) \\&\quad + 8 a_0^6 b_2^2 (89437527 a_1^6 + 38594010258 a_1^4 b_3 + 63847123071 a_1^2 b_3^2 - 59816367800 b_3^3) \\&\quad +336 a_0^7 a_1 b_2 (13826241 a_1^6 + 208531701 a_1^4 b_3 - 504056541 a_1^2 b_3^2 - 1359499225 b_3^3) \\&\quad +441 a_0^8 (706293 a_1^8 - 5704452 a_1^6 b_3 - 175464018 a_1^4 b_3^2 - 131448100 a_1^2 b_3^3 + 81493125 b_3^4)) = 0,\\&16 (27157797183599882928128 a_0 a_1 b_2^{11} + 6370472246997895348224 b_2^{12} \\&\quad +49152 a_0^3 a_1 b_2^9 (405347313952322393 a_1^2 - 2598147352384602726 b_3) \\&\quad +147456 a_0^2 b_2^10 (281784472868119963 a_1^2 - 268186906800985656 b_3) \\&\quad -1536 a_0^5 a_1 b_2^7 (16851209069176800283 a_1^4 - 3979380086262918424 a_1^2 b_3 - 97012510428436303096 b_3^2) \\&\quad -1024 a_0^4 b_2^8 (15351403088820872017 a_1^4 + 120743181355480108152 a_1^2 b_3 - 76155371083417551732 b_3^2) \\&\quad -128 a_0^7 a_1 b_2^5 (59826132685859479569 a_1^6 - 233842397088300553287 a_1^4 b_3 + 1241658449506834305102 a_1^2 b_3^2 \\&\quad -410934092963328824248 b_3^3) - 192 a_0^6 b_2^6 (83781405413462369449 a_1^6 - 372071400366966511912 a_1^4 b_3 \\&\quad +46891556191076249944 a_1^2 b_3^2 + 202893335372812712384 b_3^3) - 48 a_0^8 b_2^4 (72355429454479795701 a_1^8 \\&\quad -22201587437970313962 a_1^6 b_3 + 1282488231720067982745 a_1^4 b_3^2 - 4957211792578962761440 a_1^2 b_3^3 \\&\quad +523303618374986364300 b_3^4) - 96 a_0^9 a_1 b_2^3 (10189978402497862761 a_1^8 \\&\quad -42000151081415659614 a_1^6 b_3 - 412073512990296018156 a_1^4 b_3^2 - 1782282405723105343570 a_1^2 b_3^3 \\&\quad +1274845345921325953075 b_3^4) - 252 a_0^10 b_2^2 (252512933211941391 a_1^10 - 14993465192208803844 a_1^8 b_3 \\&\quad -93821118007359615606 a_1^6 b_3^2 - 7240136820998233340 a_1^4 b_3^3 + 516172788728252838175 a_1^2 b_3^4 \\&\quad -50378058061905315000 b_3^5) + 10584 a_0^11 a_1 b_2 (1707614860034943 a_1^10 + 69170194013267637 a_1^8 b_3 \\&\quad +75614571012035286 a_1^6 b_3^2 - 2385879536833733814 a_1^4 b_3^3 - 3465393452231732325 a_1^2 b_3^4 \\&\quad +1892755320042820625 b_3^5) + 27783 a_0^12 (48345534559593 a_1^12 + 301760670853602 a_1^10 b_3 \\&\quad -20361112706663649 a_1^8 b_3^2 - 134198879162307204 a_1^6 b_3^3 + 51156574621167575 a_1^4 b_3^4 \\&\quad +223588462531621250 a_1^2 b_3^5 - 38419614968109375 b_3^6)) = 0. \end{aligned}$$