Is there a general solution (in terms of simple known functions) for the following system of coupled non-linear EDOs ?
$$x'(t) = -a_1x^2 -bxy$$
$$y'(t) = -a_2y^2 -bxy,$$
where $a_1$, $a_2$ and $b$ are real constants.
Thank you for your help.
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6$\begingroup$ Because the system is invariant under the solvable $2$-parameter Lie group of transformations of the form $$F(x,y,t) = (x/p,y/p,pt+q)$$ for constants $q$ and $p\not=0$, it is (parametrically) solvable by quadrature, so there will be explicit formulae, and they won't be very complicated in form, but they may not be elementary functions. $\endgroup$– Robert BryantCommented Mar 8, 2023 at 13:17
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3 Answers
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with a change of variable $x = \frac{\cos \theta}{a_1r}, y = \frac{\sin \theta}{a_2r}, $ the new equations for $r \text{ and } \theta$ are:
$$\frac{dr}{dt} = f(\theta), \frac{rd\theta}{dt}=g(\theta) $$ where $f$ and $g$ are functions of $\theta$ that can be explicitly computed. the last two equations can be solves as $\ln r = \int \frac{f(\theta)}{g(\theta)}d\theta$
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$\begingroup$ Dear Abel,Thank you very much for your answer. Now I am trying to figure out how to solve the integral involving theta. $\endgroup$– silmarCommented Mar 10, 2023 at 23:54
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$\begingroup$ Abel, just one more short question: From a closer look I think that the integral in theta doesn't have a primitive. What do you think ? Thank you again. $\endgroup$– silmarCommented Mar 11, 2023 at 1:09
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$\begingroup$ @silmar I get (up to a constant)$$\ln(r)=\frac{b^2-a_1a_2}{(b-a_1)(b-a_2)}\ln\left(a_1(b-a_2)\sin(\theta)-(b-a_1)a_2\cos(\theta)\right)\qquad-\frac{a_1}{b-a_1}\ln\sin(\theta)-\frac{a_2}{b-a_2}\ln\cos(\theta)$$ $\endgroup$ Commented Mar 15, 2023 at 14:26
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$\begingroup$ Thank you. Finally I could reproduce this last result. However, when we replace the expression for r (r=h(theta)), derived from this last equation, in rd(theta)/dt = g(theta), yielding h(theta)d(theta)/dt = g(theta), we end up with an integral (in theta) which has no primitive in terms of elementary functions, right ? $\endgroup$– silmarCommented Mar 22, 2023 at 13:18
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Trying to be explicit.
Substitute $x=u^{-a_1}v^{-b_1}$, $y=v^{-a_2}u^{-b_2}$ (here $b_1=b_2=b$ but they could as well be different).
We obtain $$ \frac{du}{dt}=u^{1-a_1}v^{-b_1},\qquad\frac{dv}{dt}=v^{1-a_2}u^{-b_2}, $$ which is equivalent to $$ u^{a_1-b_2-1}du=v^{a_2-b_1-1}dv. $$ If $b_2\ne a_1$ and $b_1\ne a_2$ the general solution is given by $$ \frac{u^{a_1-b_2}}{a_1-b_2}-\frac{v^{a_2-b_1}}{a_2-b_1}=\mathrm{const.} $$ In case $b_2=a_1$ or $b_1=a_2$ or both we get logarithms instead of powers at appropriate places.
answered Mar 16, 2023 at 8:16
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$\begingroup$ +1 for your effort on this question. The OP concernes the explicite solution and your answer do that. However a geometric approach could be consideration of a metric compatible to the Lotka Volterra system. as you find in the wikipedia link I provided in my answer the system has a center so one can think to an explicite Riemannian ,metric compatible to the system $\endgroup$ Commented Mar 16, 2023 at 10:11
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1$\begingroup$ @AliTaghavi Sorry I tried various substitutions but could not reduce this system to the Lotka-Volterra form. Could you please name explicit substitution which transforms the OP system to the form $dx/dt=\alpha x-\beta xy$, $dy/dt=\delta xy-\gamma y$? $\endgroup$ Commented Mar 16, 2023 at 10:38
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1$\begingroup$ I am sorry for my mistake. misread some thing in the OP: I confused and misread $ax^2+bxy$ by $ax+bxy$. I belive that the two systems are not topological equivalent: The OP system is a Homogenous equation invariant under rescalling But the Lotka Volterra system has a center as wiikipedia indicate to. So a center in the first quadrant is not invariant under rescalling $(x,y)\mapsto (\lambda x, \lambda y)$ $\endgroup$ Commented Mar 16, 2023 at 13:58
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It is just the Lotka Volterra system
https://en.wikipedia.org/wiki/Lotka%E2%80%93Volterra_equations
The above link contains materials about this system.
I remember I learned about these material from a talk by E.C.Zeeman at 2001. I think Zeeman also devoted some researchs on this topic. So searching Lotka Voltera system +Zeeman gives you some other materials.
answered Mar 15, 2023 at 12:14
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$\begingroup$ One obtains a Lotka-Volterra like system if one replaces $x^2$ with $x$ and $y^2$ with $y$ in the OP. Do you see that they are in fact equivalent? $\endgroup$ Commented Mar 15, 2023 at 13:31
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$\begingroup$ The system in OP is itself in the Lotka Volterra form. Please see the wikipedia link $\endgroup$ Commented Mar 16, 2023 at 9:53
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$\begingroup$ @მამუკაჯიბლაძე Any way such kind of chang of coordinates is very common in ODE see for example page 2 of this paper $\pi(x,y)=(x^2,y)$ $\endgroup$ Commented Mar 16, 2023 at 10:03
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$\begingroup$ @მამუკაჯიბლაძე Any way such kind of chang of coordinates is very common in ODE see for example page 2 of this paper $\pi(x,y)=(x^2,y)$ $\endgroup$ Commented Mar 16, 2023 at 10:03
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2$\begingroup$ Let me reply here too - I am unable to find a substitution reducing the OP system to a Lotka-Volterra type system. In fact I find it unlikely since the Lotka-Volterra system contains linear terms while here everything is quadratic. I believe this means that the vector fields corresponding to these systems will be qualitatively different at the origin, no? $\endgroup$ Commented Mar 16, 2023 at 10:45