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Using various computational tools, it's possible to draw a phase plane from two first-order ODEs or a single second-order ODE. However, when there is a parameter in the equation and we don't know the value of the parameter, is there any way to draw the phase plane and see the changes with respect to the parameter? For example (e-print), if we have two first-order ODE $$ \frac{dx}{dt} = \alpha T x - \beta xy$$ $$ \frac{dy}{dt} = \alpha T y - \beta xy$$ can we draw the $x$-$y$ phase plane? We are not given any value of $\alpha$ and $\beta$, but we are given a few constraints: $$\gamma = \frac{x-y}{x+y}\;,\;\;\;\;\;\;\frac{dT}{dt} = -\left(\frac{dx}{dt}+ \frac{dy}{dt}\right)$$ $$\text{so,}\;\;\;\frac{d\gamma}{dt}=\frac{\beta}{2}(x-y)(1-\gamma^2).$$

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    $\begingroup$ As a rule, a phase portrait depends on the parameters, so without fixing them (to given values or at least ranges between bifurcations) you can't draw them. $\endgroup$
    – stafusa
    Jan 11, 2022 at 14:07
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    $\begingroup$ See page 4 - guava.physics.uiuc.edu/~nigel/REPRINTS/2017/… $\endgroup$ Jan 11, 2022 at 14:12
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    $\begingroup$ Ok, from a cursive look, the parameters are taken to be positive, but that's secondary. I'll give a quick answer shortly. BTW, I've included the reference in the body of the question. $\endgroup$
    – stafusa
    Jan 11, 2022 at 14:31

2 Answers 2

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Comment masquerading as an answer, to avoid macaronic sequences.

Looks like T is dross, to be eliminated as 𝑇=π‘βˆ’(π‘₯+𝑦), for some constant c.

You may then divide your two ODEs with each other, and get 𝑑π‘₯/𝑑𝑦 as a function of x and y, much less pretty than Lotka-Volterra, but straightforward to plot numerically for selected values of the parameters. 𝛼 may of course be absorbed into 𝛽, so it too is dross, $$ \frac{dx}{dy}=\frac{x}{y} ~~\frac{x+y-c +\frac{\beta}{\alpha} y} {x+y-c +\frac{\beta}{\alpha} x} ~. $$

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  • $\begingroup$ Eliminating T by setting T=c-(x+y) indeed did the work. But how did you come up with this expression? Why not T = x+y or something else? why T=c-(x+y)? I know the sum of the T, x and y should be a constant since we are not adding/removing any molecules from this system (here T,x,y are all concentrations of molecules, they only can convert from one to other). But how did you know that? I didn't mention any such information? did you read the original paper I linked in the comment? $\endgroup$ Jan 21, 2022 at 9:34
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    $\begingroup$ Integrate your dT/dt constraint, where you need an integration constant c. $\endgroup$ Jan 21, 2022 at 11:15
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The authors seem to consider a simple (analytic) stability analysis of the obvious equilibrium solutions (from the OP's last equation they are $x=y$ and $\gamma=\pm1$) and then to obtain the phase space not numerically, but to draw it schematically.

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  • $\begingroup$ If that's just a schematic diagram, then I understand. But the precision of that phase plane diagram made me think that it's numerically obtained and I wonder whether it can be drawn from the given information. $\endgroup$ Jan 11, 2022 at 14:43
  • $\begingroup$ @Earmen You can calculate analytically the directions of the manifolds at the fixed points and/or the scheme (if it's indeed one) can be numerically informed. $\endgroup$
    – stafusa
    Jan 11, 2022 at 15:23

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