# How to draw the phase plane of this equation?

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: $$\gamma = \frac{x-y}{x+y}\;,\;\;\;\;\;\;\frac{dT}{dt} = -\left(\frac{dx}{dt}+ \frac{dy}{dt}\right)$$ $$\text{so,}\;\;\;\frac{d\gamma}{dt}=\frac{\beta}{2}(x-y)(1-\gamma^2).$$

• 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. Jan 11, 2022 at 14:07
• See page 4 - guava.physics.uiuc.edu/~nigel/REPRINTS/2017/… Jan 11, 2022 at 14:12
• 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. Jan 11, 2022 at 14:31

You may then divide your two ODEs with each other, and get 𝑑𝑥/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, $$\frac{dx}{dy}=\frac{x}{y} ~~\frac{x+y-c +\frac{\beta}{\alpha} y} {x+y-c +\frac{\beta}{\alpha} x} ~.$$
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.