How I can get the critical points of this system?

Question

I've got this dynamic system: $$\frac{dx}{dt} = x(1-x)-\frac{\alpha xy}{x+\gamma}$$ $$\frac{dy}{dt} = \beta y\left(1-\frac{y}{x}\right)$$ Where $\alpha$, $\beta$ and $\gamma$ $>0$ .

How can I get the critical points?

I have found four: $\{{0,0}\},\{{1,0}\}$ and $$ \Bigg\{ \frac{1-\alpha-\gamma-\sqrt{(\alpha+\gamma-1)^2+4\gamma}}{2},\frac{1-\alpha-\gamma-\sqrt{(\alpha+\gamma-1)^2+4\gamma}}{2}\Bigg\} $$ $$ \Bigg\{ \frac{1-\alpha-\gamma+\sqrt{(\alpha+\gamma-1)^2+4\gamma}}{2},\frac{1-\alpha-\gamma+\sqrt{(\alpha+\gamma-1)^2+4\gamma}}{2}\Bigg\} $$

I also want to plot the equations when $\frac{dx}{dt}=0$ and $\frac{dy}{dt}=0$.

I though the critical points will be the points where the functions meet. But they aren't, furthermore $\frac{dy}{dt}=0$ is (or at least that is what i got) a straight line, thus they can only meet twice ($\frac{dx}{dy}=0$ seems to be a parabola).

Can someone confirm me if the critical points are correct and guide me to plot the functions? I usually use Maple.

Thanks!

Answer 1

Background

Prior to answering your question, I will provide my answer with a context by stating several relevant definitions. Suppose you are given a system of (autonomous) differential equations $Dy=f(y)$, where $f:U\longrightarrow\mathbb{R}^n$, $U$ is an open set such that $U\subseteq \mathbb{R}^n$ (with appropriate smoothness conditions on $f$). An equilibrium point (or a critical point) is (sometimes) defined as a function $y$, such that there exists $c \in U$, such that $y(t)=c$ for all $t\in \mathbb{R}$ and $f(c)=0$ (some general references could be "Solving Differential Equations on Manifolds" by Ernst Hairer and "Nonlinear Control Systems" by Horacio J. Marquez). For simplicity, we identify the critical point with the constant $c\in U$.

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Critical points

In your case, the dimension of the problem $n$ is 2. Also, $f=(f_1, f_2)^T$ is given by

• (1) $f_1(x,y) = x(1-x)-\frac{\alpha x y}{x + \gamma}$
• (2) $f_2(x,y) = \beta y (1 - \frac{y}{x})$

for all $(x,y)\in U$.

However, the possible domain $U$ of the definition of $f$ is implicitly restricted by the value of the parameter $\gamma\in\mathbb{R}$. In particular, $(-\gamma, y)\notin U$ for all admissible $y\in\mathbb{R}$. Furthermore, $(0,y)\notin U$ for all admissible $y\in\mathbb{R}$. Therefore, let us define $U$ as $U = \mathbb{R^2} - \{(x, y). x = -\gamma \lor x = 0\}$.

Suppose $(x,y)\in U$ is a critical point. Then, from (2), $\beta y (1 - \frac{y}{x})=0$. Thus, given that $\beta > 0$, $y = 0$ or $y = x$. Coniser these two cases explicitly:

• Case I: $y = 0$. Then, from (2), for $(x,y)$ to be a critical point we also need $x (1 - x) = 0$. Thus, $x = 0$ or $x = 1$. However, as previously established, $(0,y)\notin U$. Thus $(x_1, y_1) = (1, 0)$ is the only possible critical point in this case. It is, indeed, a critical point, which can be verified by substitution into $f(x,y)=0$.
• Case II: $y = x$. Then, from (1), $x^2 + x(\alpha+\gamma-1) - \gamma = 0$. Solving for $x$ and substituting yields the two possible critical points from your answer: $ (x_2,y_2) = \left( \frac{1-\alpha-\gamma-\sqrt{(\alpha+\gamma-1)^2+4\gamma}}{2},\frac{1-\alpha-\gamma-\sqrt{(\alpha+\gamma-1)^2+4\gamma}}{2}\right) $ and $ (x_3,y_3) =\left( \frac{1-\alpha-\gamma+\sqrt{(\alpha+\gamma-1)^2+4\gamma}}{2},\frac{1-\alpha-\gamma+\sqrt{(\alpha+\gamma-1)^2+4\gamma}}{2}\right) $. However, for these points to be, indeed, critical points, it is also necessary to ensure that $(\alpha + \gamma - 1)^2 + 4 \gamma \geq 0$. Luckily, given that $\gamma>0$, this condition is always satisfied. Furthermore, we need $x \neq -\gamma$ and $x \neq 0$. These conditions should also be satisfied, given that $\alpha$ and $\gamma$ are greater than $0$, based on my rough estimates (please double check this!). By substitution, indeed, $(x_2,y_2)$ and $(x_3,y_3)$ are critical points.

In summary, given the restrictions on the set of parameters, the system should always have exactly three critical points:

• $(1, 0)$
• $\left( \frac{1-\alpha-\gamma-\sqrt{(\alpha+\gamma-1)^2+4\gamma}}{2},\frac{1-\alpha-\gamma-\sqrt{(\alpha+\gamma-1)^2+4\gamma}}{2}\right)$
• $\left( \frac{1-\alpha-\gamma+\sqrt{(\alpha+\gamma-1)^2+4\gamma}}{2},\frac{1-\alpha-\gamma+\sqrt{(\alpha+\gamma-1)^2+4\gamma}}{2}\right)$

This is consistent with your answer, with the exception that $(0,0)$ is not a critical point based on the definitional framework that is commonly used for such problems.

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Plots

From (1), taking into account the restrictions on the values of the parameters, $\frac{dx}{dt}=0$ if and only if either $x=0$ or $y = -\frac{1}{\alpha}x^2+x\frac{1-\gamma}{\alpha}+\frac{\gamma}{\alpha}$ and $x\neq-\gamma$. If $y = -\frac{1}{\alpha}x^2+x\frac{1-\gamma}{\alpha}+\frac{\gamma}{\alpha}$ and $x=-\gamma$, then $y = 0$. Thus, technically, the plot of $\frac{dx}{dt}=0$ should consist of two lines: one is a vertical line at $x=0$ and another one is a parabola given above with the point $(-\gamma,0)$ missing. However, it is likely that you are only interested in the points $(x,y)$ such that they belong to the domain of the state function. In this case, you should omit the vertical line at $x=0$ and only leave the 'punctured' parabola.

From (2), taking into account the restrictions on the values of the parameters, $\frac{dy}{dt}=0$ if and only if either $y=0$ and $x \neq 0$ or $y = x$ and $x \neq 0$. Thus, the plot should consist of two lines: the first one is a horizontal line $y = 0$ and the other one $y = x$ with the point $(0,0)$ being omitted. This explains why there are three points of intersection.

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