Stability of equilibrium of a nonlinear system of ODE's

Question

Suppose we have the nonlinear system of ODE's $$\begin{cases} \dot{x_1} = -\beta x_1 x_2 \\ \dot{x_2} = \beta x_1 x_2 - \gamma x_2 \end{cases} $$ Where we take $\beta, \gamma > 0$ arbitrary for now. In particular I am interested in the equilibrium point $(x_1, x_2) = (1, 0)$. I first linearized the system around the point $(1, 0)$ by using the Jacobian $$J(x_1, x_2) = \begin{pmatrix} -\beta x_2 & -\beta x_1 \\ \beta x_2 & \beta x_1 - \gamma \end{pmatrix}.$$ So the linearized system around $(1, 0)$ is given by $$\begin{pmatrix} \dot{x_1} \\ \dot{x_2} \end{pmatrix} = \begin{pmatrix} 0 & -\beta \\ 0 & \beta - \gamma \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix}.$$ Hence, it follows we have eigenvalues $\lambda_1 = 0$ and $\lambda_2 = \beta - \gamma$. Now if $\beta > \gamma$ we know that the nonlinear system is unstable. However, if we let $\beta \leq \gamma$ we can not determine the stability of the nonlinear system by linearization.

The system seems relatively simple and I would expect the equilibrium to be stable or even asymptotically stable in the case $\beta \leq \gamma$, but how would one prove this when linearization fails to provide a conclusive answer? Or did I make some error in my reasoning?

Answer 1

I'd attack this system in s few steps, ending with analysis of a 2nd order ODE:

Let $u_1 = \beta x_i$ to turn the equations into $$ \left\{ \begin{array}{c} \dot{u_1} = -u_1u_2 \\ \dot{u_2} = u_1u_2-\gamma u_2\\ u_1(0) = \beta + \epsilon_1 \\ u_2(0) = \epsilon_2\end{array}\right. $$ Then get a system about $(0,0)$ by letting $y_1 = u_1-\beta, y_2 = u_2$: $$ \left\{ \begin{array}{c} \dot{y_1} = -y_2(y_1+\beta) \\ \dot{y_2} =y_2(y_1+\beta) -\gamma y_2\\ y_1(0) = \epsilon_1 \\ y_2(0) = \epsilon_2\end{array}\right. $$ Then let $z_1 = y_1+y_2, z_2=y_2$: $$ \left\{ \begin{array}{c} \dot{z_1} =-\gamma z_2\\ \dot{z_2} = z_2(z_1-z_2+\beta) \\ z_1(0) = \epsilon_1 +\epsilon_2 \equiv \epsilon_3\\ z_2(0) = \epsilon_2\end{array}\right. $$ Next, use the first equation to write $z_2 = -\dot{z_1}/\gamma$ and multiply the second line through by $-\gamma$ to get a 2nd-order ODE: $$ \left\{ \begin{array}{c} \ddot{z_1} = \dot{z_1}(z_1+\dot{z_1}/\gamma + \beta) \\ z_1(0) = \epsilon_3\\ \dot{z_1}(0) = \epsilon_2\end{array}\right. $$ The given point in the original equation is stable if and only if $z_1(0) = \dot{z_1}(0) = 0$ is a stable fixed point in this ODE. But for $\beta > 0$ and ignorably small $z_1$ and $\dot{z_1}$, $\dot{z_1}$ grows exponentially.

So it seems the system is unstable unless $\beta \leq 0$.

Answer 2

Note that since you have a $0$ eigenvalue your fixed point is not a hyperbolic fixed point and, thus, you are violating the assumptions of the Hartman–Grobman theorem. This means that the linearization will not accurately predict the behavior of your system around the fixed point.

If you plot your system, you will see that it is behaving fairly wildly, due to the fact that you have this null eigenvalue.

Note that the whole $x$ axis ($x_2 = 0$) is a fixed point in your system.