Stability proof of a seemingly simple nonlinear differential equation?

Question

How can I prove that the differential equation $\dot{x} = b(x - a x^2)$, where $a, b > 0$, has a asymptotically stable equilibirum at $x^* = 1/a$? One way to do it is via linearization, but I am interested to know whether there is a way to do it via the direct Lyapunov method?

Answer 1

Assume that $x_0>0$, then we have that $x(t)>0$ for all $t\ge0$ since 0 is an unstable equilibrium point. If $x_0<0$, then $x(t)\to-\infty$ since $\dot{x}<0$ for all $x<0$. If $x_0=0,1/a$, then $x(t)=0,1/a$ for all $t\ge0$ as those points are equilibrium points of the system.

Define then the function $V(x)=(x-1/a)^2/2$. Then, we have that

$$\dot{V}(x)=-bax(x-1/a)^2$$ along the flow of the system. Since starting from $x_0>0$, the state remains negative at all time, then the derivative of the Lyapunov function evaluated along the flow of the system is negative. This proves asymptotic stability.

We can prove exponential stability the following way. Assume that $x_0>1/a$, then we have that

$$\dot{V}(x)=-bax(x-1/a)^2\le -b(x-1/a)^2=-2bV(x)$$

and we have exponential convergence of the Lyapunov function with rate $2b$, which means that the state converges with rate $b$.

Assume now that $x<1/a$, then we have that

$$\dot{V}(x)=-bax(x-1/a)^2\le -bax_0(x-1/a)^2=-2bax_0V(x)$$

and we have exponential convergence of the Lyapunov function with rate $2abx_0$, which means that the state converges with rate $abx_0$. It is interesting to note that in this case, the rate of convergence is not uniform and depends on the initial condition.