I have this differential equation $x'=-x^3$ ,
How to study the stability and the asymtotic stability of $x=0$ ?
Please help me
Thank you .
$\endgroup$ 13 Answers
$\begingroup$Hint: Perform sign analysis. Is $x'$ positive or negative for $x>0$? What about for when $x<0$? Use this information to draw a quick sketch of some sample solution curves. Do the curves converge toward or diverge away from the equilibrium solution at $x=0$?
$\endgroup$ 2 $\begingroup$The equilibrium points are determined as follows: \begin{align} \dot{x} &= f(x), \\ 0 &= f(x_e), \\ 0 &= -x^3 \implies x_e = 0. \end{align} The equilibrium point $x_{e} = 0$ is stable or attractive because the two black arrows are heading toward the equilibrium point (i.e. the origin). The direction of the arrows are determined based on the positivity or negativity of $\dot{x}$. If the differential equation is positive as the case with this example when $x<0$, the trajectory moves to the right and vise versa. See the below picture,
In order to show whether the equilibrium point is asymptotically stable, the equilibrium point must be stable and convergent. We've just shown that the equilibrium point is stable. The equilibrium point is convergent if the trajectory goes to zero as time goes to infinity. The analytical solution for the ode is, assuming the initial time is zero (i.e. $t_0=0$):
$$ x(t) = \pm \sqrt{ \frac{x^2(0)}{1 + 2x^2(0)t}} $$
As time goes to infinity, the trajectory indeed goes to zero, therefore, the system is asymptotically stable (i.e. it is stable and convergent). The system is also globally asymptotically stable. Globally because starting from any initial value, the trajectory goes to zero as time goes to infinity. Sometimes not all initial values make the trajectory goes to zero as time goes to infinity that is the trajectory will blow with some initial values. If this is the case, the stability of the system is locally asymptotically stable.
$\endgroup$ 4 $\begingroup$You are asking how you could study the stability.
To have it visible suggest you derive the Lyapunov function which is a potential function. You have:
$$\frac{dx}{dt}=\lambda \; x^3$$
The Lyapunov function $V$ for this is a potential that can be derived form:
$$\frac{dV}{dx}=-\frac{dx}{dt}=-\lambda \;x^3$$
Hence, with $C$ a constant we integrate to:
$$V=-\lambda \frac{1}{4}x^4+C$$
Now see the graph for $\lambda<0$
You see the potential curve (assume case $C=0$), if you have a ball inside it (imagine some gravity) will oscillate until it stabilizes at $0$.
Now try the graph for $\lambda>0$ it is the same as the first graph but fliped down mirrored at the x-axis. If you have ball on it, the ball will fall either from right or left down to infinity.
So your equation is for $\lambda<0$ stable and for $\lambda>0$ instable. For a point of transition $\lambda=0$ you have stability (your ball is on a flat curve identical with x-axis). When you tune $\lambda$ from negative slowly to positive you will experience that the ball on the curve turns from stable to instable.
Hint: add a fluctuation term like white noise to the right side of your equation and see what happens. You will see transition from stability to instability in math is not the same as transition from stability to instability in real phyiscal life, there you need fluctuations.
Hope this is good answer.
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