Stability of solution

Question

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 .

Answer 1

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 $x=0$?

Answer 2

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.