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I have been trying to find a Liapunov function which would give me information about the stability of the following system of differential equations, however, I am not able to come up with any. The problem is an exercise from the book 'Differential Equations and Dynamical Systems' by Lawrence Perko (Third Edition), Chapter Two, Problem Set $9$, Question $5(c)$, so I guess, there does exist a Liapunov function for the same:

\begin{align} \frac{\mathrm{d}x_1}{\mathrm{d}t} &= -4x_2 + x_1^2 \\ \frac{\mathrm{d}x_2}{\mathrm{d}t} &= 4x_1 + x_2^2 \end{align}

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Try $V(t,x)=\frac{1}{2}\vert x\vert^2=\frac{1}{2}(x_1^2+x_2^2)$ –  smanoos Dec 9 '13 at 2:12
@smanoos If you meant $V(x) = x_1^2 + x_2^2$, then I'm not able to make it work, you get $\dot{V}(x) = 2x_1^3 + 2x_2^3$. –  Jayesh Badwaik Dec 9 '13 at 2:14
@Amzoti No, not yet I haven't tried the problem over the last week though since I was on a vacation. –  Jayesh Badwaik Dec 22 '13 at 15:09
@Amzoti If you have any solution, I will be happy to hear of it. :-) –  Jayesh Badwaik Dec 22 '13 at 15:20
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2 Answers

up vote 2 down vote accepted

I do not know which Lyapunov function will work for this example, but here is a hint how to prove that the origin is Lyapunov stable (a center). The system $$ \dot x_1=f_1(x_1,x_2),\\ \dot x_2=f_2(x_1,x_2), $$ is said to be symmetric with respect to $x_1$-axis if it is invariant with the change of the variables $(t,x_2)\to(-t,-x_2)$ (similarly, for $x_2$-axis).

Theorem: If a system symmetric with respect to $x_1$-axis, and the linear approximation is a center, then for the full nonlinear system the origin is a center.

An idea of proof is given in Perko, Section 2.10 (the very end of it).

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I have partially solved this question using the following approach. Transform the problem into polar co-ordinates, that is

\begin{align} r &= \sqrt{x_1^2 + x_2^2} \\ \theta &= \arctan \frac{x_1}{x_2} \end{align}

we get the equations \begin{align} \frac{\mathrm{d}r}{\mathrm{d}t} &= r^2 \left(\cos^3\theta + \sin^3\theta \right) \\ \frac{\mathrm{d}\theta}{\mathrm{d}t} &= -4 \end{align}

Now, we try to determine the stability of the system using this co-ordinate system, which should be somewhat easier, since $r$ is always positive and $\mathrm{d}\theta/\mathrm{d}t$ is always negative. I will work over it tomorrow morning though.

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