# Lyapunov function - stability analysis

1. Dec 10, 2012

### Liferider

1. The problem statement, all variables and given/known data
State the strongest stability property of the system (stable, asymptotically/exponentially):
\begin{align}
\dot{x_1} &= x_2 \\
\dot{x_2} &= -x_1 e^{x_1 x_2}
\end{align}
2. Relevant equations
With the Lyapunov function candidate:

V(x) = \frac{1}{2}(x_1^2 + x_2^2)

3. The attempt at a solution
V(x) is pos. def. and continously differentiable.
\begin{align}
\dot{V}(x) &= x_1 \dot{x_1} + x_2 \dot{x_2} \\
&= x_1 x_2 - x_1 x_2 e^{x_1 x_2} \\
&= x_1 x_2 (1 - e^{x_1 x_2})
\end{align}
Here, i feel like I must state that we can not conclude on the stability of the system since V-dot is not negative definite or semi-definite.

The answer is however, that it is negative semi-definite... I can not see how that is possible. Any help here please?

2. Dec 10, 2012

### BruceW

Think about the different possibilities for x1 and x2. Is it possible to make V-dot positive with any choice of x1, x2 ?

3. Dec 10, 2012

### Liferider

Well, as far as I can see, the only values of x1 and x2 that keeps V-dot negative is the solutions in the 1.st and 3.rd quadrant of the phase plane, because:
\begin{align}
x_1 x_2 (1 - e^{x_1 x_2}) &< 0 \\
1 < e^{x_1 x_2}
\end{align}
If x1 and x2 have different signs, then V-dot would become positive right? Can we prove that the solutions will never have different signs?

4. Dec 10, 2012

### BruceW

why would V-dot become positive if x1 and x2 had different signs?

5. Dec 10, 2012

### Liferider

At first, I looked at the inequality:

1 < e^{x_1 x_2}

Because it would not hold when x1x2 produced a negative number.....

However, when I tried to calculate V-dot with such a case, it still produced a negative number, it seems I forgot that I removed the x1x2 terms when I made the inequality, causing it to lose information.

What a stupid little thing.

Still, why is it only neg. semi-def. ? V-dot is allowed to be zero when either x1 or x2 is zero right? Or do they both have to be zero at the same time?

6. Dec 10, 2012

### BruceW

Only at the fixed point we don't care about the value of V-dot. This should answer your question.