Lyapunov function - stability analysis

[Liferider]

Homework Statement

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}

Homework Equations

With the Lyapunov function candidate:
\begin{equation}
V(x) = \frac{1}{2}(x_1^2 + x_2^2)
\end{equation}

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?

 

[BruceW]

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

 

[Liferider]

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

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?

 

[BruceW]

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

 

[Liferider]

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

At first, I looked at the inequality:
\begin{equation}
1 < e^{x_1 x_2}
\end{equation}
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?

 

[BruceW]

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?

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