Lyapunov function - stability analysis

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Discussion Overview

The discussion centers around the stability analysis of a dynamical system defined by a set of differential equations. Participants explore the properties of a proposed Lyapunov function and its implications for the stability of the system, particularly focusing on whether the system is stable, asymptotically stable, or exhibits other stability characteristics.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant states that the Lyapunov function candidate \( V(x) = \frac{1}{2}(x_1^2 + x_2^2) \) is positive definite and continuously differentiable, leading to the calculation of \( \dot{V}(x) = x_1 x_2 (1 - e^{x_1 x_2}) \).
  • Another participant questions whether it is possible to make \( \dot{V} \) positive with any choice of \( x_1 \) and \( x_2 \), suggesting that \( \dot{V} \) remains negative in certain quadrants of the phase plane.
  • A participant proposes that if \( x_1 \) and \( x_2 \) have different signs, \( \dot{V} \) could become positive, prompting a discussion on whether solutions can have different signs.
  • One participant reflects on the implications of the inequality \( 1 < e^{x_1 x_2} \) and acknowledges a mistake in their reasoning regarding the signs of \( x_1 \) and \( x_2 \) affecting \( \dot{V} \).
  • Another participant clarifies that \( \dot{V} \) can be zero when either \( x_1 \) or \( x_2 \) is zero, but questions whether both must be zero simultaneously for stability considerations.
  • A later reply indicates that at the fixed point, the value of \( \dot{V} \) is not of concern, which may address the previous question about conditions for \( \dot{V} \) being zero.

Areas of Agreement / Disagreement

Participants express uncertainty regarding the conditions under which \( \dot{V} \) is negative semi-definite and whether it can be positive. There is no consensus on the implications of the signs of \( x_1 \) and \( x_2 \) for the stability of the system.

Contextual Notes

Participants note that the analysis depends on the signs of \( x_1 \) and \( x_2 \) and the behavior of the exponential function in relation to the stability properties. There are unresolved questions about the conditions under which \( \dot{V} \) can be zero and the implications for stability.

Liferider
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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?
 
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Think about the different possibilities for x1 and x2. Is it possible to make V-dot positive with any choice of x1, x2 ?
 
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?
 
why would V-dot become positive if x1 and x2 had different signs?
 
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?
 
Liferider said:
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.
 

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