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Lyapunov function - stability analysis

  1. Dec 10, 2012 #1
    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:
    \begin{equation}
    V(x) = \frac{1}{2}(x_1^2 + x_2^2)
    \end{equation}
    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. jcsd
  3. Dec 10, 2012 #2

    BruceW

    User Avatar
    Homework Helper

    Think about the different possibilities for x1 and x2. Is it possible to make V-dot positive with any choice of x1, x2 ?
     
  4. Dec 10, 2012 #3
    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?
     
  5. Dec 10, 2012 #4

    BruceW

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    Homework Helper

    why would V-dot become positive if x1 and x2 had different signs?
     
  6. Dec 10, 2012 #5
    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?
     
  7. Dec 10, 2012 #6

    BruceW

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    Homework Helper

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