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Math Help - How to find a Liapunov function?

  1. #1
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    How to find a Liapunov function?

    Does anyone know in general how to find a Liapunov function when given a system of differential equations?

    For example, how would I find a strict Liapunov function for the equilibrium point (0,0) of x' = -2x - y^2 and y' = -y - x^2?
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    I don't think you will have much luck in finding a Lyapunov function for the general nonlinear case. You could try only if the known theorem about the eigenvalues of the linearized problem does not apply.



    Which is not the case here. If we call {x'=f_1,y'=f_2} and f=(f_1,f_2), the eigenvalues of Df(0,0) are negative, and so
    the origin is asymptotically stable.
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  3. #3
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    Quote Originally Posted by seadog View Post
    Does anyone know in general how to find a Liapunov function when given a system of differential equations?

    For example, how would I find a strict Liapunov function for the equilibrium point (0,0) of x' = -2x - y^2 and y' = -y - x^2?
    As Rebesques mentions, it's probably easiest to consider the linearized matrix. As this has eigenvalues of \lambda = -1, -2 (i.e., non zero real part) they are hyperbolic critical points and the Hartman-Grobman theorem gives that the stability of the origin can be obtained from the linearized system. However, you could consider the Lyapunov function

     <br />
V = \frac{1}{2} x^2 + \frac{1}{2}y^2<br />
    so

     <br />
\dot{V} = x \dot{x} + y \dot{y} = x\left(-2x - y^2\right) + y\left(-y-x^2\right)<br />

     <br />
= -x^2\left(y+2 \right) - y^2\left(x+1 \right)<br />

    and as long as |\,y\,| < 2 and |\,x\,| < 1, then you have a strict Lyapunov function and can deduce the the origin is asymptotically stable.
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