ODELF1

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Consider the system of equations \begin{cases} x'=y-xf(x,y) \\ y'=-x-yf(x,y)\end{cases}\,

Find a Lyapunov function to determine the stability of the equilibrium solution (0,0)\,. Consider the three cases in your calculations: a) f(x,y)\, positive semidefinate, b) f(x,y)\, positive definite and c) f(x,y)\, negative definate.


\begin{cases} x'=y-xf(x,y) \\ y'=-x-yf(x,y)\end{cases}\,

Let L=x^2+y^2\,.

\frac{\partial L}{\partial t} = 2x\dot{x} + 2y\dot{y}\,

=2x\left[ y-xf(x,y) \right] + 2y\left[-x-yf(x,y)\right]\,

=-2x^2f(x,y) - 2y^2f(x,y)\,

=-f(x,y)[2x^2+2y^2]\,

If f(x,y)\, is positive semidefinite, then \frac{\partial L}{\partial t} \le 0\, so the zero solution is uniformly stable.

If f(x,y)\, is positive definite, then \frac{\partial L}{\partial t} < 0\, so the zero solution is asymptotically stable.

If f(x,y)\, is negative definite, then \frac{\partial L}{\partial t} > 0\, so the zero solution is unstable.


Main Page : Ordinary Differential Equations : Lyapunov Functions

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