# ODELF1

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 $L=x^{2}+y^{2}\,$.

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

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

$=-2x^{2}f(x,y)-2y^{2}f(x,y)\,$

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

If $f(x,y)\,$ is positive semidefinite, then ${\frac {\partial L}{\partial t}}\leq 0\,$ 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.