Using Lyapunov function, prove that a critical poin (0,0) is asymptotically stable.

Question

Let a linear system $$ \ x'=-2x+x{y}^{2}\\ \ y'=-x^{3}-y\\ $$ Using Lyapunov function, prove that a critical point (0,0) is asymptotically stable.

Answer 1

The most obvious try for a Lyapunov function designed to show local stability of a differential system at $(0,0)$ might be $$ H(x,y)=x^2+y^2. $$ In the present case, $$ \frac{\mathrm d}{\mathrm dt}H(x,y)=-4x^2-2y^2+2xy(xy-x^2). $$ Using the bounds $|xy-x^2|\leqslant|xy|+x^2$ and $2|xy|\leqslant x^2+y^2$, one sees that $$ \frac{\mathrm d}{\mathrm dt}H(x,y)\leqslant-4x^2-2y^2+(x^2+y^2)\tfrac12(3x^2+y^2). $$ Let $D=\{(x,y)\mid x^2+y^2\leqslant2\}$. For every $(x,y)$ in $D$, $$ \frac{\mathrm d}{\mathrm dt}H(x,y)\leqslant-4x^2-2y^2+(3x^2+y^2)=-(x^2+y^2), $$ that is, $$ \frac{\mathrm d}{\mathrm dt}H(x,y)\leqslant-H(x,y). $$ Hence, if $(x(0),y(0))$ is in $D$, then $(x(t),y(t))$ stays in $D$ for every $t\geqslant0$ and, solving the differential inequality $u'\leqslant-u$, one sees that $$ H(x(t),y(t))\leqslant H(x(0),y(0))\mathrm e^{-t}. $$ In particular $H(x(t),y(t))\to0$ hence $(x(t),y(t))\to(0,0)$ when $t\to\infty$.