Find direct Lyapunov

Question

Show that origin is globally asymptotically stable.

\begin{align*} & \dot{x}=-x + y^2 \\ &\dot{y}=-y \end{align*}

I know to prove that $V′(x)$ has to be negative which I can prove. However, I can't seem to figure out how to get $V(x)$. Can anyone put me in right direction to how to calculate $V(x)$ for it. Thanks advanced.

Why don't you just solve the system? $$\dot{y} = -y \implies y = Ae^{-t} \implies \dot{x} + x = Ae^{-2t}$$ — Matthew Cassell, Sep 17 '19 at 06:35

score 0 · Accepted Answer · answered Sep 17 '19 at 06:50

0

For this kind of problems, I recommend you to try polynomial functions $V$ for which $$-\frac{dV(x(t), y(t))}{dt}$$ is a sum of square expressions. By inspection, you can put $V(x, y) = \frac{1}{2}x^2 + \frac{1}{4}y^4$. Then $$-\frac{dV(x(t), y(t))}{dt}=x\cdot(x-y^2)+y^3 \cdot y=x^2 -xy^2+y^4=(x-\frac{1}{2}y^2)^2 + \frac{3}{4}y^4$$ This serves as a Lyapunov function you seek.

answered Sep 17 '19 at 06:50

luxerhia

3,538

Thanks for your helping, it's so helpful for me <3 – Trọng Nhân Hồ Sep 17 '19 at 09:22

Find direct Lyapunov

1 Answers1

Linked