Suppose we have
$$\dot{x} = -\frac{x}{y+a} $$
$$\dot{y} = -y$$ for $a>0$,
Is the above system asymptotically stable?
Now, I know that we can solve for $y$ as
$$y = y(0) e^{-t}$$
and we can choose for the $x$ part of the system, a Lyapunov function as $V(x) = 0.5 x^2$ and thus obtain the derivative:
$$\dot{V} = -\frac{x^2}{y(0) e^{-t} + a}$$
which $\lim\limits_{t \rightarrow \infty} \dot{V} = -\frac{x^2}{a}$.
But, I don't think this is correct.
Is there a way to prove the $x,y$ system is stable, just by finding a positive definite Lyapunov function and proving that its derivative is negative definite?
Best Answer
We get the general solution of the DE system: $$\left\{y(t)\to e^{-t} y_0,x(t)\to x_0 (a+y_0)^{1/a} \left(a e^t+y_0\right)^{-1/a}\right\}$$
Get the limits for $t \to \infty$:
$$\underset{t\to \infty }{\text{lim}}x(t)=\fbox{$0\text{ if }\left(x_0\left|(a+y_0)^{1/a}\right|y_0\right)\in \mathbb{R}\land a>0$}$$
$$\underset{t\to \infty }{\text{lim}}y(t)=0$$
Hence the system is asymptotically stable for $a>0$ and $a+y_0>0$.