So I am trying to show that a set is forward invariant and I am not sure that this proof does the trick.
For an ODE $\dot{x} = f(x)$, a set $S$ is called forward invariant if $x_0 \in S \rightarrow \varphi(t,x_0) \in S$ for all positive $t\in I(x_0)$ where $I(x_0)$ is the maximum interval of existence for the solution to the ODE and $\varphi(t,x_0)$ is the flow of $f$ starting at $(t,x_0)$.
Consider the coupled set of differential equations on $\mathbb{R}^2$ given by:
$$\begin{matrix}\dot{x}_1=\\ \dot{x} _2=\end{matrix} \begin{matrix} -x_1 \\ (x_1x_2 -1)x_2^3 +(x_1x_2-1+x_1^2)x_2\end{matrix}$$
We would like to show that the set $\Gamma = \{x: x_1x_2\geq 2\}$ is a forward invariant set. So we define the positive definite function $V(x)= x_1x_2-2$. Note V is not positive definite everywhere but it is positive definite on $\Gamma$. Next we would like to show that the time derivative of V along all system trajectories that start in $\Gamma$ is negative. That is $\frac{dV}{dt}=L_fV = \nabla V\cdot f(x)> 0$ for all $t>0$ on $\Gamma$. Computing this quantity we get $L_fV(x)=-x_1x_2[(x_1x_2-1)x_2^2 + x_1x_2 + x_1^2]$. Since $x_1x_2 > 2$ on $\Gamma$, $[(x_1x_2-1)x_2^2+x_1x_2+x_1^2]>0$ and $-x_1x_2<0$ so $L_fV(x)<0$ on $\Gamma$ for all $t>0$. Thus the set is forward invariant. $\blacksquare$
Please let me know if I made a mistake in my reasoning. I am appealing in part to Lyapunov theory, but I do not know if it is the case that if a function $V$ that is positive everywhere on a set except the boundary, and the Lie derivative is negative on the entire set, that the set is forward invariant.
Thanks,
-akt
Best Answer
Let $f$ be the right hand side of the ODE above. Let $I$ be the maximal interval of existence with initial condition $x(0)$. Since $f$ is smooth, we see that the solution is continuous and $I$ is open.
Note that $x_1(t) = x_1(0)e^{-t}$ for all $t \ge 0$.
Also note that if $x$ solves the ODE, then so does $-x$. We see that if $x(0) \in\Gamma$, then $x_1(0),x_2(0)$ have the same parity. Hence we may assume that $x_1(0),x_2(0) > 0$. Let $\Gamma_+ = \Gamma \cap [0,\infty)^2$ (note that $\Gamma_+$ is closed).
If $x(t) \in \Gamma_+$, then by continuity of $x$ we have some $\delta>0$ such that $x_2(\tau) \ge 0$, $x_1(0)e^{-\tau} x_2(\tau) -1 \ge 0$ and $x_1(0)e^{-\tau} x_2(\tau)-2+x_1(0)^2e^{-2\tau} \ge 0$ for all $\tau \in [t,t+\delta]$. Then for $\tau \in [t,t+\delta]$ we have \begin{eqnarray} \dot{x_2}(\tau)-x_2(\tau) &=& (x_1(0)e^{-\tau} x_2(\tau) -1)x_2(\tau)^3 +(x_1(0)e^{-\tau} x_2(\tau)-2+x_1(0)^2e^{-2\tau})x_2(\tau) \\ & \ge & (x_1(0)e^{-\tau} x_2(\tau)-2+x_1(0)^2e^{-2\tau})x_2(\tau) \end{eqnarray} That is, $\dot{x_2}(\tau)-x_2(\tau) = e^\tau \frac{d}{d\tau}( e^{-\tau} x_2(\tau)) \ge 0$ for all $\tau \in [t,t+\delta]$, and so $x_2(\tau) \ge x_2(t) e^{\tau-t}$. This gives $x_1(\tau) x_2(\tau) \ge x_1(t) x_2(t)$ and hence $x(\tau) \in \Gamma_+$ for all $\tau \in [t,t+\delta]$.
To recap the above, if $t \in I$ and $x(t) \in \Gamma_+$, then there is some $\delta>0$ such that $x(\tau) \in \Gamma_+$ for all $\tau \in [t,t+\delta]$.
Now let $T = \sup I$, and let $\overline{t} = \sup \{ t \in I | x(\tau) \in \Gamma_+ \ \forall \tau \in [0,t] \}$. If $\overline{t} < T$, then since $x$ is continuous, we have $x(\overline{t}) \in \Gamma_+$, and the above shows that this leads to a contradiction. Hence $\overline{t} = T$.