The line $y=x$ is an invariant set ODE

ordinary differential equationsset-invariance

Let the system
\begin{align*}
\dot{x} &= y^3 -4x\\
\dot{y} &= y^3-y-3x
\end{align*}

(a) Prove that the line $y=x$ is an invariant set.

(b) Prove that $|x(t) – y(t)| \to 0$ when $t \to \infty$, for all other trajectories.

I have some questions of this exercise. For (a) I proposed the function $V(x,y)=x-y$, then I calculated $\dot{V}(x,y)$ and evaluate the function in the points $y=x$. The result I got was that $$\dot{V}(x,y)|_{x=y}=0$$
but then I do not know how to conclude. Is this right? For (b) I do not know well how to proceed.

Any hint?

Best Answer

First question. We subtract the two equations and we get:

$ x'(t)-y'(t)=-(x(t)-y(t))$. We define $z(t)=x(t)-y(t)$ then we get :

$z'(t)=-z(t)$ which gives $z'(t)+z(t)=0$ multiply by $e^{t}$ and get:

$(e^{t}z(t))'=0$ which gives $e^{t}z(t)=c$. If for some $t_{0}$ we have $x(t_{0})=y(t_{0})$ then $z(t_{0})=0$ and

$c=0$ and hence $z(t)=0$ for all $t$, i.e.

$x(t)=y(t)$ for all $t$. So $x=y$ is invariant.

Second question. By the same equation for $c\neq\,0$ we obtain $z(t)=ce^{-t}$ and clearly

$|z(t)|=|x(t)-y(t)|\to 0$ when $t\to +\infty$.

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