Positive invariance of a set under a system of ODEs

Question

Given the system of ODEs, $$x'=x(1-x-y)$$ $$y'=y(x-1),$$ $Q=\{(x,y):x\ge 0, y\ge 0\}$, and $S=(x,y)\in Q:x+y\le k$, $k>1$, I need to show that $S$ is invariant under this system of ODEs.

Attempted solution: $x'=x(1-(x+y))\ge x(1-k) \le 0$. Suppose $y'>0$, then $x<xk$ implies that $x-x^2-xy\ge x-xk \iff y\le (k-x)$. Hence, $y' = y(x-1) \le (k-x)(x-1)$. $y'$ changes sign at $y'=0$, thus we consider $(k-x)(x-1)=0$, whose roots are $x=k$ and $x=1$. But $x\ne k$ since $x'$ is decreasing, so $y'$ changes sign at $x=1$.

And so I don't know what to do next. I realize that we need to show that $x$ is decreasing faster than $y$ is increasing. We can solve the equation for $x(t)$ to obtain $x(t)= x_0 e^{1-k}$, and then solve for $y(t)$ and compare. Would that be practical?

Answer 1

My answer is a variation of that of D. Thomine, which I like. But it can be much shorter.

Answer: Notice first that the axes are invariant. Moreover, for $x+y\ge1$ we have $$ (x+y)' = 2x-x^2-(x+y) \le 2x-x^2-1=-(x-1)^2\le0. $$ The desired follows now immediately without the need for anything else.

Answer 2

This is not a response to your attempted solution, but since you seem stuck, may I propose a change of variables?

Like in your previous question, you can first show that any solution remains in $Q$. Then you only need to show that $x+y \leq k$ for any initial condition $(x_0,y_0) \in S$.

Consider the change of variables $v = x + y$ and $w = x - y$ and see if you can show the second requirement. Hint: Consider $\dot{v}$ and see whether it ever changes sign. Drawing might help intuition.

Answer 3

For the invariance of $Q$, see this question of yours. So what remains to prove is the boundedness of the solutions.

Let $k > 1$. We want to prove that $x(t)+y(t) \leq k$ for all $t \geq 0$ whenever this inequality is satisfied for $t = 0$. Note that we could also assume that $x>0$ and $y>0$, but that won't be useful here.

So, let's try to find a differential equation satisfied by $(x+y)$. We have:

$$(x+y)' = x-x^2-xy+yx-y = x-x^2-y.$$

It would be nice if I had a upper bound looking like $A-B(x+y)$. So, first, I want to make $x+y$ appear in the right hand side.

$$(x+y)' = 2x-x^2-(x+y) = 1-(x-1)^2-(x+y).$$

The $(x-1)^2$ term is bothersome, as it is not a function of $x+y$. Let's get rid of it.

$$(x+y)' \leq 1-(x+y).$$

Now, you may use Grönwall's lemma.

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That's the cleanest proof I can think of. Now, if you want do it your way : what you want to prove is that the vector field is pointing strictly inwards at the boundary $\{x+y=k\}$. On this line, as you noticed,

$$x' = x(1-k) \leq 0, \ \ y' = y(x-1).$$

and we want to prove that $y' < |x'| = x(k-1)$ (it is sufficient, as $x' \leq 0$). So let's make $xk$ appear in the expression of $y'$.

$$\begin{align} y' < |x'| & \Leftrightarrow y(x-1) < x(k-1) \\ & \Leftrightarrow xk-x^2-y < xk-x \\ & \Leftrightarrow x^2+y > x \\ & \Leftrightarrow x^2+k > 2x \\ & \Leftrightarrow (x-1)^2+(k-1) > 0, \end{align}$$

which is indeed true.