Global existence of solution of ODE - Gray-Scott model

I am dealing with the ODE version of Gray-Scott model: $$$$\begin{split} \dot{x} &= -xy^2 + F(1-x)\\ \dot{y} &= xy^2 - (F + k)y, \end{split}$$$$ where $$F>0$$ and $$k>0$$ are parameters and we consider $$x\geq0$$, $$y\geq 0$$. I would like to prove global existence of solution. I tried to find bound for the solution using functions of type $$V(x,y)= x^2 + y^2$$ in hopes of finding such function that has non-positive derivative. I also tried elimination of time variable to investigate the solution curves.

Does anyone have an idea how to do it? I am not sure whether global solution exists for all initial conditions. Any help or references will be greatly appreciated.

• Global as in $-\infty < t < \infty$, or are you only concerned with $0 < t < \infty$? Commented 2 days ago
• @RobertIsrael I consider $0 \leq t < \infty$. Commented 2 days ago
• As a suggestion, if this system of ODEs is based on a particular domain (e.g. chemical reaction modelling), it might be useful to try posting in chemistry stack exchange or similar also, as those communities may be more familiar with references addressing the characteristics of such equations. Commented 2 days ago
• @ad2004 Thank you for the suggestion! Commented 2 days ago

Assume by contradiction, there exists a local in time solution $$(x,y)$$, which explodes in finite time $$T$$ (this is the opposite of having a global solution by the explosion in finite time theorem).

1. Assume that $$x$$ and $$y$$ stay non negative until $$T$$. Then, considering the sum $$s = x +y$$, we have $$\dot{s} = F - Fs -ky \le F - Fs.$$ Applying the Gronwall lemma to this inequality ensures that we have a bound (which depends on T) on $$s$$: $$s \le C_T$$. Because $$x$$ and $$y$$ are non negative, we have $$0\le x,y \le s$$ so $$x$$ and $$y$$ are also bounded and cannot explode (i.e. go to the infinity) in time $$T$$, which contradicts the explosion assumption.

2. Therefore, it remains to see that $$x$$ and $$y$$ stay non negative until $$T$$. If this is not the case, introduce $$T_0< T$$ the maximal time such that $$x$$ and $$y$$ are non negative on $$[0, T_0]$$. Because $$x$$ and $$y$$ are continuous, one of them has to be $$0$$ at $$T_0$$. Let's first prove that $$y(T_0) >0$$. We eliminate the special case where $$y(0)=0$$, in which $$y=0$$ and $$x$$ solves a classical affine equation. In the other cases, since $$x$$ is non negative until $$T_0$$, we have

$$\dot{y} \ge -(F+k) y,$$ so Gronwall lemma gives $$y(T_0) \ge e^{-(F+k)T_0} y(0) >0$$ (the special case $$y(0) = 0$$ has been eliminated). Therefore, if not $$y(T_0)$$, it must be $$x(T_0)$$ that vanishes. But using the bound obtained in paragraph 1., $$y \le C_{T_0}$$ on $$[0, T_0]$$ so $$\dot{x} \ge F - (C_{T_0}^2 +F) x.$$ which by Gronwall lemma implies $$x(T_0) > 0$$. It ends the proof, hence the solutions exist globally.

Even easier/more intuitive and with some qualitative remarks on the behaviour of solutions.

It can be seen that the positive cone $$K_+:=[0,+\infty)\times [0,+\infty)$$ is forward invariant (if we start there, we stay there), simply checking the direction of the vector field at the boundary of the cone. Consider now the ''sum'' function $$V:K_+\to \mathbb{R}$$ defined by $$V(x,y)=x+y$$. It is positive definite and $$\mathcal{C}^1$$. Following a reasoning similar to the one provided in the answer of Alex440 one can prove that $$\nabla V(x,y)\cdot f(x,y)\leq -F(V(x,y)-1), \;\forall \,(x,y)\in K_+.$$ This means that $$\nabla V(x,y)\cdot f(x,y)<0,\;\;\text{if } x+y> 1.$$

This implies that the solutions are trapped in the unit triangle $$T_1:=\{(x,y)\in K_+\;\vert\; x+y\leq 1\}$$ union the triangle spanned by the initial condition $$T_{(x_0,y_0)}:=\{(x,y)\in K_+\;\vert\;x+y\leq x_0+y_0\}$$. Since this set is compact, we can conclude global existence of solutions (forward in time), for any initial condition.