Solution of non linear ODE system is always positive if its initial valus is positive

Given a system of nonlinear differential equation \begin{eqnarray}\frac{dx}{dt}=2x(3-y) \\ \frac{dy}{dt}=3y(4-x)\end{eqnarray} If $r(t)=$($x(t)$,$y(t)$) is a solution of the system with initial value $x(0)>0$ and $y(0)>0$, would you help me to prove that $x(t)>0$ and $y(t)>0$ for all real $t$.

Here is my argument: I prove it by contradiction. Let $r_1(t)=(x_1(t),x_2(t))$ be an orbit of the ODE with $x_1(0)=0$. Since, for $x=0$ we get $\frac{dx}{dt}=0$ then $x_1(t)=x_1(0)=0$. Thus, $x=0$ is an invariant manifold of the ODE. By the same argument, $y=0$ is also an invariant manifold. So the orbit that passing through a point in $x=0$ (resp $y=0$) always lies in $x=0$ (resp $y=0$). Assume there is a $t_1$ such that $x(t_1) \leq 0$ then $r(t)$ must intersect an orbit in $x=0$( or $y=0$) hence $r(t)$ must lies on $x=0$ (or $y=0$), contradicting $x(0)>0$ and $y(0)>0$.

Is there any a direct proof (or even elementary/simple proof)?

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1 Answer

It is not quite clear what exactly you mean by "direct proof", but the main idea is very simple: the boundaries of $\mathbf R^2_+$ consist of orbits and due to the uniqueness and existence theorem the orbits cannot cross, which means that there is no way to cross a boundary.

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Note that our idea is similar right, assume there is a negative x or y then the orbit intersect invariant manifold which contain different orbit. I mean direct proof is using inequality start from $x(0)>0$ and $y(0)>0$ then by invariant manfold information and some analytic process to conclude $x(t)>0$ and $y(t)>0$ –  beginner Dec 16 '12 at 22:43
@beginner Yes, of course, your idea is right. What I mean is the following: It seems to me that you overcomplicate the problem, the reasoning along the line: orbits do not cross and the boundary is composed of orbits together imply the solution will stay positive. This last sentence is a proof for me and does not require any additional arguments. –  Artem Dec 16 '12 at 23:32