Show that a solution stays in $(\Bbb R^+)^2$

Question

The problem:

We consider the following differential system:

$$ \begin{align} \dfrac{dx}{dt} & = (1-x) - \dfrac{xy}{1+x}\\ \dfrac{dy}{dt} & = (\dfrac{2x}{1+x} - 1)\times y \end{align} $$ For $t\geq 0$ and $(x(0),y(0)) \in (\Bbb R+)^2$

Show that $(x(t),y(t))\in (\Bbb R+)^2 \quad \forall t>0 $.

What I did:

Let suppose the existence of $t' >0$ such as $(x(t),y(t))\notin (\Bbb R+)^2$ Then, there are two cases:

• (1) : the trajectory cut the axis $x=0$. Then, there is $t_0$ such as $x(t_0) = 0$ and $x(t)<0$ for $t\in ]t_0,t_0 + \epsilon[$ for a certain $\epsilon$. Writing $x(t) = (t-t_0)\dfrac{dx}{dt}(t_0) + o(t-t_0) = (t-t_0) + o(t-t_0)$ we get a contradiction because $t-t_0>0$ and $x(t) <0$.
• (2): the trajectory cut the axis $y=0$. This is where I don't know how to conclude. We also have $t_0$ such as $y(t_0) = 0$. Because $\dfrac{dy}{dt} = (\dfrac{2xy}{1+x} - 1)\times y$, it is to see, by induction, that $\dfrac{d^k y}{dt^k}(t_0) = 0 \quad \forall k\in \Bbb N^*$. But that doesn't mean that the function has to be equal to zero on a neighbourhood of $t_0$, does it?

Where I do need some help:

On the second point above. I don't know where is the contradiction.

EDIT : Cauchy-Lipschitz is basically the only ODE theorem I can use here.Every common real analysis theorem can be used too.

Answer 1

Following your initial idea of all derivatives vanishing.

For general differentiable functions it is correct the concern that the vanishing of all derivatives doesn't imply that the function vanishes.

However, note that the right hand side are analytic functions away from the line $1+x=0$. I don't know which results you have at hand in your course, but Cauchy-Kowalevski's theorem tells you that the solutions of your system are analytic functions.

Therefore, the vanishing of all the derivatives of $y$ implies that it vanishes in a neighborhood.

In principle you don't need to use the full power of the theorem. You could expand in series the right hand side, and use it to show that $(x,y)$ also has a convergent power series.

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From the uniqueness theorem.

You can conclude that $y=0, x=ce^{-t}+1$, for some $c$ would be the only solution passing through that point. This function would extend to the point $t=0$, where it gives you the contradiction with $y(0)>0$.