# Solutions which start in right half plane (x>0) must remain there all time [closed]

How do I show that all solutions $$x(t)$$ and $$y(t)$$ of $$\frac{dx}{dt}=y(e^x - 1),$$ $$\frac{dy}{dt}=x + e^y$$

which start in the right half plane $$(x > 0)$$ must remain there for all time.

I thought that looking at $$\frac{dy}{dx}$$ can help but I don't know what to do with this.

For $$x>0$$, we have $$\frac{dy}{dx} = \tan\theta \in (-\frac{\pi}{2},\frac{\pi}{2})$$. Thus, the slope of the $$x-y$$ plot always points away from the $$y$$ axis. Hence, there is no way the graph can cross the $$y$$ axis i.e. it remains in the right half plane.

• Why should that be true? $y$ can be negative and so can be $dy/dx$.
– lcv
May 30 '20 at 14:38

The set $$M=\{(x,y):\; x=0\}$$ is an invariant set of the system because $$\tag{1} \left.\frac{dx}{dt}\right|_{x=0}= y(e^0-1)=0.$$ The equality (1) means that $$x$$ does not change when moving along the trajectories in $$M$$, that is, the solution can't move out or move into the set.

• As written this is of course correct. But how does that answer the question? You can have invariant set which are attractive, repulsive or neither
– lcv
May 30 '20 at 14:39
• The trajectories in $M$ are simply fixed points. You probably wanted to say that $M$ is a barrier that cannot be crossed (only, at most, reached).
– lcv
May 30 '20 at 14:45
• @lcv $M$ is an invariant set in the sense that if for some $t_0$ the point $(x(t_0),y(t_0))))\in M$, then $\forall t\in\mathbb R\; (x(t),y(t)))\in M$. The trajectory cannot cross it, because if at least one of its points belongs to M, then all its points belong to M, that is, the solution has always been contained in M and will always be contained there.
– AVK
May 30 '20 at 14:53
• @lcv The trajectories in M are not fixed points because $\left.\frac{dy}{dt}\right|_{x=0}= 0+e^{y}>0$.
– AVK
May 30 '20 at 14:56
• @AVK Thank you for the simple yet elegant solution. May 30 '20 at 15:45