Note: This proof prefaced critical points at the origin for coupled first order ODEs. It was done before showing the asymptotically stable and unstable critical points: Improper, Proper, Spiral, Center, Saddle and Degenerate.
I have read that two trajectories cannot cross at a finite value of $t$(Phase plane and nodes). The following proof was attached:
Proof: Suppose $\vec y_A(t_A) = \vec y_{P_0} = \vec y_B (t_B)$
Set $\vec y(t) = \vec y_A (t + t_A) – \vec y_B (t+t_B)$
$\Rightarrow \vec y'(t) = A \vec y(t), \vec y(0) = \vec 0$
Clearly $\vec y(t) = \vec 0$ is a solution to these equations, so uniqueness implies $\vec y(t) = \vec 0$, therefore $\vec y_A (t+t_A) = \vec y_B (t+t_b) \Rightarrow traj_A ,traj_B$ are the same near $P_0$ – but do not cross.
That is how it is written. Now I believe I am following everything correctly, but I am not sure how they reached that conclusion (in bold). Are they saying that they can't even touch? Or merely that they can't pass through each other?
(Sorry about the tags, I can't read them because the tag descriptions are mostly off the screen for some reason)
Perhaps they are saying the critical point at $t= – \infty$ is the only point where two trajectories touch, and outside of that they immediately diverge, and hence they are only every close outside of this point.
Note: I do understand that this is true. I am only confused by this proof.
Best Answer
I'm interpreting the story as follows: We are given the system of linear ODEs $$y'=Ay, \qquad A:=\left[\matrix{a&b\cr c&d\cr}\right]$$ in ${\mathbb R}^2$. Assume that we have two trajectories $$y_P:\ t\mapsto y_P(t), \quad y_P(0)=P;\qquad y_Q:\ t\mapsto y_Q(t), \quad y_Q(0)=Q$$ with initial points $P$, $Q\in{\mathbb R}^2$, found by whichever means, and assume that these trajectories have a common point $Y\in{\mathbb R}^2$. This means that there are finite $t$-values $t_P$ and $t_Q$ with $$y_P(t_P)=y_Q(t_Q)=Y\ .$$ It is claimed that in such a situation one has $$y_Q(t)=y_P(t-t_P+t_Q)\qquad(-\infty<t<\infty)\ ,\tag{1}$$ i.e., that the two trajectories differ just by a shift of the time scale. In order to prove $(1)$ one considers the auxiliary function $$\tau\mapsto z(\tau):=y_P(t_P+\tau)-y_Q(t_Q+\tau)\ .$$ This function can be viewed as the uniquely determined solution of the initial problem $$z'=Az,\quad z(0)=0\ .$$ It follows that $z(\tau)\equiv0$, first in a suitable neighborhood of $\tau=0$, and then in all of ${\mathbb R}$. For the latter a certain "local to global" argument is needed, which is proven in the general theory of ODEs.