System of ODEs: does one solution dominates the other? I am interested in the following non-linear system
\begin{align*}
\ddot{x}&=(\dot{x})^2+g(x-y+\nu)\\
\ddot{y}&=(\dot{y})^2+g(y-x)~~,\\
\end{align*}
where $g$ is a monotonic, increasing, and positive function; and $\nu>0$ is a constant. We can assume $g$ is smoother if it helps. Initial conditions are $\dot{x}(0)=x(0)=\dot{y}(0)=y(0)=0$. Both $x(t)$ and $y(t)$ are functions from the positive reals to the positive reals. I think it is not too hard to prove they are growing functions as well.
What I expect is that the answer to the question I pose in the title is "yes". That is, the initial extra "kick" on $x(t)$ will keep it above $y(t)$ at all times. Maybe $x(t)$ will blow-up before $y(t)$ (I am almost sure they both blow-up in finite time, at least when g is bounded). In fact, I think that $x(t)$ will be larger than the solution of $f''=(f')^2+g(\nu)$ and that $y(t)$ will be smaller than the solution of $f''=(f')^2+g(0)$, but I am not completly sure. This could be an approach to solve the question.
How to prove all these? (the things that are true of course). I would be very thankful if whoever kind enough to answer or comment could mention  good references to study about similar subjects. Regards.
Update (11 May 2018)
I was able to solve the equation for the particular case $g(s)=e^s$. With the change of variables $x=-\ln(p)$ and $y=-\ln(q)$, the system becomes 
\begin{align*}
-\ddot{p}&=p(t)g\left(\ln\left(\frac{q(t)}{p(t)}\right)+\nu\right)=e^\nu q(t)\\
-\ddot{q}&=q(t)g\left(\ln\left(\frac{p(t)}{q(t)}\right)\right)=p(t)~~,\\
\end{align*}
with initial conditions $p(0)=q(0)=1$ and $\dot{p}(0)=\dot{q}(0)=0$. This system is solvable, although the solutions are quite ugly so I will not put them explicitly. However, as expected, both $x$ and $y$ are increasing, and $p(t)$ becomes zero (that is, $x(t)$ blows-up) in finite time while $y(t)$ is still finite. What I postulate above about the inequalities that I expect x and y should fulfill it is also true in this particular case.
 A: For $g$ in $C^1$, define $\theta(t,\nu)=-\ln\cos(\sqrt{g(\nu)}t)$. We have that $x(t)=y(t)=\theta(t,0)$ is the solution of the system for $\nu=0$. We write $\theta$ for $\theta(t,0)$ hereafter. Let us define $h_1=x-\theta$, $h_3=\theta-y$, $h_2=\dot{h_1}$ and $h_4=\dot{h_3}$. In the new variables, the system reads
\begin{align}
\dot{h_1}&=h_2&&\tag{1}\\
\dot{h_2}&=h_2(h_2+2\dot{\theta})+g(h_1+h_3+\nu)-g(0)&&\tag{2}\\
\dot{h_3}&=h_4&&\tag{3}\\
\dot{h_4}&=h_4(2\dot{\theta}-h_4)-g\left(-(h_1+h_3)\right)+g(0)~~,&&\tag{4}\\
\end{align}
where $\dot{\theta}$ is a given function.
At this point, we cite the following result in the Encyclopedia of Mathematics, section "Differential Inequalities"

For a system of differential inequalities $$ y_i'(x)>f_i(x,y_1,...,y_n),\quad i=1,...,n,\tag{S}$$ it has been shown (Ref.1) that if each function $f_i$ is non-decreasing with respect to the arguments $y_j$ (for all $j\neq i$), the estimate $$ y_i(x)>z_i(x)\quad\text{for}\quad x_0<x\le x_2~~...$$

In the previous result, $z_i$ are the solutions of the system (S) with equality signs instead of inequalities and the same initial conditions. I assume that the result is valid with non-strict inequalities as well, and I call it hereafter "CW" theorem (of Chaplygin-Wazewski).
In order to apply CW, we define the system 
\begin{align}
\dot{z_1}&=z_2&&\tag{1'}\\
\dot{z_2}&=z_2(z_2+2\dot{\theta})+g(z_1+z_3)-g(0)&&\tag{2'}\\
\dot{z_3}&=z_4&&\tag{3'}\\
\dot{z_4}&=z_4(2\dot{\theta}-z_4)-g\left(-(z_1+z_3)\right)+g(0)~~.&&\tag{4'}\\
\end{align}
Clearly the $z$-system (1'-4') is very similar to the $h$-system (1-4), save the second equation: the right hand side of (2') is "dominated" in a certain sense by the right hand side of (2) because of the monotonicity of $g$. We can also see that the functions on the right side of equations (1-3) and (1'-3') are all growing functions of their parameters. In equation (4), the right hand side function is not a growing function but only of $h_4$. Therefore, the functions do fulfill the hypotheses of the theorem.
Solution of (1'-4') is $z_i\equiv 0$ for $i=1,2,3,4$. Therefore, by CW, $h_i\ge0$, which implies $x\ge\theta$ and $y\le\theta$, from where we conclude that $x\ge y$. Furthermore, $x(t)$ blows up in finite time if $g(0)>0$ at time $t_\text{BU}\le\frac{\pi}{2\sqrt{g(0)}}$. We also have that $y\le\theta$, but this does not guarantee that $y(t)$ would be finite when $x(t)$ blows-up: they both could blow up at the same time. 
A slightly better bound can be obtained. Now knowing that $x\ge y$ or that $h_1+h_3\ge0$, we can focus on the sub-system (1-2)
\begin{align}
\dot{h_1}&=h_2&&\tag{1}\\
\dot{h_2}&=h_2(h_2+2\theta(t,0))+g(h_1+h_3+\nu)-g(0)\ge h_2(h_2+2\theta(t,0))+g(\nu)-g(0)~~.&&\tag{2}\\
\end{align}
The solution of the "dominated" system above is $z_1=\theta(t,\nu)-\theta(t,0)$, therefore, $h_1\ge z_1=\theta(t,\nu)-\theta(t,0)$ or $x\ge\theta(t,\nu)$. That is, $t_\text{BU}\le \frac{\pi}{2\sqrt{g(\nu)}}$. If $g(\nu)>g(0)$, it is guaranteed that $x(t)$ will blow-up before $y(t)$.
Ref.1 T. Wazewski, "Systèmes des équations et des inégualités différentielles ordinaires aux deuxièmes membres monotones et leurs applications" Ann. Soc. Polon. Math. , 23 (1950) pp. 112–166
A: Hint
We have for $x \ge y$
$$
g(x-y+\nu) \ge g(x-y) \ge g(y-x)
$$
