Prove the difference of 2 solutions of different initial values problems is increasing with time

Question

I'm trying to solve this problem. In fact, I'm not really sure how to solve this, but, with guesses, I do have an attempt which is the following:

Since for the second coordinate(that is $y$ coordinate), $f_{2}$ only depends on $y$ and they have the same initial values, the second coordinate of the solution curve to be the same, so I just consider the first coordinate

Suppose the solution in the first coordinate is $x(t,x_{0})$. Here I use $x_{0}$ to denote the initial value which is a vector in $\mathbb{R}^2$and $X$ to denote this coordinate (direction). Then I have the following calculation $$\begin{align} \frac{\partial x}{\partial t}(t,x_{0})&= f_{1}(x,y) \\ \\ \frac{\partial^2 x }{\partial X\partial t}(t,x_{0}) &= \frac{\partial f}{\partial x}(x,y) \cdot \frac{\partial x}{\partial X}(t,x_{0}) \end{align}$$Then consider the ode $$\begin{align} \frac{\partial }{\partial t}\left( \frac{\partial x}{\partial X}(t,x_{0}) \right)= \frac{\partial f_{1}}{\partial x}(x,y) \cdot \frac{\partial x}{\partial X}(t,x_{0}) \end{align}$$ This is a first order linear ode with respect to $\frac{\partial x}{\partial X}$ . By solving it, I can obtain $$\begin{align} \frac{\partial x}{\partial X}(t,x_{0})=C\exp\left( \int _{0}^t \frac{\partial f_{1}}{\partial x} \ d s \right) \end{align}$$ So far, I think that if all things above are right, then as long as $C>0$ , with $g(t)=\int_{0}^t\frac{\partial f_{1}}{\partial x} \ d s>0$ , then I can say that $\frac{\partial x}{\partial X}$ is increasing as $t$ is increasing, and hence if I have 2 different solutions $$\begin{align} x(t,x_{1})< \tilde{x}(t,x_{2}) \end{align}$$ at $t=0$ . That is, $x_{1}<x_{2}$ , we can have, when $t>s$ $$\begin{align} \frac{\tilde{x}(t,x_{2})-x(t,x_{1})}{x_{2}-x_{1}}&> \frac{\tilde{x}(s,x_{2})-x(s,x_{1})}{x_{2}-x_{1}} \\ \tilde{x}(t,x_{2})-x(t,x_{1})&> \tilde{x}(s,x_{2})-x(s,x_{1}) \end{align}$$ which is done.

Now I have 2 questions:

1 I strongly doubt my attempt. If the above is wrong can anyone show me how to approach this problem?

2 If the reasoning above is right, how should I compute the constant $C$?

Thank you very much.

Answer 1

Concerning your approach:

1. Up to typos, your general approach looks fine. To make everything perfectly rigorous, you should probably say that $x_0 \in \mathbb{R}$ (and not $\mathbb{R}^2$). You don't need to keep track of the initial data for $y$, since it is fixed. Then $X \in \mathbb{R}$ and $\partial x / \partial X$ is a scalar quantity for which you can indeed solve the ODE as you did. (Beware that the exponential formula you use is only valid in the scalar case).
2. The scalar $C$ is equal to $1$. Indeed, it corresponds to the initial data at time $t = 0$ for the ODE concerning $\partial x / \partial X$. At $t = 0$, you have $x(0,x_0) = x_0$, so $\partial x / \partial X(0,x_0) = 1$. Hence $C = 1$ after substitution in your formula.

More generally, I would say that a more "straightforward'' approach is to consider the difference $z(t) := \tilde{x}(t) - x(t)$. At the initial time $z(0) = x_2 - x_1 > 0$. And, using the equality $y = \tilde{y}$ and the assumption $\partial_x f_1 > 0$ you can derive $$ \dot{z}(t) = f_1(\tilde{x}(t),\tilde{y}(t)) - f_1(x(t),y(t)) > 0 $$ as long as $z(t) = \tilde{x}(t) - x(t) > 0$, which entails the conclusion slightly more directly.