Show that for certain initial conditions the solution curves stay in $[0,\pi]\times[0,\pi]$ $$\dfrac{dx}{dt} = \sin(x)\left(-0.1\cos(x)-\cos(y)\right)$$
$$\dfrac{dy}{dt} = \sin(y)\left(\cos(x)-0.1\cos(y)\right)$$
Show that for the initial condition $(x_0,y_0)\in{[0,\pi]\times[0,\pi]}$ the solution curves are in $[0,\pi]\times[0,\pi]$ for all $t\in{\mathbb{R}}$.
I thought I could show that if the orbit of the solution enters the borders of $[0,\pi]\times[0,\pi]$. Trying a solution of the form \begin{equation}\begin{bmatrix}
x \\
0
\end{bmatrix}\end{equation} for $x\in{[0,\pi]}$. This means that
$$\dfrac{dx}{dt} = \sin(x)(-0.1\cos(x)-1)$$
But I don't know how to solve this DE. Is there a better method of solving this?
 A: You don't have to solve the DE, you merely have to bound the solutions. That's the magic of Picard-Lindelof.
Consider what Picard-Lindelof says about each equation separately: let's start with the equation for $y(t)$. Since $\sin(0) = \sin(\pi) = 0$, the equation $$\frac{dy}{dt} = \sin(y)(\cos(x) - 0.1\cos(y))$$ admits constant solutions $y(t) = 0$, $y(t) = \pi$. This property is independent of the function $x(t)$. By Picard-Lindelof, it follows that for any $C^1$ function $x(t)$ and any initial condition $y_0\in[0,\pi]$, the corresponding solution to the equation for $y$ satisfies $0 \leq y(t) \leq \pi$ for all $t$ for which $y(t)$ is defined.
We can do a similar analysis for the $dx/dt$ equation, completely independently of the $dy/dt$ equation. Again, we find that there are constant solutions $x(t) = 0$, $x(t) = \pi$, independent of the definition of $y(t)$. Therefore for any $C^1$ function $y(t)$ and any initial condition $x_0\in[0,\pi]$, the corresponding solution to $dx/dt = \sin(x)(-0.1\cos(x) - \cos(y))$ satisfies $0 \leq x(t) \leq \pi$ for all $t$ for which $x(t)$ is defined.
Creating a coupled relationship between $x(t)$ and $y(t)$ by combining the two equations into a system doesn't change these properties of the individual equations. So when we do, we find that the solution $(x(t),y(t))$ with initial conditions $(x_0,y_0)\in[0,\pi]\times[0,\pi]$ satisfies:

*

*$0 \leq x(t) \leq \pi$ since $x_0\in[0,\pi]$; remember, this property doesn't care about what exactly $y(t)$ is!

*$0 \leq y(t) \leq \pi$ since $y_0\in[0,\pi]$; similarly.

Thus the assertion is proved.
As a side note, a word of advice for studying nonlinear systems: if you ever find yourself saying "I need to solve this nonlinear system to proceed," you are likely on the wrong track. The whole point of the field is that nonlinear systems are difficult to solve analytically, but there are ways to analyze them without solving them.
A: To show that a solution curve starting in $[0, \pi] \times [0, \pi]$ always remains within that region, it suffices to show that any solution starting on the boundary of this region cannot exit the region.
Continuing from your start, we first look at the $x \in [0, \pi]$ and $y = 0$ part of the boundary. For any point here, $\frac{dy}{dt} = 0$ and $\frac{dx}{dt} \leq 0$. So any point starting on this part of the boundary will go to $(0,0)$ along this part, except for $(\pi,0)$ which remains where it is.
Repeating this analysis on the other three parts, we get the result that a point starting on the boundary will move clockwise around the boundary until it reaches a corner point where it stops (steady state). Therefore, a solution curve starting in $[0, \pi] \times [0, \pi]$ can't exit, because as soon as it touches the boundary, it gets trapped there. Moreover, a solution curve in the interior, $(0, \pi) \times (0, \pi)$, will remain in the interior, because if you reverse time for a point on the boundary, it will still be on the boundary.
