Orbits of the following system:

Question

I am asked to find orbits for the system: $$ \begin{cases} \frac{dx}{dt} = -x \\ \frac{dy}{dt} = -y \end{cases} $$ and find critic points and say whether they are stable, asymptotically stable or unstable. Moreover, I am asked to find invariant subsets. If I solve the system, how can I find the solutions for the orbits? I have the following solution for the System (with $x(0)=x_{0}$ and $y(0)=y_{0}$): $$ \begin{cases} x(t) = x_{0}e^{-t} \\ y(t) = y_{0}e^{-t} \end{cases} $$ What should I do now?

Answer 1

Hint: The orbits are given by $y_{0}x(t) = x_{0}y(t)$. Consider the cases $x_0=0$ and $y_0=0$.

Answer 2

It depends a bit on what kind of theory you have been exposed to.

You could notice that - while you have a system of two differential equations - these are really just two first order equations put next to each other. Since the equations are linear, you can simply add the solutions together to solve the system. What this implies specifically is that the $x$-axis and $y$-axis are both invariant subsets.

Regarding fixed points and stability: You can easily see that the origin is the only fixed point of the system. You can see that it is a stable equilibrium in many different ways. Again, that depends on how much theory you have been exposed to. I would just draw a phase portrait in this case.

More generaly, if you have a linear system of the form $\dot{x} = Ax$, where $x \in \mathbb{R}^2$, and $A$ is a nonsingular matrix, then the origin will be the only fixed point of the system. The directions along the eigenvectors are invariant, and the stability depends on the signs of the corresponding eigenvalues. In the above, the eigenvalues are both equal to $-1$, which implies an attractive fixed point.

(If the signs were both positive, you would have a repelling fixed point, and two different signs would imply a saddle point (unstable).)