Jacobian matrix of time dependant system

Question

Suppose we have a differential equation of the form $x^{'}(t)=(x^{'}_{1}(t),x^{'}_{2}(t))=f((x_{1}(t),x_{2}(t))$ where $f : R^2 \rightarrow R^2$

Does the Jacobian matrix of $f$ get affected by $t$?

Answer 1

The system you show

$\dot x= f(x), \quad x=[x_1,x_2] $, $x=x(t)$

is time invariant since it depends not explicitly on time. On the contrary the sysrem

$\dot x=f(x,t)$

is time variant, i.e., time dependent.

With respect to the Jacobian:

In the first time invariant case, as in your question, the Jacobian depends only on x(t).

In the second case, the Jacobian depends additionally on time dependent parameters. That means, the time dependency remains even, when you insert stationary points for $x$ in the Jacobian.

Examples:

1. Time invariant system: \begin{align} \dot x &= \begin{bmatrix} x_2\\ -x_1^3 - x_2 \end{bmatrix}\\ J &= \begin{bmatrix} 0 & 1\\ -3\,x_1^2 & - 1 \end{bmatrix} \end{align}
   Therefore, no explicit time dependence.

2. Time varying system: \begin{align} \dot x &= \begin{bmatrix} x_2\\ -x_1^3 - \sin(t)\,x_2 \end{bmatrix}\\ J &= \begin{bmatrix} 0 & 1\\ -3\,x_1^2 & - \sin(t) \end{bmatrix} \end{align} Then, the Jacobian is time dependent.

Answer 2

The Jacobian matrix contains the partial derivatives of the function $f(x_1,x_2)$ with respect to $x_1$ and $x_2$. No $t$ as far as the eye can see!

(Unless, of course, you evaluate the Jacobian at some time-dependent point $(x_1,x_2)=(g_1(t),g_2(t))$ after computing it.)