Linear time varying into linear time invariant.

Question

My original problem, is to transform Linear time varying systems of the form , for example:

$$\begin{bmatrix}\dot{x}_1 \\ \dot{x}_2 \end{bmatrix} = \begin{bmatrix} -3t^2 & 0 \\ 6t^5 & -6t^2 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2\end{bmatrix} + \begin{bmatrix} 4t^2 \\3t \end{bmatrix} u$$ into it's Linear time invariant eqivalent (i.e. constant coeffecients)

Answer 1

I do not understand what is the definition of "time invariant equivalent". Let me write $\dot X = A^* X + B^* u^*(t)$ as the "LTI" equivalent model.

The autonomous system is stable since $A$ is Hurwitz. This is because $A$ is lower diagonal and the diagonal terms are always negative. So $A^*$ is whatever Hurwitz matrix.

If all the elements of $B$ had the same dependency on $t$, then you could write $B^*$ with only constant elements, i.e. If $B = \begin{bmatrix}4t \\ 3t\end{bmatrix}$, you could define $u^*(t) =u(t) t$ and $B^* = \begin{bmatrix}4 \\ 3\end{bmatrix}$. But in your case I would say this is not possible.

Anyway, without a definition of "LTI equivalent", I can not do more precise.

Answer 2

Without context, it is not really possible to see what you want to do but one possibility is to find a bijective change of variables $z(t)=P(t)x(t)$ such that your system is equivalent to an LTI system of the form $\dot{z}(t)=\bar Az(t)+\bar Bu(t)$.

By differentiating our change of variables, we get that

$$\dot{z}(t)=(\dot{P}(t)+P(t)A(t))P(t)^{-1}z(t)+P(t)B(t)u(t).$$

So, we "just" need to find $P(t)$ such that both $(\dot{P}(t)+P(t)A(t))P(t)^{-1}$ and $P(t)B(t)$ are constant matrices. This can be reformulated as finding a solution $(Q(t),\bar A,\bar B)$ to

$$\dot{Q}(t)+A(t)Q(t)-Q(t)\bar A=0,\qquad B(t)-Q(t)\bar B=0,$$

which is a problem that is not easy to solve in general. I will be happy to add more in the case the OP provides more details about the problem.