I have the following 3-dimensional matrix initial value problem:

$$\left\{\begin{matrix}\textbf{x}'(t) &=& (A + Bt)\:\textbf{x}(t) \\ \textbf{x}(0) &=& \textbf{x}_0\end{matrix}\right.$$

The matrices $A$ and $B$ do not depend on $t$, are anti-symmetric and have the following sparsity pattern:

$$\begin{bmatrix} 0 & \times & 0 \\ \times & 0 & \times \\ 0 & \times & 0 \end{bmatrix}$$

where the crosses indicate the non-zero elements. I am wondering whether this initial value problem can be solved analytically.

I am aware that both the special cases $A = 0$ and $B = 0$ can be solved analytically, with solutions $\textbf{x}(t) = e^{\frac12Bt^2}\textbf{x}_0$ and $\textbf{x}(t) = e^{At}\textbf{x}_0$, respectively. $A$ and $B$ are very likely not to commute, so the solution $\textbf{x}(t) = e^{At + \frac12Bt^2}\textbf{x}_0$ is probably false.

Does anyone have an idea how to tackle this problem?


1 Answer 1


HERE I've proved the existence and uniqueness of such kind of problems. Define $\tilde{A}(t) := A + Bt$. Then the Dyson series is defined as

$$D(t,0):= I + \sum_{n=1}^{\infty}\int_0^{t}\int_0^{t_1}\dots\int_0^{t_{n-1}}\tilde{A}(t_1)\dots \tilde{A}(t_n)\mathrm dt_n \dots \mathrm dt_1$$

The solution will be of the Cauchy problem for the differential equation mentioned is

$$X(t) = D(t,0)X_0$$

The problem now is to find $D(t)$.

  • $\begingroup$ Thanks. :) Not exactly what I was looking for, but it is a useful result I can use. :p $\endgroup$
    – arriopolis
    Commented Oct 5, 2017 at 9:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .