Can we solve a matrix differential equation $\textbf{X}'(t)=\textbf{A}\textbf{X}(t)+\textbf{B}(t)$ by Finite Element Method?
I will be very happy if you give some special example or suggest some books about the method for solving matrix differential eauations.
In here, \begin{equation} \begin{aligned} &\textbf{X}(t)=\left( \begin{array}{cccc} x_{1}(t) \\ x_{2}(t) \\ x_{3}(t)\\ x_{4}(t) \end{array} \right), \textbf{A} = \left( \begin{array}{cccc} a_{11} & a_{12} &a_{13} & a_{14}\\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32}& a_{33} & a_{34} \\ a_{41} & a_{42} & a_{43} & a_{44} \end{array} \right),\\ &\textbf{B}(t)=\left( \begin{array}{cccc} b_{1}(t)\\ b_{2}(t)\\ b_{3}(t)\\ b_{4}(t) \end{array} \right). \end{aligned} \end{equation} initial conditions: $$x_1(0)=m_1$$ $$x_2(0)=m_2$$ $$x_3(0)=m_3$$ $$x_4(0)=m_4.$$
The differential system ${\bf X}' = {\bf A} {\bf X} + {\bf B}$ where $t \mapsto {\bf X}(t)$ is a vector-valued function of size $n=4$, ${\bf A}$ is a $n$-by-$n$ matrix and $t \mapsto {\bf B}(t)$ is a vector-valued function of size $n$, can be solved by hand in many particular cases. Indeed, this system is an order-1 linear system of ordinary differential equations (ODE). Solutions write as $$ {\bf X}(t) = \exp(t{\bf A})\, {\bf X}(0) + \int_{0}^t \exp((t-\tau) {\bf A}) \, {\bf B}(\tau)\, \mathrm{d}\tau \, , $$ where $\exp(s{\bf A})$ is a matrix exponential. Therefore, if the matrix exponential and the integral can be computed, then the solution is given above.
(1) FEM. To implement FEM, the first step is to write a weak formulation of the ODE. We write the scalar product of the ODE with the test vector field ${\bf Y}(t)$, and we integrate over $t\in \Theta$. Hence, using integration by parts, $$ \int_\Theta ({\bf Y}' + {\bf A}^\top{\bf Y})^\top\, {\bf X} = -\int_\Theta {\bf Y}^\top {\bf B}, $$ for all ${\bf Y}$ which trace equals zero on the boundaries of $\Theta$. One notes that the above weak formulation writes as $a({\bf X},{\bf Y}) = f({\bf Y})$, where $a(\cdot,\cdot)$ is a bilinear form over $H^1(\Theta)$ and $f(\cdot )$ is a linear form. It remains to discretize the above equations by considering that ${\bf X}$, ${\bf Y}$ belong to finite-dimension spaces, e.g. by using appropriate bases of Lagrange polynomials of the variable $t$.
(2) FD. The system ${\bf X}' = {\bf A} {\bf X} + {\bf B}$ can be integrated numerically by finite-difference methods. The simplest method is the first-order accurate Euler method. Here, the forward Euler method writes $$ \left\lbrace \begin{aligned} &{\bf X}_{n+1} = {\bf X}_{n} + \Delta t \left({\bf A}{\bf X}_{n} + {\bf B}(t_n)\right) , \\ &{\bf X}_{0} = {\bf X}(0) \, , \end{aligned} \right. $$ where ${\bf X}_{n} \simeq {\bf X}(t_{n})$, and $\Delta t = t_{n+1} - t_n$ denotes the time step. The method is stable provided that $ \Delta t \leq {2}/{\varrho({\bf A})} $, where $\varrho({\bf A})$ is the spectral radius of the matrix $\bf A$. The backward Euler method $$ \left\lbrace \begin{aligned} &{\bf X}_{n+1} = \left({\bf I} - \Delta t {\bf A}\right)^{-1}\, \left({\bf X}_{n} + \Delta t\, {\bf B}(t_{n+1})\right) , \\ &{\bf X}_{0} = {\bf X}(0) \, , \end{aligned} \right. $$ where ${\bf I}$ is the identity matrix, is unconditionally stable. Thus, it bypasses stability issues, but requires the computation of a matrix inverse. Many more accurate numerical methods exist, such as Runge-Kutta methods. If the system is stiff, then the time step $\Delta t$ must be chosen very small to get good accuracy, which is costly from a computational point of view. Hence, dedicated adaptive methods may be preferred. For instance, the reader is referred to the book Numerical Recipes in C: The Art of Scientific Computing for details.