System of nonlinear ODEs with symmetric coefficient matrix I am facing a problem where I have to find the solution to a nonlinearsystem of first order ODEs:
$$
\mathbf{x}'(t) = A(x_0)\mathbf{x} + \mathbf{b}(t) = (F(x_0) + B)\mathbf{x} + \mathbf{b}(t),
$$
where $x_0$ is the first component of the vector $\mathbf{x}$, and $F(x_0)$ is a matrix with only one non-zero element:
$$
F_{00} = f(x_0),\\
F_{ij} = 0,\text{ for }(i,j) \ne (0,0),
$$
and B is symmetric. A solution can always be found numerically, but first I wanted to try if I can get closer to an analytical solution, in terms of the unknown functions $f(x_0)$ and $\mathbf{b}(t)$.
If it helps, the function $f(x_0)$ will in most scenarios have the form
$$ f(x_0) = c_0\tanh(c_1(x_0 - c_2)),$$
and the function $\mathbf{b}(t)$ the form
$$\mathbf{b}(t) = \mathbf{m} + \sum_i \mathbf{n_i} e^{-\alpha_it}.$$
I am happy about any suggestions.
 A: Partial answer: Since the nonlinear equation for $x_0'(t)=f(x_0(t))+g(t)$ does not seem to have a closed form solution for your $f$, I do not expect there exists a analytic closed form solution for the system. The best I can provide you with is a partial solution:
We start by defining
$$
B=\left(\begin{array}{rrr} 
B_{00} & B_{1R} \\ 
B_{1R}^T & B_R 
\end{array}\right)\\
\mathbf{x}=\left(\begin{array}{rrr} 
x_{0}\\
\mathbf{x}_R 
\end{array}\right)\\
\mathbf{b}=\left(\begin{array}{rrr} 
b_{0}\\
\mathbf{b}_R 
\end{array}\right)\\
$$
and hence write the system of ODEs as
\begin{align}
x_0'(t)&=f(x_0(t))+B_{00}x_0(t)+B_{1R}\mathbf{x}_R(t)+b_0(t)\\
\mathbf{x}_R'(t)&=x_0(t)B_{1,R}^T+B_R\mathbf{x}_R(t)+\mathbf{b}_R (t)
\end{align}
Since $B_R$ is a symmetric real matrix it can be written as $B_R=S^{-1}DS$, with $D$ diagonal. Therefore the system for $\mathbf{x}_R$ can be solved explicitely via:
$$
\mathbf{x}_R=S^{-1}e^{tD}S\mathbf{c}+\int_0^tS^{-1}e^{(s-t)D}S\left(x_0(s)B_{1,R}^T+\mathbf{b}_R (s)\right)ds
$$
where $\mathbf{c}$ has to be fitted to the intial data. Plugging it back to the first equation yields:
$$
x_0'(t)=f(x_0(t))+B_{00}x_0(t)+B_{1R}S^{-1}e^{tD}S\mathbf{c}+B_{1R}\int_0^tS^{-1}e^{(s-t)D}S\left(x_0(s)B_{1,R}^T+\mathbf{b}_R (s)\right)ds+b_0(t)
$$
Since $D$ and $S$ can be calculated once this remaining ODE should be significantly less costly to solve numerically.
