# How to set up a time-dependant matrix for an ODE to be solved using python?

I want to solve a problem numerically in python like this:

$$y(t)' = \mathbf{M}(t)y ,\\ y(0) = (1,0,0,0 ...)$$

where $$y$$ is an $$n$$-dimensional vector and $$\mathbf{M}(t)$$ is a time-dependant $$n \times n$$ matrix.

The matrix $$\mathbf{M}(t)$$ is of the shape

$$\mathbf{M}(t)= \left[ \begin{array}{ccccc} 0 & T_1(t) & 0 & 0 & 0 & 0\\ T_1(t) & 0 & T_2(t) & 0 & 0 & 0\\ 0 & T_2(t) & 0 & T_1(t) & 0 & 0\\ 0 & 0 & T_1(t) & 0 & T_2(t) & 0\\ 0 & 0 & 0 & T_2(t) & 0 & T_1(t)\\ 0 & 0 & 0 & 0 & T_1(t) & 0\\ \end{array} \right]$$

or in numpy/scipy notation:

[[0. T1 0. 0. 0. 0.]
[T1 0. T2 0. 0. 0.]
[0. T2 0. T1 0. 0.]
[0. 0. T1 0. T2 0.]
[0. 0. 0. T2 0. T1]
[0. 0. 0. 0. T1 0.]]


where $$T_1$$ and $$T_2$$ can be arbitrarily time dependant.

My problem is, that for the routine solve_ivp in python I need to put in a function of $$y$$ into the solve_ivp routine. I don't know how to create this matrix $$\mathbf{M}(t)$$ above so that solve_ivp can use it.

Has someone some hints or suggestions?

• I edited your question for clarity and presentation. Can you check that I did not make any mistakes? Nov 27 '19 at 18:59

What about avoiding to construct the matrix by using its structure?

def odefunc(t,u):
dotu = zeros_like(u)
T1 = T1func(t)
T2 = T2func(t)
dotu[0::2] += T1*u[1::2]
dotu[1::2] += T1*u[0::2]
dotu[1:-1:2] += T2*u[2::2]
dotu[2::2] += T2*u[1:-1:2]
return dotu


This only works if the matrix size is even and not divisible by 4, like the current size 6.