13.8 Solving a system of linear ODEs using a matrix exponential
Linear systems of ODEs can be solved using a matrix exponential. This can be considerably faster than using one of the ODE solvers.
The solution to \(\frac{d}{dt} y = ay\) is \(y = y_0e^{at}\), where the constant \(y_0\) is determined by boundary conditions. We can extend this solution to the vector case: \[ \frac{d}{dt}y = A \, y \] where \(y\) is now a vector of length \(n\) and \(A\) is an \(n\) by \(n\) matrix. The solution is then given by: \[ y = e^{tA} \, y_0 \] where the matrix exponential is formally defined by the convergent power series: \[ e^{tA} = \sum_{n=0}^{\infty} \dfrac{tA^n}{n!} = I + tA + \frac{t^2A^2}{2!} + \dotsb \]
We can apply this technique to the simple harmonic oscillator example, by setting \[ y = \begin{bmatrix} y_1 \\ y_2 \end{bmatrix} \qquad A = \begin{bmatrix} 0 & 1 \\ -1 & -\theta \end{bmatrix} \]
The Stan model to simulate noisy observations using a matrix exponential function is given below.
In general, computing a matrix exponential will be more efficient than using a numerical solver. We can however only apply this technique to systems of linear ODEs.
data {
int<lower=1> T;
vector[2] y0;
array[T] real ts;
array[1] real theta;
}model {
}generated quantities {
array[T] vector[2] y_sim;
matrix[2, 2] A = [[ 0, 1],
1, -theta[1]]]
[-for (t in 1:T) {
1) * A) * y0;
y_sim[t] = matrix_exp((t -
}// add measurement error
for (t in 1:T) {
1] += normal_rng(0, 0.1);
y_sim[t, 2] += normal_rng(0, 0.1);
y_sim[t,
} }
This Stan program simulates noisy measurements from a simple harmonic
oscillator. The system of linear differential equations is coded as a
matrix. The system parameters theta
and initial state y0
are read
in as data along observation times ts
. The generated quantities
block is used to solve the ODE for the specified times and then add
random measurement error, producing observations y_sim
. Because the
ODEs are linear, we can use the matrix_exp
function to solve the
system.