- #1
member 428835
Hi PF!
Given the ODE system ##x'(t) = A(t) x(t)## where ##x## is a vector and ##A## a square matrix periodic, so that ##A(t) = A(T+t)##, would the following be a good way to solve the system's stability: fix ##t^*##. Then
$$
\int \frac{1}{x} \, dx = \int A(t^*) \, dt \implies\\
x(t) = x(0)\exp\left( A(t^*)t \right).
$$
The eigenvalues of ##A(t^*)## determine the system's stability, but by fixing ##t##, this approach assumes ##A(t)## is approximately constant for ##t##, which may not be the case. What do you think?
Given the ODE system ##x'(t) = A(t) x(t)## where ##x## is a vector and ##A## a square matrix periodic, so that ##A(t) = A(T+t)##, would the following be a good way to solve the system's stability: fix ##t^*##. Then
$$
\int \frac{1}{x} \, dx = \int A(t^*) \, dt \implies\\
x(t) = x(0)\exp\left( A(t^*)t \right).
$$
The eigenvalues of ##A(t^*)## determine the system's stability, but by fixing ##t##, this approach assumes ##A(t)## is approximately constant for ##t##, which may not be the case. What do you think?