A ODE System solution technique

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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?
 

Orodruin

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No. You essentially assumed that A is constant and even if it is you cannot divide by a vector.
 
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No. You essentially assumed that A is constant and even if it is you cannot divide by a vector.
So this technique cannot be used to determine stability of a periodic function? I know solutions won't always work, but you think this fails in general for stability for aforementioned reasons?
 

Orodruin

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You need to find the actual solution integrated over a period, which will be on the form ##x(t+T) = S x(t)##. The eigenvalues of ##S## will determine the stability.
 
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You need to find the actual solution integrated over a period, which will be on the form ##x(t+T) = S x(t)##. The eigenvalues of ##S## will determine the stability.
So you're saying something like ##B = \int_0^T A(t) \, dt## and then solve ##x' = B x##?
 
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Orodruin

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So you're saying something like ##B = \int_0^T A(t) \, dt## and then solve ##x' = B x##?
No, that will not solve the differential equation. You need the time-ordered exponential.
 
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No, that will not solve the differential equation. You need the time-ordered exponential.
Not solve, but show stability? I think you'll still say no, and thanks for the reply! :oldbiggrin:
 

Ray Vickson

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Not solve, but show stability? I think you'll still say no, and thanks for the reply! :oldbiggrin:
It will not be easy, but you need to solve ##x'(t) = A(t) x(t), \; 0 \leq t \leq T## for given ##x(0)## You can express the vector ##x(T)## as some linear transformation of ##x(0),## that is, ##x(T) = B x(0)## for some matrix ##B## that you can determine numerically. Stability issues involve ##B##.
 

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