How can the stability of an ODE system be determined without solving it?

[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?

 

[Orodruin]

No. You essentially assumed that A is constant and even if it is you cannot divide by a vector.

 

[member 428835]

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]

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.

 

[member 428835]

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 $x' = B x$?

 

[Orodruin]

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.

 

[member 428835]

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!

 

[Ray Vickson]

Not solve, but show stability? I think you'll still say no, and thanks for the reply!

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 $B$.