What is the solution for a system of ODEs with a matrix coefficient?

[kostoglotov]

I know how to solve \frac{d\vec{u}}{dt} = A\vec{u}, I was just watching a lecture, and the lecturer related that solving that equation is pretty much a direct analogy to \vec{u} = e^{At}\vec{u}(0), in so far as all we need to do after that is understand exactly what it means to take the exponential of a matrix, which is fine...

My question is this, do I solve \frac{d^2\vec{u}}{dt^2} = A\vec{u} with \vec{u}=A^{-1}e^{At}\frac{d\vec{u}}{dt}(0)?

edit:

OR do I need to go along the lines

\frac{d^2\vec{u}}{dt^2} = \lambda^2\vec{u} = A\vec{u}

and solve

(A - \lambda^2 I)\vec{u}=\vec{0}

?

 

[Svein]

One way to look at it: Introduce a new function v: \frac{d\vec{u}}{dt}= \vec{v}, \frac{d\vec{v}}{dt}=A\vec{u}.

 

[Mark44]

I know how to solve \frac{d\vec{u}}{dt} = A\vec{u}, I was just watching a lecture, and the lecturer related that solving that equation is pretty much a direct analogy to \vec{u} = e^{At}\vec{u}(0), in so far as all we need to do after that is understand exactly what it means to take the exponential of a matrix, which is fine...

My question is this, do I solve \frac{d^2\vec{u}}{dt^2} = A\vec{u} with \vec{u}=A^{-1}e^{At}\frac{d\vec{u}}{dt}(0)?

edit:

OR do I need to go along the lines

\frac{d^2\vec{u}}{dt^2} = \lambda^2\vec{u} = A\vec{u}

and solve

(A - \lambda^2 I)\vec{u}=\vec{0}

?

I haven't thought about this in quite a while.
If you had a regular ODE of the form u' = au, or u' - au = 0, the characteristic equation would be r - a = 0, so the solution would be u = Ce^at, similar to your first order example above.

If you had u'' = λ²u, or u'' - λ²u = 0, the characteristic equation would be r² - λ² = 0, or r = ±λ. The solution would be u = C₁e^λt + C₂e^-λt.

Your example, with a system of ODEs, will have a solution that is analogous to the above, with exponentials involving the matrix.