Fundamental matrix of a second order 2x2 system of ODEs

[EinsteinCross]

Let ## \mathbf{x''} = A\mathbf{x} ## be a homogenous second order system of linear differential equations where

##
A = \begin{bmatrix}
a & b\\
c & d
\end{bmatrix}
## and ##
\mathbf{x} = \begin{bmatrix}
x(t)\\ y(t))
\end{bmatrix}
##

Now to solve this equation we transform it into a 4x4 first order system ##\mathbf{X'} = M\mathbf{X} ## where

##
M = \begin{bmatrix}
0 & 1 & 0 & 0 \\
a & 0 & b & 0\\
0 & 0 & 0 & 1 \\
c & 0 & d & 0
\end{bmatrix}## and ##
\mathbf{X} = \begin{bmatrix}
x(t)\\ x'(t)
\\ y(t)
\\ y'(t)
\end{bmatrix}
##

Now solving the first order system by calculating the eigenvalues and corresponding eigenvectors gives us the fundamental matrix for the 4x4 system
##
\Phi (t) = [[\xi_{1}],[\xi_{2}],[\xi_{3}],[\xi_{4}]]
\begin{bmatrix}
e^{\lambda_{1}t }
\\ e^{\lambda_{2}t }
\\ e^{\lambda_{3}t }
\\ e^{\lambda_{4}t }
\end{bmatrix}^{T} ## where the xi brackets are the eigenvectors in column form.

But my question is this: What I am looking to do is to find the fundamental matrix of the original second order 2x2 system which is represented as ##
\Psi (t) = \begin{bmatrix}
\psi_{1}(t) & \psi_{2}(t)
\\ \psi_{3}(t)
& \psi_{4}(t)
\end{bmatrix} ## such that ##\Psi ''(t) = A\Psi (t)##. So how does one extract the solution(s) to the original 2x2 system from the fundamental matrix of the first order 4x4 system?

 

[pasmith]

It is better to construct your first order system in block matrix form as [tex]
\begin{pmatrix} \mathbf{x} \\ \mathbf{x}' \end{pmatrix}' = \begin{pmatrix} 0 & I \\ A & 0 \end{pmatrix}
\begin{pmatrix} \mathbf{x} \\ \mathbf{x}' \end{pmatrix}.[/tex] Now consider an eigenvector [itex](\mathbf{v}\, \mathbf{u})^T[/itex] of this with eigenvalue [itex]\mu[/itex]. By definition [tex]
\mu \begin{pmatrix} \mathbf{v} \\ \mathbf{u} \end{pmatrix} = \begin{pmatrix} 0 & I \\ A & 0 \end{pmatrix}
\begin{pmatrix} \mathbf{v} \\ \mathbf{u} \end{pmatrix}[/tex] so that [tex]\left. \begin{array}{c}
\mu\mathbf{v} = \mathbf{u} \\
\mu\mathbf{u} = A\mathbf{v} \end{array}\right\}\quad\Rightarrow\quad A\mathbf{v} = \mu^2 \mathbf{v}.[/tex] Hence [itex]\mu^2 = \lambda[/itex] is an eigenvalue of [itex]A[/itex] with [itex]\mathbf{v}[/itex] its corresponding eigenvector. This leads to a solution of the form [tex]
\mathbf{x}(t) = \mathbf{v}(Ae^{\sqrt{\lambda}t} + Be^{-\sqrt{\lambda}t})[/tex]

 

[jasonRF]

An equivalent approach is to leave it as a system of second-order ODEs. Start with
$$ \mathbf{x''} = A\mathbf{x} $$
and assume a solution of the form ##\mathbf{x} = \mathbf{x_0} \, e^{\lambda t}##. You then get
$$ \lambda^2 \mathbf{x_0} = A\mathbf{x_0}. $$
which is a standard eigenvalue problem. This is the approach I always use for equations of this form.

jason