Undergrad 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 <br /> \begin{pmatrix} \mathbf{x} \\ \mathbf{x}&#039; \end{pmatrix}&#039; = \begin{pmatrix} 0 &amp; I \\ A &amp; 0 \end{pmatrix}<br /> \begin{pmatrix} \mathbf{x} \\ \mathbf{x}&#039; \end{pmatrix}. Now consider an eigenvector (\mathbf{v}\, \mathbf{u})^T of this with eigenvalue \mu. By definition <br /> \mu \begin{pmatrix} \mathbf{v} \\ \mathbf{u} \end{pmatrix} = \begin{pmatrix} 0 &amp; I \\ A &amp; 0 \end{pmatrix}<br /> \begin{pmatrix} \mathbf{v} \\ \mathbf{u} \end{pmatrix} so that \left. \begin{array}{c}<br /> \mu\mathbf{v} = \mathbf{u} \\ <br /> \mu\mathbf{u} = A\mathbf{v} \end{array}\right\}\quad\Rightarrow\quad A\mathbf{v} = \mu^2 \mathbf{v}. Hence \mu^2 = \lambda is an eigenvalue of A with \mathbf{v} its corresponding eigenvector. This leads to a solution of the form <br /> \mathbf{x}(t) = \mathbf{v}(Ae^{\sqrt{\lambda}t} + Be^{-\sqrt{\lambda}t})

 

[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