# Fundamental matrix of a second order 2x2 system of ODEs

• I
• EinsteinCross

#### 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?

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

Last edited:
topsquark and EinsteinCross
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

malawi_glenn and topsquark