Convert a linear system to higher order 1D equation

Question

There are numerous explanations of converting an $n^\text{th}$ order differential equation

$$\sum_{i=0}^n a_i \frac{\partial^i x}{\partial t^i} = 0 $$

to a linear system with $n$ dimensions.

$$\frac{d \underline{x}}{dt} = \mathrm{A}\underline{x}$$

Can I perform the reverse operation and convert an arbitrary linear system back to an ode?

Obviously it is trivial to possible to perform the reverse operation when $\mathrm{A}$ has the $\begin{bmatrix}\underline{\alpha} & \beta \\ \mathrm{I} & \underline{0} \end{bmatrix}$ structure resulting from the forward operation, but my only progress for general $\mathrm{A}$ is that it is an equivalent problem to using the diagonal matrix $\Lambda$ by eigendecomposition and change of variables

$$ \frac{d \underline{u}}{dt} = \Lambda \underline{u}, \qquad \mathrm{A} = Q^{-1}\Lambda Q ,\qquad \underline{u} = Q \underline{x}$$

Answer 1

Consider the linear state system

$$\dot{x}(t) = A\,x(t) + B\,u(t)$$

$B\,u(t)$ is zero in your case. It is a time-dependent input, such as a force on a mass-spring-damper system. I'm just putting it there as it is a valid term in a linear set of equations, with the purpose of obtaining a general solution.

We can write

$$S\,I\,x(t) = A\,x(t) + B\,u(t)$$

where $S$ is the differential operator $\textrm{d}/\textrm{d}t$ and $I$ an identity matrix. Assuming that $(S\,I - A)$ is invertible (which it is unless you have two independent state systems into a single set) we can write

$$x(t) = (S\,I - A)^{-1}B\,u(t) \to \det{(S\,I - A)}\,x(t) = \mathrm{adj}\,{(S\,I - A)} B\,u(t) $$

which will give you your $n$th order differential equation. In your case, $B\,u(t)=0$, so you'll find your solution by computing $\det{(S\,I - A)}\,x(t)=0$.

Answer 2

Expanding on @JJM's response, the backward operation usually done by taking the laplace transform of the system of 1st order ODEs. Indeed if:

$$ \dot{\vec x}(t) = \Bbb{A} \vec{x}(t) + \Bbb{B}\vec{u}(t) $$

for $\vec{x}$ with $n$ components and $\vec{u}$ with $m$ components The laplace transorm of the vector equation earlier gives:

$$ \vec{X}(s) = (s\Bbb{I}_n - \Bbb{A})^{-1}\Bbb{B}\vec{U}(s) $$

Note the matrix $(s\Bbb{I}_n - \Bbb{A})^{-1}\Bbb{B}$ is a matrix of $n \times m$ tranfer functions of the general form:

$$[(s\Bbb{I}_n - \Bbb{A})^{-1}\Bbb{B}]_{ij} = \frac{\sum_{r\le{m}}{a_rs^r}}{\det((s\Bbb{I}_n - \Bbb{A})^{-1})} = \frac{X_i(s)}{U_j(s)}$$

From the transfer function(s) above we may then rearange to get the nth order ODE for any one of the state variables in $\vec{x}$ (notice how the determinat is actually an $n$th order polynomial in s.

Sorry for the vagueness, If you want more information, Springer publishing has some very good books on linear control theory. I dont have a favourite off the top of my head but I vouch for the quality of their books any day.