Ordinary Differential Equation--System with Constant Coefficients

To solve the system of differential equations

$${d{\bf x}\over dt} = {\hbox{\sf A}}{\bf x}(t) + {\bf p}(t),$$ (1)

where ${\hbox{\sf A}}$ is a Matrix and ${\bf x}$ and ${\bf p}$ are Vectors, first consider the homogeneous case with ${\bf p}={\bf0}$. Then the solutions to

$${d{\bf x}\over dt} = {\hbox{\sf A}}{\bf x}(t)$$ (2)

are given by

$${\bf x}(t)=e^{\hbox{{\sf A}}t}{\bf x(t)}.$$ (3)

But, by the Matrix Decomposition Theorem, the Matrix Exponential can be written as

$$e^{\hbox{{\sf A}}t}={\hbox{\sf u}}{\hbox{\sf D}}{\hbox{\sf u}}^{-1},$$ (4)

where the Eigenvector Matrix is

$${\hbox{\sf u}} = \left[{\matrix{{\bf u}_1 & \cdots & {\bf u}_n\cr}}\right]$$ (5)

and the Eigenvalue Matrix is

$${\hbox{\sf D}} = \left[{\matrix{e^{\lambda_1t} & 0 &\cdots &... ... & \ddots & 0\cr 0 & 0 & \cdots & e^{\lambda_nt}\cr}}\right].$$ (6)

Now consider

$$e^{\hbox{{\sf A}}t}{\hbox{\sf u}} = {\hbox{\sf u}}{\hbox{\sf D}}{\hbox{\sf u}}^{-1}{\hbox{\sf u}} = {\hbox{\sf u}}{\hbox{\sf D}}$$

$$= \left[\begin{array}{cccc} u_{11} & u_{21} & \cdots & u_{n1}\\ u_... ...dots & \vdots & \ddots & 0\\ 0 & 0 & \cdots & e^{\lambda_nt}\end{array}\right]$$

$$= \left[\begin{array}{ccc} u_{11}e^{\lambda_1t} & \cdots & u_{n1}e^... ...dots\\ u_{n1}e^{\lambda_1t} & \cdots & u_{n2}e^{\lambda_nt}\end{array}\right].$$ (7)

The individual solutions are then

$${\bf x}_i = (e^{\hbox{{\sf A}}t}{\hbox{\sf u}})\cdot \hat {\bf e}_i = {\bf u}_i e^{\lambda_i t},$$ (8)

so the homogeneous solution is

$${\bf x}=\sum_{i=1}^n c_i {\bf u}_i e^{\lambda_it},$$ (9)

where the $c_i$s are arbitrary constants.

The general procedure is therefore

1. Find the Eigenvalues of the Matrix ${\hbox{\sf A}}$ ($\lambda_1$, ..., $\lambda_n$) by solving the Characteristic Equation.

2. Determine the corresponding Eigenvectors ${\bf u}_1$, ..., ${\bf u}_n$.

3. Compute

$${\bf x}_i \equiv e^{\lambda_it}{\bf u}_i$$ (10)

for $i = 1$, ..., $n$. Then the Vectors ${\bf x}_i$ which are Real are solutions to the homogeneous equation. If ${\hbox{\sf A}}$ is a $2\times 2$ matrix, the Complex vectors ${\bf x}_j$ correspond to Real solutions to the homogeneous equation given by $\Re({\bf x}_j)$ and $\Im({\bf x}_j)$.

4. If the equation is nonhomogeneous, find the particular solution given by

$${\bf x}^*(t) = {\hbox{\sf X}}(t) \int {\hbox{\sf X}}^{-1}(t){\bf p}(t)\,dt,$$ (11)

where the Matrix ${\hbox{\sf X}}$ is defined by

$${\hbox{\sf X}}(t) \equiv \left[{\matrix{{\bf x}_1 & \cdots & {\bf x}_n\cr}}\right].$$ (12)

If the equation is homogeneous so that ${\bf p}(t)={\bf0}$, then look for a solution of the form

$${\bf x}={\boldsymbol{\xi}}e^{\lambda t}.$$ (13)

This leads to an equation

$$({\hbox{\sf A}}-\lambda {\hbox{\sf I}})\boldsymbol{\xi}={\bf0},$$ (14)

so ${\bf\xi}$ is an Eigenvector and $\lambda$ an Eigenvalue.

5. The general solution is

$${\bf x}(t) = {\bf x}^*(t) + \sum_{i=1}^n c_i {\bf x}_i.$$ (15)