Homogeneous Systems of Linear Constant-Coefficient Differential Equations

A system of linear constant-coefficient differential equations can be written in matrix form as

$X'=AX+F,$

where $X\in\mathbb{R}^n$ is an unknown vector-valued function, $A\in\mathbb{R}^{n\times n}$ is a constant coefficient matrix, and $F\in\mathbb{R}^n$ is the forcing term. When $F=0$ , we have a homogeneous system. In the homogeneous case, $X\equiv 0$ is always a solution and the set of all solutions forms an $n$ -dimensional vector space.

As in the single dimensional case, linearity leads to superposition, and specifying an initial condition $X(t_0)=X_0$ determines a unique solution.

A key feature of the system $X'=AX$ is that all solution behavior is encoded in the matrix $A$ . The eigenvalues of $A$ determine the behavior of the system, and the eigenvectors (or generalized eigenvectors) determine the directions in which these behaviors occur.

Motivation

Recall that for a Constant-Coefficient Linear ODEs, we typically seek an exponential trial solution of the form

e^{rt}

. Briefly review the derivation.

For the homogeneous system

$X'=AX,$

we use the analogous exponential ansatz

$X(t)=e^{\lambda t}\vec{v},$

where $\vec{v}$ is a nonzero constant vector. Differentiating gives

$X'(t)=\lambda \vec{v} e^{\lambda t}.$

Substituting into $X'=AX$ and cancelling the always nonzero factor $e^{\lambda t}$ yields

$\lambda v=Av \quad\Longleftrightarrow\quad (A-\lambda I)v=0.$

so nontrivial solutions exist only when

$\det(A-\lambda I)=0.$

You may recall that this is precisely the characteristic equation of the matrix $A$ . Solving the equation for $\lambda$ retrieves the eigenvalues of $A$ .

For each eigenvalue-eigenvector pair $(\lambda,\vec{v})$ , we obtain a solution $X(t)=e^{\lambda t}\vec{v}$ . The eigenvector $\vec{v}$ specifies the direction in space along which the exponential behavior $e^{\lambda t}$ occurs.

As with scalar equations, there are three important cases for systems. In each case, the solution is built from eigenvalues and (generalized) eigenvectors, in direct analogy with characteristic roots and repeated roots in the one-dimensional setting.

Distinct Real Eigenvalues

Assume $A\in\mathbb{R}^{n\times n}$ has $n$ distinct real eigenvalues $\lambda_1,\dots,\lambda_n$ with corresponding eigenvectors $\vec{v_1},\dots,\vec{v_n}$ . Then the $n$ vector functions

$X_i(t)=e^{\lambda_i t}\vec{v_i},\qquad i=1,\dots,n$

are linearly independent solutions of $X'=AX$ . The general solution is the linear combination

$X(t)=c_1 e^{\lambda_1 t}\vec{v_1}+\cdots+c_n e^{\lambda_n t}\vec{v_n},$

where the constants $c_1,\dots,c_n$ are determined by an initial condition $X(0)=X_0$ .

Example

Consider the initial value problem

$X'= \begin{pmatrix} 3 & 1 \\ 0 & 2 \end{pmatrix}X, \qquad X(0)= \begin{pmatrix} 1 \\ 1 \end{pmatrix}.$

Since $A$ is upper triangular, its eigenvalues are the diagonal entries $\lambda_1=3$ and $\lambda_2=2$ .

For $\lambda_1=3$ , we solve $(A-3I)v=0$ :

$\begin{pmatrix} 0 & 1 \\ 0 & -1 \end{pmatrix}v=0 \quad\Rightarrow\quad v_1= \begin{pmatrix} 1 \\ 0 \end{pmatrix}.$

For $\lambda_2=2$ , we solve $(A-2I)v=0$ :

$\begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}v=0 \quad\Rightarrow\quad v_2= \begin{pmatrix} 1 \\ -1 \end{pmatrix}.$

Thus the general solution is

$X(t)=c_1 e^{3t} \begin{pmatrix} 1 \\ 0 \end{pmatrix} +c_2 e^{2t} \begin{pmatrix} 1 \\ -1 \end{pmatrix}.$

Applying the initial condition and solving for $c_1$ and $c_2$

$X(0)= \begin{pmatrix} 1 \\ 1 \end{pmatrix}$

gives $c_2=-1$ and $c_1=2$ , so

$X(t)= \begin{pmatrix} 2e^{3t}-e^{2t} \\ e^{2t} \end{pmatrix}.$

Repeated Eigenvalues (Generalized Eigenvectors)

Suppose $\lambda$ is an eigenvalue of $A$ with multiplicity $m>1$ . There are two cases for the eigenvectors:

$m$ eigenvectors: we get $m$ linearly independent eigenvectors.
$n<m$ eigenvectors: we get $n$ linearly independent eigenvectors, where $n<m$ .

Recall that we will need $m$ linearly independent eigenvectors to create a basis. We first count how many eigenvectors we have, and then use generalized eigenvectors to make up the difference.

To find eigenvectors for $\lambda$ , recall that we solve the homogeneous linear system

$(A-\lambda I)v=0.$

Row-reducing $A-\lambda I$ to RREF can let us count the number of eigenvectors relatively quickly: if the RREF has $p$ pivots, then there are $n-p$ free variables, so we get $n-p$ linearly independent eigenvectors for $\lambda$ .

If there are $n-p < m$ eigenvectors, the eigenvectors will not give us a complete basis of solutions. To complete the basis for our solution space, we must find generalized eigenvectors.

Let $\vec{v}$ be an eigenvector for $\lambda$ , so $(A-\lambda I)\vec{v}=0$ . A generalized eigenvector $\vec{w}$ is a vector that solves the linear system

$(A-\lambda I)w=v.$

Once we have $\vec{v}$ and $\vec{w}$ , we obtain two independent solutions

$X_1(t)=e^{\lambda t}\vec{v}, \qquad X_2(t)=e^{\lambda t}(\vec{v}t+\vec{w}).$

Then the general solution contributed by this repeated eigenvalue is

$X(t)=c_1 X_1(t)+c_2 X_2(t) =c_1 e^{\lambda t}v+c_2 e^{\lambda t}(tv+w).$

If we still need more eigenvectors, we can continue the process by solving $(A-\lambda I)
\vec{u}=\vec{w}$, and so on. Each new generalized eigenvector produces another solution with a higher power of $t$ multiplying $e^{\lambda t}$ .

In general, if $\lambda$ has algebraic multiplicity $m$ and we can build generalized eigenvectors $\vec{v_1}, \vec{v_2}, \ldots, \vec{v_m}$ with the recurrence relation

$(A-\lambda I)\vec{v_1}=0,\qquad (A-\lambda I) \vec{v_{i}}=\vec{v_{i-1}}\ \ (i=2,\dots,m),$

then we obtain $m$ independent solutions

$X_i(t)=e^{\lambda t}\left(\frac{t^{i-1}}{(i-1)!}v_1+\frac{t^{i-2}}{(i-2)!}v_2+\cdots+t v_{i-1}+v_i\right), \qquad i=1,\dots,m,$

We can think of this as an analogy to the exponential response formula.

Example

Consider the homogeneous system

$X' = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}X.$

The characteristic polynomial is $(2-\lambda)^2$ , so $A$ has a repeated eigenvalue $\lambda=2$ .

Solving $(A-2I)v=0$ gives

$A-2I = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \qquad \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}v=0 \implies v = \begin{pmatrix} 1 \\ 0 \end{pmatrix}.$

Note that as there is 1 pivot, we have $n-p \to 2 - 1 = 1$ linearly independent eigenvectors.

To obtain a second independent solution, we find a generalized eigenvector $w$ by solving the linear system

$(A-2I)w=v.$

Here

$A-2I = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \qquad v = \begin{pmatrix} 1 \\ 0 \end{pmatrix},$

so $w$ must satisfy

$\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} w_1 \\ w_2 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}.$

From the first row we get $w_2=1$ , and $w_1$ is free, so one convenient choice is

$w = \begin{pmatrix} 0 \\ 1 \end{pmatrix}.$

Then two independent solutions are

$X_1(t) = e^{2t}v, \qquad X_2(t) = e^{2t}(tv+w),$

and therefore the general solution is

$X(t) = c_1 X_1(t) + c_2 X_2(t) = c_1 e^{2t}v + c_2 e^{2t}(tv+w).$

Complex Eigenvalues

For real matrices $A$ , nonreal eigenvalues occur in conjugate pairs $\lambda=\alpha\pm i\beta$ with $\beta\neq 0$ . If $v=p+iq$ is an eigenvector for $\alpha+i\beta$ (with $p,q\in\mathbb{R}^n$ ), then the complex solution

$X(t)=e^{(\alpha+i\beta)t}(p+iq)$

has real and imaginary parts that are real solutions. Using Euler's formula, we obtain the real solution pair

$X_1(t) = e^{\alpha t}\big(p\cos(\beta t)-q\sin(\beta t)\big),\\ X_2(t) = e^{\alpha t}\big(p\sin(\beta t)+q\cos(\beta t)\big).$

This same construction generalizes directly to $n$ dimensions: each complex conjugate eigenvalue pair produces two linearly independent real solutions obtained from the real and imaginary parts of the corresponding complex solution.

If $\alpha\pm i\beta$ is a single complex conjugate pair, then the general (real) solution contributed by this pair is

$X(t)=c_1 X_1(t)+c_2 X_2(t),$

with real constants $c_1,c_2$ .

Example

Consider

$X'= \begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}X.$

The eigenvalues are $\lambda=\pm i$ (a complex conjugate pair). Using the formulas from the complex-eigenvalue case, we obtain two real solutions

$X_1(t)=\begin{pmatrix}\cos t\\ \sin t\end{pmatrix}, \qquad X_2(t)=\begin{pmatrix}-\sin t\\ \cos t\end{pmatrix},$

and therefore the general solution is

$X(t)=c_1\begin{pmatrix}\cos t\\ \sin t\end{pmatrix}+c_2\begin{pmatrix}-\sin t\\ \cos t\end{pmatrix}$