Systems of Linear Differential Equations

Coupled equations written as $x^{'} = A x$ , solved by the eigenvalues and eigenvectors of $A$ : each eigenpair contributes a solution $e^{λ t} v$ , and the eigenvalues' signs and imaginary parts decide whether trajectories decay, grow, or circulate.

Definition

A first-order linear system with constant coefficients couples several unknown functions through their derivatives:

\begin{aligned} x_{1}^{'} & = a_{11} x_{1} + a_{12} x_{2} + \dots + a_{1 n} x_{n} \\ ⋮ \\ x_{n}^{'} & = a_{n 1} x_{1} + a_{n 2} x_{2} + \dots + a_{n n} x_{n}, \end{aligned}

which in matrix form is

x^{'} = A x, x (t) = (\begin{matrix} x_{1} (t) \\ ⋮ \\ x_{n} (t) \end{matrix}) .

Coupled means no equation can be solved alone: $x_{1}^{'}$ depends on $x_{2}$ , whose own rate depends on $x_{1}$ .

The eigenvalue solution. Try $x = e^{λ t} v$ for a constant vector $v \neq 0$ . Then $x^{'} = λ e^{λ t} v$ and $A x = e^{λ t} A v$ , so the system holds exactly when

A v = λ v .

That is the eigenvalue equation. Each eigenpair $(λ, v)$ of $A$ gives one solution $e^{λ t} v$ , and if $A$ has $n$ linearly independent eigenvectors the general solution is

x = c_{1} e^{λ_{1} t} v_{1} + \dots + c_{n} e^{λ_{n} t} v_{n},

with the $n$ constants fixed by the $n$ components of an initial vector $x (t_{0})$ .

Eigenvalues in the three shapes.

Eigenvalues of $A$	Contribution	Trajectory
real $λ$ , distinct	$e^{λ t} v$	decays toward $0$ if $λ < 0$ , grows if $λ > 0$
complex $α \pm β i$	$e^{α t} (\cos β t, \sin β t)$ combinations	circulates; spirals in if $α < 0$ , out if $α > 0$ , closed orbit if $α = 0$
repeated with too few eigenvectors	$e^{λ t} (w + t v)$	a factor of $t$ appears, as in the scalar repeated-root case

Defective matrices. A repeated eigenvalue need not supply two independent eigenvectors. When it does not, when the geometric multiplicity is less than the algebraic multiplicity, the matrix is defective, and the missing solution takes the form $e^{λ t} (w + t v)$ where $(A - λ I) w = v$ .

Every higher-order equation is a system. Setting $x_{1} = y$ , $x_{2} = y^{'}$ turns $y^{″} + p y^{'} + q y = 0$ into

x^{'} = (\begin{matrix} 0 & 1 \\ - q & - p \end{matrix}) x,

whose characteristic polynomial is the same $λ^{2} + p λ + q$ . The two theories are one theory.

Assumptions and scope

The matrix $A$ must be constant. With entries depending on $t$ , the trial $e^{λ t} v$ leaves $t$ in the equation and the eigenvalue method does not apply.
The general solution needs $n$ linearly independent solutions for an $n \times n$ system, which requires $n$ independent eigenvectors, not merely $n$ eigenvalues counted with multiplicity.
A repeated eigenvalue may or may not supply enough eigenvectors. Whether the matrix is defective must be checked by computing the rank of $A - λ I$ , never assumed from the multiplicity alone.
Complex eigenvalues of a real matrix arrive in conjugate pairs, and one pair yields two real solutions, so a conjugate pair should not be counted twice when tallying independent solutions.
Eigenvectors are determined only up to a nonzero scalar. Scaling one rescales the corresponding constant $c_{i}$ and changes no solution.
These methods solve the homogeneous system. A forcing term $x^{'} = A x + f (t)$ needs a particular solution added, as in the scalar inhomogeneous case.

Worked material

Example

A system with no real eigenvalues

The worked example had two real eigenvalues, so both solutions were exponentials along fixed directions. Here is the other kind of $2 \times 2$ system: one with no real eigenvalues at all.

Problem. Solve $x^{'} = B x$ with

B = (\begin{matrix} 0 & - 2 \\ 2 & 0 \end{matrix}),

that is $x_{1}^{'} = - 2 x_{2}$ and $x_{2}^{'} = 2 x_{1}$ .

Eigenvalues. $tr B = 0$ and $det B = 0 - (- 4) = 4$ , so the characteristic equation is

λ^{2} - 0 \cdot λ + 4 = λ^{2} + 4 = 0 ⟹ λ = \pm 2 i .

Check against the invariants. Sum $2 i + (- 2 i) = 0 = tr B$ ; product $(2 i) (- 2 i) = - 4 i^{2} = 4 = det B$ .

No real eigenvalue exists, which says something geometric: no direction is preserved. Every vector is turned by this matrix, so no solution can keep a fixed direction and merely scale, and the real-exponential solutions of the worked example have no counterpart here.

Eigenvector for $λ = 2 i$ . Solving $(B - 2 i I) v = 0$ :

(\begin{matrix} - 2 i & - 2 \\ 2 & - 2 i \end{matrix}) (\binom{v_{1}}{v_{2}}) = (\binom{0}{0}) .

The first row gives $- 2 i v_{1} - 2 v_{2} = 0$ , so $v_{2} = - i v_{1}$ . Taking $v_{1} = 1$ :

v = (\binom{1}{- i}) .

The second row is $2 v_{1} - 2 i v_{2} = 2 - 2 i (- i) = 2 - 2 = 0$ , consistent, as it must be, since $B - 2 i I$ is singular.

Real solutions. Expand $e^{2 i t} v$ by Euler's identity:

e^{2 i t} (\binom{1}{- i}) = (\cos 2 t + i \sin 2 t) (\binom{1}{- i}) = (\binom{\cos 2 t + i \sin 2 t}{\sin 2 t - i \cos 2 t}),

using $- i (\cos 2 t + i \sin 2 t) = \sin 2 t - i \cos 2 t$ . Real and imaginary parts give two real solutions, and the general solution is

x = c_{1} (\binom{\cos 2 t}{\sin 2 t}) + c_{2} (\binom{\sin 2 t}{- \cos 2 t}) .

The conjugate eigenvalue $- 2 i$ is not processed separately: it yields these same two solutions, so the pair contributes two and the $2 \times 2$ system is complete.

Verification. For the first solution, $x = (\cos 2 t, \sin 2 t)$ gives $x^{'} = (- 2 \sin 2 t, 2 \cos 2 t)$ , while

B x = (\binom{- 2 \sin 2 t}{2 \cos 2 t}),

identical. Numerically $‖ x^{'} - B x ‖$ stays below $2.6 \times 10^{- 11}$ at $t = 0, 0.5, 1$ .

What the trajectories look like. The norm is

‖ x ‖ = \sqrt{\cos^{2} 2 t + \sin^{2} 2 t} = 1

for every $t$ , confirmed as exactly $1.0$ at $t = 0, 1, 2$ . The trajectory is a circle of constant radius, traversed with angular frequency 2 and period $π$ .

That constancy is the real part being zero. The general complex case contributes $e^{α t}$ times these trigonometric terms, so:

$α = Re λ$	Trajectory
$< 0$	spirals inward to the origin
$= 0$	closed orbit, constant radius
$> 0$	spirals outward without bound

Here $α = 0$ exactly, which is the knife-edge case: the system neither gains nor loses, and circles forever. Perturbing $B$ slightly, adding $- 0.1$ to each diagonal entry, say, would move the eigenvalues to $- 0.1 \pm 2 i$ and turn every circle into an inward spiral.

Comparison with the worked example.

	$A = (\begin{matrix} 1 & 2 \\ 2 & 1 \end{matrix})$	$B = (\begin{matrix} 0 & - 2 \\ 2 & 0 \end{matrix})$
Eigenvalues	$3, - 1$ (real, distinct)	$\pm 2 i$ (pure imaginary)
Invariant directions	two lines, $(\binom{1}{1})$ and $(\binom{1}{- 1})$	none
Solutions	real exponentials	sines and cosines
Long run	escapes along $(\binom{1}{1})$	circles forever

Both were solved by the same three steps, characteristic equation, eigenvectors, combine, and the difference in behaviour was settled by the eigenvalues before any solution was written down.

Contrast

Two matrices with the same repeated eigenvalue

A repeated eigenvalue does not by itself determine the shape of the solution. These two matrices both have the single eigenvalue 3, doubled, and their solutions differ.

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Case 1 — not defective.

A_{1} = (\begin{matrix} 3 & 0 \\ 0 & 3 \end{matrix}) = 3 I .

Characteristic polynomial $(3 - λ)^{2}$ , so $λ = 3$ with algebraic multiplicity 2.

A_{1} - 3 I = (\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}), rank = 0 .

The kernel is all of $R^{2}$ , so every nonzero vector is an eigenvector and the geometric multiplicity is 2. Taking the standard basis,

x = c_{1} e^{3 t} (\binom{1}{0}) + c_{2} e^{3 t} (\binom{0}{1}) = e^{3 t} (\binom{c_{1}}{c_{2}}) .

No $t$ factor appears. Both components simply scale by $e^{3 t}$ , and any initial vector is met by reading off its components.

---

Case 2 — defective.

A_{2} = (\begin{matrix} 3 & 1 \\ 0 & 3 \end{matrix}) .

Being triangular, the characteristic polynomial is again $(3 - λ)^{2}$ . The same eigenvalue with the same multiplicity. But

A_{2} - 3 I = (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}), rank = 1,

confirmed numerically. The kernel is one-dimensional, spanned by $v = (\binom{1}{0})$ : the geometric multiplicity is 1, and the matrix is defective.

Here $c_{1} e^{3 t} v + c_{2} e^{3 t} v$ collapses to $(c_{1} + c_{2}) e^{3 t} v$ , one constant with two names, and a family confined to the line $x_{2} = 0$ , which cannot meet an initial vector such as $(\binom{1}{2})$ at all. The second solution must come from a generalised eigenvector: solving $(A_{2} - 3 I) w = v$ gives $w = (\binom{0}{1})$ , so

x = c_{1} e^{3 t} (\binom{1}{0}) + c_{2} e^{3 t} [(\binom{0}{1}) + t (\binom{1}{0})] .

With $c_{1} = 1$ , $c_{2} = 2$ this is $e^{3 t} (1 + 2 t, 2)$ , whose residual against $x^{'} = A_{2} x$ stays below $1.6 \times 10^{- 8}$ .

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What differs, and what does not.

	$A_{1} = 3 I$	$A_{2} = (\begin{matrix} 3 & 1 \\ 0 & 3 \end{matrix})$
Eigenvalue	3, twice	3, twice
Algebraic multiplicity	2	2
$rank (A - 3 I)$	0	1
Geometric multiplicity	2	1
Defective	no	yes
Solution	$e^{3 t} (\binom{c_{1}}{c_{2}})$	$e^{3 t} [c_{1} v + c_{2} (w + t v)]$
$t$ factor	none	present

The characteristic polynomial is identical; everything that distinguishes the two lives in the rank of $A - λ I$ , which the polynomial cannot see. That is why step 4 of the procedure is a computation and not an inference.

The stability reading survives either way. Both have eigenvalue 3 with positive real part, so both grow without bound, $t$ factor or not. And had the eigenvalue been $- 3$ , both would decay, since $t e^{- 3 t} \to 0$ , a polynomial factor cannot defeat a decaying exponential. Defectiveness changes the shape of the approach, never its direction.

The same distinction, one unit earlier. For scalar equations this is the difference between distinct roots and a repeated root needing $x e^{r x}$ . The companion matrix of $y^{″} - 6 y^{'} + 9 y = 0$ is defective, and its $t v$ term is that $x e^{3 x}$ .

Common errors

Common misconception

A repeated eigenvalue of a $2 \times 2$ matrix always supplies two independent eigenvectors, so the general solution is $c_{1} e^{λ t} v_{1} + c_{2} e^{λ t} v_{2}$ with no $t$ term needed.

Common misconception

The long-run behaviour of $x^{'} = A x$ can be read from the diagonal entries of $A$ , so a matrix with negative entries on the diagonal describes a system that decays.

Systems of Linear Differential Equations

Assumptions and scope

Worked material

A system with no real eigenvalues

Two matrices with the same repeated eigenvalue

Common errors

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