5.7 Applications to differential equations

Big idea

A single equation $x' = ax$ has the solution $x(t) = ce^{at}$, and that is the entire theory. A system couples several unknown functions, each derivative depending on all the functions at once, and the single-equation trick no longer applies directly because you cannot solve one equation without the others.

Eigenvectors uncouple the system. Look for a solution that keeps a fixed direction and only changes length: $\mathbf{x}(t) = e^{\lambda t}\mathbf{v}$ for a constant vector $\mathbf{v}$. Differentiating gives $\lambda e^{\lambda t}\mathbf{v}$, and substituting into $\mathbf{x}' = A\mathbf{x}$ demands $\lambda\mathbf{v} = A\mathbf{v}$. So the solutions of that shape are exactly the eigenvectors, and the exponential rates are exactly the eigenvalues.

Because the system is linear, sums of solutions are solutions. With $n$ independent eigenvectors you get $n$ independent solutions, their combinations give every solution, and the initial condition picks out the weights. Same three steps as the discrete case, with $e^{\lambda t}$ replacing $\lambda^k$.

Decoder

The functions $\mathbf{x}(t) = e^{\lambda t}\mathbf{v}$ solve $\mathbf{x}' = A\mathbf{x}$ precisely when $\lambda$ is an eigenvalue of $A$ and $\mathbf{v}$ a corresponding eigenvector.

Read it as a search rather than a claim. You guess that some solution keeps a fixed direction, you plug the guess in, and the equation you are left with is the eigenvalue equation. Nothing has been assumed about $A$; the guess either produces eigenvectors or produces nothing.

Definitions and results

The system. A system of first-order linear differential equations with constant coefficients, written $\mathbf{x}'(t) = A\mathbf{x}(t)$, where $\mathbf{x}(t)$ is a vector of unknown functions and $\mathbf{x}'(t)$ is the vector of their derivatives. An initial value problem adds a required value $\mathbf{x}(0)$.

Eigen-solutions. For each eigenvalue $\lambda$ with eigenvector $\mathbf{v}$, the function $e^{\lambda t}\mathbf{v}$ solves the system. Along it the state stays on the line through $\mathbf{v}$ forever, growing when $\lambda > 0$, decaying when $\lambda < 0$, and standing still when $\lambda = 0$.

General solution, distinct real eigenvalues. If $A$ is $n \times n$ with $n$ independent eigenvectors $\mathbf{v}_1, \dots, \mathbf{v}_n$ and real eigenvalues $\lambda_1, \dots, \lambda_n$, every solution has the form

$$ \mathbf{x}(t) = c_1e^{\lambda_1 t}\mathbf{v}_1 + \cdots + c_ne^{\lambda_n t}\mathbf{v}_n $$

for constants $c_1, \dots, c_n$. The solution set is a vector space of dimension $n$, and those $n$ functions are a basis for it.

Fitting the initial condition. Setting $t = 0$ makes every exponential equal to $1$, so the requirement becomes $c_1\mathbf{v}_1 + \cdots + c_n\mathbf{v}_n = \mathbf{x}(0)$. That is one linear system in the weights, solved once. An initial value problem with $n$ independent eigenvectors has exactly one solution.

Decoupling. With $A = PDP^{-1}$, substitute $\mathbf{x} = P\mathbf{y}$. Then $P\mathbf{y}' = AP\mathbf{y}$, and multiplying on the left by $P^{-1}$ gives $\mathbf{y}' = D\mathbf{y}$. The rows of that system are separate one-variable equations $y_i' = \lambda_i y_i$, each solved by $y_i = c_ie^{\lambda_i t}$. Converting back with $\mathbf{x} = P\mathbf{y}$ reproduces the general solution. This is what “the eigenvector basis uncouples the system” means concretely.

Classifying the origin. The zero function always solves the system, so the origin is an equilibrium. With real eigenvalues, it is an attractor when all are negative, since every term decays; a repeller when all are positive; and a saddle point when the signs are mixed, with the negative directions feeding in and the positive ones carrying trajectories out.

Which direction dominates. For large positive $t$, the term with the largest eigenvalue dominates and the trajectory lines up with its eigenvector. For large negative $t$, the smallest eigenvalue dominates. In a decaying system all terms vanish, but the least negative eigenvalue decays slowest, so the approach to the origin happens along its eigenvector.

Complex eigenvalues. A conjugate pair $a \pm bi$ produces solutions that spiral: the real and imaginary parts of $e^{\lambda t}\mathbf{v}$ give two real solutions, built from $e^{at}\cos bt$ and $e^{at}\sin bt$. The sign of the real part $a$ decides inward, outward, or a closed orbit, exactly as the modulus did in the discrete case.

Worked examples

A saddle, with an initial condition

Let

$$ A = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}, \qquad \mathbf{x}(0) = \begin{bmatrix} 5 \\ 1 \end{bmatrix} $$

The trace is $2$ and the determinant is $1 - 4 = -3$, so the characteristic polynomial is $\lambda^2 - 2\lambda - 3 = (\lambda - 3)(\lambda + 1)$. For $\lambda = 3$ the shifted matrix has rows $(-2, 2)$, giving the eigenvector $(1,1)$; check $A(1,1) = (3,3)$. For $\lambda = -1$ the rows are $(2,2)$, giving $(1,-1)$; check $A(1,-1) = (1 - 2, \; 2 - 1) = (-1, 1)$, which is $-1$ times the vector.

The general solution is

$$ \mathbf{x}(t) = c_1e^{3t}\begin{bmatrix} 1 \\ 1 \end{bmatrix} + c_2e^{-t}\begin{bmatrix} 1 \\ -1 \end{bmatrix} $$

Set $t = 0$: $c_1 + c_2 = 5$ and $c_1 - c_2 = 1$, so $c_1 = 3$ and $c_2 = 2$.

Check the answer twice. At $t = 0$ it gives $3(1,1) + 2(1,-1) = (5,1)$, the required starting point. Differentiating and setting $t = 0$ gives $9(1,1) - 2(1,-1) = (7, 11)$, while $A\mathbf{x}(0) = (5 + 2, \; 10 + 1) = (7,11)$. The differential equation holds at the start, and by construction it holds everywhere.

The signs are mixed, so the origin is a saddle point. The $e^{-t}$ term dies and the $e^{3t}$ term grows, so the trajectory swings toward the line through $(1,1)$ and runs out along it.

Everything decays

Let

$$ B = \begin{bmatrix} -3 & 1 \\ 1 & -3 \end{bmatrix}, \qquad \mathbf{x}(0) = \begin{bmatrix} 6 \\ 2 \end{bmatrix} $$

The trace is $-6$ and the determinant is $9 - 1 = 8$, so the characteristic polynomial is $\lambda^2 + 6\lambda + 8 = (\lambda + 2)(\lambda + 4)$. For $\lambda = -2$, the shifted matrix has rows $(-1,1)$ and the eigenvector is $(1,1)$; check $B(1,1) = (-2,-2)$. For $\lambda = -4$, the rows are $(1,1)$ and the eigenvector is $(1,-1)$; check $B(1,-1) = (-3 - 1, \; 1 + 3) = (-4, 4)$.

So

$$ \mathbf{x}(t) = c_1e^{-2t}\begin{bmatrix} 1 \\ 1 \end{bmatrix} + c_2e^{-4t}\begin{bmatrix} 1 \\ -1 \end{bmatrix} $$

and the initial condition gives $c_1 + c_2 = 6$, $c_1 - c_2 = 2$, so $c_1 = 4$ and $c_2 = 2$.

Check at $t = 0$: $4(1,1) + 2(1,-1) = (6,2)$. Check the derivative at $t = 0$: $-8(1,1) - 8(1,-1) = (-16, 0)$, while $B\mathbf{x}(0) = (-18 + 2, \; 6 - 6) = (-16, 0)$. Both match.

Both eigenvalues are negative, so the origin is an attractor and $\mathbf{x}(t) \to \mathbf{0}$. The $e^{-4t}$ term vanishes faster, so late in the decay the state is essentially $4e^{-2t}(1,1)$ and the trajectory comes into the origin along the line through $(1,1)$.

The same problem by decoupling

Take $A$ and $\mathbf{x}(0) = (5,1)$ from the first example and set

$$ P = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}, \qquad D = \begin{bmatrix} 3 & 0 \\ 0 & -1 \end{bmatrix} $$

Check the factorization without inverting: $AP$ has columns $A(1,1) = (3,3)$ and $A(1,-1) = (-1,1)$, while $PD$ has columns $3(1,1) = (3,3)$ and $-1(1,-1) = (-1,1)$. So $AP = PD$ and $A = PDP^{-1}$.

Substituting $\mathbf{x} = P\mathbf{y}$ turns the system into $\mathbf{y}' = D\mathbf{y}$, which is two separate equations, $y_1' = 3y_1$ and $y_2' = -y_2$. Their solutions are $y_1 = c_1e^{3t}$ and $y_2 = c_2e^{-t}$, with no coupling left to worry about.

The new initial condition is $\mathbf{y}(0) = P^{-1}\mathbf{x}(0)$. Since $\det P = -2$,

$$ P^{-1} = \begin{bmatrix} 0.5 & 0.5 \\ 0.5 & -0.5 \end{bmatrix}, \qquad \mathbf{y}(0) = \begin{bmatrix} 0.5(5) + 0.5(1) \\ 0.5(5) - 0.5(1) \end{bmatrix} = \begin{bmatrix} 3 \\ 2 \end{bmatrix} $$

So $c_1 = 3$ and $c_2 = 2$, and converting back with $\mathbf{x} = P\mathbf{y}$ gives $3e^{3t}(1,1) + 2e^{-t}(1,-1)$, the answer from the first example. The two routes are the same computation in a different order: decoupling does the change of coordinates up front, the eigen-solution route does it implicitly when solving for the weights.

Practice

First the eigenvalues and eigenvectors of the coefficient matrix, since nothing else can start until those are in hand.

Then the full solution: combine the eigen-solutions and fit the initial condition.

Quiz

Five items on eigenvalues of a coefficient matrix and solutions of systems with distinct real eigenvalues.