Week 11: Differential Equations

Systems of Differential Equations#

A single scalar equation $\frac{du}{dt} = a u$ has the familiar solution $u(t) = c\,e^{a t}$. The constant $a$ is a growth rate: negative means decay to $0$, positive means explosive growth. Many models couple several unknowns at once — concentrations of chemicals, positions of masses on springs, currents in a circuit, or components of a robot state. Those systems are still first-order and linear, but the right-hand side is a matrix times a vector:

Here $u(t) \in \mathbb{R}^n$ is the state at time $t$, and $A$ is a fixed $n \times n$ matrix of constant coefficients. Component-wise this is $n$ equations

so each rate depends on every component when $A$ is dense. The equations are coupled: you cannot solve for $u_1$ without knowing $u_2$, and vice versa.

Why eigenvalues enter. Week 10 diagonalized $A$ as $A = S\Lambda S^{-1}$ whenever $A$ has $n$ independent eigenvectors. Change variables to the eigen-coordinates $v = S^{-1}u$ (so $u = Sv$). Differentiating and substituting:

Because $\Lambda$ is diagonal, the system for $v$ decouples into $n$ independent scalar ODEs:

Each eigen-direction evolves on its own with its own exponential rate $\lambda_i$. Transforming back with $u = Sv$ reassembles the solution in the original coordinates. That is the whole strategy of this week: diagonalize, solve $n$ scalar problems, change back.

Initial condition. The system is first-order, so one vector $u(0) = u_0$ pins down the solution completely. In eigen-coordinates that means $v(0) = S^{-1}u_0 = c$, the vector of coefficients $c_i$.

The Solution Formula#

Suppose $A$ is diagonalizable with eigenvalues $\lambda_1,\ldots,\lambda_n$ and corresponding independent eigenvectors $x_1,\ldots,x_n$ (columns of $S$). The general solution of $\frac{du}{dt} = Au$ is

Each term is an eigenvector times a scalar exponential whose rate is the matching eigenvalue. The constants $c_1,\ldots,c_n$ are fixed by the initial condition:

In matrix form that is $S c = u_0$, so $c = S^{-1}u_0$ — solve one linear system once, then the solution is known for every $t$.

Matrix exponential view. The same formula can be written compactly as $u(t) = e^{At}\, u_0$. When $A = S\Lambda S^{-1}$,

Expanding $S e^{\Lambda t} S^{-1} u_0$ recovers the sum of $c_i e^{\lambda_i t} x_i$ above. You do not need the matrix-exponential series for this course — the eigen-sum is the practical formula — but it explains why diagonalization is the natural tool for continuous-time dynamics just as it was for discrete powers $A^k$ in Week 10.

Complex eigenvalues. If a real matrix $A$ has a complex conjugate pair $\alpha \pm i\omega$, the corresponding complex exponential $e^{(\alpha\pm i\omega)t} = e^{\alpha t}(\cos\omega t \pm i\sin\omega t)$ combines into real solutions that oscillate at frequency $\omega$ while growing or decaying with envelope $e^{\alpha t}$. The real part $\alpha$ still governs stability; the imaginary part sets the oscillation rate.

Stability#

Long-run behavior of every solution is read from the eigenvalues of $A$. Write each eigenvalue as $\lambda = \alpha + i\omega$ with real part $\alpha = \operatorname{Re}(\lambda)$. Then $|e^{\lambda t}| = e^{\alpha t}$, so:

| Eigenvalues of $A$ | Long-run behavior of $u(t)$ | Classification | |---|---|---| | Every $\operatorname{Re}(\lambda) < 0$ | All terms $e^{\lambda t}x$ decay to $0$ | Stable (asymptotically stable origin) | | Some $\operatorname{Re}(\lambda) > 0$ | At least one term grows without bound | Unstable | | Some $\operatorname{Re}(\lambda) = 0$, none positive | Oscillation or constants (no decay, no growth from those modes) | Marginally stable / neutrally stable (with care) |

The decisive rule for this course:

• Stable if every eigenvalue has negative real part — every solution $\to 0$ as $t \to \infty$.
• Unstable if any eigenvalue has positive real part — generic solutions explode.
• Purely imaginary eigenvalues ($\operatorname{Re}(\lambda) = 0$, $\operatorname{Im}(\lambda) \neq 0$) produce sustained oscillation: $e^{i\omega t}$ neither grows nor decays. Pure zero eigenvalues produce constant modes along the corresponding eigenvectors.

A single unstable eigenvalue ruins the system: even if ten modes decay, the one growing mode dominates for large $t$. Conversely, stability requires all eigenvalues to lie in the open left half-plane.

Intuition from the scalar case. For $\frac{du}{dt} = a u$, stability is simply $a < 0$. The matrix case is the same test applied to every eigen-direction independently — that is why diagonalization converts a hard coupled question into $n$ easy sign checks.

Preview of Week 12. When $A$ is symmetric (real), every eigenvalue is real, so there is no oscillatory pair. Stability then reduces to “are all eigenvalues negative?” Symmetric matrices with nonpositive eigenvalues are a clean special case of the table above; Week 12 develops the spectral theory that makes those eigenvalues especially well-behaved.

Example: A Two-Component System#

Work a full $2 \times 2$ example end to end. Let

Step 1 — eigenvalues. Trace $\operatorname{tr}(A) = -6$ and $\det(A) = 9 - 1 = 8$, so

Eigenvalues: $\lambda_1 = -2$, $\lambda_2 = -4$. Both are real and negative, so the origin is stable — every solution will decay to $0$.

Step 2 — eigenvectors. For $\lambda_1 = -2$:

For $\lambda_2 = -4$:

So

Step 3 — coefficients from $u(0)$.

Step 4 — solution formula.

Check $u(0) = (1,0)^\top$: both components match. As $t \to \infty$, both exponentials vanish, so $u(t) \to 0$, confirming stability from the signs of $\lambda$.

Contrast — an unstable matrix. Replace $A$ by $-A$:

has eigenvalues $+2$ and $+4$. The same algebra with $e^{+2t}$ and $e^{+4t}$ produces solutions that grow without bound — unstable. The eigenvectors stay the same; only the signs of the rates flip.

Contrast — pure oscillation. The rotation generator

has eigenvalues $\pm i$. Solutions are rotations: $u(t)$ circles the origin at constant radius (sustained oscillation, neither decay nor growth). Purely imaginary eigenvalues sit on the stability boundary.

Knowledge check#

Browser lab#

Simulate a stable and an unstable system side by side.

import numpy as np
import matplotlib.pyplot as plt

A_stable = np.array([[-1., 0.], [0., -3.]])
A_unstable = np.array([[1., 0.], [0., 2.]])

def simulate(A, u0, t):
    eigvals, eigvecs = np.linalg.eig(A)
    c = np.linalg.solve(eigvecs, u0)
    U = np.zeros((len(t), 2))
    for i, ti in enumerate(t):
        U[i] = (eigvecs @ (c * np.exp(eigvals * ti))).real
    return U

t = np.linspace(0, 3, 100)
u0 = np.array([1.0, 1.0])

U_stable = simulate(A_stable, u0, t)
U_unstable = simulate(A_unstable, u0, t)

fig, axes = plt.subplots(1, 2, figsize=(10, 4))
axes[0].plot(t, U_stable)
axes[0].set_title("Stable: eigenvalues -1, -3")
axes[1].plot(t, U_unstable)
axes[1].set_title("Unstable: eigenvalues 1, 2")
plt.tight_layout()
plt.show()

The left panel decays to zero (negative eigenvalues). The right panel explodes (positive eigenvalues). The lab uses the eigen-sum formula from this lesson: solve $S c = u_0$, then plot $S(c \odot e^{\lambda t})$ at each $t$.

Quiz#