The Eigenvalue Problem

Why Exponentials?

We want to solve \(\mathbf{x}' = A\mathbf{x}\), where \(A\) is a constant \(2\times 2\) matrix.

Key observation: for all terms to cancel when substituted into the ODE, the solution and its derivative must have the same form. Exponential functions satisfy this.

Example 4.27 — Finding Eigenpairs

\[A = \begin{pmatrix}1&1\\4&1\end{pmatrix}, \quad \text{tr}\,A=2, \quad \det A = -3.\]

Characteristic equation: \(\lambda^2 - 2\lambda - 3 = (\lambda+1)(\lambda-3) = 0\).

Eigenvalues: \(\lambda_1 = -1\), \(\lambda_2 = 3\).

s = sym.Symbol('lam')
A = sym.Matrix([[1,1],[4,1]])
display(Math(r"\text{Char. poly: }" + sym.latex(A.charpoly(s).as_expr())))
for val, mult, vecs in A.eigenvects():
    display(Math(rf"\lambda={val},\;\mathbf{{v}}={sym.latex(vecs[0])}"))

Example 4.28 — Complex Eigenvalues

\[A = \begin{pmatrix}-2 & -3\\3 & -2\end{pmatrix}, \quad \text{tr}\,A=-4, \quad \det A = 13.\]

Characteristic equation: \(\lambda^2 + 4\lambda + 13 = 0 \implies \lambda = -2 \pm 3i\).

Solving Linear Systems

The Master Strategy

Each eigenpair \(\lambda,\,\mathbf{v}\) gives a solution \(\mathbf{x}(t) = \mathbf{v}e^{\lambda t}\).

The structure of the solution — and the phase portrait — depends entirely on the nature of the eigenvalues:

General Solution and Linear Orbits

If \(\lambda_1 \neq \lambda_2\) are real with eigenvectors \(\mathbf{v}_1,\,\mathbf{v}_2\) (independent), the general solution is

\[\mathbf{x}(t) = c_1\mathbf{v}_1 e^{\lambda_1 t} + c_2\mathbf{v}_2 e^{\lambda_2 t}. \tag{4.34}\]

Saddle Points — Opposite Signs

When \(\lambda_1 < 0 < \lambda_2\):

• Linear orbits along \(\mathbf{v}_1\): approach origin (\(\lambda_1 < 0\))
• Linear orbits along \(\mathbf{v}_2\): leave origin (\(\lambda_2 > 0\))
• All other orbits: hyperbolic curves

Example 4.30 — Saddle Point

\[\mathbf{x}' = \begin{pmatrix}1&1\\4&1\end{pmatrix}\mathbf{x}, \quad \lambda_1=-1,\;\mathbf{v}_1=\begin{pmatrix}1\\-2\end{pmatrix},\quad \lambda_2=3,\;\mathbf{v}_2=\begin{pmatrix}1\\2\end{pmatrix}.\]

\[\mathbf{x}(t) = c_1\begin{pmatrix}1\\-2\end{pmatrix}e^{-t} + c_2\begin{pmatrix}1\\2\end{pmatrix}e^{3t}. \tag{4.35}\]

Stable and Unstable Nodes

Both \(\lambda_1 < \lambda_2 < 0\) (stable node):

• All orbits enter origin as \(t\to+\infty\)
• Orbits tangent to slow eigendirection (\(\mathbf{v}_2\), eigenvalue closer to 0) as \(t\to+\infty\)
• Orbits parallel to fast eigendirection (\(\mathbf{v}_1\)) as \(t\to-\infty\)

\(\det A = 0\), \(\lambda=0\):

• Non-isolated equilibria along entire line \(ax+by=0\)
• Other orbits are parallel straight lines \(y=mx+C\)

From Complex Eigenpairs to Real Solutions

If \(\lambda = a \pm bi\) (complex), write the eigenvector as \(\mathbf{v} = \mathbf{w} \pm i\mathbf{z}\).

Expand one complex solution via Euler’s formula \(e^{ibt} = \cos bt + i\sin bt\):

\[(\mathbf{w}+i\mathbf{z})e^{(a+bi)t} = e^{at}(\mathbf{w}\cos bt - \mathbf{z}\sin bt) + i\,e^{at}(\mathbf{w}\sin bt + \mathbf{z}\cos bt).\]

Orbital Behavior: Spirals and Centers

The factor \(e^{at}\) controls amplitude; the trig terms produce rotation:

Example 4.37 — Asymptotically Stable Spiral

\[\mathbf{x}' = \begin{pmatrix}-2&-3\\3&-2\end{pmatrix}\mathbf{x}, \quad \lambda=-2\pm 3i, \quad \mathbf{w}=\begin{pmatrix}-1\\0\end{pmatrix},\; \mathbf{z}=\begin{pmatrix}0\\1\end{pmatrix}.\]

Two real independent solutions:

\[\mathbf{x}_1(t) = e^{-2t}\begin{pmatrix}-\cos 3t\\-\sin 3t\end{pmatrix}, \quad \mathbf{x}_2(t) = e^{-2t}\begin{pmatrix}-\sin 3t\\-\cos 3t\end{pmatrix}.\]

Since \(a = -2 < 0\): asymptotically stable spiral. At \((1,1)\): \((x',y')=(-5,1)\) — spirals are counterclockwise. \(\square\)

Two Subcases

Suppose \(\lambda_1 = \lambda_2 \equiv \lambda \neq 0\). There are two cases:

Case 1 — Non-deficient: \(A\) has two independent eigenvectors \(\mathbf{v}_1,\,\mathbf{v}_2\).

\[\mathbf{x}(t) = c_1\mathbf{v}_1 e^{\lambda t} + c_2\mathbf{v}_2 e^{\lambda t}.\]

Every direction through the origin is a linear orbit — a star node.

Deficient Case: Why \(t\mathbf{v}e^{\lambda t}\) Alone Fails

Intuition from Chapter 2: for scalar equations with repeated roots, the second solution is \(te^{\lambda t}\). For systems, the direct analog \(t\mathbf{v}e^{\lambda t}\) does not work.

Example 4.39 — Deficient Matrix

\[\mathbf{x}' = \begin{pmatrix}2&1\\-1&4\end{pmatrix}\mathbf{x}, \quad \lambda=3,3, \quad \mathbf{v}=\begin{pmatrix}1\\1\end{pmatrix}.\]

First solution: \(\mathbf{x}_1(t) = \begin{pmatrix}1\\1\end{pmatrix}e^{3t}\) (linear orbit along \(y=x\)).

Find \(\mathbf{w}\): solve \((A-3I)\mathbf{w} = \mathbf{v}\):

\[\begin{pmatrix}-1&1\\-1&1\end{pmatrix}\mathbf{w} = \begin{pmatrix}1\\1\end{pmatrix} \implies -w_1+w_2=1.\]

Choose \(\mathbf{w} = (0,1)^T\).

Example 4.39 — Phase Portrait

Key Takeaways

• The eigenvalue ansatz \(\mathbf{x} = \mathbf{v}e^{\lambda t}\) converts \(\mathbf{x}' = A\mathbf{x}\) into the algebraic problem \(A\mathbf{v} = \lambda\mathbf{v}\). Every eigenpair gives a solution.

• Eigenvalues come from the characteristic equation \(\lambda^2 - (\text{tr}\,A)\lambda + \det A = 0\); eigenvectors from solving \((A-\lambda I)\mathbf{v}=\mathbf{0}\).

• Real unequal eigenvalues give linear orbits (rays) and either a saddle (opposite signs) or a node (same sign). Orbits of a node enter/exit the origin tangent to the slow eigenvector.

• Complex eigenvalues \(\lambda=a\pm bi\) give oscillatory solutions via Euler’s formula. The sign of \(a\) determines whether orbits spiral in (stable), spiral out (unstable), or close (center).

• Repeated eigenvalues with two independent eigenvectors give a star node. If the matrix is deficient (one eigenvector), a generalized eigenvector \(\mathbf{w}\) satisfying \((A-\lambda I)\mathbf{w}=\mathbf{v}\) is needed, giving the second solution \(e^{\lambda t}(t\mathbf{v}+\mathbf{w})\) — a degenerate node.

• In all cases, stability is determined by the sign of the real part of the eigenvalues: negative real parts → stable; any positive real part → unstable; zero real part → inconclusive (or center).