From 42d59b1a1a4a8441e157c862ee73031187fc57cc Mon Sep 17 00:00:00 2001
From: Worker Agent <worker@diff-eq-handbook.local>
Date: Thu, 4 Jun 2026 18:44:45 -0500
Subject: [PATCH] ch14: nonlinear systems, linearization, Lotka-Volterra

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+% =============================================================================
+% ch14_nonlinear_systems.tex
+% Chapter 14: Nonlinear Systems
+% =============================================================================
+
 \section{Nonlinear Systems}
 \label{ch:nonlinear_systems}
 
-\subsection{Equilibrium Points}
-\label{sec:ch14_equilibrium}
+Linear systems admit a powerful theory built on superposition and eigenvalue analysis
+(\cref{ch:systems}). When the equations become nonlinear, the principle of superposition
+breaks down: the sum of two solutions is generally \emph{not} a solution. No universal
+closed-form solution formula exists for nonlinear systems. Instead, we rely on
+\textbf{qualitative methods} --- studying the structure of solutions without finding
+explicit formulas.
 
-% Content goes here
+\subsection{Nonlinear Autonomous Systems}
+\label{sec:ch14_autonomous}
+
+A \textbf{nonlinear autonomous system} in two variables takes the form
+\begin{equation}
+  \label{eq:nonlinear_autonomous}
+  \frac{\mathrm{d}x}{\mathrm{d}t} = f(x,y), \qquad
+  \frac{\mathrm{d}y}{\mathrm{d}t} = g(x,y),
+\end{equation}
+where $f$ and $g$ are nonlinear functions. The system is \emph{autonomous} because
+$f$ and $g$ do not depend explicitly on $t$.
+
+\paragraph{Why analytical methods fail.}
+For linear systems $\mathbf{x}' = A\mathbf{x}$, we can construct the general solution
+from eigenvalues and eigenvectors. For nonlinear systems:
+\begin{itemize}
+  \item \textbf{No superposition:} If $\mathbf{x}_1(t)$ and $\mathbf{x}_2(t)$ are solutions,
+        $c_1\mathbf{x}_1(t) + c_2\mathbf{x}_2(t)$ is \emph{not} generally a solution.
+  \item \textbf{No general formula:} There is no algorithm that produces a closed-form
+        solution for an arbitrary nonlinear system.
+  \item \textbf{Rich behavior:} Nonlinear systems exhibit phenomena absent in linear
+        systems, including multiple equilibria, limit cycles, bifurcations, and chaos.
+\end{itemize}
+
+The strategy is to understand the system's \textbf{phase portrait} --- a qualitative
+map of all possible solution trajectories in the $(x,y)$-plane --- by combining
+local linear analysis near equilibria with global theorems.
+
+\paragraph{Equilibrium points.}
+As in \cref{sec:ch03_autonomous} and \cref{sec:ch08_phase_plane}, equilibrium
+(singular) points are constant solutions where the system comes to rest.
+
+\begin{definition}[Equilibrium point]
+  An \textbf{equilibrium point} (or \textbf{fixed point}, or \textbf{critical point})
+  of \cref{eq:nonlinear_autonomous} is a point $(x^*,y^*)$ satisfying
+  \[
+    f(x^*,y^*) = 0 \quad\text{and}\quad g(x^*,y^*) = 0.
+  \]
+  At an equilibrium, both derivatives vanish and the solution is stationary:
+  $x(t) \equiv x^*$, $y(t) \equiv y^*$.
+\end{definition}
+
+Finding equilibria is an algebraic problem: solve the system of two nonlinear
+equations $f = 0$, $g = 0$. Unlike the one-dimensional case (\cref{ch:qualitative}),
+there may be zero, one, or many equilibrium points.
+
+\begin{workedexample}
+  \textbf{Find all equilibrium points of the system}
+  \[
+    \frac{\mathrm{d}x}{\mathrm{d}t} = x - x^2 - xy, \qquad
+    \frac{\mathrm{d}y}{\mathrm{d}t} = y - y^2 - xy.
+  \]
+
+  \textbf{Solution.} Set both equations to zero simultaneously:
+  \begin{align}
+    x(1 - x - y) &= 0, \label{eq:ex_eq1} \\
+    y(1 - x - y) &= 0. \label{eq:ex_eq2}
+  \end{align}
+  From \cref{eq:ex_eq1}, either $x = 0$ or $1 - x - y = 0$.
+
+  \textit{Case 1:} $x = 0$. Substituting into \cref{eq:ex_eq2}: $y(1 - 0 - y) = 0$,
+  so $y = 0$ or $y = 1$. This gives equilibria $(0,0)$ and $(0,1)$.
+
+  \textit{Case 2:} $1 - x - y = 0$, i.e.\ $y = 1 - x$. Substituting into
+  \cref{eq:ex_eq2}: $y(0) = 0$, which is satisfied for any $y$. But we must
+  also have $x = 1 - y$ from the same relation. Setting $x = 0$ gives $y = 1$
+  (already found), and setting $y = 0$ gives $x = 1$, yielding the equilibrium $(1,0)$.
+
+  \textit{Verification:} Check $(1,0)$: $f(1,0) = 1 - 1 - 0 = 0$, $g(1,0) = 0 - 0 - 0 = 0$. $\checkmark$
+
+  The system has three equilibrium points: $(0,0)$, $(0,1)$, and $(1,0)$.
+\end{workedexample}
 
 \subsection{Jacobian Linearization}
 \label{sec:ch14_jacobian}
 
-% Content goes here
+The central technique for analyzing nonlinear systems near an equilibrium is
+\textbf{linearization}. Just as in \cref{sec:ch03_stability}, we approximate the
+nonlinear functions by their first-order Taylor expansions near the equilibrium.
+This reduces the nonlinear system to a linear one, whose behavior we understand
+completely from \cref{ch:systems}.
+
+\paragraph{Derivation via Taylor expansion.}
+Let $(x^*,y^*)$ be an equilibrium. Introduce deviation variables
+$u = x - x^*$ and $v = y - y^*$. Expanding $f$ and $g$ in a
+two-variable Taylor series about $(x^*,y^*)$:
+\begin{align*}
+  f(x,y) &= f(x^*,y^*) + \pd{f}{x}\Big|_{(x^*,y^*)} (x-x^*)
+           + \pd{f}{y}\Big|_{(x^*,y^*)} (y-y^*) + O\!\bigl(|u|^2 + |v|^2\bigr), \\
+  g(x,y) &= g(x^*,y^*) + \pd{g}{x}\Big|_{(x^*,y^*)} (x-x^*)
+           + \pd{g}{y}\Big|_{(x^*,y^*)} (y-y^*) + O\!\bigl(|u|^2 + |v|^2\bigr).
+\end{align*}
+Since $(x^*,y^*)$ is an equilibrium, $f(x^*,y^*) = g(x^*,y^*) = 0$.
+Discarding the higher-order terms gives the \textbf{linearized system}
+\begin{equation}
+  \label{eq:linearized_system}
+  \begin{pmatrix} u' \\ v' \end{pmatrix}
+  =
+  \underbrace{
+    \begin{pmatrix}
+      \pd{f}{x}\big|_{(x^*,y^*)} & \pd{f}{y}\big|_{(x^*,y^*)} \\[6pt]
+      \pd{g}{x}\big|_{(x^*,y^*)} & \pd{g}{y}\big|_{(x^*,y^*)}
+    \end{pmatrix}
+  }_{\displaystyle = \; J}
+  \begin{pmatrix} u \\ v \end{pmatrix}.
+\end{equation}
+
+\begin{keyresult}
+  \textbf{Jacobian matrix and linearization.}
+  The \textbf{Jacobian matrix} of the system $x' = f(x,y)$, $y' = g(x,y)$ is
+  \[
+    J(x,y) =
+    \begin{pmatrix}
+      \pd{f}{x} & \pd{f}{y} \\[6pt]
+      \pd{g}{x} & \pd{g}{y}
+    \end{pmatrix}.
+  \]
+  Evaluated at an equilibrium $(x^*,y^*)$, the linearized system is
+  $\begin{pmatrix} u' \\ v' \end{pmatrix} = J(x^*,y^*) \begin{pmatrix} u \\ v \end{pmatrix}$,
+  where $u = x-x^*$, $v = y-y^*$. Classify $(x^*,y^*)$ by the eigenvalues
+  of $J(x^*,y^*)$ using the phase plane classification from \cref{ch:systems}.
+\end{keyresult}
+
+\begin{keyresult}
+  \textbf{Eigenvalue classification at an equilibrium of a nonlinear system.}
+
+  Let $\lambda_1, \lambda_2$ be the eigenvalues of $J(x^*,y^*)$.
+
+  \begin{center}
+  \begin{tabular}{l l p{5cm}}
+    \toprule
+    \textbf{Eigenvalues} & \textbf{Type} & \textbf{Stability} \\
+    \midrule
+    Real, both $> 0$ & Unstable node (source) & Unstable \\[4pt]
+    Real, both $< 0$ & Stable node (sink) & Asymptotically stable \\[4pt]
+    Real, opposite signs & Saddle point & Unstable \\[4pt]
+    Complex $\alpha \pm i\beta$, $\alpha > 0$ & Unstable spiral & Unstable \\[4pt]
+    Complex $\alpha \pm i\beta$, $\alpha < 0$ & Stable spiral & Asymptotically stable \\[4pt]
+    Purely imaginary $\pm i\beta$ & Center (inconclusive) & \textit{Linearization inconclusive} \\
+    \bottomrule
+  \end{tabular}
+  \end{center}
+
+  The center case is the only one where the linearization does \emph{not}
+  determine the true nonlinear behavior. See \cref{sec:ch14_hartman_grobman} for details.
+\end{keyresult}
+
+\begin{workedexample}
+  \textbf{Linearize and classify the equilibria of}
+  \[
+    \frac{\mathrm{d}x}{\mathrm{d}t} = y, \qquad
+    \frac{\mathrm{d}y}{\mathrm{d}t} = x - x^2.
+  \]
+  \textit{(This is the undamped nonlinear pendulum in phase-plane form.)}
+
+  \textbf{Step 1: Find equilibria.}
+  Set $y = 0$ and $x - x^2 = 0 \implies x(1-x) = 0$.
+  \[
+    \text{Equilibria: } (0,0) \text{ and } (1,0).
+  \]
+
+  \textbf{Step 2: Jacobian.}
+  With $f(x,y) = y$ and $g(x,y) = x - x^2$:
+  \[
+    \pd{f}{x} = 0, \quad \pd{f}{y} = 1, \quad
+    \pd{g}{x} = 1 - 2x, \quad \pd{g}{y} = 0.
+  \]
+  \[
+    J(x,y) =
+    \begin{pmatrix}
+      0 & 1 \\
+      1 - 2x & 0
+    \end{pmatrix}.
+  \]
+
+  \textbf{Step 3: Classify $(0,0)$.}
+  \[
+    J(0,0) = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.
+  \]
+  Characteristic equation: $\lambda^2 - 1 = 0 \implies \lambda = \pm 1$.
+  Opposite signs $\implies$ \textbf{saddle point} (unstable).
+
+  \textbf{Step 4: Classify $(1,0)$.}
+  \[
+    J(1,0) = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}.
+  \]
+  Characteristic equation: $\lambda^2 + 1 = 0 \implies \lambda = \pm i$.
+  Purely imaginary eigenvalues $\implies$ \textbf{center by linearization}.
+  The linearization predicts closed orbits, but the nonlinear behavior
+  requires further analysis (confirmed below to be a true center).
+\end{workedexample}
+
+\begin{workedexample}
+  \textbf{Linearize and classify the equilibria of the competing species model}
+  \[
+    \frac{\mathrm{d}x}{\mathrm{d}t} = 2x - x^2 - xy, \qquad
+    \frac{\mathrm{d}y}{\mathrm{d}t} = y - y^2 - xy.
+  \]
+
+  \textbf{Step 1: Find equilibria.}
+  \begin{align}
+    x(2 - x - y) &= 0, \label{eq:comp1} \\
+    y(1 - x - y) &= 0. \label{eq:comp2}
+  \end{align}
+  From \cref{eq:comp1}, either $x = 0$ or $x + y = 2$.
+
+  \textit{Case 1:} $x = 0$. From \cref{eq:comp2}: $y(1-y) = 0$, giving $y = 0$ or $y = 1$.
+  Equilibria: $(0,0)$ and $(0,1)$.
+
+  \textit{Case 2:} $x + y = 2$, so $y = 2 - x$. From \cref{eq:comp2}: $y(1 - x - y) = y(1 - 2) = -y = 0$,
+  so $y = 0$, which gives $x = 2$. Equilibrium: $(2,0)$.
+
+  The three equilibria are $(0,0)$, $(0,1)$, and $(2,0)$.
+
+  \textbf{Step 2: Jacobian.}
+  With $f(x,y) = 2x - x^2 - xy$ and $g(x,y) = y - y^2 - xy$:
+  \[
+    J(x,y) =
+    \begin{pmatrix}
+      2 - 2x - y & -x \\[4pt]
+      -y & 1 - 2y - x
+    \end{pmatrix}.
+  \]
+
+  \textbf{Step 3: Classify $(0,0)$.}
+  \[
+    J(0,0) = \begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix},
+    \quad \lambda_1 = 2,\; \lambda_2 = 1.
+  \]
+  Both positive $\implies$ \textbf{unstable node (source)}. Both species die out is unstable:
+  a small introduction of either population grows.
+
+  \textbf{Step 4: Classify $(0,1)$.}
+  \[
+    J(0,1) = \begin{pmatrix} 2-1 & 0 \\ -1 & 1-2 \end{pmatrix}
+    = \begin{pmatrix} 1 & 0 \\ -1 & -1 \end{pmatrix}.
+  \]
+  Since $J$ is lower triangular, eigenvalues are the diagonal entries:
+  $\lambda_1 = 1$, $\lambda_2 = -1$.
+  Opposite signs $\implies$ \textbf{saddle point} (unstable).
+  Species~$y$ alone at carrying capacity is unstable to invasion by species~$x$.
+
+  \textbf{Step 5: Classify $(2,0)$.}
+  \[
+    J(2,0) = \begin{pmatrix} 2-4 & -2 \\ 0 & 1-2 \end{pmatrix}
+    = \begin{pmatrix} -2 & -2 \\ 0 & -1 \end{pmatrix}.
+  \]
+  Upper triangular: $\lambda_1 = -2$, $\lambda_2 = -1$.
+  Both negative $\implies$ \textbf{stable node (sink)}.
+  Species~$x$ alone at carrying capacity is stable; species~$y$ cannot invade.
+\end{workedexample}
 
 \subsection{Trace-Determinant Classification}
 \label{sec:ch14_trace_determinant}
 
-% Content goes here
+Just as for linear systems (\cref{sec:ch08_trace_det}), the trace and determinant
+of the Jacobian at an equilibrium provide a quick classification without computing
+eigenvalues explicitly.
+
+For $J = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ at an equilibrium:
+\[
+  \tau = \tr(J) = a + d, \qquad
+  \Delta = \det(J) = ad - bc.
+\]
+The eigenvalues satisfy $\lambda^2 - \tau\lambda + \Delta = 0$ with
+discriminant $D = \tau^2 - 4\Delta$.
+
+\begin{keyresult}
+  \textbf{Trace-determinant classification for nonlinear equilibria.}
+  Evaluate $\tau = \tr(J)$ and $\Delta = \det(J)$ at the equilibrium point.
+
+  \begin{center}
+  \begin{tabular}{l l l}
+    \toprule
+    \textbf{Region} & \textbf{Conditions} & \textbf{Classification} \\
+    \midrule
+    $\Delta < 0$ & Below $\tau$-axis & Saddle (unstable) \\[4pt]
+    $\Delta > 0$, $\tau > 0$, $D > 0$ & Right, above parabola & Unstable node \\[4pt]
+    $\Delta > 0$, $\tau < 0$, $D > 0$ & Left, above parabola & Stable node \\[4pt]
+    $\Delta > 0$, $\tau > 0$, $D < 0$ & Right, below parabola & Unstable spiral \\[4pt]
+    $\Delta > 0$, $\tau < 0$, $D < 0$ & Left, below parabola & Stable spiral \\[4pt]
+    $\Delta > 0$, $\tau = 0$ & On positive $\Delta$-axis & Center (inconclusive) \\
+    \bottomrule
+  \end{tabular}
+  \end{center}
+\end{keyresult}
+
+\paragraph{The trace-determinant plane.}
+
+\begin{center}
+\begin{tikzpicture}[scale=0.85]
+  % Axes
+  \draw[->, thick] (-5,0) -- (5,0) node[right] {$\tau$ (trace)};
+  \draw[->, thick] (0,-1) -- (0,5) node[above] {$\Delta$ (det)};
+
+  % Parabola \Delta = \tau^2 / 4
+  \draw[thick, dashed, red!80] plot[domain=-4.5:4.5, samples=100, smooth, variable=\t]
+    ({\t}, {\t*\t/4});
+  \node[red!80, font=\footnotesize, anchor=west] at (3.8, 4.3) {$\Delta = \tau^2/4$};
+
+  % Fill saddle region (below \tau-axis)
+  \fill[pattern=north west lines, pattern color=red!25] (-5,0) rectangle (5,-0.45);
+
+  % Stability boundary (\tau = 0 line)
+  \draw[dashed, thick, blue!60] (0,0) -- (0,5);
+
+  % Labels for regions
+  \node[font=\footnotesize, align=center] at (0,-0.25) {\textbf{Saddle}\\$(\Delta < 0)$};
+
+  \node[font=\footnotesize, align=center] at (-3,3.5) {\textbf{Stable node}\\$(\tau < 0,\; D > 0)$};
+  \node[font=\footnotesize, align=center] at (3,3.5) {\textbf{Unstable node}\\$(\tau > 0,\; D > 0)$};
+
+  \node[font=\footnotesize, align=center] at (-2,1.1) {\textbf{Stable spiral}\\$(\tau < 0,\; D < 0)$};
+  \node[font=\footnotesize, align=center] at (2,1.1) {\textbf{Unstable spiral}\\$(\tau > 0,\; D < 0)$};
+
+  \node[font=\footnotesize, align=center] at (0,4.5) {\textbf{Center}\\$(\tau = 0,\; \Delta > 0)$};
+
+  % Origin
+  \filldraw[black] (0,0) circle (1.5pt);
+  \node[font=\scriptsize, below left] at (0,0) {$(0,0)$};
+
+  % Parabola tip label
+  \node[font=\scriptsize, red!80, align=center] at (0,0.15) {parabola};
+\end{tikzpicture}
+\end{center}
+
+The parabola $\Delta = \tau^2/4$ separates real from complex eigenvalues.
+The $\Delta$-axis ($\tau = 0$) separates stable from unstable systems.
+The $\tau$-axis ($\Delta = 0$) separates saddles from nodes/spirals/centers.
+
+\begin{workedexample}
+  \textbf{Classify the equilibria of the competing species model from
+  the previous section using the trace-determinant plane.}
+
+  Recall the Jacobian:
+  \[
+    J(x,y) =
+    \begin{pmatrix}
+      2 - 2x - y & -x \\[4pt]
+      -y & 1 - 2y - x
+    \end{pmatrix}.
+  \]
+
+  \textbf{At $(0,0)$:}
+  \[
+    \tau = 2 + 1 = 3 > 0, \qquad \Delta = (2)(1) - (0) = 2 > 0.
+  \]
+  Discriminant: $D = 9 - 8 = 1 > 0$.
+  Region: $\tau > 0$, $\Delta > 0$, $D > 0$ $\implies$ \textbf{unstable node}. Consistent with $\lambda = 2, 1$.
+
+  \textbf{At $(0,1)$:}
+  \[
+    \tau = 1 + (-1) = 0, \qquad \Delta = (1)(-1) - (0) = -1 < 0.
+  \]
+  Region: $\Delta < 0$ $\implies$ \textbf{saddle point}. Consistent with $\lambda = \pm 1$.
+
+  \textbf{At $(2,0)$:}
+  \[
+    \tau = (-2) + (-1) = -3 < 0, \qquad \Delta = (-2)(-1) - (0) = 2 > 0.
+  \]
+  Discriminant: $D = 9 - 8 = 1 > 0$.
+  Region: $\tau < 0$, $\Delta > 0$, $D > 0$ $\implies$ \textbf{stable node}. Consistent with $\lambda = -2, -1$.
+\end{workedexample}
 
 \subsection{Hartman--Grobman Theorem}
 \label{sec:ch14_hartman_grobman}
 
-% Content goes here
+The linearization tells us about the behavior of the nonlinear system in a
+\emph{small neighborhood} of the equilibrium. The Hartman--Grobman theorem
+provides the rigorous justification for this connection.
+
+\begin{theorem}[Hartman--Grobman]
+  \label{thm:hartman_grobman}
+  Let $(x^*,y^*)$ be an equilibrium point of the autonomous system
+  $x' = f(x,y)$, $y' = g(x,y)$, where $f$ and $g$ are $C^1$ (continuously differentiable).
+  Let $J$ be the Jacobian matrix evaluated at $(x^*,y^*)$.
+
+  If $J$ has \textbf{no eigenvalues with zero real part} (i.e., the equilibrium is
+  \textbf{hyperbolic}), then there exists a neighborhood of $(x^*,y^*)$ in which
+  the nonlinear system is \textbf{topologically conjugate} to its linearization
+  $u' = a u + b v$, $v' = c u + d v$.
+
+  In particular, the local phase portrait of the nonlinear system near $(x^*,y^*)$
+  has the same topological structure as that of the linearized system.
+\end{theorem}
+
+\paragraph{What topological conjugacy means.}
+Two systems are topologically conjugate if there exists a continuous, invertible
+change of coordinates (a homeomorphism) that maps the trajectories of one system
+onto the trajectories of the other, preserving the direction of time. Practically,
+this means:
+\begin{itemize}
+  \item A stable node of the linearization corresponds to a stable node of the nonlinear system.
+  \item A saddle of the linearization corresponds to a saddle of the nonlinear system.
+  \item The qualitative flow pattern (arrows, attraction, repulsion) is preserved.
+\end{itemize}
+
+The actual shapes of trajectories may differ --- a nonlinear trajectory near
+a stable spiral may not be a perfect logarithmic spiral --- but the
+essential features (spiraling inward, counterclockwise direction, etc.) are identical.
+
+\paragraph{Limitations: non-hyperbolic equilibria.}
+The theorem \textbf{does not apply} when the Jacobian has eigenvalues with zero real part:
+\begin{itemize}
+  \item \textbf{Centers} ($\lambda = \pm i\beta$): The nonlinear system near a center
+        can be a center, a stable spiral, or an unstable spiral. The linearization
+        alone cannot distinguish between these cases. Higher-order terms in the
+        Taylor expansion must be examined.
+  \item \textbf{Non-hyperbolic points} with real zero eigenvalues ($\lambda = 0$):
+        The dynamics can be extremely sensitive to nonlinear terms.
+        Examples include semi-stable equilibria and bifurcation points.
+\end{itemize}
+
+\begin{workedexample}
+  \textbf{Hartman--Grobman applies: stable spiral.}
+
+  Consider $x' = -x + y + x(x^2+y^2)$, $y' = -x - y + y(x^2+y^2)$.
+
+  \textit{Equilibrium:} $(0,0)$ (the only one easily found).
+
+  \textit{Jacobian at $(0,0)$:}
+  \[
+    J(0,0) = \begin{pmatrix} -1 & 1 \\ -1 & -1 \end{pmatrix},
+    \quad \tau = -2,\; \Delta = 2,\; D = 4-8 = -4 < 0.
+  \]
+  Eigenvalues: $\lambda = -1 \pm i$. Both have nonzero real part
+  ($\Re(\lambda) = -1 \neq 0$).
+
+  \textbf{Conclusion:} The equilibrium is hyperbolic. By the Hartman--Grobman
+  theorem, the nonlinear system near $(0,0)$ is a \textbf{stable spiral},
+  topologically equivalent to its linearization.
+
+  (In fact, the nonlinear terms $x(x^2+y^2)$ and $y(x^2+y^2)$ cause
+  outward spiraling for large radii, creating a stable limit cycle ---
+  but near the origin, the local portrait is exactly a stable spiral.)
+\end{workedexample}
+
+\begin{workedexample}
+  \textbf{Hartman--Grobman does not apply: center vs.\ spiral.}
+
+  Consider the two systems:
+  \begin{align}
+    \text{(A)} \quad x' &= y, \quad y' = -x, \\
+    \text{(B)} \quad x' &= y + x(x^2+y^2), \quad y' = -x + y(x^2+y^2).
+  \end{align}
+
+  \textit{Jacobian at $(0,0)$ for both systems:}
+  \[
+    J(0,0) = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix},
+    \quad \lambda = \pm i.
+  \]
+  Purely imaginary eigenvalues $\implies$ the equilibrium is \textbf{not hyperbolic}.
+  Hartman--Grobman \textbf{does not apply}.
+
+  \textit{System (A):} The linear system. Trajectories are circles $x^2 + y^2 = C$.
+  This is a true \textbf{center}.
+
+  \textit{System (B):} Convert to polar coordinates ($x = r\cos\theta$, $y = r\sin\theta$):
+  \[
+    r' = r^3, \quad \theta' = -1.
+  \]
+  Since $r' = r^3 > 0$ for $r > 0$, the radius strictly increases.
+  The origin is actually an \textbf{unstable spiral}, \emph{not} a center,
+  despite the linearization suggesting a center.
+
+  \textbf{Lesson:} When $\lambda = \pm i\beta$, the nonlinear terms determine
+  the true behavior. Always use additional tools (Lyapunov functions,
+  polar coordinates, first integrals) to resolve non-hyperbolic equilibria.
+\end{workedexample}
 
 \subsection{Limit Cycles}
 \label{sec:ch14_limit_cycles}
 
-% Content goes here
+\begin{definition}[Limit cycle]
+  A \textbf{limit cycle} is a \textbf{closed, isolated periodic orbit}
+  of an autonomous system in the phase plane.
+\end{definition}
 
-\subsection{Lotka--Volterra Predator-Prey}
+The key word is \emph{isolated}: a limit cycle is a single closed trajectory
+that is not part of a continuous family of closed orbits. Nearby trajectories
+either spiral toward the limit cycle (stable limit cycle) or spiral away from
+it (unstable limit cycle).
+
+This contrasts with the center in linear systems, where every trajectory near
+the equilibrium is a closed orbit forming a continuous family.
+
+\begin{theorem}[Poincar\'{e}--Bendixson]
+  \label{thm:poincare_bendixson}
+  Let $R$ be a closed, bounded (compact) region in the plane containing no
+  equilibrium points, and let a trajectory enter $R$ and remain there for
+  all future time $t > 0$. Then the trajectory either:
+  \begin{enumerate}
+    \item approaches a periodic orbit (a closed trajectory) as $t \to \infty$, or
+    \item is itself a periodic orbit.
+  \end{enumerate}
+\end{theorem}
+
+\paragraph{Implications.}
+The Poincar\'{e}--Bendixson theorem is a powerful tool for proving the
+existence of limit cycles in two-dimensional systems:
+\begin{itemize}
+  \item If you can construct a trapping region $R$ that contains no equilibria,
+        any trajectory entering $R$ must approach a periodic orbit.
+  \item In the plane, the only possible long-term behaviors of bounded
+        trajectories are: equilibrium points, periodic orbits (limit cycles),
+        or trajectories that approach limit cycles.
+  \item \textbf{No chaos in 2D autonomous systems:} Strange attractors and
+        chaotic behavior require at least three dimensions.
+\end{itemize}
+
+\paragraph{Physical examples.}
+Limit cycles model sustained oscillations in physical systems:
+\begin{itemize}
+  \item \textbf{Electrical circuits:} The van der Pol oscillator models self-sustained
+        oscillations in vacuum tube circuits.
+  \item \textbf{Chemical reactions:} The Belousov--Zhabotinsky reaction exhibits
+        periodic color changes due to a chemical limit cycle.
+  \item \textbf{Biological rhythms:} Heartbeat, neural firing, and circadian
+        rhythms can be modeled as limit cycles.
+  \item \textbf{Mechanical oscillators:} A clock pendulum with a periodic
+        driving force (the escapement mechanism) exhibits a limit cycle.
+\end{itemize}
+
+\subsection{Lotka--Volterra Predator--Prey Model}
 \label{sec:ch14_lotka_volterra}
 
-% Content goes here
+The Lotka--Volterra model is the classic example of a nonlinear system with
+closed orbits, first introduced in the 1920s to model predator--prey
+population dynamics.
+
+\paragraph{Biological motivation and model derivation.}
+Consider two interacting populations:
+\begin{itemize}
+  \item $x(t)$: number of \textbf{prey} (e.g., rabbits).
+  \item $y(t)$: number of \textbf{predators} (e.g., foxes).
+\end{itemize}
+
+We make four biological assumptions:
+\begin{enumerate}
+  \item In the absence of predators, prey grow exponentially at rate $\alpha$:
+        $x' = \alpha x$.
+  \item Predators eat prey at a rate proportional to encounters $\alpha xy$.
+        Combined with assumption~1: $x' = \alpha x - \beta xy$.
+  \item In the absence of prey, predators die off at rate $\gamma$:
+        $y' = -\gamma y$.
+  \item Predators reproduce at a rate proportional to food consumption $\beta xy$.
+        Combined with assumption~3: $y' = \delta xy - \gamma y$.
+\end{enumerate}
+
+This yields the \textbf{Lotka--Volterra predator--prey equations}:
+\begin{equation}
+  \label{eq:lotka_volterra}
+  \boxed{
+    \frac{\mathrm{d}x}{\mathrm{d}t} = \alpha x - \beta xy, \qquad
+    \frac{\mathrm{d}y}{\mathrm{d}t} = \delta xy - \gamma y,
+  }
+\end{equation}
+where $\alpha, \beta, \gamma, \delta > 0$ are parameters.
+
+\begin{definition}[Lotka--Volterra model]
+  The Lotka--Volterra predator--prey system is the autonomous nonlinear
+  system \cref{eq:lotka_volterra} with $\alpha, \beta, \gamma, \delta > 0$.
+  It has two equilibrium points:
+  \[
+    (0,0) \text{ (extinction)}, \qquad
+    \left(\frac{\gamma}{\delta},\, \frac{\alpha}{\beta}\right) \text{ (coexistence)}.
+  \]
+\end{definition}
+
+\paragraph{Equilibrium analysis.}
+Setting the right-hand sides to zero:
+\begin{align}
+  x(\alpha - \beta y) &= 0, \label{eq:lv_eq1} \\
+  y(\delta x - \gamma) &= 0. \label{eq:lv_eq2}
+\end{align}
+From \cref{eq:lv_eq1}, $x = 0$ or $y = \alpha/\beta$.
+
+\textit{Case 1:} $x = 0$. From \cref{eq:lv_eq2}, $y(\,-\gamma) = 0 \implies y = 0$.
+Equilibrium: $(0,0)$.
+
+\textit{Case 2:} $y = \alpha/\beta$. From \cref{eq:lv_eq2},
+$\delta x - \gamma = 0 \implies x = \gamma/\delta$.
+Equilibrium: $(\gamma/\delta, \alpha/\beta)$.
+
+\paragraph{Jacobian and classification.}
+The Jacobian of the system is
+\[
+  J(x,y) =
+  \begin{pmatrix}
+    \alpha - \beta y & -\beta x \\[6pt]
+    \delta y & \delta x - \gamma
+  \end{pmatrix}.
+\]
+
+\textbf{At $(0,0)$:}
+\[
+  J(0,0) = \begin{pmatrix} \alpha & 0 \\ 0 & -\gamma \end{pmatrix},
+  \quad \lambda_1 = \alpha > 0,\; \lambda_2 = -\gamma < 0.
+\]
+Opposite signs $\implies$ \textbf{saddle point} (unstable).
+Both populations at zero is unstable: a small number of prey grows,
+triggering predator response.
+
+\textbf{At $(\gamma/\delta, \alpha/\beta)$:}
+\[
+  J\!\left(\frac{\gamma}{\delta}, \frac{\alpha}{\beta}\right)
+  =
+  \begin{pmatrix}
+    0 & -\beta\gamma/\delta \\[6pt]
+    \delta\alpha/\beta & 0
+  \end{pmatrix}
+  =
+  \begin{pmatrix}
+    0 & -\dfrac{\beta\gamma}{\delta} \\[8pt]
+    \dfrac{\delta\alpha}{\beta} & 0
+  \end{pmatrix}.
+\]
+Trace: $\tau = 0$. Determinant: $\Delta = (0) - (-\beta\gamma/\delta)(\delta\alpha/\beta) = \alpha\gamma > 0$.
+Characteristic equation: $\lambda^2 + \alpha\gamma = 0$, so
+\[
+  \lambda = \pm i\sqrt{\alpha\gamma}.
+\]
+Purely imaginary eigenvalues $\implies$ \textbf{center} by linearization.
+Hartman--Grobman does not apply. We must analyze the nonlinear system.
+
+\paragraph{First integral (conservation law).}
+The Lotka--Volterra system admits an exact first integral, proving that
+the orbits are indeed closed. Divide the two equations:
+\[
+  \frac{\mathrm{d}y}{\mathrm{d}x}
+  = \frac{\delta xy - \gamma y}{\alpha x - \beta xy}
+  = \frac{y(\delta x - \gamma)}{x(\alpha - \beta y)}.
+\]
+Separate variables:
+\[
+  \frac{\alpha - \beta y}{y}\,\mathrm{d}y
+  = \frac{\delta x - \gamma}{x}\,\mathrm{d}x.
+\]
+Rewrite and integrate:
+\[
+  \left(\frac{\alpha}{y} - \beta\right)\mathrm{d}y
+  = \left(\delta - \frac{\gamma}{x}\right)\mathrm{d}x.
+\]
+\[
+  \alpha \ln y - \beta y = \delta x - \gamma \ln x + C.
+\]
+Rearranging gives the \textbf{first integral}:
+\begin{equation}
+  \label{eq:lv_first_integral}
+  V(x,y) = \delta x - \gamma \ln x + \beta y - \alpha \ln y = \text{constant}.
+\end{equation}
+Each value of $C$ defines a closed orbit in the phase plane.
+The function $V(x,y)$ has a strict global minimum at the coexistence equilibrium
+$(\gamma/\delta, \alpha/\beta)$, and level curves $V(x,y) = C$ for
+$C > V_{\min}$ are closed curves surrounding this point.
+
+\begin{keyresult}
+  \textbf{Lotka--Volterra closed orbits.}
+  The coexistence equilibrium $(\gamma/\delta, \alpha/\beta)$ is a true
+  center for the nonlinear system. Every solution starting in the
+  first quadrant $(x > 0, y > 0)$ traces a closed orbit around
+  the coexistence point, determined by the initial conditions through
+  the first integral \cref{eq:lv_first_integral}.
+\end{keyresult}
+
+\paragraph{Nullclines.}
+The \textbf{nullclines} are curves in the phase plane where one of the
+derivatives vanishes:
+\begin{itemize}
+  \item \textbf{$x$-nullcline} ($x' = 0$): $x = 0$ or $y = \alpha/\beta$.
+        On $x = 0$, the prey population is zero.
+        On $y = \alpha/\beta$ (horizontal line), the prey population
+        is momentarily stationary.
+  \item \textbf{$y$-nullcline} ($y' = 0$): $y = 0$ or $x = \gamma/\delta$.
+        On $y = 0$, the predator population is zero.
+        On $x = \gamma/\delta$ (vertical line), the predator population
+        is momentarily stationary.
+\end{itemize}
+The two nontrivial nullclines intersect at the coexistence equilibrium
+$(\gamma/\delta, \alpha/\beta)$. The nullclines divide the first quadrant
+into four regions, each with a characteristic direction of motion:
+\begin{itemize}
+  \item Region I ($x > \gamma/\delta, y < \alpha/\beta$): $x' > 0$, $y' > 0$
+        $\implies$ motion up and right.
+  \item Region II ($x < \gamma/\delta, y < \alpha/\beta$): $x' > 0$, $y' < 0$
+        $\implies$ motion down and right.
+  \item Region III ($x < \gamma/\delta, y > \alpha/\beta$): $x' < 0$, $y' < 0$
+        $\implies$ motion down and left.
+  \item Region IV ($x > \gamma/\delta, y > \alpha/\beta$): $x' < 0$, $y' > 0$
+        $\implies$ motion up and left.
+\end{itemize}
+The resulting flow is counterclockwise around the equilibrium.
+
+\begin{workedexample}
+  \textbf{Full analysis of the Lotka--Volterra system with parameters}
+  $\alpha = 1.5$, $\beta = 1$, $\delta = 3$, $\gamma = 1$.
+
+  \textbf{System:}
+  \[
+    \frac{\mathrm{d}x}{\mathrm{d}t} = 1.5\,x - x y, \qquad
+    \frac{\mathrm{d}y}{\mathrm{d}t} = 3\,x y - y.
+  \]
+
+  \textbf{Equilibria:}
+  \[
+    (0,0) \text{ and } \left(\frac{\gamma}{\delta}, \frac{\alpha}{\beta}\right)
+    = \left(\frac{1}{3}, \frac{1.5}{1}\right) = \left(\frac{1}{3}, 1.5\right).
+  \]
+
+  \textbf{Jacobian:}
+  \[
+    J(x,y) =
+    \begin{pmatrix}
+      1.5 - y & -x \\[6pt]
+      3y & 3x - 1
+    \end{pmatrix}.
+  \]
+
+  \textbf{At $(0,0)$:}
+  \[
+    J(0,0) = \begin{pmatrix} 1.5 & 0 \\ 0 & -1 \end{pmatrix},
+    \quad \lambda_1 = 1.5 > 0,\; \lambda_2 = -1 < 0.
+  \]
+  \textbf{Saddle point} (unstable).
+
+  \textbf{At $(1/3, 1.5)$:}
+  \[
+    J\!\left(\tfrac{1}{3}, 1.5\right)
+    = \begin{pmatrix} 0 & -1/3 \\ 4.5 & 0 \end{pmatrix}.
+  \]
+  \[
+    \tau = 0, \quad \Delta = (0) - (-1/3)(4.5) = 1.5.
+  \]
+  \[
+    \lambda = \pm i\sqrt{1.5} \approx \pm 1.225\,i.
+  \]
+  Purely imaginary eigenvalues. The coexistence point is a \textbf{center}
+  (confirmed by the first integral, not just linearization).
+
+  \textbf{Nullclines:}
+  \begin{itemize}
+    \item $x$-nullcline: $x = 0$ or $y = 1.5$ (horizontal line).
+    \item $y$-nullcline: $y = 0$ or $x = 1/3$ (vertical line).
+  \end{itemize}
+
+  \textbf{First integral:}
+  \[
+    V(x,y) = 3x - \ln x + y - 1.5\,\ln y = \text{constant}.
+  \]
+
+  \textbf{Period of oscillation:} Near the equilibrium, the period is
+  approximately $T \approx 2\pi/\sqrt{\alpha\gamma} = 2\pi/\sqrt{1.5}
+  \approx 5.13$ time units.
+\end{workedexample}
+
+\paragraph{Phase portrait.}
+
+\begin{center}
+\begin{tikzpicture}[scale=0.9]
+  % Axes
+  \draw[->, thick] (-0.3,0) -- (5,0) node[right] {$x$ (prey)};
+  \draw[->, thick] (0,-0.3) -- (0,5) node[above] {$y$ (predator)};
+
+  % Nullclines
+  \draw[dashed, thick, blue!60] (1,0) -- (1,4.5) node[above, font=\scriptsize, blue!60] {$x=1/3$};
+  \draw[dashed, thick, red!60] (0,1.875) -- (4.8,1.875) node[right, font=\scriptsize, red!60] {$y=1.5$};
+
+  % Closed orbits (ellipses centered at (1/3, 1.5))
+  % Inner orbit
+  \draw[thick, purple!80] (1,1.875) ellipse (0.6 and 0.6);
+  % Middle orbit
+  \draw[thick, purple!80] (1,1.875) ellipse (1.2 and 1.2);
+  % Outer orbit
+  \draw[thick, purple!80] (1,1.875) ellipse (1.9 and 1.8);
+
+  % Arrows on orbits (counterclockwise)
+  % Inner orbit arrows
+  \draw[thick, purple!80, ->] (1.6,1.875) -- (1.65,2.05);
+  \draw[thick, purple!80, ->] (1,2.475) -- (0.83,2.45);
+  \draw[thick, purple!80, ->] (0.4,1.875) -- (0.4,1.65);
+  \draw[thick, purple!80, ->] (1,1.275) -- (1.17,1.295);
+
+  % Middle orbit arrows
+  \draw[thick, purple!80, ->] (2.2,1.875) -- (2.28,2.12);
+  \draw[thick, purple!80, ->] (1,3.075) -- (0.76,3.05);
+  \draw[thick, purple!80, ->] (-0.2,1.875) -- (-0.2,1.55);
+  \draw[thick, purple!80, ->] (1,0.675) -- (1.24,0.70);
+
+  % Equilibrium points
+  \filldraw[black] (0,0) circle (2.5pt) node[below left=1pt] {$(0,0)$};
+  \filldraw[black] (1,1.875) circle (2.5pt) node[below right=1pt] {$(\gamma/\delta,\,\alpha/\beta)$};
+
+  % Region labels
+  \node[font=\scriptsize, align=center, blue!60] at (2.5,2.8) {Region I\\($x' > 0,\; y' > 0$)};
+  \node[font=\scriptsize, align=center, blue!60] at (0.4,0.9) {Region II\\($x' > 0,\; y' < 0$)};
+  \node[font=\scriptsize, align=center, blue!60] at (-0.5,2.5) {Region III\\($x' < 0,\; y' < 0$)};
+  \node[font=\scriptsize, align=center, blue!60] at (2.5,3.5) {Region IV\\($x' < 0,\; y' > 0$)};
+
+  % Legend
+  \node[font=\footnotesize, align=left, anchor=west] at (2.8,4.2) {
+    \textcolor{blue!60}{\rule{6pt}{1pt}} $x$-nullcline \\
+    \textcolor{red!60}{\rule{6pt}{1pt}} $y$-nullcline \\
+    \textcolor{purple!80}{\rule{6pt}{1pt}} Closed orbits
+  };
+\end{tikzpicture}
+\end{center}
+
+\paragraph{Interpretation of the oscillations.}
+The closed orbits correspond to periodic predator--prey cycles:
+\begin{enumerate}
+  \item Prey population grows (few predators) $\to$ predators have abundant food.
+  \item Predator population grows $\to$ predation pressure increases.
+  \item Prey population declines $\to$ predators face food shortage.
+  \item Predator population declines $\to$ prey can recover.
+  \item Cycle repeats.
+\end{enumerate}
+The amplitude of the oscillation is determined by the initial conditions:
+the further the initial point from the equilibrium, the larger the orbit.
+
+\paragraph{Limitations of the model.}
+The Lotka--Volterra model has several simplifications:
+\begin{itemize}
+  \item Unlimited prey growth in the absence of predators (no carrying capacity).
+  \item Linear functional response (predation rate proportional to $xy$).
+  \item No time delays in predator reproduction.
+  \item Homogeneous mixing (well-mixed populations).
+\end{itemize}
+Realistic extensions include logistic prey growth, Holling-type functional
+responses, and more complex age-structured models. These modifications can
+create stable limit cycles (rather than neutral centers) and richer dynamics.
 
 \subsection{Summary}
 \label{sec:ch14_summary}
 
-% Summary table at end of chapter
+Nonlinear systems cannot be solved by superposition, but their qualitative
+behavior near equilibria is accessible through linearization. The Jacobian
+matrix, combined with the trace-determinant classification, provides a
+systematic framework for understanding local dynamics. The Hartman--Grobman
+theorem guarantees that this linear picture is topologically correct for
+hyperbolic equilibria. Global phenomena like limit cycles require additional
+tools: the Poincar\'{e}--Bendixson theorem, first integrals, and Lyapunov
+functions.
+
 \begin{table}[htbp]
 \centering
-\caption{Chapter Summary}
+\caption{Chapter 14 Summary: Nonlinear Systems}
 \label{tab:ch14_summary}
-\begin{tabular}{l l}
+\begin{tabular}{l p{8.5cm}}
 \toprule
 \textbf{Concept} & \textbf{Key Formula/Method} \\
 \midrule
-TBD & TBD \\
+Nonlinear autonomous system & $\displaystyle x' = f(x,y), \;\; y' = g(x,y)$, no superposition \\
+Equilibrium point & Solve $f(x^*,y^*) = 0$, $g(x^*,y^*) = 0$ \\
+Jacobian matrix & $J = \begin{pmatrix} \pd{f}{x} & \pd{f}{y} \\[4pt] \pd{g}{x} & \pd{g}{y} \end{pmatrix}$ \\
+Linearized system & $\begin{pmatrix} u' \\ v' \end{pmatrix} = J(x^*,y^*) \begin{pmatrix} u \\ v \end{pmatrix}$, $u=x-x^*$, $v=y-y^*$ \\
+Trace-determinant & $\tau = \tr(J)$, $\Delta = \det(J)$, $D = \tau^2 - 4\Delta$ \\
+Hartman--Grobman theorem & Hyperbolic equilibria ($\Re(\lambda) \neq 0$): nonlinear $\cong$ linearization \\
+Non-hyperbolic case & $\Re(\lambda) = 0$: linearization inconclusive; use first integrals, Lyapunov functions \\
+Limit cycle & Closed, isolated periodic orbit \\
+Poincar\'{e}--Bendixson theorem & Bounded trajectory in 2D with no equilibria $\implies$ periodic orbit \\
+Lotka--Volterra model & $x' = \alpha x - \beta xy$, $y' = \delta xy - \gamma y$ \\
+LV equilibria & $(0,0)$ saddle; $(\gamma/\delta, \alpha/\beta)$ center (closed orbits) \\
+LV first integral & $V(x,y) = \delta x - \gamma\ln x + \beta y - \alpha\ln y = C$ \\
+LV nullclines & $x$-nullcline: $y = \alpha/\beta$; $y$-nullcline: $x = \gamma/\delta$ \\
 \bottomrule
 \end{tabular}
 \end{table}
+
+\begin{hintbox}
+  \textbf{Problem-solving workflow for nonlinear autonomous systems.}
+  \begin{enumerate}
+    \item Find all equilibrium points by solving $f(x,y) = 0$, $g(x,y) = 0$.
+    \item Compute the Jacobian matrix $J(x,y)$ and evaluate it at each equilibrium.
+    \item Classify each equilibrium using eigenvalues or the trace-determinant plane.
+    \item Apply the Hartman--Grobman theorem: if the equilibrium is hyperbolic,
+          the local phase portrait matches the linear classification.
+    \item For non-hyperbolic equilibria ($\Re(\lambda) = 0$), use additional tools:
+          first integrals, Lyapunov functions, or polar coordinates.
+    \item Draw nullclines to understand the flow direction in each region.
+    \item Look for limit cycles using the Poincar\'{e}--Bendixson theorem
+          (requires constructing a trapping region).
+    \item Sketch the full phase portrait combining local and global information.
+  \end{enumerate}
+\end{hintbox}