fix(ch8): correct eigenvalue example, add missing phase portraits, clean stability example

chapters/ch08_systems.tex

@@ -681,6 +681,74 @@ Trajectories spiral toward the origin. The number of complete rotations depends
 
 Trajectories are closed orbits (circles or ellipses). The equilibrium is \textbf{Lyapunov stable} but not asymptotically stable: nearby solutions stay nearby but do not converge to the origin.
 
+\subsubsection{Unstable Node (Source)}
+
+\begin{center}
+\begin{tikzpicture}[scale=1.0]
+  % Axes
+  \draw[->, thick] (-3.2,0) -- (3.2,0) node[right] {$x$};
+  \draw[->, thick] (0,-3.2) -- (0,3.2) node[above] {$y$};
+
+  % Eigenvector lines (dashed)
+  \draw[dashed, thick, blue!50] (-0.6,-1.8) -- (0.6,1.8);
+  \draw[dashed, thick, red!50] (-2,-1) -- (2,1);
+
+  % Trajectories (all outward)
+  \draw[thick, ->] (0.1,0.04) .. controls (0.6,0.7) and (1.5,1.8) .. (2.5,3);
+  \draw[thick, ->] (0.1,-0.04) .. controls (0.6,-0.7) and (1.5,-1.8) .. (2.5,-3);
+  \draw[thick, ->] (-0.1,0.04) .. controls (-0.6,0.7) and (-1.5,1.8) .. (-2.5,3);
+  \draw[thick, ->] (-0.1,-0.04) .. controls (-0.6,-0.7) and (-1.5,-1.8) .. (-2.5,-3);
+
+  \draw[thick, ->] (0.1,0.005) .. controls (0.5,0.05) and (1.8,0.2) .. (3,0.5);
+  \draw[thick, ->] (-0.1,0.005) .. controls (-0.5,0.05) and (-1.8,0.2) .. (-3,0.5);
+  \draw[thick, ->] (0.1,-0.005) .. controls (0.5,-0.05) and (1.8,-0.2) .. (3,-0.5);
+  \draw[thick, ->] (-0.1,-0.005) .. controls (-0.5,-0.05) and (-1.8,-0.2) .. (-3,-0.5);
+
+  \draw[thick, ->] (0.005,0.05) .. controls (0.02,0.3) and (0.1,1.5) .. (0,3);
+  \draw[thick, ->] (-0.005,-0.05) .. controls (-0.02,-0.3) and (-0.1,-1.5) .. (0,-3);
+
+  % Equilibrium point
+  \filldraw[black] (0,0) circle (2pt) node[below left=1pt] {$\mathbf{0}$};
+
+  % Label
+  \node[font=\footnotesize, align=center] at (2.5,2.7) {Unstable node (source)\\$(\lambda_1 > \lambda_2 > 0)$};
+\end{tikzpicture}
+\end{center}
+
+All trajectories diverge from the origin. Far from the origin, trajectories become tangent to the eigenvector associated with the larger eigenvalue (the ``fast'' direction).
+
+\subsubsection{Unstable Spiral (Source Spiral)}
+
+\begin{center}
+\begin{tikzpicture}[scale=1.0]
+  % Axes
+  \draw[->, thick] (-3.2,0) -- (3.2,0) node[right] {$x$};
+  \draw[->, thick] (0,-3.2) -- (0,3.2) node[above] {$y$};
+
+  % Spiral trajectory (outward)
+  \draw[thick, purple!80] plot[domain=-20:0, samples=200, smooth, variable=\t]
+    ({0.15*\t*exp(0.25*\t)*cos(1.2*\t)}, {0.15*\t*exp(0.25*\t)*sin(1.2*\t)});
+
+  % Arrows along the spiral (outward)
+  \draw[thick, purple!80, ->] (0.5,-0.35) -- (1.1,-0.2);
+  \draw[thick, purple!80, ->] (1.6,0.9) -- (2.0,1.5);
+  \draw[thick, purple!80, ->] (-1.8,1.7) -- (-2.3,1.2);
+  \draw[thick, purple!80, ->] (-1.0,-1.4) -- (-1.5,-1.9);
+
+  % Second spiral
+  \draw[thick, orange!60] plot[domain=-18:-3, samples=200, smooth, variable=\t]
+    ({0.25*\t*exp(0.25*\t)*cos(1.2*\t + 1)}, {0.25*\t*exp(0.25*\t)*sin(1.2*\t + 1)});
+
+  % Equilibrium point
+  \filldraw[black] (0,0) circle (2pt) node[below left=1pt] {$\mathbf{0}$};
+
+  % Label
+  \node[font=\footnotesize, align=center] at (2.3,2.7) {Unstable spiral\\$(\alpha \pm i\beta,\;\alpha > 0)$};
+\end{tikzpicture}
+\end{center}
+
+Trajectories spiral outward from the origin. The rate of divergence is governed by the real part $\alpha$ of the eigenvalues, while the frequency of rotation depends on $\beta$.
+
 \subsubsection{Classifying Equilibria: Worked Examples}
 
 \begin{workedexample}
@@ -690,19 +758,30 @@ Trajectories are closed orbits (circles or ellipses). The equilibrium is \textbf
 \end{workedexample}
 
 \begin{workedexample}
-  Classify the equilibrium of $\mathbf{x}' = \begin{pmatrix} -1 & -2 \\ -2 & -4 \end{pmatrix}\mathbf{x}$.
+  Classify the equilibrium of $\mathbf{x}' = \begin{pmatrix} -2 & 1 \\ 0 & -3 \end{pmatrix}\mathbf{x}$.
 
-  \textbf{Solution.} The characteristic polynomial is
+  \textbf{Solution.} Since $A$ is upper triangular, the eigenvalues are the diagonal entries:
+  $\lambda_1 = -2$ and $\lambda_2 = -3$. Both are negative $\implies$ \textbf{stable node (sink)}.
+
+  \textbf{Eigenvectors.} For $\lambda_1 = -2$, solve $(A + 2I)\mathbf{v} = \mathbf{0}$:
   \[
-    \det\!\begin{pmatrix} -1-\lambda & -2 \\ -2 & -4-\lambda \end{pmatrix}
+    \begin{pmatrix} 0 & 1 \\ 0 & -1 \end{pmatrix}
-    = (-1-\lambda)(-4-\lambda) - 4
+    \begin{pmatrix} v_1 \\ v_2 \end{pmatrix}
-    = \lambda^2 + 5\lambda + 4
+    =
-    = (\lambda+1)(\lambda+4).
+    \begin{pmatrix} 0 \\ 0 \end{pmatrix},
   \]
-  Eigenvalues: $\lambda_1 = -1$, $\lambda_2 = -4$. Both negative $\implies$ \textbf{stable node (sink)}.
+  giving $v_2 = 0$ and $\mathbf{v}_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ (fast eigendirection, along the $x$-axis).
 
-  Eigenvectors: For $\lambda_1 = -1$, $(A+I)\mathbf{v} = \mathbf{0}$ gives $\begin{pmatrix} 0 & -2 \\ -2 & -3 \end{pmatrix}\mathbf{v} = \mathbf{0}$, so $v_2 = 0$ and $\mathbf{v}_1 = (1,0)^\top$.
+  For $\lambda_2 = -3$, solve $(A + 3I)\mathbf{v} = \mathbf{0}$:
-  For $\lambda_2 = -4$, $(A+4I)\mathbf{v} = \mathbf{0}$ gives $\begin{pmatrix} 3 & -2 \\ -2 & 0 \end{pmatrix}\mathbf{v} = \mathbf{0}$, so $v_2 = 0$ and $3v_1 = 0$, hence $\mathbf{v}_2 = (0,1)^\top$ is incorrect; instead $-2v_1 = 0 \implies v_1 = 0$, so $\mathbf{v}_2 = (0,1)^\top$.
+  \[
+    \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}
+    \begin{pmatrix} v_1 \\ v_2 \end{pmatrix}
+    =
+    \begin{pmatrix} 0 \\ 0 \end{pmatrix},
+  \]
+  giving $v_1 + v_2 = 0$, so $\mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$ (slow eigendirection).
+
+  Since $\lvert\lambda_1\rvert > \lvert\lambda_2\rvert$, trajectories approach the origin tangent to the slow eigenvector $\mathbf{v}_2$.
 \end{workedexample}
 
 \subsection{Trace-Determinant Classification}
@@ -874,19 +953,21 @@ Stability analysis for linear systems follows directly from the eigenvalue struc
 \end{workedexample}
 
 \begin{workedexample}
-  Show that $\mathbf{x}' = \begin{pmatrix} 1 & -4 \\ 1 & -3 \end{pmatrix}\mathbf{x}$ is unstable.
+  Show that $\mathbf{x}' = \begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix}\mathbf{x}$ is unstable.
 
   \textbf{Solution.}
   \[
-    \tau = 1 + (-3) = -2, \qquad \Delta = (1)(-3) - (-4)(1) = -3 + 4 = 1.
+    \tau = \tr(A) = 3 + 3 = 6, \qquad
+    \Delta = \det(A) = (3)(3) - (1)(1) = 9 - 1 = 8.
   \]
-  Wait --- let me recompute: $\Delta = -3 - (-4) = -3 + 4 = 1 > 0$ and $\tau = -2 < 0$. By the criterion, this would be stable. Let me check the eigenvalues:
+  Discriminant: $D = \tau^2 - 4\Delta = 36 - 32 = 4 > 0$. Since $\tau = 6 > 0$ and $\Delta = 8 > 0$ with $D > 0$, the point $(6, 8)$ lies in the \textbf{unstable node} region (right of $\Delta$-axis, above the parabola).
-  \[
-    \lambda^2 + 2\lambda + 1 = (\lambda + 1)^2 = 0, \quad\text{so } \lambda = -1 \text{ (double)}.
-  \]
-  Since $\lambda = -1 < 0$ and the repeated eigenvalue has negative real part, the system is asymptotically stable (degenerate stable node).
 
-  For an unstable example, consider $A = \begin{pmatrix} 2 & 1 \\ 0 & -1 \end{pmatrix}$. Here $\tau = 1 > 0$ and $\Delta = -2 < 0$. The eigenvalues are $\lambda_1 = 2$ and $\lambda_2 = -1$ (opposite signs), so the system is a \textbf{saddle} --- unstable.
+  Verification: eigenvalues satisfy $\lambda^2 - 6\lambda + 8 = 0$, giving
+  \[
+    \lambda = \frac{6 \pm \sqrt{36-32}}{2} = \frac{6 \pm 2}{2},
+    \quad\text{so}\quad \lambda_1 = 4, \;\; \lambda_2 = 2.
+  \]
+  Both eigenvalues are positive, confirming an \textbf{unstable node (source)}. All nonzero solutions diverge from the origin as $t \to \infty$.
 \end{workedexample}
 
 \subsection{Matrix Exponential}