Chapter 1 • Theory & Derivations

Systems of Linear First-Order ODEs: Foundations, Operator Elimination & Phase Plane Topology

Transformation of higher-order differential equations into first-order dynamical systems, algebraic operator elimination, autonomous phase plane topology, and complete classification of equilibrium points.

§1.1Conversion of n-th Order Linear ODEs into First-Order Vector Systems & Fundamental Existence-Uniqueness Theorem

1. The Canonical Vector Formulation

In analytical dynamics, control theory, and multivariable physics, physical systems rarely present themselves as single isolated scalar equations. Instead, systems of interconnected components are modeled by coupled differential equations. Consider the general $n$-th order linear ordinary differential equation:

$$y^{(n)}(t) + p_{n-1}(t) y^{(n-1)}(t) + \dots + p_1(t) y'(t) + p_0(t) y(t) = g(t)$$

We convert this single $n$-th order equation into an equivalent system of $n$ first-order differential equations by introducing state variables:

$$\begin{aligned} x_1(t) &= y(t) \\ x_2(t) &= y'(t) = x_1'(t) \\ x_3(t) &= y''(t) = x_2'(t) \\ &\;\;\vdots \\ x_n(t) &= y^{(n-1)}(t) = x_{n-1}'(t) \end{aligned}$$

Differentiating the state vector $\vec{x}(t) = \begin{pmatrix} x_1(t) & x_2(t) & \dots & x_n(t) \end{pmatrix}^T$, the final derivative is:

$$x_n'(t) = y^{(n)}(t) = -p_0(t) x_1(t) - p_1(t) x_2(t) - \dots - p_{n-1}(t) x_n(t) + g(t)$$

In compact matrix-vector notation, this yields the standard linear dynamical system:

$$\frac{d\vec{x}}{dt} = \mathbf{A}(t) \vec{x}(t) + \vec{g}(t)$$

where the system companion matrix $\mathbf{A}(t) \in \mathbb{R}^{n \times n}$ and nonhomogeneous source vector $\vec{g}(t)$ are:

$$\mathbf{A}(t) = \begin{pmatrix} 0 & 1 & 0 & \dots & 0 \\ 0 & 0 & 1 & \dots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \dots & 1 \\ -p_0(t) & -p_1(t) & -p_2(t) & \dots & -p_{n-1}(t) \end{pmatrix}, \quad \vec{g}(t) = \begin{pmatrix} 0 \\ 0 \\ \vdots \\ 0 \\ g(t) \end{pmatrix}$$

2. The Picard-Lindelöf Theorem for Vector Systems

Theorem 1.1 (Existence and Uniqueness for Linear Systems): Let the matrix function $\mathbf{A}(t)$ and the vector function $\vec{g}(t)$ be continuous on an open interval $I = (\alpha, \beta) \subseteq \mathbb{R}$. Let $t_0 \in I$, and let $\vec{x}_0 \in \mathbb{R}^n$ be an arbitrary prescribed initial vector. Then the initial value problem (IVP): $$\frac{d\vec{x}}{dt} = \mathbf{A}(t) \vec{x}(t) + \vec{g}(t), \quad \vec{x}(t_0) = \vec{x}_0$$ possesses a unique continuously differentiable solution $\vec{x}(t)$ defined across the entire interval $I$.

The proof relies on constructing the vector Picard iteration operator $\mathcal{T}[\vec{x}](t) = \vec{x}_0 + \int_{t_0}^t [\mathbf{A}(s)\vec{x}(s) + \vec{g}(s)]\,ds$ and proving it is a contraction mapping on the Banach space $C(J, \mathbb{R}^n)$ under the supremum norm $\|\vec{x}\|_\infty = \sup_{t \in J} \|\vec{x}(t)\|$.

§1.2The Method of Elimination & Differential Polynomial Determinants

1. Differential Operator Notation

Let $D = \frac{d}{dt}$ denote the differential operator. A linear coupled system of two differential equations in dependent variables $x(t)$ and $y(t)$ can be expressed algebraically as:

$$\begin{aligned} L_1[x] + L_2[y] &= f_1(t) \\ L_3[x] + L_4[y] &= f_2(t) \end{aligned}$$

where $L_1, L_2, L_3, L_4 \in \mathbb{R}[D]$ are polynomial differential operators with constant coefficients. Written in operator matrix form:

$$\begin{pmatrix} L_1(D) & L_2(D) \\ L_3(D) & L_4(D) \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} f_1(t) \\ f_2(t) \end{pmatrix}$$

2. Elimination via Cramer's Operator Rule

Applying the operator determinant $\Delta(D) = L_1(D) L_4(D) - L_2(D) L_3(D)$ eliminates variables systematically:

$$\begin{aligned} \Delta(D) x(t) &= L_4(D) f_1(t) - L_2(D) f_2(t) \\ \Delta(D) y(t) &= L_1(D) f_2(t) - L_3(D) f_1(t) \end{aligned}$$

Caution on Arbitrary Constants: If $\Delta(D)$ has degree $m$, each scalar equation produces $m$ arbitrary constants. However, the original system requires only $m$ independent constants in total! Substituting the solutions back into the original coupled equations establishes the algebraic constraints connecting the constant pairs $(c_1, \dots, c_m)$ and $(k_1, \dots, k_m)$.

§1.3Autonomous 2D Linear Systems, Phase Portraits & Classification of Critical Equilibrium Points

1. Autonomous 2D Linear Systems

Consider the planar autonomous linear system:

$$\frac{d\vec{x}}{dt} = \mathbf{A} \vec{x}, \quad \mathbf{A} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \mathbb{R}^{2 \times 2}$$

The origin $\vec{x}^* = \begin{pmatrix} 0 & 0 \end{pmatrix}^T$ is the unique equilibrium point whenever $\det(\mathbf{A}) \ne 0$. The characteristic equation governing the eigenvalues $\lambda$ is:

$$\det(\mathbf{A} - \lambda \mathbf{I}) = \lambda^2 - \tau \lambda + \Delta = 0$$

where $\tau = \text{Tr}(\mathbf{A}) = a + d$ is the matrix trace, and $\Delta = \det(\mathbf{A}) = ad - bc$ is the determinant. The discriminant is:

$$\mathcal{D} = \tau^2 - 4\Delta$$

2. The Poincaré Trace-Determinant Classification

Classification of Critical Points:

$\Delta < 0$: Eigenvalues $\lambda_1 < 0 < \lambda_2$ are real with opposite signs. The origin is a Saddle Point (always asymptotically unstable). Orbits approach along the stable eigenvector and depart along the unstable eigenvector.
$\Delta > 0$ and $\mathcal{D} > 0$: Real distinct eigenvalues of the same sign.
- $\tau < 0 \implies \lambda_1, \lambda_2 < 0$: Stable Node (asymptotically stable attractor).
- $\tau > 0 \implies \lambda_1, \lambda_2 > 0$: Unstable Node (repeller).
$\Delta > 0$ and $\mathcal{D} < 0$: Complex conjugate eigenvalues $\lambda = \alpha \pm i\beta$ where $\alpha = \frac{\tau}{2}$ and $\beta = \frac{\sqrt{4\Delta - \tau^2}}{2}$.
- $\tau < 0 \implies \alpha < 0$: Stable Spiral (Focus). Orbits spiral inward toward the origin.
- $\tau > 0 \implies \alpha > 0$: Unstable Spiral (Focus). Orbits spiral outward to infinity.
- $\tau = 0 \implies \alpha = 0$: Center (Vortex). Eigenvalues are purely imaginary $\lambda = \pm i\beta$. Orbits form closed concentric ellipses (neutrally stable, non-isolated periodic orbits).
$\mathcal{D} = 0$: Repeated eigenvalue $\lambda_1 = \lambda_2 = \frac{\tau}{2}$.
- If $\mathbf{A} = \lambda \mathbf{I}$: Proper Star Node (every radial direction is an eigenvector trajectory).
- If $\mathbf{A} \ne \lambda \mathbf{I}$ (defective matrix): Degenerate / Improper Node (trajectories tangent to the single eigenvector span).

TIERED UNIVERSITY HONORS PROBLEMS

Step-by-Step Solved Examination Problems

Comprehensive analytical derivations, multi-tier solutions (Foundational, Intermediate Exam, and Honors/Proof Challenge) with complete line-by-line verification.

Tier 1 • Foundational Homogeneous 2x2 System with Real Distinct Eigenvalues

Solve the initial value problem for the linear system:
$$\frac{d\vec{x}}{dt} = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix} \vec{x}, \quad \vec{x}(0) = \begin{pmatrix} 3 \\ 1 \end{pmatrix}$$

Tier 2 • Intermediate Exam 3x3 System with Complex Eigenvalues & Invariant Orbital Manifold

Solve the $3 \times 3$ linear differential system:
$$\vec{x}' = \begin{pmatrix} 0 & 1 & 0 \\ -4 & 0 & 0 \\ 0 & 0 & -2 \end{pmatrix} \vec{x}, \quad \vec{x}(0) = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}$$

Tier 3 • Honors Challenge Matrix Riccati Transformation & Hamiltonian Linear System

Convert the nonlinear matrix Riccati differential equation $\mathbf{R}' = \mathbf{B} - \mathbf{R} \mathbf{D} \mathbf{R}$ into an equivalent $2n \times 2n$ linear Hamiltonian dynamical system $\vec{z}' = \mathbf{H}\vec{z}$.

Step 1: Fractional Linear Transformation
Let $\mathbf{R}(t) = \mathbf{X}(t) \mathbf{Y}^{-1}(t)$ where $\mathbf{X}(t), \mathbf{Y}(t) \in \mathbb{R}^{n \times n}$ and $\det \mathbf{Y}(t) \ne 0$.

Step 2: Differentiate the quotient
Using the matrix product rule and derivative of inverse $\frac{d}{dt}[\mathbf{Y}^{-1}] = -\mathbf{Y}^{-1} \mathbf{Y}' \mathbf{Y}^{-1}$: $$\mathbf{R}' = \mathbf{X}' \mathbf{Y}^{-1} + \mathbf{X} (\mathbf{Y}^{-1})' = \mathbf{X}' \mathbf{Y}^{-1} - \mathbf{X} \mathbf{Y}^{-1} \mathbf{Y}' \mathbf{Y}^{-1} = (\mathbf{X}' - \mathbf{R} \mathbf{Y}') \mathbf{Y}^{-1}$$
Step 3: Equate to Riccati expression
$$\mathbf{R}' = \mathbf{B} - \mathbf{R} \mathbf{D} \mathbf{R} = \mathbf{B} - \mathbf{R} \mathbf{D} \mathbf{X} \mathbf{Y}^{-1}$$ Equating both expressions and post-multiplying by $\mathbf{Y}$: $$\mathbf{X}' - \mathbf{R} \mathbf{Y}' = \mathbf{B} \mathbf{Y} - \mathbf{R} \mathbf{D} \mathbf{X}$$ Rearranging terms by isolating $\mathbf{R}$: $$\mathbf{X}' - \mathbf{B} \mathbf{Y} = \mathbf{R} (\mathbf{Y}' - \mathbf{D} \mathbf{X})$$
Step 4: Decouple into Linear System
This identity holds identically if both parenthetical sides vanish: $$\mathbf{X}' = \mathbf{B} \mathbf{Y}, \quad \mathbf{Y}' = \mathbf{D} \mathbf{X}$$ Stacking state vectors $\vec{z} = \begin{pmatrix} \mathbf{X} \\ \mathbf{Y} \end{pmatrix}$: $$\frac{d}{dt} \begin{pmatrix} \mathbf{X} \\ \mathbf{Y} \end{pmatrix} = \begin{pmatrix} \mathbf{0} & \mathbf{B} \\ \mathbf{D} & \mathbf{0} \end{pmatrix} \begin{pmatrix} \mathbf{X} \\ \mathbf{Y} \end{pmatrix}$$ This establishes that the solution of the nonlinear Riccati equation is $\mathbf{R}(t) = \mathbf{X}(t)\mathbf{Y}^{-1}(t)$ generated by the linear flow map $e^{\mathbf{H}t}$.

Final Answer & Analytical Insight

$\mathbf{R}(t) = \mathbf{X}(t)\mathbf{Y}(t)^{-1}$ where $\frac{d}{dt}\begin{pmatrix} \mathbf{X} \\ \mathbf{Y} \end{pmatrix} = \begin{pmatrix} \mathbf{0} & \mathbf{B} \\ \mathbf{D} & \mathbf{0} \end{pmatrix} \begin{pmatrix} \mathbf{X} \\ \mathbf{Y} \end{pmatrix}$.