Chapter 3 • Theory & Derivations

The Fundamental Matrix, Matrix Exponential & Nonhomogeneous Systems

Fundamental matrix manifolds, the matrix exponential operator e^(At) via Putzer's algorithm, and nonhomogeneous systems via matrix variation of parameters.

§3.1 The Fundamental Matrix Φ(t), System Wronskian & Abel-Liouville Identity

1. The Fundamental Matrix

Let $\vec{x}_1(t), \vec{x}_2(t), \dots, \vec{x}_n(t)$ be $n$ linearly independent solutions of $\vec{x}' = \mathbf{A}(t)\vec{x}$ on an interval $I$. The $n \times n$ matrix whose columns are these solution vectors:

$$\mathbf{\Phi}(t) = \begin{pmatrix} \vec{x}_1(t) & \vec{x}_2(t) & \dots & \vec{x}_n(t) \end{pmatrix}$$

is called a fundamental matrix of the system. By construction, it satisfies the matrix differential equation:

$$\frac{d\mathbf{\Phi}}{dt} = \mathbf{A}(t) \mathbf{\Phi}(t)$$

The general homogeneous solution is simply $\vec{x}(t) = \mathbf{\Phi}(t)\vec{c}$ where $\vec{c} \in \mathbb{R}^n$ is a constant vector.

2. The System Wronskian & Abel-Liouville Formula

The Wronskian determinant of the system is $W(t) = \det \mathbf{\Phi}(t)$. Differentiating the determinant:

$$\frac{dW}{dt} = \text{Tr}(\mathbf{A}(t)) W(t)$$

Theorem 3.1 (Abel-Liouville Identity for Systems): For any fundamental matrix $\mathbf{\Phi}(t)$ on $I$, and any $t_0 \in I$: $$W(t) = W(t_0) \exp\left( \int_{t_0}^t \text{Tr}(\mathbf{A}(s))\,ds \right)$$ Consequently, $W(t) \ne 0$ for all $t \in I$ if and only if $W(t_0) \ne 0$.

§3.2 The Matrix Exponential e^(At): Putzer's Algorithm, Cayley-Hamilton & Series

1. The Matrix Exponential Operator

For any constant square matrix $\mathbf{A} \in \mathbb{C}^{n \times n}$, the matrix exponential is defined by the absolutely convergent power series:

$$e^{\mathbf{A}t} = \sum_{k=0}^\infty \frac{t^k}{k!} \mathbf{A}^k = \mathbf{I} + t\mathbf{A} + \frac{t^2}{2!} \mathbf{A}^2 + \frac{t^3}{3!} \mathbf{A}^3 + \dots$$

The matrix exponential $e^{\mathbf{A}t}$ is the unique fundamental matrix that equals the identity at $t = 0$: $\mathbf{\Phi}(0) = \mathbf{I}$. The unique solution to the IVP $\vec{x}' = \mathbf{A}\vec{x}, \vec{x}(0) = \vec{x}_0$ is:

$$\vec{x}(t) = e^{\mathbf{A}t} \vec{x}_0$$

2. Putzer's Algorithm

Putzer's algorithm computes $e^{\mathbf{A}t}$ without matrix inversions or Jordan canonical forms. Let $\lambda_1, \lambda_2, \dots, \lambda_n$ be the eigenvalues of $\mathbf{A}$ in any order. Define the polynomial sequence of matrices:

$$\mathbf{P}_0 = \mathbf{I}, \quad \mathbf{P}_k = \prod_{j=1}^k (\mathbf{A} - \lambda_j \mathbf{I}) = (\mathbf{A} - \lambda_k \mathbf{I}) \mathbf{P}_{k-1}, \quad k = 1, \dots, n-1$$

Then the matrix exponential is expressed as:

$$e^{\mathbf{A}t} = \sum_{k=0}^{n-1} r_{k+1}(t) \mathbf{P}_k$$

where the scalar functions $r_1(t), \dots, r_n(t)$ satisfy the triangular scalar differential system:

$$\begin{aligned} r_1'(t) &= \lambda_1 r_1(t), \quad r_1(0) = 1 \\ r_k'(t) &= \lambda_k r_k(t) + r_{k-1}(t), \quad r_k(0) = 0 \quad (k = 2, \dots, n) \end{aligned}$$

§3.3 Nonhomogeneous Systems: Undetermined Coefficients & Variation of Parameters

1. Matrix Variation of Parameters

Consider the nonhomogeneous linear system $\vec{x}'(t) = \mathbf{A}(t)\vec{x}(t) + \vec{g}(t)$. Let $\mathbf{\Phi}(t)$ be a fundamental matrix for the associated homogeneous system $\vec{x}' = \mathbf{A}(t)\vec{x}$. We seek a particular solution of the form:

$$\vec{x}_p(t) = \mathbf{\Phi}(t) \vec{u}(t)$$

Differentiating using the product rule:

$$\vec{x}_p'(t) = \mathbf{\Phi}'(t) \vec{u}(t) + \mathbf{\Phi}(t) \vec{u}'(t) = \mathbf{A}(t) \mathbf{\Phi}(t) \vec{u}(t) + \mathbf{\Phi}(t) \vec{u}'(t)$$

Substituting into the nonhomogeneous equation:

$$\mathbf{A}(t) \mathbf{\Phi}(t) \vec{u}(t) + \mathbf{\Phi}(t) \vec{u}'(t) = \mathbf{A}(t) \mathbf{\Phi}(t) \vec{u}(t) + \vec{g}(t)$$ $$\mathbf{\Phi}(t) \vec{u}'(t) = \vec{g}(t) \implies \vec{u}'(t) = \mathbf{\Phi}^{-1}(t) \vec{g}(t)$$

Theorem 3.2 (Variation of Parameters Formula for Systems): Integrating $\vec{u}'(t)$ from $t_0$ to $t$ yields the complete general solution: $$\vec{x}(t) = \mathbf{\Phi}(t)\mathbf{\Phi}^{-1}(t_0) \vec{x}_0 + \mathbf{\Phi}(t) \int_{t_0}^t \mathbf{\Phi}^{-1}(s) \vec{g}(s)\,ds$$ For constant matrix $\mathbf{A}$, where $\mathbf{\Phi}(t) = e^{\mathbf{A}t}$, this simplifies to Duhamel's convolution integral: $$\vec{x}(t) = e^{\mathbf{A}(t - t_0)} \vec{x}_0 + \int_{t_0}^t e^{\mathbf{A}(t - s)} \vec{g}(s)\,ds$$

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 Example 3.1: Variation of Parameters for Nonhomogeneous 2x2 System

Find a particular solution using Variation of Parameters for the system:

\vec{x}' = \begin{pmatrix} 0 & 1 \\ -2 & 3 \end{pmatrix} \vec{x} + \begin{pmatrix} 0 \\ e^t \end{pmatrix}

Step 1: Homogeneous eigenvalues and eigenvectors
$\det(\mathbf{A} - \lambda \mathbf{I}) = \lambda^2 - 3\lambda + 2 = (\lambda - 1)(\lambda - 2) = 0 \implies \lambda_1 = 1, \lambda_2 = 2$.
For $\lambda_1 = 1$: $(\mathbf{A} - \mathbf{I})\vec{v}_1 = \begin{pmatrix} -1 & 1 \\ -2 & 2 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \vec{0} \implies \vec{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$.
For $\lambda_2 = 2$: $(\mathbf{A} - 2\mathbf{I})\vec{v}_2 = \begin{pmatrix} -2 & 1 \\ -2 & 1 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \vec{0} \implies \vec{v}_2 = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$.

Step 2: Fundamental Matrix $\mathbf{\Phi}(t)$

\mathbf{\Phi}(t) = \begin{pmatrix} e^t & e^{2t} \\ e^t & 2e^{2t} \end{pmatrix}, \quad \det \mathbf{\Phi}(t) = 2e^{3t} - e^{3t} = e^{3t}

\mathbf{\Phi}^{-1}(t) = \frac{1}{e^{3t}} \begin{pmatrix} 2e^{2t} & -e^{2t} \\ -e^t & e^t \end{pmatrix} = \begin{pmatrix} 2e^{-t} & -e^{-t} \\ -e^{-2t} & e^{-2t} \end{pmatrix}

Step 3: Integrate $\vec{u}'(t) = \mathbf{\Phi}^{-1}(t) \vec{g}(t)$

\vec{u}'(t) = \begin{pmatrix} 2e^{-t} & -e^{-t} \\ -e^{-2t} & e^{-2t} \end{pmatrix} \begin{pmatrix} 0 \\ e^t \end{pmatrix} = \begin{pmatrix} -1 \\ e^{-t} \end{pmatrix}

Integrating with respect to $t$:

\vec{u}(t) = \begin{pmatrix} -t \\ -e^{-t} \end{pmatrix}

Step 4: Form Particular Solution $\vec{x}_p(t) = \mathbf{\Phi}(t) \vec{u}(t)$

\vec{x}_p(t) = \begin{pmatrix} e^t & e^{2t} \\ e^t & 2e^{2t} \end{pmatrix} \begin{pmatrix} -t \\ -e^{-t} \end{pmatrix} = \begin{pmatrix} -t e^t - e^t \\ -t e^t - 2e^t \end{pmatrix}

Final Answer & Physical Insight

$\vec{x}_p(t) = \begin{pmatrix} -(t + 1)e^t \\ -(t + 2)e^t \end{pmatrix}$

Tier 2 • Intermediate Exam Example 3.2: Bessel Differential Recurrence & Integral Evaluation

Prove the recurrence identity $\frac{d}{dx}[x^\nu J_\nu(x)] = x^\nu J_{\nu-1}(x)$ and use it to evaluate $\int x^3 J_0(x)\,dx$.

Step 1: Proof of identity
From the series definition:

x^\nu J_\nu(x) = \sum_{m=0}^\infty \frac{(-1)^m}{m!\, \Gamma(m + \nu + 1)} \frac{x^{2m + 2\nu}}{2^{2m + \nu}}

Differentiating with respect to $x$:

\frac{d}{dx}[x^\nu J_\nu(x)] = \sum_{m=0}^\infty \frac{(-1)^m (2m + 2\nu)}{m!\, \Gamma(m + \nu + 1)} \frac{x^{2m + 2\nu - 1}}{2^{2m + \nu}}

Since $2m + 2\nu = 2(m + \nu)$ and $\Gamma(m + \nu + 1) = (m + \nu)\Gamma(m + \nu)$:

= x^\nu \sum_{m=0}^\infty \frac{(-1)^m}{m!\, \Gamma(m + \nu)} \left(\frac{x}{2}\right)^{2m + \nu - 1} = x^\nu J_{\nu-1}(x)

Step 2: Integration of $\int x^3 J_0(x)\,dx$
Rewrite integrand as $x^2 \cdot [x J_0(x)]$. From identity with $\nu = 1$: $\frac{d}{dx}[x J_1(x)] = x J_0(x) \implies \int x J_0(x)\,dx = x J_1(x)$.
Integrate by parts:

u = x^2 \implies du = 2x\,dx, \quad dv = x J_0(x)\,dx \implies v = x J_1(x)

\int x^3 J_0(x)\,dx = x^2 [x J_1(x)] - \int [x J_1(x)](2x)\,dx = x^3 J_1(x) - 2 \int x^2 J_1(x)\,dx

Now apply identity with $\nu = 2$: $\frac{d}{dx}[x^2 J_2(x)] = x^2 J_1(x) \implies \int x^2 J_1(x)\,dx = x^2 J_2(x)$.

$$= x^3 J_1(x) - 2 x^2 J_2(x) + C$$

Using recurrence $J_2(x) = \frac{2}{x} J_1(x) - J_0(x)$:

x^3 J_1(x) - 2x^2 \left( \frac{2}{x} J_1(x) - J_0(x) \right) = (x^3 - 4x) J_1(x) + 2x^2 J_0(x) + C

Final Answer & Physical Insight

$\int x^3 J_0(x)\,dx = x^3 J_1(x) - 2x^2 J_2(x) + C = (x^3 - 4x)J_1(x) + 2x^2 J_0(x) + C$

Tier 3 • Honors Challenge Example 3.3: Associated Laguerre Transformation of Hydrogen Radial Schrödinger Equation

Transform the radial hydrogenic Schrödinger equation $\frac{d^2 R}{dr^2} + \frac{2}{r} \frac{dR}{dr} + \left[ \frac{2Z}{r} - \frac{l(l+1)}{r^2} - \kappa^2 \right] R = 0$ into the confluent hypergeometric Associated Laguerre equation.

Step 1: Asymptotic Behavior
As $r \to \infty$: $R'' - \kappa^2 R \approx 0 \implies R(r) \sim e^{-\kappa r}$.
As $r \to 0$: $R'' + \frac{2}{r} R' - \frac{l(l+1)}{r^2} R \approx 0 \implies R(r) \sim r^l$.

Step 2: Dimensionless variable and ansatz
Define dimensionless coordinate $\rho = 2\kappa r$.
Let $R(\rho) = \rho^l e^{-\rho/2} v(\rho)$.

Step 3: Derivatives of the ansatz
Compute $R'$ and $R''$ in terms of $\rho$ and substitute into the radial equation. After factorizing $\rho^l e^{-\rho/2}$, the equation reduces to:

\rho \frac{d^2 v}{d\rho^2} + [2(l + 1) - \rho] \frac{dv}{d\rho} + (n - l - 1) v = 0

where $n = \frac{Z}{\kappa}$ is the principal quantum number.

Step 4: Associated Laguerre Polynomials
Let $k = 2l + 1$ and $p = n - l - 1 \ge 0$. The equation becomes:

\rho v'' + (k + 1 - \rho) v' + p v = 0

which is precisely the Associated Laguerre differential equation! The solutions are $v(\rho) = L_{n - l - 1}^{2l + 1}(\rho)$, proving that the bound state energy levels are quantized:

E_n = -\frac{\hbar^2 \kappa^2}{2\mu} = -\frac{\mu Z^2 e^4}{2\hbar^2 n^2}, \quad n = l + 1, l + 2, \dots

Final Answer & Physical Insight

$\rho v'' + (2l + 2 - \rho)v' + (n - l - 1)v = 0$ with solution $v(\rho) = L_{n-l-1}^{2l+1}(2\kappa r)$.