Sturm-Liouville Theory, Self-Adjoint Operators & Oscillation Theorems
Formal self-adjoint differential operators, Lagrange's identity, regular Sturm-Liouville boundary value problems, reality of eigenvalues, eigenfunction orthogonality, and Sturm oscillation theorems.
§7.1 Formal Self-Adjoint Differential Operators & Lagrange's Identity
1. The Sturm-Liouville Differential Operator
Consider the second-order linear differential operator acting on $C^2[a, b]$:
$$L[y] = -\frac{d}{dx}\left[ p(x) \frac{dy}{dx} \right] + q(x) y$$where $p(x) > 0$, $p'(x)$, $q(x)$, and weight $w(x) > 0$ are continuous on $[a, b]$. Any linear equation $a_2(x)y'' + a_1(x)y' + a_0(x)y = 0$ with $a_2(x) > 0$ can be transformed into Sturm-Liouville form by multiplying by the integrating factor:
$$\mu(x) = \frac{1}{a_2(x)} \exp\left( \int \frac{a_1(x)}{a_2(x)}\,dx \right)$$2. Lagrange's Identity & Green's Formula
§7.2 Regular Sturm-Liouville Problems: Reality of Eigenvalues & Orthogonality
1. The Regular Sturm-Liouville Problem
The eigenvalue problem consists of $L[y] = \lambda w(x) y$ subject to separated boundary conditions:
$$\begin{aligned} \alpha_1 y(a) + \alpha_2 y'(a) &= 0 \quad (|\alpha_1| + |\alpha_2| > 0) \\ \beta_1 y(b) + \beta_2 y'(b) &= 0 \quad (|\beta_1| + |\beta_2| > 0) \end{aligned}$$Under these boundary conditions, the boundary term in Green's formula vanishes: $[-p(x)(u v' - v u')]_a^b = 0$. Hence $L$ is self-adjoint (Hermitian) on the domain of admissible boundary functions.
- All eigenvalues $\lambda_n$ are strictly real.
- The eigenvalues are countably infinite, discrete, bounded below, and can be ordered: $$\lambda_1 < \lambda_2 < \lambda_3 < \dots < \lambda_n \to \infty$$
- To each eigenvalue $\lambda_n$, there corresponds a unique eigenfunction $y_n(x)$ (up to a scalar multiple). There is no degeneracy: all eigenspaces are one-dimensional.
- Eigenfunctions corresponding to distinct eigenvalues are orthogonal with respect to $w(x)$: $$\int_a^b y_n(x) y_m(x) w(x)\,dx = 0 \quad (n \ne m)$$
§7.3 Sturm Oscillation Theorem & Sturm Comparison/Separation Theorems
1. The Sturm Oscillation Theorem
For example, the fundamental mode $y_1(x)$ has 0 interior zeros (does not change sign), $y_2(x)$ has 1 zero, $y_3(x)$ has 2 zeros, etc.
2. Sturm Separation and Comparison Theorems
Sturm's Comparison Theorem: Let $u'' + q_1(x)u = 0$ and $v'' + q_2(x)v = 0$ where $q_2(x) \ge q_1(x)$. If $x_1, x_2$ are consecutive zeros of $u(x)$, then $v(x)$ must have at least one zero in $[x_1, x_2]$.
§7.4 Generalized Fourier Series & Eigenfunction Completeness in L^2_w[a, b]
1. Generalized Fourier Series Expansions
The normalized eigenfunctions $\phi_n(x) = \frac{y_n(x)}{\sqrt{\int_a^b y_n^2 w\,dx}}$ form an orthonormal basis for the Hilbert space $L^2_w[a, b]$. Any piecewise smooth function $f(x)$ on $[a, b]$ can be expanded in a generalized Fourier series:
$$f(x) \sim \sum_{n=1}^\infty c_n \phi_n(x), \quad c_n = \langle f, \phi_n \rangle_w = \int_a^b f(x) \phi_n(x) w(x)\,dx$$The series converges pointwise to $\frac{f(x^+) + f(x^-)}{2}$ at all interior points $x \in (a, b)$, and satisfies Parseval's identity:
$$\|f\|_w^2 = \int_a^b [f(x)]^2 w(x)\,dx = \sum_{n=1}^\infty c_n^2$$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.
Find all eigenvalues and eigenfunctions of the regular Sturm-Liouville problem:
Case 1: $\lambda < 0$ ($\lambda = -\mu^2, \mu > 0$)
$y(x) = c_1 \cosh(\mu x) + c_2 \sinh(\mu x)$.
$y(0) = c_1 = 0 \implies y(x) = c_2 \sinh(\mu x)$.
$y'(\pi) = c_2 \mu \cosh(\mu \pi) = 0$. Since $\mu \ne 0$ and $\cosh(\mu \pi) \ge 1$, $c_2 = 0$. No non-trivial solutions.
Case 2: $\lambda = 0$
$y(x) = c_1 x + c_2$. $y(0) = c_2 = 0$. $y'(x) = c_1 \implies y'(\pi) = c_1 = 0$. Only trivial solution.
Case 3: $\lambda > 0$ ($\lambda = k^2, k > 0$)
$y(x) = c_1 \cos(kx) + c_2 \sin(kx)$.
$y(0) = c_1 = 0 \implies y(x) = c_2 \sin(kx)$.
$y'(x) = c_2 k \cos(kx) \implies y'(\pi) = c_2 k \cos(k\pi) = 0$.
For non-trivial solutions ($c_2 \ne 0$), we require:
Step 4: Eigenvalues & Orthonormal Eigenfunctions
Norm: $\int_0^\pi \sin^2\left(\frac{2n - 1}{2} x\right) dx = \frac{\pi}{2}$. Orthonormal set: $\phi_n(x) = \sqrt{\frac{2}{\pi}} \sin\left(\frac{2n - 1}{2} x\right)$.
$\lambda_n = \frac{(2n - 1)^2}{4}, \quad y_n(x) = \sin\left(\frac{2n - 1}{2} x\right), \quad n = 1, 2, 3, \dots$
Prove using Sturm's Separation/Comparison Theorem that between any two consecutive positive zeros of $J_0(x)$, there is exactly one zero of $J_1(x)$.
Step 1: Recurrence relation between $J_0$ and $J_1$
Recall from Bessel recurrence relations:
Step 2: Apply Rolle's Theorem
Let $0 < x_1 < x_2$ be two consecutive positive zeros of $J_0(x)$, so $J_0(x_1) = 0$ and $J_0(x_2) = 0$, and $J_0(x) \ne 0$ for all $x \in (x_1, x_2)$.
Since $J_0(x)$ is continuously differentiable on $[x_1, x_2]$, Rolle's theorem guarantees that there exists at least one point $\xi \in (x_1, x_2)$ such that $J_0'(\xi) = 0$.
Since $J_0'(x) = -J_1(x)$, this means $J_1(\xi) = 0$. Thus $J_1$ has at least one zero in $(x_1, x_2)$.
Step 3: Uniqueness via Sturm Separation
Now consider the identity:
Suppose $J_1(x)$ had two zeros $\xi_1 < \xi_2$ in $(x_1, x_2)$. Then applying Rolle's theorem to $g(x) = x J_1(x)$ on $[\xi_1, \xi_2]$ would imply $g'(\eta) = \eta J_0(\eta) = 0$ for some $\eta \in (\xi_1, \xi_2) \subset (x_1, x_2)$.
Since $\eta > 0$, this would force $J_0(\eta) = 0$, directly contradicting that $x_1$ and $x_2$ are consecutive zeros of $J_0(x)$!
Conclusion: There is strictly one zero of $J_1(x)$ between any two consecutive zeros of $J_0(x)$. The zeros strictly interlace: $0 < j_{0, 1} < j_{1, 1} < j_{0, 2} < j_{1, 2} < \dots$.
Proved: zeros of $J_0(x)$ and $J_1(x)$ strictly interlace.
Establish the self-adjointness, non-negative spectrum, and double degeneracy for the periodic Sturm-Liouville problem:
Step 1: Self-Adjointness via Green's Formula
For $L[y] = -y''$:
Using periodic conditions $u(-\pi) = u(\pi)$ and $u'(-\pi) = u'(\pi)$:
The periodic boundary conditions make the boundary term vanish. Hence $L$ is self-adjoint.
Step 2: Non-negative eigenvalues ($\lambda \ge 0$)
Multiply by $y$ and integrate by parts:
Hence $\lambda \ge 0$.
Step 3: Eigenvalues & Degeneracy
For $\lambda_0 = 0$: $y_0(x) = 1$ (non-degenerate, dimension 1).
For $\lambda_n = n^2 > 0$ ($n = 1, 2, 3, \dots$): Both $y_{n, 1}(x) = \cos(nx)$ and $y_{n, 2}(x) = \sin(nx)$ satisfy the periodic boundary conditions! Thus every eigenvalue $\lambda_n = n^2$ has multiplicity 2 (doubly degenerate eigenspace spanned by $\{\cos nx, \sin nx\}$), generating the classical full Fourier series!
$\lambda_0 = 0$ (simple), $\lambda_n = n^2$ ($n \ge 1$, doubly degenerate with basis $\{\cos nx, \sin nx\}$).