Chapter 8 • Theory & Derivations

Nonhomogeneous Boundary Value Problems, The Fredholm Alternative & Green's Functions

Solvability criteria via the Fredholm Alternative, the Dirac delta distribution, jump conditions, and explicit closed-form construction of Green's functions.

§8.1 Nonhomogeneous Boundary Value Problems & Solvability via The Fredholm Alternative

1. Inhomogeneous Boundary Value Problems

Consider the nonhomogeneous boundary value problem:

$$L[y] = f(x), \quad x \in [a, b], \quad B_1[y] = 0, \quad B_2[y] = 0$$

where $L$ is a self-adjoint differential operator with homogeneous boundary operators $B_1, B_2$.

Theorem 8.1 (The Fredholm Alternative for Differential Operators):

Case I: If the homogeneous problem $L[y] = 0, B_1[y] = 0, B_2[y] = 0$ has only the trivial solution $y \equiv 0$, then for any continuous function $f(x)$, the nonhomogeneous problem has a unique solution $y(x)$.
Case II: If the homogeneous problem has non-trivial solutions $\phi_1(x), \dots, \phi_k(x)$, then the nonhomogeneous problem has solutions if and only if $f(x)$ is orthogonal to every homogeneous solution: $$\int_a^b f(x) \phi_j(x)\,dx = 0 \quad \text{for all } j = 1, \dots, k$$ If this orthogonality condition holds, there exists an infinite family of solutions.

§8.2 Green's Functions: Definition, Dirac Delta Distribution & Jump Discontinuity Conditions

1. The Green's Function as Impulse Response

The Green's function $G(x, \xi)$ is the kernel representing the physical response at point $x$ produced by an idealized unit point-source impulse $\delta(x - \xi)$ applied at $\xi \in (a, b)$:

$$L[G(x, \xi)] = \delta(x - \xi), \quad B_1[G] = 0, \quad B_2[G] = 0$$

By linearity and superposition, once $G(x, \xi)$ is known, the solution for any distributed source $f(x)$ is given by the integral convolution:

$$y(x) = \int_a^b G(x, \xi) f(\xi)\,d\xi$$

2. The Four Defining Properties of G(x, ξ)

Axiomatic Properties of Green's Function:

Differential Equation: For $x \ne \xi$, $L[G(x, \xi)] = 0$.
Boundary Conditions: $G(x, \xi)$ satisfies the homogeneous boundary conditions $B_1[G] = 0$ at $x = a$ and $B_2[G] = 0$ at $x = b$.
Continuity at $x = \xi$: $G(x, \xi)$ is continuous across the point source: $$\lim_{x \to \xi^+} G(x, \xi) = \lim_{x \to \xi^-} G(x, \xi)$$
Jump Discontinuity in Derivative: Integrating $L[G] = -\frac{d}{dx}[p(x) G'] + q(x)G = \delta(x - \xi)$ over $[\xi - \epsilon, \xi + \epsilon]$ as $\epsilon \to 0$ yields the jump condition: $$\left. \frac{\partial G}{\partial x} \right|_{x = \xi^+} - \left. \frac{\partial G}{\partial x} \right|_{x = \xi^-} = -\frac{1}{p(\xi)}$$

§8.3 Explicit Closed-Form Construction of Green's Functions for Second-Order Operators

1. Closed-Form Construction Recipe

Let $y_1(x)$ be a non-trivial solution of $L[y] = 0$ satisfying the boundary condition at $x = a$, and let $y_2(x)$ be a non-trivial solution satisfying the boundary condition at $x = b$.

Since $G(x, \xi)$ must satisfy $B_1[G] = 0$ for $x < \xi$ and $B_2[G] = 0$ for $x > \xi$:

$$G(x, \xi) = \begin{cases} c_1(\xi) y_1(x), & a \le x \le \xi \\ c_2(\xi) y_2(x), & \xi \le x \le b \end{cases}$$

Applying the continuity condition $c_2(\xi) y_2(\xi) - c_1(\xi) y_1(\xi) = 0$ and the jump condition $c_2(\xi) y_2'(\xi) - c_1(\xi) y_1'(\xi) = -\frac{1}{p(\xi)}$ gives a $2 \times 2$ linear system for $c_1, c_2$ with determinant $W(y_1, y_2)(\xi)$:

$$c_1(\xi) = -\frac{y_2(\xi)}{p(\xi) W(\xi)}, \quad c_2(\xi) = -\frac{y_1(\xi)}{p(\xi) W(\xi)}$$

Theorem 8.2 (Universal Green's Function Formula): $$G(x, \xi) = \begin{cases} -\frac{y_1(x) y_2(\xi)}{p(\xi) W(\xi)}, & a \le x \le \xi \\ -\frac{y_1(\xi) y_2(x)}{p(\xi) W(\xi)}, & \xi \le x \le b \end{cases}$$ Because $p(\xi) W(\xi) = \text{const}$ by Abel's identity, $G(x, \xi)$ is perfectly symmetric: $G(x, \xi) = G(\xi, x)$.

§8.4 Green's Function Representation of Inhomogeneous Solutions & Physical Applications

1. Physical Example: Deflection of a Taut Elastic String

A taut string of length $L$ under uniform tension $T$ clamped at both ends ($y(0) = 0, y(L) = 0$) subject to transverse load density $f(x)$ satisfies:

$$-T y''(x) = f(x), \quad y(0) = 0, \quad y(L) = 0$$

Here $p(x) = T$. The solutions satisfying the boundary conditions are $y_1(x) = x$ and $y_2(x) = L - x$. The Wronskian is $W(y_1, y_2) = x(-1) - (L - x)(1) = -L$. Hence $p W = -TL$. The Green's function is:

$$G(x, \xi) = \begin{cases} \frac{x(L - \xi)}{TL}, & 0 \le x \le \xi \\ \frac{\xi(L - x)}{TL}, & \xi \le x \le L \end{cases}$$

The deflection caused by a uniform distributed gravitational load $f(x) = \rho g$ is:

$$y(x) = \int_0^L G(x, \xi) \rho g\,d\xi = \frac{\rho g}{2T} x(L - x)$$

which matches the classical parabolic profile of hanging cables and beam theory.

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 8.1: Explicit Construction of Green's Function for y'' = f(x)

Construct the Green's function for the Dirichlet boundary value problem:

y'' = f(x), \quad y(0) = 0, \quad y(1) = 0

and find the solution for $f(x) = x$.

Step 1: Linearly independent solutions satisfying boundary conditions
Homogeneous ODE: $y'' = 0 \implies y(x) = c_1 x + c_2$.
Left solution satisfying $y_1(0) = 0$: $y_1(x) = x$.
Right solution satisfying $y_2(1) = 0$: $y_2(x) = 1 - x$.

Step 2: Wronskian computation

$$W(y_1, y_2) = y_1 y_2' - y_2 y_1' = x(-1) - (1 - x)(1) = -x - 1 + x = -1$$

In $L[y] = -y''$, $p(x) = 1$. So $-p W = -1(-1) = 1$.

Step 3: Green's Function Formulation
For $-y'' = -f(x)$, or directly:

G(x, \xi) = \begin{cases} x(1 - \xi), & 0 \le x \le \xi \\ \xi(1 - x), & \xi \le x \le 1 \end{cases}

Step 4: Solution for $f(x) = x$ (where $-y'' = -x \implies y'' = x$)

y(x) = \int_0^1 G(x, \xi) (-\xi)\,d\xi = -\int_0^x \xi(1 - x) \xi\,d\xi - \int_x^1 x(1 - \xi) \xi\,d\xi

First integral:

-(1 - x) \int_0^x \xi^2 d\xi = -(1 - x) \frac{x^3}{3} = -\frac{x^3}{3} + \frac{x^4}{3}

Second integral:

-x \int_x^1 (\xi - \xi^2) d\xi = -x \left[ \frac{\xi^2}{2} - \frac{\xi^3}{3} \right]_x^1 = -x \left( \frac{1}{6} - \frac{x^2}{2} + \frac{x^3}{3} \right) = -\frac{x}{6} + \frac{x^3}{2} - \frac{x^4}{3}

Summing both:

y(x) = -\frac{x^3}{3} + \frac{x^4}{3} - \frac{x}{6} + \frac{x^3}{2} - \frac{x^4}{3} = \frac{x^3}{6} - \frac{x}{6} = \frac{x(x^2 - 1)}{6}

Verify: $y'' = x, y(0) = 0, y(1) = 0$. Matches!

Final Answer & Physical Insight

$G(x, \xi) = \begin{cases} x(1-\xi), & x \le \xi \\ \xi(1-x), & x > \xi \end{cases}$ and $y(x) = \frac{x(x^2 - 1)}{6}$

Tier 2 • Intermediate Exam Example 8.2: Solvability Condition via the Fredholm Alternative

Determine the exact condition on the function $h(x)$ for the boundary value problem to possess a solution:

y'' + \pi^2 y = h(x), \quad y(0) = 0, \quad y(1) = 0

Step 1: Check homogeneous boundary value problem

y_h'' + \pi^2 y_h = 0 \implies y_h(x) = c_1 \cos(\pi x) + c_2 \sin(\pi x)

Boundary conditions:

y_h(0) = c_1 = 0 \implies y_h(x) = c_2 \sin(\pi x)

y_h(1) = c_2 \sin(\pi) = 0 \quad (\text{satisfied for any } c_2)

The homogeneous problem has non-trivial solution $\phi(x) = \sin(\pi x)$.

Step 2: Self-adjointness of operator
The operator $L[y] = y'' + \pi^2 y$ with Dirichlet conditions $y(0) = y(1) = 0$ is formally self-adjoint:

\int_0^1 (u L[v] - v L[u])\,dx = [u v' - v u']_0^1 = 0

Step 3: Apply the Fredholm Alternative (Case II)
By Theorem 8.1, the nonhomogeneous problem has a solution if and only if $h(x)$ is orthogonal to the null space of the adjoint operator:

\int_0^1 h(x) \phi(x)\,dx = 0 \implies \int_0^1 h(x) \sin(\pi x)\,dx = 0

If this integral is zero, an infinite family of solutions exists ($y(x) = y_p(x) + c \sin(\pi x)$). If non-zero, NO solution exists.

Final Answer & Physical Insight

Solvability condition: $\int_0^1 h(x) \sin(\pi x)\,dx = 0$.

Tier 3 • Honors Challenge Example 8.3: Bilinear Mercer Eigenfunction Expansion of Green's Function

Prove the bilinear eigenfunction expansion for the Green's function of a regular Sturm-Liouville problem:

G(x, \xi) = \sum_{n=1}^\infty \frac{\phi_n(x) \phi_n(\xi)}{\lambda_n}

Step 1: Expand Green's function in orthonormal eigenfunctions
Let $\{\phi_n(x)\}$ be the complete orthonormal eigenfunctions of $L[\phi_n] = \lambda_n w(x) \phi_n$ with separated boundary conditions. For any fixed $\xi \in (a, b)$, $G(x, \xi)$ satisfies the boundary conditions in $x$. Expand $G(x, \xi)$ in terms of $\phi_n(x)$:

G(x, \xi) = \sum_{n=1}^\infty c_n(\xi) \phi_n(x)

Step 2: Determine expansion coefficients $c_n(\xi)$
By orthonormality $\int_a^b \phi_n(x) \phi_m(x) w(x)\,dx = \delta_{nm}$:

c_n(\xi) = \int_a^b G(x, \xi) \phi_n(x) w(x)\,dx

Step 3: Exploit self-adjointness and eigenvalue equation
Since $\phi_n(x) = \frac{1}{\lambda_n w(x)} L[\phi_n](x)$:

c_n(\xi) = \frac{1}{\lambda_n} \int_a^b G(x, \xi) L[\phi_n](x)\,dx

Using Green's formula $\int_a^b (u L[v] - v L[u])\,dx = 0$:

c_n(\xi) = \frac{1}{\lambda_n} \int_a^b \phi_n(x) L_x[G(x, \xi)]\,dx

Since $L_x[G(x, \xi)] = \delta(x - \xi)$:

c_n(\xi) = \frac{1}{\lambda_n} \int_a^b \phi_n(x) \delta(x - \xi)\,dx = \frac{\phi_n(\xi)}{\lambda_n}

Step 4: Substitute $c_n(\xi)$ back

G(x, \xi) = \sum_{n=1}^\infty \frac{\phi_n(x) \phi_n(\xi)}{\lambda_n}

This Mercer series proves the exact symmetry $G(x, \xi) = G(\xi, x)$ and convergence in $L^2$.

Final Answer & Physical Insight

$G(x, \xi) = \sum_{n=1}^\infty \frac{\phi_n(x) \phi_n(\xi)}{\lambda_n}$ (Mercer's Bilinear Expansion).