Fourier Coefficients: Why They're Symbolic

Topic: advanced.pde.fourier_coefficients

Explanation of why Fourier coefficients in PDE solutions are returned as symbolic expressions rather than numerical values. Covers the orthogonality principle, symbolic integration requirements, and workarounds for computing coefficients manually.

Mathematical Definition

For any PDE solution via separation of variables: $$u(x,t)=\sum _{n=1}^{\infty }{A}_n{X}_n(x)T_n(t)$$

Coefficients from initial conditions: $$u(x,0)=f(x)=\sum _{n=1}^{\infty }{A}_n{X}_n(x)$$

Orthogonality gives: $$A_n=\frac{⟨f,X_n⟩}{⟨X_n,X_n⟩}=\frac{{\int }_0^{L}f(x)X_n(x)dx}{{\int }_0^L{X}_n^2(x)dx}$$

Heat Equation (Dirichlet BCs): $$X_n(x)=\sin \left(\frac{n\pi x}{L}\right)$$

$$A_n=\frac{2}{L}{\int }_0^{L}f(x)\sin \left(\frac{n\pi x}{L}\right)dx$$

Constant Initial Condition ($f(x)=c$): $$A_n=\frac{2c}{n\pi }[1-(-1{)}^n]=\{\begin{array}{ll}\frac{4c}{n\pi }& n\text{ odd}\\ 0& n\text{ even}\end{array}$$

Fourier Coefficients: Why They're Symbolic

The Coefficient Problem

All PDE solutions via separation of variables involve Fourier series coefficients that must be computed from initial/boundary conditions.

General Form

For any PDE solution:

$$u(x,t)=\sum _{n=1}^{\infty }{A}_n{X}_n(x)T_n(t)$$

The coefficients $A_n$ come from matching initial conditions:

$$u(x,0)=f(x)=\sum _{n=1}^{\infty }{A}_n{X}_n(x)$$

Orthogonality of eigenfunctions $X_n(x)$ gives:

$$A_n=\frac{⟨f,X_n⟩}{⟨X_n,X_n⟩}=\frac{{\int }_0^{L}f(x)X_n(x)dx}{{\int }_0^L{X}_n^2(x)dx}$$

This requires symbolic integration.

Heat Equation Example

For heat equation with Dirichlet BCs on $[0,L]$:

$$X_n(x)=\sin \left(\frac{n\pi x}{L}\right)$$

Orthogonality:

$${\int }_0^L\sin \left(\frac{n\pi x}{L}\right)\sin \left(\frac{m\pi x}{L}\right)dx=\{\begin{array}{ll}L/2& n=m\\ 0& n\ne m\end{array}$$

Fourier coefficients:

$$A_n=\frac{2}{L}{\int }_0^{L}f(x)\sin \left(\frac{n\pi x}{L}\right)dx$$

Requires symbolic integration of $f(x)\sin (n\pi x/L)$.

Simple Case: Constant Initial Condition

$$f(x)=c\text{(constant)}$$

$$A_n=\frac{2c}{L}{\int }_0^L\sin \left(\frac{n\pi x}{L}\right)dx=\frac{2c}{L}{\left[-\frac{L}{n\pi }\cos \left(\frac{n\pi x}{L}\right)\right]}_0^L$$

$$=\frac{2c}{n\pi }[1-\cos (n\pi )]=\frac{2c}{n\pi }[1-(-1{)}^n]$$

$$=\{\begin{array}{ll}\frac{4c}{n\pi }& n\text{ odd}\\ 0& n\text{ even}\end{array}$$

MathHook can compute this (Phase 2) with symbolic integration.

Complex Case: Arbitrary Function

$$f(x)=x(L-x)\text{(parabola)}$$

$$A_n=\frac{2}{L}{\int }_0^{L}x(L-x)\sin \left(\frac{n\pi x}{L}\right)dx$$

Requires integration by parts twice:

$$=\frac{2}{L}\left[\frac{4L^3}{n^3{\pi }^3}[1-(-1{)}^n]\right]=\frac{8L^2}{n^3{\pi }^3}[1-(-1{)}^n]$$

MathHook needs symbolic integration for this.

Why MathHook Returns Symbolic Coefficients

Current Implementation

MathHook solvers return:

#![allow(unused)]
fn main() {
pub struct HeatSolution {
    pub solution: Expression,     // Σ A_n sin(λₙx) exp(-λₙαt)
    pub eigenvalues: Vec<Expression>,  // [λ₁, λ₂, λ₃, ...] ✅ COMPUTED
    pub coefficients: Vec<Expression>, // [A_1, A_2, A_3, ...] ⚠️ SYMBOLIC
}
}

The coefficients are symbolic symbols $A_1,A_2,A_3,\dots$ (not numerical values).

Why?

Computing numerical $A_n$ requires:

$$A_n=\frac{2}{L}{\int }_0^{L}f(x)\sin \left(\frac{n\pi x}{L}\right)dx$$

This is symbolic integration of a user-provided function $f(x)$.

MathHook's integration module (Phase 1) focuses on:

• Standard integrals ($\int x^{n}dx$, $\int \sin (x)dx$, etc.)
• Integration by substitution
• Integration by parts

NOT YET IMPLEMENTED:

• Definite integral evaluation with symbolic limits
• Fourier sine/cosine integral tables
• Automated integration strategy selection

Phase 2 Roadmap

Goal: Automatically compute Fourier coefficients for common initial conditions.

Requirements:

1. Symbolic definite integration

2. Fourier integral table:

   • ${\int }_0^{L}1\cdot \sin (n\pi x/L)dx$
   • ${\int }_0^{L}x\cdot \sin (n\pi x/L)dx$
   • ${\int }_0^L{x}^2\cdot \sin (n\pi x/L)dx$
   • ${\int }_0^L{e}^{ax}\cdot \sin (n\pi x/L)dx$

3. Pattern matching for common forms:

   • Polynomial × trig
   • Exponential × trig
   • Piecewise functions

Validation: SymPy Also Returns Symbolic

Important: SymPy's pdsolve() ALSO returns symbolic coefficients.

Why? SymPy separates:

1. Solving PDE → symbolic solution structure
2. Matching ICs → separate fourier_series() function

MathHook follows the same philosophy.

Examples: Computing Coefficients

Heat Equation: Constant Initial Temp

Initial condition: $u(x,0)=T_0$

Analytical:

$$A_n=\frac{2T_0}{L}{\int }_0^L\sin \left(\frac{n\pi x}{L}\right)dx=\frac{2T_0}{n\pi }[1-(-1{)}^n]$$

Numerical (for $T_0=100$, $L=1$):

• $A_1=\frac{200}{\pi }\approx 63.66$
• $A_2=0$
• $A_3=\frac{200}{3\pi }\approx 21.22$
• $A_4=0$
• $A_5=\frac{200}{5\pi }\approx 12.73$

Wave Equation: Triangular Pluck

Initial position: Triangular (plucked at center)

Analytical:

$$A_n=\frac{8h}{n^2{\pi }^2}\sin \left(\frac{n\pi }{2}\right)$$

Numerical (for $h=0.005$ m):

• $A_1=\frac{8\times 0.005}{{\pi }^2}\approx 0.00405$ m
• $A_2=0$ (sin(π) = 0)
• $A_3=-\frac{8\times 0.005}{9{\pi }^2}\approx -0.00045$ m
• $A_4=0$

Laplace Equation: Fixed Top Edge

Boundary condition: $u(x,b)=V_0$

Analytical:

$$C_n=\frac{2V_0}{n\pi \sinh (n\pi b/a)}[1-(-1{)}^n]$$

Numerical (for $V_0=100$, $a=0.1$, $b=0.05$):

• $C_1=\frac{200}{\pi \sinh (\pi /2)}\approx 27.67$
• $C_2=0$
• $C_3=\frac{200}{3\pi \sinh (3\pi /2)}\approx 0.83$

Examples

Manual Coefficient Computation for Constant Initial Condition

Computing Fourier coefficients manually for heat equation with constant initial temperature

#![allow(unused)]
fn main() {
use mathhook_core::pde::standard::heat::HeatEquationSolver;
use mathhook_core::{symbol, expr};

// Setup PDE, BCs, IC...
let result = solver.solve_heat_equation_1d(&pde, &alpha, &bcs, &ic)?;

// Coefficients are symbolic
println!("Symbolic: {:?}", result.coefficients);  // [A_1, A_2, A_3, ...]

// Manually compute for f(x) = 100 (constant)
let mut numerical_coeffs = Vec::new();
for n in 1..=10 {
    let a_n = if n % 2 == 1 {
        // Odd n: A_n = 400/(nπ)
        expr!(400.0 / ((n as f64) * std::f64::consts::PI))
    } else {
        // Even n: A_n = 0
        expr!(0)
    };
    numerical_coeffs.push(a_n);
}

}

from mathhook.pde.heat import HeatEquationSolver
from mathhook import symbol, expr
import math

# Setup PDE, BCs, IC...
result = solver.solve_heat_equation_1d(pde, alpha, bcs, ic)

# Coefficients are symbolic
print("Symbolic:", result.coefficients)  # [A_1, A_2, A_3, ...]

# Manually compute for f(x) = 100 (constant)
numerical_coeffs = []
for n in range(1, 11):
    if n % 2 == 1:
        # Odd n: A_n = 400/(nπ)
        a_n = expr(400.0 / (n * math.pi))
    else:
        # Even n: A_n = 0
        a_n = expr(0)
    numerical_coeffs.append(a_n)

const { HeatEquationSolver } = require('mathhook/pde/heat');
const { symbol, expr } = require('mathhook');

// Setup PDE, BCs, IC...
const result = solver.solveHeatEquation1d(pde, alpha, bcs, ic);

// Coefficients are symbolic
console.log("Symbolic:", result.coefficients);  // [A_1, A_2, A_3, ...]

// Manually compute for f(x) = 100 (constant)
const numericalCoeffs = [];
for (let n = 1; n <= 10; n++) {
    let aN;
    if (n % 2 === 1) {
        // Odd n: A_n = 400/(nπ)
        aN = expr(400.0 / (n * Math.PI));
    } else {
        // Even n: A_n = 0
        aN = expr(0);
    }
    numericalCoeffs.push(aN);
}

Performance

Time Complexity: O(n)

API Reference

• Rust: mathhook_core::pde::fourier::coefficients
• Python: mathhook.pde.fourier.coefficients
• JavaScript: mathhook.pde.fourier.coefficients

See Also

• advanced.pde.boundary_conditions

• advanced.pde.classification

• calculus.integration

• calculus.heat_equation

• calculus.wave_equation