Fourier Series Solutions of Boundary Value Problems

The orthogonality-and-projection technique developed here is not confined to the heat equation. In acoustics and structural engineering, the same sine and cosine eigenfunctions describe vibration modes of strings and beams, and their coefficients determine how a given pluck or impact excites each mode. In electrical engineering and signal processing, Fourier series and their discrete analogues (the DFT and FFT) decompose periodic signals into frequency components, using exactly the same orthogonality of sinusoids over a period. In quantum mechanics, the coefficient-extraction idea generalizes directly: wavefunctions are expanded in eigenfunctions of the Hamiltonian, and the "Fourier coefficients" become probability amplitudes for measuring definite energy states. In numerical analysis, truncated Fourier expansions underlie spectral methods for solving PDEs, prized for their rapid convergence on smooth periodic data. In statistics and cryptography, orthogonal function expansions and their discrete cousins appear in signal reconstruction, compressed sensing, and the algebraic structure exploited by number-theoretic transforms.

Ask the tutor: How does this coefficient-extraction method generalize when the boundary value problem produces eigenfunctions that are not sines or cosines, such as Bessel functions or Legendre polynomials?

:::definition[Fourier Sine and Cosine Series on $[0,L]$] Let $f:[0,L]\to\mathbb{R}$ be piecewise continuous. Define the Fourier sine series of $f$ as

and the Fourier cosine series of $f$ as

These are, respectively, the ordinary Fourier series of the odd and even $2L$-periodic extensions of $f$ from $[0,L]$ to $\mathbb{R}$. The symbol $\sim$ denotes formal correspondence; conditions under which the series converges pointwise to $f$ are addressed separately (see Convergence of Fourier Series Solutions).

Think of piecewise continuous functions on $[0,L]$ as vectors in an infinite-dimensional space, equipped with the inner product $\langle f,g\rangle = \int_0^L f(x)g(x)\,dx$. In ordinary Euclidean space $\mathbb{R}^n$, if $\{e_1,\dots,e_n\}$ is an orthogonal basis, any vector $v$ decomposes as $v=\sum c_i e_i$ with $c_i = \langle v,e_i\rangle/\langle e_i,e_i\rangle$ — you find each component by projecting onto the corresponding axis, and the other axes contribute nothing because they are perpendicular. The functions $\sin(n\pi x/L)$ for $n=1,2,3,\dots$ play the role of orthogonal axes in this function space. When you "dot" $f$ with $\sin(m\pi x/L)$, every term in the series except the $m$-th integrates to zero, and the $m$-th term integrates to a computable nonzero number. That is the entire mechanism: orthogonality annihilates cross terms and isolates one coefficient at a time. The reason sines (rather than cosines, or some other family) appear is dictated by the boundary conditions of the physical problem. A rod with both ends held at zero temperature (Dirichlet conditions $u(0,t)=u(L,t)=0$) forces the spatial eigenfunctions to vanish at $x=0,L$, which only $\sin(n\pi x/L)$ does among $\{\sin,\cos\}$. A rod with insulated ends (Neumann conditions $u_x(0,t)=u_x(L,t)=0$) forces zero slope at the ends, which selects $\cos(n\pi x/L)$ instead (including the constant function, $n=0$).

:::svg[Orthogonal sine eigenfunctions on $[0,L]$ and their role as "axes"]

Consider the standard heat conduction boundary value problem on a rod of length $L$ with fixed zero end temperatures:

Separation of variables (assumed known) produces the elementary product solutions

each of which satisfies the PDE and both boundary conditions. Because the PDE and boundary conditions are linear and homogeneous, any finite linear combination of the $u_n$ is again a solution, and — accepting termwise differentiability of the infinite sum, a technical point addressed in the study of convergence — so is the infinite superposition

The only remaining requirement is the initial condition:

This is exactly a Fourier sine series matching problem. The theorem below shows how orthogonality determines the $b_n$ uniquely.

:::theorem[Orthogonality of the Sine and Cosine Eigenfunctions on $[0,L]$] Let $L>0$ and let $m,n$ be positive integers. Then

and for nonnegative integers $m,n$,

Moreover, for all nonnegative integers $m,n$,

need not vanish in general (sine and cosine bases are not mutually compared here; each family is used separately, so this mixed integral plays no role in what follows).

Example 1 — Dirichlet rod with an asymmetric initial temperature. Solve

Here $\alpha^2=4$, $L=3$. Note $f$ is deliberately asymmetric (a triangular pulse skewed to one side, not centered at $x=L/2$), and its two pieces have different slopes ($1$ and $-\tfrac12$), so no shortcut from symmetry is available. By separation of variables, the solution has the form

and matching $t=0$ requires $f(x) = \sum b_n \sin(n\pi x/3)$ on $(0,3)$. By the Corollary,

First integral. Let $k=n\pi/3$. Using integration by parts, $\int x\sin(kx)\,dx = -\frac{x}{k}\cos(kx) + \frac{1}{k^2}\sin(kx)$, so

With $k = n\pi/3$:

Second integral. Write $\frac{3-x}{2} = \frac{3}{2} - \frac{x}{2}$, so

First piece:

With $k=n\pi/3$, $3k = n\pi$, so $\cos(3k)=\cos(n\pi) = (-1)^n$. Thus this piece contributes

Second piece, using the antiderivative from above:

Since $3k=n\pi$, $\sin(3k)=0$ and $\cos(3k)=(-1)^n$, so this equals

Dividing by $2$ (from the $-x/2$ coefficient) and recalling $k=n\pi/3$ gives, after substitution,

Combining, $I_2$ equals the first piece minus this:

Simplifying the $\cos(n\pi/3)$ and $(-1)^n$ coefficients:

So remarkably the $(-1)^n$ terms cancel and

Total. Adding $I_1 + I_2$:

Finally,

As a sanity check, take $n=3$: $\sin(\pi)=0$, so $b_3=0$; take $n=1$: $\sin(\pi/3)=\tfrac{\sqrt3}{2}$, so $b_1 = \frac{15\sqrt3}{4\pi^2}\approx \frac{25.98}{39.48}\approx 0.658$, a sensible magnitude comparable to the peak value $f(1)=1$. The full solution is

Example 2 — Neumann (insulated) rod with an asymmetric initial temperature. Solve

Insulated ends ($\alpha^2=1$, $L=2$) force the spatial eigenfunctions to satisfy $X'(0)=X'(2)=0$, which selects $\cos(n\pi x/2)$, $n=0,1,2,\dots$, so the solution has the form

Matching $t=0$ requires $f(x) = x^2 = \frac{a_0}{2}+\sum_{n\ge1} a_n\cos(n\pi x/2)$ on $(0,2)$, an asymmetric (non-even-about-midpoint, genuinely quadratic) target function. Computing $a_0$.

So the constant term is $a_0/2 = 4/3$, the average value of $x^2$ over $[0,2]$ — a useful sanity check since $\frac{1}{2}\int_0^2 x^2\,dx = \frac{1}{2}\cdot \frac83 = \frac43$. Computing $a_n$ for $n\ge 1$. With $k=n\pi/2$,

Integrate by parts twice. First, $\int x^2\cos(kx)\,dx = \frac{x^2}{k}\sin(kx) + \frac{2x}{k^2}\cos(kx) - \frac{2}{k^3}\sin(kx)$ (differentiate to check: $\frac{d}{dx}$ of this expression is

confirming the antiderivative is correct). Evaluate from $0$ to $2$, noting $2k = n\pi$ so $\sin(2k) = \sin(n\pi)=0$ and $\cos(2k)=\cos(n\pi)=(-1)^n$:

Substituting $k=n\pi/2$, so $k^2 = n^2\pi^2/4$:

As a check, $n=1$ gives $a_1 = -16/\pi^2 \approx -1.621$, a plausible-sized correction to the mean value $4/3\approx 1.333$; $n=2$ gives $a_2 = 16/(4\pi^2)=4/\pi^2\approx 0.405$. The full solution is

Note that as $t\to\infty$, $u(x,t)\to 4/3$, the spatial average of the initial data — exactly as expected physically for an insulated rod with no heat loss, where total thermal energy is conserved and temperature equilibrates to the mean.