Fourier Series and Convergence

A function defined on an interval can be written as an infinite sum of sines and cosines. That expansion turns a partial differential equation into a family of ordinary ones, one per term. The sines and cosines themselves come from a boundary value problem posed on the interval, and an algebraic structure on that problem selects which of them appear.

Two-point boundary value problems

Until now the data for a second-order equation has been given at a single point: the initial value problem prescribes $y (t_{0})$ and $y^{'} (t_{0})$ and asks for the solution afterward. Physical problems in space instead prescribe the solution at two different points, the ends of an interval.

Moving from one point to two changes the question of existence. An initial value problem with continuous coefficients has exactly one solution. A boundary value problem may have one solution, none, or infinitely many. That trichotomy is the same one that governs the linear algebraic system $A x = b$ :¹

Unique solution. If $A$ is nonsingular, $A x = b$ has one solution for every $b$ , and the homogeneous system $A x = 0$ has only $x = 0$ .
None or infinitely many. If $A$ is singular, $A x = b$ has no solution unless $b$ meets a compatibility condition, in which case it has infinitely many; the homogeneous system then has nontrivial solutions.

The nonhomogeneous boundary value problem inherits this dichotomy from its associated homogeneous problem, exactly as a matrix system does.

For example, $y^{''} + 2 y = 0$ with $y (0) = 1$ , $y (π) = 0$ has the single solution $y = cos (2 x) - cot (2 π) sin (2 x)$ : a unique answer, matching a nonsingular $A$ . But $y^{''} + y = 0$ with $y (0) = 1$ , $y (π) = a$ demands $c_{1} = 1$ from the left condition and $- c_{1} = a$ from the right; these are compatible only when $a = - 1$ , and then $y = cos x + c_{2} sin x$ leaves $c_{2}$ free. The same equation has no solution for $a \neq = - 1$ and infinitely many for $a = - 1$ .

The eigenvalue problem

The homogeneous problem that drives this module attaches a parameter to the equation. Consider

y^{''} + λ y = 0, y (0) = 0, y (π) = 0.

For every $λ$ this has the trivial solution $y \equiv 0$ . The question is for which $λ$ it has others. This is the differential analogue of $A x = λ x$ , and the vocabulary carries over.

The form of the general solution differs for positive, zero, and negative $λ$ , so the three cases are treated separately.

Case	Equation	General solution	Boundary conditions force
$λ = μ^{2} > 0$	$y^{''} + μ^{2} y = 0$	$c_{1} cos (μx) + c_{2} sin (μx)$	$c_{1} = 0$ , then $sin (μ π) = 0$
$λ = 0$	$y^{''} = 0$	$c_{1} x + c_{2}$	$c_{1} = c_{2} = 0$ : trivial only
$λ = - μ^{2} < 0$	$y^{''} - μ^{2} y = 0$	$c_{1} cosh (μx) + c_{2} sinh (μx)$	$c_{1} = c_{2} = 0$ : trivial only

Only the positive case gives nontrivial solutions. There $sin (μ π) = 0$ requires $μ$ to be a positive integer, so the eigenvalues and eigenfunctions are

λ_{n} = n^{2}, y_{n} (x) = sin (n x), n = 1, 2, 3, \dots

Replacing the endpoint $π$ by an arbitrary length $L$ shifts the condition to $sin (μL) = 0$ , giving $μ = nπ / L$ and

λ_{n} = \frac{n ^{2} π ^{2}}{L ^{2}}, y_{n} (x) = sin (\frac{nπ x}{L}) .

These eigenfunctions are the standing shapes that a fixed-end string or a zero-temperature rod can hold. Each is a half-integer number of arches fitted into the interval.

The first three eigenfunctions of

y^{''} + λ y = 0

with zero endpoints: mode

n

fits

n

half-waves into

[0, L]

and vanishes at both ends.

Orthogonality

The sines and cosines from these problems are mutually orthogonal, and that orthogonality supports the whole theory. Generalize the dot product of vectors to functions.

The functions $sin (mπ x / L)$ and $cos (mπ x / L)$ for $m = 1, 2, \dots$ are mutually orthogonal on $- L \leq x \leq L$ . Direct integration, using the product-to-sum identities, gives the orthogonality relations

- L \int L cos (\frac{mπ x}{L}) cos (\frac{nπ x}{L}) d x = {0, L, m \neq = n, m = n,

- L \int L sin (\frac{mπ x}{L}) sin (\frac{nπ x}{L}) d x = {0, L, m \neq = n, m = n,

- L \int L cos (\frac{mπ x}{L}) sin (\frac{nπ x}{L}) d x = 0 for all m, n .

Sines are orthogonal to cosines; distinct sines are orthogonal to each other, as are distinct cosines; and a function integrated against itself returns its squared length $L$ . Integration plays the role of the dot product, and the trigonometric functions form an orthogonal basis in the space of functions on $[- L, L]$ .

Orthogonality as a Gram matrix: the inner product of any two distinct basis functions over a full period is zero, so only the diagonal survives.

The Euler-Fourier formulas

Suppose a function $f$ on $- L \leq x \leq L$ can be written as the trigonometric series

f (x) = \frac{a _{0}}{2} + m = 1 \sum \infty [a_{m} cos (\frac{mπ x}{L}) + b_{m} sin (\frac{mπ x}{L})] .

Orthogonality recovers every coefficient. Multiply both sides by $cos (nπ x / L)$ and integrate from $- L$ to $L$ . On the right, every product integrates to zero except the one term with $m = n$ in the cosine sum, which contributes $a_{n} L$ . Hence $\int_{- L}^{L} f (x) cos (nπ x / L) d x = a_{n} L$ . The same maneuver with $sin (nπ x / L)$ isolates $b_{n}$ .

Writing the constant term as $a_{0} /2$ lets the single formula for $a_{n}$ cover $n = 0$ as well: $a_{0} = \frac{1}{L} \int_{- L}^{L} f d x$ is twice the average of $f$ . Each coefficient is computed independently of the others, and each depends only on the values of $f$ on one period. The coefficient $a_{n}$ is the projection of $f$ onto the $n$ -th cosine, divided by that cosine's squared length, precisely the vector formula $a_{i} = (u \cdot v_{i}) / (v_{i} \cdot v_{i})$ read one dimension at a time.²

The convergence theorem

The Euler-Fourier formulas manufacture coefficients from any integrable $f$ , but the resulting series need not converge to $f$ , and constructed examples show it can fail at isolated points. Convergence requires a mild smoothness hypothesis, the same piecewise continuity used to define the Laplace transform: $f$ is piecewise continuous on an interval if the interval splits into finitely many pieces on each of which $f$ is continuous and has finite one-sided limits. Write $f (c^{+})$ and $f (c^{-})$ for the right- and left-hand limits at $c$ .

The hypotheses are sufficient, not necessary, and the proof is beyond scope, but the statement gives the applications everything they need. At a jump the series splits the difference: for the square wave, the endpoints $x = 0, \pm L$ carry a jump between $0$ and $L$ , and every sine term vanishes there, leaving the sum $L /2$ , the midpoint.

A Fourier series of infinitely differentiable terms can converge to a discontinuous sum, and the convergence is not uniform.

A five-term partial sum of the square wave. The approximation tracks the flat sections but overshoots at each jump, the Gibbs phenomenon, and the overshoot height persists as terms are added.

The rate of decay of the coefficients governs the speed of convergence. Smoother functions decay faster: the triangular wave, whose coefficients drop like $1/ m^{2}$ , needs about $21$ terms for uniform error below $0.01$ , while the square wave's $1/ m$ coefficients converge far more slowly. In general, the more continuous derivatives $f$ has, the faster its series converges.

Even and odd symmetry

Symmetry cuts the work in half and dictates which functions appear. Recall $f$ is even when $f (- x) = f (x)$ and odd when $f (- x) = - f (x)$ . Products follow sign rules: even times even is even, odd times odd is even, even times odd is odd. The integral consequences are what matter:

- L \int L f d x = 2 0 \int L f d x (f even), - L \int L f d x = 0 (f odd) .

Apply these to the Euler-Fourier formulas. If $f$ is even, then $f (x) sin (nπ x / L)$ is odd and integrates to zero, so every $b_{n}$ vanishes. If $f$ is odd, every $a_{n}$ vanishes.

Symmetry of $f$	Series	Nonzero coefficients	Integral range
Even	cosine only	$a_{n}$	$\int_{0}^{L}$ , doubled
Odd	sine only	$b_{n}$	$\int_{0}^{L}$ , doubled
Neither	both	$a_{n}$ and $b_{n}$	$\int_{- L}^{L}$

Half-range expansions

A function given only on $0 \leq x \leq L$ can be expanded as either a sine or a cosine series by first extending it to $[- L, L]$ with the symmetry you want, then extending periodically.⁴

Even (cosine) extension. Reflect across the $y$ -axis: $g (x) = f (- x)$ for $- L < x < 0$ . Its cosine series represents $f$ on $[0, L]$ .
Odd (sine) extension. Reflect through the origin: $h (x) = - f (- x)$ for $- L < x < 0$ , with $h (0) = 0$ . Its sine series represents $f$ on $(0, L)$ .

The same $f (x) = x$ on $[0, L]$ becomes a triangular wave under even extension and a sawtooth under odd extension, with different series but the same values on the original interval. The choice is dictated by the boundary conditions of the problem being solved: a rod held at zero temperature at both ends calls for a sine series, because sines vanish at the ends.

A function on the half-interval

[0, L]

(solid) extended two ways: the even reflection (left, dashed) builds a cosine series, the odd reflection (right, dashed) builds a sine series.

The same $f$ expanded both ways gives two series with the same values on $[0, L]$ but different smoothness off it.

Worked example. Take $f (x) = x$ on $0 \leq x \leq L$ .

Even (cosine) extension. The even extension is $∣ x ∣$ on $[- L, L]$ , a triangular wave. All $b_{n}$ vanish, and

a_{0} = \frac{2}{L} 0 \int L x d x = L, a_{m} = \frac{2}{L} 0 \int L x cos (\frac{mπ x}{L}) d x = \frac{2 L}{m ^{2} π ^{2}} (cos mπ - 1),

so $a_{m} = - 4 L / (m^{2} π^{2})$ for $m$ odd and $0$ for $m$ even, giving

x = \frac{L}{2} - \frac{4 L}{π ^{2}} n = 1 \sum \infty \frac{1}{( 2 n - 1 ) ^{2}} cos (\frac{( 2 n - 1 ) π x}{L}) .

Odd (sine) extension. The odd extension is $x$ on $(- L, L)$ , a sawtooth. All $a_{n}$ vanish, and integration by parts gives

b_{m} = \frac{2}{L} 0 \int L x sin (\frac{mπ x}{L}) d x = \frac{2 L}{mπ} (- 1)^{m + 1}, x = \frac{2 L}{π} m = 1 \sum \infty \frac{( - 1 ) ^{m + 1}}{m} sin (\frac{mπ x}{L}) .

The cosine coefficients decay like $1/ m^{2}$ and the series is continuous; the sine coefficients decay only like $1/ m$ , and the sawtooth jumps by $2 L$ at each odd multiple of $L$ , where the series returns the midpoint $0$ and overshoots by the Gibbs amount on either side.

The two periodic extensions of

f (x) = x

[0, L]

: the even extension (top) is a continuous triangular wave, the odd extension (bottom) a sawtooth with a jump at each end of the base interval.

Boyce, Elementary Differential Equations and Boundary Value Problems, §10.1 — two-point boundary value problems, the parallel with singular and nonsingular linear systems $A x = b$ , and the eigenvalue problem $y^{''} + λ y = 0$ with $λ_{n} = n^{2} π^{2} / L^{2}$ . ↩
Simmons, Differential Equations with Applications and Historical Notes, §33, §37 — the Fourier coefficients derived from orthogonality, and orthogonal functions as an inner-product geometry with coefficients read as projections. ↩
Boyce, §10.3 — the Fourier Convergence Theorem for piecewise continuous $f$ and $f^{'}$ , convergence to the mean of one-sided limits at jumps, and the Gibbs phenomenon at discontinuities. ↩
Simmons, §35 — even and odd functions, the vanishing of the complementary coefficients, and half-range sine and cosine expansions via symmetric extension. ↩