Partial differential equations and Fourier series

MotivationH

Three equations that run the world

Heat a metal rod at its middle and hold its ends in ice water. How does the temperature evolve? Pluck a string on a kora in Bamako, a sitar in Varanasi or a guitar in Seville: why does it sound a note and its overtones, and nothing in between? Dip a bent wire loop into soapy water: what shape does the film take? Each question is answered by a partial differential equation (PDE) — an equation linking the rates of change of a function of several variables:

u_t = k\,u_{xx} \;\;\text{(heat)}, \qquad u_{tt} = c^2 u_{xx} \;\;\text{(wave)}, \qquad u_{xx} + u_{yy} = 0 \;\;\text{(Laplace)}.

In 1807 Joseph Fourier claimed something outrageous about the first one. Every initial temperature profile, he said, even one with corners and jumps, can be written as a sum of sine waves, and each sine wave simply decays at its own rate. The leading analysts of the day, Lagrange among them, did not believe that a discontinuous function could be a sum of smooth ones. Making sense of Fourier's claim took a century and created much of modern analysis: the Riemann and Lebesgue integrals, uniform convergence, set theory (Cantor began with the uniqueness of trigonometric series), and Hilbert spaces.

Today Fourier's idea is everywhere. MP3 audio and JPEG images store a signal by its frequency content, an MRI scanner measures the Fourier transform of your body's tissue directly, and quantum mechanics computes atomic energy levels as the frequencies of a wave equation. This lesson solves the three classical equations on an interval and develops the theory of Fourier series that makes the solutions rigorous. It also proves that the solutions are unique.

IntuitionH

Pure tones, and averages

Heat flows from hot to cold at a rate proportional to the temperature gradient (Fourier's law). In a thin rod, the heat in a small piece $[x, x + \Delta x]$ changes by the flux in at one end minus the flux out at the other. Dividing by $\Delta x$ and letting $\Delta x \to 0$ gives $u_t = k u_{xx}$ . The sign tells the story: where the temperature profile is curved upwards ( $u_{xx} > 0$ , a cold dip), the temperature rises. Heat flattens bumps.

Sine waves are the natural modes. If the initial profile is $\sin(nx)$ , the solution keeps its shape and just shrinks: $u = e^{-kn^2t}\sin(nx)$ . The decay rate grows like $n^2$ , so wiggly components die almost instantly and the profile quickly smooths out. If the initial profile is a sum of sines, each one decays independently, because the equation is linear. So everything hinges on Fourier's claim that every reasonable profile is such a sum.

Waves keep their shape and travel. A string's displacement satisfies $u_{tt} = c^2 u_{xx}$ (Newton's law for each small piece, with tension $\tau$ and density $\rho$ , $c^2 = \tau/\rho$ ). Any shape $f$ moving right at speed $c$ , $u = f(x - ct)$ , is a solution, and so is any shape moving left.

Harmonic functions — solutions of Laplace's equation — are steady states of heat flow. They have no bumps at all: the value at each point equals the average over any circle around it. So a harmonic function cannot have a maximum inside its domain.

Separation of variables on a string fixed at both ends: the first three modes $\sin x$, $\sin 2x$, $\sin 3x$ on $[0, \pi]$. Each oscillates (wave equation) or decays (heat equation) at its own rate, $n$ or $n^2$, without changing shape. A general solution is a superposition of them.

DefinitionsH

Definitions

Definition— the three classical equations

For $u = u(t, x)$ (or $u(x, y)$ ) and constants $k, c > 0$ : the heat equation $u_t = k\,u_{xx}$ , the wave equation $u_{tt} = c^2 u_{xx}$ , and Laplace's equation $\Delta u = u_{xx} + u_{yy} = 0$ . A solution of Laplace's equation is harmonic. All three are linear: sums and multiples of solutions are solutions (the superposition principle).

A problem consists of an equation plus side conditions. On a rod $[0, L]$ we impose initial conditions — $u(0, x) = \varphi(x)$ for heat, and also $u_t(0, x) = \psi(x)$ for waves — and boundary conditions such as the Dirichlet conditions $u(t, 0) = u(t, L) = 0$ . Following Hadamard (1902), a problem is well posed if a solution exists, is unique, and depends continuously on the data.

Definition— Fourier series

Let $f$ be $2\pi$ -periodic and Riemann integrable on $[-\pi, \pi]$ . Its Fourier coefficients are

\hat f(n) = c_n = \frac{1}{2\pi}\int_{-\pi}^{\pi} f(x)\,e^{-inx}\,dx \quad (n \in \Z),

and its Fourier series is $\sum_{n \in \Z} c_n e^{inx}$ , with partial sums $S_N f(x) = \sum_{|n| \le N} c_n e^{inx}$ . For real $f$ this is $\frac{a_0}{2} + \sum_{n \ge 1}(a_n \cos nx + b_n \sin nx)$ with $a_n = \frac1\pi \int_{-\pi}^{\pi} f(x) \cos nx\,dx$ and $b_n = \frac1\pi\int_{-\pi}^{\pi} f(x)\sin nx\,dx$ .

Definition— Fourier sine series on [0, L]

For $f$ integrable on $[0, L]$ , the sine coefficients are $b_n = \frac{2}{L}\int_0^L f(x)\sin\frac{n\pi x}{L}\,dx$ and the sine series is $\sum_{n \ge 1} b_n \sin\frac{n\pi x}{L}$ . It is the full Fourier series of the odd $2L$ -periodic extension of $f$ .

Definition— inner product and L² norm

For complex-valued integrable functions on $[-\pi, \pi]$ , put $\langle f, g \rangle = \frac{1}{2\pi}\int_{-\pi}^{\pi} f(x)\overline{g(x)}\,dx$ and $\|f\|_2 = \langle f, f\rangle^{1/2}$ . The functions $e_n(x) = e^{inx}$ are orthonormal: $\langle e_m, e_n\rangle = 1$ if $m = n$ and $0$ otherwise, and $c_n = \langle f, e_n\rangle$ .

A function on $[a, b]$ is piecewise $C^1$ if it is continuously differentiable except at finitely many points, where $f$ and $f'$ have one-sided limits.

Worked examplesH

Worked examples

Example— The square wave

Let $f = \operatorname{sign}(\sin x)$ : $f = 1$ on $(0, \pi)$ and $-1$ on $(-\pi, 0)$ . It is odd, so $a_n = 0$ , and

b_n = \frac{2}{\pi}\int_0^\pi \sin nx\,dx = \frac{2}{\pi}\cdot\frac{1 - (-1)^n}{n} = \begin{cases} \frac{4}{\pi n} & n \text{ odd}, \\ 0 & n \text{ even}. \end{cases}

So $f \sim \frac{4}{\pi}\left(\sin x + \frac{\sin 3x}{3} + \frac{\sin 5x}{5} + \cdots\right)$ . At $x = \pi/2$ this gives Leibniz's series $\frac\pi4 = 1 - \frac13 + \frac15 - \cdots$ .

Example— The sawtooth and the triangle

For $f(x) = x$ on $(-\pi, \pi)$ , integration by parts gives $b_n = \frac{2(-1)^{n+1}}{n}$ , so $x \sim 2\left(\sin x - \frac{\sin 2x}{2} + \frac{\sin 3x}{3} - \cdots\right)$ . The coefficients decay like $1/n$ : the periodic extension jumps at $\pm\pi$ . For the continuous triangle wave $f(x) = |x|$ , $a_0 = \pi$ and $a_n = -\frac{4}{\pi n^2}$ for odd $n$ (zero for even $n \ge 2$ ). The coefficients decay like $1/n^2$ , one power faster, because one more derivative is continuous. Smoothness of $f$ is decay of $\hat f$ .

Example— The heat equation with initial profile x(π − x)

On $[0, \pi]$ , $\varphi(x) = x(\pi - x)$ has sine coefficients $b_n = \frac{2}{\pi}\int_0^\pi x(\pi - x)\sin nx\,dx = \frac{4(1 - (-1)^n)}{\pi n^3}$ , that is $\frac{8}{\pi n^3}$ for odd $n$ . So

u(t, x) = \frac{8}{\pi}\sum_{n \text{ odd}} \frac{e^{-n^2 t}\sin nx}{n^3}.

By $t = 1$ the second term is $e^{-8}/27 \approx 10^{-5}$ times the first: the profile is essentially $\frac{8}{\pi}e^{-t}\sin x$ .

Example— An ill-posed problem (Hadamard)

Pose Laplace's equation as an initial-value problem: $u_{xx} + u_{yy} = 0$ for $y > 0$ with $u(x, 0) = 0$ and $u_y(x, 0) = \frac{\sin nx}{n}$ . The solution is $u_n = \frac{\sin nx \sinh ny}{n^2}$ . As $n \to \infty$ the data tend to $0$ uniformly, but at $y = 1$ the solution has size $\frac{\sinh n}{n^2} \to \infty$ . Tiny changes in the data cause enormous changes in the solution, so the problem is not well posed — even though a solution exists and is unique. Laplace's equation wants boundary data, not initial data.

ProofH

Separation of variables

We solve the heat equation on $[0, \pi]$ with $u(t, 0) = u(t, \pi) = 0$ and $u(0, x) = \varphi(x)$ . (Rescaling $x$ handles any interval $[0, L]$ .)

Step 1: separated solutions. Look for $u = T(t)X(x)$ . Then $T'X = kTX''$ , so $\frac{T'}{kT} = \frac{X''}{X}$ wherever these are defined. The left side depends only on $t$ and the right only on $x$ , so both equal a constant $-\lambda$ . The problem $X'' = -\lambda X$ with $X(0) = X(\pi) = 0$ has a non-zero solution only when $\lambda = n^2$ , $n = 1, 2, \dots$ , with $X = \sin nx$ . (For $\lambda \le 0$ the solutions are $A + Bx$ or combinations of $e^{\pm\sqrt{-\lambda}x}$ , and the boundary conditions force them to vanish; for $\lambda > 0$ , $X = A\cos\sqrt\lambda x + B\sin\sqrt\lambda x$ , and $X(0) = 0$ , $X(\pi) = 0$ force $A = 0$ and $\sqrt\lambda \in \Z$ .) Then $T = e^{-kn^2t}$ .

Step 2: superposition. Formally, $u(t, x) = \sum_{n \ge 1} b_n e^{-kn^2t}\sin nx$ with $b_n$ the sine coefficients of $\varphi$ , so that $u(0, x) = \varphi(x)$ .

Theorem— Theorem 1 (the series solves the heat equation)

Let $\varphi$ be integrable on $[0, \pi]$ with sine coefficients $b_n$ . Then $u(t, x) = \sum_{n \ge 1} b_n e^{-kn^2t}\sin nx$ converges for $t > 0$ , together with all its derivatives, uniformly on $[t_0, \infty) \times [0, \pi]$ for each $t_0 > 0$ . The function $u$ is infinitely differentiable for $t > 0$ , satisfies $u_t = k u_{xx}$ and $u(t, 0) = u(t, \pi) = 0$ , and if $\varphi$ is continuous and piecewise $C^1$ with $\varphi(0) = \varphi(\pi) = 0$ , then $u(t, x) \to \varphi(x)$ uniformly as $t \to 0^+$ .

Proof

$|b_n| \le \frac2\pi\int_0^\pi|\varphi| =: B$ . A derivative $\partial_t^i\partial_x^j$ of the $n$ -th term is bounded by $B (kn^2)^i n^j e^{-kn^2t_0}$ for $t \ge t_0$ , and $\sum_n n^{2i + j}e^{-kn^2t_0} < \infty$ . By the Weierstrass M-test every differentiated series converges uniformly, so (by the theorem on derivatives of uniform limits in real analysis) we may differentiate term by term. Each term solves the equation and the boundary conditions, so $u$ does too. For the last claim, the hypotheses give $\sum|b_n| < \infty$ (Theorem 4 below, applied to the odd extension of $\varphi$ , which is continuous and piecewise $C^1$ ). Then $|u(t, x) - \varphi(x)| \le \sum_n |b_n|(1 - e^{-kn^2t})$ , which tends to $0$ uniformly by dominated convergence for series (or directly: split the sum at a large $N$ ).

∎

Theorem 1 shows the smoothing effect of the heat equation. Even a discontinuous initial profile becomes infinitely differentiable instantly. Running time backwards is the opposite. The backward heat equation multiplies the $n$ -th mode by $e^{+kn^2t}$ , which explodes, so it is ill posed: you cannot un-mix milk from coffee.

The vibrating string. The same method for $u_{tt} = c^2u_{xx}$ on $[0, \pi]$ with fixed ends gives $T'' = -c^2n^2T$ , so

u(t, x) = \sum_{n \ge 1}\big(a_n\cos nct + b_n\sin nct\big)\sin nx,

with $a_n$ the sine coefficients of $u(0, \cdot)$ and $nc\,b_n$ those of $u_t(0, \cdot)$ . The $n$ -th mode vibrates at frequency $n$ times the fundamental: the harmonic series of overtones that gives a string its timbre. Here nothing decays, so convergence requires smoother data, and on the whole line d'Alembert's formula below is cleaner.

ProofH

Fourier series: Bessel, Parseval and convergence

Theorem— Theorem 2 (Bessel's inequality; best approximation)

For $f$ integrable on $[-\pi, \pi]$ and every $N$ : $\|f - S_Nf\|_2^2 = \|f\|_2^2 - \sum_{|n| \le N}|c_n|^2$ . Hence $\sum_{n \in \Z}|c_n|^2 \le \|f\|_2^2$ , and in particular $c_n \to 0$ (the Riemann–Lebesgue lemma for square-integrable $f$ ). Moreover $S_Nf$ is the closest trigonometric polynomial of degree $\le N$ to $f$ in the $L^2$ norm.

Proof

By orthonormality, $\langle f - S_Nf, e_n\rangle = c_n - c_n = 0$ for $|n| \le N$ , so $f - S_Nf$ is orthogonal to every trigonometric polynomial $P$ of degree $\le N$ . Pythagoras gives $\|f - P\|_2^2 = \|f - S_Nf\|_2^2 + \|S_Nf - P\|_2^2 \ge \|f - S_Nf\|_2^2$ (best approximation). Taking $P = 0$ : $\|f\|_2^2 = \|f - S_Nf\|_2^2 + \|S_Nf\|_2^2$ , and $\|S_Nf\|_2^2 = \sum_{|n|\le N}|c_n|^2$ .

∎

Theorem— Theorem 3 (Parseval; stated)

For every square-integrable $f$ on $[-\pi, \pi]$ , $\|f - S_Nf\|_2 \to 0$ , and therefore

\sum_{n \in \Z}|c_n|^2 = \frac{1}{2\pi}\int_{-\pi}^{\pi}|f(x)|^2\,dx.

By Theorem 2, Parseval is equivalent to the density of trigonometric polynomials in $L^2$ . For continuous $f$ , density follows from Fejér's theorem: the averages $\sigma_N f = \frac{1}{N+1}\sum_{m=0}^N S_mf$ converge to $f$ uniformly, because they are convolutions with the positive kernel $F_N(t) = \frac{1}{N+1}\left(\frac{\sin((N+1)t/2)}{\sin(t/2)}\right)^2$ , which concentrates at $0$ . Continuous functions are in turn dense in $L^2$ . Full proofs are in Stein and Shakarchi (2003), chapters 2–3, and Mathlib proves the Hilbert-space version (see the formal section).

Example— Basel from Parseval

For $f(x) = x$ , $|c_n| = |b_n|/2 = \frac1{|n|}$ for $n \ne 0$ , and $\frac{1}{2\pi}\int_{-\pi}^{\pi}x^2\,dx = \frac{\pi^2}{3}$ . Parseval gives $2\sum_{n\ge1}\frac{1}{n^2} = \frac{\pi^2}{3}$ , that is, $\sum_{n\ge1}\frac{1}{n^2} = \frac{\pi^2}{6}$ , Euler's 1734 answer to the Basel problem.

$L^2$ convergence says nothing about any particular point. Pointwise convergence needs the Dirichlet kernel: substituting the coefficients into $S_Nf$ and summing the geometric series gives

S_Nf(x) = \frac{1}{2\pi}\int_{-\pi}^{\pi}f(x - t)\,D_N(t)\,dt, \qquad D_N(t) = \sum_{|n|\le N}e^{int} = \frac{\sin\big((N + \frac12)t\big)}{\sin(t/2)},

with $\frac{1}{2\pi}\int_{-\pi}^\pi D_N = 1$ .

Theorem— Theorem 4 (Dirichlet; uniform convergence)

Let $f$ be $2\pi$ -periodic and integrable.

(a) If $f$ is differentiable at $x$ (or merely Lipschitz near $x$ ), then $S_Nf(x) \to f(x)$ .

(b) If $f$ is piecewise $C^1$ , then $S_Nf(x) \to \frac12\big(f(x^+) + f(x^-)\big)$ at every $x$ .

(c) If $f$ is continuous and piecewise $C^1$ , then $\sum_n|c_n| < \infty$ , and $S_Nf \to f$ uniformly.

Proof

(a) Since $\frac{1}{2\pi}\int D_N = 1$ ,

S_Nf(x) - f(x) = \frac{1}{2\pi}\int_{-\pi}^{\pi}g(t)\sin\big((N + \tfrac12)t\big)\,dt, \qquad g(t) = \frac{f(x - t) - f(x)}{\sin(t/2)}.

Near $t = 0$ , $|f(x - t) - f(x)| \le C|t|$ and $|\sin(t/2)| \ge |t|/\pi$ , so $g$ is bounded, hence square integrable. Expand $\sin((N + \frac12)t) = \sin Nt\cos\frac t2 + \cos Nt\sin\frac t2$ . The integral becomes a combination of Fourier coefficients of the square-integrable functions $g(t)\cos\frac t2$ and $g(t)\sin\frac t2$ , which tend to $0$ by the Riemann–Lebesgue lemma (Theorem 2).

(b) Split the integral into $t > 0$ and $t < 0$ . Since $D_N$ is even, $\frac{1}{2\pi}\int_0^\pi D_N = \frac12$ . Repeat (a) on each half with $f(x^\mp)$ in place of $f(x)$ , using the one-sided Lipschitz bounds that piecewise $C^1$ provides.

(c) Integration by parts (valid because $f$ is continuous and periodic) gives $\widehat{f'}(n) = in\,c_n$ . By Cauchy–Schwarz and Bessel for $f'$ ,

\sum_{n \ne 0}|c_n| = \sum_{n\ne0}\frac{|\widehat{f'}(n)|}{|n|} \le \Big(\sum_{n\ne0}\frac1{n^2}\Big)^{1/2}\Big(\sum_n|\widehat{f'}(n)|^2\Big)^{1/2} < \infty.

By the M-test, $S_Nf$ converges uniformly to some continuous $g$ , and by (a) $g = f$ .

∎

The Gibbs phenomenon. Near a jump, convergence cannot be uniform: a uniform limit of continuous functions is continuous. But the failure has a precise, universal size. For the square wave, $S_n$ (with $n$ non-zero terms) has derivative $\frac{4}{\pi}\sum_{j=1}^n\cos((2j - 1)x) = \frac{2}{\pi}\frac{\sin 2nx}{\sin x}$ . So its first maximum to the right of $0$ is at $x = \frac{\pi}{2n}$ , where, with $t_j = \frac{(2j - 1)\pi}{2n}$ ,

S_n\Big(\frac{\pi}{2n}\Big) = \frac{4}{\pi}\sum_{j=1}^{n}\frac{\sin t_j}{2j - 1} = \frac{2}{\pi}\sum_{j=1}^n\frac{\pi}{n}\cdot\frac{\sin t_j}{t_j} \;\longrightarrow\; \frac{2}{\pi}\int_0^\pi\frac{\sin t}{t}\,dt = \frac{2}{\pi}\mathrm{Si}(\pi) \approx 1.17898,

because the middle sum is a midpoint Riemann sum. The partial sums overshoot the value $1$ by about $9\%$ of the jump $2$ , ever closer to the discontinuity, and the overshoot never shrinks. Every jump discontinuity of a piecewise $C^1$ function shows the same proportional overshoot.

What can go wrong. Continuity alone is not enough for pointwise convergence. Paul du Bois-Reymond (1873) constructed a continuous periodic function whose Fourier series diverges at a point, and Kolmogorov (1926) found an integrable function whose Fourier series diverges everywhere. In the positive direction, Lennart Carleson (1966) proved that the Fourier series of every square-integrable function converges almost everywhere — one of the deepest theorems of twentieth-century analysis.

Term-by-term differentiation can fail. Differentiating $x = 2\sum(-1)^{n+1}\frac{\sin nx}{n}$ term by term gives $2\sum(-1)^{n+1}\cos nx$ , whose terms do not even tend to $0$ . The problem is the jump of the periodic sawtooth at $\pm\pi$ . For a continuous, piecewise $C^1$ periodic function, $\widehat{f'}(n) = in\,c_n$ , so the differentiated series is the Fourier series of $f'$ . That series converges to $f'$ wherever Theorem 4 applies to $f'$ .

ProofH

d'Alembert's formula

Theorem— Theorem 5 (d'Alembert)

Every $C^2$ solution of $u_{tt} = c^2u_{xx}$ on $\R^2$ has the form $u(t, x) = F(x - ct) + G(x + ct)$ with $F, G \in C^2(\R)$ . The solution with $u(0, x) = \varphi(x)$ and $u_t(0, x) = \psi(x)$ ( $\varphi \in C^2$ , $\psi \in C^1$ ) is unique and given by

u(t, x) = \frac{\varphi(x - ct) + \varphi(x + ct)}{2} + \frac{1}{2c}\int_{x - ct}^{x + ct}\psi(s)\,ds.

Proof

Change to characteristic coordinates $\xi = x - ct$ , $\eta = x + ct$ , and write $u(t, x) = w(\xi, \eta)$ . The chain rule gives $u_x = w_\xi + w_\eta$ , $u_t = c(w_\eta - w_\xi)$ , and

u_{tt} - c^2u_{xx} = c^2(w_{\xi\xi} - 2w_{\xi\eta} + w_{\eta\eta}) - c^2(w_{\xi\xi} + 2w_{\xi\eta} + w_{\eta\eta}) = -4c^2 w_{\xi\eta}.

So $w_{\xi\eta} = 0$ . Then $w_\xi$ does not depend on $\eta$ , say $w_\xi = F'(\xi)$ , and $w - F(\xi)$ has zero $\xi$ -derivative, so it is a function $G(\eta)$ . Conversely, every such sum solves the equation (proved in Lean below).

For the initial values: $F + G = \varphi$ and $c(G' - F') = \psi$ . Integrating the second, $G - F = \frac1c\int_0^x\psi + C$ . Solving, $F(x) = \frac12\varphi(x) - \frac1{2c}\int_0^x\psi - \frac C2$ and $G(x) = \frac12\varphi(x) + \frac1{2c}\int_0^x\psi + \frac C2$ . Substituting gives the formula, and since $F, G$ were forced, the solution is unique.

∎

The formula shows two physical facts. Finite speed of propagation: $u(t, x)$ depends only on the data in $[x - ct, x + ct]$ , the domain of dependence, so disturbances travel at speed exactly $c$ . And no smoothing: a corner in $\varphi$ travels along the lines $x \pm ct$ forever, unlike for the heat equation.

ProofH

Maximum principles and energy

Theorem— Theorem 6 (weak maximum principle for the heat equation)

Let $u$ be continuous on the rectangle $R = [0, T] \times [0, L]$ , with $u_t, u_{xx}$ continuous for $0 < t \le T$ , $0 < x < L$ , and $u_t = ku_{xx}$ there. Then the maximum of $u$ over $R$ is attained on the parabolic boundary $\Gamma$ : the bottom $t = 0$ or the sides $x = 0$ , $x = L$ .

Proof

Let $M = \max_\Gamma u$ , and for $\varepsilon > 0$ put $v = u + \varepsilon x^2$ , so that $v_t - kv_{xx} = -2k\varepsilon < 0$ inside. Suppose $v$ attains its maximum over $R$ at a point $(t_0, x_0)$ not on $\Gamma$ , so $0 < t_0 \le T$ and $0 < x_0 < L$ . In the $x$ -direction this is an interior maximum, so $v_{xx}(t_0, x_0) \le 0$ . In the $t$ -direction it is a maximum at an interior point or at the top edge $t_0 = T$ , so $v_t(t_0, x_0) \ge 0$ . Then $v_t - kv_{xx} \ge 0$ , a contradiction. So $\max_R v = \max_\Gamma v \le M + \varepsilon L^2$ , hence $u \le v \le M + \varepsilon L^2$ on $R$ . Let $\varepsilon \to 0$ .

∎

(Without the $\varepsilon x^2$ trick the argument fails: at an interior maximum of $u$ we only get $u_t \ge 0 \ge ku_{xx}$ , which is consistent with $u_t = ku_{xx}$ .)

Theorem— Corollary 7 (uniqueness and stability for the heat equation)

The Dirichlet problem $u_t = ku_{xx}$ on $(0, T] \times (0, L)$ , $u(0, x) = \varphi(x)$ , $u(t, 0) = g(t)$ , $u(t, L) = h(t)$ has at most one solution continuous on $R$ . If $u_1, u_2$ solve it with initial data $\varphi_1, \varphi_2$ and the same boundary data, then $\max_R|u_1 - u_2| \le \max|\varphi_1 - \varphi_2|$ .

Proof

$w = u_1 - u_2$ solves the heat equation with zero boundary data. Apply Theorem 6 to $w$ and to $-w$ .

∎

The energy method gives uniqueness by integration instead.

Theorem— Theorem 8 (energy identities)

(a) If $w$ is a smooth solution of $w_t = kw_{xx}$ on $[0, L]$ with $w(t, 0) = w(t, L) = 0$ , then $E(t) = \int_0^Lw^2\,dx$ is non-increasing. (b) If $w$ is a smooth solution of $w_{tt} = c^2w_{xx}$ with the same boundary conditions, then $E(t) = \frac12\int_0^L\big(w_t^2 + c^2w_x^2\big)\,dx$ is constant. Consequently both Dirichlet problems have at most one smooth solution.

Proof

(a) $E'(t) = 2\int_0^Lww_t = 2k\int_0^Lww_{xx} = 2k\big[ww_x\big]_0^L - 2k\int_0^Lw_x^2 = -2k\int_0^Lw_x^2 \le 0$ . The boundary term vanishes because $w = 0$ at both ends.

(b) $E'(t) = \int_0^L(w_tw_{tt} + c^2w_xw_{xt}) = c^2\int_0^L(w_tw_{xx} + w_xw_{xt}) = c^2\big[w_tw_x\big]_0^L = 0$ , because $w_t = 0$ at the fixed ends.

For uniqueness, the difference of two solutions has zero initial data, so $E(0) = 0$ . In (a) this gives $E \equiv 0$ and $w \equiv 0$ . In (b) it gives $w_t = w_x = 0$ , so $w$ is constant, hence $0$ .

∎

For Laplace's equation the corresponding facts are the mean value property and the maximum principle.

Theorem— Theorem 9 (harmonic functions)

Let $u$ be harmonic ( $C^2$ with $\Delta u = 0$ ) on an open set $U \subseteq \R^2$ . (a) If the closed disc $\overline{B(p, r)}$ lies in $U$ , then $u(p) = \frac{1}{2\pi}\int_0^{2\pi}u(p + re^{i\theta})\,d\theta$ . (b) If $U$ is connected and $u$ attains its maximum at a point of $U$ , then $u$ is constant. Hence on a bounded domain, a harmonic function continuous up to the boundary attains its maximum and minimum on the boundary, and the Dirichlet problem for $\Delta u = 0$ has at most one solution.

Proof

(a) Let $m(s) = \frac{1}{2\pi}\int_0^{2\pi}u(p + se^{i\theta})\,d\theta$ . Differentiating under the integral, $m'(s) = \frac{1}{2\pi}\int_0^{2\pi}\nabla u \cdot e^{i\theta}\,d\theta = \frac{1}{2\pi s}\oint_{|z-p|=s}\frac{\partial u}{\partial n}\,d\ell$ . By the divergence theorem this equals $\frac{1}{2\pi s}\iint_{B(p,s)}\Delta u = 0$ . So $m$ is constant, and $m(s) \to u(p)$ as $s \to 0$ by continuity.

(b) The set where $u$ equals its maximum $M$ is closed in $U$ . It is also open: if $u(p) = M$ , the mean value property on every small circle around $p$ forces $u = M$ on those circles (an average of values $\le M$ equals $M$ only if all are $M$ , by continuity). By connectedness it is all of $U$ . For uniqueness, apply this to the difference of two solutions and its negative.

∎

Formal mathematicsF

PDEs and Fourier series in Lean and Mathlib

Mathlib has a substantial theory of Fourier series on the circle $\R/T\Z$ (the type AddCircle T) but, at this version, no library of partial differential equations. So the PDE results below are proved directly, by computing derivatives with the chain rule. Partial derivatives are written as ordinary derivatives of one-variable slices: $u_t$ at $(t, x)$ is deriv (fun s => u s x) t.

The separated solutions really solve their equations, and satisfy the boundary conditions:

heat_modeLean 4Proved here from basic factsFormally verified

/-- **A separated solution of the heat equation.** `u(t, x) = e^{-k n² t} sin(n x)` satisfies
`u_t = k u_xx`: differentiating in `t` brings down `-k n²`, and differentiating twice in `x`
brings down `-n²`. Both sides equal `-k n² e^{-k n² t} sin(n x)`. -/
theorem heat_mode (k n t x : ℝ) :
    deriv (fun s => exp (-k * n ^ 2 * s) * sin (n * x)) t =
      k * deriv (fun y => deriv (fun z => exp (-k * n ^ 2 * t) * sin (n * z)) y) x := by
  -- First and second derivatives in `x`.
  have hx1 : ∀ y, HasDerivAt (fun z => exp (-k * n ^ 2 * t) * sin (n * z))
      (exp (-k * n ^ 2 * t) * (cos (n * y) * n)) y := fun y =>
    (((hasDerivAt_id' y).const_mul n).sin.congr_deriv (by simp)).const_mul _
  have hfun : deriv (fun z => exp (-k * n ^ 2 * t) * sin (n * z)) =
      fun y => exp (-k * n ^ 2 * t) * (cos (n * y) * n) := funext fun y => (hx1 y).deriv
  have hx2 : HasDerivAt (fun y => exp (-k * n ^ 2 * t) * (cos (n * y) * n))
      (exp (-k * n ^ 2 * t) * (-sin (n * x) * n * n)) x := by
    have := ((((hasDerivAt_id' x).const_mul n).cos).mul_const n).const_mul (exp (-k * n ^ 2 * t))
    convert this using 1; ring
  -- The derivative in `t`.
  have ht : HasDerivAt (fun s => exp (-k * n ^ 2 * s) * sin (n * x))
      (exp (-k * n ^ 2 * t) * (-k * n ^ 2) * sin (n * x)) t := by
    have := (((hasDerivAt_id' t).const_mul (-k * n ^ 2)).exp).mul_const (sin (n * x))
    convert this using 1; ring
  rw [ht.deriv, hfun, hx2.deriv]
  ring