5.9 Partial differential equations

A partial differential equation relates the partial derivatives of an unknown function of several variables, and its solutions contain whole arbitrary functions instead of constants. A first-order linear equation with constant coefficients says the unknown is constant along straight characteristic lines. The wave equation splits into two such transport equations (d'Alembert), the heat equation smooths its data mode by mode (separation of variables), and the solutions of Laplace's equation obey a mean value property that forbids interior maxima.

In one variable a differential equation such as $y' = 3y$ is solved by a function that depends on one unknown constant. Most of physics is about quantities that depend on position and time at once: the height of a string, the temperature in a rod, the potential around a charge. Their laws say how the rates of change in different directions are related, and the unknown is a whole function of several variables. The equation $u_t + 2u_x = 0$ is solved by $g(x - 2t)$ for any function $g$ at all, so the solution space is not a few-dimensional space of formulas but a space of functions (Apostol 9.1: it is infinite-dimensional). This chapter does the three famous second-order equations, plus the first-order one that sits underneath the wave equation, with the tools of Unit 5: the chain rule of 5.4 to check and to change variables, equality of mixed partials of 5.5, and the gradient picture of 5.3.

Before you start

5.4: the chain rule for a composite $f(\text{linear change of variables})$ . 5.3: the gradient is perpendicular to level sets, and the derivative of a field along a path. 2.9: inner products of functions and orthogonality.

Quick checkLet

f(x, y) = g(2x - 3y)

with

g

differentiable. What is

3f_x + 2f_y

Quick checkFor a

C^2

function

u(x, t)

, which of these is always true?

Quick checkWhy are the functions

\sin(n\pi x)

and

\sin(m\pi x)

for

n \ne m

orthogonal on

[0, 1]

The picture: standing waves and travelling pulses

Two pictures of the same equation. In the first, the string vibrates as a sum of standing waves, each of which keeps its shape and only changes its height: this is what separation of variables finds. Press Play, then turn the second and third modes on. In the second picture (choose it in the menu), an initial bump splits into two copies that run apart at the wave speed: this is d’Alembert’s solution. Tick ‘fixed ends’ and watch what a reflection does to the pulse. The two descriptions are the same solution: $\sin a\cos b = \frac12[\sin(a + b) + \sin(a - b)]$ says that a standing wave is a right-moving and a left-moving wave added together.

The idea, in words

An ordinary differential equation pins down a function of one variable up to a few constants. A partial differential equation relates rates of change in several directions, and such a relation constrains the function only along certain curves, leaving freedom across them. For $u_t + 2u_x = 0$ it says that the rate of change of $u$ in the direction $(2, 1)$ of the $(x, t)$ plane is zero: $u$ is constant along every line $x - 2t = \text{const}$ , and what $u$ does from one such line to the next is free. That freedom is an arbitrary function. Initial and boundary conditions (the shape of the string at $t = 0$ , the temperature of the ends) are what select one function from the freedom.

Where this came from· d'Alembert's strings, Bernoulli's harmonics, and Fourier's heat

In 1747 Jean le Rond d’Alembert wrote down the equation for the vibrating string and found that its solutions are sums of a wave running left and a wave running right, each of an arbitrary shape. Leonhard Euler took up the problem within a year or two and argued that the shapes need not be given by a single formula, such as a plucked string with a corner, which d’Alembert disputed. Around 1753 Daniel Bernoulli proposed that the string’s motion is an infinite sum of a fundamental tone and its harmonics, the standing waves of separation of variables, without a method to find the coefficients. The three men disagreed for decades on which shapes counted as solutions and whether trigonometric sums were general enough. Joseph Fourier, who presented his first memoir on heat in 1807 and published the full Théorie analytique de la chaleur in 1822, used such sums for heat and claimed that every function could be expanded; Dirichlet in 1829 proved that Fourier series converge for a wide class of functions. Both descriptions in the picture above are therefore correct, and both are older than the language (function, continuity, convergence) needed to say why.

Definitions

Definition A

Partial differential equation

An equation involving a scalar field $u$ and its partial derivatives is a partial differential equation (PDE). Its order is that of the highest derivative present. It is linear if it has the form $L(u) = s$ where $L$ is a linear combination of $u$ and its partial derivatives (with coefficients that may depend on the variables, but not on $u$ ) and homogeneous if $s = 0$ . Examples with $u = u(x, y)$ or $u(x, t)$ : $u_x = 0$ (order 1), $u_t + 2u_x = 0$ (order 1), $u_{xx} + u_{yy} = 0$ (order 2). Apostol 9.1.

In one lineAn equation involving an unknown function of two or more variables and some of its partial derivatives.

Why it existsIt is the natural law for any quantity that depends on position and time: it says how the change in each direction is tied to the changes in the others.

Proposition A Apostol 9.1

Superposition for linear homogeneous equations

If $L$ is a linear differential operator and $L(u_1) = L(u_2) = 0$ , then $L(au_1 + bu_2) = 0$ for all real $a, b$ .

In one lineThe solutions of a linear homogeneous PDE form a vector space: sums and multiples of solutions are solutions.

Why it is trueIt is why one can build solutions from simple ones (standing waves, plane waves) and why a general solution is a sum.

ProofApostol 9.1

The idea: The operator is a linear map on a space of functions; the null space of a linear map is a subspace.

L(au_1 + bu_2) = aL(u_1) + bL(u_2)

Derivatives are linear:

(au_1 + bu_2)_x = au_{1x} + bu_{2x}

, and likewise for every partial derivative.

L(au_1 + bu_2) = 0

Both terms are zero by hypothesis. This is the null space of the linear map

L

(2.3).

\blacksquare

The null space can be infinite-dimensional: every function $g(y)$ solves $u_x = 0$ (Apostol 9.1), because $u$ may change with $y$ but not with $x$ . The functions $y, y^2, y^3, \dots$ are infinitely many solutions that are linearly independent.

First-order equations: constant along characteristic lines

Theorem A Apostol Thm 9.1 (9.2)

First-order equation with constant coefficients

Let $a, b$ be constants, not both zero. A differentiable $f$ on $\mathbb R^2$ satisfies $a\,f_x + b\,f_y = 0$ everywhere if and only if $f(x, y) = g(bx - ay)$ for some differentiable $g: \mathbb R \to \mathbb R$ .

In one lineIf $a, b$ are constants, not both $0$ , then $f$ satisfies $af_x + bf_y = 0$ if and only if $f(x, y) = g(bx - ay)$ for some differentiable function $g$ of one variable.

Why it is trueIt names every solution of the simplest PDE: constant on straight lines in the direction $(a, b)$ , free from line to line. Most of the later equations factor into equations of this type.

ProofApostol 9.2 (the geometric argument), made precise with characteristics

The idea: Write the equation as a dot product: the derivative of f in the direction (a, b) is zero, so f is constant on every line in that direction. The lines are bx - ay = const, and one line is a single value of that expression. Conversely the chain rule checks any g.

Fix a point

(x_0, y_0)

and put

\phi(s) = f(x_0 + as, y_0 + bs)

. Then

\phi'(s) = af_x + bf_y = 0

\frac{d}{ds}f(x_0 + as,\ y_0 + bs) = af_x + bf_y

by the path theorem of 5.3.

f

is constant on the line through

(x_0, y_0)

with direction

(a, b)

. This line is the curve

bx - ay = bx_0 - ay_0

(a characteristic).

A function with zero derivative on an interval is constant (mean value theorem, 1.4).

Define

g(k) = f(p(k))

with

p(k) = \dfrac{k}{a^2 + b^2}(b, -a)

. It is differentiable, as a composite of

f

with a linear map, and the line

bx - ay = k

passes through

p(k)

Every point lies on exactly one such line. The point

p(k)

lies on the line

bx - ay = k

, because

b\cdot\dfrac{kb}{a^2 + b^2} - a\cdot\dfrac{-ka}{a^2 + b^2} = k

. So the lines are labelled by

k

, and

p(k)

is one chosen point on each.

Hence

f(x, y) = g(bx - ay)

f

is constant on the line, and

bx - ay = k

at the point

(x, y)

itself. So

f(x, y) = f(p(k)) = g(k)

with

k = bx - ay

Every such

f

is a solution.

Conversely, chain rule:

f_x = b\,g'

f_y = -a\,g'

, so

af_x + bf_y = abg' - abg' = 0

\blacksquare

Definition A

Characteristic curve

For the first-order linear equation $a(x, y)u_x + b(x, y)u_y + c(x, y)u = f(x, y)$ , a characteristic curve is a solution of the system $\dfrac{dx}{ds} = a(x, y)$ , $\dfrac{dy}{ds} = b(x, y)$ . Along it the field $U(s) = u(x(s), y(s))$ has $U'(s) = u_xx' + u_yy' = au_x + bu_y = f - cu$ . So the PDE becomes the ODE $U' + cU = f$ for $U(s)$ . For constant $a, b$ the characteristics are the straight lines of the theorem. The method of characteristics is: find the curves, solve the ODE along each, and label the curves by where they cross the line (or curve) where the initial data is given.

In one lineA curve $(x(s), y(s))$ along which the PDE $a\,u_x + b\,u_y + c\,u = f$ turns into an ordinary differential equation: $x' = a$ , $y' = b$ .

Why it existsIt is the idea that reduces a PDE to ODEs: follow the unknown along these curves and it changes by a rule in one variable.

Worked example Apostol 9.3 #1

Find the solution of

4f_x + 3f_y = 0

with

f(x, 0) = \sin x

for all

x

Solution.

General solution. With $a = 4$ , $b = 3$ the theorem gives $f(x, y) = g(3x - 4y)$ . Characteristics are the lines $3x - 4y = \text{const}$ with direction $(4, 3)$ .
Use the condition. At $y = 0$ : $f(x, 0) = g(3x) = \sin x$ for all $x$ . Put $s = 3x$ : $g(s) = \sin\dfrac s3$ . The condition determines $g$ on the whole real line, since $3x$ takes every value.
Assemble. $f(x, y) = \sin\dfrac{3x - 4y}{3} = \sin\big(x - \tfrac43y\big)$ . Substitute $s = 3x - 4y$ .
Check. $f_x = \cos(x - \frac43y)$ , $f_y = -\frac43\cos(x - \frac43y)$ , so $4f_x + 3f_y = 4\cos - 4\cos = 0$ and $f(x, 0) = \sin x$ . Always verify.

Answer. $f(x, y) = \sin\big(x - \frac43y\big)$ : the sine wave $\sin x$ at $y = 0$ is carried along the direction $(4, 3)$ .

Worked example Apostol 9.3 #3

(a) Show that

u(x, y) = f(xy)

satisfies

xu_x - yu_y = 0

, and find the solution with

u(x, x) = x^4e^{x^2}

. (b) Show that

v = f(x/y)

satisfies

xv_x + yv_y = 0

, and find the one with

v(1, 1) = 2

and

D_1v(x, 1/x) = 1/x

Solution.

(a) Verify. $u_x = yf'(xy)$ and $u_y = xf'(xy)$ , so $xu_x - yu_y = xyf' - xyf' = 0$ . Chain rule. The characteristics here solve $x' = x$ , $y' = -y$ , which are the hyperbolas $xy = \text{const}$ , and $u$ is constant on each.
(a) Solve. On the diagonal, $u(x, x) = f(x^2) = x^4e^{x^2}$ . Put $t = x^2$ : $f(t) = t^2e^t$ for $t \ge 0$ . So $u(x, y) = (xy)^2e^{xy}$ . The data on $y = x$ determines $f$ at all values $t = x^2 \ge 0$ , which is the range of $xy$ on the diagonal; the formula extends naturally.
(b) Verify. $v_x = f'(x/y)\cdot\frac1y$ , $v_y = -f'(x/y)\cdot\frac x{y^2}$ , so $xv_x + yv_y = \frac xyf' - \frac xyf' = 0$ . Characteristics $x' = x$ , $y' = y$ are the rays through the origin, along which $x/y$ is constant.
(b) Use $D_1v(x, 1/x) = 1/x$ . $D_1v(x, y) = f'(x/y)/y$ ; at $y = 1/x$ this is $xf'(x^2) = \frac1x$ , so $f'(x^2) = \frac1{x^2}$ , i.e. $f'(t) = \frac1t$ and $f(t) = \ln t + C$ for $t \gt 0$ . Put $t = x^2$ and use $x \gt 0$ .
(b) Use $v(1, 1) = 2$ . $v(1, 1) = f(1) = C = 2$ . $\ln1 = 0$ .

Answer. (a) $u = (xy)^2e^{xy}$ . (b) $v(x, y) = \ln\dfrac xy + 2$ .

Worked example new (method of characteristics)

Solve the transport equation with decay,

u_t + 2u_x + u = 0

u(x, 0) = e^{-x^2}

Solution.

Characteristics. $\dfrac{dx}{ds} = 2$ , $\dfrac{dt}{ds} = 1$ : the lines $x = x_0 + 2t$ (take $s = t$ ). Coefficients of $u_x$ and $u_t$ in the PDE.
The ODE along a characteristic. Put $U(t) = u(x_0 + 2t, t)$ . Then $U' = u_x\cdot2 + u_t = -u = -U$ , so $U(t) = U(0)e^{-t}$ . The PDE says $u_t + 2u_x = -u$ .
Use the data. $U(0) = u(x_0, 0) = e^{-x_0^2}$ . At a point $(x, t)$ the characteristic through it started at $x_0 = x - 2t$ . Label each line by where it crosses $t = 0$ .
Assemble. $u(x, t) = e^{-t}\,e^{-(x - 2t)^2}$ . Substitute $x_0 = x - 2t$ .
Check. $u_t = -u + 4(x - 2t)u$ and $2u_x = -4(x - 2t)u$ , so $u_t + 2u_x + u = 0$ . ✓

Answer. $u(x, t) = e^{-t}e^{-(x - 2t)^2}$ : the bump moves right at speed $2$ while its height decays like $e^{-t}$ . The same recipe gives, for $u_t + au_x + cu = 0$ , the solution $u(x, t) = e^{-ct}u_0(x - at)$ .

Worked example Apostol 9.3 #8 and #9 (the converse of Euler's theorem)

Euler's theorem by characteristics: if

f

satisfies

x\cdot\nabla f(x) = pf(x)

\mathbb R^2\setminus\{O\}

, show that

f(sx) = s^pf(x)

for

s \gt 0

Solution.

The characteristics. The coefficients of $f_x, f_y$ are $x, y$ , so the characteristics solve $x' = x$ , $y' = y$ : the rays $\sigma \mapsto e^\sigma(x_0, y_0)$ . The field of directions $(x, y)$ points radially outward.
The ODE along a ray. $F(\sigma) = f(e^\sigma x_0, e^\sigma y_0)$ has $F'(\sigma) = \nabla f\cdot(e^\sigma x_0, e^\sigma y_0) = p\,F(\sigma)$ by the equation. So $F(\sigma) = F(0)e^{p\sigma}$ . Path theorem of 5.3, then the equation $x\cdot\nabla f = pf$ at the point $e^\sigma x_0$ .
Translate. With $s = e^\sigma$ : $f(sx_0) = s^pf(x_0)$ . That is the definition of homogeneous of degree $p$ .

Answer. $f$ is homogeneous of degree $p$ . Conversely, differentiating $f(sx) = s^pf(x)$ in $s$ at $s = 1$ gives $x\cdot\nabla f = pf$ (Euler’s theorem). Example: $f = x^2 + 3xy$ has $p = 2$ and $x f_x + yf_y = 2x^2 + 3xy + 3xy = 2f$ . ✓

Quick checkWhich of these functions solves

u_t + 3u_x = 0

The wave equation

Definition A

Wave equation

The wave equation in one space variable is $u_{tt} = c^2u_{xx}$ , where $c \gt 0$ is a constant. In two or three space variables $u_{tt} = c^2\nabla^2u$ . Apostol 9.4.

In one lineThe equation $u_{tt} = c^2u_{xx}$ for the displacement $u(x, t)$ of a vibrating string, with $c$ the wave speed.

Why it existsIt is Newton's second law for a thin string under tension, and the model of every one-dimensional wave: sound in a pipe, light in a cable, a vibrating guitar string.

Why· Where the equation comes from: Newton's law for a small piece of string

A string with mass $\rho$ per unit length is pulled with tension $T$ (assumed constant). Let $u(x, t)$ be the height at $x$ . For a small piece between $x$ and $x + dx$ , the tension pulls up on the right end with vertical component $T\sin\theta(x + dx)$ and down on the left with $T\sin\theta(x)$ , where $\theta$ is the angle of the string. For small displacements $\sin\theta \approx \tan\theta = u_x$ , so the net force is $T\big(u_x(x + dx, t) - u_x(x, t)\big) \approx T\,u_{xx}\,dx$ . The mass is $\rho\,dx$ and the acceleration $u_{tt}$ , so Newton’s law gives $\rho\,u_{tt}\,dx = T\,u_{xx}\,dx$ . Hence $u_{tt} = c^2u_{xx}$ with $c = \sqrt{T/\rho}$ : the wave speed grows with tension and falls with mass. A string of length $0.65$ m and linear density $0.5$ g/m tuned to the note E4, $329.63$ Hz, has (from the formula $c = 2Lf$ derived below) $c = 2\cdot0.65\cdot329.63 = 428.5$ m/s and tension $T = \rho c^2 = 0.0005\cdot428.5^2 = 91.8$ N, about the weight of 9.4 kg.

Theorem A Apostol Thm 9.2 (9.4); d'Alembert 1747

d'Alembert's solution of the wave equation

Let $F \in C^2(\mathbb R)$ and $G \in C^1(\mathbb R)$ . Then $u(x, t) = \tfrac12\big[F(x + ct) + F(x - ct)\big] + \dfrac1{2c}\displaystyle\int_{x - ct}^{x + ct}G(s)\,ds$ satisfies $u_{tt} = c^2u_{xx}$ , $u(x, 0) = F(x)$ , $u_t(x, 0) = G(x)$ . Conversely, if $u$ is $C^2$ on $\mathbb R^2$ and solves the wave equation with these initial data, then $u$ is given by this formula.

In one lineIf $F$ is twice differentiable and $G$ is differentiable, then $u(x, t) = \tfrac12\big[F(x + ct) + F(x - ct)\big] + \dfrac1{2c}\displaystyle\int_{x - ct}^{x + ct}G(s)\,ds$ solves $u_{tt} = c^2u_{xx}$ with $u(x, 0) = F(x)$ and $u_t(x, 0) = G(x)$ ; conversely every $C^2$ solution on $\mathbb R^2$ has this form.

Why it is trueIt solves the initial value problem for the string in closed form: the motion is a left-moving and a right-moving wave. It shows information travels at speed $c$ .

ProofApostol 9.4 (this is his 'one method', via 9.3 #4)

The idea: Change variables to xi = x + ct and eta = x - ct (the two characteristic directions). Then the wave equation becomes v_xi_eta = 0 by the chain rule, so v is a function of xi plus a function of eta. The initial conditions fix the two functions.

Put

\xi = x + ct

\eta = x - ct

and

u(x, t) = v(\xi, \eta)

Define

v(\xi, \eta) = u\big(\frac{\xi + \eta}2,\ \frac{\xi - \eta}{2c}\big)

, the same field in the new coordinates, which is

C^2

because the change of variables is linear.

u_x = v_\xi + v_\eta

and

u_t = c(v_\xi - v_\eta)

Chain rule (5.4):

\xi_x = \eta_x = 1

\xi_t = c

\eta_t = -c

u_{xx} = v_{\xi\xi} + 2v_{\xi\eta} + v_{\eta\eta}

and

u_{tt} = c^2(v_{\xi\xi} - 2v_{\xi\eta} + v_{\eta\eta})

Apply the chain rule again, using

v_{\xi\eta} = v_{\eta\xi}

(5.5).

u_{tt} - c^2u_{xx} = -4c^2v_{\xi\eta}

. So the wave equation holds if and only if

v_{\xi\eta} = 0

Subtract: the

v_{\xi\xi}

and

v_{\eta\eta}

terms cancel.

Hence

v(\xi, \eta) = \varphi(\xi) + \psi(\eta)

, that is

u(x, t) = \varphi(x + ct) + \psi(x - ct)

\partial_\eta(v_\xi) = 0

means

v_\xi

does not depend on

\eta

(mean value theorem in

\eta

for fixed

\xi

, on the whole plane):

v_\xi = \varphi'(\xi)

. Then

v - \varphi(\xi)

has zero

\xi

-derivative, so depends on

\eta

only.

Solving the two linear equations:

\varphi' = \frac12F' + \frac1{2c}G

\psi' = \frac12F' - \frac1{2c}G

Initial data:

u(x, 0) = \varphi(x) + \psi(x) = F(x)

and

u_t(x, 0) = c\big(\varphi'(x) - \psi'(x)\big) = G(x)

. Differentiate the first:

\varphi' + \psi' = F'

\varphi(x) = \frac12F(x) + \frac1{2c}\int_0^xG + C

\psi(x) = \frac12F(x) - \frac1{2c}\int_0^xG - C

, which gives the formula.

Integrate from

0

, with

\varphi + \psi = F

0

giving the constants (they cancel). Then

u = \varphi(x + ct) + \psi(x - ct)

and the two integrals combine:

\int_0^{x + ct}G - \int_0^{x - ct}G = \int_{x - ct}^{x + ct}G

The formula is a solution.

Conversely, for the formula

u = \varphi(x + ct) + \psi(x - ct)

with

\varphi, \psi \in C^2

(which they are, by the hypotheses on

F

and

G

u_{tt} = c^2\varphi'' + c^2\psi''= c^2u_{xx}

by the chain rule, and the initial conditions are checked as above.

\blacksquare

Proof workshopGive the reason for each line of d'Alembert's derivationApostol 9.4

Each line of the proof is shown in order. Choose the reason that justifies it, then press Check.

Put $\xi = x + ct$ , $\eta = x - ct$ , $u(x, t) = v(\xi, \eta)$ .
$u_{tt} - c^2u_{xx} = -4c^2v_{\xi\eta}$ .
$v_{\xi\eta} = 0$ , so $v = \varphi(\xi) + \psi(\eta)$ .
$\varphi + \psi = F$ and $c(\varphi' - \psi') = G$ at $t = 0$ .
$\varphi' = \frac12F' + \frac1{2c}G$ and $\psi' = \frac12F' - \frac1{2c}G$ .
$u = \frac12[F(x + ct) + F(x - ct)] + \frac1{2c}\int_{x - ct}^{x + ct}G$ .

Worked example new (compare Apostol 9.4, Example)

Wave speed 2, initial shape

F(x) = e^{-x^2}

, initially at rest (

G = 0

). Describe the solution and find

u(3, 1.5)

Solution.

Formula. With $G = 0$ : $u(x, t) = \frac12\big[e^{-(x + 2t)^2} + e^{-(x - 2t)^2}\big]$ . d’Alembert with $c = 2$ .
Interpret. The first term is a bump of height $\frac12$ moving left at speed $2$ ; the second is a bump of height $\frac12$ moving right at speed $2$ . $F(x - ct)$ is $F$ shifted right by $ct$ .
Evaluate. At $x = 3$ , $t = 1.5$ : $x - 2t = 0$ , $x + 2t = 6$ . So $u = \frac12\big[e^{-36} + 1\big] \approx 0.5$ . The right-moving bump is exactly over $x = 3$ ; the left-moving one is $6$ units away and negligible.

Answer. Two bumps of half height run apart at speed $2$ ; $u(3, 1.5) = \frac12(1 + e^{-36}) \approx 0.5$ . Apostol’s own example is the same with the bump $1 + \cos\pi x$ on $(-1, 1)$ (which has only one continuous derivative at its edges, so it is a solution only in a generalised sense, but the formula still describes the motion).

Your answerFor the wave equation with

c = 2

F = e^{-x^2}

G = 0

, what is

u(3, 1.5)

Worked example new (the velocity term of d'Alembert's formula)

With

F = 0

and

G(x) = \cos x

, find the solution of the wave equation and its value at

x = 0

t = \pi/6

for

c = 3

Solution.

Formula. $u = \dfrac1{2c}\int_{x - ct}^{x + ct}\cos s\,ds = \dfrac1{2c}\big[\sin(x + ct) - \sin(x - ct)\big]$ . $F = 0$ , so only the integral term survives.
Simplify. $\sin(x + ct) - \sin(x - ct) = 2\cos x\sin(ct)$ , so $u = \dfrac1c\cos x\sin(ct)$ . Sum-to-product identity: $\sin A - \sin B = 2\cos\frac{A + B}2\sin\frac{A - B}2$ .
Check. $u_{tt} = -c\cos x\sin(ct)$ and $c^2u_{xx} = c^2\cdot\big(-\frac1c\cos x\sin ct\big) = -c\cos x\sin ct$ . ✓ And $u_t(x, 0) = \cos x = G$ . It solves the equation and the initial data.
Evaluate. For $c = 3$ , $x = 0$ , $t = \frac\pi6$ : $u = \frac13\cdot1\cdot\sin\frac\pi2 = \frac13$ . $ct = \frac\pi2$ .

Answer. $u = \frac1c\cos x\sin(ct)$ ; $u(0, \pi/6) = \frac13$ . This is a standing wave: the profile $\cos x$ stays in place and oscillates in time with angular frequency $c$ (the factor $\sin ct$ ).

Your answerFor

c = 3

F = 0

G = \cos x

, find

u(0, \pi/6)

Beyond the course: for understanding· A finite string: reflections by an odd extension

For a string on $[0, L]$ fixed at both ends, $u(0, t) = u(L, t) = 0$ , extend the initial shape $F$ to an odd function of period $2L$ and use the same formula. Then $u(0, t) = \frac12[F(ct) + F(-ct)] = 0$ because $F$ is odd, and $u(L, t) = \frac12[F(L + ct) + F(L - ct)] = \frac12[F(-L + ct) + F(L - ct)] = \frac12[-F(L - ct) + F(L - ct)] = 0$ by periodicity and oddness. The pulses of the formula pass through the walls as upside-down mirror images: a reflection at a fixed end flips the sign, as in the second picture of the visualiser. The motion repeats every $2L/c$ , which gives the fundamental frequency $f_1 = \frac{c}{2L}$ , and the standing waves $\sin(n\pi x/L)$ have frequencies $f_n = n\,\frac{c}{2L}$ : a note and its harmonics. For the E4 string above, $c = 2Lf_1 = 428.5$ m/s and the harmonics are $329.6$ , $659.3$ , $988.9$ Hz, and so on.

Your answerA string of length

0.65

m has fundamental frequency

329.63

Hz. What is the wave speed

c = 2Lf_1

in m/s? (Give a decimal.)

Quick checkA pulse travelling to the right along a string reaches a fixed end. What does the reflected pulse look like?

The heat equation

Definition A

Heat equation

The heat equation (or diffusion equation) is $u_t = \alpha u_{xx}$ for a rod, $u_t = \alpha(u_{xx} + u_{yy})$ for a plate, and $u_t = \alpha\nabla^2u$ in general, with constant $\alpha \gt 0$ . The first-order time derivative is the difference from the wave equation.

In one lineThe equation $u_t = \alpha u_{xx}$ (or $u_t = \alpha\nabla^2u$ in several space variables) for the temperature $u$ of a conducting body, with $\alpha \gt 0$ the diffusivity.

Why it existsIt is energy conservation plus Fourier's law, and the model of every diffusion: heat, dissolved salt, the spread of a probability distribution.

Why· Where the equation comes from: conservation of energy and Fourier's law

Take a rod of cross-section $A$ , density $\rho$ , specific heat $c_p$ and conductivity $k$ , with temperature $u(x, t)$ . The heat stored between $x$ and $x + dx$ is $\rho c_pA\,u\,dx$ . Fourier’s law (5.2) says heat flows from hot to cold at the rate $q = -k\,u_x$ per unit area. In time $dt$ the piece gains $A\big(q(x) - q(x + dx)\big)dt = Ak\big(u_x(x + dx) - u_x(x)\big)dt \approx Ak\,u_{xx}\,dx\,dt$ . Equating with the rise in stored heat, $\rho c_pA\,u_t\,dx\,dt$ , gives $\rho c_pu_t = ku_{xx}$ , that is $u_t = \alpha u_{xx}$ with $\alpha = \dfrac{k}{\rho c_p}$ . For copper, $k = 400$ W/(m K) (the value of 5.2), $\rho = 8960$ kg/m $^3$ and $c_p = 385$ J/(kg K) give $\alpha = \dfrac{400}{8960\cdot385} = 1.16\times10^{-4}$ m $^2$ /s. The time for heat to cross a length $L$ is of order $L^2/\alpha$ : for a $0.2$ m rod, $0.04/(1.16\times10^{-4}) \approx 345$ s, about six minutes.

Worked example MATH 375 Homework 11 style (Apostol 8.9 #18); related to Apostol 9.5 #1, whose solution is an integral of this kernel

Show that

u(x, t) = \dfrac1{\sqrt{4\pi\alpha t}}\,e^{-x^2/(4\alpha t)}

solves

u_t = \alpha u_{xx}

for

t \gt 0

, and say what it describes.

Solution.

Logarithmic derivative in $x$ . $u_x = -\dfrac{x}{2\alpha t}\,u$ . Chain rule on the exponent $-x^2/(4\alpha t)$ .
Second derivative. $u_{xx} = -\dfrac{1}{2\alpha t}u - \dfrac{x}{2\alpha t}u_x = \Big(-\dfrac{1}{2\alpha t} + \dfrac{x^2}{4\alpha^2t^2}\Big)u$ . Product rule, then substitute $u_x$ .
Derivative in $t$ . $u_t = \Big(-\dfrac1{2t} + \dfrac{x^2}{4\alpha t^2}\Big)u$ . The prefactor $t^{-1/2}$ gives $-\frac1{2t}$ , and the exponent $-\frac{x^2}{4\alpha}t^{-1}$ has $t$ -derivative $+\frac{x^2}{4\alpha t^2}$ .
Compare. $\alpha u_{xx} = \Big(-\dfrac1{2t} + \dfrac{x^2}{4\alpha t^2}\Big)u = u_t$ . ✓

Answer. It is a solution. At each time it is a Gaussian (bell curve) with variance $2\alpha t$ and total area $1$ : the temperature after a unit of heat is released at the single point $x = 0$ at time $0$ . The width grows like $\sqrt t$ : diffusion spreads like the square root of time, not linearly. This is the one-dimensional cousin of the function $t^{-3/2}e^{-r^2/4t}$ of 5.2. Convolving any initial temperature with it gives the solution, which is also why the heat equation is Gaussian blurring.

Separation of variables

Before the method, one check that every PDE student must be able to do: test a candidate.

Worked example new (checking a candidate solution)

u = e^{-4t}\sin x

a solution of

u_t = 4u_{xx}

? Of

u_{tt} = 4u_{xx}

Solution.

Differentiate in $t$ . $u_t = -4e^{-4t}\sin x = -4u$ and $u_{tt} = 16e^{-4t}\sin x = 16u$ . Treat $x$ as a constant; only the factor $e^{-4t}$ depends on $t$ .
Differentiate in $x$ . $u_x = e^{-4t}\cos x$ and $u_{xx} = -e^{-4t}\sin x = -u$ . Treat $t$ as a constant.
Heat equation. $4u_{xx} = -4u = u_t$ . Both sides are the same function.
Wave equation. $4u_{xx} = -4u$ but $u_{tt} = 16u$ , and $-4u \ne 16u$ wherever $u \ne 0$ . One point where the two sides differ is enough to fail.

Answer. Yes for the heat equation (the shape $\sin x$ decays at rate $4$ ), no for the wave equation (a wave equation would need an oscillating time factor, not a decaying one).

Definition A

Separation of variables

To solve the heat equation on the rod $0 \le x \le L$ with the ends held at temperature $0$ , $u(0, t) = u(L, t) = 0$ , try $u = X(x)T(t)$ . Then $u_t = XT'$ and $u_{xx} = X''T$ , so the equation is $XT' = \alpha X''T$ . Dividing by $\alpha XT$ (where nonzero): $\dfrac{T'}{\alpha T} = \dfrac{X''}{X}$ . The left side depends only on $t$ and the right side only on $x$ ; two functions of different variables that are equal for all $x, t$ are both equal to one constant, call it $-\kappa$ . So $X'' = -\kappa X$ and $T' = -\alpha\kappa T$ . The conditions $X(0) = X(L) = 0$ force $\kappa = (n\pi/L)^2$ and $X = \sin(n\pi x/L)$ , $n = 1, 2, 3, \dots$ (for $\kappa \le 0$ the only solution of $X'' = -\kappa X$ with both ends zero is $X = 0$ ; for $\kappa \gt 0$ the solutions are $A\cos(\sqrt\kappa x) + B\sin(\sqrt\kappa x)$ , which vanish at $L$ only when $\sqrt\kappa L$ is a multiple of $\pi$ ). Then $T = e^{-\alpha(n\pi/L)^2t}$ .

In one lineThe method of looking for solutions of a PDE that are products of functions of one variable each, such as $u = X(x)T(t)$ , which turns the PDE into ordinary differential equations.

Why it existsIt replaces one PDE by several equations in one variable, which can be solved, and the linearity of the PDE then lets us add the product solutions.

So the modes $u_n(x, t) = e^{-\alpha(n\pi/L)^2t}\sin\dfrac{n\pi x}L$ are solutions with zero ends, and by superposition any sum $\sum b_nu_n$ is too. The decay rate grows like $n^2$ : the wiggly modes die first, which is why heat smooths. To match an initial temperature $f$ we need $\sum b_n\sin(n\pi x/L) = f(x)$ and that needs the coefficients.

Definition A

Fourier sine series

For a function $f$ on $[0, L]$ , its Fourier sine series is $\sum_{n \ge 1}b_n\sin\frac{n\pi x}L$ with $b_n = \dfrac2L\displaystyle\int_0^Lf(x)\sin\dfrac{n\pi x}L\,dx$ . For piecewise smooth $f$ the series converges to $f$ at every point of continuity inside $(0, L)$ (at the ends it converges to $0$ , the value of every sine, so it matches $f$ there only if $f(0) = f(L) = 0$ ), a theorem of Dirichlet and Fourier analysis (B, MATH 521). Here we prove the formula for $b_n$ when $f$ is a finite sum of sines, and use it as the recipe for the infinite series.

In one lineThe expansion $f(x) = \sum_{n \ge 1}b_n\sin\dfrac{n\pi x}L$ on $[0, L]$ , with $b_n = \dfrac2L\int_0^Lf(x)\sin\dfrac{n\pi x}L\,dx$ .

Why it existsIt writes an arbitrary profile as a sum of the modes that separation of variables produced; the coefficient formula is the projection onto an orthogonal basis from 2.9.

Why· Fourier sine series from scratch: why this formula, and why sines

Separation of variables left us with a ladder of solutions $u_n$ , one for each whole number $n$ . The starting temperature is given as some profile $f(x)$ , so we need numbers $b_n$ with $f(x) = b_1\sin\frac{\pi x}L + b_2\sin\frac{2\pi x}L + \cdots$ . This is the vector problem of 2.9 in disguise.

In $\mathbb R^3$ , suppose $e_1, e_2, e_3$ are perpendicular and $v = c_1e_1 + c_2e_2 + c_3e_3$ . Take the dot product with $e_1$ : the other two terms drop out because they are perpendicular to $e_1$ , leaving $v\cdot e_1 = c_1\,(e_1\cdot e_1)$ , so $c_1 = \dfrac{v\cdot e_1}{e_1\cdot e_1}$ . Now let the “vectors” be functions on $[0, L]$ , with the inner product $\langle f, g\rangle = \int_0^Lf(x)g(x)\,dx$ (the dot product of two functions: multiply value by value at every $x$ and add up). The “axes” are the sines $s_n(x) = \sin\frac{n\pi x}L$ . The lemma below shows they are perpendicular, $\langle s_n, s_m\rangle = 0$ for $n \ne m$ , and that each has squared length $\langle s_n, s_n\rangle = \frac L2$ . The recipe is the same: $b_n = \dfrac{\langle f, s_n\rangle}{\langle s_n, s_n\rangle} = \dfrac2L\int_0^Lf\,s_n\,dx$ .

Why sines and not cosines? The ends are held at $0$ , and every $s_n$ is $0$ at $x = 0$ and at $x = L$ , so every sum of them also has zero ends. A cosine is not $0$ at $x = 0$ . Equivalently: extend $f$ to $[-L, L]$ as an odd function, $f(-x) = -f(x)$ ; odd functions are built from sines alone.

What is different from $\mathbb R^3$ ? There are infinitely many axes, so ” $f$ equals the sum” needs a meaning of convergence for an infinite sum of functions. Proving that it holds is a theorem of analysis (Dirichlet; MATH 521). For the heat equation the risk is small: for $t \gt 0$ the factors $e^{-\alpha(n\pi/L)^2t}$ shrink the high terms so fast that the series can be differentiated term by term (MATH 521 proves this), and each term solves the equation. At $t = 0$ the series may converge slowly, or to the wrong value at the ends (every sine is $0$ there), which is why the example below tests it at an interior point.

Lemma A Apostol 1.12 (orthogonal families); Fourier 1822

Orthogonality of the sine modes

For positive integers $n, m$ : $\langle\sin\frac{n\pi x}L, \sin\frac{m\pi x}L\rangle = \int_0^L\sin\frac{n\pi x}L\sin\frac{m\pi x}L\,dx$ equals $0$ if $n \ne m$ and $\frac L2$ if $n = m$ .

In one line $\displaystyle\int_0^L\sin\frac{n\pi x}L\sin\frac{m\pi x}L\,dx = 0$ for $n \ne m$ and $= \frac L2$ for $n = m$ ; hence if $f = \sum b_m\sin\frac{m\pi x}L$ (finite sum) then $b_n = \frac2L\int_0^Lf\sin\frac{n\pi x}L$ .

Why it is trueIt is what makes the coefficient formula work: take the inner product with one mode and every other term vanishes (2.9).

Proofstandard

The idea: Turn the product of sines into a sum of cosines and integrate; the integrals of cosines over whole periods vanish.

\sin\frac{n\pi x}L\sin\frac{m\pi x}L = \frac12\Big[\cos\frac{(n - m)\pi x}L - \cos\frac{(n + m)\pi x}L\Big]

Product-to-sum identity

\sin A\sin B = \frac12[\cos(A - B) - \cos(A + B)]

So for

n \ne m

both integrals are

0

, and for

n = m

the integral is

\frac12(L - 0) = \frac L2

For an integer

k \ne 0

\int_0^L\cos\frac{k\pi x}L\,dx = \frac{L}{k\pi}\sin(k\pi) = 0

. For

k = 0

the integral is

L

. If

n \ne m

both

n - m

and

n + m

are nonzero; if

n = m

only

n + m

is.

Hence

b_n = \frac2L\int_0^Lf\sin\frac{n\pi x}L\,dx

Multiply

f = \sum_mb_m\sin\frac{m\pi x}L

\sin\frac{n\pi x}L

and integrate: only

m = n

survives, giving

b_n\frac L2

\blacksquare

Worked example new (separation of variables)

A rod of length

1

with diffusivity

\alpha = 1

and ends at

0

starts at the temperature

f(x) = \sin\pi x + \frac12\sin3\pi x

. Find

u(x, t)

and

u\big(\frac12, 0.05\big)

Solution.

Identify the modes. $f$ is already a combination of the modes $n = 1$ and $n = 3$ with $b_1 = 1$ , $b_3 = \frac12$ , all other $b_n = 0$ . By orthogonality the coefficients are unique.
Let each mode decay. $u(x, t) = e^{-\pi^2t}\sin\pi x + \frac12e^{-9\pi^2t}\sin3\pi x$ . Mode $n$ decays at the rate $\alpha n^2\pi^2/L^2 = n^2\pi^2$ .
Check. Each term has $u_t = -n^2\pi^2u_n = u_{xx}$ , and $u(0, t) = u(1, t) = 0$ , $u(x, 0) = f$ . All three requirements.
Evaluate at $x = \frac12$ , $t = 0.05$ . $\sin\frac\pi2 = 1$ and $\sin\frac{3\pi}2 = -1$ , so $u = e^{-0.05\pi^2} - \frac12e^{-0.45\pi^2} = 0.6105 - 0.0059 = 0.6046$ . $e^{-0.4935} = 0.6105$ and $e^{-4.441} = 0.01178$ .

Answer. $u(x, t) = e^{-\pi^2t}\sin\pi x + \frac12e^{-9\pi^2t}\sin3\pi x$ ; $u(\frac12, 0.05) \approx 0.6046$ . By $t = 0.05$ the third mode has fallen to $e^{-9\pi^2\cdot0.05} = 1.2\%$ of its starting size and the first to $61\%$ : only the smoothest part of the profile survives.

Your answerFor the rod above, what is

u(\tfrac12, 0.05)

? (Give a decimal.)

Worked example new (the Fourier series of 2.9 in action)

Find the Fourier sine coefficients of

f(x) = x(1 - x)

[0, 1]

and write the solution of the heat equation with

\alpha = 1

Solution.

Coefficient formula. $b_n = 2\displaystyle\int_0^1x(1 - x)\sin n\pi x\,dx$ . $L = 1$ .
Integrate by parts twice. With $p = x - x^2$ and $k = n\pi$ : $\displaystyle\int_0^1p\sin kx\,dx = \Big[-\frac{p\cos kx}k\Big]_0^1 + \frac1k\int_0^1p'\cos kx\,dx = 0 + \frac1{k^2}\Big[p'\sin kx\Big]_0^1 - \frac1{k^2}\int_0^1p''\sin kx\,dx = \frac2{k^2}\int_0^1\sin kx\,dx = \frac{2(1 - \cos k)}{k^3}$ , which is $\dfrac{2\big(1 - (-1)^n\big)}{(n\pi)^3}$ . The boundary terms vanish because $p$ is $0$ at both ends and $\sin k = \sin0 = 0$ ; and $p'' = -2$ .
Result. $b_n = \dfrac{4\big(1 - (-1)^n\big)}{(n\pi)^3}$ : $b_n = \dfrac{8}{n^3\pi^3}$ for odd $n$ and $0$ for even $n$ . For example $b_1 = \dfrac8{\pi^3} = 0.258$ , $b_3 = 0.00956$ . The profile is symmetric about $\frac12$ , so even modes (antisymmetric) are absent.
The solution. $u(x, t) = \displaystyle\sum_{n\text{ odd}}\frac{8}{n^3\pi^3}\,e^{-n^2\pi^2t}\sin n\pi x$ .
Test at $t = 0$ , $x = \frac12$ . $\sin\frac{n\pi}2 = (-1)^{(n - 1)/2}$ for odd $n$ , so $\sum_k\frac{8(-1)^k}{(2k + 1)^3\pi^3} = \frac{8}{\pi^3}\cdot\frac{\pi^3}{32} = \frac14 = f(\frac12)$ . ✓ Uses the known sum $1 - \frac1{27} + \frac1{125} - \dots = \frac{\pi^3}{32}$ . The partial sums show it happening: $0.2580$ after one term, $0.2485$ after two, $0.2505$ after three, against $f(\frac12) = 0.25$ .

Answer. $b_n = 8/(n\pi)^3$ for odd $n$ . The coefficients fall like $n^{-3}$ because the odd periodic extension of $f$ is continuous with a continuous first derivative but its second derivative jumps (from $-2$ to $+2$ ) at the ends; each further derivative of smoothness gives another factor $1/n$ , which is why the series converges quickly.

Theorem B the energy method; Evans 2.3, Strauss 2.3

Uniqueness for the heat equation

Suppose $u_1, u_2$ are $C^2$ in $x$ and $C^1$ in $t$ on $[0, L]\times[0, T]$ , both solve $u_t = \alpha u_{xx}$ , both have the same values at $t = 0$ and the same values at $x = 0$ and $x = L$ for all $t$ . Then $u_1 = u_2$ .

In one lineA $C^2$ solution of $u_t = \alpha u_{xx}$ on $[0, L]\times[0, T]$ with given initial temperature and given end temperatures is unique.

Why it is trueIt says that separation of variables does not miss anything: once you have found a solution of the problem, it is the solution.

ProofEvans 2.3

The idea: The difference w solves the same equation with zero data. Its total squared temperature E(t) = the integral of w squared never increases, because its derivative is minus a sum of squares; it starts at zero, so it stays zero.

Put

w = u_1 - u_2

and

E(t) = \int_0^Lw(x, t)^2\,dx \ge 0

The equation is linear (superposition), so

w = u_1 - u_2

solves it, with

w(x, 0) = 0

and

w(0, t) = w(L, t) = 0

E'(t) = 2\int_0^Lww_t\,dx = 2\alpha\int_0^Lww_{xx}\,dx

Differentiate under the integral sign (the integrand is

C^1

on a compact rectangle, hence

w_t

is uniformly continuous and the integral may be differentiated), then use the equation.

E'(t) = -2\alpha\int_0^Lw_x^2\,dx \le 0

Integration by parts:

\int_0^Lww_{xx}\,dx = [ww_x]_0^L - \int_0^Lw_x^2\,dx

, and

w = 0

x = 0

and

L

kills the bracket.

Then

\int_0^Lw^2\,dx = 0

for all

t

; as

w^2

is continuous and non-negative,

w \equiv 0

\blacksquare

E(0) = 0

since

w(\cdot, 0) = 0

E \ge 0

and

E

is non-increasing. So

E \equiv 0

Quick checkWhy can the heat equation not be run backwards in time?

Python labHeat equation by finite differences, and when it blows up

import numpy as np
import matplotlib
matplotlib.use("Agg")
import matplotlib.pyplot as plt

N = 50; dx = 1 / N                       # 50 intervals on a rod of length 1, diffusivity alpha = 1
xs = np.linspace(0, 1, N + 1)
f = lambda x: np.sin(np.pi * x) + 0.5 * np.sin(3 * np.pi * x)       # initial temperature
def exact(x, t):                          # separation of variables: each mode decays like exp(-(n pi)^2 t)
    return np.exp(-np.pi**2 * t) * np.sin(np.pi * x) + 0.5 * np.exp(-9 * np.pi**2 * t) * np.sin(3 * np.pi * x)

def solve(r, T):
    """Explicit scheme u_j <- u_j + r (u_{j+1} - 2 u_j + u_{j-1}) with r = dt/dx^2; the ends stay at 0."""
    dt = r * dx**2; steps = int(round(T / dt)); u = f(xs)
    for _ in range(steps):
        u[1:-1] = u[1:-1] + r * (u[2:] - 2 * u[1:-1] + u[:-2])
    return u, steps * dt

print(" r     steps   max|u|        max error vs exact")
for r in (0.4, 0.5, 0.6):
    u, T = solve(r, 0.05)
    print(f"{r:.1f}   {int(round(T / (r * dx**2))):5d}   {np.abs(u).max():.3e}   {np.abs(u - exact(xs, T)).max():.3e}")

plt.figure(figsize=(6, 3.2))
for T in (0.0, 0.005, 0.02, 0.05):
    u, T2 = solve(0.4, T) if T > 0 else (f(xs), 0)
    plt.plot(xs, u, label=f"t = {T}")
plt.plot(xs, exact(xs, 0.05), "k--", lw=1, label="exact, t = 0.05")
plt.legend(); plt.xlabel("x"); plt.ylabel("temperature")

Try this. Run it. For r = 0.4 and 0.5 the numerical solution agrees with the exact two-mode solution to about 2e-4 (the plot shows the profile flattening and the third mode vanishing first). For r = 0.6 the same code gives a number near 3e13: the explicit scheme is unstable when r = alpha dt / dx^2 exceeds 1/2, because the highest-frequency grid mode is then multiplied by a factor of absolute value above 1 at every step. Change N to 100 with r = 0.5 and check that the number of steps quadruples (the stable step is proportional to dx squared). Change f to a single spike (u = 1 at x = 0.5, 0 elsewhere) and watch it become a bell curve of width about sqrt(2 t).

Standing waves from the same method

For the wave equation on $[0, L]$ with fixed ends, $u = X(x)T(t)$ gives $\dfrac{T''}{c^2T} = \dfrac{X''}{X} = -\kappa$ with the same $X = \sin\frac{n\pi x}L$ , and $T'' = -c^2\big(\frac{n\pi}L\big)^2T$ , so $T = A\cos\frac{n\pi ct}L + B\sin\frac{n\pi ct}L$ . The modes $\sin\frac{n\pi x}L\cos\frac{n\pi ct}L$ are the standing waves of the first picture: mode $n$ vibrates at frequency $n\,c/(2L)$ . The only difference from the heat case is the sign in the time equation: oscillation instead of decay. That one sign separates a string, which rings forever (in the model), from a rod, which cools.

Laplace’s equation

Definition A

Laplace equation

Laplace’s equation is $\nabla^2u = 0$ , where $\nabla^2u = \sum_ku_{x_kx_k}$ is the Laplacian, the sum of the pure second partial derivatives (the trace of the Hessian, 5.6). If $u_t = \alpha\nabla^2u$ and $u$ does not depend on $t$ , then $\nabla^2u = 0$ . Apostol 9.1 (equation 9.2).

In one lineThe equation $\nabla^2u = u_{xx} + u_{yy} = 0$ (or $u_{xx} + u_{yy} + u_{zz} = 0$ ): the steady state of the heat equation.

Why it existsIt is what remains when nothing changes in time: the temperature of a plate once it has settled, the electrostatic potential in empty space.

Definition A

Harmonic function

A harmonic function on an open set $\Omega$ is a $C^2$ function with $\nabla^2u = 0$ throughout $\Omega$ . Examples in the plane: constants, $x$ , $y$ , $xy$ , $x^2 - y^2$ , $x^3 - 3xy^2$ , $e^x\cos y$ , $\ln\sqrt{x^2 + y^2}$ (away from the origin). Not harmonic: $x^2 + y^2$ (its Laplacian is $4$ ). In space, $1/\sqrt{x^2 + y^2 + z^2}$ is harmonic away from the origin (the potential of a point charge or a point mass).

In one lineA $C^2$ function $u$ on an open set that satisfies $\nabla^2u = 0$ there.

Why it existsHarmonic functions are the solutions of Laplace's equation; they have remarkable rigidity (the mean value property) that makes them the model of 'no sources and no sinks'.

Lemma A Apostol 9.5 #4

Polar form of the Laplacian

For $u \in C^2$ and $x = r\cos\theta$ , $y = r\sin\theta$ , $r \gt 0$ : $u_{xx} + u_{yy} = U_{rr} + \dfrac1rU_r + \dfrac1{r^2}U_{\theta\theta}$ .

In one lineIf $U(r, \theta) = u(r\cos\theta, r\sin\theta)$ with $u$ of class $C^2$ , then for $r \gt 0$ : $u_{xx} + u_{yy} = U_{rr} + \dfrac1rU_r + \dfrac1{r^2}U_{\theta\theta}$ .

Why it is trueIt is the form of Laplace's equation that suits discs and circles, and it is the key to the mean value property and to separation of variables in polar coordinates.

ProofApostol 9.5 #4

The idea: Compute U_r, U_rr, U_theta and U_theta theta with the chain rule. The cross terms cancel and the first-order terms in U_theta theta combine with U_r.

U_r = u_x\cos\theta + u_y\sin\theta

and

U_{rr} = u_{xx}\cos^2\theta + 2u_{xy}\sin\theta\cos\theta + u_{yy}\sin^2\theta

Chain rule (5.4) with

x_r = \cos\theta

y_r = \sin\theta

U_\theta = -r\sin\theta\,u_x + r\cos\theta\,u_y

Chain rule with

x_\theta = -r\sin\theta

y_\theta = r\cos\theta

U_{\theta\theta} = -r\big(u_x\cos\theta + u_y\sin\theta\big) + r^2\big(u_{xx}\sin^2\theta - 2u_{xy}\sin\theta\cos\theta + u_{yy}\cos^2\theta\big)

Differentiate

U_\theta

again: the product rule on the factors

-r\sin\theta

r\cos\theta

gives

-r\cos\theta\,u_x - r\sin\theta\,u_y

, and the chain rule on

u_x, u_y

gives the second-order terms (with

u_{xy} = u_{yx}

by 5.5).

U_{rr} + \frac1rU_r + \frac1{r^2}U_{\theta\theta} = u_{xx} + u_{yy}

The first bracket is

U_r

, so

U_{\theta\theta} = -rU_r + r^2\big(\ldots\big)

. Divide by

r^2

and add

U_{rr}

and

\frac1rU_r

: the

U_r

terms cancel, the

u_{xy}

terms cancel, and

\cos^2 + \sin^2 = 1

multiplies

u_{xx}

and

u_{yy}

\blacksquare

Worked example Apostol 9.5 #2 and #3

Find all radial harmonic functions

u = g(r)

in the plane away from the origin, and in

\mathbb R^3

Solution.

Plane. For $u = g(r)$ the lemma gives $\nabla^2u = g'' + \dfrac1rg' = \dfrac1r(rg')'$ . $U$ does not depend on $\theta$ , so $U_{\theta\theta} = 0$ .
Solve. $\nabla^2u = 0$ means $(rg')' = 0$ , so $rg' = a$ and $g = a\ln r + b$ . Integrate twice.
Space. In $\mathbb R^n$ the same computation gives $\nabla^2g(r) = g'' + \dfrac{n - 1}rg'$ , which vanishes when $r^{n - 1}g'$ is constant: $g = ar^{2 - n} + b$ for $n \ge 3$ . For $n = 3$ , $g = a/r + b$ . Check $n = 3$ directly: $g = r^{-1}$ has $g'' = 2r^{-3}$ and $\frac2rg' = -2r^{-3}$ .

Answer. $a\ln\sqrt{x^2 + y^2} + b$ in the plane; $a/\sqrt{x^2 + y^2 + z^2} + b$ in space. These are the potentials of a line charge and of a point charge (Apostol’s answer for the plane is written $a\log(x^2 + y^2) + b$ ).

Theorem B Gauss; Evans 2.2.2; Strauss 6.3

Mean value property of harmonic functions

Let $u$ be harmonic on an open set containing the closed disc $\bar B(a, R)$ of the plane. Then

u(a) = \frac1{2\pi}\int_0^{2\pi}u\big(a_1 + R\cos\theta,\ a_2 + R\sin\theta\big)\,d\theta.

In one lineIf $u$ is harmonic on an open set containing the closed disc of radius $R$ about $a$ , then $u(a) = \dfrac1{2\pi}\displaystyle\int_0^{2\pi}u\big(a_1 + R\cos\theta,\ a_2 + R\sin\theta\big)\,d\theta$ : the value at the centre is the average over the circle.

Why it is trueIt says a harmonic function has no preferred point: its value at any point is the average of its neighbours. That one fact gives the maximum principle, smoothness, and the uniqueness of the temperature or potential with given boundary values.

ProofEvans 2.2.2, in polar form

The idea: Let M(r) be the average of u over the circle of radius r. Show its derivative is zero by using Laplace's equation in polar form and the fact that the circle has no end points, so a derivative with respect to the angle integrates to zero. Then M(R) = M(0) = u(a).

Define

M(r) = \dfrac1{2\pi}\displaystyle\int_0^{2\pi}U(r, \theta)\,d\theta

. Then

M(0) = u(a)

and we want

M(R) = M(0)

Write

U(r, \theta) = u(a_1 + r\cos\theta, a_2 + r\sin\theta)

for

0 \le r \le R

; it is

C^2

and

2\pi

-periodic in

\theta

. At

r = 0

it equals

u(a)

for every

\theta

rM'(r) = \dfrac1{2\pi}\displaystyle\int_0^{2\pi}rU_r\,d\theta =: N(r)

Differentiate under the integral sign:

U_r

is continuous on the compact set

[0, R]\times[0, 2\pi]

, hence uniformly continuous (4.3, 1.3), which lets the difference quotients converge uniformly.

N(r) \to 0

r \to 0^+

\lvert N(r)\rvert \le r\max\lvert\nabla u\rvert

, since

\lvert U_r\rvert = \lvert\nabla u\cdot(\cos\theta, \sin\theta)\rvert \le \lVert\nabla u\rVert

(Cauchy–Schwarz).

N'(r) = \dfrac1{2\pi}\displaystyle\int_0^{2\pi}r\Big(U_{rr} + \frac1rU_r\Big)d\theta

Differentiate again under the integral:

(rU_r)_r = rU_{rr} + U_r = r\big(U_{rr} + \frac1rU_r\big)

N'(r) = -\dfrac1{2\pi r}\displaystyle\int_0^{2\pi}U_{\theta\theta}\,d\theta

Laplace's equation in polar form (the lemma):

U_{rr} + \frac1rU_r = -\frac1{r^2}U_{\theta\theta}

N'(r) = 0

for

0 \lt r \le R

Fundamental theorem of calculus:

\int_0^{2\pi}U_{\theta\theta}\,d\theta = U_\theta(r, 2\pi) - U_\theta(r, 0) = 0

because

U_\theta

2\pi

-periodic.This is the step where the circle being closed matters: a derivative integrated around a closed loop gives zero.

N \equiv 0

, hence

M

is constant.

N

is constant on

(0, R]

and tends to

0

0

, so

N \equiv 0

. Then

M'(r) = N(r)/r = 0

(0, R]

, so

M

is constant on

[0, R]

(mean value theorem, continuity at

0

), and

M(R) = M(0) = u(a)

\blacksquare

Proof workshopRebuild the proof of the mean value propertyEvans 2.2.2

Tap the lines in the order a proof would write them. Each line can only use what is already above it.

Corollary B Evans 2.2.3; Strauss 6.1

Maximum principle for harmonic functions

Let $\Omega \subseteq \mathbb R^2$ be open and connected and $u$ harmonic on $\Omega$ . If $u(p) \ge u(x)$ for all $x \in \Omega$ for some $p \in \Omega$ , then $u$ is constant on $\Omega$ . The same for a minimum.

In one lineIf $u$ is harmonic on a connected open set $\Omega$ and has a maximum (or a minimum) at a point of $\Omega$ , then $u$ is constant. So a non-constant harmonic function on a bounded region, continuous up to the boundary, attains its largest and smallest values on the boundary only.

Why it is trueIt says harmonic functions have no interior peaks or valleys, which gives the uniqueness of the steady-state temperature or potential with given boundary values.

ProofEvans 2.2.3

The idea: The set where u equals its maximum value M is closed by continuity and open by the mean value property, so it is everything because the region is connected.

A

is a non-empty closed subset of

\Omega

Let

M = u(p)

and

A = \{x \in \Omega : u(x) = M\}

. It is non-empty (it contains

p

) and closed in

\Omega

, as the preimage of the closed set

\{M\}

under the continuous

u

u = M

on every circle of radius

r \lt \rho

about

x

, so

u = M

B(x, \rho)

and

B(x, \rho) \subseteq A

A

is open.

Let

x \in A

and a ball

B(x, \rho)

inside

\Omega

. For

0 \lt r \lt \rho

the mean value property gives

M = u(x) = \frac1{2\pi}\int_0^{2\pi}u(x + r(\cos\theta, \sin\theta))\,d\theta

. The integrand is at most

M

, and an average of numbers at most

M

equals

M

only if all of them are

M

(continuity of the integrand).

A = \Omega

, that is

u \equiv M

A set that is open, closed and non-empty in a connected set is the whole set (4.4). For a minimum apply the result to

-u

, which is also harmonic.

\blacksquare

Quick checkA bounded function

u

is harmonic on the open unit disc, with boundary values tending to

1

on the upper half of the boundary circle and to

0

on the lower half (except at the two points where the data jumps). Which statement does the maximum principle give?

Your answerThe function

u = x^3 - 3xy^2

is harmonic. What is its average value over the circle of radius

2

centred at

(1, 1)

Your answerRod of length

1

, diffusivity

1

, ends at

0

: after how long does the first mode

e^{-\pi^2t}\sin\pi x

fall to half its initial size? (Give a decimal.)

Where it is used

Used inside maths· Inside maths: complex analysis, saddle points, and Fourier series

Harmonic functions and complex numbers. If $f = u + iv$ is a differentiable function of $z = x + iy$ (the complex analysis of MATH 521), then $u$ and $v$ satisfy the Cauchy–Riemann equations $u_x = v_y$ and $u_y = -v_x$ . Differentiating the first in $x$ and the second in $y$ and using the equality of mixed partials (5.5): $u_{xx} + u_{yy} = v_{yx} - v_{xy} = 0$ . So the real and imaginary parts of $z^2$ , namely $x^2 - y^2$ and $2xy$ , and of $z^3$ , namely $x^3 - 3xy^2$ and $3x^2y - y^3$ , are harmonic; every conformal picture in the plane is a pair of harmonic functions. No interior extremes. $\nabla^2u$ is the trace of the Hessian (5.6), so at a critical point of a harmonic function the Hessian has trace $0$ and cannot be positive or negative definite: the second-derivative test can only report a saddle or a degenerate point, which is the maximum principle seen through the Hessian. Fourier series. The sine modes are an orthogonal family in the inner product of 2.9, and the coefficient formula is the orthogonal projection of $f$ onto each: the same formula that gives the best approximation in 2.9. The convergence of the series is the subject of Fourier analysis, and the reason Dirichlet’s 1829 theorem was needed.

Used in the real world· Guitar strings, copper rods, blurred images and potentials

A guitar string. With $L = 0.65$ m, density $0.5$ g/m and tension $91.8$ N, the wave speed is $c = \sqrt{T/\rho} = 428.5$ m/s, the fundamental is $f_1 = c/(2L) = 329.6$ Hz (the note E4) and the harmonics are $659.3$ and $988.9$ Hz. Plucking the string at a third of its length leaves out mode $3$ (and every multiple of $3$ ), because $\sin(n\pi x/L)$ is $0$ at $x = L/3$ when $n = 3$ : the coefficients of a plucked string are computed with the sine formula above. Doubling the tension multiplies the pitch by $\sqrt2$ , an interval of six semitones (a tritone).

A copper rod. From the heat equation, a rod of length $0.2$ m with ends held at the ambient temperature and an initial half-sine profile of excess temperature loses half of that excess every $24$ s; the second mode decays four times faster. A hot-spot of heat at the centre spreads as a bell curve of width $\sqrt{2\alpha t}$ : at $t = 60$ s the standard deviation is $\sqrt{2\cdot1.16\times10^{-4}\cdot60} = 0.118$ m.

Image blurring and diffusion models. The convolution of an image with a Gaussian of standard deviation $\sigma$ pixels is the solution of the heat equation (with $\alpha = 1$ ) at time $t = \sigma^2/2$ : a blur with $\sigma = 3$ pixels is $4.5$ time units of diffusion, and edge detection by ‘blur, then differentiate’ (the Laplacian of Gaussian) uses that the Gaussian solves the equation. In the simplest diffusion model for images, the noise added during the forward process is Gaussian with growing variance, so the density of the noisy data evolves by the heat equation; generating an image runs this process backwards, and running the heat equation backwards is ill-posed (see the quiz), so in practice the reverse process is steered by the gradient of the logarithm of the density, which a neural network learns.

Potentials. The electrostatic potential of a point charge $q$ in vacuum is $V = \dfrac{q}{4\pi\varepsilon_0r}$ , harmonic away from the origin, and by the mean value property the potential at the centre of any empty sphere equals its average over the sphere: the potential has no local minimum in empty space, which is the idea behind Earnshaw’s theorem that charges cannot be held in stable static equilibrium by electrostatic forces alone.

Traps

Watch out· The solution of a first-order PDE contains a function, not a constant

$u_t + 2u_x = 0$ has the general solution $g(x - 2t)$ for any differentiable $g$ . Writing ' $u = C$ ' or ’ $u = e^{x - 2t}$ times a constant’ misses all other solutions. The function $g$ is selected by the initial data, and the sign in $x - 2t$ matters: $g(x + 2t)$ moves the other way.

Watch out· Separation of variables finds special solutions only

A product $X(x)T(t)$ is a special solution. A general solution is a sum (or series) of them, with coefficients from the initial data. And dividing by $XT$ assumes it is nonzero: the constant $-\kappa$ is a separation constant and must be the same on both sides. Check the boundary conditions before you pick the sign of $\kappa$ .

Watch out· Count the conditions

Wave equation: two conditions in time ( $u$ and $u_t$ at $t = 0$ ) and conditions at the boundary. Heat equation: one condition in time. Laplace: none in time, only boundary values. Giving too few leaves a function free; giving too many may leave no solution. And the heat equation cannot be solved backwards.

Watch out· Smoothness hypotheses are real

d’Alembert’s formula needs $F \in C^2$ to be a classical solution. For a plucked string with a corner, the formula still describes the motion but $u_{xx}$ does not exist at the corner. The mean value property needs the closed disc inside the domain of harmonicity.

Watch out· Numerical stability is not optional

The explicit finite-difference scheme for the heat equation needs $\alpha\,dt/dx^2 \le \frac12$ . Halving the grid spacing forces the time step to a quarter, so a four-times finer grid in one dimension costs 64 times the work. See the lab above, where $r = 0.6$ explodes.

Problems

check

Is $e^{-9t}\sin3x$ a solution of $u_t = u_{xx}$ ? Of $u_{tt} = 9u_{xx}$ ?

Solution

$u_t = -9u$ and $u_{xx} = -9u$ , so $u_t = u_{xx}$ : yes for the heat equation. For the wave equation, $u_{tt} = 81u$ while $9u_{xx} = -81u$ : no.

checkApostol 9.2

Find the general solution of $5f_x - 2f_y = 0$ .

Solution

With $a = 5$ , $b = -2$ : $f(x, y) = g(bx - ay) = g(-2x - 5y)$ , which is the same family as $h(2x + 5y)$ for an arbitrary differentiable $h$ . Check: $f_x = 2h'$ , $f_y = 5h'$ for $f = h(2x + 5y)$ , so $5f_x - 2f_y = 10h' - 10h' = 0$ .

checknew (compare Apostol 9.5 #2)

Which of $x^2 + y^2$ , $x^2 - y^2$ , $xy$ , $\ln(x^2 + y^2)$ , $e^x\sin y$ are harmonic in the plane (away from the origin for the logarithm)?

Solution

$x^2 + y^2$ : $\nabla^2 = 4$ , no. $x^2 - y^2$ : $2 - 2 = 0$ , yes. $xy$ : $0 + 0 = 0$ , yes. $\ln(x^2 + y^2) = 2\ln r$ : yes (the radial formula). $e^x\sin y$ : $e^x\sin y - e^x\sin y = 0$ , yes.

check

How many initial conditions does each of the wave, heat and Laplace equations take, and why?

Solution

Wave: two ( $u$ and $u_t$ at $t = 0$ ), since it is second order in time. Heat: one ( $u$ at $t = 0$ ), first order in time. Laplace: none in time (no time variable); the solution is determined by its values on the boundary of the region.

usenew (first-order PDE)

Solve $3u_x + 5u_y = 0$ with $u(0, y) = y^2$ .

Solution

General solution $u = g(5x - 3y)$ (here $a = 3$ , $b = 5$ ). At $x = 0$ : $g(-3y) = y^2$ , so $g(s) = \frac{s^2}9$ . Hence $u(x, y) = \dfrac{(5x - 3y)^2}9$ . Check: $u_x = \frac{10(5x - 3y)}{9}$ , $u_y = -\frac{6(5x - 3y)}{9}$ , so $3u_x + 5u_y = \frac{30 - 30}{9}(5x - 3y) = 0$ .

usenew (a velocity impulse; d'Alembert with G nonzero)

For $c = 1$ , $F = 0$ and $G(x) = 1$ for $\lvert x\rvert \lt 1$ , $G(x) = 0$ otherwise (a string struck along an interval), find $u(0, t)$ from d’Alembert’s formula.

Solution

$u(0, t) = \frac1{2}\int_{-t}^{t}G(s)\,ds$ (with $c = 1$ , $x = 0$ ). For $0 \le t \le 1$ the interval $[-t, t]$ lies inside $(-1, 1)$ , so the integral is $2t$ and $u(0, t) = t$ . For $t \ge 1$ the integral is $2$ (the whole interval of $G$ ) and $u(0, t) = 1$ . So the midpoint rises linearly and then stays at height $1$ : the wave leaves behind a raised plateau. (The formula applies even though $G$ is not continuous, as an integral.)

usecopper rod

A copper rod of length $0.2$ m ( $\alpha = 1.16\times10^{-4}$ m $^2$ /s) has ends held at the ambient temperature, and the excess temperature starts as $\sin(\pi x/L)$ . After how many seconds does the excess at the centre fall to $10\%$ of its start?

Solution

Mode $1$ decays as $e^{-\alpha\pi^2t/L^2}$ . Setting $e^{-\alpha\pi^2t/L^2} = 0.1$ : $t = \dfrac{L^2\ln10}{\alpha\pi^2} = \dfrac{0.04\cdot2.3026}{1.16\times10^{-4}\cdot9.8696} = \dfrac{0.092103}{0.0011449} = 80.4$ s.

useApostol 9.5 #3

Find all harmonic functions of the form $g(r)$ , $r = \lVert x\rVert$ , in $\mathbb R^4$ away from the origin.

Solution

In $\mathbb R^n$ : $\nabla^2g(r) = g'' + \frac{n - 1}rg'$ . For $n = 4$ this is $g'' + \frac3rg' = r^{-3}(r^3g')'$ , zero when $r^3g' = a$ , so $g = -\frac a{2r^2} + b$ , that is $g = \alpha r^{-2} + b$ . (This is $r^{2 - n}$ for $n = 4$ .)

provenew (energy conservation of the string)

Show that a $C^2$ solution of the wave equation on $[0, L]$ with $u(0, t) = u(L, t) = 0$ conserves the energy $E(t) = \frac12\int_0^L\big(u_t^2 + c^2u_x^2\big)\,dx$ .

Solution

Plan. Differentiate $E$ , use the equation, and integrate by parts.

$E'(t) = \int_0^L\big(u_tu_{tt} + c^2u_xu_{xt}\big)dx$ (differentiating under the integral sign; the integrand is $C^1$ on a compact set). With $u_{tt} = c^2u_{xx}$ : $E' = c^2\int_0^L\big(u_tu_{xx} + u_xu_{xt}\big)dx = c^2\int_0^L\frac{\partial}{\partial x}\big(u_tu_x\big)dx$ (product rule, using $u_{xt} = u_{tx}$ from 5.5) $= c^2\big[u_tu_x\big]_{x = 0}^{x = L}$ . Since $u(0, t) = 0$ for all $t$ , $u_t(0, t) = 0$ , and likewise at $L$ , so the bracket is $0$ . Hence $E' = 0$ and $E$ is constant. $\blacksquare$ ( $\frac12\int u_t^2$ is kinetic and $\frac{c^2}{2}\int u_x^2$ is potential energy; the string never loses energy in this model.)

provenew (orthogonality used for coefficients)

Prove that $\int_0^L\sin\frac{n\pi x}L\sin\frac{m\pi x}L\,dx = 0$ for integers $n \ne m$ , both positive, and compute the value for $n = m$ .

Solution

Plan. Product-to-sum, then integrate.

$\sin A\sin B = \frac12[\cos(A - B) - \cos(A + B)]$ , so the integral is $\frac12\int_0^L\big[\cos\frac{(n - m)\pi x}L - \cos\frac{(n + m)\pi x}L\big]dx$ . For a nonzero integer $k$ , $\int_0^L\cos\frac{k\pi x}L\,dx = \frac L{k\pi}\sin k\pi = 0$ . When $n \ne m$ both $n - m$ and $n + m$ are nonzero integers, so the integral is $0$ . When $n = m$ the first cosine is $\cos0 = 1$ with integral $L$ and the second integrates to $0$ : the value is $\frac L2$ . $\blacksquare$

provenew (uniqueness of the Dirichlet problem)

Let $\Omega$ be a bounded connected open set of the plane, and $u_1, u_2$ functions that are continuous on $\bar\Omega$ and harmonic in $\Omega$ and agree on the boundary $\partial\Omega$ . Prove $u_1 = u_2$ .

Solution

Plan. The difference is harmonic and zero on the boundary; apply the maximum principle on a compact set.

Put $w = u_1 - u_2$ , harmonic in $\Omega$ (linearity), continuous on the compact set $\bar\Omega$ (closed and bounded), and $w = 0$ on $\partial\Omega$ . By the extreme value theorem (4.3) $w$ has a maximum at some point of $\bar\Omega$ . If it is in $\Omega$ , then by the maximum principle $w$ is constant on $\Omega$ , hence (by continuity) equal to its boundary value $0$ on $\bar\Omega$ . Otherwise it is on $\partial\Omega$ , where $w = 0$ , so $w \le 0$ everywhere. The same argument for $-w$ gives $w \ge 0$ . So $w \equiv 0$ . $\blacksquare$

stretchnew (Cauchy–Riemann and harmonic functions)

Let $u$ and $v$ be $C^2$ with $u_x = v_y$ and $u_y = -v_x$ . Prove that $u$ and $v$ are harmonic, and that $\nabla u\perp\nabla v$ . Find $v$ when $u = x^2 - y^2$ , and verify the perpendicularity.

Solution

Differentiate: $u_{xx} = v_{yx}$ and $u_{yy} = -v_{xy}$ , so $u_{xx} + u_{yy} = v_{yx} - v_{xy} = 0$ by equality of mixed partials (5.5). Likewise $v_{xx} = -u_{yx}$ and $v_{yy} = u_{xy}$ , so $v_{xx} + v_{yy} = 0$ . And $\nabla u\cdot\nabla v = u_xv_x + u_yv_y = u_x(-u_y) + u_y(u_x) = 0$ . For $u = x^2 - y^2$ : $u_x = 2x = v_y$ gives $v = 2xy + h(x)$ , and $u_y = -2y = -v_x = -2y - h'(x)$ gives $h' = 0$ : $v = 2xy + C$ . Then $\nabla u = (2x, -2y)$ , $\nabla v = (2y, 2x)$ , and $\nabla u\cdot\nabla v = 4xy - 4xy = 0$ . ✓ (The level curves of $u$ and $v$ are two orthogonal families of hyperbolas.)

stretchnew (ill-posedness of the backward heat equation)

Solve the backward heat equation $u_t = -u_{xx}$ on $[0, 1]$ with zero ends and $u(x, 0) = \frac1n\sin n\pi x$ . Compare the size of the initial data with the size of the solution at $t = 0.1$ for $n = 10$ .

Solution

Separation of variables for $u_t = -u_{xx}$ : $T' = +(n\pi)^2T$ , so $u(x, t) = \frac1ne^{n^2\pi^2t}\sin n\pi x$ . For $n = 10$ : the initial data has maximum $\frac1{10} = 0.1$ , and at $t = 0.1$ the solution has maximum $\frac1{10}e^{100\pi^2\cdot0.1} = \frac1{10}e^{98.7} \approx 7\times10^{41}$ . A data error of size $0.1$ becomes an error of size $10^{41}$ in a tenth of a time unit, and larger $n$ is worse without limit: the solution does not depend continuously on the data, so the problem is ill-posed.

Summary

A PDE relates the partial derivatives of an unknown function of several variables. For linear homogeneous equations the solutions form a vector space, which is usually infinite-dimensional because arbitrary functions appear.
First order, constant coefficients: $af_x + bf_y = 0$ has the general solution $f = g(bx - ay)$ : constant along the characteristic lines in the direction $(a, b)$ , with an arbitrary profile $g$ . The method of characteristics follows the unknown along the curves $x' = a$ , $y' = b$ , where the PDE becomes an ODE ( $u_t + au_x + cu = 0$ gives $e^{-ct}u_0(x - at)$ ).
Wave equation $u_{tt} = c^2u_{xx}$ (Newton’s law for a string, $c = \sqrt{T/\rho}$ ): in the variables $\xi = x + ct$ , $\eta = x - ct$ it becomes $v_{\xi\eta} = 0$ . d’Alembert: $u = \frac12[F(x + ct) + F(x - ct)] + \frac1{2c}\int_{x - ct}^{x + ct}G$ . A pulse splits into two half-height pulses; a fixed end reflects with a sign change.
Heat equation $u_t = \alpha u_{xx}$ (energy conservation plus Fourier’s law, $\alpha = k/(\rho c_p)$ ). Separation of variables $u = X(x)T(t)$ gives the modes $e^{-\alpha(n\pi/L)^2t}\sin\frac{n\pi x}L$ ; orthogonality of the sines gives the coefficients $b_n = \frac2L\int f\sin\frac{n\pi x}L$ ; high modes die like $n^2$ , so heat smooths and cannot be run backwards. Uniqueness by the energy method.
Laplace’s equation $\nabla^2u = 0$ (steady state); in polar form $U_{rr} + \frac1rU_r + \frac1{r^2}U_{\theta\theta} = 0$ . Radial solutions: $\ln r$ in the plane, $1/r$ in space.
Mean value property: a harmonic function’s value at a point is its average over any circle around it (proof: the circle average has zero derivative). Maximum principle: no interior extremes unless constant; hence the Dirichlet problem has at most one solution.
Numerically, the explicit scheme for the heat equation needs $\alpha\,dt/dx^2 \le \frac12$ .

Watch and read

The same ideas explained by someone else. Use these when an explanation here does not land.

The picture: standing waves and travelling pulses

The idea, in words

Definitions

Partial differential equation

Superposition for linear homogeneous equations

First-order equations: constant along characteristic lines

First-order equation with constant coefficients

Characteristic curve

The wave equation

Wave equation

d'Alembert's solution of the wave equation

The heat equation

Heat equation

Separation of variables

Separation of variables

Fourier sine series

Orthogonality of the sine modes

Uniqueness for the heat equation

Standing waves from the same method

Laplace’s equation

Laplace equation

Harmonic function

Polar form of the Laplacian

Mean value property of harmonic functions

Maximum principle for harmonic functions

Where it is used

Traps

Problems

Summary

Watch and read

Where this sits

Comes from

Leads to

Same idea elsewhere