Solving wave equation — D'Alembert's solution: Question bank

Reminder of the cast, so no symbol is unearned:

$u (x, t)$ — displacement of the string at position $x$ , time $t$ .
$c > 0$ — the fixed wave speed.
$f (x) = u (x, 0)$ — the initial shape; $g (x) = u_{t} (x, 0)$ — the initial velocity.
$F, G$ — the two travelling profiles: the general solution is $u = F (x - c t) + G (x + c t)$ , a right-mover $F$ plus a left-mover $G$ .
$ξ = x - c t$ , $η = x + c t$ — the characteristic coordinates (lines along which info travels).
The formula under test: $u (x, t) = \frac{1}{2} [f (x - c t) + f (x + c t)] + \frac{1}{2 c} \int_{x - c t}^{x + c t} g (s) d s$ .

Picture 1 — a bump splits into two half-height copies

Figure — Solving wave equation — D'Alembert's solution

Picture 2 — characteristics and the domain of dependence

Picture 3 — solving E1 and E2 to see where $\frac{1}{2}$ and $\frac{1}{2 c}$ come from

True or false — justify

Each answer must give the reason, never a bare verdict.

The general solution

F (x - c t) + G (x + c t)

needs

F

and

G

to be the same function.

False — they generally differ by the velocity integral;

F = G

only in the special case

g = 0

, where both equal

\frac{1}{2} f

(see Picture 1).

f (x - c t)

describes a wave moving in the positive

x

-direction.

True — to keep

x - c t

constant as

t

grows,

x

must grow, so the profile slides toward

+ x

at speed

c

(orange copy in Picture 1).

If the initial velocity

g \equiv 0

, the string never moves.

False — a nonzero initial shape

f

still splits into two half-height copies that travel apart; "zero velocity" only means the instantaneous speed at

t = 0

is zero.

D'Alembert's formula requires

f

to be smooth (infinitely differentiable).

False — the formula only shifts and averages

f

, so even a corner (kink) travels along the characteristics; it is a valid "weak" solution though

u_{xx}

may not exist at the kink.

The wave equation smooths out sharp corners over time, like the heat equation does.

False — the Heat Equation smooths instantly, but the wave equation transports corners unchanged along

x \pm c t

; a kink stays a kink forever.

Doubling the wave speed

c

halves the coefficient in front of the velocity integral.

True — the coefficient is

\frac{1}{2 c}

, so

c \to 2 c

turns

\frac{1}{2 c}

into

\frac{1}{4 c}

, i.e. half — though the interval

[x - c t, x + c t]

also widens.

Information from the initial data can reach every point

x

instantly.

False — finite speed

c

means

(x, t)

only "sees" the initial line inside

[x - c t, x + c t]

; points outside cannot influence it yet (the triangle base in Picture 2).

The value

u (x_{0}, t_{0})

depends on

g

at just the two endpoints

x_{0} \pm c t_{0}

False —

g

is integrated over the whole interval

[x_{0} - c t_{0}, x_{0} + c t_{0}]

; it is

f

that enters only at the two endpoints.

Adding a constant to

g

leaves the solution unchanged.

False —

\int_{x - c t}^{x + c t} (g + a) d s

adds

\frac{1}{2 c} \cdot a \cdot 2 c t = a t

, a term growing linearly in time (a steady drift).

The two characteristic families

ξ =

const and

η =

const have opposite slopes in the

x

–

t

plane.

True —

x - c t =

const has slope

\frac{d t}{d x} = \frac{1}{c}

while

x + c t =

const has slope

- \frac{1}{c}

; they form the "light-cone" cross through each point (Picture 2).

Why questions

Why do we get two arbitrary functions rather than two arbitrary constants?

A second-order PDE in two independent variables integrates "twice," but each integration in

ξ

η

leaves a free function of the other variable — not a mere constant.

Why do the characteristic coordinates

ξ, η

make the messy

u_{xx}, u_{tt}

terms cancel?

In the new variables

u_{tt} - c^{2} u_{xx} = - 4 c^{2} u_{ξ η}

; the like terms

u_{ξ ξ}

and

u_{η η}

appear with equal magnitude and opposite sign, so only the cross term survives.

Why is D'Alembert's the natural method for the wave equation but not the Heat Equation?

The wave equation is hyperbolic with two real characteristic families to travel along; the heat equation is parabolic with no such real characteristics, so it has no finite propagation speed and no travelling-wave factorisation.

Why does the domain of dependence shrink to a point as

t \to 0

The interval

[x - c t, x + c t]

has width

2 c t

, which vanishes as

t \to 0

(the triangle collapses in Picture 2), recovering

u (x, 0) = f (x)

— the point sees only itself at the initial instant.

Why does a constant initial velocity

g \equiv 1

make the string drift upward linearly?

\frac{1}{2 c} \int_{x - c t}^{x + c t} 1 d s = \frac{1}{2 c} (2 c t) = t

, a term independent of

x

that grows steadily in time.

Why must we integrate the velocity condition

(E2)

from

0

x

(introducing a constant

K

) rather than just antidifferentiate freely?

We need a definite relation between

F

and

G

(equation

(E3)

); the constant

K

then provably cancels when reassembled into

u = F (x - c t) + G (x + c t)

, so the answer is well-defined.

Why is D'Alembert's construction linked to the Transport Equation?

Each first-order factor

(\partial_{t} \pm c \partial_{x}) w = 0

is a transport equation whose solution is constant along

x \mp c t

; chaining the two builds the full wave solution.

Edge cases

What is

u (x, 0)

from the formula, and does it match the data?

The interval collapses so the integral is

0

and

\frac{1}{2} [f (x) + f (x)] = f (x)

— it correctly returns the initial shape.

What happens to the formula if

c \to 0^{+}

The characteristics flatten (

ξ, η \to x

) and the

\frac{1}{2 c}

coefficient blows up; the wave equation degenerates (

u_{tt} = 0

), so D'Alembert's finite-speed picture no longer applies.

f

and

g

are both zero everywhere, what is

u

Identically

u \equiv 0

— no shape, no push means no motion; both terms of the formula vanish for all

t

g

is nonzero only on a tiny interval

[a, b]

, when does a far point

x_{0}

first move?

When the interval

[x_{0} - c t, x_{0} + c t]

first overlaps

[a, b]

, i.e. at

t = dist (x_{0}, [a, b]) / c

— before that the far point is still at rest (finite speed).

f

has a jump discontinuity, what does the solution do?

The jump splits into two half-jumps that travel along

x \pm c t

unchanged; the wave equation neither heals nor spreads the discontinuity, unlike diffusion.

What is the solution if

f

is a pure right-mover, e.g.

f = 0

but you want only

F

You cannot force a pure right-mover from shape data alone; a nonzero

f

always contributes

\frac{1}{2} f

to both directions unless the velocity

g = - c f^{'}

is tuned via

(E2)

to cancel the left-mover.

Does the formula still work for

t < 0

(running the wave backward)?

Yes — the wave equation is time-reversible, so replacing

t

- t

still solves it; the two travellers simply swap directions and reconverge.

Recall One-line self-test

If someone hands you $f$ , $g$ , $c$ and asks for $u (x_{0}, t_{0})$ , name the exact pieces of the initial line you must inspect. Answer ::: $f$ at the two endpoints $x_{0} \pm c t_{0}$ , and $g$ over the whole interval between them — nothing else (the base of the triangle in Picture 2).

Connections

Parent: D'Alembert's solution — the derivation these traps probe; Picture 3 is its data-fitting step.
Domain of Dependence and Influence — literally Picture 2: the shaded triangle and its base.
Transport Equation — the single characteristic lines of Picture 2, taken one at a time ( $\partial_{t} \pm c \partial_{x}$ ).
Method of Characteristics — where the $ξ, η$ lines drawn in Picture 2 come from.
Classification of Second-Order PDEs — why "hyperbolic" gives the two real characteristic slopes $\pm \frac{1}{c}$ .
Heat Equation — the contrast case: no straight characteristics, so no triangle, no finite speed.

Question bank — Solving wave equation — D'Alembert's solution