Solving wave equation — D'Alembert's solution: Worked examples

If any symbol here feels unfamiliar, the parent note builds it: parent topic. The characteristic lines $x - c t$ , $x + c t$ come from Method of Characteristics and the two first-order factors are Transport Equations.

The scenario matrix

Every wave problem on the infinite line is built from three switches. Below, ✓ means "present / non-trivial", $0$ means "switched off".

Cell	Initial shape $f$	Initial push $g$	Special feature	Example
A	✓	$0$	smooth bump — shape splits	Ex 1
B	$0$	✓	pure velocity kick — push spreads	Ex 2
C	✓	✓	both together (superpose)	Ex 3
D	✓	$0$	corner / non-smooth shape (triangle)	Ex 4
E	$0$	✓ (bounded lump)	finite-width kick → domain of influence	Ex 5
F	✓	✓	sign / direction check ( $c$ vs $- c$ , left vs right)	Ex 6
G	—	—	degenerate limits: $c \to 0$ , $t = 0$ , $t \to \infty$	Ex 7
H	✓	✓	word problem (guitar string flick)	Ex 8

The switches:

$f \neq = 0$ ? → first bracket lights up.
$g \neq = 0$ ? → integral term lights up.
Is $f$ smooth or cornered? → corners travel, they don't smooth out (no diffusion here — contrast Heat Equation).
Where is $g$ nonzero? → decides the Domain of Dependence and Influence.

Case A — shape only, no push

Worked example Ex 1 · Cell A ·

$f (x) = e^{- x^{2}}$ , $g = 0$ , $c = 1$ Forecast: with no starting push, guess what one Gaussian bump does after a moment. Split? Stay? Grow?

Step 1 — switch off the integral. $g = 0$ , so $\int_{x - c t}^{x + c t} g d s = 0$ . Only the shape term survives. Why this step? The formula has two independent contributions; killing $g$ removes one entirely, so we never even compute the integral.

Step 2 — plug $f$ into the shape term with $c = 1$ . $u (x, t) = \frac{1}{2} e^{- (x - t)^{2}} + \frac{1}{2} e^{- (x + t)^{2}} .$ Why this step? $f (x - c t)$ is the same Gaussian with $x$ replaced by $x - t$ : a copy that has slid right by $t$ . $f (x + t)$ slides left. Each is halved because the formula puts $\frac{1}{2}$ in front.

Step 3 — read the picture (figure below). Two half-height bumps march apart at speed $1$ .

Verify: at $t = 0$ , both exponents become $- x^{2}$ , so $u = \frac{1}{2} e^{- x^{2}} + \frac{1}{2} e^{- x^{2}} = e^{- x^{2}} = f (x)$ ✓. Height at the centre $x = 0$ : $u (0, 1) = \frac{1}{2} e^{- 1} + \frac{1}{2} e^{- 1} = e^{- 1} \approx 0.3679$ .

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

Case B — push only, no shape

Worked example Ex 2 · Cell B ·

$f = 0$ , $g (x) = cos x$ , $c = 2$ Forecast: the string starts flat but every point is moving. Will it oscillate in place, travel, or both?

Step 1 — switch off the shape term. $f = 0$ , so $\frac{1}{2} [f (x - c t) + f (x + c t)] = 0$ . Only the integral survives. Why this step? Same logic as before — a flat start ( $f = 0$ ) means the first bracket contributes nothing.

Step 2 — write the integral with $c = 2$ . $u (x, t) = \frac{1}{2 \cdot 2} \int_{x - 2 t}^{x + 2 t} cos s d s = \frac{1}{4} [sin s]_{x - 2 t}^{x + 2 t} = \frac{1}{4} [sin (x + 2 t) - sin (x - 2 t)] .$ Why this step? The antiderivative of $cos$ is $sin$ ; the domain of dependence is $[x - 2 t, x + 2 t]$ because $c t = 2 t$ .

Step 3 — simplify with the sum-to-product identity. $sin A - sin B = 2 cos \frac{A + B}{2} sin \frac{A - B}{2}, A = x + 2 t, B = x - 2 t .$ Then $\frac{A + B}{2} = x$ , $\frac{A - B}{2} = 2 t$ , so $u (x, t) = \frac{1}{4} \cdot 2 cos x sin 2 t = \frac{1}{2} cos x sin 2 t .$ Why this step? This turns a travelling form into a standing-wave form: a fixed spatial shape $cos x$ breathing in time via $sin 2 t$ .

Verify: at $t = 0$ , $sin 0 = 0 \Rightarrow u = 0 = f$ ✓. Velocity: $u_{t} = \frac{1}{2} cos x \cdot 2 cos 2 t = cos x cos 2 t$ ; at $t = 0$ this is $cos x = g$ ✓. Sample: $u (0, π /4) = \frac{1}{2} cos 0 sin (π /2) = \frac{1}{2}$ .

Case C — both shape and push

Worked example Ex 3 · Cell C ·

$f (x) = sin x$ , $g (x) = 1$ , $c = 1$ Forecast: an oscillating start with a constant upward push everywhere. Does it just wobble, or also drift up forever?

Step 1 — shape term. $\frac{1}{2} [sin (x - t) + sin (x + t)] = sin x cos t$ via sum-to-product ( $\frac{A + B}{2} = x$ , $\frac{A - B}{2} = t$ , factor $\frac{1}{2} \cdot 2$ ). Why this step? Collapses two travellers into one clean standing oscillation.

Step 2 — push term. $\frac{1}{2} \int_{x - t}^{x + t} 1 d s = \frac{1}{2} \cdot (interval length) = \frac{1}{2} \cdot 2 t = t .$ Why this step? Integrating the constant $1$ just gives the length of $[x - t, x + t]$ , which is $2 t$ ; the $\frac{1}{2 c} = \frac{1}{2}$ halves it.

Step 3 — add. $u (x, t) = sin x cos t + t .$ Why this step? Superposition: shape-part and push-part are added because the wave equation is linear.

Verify: $t = 0 \Rightarrow u = sin x + 0 = sin x = f$ ✓. $u_{t} = - sin x sin t + 1$ ; at $t = 0$ , $u_{t} = 1 = g$ ✓. The $+ t$ is a genuine upward drift (constant velocity input never stops).

Case D — a cornered shape (non-smooth)

Worked example Ex 4 · Cell D · triangular pluck,

$g = 0$ , $c = 1$ Let $f$ be a tent: $f (x) = 1 - ∣ x ∣$ for $∣ x ∣ \leq 1$ , else $0$ . Forecast: in the Heat Equation the corner would instantly smooth out. Here — does the corner survive?

Step 1 — switch off the integral ( $g = 0$ ), as in Ex 1. $u (x, t) = \frac{1}{2} f (x - t) + \frac{1}{2} f (x + t) .$ Why this step? Same recipe — a shape-only problem.

Step 2 — locate the two half-tents. The right tent lives where $∣ x - t ∣ \leq 1$ , i.e. centred at $x = t$ ; the left tent centred at $x = - t$ . Each has height $\frac{1}{2}$ . Why this step? Replacing $x \to x - t$ shifts the tent's centre from $0$ to $t$ .

Step 3 — evaluate at a chosen point. Take $x = 0, t = 2$ : the right tent needs $∣0 - 2∣ = 2 \leq 1$ ? No. Left tent $∣0 + 2∣ = 2 \leq 1$ ? No. So $u (0, 2) = 0$ : both corners have already marched past the origin. Why this step? Confirms corners travel undistorted — they do not diffuse. The corner is a feature riding along the characteristic $x = t$ .

Verify: at $t = 0$ , $u = \frac{1}{2} f + \frac{1}{2} f = f$ ✓ (tent restored). At $x = 0.5, t = 0.5$ : right tent value $f (0) = 1$ , left tent $f (1) = 0$ , so $u = \frac{1}{2} (1) + \frac{1}{2} (0) = 0.5$ .

Case E — a finite-width kick (domain of influence)

Worked example Ex 5 · Cell E ·

$f = 0$ , $g (x) = 1$ on $[- 1, 1]$ else $0$ , $c = 1$ Forecast: you smack only the middle metre of the rope. Which points ever feel it? All of them, or only some?

Step 1 — push-only formula. $u (x, t) = \frac{1}{2} \int_{x - t}^{x + t} g (s) d s = \frac{1}{2} \cdot (length of overlap of [x - t, x + t] with [- 1, 1]) .$ Why this step? $g$ is $1$ only inside $[- 1, 1]$ , so the integral just measures how much of the dependence interval lands on the struck region.

Step 2 — a point far away, $x = 5$ . For small $t$ , $[5 - t, 5 + t]$ is nowhere near $[- 1, 1]$ , so $u = 0$ . The signal reaches $x = 5$ only once $5 - t \leq 1$ , i.e. $t \geq 4$ . Why this step? Shows finite propagation speed: the Domain of Dependence and Influence of the kick is a widening cone, not the whole line at once.

Step 3 — a point that fully engulfs the kick. Once $x - t \leq - 1$ and $x + t \geq 1$ , the whole struck region is inside, integral $=$ its length $= 2$ , so $u = \frac{1}{2} \cdot 2 = 1$ . Take $x = 0, t = 3$ : interval $[- 3, 3] \supset [- 1, 1]$ , so $u (0, 3) = 1$ . Why this step? After the wave has fully swept over, the displacement settles to a permanent value (residual push).

Verify: $x = 0, t = 0.5$ : interval $[- 0.5, 0.5] \subset [- 1, 1]$ , overlap length $= 1$ , so $u = \frac{1}{2} \cdot 1 = 0.5$ . And $u (0, 3) = 1$ , $u (5, 1) = 0$ . All ✓.

Case F — signs and direction (the trap)

Worked example Ex 6 · Cell F · which way does

$f (x - c t)$ go? Let $f (x) = H (x)$ (a step: $0$ for $x < 0$ , $1$ for $x \geq 0$ ), $g = 0$ , $c = 3$ . Forecast: does the piece $f (x - c t)$ move left or right as $t$ grows?

Step 1 — track a fixed value. The jump of $f (x - c t)$ sits where the argument $= 0$ : $x - 3 t = 0 \Rightarrow x = 3 t$ . Why this step? A travelling wave keeps its features where its argument is constant. Follow the argument, not $x$ alone.

Step 2 — read the motion. As $t ↑$ , $x = 3 t ↑$ : the jump moves in the $+ x$ direction at speed $3$ . So minus sign ⇒ rightward. Why this step? This is the single most-confused sign in the subject. $f (x - c t)$ = right-mover, $f (x + c t)$ = left-mover.

Step 3 — full solution. $u (x, t) = \frac{1}{2} H (x - 3 t) + \frac{1}{2} H (x + 3 t)$ : a right step at $x = 3 t$ , a left step at $x = - 3 t$ , each height $\frac{1}{2}$ . Why this step? Splits the unit step into two half-steps sprinting apart.

Verify: at $t = 0$ both are $H (x)$ , sum $= H (x) = f$ ✓. At $x = 0, t = 1$ : $H (- 3) = 0$ , $H (3) = 1$ , so $u = \frac{1}{2} (0) + \frac{1}{2} (1) = 0.5$ .

Case G — degenerate limits

Worked example Ex 7 · Cell G · the edge cases every formula must survive

Take $f (x) = cos x$ , $g (x) = 0$ and probe three limits.

(i) $t = 0$ (no time has passed). $u (x, 0) = \frac{1}{2} [cos x + cos x] + 0 = cos x = f$ . Why this step? Any correct solution must reproduce the initial shape at $t = 0$ ; this is the first sanity gate.

(ii) $c \to 0$ (speed vanishes). Shape term $\to \frac{1}{2} [cos x + cos x] = cos x$ (frozen). Integral: with $g = 0$ it's $0$ anyway, so $u (x, t) = cos x$ for all $t$ — nothing moves. Why this step? $c = 0$ removes travel; the wave equation degenerates to $u_{tt} = 0$ , meaning $u$ is at most linear in $t$ . With $g = 0$ it's simply constant in time. (Careful: for $g \neq = 0$ the $\frac{1}{2 c}$ would blow up — a warning that $c = 0$ is genuinely a different, non-hyperbolic regime; see Classification of Second-Order PDEs.)

(iii) $t \to \infty$ . $u = \frac{1}{2} cos (x - c t) + \frac{1}{2} cos (x + c t) = cos x cos c t$ . This does not settle — it oscillates forever between $\pm cos x$ (no damping in the ideal wave equation). Why this step? Contrasts with the Heat Equation, whose solutions decay to a steady state. Waves conserve energy; they ring forever.

Verify: (i) $u (x, 0) = cos x$ ✓. (iii) at $x = 0$ : $u (0, t) = cos c t$ , which at $t = π / c$ equals $cos π = - 1$ ✓ (full oscillation, no decay).

Case H — a real-world word problem

Worked example Ex 8 · Cell H · flicking a long string

A very long horizontal string ( $c = 1$ m/s) is flat ( $f = 0$ ). You give the segment $[- a, a]$ an upward velocity $v_{0}$ (m/s), zero elsewhere: $g (x) = v_{0}$ for $∣ x ∣ \leq a$ , else $0$ . Find the displacement of the midpoint $x = 0$ at time $t$ , for $t > a$ (after the edges have run past the origin). Take $a = 1, v_{0} = 2$ .

Forecast: does the midpoint keep rising forever, or freeze at some height once the kick has swept past?

Step 1 — push-only formula (as Ex 5), with $c = 1$ . $u (0, t) = \frac{1}{2} \int_{- t}^{t} g (s) d s .$ Why this step? $f = 0$ kills the shape term; dependence interval at $x = 0$ is $[- t, t]$ .

Step 2 — for $t > a$ , the whole struck region is inside $[- t, t]$ . $u (0, t) = \frac{1}{2} \int_{- a}^{a} v_{0} d s = \frac{1}{2} v_{0} (2 a) = v_{0} a .$ Why this step? Beyond $t = a$ the integral no longer grows — the interval already contains all the pushed points, so displacement saturates.

Step 3 — numbers. $v_{0} a = 2 \cdot 1 = 2$ m. So for all $t > 1$ s the midpoint sits at a constant height $2$ m. Why this step? Physically: the finite total impulse gives a permanent net lift; the "edges" of the disturbance keep racing outward but the centre is done moving.

Verify: dimension check — $[v_{0}] [a] = (m/s) (s used implicitly? no)$ … units: $v_{0}$ (m/s) $\times$ $\frac{a}{c}$ ? Cleanly, $u = \frac{1}{2 c} \int g$ has units $\frac{1}{m/s} \cdot (m/s \cdot m) = m$ ✓. Numeric: $u (0, 2) = \frac{1}{2} \int_{- 1}^{1} 2 d s = \frac{1}{2} \cdot 4 = 2$ ✓. At $t = 0.5 < a$ : interval $[- 0.5, 0.5]$ , $u = \frac{1}{2} \int_{- 0.5}^{0.5} 2 d s = \frac{1}{2} \cdot 2 = 1$ .

Recall Quick self-test

Which cell does " $f = 0$ , $g = sin x$ " fall in? ::: Cell B — push only; use the integral term. A cornered initial shape with $g = 0$ — does the corner smooth out? ::: No. It travels undistorted along $x = \pm c t$ ; waves don't diffuse. Direction of $f (x - c t)$ ? ::: Rightward (+x), because its features sit where $x - c t =$ const, so $x$ grows with $t$ . Ex 8 midpoint height for $t > a$ ? ::: $v_{0} a$ (here $2$ m), and it saturates — stops rising. Why does $c \to 0$ break the integral term when $g \neq = 0$ ? ::: The $\frac{1}{2 c}$ factor blows up; the PDE stops being hyperbolic.

Connections

Parent topic — the formula these examples exercise.
Method of Characteristics — the $x \pm c t$ lines every example rides.
Transport Equation — the first-order right/left movers behind Ex 6's sign.
Domain of Dependence and Influence — Ex 5 and Ex 8 are it in action.
Separation of Variables — Wave Equation — how these look on a bounded string.
Classification of Second-Order PDEs — why $c \to 0$ (Ex 7) leaves the hyperbolic world.
Heat Equation — the contrast: corners smooth, solutions decay (Ex 4, Ex 7).