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

4.7.11 · D3 · Maths › Partial Differential Equations › Solving wave equation — D'Alembert's solution

Agar koi symbol yahan unfamiliar lage, toh parent note use build karta hai: parent topic. Characteristic lines $x - c t$ , $x + c t$ Method of Characteristics se aati hain aur do first-order factors Transport Equations hain.

Scenario matrix

Infinite line pe har wave problem teen switches se bana hota hai. Neeche, ✓ ka matlab "present / non-trivial" hai, $0$ ka matlab "switched off" hai.

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	✓	✓	dono saath (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

Switches:

$f \neq = 0$ ? → pehla bracket light up hota hai.
$g \neq = 0$ ? → integral term light up hoti hai.
Kya $f$ smooth hai ya cornered? → corners travel karte hain, smooth nahi hote (yahan koi diffusion nahi — Heat Equation se contrast karo).
$g$ kahan nonzero hai? → Domain of Dependence and Influence decide hota hai.

Case A — shape only, no push

Worked example Ex 1 · Cell A ·

$f (x) = e^{- x^{2}}$ , $g = 0$ , $c = 1$ Forecast: bina starting push ke, socho ek Gaussian bump thodi der baad kya karta hai. Split hoga? Rukega? Badhega?

Step 1 — integral ko switch off karo. $g = 0$ , toh $\int_{x - c t}^{x + c t} g d s = 0$ . Sirf shape term bachta hai. Yeh step kyun? Formula mein do independent contributions hain; $g$ ko khatam karne se ek poori tarah remove ho jaati hai, toh integral compute hi nahi karna.

Step 2 — $f$ ko shape term mein $c = 1$ ke saath plug karo. $u (x, t) = \frac{1}{2} e^{- (x - t)^{2}} + \frac{1}{2} e^{- (x + t)^{2}} .$ Yeh step kyun? $f (x - c t)$ wahi Gaussian hai jisme $x$ ki jagah $x - t$ hai: ek copy jo right mein $t$ khisak gayi hai. $f (x + t)$ left mein khisakti hai. Dono aadhi hain kyunki formula mein $\frac{1}{2}$ lagta hai.

Step 3 — picture padho (neeche figure). Do half-height bumps speed $1$ pe alag-alag march karte hain.

Verify: $t = 0$ pe, dono exponents $- x^{2}$ ban jaate hain, toh $u = \frac{1}{2} e^{- x^{2}} + \frac{1}{2} e^{- x^{2}} = e^{- x^{2}} = f (x)$ ✓. Centre $x = 0$ pe height: $u (0, 1) = \frac{1}{2} e^{- 1} + \frac{1}{2} e^{- 1} = e^{- 1} \approx 0.3679$ .

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: string flat start karti hai lekin har point move kar raha hai. Kya yeh in place oscillate karega, travel karega, ya dono?

Step 1 — shape term ko switch off karo. $f = 0$ , toh $\frac{1}{2} [f (x - c t) + f (x + c t)] = 0$ . Sirf integral bachti hai. Yeh step kyun? Wahi logic — flat start ( $f = 0$ ) ka matlab hai pehla bracket kuch contribute nahi karta.

Step 2 — $c = 2$ ke saath integral likho. $u (x, t) = \frac{1}{2 \cdot 2} \int_{x - 2 t}^{x + 2 t} cos s d s = \frac{1}{4} [sin s]_{x - 2 t}^{x + 2 t} = \frac{1}{4} [sin (x + 2 t) - sin (x - 2 t)] .$ Yeh step kyun? $cos$ ka antiderivative $sin$ hai; domain of dependence $[x - 2 t, x + 2 t]$ hai kyunki $c t = 2 t$ .

Step 3 — sum-to-product identity se simplify karo. $sin A - sin B = 2 cos \frac{A + B}{2} sin \frac{A - B}{2}, A = x + 2 t, B = x - 2 t .$ Phir $\frac{A + B}{2} = x$ , $\frac{A - B}{2} = 2 t$ , toh $u (x, t) = \frac{1}{4} \cdot 2 cos x sin 2 t = \frac{1}{2} cos x sin 2 t .$ Yeh step kyun? Yeh ek travelling form ko standing-wave form mein badal deta hai: ek fixed spatial shape $cos x$ jo time mein $sin 2 t$ ke zariye breathe karta hai.

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

Case C — shape aur push dono

Worked example Ex 3 · Cell C ·

$f (x) = sin x$ , $g (x) = 1$ , $c = 1$ Forecast: oscillating start ke saath har jagah constant upward push. Kya yeh sirf wobble karega, ya upar forever drift bhi karega?

Step 1 — shape term. $\frac{1}{2} [sin (x - t) + sin (x + t)] = sin x cos t$ sum-to-product se ( $\frac{A + B}{2} = x$ , $\frac{A - B}{2} = t$ , factor $\frac{1}{2} \cdot 2$ ). Yeh step kyun? Do travellers ko ek clean standing oscillation mein collapse karta hai.

Step 2 — push term. $\frac{1}{2} \int_{x - t}^{x + t} 1 d s = \frac{1}{2} \cdot (interval length) = \frac{1}{2} \cdot 2 t = t .$ Yeh step kyun? Constant $1$ ko integrate karna sirf $[x - t, x + t]$ ki length deta hai, jo $2 t$ hai; $\frac{1}{2 c} = \frac{1}{2}$ use aadha kar deta hai.

Step 3 — add karo. $u (x, t) = sin x cos t + t .$ Yeh step kyun? Superposition: shape-part aur push-part add hote hain kyunki wave equation linear hai.

Verify: $t = 0 \Rightarrow u = sin x + 0 = sin x = f$ ✓. $u_{t} = - sin x sin t + 1$ ; $t = 0$ pe, $u_{t} = 1 = g$ ✓. $+ t$ ek genuine upward drift hai (constant velocity input kabhi rukta nahi).

Case D — ek cornered shape (non-smooth)

Worked example Ex 4 · Cell D · triangular pluck,

$g = 0$ , $c = 1$ Maano $f$ ek tent hai: $f (x) = 1 - ∣ x ∣$ for $∣ x ∣ \leq 1$ , baaki $0$ . Forecast: Heat Equation mein corner instantly smooth ho jaata. Yahan — kya corner survive karta hai?

Step 1 — integral switch off karo ( $g = 0$ ), jaise Ex 1 mein. $u (x, t) = \frac{1}{2} f (x - t) + \frac{1}{2} f (x + t) .$ Yeh step kyun? Wahi recipe — ek shape-only problem.

Step 2 — do half-tents locate karo. Right tent wahan rehta hai jahan $∣ x - t ∣ \leq 1$ , yaani $x = t$ pe centred; left tent $x = - t$ pe centred. Har ek ki height $\frac{1}{2}$ hai. Yeh step kyun? $x \to x - t$ replace karna tent ka centre $0$ se $t$ pe shift kar deta hai.

Step 3 — ek chosen point pe evaluate karo. Lo $x = 0, t = 2$ : right tent ko chahiye $∣0 - 2∣ = 2 \leq 1$ ? Nahi. Left tent $∣0 + 2∣ = 2 \leq 1$ ? Nahi. Toh $u (0, 2) = 0$ : dono corners origin se march karke nikal chuke hain. Yeh step kyun? Confirm karta hai ki corners undistorted travel karte hain — diffuse nahi hote. Corner ek feature hai jo characteristic $x = t$ pe ride karta hai.

Verify: $t = 0$ pe, $u = \frac{1}{2} f + \frac{1}{2} f = f$ ✓ (tent restore). $x = 0.5, t = 0.5$ pe: right tent value $f (0) = 1$ , left tent $f (1) = 0$ , toh $u = \frac{1}{2} (1) + \frac{1}{2} (0) = 0.5$ .

Case E — ek 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: tum sirf rope ka beech wala ek metre maro. Kaunse points kabhi bhi isko feel karte hain? Sab, ya kuch hi?

Step 1 — push-only formula. $u (x, t) = \frac{1}{2} \int_{x - t}^{x + t} g (s) d s = \frac{1}{2} \cdot (length of overlap of [x - t, x + t] with [- 1, 1]) .$ Yeh step kyun? $g$ sirf $[- 1, 1]$ ke andar $1$ hai, toh integral sirf measure karta hai ki dependence interval ka kitna hissa struck region pe land karta hai.

Step 2 — ek door wala point, $x = 5$ . Chote $t$ ke liye, $[5 - t, 5 + t]$ $[- 1, 1]$ ke paas kahin nahi hai, toh $u = 0$ . Signal $x = 5$ tak pahunchta hai tabhi jab $5 - t \leq 1$ , yaani $t \geq 4$ . Yeh step kyun? Finite propagation speed dikhata hai: kick ka Domain of Dependence and Influence ek widening cone hai, na ki poori line ek saath.

Step 3 — ek aisa point jo kick ko poora engulf kare. Jab ek baar $x - t \leq - 1$ aur $x + t \geq 1$ ho, poora struck region andar hai, integral $=$ uski length $= 2$ , toh $u = \frac{1}{2} \cdot 2 = 1$ . Lo $x = 0, t = 3$ : interval $[- 3, 3] \supset [- 1, 1]$ , toh $u (0, 3) = 1$ . Yeh step kyun? Wave ke poori tarah sweep karne ke baad, displacement ek permanent value pe settle ho jaata hai (residual push).

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

Case F — signs aur direction (the trap)

Worked example Ex 6 · Cell F ·

$f (x - c t)$ kis taraf jaata hai? Maano $f (x) = H (x)$ (ek step: $x < 0$ ke liye $0$ , $x \geq 0$ ke liye $1$ ), $g = 0$ , $c = 3$ . Forecast: kya piece $f (x - c t)$ left jaata hai ya right jab $t$ barhta hai?

Step 1 — ek fixed value track karo. $f (x - c t)$ ka jump wahan baithta hai jahan argument $= 0$ ho: $x - 3 t = 0 \Rightarrow x = 3 t$ . Yeh step kyun? Ek travelling wave apni features wahan rakhta hai jahan uska argument constant ho. Argument follow karo, sirf $x$ nahi.

Step 2 — motion padho. Jab $t ↑$ , $x = 3 t ↑$ : jump $+ x$ direction mein speed $3$ se move karta hai. Toh minus sign ⇒ rightward. Yeh step kyun? Yeh is subject ka sabse zyada confused sign hai. $f (x - c t)$ = right-mover, $f (x + c t)$ = left-mover.

Step 3 — full solution. $u (x, t) = \frac{1}{2} H (x - 3 t) + \frac{1}{2} H (x + 3 t)$ : ek right step $x = 3 t$ pe, ek left step $x = - 3 t$ pe, dono height $\frac{1}{2}$ . Yeh step kyun? Unit step ko do half-steps mein split karta hai jo alag-alag sprint karte hain.

Verify: $t = 0$ pe dono $H (x)$ hain, sum $= H (x) = f$ ✓. $x = 0, t = 1$ pe: $H (- 3) = 0$ , $H (3) = 1$ , toh $u = \frac{1}{2} (0) + \frac{1}{2} (1) = 0.5$ .

Case G — degenerate limits

Worked example Ex 7 · Cell G · edge cases jo har formula ko survive karne chahiye

Lo $f (x) = cos x$ , $g (x) = 0$ aur teen limits probe karo.

(i) $t = 0$ (koi time nahi guzra). $u (x, 0) = \frac{1}{2} [cos x + cos x] + 0 = cos x = f$ . Yeh step kyun? Kisi bhi correct solution ko $t = 0$ pe initial shape reproduce karni chahiye; yeh pehla sanity gate hai.

(ii) $c \to 0$ (speed khatam hoti hai). Shape term $\to \frac{1}{2} [cos x + cos x] = cos x$ (frozen). Integral: $g = 0$ ke saath yeh $0$ hai waise bhi, toh $u (x, t) = cos x$ sab $t$ ke liye — kuch move nahi karta. Yeh step kyun? $c = 0$ travel ko remove karta hai; wave equation $u_{tt} = 0$ mein degenerate ho jaati hai, matlab $u$ $t$ mein zyada se zyada linear hai. $g = 0$ ke saath yeh time mein simply constant hai. (Dhyan raho: $g \neq = 0$ ke liye $\frac{1}{2 c}$ blow up ho jaata — ek warning ki $c = 0$ genuinely ek alag, non-hyperbolic regime hai; dekho Classification of Second-Order PDEs.)

(iii) $t \to \infty$ . $u = \frac{1}{2} cos (x - c t) + \frac{1}{2} cos (x + c t) = cos x cos c t$ . Yeh settle nahi hota — yeh forever $\pm cos x$ ke beech oscillate karta rehta hai (ideal wave equation mein koi damping nahi). Yeh step kyun? Heat Equation se contrast karta hai, jiske solutions steady state mein decay karte hain. Waves energy conserve karti hain; woh forever ring karti hain.

Verify: (i) $u (x, 0) = cos x$ ✓. (iii) $x = 0$ pe: $u (0, t) = cos c t$ , jo $t = π / c$ pe $cos π = - 1$ hai ✓ (full oscillation, koi decay nahi).

Case H — ek real-world word problem

Worked example Ex 8 · Cell H · ek lambi string ko flick karna

Ek bahut lambi horizontal string ( $c = 1$ m/s) flat hai ( $f = 0$ ). Tum segment $[- a, a]$ ko upward velocity $v_{0}$ (m/s) dete ho, baaki jagah zero: $g (x) = v_{0}$ for $∣ x ∣ \leq a$ , else $0$ . Midpoint $x = 0$ ka displacement time $t$ pe dhundho, $t > a$ ke liye (edges ke origin se guzarne ke baad). Lo $a = 1, v_{0} = 2$ .

Forecast: kya midpoint forever uthta rehta hai, ya kisi height pe freeze ho jaata hai jab kick guzar jaati hai?

Step 1 — push-only formula (jaise Ex 5), $c = 1$ ke saath. $u (0, t) = \frac{1}{2} \int_{- t}^{t} g (s) d s .$ Yeh step kyun? $f = 0$ shape term ko khatam karta hai; $x = 0$ pe dependence interval $[- t, t]$ hai.

Step 2 — $t > a$ ke liye, poora struck region $[- t, t]$ ke andar hai. $u (0, t) = \frac{1}{2} \int_{- a}^{a} v_{0} d s = \frac{1}{2} v_{0} (2 a) = v_{0} a .$ Yeh step kyun? $t = a$ se aage integral aur nahi barhta — interval mein pehle se sab pushed points hain, toh displacement saturate ho jaata hai.

Step 3 — numbers. $v_{0} a = 2 \cdot 1 = 2$ m. Toh sab $t > 1$ s ke liye midpoint constant height $2$ m pe baitha rehta hai. Yeh step kyun? Physically: finite total impulse ek permanent net lift deta hai; disturbance ke "edges" bahar race karte rehte hain lekin centre ka kaam khatam ho chuka hai.

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

" $f = 0$ , $g = sin x$ " — yeh kaun se cell mein aata hai? ::: Cell B — push only; integral term use karo. Cornered initial shape $g = 0$ ke saath — kya corner smooth ho jaata hai? ::: Nahi. Yeh $x = \pm c t$ ke saath undistorted travel karta hai; waves diffuse nahi karti. $f (x - c t)$ ki direction? ::: Rightward (+x), kyunki iske features wahan hain jahan $x - c t =$ const hai, toh $x$ $t$ ke saath barhta hai. Ex 8 midpoint height $t > a$ ke liye? ::: $v_{0} a$ (yahan $2$ m), aur yeh saturate hota hai — uthna band ho jaata hai. $g \neq = 0$ ke liye $c \to 0$ integral term kyun tod deta hai? ::: $\frac{1}{2 c}$ factor blow up ho jaata hai; PDE hyperbolic rehna band ho jaati hai.

Connections

Parent topic — woh formula jo yeh examples exercise karte hain.
Method of Characteristics — $x \pm c t$ lines jis pe har example ride karta hai.
Transport Equation — Ex 6 ke sign ke peeche first-order right/left movers.
Domain of Dependence and Influence — Ex 5 aur Ex 8 iska action mein demonstration hain.
Separation of Variables — Wave Equation — bounded string pe yeh kaisa dikhta hai.
Classification of Second-Order PDEs — kyun $c \to 0$ (Ex 7) hyperbolic world se bahar jaata hai.
Heat Equation — contrast: corners smooth hote hain, solutions decay karte hain (Ex 4, Ex 7).