Wave equation

At a Glance

• Frequency: 7 sub-parts across 6 of 13 years (2013, 2014, 2017, 2020, 2021, 2023)
• Priority tier: T2
• Marks (count): 15 (2), 20 (5)
• Average solve time: ~15 min
• Difficulty mix: medium 5, hard 2
• Section: B | Dominant type: computation

Why This Chapter Matters

The wave equation appears in 6 of the last 11 years and is essentially always a 15- or 20-mark question — the highest mark range in Paper 2’s PDE cluster. Every question uses the same algorithm: separate variables, quantise the spatial modes using the fixed-end boundary conditions, apply the two initial conditions to determine the Fourier coefficients, and write the series. The variation between questions lies entirely in the initial conditions (zero vs. non-zero displacement, zero vs. non-zero velocity, polynomial vs. trigonometric IC), which determines which coefficients survive and how much integration is required. Mastering the algorithm once covers all seven appearances.

Minimum Theory

Wave equation. The transverse displacement $u(x,t)$ of a tightly stretched string of length $L$ satisfies $u_{tt}=c^2 u_{xx},\qquad 0<x<L,\;t>0,$ with $c^2=T/\mu$ (tension/mass-per-unit-length). Boundary conditions $u(0,t)=u(L,t)=0$ (fixed ends). Initial conditions: $u(x,0)=f(x)$ (initial shape) and $u_t(x,0)=g(x)$ (initial velocity).

Separation of variables. Set $u=X(x)T(t)$. The BCs force $X(0)=X(L)=0$ and the separation constant must be negative (call it $-\lambda^2$) to get oscillatory solutions. The normalised modes are $X_n(x)=\sin(n\pi x/L)$, $n=1,2,\ldots$, with corresponding time factors $T_n(t)=A_n\cos(n\pi ct/L)+B_n\sin(n\pi ct/L)$. General solution: $u(x,t)=\sum_{n=1}^\infty\sin\frac{n\pi x}{L}\!\left[A_n\cos\frac{n\pi ct}{L}+B_n\sin\frac{n\pi ct}{L}\right].$

IC rule. Apply $u(x,0)=f(x)$: $A_n=\dfrac{2}{L}\displaystyle\int_0^L f(x)\sin\frac{n\pi x}{L}\,dx$.

Apply $u_t(x,0)=g(x)$: $\dfrac{n\pi c}{L}B_n=\dfrac{2}{L}\displaystyle\int_0^L g(x)\sin\frac{n\pi x}{L}\,dx$, i.e. $B_n=\dfrac{2}{n\pi c}\displaystyle\int_0^L g(x)\sin\frac{n\pi x}{L}\,dx$.

Zero IC simplifications: If $f(x)=0$ then $A_n=0$ for all $n$ (pure $\sin$ in time). If $g(x)=0$ then $B_n=0$ (pure $\cos$ in time). If $f$ is already one eigenmode $a\sin(n_0\pi x/L)$, no series is needed — only $A_{n_0}=a$ survives.

IBP for polynomial ICs. For $f(x)=x(L-x)$ on $(0,L)$: $\int_0^L x(L-x)\sin(n\pi x/L)\,dx=\dfrac{2L^3}{n^3\pi^3}[1-(-1)^n]$, so only odd $n$ contribute ($1/n^3$ decay). For $g(x)=\lambda x(L-x)$: the $B_n$ coefficient is $\dfrac{8\lambda L^3}{n^4\pi^4 c}$ for odd $n$. For $f(x)=x$ on $(0,L)$: $A_n=\dfrac{2L(-1)^{n+1}}{n\pi}$ (slow $1/n$ decay, alternating). For constant $g=1$: $B_n=\dfrac{4L}{n^2\pi^2 c}$ for odd $n$ only.

Triangular IC. For a string plucked at $x=a$ to height $h$: $b_n=\dfrac{2hL^2}{\pi^2 n^2 a(L-a)}\sin\dfrac{n\pi a}{L}$. Harmonics for which $\sin(n\pi a/L)=0$ are absent (nodes at the pluck point).

Question Archetypes

All wave-equation questions in the corpus share the same archetype.

vibrating-string (7 question(s); 2013, 2014, 2014, 2017, 2020, 2021, 2023)

Recognition Cues

• “A tightly stretched string of length $L$ with both ends fixed; find the displacement at any time.”
• One or both initial conditions given: initial shape $f(x)$ and/or initial velocity $g(x)$.
• Variants: a single eigenmode IC (no series needed); triangular (plucked) IC; polynomial IC (requires IBP twice).

Solution Template

1. Write the general sine-series solution $u=\sum_{n\ge1}\sin(n\pi x/L)[A_n\cos\omega_n t+B_n\sin\omega_n t]$ where $\omega_n=n\pi c/L$.
2. Set $t=0$: $\sum A_n\sin(n\pi x/L)=f(x)$. Compute $A_n$ by Fourier projection or by inspection if $f$ is already a finite sum of modes.
3. Differentiate in $t$ and set $t=0$: $\sum B_n\omega_n\sin(n\pi x/L)=g(x)$. Compute $B_n=\frac{2}{\omega_n L}\int_0^L g\sin(n\pi x/L)\,dx$.
4. For polynomial ICs: use two IBPs (bracket terms vanish at fixed ends); collect odd/even $n$ cases.
5. Write the final boxed series. Note which harmonics vanish (e.g. $n$ divisible by 3 if plucked at $L/3$).

Worked Example(s)

2013 Paper 2, 2013-P2-Q6c (20 marks)

String length $l$, fixed ends, released from rest; initial velocity $\lambda x(l-x)$.

$f(x)=0$ → all $A_n=0$. Apply $g(x)=\lambda x(l-x)$: $B_n\cdot\frac{n\pi c}{l}=\frac{2\lambda}{l}\int_0^l x(l-x)\sin\frac{n\pi x}{l}\,dx=\frac{2\lambda}{l}\cdot\frac{2l^3\cdot[1-(-1)^n]}{n^3\pi^3}.$

For even $n$: $B_n=0$. For odd $n$: $B_n=\dfrac{8\lambda l^3}{n^4\pi^4 c}$.

$\boxed{\;u(x,t)=\frac{8\lambda l^3}{\pi^4 c}\sum_{n\text{ odd}}\frac{1}{n^4}\sin\frac{n\pi x}{l}\sin\frac{n\pi ct}{l}.\;}$

The $1/n^4$ decay signals a quadratic IC; the odd-harmonics-only pattern follows from $x(l-x)$ being even about $l/2$.

2014 Paper 2, 2014-P2-Q7a (15 marks)

String length $\pi$, $c=1$; zero velocity; initial shape $k(\sin x-\sin 2x)$.

$g(x)=0$ → all $B_n=0$. The IC $f(x)=k\sin x-k\sin 2x$ is already a finite sum of eigenmodes ($L=\pi$, so basis is $\sin nx$): by inspection $A_1=k$, $A_2=-k$, all others zero.

$\boxed{\;u(x,t)=k\sin x\cos t-k\sin 2x\cos 2t.\;}$

No Fourier integral required; the solution is a finite sum and stays finite for all $t$.

2014 Paper 2, 2014-P2-Q8a (15 marks)

$u_{tt}=u_{xx}$ on $(0,1)$; $u(x,0)=0$; $u_t(x,0)=x^2$.

$L=c=1$. $A_n=0$. Compute $B_n$ from $x^2$ via two IBPs (using $\int x^2\sin(n\pi x)\,dx=-x^2\cos(n\pi x)/(n\pi)+2x\sin(n\pi x)/(n\pi)^2+2\cos(n\pi x)/(n\pi)^3$):

• Odd $n$: $B_n=\dfrac{2[(n\pi)^2-4]}{(n\pi)^4}$.
• Even $n$: $B_n=-\dfrac{2}{(n\pi)^2}$.

$\boxed{\;u(x,t)=\sum_{n=1}^\infty B_n\sin(n\pi t)\sin(n\pi x),\;}$

with the $B_n$ as above. Unlike the $x(l-x)$ case, both odd and even harmonics contribute (because $x^2$ lacks the midpoint symmetry).

2017 Paper 2, 2017-P2-Q8a (20 marks)

(i) Derive the general solution by separation of variables. (ii) Boundary conditions $y(0,t)=y(l,t)=0$, $y_t(x,0)=0$, $y(x,0)=a\sin(\pi x/l)$.

Part (i). Set $y=X(x)G(t)$; show the separation constant must be negative $-k^2$; write $y=(A\cos kx+B\sin kx)(C\cos ckt+E\sin ckt)$. State why $+k^2$ (exponential in $x$) is rejected. The appropriate solution is the oscillatory form with negative constant.

Part (ii). BCs: $A=0$, $k=n\pi/l$. IC $y_t=0$: $E_n=0$. IC $y(x,0)=a\sin(\pi x/l)$ is already the $n=1$ mode: $C_1=a$, all others zero.

$\boxed{\;y(x,t)=a\sin\frac{\pi x}{l}\cos\frac{\pi ct}{l}.\;}$

The initial shape is a single eigenmode so no Fourier series arises — only the fundamental survives.

2020 Paper 2, 2020-P2-Q8a (20 marks)

String length $l$, plucked at $x=l/3$ to height $h$, released from rest.

$g(x)=0$ → $B_n=0$. The initial shape is the triangular function: $f(x)=\begin{cases}3hx/l,&0\le x\le l/3,\\3h(l-x)/(2l),&l/3\le x\le l.\end{cases}$

Fourier sine coefficients (integrate piecewise + IBP on each segment): $b_n=\frac{9h}{\pi^2 n^2}\sin\frac{n\pi}{3}.$

Since $\sin(n\pi/3)=0$ for $n=3,6,9,\ldots$, every third harmonic vanishes (nodes at the pluck point).

$\boxed{\;y(x,t)=\frac{9h}{\pi^2}\sum_{n=1}^\infty\frac{1}{n^2}\sin\frac{n\pi}{3}\sin\frac{n\pi x}{l}\cos\frac{n\pi ct}{l}.\;}$

2021 Paper 2, 2021-P2-Q6a (20 marks)

$a^2u_{xx}=u_{tt}$ on $(0,L)$; $u(x,0)=x(L-x)/4$; $u_t(x,0)=0$.

$B_n=0$ (zero velocity). Compute $A_n$ from $x(L-x)/4$; using the standard integral $\int_0^L x(L-x)\sin(n\pi x/L)\,dx=2L^3[1-(-1)^n]/(n^3\pi^3)$, only odd $n$ contribute: $A_n=\frac{2}{L}\cdot\frac{1}{4}\cdot\frac{4L^3}{n^3\pi^3}=\frac{2L^2}{n^3\pi^3}\quad(n\text{ odd}).$

$\boxed{\;u(x,t)=\sum_{n\text{ odd}}\frac{2L^2}{n^3\pi^3}\sin\!\frac{n\pi x}{L}\cos\!\frac{an\pi t}{L}.\;}$

Same integral structure as 2013-P2-Q6c (quadratic IC, odd harmonics, $1/n^3$ decay from shape — noting here it’s displacement, not velocity, so decay is $1/n^3$ not $1/n^4$).

2023 Paper 2, 2023-P2-Q7c (20 marks)

$a^2u_{xx}=u_{tt}$ on $(0,L)$; $u(x,0)=x$; $u_t(x,0)=1$.

Both ICs non-zero; both $A_n$ and $B_n$ survive.

$A_n$: Fourier sine series of $x$ on $(0,L)$: $A_n=\dfrac{2L(-1)^{n+1}}{n\pi}$ (slow $1/n$ decay, alternating sign — because $f(L)=L\ne0$).

$B_n$: Fourier sine series of constant $1$: $\dfrac{n\pi a}{L}B_n=\dfrac{2[1-(-1)^n]}{n\pi}$, giving $B_n=\dfrac{4L}{n^2\pi^2 a}$ for odd $n$, $0$ for even.

$\boxed{\;u=\sum_{n=1}^\infty\frac{2L(-1)^{n+1}}{n\pi}\sin\frac{n\pi x}{L}\cos\frac{n\pi at}{L}+\sum_{n\text{ odd}}\frac{4L}{n^2\pi^2 a}\sin\frac{n\pi x}{L}\sin\frac{n\pi at}{L}.\;}$

Common Traps

• Sign of separation constant. For a vibrating string the separation constant must be $-k^2<0$; choosing $+k^2$ gives exponentials in $x$ that cannot satisfy the fixed-end BCs non-trivially. State the reason.
• Which IC kills which coefficient. Zero initial displacement → $A_n=0$, pure $\sin ct$ in time. Zero initial velocity → $B_n=0$, pure $\cos ct$ in time. Mixing these up is the most common source of errors.
• The factor $L/(n\pi c)$ from the $B_n$ formula. $B_n\omega_n=$ (Fourier coefficient of $g$) but $\omega_n=n\pi c/L$, so $B_n=(L/(n\pi c))\times$(Fourier coefficient). Forgetting this factor is frequent.
• Triangular IC (2020). The two-segment integral requires careful setup of limits and slopes; the result $b_n=\frac{9h}{\pi^2 n^2}\sin(n\pi/3)$ can be verified at $n=3$: $\sin(\pi)=0$ ✓ (no third harmonic when plucked at $L/3$).
• The $x(L-x)$ integral. Commit this to memory: $\int_0^L x(L-x)\sin(n\pi x/L)\,dx=\frac{2L^3}{n^3\pi^3}[1-(-1)^n]$. It appears in 2013, 2021 (and analogously in heat-equation chapters). Odd $n$ only.

Marks-Aware Writing

15-mark questions (2014-Q7a, 2014-Q8a): Four steps — write the general solution, apply the zero-velocity or zero-displacement IC (one sentence), compute the Fourier coefficient integral (show the IBP setup), write the final series. For Q7a: identify that the IC is already a finite sum; no integral needed (2 marks saved).

20-mark questions (2013, 2017, 2020, 2021, 2023): The mark scheme expects all five components: PDE + general solution derivation (step 1), BC application (step 2), first IC (step 3), second IC (step 4), assembled answer (step 5). For the 2017 two-part question, Part (i) derivation is worth ~8 marks on its own; state the rejection of the positive constant explicitly.

Practice Set