Heat equation

At a Glance

• Frequency: 3 sub-parts across 3 of 13 years (2015, 2016, 2022)
• Priority tier: T3
• Marks (count): 15 (1), 20 (2)
• Average solve time: ~12 min
• Difficulty mix: medium 3
• Section: B | Dominant type: computation

Why This Chapter Matters

The heat equation recurs at intervals of roughly three years and always commands 15 or 20 marks, making it one of the higher-yield single algorithms in Section B. All three questions use the same separation-of-variables pipeline ending in a Fourier sine series; the variation is entirely in the initial condition (quadratic $x(l-x)$ vs.\ a single eigenmode vs.\ a modified PDE with extra $+u$ term). Recognising which IC triggers a full series vs.\ a one-term answer can save 6–8 minutes. The 2016 question adds a physical setup with given material constants — diffusivity must be computed first.

Minimum Theory

Heat equation on a bar. For a laterally insulated bar of length $L$ with Dirichlet ends:

$u_t = c^2 u_{xx},\quad 0<x<L,\;t>0,\quad u(0,t)=u(L,t)=0,\quad u(x,0)=f(x).$

The diffusivity $c^2 = K/(\rho\sigma)$ where $K$ is thermal conductivity, $\rho$ density, $\sigma$ specific heat.

Separation of variables. Set $u=X(x)T(t)$. Substituting and separating gives

$\frac{T'}{c^2 T}=\frac{X''}{X}=-\lambda^2\quad(\lambda>0\text{ for oscillatory }X).$

Spatial: $X''+\lambda^2 X=0$ with $X(0)=X(L)=0$ gives $X_n=\sin(n\pi x/L)$, $\lambda_n=n\pi/L$.

Temporal: $T'+c^2\lambda_n^2 T=0$ gives $T_n=e^{-c^2 n^2\pi^2 t/L^2}$.

General solution:

$u(x,t)=\sum_{n=1}^{\infty}B_n\sin\!\left(\frac{n\pi x}{L}\right)e^{-c^2 n^2\pi^2 t/L^2},\quad B_n=\frac{2}{L}\int_0^L f(x)\sin\frac{n\pi x}{L}\,dx.$

Modified equation $u_t - u_{xx} + u = 0$. The $+u$ term modifies only the temporal decay: the eigenfunction equation for $X$ is unchanged; the $T$-equation becomes $T'=-(1+\lambda_n^2)T$, so the decay exponent is replaced by $1+n^2\pi^2/l^2$.

Key integral (quadratic IC). For $f(x)=x(l-x)$ on $[0,l]$:

$B_n=\frac{2}{l}\int_0^l x(l-x)\sin\frac{n\pi x}{l}\,dx=\begin{cases}8l^2/(n^3\pi^3)&n\text{ odd},\\0&n\text{ even}.\end{cases}$

(Only odd harmonics survive; the $1/n^3$ decay comes from the $x^2$ factor in the IC.)

Question Archetypes

heat-equation (3 question(s); 2015, 2016, 2022)

Recognition Cues

• “Solve $u_t=u_{xx}$” or “find the temperature $u(x,t)$” with $u(0,t)=u(L,t)=0$ and $u(x,0)=f(x)$.
• Physical variants: bar with given $\rho,K,\sigma$ — compute $c^2$ first.
• Modified PDE: $u_t-u_{xx}+u=0$ — same spatial eigenfunctions, shifted decay.
• Shortcut: if $f(x)$ is already $\sin(n_0\pi x/L)$, only one mode survives.

Solution Template

1. Identify $L$, $c^2$ (compute from material constants if needed), and $f(x)$.
2. Assume $u=X(x)T(t)$; write the separated equations; state the eigenvalue problem.
3. Solve: $X_n=\sin(n\pi x/L)$, $T_n=e^{-c^2(n\pi/L)^2 t}$ (adjust decay for any extra linear terms in the PDE).
4. Write the general series $u=\sum B_n\sin(n\pi x/L)\,T_n(t)$.
5. Apply $u(x,0)=f(x)$: compute $B_n$ by Fourier projection. For $f(x)=x(l-x)$ use the standard result; for $f(x)=\sin(k\pi x/L)$ read off $B_k=1$ by inspection.
6. Write the final boxed series; for single-mode IC write the single term explicitly.

Worked Examples

2015 Paper 2, 2015-P2-Q7a (15 marks)

Solve $u_t - u_{xx} + u = 0$, $0<x<l$, $u(0,t)=u(l,t)=0$, $u(x,0)=x(l-x)$.

Set $u=X(x)T(t)$. Substituting: $XT'-X''T+XT=0$, so

$\frac{T'+T}{T}=\frac{X''}{X}=-\lambda^2.$

Spatial: $X''+\lambda^2 X=0$ with $X(0)=X(l)=0$ gives $X_n=\sin(n\pi x/l)$, $\lambda_n=n\pi/l$.

Temporal: $T'+T=-\lambda_n^2 T$, i.e.\ $T'=-(1+\lambda_n^2)T$, giving

$T_n=e^{-(1+n^2\pi^2/l^2)t}.$

General solution: $u(x,t)=\displaystyle\sum_{n=1}^\infty B_n\sin\!\left(\dfrac{n\pi x}{l}\right)e^{-(1+n^2\pi^2/l^2)t}$.

Apply $u(x,0)=x(l-x)$: $B_n=\dfrac{2}{l}\displaystyle\int_0^l x(l-x)\sin\!\left(\dfrac{n\pi x}{l}\right)dx$. Using the standard quadratic-IC result:

• $n$ odd: $B_n=\dfrac{8l^2}{n^3\pi^3}$.
• $n$ even: $B_n=0$.

$\boxed{u(x,t)=\sum_{\substack{n=1\\n\;\text{odd}}}^{\infty}\frac{8l^2}{n^3\pi^3}\sin\!\left(\frac{n\pi x}{l}\right)e^{-(1+n^2\pi^2/l^2)t}}$

The extra $+u$ term shifts the decay exponent from $n^2\pi^2/l^2$ to $1+n^2\pi^2/l^2$, giving slightly faster decay.

2016 Paper 2, 2016-P2-Q8a (20 marks)

Find $u(x,t)$ in a 10 cm silver bar with $\rho=10.6$ g/cm$^3$, $K=1.04$ cal/(cm s $^\circ$C), $\sigma=0.056$ cal/(g $^\circ$C), ends at $0^\circ$C, and initial temperature $f(x)=\sin(0.1\pi x)$.

PDE: $u_t=c^2 u_{xx}$ on $0\le x\le 10$, $u(0,t)=u(10,t)=0$, $u(x,0)=\sin(0.1\pi x)$.

Step 1 — Diffusivity:

$c^2=\frac{K}{\rho\sigma}=\frac{1.04}{10.6\times0.056}=\frac{1.04}{0.5936}=1.752\ \text{cm}^2/\text{s}.$

Step 2 — General series (with $L=10$):

$u(x,t)=\sum_{n=1}^{\infty}B_n\sin\!\frac{n\pi x}{10}\,e^{-c^2(n\pi/10)^2 t}.$

Step 3 — Match IC. The initial profile $f(x)=\sin(0.1\pi x)=\sin(\pi x/10)$ is exactly the $n=1$ eigenfunction. By inspection $B_1=1$ and $B_n=0$ for $n\ge2$.

Step 4 — Decay rate:

$c^2\!\left(\frac{\pi}{10}\right)^2=1.752\times\frac{\pi^2}{100}=1.752\times0.09870=0.1729\ \text{s}^{-1}.$

$\boxed{u(x,t)=\sin\!\left(\frac{\pi x}{10}\right)e^{-0.1729\,t}}$

The bar keeps its initial half-sine shape and decays with time constant $1/0.1729\approx5.78$ s. No Fourier series integration is needed because the initial profile is a single eigenmode.

2022 Paper 2, 2022-P2-Q6a (20 marks)

Solve $u_t=u_{xx}$, $0<x<l$, $t>0$, with $u(0,t)=u(l,t)=0$ and $u(x,0)=x(l-x)$.

Separate $u=X(x)T(t)$: $X''+\lambda^2 X=0$, $X(0)=X(l)=0$ gives $X_n=\sin(n\pi x/l)$; $T'+\lambda_n^2 T=0$ gives $T_n=e^{-n^2\pi^2 t/l^2}$.

General solution:

$u(x,t)=\sum_{n=1}^{\infty}B_n\sin\!\left(\frac{n\pi x}{l}\right)e^{-n^2\pi^2 t/l^2}.$

Apply $u(x,0)=x(l-x)$: $B_n=\dfrac{2}{l}\displaystyle\int_0^l x(l-x)\sin\!\left(\dfrac{n\pi x}{l}\right)dx$. Standard quadratic-IC integral gives $B_n=8l^2/(n^3\pi^3)$ for odd $n$, and $0$ for even $n$.

$\boxed{u(x,t)=\sum_{\substack{n=1\\n\;\text{odd}}}^{\infty}\frac{8l^2}{n^3\pi^3}\sin\!\left(\frac{n\pi x}{l}\right)e^{-n^2\pi^2 t/l^2}}$

After short time the dominant mode ($n=1$) approximates $u\approx\dfrac{8l^2}{\pi^3}\sin(\pi x/l)\,e^{-\pi^2 t/l^2}$.

Common Traps

• The $+u$ term (2015). Do not treat it as a forcing term. It modifies only the time decay: each mode gains an extra factor $e^{-t}$, shifting the exponent from $n^2\pi^2/l^2$ to $1+n^2\pi^2/l^2$. The spatial eigenfunctions and Fourier coefficients are unchanged.
• Single-mode IC (2016). When $f(x)=\sin(k\pi x/L)$ is an eigenfunction, the answer is a single term. Computing the full Fourier integral wastes time (the integral gives $B_k=1$, all others $0$ anyway — but state that explicitly to secure the marks).
• Material-constant arithmetic (2016). $\rho\sigma=10.6\times0.056=0.5936$; $c^2=1.04/0.5936=1.752$. The decay exponent is $c^2(\pi/L)^2$ with $L=10$, giving $\pi^2/100\approx0.09870$ — a common error is using $L=1$ or omitting the $(1/L)^2$ factor.
• Odd-modes-only pattern. For $f(x)=x(l-x)$, the IC is symmetric about $x=l/2$; even harmonics integrate to zero. The coefficient $8l^2/(n^3\pi^3)$ for odd $n$ should be memorised — it reappears whenever a quadratic IC occurs.

Marks-Aware Writing

15-mark questions (2015). Show: (1) the separation and eigenvalue problem for $X$ and $T$ with the modified decay; (2) the Fourier coefficient integral setup; (3) the IBP to evaluate $\int x(l-x)\sin(n\pi x/l)\,dx$; (4) odd-only conclusion; (5) boxed series. State “the $+u$ term shifts the time decay by $+1$” for a line that secures method marks.

20-mark questions (2016, 2022). The extra 5 marks reward: full derivation of the eigenvalue problem (not just quoting the answer), the explicit series formula before applying the IC, computation of $c^2$ with units (2016), and the final one-term vs.\ full-series conclusion with justification. For 2022, write out the IBP explicitly: at least two integration-by-parts steps must be shown.

Practice Set