Laplace equation: Dirichlet/Neumann, separation of variables

At a Glance

• Frequency: 5 sub-parts across 5 of 14 years (2017, 2018, 2024, 2025, 2026)
• Priority tier: T3
• Marks (count): 10 (1), 20 (3)
• Average solve time: ~17 min
• Difficulty mix: hard 2, medium 1, easy 1
• Section: B | Dominant type: proof

Why This Chapter Matters

Every Laplace equation question carries 10–20 marks and involves one of three techniques: the energy-integral uniqueness proof, separation of variables on a rectangle or annulus, or Fourier transform derivation of the Poisson kernel. The 2024 and 2025 papers together accounted for 40 marks from this one atom. The hardest trap — encountered in 2018 — is that the boundary datum $\cos(\theta/2)$ has period $4\pi$, not $2\pi$, so it must be expanded in a Fourier series rather than matched to a single harmonic. Knowing this trap is worth 10 marks on its own.

Minimum Theory

Laplace’s equation. $\nabla^2 u = u_{xx} + u_{yy} = 0$ (or in polar: $u_{rr} + \tfrac{1}{r}u_r + \tfrac{1}{r^2}u_{\theta\theta} = 0$). A solution is called harmonic. Harmonic functions satisfy the maximum principle: on a bounded domain $\Omega$, $u$ attains its max and min on the boundary $\partial\Omega$.

Uniqueness via Green’s identity. Green’s first identity: $\iint_S \left(u\,\nabla^2 u + |\nabla u|^2\right)dA = \oint_\Gamma u\,\frac{\partial u}{\partial n}\,ds.$ For the Dirichlet problem ($u$ prescribed on $\Gamma$): if two solutions $u_1, u_2$ both satisfy the same BVP, their difference $v = u_1 - u_2$ is harmonic with $v = 0$ on $\Gamma$. Applying the identity to $v$ gives $\iint_S |\nabla v|^2\,dA = 0$, hence $v \equiv 0$.

Separation of variables on a rectangle. For $u(x,y) = X(x)Y(y)$:

• $X'' + \lambda X = 0$ with BCs on $x$ gives the Sturm–Liouville problem; the eigenvalues are $\lambda_n = (n\pi/a)^2$.
• Homogeneous edges (zero) give sine eigenfunctions $X_n = \sin(n\pi x/a)$.
• The $Y$-equation $Y'' - \lambda_n Y = 0$ with $Y(0) = 0$ gives $Y_n = \sinh(n\pi y/a)$.
• The non-homogeneous top edge $u(x,b) = f(x)$ is matched by a Fourier sine series.

Question Archetypes

uniqueness-proof (1 question; 2017)

Recognition Cues

• “Prove that any solution satisfying these conditions is unique.”
• The setup is the Dirichlet problem for Poisson’s equation: $\nabla^2 w = f$ with $w$ prescribed on $\Gamma$.

Solution Template

1. Suppose two solutions $w_1$, $w_2$. Define $u = w_1 - w_2$.
2. Establish the homogeneous problem: $\nabla^2 u = 0$ in $S$; $u = 0$ on $\Gamma$.
3. Apply Green’s first identity to $u$: $\iint_S |\nabla u|^2\,dA = \oint_\Gamma u\,\partial u/\partial n\,ds = 0$ (boundary term vanishes since $u = 0$ on $\Gamma$).
4. Conclude: $|\nabla u|^2 \ge 0$ and its integral is 0, so $|\nabla u|^2 \equiv 0 \Rightarrow u = \text{const}$. Boundary condition $u = 0$ on $\Gamma$ forces $u \equiv 0$.

Worked Example

2017 Paper 2, 2017-P2-Q5d (10 marks)

Prove uniqueness for $\nabla^2 w = f$ in $S$ with $w = g$ on $\Gamma$.

Step 1. Suppose $w_1$ and $w_2$ both satisfy the problem. Let $u = w_1 - w_2$. By linearity, $\nabla^2 u = 0$ in $S$ and $u = 0$ on $\Gamma$.

Step 2. Apply Green’s first identity to $u$ (with $u$ in both positions): $\iint_S \left(u\,\nabla^2 u + |\nabla u|^2\right)dA = \oint_\Gamma u\,\frac{\partial u}{\partial n}\,ds.$ The right side is 0 (since $u = 0$ on $\Gamma$). The first term on the left is 0 (since $\nabla^2 u = 0$). Hence: $\iint_S |\nabla u|^2\,dA = 0.$

Step 3. Since $|\nabla u|^2 = u_x^2 + u_y^2 \ge 0$ everywhere and integrates to 0, we have $|\nabla u|^2 \equiv 0$, so $u$ is constant in $S$. By continuity and $u = 0$ on $\Gamma$, this constant is 0. Hence $u \equiv 0$, i.e. $w_1 \equiv w_2$. $\square$

Alternative. By the maximum principle, a harmonic function on $\bar S$ attains its max and min on $\Gamma$. Since $u = 0$ on $\Gamma$: $0 \le u \le 0$, so $u \equiv 0$.

Common Traps

• The difference $u$ satisfies the homogeneous Laplace equation ($\nabla^2 u = 0$, not $f$); both the $f$ terms cancel. This reduction is the first step and must appear explicitly.
• For Neumann BCs ($\partial u/\partial n = 0$ on $\Gamma$): the boundary integral still vanishes, and the same Green’s argument gives $u =$ const — but only uniqueness up to an additive constant.

laplace-rectangle (1 question; 2025)

Recognition Cues

• “Solve $u_{xx}+u_{yy}=0$ on a rectangle $[0,a]\times[0,b]$.”
• Three zero BCs ($u=0$ on three edges) and one data edge ($u=f(x)$ on the top).

Solution Template

1. Assume $u(x,y) = X(x)Y(y)$. Substitute, separate: $X''/X = -Y''/Y = -\lambda$.
2. Solve the $x$-problem with homogeneous BCs ($X(0) = X(a) = 0$): $\lambda_n = (n\pi/a)^2$, $X_n = \sin(n\pi x/a)$.
3. Solve the $y$-problem with $Y(0) = 0$: $Y_n = \sinh(n\pi y/a)$ (not $\cosh$ — that would fail $Y(0) = 0$).
4. Superpose: $u = \sum_{n=1}^\infty B_n \sin(n\pi x/a)\sinh(n\pi y/a)$.
5. Apply top BC $u(x,b) = f(x)$: Fourier sine series gives $B_n = \dfrac{2}{a\,\sinh(n\pi b/a)}\int_0^a f(\xi)\sin(n\pi\xi/a)\,d\xi$.

Worked Example

2025 Paper 2, 2025-P2-Q6a (20 marks)

Solve $\nabla^2 u = 0$ on $[0,a]\times[0,b]$ with $u(0,y)=u(a,y)=u(x,0)=0$ and $u(x,b)=f(x)$.

Following the template: the three homogeneous BCs force sine eigenfunctions and $\sinh$ in $y$. Superposition then gives:

$\boxed{u(x,y) = \sum_{n=1}^\infty B_n\,\sin\frac{n\pi x}{a}\,\sinh\frac{n\pi y}{a},\quad B_n = \frac{2}{a\,\sinh(n\pi b/a)}\int_0^a f(\xi)\sin\frac{n\pi\xi}{a}\,d\xi.}$

Common Traps

• The wrong sign of the separation constant ($\lambda < 0$) gives exponentials/hyperbolics in $x$ and sinusoids in $y$, which cannot satisfy $X(0)=X(a)=0$ nontrivially.
• Use $\sinh(n\pi y/a)$, not $\cosh(n\pi y/a)$: $\sinh(0) = 0$ satisfies $Y(0) = 0$, but $\cosh(0) = 1 \ne 0$.
• The denominator of $B_n$ is $\sinh(n\pi b/a)$, not 1. Omitting it loses the final marks.

laplace-polar (1 question; 2018)

Recognition Cues

• “Steady temperature in an annulus $a \le r \le b$” with a trigonometric outer boundary condition.
• The boundary condition contains $\cos(\theta/2)$ or $\sin(\theta/k)$ for non-integer $k$ — a period trap.

Key trap. The general Laplace solution in an annulus is $2\pi$-periodic in $\theta$ (since $T$ must be single-valued). A datum like $\cos(\theta/2)$ has period $4\pi$ — it is not a single $2\pi$-periodic harmonic. It must be expanded in a Fourier series in $\{1, \cos n\theta, \sin n\theta\}$ on $[0, 2\pi]$.

Solution Template

1. General solution: $T = A_0 + B_0\ln r + \sum_{n=1}^\infty (A_n r^n + B_n r^{-n})\cos n\theta + (C_n r^n + D_n r^{-n})\sin n\theta$.
2. Fourier-expand the outer BC on $[0, 2\pi]$. For $K\cos(\theta/2)$, only sine terms survive (it’s odd about $\theta = \pi$): $b_n = \frac{8Kn}{\pi(4n^2-1)}$.
3. Match modes. For each $n$: apply inner BC ($T(a,\theta) = 0$) and outer BC mode coefficients. Solve the $2\times2$ system for $C_n, D_n$ (sine family only).

Worked Example

2018 Paper 2, 2018-P2-Q8c (20 marks)

Annulus $a \le r \le b$; $T(a,\theta) = 0$; $T(b,\theta) = K\cos(\theta/2)$. Find the temperature distribution.

Fourier expansion of $K\cos(\theta/2)$ on $[0, 2\pi]$: all cosine and constant terms vanish (odd symmetry about $\pi$); the sine coefficients are: $b_n = \frac{1}{\pi}\int_0^{2\pi} K\cos\frac{\theta}{2}\sin n\theta\,d\theta = \frac{8Kn}{\pi(4n^2-1)}.$

So $K\cos(\theta/2) = \sum_{n=1}^\infty \dfrac{8Kn}{\pi(4n^2-1)}\sin n\theta$ for $0 < \theta < 2\pi$.

Match modes (only sine family survives). For each $n$, the radial function $R_n(r) = C_n r^n + D_n r^{-n}$ satisfies: $R_n(a) = 0 \;\Rightarrow\; D_n = -C_n a^{2n}. \qquad R_n(b) = \frac{8Kn}{\pi(4n^2-1)}.$

Solving: $\boxed{T(r,\theta) = \sum_{n=1}^\infty \frac{8Kn}{\pi(4n^2-1)}\cdot\frac{(r/a)^n - (a/r)^n}{(b/a)^n - (a/b)^n}\,\sin n\theta.}$

Common Traps

• Treating $\cos(\theta/2)$ as a single Fourier mode and writing $T = C(r^{1/2} + Dr^{-1/2})\cos(\theta/2)$ is wrong: $r^{1/2}$ is not single-valued on the annulus.
• All cosine and constant Fourier coefficients of $\cos(\theta/2)$ on $[0, 2\pi]$ are zero — only sine terms appear.

laplace-fourier-transform (1 question; 2024)

Recognition Cues

• “Show that the solution of Laplace on the upper half-plane with $\phi(x,0) = f(x)$ is given by $\phi(x,y) = \dfrac{y}{\pi}\int_{-\infty}^\infty \dfrac{f(\xi)\,d\xi}{y^2+(x-\xi)^2}$.”
• This is a derivation proof, not a computation. The steps are: Fourier transform in $x$ → ODE in $y$ → apply decay condition → invert (Poisson kernel).

Solution Template

1. Fourier transform in $x$: $\Phi(k,y) = \widehat\phi$. The PDE becomes $\Phi_{yy} - k^2\Phi = 0$.
2. Solve: $\Phi = A(k)e^{-|k|y} + B(k)e^{|k|y}$. Decay condition ($\phi \to 0$ as $y \to \infty$) forces $B = 0$.
3. Apply BC: $\Phi(k,0) = F(k) = \widehat f$, so $\Phi(k,y) = F(k)e^{-|k|y}$.
4. Invert by convolution: $\phi = f * P_y$ where $P_y(x) = \mathcal{F}^{-1}(e^{-|k|y}) = \dfrac{y}{\pi(y^2+x^2)}$.
5. Compute $P_y(x)$: split the integral at $k=0$, evaluate each half as a standard exponential, add.

Worked Example

2024 Paper 2, 2024-P2-Q6a (20 marks)

Show $\phi(x,y) = \dfrac{y}{\pi}\int_{-\infty}^\infty \dfrac{f(\xi)\,d\xi}{y^2+(x-\xi)^2}$.

Steps 1–3 give $\Phi(k,y) = F(k)\,e^{-|k|y}$.

Step 4 (Poisson kernel): $P_y(x) = \frac{1}{2\pi}\int_{-\infty}^\infty e^{-|k|y}e^{ikx}\,dk = \frac{1}{2\pi}\left[\frac{1}{y-ix} + \frac{1}{y+ix}\right] = \frac{1}{2\pi}\cdot\frac{2y}{y^2+x^2} = \frac{y}{\pi(y^2+x^2)}.$

Step 5 (Convolution): $\phi(x,y) = \int_{-\infty}^\infty f(\xi)\,P_y(x-\xi)\,d\xi = \frac{y}{\pi}\int_{-\infty}^\infty \frac{f(\xi)\,d\xi}{y^2+(x-\xi)^2}. \quad\square$

Common Traps

• The decay condition at $y = \infty$ is essential — without it, you cannot choose between $e^{-|k|y}$ and $e^{+|k|y}$.
• The two halves of the Poisson kernel integral ($k < 0$ and $k > 0$) both converge to $\dfrac{1}{y \pm ix}$; adding them gives the $2y$ numerator.
• Convolution theorem direction: $\widehat{f * g} = \hat f \cdot \hat g$, so $\Phi = F \cdot e^{-|k|y}$ inverts to a convolution.

Marks-Aware Writing

10-mark uniqueness: Three steps: difference is harmonic with zero boundary data; apply Green’s identity (write it out); conclude gradient vanishes and boundary forces $u=0$. State both hypotheses in step 3 explicitly ($|\nabla u|^2 \ge 0$ and integral = 0 implies identically 0).

20-mark separation of variables: Write out the full separation (all three steps: ODE for $X$, eigenvalue problem, ODE for $Y$). Justify why $\cosh$ is excluded ($Y(0)\ne0$). Write the superposition. Derive the Fourier coefficient formula with the correct $\sinh$ denominator.

20-mark Fourier transform (2024): Show each step: FT of PDE, ODE solution, decay argument, application of BC, convolution theorem, Poisson kernel computation. Don’t skip the kernel calculation — it’s worth multiple marks.

Practice Set

(No additional practice items in the scaffold — the four worked examples cover all archetypes.)