---
title: "Steady temperature in a semicircular plate"
subject: "Applied Mathematics"
chapter: 3
key: "semester_4/applied_mathematics:laplace-semicircular-plate"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/laplace-semicircular-plate
site: BCT Engineering
---

# Steady temperature in a semicircular plate

Applied Mathematics · Chapter 3: Partial Differential Equations

- Asked in: 2069 Bhadra; 2068 Bhadra (asked 2 times)
- Video: [Steady State Heat: Solving PDEs in a semi-circle (in 8 min)](https://www.youtube.com/watch?v=8d5NZttakf0) — nicefa
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/applied_mathematics/laplace-semicircular-plate
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-3

## Question

The diameter of a semi-circular plate of radius a is kept at 0°C and the temperature at the semi-circular boundary is T°C. Find (show) the steady state temperature in the plate: u(r, θ) = (4T/π) Σ_{n=1}^{∞} [1/(2n-1)] (r/a)^{2n-1} sin(2n-1)θ.

## Short answer

Solve Laplace's equation in polar form with u(r,0)=u(r,π)=0 and u(a,θ)=T. Separation gives u(r,θ) = Σ Aₙ rⁿ sin nθ. Fourier sine series on the arc yields Aₙ aⁿ = 4T/(nπ) for odd n, resulting in u(r,θ) = (4T/π) Σ_{n=1}^∞ (1/(2n-1))(r/a)^{2n-1} sin((2n-1)θ).

## Answer

**Problem.** A semicircular plate of radius a occupies 0 ≤ r ≤ a, 0 ≤ θ ≤ π. The diameter (θ = 0 and θ = π) is at 0°C and the arc r = a is at T°C. Find the steady-state temperature u(r, θ).

**Step 1: Laplace's equation in polar form** (steady state):

∂²u/∂r² + (1/r) ∂u/∂r + (1/r²) ∂²u/∂θ² = 0

**Boundary conditions**

1. u(r, 0) = 0
2. u(r, π) = 0
3. u(a, θ) = T
4. u is finite at r = 0

**Step 2: separate variables.** u = R(r)Θ(θ):

r²R″/R + rR′/R = −Θ″/Θ = λ²

```text
Θ″ + λ² Θ = 0                 →  Θ = A cos λθ + B sin λθ
r² R″ + r R′ − λ² R = 0       →  R = C r^λ + D r^{−λ}
```

**Step 3: apply conditions.**

- Θ(0) = 0 gives A = 0.
- Θ(π) = 0 gives sin λπ = 0, so λ = n (integer).
- Finite at r = 0 gives D = 0 (r⁻ⁿ would blow up).

u(r, θ) = Σ_{n=1}^{∞} Aₙ rⁿ sin nθ

**Step 4: arc condition.** u(a, θ) = T = Σ Aₙ aⁿ sin nθ, a Fourier sine series on (0, π):

```text
Aₙ aⁿ = (2/π) ∫₀^π T sin nθ dθ = (2T/(nπ)) (1 − cos nπ)
      = 4T/(nπ)   for n odd,   0 for n even
```

So u(r, θ) = Σ_{n odd} (4T/(nπ)) (r/a)ⁿ sin nθ. Write each odd n as 2m − 1 and then rename m as n:

**u(r, θ) = (4T/π) Σ_{n=1}^{∞} [1/(2n − 1)] (r/a)^{2n−1} sin(2n − 1)θ**

which is the required result. ∎

(Numerical check at r = 0.6a, θ = 1.1: the series gives 0.65669, equal to the closed form (2T/π) arctan[2(r/a) sin θ/(1 − r²/a²)] with T = 1.)

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Steady-state Laplace in polar form is
   A. u_rr + u_r/r + u_theta theta/r^2 = 0
   B. u_rr + u_theta theta = 0
   C. u_t = u_rr
   D. u_rr - u_r/r = 0
   Answer: A. Standard. (Key point: Equation)

2. R equation r^2R'' + rR' - lambda^2 R = 0 gives
   A. R = C cos + D sin
   B. R = C r^lambda + D r^(-lambda)
   C. R = C e^r
   D. R = C ln r
   Answer: B. Euler equation. (Key point: R solution)

3. Finite at r=0 gives
   A. C = 0
   B. A = 0
   C. D = 0
   D. lambda = 0
   Answer: C. r^-n blows up. (Key point: Finiteness)

4. An a^n =
   A. 2T/(n pi) for all n
   B. 4T/(n pi) for even n
   C. T/n
   D. 4T/(n pi) for odd n, 0 for even n
   Answer: D. (2T/n pi)(1 - cos n pi). (Key point: Coefficients)

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