← 2018 Paper 2

UPSC 2018 Maths Optional Paper 2 Q8c — Step-by-Step Solution

20 marks · Section B

Laplace equation: Dirichlet/Neumann, separation of variables · PDEs · asked 4× in 13 yrs · Read the full method →

Question

A thin annulus occupies the region 0<a≤r≤b, 0≤θ≤2π0<a\le r\le b,\ 0\le\theta\le2\pi. The faces are insulated. Along the inner edge the temperature is maintained at 0°0°, while along the outer edge the temperature is held at T=Kcos⁡θ2T=K\cos\dfrac{\theta}{2}, where KK is a constant. Determine the temperature distribution in the annulus.

Technique

Separation of variables in polar coordinates; expand the non-harmonic datum Kcos⁡(θ/2)K\cos(\theta/2) in a full Fourier (sine) series on [0,2π][0,2\pi]; solve each nn-mode radial ODE with rn,r−nr^n,r^{-n} matched to T(a)=0, T(b)=T(a)=0,\ T(b)= Fourier coefficient.

Solution

Setup. Steady temperature T(r,θ)T(r,\theta) satisfies Laplace’s equation in polar coordinates,

∇2T=∂2T∂r2+1r∂T∂r+1r2∂2T∂θ2=0,\nabla^2T=\frac{\partial^2T}{\partial r^2}+\frac1r\frac{\partial T}{\partial r}+\frac{1}{r^2}\frac{\partial^2T}{\partial\theta^2}=0,

on the annulus a≤r≤ba\le r\le b, with boundary conditions

T(a,θ)=0,T(b,θ)=Kcos⁡θ2.T(a,\theta)=0,\qquad T(b,\theta)=K\cos\frac\theta2.

Periodicity subtlety (the heart of the problem). Physically TT must be single-valued and 2π2\pi-periodic in θ\theta. But the boundary datum cos⁡(θ/2)\cos(\theta/2) has period 4π4\pi, not 2π2\pi — it is not a single Fourier harmonic of the annulus. (Indeed cos⁡(θ/2)\cos(\theta/2) is discontinuous when wrapped to period 2π2\pi: it equals +1+1 at θ=0+\theta=0^+ and −1-1 at θ=2π−\theta=2\pi^-.) So we cannot match it to a single cos⁡nθ\cos n\theta. Instead we must expand cos⁡(θ/2)\cos(\theta/2) as a Fourier series in the admissible 2π2\pi-periodic harmonics {1,cos⁡nθ,sin⁡nθ}\{1,\cos n\theta,\sin n\theta\} on [0,2π][0,2\pi] and solve mode by mode.

Step 1 — Separated (admissible) solutions of Laplace’s equation

Single-valued 2π2\pi-periodic separated solutions are, for integer n≥1n\ge1,

1, ln⁡r,(rn, r−n){cos⁡nθ,sin⁡nθ}.1,\ \ln r,\qquad (r^n,\ r^{-n})\{\cos n\theta,\sin n\theta\}.

So

T(r,θ)=A0+B0ln⁡r+∑n=1∞(Anrn+Bnr−n)cos⁡nθ+∑n=1∞(Cnrn+Dnr−n)sin⁡nθ.T(r,\theta)=A_0+B_0\ln r+\sum_{n=1}^\infty\big(A_n r^n+B_n r^{-n}\big)\cos n\theta+\sum_{n=1}^\infty\big(C_n r^n+D_n r^{-n}\big)\sin n\theta.

Step 2 — Fourier expansion of the boundary datum on [0,2π][0,2\pi]

Expand f(θ)=Kcos⁡(θ/2)f(\theta)=K\cos(\theta/2) on 0≤θ≤2π0\le\theta\le2\pi (period 2π2\pi):

a0=12π∫02πKcos⁡θ2 dθ=K2π[2sin⁡θ2]02π=K2π (2⋅0−0)=0,a_0=\frac{1}{2\pi}\int_0^{2\pi}K\cos\frac\theta2\,d\theta=\frac{K}{2\pi}\Big[2\sin\frac\theta2\Big]_0^{2\pi}=\frac{K}{2\pi}\,(2\cdot0-0)=0, an=1π∫02πKcos⁡θ2cos⁡nθ dθ=0(n≥1),a_n=\frac1\pi\int_0^{2\pi}K\cos\frac\theta2\cos n\theta\,d\theta=0\quad(n\ge1), bn=1π∫02πKcos⁡θ2sin⁡nθ dθ=8Knπ(4n2−1).b_n=\frac1\pi\int_0^{2\pi}K\cos\frac\theta2\sin n\theta\,d\theta=\frac{8Kn}{\pi(4n^2-1)}.

(The cosine and constant coefficients vanish because cos⁡(θ/2)\cos(\theta/2) is odd about θ=π\theta=\pi on [0,2π][0,2\pi].) Thus

Kcos⁡θ2=∑n=1∞8Knπ(4n2−1)sin⁡nθ,0<θ<2π.K\cos\frac\theta2=\sum_{n=1}^\infty\frac{8Kn}{\pi(4n^2-1)}\sin n\theta,\qquad 0<\theta<2\pi.

Step 3 — Match modes; only sine terms survive

Since the boundary data are pure sines, set A0=B0=0A_0=B_0=0 and all An=Bn=0A_n=B_n=0; keep only the sin⁡nθ\sin n\theta family Rn(r)=Cnrn+Dnr−nR_n(r)=C_nr^n+D_nr^{-n}.

Inner edge T(a,θ)=0T(a,\theta)=0: for every nn, Cnan+Dna−n=0C_na^n+D_na^{-n}=0. Outer edge T(b,θ)=Kcos⁡(θ/2)T(b,\theta)=K\cos(\theta/2): for every nn, Cnbn+Dnb−n=8Knπ(4n2−1)C_nb^n+D_nb^{-n}=\dfrac{8Kn}{\pi(4n^2-1)}.

Solving this 2×22\times2 system, the radial factor simplifies to

Rn(r)=8Knπ(4n2−1)⋅(r/a)n−(a/r)n(b/a)n−(a/b)n.R_n(r)=\frac{8Kn}{\pi(4n^2-1)}\cdot\frac{(r/a)^n-(a/r)^n}{(b/a)^n-(a/b)^n}.

(Equivalently Rn(r)=8Knπ(4n2−1)⋅bn (r2n−a2n)rn (b2n−a2n)R_n(r)=\dfrac{8Kn}{\pi(4n^2-1)}\cdot\dfrac{b^n\,(r^{2n}-a^{2n})}{r^n\,(b^{2n}-a^{2n})}.) This satisfies Rn(a)=0R_n(a)=0 and Rn(b)=8Knπ(4n2−1)R_n(b)=\dfrac{8Kn}{\pi(4n^2-1)}.

Step 4 — Temperature distribution

Answer

  T(r,θ)=∑n=1∞8Knπ(4n2−1) (r/a)n−(a/r)n(b/a)n−(a/b)n sin⁡nθ.  \boxed{\;T(r,\theta)=\sum_{n=1}^\infty\frac{8Kn}{\pi(4n^2-1)}\,\frac{(r/a)^n-(a/r)^n}{(b/a)^n-(a/b)^n}\,\sin n\theta.\;}
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