← 2016 Paper 2

UPSC 2016 Maths Optional Paper 2 Q8a — Step-by-Step Solution

20 marks · Section B

Heat equation · PDEs · asked 3× in 13 yrs · Read the full method →

Question

Find the temperature u(x,t)u(x,t) in a bar of silver of length 10 cm and constant cross-section of area 1 cm². Let density ρ=10.6\rho=10.6 g/cm³, thermal conductivity K=1.04K=1.04 cal/(cm sec °C) and specific heat σ=0.056\sigma=0.056 cal/g °C. The bar is perfectly isolated laterally, with ends kept at 0°C and initial temperature f(x)=sin(0.1πx)f(x)=\sin(0.1\pi x) °C. Note that u(x,t)u(x,t) follows the heat equation ut=c2uxxu_t=c^2 u_{xx}, where c2=K/(ρσ)c^2=K/(\rho\sigma).

Technique

Separation of variables for the heat equation with homogeneous Dirichlet ends; eigenfunctions sin(nπx/10)\sin(n\pi x/10) with decay ec2(nπ/10)2te^{-c^2(n\pi/10)^2 t}; the initial condition is the n=1n=1 eigenfunction, so only that single mode survives — no Fourier coefficients to integrate.

Solution

Setup. One-dimensional heat equation on 0x100\le x\le 10:

ut=c2uxx,u(0,t)=u(10,t)=0,u(x,0)=sin(0.1πx),u_t=c^2u_{xx},\qquad u(0,t)=u(10,t)=0,\qquad u(x,0)=\sin(0.1\pi x),

with c2=Kρσc^2=\dfrac{K}{\rho\sigma}.

Step 1 — Diffusivity

c2=Kρσ=1.0410.6×0.056=1.040.5936=1.7520 cm2/s.c^2=\frac{K}{\rho\sigma}=\frac{1.04}{10.6\times0.056}=\frac{1.04}{0.5936}=1.7520\ \text{cm}^2/\text{s}.

Step 2 — Separation of variables (general)

Seeking u=X(x)T(t)u=X(x)T(t) with u(0)=u(L)=0u(0)=u(L)=0, L=10L=10, gives the eigenfunctions and modes

Xn(x)=sinnπxL=sinnπx10,Tn(t)=ec2(nπ/L)2t,X_n(x)=\sin\frac{n\pi x}{L}=\sin\frac{n\pi x}{10},\qquad T_n(t)=e^{-c^2(n\pi/L)^2 t},

so the general solution is the Fourier sine series

u(x,t)=n=1Bnsinnπx10ec2(nπ/10)2t.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 the initial condition

The initial profile is f(x)=sin(0.1πx)=sinπx10f(x)=\sin(0.1\pi x)=\sin\dfrac{\pi x}{10} — exactly the first eigenfunction (n=1n=1). Hence B1=1B_1=1 and all other Bn=0B_n=0; no series summation is needed.

Step 4 — Temperature

With n=1n=1, the decay rate is

c2(π10)2=1.7520×π2100=1.7520×0.098696=0.17292 s1.c^2\Big(\frac{\pi}{10}\Big)^2=1.7520\times\frac{\pi^2}{100}=1.7520\times0.098696=0.17292\ \text{s}^{-1}.

Answer

  u(x,t)=sin ⁣(πx10)e0.1729t=sin(0.1πx)ec2(π/10)2t,c2=1.752 cm2/s.  \boxed{\;u(x,t)=\sin\!\Big(\frac{\pi x}{10}\Big)\,e^{-0.1729\,t} =\sin(0.1\pi x)\,e^{-\,c^2(\pi/10)^2\,t},\quad c^2=1.752\ \text{cm}^2/\text{s}.\;}
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