A thin rod of length has its two ends held at

PHYSICS · Multi-part · Fermi University Physics · Mathematical Methods for Physics · Numerical and computational mathematics · Numerical and computational mathematics

Error, Chance and the Machine, Ch Linear systems, least squares and the singular value decomposition

Problem

A thin rod of length 5 Δx5\,\Delta x has its two ends held at 0 °C0\,\text{°C}. One implicit time step of the diffusion equation with μ=D Δt/Δx2=1.5\mu = D\,\Delta t/\Delta x^2 = 1.5 gives, for the four interior nodal temperatures T1,T2,T3,T4T_1, T_2, T_3, T_4 (in °C\text{°C}), the tridiagonal system −1.5 Ti−1+4Ti−1.5 Ti+1=Tiold,i=1,…,4,-1.5\,T_{i-1} + 4T_i - 1.5\,T_{i+1} = T_i^{\rm old}, \qquad i = 1,\dots,4, with T0=T5=0T_0 = T_5 = 0 and old (start-of-step) temperatures T1old=100T_1^{\rm old} = 100, T2old=80T_2^{\rm old} = 80, T3old=50T_3^{\rm old} = 50, T4old=20T_4^{\rm old} = 20 (all in °C\text{°C}). ((a)) Write down the coefficient matrix and check whether it is diagonally dominant, row by row. ((b)) Use the tridiagonal (Thomas) elimination, which needs no pivoting, to solve for T1,T2,T3,T4T_1, T_2, T_3, T_4 to three significant figures.

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