A metal rod of length has both ends held at zero excess temperature

PHYSICS · Subjective · Fermi University Physics · Mathematical Methods for Physics · Linear algebra and vector spaces · Function spaces and operator theory

Mathematical Methods for Physics, Ch Vector spaces, bases, linear maps and dual vectors

Problem

A metal rod of length L=0.80 mL = 0.80\,\mathrm{m} has both ends held at zero excess temperature. At time t=0t=0 its middle half has been heated, and its excess-temperature profile is idealised as the step f(x)={T0,L/4≤x≤3L/40,elsewhere on [0,L],T0=10.0 K,f(x) = \begin{cases} T_0, & L/4 \le x \le 3L/4 \\ 0, & \text{elsewhere on } [0,L] \end{cases}, \qquad T_0 = 10.0\,\mathrm{K}, which is zero at both ends. The rod's natural shapes are sin⁡n(x)≡sin⁡(nπx/L)\sin_n(x) \equiv \sin(n\pi x/L) for positive integers nn, orthogonal on [0,L][0,L] with ⟨sin⁡n,sin⁡n⟩=L/2\langle\sin_n,\sin_n\rangle = L/2 for every nn. Take the share of ff along each shape from the projection formula, bn=⟨sin⁡n,f⟩/⟨sin⁡n,sin⁡n⟩b_n = \langle\sin_n, f\rangle/\langle\sin_n,\sin_n\rangle. (a) Show that the remainder r=f−b1sin⁡1−b2sin⁡2r = f - b_1\sin_1 - b_2\sin_2 is orthogonal to both sin⁡1\sin_1 and sin⁡2\sin_2, and deduce that (b12+b22)(L/2)≤∥f∥2(b_1^2+b_2^2)(L/2) \le \|f\|^2, where ∥f∥2=∫0Lf(x)2 dx\|f\|^2 = \int_0^L f(x)^2\,\mathrm{d}x. (b) Before integrating, decide whether b2b_2 can be non-zero for this profile, and give the reason. Then evaluate b1b_1 and b2b_2 to three significant figures. (c) Compute (b12+b22)(L/2)(b_1^2+b_2^2)(L/2) and ∥f∥2\|f\|^2, and state what fraction of ∥f∥2\|f\|^2 the first two shares carry.

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