Question

This is a question about Partial differential equation - Heat equation. Please help solving part (a) and show clear explanations. Thanks!

=K х 7. The temperature T(2,t) in an insulated rod of length L and diffusivity k is given by the heat equation ОТ 22T 0 < x < L. at Əx2' Initially this rod is at constant temperature To, and immediately after t=0 the temperature at x = L is suddenly increased to T1. The temperature at x = 0 is kept at To. (a) Let T(x,t) = To + (T1 – To) ( + u(x,t)) L Show that u(x, t) also satisfies the heat equation, and that u(0,t) = u(L,t) = 0. Given that T(x,0) To what is the initial condition for u(x,0)? [4 marks] (b) Use the method of separation of variables to derive the general solution u(x, t) from (a) which satisfies the given boundary conditions. You are required to analyze all cases of the separation constant, i.e., I = 0,1> 0 and < 0. All the steps must be shown in detail. (18 marks] (c) For the solution derived in (b) write down expressions for the Fourier coefficients in terms of an arbitrary initial condition u(x,0) = u(x). [2 marks]

Answer 1

Soln T(2,4) = To +(Ti-to)( + 4(xst)) T(x, t) satisfy heat equation भ ชา 2 ( to Hi-T ulx,t) =K (T6 +(T+1)(*44(8,9) K at Ox? a2 + 5x² at = 0 + TT, -T).0 + 2(x, t) K.0 + +T-To):vorkozy(2,4) ot 04(x, t) at au (x,t) = Қ 2 Ox = To x=0 it's satisfy heat equation. T(x, t) = to 7(7.-T.) ( + 4(e,t)) as given Tlo,t) to = Hat(Ti-To) (+4(0,t)) 0 = (1,7o) 4(0,t) =) u 10,t) = x=L T(5,t) = T, Ti = to +(T. -To) u(4,4)) 1+ ull,t) x =X+u(4,t) u(4,t) u(at) = ull, t) = 0 +(x,o) To = To +(-)( + ulæst)) 4(x,0) = -2 s ) - Ti-To T-to -) =0 =) = To tso s )

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