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5. A problem with odd harmonics only. Show that the solution of the heat equation (1), subject to the boundary conditions u(0, t) = 0 and ux(L, t) = 0, and the initial condition u(x, 0) = f(x), is where ㅠ u(x, t) = Σ Bn sin [77 (2n + 1)x]e¯[cz(²n+1)]²t -2L n=0 Bn - 2 L S 0 f(x) sin [27 (2n +1)x] dx. -2L

5. A problem with odd harmonics only. Show that the solution of the heat equation (1), subject to the boundary conditions u(0, t) = 0 and ux(L, t) = 0, and the initial condition u(x, 0) = f(x), is where ㅠ u(x, t) = Σ Bn sin [77 (2n + 1)x]e¯[cz(²n+1)]²t -2L n=0 Bn - 2 L S 0 f(x) sin [27 (2n +1)x] dx. -2L

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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5. A problem with odd harmonics only. Show that the solution of the heat equation (1), subject to the boundary conditions u(0, t) = 0 and ux(L, t) = 0, and the initial condition u(x, 0) = f(x), is where ㅠ = Σ B, sia [27 2L n=0 u(x, t) => B₁ sin Bn = 2 ²/² (2n +1)x]e¯ [cz (²n+1)] ²t f(x) sin [277 (2n + 1)x] dx. 2L
Transcribed Image Text:5. A problem with odd harmonics only. Show that the solution of the heat equation (1), subject to the boundary conditions u(0, t) = 0 and ux(L, t) = 0, and the initial condition u(x, 0) = f(x), is where ㅠ = Σ B, sia [27 2L n=0 u(x, t) => B₁ sin Bn = 2 ²/² (2n +1)x]e¯ [cz (²n+1)] ²t f(x) sin [277 (2n + 1)x] dx. 2L
The problem of heat transfer in a bar of length L with initial heat distribution f(x) and no heat loss at either end (Figure 1) is modeled by the heat equation (1) du Ət = C² J²u U Əx²¹ 0 < x < L, t>0,
Transcribed Image Text:The problem of heat transfer in a bar of length L with initial heat distribution f(x) and no heat loss at either end (Figure 1) is modeled by the heat equation (1) du Ət = C² J²u U Əx²¹ 0 < x < L, t>0,
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