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the resulting zero-ends (homogeneous) heat boundary value problem c² a²u (10) 0 < x 0, Ət (11) u(0,t) = 0_and_u(L,t)=0, t>0, u(x, 0) = f(x) — u₁(x), 0 < x < L. (12) Let u₂(x, t) be the solution of (10)-(12). According to (4) and (5), we have (13) where An ди = (15) cn and L' - Əx², u₂(x, t): = 2 bn = f(f(x) - ∞ n=1 bne -X²/2t u1 (r) T2 - T₁ L NTT sin -X, L (14) Now the solution of (6)–(8) is obtained by adding to u₂(x, t) the steady-state solution u₁(x) as follows: -x+ T₁)) sin x dx. u(x, t) = u₁(x) + u₂(x, t).

the resulting zero-ends (homogeneous) heat boundary value problem c² a²u (10) 0 < x 0, Ət (11) u(0,t) = 0_and_u(L,t)=0, t>0, u(x, 0) = f(x) — u₁(x), 0 < x < L. (12) Let u₂(x, t) be the solution of (10)-(12). According to (4) and (5), we have (13) where An ди = (15) cn and L' - Əx², u₂(x, t): = 2 bn = f(f(x) - ∞ n=1 bne -X²/2t u1 (r) T2 - T₁ L NTT sin -X, L (14) Now the solution of (6)–(8) is obtained by adding to u₂(x, t) the steady-state solution u₁(x) as follows: -x+ T₁)) sin x dx. u(x, t) = u₁(x) + u₂(x, t).

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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please also show solution satisfies boundry conditions, intial conditions, and the PDE. Thanks!

the resulting zero-ends (homogeneous) heat boundary value problem = c² 0²u Əx², (10) du Ət 0 < x < L, t> 0, (11) u(0, t) = 0 and u(L, t) = 0, t> 0, (12) u(x, 0) = f(x) — u₁(x), 0<x< L. Let u₂(x, t) be the solution of (10)-(12). According to (4) and (5), we have (13) where An = cn and L' u₂(x, t) = n=1 bne -X²2/t u₁(x) Nπ sin -X, L T₂ - T₁ bn = ²/ fő (f(x) - ( -x+ T₁)) sin x dx. L (14) Now the solution of (6)-(8) is obtained by adding to u₂(x, t) the steady-state solution u₁(x) as follows: (15) u(x, t) = u₁(x) + u₂(x, t).
Transcribed Image Text:the resulting zero-ends (homogeneous) heat boundary value problem = c² 0²u Əx², (10) du Ət 0 < x < L, t> 0, (11) u(0, t) = 0 and u(L, t) = 0, t> 0, (12) u(x, 0) = f(x) — u₁(x), 0<x< L. Let u₂(x, t) be the solution of (10)-(12). According to (4) and (5), we have (13) where An = cn and L' u₂(x, t) = n=1 bne -X²2/t u₁(x) Nπ sin -X, L T₂ - T₁ bn = ²/ fő (f(x) - ( -x+ T₁)) sin x dx. L (14) Now the solution of (6)-(8) is obtained by adding to u₂(x, t) the steady-state solution u₁(x) as follows: (15) u(x, t) = u₁(x) + u₂(x, t).
In Exercises 11-14, solve the nonhomogeneous boundary value problem (6)–(8) for the given data. 11. u(0, t) = 100, u(1, t) = 0, ƒ(x) = 30 sin(πx), L = 1, c = 1.
Transcribed Image Text:In Exercises 11-14, solve the nonhomogeneous boundary value problem (6)–(8) for the given data. 11. u(0, t) = 100, u(1, t) = 0, ƒ(x) = 30 sin(πx), L = 1, c = 1.
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