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find u (x, t) for 0 < x < 1 and t> 0 which solves Ut - Uxx = et, u (x,0) = 1, u (1,t) = u (0,t) = 0.

find u (x, t) for 0 < x < 1 and t> 0 which solves Ut - Uxx = et, u (x,0) = 1, u (1,t) = u (0,t) = 0.

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter3: Functions And Graphs
Section3.5: Graphs Of Functions
Problem 58E
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[Second Order Equations] How do you solve 2? Use the expansion method in the second picture

1. 2. find u (x, t) for 0 < x < 1 and t> 0 which solves UtUxx = 0, u (x,0) = 1, ux (1,t) = t, find u (x, t) for 0 < x < 1 and t > 0 which solves Ut - Uxx = et, u (x,0) = 1, ux (0, t) = 0. u (1, t) = u (0,t) = 0.
Transcribed Image Text:1. 2. find u (x, t) for 0 < x < 1 and t> 0 which solves UtUxx = 0, u (x,0) = 1, ux (1,t) = t, find u (x, t) for 0 < x < 1 and t > 0 which solves Ut - Uxx = et, u (x,0) = 1, ux (0, t) = 0. u (1, t) = u (0,t) = 0.
EXPANSION METHOD We already know that for the corresponding homogeneous problem the correct expansion is the Fourier sine series. For each t, we certainly can expand u(x, t) = Σun(t) sin n=1 2 ²7 [u for some coefficients un(t), because the completeness theorems guarantee that any function in (0, 1) can be so expanded. The coefficients are necessarily given by un(t) = u(x, t) sin ηπ 1 0 = u₁ - kuxx = Σ ηπχ 1 dt dx. (3) You may object that each term in the series vanishes at both endpoints and thereby violates the boundary conditions. The answer is that we simply do not insist that the series converge at the endpoints but only inside the interval. In fact, we are exactly in the situation of Theorems 3 and 4 but not of Theorem 2 of Section 5.4. Now differentiating the series (2) term by term, we get =Σ [dun+ku,((77)²] sin (2) ηπ 1
Transcribed Image Text:EXPANSION METHOD We already know that for the corresponding homogeneous problem the correct expansion is the Fourier sine series. For each t, we certainly can expand u(x, t) = Σun(t) sin n=1 2 ²7 [u for some coefficients un(t), because the completeness theorems guarantee that any function in (0, 1) can be so expanded. The coefficients are necessarily given by un(t) = u(x, t) sin ηπ 1 0 = u₁ - kuxx = Σ ηπχ 1 dt dx. (3) You may object that each term in the series vanishes at both endpoints and thereby violates the boundary conditions. The answer is that we simply do not insist that the series converge at the endpoints but only inside the interval. In fact, we are exactly in the situation of Theorems 3 and 4 but not of Theorem 2 of Section 5.4. Now differentiating the series (2) term by term, we get =Σ [dun+ku,((77)²] sin (2) ηπ 1
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