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Given: Pu The given boundary value problem is a² и (0, г) %3 dx2 0, t> 0, и (х, 0) —D f (x) and ди dt It=0 0, х Calculation: The given boundary value problem is, a² dx2 dt2 Take Fourier transform of both sides of the abo a²F { #u } = F{ #u} dx² dt2 d²U -a?a? U (a, t) + a² au (0, t) dt2 -a?a?U (a, t) = U (α, d²U dt2 Therefore, the equation is, d²U + a²a?U (a, t) = 0 dt2

Given: Pu The given boundary value problem is a² и (0, г) %3 dx2 0, t> 0, и (х, 0) —D f (x) and ди dt It=0 0, х Calculation: The given boundary value problem is, a² dx2 dt2 Take Fourier transform of both sides of the abo a²F { #u } = F{ #u} dx² dt2 d²U -a?a? U (a, t) + a² au (0, t) dt2 -a?a?U (a, t) = U (α, d²U dt2 Therefore, the equation is, d²U + a²a?U (a, t) = 0 dt2

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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I don't understand why the Fourier transform is -a^2a^2U(a,t)+a^2aU(0,t). Where did aU(0,t) come from? Please let me know. Thank you 

3:19 PM Mon May 17 * 42% A bartleby.com Given: The given boundary value problem is a² u u (0, t) = 0, t > 0, u (x, 0) = f (x) and 0 dx2 ди = 0, x > dt Calculation: The given boundary value problem is, q² ®u dx2 dt2 Take Fourier transform of both sides of the abov { } -r {) {2} Pu Pu a²F dx2 = F dt2 -a?a?U (a, t) + a²au (0, t) = dU dt2 d?U -a?a?U (a, t) = dt2 Therefore, the equation is, d²U + a²a²U (a, t) = 0 dt2
Transcribed Image Text:3:19 PM Mon May 17 * 42% A bartleby.com Given: The given boundary value problem is a² u u (0, t) = 0, t > 0, u (x, 0) = f (x) and 0 dx2 ди = 0, x > dt Calculation: The given boundary value problem is, q² ®u dx2 dt2 Take Fourier transform of both sides of the abov { } -r {) {2} Pu Pu a²F dx2 = F dt2 -a?a?U (a, t) + a²au (0, t) = dU dt2 d?U -a?a?U (a, t) = dt2 Therefore, the equation is, d²U + a²a²U (a, t) = 0 dt2
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