Solve Completely to the final form of the 200 Solution as a Fourier Series the heat equation boundary value problem. (homogeneous PDĚ, Dirichlet, but constant B.C.) X€ [0₂L] L= 4, k = 3 +₁²0A B.C. IC. O 2u 24 1 non-homogeneous sol sy/ad k 2²u = 0 2x² : 10 xut food go wol u(0₁t) = 5 $ vot stand on : You 5 die u(L₁Z) = -5 u(x₂0.) = x DR

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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Solve Completely to the final form of the 200
Solution as a Fourier Series the heat
equation boundary value problem. A
(homogeneous PDĚ, Dirichlet, non-homogeneous
but constant B.C.)
xe LOL] L: 4, kẸ 3 t 201
R 2²u = 0
B.C.
IC
2u
24
2x² 10 xut food
go
u(0₁t) = 5 $ roz sand on : You alia
u(L₁Z) = -5
A
A
u(x₂0) = x
9
wole
9
Transcribed Image Text:Solve Completely to the final form of the 200 Solution as a Fourier Series the heat equation boundary value problem. A (homogeneous PDĚ, Dirichlet, non-homogeneous but constant B.C.) xe LOL] L: 4, kẸ 3 t 201 R 2²u = 0 B.C. IC 2u 24 2x² 10 xut food go u(0₁t) = 5 $ roz sand on : You alia u(L₁Z) = -5 A A u(x₂0) = x 9 wole 9
Expert Solution
Step 1: Heat equation solve

Given Heat equation is the following :

fraction numerator partial differential u over denominator partial differential t end fraction equals 3 fraction numerator partial differential squared u over denominator partial differential x squared end fraction comma 0 less or equal than x less or equal than 4 comma t greater than 0

u left parenthesis 0 comma t right parenthesis equals 5 comma u left parenthesis 4 comma t right parenthesis equals negative 5 comma t greater than 0 space a n d space

u left parenthesis x comma 0 right parenthesis equals x comma 0 less or equal than x less or equal than 4

Let, v left parenthesis x comma t right parenthesis equals u left parenthesis x comma t right parenthesis minus open curly brackets 5 plus x over 4 open parentheses negative 5 minus 5 close parentheses close curly brackets equals u left parenthesis x comma t right parenthesis plus 5 over 2 x minus 5

so we get v subscript t equals u subscript t space a n d space v subscript x x end subscript equals u subscript x x end subscript

By substitution ,we get v subscript t equals 3 v subscript x x end subscript

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