We will solve the heat equation=2ur2 Usc0 0with boundary/initial conditions:(0, t) = 0,2(6, t) = 0,2, 0 Find Eigenfunctions for ).The problem splits into cases based on the sign of A(Notation: For the cases below, use constants a and b)• Case 1: A = 0X(x)Plugging the boundary values into this formula gives0 = X(0)%3D0 = X(6)So X(x) =which means u(x, t) =We can ignore this case.• Case 2: A =-y² <0(In your answers below use gamma instead of lambda)X(x) =Plugging the boundary values into this formula gives0 – X(0)0 = X(6) =So X(x) =which means u(x, t) =We can ingore this case.• Case 3: A= >X(x) =(In your answers below use gamma instead of lambda)Plugging in the boundary values into this formula gives0 = X(0) =%3D0 = X(6)Which leads us to the eigenvalues A, = Yn where Yn =and eigenfunctions X,(z) =(Notation: Eigenfunctions should not include any constants a or b.)

Given that, ut=2uxx, 0<x<6, t≥0⋯⋯(1) BC: Ux(0,t)=0, ux6,t=0 IC: ux,0=2, 0<x≤30,…

Answered: u = 2 uzz) 0 < < 6, t >0 with…

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u = 2 uzz) 0 < < 6, t >0 with boundary/initial conditions: u(0, t) = 0, u(6, t) = 0, S 2, 0

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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Question

We will solve the heat equation
=2ur
2 Usc
0<r<6, t> 0
with boundary/initial conditions:
(0, t) = 0,
2(6, t) = 0,
2, 0<x<3
and u(x, 0) = {
0, 3<x < 6
This models temperature in a thin rod of length L = 6 with thermal diffusivity a- 2 where the temperature at the ends is fixed at 0 and the initial temperature
distribution is u(x, 0).
For extra practice we will solve this problem from scratch.

Find Eigenfunctions for ).
The problem splits into cases based on the sign of A
(Notation: For the cases below, use constants a and b)
• Case 1: A = 0
X(x)
Plugging the boundary values into this formula gives
0 = X(0)
%3D
0 = X(6)
So X(x) =
which means u(x, t) =
We can ignore this case.
• Case 2: A =-y² <0
(In your answers below use gamma instead of lambda)
X(x) =
Plugging the boundary values into this formula gives
0 – X(0)
0 = X(6) =
So X(x) =
which means u(x, t) =
We can ingore this case.
• Case 3: A= >
X(x) =
(In your answers below use gamma instead of lambda)
Plugging in the boundary values into this formula gives
0 = X(0) =
%3D
0 = X(6)
Which leads us to the eigenvalues A, = Yn where Yn =
and eigenfunctions X,(z) =
(Notation: Eigenfunctions should not include any constants a or b.)

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