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4. Consider the following 1D heat equation: U = a 00 t where a is a real positive constant, subject to the initial condition: 1 0 0 is a good one. (b) The resulting ODE system for X(x) is given by: d²X +X²X = 0, dr? X'(0) = 0, X'(L) = 0 solve this and obtain: X (r) = B cos n = 0,1,2,3, . %3D with B some constant. (c) The T solution is given by T(t) = Ae-ana*t/L*, for some constant A. Show that the initial condition (5) reduces down to: %3D L/2 COS Cos COS dr n=0 where D, is a constant which depends on n.

4. Consider the following 1D heat equation: U = a 00 t where a is a real positive constant, subject to the initial condition: 1 0 0 is a good one. (b) The resulting ODE system for X(x) is given by: d²X +X²X = 0, dr? X'(0) = 0, X'(L) = 0 solve this and obtain: X (r) = B cos n = 0,1,2,3, . %3D with B some constant. (c) The T solution is given by T(t) = Ae-ana*t/L*, for some constant A. Show that the initial condition (5) reduces down to: %3D L/2 COS Cos COS dr n=0 where D, is a constant which depends on n.

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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part C and D solutio needed urgenty 

 

4. Consider the following 1D heat equation: = a 0<r < L,t > 0 at where a is a real positive constant, subject to the initial condition: 1 0<r< L/2 U(1,0) = {0 L/2 <SL and boundary conditions: (0,t) = 0, (L,t) = 0. (a) Use the method of separation of variables: U(r, t) = X(x)T(t) and show that the heat equation (4) can be reduced to the following equation: %3D 1 dT 1 d²X aT dt X da? where 3 is a real constant. Explain why these ODES must be equal to a constant and why (in this case) this constant should be purely negative so that a choice of 3 = -X, for A > 0 is a good one. (b) The resulting ODE system for X(x) is given by: d²X +X²X = 0, dr2 x'(0) = 0, X'(L) = 0 solve this and obtain: X (r) = B cos n = 0,1, 2,3, . %3D with B some constant. (c) The T solution is given by T(t) = Ae-an*a*t/L*, for some constant A. Show that the initial condition (5) reduces down to: L/2 COS Cos COS da n=0 where D, is a constant which depends on n. (d) Using the result in Part (c) and the fact that: { n+ m L/2 n= m+0 n = m = 0 Cos cos dx = and sin(r) lim 1, show that the final solution U(r, t) of the problem is U (r,t): sin cos n=1
Transcribed Image Text:4. Consider the following 1D heat equation: = a 0<r < L,t > 0 at where a is a real positive constant, subject to the initial condition: 1 0<r< L/2 U(1,0) = {0 L/2 <SL and boundary conditions: (0,t) = 0, (L,t) = 0. (a) Use the method of separation of variables: U(r, t) = X(x)T(t) and show that the heat equation (4) can be reduced to the following equation: %3D 1 dT 1 d²X aT dt X da? where 3 is a real constant. Explain why these ODES must be equal to a constant and why (in this case) this constant should be purely negative so that a choice of 3 = -X, for A > 0 is a good one. (b) The resulting ODE system for X(x) is given by: d²X +X²X = 0, dr2 x'(0) = 0, X'(L) = 0 solve this and obtain: X (r) = B cos n = 0,1, 2,3, . %3D with B some constant. (c) The T solution is given by T(t) = Ae-an*a*t/L*, for some constant A. Show that the initial condition (5) reduces down to: L/2 COS Cos COS da n=0 where D, is a constant which depends on n. (d) Using the result in Part (c) and the fact that: { n+ m L/2 n= m+0 n = m = 0 Cos cos dx = and sin(r) lim 1, show that the final solution U(r, t) of the problem is U (r,t): sin cos n=1
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