Explanation Given: The boundary value problem is ∂ 2 u ∂ r 2 + 2 r ∂ u ∂ r = ∂ u ∂ t , 0 < r < 1 , t > 0 and the boundary conditions are u ( 1 , t ) = 100 and u ( r , 0 ) = 0 . Calculation: The given boundary value problem is, ∂ 2 u ∂ r 2 + 2 r ∂ u ∂ r = ∂ u ∂ t (1) Firstly verify the left hand side of the given boundary value problem is equals to given identity, 1 r ∂ 2 ∂ r 2 ( r u ) = 1 r ∂ ∂ r [ r ∂ u ∂ r + u ] = 1 r [ r ∂ 2 u ∂ r 2 + ∂ u ∂ r + ∂ u ∂ r ] = ∂ 2 u ∂ r 2 + 2 r ∂ u ∂ r Therefore, the above equation can be written as, 1 r ∂ 2 ∂ r 2 ( r u ) = ∂ u ∂ t Consider the equation, r u ( r , t ) = v ( r , t ) + ψ ( r ) (2) Partially differentiate both sides of the equation (2) with respect to r , r ∂ u ∂ r + u ( r , t ) = ∂ v ∂ r + ψ ′ ( r ) Partially differentiate both sides of the above equation with respect to r , r ∂ 2 u ∂ r 2 + 2 ∂ u ∂ r = ∂ 2 v ∂ r 2 + ψ ″ ( r ) Substitute r ∂ 2 u ∂ r 2 + 2 ∂ u ∂ r = ∂ 2 v ∂ r 2 + ψ ″ ( r ) and ∂ u ∂ t = ∂ v ∂ t in equation (1), ∂ 2 v ∂ r 2 + ψ ″ ( r ) = ∂ v ∂ t (3) Second derivative of ψ is zero in the equation (3) by homogeneous system, ψ ″ ( r ) = 0 Integrate the above equation, ψ ′ ( r ) = c 1 Integrate the above equation, ψ ( r ) = c 1 r + c 2 (4) At boundary conditions r = 0 and u ( 1 , t ) = 100 , c 1 = 100 and c 2 = 0 Substitute the value of c 1 and c 2 in equation (4), ψ ( r ) = 100 r (5) Take the equation (1) in the form of v , ∂ 2 v ∂ r 2 = ∂ v ∂ t In form of v the boundary conditions are v ( 1 , t ) = 0 , lim r → 0 1 r v ( r , t )

6th Edition

ISBN: 9781284127003

Author: ZILL

Publisher: JONES+BARTLETT LEARNING,LLC (CC)

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Chapter 14.3, Problem 9E

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Ch. 14.1 - Prob. 1E Ch. 14.1 - Prob. 2E Ch. 14.1 - Prob. 3E Ch. 14.1 - Prob. 4E Ch. 14.1 - Prob. 5E Ch. 14.1 - Prob. 6E Ch. 14.1 - Prob. 7E Ch. 14.1 - Prob. 8E Ch. 14.1 - Prob. 10E Ch. 14.1 - Prob. 11E

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