Explanation Given: The boundary conditions are as follows: ∂ 2 u ∂ r 2 + 1 r ∂ u ∂ r + ∂ 2 u ∂ z 2 = 0 0 < r < c , 0 < z < L u ( c , z ) = h ( z ) u ( r , L ) = f ( r ) u ( r , 0 ) = g ( r ) Calculation: Consider the steady state temperature of a finite cylinder as u ( r , z ) . As per the superposition principle of solution of homogenous partial differential equation the steady state temperature u ( r , z ) is expressed as follows: u ( r , z ) = u 1 ( r , z ) + u 2 ( r , z ) … … ( 1 ) The boundary value conditions for u 1 ( r , z ) are as follows: ∂ 2 u 1 ∂ r 2 + 1 r ∂ u 1 ∂ r + ∂ 2 u 1 ∂ z 2 = 0 0 < r < c , 0 < z < L u 1 ( r , 0 ) = 0 u 1 ( r , L ) = 0 u 1 ( c , z ) = h ( z ) Separate the variables with the help of separation constant and take the separation constant λ as follows: R 1 ′ ′ + 1 r R 1 ′ R 1 = − Z 1 ′ ′ Z 1 = λ The separated equations are written as follows: r R 1 ′ ′ + R 1 ′ − r λ R 1 = 0 … … ( 2 ) Z 1 ′ ′ + λ Z 1 = 0 … … ( 3 ) The product solution is as follows: u 1 ( r , z ) = R 1 ( r ) Z 1 ( z ) … … ( 4 ) Substitute 0 for z in equation ( 4 ) . u 1 ( r , 0 ) = R 1 ( r ) Z 1 ( 0 ) R 1 ( r ) Z 1 ( 0 ) = 0 ( u 1 ( r , 0 ) = 0 ) Z 1 ( 0 ) = 0 Substitute L for z in equation ( 4 ) . u 1 ( r , L ) = R 1 ( r ) Z 1 ( L ) R 1 ( r ) Z 1 ( L ) = 0 ( u 1 ( r , L ) = 0 ) Z 1 ( L ) = 0 There are three cases for the value of λ . First case is λ = 0 . Substitute 0 for λ in equation ( 3 ) . Z 1 ′ ′ = 0 The general solution of the above equation is as follows: Z 1 ( z ) = c 1 + c 2 z … … ( 5 ) Substitute 0 for z in equation ( 5 ) . Z 1 ( 0 ) = c 1 + c 2 ⋅ 0 c 1 = 0 ( Z 1 ( 0 ) = 0 ) Substitute 0 for c 1 in equation ( 5 ) . Z 1 ( z ) = c 2 z Substitute L for z in the above equation. Z 1 ( L ) = c 2 L c 2 L = 0 ( Z 1 ( L ) = 0 ) c 2 = 0 Since, the value of c 1 and c 2 is 0 so, for λ = 0 non-trivial solution does not exist. Second case is λ = − α 2 ( λ < 0 ) . Substitute λ = − α 2 in equation ( 3 ) . Z 1 ′ ′ − α 2 Z 1 = 0 The general solution of the above equation is as follows: Z 1 ( z ) = c 3 cos h ( α z ) + c 4 sinh ( α z ) … … ( 6 ) Substitute 0 for z in equation ( 6 ) . Z 1 ( 0 ) = c 3 cos h ( 0 ) + c 4 sinh ( 0 ) c 3 = 0 ( Z 1 ( 0 ) = 0 ) Substitute 0 for c 3 in equation ( 6 ) . Z 1 ( z ) = c 4 sinh ( α z ) Substitute L for z in the above equation. Z 1 ( L ) = c 4 sinh ( α L ) c 4 sinh ( α L ) = 0 ( Z 1 ( L ) = 0 ) c 4 = 0 Since, the value of c 3 and c 4 is 0 so, for λ = − α 2 ( λ < 0 ) non-trivial solution does not exist. Third case is λ = α 2 ( λ > 0 ) . Substitute λ = α 2 in equation ( 3 ) . Z 1 ′ ′ + α 2 Z 1 = 0 The general solution of the above equation is as follows: Z 1 ( z ) = c 5 cos ( α z ) + c 6 sin ( α z ) … … ( 7 ) Substitute 0 for z in equation ( 7 ) . Z 1 ( 0 ) = c 5 cos ( 0 ) + c 6 sin ( 0 ) c 5 = 0 ( Z 1 ( 0 ) = 0 ) Substitute 0 for c 5 in equation ( 7 ) . Z 1 ( z ) = c 6 sin ( α z ) Substitute L for z in the above equation. Z 1 ( L ) = c 6 sin ( α L ) c 6 sin ( α L ) = 0 ( Z 1 ( L ) = 0 ) sin ( α L ) = 0 α L = n π Further solve the above equation to obtain the value of α . α L = n π α = n π L The eigenvalues are as follows: λ n = n 2 π 2 L 2 Substitute λ = α 2 in equation ( 2 ) . r R 1 ′ ′ + R 1 ′ − r 1 α 2 R 1 = 0 The general solution of the above equation is as follows: R 1 ( r ) = c 7 I 0 ( α r ) + c 8 K 0 ( α r ) … … ( 8 ) The value of c 8 in necessarily 0 because the Bessel function of second kind ( K 0 ( α r ) ) becomes unbounded as r → 0 . Substitute 0 for c 8 in equation ( 8 ) . R 1 ( r ) = c 7 I 0 ( α r ) Substitute n π L for α in the above equation. R 1 ( r ) = c 7 I 0 ( n π r L ) The product solution u 1 ( r , z ) is expressed as follows: u 1 ( r , z ) = c 6 sin ( n π z L ) ( c 7 I 0 ( n π r L ) ) = sin ( n π z L ) ( c 6 c 7 I 0 ( n π r L ) ) As per the superposition principle the above equation in written as follows: u 1 ( r , z ) = ∑ n = 1 ∞ sin ( n π z L ) ( c 6 c 7 I 0 ( n π r L ) ) Replace the term c 6 c 7 by A n respectively in the above equation

6th Edition

ISBN: 9781284127003

Author: ZILL

Publisher: JONES+BARTLETT LEARNING,LLC (CC)

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Chapter 14, Problem 19CR

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The steady state temperature u(r,z) of a finite cylinder of length L such that 0<r<c and 0<z<L.

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Chapter 14, Problem 18CR

Chapter 14, Problem 20CR

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Chapter 14 Solutions

EBK ADVANCED ENGINEERING MATHEMATICS

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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The steady state temperature u ( r , z ) of a finite cylinder of length L such that 0 &lt; r &lt; c and 0 &lt; z &lt; L .

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The steady state temperature u(r,z) of a finite cylinder of length L such that 0<r<c and 0<z<L.

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Chapter 14 Solutions

The steady state temperature u ( r , z ) of a finite cylinder of length L such that 0 < r < c and 0 < z < L .