Explanation Given: The boundary conditions are as follows: ∂ 2 u ∂ r 2 + 1 r ∂ u ∂ r − u r 2 + ∂ 2 u ∂ z 2 = 0 0 < r < 1 , 0 < z < 1 u ( 1 , z ) = 0 u ( r , 0 ) = 0 u ( r , 1 ) = r − r 3 Calculation: The boundary conditions are given as follows: ∂ 2 u ∂ r 2 + 1 r ∂ u ∂ r − u r 2 + ∂ 2 u ∂ z 2 = 0 0 < r < 1 , 0 < z < 1 u ( 1 , z ) = 0 u ( r , 0 ) = 0 u ( r , 1 ) = 1 − r 3 Separate the variables with the help of separation constant and take the separation constant − λ as follows: R ″ R + 1 r R ′ R − 1 r 2 = − Z ″ Z = − λ The separated equations are written as follows: r R ″ + r R ′ + ( r 2 λ − 1 ) R = 0 … … ( 1 ) Z ″ − λ Z = 0 … … ( 2 ) The product solution is as follows: u ( r , z ) = R ( r ) Z ( z ) … … ( 3 ) Substitute 1 for r in equation ( 3 ) . u ( 1 , z ) = R ( 1 ) Z ( z ) R ( 1 ) Z ( z ) = 0 ( u ( 1 , z ) = 0 ) R ( 1 ) = 0 Substitute 0 for z in equation ( 3 ) . u ( r , 0 ) = R ( r ) Z ( 0 ) R ( r ) Z ( 0 ) = 0 ( u ( r , 0 ) = 0 ) Z ( 0 ) = 0 There are three cases for the value of λ . First case is λ = 0 . Substitute 0 for λ in equation ( 1 ) . r R ″ + r R ′ − R = 0 Above equation is in the Cauchy-Euler form and its general solution is as follows: R ( r ) = c 1 r + c 2 r … … ( 4 ) The value of c 2 is necessarily 0 because the function is bounded at r = 0 . Substitute 0 for c 2 in equation ( 4 ) . R ( r ) = c 1 r Substitute 1 for r in the above equation. R ( 1 ) = c 1 c 1 = 0 ( R ( 1 ) = 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 ( 1 ) . r R ″ + r R ′ − ( r 2 α 2 − 1 ) R = 0 The general solution of the above equation is as follows: R ( r ) = c 3 I 1 ( α r ) + c 4 K 1 ( α r ) … … ( 5 ) The value of c 4 is necessarily 0 because the Bessel function of second kind K 1 ( α r ) becomes unbounded as r → 0 . Substitute 0 for c 4 in equation ( 5 ) . R ( r ) = c 3 I 1 ( α r ) Substitute 1 for r in the above equation. R ( 1 ) = c 3 I 1 ( α ) c 3 I 1 ( α ) = 0 ( R ( 1 ) = 0 ) c 3 = 0 Substitute 0 for c 3 in R ( r ) = c 3 I 1 ( α r ) . R ( r ) = 0 Since, the value of c 3 and c 4 is o 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 ( 1 ) . r R ″ + r R ′ + ( r 2 α 2 − 1 ) R = 0 The general solution of the above equation is as follows: R ( r ) = c 5 J 1 ( α r ) + c 6 Y 1 ( α r ) … … ( 6 ) The value of c 6 is necessarily 0 because the function becomes unbounded as r → 0 . Substitute 0 for c 6 in equation ( 6 ) . R ( r ) = c 5 J 1 ( α r ) Substitute λ = α 2 in equation ( 2 )

Bundle: Differential Equations with Boundary-Value Problems, 9th + WebAssign Printed Access Card for Zill's Differential Equations with Boundary-Value Problems, 9th Edition, Single-Term

9th Edition

ISBN: 9781337604918

Author: Dennis G. Zill

Publisher: Cengage Learning

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Chapter 13, Problem 20RE

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The steady state temperature u(r,z) such that ∂2u∂r2+1r∂u∂r−ur2+∂2u∂z2=0, u(1,z)=0, u(r,0)=0 and u(r,1)=r−r3.

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Chapter 13, Problem 19RE

Chapter 13, Problem 21RE

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Chapter 13 Solutions

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The steady state temperature u ( r , z ) such that ∂ 2 u ∂ r 2 + 1 r ∂ u ∂ r − u r 2 + ∂ 2 u ∂ z 2 = 0 , u ( 1 , z ) = 0 , u ( r , 0 ) = 0 and u ( r , 1 ) = r − r 3 .

Solution Summary: The author explains the steady state temperature u(r,z) and the boundary conditions.

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The steady state temperature u(r,z) such that ∂2u∂r2+1r∂u∂r−ur2+∂2u∂z2=0, u(1,z)=0, u(r,0)=0 and u(r,1)=r−r3.

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Chapter 13 Solutions