Explanation Given: The boundary conditions are u ( 1 2 , θ ) = 100 ( 1 + 0.5 cos θ ) and u ( 1 , θ ) = 200 , 0 ≤ θ ≤ 2 π . Calculation: The boundary value problem is as follows: ∂ 2 u ∂ r 2 + 1 r ∂ u ∂ r + 1 r 2 ∂ 2 u ∂ θ 2 = 0 Take separation of the variables of the above equation and let the separation constant is λ . r 2 R ″ + r R ′ R = λ and − Θ ″ Θ = λ The equation of variable R is as follows: r 2 R ″ + r R ′ − λ R = 0 … … ( 1 ) The equation of variable θ is as follows: Θ ″ + λ Θ = 0 … … ( 2 ) Consider the equation as follows: u ( r , θ ) = R ( r ) Θ ( θ ) … … ( 3 ) There are three possible cases for λ . First case is λ = 0 . Substitute 0 for λ in equation ( 2 ) . Θ ″ = 0 The general solution of the above equation is as follows: Θ ( θ ) = c 1 + c 2 θ … … ( 4 ) For the boundary conditions Θ ( 0 ) = 0 . Θ ( 0 ) = c 1 + c 2 ( 0 ) c 1 = 0 Substitute 0 for c 1 in equation ( 4 ) . Θ ( θ ) = c 2 θ For the boundary conditions Θ ( 2 π ) = 0 . Θ ( 2 π ) = c 2 ( 2 π ) c 2 ( 2 π ) = 0 c 2 = 0 Since, c 1 = 0 and c 2 = 0 , so for the case λ = 0 non-trivial solution does not exist. Second case is λ = − α 2 , such that λ < 0 . Substitute − α 2 for λ in equation ( 2 ) . Θ ″ − α 2 Θ = 0 The general solution of the above differential equation is as follows: Θ ( θ ) = c 3 cos h ( α θ ) + c 4 sin h ( α θ ) … … ( 5 ) For the boundary conditions Θ ( 0 ) = 0 . Θ ( 0 ) = c 3 cos h ( 0 ) + c 4 sin h ( 0 ) c 3 = 0 Substitute 0 for c 3 in equation ( 5 ) . Θ ( θ ) = c 4 sin h ( α θ ) For the boundary conditions Θ ( 2 π ) = 0 . Θ ( 2 π ) = c 4 sin h ( 2 π α ) c 4 sin h ( 2 π α ) = 0 c 4 = 0 Since, c 3 = 0 and c 4 = 0 , so for the case λ = − α 2 non-trivial solution does not exist. Third case is λ = α 2 such that λ > 0 . Substitute α 2 for λ in equation ( 2 ) . Θ ″ + α 2 Θ = 0 The general solution of the above differential equation is as follows: Θ ( θ ) = c 5 cos ( α θ ) + c 6 sin ( α θ ) … … ( 6 ) For the boundary conditions Θ ( 0 ) = 0 . Θ ( 0 ) = c 5 cos ( 0 ) + c 6 sin ( 0 ) c 5 = 0 Substitute 0 for c 5 in equation ( 6 ) . Θ ( θ ) = c 6 sin ( α θ ) For the boundary conditions Θ ( 2 π ) = 0 . Θ ( 2 π ) = c 6 sin ( 2 π α ) c 6 sin ( 2 π α ) = 0 sin ( 2 π α ) = 0 Further solve the above equation. 2 π α = n π α = n 2 Substitute n 2 for α in Θ ( θ ) = c 6 sin ( α θ )

Differential Equations with Boundary-Value Problems

9th Edition

ISBN: 9781337632515

Author: Dennis G. Zill

Publisher: Cengage Learning US

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Chapter 13.1, Problem 20E

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The steady state temperature of the annular plate with the boundary condition as u(12,θ)=100(1+0.5cosθ) and u(1,θ)=200,0≤θ≤2π.

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Chapter 13.1, Problem 19E

Chapter 13.1, Problem 21E

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Differential Equations with Boundary-Value Problems

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The steady state temperature of the annular plate with the boundary condition as u ( 1 2 , θ ) = 100 ( 1 + 0.5 cos θ ) and u ( 1 , θ ) = 200 , 0 ≤ θ ≤ 2 π .

Solution Summary: The author explains the steady state temperature of the annular plate with the boundary condition as u(12,theta )=200,0le

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The steady state temperature of the annular plate with the boundary condition as u(12,θ)=100(1+0.5cosθ) and u(1,θ)=200,0≤θ≤2π.

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