Explanation Given: The given Equations are, ∂ 2 u ∂ r 2 + 1 r ∂ u ∂ r + 1 r 2 ∂ 2 u ∂ θ 2 = 0 with π < r < 2 π , 0 < θ < π And u ( r , 0 ) = sin r , π ≤ r ≤ 2 π u ( r , π ) = 0 , π ≤ r ≤ 2 π u ( π , θ ) = u ( 2 π , θ ) = 0 , 0 ≤ θ ≤ π Calculation: Consider the given equation, ∂ 2 u ∂ r 2 + 1 r ∂ u ∂ r + 1 r 2 ∂ 2 u ∂ θ 2 = 0 … ( 1 ) Consider the relation, u ( r , θ ) = R ( r ) T ( θ ) … ( 2 ) Differentiate Equation (2) with respect to r . ∂ u ∂ r = R ′ ( r ) T ( θ ) … ( 3 ) Differentiate Equation (3) with respect to r . ∂ 2 u ∂ r 2 = R ″ ( r ) T ( θ ) Differentiate Equation (2) with respect to θ . ∂ u ∂ θ = R ( r ) T ′ ( θ ) … ( 4 ) Differentiate Equation (4) with respect to r . ∂ 2 u ∂ θ 2 = R ( r ) T ″ ( θ ) Substitute R ( r ) T ″ ( θ ) for ∂ 2 u ∂ θ 2 , R ″ ( r ) T ( θ ) for ∂ 2 u ∂ r 2 and R ′ ( r ) T ( θ ) for in ∂ u ∂ r Equation (1). R ″ ( r ) T ( θ ) + 1 r R ′ ( r ) T ( θ ) + 1 r 2 R ( r ) T ″ ( θ ) = 0 r 2 R ″ ( r ) + r R ′ ( r ) R ( r ) = − T ″ ( θ ) T ( θ ) … ( 5 ) Equate the ratios of Equation (5) to λ . r 2 R ″ ( r ) + r R ′ ( r ) − λ R ( r ) = 0 … ( 6 ) T ″ ( θ ) + λ T ( θ ) = 0 … ( 7 ) The auxiliary equation is, r 2 + λ = 0 r = ± i λ The solution for Equation (7) is given as,

Fundamentals of Differential Equations and Boundary Value Problems

7th Edition

ISBN: 9780321977106

Author: Nagle, R. Kent

Publisher: Pearson Education, Limited

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Chapter 10.7, Problem 16E

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The solution of the following Dirichlet problem for a half annulus.

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Fundamentals of Differential Equations and Boundary Value Problems

Ch. 10.2 - In Problems 1-8, determine all the solutions, if...Ch. 10.2 - Prob. 2E Ch. 10.2 - Prob. 3E Ch. 10.2 - Prob. 4E Ch. 10.2 - In Problems 1-8, determine all the solutions, if...Ch. 10.2 - Prob. 6E Ch. 10.2 - In Problems 1-8, determine all the solutions, if...Ch. 10.2 - In Problems 1-8, determine all the solutions, if...Ch. 10.2 - In Problems 9-14, find the values of eigenvalues...Ch. 10.2 - In Problems 9-14, find the values of eigenvalues...

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The solution of the following Dirichlet problem for a half annulus.

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