Explanation Given: The boundary value problem is as follows: ∂ 2 u ∂ r 2 + 1 r ∂ u ∂ r + 1 r 2 ∂ 2 u ∂ θ 2 = 0 u ( 1 , θ ) = sin 2 θ ∂ u ∂ r | r = 2 = 0 0 < θ < 2 π , 1 < r < 2 Calculation: The boundary value problem is given as follows: ∂ 2 u ∂ r 2 + 1 r ∂ u ∂ r + 1 r 2 ∂ 2 u ∂ θ 2 = 0 u ( 1 , θ ) = sin 2 θ ∂ u ∂ r | r = 2 = 0 0 < θ < 2 π , 1 < r < 2 Take separation of the variables of the above equation and 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 product solution as follows: u ( r , θ ) = R ( r ) Θ ( θ ) … … ( 3 ) Differentiate equation ( 3 ) with respect to r . u ′ ( r , θ ) = R ′ ( r ) Θ ( θ ) Substitute 2 for r in the above equation. u ′ ( 2 , θ ) = R ′ ( 2 ) Θ ( θ ) R ′ ( 2 ) Θ ( θ ) = 0 R ′ ( 2 ) = 0 The function u ( r , θ ) is periodic in θ with the period 2 π , so u ( r , θ ) = u ( r , 2 π + θ ) . There are three possible cases for λ . First case is λ = 0 . Substitute λ = 0 in equation ( 1 ) . r 2 R ″ + r R ′ = 0 The general solution of the above equation is as follows: R ( r ) = c 1 + c 2 ln r … … ( 4 ) Differentiate equation ( 4 ) with respect to r . R ′ ( r ) = c 2 r Substitute 2 for r in the above equation. R ′ ( 2 ) = c 2 2 c 2 2 = 0 ( R ′ ( 2 ) = 0 ) c 2 = 0 Substitute 0 for c 2 in equation ( 4 ) . R ( r ) = c 1 The general solution of equation ( 2 ) for λ = 0 is as follows: Θ ( θ ) = c 3 + c 4 θ … … ( 5 ) Substitute the value of θ as θ + 2 π in the above equation. Θ ( θ + 2 π ) = c 3 + c 4 ( θ + 2 π ) Substitute the value of Θ ( θ ) and Θ ( θ + 2 π ) in Θ ( θ + 2 π ) = Θ ( θ ) . c 3 + c 4 ( θ + 2 π ) = c 3 + c 4 θ c 4 θ = c 4 θ + 2 π c 4 2 π c 4 = 0 c 4 = 0 Substitute the value of c 4 in equation ( 5 ) . Θ ( θ ) = c 3 Second case is λ = − α 2 ( λ < 0 ) . The general solution of equation ( 2 ) for λ = − α 2 ( λ < 0 ) is as follows: Θ ( θ ) = c 5 cos h ( α θ ) + c 6 sinh ( α θ ) … … ( 6 ) The function u ( r , θ ) is periodic in θ with the period 2 π . Since the function Θ ( θ ) in equation ( 6 ) is not periodic because hyperbolic functions are not periodic. Therefore, 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 2 R ″ + r R ′ − λ R = 0 The general solution of the above equation is as follows: R ( r ) = c 7 r α + c 8 r − α … … ( 7 ) Differentiate the above equation with respect to r . R ′ ( r ) = c 7 α r α − 1 + c 8 α r − α − 1 Substitute 2 for r in the above equation. R ′ ( 2 ) = c 7 α 2 α − 1 + c 8 α 2 − α − 1 α ( c 7 2 α − 1 + c 8 2 − α − 1 ) = 0 ( R ′ ( 2 ) = 0 ) c 7 2 α − 1 = c 8 2 α + 1 c 8 = c 7 2 2 α Substitute the value of c 8 in equation ( 7 ) . R ( r ) = c 7 r α + c 7 2 2 α r − α = c 7 ( r α + 4 α r − α ) For α = n the above equation is written as follows: R ( r ) = c 7 ( r n + 4 n r − n ) 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 21RE

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To show: The steady state temperature u(r,θ) in the plate is u(r,θ)=12−134r2cos(2θ)−817r−2cos(2θ) such that u(1,θ)=sin2θ, ∂u∂r|r=2=0 and 0<θ<2π, 1<r<2.

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Chapter 13, Problem 20RE

Chapter 14.1, Problem 1E

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Chapter 13 Solutions

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The steady state temperature u ( r , θ ) in the plate is u ( r , θ ) = 1 2 − 1 34 r 2 cos ( 2 θ ) − 8 17 r − 2 cos ( 2 θ ) such that u ( 1 , θ ) = sin 2 θ , ∂ u ∂ r | r = 2 = 0 and 0 < θ < 2 π , 1 < r < 2 .

Solution Summary: The author shows the steady state temperature u(r,theta ) in the plate.

Videos

To show: The steady state temperature u(r,θ) in the plate is u(r,θ)=12−134r2cos(2θ)−817r−2cos(2θ) such that u(1,θ)=sin2θ, ∂u∂r|r=2=0 and 0<θ<2π, 1<r<2.

Want to see the full answer?

Chapter 13 Solutions

The steady state temperature u ( r , θ ) in the plate is u ( r , θ ) = 1 2 − 1 34 r 2 cos ( 2 θ ) − 8 17 r − 2 cos ( 2 θ ) such that u ( 1 , θ ) = sin 2 θ , ∂ u ∂ r | r = 2 = 0 and 0 &lt; θ &lt; 2 π , 1 &lt; r &lt; 2 .

Solution Summary: The author shows the steady state temperature u(r,theta ) in the plate.

Videos

To show: The steady state temperature u(r,θ) in the plate is u(r,θ)=12−134r2cos(2θ)−817r−2cos(2θ) such that u(1,θ)=sin2θ, ∂u∂r|r=2=0 and 0<θ<2π, 1<r<2.

Want to see the full answer?

Chapter 13 Solutions

The steady state temperature u ( r , θ ) in the plate is u ( r , θ ) = 1 2 − 1 34 r 2 cos ( 2 θ ) − 8 17 r − 2 cos ( 2 θ ) such that u ( 1 , θ ) = sin 2 θ , ∂ u ∂ r | r = 2 = 0 and 0 < θ < 2 π , 1 < r < 2 .