# Chapt_14 - 14 14 1 We have Boundary-Value Problems in Other...

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14 14 Boundary-Value Problems in Other Coordinate Systems EXERCISES 14.1 Problems in Polar Coordinates 1. We have A 0 = 1 2 π Z π 0 u 0 = u 0 2 A n = 1 π Z π 0 u 0 cos nθ dθ =0 B n = 1 π Z π 0 u 0 sin nθ dθ = u 0 [1 ( 1) n ] and so u ( r, θ )= u 0 2 + u 0 π X n =1 1 ( 1) n n r n sin nθ. 2. A 0 = 1 2 π Z π 0 θdθ + 1 2 π Z 2 π π ( π θ ) A n = 1 π Z π 0 θ cos nθ dθ + 1 π Z 2 π π ( π θ )cos nθ dθ = 2 n 2 π [( 1) n 1] B n = 1 π Z π 0 θ sin nθ dθ + 1 π Z 2 π π ( π θ )sin nθ dθ = 1 n [1 ( 1) n ] and so u ( r, θ X n =1 r n · ( 1) n 1 n 2 π cos + 1 ( 1) n n sin ¸ . 3. A 0 = 1 2 π Z 2 π 0 (2 πθ θ 2 ) = 2 π 2 3 A n = 1 π Z 2 π 0 (2 θ 2 nθ dθ = 4 n 2 B n = 1 π Z 2 π 0 (2 θ 2 nθ dθ and so u ( r, θ 2 π 2 3 4 X n =1 r n n 2 cos nθ. 755

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14.1 Problems in Polar Coordinates 4. We have A 0 = 1 2 π Z 2 π 0 θdθ = π A n = 1 π Z 2 π 0 θ cos nθ dθ =0 B n = 1 π Z 2 π 0 θ sin nθ dθ = 2 n and so u ( r, θ )= π 2 X n =1 r n n sin nθ. 5. As in Example 1 in the text we have R ( r c 3 r n + c 4 r n . In order that the solution be bounded as r →∞ we must defne c 3 = 0. Hence u ( r, θ A 0 + X n =1 r n ( A n cos + B n sin ) A 0 = 1 2 π Z 2 π 0 f ( θ ) where A n = c n π Z 2 π 0 f ( θ )cos nθ dθ B n = c n π Z 2 π 0 f ( θ )sin nθ dθ. 6. Using the same reasoning as in Example 1 in the text we obtain u ( r, θ A 0 + X n =1 r n ( A n cos + B n sin ) . The boundary condition at r = c implies f ( θ X n =1 nc n 1 ( A n cos + B n sin ) . Since this condition does not determine A 0 , it is an arbitrary constant. However, to be a Full ±ourier series on [0 , 2 π ] we must require that f ( θ ) satisFy the condition A 0 = a 0 / 2=0or R 2 π 0 f ( θ ) = 0. IF this integral were not 0, then the series For f ( θ ) would contain a nonzero constant, which it obviously does not. With this as a necessary compatibility condition we can then make the identifcations nc n 1 A n = a n and nc n 1 B n = b n or A n = 1 nc n 1 π Z 2 π 0 f ( θ nθ dθ and B n = 1 nc n 1 π Z 2 π 0 f ( θ nθ dθ. 7. Proceeding as in Example 1 in the text and again using the periodicity oF u ( r, θ ), we have Θ( θ c 1 cos αθ + c 2 sin αθ where α = n For n ,1 ,2 , ... . Then R ( r c 3 r n + c 4 r n . 756
14.1 Problems in Polar Coordinates [We do not have c 4 = 0 in this case since 0 <a r .] Since u ( b, θ )=0wehave u ( r, θ )= A 0 ln r b + X n =1 ·µ b r n ³ r b ´ n ¸ [ A n cos + B n sin ] . From u ( a, θ f ( θ A 0 ln a b + X n =1 ·µ b a n ³ a b ´ n ¸ [ A n cos + B n sin ] we fnd A 0 ln a b = 1 2 π Z 2 π 0 f ( θ ) dθ, ·µ b a n ³ a b ´ n ¸ A n = 1 π Z 2 π 0 f ( θ )cos nθ dθ, and ·µ b a n ³ a b ´ n ¸ B n = 1 π Z 2 π 0 f ( θ )sin nθ dθ. 8. Substituting u ( r, θ v ( r, θ )+ ψ ( r ) into the partial di±erential equation we obtain 2 v ∂r 2 + ψ 0 ( r 1 r · ∂v + ψ 0 ( r ) ¸ + 1 r 2 2 v ∂θ 2 =0 .

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Chapt_14 - 14 14 1 We have Boundary-Value Problems in Other...

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