AEM_3e_Chapter_14

# AEM_3e_Chapter_14 - 14 14 3 We have Boundary-Value Problems...

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14 14 Boundary-Value Problems in Other Coordinate Systems EXERCISES 14.1 Problems in Polar Coordinates 3. We have 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 )cos nθ dθ = 4 n 2 B n = 1 π Z 2 π 0 (2 θ 2 )sin nθ dθ =0 and so u ( r, θ )= 2 π 2 3 4 X n =1 r n n 2 cos nθ. 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 Fourier 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 identi±cations 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. 9. We proceed as in Example 1 in the text and use the periodicity of u ( r, θ ). When n ,Θ( θ c 5 θ + c 6 and R ( r c 7 + c 8 ln r . Periodicity implies c 5 = 0 and the insulation condition at r = a implies c 8 = 0. Thus, we take u 0 = A 0 = c 6 c 7 . Then, for n =1 , 2 , 3 ,... ( θ c 1 cos + c 2 sin and R ( r c 3 r n + c 4 r n . From R 0 ( a ) = 0 we get c 3 na n 1 c 4 na n 1 = 0, which implies c 4 = c 3 a 2 n . Then R ( r c 3 ( r n + a 2 n r n c 3 r 2 n + a 2 n r n 244

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14.1 Problems in Polar Coordinates and u ( r, θ )= A 0 + X n =1 r 2 n + a 2 n r n ( A n cos + B n sin ) . Taking r = b we have f ( θ A 0 + X n =1 b 2 n + a 2 n b n ( A n cos + B n sin ) , which implies A 0 = a 0 2 = 1 2 π Z 2 π 0 f ( θ ) and b 2 n + a 2 n b n A n = 1 π Z 2 π 0 f ( θ )cos nθ dθ and b 2 n + a 2 n b n B n = 1 π Z 2 π 0 f ( θ )sin2 nθ dθ. Hence A n = b n π ( a 2 n + b 2 n ) Z 2 π 0 f ( θ nθ dθ and B n = b n π ( a 2 n + b 2 n ) Z 2 π 0 f ( θ )sin nθ dθ. 12. We solve 2 u ∂r 2 + 1 r ∂u + 1 r 2 2 u ∂θ 2 =0 , 0 <θ<π/ 4 ,r > 0 u ( r, 0) = 0 > 0 u ( r, π/ 4) = 30 > 0 . Proceeding as in Example 1 in the text we fnd the separated ordinary diFerential equations to be r 2 R 0 + rR 0 λR Θ 0 + λ Θ=0 . With λ = α 2 > 0 the corresponding general solutions are R ( r c 1 r α + c 2 r α Θ( θ c 3 cos αθ + c 4 sin αθ. The condition Θ(0) = 0 implies c 3 = 0 so that Θ = c 4 sin αθ . Now, in order that the temperature be bounded as r →∞ we defne c 1 = 0. Similarly, in order that the temperature be bounded as r 0 we are ±orced to defne c 2 =0.Thus R ( r ) = 0 and so no nontrivial solution exists ±or λ> 0. ²or λ = 0 the separated diFerential equations are r 2 R 0 + 0 = 0 and Θ 0 . Solutions o± these latter equations are R ( r c 1 + c 2 ln r and Θ( θ c 3 θ + c 4 . Θ(0) = 0 still implies c 4 = 0, whereas boundedness as r 0 demands c 2 = 0. Thus, a product solution is u = c 1 c 3 θ = Aθ.
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## This note was uploaded on 01/03/2011 for the course BIS 511 at Yale.

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AEM_3e_Chapter_14 - 14 14 3 We have Boundary-Value Problems...

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