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Chapter 14

# Chapter 14 - 14 Boundary-Value Problems in Other Coordinate...

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14 Boundary-Value Problems in Other Coordinate Systems Exercises 14.1 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. 4. A 0 = 1 2 π Z 2 π 0 = π A n = 1 π Z 2 π 0 θ cos nθ dθ B n = 1 π Z 2 π 0 θ sin nθ dθ = 2 n 670

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Exercises 14.1 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. We solve 2 u ∂r 2 + 1 r ∂u + 1 r 2 2 u ∂θ 2 =0 , 0 <θ< π 2 , 0 <r<c, u ( c,θ f ( θ ) , 0 π 2 , u ( r, 0) = 0 ,u ( r, π/ 2) = 0 , 0 <r<c. Proceeding as in Example 1 in the text we obtain the separated diFerential equations r 2 R 0 + rR 0 λ 2 R Θ 0 + λ 2 Θ=0 with solutions Θ( θ c 1 cos λθ + c 2 sin λθ R ( r c 3 r λ + c 4 r λ . Since we want R ( r ) to be bounded as r 0 we require c 4 = 0. Applying the boundary conditions Θ(0) = 0 and Θ( π/ 2) = 0 we fnd that c 1 = 0 and λ =2 n ±or n =1 ,2 ,3 , ... . There±ore u ( r, θ X n =1 A n r 2 n sin2 nθ. ²rom u ( f ( θ X n =1 A n c n we fnd A n = 4 πc 2 n Z 2 0 f ( θ )sin2 nθ dθ. 7. Re±erring to the solution o± Problem 6 above we have Θ( θ c 1 cos λθ + c 2 sin λθ R ( r c 3 r n . Applying the boundary conditions Θ 0 (0) = 0 and Θ 0 ( 2) = 0 we fnd that c 2 = 0 and λ n ±or 671
Exercises 14.1 n =0 ,1 ,2 , ... . Therefore u ( r, θ )= A 0 + X n =1 A n r 2 n cos2 nθ. From u ( c,θ ½ 1 , 0 <θ<π/ 4 0 / 4 2 = A 0 + X n =1 A n c 2 n we ±nd A 0 = 1 π/ 2 Z 4 0 = 1 2 and c 2 n A n = 2 2 Z 4 0 nθ dθ = 2 sin 2 . Thus u ( r, θ 1 2 + 2 π X n =1 1 n sin 2 ³ r c ´ 2 n nθ. 8. We solve 2 u ∂r 2 + 1 r ∂u + 1 r 2 2 u ∂θ 2 , 0 4 ,r > 0 u ( r, 0) = 0 > 0 u ( r, π/ 4) = 30 > 0 . Proceeding as in Example 1 in the text we ±nd the separated ordinary di²erential equations to be r 2 R 0 + rR 0 λ 2 R Θ 0 + λ 2 Θ=0 . The corresponding general solutions are R ( r c 1 r λ + c 2 r λ Θ( θ c 3 cos λθ + c 4 sin λθ. The condition Θ(0) = 0 implies c 3 = 0 so that Θ = c 4 sin λθ . Now, in order that the temperature be bounded as r →∞ we de±ne c 1 = 0. Similarly, in order that the temperature be bounded as r 0 we are forced to de±ne c 2 = 0. Thus R ( r ) = 0 and so no nontrivial solution exists for λ> 0. For λ = 0 the separated di²erential equations are r 2 R 0 + 0 = 0 and Θ 0 .

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Chapter 14 - 14 Boundary-Value Problems in Other Coordinate...

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