Chapter 14

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

• Notes
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14 Boundary-Value Problems in Other Coordinate Systems Exercises 14.1 1. We have A 0 = 1 2 π π 0 u 0 = u 0 2 A n = 1 π π 0 u 0 cos nθ dθ = 0 B n = 1 π π 0 u 0 sin nθ dθ = u 0 [1 ( 1) n ] and so u ( r, θ ) = u 0 2 + u 0 π n =1 1 ( 1) n n r n sin nθ. 2. We have A 0 = 1 2 π π 0 θ dθ + 1 2 π 2 π π ( π θ ) = 0 A n = 1 π π 0 θ cos nθ dθ + 1 π 2 π π ( π θ )cos nθ dθ = 2 n 2 π [( 1) n 1] B n = 1 π π 0 θ sin nθ dθ + 1 π 2 π π ( π θ )sin nθ dθ = 1 n [1 ( 1) n ] and so u ( r, θ ) = n =1 r n ( 1) n 1 n 2 π cos + 1 ( 1) n n sin . 3. We have A 0 = 1 2 π 2 π 0 (2 πθ θ 2 ) = 2 π 2 3 A n = 1 π 2 π 0 (2 πθ θ 2 )cos nθ dθ = 4 n 2 B n = 1 π 2 π 0 (2 πθ θ 2 )sin nθ dθ = 0 and so u ( r, θ ) = 2 π 2 3 4 n =1 r n n 2 cos nθ. 4. We have A 0 = 1 2 π 2 π 0 θ dθ = π A n = 1 π 2 π 0 θ cos nθ dθ = 0 B n = 1 π 2 π 0 θ sin nθ dθ = 2 n 670

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Exercises 14.1 and so u ( r, θ ) = π 2 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 define c 3 = 0. Hence u ( r, θ ) = A 0 + n =1 r n ( A n cos + B n sin ) A 0 = 1 2 π 2 π 0 f ( θ ) where A n = c n π 2 π 0 f ( θ )cos nθ dθ B n = c n π 2 π 0 f ( θ )sin nθ dθ. 6. We solve 2 u ∂r 2 + 1 r ∂u ∂r + 1 r 2 2 u ∂θ 2 = 0 , 0 < θ < π 2 , 0 < r < c, u ( c, θ ) = f ( θ ) , 0 < θ < 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 differential equations r 2 R + rR λ 2 R = 0 Θ + 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 find that c 1 = 0 and λ = 2 n for n = 1, 2, 3, . . . . Therefore u ( r, θ ) = n =1 A n r 2 n sin2 nθ. From u ( c, θ ) = f ( θ ) = n =1 A n c n sin2 we find A n = 4 πc 2 n π/ 2 0 f ( θ )sin2 nθ dθ. 7. Referring to the solution of 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 and Θ ( π/ 2) = 0 we find that c 2 = 0 and λ = 2 n for 671
Exercises 14.1 n = 0, 1, 2, . . . . Therefore u ( r, θ ) = A 0 + n =1 A n r 2 n cos2 nθ. From u ( c, θ ) = 1 , 0 < θ < π/ 4 0 , π/ 4 < θ < π/ 2 = A 0 + n =1 A n c 2 n cos2 we find A 0 = 1 π/ 2 π/ 4 0 = 1 2 and c 2 n A n = 2 π/ 2 π/ 4 0 cos2 nθ dθ = 2 sin 2 . Thus u ( r, θ ) = 1 2 + 2 π n =1 1 n sin 2 r c 2 n cos2 nθ. 8. We solve 2 u ∂r 2 + 1 r ∂u ∂r + 1 r 2 2 u ∂θ 2 = 0 , 0 < θ < π/ 4 , r > 0 u ( r, 0) = 0 , r > 0 u ( r, π/ 4) = 30 , r > 0 . Proceeding as in Example 1 in the text we find the separated ordinary differential equations to be r 2 R + rR λ 2 R = 0 Θ + 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 define c 1 = 0. Similarly, in order that the temperature be bounded as r 0 we are forced to define c 2 = 0. Thus R ( r ) = 0 and so no nontrivial solution exists for λ > 0. For λ = 0 the separated differential equations are r 2 R + rR = 0 and Θ = 0 .

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• Fall '07
• Delahanty
• Sin, Cos, Trigraph, Boundary value problem

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