Page 878 - Advanced_Engineering_Mathematics o'neil

858 Answers to Selected Problems

1 n 4(−1) n+1
r
2
7. u(r,θ) = 1 − π + ∞ cos(nθ)
n=1
3 8 n 2
9. In polar coordinates, the problem is to solve
2
2
∇ U(r,θ) = 0for r < 4,U(4,θ) = 16cos (θ).
This has solution
1

2
U(r,θ) = 8 +r 2 cos (θ) − ,
2
so
1
2
2
u(x, y) = (x − y ) + 8.
2
2
2
2
2
11. In polar coordinates, U(r,θ) =r (2cos (θ) − 1),so u(x, y) = x − y .
Section 18.4 Poisson’s Integral Formula
ξ
π
3
1. u(1/2,π) = dξ = 0 (odd integrand)
8π −π 5/4 − cos(ξ − π)
u(3/4,π/3) ≈ 0.88261, u(0.2,π/4) ≈ 0.024076
3. u(4,π) = 0, u(12,3π/2) ≈−2.571176,
u(8,π/4) ≈ 0.59705, u(7,0) ≈ 0
Section 18.5 Dirichlet Problem for Unbounded Regions
1
4 + x 4 − x
x
1. u(x, y) = 2arctan − arctan + arctan
π y y y

1 1
∞ −ξ
2
3. u(x, y) = − e cos(ξ)dξ
2
π 0 y + (ξ − x) 2 y + (ξ + x) 2
2
5. 2 ∞ ∞ −ωy
u(x, y) = 0 0 f (ξ)sin(ωξ)dξ sin(ωx)e dω
π
∞ ∞ −ωx
2
+ g(ξ)sin(ωξ)dξ sin(ωy)e dω
π 0 0
1

2 ∞
7. u(x, y) = sin(ωy)e −ωx dω
π 0 1 + ω 2

y 8 A A x − 8 x − 4
9. u(x, y) = dξ = −arctan + arctan
π 4 y + (ξ − x) 2 π y y
2
1
1 − y 1 + y
11. u(x, y) = arctan + arctan
π x x
Section 18.6 Dirichlet Problem for a Cube
∞ ∞ 4(−1) n+m √
2
2
1. u(x, y, z) = √ sin(nπx)sin(mπy)sinh(π n + m z)
n=1 m=1
nmπ 2 sinh(π n 2 +m 2 )
3.
∞ ∞

2
2
2
u(x, y, z) = a nm sin(nπx)sin(mπy/2)sinh( n π + m /4z)
n=1 m=1
∞ ∞
√

2
2
2
+ c nm sin(nπx)sin(mz)sinh( n π + m z)
n=1 m=1
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October 14, 2010 17:50 THM/NEIL Page-858 27410_25_Ans_p801-866