Page 685 - Advanced_Engineering_Mathematics o'neil

18.8 The Neumann Problem 665

Then v is harmonic wherever u is. Furthermore,
∂u
v(x,0) = (x,0) = f (x)
∂y
on the x-axis. Therefore v satisfies a Dirichlet problem for the upper half-plane. We know the
solution of this problem is
y ∞ f (ξ)
v(x, y) = dξ.
2
π −∞ y + (ξ − x) 2
Now integrate to recover u from v. To within an arbitrary constant,
∂u y f (ξ)
∞
u(x, y) = dy = dξ dy
2
∂y π −∞ y + (ξ − x) 2
1 ∞ ∞ y
= dy f (ξ)dξ
2
π −∞ −∞ y + (ξ − x) 2
1 ∞
2
2
= ln(y + (ξ − x) ) f (ξ)dξ.
2π
−∞

SECTION 18.8 PROBLEMS

2
1. ∇ u(x, y)= 0for0 < x < 1,0 < y < 1, 5. Attempt a separation of variables to solve
∂u ∂u ∇ u(x, y) = 0for0 < x < 1,0 < y < 1
2
(x,0)= 4cos(πx), (x,1) = 0for 0 ≤ x ≤ 1,
∂y ∂y
∂u ∂u u(x,0) = u(x,1) = 0for 0 ≤ x ≤ 1
(0, y)= (1, y) = 0for0 ≤ y ≤ 1
∂x ∂x ∂u (0, y) = 3y − 2y, ∂u (1, y) = 0for 0 ≤ y ≤ 1
2
2
2. ∇ u(x, y)= 0for0 < x < 1,0 < y <π, ∂x ∂x
∂u ∂u 6. Write a series solution for
(x,0)= (x,π) = 0for 0 ≤ x ≤ 1,
2
∂y ∂y ∇ u(r,θ) = 0for0 ≤r < R,−π ≤ θ ≤ π
∂u π ∂u
(0, y)= y − , (π, y) ∂u
∂x 2 ∂x (R,θ) = sin(3θ) for − π ≤ θ ≤ π
= cos(y) for 0 ≤ y ≤ π ∂r
7. Write a series solution for
2
3. ∇ u(x, y)= 0for0 < x <π,0 < y <π,
2
∇ u(r,θ) = 0for0 ≤r < R,−π ≤ θ ≤ π
∂u
(x,0)= cos(3x) for 0 ≤ x ≤ π
∂y ∂u (R,θ) = cos(2θ) for − π ≤ θ ≤ π
∂u ∂r
(x,π)= 6x − 3π for 0 ≤ x ≤ π
∂y 8. Solve the following Neumann problem for the upper
∂u ∂u half-plane:
(0, y)= (π, y) = 0for0 ≤ y ≤ π
∂x ∂x
2
∇ u(x, y) = 0for −∞ < x < ∞, y > 0
4. Use separation of variables to write an expres-
sion for the solution of the mixed boundary value ∂u (x,0) = xe −|x| for −∞ < x < ∞
problem ∂y
2
∇ u(x, y) = 0for0 < x <π,0 < y <π 9. Solve the following Neumann problem for the upper
half-plane:
u(x,0) = f (x),u(x,π) = 0for0 ≤ x ≤ π
2
∇ u(x, y) = 0for −∞ < x < ∞, y > 0
∂u ∂u
(0, y) = (π, y) = 0for 0 ≤ y ≤ π
∂x ∂x ∂u (x,0) = e −|x| sin(x) for −∞ < x < ∞
Does this problem have a unique solution? ∂y
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October 14, 2010 15:27 THM/NEIL Page-665 27410_18_ch18_p641-666