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652 CHAPTER 18 The Potential Equation

18.5.2 The Right Quarter-Plane
The right quarter-plane has the nonnegative horizontal and vertical axes as boundary. The
Dirichlet problem for this region is
2
∇ u(x, y) = 0for x > 0, y > 0,
u(x,0) = f (x) for x ≥ 0,

and
u(0, y) = g(y) for y ≥ 0.
This problem can be treated by solving separately the cases that either f (x) or g(y) is iden-
tically zero. Separation of variables applies to both cases, and the solution of the given problem
is the sum of the solutions of these simpler problems.
We will demonstrate a different method for the case that g(y) = 0. Notice that if we fold
the upper half-plane across the vertical axis we obtain the right quarter-plane. This suggests that
we might be able to use the solution for the upper half-plane to solve the problem for the right
quarter-plane. To do this, let

f (x) for x ≥ 0
w(x) =
anything for x < 0.
where by “anything” we mean we will fill in this part shortly. We now have a Dirichlet problem
for the upper half-plane with the data function u(x,0) = w(x). We know the solution u hp of this
problem for the upper half-plane:
y ∞ w(ξ)
u hp (x, y) = dξ.
π −∞ y + (ξ − x) 2
2
Write this as
y 0 w(ξ) ∞ w(ξ)
u hp (x, y) = dξ + dξ .
2
2
π −∞ y + (ξ − x) 2 0 y + (ξ − x) 2
Change variables in the first integral on the right by letting ζ =−ξ:
w(ξ) w(−ζ)
0 ∞
dξ = (−1)dζ.
2
2
−∞ y + (ξ − x) 2 0 y + (ζ + x) 2
For uniformity in notation, replace the dummy variable of integration on the right with ξ to write
y 0 w(−ξ) ∞ w(ξ)
u hp (x, y) = (−1)dξ + dξ
2
2
π ∞ y + (ξ + x) 2 0 y + (ξ − x) 2
y ∞ w(−ξ) f (ξ)
= + dξ.
2
2
π 0 y + (ξ + x) 2 y + (ξ − x) 2
In the last integral, we used the fact that w(ξ) = f (ξ) for ξ ≥ 0. Now fill in the “anything” in the
definition of w. Notice that the last integral will vanish at points (0, y) on the positive y-axis if
f (ξ) + w(−ξ) = 0for ξ ≥ 0. This will occur if w(−ξ) =− f (ξ).Make w the odd extension of
f to the entire line. In this way, we obtain the solution for the upper half-plane:
y ∞ 1 1
u hp (x, y) = − f (ξ)dξ.
2
π 0 y + (ξ − x) 2 y + (ξ + x) 2
2
But this function is also harmonic on the right quarter-plane, vanishes when x = 0, and equals
f (x) if x ≥ 0 and y = 0. Therefore, this function is also the solution of the problem for the right
quarter-plane (in this case of zero data along the positive y-axis).

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October 14, 2010 15:27 THM/NEIL Page-652 27410_18_ch18_p641-666