Differentiation 323
so that Uxx + Uyy = O, and u is harmonic (in C). Since Vx = -Uy = 2y,
we get
v = 2xy + F(y) (6.6-6)
where F(y) is some differentiable function of y to be determined.
Differentiating with respect to y, we obtain
Vy= 2x + F'(y) = Ux = 2x + 1
Hence, F'(y) = 1, so that F(y) = y + C. Replacing this result in (6.6-6)
we find that
v = 2xy+y+C
where C is an arbitrary constant. Therefore, the analytic function with u
and v as components is
f(z) = (x^2 + x - y^2 ) + i(2xy + y + C)
= (x + iy)^2 + (x + iy) + iC
= z^2 + z + iCAlternatively, we may compute v by using (6.6-5). We obtain
as before.
1
(x,y)v(x,y)= 2ydx+(2x+l)dy
(xo,Yo)= r 2yo dx + r (2x + 1) dy
lxo }yo
= 2yo(x - xo) + (2x + l)(y - Yo)= 2xy +y+ C
- Let u = % ln(x^2 + y^2 ) be defined in R = {(x, y): x^2 + y^2 #-O}. We
have
x
Ux= ---x2 + y2'
y2 -x2
Uxx = (x2 + y2)2'y
Uy= x2 + y2x2 -y2
u -
YY - (x2 + y2)2These derivatives exist, are continuous in R, and Uxx + Uyy = 0, so that u
is harmonic in R. Equation (6.6-4) gives
-y d x d x dy - y dx d ( 1 y )
dv = x2 + y2 x + x2 + y2 y = x2 + y2 = tan - -xv = tan-^1 '#.. + C = argz + C
x