File: Ex19
Solution for Ex. 19 in Derivative.pdf
September 28, 2006 5:50 pm
Prof. W. Kahan Math. 185 Page 1/1
Exercise 19:
By integrating derivatives of a harmonic
function
g(x, y) we recover one of its
harmonic conjugates h(x, y) and then set ƒ(x +
ı
y) := g(x, y) +
ı
h(x, y) to recover an analytic
function ƒ(z)
. The recovery of an analytic
expression
ƒ from a harmonic
expression
g that
is also a real analytic function of each of its arguments can be achieved more easily by setting
ƒ(z) := 2g((z
o
+z)/2,
ı
(z
o
–z)/2) for any z
o
= x
o
+
ı
y
o
inside the domain of g
; explain why. If
this procedure fails to recover a
function
ƒ analytic throughout the domain of g
, explain why.
Explanation:
A well-formed
expression
provides a formula to evaluate an analytic function by
means of finitely many operations of addition, subtraction, multiplication, division, roots, log,
cos, tanh,
etc
. For example, ƒ(z) := z
2
–
ı
/
z leads to a formula g(x, y) := x
2
– y
2
– y
/
(x
2
+ y
2
)
for the harmonic function g that satisfies ƒ(x
+
ı
y) = g(x, y) +
ı
h(x, y) when x and y are real.
This g(x, y) is a formula also for a real analytic function of each of x and y when they are
reinterpreted as complex variables. And after suitable complex expressions are substituted for x
and y in the formula 2g((w+z)/2,
ı
(w–z)/2) it simplifies to ƒ(z) + (an expression in w)
, which
recovers ƒ from g within a “constant” without integrating derivatives. Does this always work?
The complex conjugate of ƒ(x +
ı
y) := g(x, y) +
ı
h(x, y) is ƒ(x +
ı
y) := g(x, y) –
ı
h(x, y) when x
and y are real. This ƒ(z) is generally not an analytic function of z because g and –h violate
the Cauchy-Riemann conditions. However
Φ
(z) := ƒ(z)
is
an analytic function of z because
Φ
(x +
ı
y) = g(x, –y) –
ı
h(x, –y) for real x and y has real and imaginary parts g(x, –y) and
–h(x, –y) that inherit satisfaction of the Cauchy-Riemann conditions from g(x, y) and h(x, y)
.
For later reference note that the domain of
Φ
(z) is the complex conjugate of the domain of ƒ
.
We are given a harmonic g and wish to recover ƒ (within a constant) without first constructing
g
’s harmonic conjugate h (within a constant). Eliminating h from foregoing formulas yields
2g(x, y) = ƒ(x +
ı
y) +
Φ
(x –
ı
y)
. This relation among functions is also, presumably, a relation
among expressions (formulas), and as such is a relation among formulas into which independent
complex variables may be substituted for x and y because all three functions are analytic in their
arguments no matter whether real or complex. An apt substitution is suggested by the equations
x = (z+z)/2 and y =
ı
(z–z)/2 in which z is replaced by an independent complex variable w
,
say. Thus 2g((w+z)/2,
ı
(w–z)/2) = ƒ(z) +
Φ
(w) = ƒ(z) + ƒ(w) provided w is chosen in the
domain of
Φ
, which puts w in the domain of ƒ and of ƒ
. The last equation delivers a formula
for ƒ(z) within a constant when w is held constant. Almost.
This formula for ƒ can malfunction in two ways. One way occurs when w does not lie in the
domain of ƒ
; for example when g(x, y) = x
2
– y
2
– y
/
(x
2
+ y
2
) the choice w := 0
±
ı
0 makes
2g((w+z)/2,
ı
(w–z)/2) =
∞
instead of z
2
–
ı
/
z + (finite constant)
. This malfunction is predictable
because g(x, y) misbehaves around (0, 0) in the (x, y)-plane, so ƒ(z) must be expected to have
a singularity at z = 0
±
ı
0 in the z-plane. The second kind of malfunction is more subtle.
If not simply connected, the domain in the (x, y)-plane whereon g(x, y) is harmonic may extend
beyond the domain in the z-plane whereon ƒ(z) is differentiable. Try the Principal Value of
ƒ(z) := log(z) whose real part g(x, y) = log(x
2
+ y
2
)
/
2 is harmonic everywhere but at (0, 0)
; its
2g((w+z)/2,
ı
(w–z)/2) = log(w·z) is discontinuous across the ray z/w < 0 instead of z < 0 .
In general 2g((w+z)/2,
ı
(w–z)/2) – ƒ(z) may take several “constant” values. Try ƒ(z) :=
√
(2z)
.