More on Harmonic Functions · harmonic functions are real parts of analytic functions. Bernd...

Constructing Harmonic Conjugates Transforming Domains

More on Harmonic Functions

Bernd Schroder

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Introduction

1. The problem of finding a harmonic function with prescribedvalues on the boundary of its domain is called the Dirichletproblem.

2. Think of a drum membrane or a soap film that is stretched in aframe that has been bent to a certain shape. The way the frame isbent gives the boundary values. The shape of the membrane/filmis a solution of the Dirichlet problem.

3. The problem of finding a harmonic function with prescribedvalues for the normal derivative along the boundary of its domainis called the Neumann problem.

4. We will investigate how to solve the Dirichlet and Neumannproblems for certain simply connected domains.

5. The first step is to realize that, on simply connected domains,harmonic functions are real parts of analytic functions.

Introduction1. The problem of finding a harmonic function with prescribed

values on the boundary of its domain is called the Dirichletproblem.

2. Think of a drum membrane or a soap film that is stretched in aframe that has been bent to a certain shape.

The way the frame isbent gives the boundary values. The shape of the membrane/filmis a solution of the Dirichlet problem.

2. Think of a drum membrane or a soap film that is stretched in aframe that has been bent to a certain shape. The way the frame isbent gives the boundary values.

The shape of the membrane/filmis a solution of the Dirichlet problem.

Theorem.

Let D be a simply connected domain and let P(s, t) andQ(s, t) have continuous first order partial derivatives. If Pt = Qs, thenfor any simple closed curve B we have∫

BP(s, t) ds+Q(s, t) dt

interiorPt−Qs dA

Hence, in particular, if (x0,y0) and (x,y) are in D, then the integral∫C

P(s, t) ds+Q(s, t) dt (which is the line integral of the vector field(PQ

)along C) is the same for any simple curve C from (x0,y0) to

(x,y).

Proof. For a proof of Green’s Theorem (that’s the step in thebrackets), see (advanced) calculus.

Theorem. Let D be a simply connected domain and let P(s, t) andQ(s, t) have continuous first order partial derivatives.

If Pt = Qs, thenfor any simple closed curve B we have∫

interiorPt−Qs dA

(x,y).

Theorem. Let D be a simply connected domain and let P(s, t) andQ(s, t) have continuous first order partial derivatives. If Pt = Qs, thenfor any simple closed curve B we have∫

interiorPt−Qs dA

(x,y).

interiorPt−Qs dA

(x,y).

interiorPt−Qs dA

(x,y).

interiorPt−Qs dA

P(s, t) ds+Q(s, t) dt

(which is the line integral of the vector field(PQ

(x,y).

interiorPt−Qs dA

)along C)

is the same for any simple curve C from (x0,y0) to

(x,y).

interiorPt−Qs dA

(x,y).

interiorPt−Qs dA

(x,y).

Proof.

For a proof of Green’s Theorem (that’s the step in thebrackets), see (advanced) calculus.

interiorPt−Qs dA

(x,y).

Proof. For a proof of Green’s Theorem

(that’s the step in thebrackets), see (advanced) calculus.

interiorPt−Qs dA

(x,y).

Proof. For a proof of Green’s Theorem (that’s the step in thebrackets)

, see (advanced) calculus.

interiorPt−Qs dA

(x,y).

interiorPt−Qs dA

(x,y).

Theorem.

Let D be a simply connected domain, let u be a harmonicfunction on D and let (x0,y0) be a fixed point in D. Then

v(x,y) :=∫ (x,y)

(x0,y0)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

defines a harmonic conjugate of u on D.

Theorem. Let D be a simply connected domain, let u be a harmonicfunction on D and let (x0,y0) be a fixed point in D.

v(x,y) :=∫ (x,y)

(x0,y0)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

Theorem. Let D be a simply connected domain, let u be a harmonicfunction on D and let (x0,y0) be a fixed point in D. Then

v(x,y) :=∫ (x,y)

(x0,y0)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

Proof.

Because(−∂u

)t=−∂ 2u

∂ t2 =∂ 2u∂ s2 =

(∂u∂ s

)s, the function

v(x,y) =∫ (x,y)

(x0,y0)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

is well-defined by the previous theorem. Moreover,

∂xv(x,y) =

∫ (x,y)

(x0,y0)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds

= −∂u∂ t

(x,y) =−∂u∂y

and similarly we prove that∂v∂y

=∂u∂x

Proof. Because(−∂u

=−∂ 2u∂ t2 =

∂ 2u∂ s2 =

(∂u∂ s

)s, the function

v(x,y) =∫ (x,y)

(x0,y0)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

∂xv(x,y) =

∫ (x,y)

(x0,y0)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds

= −∂u∂ t

(x,y) =−∂u∂y

=∂u∂x

)t=−∂ 2u

∂ t2

=∂ 2u∂ s2 =

(∂u∂ s

)s, the function

v(x,y) =∫ (x,y)

(x0,y0)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

∂xv(x,y) =

∫ (x,y)

(x0,y0)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds

= −∂u∂ t

(x,y) =−∂u∂y

=∂u∂x

)t=−∂ 2u

∂ t2 =∂ 2u∂ s2

∂u∂ s

)s, the function

v(x,y) =∫ (x,y)

(x0,y0)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

∂xv(x,y) =

∫ (x,y)

(x0,y0)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds

= −∂u∂ t

(x,y) =−∂u∂y

=∂u∂x

)t=−∂ 2u

∂ t2 =∂ 2u∂ s2 =

(∂u∂ s

, the function

v(x,y) =∫ (x,y)

(x0,y0)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

∂xv(x,y) =

∫ (x,y)

(x0,y0)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds

= −∂u∂ t

(x,y) =−∂u∂y

=∂u∂x

)t=−∂ 2u

∂ t2 =∂ 2u∂ s2 =

(∂u∂ s

)s, the function

v(x,y) =∫ (x,y)

(x0,y0)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

is well-defined by the previous theorem.

Moreover,

∂xv(x,y) =

∫ (x,y)

(x0,y0)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds

= −∂u∂ t

(x,y) =−∂u∂y

=∂u∂x

)t=−∂ 2u

∂ t2 =∂ 2u∂ s2 =

(∂u∂ s

)s, the function

v(x,y) =∫ (x,y)

(x0,y0)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

∂xv(x,y)

∫ (x,y)

(x0,y0)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds

= −∂u∂ t

(x,y) =−∂u∂y

=∂u∂x

)t=−∂ 2u

∂ t2 =∂ 2u∂ s2 =

(∂u∂ s

)s, the function

v(x,y) =∫ (x,y)

(x0,y0)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

∂xv(x,y) =

∫ (x,y)

(x0,y0)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds

= −∂u∂ t

(x,y) =−∂u∂y

=∂u∂x

)t=−∂ 2u

∂ t2 =∂ 2u∂ s2 =

(∂u∂ s

)s, the function

v(x,y) =∫ (x,y)

(x0,y0)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

∂xv(x,y) =

∫ (x,y)

(x0,y0)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds

= −∂u∂ t

(x,y) =−∂u∂y

=∂u∂x

)t=−∂ 2u

∂ t2 =∂ 2u∂ s2 =

(∂u∂ s

)s, the function

v(x,y) =∫ (x,y)

(x0,y0)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

∂xv(x,y) =

∫ (x,y)

(x0,y0)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds

= −∂u∂ t

(x,y) =−∂u∂y

=∂u∂x

)t=−∂ 2u

∂ t2 =∂ 2u∂ s2 =

(∂u∂ s

)s, the function

v(x,y) =∫ (x,y)

(x0,y0)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

∂xv(x,y) =

∫ (x,y)

(x0,y0)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds

= −∂u∂ t

=−∂u∂y

=∂u∂x

)t=−∂ 2u

∂ t2 =∂ 2u∂ s2 =

(∂u∂ s

)s, the function

v(x,y) =∫ (x,y)

(x0,y0)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

∂xv(x,y) =

∫ (x,y)

(x0,y0)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds

= −∂u∂ t

(x,y) =−∂u∂y

=∂u∂x

)t=−∂ 2u

∂ t2 =∂ 2u∂ s2 =

(∂u∂ s

)s, the function

v(x,y) =∫ (x,y)

(x0,y0)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

∂xv(x,y) =

∫ (x,y)

(x0,y0)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds

= −∂u∂ t

(x,y) =−∂u∂y

=∂u∂x

.Bernd Schroder Louisiana Tech University, College of Engineering and Science

)t=−∂ 2u

∂ t2 =∂ 2u∂ s2 =

(∂u∂ s

)s, the function

v(x,y) =∫ (x,y)

(x0,y0)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

∂xv(x,y) =

∫ (x,y)

(x0,y0)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds

= −∂u∂ t

(x,y) =−∂u∂y

=∂u∂x

.Bernd Schroder Louisiana Tech University, College of Engineering and Science

Example.

Find a harmonic conjugate for u(x,y) = x2− y2 in theplane.

u(s, t) = s2− t2, so∂u∂ s

= 2s and −∂u∂ t

= 2t. Now consider

v(x,y) =∫ (x,y)

(0,0)2t ds+2s dt

=∫ (x,0)

(0,0)2t ds+2s dt +

∫ (x,y)

(x,0)2t ds+2s dt

= 2ts∣∣∣(s,t)=(x,0)

(s,t)=(0,0)+2st

∣∣∣(s,t)=(x,y)

(s,t)=(x,0)= 0−0+2xy−0 = 2xy

and we can easily verify via the Cauchy-Riemann equations that thisfunction is a harmonic conjugate of u.

Example. Find a harmonic conjugate for u(x,y) = x2− y2 in theplane.

u(s, t) = s2− t2, so∂u∂ s

= 2t. Now consider

v(x,y) =∫ (x,y)

(0,0)2t ds+2s dt

=∫ (x,0)

(0,0)2t ds+2s dt +

∫ (x,y)

(x,0)2t ds+2s dt

= 2ts∣∣∣(s,t)=(x,0)

(s,t)=(0,0)+2st

∣∣∣(s,t)=(x,y)

(s,t)=(x,0)= 0−0+2xy−0 = 2xy

u(s, t) = s2− t2

, so∂u∂ s

= 2t. Now consider

v(x,y) =∫ (x,y)

(0,0)2t ds+2s dt

=∫ (x,0)

(0,0)2t ds+2s dt +

∫ (x,y)

(x,0)2t ds+2s dt

= 2ts∣∣∣(s,t)=(x,0)

(s,t)=(0,0)+2st

∣∣∣(s,t)=(x,y)

(s,t)=(x,0)= 0−0+2xy−0 = 2xy

u(s, t) = s2− t2, so∂u∂ s

and −∂u∂ t

= 2t. Now consider

v(x,y) =∫ (x,y)

(0,0)2t ds+2s dt

=∫ (x,0)

(0,0)2t ds+2s dt +

∫ (x,y)

(x,0)2t ds+2s dt

= 2ts∣∣∣(s,t)=(x,0)

(s,t)=(0,0)+2st

∣∣∣(s,t)=(x,y)

(s,t)=(x,0)= 0−0+2xy−0 = 2xy

u(s, t) = s2− t2, so∂u∂ s

Now consider

v(x,y) =∫ (x,y)

(0,0)2t ds+2s dt

=∫ (x,0)

(0,0)2t ds+2s dt +

∫ (x,y)

(x,0)2t ds+2s dt

= 2ts∣∣∣(s,t)=(x,0)

(s,t)=(0,0)+2st

∣∣∣(s,t)=(x,y)

(s,t)=(x,0)= 0−0+2xy−0 = 2xy

u(s, t) = s2− t2, so∂u∂ s

= 2t. Now consider

v(x,y)

=∫ (x,y)

(0,0)2t ds+2s dt

=∫ (x,0)

(0,0)2t ds+2s dt +

∫ (x,y)

(x,0)2t ds+2s dt

= 2ts∣∣∣(s,t)=(x,0)

(s,t)=(0,0)+2st

∣∣∣(s,t)=(x,y)

(s,t)=(x,0)= 0−0+2xy−0 = 2xy

u(s, t) = s2− t2, so∂u∂ s

= 2t. Now consider

v(x,y) =∫ (x,y)

(0,0)2t ds+2s dt

=∫ (x,0)

(0,0)2t ds+2s dt +

∫ (x,y)

(x,0)2t ds+2s dt

= 2ts∣∣∣(s,t)=(x,0)

(s,t)=(0,0)+2st

∣∣∣(s,t)=(x,y)

(s,t)=(x,0)= 0−0+2xy−0 = 2xy

u(s, t) = s2− t2, so∂u∂ s

= 2t. Now consider

v(x,y) =∫ (x,y)

(0,0)2t ds+2s dt

=∫ (x,0)

(0,0)2t ds+2s dt

+∫ (x,y)

(x,0)2t ds+2s dt

= 2ts∣∣∣(s,t)=(x,0)

(s,t)=(0,0)+2st

∣∣∣(s,t)=(x,y)

(s,t)=(x,0)= 0−0+2xy−0 = 2xy

u(s, t) = s2− t2, so∂u∂ s

= 2t. Now consider

v(x,y) =∫ (x,y)

(0,0)2t ds+2s dt

=∫ (x,0)

(0,0)2t ds+2s dt +

∫ (x,y)

(x,0)2t ds+2s dt

= 2ts∣∣∣(s,t)=(x,0)

(s,t)=(0,0)+2st

∣∣∣(s,t)=(x,y)

(s,t)=(x,0)= 0−0+2xy−0 = 2xy

u(s, t) = s2− t2, so∂u∂ s

= 2t. Now consider

v(x,y) =∫ (x,y)

(0,0)2t ds+2s dt

=∫ (x,0)

(0,0)2t ds+2s dt +

∫ (x,y)

(x,0)2t ds+2s dt

= 2ts∣∣∣(s,t)=(x,0)

(s,t)=(0,0)

+2st∣∣∣(s,t)=(x,y)

(s,t)=(x,0)= 0−0+2xy−0 = 2xy

u(s, t) = s2− t2, so∂u∂ s

= 2t. Now consider

v(x,y) =∫ (x,y)

(0,0)2t ds+2s dt

=∫ (x,0)

(0,0)2t ds+2s dt +

∫ (x,y)

(x,0)2t ds+2s dt

= 2ts∣∣∣(s,t)=(x,0)

(s,t)=(0,0)+2st

∣∣∣(s,t)=(x,y)

(s,t)=(x,0)

= 0−0+2xy−0 = 2xy

u(s, t) = s2− t2, so∂u∂ s

= 2t. Now consider

v(x,y) =∫ (x,y)

(0,0)2t ds+2s dt

=∫ (x,0)

(0,0)2t ds+2s dt +

∫ (x,y)

(x,0)2t ds+2s dt

= 2ts∣∣∣(s,t)=(x,0)

(s,t)=(0,0)+2st

∣∣∣(s,t)=(x,y)

(s,t)=(x,0)= 0−0

+2xy−0 = 2xy

u(s, t) = s2− t2, so∂u∂ s

= 2t. Now consider

v(x,y) =∫ (x,y)

(0,0)2t ds+2s dt

=∫ (x,0)

(0,0)2t ds+2s dt +

∫ (x,y)

(x,0)2t ds+2s dt

= 2ts∣∣∣(s,t)=(x,0)

(s,t)=(0,0)+2st

∣∣∣(s,t)=(x,y)

(s,t)=(x,0)= 0−0+2xy−0

u(s, t) = s2− t2, so∂u∂ s

= 2t. Now consider

v(x,y) =∫ (x,y)

(0,0)2t ds+2s dt

=∫ (x,0)

(0,0)2t ds+2s dt +

∫ (x,y)

(x,0)2t ds+2s dt

= 2ts∣∣∣(s,t)=(x,0)

(s,t)=(0,0)+2st

∣∣∣(s,t)=(x,y)

(s,t)=(x,0)= 0−0+2xy−0 = 2xy

u(s, t) = s2− t2, so∂u∂ s

= 2t. Now consider

v(x,y) =∫ (x,y)

(0,0)2t ds+2s dt

=∫ (x,0)

(0,0)2t ds+2s dt +

∫ (x,y)

(x,0)2t ds+2s dt

= 2ts∣∣∣(s,t)=(x,0)

(s,t)=(0,0)+2st

∣∣∣(s,t)=(x,y)

(s,t)=(x,0)= 0−0+2xy−0 = 2xy

Theorem.

Let f = u+ iv be an analytic function that maps a domainDz to a domain Dw. If h(u,v) is a harmonic function on the domainDw, then H(x,y) := h

(u(x,y),v(x,y)

)is harmonic on Dz.

Proof.∂H∂x

=∂H∂u

∂u∂x

+∂H∂v

∂v∂x

∂ 2H∂x2 =

(∂H∂u

∂u∂x

+∂H∂v

∂v∂x

(∂H∂u

)∂u∂x

+∂H∂u

∂ 2u∂x2 +

(∂H∂v

)∂v∂x

+∂H∂v

∂ 2v∂x2

∂ 2H∂u2

∂u∂x

+∂ 2H∂u∂v

∂v∂x

)∂u∂x

+∂H∂u

∂ 2u∂x2

∂ 2H∂u∂v

∂u∂x

+∂ 2H∂v2

∂v∂x

)∂v∂x

+∂H∂v

∂ 2v∂x2

Theorem. Let f = u+ iv be an analytic function that maps a domainDz to a domain Dw.

If h(u,v) is a harmonic function on the domainDw, then H(x,y) := h

(u(x,y),v(x,y)

)is harmonic on Dz.

Proof.∂H∂x

=∂H∂u

∂u∂x

+∂H∂v

∂v∂x

∂ 2H∂x2 =

(∂H∂u

∂u∂x

+∂H∂v

∂v∂x

(∂H∂u

)∂u∂x

+∂H∂u

∂ 2u∂x2 +

(∂H∂v

)∂v∂x

+∂H∂v

∂ 2v∂x2

∂ 2H∂u2

∂u∂x

+∂ 2H∂u∂v

∂v∂x

)∂u∂x

+∂H∂u

∂ 2u∂x2

∂ 2H∂u∂v

∂u∂x

+∂ 2H∂v2

∂v∂x

)∂v∂x

+∂H∂v

∂ 2v∂x2

Theorem. Let f = u+ iv be an analytic function that maps a domainDz to a domain Dw. If h(u,v) is a harmonic function on the domainDw, then H(x,y) := h

(u(x,y),v(x,y)

)is harmonic on Dz.

Proof.∂H∂x

=∂H∂u

∂u∂x

+∂H∂v

∂v∂x

∂ 2H∂x2 =

(∂H∂u

∂u∂x

+∂H∂v

∂v∂x

(∂H∂u

)∂u∂x

+∂H∂u

∂ 2u∂x2 +

(∂H∂v

)∂v∂x

+∂H∂v

∂ 2v∂x2

∂ 2H∂u2

∂u∂x

+∂ 2H∂u∂v

∂v∂x

)∂u∂x

+∂H∂u

∂ 2u∂x2

∂ 2H∂u∂v

∂u∂x

+∂ 2H∂v2

∂v∂x

)∂v∂x

+∂H∂v

∂ 2v∂x2

(u(x,y),v(x,y)

)is harmonic on Dz.

Proof.

∂H∂x

=∂H∂u

∂u∂x

+∂H∂v

∂v∂x

∂ 2H∂x2 =

(∂H∂u

∂u∂x

+∂H∂v

∂v∂x

(∂H∂u

)∂u∂x

+∂H∂u

∂ 2u∂x2 +

(∂H∂v

)∂v∂x

+∂H∂v

∂ 2v∂x2

∂ 2H∂u2

∂u∂x

+∂ 2H∂u∂v

∂v∂x

)∂u∂x

+∂H∂u

∂ 2u∂x2

∂ 2H∂u∂v

∂u∂x

+∂ 2H∂v2

∂v∂x

)∂v∂x

+∂H∂v

∂ 2v∂x2

(u(x,y),v(x,y)

)is harmonic on Dz.

Proof.∂H∂x

=∂H∂u

∂u∂x

+∂H∂v

∂v∂x

∂ 2H∂x2 =

(∂H∂u

∂u∂x

+∂H∂v

∂v∂x

(∂H∂u

)∂u∂x

+∂H∂u

∂ 2u∂x2 +

(∂H∂v

)∂v∂x

+∂H∂v

∂ 2v∂x2

∂ 2H∂u2

∂u∂x

+∂ 2H∂u∂v

∂v∂x

)∂u∂x

+∂H∂u

∂ 2u∂x2

∂ 2H∂u∂v

∂u∂x

+∂ 2H∂v2

∂v∂x

)∂v∂x

+∂H∂v

∂ 2v∂x2

(u(x,y),v(x,y)

)is harmonic on Dz.

Proof.∂H∂x

=∂H∂u

∂u∂x

+∂H∂v

∂v∂x

∂ 2H∂x2 =

(∂H∂u

∂u∂x

+∂H∂v

∂v∂x

(∂H∂u

)∂u∂x

+∂H∂u

∂ 2u∂x2 +

(∂H∂v

)∂v∂x

+∂H∂v

∂ 2v∂x2

∂ 2H∂u2

∂u∂x

+∂ 2H∂u∂v

∂v∂x

)∂u∂x

+∂H∂u

∂ 2u∂x2

∂ 2H∂u∂v

∂u∂x

+∂ 2H∂v2

∂v∂x

)∂v∂x

+∂H∂v

∂ 2v∂x2

(u(x,y),v(x,y)

)is harmonic on Dz.

Proof.∂H∂x

=∂H∂u

∂u∂x

+∂H∂v

∂v∂x

∂ 2H∂x2

(∂H∂u

∂u∂x

+∂H∂v

∂v∂x

(∂H∂u

)∂u∂x

+∂H∂u

∂ 2u∂x2 +

(∂H∂v

)∂v∂x

+∂H∂v

∂ 2v∂x2

∂ 2H∂u2

∂u∂x

+∂ 2H∂u∂v

∂v∂x

)∂u∂x

+∂H∂u

∂ 2u∂x2

∂ 2H∂u∂v

∂u∂x

+∂ 2H∂v2

∂v∂x

)∂v∂x

+∂H∂v

∂ 2v∂x2

(u(x,y),v(x,y)

)is harmonic on Dz.

Proof.∂H∂x

=∂H∂u

∂u∂x

+∂H∂v

∂v∂x

∂ 2H∂x2 =

(∂H∂u

∂u∂x

+∂H∂v

∂v∂x

(∂H∂u

)∂u∂x

+∂H∂u

∂ 2u∂x2 +

(∂H∂v

)∂v∂x

+∂H∂v

∂ 2v∂x2

∂ 2H∂u2

∂u∂x

+∂ 2H∂u∂v

∂v∂x

)∂u∂x

+∂H∂u

∂ 2u∂x2

∂ 2H∂u∂v

∂u∂x

+∂ 2H∂v2

∂v∂x

)∂v∂x

+∂H∂v

∂ 2v∂x2

(u(x,y),v(x,y)

)is harmonic on Dz.

Proof.∂H∂x

=∂H∂u

∂u∂x

+∂H∂v

∂v∂x

∂ 2H∂x2 =

(∂H∂u

∂u∂x

+∂H∂v

∂v∂x

(∂H∂u

)∂u∂x

+∂H∂u

∂ 2u∂x2

(∂H∂v

)∂v∂x

+∂H∂v

∂ 2v∂x2

∂ 2H∂u2

∂u∂x

+∂ 2H∂u∂v

∂v∂x

)∂u∂x

+∂H∂u

∂ 2u∂x2

∂ 2H∂u∂v

∂u∂x

+∂ 2H∂v2

∂v∂x

)∂v∂x

+∂H∂v

∂ 2v∂x2

(u(x,y),v(x,y)

)is harmonic on Dz.

Proof.∂H∂x

=∂H∂u

∂u∂x

+∂H∂v

∂v∂x

∂ 2H∂x2 =

(∂H∂u

∂u∂x

+∂H∂v

∂v∂x

(∂H∂u

)∂u∂x

+∂H∂u

∂ 2u∂x2 +

(∂H∂v

)∂v∂x

+∂H∂v

∂ 2v∂x2

∂ 2H∂u2

∂u∂x

+∂ 2H∂u∂v

∂v∂x

)∂u∂x

+∂H∂u

∂ 2u∂x2

∂ 2H∂u∂v

∂u∂x

+∂ 2H∂v2

∂v∂x

)∂v∂x

+∂H∂v

∂ 2v∂x2

(u(x,y),v(x,y)

)is harmonic on Dz.

Proof.∂H∂x

=∂H∂u

∂u∂x

+∂H∂v

∂v∂x

∂ 2H∂x2 =

(∂H∂u

∂u∂x

+∂H∂v

∂v∂x

(∂H∂u

)∂u∂x

+∂H∂u

∂ 2u∂x2 +

(∂H∂v

)∂v∂x

+∂H∂v

∂ 2v∂x2

∂ 2H∂u2

∂u∂x

+∂ 2H∂u∂v

∂v∂x

)∂u∂x

+∂H∂u

∂ 2u∂x2

∂ 2H∂u∂v

∂u∂x

+∂ 2H∂v2

∂v∂x

)∂v∂x

+∂H∂v

∂ 2v∂x2

(u(x,y),v(x,y)

)is harmonic on Dz.

Proof.∂H∂x

=∂H∂u

∂u∂x

+∂H∂v

∂v∂x

∂ 2H∂x2 =

(∂H∂u

∂u∂x

+∂H∂v

∂v∂x

(∂H∂u

)∂u∂x

+∂H∂u

∂ 2u∂x2 +

(∂H∂v

)∂v∂x

+∂H∂v

∂ 2v∂x2

∂ 2H∂u2

∂u∂x

+∂ 2H∂u∂v

∂v∂x

)∂u∂x

+∂H∂u

∂ 2u∂x2

∂ 2H∂u∂v

∂u∂x

+∂ 2H∂v2

∂v∂x

)∂v∂x

+∂H∂v

∂ 2v∂x2

∂ 2H∂x2 =

(∂ 2H∂u2

∂u∂x

+∂ 2H∂u∂v

∂v∂x

)∂u∂x

+∂H∂u

∂ 2u∂x2

∂ 2H∂u∂v

∂u∂x

+∂ 2H∂v2

∂v∂x

)∂v∂x

+∂H∂v

∂ 2v∂x2

∂ 2H∂y2 =

(∂ 2H∂u2

∂u∂y

+∂ 2H∂u∂v

∂v∂y

)∂u∂y

+∂H∂u

∂ 2u∂y2

∂ 2H∂u∂v

∂u∂y

+∂ 2H∂v2

∂v∂y

)∂v∂y

+∂H∂v

∂ 2v∂y2

∂v∂y

∂u∂y

=∂u∂x

(−∂v

)∂ 2H∂x2 +

∂ 2H∂y2 =

∂ 2H∂u2

(∂u∂x

+∂ 2H∂v2

(∂v∂x

+∂ 2H∂u2

(∂u∂y

+∂ 2H∂v2

(∂v∂y

=∂ 2H∂u2

(∂u∂x

+∂ 2H∂v2

(∂u∂y

+∂ 2H∂u2

(∂u∂y

+∂ 2H∂v2

(∂u∂x

∂ 2H∂x2 =

(∂ 2H∂u2

∂u∂x

+∂ 2H∂u∂v

∂v∂x

)∂u∂x

+∂H∂u

∂ 2u∂x2

∂ 2H∂u∂v

∂u∂x

+∂ 2H∂v2

∂v∂x

)∂v∂x

+∂H∂v

∂ 2v∂x2

∂ 2H∂y2 =

(∂ 2H∂u2

∂u∂y

+∂ 2H∂u∂v

∂v∂y

)∂u∂y

+∂H∂u

∂ 2u∂y2

∂ 2H∂u∂v

∂u∂y

+∂ 2H∂v2

∂v∂y

)∂v∂y

+∂H∂v

∂ 2v∂y2

∂v∂y

∂u∂y

=∂u∂x

(−∂v

)∂ 2H∂x2 +

∂ 2H∂y2 =

∂ 2H∂u2

(∂u∂x

+∂ 2H∂v2

(∂v∂x

+∂ 2H∂u2

(∂u∂y

+∂ 2H∂v2

(∂v∂y

=∂ 2H∂u2

(∂u∂x

+∂ 2H∂v2

(∂u∂y

+∂ 2H∂u2

(∂u∂y

+∂ 2H∂v2

(∂u∂x

∂ 2H∂x2 =

(∂ 2H∂u2

∂u∂x

+∂ 2H∂u∂v

∂v∂x

)∂u∂x

+∂H∂u

∂ 2u∂x2

∂ 2H∂u∂v

∂u∂x

+∂ 2H∂v2

∂v∂x

)∂v∂x

+∂H∂v

∂ 2v∂x2

∂ 2H∂y2 =

(∂ 2H∂u2

∂u∂y

+∂ 2H∂u∂v

∂v∂y

)∂u∂y

+∂H∂u

∂ 2u∂y2

∂ 2H∂u∂v

∂u∂y

+∂ 2H∂v2

∂v∂y

)∂v∂y

+∂H∂v

∂ 2v∂y2

∂v∂y

∂u∂y

=∂u∂x

(−∂v

)∂ 2H∂x2 +

∂ 2H∂y2 =

∂ 2H∂u2

(∂u∂x

+∂ 2H∂v2

(∂v∂x

+∂ 2H∂u2

(∂u∂y

+∂ 2H∂v2

(∂v∂y

=∂ 2H∂u2

(∂u∂x

+∂ 2H∂v2

(∂u∂y

+∂ 2H∂u2

(∂u∂y

+∂ 2H∂v2

(∂u∂x

∂ 2H∂x2 =

(∂ 2H∂u2

∂u∂x

+∂ 2H∂u∂v

∂v∂x

)∂u∂x

+∂H∂u

∂ 2u∂x2

∂ 2H∂u∂v

∂u∂x

+∂ 2H∂v2

∂v∂x

)∂v∂x

+∂H∂v

∂ 2v∂x2

∂ 2H∂y2 =

(∂ 2H∂u2

∂u∂y

+∂ 2H∂u∂v

∂v∂y

)∂u∂y

+∂H∂u

∂ 2u∂y2

∂ 2H∂u∂v

∂u∂y

+∂ 2H∂v2

∂v∂y

)∂v∂y

+∂H∂v

∂ 2v∂y2

∂v∂y

∂u∂y

=∂u∂x

(−∂v

)∂ 2H∂x2 +

∂ 2H∂y2 =

∂ 2H∂u2

(∂u∂x

+∂ 2H∂v2

(∂v∂x

+∂ 2H∂u2

(∂u∂y

+∂ 2H∂v2

(∂v∂y

=∂ 2H∂u2

(∂u∂x

+∂ 2H∂v2

(∂u∂y

+∂ 2H∂u2

(∂u∂y

+∂ 2H∂v2

(∂u∂x

∂ 2H∂x2 =

(∂ 2H∂u2

∂u∂x

+∂ 2H∂u∂v

∂v∂x

)∂u∂x

+∂H∂u

∂ 2u∂x2

∂ 2H∂u∂v

∂u∂x

+∂ 2H∂v2

∂v∂x

)∂v∂x

+∂H∂v

∂ 2v∂x2

∂ 2H∂y2 =

(∂ 2H∂u2

∂u∂y

+∂ 2H∂u∂v

∂v∂y

)∂u∂y

+∂H∂u

∂ 2u∂y2

∂ 2H∂u∂v

∂u∂y

+∂ 2H∂v2

∂v∂y

)∂v∂y

+∂H∂v

∂ 2v∂y2

∂v∂y

∂u∂y

=∂u∂x

(−∂v

∂ 2H∂x2 +

∂ 2H∂y2 =

∂ 2H∂u2

(∂u∂x

+∂ 2H∂v2

(∂v∂x

+∂ 2H∂u2

(∂u∂y

+∂ 2H∂v2

(∂v∂y

=∂ 2H∂u2

(∂u∂x

+∂ 2H∂v2

(∂u∂y

+∂ 2H∂u2

(∂u∂y

+∂ 2H∂v2

(∂u∂x

∂ 2H∂x2 =

(∂ 2H∂u2

∂u∂x

+∂ 2H∂u∂v

∂v∂x

)∂u∂x

+∂H∂u

∂ 2u∂x2

∂ 2H∂u∂v

∂u∂x

+∂ 2H∂v2

∂v∂x

)∂v∂x

+∂H∂v

∂ 2v∂x2

∂ 2H∂y2 =

(∂ 2H∂u2

∂u∂y

+∂ 2H∂u∂v

∂v∂y

)∂u∂y

+∂H∂u

∂ 2u∂y2

∂ 2H∂u∂v

∂u∂y

+∂ 2H∂v2

∂v∂y

)∂v∂y

+∂H∂v

∂ 2v∂y2

∂v∂y

∂u∂y

=∂u∂x

(−∂v

)∂ 2H∂x2 +

∂ 2H∂y2

=∂ 2H∂u2

(∂u∂x

+∂ 2H∂v2

(∂v∂x

+∂ 2H∂u2

(∂u∂y

+∂ 2H∂v2

(∂v∂y

=∂ 2H∂u2

(∂u∂x

+∂ 2H∂v2

(∂u∂y

+∂ 2H∂u2

(∂u∂y

+∂ 2H∂v2

(∂u∂x

∂ 2H∂x2 =

(∂ 2H∂u2

∂u∂x

+∂ 2H∂u∂v

∂v∂x

)∂u∂x

+∂H∂u

∂ 2u∂x2

∂ 2H∂u∂v

∂u∂x

+∂ 2H∂v2

∂v∂x

)∂v∂x

+∂H∂v

∂ 2v∂x2

∂ 2H∂y2 =

(∂ 2H∂u2

∂u∂y

+∂ 2H∂u∂v

∂v∂y

)∂u∂y

+∂H∂u

∂ 2u∂y2

∂ 2H∂u∂v

∂u∂y

+∂ 2H∂v2

∂v∂y

)∂v∂y

+∂H∂v

∂ 2v∂y2

∂v∂y

∂u∂y

=∂u∂x

(−∂v

)∂ 2H∂x2 +

∂ 2H∂y2 =

∂ 2H∂u2

(∂u∂x

+∂ 2H∂v2

(∂v∂x

+∂ 2H∂u2

(∂u∂y

+∂ 2H∂v2

(∂v∂y

=∂ 2H∂u2

(∂u∂x

+∂ 2H∂v2

(∂u∂y

+∂ 2H∂u2

(∂u∂y

+∂ 2H∂v2

(∂u∂x

∂ 2H∂x2 =

(∂ 2H∂u2

∂u∂x

+∂ 2H∂u∂v

∂v∂x

)∂u∂x

+∂H∂u

∂ 2u∂x2

∂ 2H∂u∂v

∂u∂x

+∂ 2H∂v2

∂v∂x

)∂v∂x

+∂H∂v

∂ 2v∂x2

∂ 2H∂y2 =

(∂ 2H∂u2

∂u∂y

+∂ 2H∂u∂v

∂v∂y

)∂u∂y

+∂H∂u

∂ 2u∂y2

∂ 2H∂u∂v

∂u∂y

+∂ 2H∂v2

∂v∂y

)∂v∂y

+∂H∂v

∂ 2v∂y2

∂v∂y

∂u∂y

=∂u∂x

(−∂v

)∂ 2H∂x2 +

∂ 2H∂y2 =

∂ 2H∂u2

(∂u∂x

+∂ 2H∂v2

(∂v∂x

+∂ 2H∂u2

(∂u∂y

+∂ 2H∂v2

(∂v∂y

=∂ 2H∂u2

(∂u∂x

+∂ 2H∂v2

(∂u∂y

+∂ 2H∂u2

(∂u∂y

+∂ 2H∂v2

(∂u∂x

∂ 2H∂x2 =

(∂ 2H∂u2

∂u∂x

+∂ 2H∂u∂v

∂v∂x

)∂u∂x

+∂H∂u

∂ 2u∂x2

∂ 2H∂u∂v

∂u∂x

+∂ 2H∂v2

∂v∂x

)∂v∂x

+∂H∂v

∂ 2v∂x2

∂ 2H∂y2 =

(∂ 2H∂u2

∂u∂y

+∂ 2H∂u∂v

∂v∂y

)∂u∂y

+∂H∂u

∂ 2u∂y2

∂ 2H∂u∂v

∂u∂y

+∂ 2H∂v2

∂v∂y

)∂v∂y

+∂H∂v

∂ 2v∂y2

∂v∂y

∂u∂y

=∂u∂x

(−∂v

)∂ 2H∂x2 +

∂ 2H∂y2 =

∂ 2H∂u2

(∂u∂x

+∂ 2H∂v2

(∂v∂x

+∂ 2H∂u2

(∂u∂y

+∂ 2H∂v2

(∂v∂y

=∂ 2H∂u2

(∂u∂x

+∂ 2H∂v2

(∂u∂y

+∂ 2H∂u2

(∂u∂y

+∂ 2H∂v2

(∂u∂x

∂ 2H∂x2 =

(∂ 2H∂u2

∂u∂x

+∂ 2H∂u∂v

∂v∂x

)∂u∂x

+∂H∂u

∂ 2u∂x2

∂ 2H∂u∂v

∂u∂x

+∂ 2H∂v2

∂v∂x

)∂v∂x

+∂H∂v

∂ 2v∂x2

∂ 2H∂y2 =

(∂ 2H∂u2

∂u∂y

+∂ 2H∂u∂v

∂v∂y

)∂u∂y

+∂H∂u

∂ 2u∂y2

∂ 2H∂u∂v

∂u∂y

+∂ 2H∂v2

∂v∂y

)∂v∂y

+∂H∂v

∂ 2v∂y2

∂v∂y

∂u∂y

=∂u∂x

(−∂v

)∂ 2H∂x2 +

∂ 2H∂y2 =

∂ 2H∂u2

(∂u∂x

+∂ 2H∂v2

(∂v∂x

+∂ 2H∂u2

(∂u∂y

+∂ 2H∂v2

(∂v∂y

=∂ 2H∂u2

(∂u∂x

+∂ 2H∂v2

(∂u∂y

+∂ 2H∂u2

(∂u∂y

+∂ 2H∂v2

(∂u∂x

Example.

h(u,v) = eu cos(v) is harmonic on the disk u2 + v2 ≤ 1 and

f (z) =z−1z+1

is an analytic function that maps the right half plane

x > 0 to u2 + v2 ≤ 1. So, with

u = ℜ

(z−1z+1

)= ℜ

(z−1z+1

z+1z+1

)= ℜ

(|z|2− z+ z−1|z+1|2

)= ℜ

(x2 + y2−1+2iy

(1+ x)2 + y2

x2 + y2−1(1+ x)2 + y2 and

v = ℑ

(x2 + y2−1+2iy

(1+ x)2 + y2

2y(1+ x)2 + y2 the function

H(x,y) = h(u(x,y),v(x,y)

x2+y2−1(1+x)2+y2 cos

(1+ x)2 + y2

)is harmonic on the right half plane x > 0.

Example. h(u,v) = eu cos(v) is harmonic on the disk u2 + v2 ≤ 1

f (z) =z−1z+1

x > 0 to u2 + v2 ≤ 1. So, with

u = ℜ

(z−1z+1

)= ℜ

(z−1z+1

z+1z+1

)= ℜ

(|z|2− z+ z−1|z+1|2

)= ℜ

(x2 + y2−1+2iy

(1+ x)2 + y2

x2 + y2−1(1+ x)2 + y2 and

v = ℑ

(x2 + y2−1+2iy

(1+ x)2 + y2

x2+y2−1(1+x)2+y2 cos

(1+ x)2 + y2

Example. h(u,v) = eu cos(v) is harmonic on the disk u2 + v2 ≤ 1 and

f (z) =z−1z+1

x > 0 to u2 + v2 ≤ 1.

So, with

u = ℜ

(z−1z+1

)= ℜ

(z−1z+1

z+1z+1

)= ℜ

(|z|2− z+ z−1|z+1|2

)= ℜ

(x2 + y2−1+2iy

(1+ x)2 + y2

x2 + y2−1(1+ x)2 + y2 and

v = ℑ

(x2 + y2−1+2iy

(1+ x)2 + y2

x2+y2−1(1+x)2+y2 cos

(1+ x)2 + y2

f (z) =z−1z+1

x > 0 to u2 + v2 ≤ 1. So, with

(z−1z+1

)= ℜ

(z−1z+1

z+1z+1

)= ℜ

(|z|2− z+ z−1|z+1|2

)= ℜ

(x2 + y2−1+2iy

(1+ x)2 + y2

x2 + y2−1(1+ x)2 + y2 and

v = ℑ

(x2 + y2−1+2iy

(1+ x)2 + y2

x2+y2−1(1+x)2+y2 cos

(1+ x)2 + y2

f (z) =z−1z+1

x > 0 to u2 + v2 ≤ 1. So, with

u = ℜ

(z−1z+1

z+1z+1

)= ℜ

(|z|2− z+ z−1|z+1|2

)= ℜ

(x2 + y2−1+2iy

(1+ x)2 + y2

x2 + y2−1(1+ x)2 + y2 and

v = ℑ

(x2 + y2−1+2iy

(1+ x)2 + y2

x2+y2−1(1+x)2+y2 cos

(1+ x)2 + y2

f (z) =z−1z+1

x > 0 to u2 + v2 ≤ 1. So, with

u = ℜ

(z−1z+1

)= ℜ

(z−1z+1

z+1z+1

(|z|2− z+ z−1|z+1|2

)= ℜ

(x2 + y2−1+2iy

(1+ x)2 + y2

x2 + y2−1(1+ x)2 + y2 and

v = ℑ

(x2 + y2−1+2iy

(1+ x)2 + y2

x2+y2−1(1+x)2+y2 cos

(1+ x)2 + y2

f (z) =z−1z+1

x > 0 to u2 + v2 ≤ 1. So, with

u = ℜ

(z−1z+1

)= ℜ

(z−1z+1

z+1z+1

)= ℜ

(|z|2− z+ z−1|z+1|2

(x2 + y2−1+2iy

(1+ x)2 + y2

x2 + y2−1(1+ x)2 + y2 and

v = ℑ

(x2 + y2−1+2iy

(1+ x)2 + y2

x2+y2−1(1+x)2+y2 cos

(1+ x)2 + y2

f (z) =z−1z+1

x > 0 to u2 + v2 ≤ 1. So, with

u = ℜ

(z−1z+1

)= ℜ

(z−1z+1

z+1z+1

)= ℜ

(|z|2− z+ z−1|z+1|2

)= ℜ

(x2 + y2−1+2iy

(1+ x)2 + y2

=x2 + y2−1

(1+ x)2 + y2 and

v = ℑ

(x2 + y2−1+2iy

(1+ x)2 + y2

x2+y2−1(1+x)2+y2 cos

(1+ x)2 + y2

f (z) =z−1z+1

x > 0 to u2 + v2 ≤ 1. So, with

u = ℜ

(z−1z+1

)= ℜ

(z−1z+1

z+1z+1

)= ℜ

(|z|2− z+ z−1|z+1|2

)= ℜ

(x2 + y2−1+2iy

(1+ x)2 + y2

x2 + y2−1(1+ x)2 + y2

v = ℑ

(x2 + y2−1+2iy

(1+ x)2 + y2

x2+y2−1(1+x)2+y2 cos

(1+ x)2 + y2

f (z) =z−1z+1

x > 0 to u2 + v2 ≤ 1. So, with

u = ℜ

(z−1z+1

)= ℜ

(z−1z+1

z+1z+1

)= ℜ

(|z|2− z+ z−1|z+1|2

)= ℜ

(x2 + y2−1+2iy

(1+ x)2 + y2

x2 + y2−1(1+ x)2 + y2 and

(x2 + y2−1+2iy

(1+ x)2 + y2

x2+y2−1(1+x)2+y2 cos

(1+ x)2 + y2

f (z) =z−1z+1

x > 0 to u2 + v2 ≤ 1. So, with

u = ℜ

(z−1z+1

)= ℜ

(z−1z+1

z+1z+1

)= ℜ

(|z|2− z+ z−1|z+1|2

)= ℜ

(x2 + y2−1+2iy

(1+ x)2 + y2

x2 + y2−1(1+ x)2 + y2 and

v = ℑ

(x2 + y2−1+2iy

(1+ x)2 + y2

(1+ x)2 + y2 the function

x2+y2−1(1+x)2+y2 cos

(1+ x)2 + y2

f (z) =z−1z+1

x > 0 to u2 + v2 ≤ 1. So, with

u = ℜ

(z−1z+1

)= ℜ

(z−1z+1

z+1z+1

)= ℜ

(|z|2− z+ z−1|z+1|2

)= ℜ

(x2 + y2−1+2iy

(1+ x)2 + y2

x2 + y2−1(1+ x)2 + y2 and

v = ℑ

(x2 + y2−1+2iy

(1+ x)2 + y2

2y(1+ x)2 + y2

the function

x2+y2−1(1+x)2+y2 cos

(1+ x)2 + y2

f (z) =z−1z+1

x > 0 to u2 + v2 ≤ 1. So, with

u = ℜ

(z−1z+1

)= ℜ

(z−1z+1

z+1z+1

)= ℜ

(|z|2− z+ z−1|z+1|2

)= ℜ

(x2 + y2−1+2iy

(1+ x)2 + y2

x2 + y2−1(1+ x)2 + y2 and

v = ℑ

(x2 + y2−1+2iy

(1+ x)2 + y2

H(x,y)

= h(u(x,y),v(x,y)

x2+y2−1(1+x)2+y2 cos

(1+ x)2 + y2

f (z) =z−1z+1

x > 0 to u2 + v2 ≤ 1. So, with

u = ℜ

(z−1z+1

)= ℜ

(z−1z+1

z+1z+1

)= ℜ

(|z|2− z+ z−1|z+1|2

)= ℜ

(x2 + y2−1+2iy

(1+ x)2 + y2

x2 + y2−1(1+ x)2 + y2 and

v = ℑ

(x2 + y2−1+2iy

(1+ x)2 + y2

= ex2+y2−1

(1+x)2+y2 cos(

2y(1+ x)2 + y2

f (z) =z−1z+1

x > 0 to u2 + v2 ≤ 1. So, with

u = ℜ

(z−1z+1

)= ℜ

(z−1z+1

z+1z+1

)= ℜ

(|z|2− z+ z−1|z+1|2

)= ℜ

(x2 + y2−1+2iy

(1+ x)2 + y2

x2 + y2−1(1+ x)2 + y2 and

v = ℑ

(x2 + y2−1+2iy

(1+ x)2 + y2

x2+y2−1(1+x)2+y2 cos

(1+ x)2 + y2

is harmonic on the right half plane x > 0.

f (z) =z−1z+1

x > 0 to u2 + v2 ≤ 1. So, with

u = ℜ

(z−1z+1

)= ℜ

(z−1z+1

z+1z+1

)= ℜ

(|z|2− z+ z−1|z+1|2

)= ℜ

(x2 + y2−1+2iy

(1+ x)2 + y2

x2 + y2−1(1+ x)2 + y2 and

v = ℑ

(x2 + y2−1+2iy

(1+ x)2 + y2

x2+y2−1(1+x)2+y2 cos

(1+ x)2 + y2

Theorem.

Let f (x,y) = u(x,y)+ iv(x,y) be a conformal map on aneighborhood of a smooth arc C and let Γ := f [C]. If, along Γ, afunction h(u,v) satisfies the condition h = h0, where h0 is a realnumber, then H(x,y) := h

(u(x,y),v(x,y)

)satisfies the condition

H = h0 along C.

Proof. For all (x,y) on C we have that(u(x,y),v(x,y)

)is on Γ.

If h(u,v) = h0 for all (u,v) on Γ, then for all (x,y) on C we haveH(x,y) = h

(u(x,y),v(x,y)

)= h0.

Theorem. Let f (x,y) = u(x,y)+ iv(x,y) be a conformal map on aneighborhood of a smooth arc C and let Γ := f [C].

If, along Γ, afunction h(u,v) satisfies the condition h = h0, where h0 is a realnumber, then H(x,y) := h

(u(x,y),v(x,y)

H = h0 along C.

)is on Γ.

(u(x,y),v(x,y)

)= h0.

Theorem. Let f (x,y) = u(x,y)+ iv(x,y) be a conformal map on aneighborhood of a smooth arc C and let Γ := f [C]. If, along Γ, afunction h(u,v) satisfies the condition h = h0

, where h0 is a realnumber, then H(x,y) := h

(u(x,y),v(x,y)

H = h0 along C.

)is on Γ.

(u(x,y),v(x,y)

)= h0.

Theorem. Let f (x,y) = u(x,y)+ iv(x,y) be a conformal map on aneighborhood of a smooth arc C and let Γ := f [C]. If, along Γ, afunction h(u,v) satisfies the condition h = h0, where h0 is a realnumber

, then H(x,y) := h(u(x,y),v(x,y)

H = h0 along C.

)is on Γ.

(u(x,y),v(x,y)

)= h0.

Theorem. Let f (x,y) = u(x,y)+ iv(x,y) be a conformal map on aneighborhood of a smooth arc C and let Γ := f [C]. If, along Γ, afunction h(u,v) satisfies the condition h = h0, where h0 is a realnumber, then H(x,y) := h

(u(x,y),v(x,y)

H = h0 along C.

)is on Γ.

(u(x,y),v(x,y)

)= h0.

(u(x,y),v(x,y)

H = h0 along C.

Proof.

For all (x,y) on C we have that(u(x,y),v(x,y)

)is on Γ.

(u(x,y),v(x,y)

)= h0.

(u(x,y),v(x,y)

H = h0 along C.

)is on Γ.

(u(x,y),v(x,y)

)= h0.

(u(x,y),v(x,y)

H = h0 along C.

)is on Γ.

(u(x,y),v(x,y)

)= h0.

(u(x,y),v(x,y)

H = h0 along C.

)is on Γ.

(u(x,y),v(x,y)

)= h0.

Theorem.

Let f (x,y) = u(x,y)+ iv(x,y) be a conformal map on aneighborhood of a smooth arc C and let Γ := f [C]. If, along Γ, a

function h(u,v) satisfies the condition∂h∂n

= 0, where∂

∂ndenotes the

normal derivative, then H(x,y) := h(u(x,y),v(x,y)

)satisfies the

condition∂H∂N

= 0 along C.

Proof. Note that∂ f∂n

= ∇f ·~n, where~n is a unit normal vector to the

curve. Moreover, if a curve is parametrized by a(t) and b(t), then(dadtdbdt

)is tangential to the curve and

(−db

dtdadt

)is normal to the curve.

Theorem. Let f (x,y) = u(x,y)+ iv(x,y) be a conformal map on aneighborhood of a smooth arc C and let Γ := f [C].

If, along Γ, a

= 0, where∂

∂ndenotes the

)satisfies the

condition∂H∂N

= 0 along C.

(−db

dtdadt

Theorem. Let f (x,y) = u(x,y)+ iv(x,y) be a conformal map on aneighborhood of a smooth arc C and let Γ := f [C]. If, along Γ, a

, where∂

∂ndenotes the

)satisfies the

condition∂H∂N

= 0 along C.

(−db

dtdadt

= 0, where∂

∂ndenotes the

normal derivative

, then H(x,y) := h(u(x,y),v(x,y)

)satisfies the

condition∂H∂N

= 0 along C.

(−db

dtdadt

= 0, where∂

∂ndenotes the

)satisfies the

condition∂H∂N

along C.

(−db

dtdadt

= 0, where∂

∂ndenotes the

)satisfies the

condition∂H∂N

= 0 along C.

(−db

dtdadt

= 0, where∂

∂ndenotes the

)satisfies the

condition∂H∂N

= 0 along C.

Proof.

Note that∂ f∂n

(−db

dtdadt

= 0, where∂

∂ndenotes the

)satisfies the

condition∂H∂N

= 0 along C.

curve.

Moreover, if a curve is parametrized by a(t) and b(t), then(dadtdbdt

(−db

dtdadt

= 0, where∂

∂ndenotes the

)satisfies the

condition∂H∂N

= 0 along C.

curve. Moreover, if a curve is parametrized by a(t) and b(t)

, then(dadtdbdt

(−db

dtdadt

= 0, where∂

∂ndenotes the

)satisfies the

condition∂H∂N

= 0 along C.

)is tangential to the curve

and(−db

dtdadt

= 0, where∂

∂ndenotes the

)satisfies the

condition∂H∂N

= 0 along C.

(−db

dtdadt

= 0, where∂

∂ndenotes the

)satisfies the

condition∂H∂N

= 0 along C.

(−db

dtdadt

Proof.

With c :=

√(dxdt

)2we have

c∂H∂N

= c∇h(u(x,y),v(x,y)

)·~N =

∂x h(u(x,y),v(x,y)

∂y h(u(x,y),v(x,y)

)) ·(−dydt

(∂h∂u

(u(x,y),v(x,y)

)∂u∂x (x,y)+ ∂h

(u(x,y),v(x,y)

)∂v∂x(x,y)

∂h∂u

(u(x,y),v(x,y)

)∂u∂y (x,y)+ ∂h

(u(x,y),v(x,y)

)∂v∂y(x,y)

)·(−dy

dtdxdt

∂h∂u

(u(x,y),v(x,y)

)∂u∂x

(x,y)(−dy

∂h∂v

(u(x,y),v(x,y)

)∂v∂x

(x,y)(−dy

∂h∂u

(u(x,y),v(x,y)

)∂u∂y

(x,y)(

∂h∂v

(u(x,y),v(x,y)

)∂v∂y

(x,y)(

∂h∂u

(u(x,y),v(x,y)

)∂v∂y

(−dy

∂h∂v

(u(x,y),v(x,y)

)(−∂u

)(−dy

∂h∂u

(u(x,y),v(x,y)

)(−∂v

)(dxdt

∂h∂v

(u(x,y),v(x,y)

)∂u∂x

Proof. With c :=

√(dxdt

)2we have

c∂H∂N

)·~N =

)) ·(−dydt

(∂h∂u

(u(x,y),v(x,y)

)∂u∂x (x,y)+ ∂h

(u(x,y),v(x,y)

)∂v∂x(x,y)

∂h∂u

(u(x,y),v(x,y)

)∂u∂y (x,y)+ ∂h

(u(x,y),v(x,y)

)∂v∂y(x,y)

)·(−dy

dtdxdt

∂h∂u

(u(x,y),v(x,y)

)∂u∂x

(x,y)(−dy

∂h∂v

(u(x,y),v(x,y)

)∂v∂x

(x,y)(−dy

∂h∂u

(u(x,y),v(x,y)

)∂u∂y

(x,y)(

∂h∂v

(u(x,y),v(x,y)

)∂v∂y

(x,y)(

∂h∂u

(u(x,y),v(x,y)

)∂v∂y

(−dy

∂h∂v

(u(x,y),v(x,y)

)(−∂u

)(−dy

∂h∂u

(u(x,y),v(x,y)

)(−∂v

)(dxdt

∂h∂v

(u(x,y),v(x,y)

)∂u∂x

Proof. With c :=

√(dxdt

)2we have

c∂H∂N

)·~N =

)) ·(−dydt

(∂h∂u

(u(x,y),v(x,y)

)∂u∂x (x,y)+ ∂h

(u(x,y),v(x,y)

)∂v∂x(x,y)

∂h∂u

(u(x,y),v(x,y)

)∂u∂y (x,y)+ ∂h

(u(x,y),v(x,y)

)∂v∂y(x,y)

)·(−dy

dtdxdt

∂h∂u

(u(x,y),v(x,y)

)∂u∂x

(x,y)(−dy

∂h∂v

(u(x,y),v(x,y)

)∂v∂x

(x,y)(−dy

∂h∂u

(u(x,y),v(x,y)

)∂u∂y

(x,y)(

∂h∂v

(u(x,y),v(x,y)

)∂v∂y

(x,y)(

∂h∂u

(u(x,y),v(x,y)

)∂v∂y

(−dy

∂h∂v

(u(x,y),v(x,y)

)(−∂u

)(−dy

∂h∂u

(u(x,y),v(x,y)

)(−∂v

)(dxdt

∂h∂v

(u(x,y),v(x,y)

)∂u∂x

Proof. With c :=

√(dxdt

)2we have

c∂H∂N

)) ·(−dydt

(∂h∂u

(u(x,y),v(x,y)

)∂u∂x (x,y)+ ∂h

(u(x,y),v(x,y)

)∂v∂x(x,y)

∂h∂u

(u(x,y),v(x,y)

)∂u∂y (x,y)+ ∂h

(u(x,y),v(x,y)

)∂v∂y(x,y)

)·(−dy

dtdxdt

∂h∂u

(u(x,y),v(x,y)

)∂u∂x

(x,y)(−dy

∂h∂v

(u(x,y),v(x,y)

)∂v∂x

(x,y)(−dy

∂h∂u

(u(x,y),v(x,y)

)∂u∂y

(x,y)(

∂h∂v

(u(x,y),v(x,y)

)∂v∂y

(x,y)(

∂h∂u

(u(x,y),v(x,y)

)∂v∂y

(−dy

∂h∂v

(u(x,y),v(x,y)

)(−∂u

)(−dy

∂h∂u

(u(x,y),v(x,y)

)(−∂v

)(dxdt

∂h∂v

(u(x,y),v(x,y)

)∂u∂x

Proof. With c :=

√(dxdt

)2we have

c∂H∂N

)·~N =

)) ·(−dydt

(∂h∂u

(u(x,y),v(x,y)

)∂u∂x (x,y)+ ∂h

(u(x,y),v(x,y)

)∂v∂x(x,y)

∂h∂u

(u(x,y),v(x,y)

)∂u∂y (x,y)+ ∂h

(u(x,y),v(x,y)

)∂v∂y(x,y)

)·(−dy

dtdxdt

∂h∂u

(u(x,y),v(x,y)

)∂u∂x

(x,y)(−dy

∂h∂v

(u(x,y),v(x,y)

)∂v∂x

(x,y)(−dy

∂h∂u

(u(x,y),v(x,y)

)∂u∂y

(x,y)(

∂h∂v

(u(x,y),v(x,y)

)∂v∂y

(x,y)(

∂h∂u

(u(x,y),v(x,y)

)∂v∂y

(−dy

∂h∂v

(u(x,y),v(x,y)

)(−∂u

)(−dy

∂h∂u

(u(x,y),v(x,y)

)(−∂v

)(dxdt

∂h∂v

(u(x,y),v(x,y)

)∂u∂x

Proof. With c :=

√(dxdt

)2we have

c∂H∂N

)·~N =

)) ·(−dydt

(∂h∂u

(u(x,y),v(x,y)

)∂u∂x (x,y)+ ∂h

(u(x,y),v(x,y)

)∂v∂x(x,y)

∂h∂u

(u(x,y),v(x,y)

)∂u∂y (x,y)+ ∂h

(u(x,y),v(x,y)

)∂v∂y(x,y)

)·(−dy

dtdxdt

=∂h∂u

(u(x,y),v(x,y)

)∂u∂x

(x,y)(−dy

∂h∂v

(u(x,y),v(x,y)

)∂v∂x

(x,y)(−dy

∂h∂u

(u(x,y),v(x,y)

)∂u∂y

(x,y)(

∂h∂v

(u(x,y),v(x,y)

)∂v∂y

(x,y)(

∂h∂u

(u(x,y),v(x,y)

)∂v∂y

(−dy

∂h∂v

(u(x,y),v(x,y)

)(−∂u

)(−dy

∂h∂u

(u(x,y),v(x,y)

)(−∂v

)(dxdt

∂h∂v

(u(x,y),v(x,y)

)∂u∂x

Proof. With c :=

√(dxdt

)2we have

c∂H∂N

)·~N =

)) ·(−dydt

(∂h∂u

(u(x,y),v(x,y)

)∂u∂x (x,y)+ ∂h

(u(x,y),v(x,y)

)∂v∂x(x,y)

∂h∂u

(u(x,y),v(x,y)

)∂u∂y (x,y)+ ∂h

(u(x,y),v(x,y)

)∂v∂y(x,y)

)·(−dy

dtdxdt

∂h∂u

(u(x,y),v(x,y)

)∂u∂x

(x,y)(−dy

∂h∂v

(u(x,y),v(x,y)

)∂v∂x

(x,y)(−dy

+∂h∂u

(u(x,y),v(x,y)

)∂u∂y

(x,y)(

∂h∂v

(u(x,y),v(x,y)

)∂v∂y

(x,y)(

∂h∂u

(u(x,y),v(x,y)

)∂v∂y

(−dy

∂h∂v

(u(x,y),v(x,y)

)(−∂u

)(−dy

∂h∂u

(u(x,y),v(x,y)

)(−∂v

)(dxdt

∂h∂v

(u(x,y),v(x,y)

)∂u∂x

Proof. With c :=

√(dxdt

)2we have

c∂H∂N

)·~N =

)) ·(−dydt

(∂h∂u

(u(x,y),v(x,y)

)∂u∂x (x,y)+ ∂h

(u(x,y),v(x,y)

)∂v∂x(x,y)

∂h∂u

(u(x,y),v(x,y)

)∂u∂y (x,y)+ ∂h

(u(x,y),v(x,y)

)∂v∂y(x,y)

)·(−dy

dtdxdt

∂h∂u

(u(x,y),v(x,y)

)∂u∂x

(x,y)(−dy

∂h∂v

(u(x,y),v(x,y)

)∂v∂x

(x,y)(−dy

∂h∂u

(u(x,y),v(x,y)

)∂u∂y

(x,y)(

∂h∂v

(u(x,y),v(x,y)

)∂v∂y

(x,y)(

=∂h∂u

(u(x,y),v(x,y)

)∂v∂y

(−dy

∂h∂v

(u(x,y),v(x,y)

)(−∂u

)(−dy

∂h∂u

(u(x,y),v(x,y)

)(−∂v

)(dxdt

∂h∂v

(u(x,y),v(x,y)

)∂u∂x

Proof. With c :=

√(dxdt

)2we have

c∂H∂N

)·~N =

)) ·(−dydt

(∂h∂u

(u(x,y),v(x,y)

)∂u∂x (x,y)+ ∂h

(u(x,y),v(x,y)

)∂v∂x(x,y)

∂h∂u

(u(x,y),v(x,y)

)∂u∂y (x,y)+ ∂h

(u(x,y),v(x,y)

)∂v∂y(x,y)

)·(−dy

dtdxdt

∂h∂u

(u(x,y),v(x,y)

)∂u∂x

(x,y)(−dy

∂h∂v

(u(x,y),v(x,y)

)∂v∂x

(x,y)(−dy

∂h∂u

(u(x,y),v(x,y)

)∂u∂y

(x,y)(

∂h∂v

(u(x,y),v(x,y)

)∂v∂y

(x,y)(

∂h∂u

(u(x,y),v(x,y)

)∂v∂y

(−dy

∂h∂v

(u(x,y),v(x,y)

)(−∂u

)(−dy

+∂h∂u

(u(x,y),v(x,y)

)(−∂v

)(dxdt

∂h∂v

(u(x,y),v(x,y)

)∂u∂x

Proof. With c :=

√(dxdt

)2we have

c∂H∂N

)·~N =

)) ·(−dydt

(∂h∂u

(u(x,y),v(x,y)

)∂u∂x (x,y)+ ∂h

(u(x,y),v(x,y)

)∂v∂x(x,y)

∂h∂u

(u(x,y),v(x,y)

)∂u∂y (x,y)+ ∂h

(u(x,y),v(x,y)

)∂v∂y(x,y)

)·(−dy

dtdxdt

∂h∂u

(u(x,y),v(x,y)

)∂u∂x

(x,y)(−dy

∂h∂v

(u(x,y),v(x,y)

)∂v∂x

(x,y)(−dy

∂h∂u

(u(x,y),v(x,y)

)∂u∂y

(x,y)(

∂h∂v

(u(x,y),v(x,y)

)∂v∂y

(x,y)(

∂h∂u

(u(x,y),v(x,y)

)∂v∂y

(−dy

∂h∂v

(u(x,y),v(x,y)

)(−∂u

)(−dy

∂h∂u

(u(x,y),v(x,y)

)(−∂v

)(dxdt

∂h∂v

(u(x,y),v(x,y)

)∂u∂x

)Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proof.

=∂h∂u

(u(x,y),v(x,y)

)∂v∂y

(−dy

∂h∂v

(u(x,y),v(x,y)

)(−∂u

)(−dy

∂h∂u

(u(x,y),v(x,y)

)(−∂v

)(dxdt

∂h∂v

(u(x,y),v(x,y)

)∂u∂x

∂h∂u

(u(x,y),v(x,y)

)(−dv

∂h∂v

(u(x,y),v(x,y)

∂h∂u

(u(x,y),v(x,y)

)∂h∂v

(u(x,y),v(x,y)

)) ·(−dvdt

∂h∂u

(u(x,y),v(x,y)

)∂h∂v

(u(x,y),v(x,y)

)) ·~n√(

Proof.

=∂h∂u

(u(x,y),v(x,y)

)∂v∂y

(−dy

∂h∂v

(u(x,y),v(x,y)

)(−∂u

)(−dy

∂h∂u

(u(x,y),v(x,y)

)(−∂v

)(dxdt

∂h∂v

(u(x,y),v(x,y)

)∂u∂x

=∂h∂u

(u(x,y),v(x,y)

)(−dv

∂h∂v

(u(x,y),v(x,y)

∂h∂u

(u(x,y),v(x,y)

)∂h∂v

(u(x,y),v(x,y)

)) ·(−dvdt

∂h∂u

(u(x,y),v(x,y)

)∂h∂v

(u(x,y),v(x,y)

)) ·~n√(

Proof.

=∂h∂u

(u(x,y),v(x,y)

)∂v∂y

(−dy

∂h∂v

(u(x,y),v(x,y)

)(−∂u

)(−dy

∂h∂u

(u(x,y),v(x,y)

)(−∂v

)(dxdt

∂h∂v

(u(x,y),v(x,y)

)∂u∂x

∂h∂u

(u(x,y),v(x,y)

)(−dv

∂h∂v

(u(x,y),v(x,y)

∂h∂u

(u(x,y),v(x,y)

)∂h∂v

(u(x,y),v(x,y)

)) ·(−dvdt

∂h∂u

(u(x,y),v(x,y)

)∂h∂v

(u(x,y),v(x,y)

)) ·~n√(

Proof.

=∂h∂u

(u(x,y),v(x,y)

)∂v∂y

(−dy

∂h∂v

(u(x,y),v(x,y)

)(−∂u

)(−dy

∂h∂u

(u(x,y),v(x,y)

)(−∂v

)(dxdt

∂h∂v

(u(x,y),v(x,y)

)∂u∂x

∂h∂u

(u(x,y),v(x,y)

)(−dv

∂h∂v

(u(x,y),v(x,y)

∂h∂u

(u(x,y),v(x,y)

)∂h∂v

(u(x,y),v(x,y)

)) ·(−dvdt

∂h∂u

(u(x,y),v(x,y)

)∂h∂v

(u(x,y),v(x,y)

)) ·~n√(

Proof.

=∂h∂u

(u(x,y),v(x,y)

)∂v∂y

(−dy

∂h∂v

(u(x,y),v(x,y)

)(−∂u

)(−dy

∂h∂u

(u(x,y),v(x,y)

)(−∂v

)(dxdt

∂h∂v

(u(x,y),v(x,y)

)∂u∂x

∂h∂u

(u(x,y),v(x,y)

)(−dv

∂h∂v

(u(x,y),v(x,y)

∂h∂u

(u(x,y),v(x,y)

)∂h∂v

(u(x,y),v(x,y)

)) ·(−dvdt

∂h∂u

(u(x,y),v(x,y)

)∂h∂v

(u(x,y),v(x,y)

)) ·~n√(

Proof.

=∂h∂u

(u(x,y),v(x,y)

)∂v∂y

(−dy

∂h∂v

(u(x,y),v(x,y)

)(−∂u

)(−dy

∂h∂u

(u(x,y),v(x,y)

)(−∂v

)(dxdt

∂h∂v

(u(x,y),v(x,y)

)∂u∂x

∂h∂u

(u(x,y),v(x,y)

)(−dv

∂h∂v

(u(x,y),v(x,y)

∂h∂u

(u(x,y),v(x,y)

)∂h∂v

(u(x,y),v(x,y)

)) ·(−dvdt

∂h∂u

(u(x,y),v(x,y)

)∂h∂v

(u(x,y),v(x,y)

)) ·~n√(

Proof.

=∂h∂u

(u(x,y),v(x,y)

)∂v∂y

(−dy

∂h∂v

(u(x,y),v(x,y)

)(−∂u

)(−dy

∂h∂u

(u(x,y),v(x,y)

)(−∂v

)(dxdt

∂h∂v

(u(x,y),v(x,y)

)∂u∂x

∂h∂u

(u(x,y),v(x,y)

)(−dv

∂h∂v

(u(x,y),v(x,y)

∂h∂u

(u(x,y),v(x,y)

)∂h∂v

(u(x,y),v(x,y)

)) ·(−dvdt

∂h∂u

(u(x,y),v(x,y)

)∂h∂v

(u(x,y),v(x,y)

)) ·~n√(

Example.

h(u,v) = v is harmonic on the right half plane u≥ 0 with∂h∂n

=∂v∂u

= 0 and f (z) =z+11− z

is an analytic function that maps the

unit disk x2 + y2 ≤ 1 to the right half plane u > 0. So,

(z+11− z

)= ℑ

(z+11− z

1− z1− z

)= ℑ

(z+1− z−|z|2

(1− x)2 + y2

)= ℑ

(1+2iy− x2− y2

(1− x)2 + y2

2y(1− x)2 + y2

is harmonic on the unit disk and it satisfies∂H∂N

= 0 on the boundary.

Example. h(u,v) = v is harmonic on the right half plane u≥ 0

with∂h∂n

=∂v∂u

= 0 and f (z) =z+11− z

(z+11− z

)= ℑ

(z+11− z

1− z1− z

)= ℑ

(z+1− z−|z|2

(1− x)2 + y2

)= ℑ

(1+2iy− x2− y2

(1− x)2 + y2

2y(1− x)2 + y2

Example. h(u,v) = v is harmonic on the right half plane u≥ 0 with∂h∂n

=∂v∂u

= 0 and f (z) =z+11− z

(z+11− z

)= ℑ

(z+11− z

1− z1− z

)= ℑ

(z+1− z−|z|2

(1− x)2 + y2

)= ℑ

(1+2iy− x2− y2

(1− x)2 + y2

2y(1− x)2 + y2

=∂v∂u

= 0 and f (z) =z+11− z

(z+11− z

)= ℑ

(z+11− z

1− z1− z

)= ℑ

(z+1− z−|z|2

(1− x)2 + y2

)= ℑ

(1+2iy− x2− y2

(1− x)2 + y2

2y(1− x)2 + y2

=∂v∂u

and f (z) =z+11− z

(z+11− z

)= ℑ

(z+11− z

1− z1− z

)= ℑ

(z+1− z−|z|2

(1− x)2 + y2

)= ℑ

(1+2iy− x2− y2

(1− x)2 + y2

2y(1− x)2 + y2

=∂v∂u

= 0 and f (z) =z+11− z

unit disk x2 + y2 ≤ 1 to the right half plane u > 0.

(z+11− z

)= ℑ

(z+11− z

1− z1− z

)= ℑ

(z+1− z−|z|2

(1− x)2 + y2

)= ℑ

(1+2iy− x2− y2

(1− x)2 + y2

2y(1− x)2 + y2

=∂v∂u

= 0 and f (z) =z+11− z

(z+11− z

1− z1− z

)= ℑ

(z+1− z−|z|2

(1− x)2 + y2

)= ℑ

(1+2iy− x2− y2

(1− x)2 + y2

2y(1− x)2 + y2

=∂v∂u

= 0 and f (z) =z+11− z

(z+11− z

)= ℑ

(z+11− z

1− z1− z

(z+1− z−|z|2

(1− x)2 + y2

)= ℑ

(1+2iy− x2− y2

(1− x)2 + y2

2y(1− x)2 + y2

=∂v∂u

= 0 and f (z) =z+11− z

(z+11− z

)= ℑ

(z+11− z

1− z1− z

)= ℑ

(z+1− z−|z|2

(1− x)2 + y2

(1+2iy− x2− y2

(1− x)2 + y2

2y(1− x)2 + y2

=∂v∂u

= 0 and f (z) =z+11− z

(z+11− z

)= ℑ

(z+11− z

1− z1− z

)= ℑ

(z+1− z−|z|2

(1− x)2 + y2

)= ℑ

(1+2iy− x2− y2

(1− x)2 + y2

=∂v∂u

= 0 and f (z) =z+11− z

(z+11− z

)= ℑ

(z+11− z

1− z1− z

)= ℑ

(z+1− z−|z|2

(1− x)2 + y2

)= ℑ

(1+2iy− x2− y2

(1− x)2 + y2

2y(1− x)2 + y2

=∂v∂u

= 0 and f (z) =z+11− z

(z+11− z

)= ℑ

(z+11− z

1− z1− z

)= ℑ

(z+1− z−|z|2

(1− x)2 + y2

)= ℑ

(1+2iy− x2− y2

(1− x)2 + y2

2y(1− x)2 + y2

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