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

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Constructing Harmonic Conjugates Transforming Domains

More on Harmonic Functions

Bernd Schroder

Bernd Schroder Louisiana Tech University, College of Engineering and Science


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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.



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Introduction1. The problem of finding a harmonic function with prescribed

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







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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.






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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.






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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

]= 0.

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.



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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

]= 0.




(x,y).




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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

]= 0.




(x,y).




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[=∫

interiorPt−Qs dA

]

= 0.




(x,y).




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[=∫

interiorPt−Qs dA

]= 0.




(x,y).




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[=∫

interiorPt−Qs dA

]= 0.


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

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


(x,y).




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[=∫

interiorPt−Qs dA

]= 0.



)along C)

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

(x,y).




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[=∫

interiorPt−Qs dA

]= 0.




(x,y).




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[=∫

interiorPt−Qs dA

]= 0.




(x,y).

Proof.

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



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[=∫

interiorPt−Qs dA

]= 0.




(x,y).

Proof. For a proof of Green’s Theorem

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



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[=∫

interiorPt−Qs dA

]= 0.




(x,y).

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

, see (advanced) calculus.



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[=∫

interiorPt−Qs dA

]= 0.




(x,y).




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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.



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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




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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




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Proof.

Because(−∂u

∂ t

)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

∫ (x,y)

(x0,y0)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

1h

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

1h

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds

= −∂u∂ t

(x,y) =−∂u∂y

(x,y)

and similarly we prove that∂v∂y

=∂u∂x

.



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Proof. Because(−∂u

∂ t

)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

∫ (x,y)

(x0,y0)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

1h

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

1h

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds

= −∂u∂ t

(x,y) =−∂u∂y

(x,y)


=∂u∂x

.



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∂ t

)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

∫ (x,y)

(x0,y0)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

1h

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

1h

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds

= −∂u∂ t

(x,y) =−∂u∂y

(x,y)


=∂u∂x

.



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∂ t

)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

∫ (x,y)

(x0,y0)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

1h

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

1h

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds

= −∂u∂ t

(x,y) =−∂u∂y

(x,y)


=∂u∂x

.



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∂ t

)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

∫ (x,y)

(x0,y0)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

1h

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

1h

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds

= −∂u∂ t

(x,y) =−∂u∂y

(x,y)


=∂u∂x

.



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∂ t

)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

∫ (x,y)

(x0,y0)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

1h

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

1h

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds

= −∂u∂ t

(x,y) =−∂u∂y

(x,y)


=∂u∂x

.



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∂ t

)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

∫ (x,y)

(x0,y0)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

1h

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

1h

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds

= −∂u∂ t

(x,y) =−∂u∂y

(x,y)


=∂u∂x

.



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∂ t

)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

∫ (x,y)

(x0,y0)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

1h

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

1h

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds

= −∂u∂ t

(x,y) =−∂u∂y

(x,y)


=∂u∂x

.



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∂ t

)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

∫ (x,y)

(x0,y0)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

1h

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

1h

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds

= −∂u∂ t

(x,y)

=−∂u∂y

(x,y)


=∂u∂x

.



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∂ t

)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

∫ (x,y)

(x0,y0)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

1h

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

1h

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds

= −∂u∂ t

(x,y) =−∂u∂y

(x,y)


=∂u∂x

.



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∂ t

)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

∫ (x,y)

(x0,y0)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

1h

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds+

∂u∂ s

(s, t) dt

= limh→0

1h

∫ (x+h,y)

(x,y)−∂u

∂ t(s, t) ds

= −∂u∂ t

(x,y) =−∂u∂y

(x,y)


=∂u∂x

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


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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.



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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




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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




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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




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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




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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




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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




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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




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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




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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




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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




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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




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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




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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




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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 =

∂

∂x

(∂H∂u

∂u∂x

+∂H∂v

∂v∂x

)=

∂

∂x

(∂H∂u

)∂u∂x

+∂H∂u

∂ 2u∂x2 +

∂

∂x

(∂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



logo1


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 =

∂

∂x

(∂H∂u

∂u∂x

+∂H∂v

∂v∂x

)=

∂

∂x

(∂H∂u

)∂u∂x

+∂H∂u

∂ 2u∂x2 +

∂

∂x

(∂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



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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 =

∂

∂x

(∂H∂u

∂u∂x

+∂H∂v

∂v∂x

)=

∂

∂x

(∂H∂u

)∂u∂x

+∂H∂u

∂ 2u∂x2 +

∂

∂x

(∂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



logo1



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

)is harmonic on Dz.

Proof.

∂H∂x

=∂H∂u

∂u∂x

+∂H∂v

∂v∂x

∂ 2H∂x2 =

∂

∂x

(∂H∂u

∂u∂x

+∂H∂v

∂v∂x

)=

∂

∂x

(∂H∂u

)∂u∂x

+∂H∂u

∂ 2u∂x2 +

∂

∂x

(∂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



logo1



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

)is harmonic on Dz.

Proof.∂H∂x

=∂H∂u

∂u∂x

+∂H∂v

∂v∂x

∂ 2H∂x2 =

∂

∂x

(∂H∂u

∂u∂x

+∂H∂v

∂v∂x

)=

∂

∂x

(∂H∂u

)∂u∂x

+∂H∂u

∂ 2u∂x2 +

∂

∂x

(∂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



logo1



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

)is harmonic on Dz.

Proof.∂H∂x

=∂H∂u

∂u∂x

+∂H∂v

∂v∂x

∂ 2H∂x2

=∂

∂x

(∂H∂u

∂u∂x

+∂H∂v

∂v∂x

)=

∂

∂x

(∂H∂u

)∂u∂x

+∂H∂u

∂ 2u∂x2 +

∂

∂x

(∂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



logo1



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

)is harmonic on Dz.

Proof.∂H∂x

=∂H∂u

∂u∂x

+∂H∂v

∂v∂x

∂ 2H∂x2 =

∂

∂x

(∂H∂u

∂u∂x

+∂H∂v

∂v∂x

)

=∂

∂x

(∂H∂u

)∂u∂x

+∂H∂u

∂ 2u∂x2 +

∂

∂x

(∂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



logo1



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

)is harmonic on Dz.

Proof.∂H∂x

=∂H∂u

∂u∂x

+∂H∂v

∂v∂x

∂ 2H∂x2 =

∂

∂x

(∂H∂u

∂u∂x

+∂H∂v

∂v∂x

)=

∂

∂x

(∂H∂u

)∂u∂x

+∂H∂u

∂ 2u∂x2

+∂

∂x

(∂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



logo1



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

)is harmonic on Dz.

Proof.∂H∂x

=∂H∂u

∂u∂x

+∂H∂v

∂v∂x

∂ 2H∂x2 =

∂

∂x

(∂H∂u

∂u∂x

+∂H∂v

∂v∂x

)=

∂

∂x

(∂H∂u

)∂u∂x

+∂H∂u

∂ 2u∂x2 +

∂

∂x

(∂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



logo1


∂ 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

∂x

)∂ 2H∂x2 +

∂ 2H∂y2 =

∂ 2H∂u2

(∂u∂x

)2

+∂ 2H∂v2

(∂v∂x

)2

+∂ 2H∂u2

(∂u∂y

)2

+∂ 2H∂v2

(∂v∂y

)2

=∂ 2H∂u2

(∂u∂x

)2

+∂ 2H∂v2

(∂u∂y

)2

+∂ 2H∂u2

(∂u∂y

)2

+∂ 2H∂v2

(∂u∂x

)2

= 0



logo1


∂ 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

∂x

)

∂ 2H∂x2 +

∂ 2H∂y2 =

∂ 2H∂u2

(∂u∂x

)2

+∂ 2H∂v2

(∂v∂x

)2

+∂ 2H∂u2

(∂u∂y

)2

+∂ 2H∂v2

(∂v∂y

)2

=∂ 2H∂u2

(∂u∂x

)2

+∂ 2H∂v2

(∂u∂y

)2

+∂ 2H∂u2

(∂u∂y

)2

+∂ 2H∂v2

(∂u∂x

)2

= 0



logo1


∂ 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

∂x

)∂ 2H∂x2 +

∂ 2H∂y2

=∂ 2H∂u2

(∂u∂x

)2

+∂ 2H∂v2

(∂v∂x

)2

+∂ 2H∂u2

(∂u∂y

)2

+∂ 2H∂v2

(∂v∂y

)2

=∂ 2H∂u2

(∂u∂x

)2

+∂ 2H∂v2

(∂u∂y

)2

+∂ 2H∂u2

(∂u∂y

)2

+∂ 2H∂v2

(∂u∂x

)2

= 0



logo1


∂ 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

∂x

)∂ 2H∂x2 +

∂ 2H∂y2 =

∂ 2H∂u2

(∂u∂x

)2

+∂ 2H∂v2

(∂v∂x

)2

+∂ 2H∂u2

(∂u∂y

)2

+∂ 2H∂v2

(∂v∂y

)2

=∂ 2H∂u2

(∂u∂x

)2

+∂ 2H∂v2

(∂u∂y

)2

+∂ 2H∂u2

(∂u∂y

)2

+∂ 2H∂v2

(∂u∂x

)2

= 0



logo1


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)

)= e

x2+y2−1(1+x)2+y2 cos

(2y

(1+ x)2 + y2

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



logo1


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

)=



)= e

x2+y2−1(1+x)2+y2 cos

(2y

(1+ x)2 + y2




logo1


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

)=



)= e

x2+y2−1(1+x)2+y2 cos

(2y

(1+ x)2 + y2




logo1



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

)=



)= e

x2+y2−1(1+x)2+y2 cos

(2y

(1+ x)2 + y2




logo1



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

)=



)= e

x2+y2−1(1+x)2+y2 cos

(2y

(1+ x)2 + y2




logo1



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

)=



)= e

x2+y2−1(1+x)2+y2 cos

(2y

(1+ x)2 + y2




logo1



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

)=



)= e

x2+y2−1(1+x)2+y2 cos

(2y

(1+ x)2 + y2




logo1



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

)=



)= e

x2+y2−1(1+x)2+y2 cos

(2y

(1+ x)2 + y2




logo1



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

)=



)= e

x2+y2−1(1+x)2+y2 cos

(2y

(1+ x)2 + y2




logo1



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

)=



)= e

x2+y2−1(1+x)2+y2 cos

(2y

(1+ x)2 + y2




logo1



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


)= e

x2+y2−1(1+x)2+y2 cos

(2y

(1+ x)2 + y2




logo1



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


)= e

x2+y2−1(1+x)2+y2 cos

(2y

(1+ x)2 + y2




logo1



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)

)= e

x2+y2−1(1+x)2+y2 cos

(2y

(1+ x)2 + y2




logo1



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




logo1



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

)=



)= e

x2+y2−1(1+x)2+y2 cos

(2y

(1+ x)2 + y2

)

is harmonic on the right half plane x > 0.



logo1



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

)=



)= e

x2+y2−1(1+x)2+y2 cos

(2y

(1+ x)2 + y2




logo1


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.



logo1


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.



logo1


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.



logo1


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.



logo1


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.



logo1



(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.



logo1



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


H = h0 along C.


)is on Γ.


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

)= h0.



logo1


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.

Now



logo1


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


Now



logo1


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


Now



logo1




= 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


Now



logo1




= 0, where∂

∂ndenotes the


)satisfies the

condition∂H∂N

= 0

along C.





(−db

dtdadt


Now



logo1




= 0, where∂

∂ndenotes the


)satisfies the

condition∂H∂N

= 0 along C.





(−db

dtdadt


Now



logo1




= 0, where∂

∂ndenotes the


)satisfies the

condition∂H∂N

= 0 along C.

Proof.

Note that∂ f∂n




(−db

dtdadt


Now



logo1




= 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


Now



logo1




= 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


Now



logo1




= 0, where∂

∂ndenotes the


)satisfies the

condition∂H∂N

= 0 along C.




)is tangential to the curve

and(−db

dtdadt


Now



logo1




= 0, where∂

∂ndenotes the


)satisfies the

condition∂H∂N

= 0 along C.





(−db

dtdadt


Now



logo1


Proof.

With c :=

√(dxdt

)2+(

dydt

)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

dxdt

)

=

(∂h∂u

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

)∂u∂x (x,y)+ ∂h

∂v

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

)∂v∂x(x,y)

∂h∂u

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

)∂u∂y (x,y)+ ∂h

∂v

(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

dt

)+

∂h∂v

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

)∂v∂x

(x,y)(−dy

dt

)+

∂h∂u

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

)∂u∂y

(x,y)(

dxdt

)+

∂h∂v

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

)∂v∂y

(x,y)(

dxdt

)=

∂h∂u

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

)∂v∂y

(−dy

dt

)+

∂h∂v

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

)(−∂u

∂y

)(−dy

dt

)+

∂h∂u

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

)(−∂v

∂x

)(dxdt

)+

∂h∂v

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

)∂u∂x

(dxdt

)



logo1


Proof. With c :=

√(dxdt

)2+(

dydt

)2we have

c∂H∂N


)·~N =

(∂


)∂


)) ·(−dydt

dxdt

)

=

(∂h∂u

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

)∂u∂x (x,y)+ ∂h

∂v

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

)∂v∂x(x,y)

∂h∂u

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

)∂u∂y (x,y)+ ∂h

∂v

(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

dt

)+

∂h∂v

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

)∂v∂x

(x,y)(−dy

dt

)+

∂h∂u

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

)∂u∂y

(x,y)(

dxdt

)+

∂h∂v

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

)∂v∂y

(x,y)(

dxdt

)=

∂h∂u

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

)∂v∂y

(−dy

dt

)+

∂h∂v

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

)(−∂u

∂y

)(−dy

dt

)+

∂h∂u

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

)(−∂v

∂x

)(dxdt

)+

∂h∂v

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

)∂u∂x

(dxdt

)



logo1


Proof. With c :=

√(dxdt

)2+(

dydt

)2we have

c∂H∂N


)·~N

=

(∂


)∂


)) ·(−dydt

dxdt

)

=

(∂h∂u

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

)∂u∂x (x,y)+ ∂h

∂v

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

)∂v∂x(x,y)

∂h∂u

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

)∂u∂y (x,y)+ ∂h

∂v

(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

dt

)+

∂h∂v

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

)∂v∂x

(x,y)(−dy

dt

)+

∂h∂u

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

)∂u∂y

(x,y)(

dxdt

)+

∂h∂v

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

)∂v∂y

(x,y)(

dxdt

)=

∂h∂u

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

)∂v∂y

(−dy

dt

)+

∂h∂v

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

)(−∂u

∂y

)(−dy

dt

)+

∂h∂u

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

)(−∂v

∂x

)(dxdt

)+

∂h∂v

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

)∂u∂x

(dxdt

)



logo1


Proof. With c :=

√(dxdt

)2+(

dydt

)2we have

c∂H∂N


)·~N =

(∂


)∂


)) ·(−dydt

dxdt

)

=

(∂h∂u

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

)∂u∂x (x,y)+ ∂h

∂v

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

)∂v∂x(x,y)

∂h∂u

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

)∂u∂y (x,y)+ ∂h

∂v

(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

dt

)+

∂h∂v

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

)∂v∂x

(x,y)(−dy

dt

)+

∂h∂u

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

)∂u∂y

(x,y)(

dxdt

)+

∂h∂v

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

)∂v∂y

(x,y)(

dxdt

)=

∂h∂u

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

)∂v∂y

(−dy

dt

)+

∂h∂v

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

)(−∂u

∂y

)(−dy

dt

)+

∂h∂u

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

)(−∂v

∂x

)(dxdt

)+

∂h∂v

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

)∂u∂x

(dxdt

)



logo1


Proof. With c :=

√(dxdt

)2+(

dydt

)2we have

c∂H∂N


)·~N =

(∂


)∂


)) ·(−dydt

dxdt

)

=

(∂h∂u

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

)∂u∂x (x,y)+ ∂h

∂v

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

)∂v∂x(x,y)

∂h∂u

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

)∂u∂y (x,y)+ ∂h

∂v

(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

dt

)+

∂h∂v

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

)∂v∂x

(x,y)(−dy

dt

)+

∂h∂u

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

)∂u∂y

(x,y)(

dxdt

)+

∂h∂v

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

)∂v∂y

(x,y)(

dxdt

)=

∂h∂u

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

)∂v∂y

(−dy

dt

)+

∂h∂v

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

)(−∂u

∂y

)(−dy

dt

)+

∂h∂u

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

)(−∂v

∂x

)(dxdt

)+

∂h∂v

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

)∂u∂x

(dxdt

)



logo1


Proof. With c :=

√(dxdt

)2+(

dydt

)2we have

c∂H∂N


)·~N =

(∂


)∂


)) ·(−dydt

dxdt

)

=

(∂h∂u

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

)∂u∂x (x,y)+ ∂h

∂v

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

)∂v∂x(x,y)

∂h∂u

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

)∂u∂y (x,y)+ ∂h

∂v

(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

dt

)+

∂h∂v

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

)∂v∂x

(x,y)(−dy

dt

)

+∂h∂u

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

)∂u∂y

(x,y)(

dxdt

)+

∂h∂v

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

)∂v∂y

(x,y)(

dxdt

)=

∂h∂u

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

)∂v∂y

(−dy

dt

)+

∂h∂v

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

)(−∂u

∂y

)(−dy

dt

)+

∂h∂u

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

)(−∂v

∂x

)(dxdt

)+

∂h∂v

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

)∂u∂x

(dxdt

)



logo1


Proof. With c :=

√(dxdt

)2+(

dydt

)2we have

c∂H∂N


)·~N =

(∂


)∂


)) ·(−dydt

dxdt

)

=

(∂h∂u

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

)∂u∂x (x,y)+ ∂h

∂v

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

)∂v∂x(x,y)

∂h∂u

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

)∂u∂y (x,y)+ ∂h

∂v

(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

dt

)+

∂h∂v

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

)∂v∂x

(x,y)(−dy

dt

)+

∂h∂u

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

)∂u∂y

(x,y)(

dxdt

)+

∂h∂v

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

)∂v∂y

(x,y)(

dxdt

)

=∂h∂u

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

)∂v∂y

(−dy

dt

)+

∂h∂v

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

)(−∂u

∂y

)(−dy

dt

)+

∂h∂u

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

)(−∂v

∂x

)(dxdt

)+

∂h∂v

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

)∂u∂x

(dxdt

)



logo1


Proof. With c :=

√(dxdt

)2+(

dydt

)2we have

c∂H∂N


)·~N =

(∂


)∂


)) ·(−dydt

dxdt

)

=

(∂h∂u

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

)∂u∂x (x,y)+ ∂h

∂v

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

)∂v∂x(x,y)

∂h∂u

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

)∂u∂y (x,y)+ ∂h

∂v

(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

dt

)+

∂h∂v

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

)∂v∂x

(x,y)(−dy

dt

)+

∂h∂u

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

)∂u∂y

(x,y)(

dxdt

)+

∂h∂v

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

)∂v∂y

(x,y)(

dxdt

)=

∂h∂u

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

)∂v∂y

(−dy

dt

)+

∂h∂v

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

)(−∂u

∂y

)(−dy

dt

)

+∂h∂u

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

)(−∂v

∂x

)(dxdt

)+

∂h∂v

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

)∂u∂x

(dxdt

)



logo1


Proof. With c :=

√(dxdt

)2+(

dydt

)2we have

c∂H∂N


)·~N =

(∂


)∂


)) ·(−dydt

dxdt

)

=

(∂h∂u

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

)∂u∂x (x,y)+ ∂h

∂v

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

)∂v∂x(x,y)

∂h∂u

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

)∂u∂y (x,y)+ ∂h

∂v

(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

dt

)+

∂h∂v

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

)∂v∂x

(x,y)(−dy

dt

)+

∂h∂u

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

)∂u∂y

(x,y)(

dxdt

)+

∂h∂v

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

)∂v∂y

(x,y)(

dxdt

)=

∂h∂u

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

)∂v∂y

(−dy

dt

)+

∂h∂v

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

)(−∂u

∂y

)(−dy

dt

)+

∂h∂u

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

)(−∂v

∂x

)(dxdt

)+

∂h∂v

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

)∂u∂x

(dxdt

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


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Proof.

=∂h∂u

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

)∂v∂y

(−dy

dt

)+

∂h∂v

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

)(−∂u

∂y

)(−dy

dt

)+

∂h∂u

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

)(−∂v

∂x

)(dxdt

)+

∂h∂v

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

)∂u∂x

(dxdt

)=

∂h∂u

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

)(−dv

dt

)+

∂h∂v

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

)dudt

=(

∂h∂u

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

)∂h∂v

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

)) ·(−dvdt

dudt

)

=(

∂h∂u

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

)∂h∂v

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

)) ·~n√(

dudt

)2

+(

dvdt

)2

= 0



logo1


Proof.

=∂h∂u

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

)∂v∂y

(−dy

dt

)+

∂h∂v

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

)(−∂u

∂y

)(−dy

dt

)+

∂h∂u

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

)(−∂v

∂x

)(dxdt

)+

∂h∂v

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

)∂u∂x

(dxdt

)

=∂h∂u

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

)(−dv

dt

)+

∂h∂v

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

)dudt

=(

∂h∂u

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

)∂h∂v

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

)) ·(−dvdt

dudt

)

=(

∂h∂u

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

)∂h∂v

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

)) ·~n√(

dudt

)2

+(

dvdt

)2

= 0



logo1


Proof.

=∂h∂u

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

)∂v∂y

(−dy

dt

)+

∂h∂v

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

)(−∂u

∂y

)(−dy

dt

)+

∂h∂u

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

)(−∂v

∂x

)(dxdt

)+

∂h∂v

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

)∂u∂x

(dxdt

)=

∂h∂u

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

)(−dv

dt

)+

∂h∂v

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

)dudt

=(

∂h∂u

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

)∂h∂v

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

)) ·(−dvdt

dudt

)

=(

∂h∂u

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

)∂h∂v

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

)) ·~n√(

dudt

)2

+(

dvdt

)2

= 0



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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,

ℑ

(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.



logo1


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



ℑ

(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





logo1


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



ℑ

(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





logo1



=∂v∂u

= 0

and f (z) =z+11− z



ℑ

(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





logo1



=∂v∂u

= 0 and f (z) =z+11− z


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

So,

ℑ

(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





logo1



=∂v∂u

= 0 and f (z) =z+11− z



ℑ

(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





logo1



=∂v∂u

= 0 and f (z) =z+11− z



ℑ

(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





logo1



=∂v∂u

= 0 and f (z) =z+11− z



ℑ

(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





logo1



=∂v∂u

= 0 and f (z) =z+11− z



ℑ

(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





logo1



=∂v∂u

= 0 and f (z) =z+11− z



ℑ

(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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