# Potential Theory - notes

```Potential Theory - Glossary
Analytic function A complex function ๐ค = ๐(๐ง) is said to be analytic at a point ๐ง0 if ๐ is
differentiable at ๐ง0 and at every point in some neighbourhood of ๐ง0 . A function is analytic in a
domain ๐ท if it is analytic at every point in ๐ท.
Note: the term analytic has been used to refer to a function that is in fact holomorphic (i.e.
differentiable) rather than expressible as a power series. However, this is not a problem since
a complex function is holomorphic if, and only if it is analytic.
Complex potential Let ๐(๐ฅ, ๐ฆ) be a harmonic conjugate of a potential function ๐(๐ฅ, ๐ฆ).
Then the analytic function
ฮฉ(๐ง) = ๐(๐ฅ, ๐ฆ) + ๐๐(๐ฅ, ๐ฆ)
is called the complex potential corresponding to the real potential ๐. The function ๐ is called
the stream function.
The level curves
๐(๐ฅ, ๐ฆ) = ๐1
and ๐(๐ฅ, ๐ฆ) = ๐2
of ๐ and ๐, respectively, are orthogonal families.
Conformal Mapping Let ๐ค = ๐(๐ง) be a complex mapping defined in a domain ๐ท, and let ๐ง0
be a point in ๐ท. Then ๐ค is a conformal mapping (or conformal) at ๐ง0 if for every pair of smooth
oriented curves ๐ถ1 and ๐ถ2 in ๐ท intersecting at ๐ง0 , the angle between ๐ถ1 and ๐ถ2 is equal to the
angle between the image curves ๐ถ1โฒ and ๐ถ2โฒ at ๐(๐ง0 ) in both magnitude and sense (direction).
If ๐ค = ๐(๐ง) maps a domain ๐ท onto a domain ๐ทโฒ and if ๐ค is conformal at all points in ๐ท, then ๐ค
is a conformal mapping of ๐ท onto ๐ทโฒ .
Critical Point Let the complex function ๐(๐ง) be analytic in a domain ๐ท containing the point
๐ง0 . If ๐ โฒ (๐ง0 ) = 0, then the mapping ๐ค = ๐(๐ง) is not conformal at ๐ง0 , and ๐ง0 is called a critical
point of ๐.
Dipole
A pair consisting of a source and a sink is called a dipole (or doublet).
Dirichlet's principle Let ๐ท be a domain in the plane bounded by a simple closed curve ๐ถ.
Let ๐ be a real-valued smooth function defined on ๐ถ, and let ๐น be the family of real-valued
functions with smooth derivatives on ๐ท and ๐ถ and boundary values ๐. If
inf {โฌ (๐ข๐ฅ2 + ๐ข๐ฆ2 ) ๐๐ด: ๐ข โ ๐น}
๐ท
is attained by ๐ โ ๐น, then ๐ is the solution of the Dirichlet problem for ๐ท with boundary
values ๐.
Dirichlet problem Let ๐ท be a domain in the plane bounded by a finite number of nonintersecting simple closed contours ๐ถ๐ , and let ๐๐ be ๐ continuous functions defined on the ๐
contours. The problem of finding a function ๐ข(๐ฅ, ๐ฆ), which satisfies Laplace's equation in ๐ท
and which takes on the values ๐๐ on ๐ถ๐ is called a Dirichlet problem.
Greenโs function Let ๐ง0 = ๐ฅ0 + ๐๐ฆ0 be a point in a region of the plane bounded by a circle ๐ถ
of radius ๐. Then a function of the form
1
ln|๐ง โ ๐ง0 | + ๐ป(๐ฅ, ๐ฆ, ๐ฅ0 , ๐ฆ0 )
2๐
where ๐ป is harmonic for |๐ง| < ๐ and
๐บ(๐ฅ, ๐ฆ, ๐ฅ0 , ๐ฆ0 ) = โ
๐ป(๐ cos ๐ , ๐ sin ๐ , ๐ฅ0 , ๐ฆ0 ) =
1
ln|(๐ cos ๐ โ ๐ฅ0 ) + ๐(๐ sin ๐ โ ๐ฆ0 )|
2๐
Potential Theory - Glossary
is called a Greenโs function.
The essential feature of a Greenโs function ๐บ is the singular behaviour specified by
1
ln|๐ง โ ๐ง0 |.
2๐
The behaviour on ๐ถ is specified so that an application of Greenโs theorem leads to the desired
result.
โ
Green's Theorem in the Plane Let ๐ be a region in the plane whose boundary ๐ถ consists of
one, or more smooth, non-self-intersecting, closed curves that are positively oriented with
respect to ๐. If ๐(๐ฅ, ๐ฆ)i + ๐(๐ฅ, ๐ฆ)j is a smooth vector field on ๐, then
โฌ(
๐
๐๐ ๐๐
โ ) ๐๐ฅ ๐๐ฆ = โฎ๐(๐ฅ, ๐ฆ)๐๐ฅ + ๐(๐ฅ, ๐ฆ)๐๐ฆ.
๐๐ฅ ๐๐ฆ
๐ถ
Harmonic conjugate Suppose that the real function ๐ข(๐ฅ, ๐ฆ) is harmonic on a domain ๐ท, and
there exists a real function ๐ฃ(๐ฅ, ๐ฆ) such that the partial derivatives of ๐ข and ๐ฃ satisfy the
Cauchy-Riemann equations throughout ๐ท. Then ๐ฃ is said to be a harmonic conjugate of ๐ข.
Furthermore, it follows that ๐(๐ง) = ๐ข(๐ฅ, ๐ฆ) + ๐๐ฃ(๐ฅ, ๐ฆ) is analytic on ๐ท.
Harmonic function A real valued function ๐(๐ฅ, ๐ฆ) that has continuous first and secondorder partial derivatives in a domain ๐ท, and satisfies Laplaceโs equation in two dimensions
โ2 ๐ =
๐ 2๐ ๐ 2๐
+
=0
๐๐ฅ 2 ๐๐ฆ 2
is said to be harmonic in ๐ท. The function ๐ is called a harmonic (or potential) function.
Laplace's equation (in two dimensions) Let ฮฆ(๐ฅ, ๐ฆ) be a real valued function of two real
variables ๐ฅ and ๐ฆ. The partial differential equation
ฮฆ๐ฅ๐ฅ (๐ฅ, ๐ฆ) + ฮฆ๐ฆ๐ฆ (๐ฅ, ๐ฆ) = 0
or alternatively
๐ 2ฮฆ ๐ 2ฮฆ
+
=0
๐๐ฅ 2 ๐๐ฆ 2
is known as Laplace's equation, or the potential equation. The polar form of Laplace's
equation is given by
๐ 2 ฮฆ๐๐ (๐, ๐) + ๐ฮฆ๐ (๐, ๐) + ฮฆ๐๐ (๐, ๐) = 0
or alternatively
๐ 2ฮฆ
๐ฮฆ ๐ 2 ฮฆ
+
๐
+
= 0.
๐๐ 2
๐๐ ๐๐ 2
Level curves Suppose the function ๐(๐ง) = ๐ข(๐ฅ, ๐ฆ) + ๐๐ฃ(๐ฅ, ๐ฆ) is analytic in a domain ๐ท. Then
the real and imaginary parts of ๐ can be used to define two families of curves in ๐ท. The
equations
๐2
๐ข(๐ฅ, ๐ฆ) = ๐1
and ๐ฃ(๐ฅ, ๐ฆ) = ๐2
where ๐1 and ๐2 are arbitrary real constants, are called level curves of ๐ข and ๐ฃ, respectively.
The level curves are orthogonal families.
Neumann problem Let ๐ท be a simply connected domain in the plane bounded by a simple
closed smooth contour ๐ถ, and let ๐ be a function defined on ๐ถ. The problem of finding a
function ๐ข(๐ฅ, ๐ฆ), which satisfies Laplace's equation in ๐ท and such that โ๐ข โ N = ๐ where N is
the unit outward normal on ๐ถ is called a Neumann problem.
Potential Theory - Glossary
Poisson integral formulae The Poisson integral formulae are general solutions to the
Dirichlet problem in (i) the upper half plane ๐ฆ > 0, and (ii) the unit open disk |๐ง| < 1.
(i) Poisson integral formula for the upper half plane.
Let ๐(๐ฅ) be a piecewise continuous and bounded function on โโ < ๐ฅ < โ. Then the function
defined by
๐(๐ฅ, ๐ฆ) =
๐ฆ โ
๐(๐ )
โซ
๐๐
๐ โโ (๐ฅ โ ๐ )2 + ๐ฆ 2
is a solution of the Dirichlet problem in the upper half plane ๐ฆ > 0 with boundary condition
๐(๐ฅ, 0) = ๐(๐ฅ) at all points of continuity of ๐.
(ii) Poisson integral formula for the unit disk.
Cartesian form. Let ๐(๐ง) be a complex function for which the values ๐(๐ ๐๐ ) on the unit circle
๐ง = ๐ ๐๐ give a piecewise continuous and bounded function for โ๐ โค ๐ โค ๐. Then the function
defined by
๐(๐ฅ, ๐ฆ) =
1 ๐ ๐(๐ ๐๐  )(1 โ |๐ง|2 )
โซ
๐๐
|๐ ๐๐  โ ๐ง|2
2๐ โ๐
is a solution of the Dirichlet problem in the open unit disk |๐ง| < 1 with boundary condition
๐(cos ๐ , sin ๐) = ๐(๐ ๐๐ ) at all points of continuity of ๐.
Polar form. Let ๐(๐) be a piecewise continuous and bounded function on โ๐ < ๐ < ๐. Then
the function defined by
๐(๐, ๐) =
1 ๐
๐(๐ )(1 โ ๐ 2 )
โซ
๐๐
2๐ โ๐ 1 โ 2๐ cos(๐ โ ๐ ) + ๐ 2
is a solution of the Dirichlet problem in the open unit disk |๐ง| < 1 with boundary condition
๐(๐ = 1, ๐) = ๐(๐) at all points of continuity of ๐.
Poisson kernel
The function ๐: โ โ โ defined by
๐2 โ ๐ 2
๐(๐๐ ) = 2
๐ โ 2๐๐ cos ๐ + ๐ 2
is known as the Poisson kernel for the disk {๐ง: |๐ง| < ๐}. It vanishes at every point of
{๐ง: |๐ง| = ๐} except at ๐ง = ๐, where it has a singularity, and has the property of being positive
for 0 < ๐ < ๐.
๐๐
Polygonal region A polygonal region in the complex plane is a region that is bounded by a
simple, connected, piecewise smooth curve consisting of a finite number of line segments. The
boundary curve of a polygonal region is called a polygon and the endpoints of the line
segments in the polygon are called vertices of the polygon. If a polygon is a closed curve, then
the region enclosed by the polygon is called a bounded polygonal region, and a polygonal
region that is not bounded is called an unbounded polygonal region. In the case of an
unbounded polygonal region, the ideal point โ is also called a vertex of the polygon.
Simple function A function ๐ค = ๐(๐ง) is said to be simple (or univalent) in a domain ๐ท if it
is analytic and one-to-one in ๐ท.
Sources and Sinks A source is a point at which fluid is produced (or introduced), whereas,
a sink is a point at which fluid disappears (or is removed).
Potential Theory - Glossary
If the motion of an ideal fluid consists of an outward radial flow from a point and is
symmetrical in all directions, then the point is called a simple source. The point is called a
simple sink if the flow is inward rather than outward.
Streamlining The process of constructing a flow of an ideal fluid that remains inside a given
domain ๐ท is called streamlining.
Potential Theory - Notes
Harmonic Functions and Conjugates
Theorem Suppose the complex function ๐(๐ง) = ๐ข(๐ฅ, ๐ฆ) + ๐๐ฃ(๐ฅ, ๐ฆ) is analytic in a domain ๐ท.
Then the functions ๐ข(๐ฅ, ๐ฆ) and ๐ฃ(๐ฅ, ๐ฆ) are harmonic in ๐ท.
Theorem Suppose that the real function ๐ข(๐ฅ, ๐ฆ) is harmonic in an ๐-neighbourhood of the
point (๐ฅ0 , ๐ฆ0 ). Then there exists a conjugate harmonic function
๐ฃ(๐ฅ, ๐ฆ) = โซ ๐ข๐ฅ (๐ฅ, ๐ฆ) ๐๐ฆ โ โซ ๐ข๐ฆ (๐ฅ, ๐ฆ) ๐๐ฅ โ โฌ ๐ข๐ฅ๐ฅ (๐ฅ, ๐ฆ) ๐๐ฆ๐๐ฅ
defined in this neighbourhood such that ๐(๐ง) = ๐ข(๐ฅ, ๐ฆ) + ๐๐ฃ(๐ฅ, ๐ฆ) is an analytic function.
A harmonic conjugate can be constructed by using the following method.
First confirm that ๐ข(๐ฅ, ๐ฆ) is harmonic by verifying that
๐ข๐ฅ๐ฅ (๐ฅ, ๐ฆ) + ๐ข๐ฆ๐ฆ (๐ฅ, ๐ฆ) = 0.
Then, using the Cauchy-Riemann equation ๐ฃ๐ฆ (๐ฅ, ๐ฆ) = ๐ข๐ฅ (๐ฅ, ๐ฆ) gives
๐ฃ(๐ฅ, ๐ฆ) = โซ ๐ฃ๐ฆ (๐ฅ, ๐ฆ) ๐๐ฆ + ๐ถ(๐ฅ) = โซ ๐ข๐ฅ (๐ฅ, ๐ฆ) ๐๐ฆ + ๐ถ(๐ฅ)
where ๐ถ(๐ฅ) is a function of ๐ฅ alone.
Differentiating both sides with respect to ๐ฅ and using the Cauchy-Riemann equation
๐ฃ๐ฅ (๐ฅ, ๐ฆ) = โ๐ข๐ฆ (๐ฅ, ๐ฆ) gives
โ๐ข๐ฆ (๐ฅ, ๐ฆ) =
๐
(โซ ๐ข๐ฅ (๐ฅ, ๐ฆ) ๐๐ฆ) + ๐ถ โฒ (๐ฅ).
๐๐ฅ
All terms except those involving ๐ฅ will cancel leaving a formula for ๐ถ โฒ (๐ฅ) in terms of ๐ฅ alone,
say ๐ถ โฒ (๐ฅ) = ๐(๐ฅ).
Integrating both sides with respect to ๐ฅ gives
๐ถ(๐ฅ) = ๐น(๐ฅ) + ๐
where ๐น is an antiderivative of ๐ and ๐ is an arbitrary constant.
Finally, letting ๐ = 0 gives
๐ฃ(๐ฅ, ๐ฆ) = โซ ๐ข๐ฅ (๐ฅ, ๐ฆ) ๐๐ฆ + ๐น(๐ฅ).
A similar method can be used if we are given ๐ฃ(๐ฅ, ๐ฆ) and wish to find ๐ข(๐ฅ, ๐ฆ).
Laplaceโs Equation in Physics
Electromagnetism
For an electrostatic field, the electric field E = (๐ธ๐ฅ , ๐ธ๐ฆ , ๐ธ๐ง ), where ๐ธ๐ฅ , ๐ธ๐ฆ , ๐ธ๐ง are the Cartesian
components of E, is given by
๐๐
๐๐
๐๐
i+
j+
k,
๐๐ฅ
๐๐ฆ
๐๐ง
where ๐ is an arbitrary function of position known as the electrostatic potential.
The electrostatic potential function ๐ satisfies Laplace's equation and is therefore harmonic in
some domain ๐ท. If we restrict our attention to two dimensions, it follows that there exists a
Potential Theory - Notes
harmonic conjugate function ๐(๐ฅ, ๐ฆ) defined in some domain ๐ธ in โ2 , so that the complex
potential function ฮฉ(๐ง) = ๐(๐ฅ, ๐ฆ) + ๐๐(๐ฅ, ๐ฆ) is analytic in ๐ธ.
The electric field E (i.e. the electron flow) is given by
๐ฮฉ
= ๐ธ๐ฅ + ๐๐ธ๐ฆ .
๐๐ง
The level curves ๐(๐ฅ, ๐ฆ) = ๐1 are called equipotential curves, that is, curves along which the
electrostatic potential is constant. Whereas, the level curves ๐(๐ฅ, ๐ฆ) = ๐2 , curves that are
orthogonal to the family ๐(๐ฅ, ๐ฆ) = ๐1, are called lines of force and are the paths along which a
charged particle will move in the electrostatic field.
Fluid Mechanics
For an incompressible fluid, the velocity field v = (๐ฃ๐ฅ , ๐ฃ๐ฆ , ๐ฃ๐ง ), where ๐ฃ๐ฅ , ๐ฃ๐ฆ , ๐ฃ๐ง are the
Cartesian components of v, is given by
๐๐
๐๐
๐๐
i+
j+
k,
๐๐ฅ
๐๐ฆ
๐๐ง
where ๐ is an arbitrary function of position known as the velocity potential.
The velocity potential function ๐ satisfies Laplace's equation and is therefore harmonic in
some domain ๐ท. If we restrict our attention to two dimensions, it follows that there exists a
harmonic conjugate function ๐(๐ฅ, ๐ฆ) defined in some domain ๐ธ in โ2 , so that the complex
potential function ฮฉ(๐ง) = ๐(๐ฅ, ๐ฆ) + ๐๐(๐ฅ, ๐ฆ) is analytic in ๐ธ. In this case, the complex potential
function ฮฉ(๐ง) is called the complex velocity potential of the flow.
The velocity field v (i.e. the fluid flow) is given by
๐ฮฉ
= ๐ฃ๐ฅ + ๐๐ฃ๐ฆ .
๐๐ง
The level curves ๐(๐ฅ, ๐ฆ) = ๐1 are called equipotential curves, that is, curves along which the
velocity potential is constant. Whereas, the level curves ๐(๐ฅ, ๐ฆ) = ๐2 , curves that are
orthogonal to the family ๐(๐ฅ, ๐ฆ) = ๐1, are called streamlines and represent the actual paths
along which particles in the fluid will move.
The Dirichlet Problem
Theorem 6.2.1 If the Dirichlet problem has a solution ๐ for a given domain ๐ท and set of
functions ๐๐ , then the solution is unique.
Theorem 6.2.2
The Dirichlet problem is solvable for the disk |๐ง| < ๐.
Corollary 6.2.1 Let ๐(๐) be a piecewise continuous function on 0 โค ๐ โค 2๐. Then the
function defined by
1 2๐
๐(๐ )(๐ 2 โ ๐ 2 )
๐(๐, ๐) =
โซ
๐๐
2๐ 0 ๐ 2 โ 2๐๐ cos(๐ โ ๐ ) + ๐ 2
is harmonic in the open disk |๐ง| < ๐, and lim ๐(๐, ๐) = ๐(๐) for all but a finite number of
values of ๐.
๐โ๐
Theorem 6.2.3 Let ๐ท be a simply connected domain with boundary ๐ถ. Let ๐ค = ๐๐ ๐๐ = ๐(๐ง)
be analytic in ๐ท, and map ๐ท โช ๐ถ in a one to one fashion onto the closed disk |๐ค| โค 1 so that ๐ท
maps onto the interior of the disk |๐ค| < 1 and ๐ถ maps onto its boundary |๐ค| = 1. Then
Potential Theory - Notes
๐ โฒ (๐ข, ๐ฃ) =
1 2๐
๐(๐ )(1 โ ๐2 )
โซ
๐๐
2๐ 0 1 โ 2๐ cos(๐ โ ๐ ) + ๐2
where ๐ข = Re(๐ค) = ๐ cos ๐ and ๐ฃ = Im(๐ค) = ๐ sin ๐, solves the Dirichlet problem in the
sense that ๐(๐ฅ, ๐ฆ) = ๐ โฒ (๐ข, ๐ฃ) at corresponding points under the mapping.
Steps for Solving a Dirichlet Problem
(i)
Find an analytic function ๐(๐ง) = ๐ข(๐ฅ, ๐ฆ) + ๐๐ฃ(๐ฅ, ๐ฆ) that maps the domain ๐ท in the ๐งplane onto a simpler domain ๐ทโฒ in the ๐ค-plane, and that maps the boundary curves
๐ถ1 , ๐ถ2 , โฆ , ๐ถ๐ onto the curves ๐ถ1โฒ , ๐ถ2โฒ , โฆ , ๐ถ๐โฒ , respectively.
(ii)
Transform the boundary conditions on ๐ถ1 , ๐ถ2 , โฆ , ๐ถ๐ to boundary conditions on
๐ถ1โฒ , ๐ถ2โฒ , โฆ , ๐ถ๐โฒ .
(iii) Solve this new (and easier ) Dirichlet problem in ๐ทโฒ to obtain a harmonic function
ฮฆ(๐ข, ๐ฃ).
(iv) Substitute the real and imaginary parts ๐ข(๐ฅ, ๐ฆ) and ๐ฃ(๐ฅ, ๐ฆ) of ๐ for the variables ๐ข and ๐ฃ
in ฮฆ(๐ข, ๐ฃ). Then the function ๐(๐ฅ, ๐ฆ) = ฮฆ(๐ข(๐ฅ, ๐ฆ), ๐ฃ(๐ฅ, ๐ฆ)) is a solution to the Dirichlet
problem in ๐ท.
Greenโs Functions
Theorem
Let ๐ถ be a simple closed smooth contour with interior ๐, then
โฌ (๐ขโ2 ๐ฃ + โ๐ข โ โ๐ฃ) ๐๐ฅ ๐๐ฆ = โซ๐ขโ๐ฃ โ N ๐๐
๐
๐ถ
and
โฌ (๐ขโ2 ๐ฃ โ ๐ฃโ2 ๐ข) ๐๐ฅ ๐๐ฆ = โซ(๐ขโ๐ฃ โ N โ ๐ฃโ๐ข โ N) ๐๐ .
๐
๐ถ
where ๐ข and ๐ฃ are functions with continuous first and second order partial derivatives within,
and on ๐ถ, N is the unit outward normal on ๐ถ, and ๐๐  is the element of arc length on ๐ถ.
Greenโs function for the Dirichlet problem Let ๐ท be a bounded simply connected domain
bounded by a simple closed smooth contour ๐ถ, and let ๐ง0 = ๐ฅ0 + ๐๐ฆ0 be a point in ๐ท. If
๐บ(๐ฅ, ๐ฆ, ๐ฅ0 , ๐ฆ0 ) is a function satisfying the following properties:
1
(i)
๐บ(๐ฅ, ๐ฆ, ๐ฅ0 , ๐ฆ0 ) = โ 2๐ ln|๐ง โ ๐ง0 | + ๐ป(๐ฅ, ๐ฆ, ๐ฅ0 , ๐ฆ0 ) where ๐ป is harmonic inside ๐ถ;
(ii)
๐บ is continuous in ๐ท โช ๐ถ, except at (๐ฅ0 , ๐ฆ0 ) in ๐ท;
(iii) ๐บ(๐ฅ, ๐ฆ, ๐ฅ0 , ๐ฆ0 ) = 0 when (๐ฅ, ๐ฆ) is on ๐ถ.
Then ๐บ is a Greenโs function associated with the Dirichlet problem for the region ๐ = ๐ท โช ๐ถ.
Theorem 6.3.1
If a Greenโs function (for the Dirichlet problem) exists, then it is unique.
Theorem 6.3.2
positive in ๐ท.
The Greenโs function (for the Dirichlet problem) for the region ๐ = ๐ท โช ๐ถ is
Theorem 6.3.3 If ๐บ is the Greenโs function (for the Dirichlet problem) for the region ๐ =
๐ท โช ๐ถ, then ๐บ(๐ฅ, ๐ฆ, ๐ฅ1 , ๐ฆ1 ) = ๐บ(๐ฅ, ๐ฆ, ๐ฅ2 , ๐ฆ2 ) for any pair of distinct points (๐ฅ1 , ๐ฆ1 ) and (๐ฅ2 , ๐ฆ2 ) in
๐ท.
Theorem 6.3.4 If the Greenโs function (for the Dirichlet problem) exists for the region ๐ =
๐ท โช ๐ถ, then for ๐(๐ ) continuous on ๐ถ
Potential Theory - Notes
๐ข(๐ฅ0 , ๐ฆ0 ) = โ โซ(๐(๐ )โ๐บ โ N) ๐๐
๐ถ
where N is the unit outward normal on ๐ถ and ๐๐  is the element of arc length on ๐ถ, solves the
Dirichlet problem โ2 ๐ข = 0 in ๐ท with ๐ข = ๐ on ๐ถ.
Theorem 6.3.5 Let ๐ค = ๐(๐ง) map the region ๐ onto the unit closed disk |๐ง| โค 1 so that ๐ท,
the interior of ๐, maps onto the open disk |๐ง| < 1 and ๐ถ, the boundary of ๐, maps onto the
unit circle |๐ง| = 1. Furthermore, let ๐ โฒ (๐ง) exist and never vanish in ๐ท, and let ๐ง0 map into the
1
origin. Then ๐บ(๐ฅ, ๐ฆ, ๐ฅ1 , ๐ฆ1 ) = โ 2๐ ln|๐(๐ง)| is the Greenโs function (for the Dirichlet problem)
for ๐.
The Neumann Problem
An insulated boundary curve in a heat flow problem corresponds to a boundary condition of
the form โ๐ข โ N = 0, and so is an example of a Neumann problem.
The solution ๐ข of a Neumann problem is not unique since ๐ข + ๐ is also a solution for any
constant ๐.
Theorem 6.3.6
The solution ๐ข of the Neumann problem such that โซ๐ถ ๐ข ๐๐  = 0 is unique.
Theorem 6.3.7
โซ๐ถ ๐(๐ ) ๐๐  = 0.
A necessary condition for the Neumann problem to have a solution is that
Greenโs function for the Neumann problem Let ๐ท be a bounded simply connected domain
bounded by a simple closed smooth contour ๐ถ, and let ๐ง0 = ๐ฅ0 + ๐๐ฆ0 be a point in ๐ท. If
๐บ(๐ฅ, ๐ฆ, ๐ฅ0 , ๐ฆ0 ) is a function satisfying the following properties:
1
(i)
๐บ(๐ฅ, ๐ฆ, ๐ฅ0 , ๐ฆ0 ) = โ
(ii)
๐บ and its first partial derivatives are continuous in ๐ท โช ๐ถ, except at (๐ฅ0 , ๐ฆ0 ) in ๐ท;
2๐
ln|๐ง โ ๐ง0 | + ๐ป(๐ฅ, ๐ฆ, ๐ฅ0 , ๐ฆ0 ) where ๐ป is harmonic in ๐ท;
(iii) โ๐บ โ N = โ 1โ๐ฟ where ๐ฟ is the length of ๐ถ, when (๐ฅ, ๐ฆ) is on ๐ถ;
(iv) โซ๐ถ ๐บ ๐๐  = 0.
Then ๐บ is a Greenโs function associated with the Neumann problem for the region ๐ = ๐ท โช ๐ถ.
Theorem
If a Greenโs function (for the Neumann problem) exists, then it is unique.
Theorem 6.3.9 If the Greenโs function (for the Neumann problem) exists for the region ๐ =
๐ท โช ๐ถ, then for ๐(๐ ) continuous on ๐ถ
๐ข(๐ฅ0 , ๐ฆ0 ) = โซ๐(๐ )๐บ ๐๐
๐ถ
where ๐ข satisfies Theorems 6.3.6 and 6.3.7 and ๐๐  is the element of arc length on ๐ถ, solves the
Neumann problem โ2 ๐ข = 0 in ๐ท with โ๐ข โ N = ๐ on ๐ถ.
Conformal Mappings
Theorem Let the complex function ๐(๐ง) be analytic in a domain ๐ท containing a point ๐ง0 . If
๐ โฒ (๐ง0 ) โ  0, then ๐ค = ๐(๐ง) is conformal at ๐ง0 .
Theorem Let the complex function ๐(๐ง) be analytic at a critical point ๐ง0 . If ๐ > 1 is an
integer such that ๐ โฒ (๐ง0 ) = ๐ โฒโฒ (๐ง0 ) = โฏ = ๐ (๐โ1) (๐ง0 ) = 0, and ๐ (๐) (๐ง0 ) โ  0, then the angle
between any two smooth curves intersecting at ๐ง0 is increased by a factor ๐ by the complex
mapping ๐ค = ๐(๐ง). In particular, ๐ค = ๐(๐ง) is not a conformal mapping at ๐ง0 .
Potential Theory - Notes
Theorem Laplace's equation is preserved under a mapping ๐ค = ๐(๐ง) provided that ๐ โฒ (๐ง0 ) โ
0.
Theorem 6.4.1
If ๐(๐ง) is simple in a domain ๐ท, then ๐ โฒ (๐ง0 ) โ  0 in ๐ท.
Corollary 6.4.1
If ๐(๐ง) is simple in a domain ๐ท, then it is conformal in ๐ท.
Simple functions have the following desirable properties as mappings:
(i)
they are conformal, i.e. they preserve angles in both magnitude and direction;
(ii)
they preserve Laplace's equation;
(iii) they have a unique inverse which is also a simple function and at every point is the same
as the local inverse.
Theorem 6.4.2 If ๐ค = ๐(๐ง) has range ๐ทโฒ and is simple in a domain ๐ท, and ๐น(๐ค) is simple in
๐ทโฒ , then the composite (๐น โ ๐)(๐ง) = ๐น[๐(๐ง)] is simple in ๐ท.
Theorem 6.4.3 Let ๐ถ be a simple closed contour with interior ๐ท. Furthermore, let ๐(๐ง) be
analytic within and on ๐ถ, and take no value more than once on ๐ถ. Then ๐(๐ง) is simple in ๐ท.
Theorem 6.4.4 Let ๐๐ (๐ง) be simple in a domain ๐ท for all ๐ = 1,2,3, โฆ. Furthermore, let ๐๐ (๐ง)
converge uniformly in compact subsets of ๐ท. Then ๐๐ (๐ง) is either constant, or simple in ๐ท.
A fundamental problem in the theory of conformal mappings is to find a simple function ๐ค =
๐(๐ง) which maps a given simply connected domain onto the open unit disk |๐ค| < 1.
Theorem 6.4.7 (Riemann Mapping Theorem) Let ๐ท be a simply connected domain in the
complex plane โ with at least two boundary points. Then there exists a simple function ๐ค =
๐(๐ง) which maps ๐ท onto the open unit disk |๐ค| < 1 in the ๐ค-plane. If we specify that a given
point ๐ง0 in ๐ท maps into the origin ๐ค = 0, and a given direction at ๐ง0 is mapped into a given
direction at the origin, then the mapping is unique.
Let ๐ท and ๐ทโฒ be simply connected domains in โ. We can use the Riemann mapping theorem to
establish the existence of a simple mapping from ๐ท onto ๐ทโฒ . According to the theorem there is
a simple mapping ๐ from ๐ท onto the open unit disk |๐ค| < 1. Similarly, there is a simple
mapping ๐ from ๐ทโฒ onto the open unit disk |๐ค| < 1. Since ๐ is simple, it has a well defined
inverse function ๐โ1 that maps the open unit disk |๐ค| < 1 onto ๐ทโฒ . The desired mapping from
๐ท onto ๐ทโฒ is then given by the composition ๐ค = (๐โ1 โ ๐)(๐ง).
The Riemann mapping theorem is inadequate in two ways:
(i)
it is strictly an existence theorem, which assert the existence of a mapping function but
does not show how it can be constructed;
(ii)
it does not show how boundary points of ๐ท correspond to boundary points of ๐ทโฒ .
Theorem 6.4.8 Let ๐ท and ๐ทโฒ be domains bounded by simple closed contours ๐ถ and ๐ถ โฒ ,
respectively. Then the conformal map ๐: ๐ท โ ๐ทโฒ is continuous in ๐ท โช ๐ถ and establishes a oneto-one correspondence between ๐ถ and ๐ถ โฒ .
The Schwarz-Christoffel Transformation
The Schwarz-Christoffel transformation provides a recipe for a conformal mapping of the
upper half of the ๐ง-plane to a polygonal region in the ๐ค-plane.
Theorem (Schwarz-Christoffel formula)
domain ๐ฆ > 0 and has the derivative
Let ๐ be a complex function that is analytic in the
๐ โฒ (๐ง) = ๐ด(๐ง โ ๐ฅ1 )(๐ผ1 โ๐)โ1 × (๐ง โ ๐ฅ2 )(๐ผ2 โ๐)โ1 × โฏ × (๐ง โ ๐ฅ๐ )(๐ผ๐โ๐)โ1
Potential Theory - Notes
where ๐ฅ1 < ๐ฅ2 < โฏ < ๐ฅ๐ are arbitrary points on the ๐ฅ-axis, 0 < ๐ผ๐ < 2๐ for 1 โค ๐ โค ๐, and ๐ด
is a complex constant. Then the upper half plane ๐ฆ โฅ 0 is mapped by ๐ค = ๐(๐ง) onto an
unbounded polygonal region with interior angles ๐ผ1 , ๐ผ2 , โฆ , ๐ผ๐ .
Since ๐ โฒ (๐ฅ + ๐๐ฆ) โ  0 when ๐ฆ > 0, it follows that the function given by the Schwarz-Christoffel
formula is a conformal mapping in the domain ๐ฆ > 0. Furthermore, although the mapping is
defined on the upper half plane ๐ฆ โฅ 0, it is only conformal in the domain ๐ฆ > 0.
The above theorem provides a formula for the derivative of ๐, which may in principle be
integrated to obtain ๐ค as follows
๐ค = ๐(๐ง) = ๐ด โซ(๐ง โ ๐ฅ1 )(๐ผ1 โ๐)โ1 × (๐ง โ ๐ฅ2 )(๐ผ2 โ๐)โ1 × โฏ × (๐ง โ ๐ฅ๐ )(๐ผ๐โ๐)โ1 ๐๐ง + ๐ต
where ๐ด and ๐ต are complex constants. Thus ๐ is the composition of the function
๐(๐ง) = โซ(๐ง โ ๐ฅ1 )(๐ผ1 โ๐)โ1 × (๐ง โ ๐ฅ2 )(๐ผ2 โ๐)โ1 × โฏ × (๐ง โ ๐ฅ๐ )(๐ผ๐โ๐)โ1 ๐๐ง
and the linear mapping โ(๐ง) = ๐ด๐ง + ๐ต. The linear mapping โ allows us to rotate, scale, and
translate the polygonal region produced by ๐.
In practice we usually have a certain amount of freedom in the selection of the points ๐ฅ๐ on
the ๐ฅ-axis. A judicious choice can simplify the computation of ๐(๐ง).
The Schwarz-Christoffel formula may be used to construct a mapping of the upper half plane
๐ฆ โฅ 0 onto a bounded polygonal region. To do so, we apply the formula using only ๐ โ 1 on
the ๐ interior angles of the bounded polygonal region. Essentially, we are choosing the point
๐ฅ๐ on the ๐ฅ-axis to lie at infinity.
Flows with Sources and Sinks
Streamlining
Suppose that the complex velocity potential ฮฉ(๐ง) = ๐(๐ฅ, ๐ฆ) + ๐๐(๐ฅ, ๐ฆ) is analytic in a domain
๐ท and that ๐ is constant on the boundary of ๐ท. Then ฮฉโฒ (๐ง) is a complex representation of the
velocity field of a flow of an ideal fluid in ๐ท. Moreover, if a particle is placed in ๐ท and allowed
to flow with the fluid, then its path ๐ง(๐ก) remains in ๐ท.
Solving a streamlining problem If ๐ค = ๐(๐ง) is a simple mapping of the domain ๐ท onto a
domain ๐ทโฒ in the ๐ค-plane such that the image of the boundary ๐ถ of ๐ท is a horizontal line in the
๐ค-plane, then ๐ค = ฮฉ(๐ง) is a complex velocity potential of the flow of an ideal fluid in ๐ท and
ฮฉโฒ (๐ง) is a complex representation of its velocity field.
An alternative method is to apply the mapping ๐ค = ๐(๐ง) directly to the complex velocity
potential of a uniform flow in the ๐ค-plane
ฮฉ๐ทโฒ = ๐ฃ0 ๐ค
โ
ฮฉ๐ท = ๐ฃ0 ๐(๐ง).
Sources, Sinks and Dipioles
The complex velocity potential of a (simple) source of strength ๐ด at the point ๐ง0 is
ฮฉ(๐ง) = ๐ด log(๐ง โ ๐ง0 )
where ๐ด is a positive real constant.
The complex velocity potential of a (simple) sink of strength ๐ด at the point ๐ง0 is
ฮฉ(๐ง) = โ๐ด log(๐ง โ ๐ง0 )
Potential Theory - Notes
where ๐ด is a positive real constant.
A flow containing both sources and sinks can be described by adding together their respective
complex velocity potentials.
Suppose that there is a source of strength ๐ด at the point ๐ง1 = ๐๐ ๐๐ and a sink of the same
strength at the point ๐ง0 . Then the complex velocity potential of the dipole in the direction ๐ at
๐ง0 is
ฮฉ(๐ง) = โ๐ด๐
๐ ๐๐
๐ง โ ๐ง0
where ๐ = ๐ด๐ is strength of the dipole.
The velocity potential is
๐
[log|๐ง โ ๐ง0 |]
๐๐
where ๐โ๐๐ is the directional derivative in the direction ๐.
๐(๐ฅ, ๐ฆ) = โ๐
Determining fluid flows produced by sources and sinks If ๐ค = ๐(๐ง) is a simple mapping of
the domain ๐ท onto a domain ๐ทโฒ in the ๐ค-plane such that the image of boundary curves,
sources, and sinks for the flow in ๐ทare mapped onto boundary curves, sources, and sinks in ๐ทโฒ
where the complex potential is known to be ฮฉ๐ทโฒ (๐ค). Then the complex potential in ๐ท is given
by ฮฉ๐ท (๐ง) = ฮฉ๐ทโฒ (๐(๐ง)).
Sources, sinks and dipoles, and their potentials, have their counterparts in electrostatics,
magnetostatics, heat transfer, and so on.
Potential Theory - SAQโs
SAQ 1
Which flow is given by the velocity potential
๐(๐ฅ, ๐ฆ) = ๐ฅ +
๐ฅ2
2๐ฅ
?
+ ๐ฆ2
For very large ๐ฅ 2 + ๐ฆ 2 , ๐ is approximately ๐ฅ, and
๐ฃ๐ฅ =
๐๐
=1
๐๐ฅ
and ๐ฃ๐ฆ =
๐๐
= 0.
๐๐ฆ
Hence, at large distances from the origin, the flow is that of a uniform stream with velocity 1
to the right.
Consider
๐ง+
2
2
2(๐ฅ โ ๐๐ฆ)
2(๐ฅ โ ๐๐ฆ)
= ๐ฅ + ๐๐ฆ +
= ๐ฅ + ๐๐ฆ +
= ๐ฅ + ๐๐ฆ + 2
(๐ฅ + ๐๐ฆ)(๐ฅ โ ๐๐ฆ)
๐ง
๐ฅ + ๐๐ฆ
๐ฅ + ๐ฆ2
2๐ฅ
2๐ฆ
=๐ฅ+ 2
+ ๐ (๐ฆ โ 2
).
2
๐ฅ +๐ฆ
๐ฅ + ๐ฆ2
Let
๐(๐ฅ, ๐ฆ) = ๐ฆ โ
๐ฅ2
2๐ฆ
.
+ ๐ฆ2
Now
๐๐ฅ = 1 +
2(๐ฅ 2 + ๐ฆ 2 ) โ 4๐ฅ 2
2(๐ฅ 2 + ๐ฆ 2 ) โ 4๐ฆ 2
=
1
โ
= ๐๐ฆ
(๐ฅ 2 + ๐ฆ 2 )2
(๐ฅ 2 + ๐ฆ 2 )2
and
๐๐ฆ =
โ4๐ฅ๐ฆ
= โ๐๐ฅ .
+ ๐ฆ 2 )2
(๐ฅ 2
Hence, the Cauchy-Riemann equations are satisfied, and ๐ is a harmonic conjugate of ๐. It
follows that
ฮฉ(๐ง) = ๐ง +
2
2๐ฅ
2๐ฆ
=๐ฅ+ 2
+
๐
(๐ฆ
โ
)
๐ง
๐ฅ + ๐ฆ2
๐ฅ2 + ๐ฆ2
is the complex potential corresponding to ๐, and the streamlines of the flow are given by the
level curves of ๐.
Now
๐ฆโ
๐ฅ2
2๐ฆ
= 0 โ ๐ฆ 3 + ๐ฅ 2 ๐ฆ โ 2๐ฆ = 0 โ ๐ฅ 2 + ๐ฆ 2 = 2,
+ ๐ฆ2
which is the equation of a circle of radius โ2 centred at the origin. Since the flow is tangential
to this circle it can be considered as a rigid obstacle past which the stream is flowing.
Therefore the flow given by the velocity potential ๐ is that of a uniform stream past a circular
SAQ 2 Express โ2 ๐ข = 0 in two dimensions in polar coordinates by considering ๐ข as the real
part of an analytic function and using the Cauchy-Riemann equations in polar form.
Potential Theory - SAQโs
Let ๐(๐ง) = ๐ข(๐, ๐) + ๐๐ฃ(๐, ๐) be an analytic function, then the polar form of the CauchyRiemann equations give
๐๐ข ๐๐ฃ
๐๐ฃ
๐๐ข
=
and ๐
=โ .
๐๐ ๐๐
๐๐
๐๐
Since ๐ข and ๐ฃ are analytic,
๐
๐
๐๐ข
๐ ๐๐ฃ
๐ ๐๐ฃ
๐ 1 ๐๐ข
(๐ ) = ( ) =
( )=โ (
).
๐๐ ๐๐
๐๐ ๐๐
๐๐ ๐๐
๐๐ ๐ ๐๐
Hence
๐
๐๐ข
๐ 1 ๐๐ข
(๐ ) +
(
) = 0.
๐๐ ๐๐
๐๐ ๐ ๐๐
Using the chain rule gives
๐
๐๐ข
๐ 1 ๐๐ข
๐ 2 ๐ข ๐๐ข 1 ๐ 2 ๐ข
(๐ ) +
(
)=๐ 2+
+
= 0.
๐๐ ๐๐
๐๐ ๐ ๐๐
๐๐
๐๐ ๐ ๐๐ 2
Which gives
๐2
๐ 2๐ข
๐๐ข ๐ 2 ๐ข
+
๐
+
= 0.
๐๐ 2
๐๐ ๐๐ 2
SAQ 3 Find the potential inside a capacitor consisting of an infinitely long circular cylinder
of radius ๐ charged to a potential of 10 volts, and an infinitely long wire of radius ๐ along the
axis of the cylinder at a potential of zero.
Hint: by symmetry, the solution must be independent of the polar angle ๐.
The electrostatic potential ๐(๐) inside the capacitor is independent of the polar angle ๐ and so
satisfies the polar form of Laplace's equation
๐2 ๐
๐๐
+๐
= 0, ๐ < ๐ < ๐
2
๐๐
๐๐
with boundary conditions
๐2
๐(๐) = 0, ๐(๐) = 10.
The differential equation above is a Cauchy-Euler equation.
Indicial equation:
๐2 = 0 โ ๐ = 0.
General solution:
๐(๐) = ๐ด + ๐ต ln ๐.
Applying the boundary conditions gives
๐(๐) = ๐ด + ๐ต ln ๐ = 0 โ ๐ด = โ๐ต ln ๐
and
๐(๐) = ๐ด + ๐ต ln ๐ = 10 โ โ๐ต ln ๐ + ๐ต ln ๐ = 10 โ ๐ต =
Hence, the required potential equation is
10
.
ln ๐ โ ln ๐
Potential Theory - SAQโs
๐(๐) = โ๐ต ln ๐ +
10
10(ln ๐ โ ln ๐)
ln ๐ =
.
ln ๐ โ ln ๐
ln ๐ โ ln ๐
SAQ 4 Find the potential in an annulus bounded by circles of radii ๐ < ๐ if the inner circle
has potential ๐ผ and the outer circle has potential ๐ฝ.
The electrostatic potential ๐(๐) inside the capacitor is independent of the polar angle ๐ and so
satisfies the polar form of Laplace's equation
๐2 ๐
๐๐
+๐
= 0, ๐ < ๐ < ๐
2
๐๐
๐๐
with boundary conditions
๐2
๐(๐) = ๐ผ,
๐(๐) = ๐ฝ.
The differential equation above is a Cauchy-Euler equation.
Indicial equation:
๐2 = 0 โ ๐ = 0.
General solution:
๐(๐) = ๐ด + ๐ต ln ๐.
Applying the boundary conditions gives
๐(๐) = ๐ด + ๐ต ln ๐ = ๐ผ โ ๐ด = ๐ผ โ ๐ต ln ๐
and
๐(๐) = ๐ด + ๐ต ln ๐ = ๐ฝ โ ๐ผ โ ๐ต ln ๐ + ๐ต ln ๐ = ๐ฝ โ ๐ต =
๐ฝโ๐ผ
.
ln ๐ โ ln ๐
Hence, the required potential equation is
๐(๐) = ๐ผ โ ๐ต ln ๐ +
SAQ 5
๐ฝโ๐ผ
๐ผ ln ๐ โ ๐ฝ ln ๐ + (๐ฝ โ ๐ผ) ln ๐
ln ๐ =
.
ln ๐ โ ln ๐
ln ๐ โ ln ๐
What conditions on the real numbers ๐, ๐, ๐, ๐ ensure that
๐๐ฅ 3 + ๐๐ฅ 2 ๐ฆ + ๐๐ฅ๐ฆ 2 + ๐๐ฆ 3
is harmonic.
Let
๐(๐ฅ, ๐ฆ) = ๐๐ฅ 3 + ๐๐ฅ 2 ๐ฆ + ๐๐ฅ๐ฆ 2 + ๐๐ฆ 3 .
Then ๐ is harmonic if it has continuous first and second-order partial derivatives in a domain
๐ท, and satisfies Laplaceโs equation
โ2 ๐ =
Now
๐ 2๐ ๐ 2๐
+
= 0.
๐๐ฅ 2 ๐๐ฆ 2
Potential Theory - SAQโs
๐๐
= 3๐๐ฅ 2 + 2๐๐ฅ๐ฆ + ๐๐ฆ 2
๐๐ฅ
and
๐ 2๐
= 6๐๐ฅ + 2๐๐ฆ
๐๐ฅ 2
๐๐
= 2๐๐ฅ + 2๐๐ฅ๐ฆ + 3๐๐ฆ 2
๐๐ฆ
and
๐ 2๐
= 2๐๐ฅ + 6๐๐ฆ.
๐๐ฆ 2
and
So ๐ has continuous first and second-order partial derivatives in the complex plane.
Substituting the second-order partial derivatives into Laplace's equation gives
6๐๐ฅ + 2๐๐ฆ + 2๐๐ฅ + 6๐๐ฆ = 2(3๐ + ๐)๐ฅ + 2(๐ + 3๐)๐ฆ = 0
which is satisfied when
๐ = โ3๐
and ๐ = โ3๐ .
Hence
๐(๐ฅ, ๐ฆ) = ๐๐ฅ 3 โ 3๐๐ฅ 2 ๐ฆ โ 3๐๐ฅ๐ฆ 2 + ๐๐ฆ 3 = ๐(๐ฅ 3 โ 3๐ฅ๐ฆ 2 ) + ๐(๐ฆ 3 โ 3๐ฅ 2 ๐ฆ).
Now
๐ง 3 = (๐ฅ + ๐๐ฆ)3 = ๐ฅ 3 โ 3๐ฅ๐ฆ 2 + ๐(3๐ฅ 2 ๐ฆ โ ๐ฆ 3 ).
Which gives
๐(๐ฅ, ๐ฆ) = ๐ Re(๐ง 3 ) โ ๐ Im(๐ง 3 )
for all complex ๐ง.
SAQ 6
๐ง+1
By considering arg (๐งโ1), find a function ๐ harmonic in the unit disk such that
๐(๐ ๐๐ ) = {
1
0<๐<๐
โ1 โ๐ < ๐ < 0.
Let ๐ง = ๐ ๐๐ , then |๐ง| < 1 and
(cos ๐ + 1) + ๐ sin ๐
๐ ๐๐ + 1
arg ( ๐๐
) = arg (
)
(cos ๐ โ 1) + ๐ sin ๐
๐ โ1
[(cos ๐ + 1) + ๐ sin ๐][(cos ๐ โ 1) โ ๐ sin ๐]
= arg (
)
(cos ๐ โ 1)2 + sin2 ๐
โ sin ๐
= arg (
๐)
1 โ cos ๐
๐
โ
0<๐<๐
2
={ ๐
โ๐ < ๐ < 0.
2
So, we define
2
๐ง+1
๐(๐ง) = โ arg (
).
๐
๐งโ1
Let
๐ง+1
๐ง+1
๐(๐ง) = log (
)=|
| + ๐๐(๐ง).
๐งโ1
๐งโ1
Potential Theory - SAQโs
Since ๐ is analytic in the unit open disk |๐ง| < 1, it follows that ๐(๐ง) = Im(๐(๐ง)) is harmonic
there.
SAQ 7 By considering the definition of a Green's function on page 245 and Theorem 6.3.4
prove the result given in Exercises 6.3 #2.
We must show that
1
1
ln โ(๐ฅ โ ๐ฅ0 )2 + (๐ฆ โ ๐ฆ0 )2 +
ln โ(๐ฅ โ ๐ฅ0 )2 + (๐ฆ + ๐ฆ0 )2
2๐
2๐
is a Green's function for the upper half-plane problem in the sense that the solution, if it exists,
can be expressed as
๐บ(๐ฅ, ๐ฆ, ๐ฅ0 , ๐ฆ0 ) = โ
โ
๐ฆ0 โ
๐(๐ฅ)
๐ข(๐ฅ0 , ๐ฆ0 ) = โซ ๐(๐ฅ)๐บ๐ฆ (๐ฅ, 0, ๐ฅ0 , ๐ฆ0 ) ๐๐ฅ = โซ
๐๐ฅ.
๐ โโ (๐ฅ โ ๐ฅ0 )2 + ๐ฆ02
โโ
Let let ๐ง0 = ๐ฅ0 + ๐๐ฆ0 be a point in the open upper half-plane๐ท, and let ๐ = ๐ท โช ๐ถ where ๐ถ is
the ๐ฅ-axis. Then
1
1
ln|๐ง โ ๐ง0 | +
ln|๐ง โ ๐ง0 |
2๐
2๐
1
1
=โ
Re(log(๐ง โ ๐ง0 )) +
Re(log(๐ง โ ๐ง0 )).
2๐
2๐
๐บ(๐ฅ, ๐ฆ, ๐ฅ0 , ๐ฆ0 ) = โ
Hence ๐บ is analytic in ๐ท. It follows that
1
1
Re(log(๐ง โ ๐ง0 )) =
ln โ(๐ฅ โ ๐ฅ0 )2 + (๐ฆ + ๐ฆ0 )2
2๐
2๐
is harmonic in ๐ท.
Also, ๐บ is continuous in ๐ท โช ๐ถ, except at (๐ฅ0 , ๐ฆ0 ) in ๐ท.
Finally, since ๐บ(๐ฅ, 0, ๐ฅ0 , ๐ฆ0 ) = 0, we have ๐บ(๐ฅ, ๐ฆ, ๐ฅ0 , ๐ฆ0 ) = 0 when (๐ฅ, ๐ฆ) is on ๐ถ.
Hence, ๐บ is a Greenโs function associated with the Dirichlet problem for the closed upper halfplane.
By Theorem 6.3.4, the solution to the Dirichlet problem is given by
๐ข(๐ฅ0 , ๐ฆ0 ) = โ โซ(๐(๐ )โ๐บ โ N) ๐๐
๐ถ
where N is the unit outward normal on ๐ถ and ๐๐  is the element of arc length on ๐ถ, solves the
Dirichlet problem โ2 ๐ข = 0 in ๐ท with ๐ข = ๐ on ๐ถ.
In this case
N = โj and ๐๐  = ๐๐ฅ.
Hence, integrating along the ๐ฅ-axis gives
โ
โ
๐ข(๐ฅ0 , ๐ฆ0 ) = โซ (๐(๐ฅ)โ๐บ โ j) ๐๐ฅ = โซ (๐(๐ฅ)๐บ๐ฆ (๐ฅ, ๐ฆ, ๐ฅ0 , ๐ฆ0 )|๐ฆ=0 ) ๐๐ฅ
โโ
โโ
Potential Theory - SAQโs
โ
= โซ (๐(๐ฅ)
โโ
+
1
ln โ(๐ฅ โ ๐ฅ0 )2 + (๐ฆ + ๐ฆ0 )2 ] ) ๐๐ฅ
2๐
๐ฆ=0
โ
= โซ (๐(๐ฅ)
โโ
+
๐
1
[โ ln โ(๐ฅ โ ๐ฅ0 )2 + (๐ฆ โ ๐ฆ0 )2
๐๐ฆ 2๐
๐
1
[โ ln((๐ฅ โ ๐ฅ0 )2 + (๐ฆ โ ๐ฆ0 )2 )
๐๐ฆ 4๐
1
ln((๐ฅ โ ๐ฅ0 )2 + (๐ฆ + ๐ฆ0 )2 )] ) ๐๐ฅ
4๐
๐ฆ=0
=โ
1 โ
2(๐ฆ โ ๐ฆ0 )
2(๐ฆ + ๐ฆ0 )
โซ (๐(๐ฅ) [
โ
] ) ๐๐ฅ
2
2
(๐ฅ โ ๐ฅ0 ) + (๐ฆ โ ๐ฆ0 )
(๐ฅ โ ๐ฅ0 )2 + (๐ฆ + ๐ฆ0 )2
4๐ โโ
๐ฆ=0
=โ
1 โ
โ2๐ฆ0
2๐ฆ0
โซ (๐(๐ฅ) [
โ
]) ๐๐ฅ
2
2
(๐ฅ โ ๐ฅ0 ) + ๐ฆ0 (๐ฅ โ ๐ฅ0 )2 + ๐ฆ02
4๐ โโ
๐ฆ0 โ
๐(๐ฅ)
= โซ
๐๐ฅ.
๐ โโ (๐ฅ โ ๐ฅ0 )2 + ๐ฆ02
SAQ 8 Prove that the function ๐(๐ง) = ๐ง + ๐2 ๐ง 2 + ๐3 ๐ง 3 + โฏ is simple for |๐ง| < 1 if
โโ
๐=2 ๐|๐๐ | โค 1.
Since
โ
โ
๐ด โ ๐ โค โ ๐|๐๐ | โค 1
๐=2
๐=2
where ๐ด = min{|๐2 |, |๐3 |, โฆ }, we have
๐ดโค
1
โโ
๐=2 ๐
which implies that |๐๐ | โ 0 as ๐ โ โ.
๐
Hence, by the root test, ๐(๐ง) = โโ
๐=2 ๐๐ ๐ง defines an analytic function on |๐ง| < โ.
Suppose that ๐ง1 โ  ๐ง2 , then
๐(๐ง1 ) โ ๐(๐ง2 ) = (๐ง1 + ๐2 ๐ง12 + ๐3 ๐ง13 + โฏ ) โ (๐ง2 + ๐2 ๐ง22 + ๐3 ๐ง23 + โฏ )
= (๐ง1 โ ๐ง2 + ๐2 (๐ง12 โ ๐ง22 ) + ๐3 (๐ง13 โ ๐ง23 ) + โฏ )
= (๐ง1 โ ๐ง2 )(1 + ๐2 (๐ง1 + ๐ง2 ) + ๐3 (๐ง12 + ๐ง1 ๐ง2 + ๐ง22 ) + โฏ ).
Hence, by the reverse triangle inequality
|๐(๐ง1 ) โ ๐(๐ง2 )| โฅ ||๐ง1 + ๐2 ๐ง12 + ๐3 ๐ง13 + โฏ | โ |๐ง2 + ๐2 ๐ง22 + ๐3 ๐ง23 + โฏ ||
= |๐ง1 โ ๐ง2 |(1 + |๐2 |(|๐ง1 | + |๐ง2 |) + |๐3 |(|๐ง12 | + |๐ง1 ๐ง2 | + |๐ง22 |) + โฏ )
โฅ |๐ง1 โ ๐ง2 |(1 + 2|๐2 | + 3|๐3 | + โฏ ) > 0.
Potential Theory - SAQโs
So, ๐ง1 โ  ๐ง2 implies that ๐(๐ง1 ) โ  ๐(๐ง2 ), which is the contrapositive of ๐(๐ง1 ) = ๐(๐ง2 ) โ ๐ง1 = ๐ง2 ;
the necessary condition for a one-to-one function.
It follows that ๐(๐ง) is analytic and one-to-one on |๐ง| < 1, and so is simple there.
SAQ 9 Let ๐(๐ง) be simple within and on a simple closed contour ๐ถ. Let ๐ท the interior of ๐ถ,
map onto ๐ทโฒ , and ๐ถ map onto ๐ถ โฒ , the boundary of ๐ทโฒ .
(i)
Prove that the length of ๐ถ โฒ is โซ๐ถ|๐ โฒ (๐ง)| |๐๐ง|;
(ii)
Prove that the area of ๐ทโฒ is โฌ๐ท|๐ โฒ (๐ง)|2 ๐๐ฅ๐๐ฆ;
(iii) If ๐ถ is a circle of radius ๐ centred at the origin, prove that the area of ๐ท โฒ is greater than,
or equal to ๐|๐ โฒ (0)|2 ๐ 2.
(i) Since ๐ is simple, the curve ๐ถ is mapped one-to-one onto the curve ๐ถ โฒ . Hence, ๐ถ โฒ is smooth
and has a length ๐ฟ given by
๐ฟ = โซ |๐๐ค|.
๐ถโฒ
Now ๐๐ค = ๐ โฒ (๐ง)๐๐ง, which gives
๐ฟ = โซ |๐ โฒ (๐ง)๐๐ง| = โซ|๐ โฒ (๐ง)||๐๐ง|.
๐ถโฒ
๐ถ
(ii) Let ๐ง = ๐ฅ + ๐๐ฆ, then the area element ๐๐ด at ๐ง is ๐๐ฅ๐๐ฆ. The mapping ๐ค = ๐(๐ง) changes
small distances near ๐ง0 by the scale factor |๐ โฒ (๐ง0 )|. Therefore, ๐๐ด is transformed by ๐ into an
element of area of ๐ทโฒ given by |๐ โฒ (๐ง)|2 ๐๐ฅ๐๐ฆ at ๐(๐ง) โ ๐ทโฒ . Hence, ๐ทโฒ has an area ๐ด given by
๐ด = โฌ |๐ โฒ (๐ง)|2 ๐๐ฅ๐๐ฆ.
๐ท
(iii) Too difficult to waste time on.
SAQ 10
Prove that ๐ค = 3๐ง + ๐ง 2 is simple on the disk |๐ง| < 1.
Let ๐ค = ๐(๐ง), then ๐(๐ง) is analytic on |๐ง| < 1.
Suppose that ๐ง1 โ  ๐ง2 , then
๐(๐ง1 ) โ ๐(๐ง2 ) = (3๐ง1 + ๐ง12 ) โ (3๐ง2 + ๐ง22 ) = (3(๐ง1 โ ๐ง2 ) + (๐ง12 โ ๐ง22 ))
= (๐ง1 โ ๐ง2 )(3 + ๐ง1 + ๐ง2 ).
Hence, by the reverse triangle inequality
|๐(๐ง1 ) โ ๐(๐ง2 )| โฅ ||3๐ง1 + ๐ง12 | โ |3๐ง2 + ๐ง22 || = |๐ง1 โ ๐ง2 |(3 + |๐ง1 | + |๐ง2 |)
โฅ 5|๐ง1 โ ๐ง2 | > 0.
So, ๐ง1 โ  ๐ง2 implies that ๐(๐ง1 ) โ  ๐(๐ง2 ), which is the contrapositive of ๐(๐ง1 ) = ๐(๐ง2 ) โ ๐ง1 = ๐ง2 ;
the necessary condition for a one-to-one function.
It follows that ๐(๐ง) is analytic and one-to-one on |๐ง| < 1, and so is simple there.
Potential Theory - Problems
P1 Consider a 2-dimensional flow outside a simple curve ๐ถ in the (๐ฅ, ๐ฆ)-plane. If the
velocity v = (๐ฃ๐ฅ , ๐ฃ๐ฆ ) is such that ๐ฃ๐ฅ โ ๐๐ฃ๐ฆ is analytic outside ๐ถ, i.e. ๐ฃ๐ฅ โ ๐๐ฃ๐ฆ = ๐(๐ง) with ๐ง =
๐ฅ + ๐๐ฆ and ๐(๐ง) analytic, show that the circulation is the same round any simple curve
surrounding ๐ถ.
Cannot answer this problem. There is nothing in the text book or course notes that suggests a
P2 Consider the hypothesis of the preceding problem with ๐ถ the unit circle {|๐ง| = 1}. Prove
๐
that if the circulation is non-zero, then ๐(๐ง) is of the form โโ
โโ ๐๐ ๐ง with Im(๐โ1 ) = 0.
Since ๐ is analytic on the open annulus ๐ด = {1 < |๐ง| < โ}, it has the Laurent series
representation
โ
๐(๐ง) = โ ๐๐ ๐ง ๐
๐=โโ
for all ๐ง โ ๐ด.
The velocity field v is given by
๐ฮฉ
= ๐ฃ๐ฅ + ๐๐ฃ๐ฆ .
๐๐ง
Hence
๐ฮฉ
= ๐ฃ๐ฅ โ ๐๐ฃ๐ฆ = ๐(๐ง) โ
๐๐ง
ฮฉ(๐ง) = โซ๐(๐ง) ๐๐ง.
๐ถ
Which gives, for the circle |๐ง| = 1
โ
๐(๐ฅ, ๐ฆ) = Re (โซ
โ ๐๐ ๐ง ๐ ๐๐ง).
๐(๐ง) ๐๐ง) = Re (โซ
|๐ง|=1
|๐ง|=1
๐=โโ
On the unit circle ๐ง = ๐ , which gives
๐๐
โ
2๐
๐(๐ฅ, ๐ฆ) = Re (โซ
(๐๐
๐๐
0
โ ๐๐ ๐ ๐๐๐ ) ๐๐)
๐=โโ
โ
2๐
= Re (๐ โ ๐๐ โซ ๐ ๐(๐+1)๐ ๐๐)
0
๐=โโ
2๐
= Re (๐๐โ1 โซ ๐๐) = Re(๐2๐๐โ1 )
0
since
2๐
โซ ๐ ๐๐๐ ๐๐ = 0
0
for any non-zero integer ๐.
Consequently, ๐(๐ฅ, ๐ฆ) is non-zero if Im(๐โ1 ) = 0.
Potential Theory - Problems
P3 Find a steady state temperature for a thin metal infinite strip {(๐ฅ, ๐ฆ): โโ < ๐ฅ < โ, 0 โค
๐ฆ โค 1} when the upper edge is perfectly insulated and ๐ข(โ1,0) = 3 and ๐ข(1,0) = 2. How
many solutions are there for ๐ข(๐ฅ, ๐ฆ)?
The steady-state temperature ๐ข is a solution of Laplace's equation in the domain ๐ท defined by
โโ < ๐ฅ < โ, 0 โค ๐ฆ โค 1, which satisfies the boundary conditions
๐ข(โ1,0) = 3;
๐ข(1,0) = 2;
โ๐ข โ N = 0
for
โโ < ๐ฅ < โ, ๐ฆ = 1.
In this case, the unit outward facing normal is N = j, hence
(โ๐ข โ N)|๐ฆ=1 =
๐๐ข
|
= 0.
๐๐ฆ ๐ฆ=1
We seek a harmonic function ๐ข, which behaves as required on the boundary, and in particular
satisfies the condition (โ๐ข โ N)|๐ฆ=1 = 0.
One possibility is that ๐ข(๐ฅ, ๐ฆ) is a function of ๐ฅ alone, i.e.
๐ข(๐ฅ, ๐ฆ) = ๐(๐ฅ)
for some function ๐ of ๐ฅ.
In order to satisfy Laplace's equation we must have
๐2 ๐ข ๐2๐ข ๐2๐
+
=
= 0.
๐๐ฅ 2 ๐๐ฆ 2 ๐๐ฅ 2
Integrating twice gives
๐(๐ฅ) = ๐ด๐ฅ + ๐ต
for arbitrary constants ๐ด, ๐ต.
Applying the remaining boundary conditions gives
๐ข(โ1,0) = ๐(โ1) = โ๐ด + ๐ต = 3
and
๐ข(1,0) = ๐(1) = ๐ด + ๐ต = 2.
Hence
1
5
and ๐ต =
2
2
which gives a solution
๐ด=โ
5 ๐ฅ
โ .
2 2
Another possibility is ๐ข(๐ฅ, ๐ฆ) = ๐(๐ฅ)โ(๐ฆ) where โโฒ (1) = 0.
๐ข(๐ฅ, ๐ฆ) =
Consider the function ๐(๐ง) = ๐ ๐๐งโ๐๐ for real ๐. Since ๐ is analytic in ๐ท, it follows that
Re(๐ ๐๐งโ๐๐ ) = Re(๐ ๐๐ฅ ๐ ๐(๐ฆโ1)๐ ) = ๐ ๐๐ฅ cos(๐(๐ฆ โ 1))
Potential Theory - Problems
is harmonic in ๐ท.
Now
๐ ๐๐ฅ
[๐ cos(๐(๐ฆ โ 1))]|
= โ๐ ๐๐ฅ ๐ sin(0) = 0.
๐๐ฆ
๐ฆ=1
Choosing ๐ = ๐โ2 gives
๐
๐ ๐๐ฅโ2 cos ( (๐ฆ โ 1))
2
which is equal to zero when ๐ฆ = 0.
Therefore
๐ข(๐ฅ, ๐ฆ) =
5 ๐ฅ
๐
โ + ๐ถ๐ ๐๐ฅโ2 cos ( (๐ฆ โ 1))
2 2
2
solves the problem for any constant ๐ถ, which means there are infinitely many solutions.
P4
Prove the assertion on p.235 that
๐พ๐ 2
โซ 3 (๐ฅ๐๐ฅ + ๐ฆ๐๐ฆ + ๐ง๐๐ง)
๐ถ ๐
is independent of the path assuming that it is smooth.
Note: the quantity represented by ๐ 3 is the distance from the origin.
Let the path ๐ถ be parameterised by
(๐ฅ(๐ก), ๐ฆ(๐ก), ๐ง(๐ก))
for
๐โค๐กโค๐
where ๐ and ๐ are the left and right endpoints of ๐ถ, respectively.
Since ๐ถ is smooth, the derivatives ๐ฅ โฒ (๐ก), ๐ฆ โฒ (๐ก), ๐ง โฒ (๐ก) exist. Which gives
๐ฅ๐๐ฅ + ๐ฆ๐๐ฆ + ๐ง๐๐ง = ๐ฅ(๐ก)
๐๐ฅ
๐๐ฆ
๐๐ง
๐๐ก + ๐ฆ(๐ก) ๐๐ก + ๐ง(๐ก) ๐๐ก
๐๐ก
๐๐ก
๐๐ก
on ๐ถ.
Hence
๐
๐พ๐ 2
๐ฅ(๐ก)๐ฅ โฒ (๐ก) + ๐ฆ(๐ก)๐ฆ โฒ (๐ก) + ๐ง(๐ก)๐ง โฒ (๐ก)
2
(๐ฅ๐๐ฅ
+
๐ฆ๐๐ฆ
+
๐ง๐๐ง)
=
๐พ๐
โซ
๐๐ก
3
3โ2
๐ถ ๐
๐
(๐ฅ 2 (๐ก) + ๐ฆ 2 (๐ก) + ๐ง 2 (๐ก))
โซ
= โ๐พ๐
2
๐
1
(๐ฅ 2 (๐ก) + ๐ฆ 2 (๐ก) + ๐ง 2 (๐ก))
1โ2
|
๐
since
๐ 2
2
2
๐
1
1 ๐๐ก [๐ฅ (๐ก) + ๐ฆ (๐ก) + ๐ง (๐ก)]
[
]=โ
๐๐ก (๐ฅ 2 (๐ก) + ๐ฆ 2 (๐ก) + ๐ง 2 (๐ก))1โ2
2 (๐ฅ 2 (๐ก) + ๐ฆ 2 (๐ก) + ๐ง 2 (๐ก))3โ2
Potential Theory - Problems
=โ
๐ฅ(๐ก)๐ฅ โฒ (๐ก) + ๐ฆ(๐ก)๐ฆ โฒ (๐ก) + ๐ง(๐ก)๐ง โฒ (๐ก)
(๐ฅ 2 (๐ก) + ๐ฆ 2 (๐ก) + ๐ง 2 (๐ก))
3โ2
.
So the line integral depends only on the endpoints of ๐ถ, which are independent of its actual
path.
P5 If ๐ข(๐ง) is harmonic and positive in the unit disk and continuous in the closed unit disk
show that
1โ๐
1+๐
โค ๐ข(๐๐ ๐๐ ) โค ๐ข(0)
.
1+๐
1โ๐
Hint: first find upper and lower bounds for the Poisson kernel of the unit disk.
๐ข(0)
As shown in the proof of Theorem 6.2.2, for the unit disk {๐ง: |๐ง| < ๐ = 1}
(1 โ ๐ 2 )๐ข(๐ ๐๐ )
1 2๐
๐ข(๐๐ ) =
โซ
๐๐
2๐ 0 1 โ 2๐ cos(๐ โ ๐) + ๐ 2
๐๐
1 2๐
1 โ ๐2
=
โซ (
) ๐ข(๐ ๐๐ ) ๐๐
2
2๐ 0
1 โ 2๐ cos(๐ก) + ๐
with ๐ก = ๐ โ ๐.
Hence
๐ข(0) =
1 2๐
โซ ๐ข(๐ ๐๐ ) ๐๐.
2๐ 0
Since both the Poisson kernel and ๐ข(๐ ๐๐ ) are positive
1 2๐
1 โ ๐2
โซ min
๐ข(๐ ๐๐ ) ๐๐ โค ๐ข(๐๐ ๐๐ )
2
0โค๐กโค2๐
2๐ 0
1 โ 2๐ cos(๐ก) + ๐
โค
1 2๐
1 โ ๐2
โซ max
๐ข(๐ ๐๐ ) ๐๐.
2๐ 0 0โค๐กโค2๐ 1 โ 2๐ cos(๐ก) + ๐ 2
Since
1 โ ๐2
1 โ ๐2
1 โ ๐2
=
=
0โค๐กโค2๐ 1 โ 2๐ cos(๐ก) + ๐ 2
1 + 2๐ + ๐ 2 (1 + ๐)2
min
and
1 โ ๐2
1 โ ๐2
1 โ ๐2
=
=
.
0โค๐กโค2๐ 1 โ 2๐ cos(๐ก) + ๐ 2
1 โ 2๐ + ๐ 2 (1 โ ๐)2
max
We have
1 โ ๐ 2 1 2๐
1 โ ๐ 2 1 2๐
๐๐
๐๐
โซ ๐ข(๐ ) ๐๐ โค ๐ข(๐๐ ) โค
โซ ๐ข(๐ ๐๐ ) ๐๐.
(1 + ๐)2 2๐ 0
(1 โ ๐)2 2๐ 0
Which gives, since ๐ < 1
1โ๐
1+๐
๐ข(0) โค ๐ข(๐๐ ๐๐ ) โค
๐ข(0).
1+๐
1โ๐
Potential Theory - Problems
P6 Let ๐ท be a domain in the plane bounded by a simple closed curve ๐ถ. Let ๐ be a realvalued smooth function defined on ๐ถ, and let ๐น be the family of real-valued functions with
smooth derivatives on ๐ท and ๐ถ and boundary values ๐. If
inf {โฌ (๐ข๐ฅ2 + ๐ข๐ฆ2 ) ๐๐ด: ๐ข โ ๐น}
๐ท
is attained by ๐ โ ๐น, show that ๐ is the solution of the Dirichlet problem for ๐ท with boundary
values ๐.
A long and complicated proof is the required answer. Not worth wasting time on.
P7
Find a univalent mapping of the half disk {๐ง: |๐ง| < 1, Im(๐ง) > 0} onto the unit disk.
Let ๐ท = {๐ง: |๐ง| < 1, Im(๐ง) > 0}. According to the Riemann mapping theorem, there exists a
simple function ๐ค = ๐(๐ง) which maps ๐ท onto the open unit disk |๐ค| < 1 in the ๐ค-plane.
The boundary of the domain ๐ท consists of two circular arcs
(i)
the upper semicircle {๐ง: |๐ง| = 1, Im(๐ง) > 0};
(ii)
the ๐ฅ-axis as a circle through ๐ง = 0 and ๐ง = โ.
Consider the linear fractional transformation
1+๐ง
.
1โ๐ง
The images of the points โ1, ๐, and 1 under this transformation are
๐ค=
๐ค(โ1) = 0, ๐ค(๐) = ๐,
and ๐ค(1) = โ.
Therefore, the upper semicircle is mapped to the positive imaginary axis.
The images of the points โ1,0, and 1 under this transformation are
๐ค(โ1) = 0, ๐ค(0) = 1,
and ๐ค(1) = โ.
Therefore, the ๐ฅ-axis is mapped to the positive real axis.
The test point ๐ง = 1โ2 is mapped to
๐ค(1 + ๐) =
1 + ๐ โ2 3 4
= + ๐.
1 โ ๐ โ2 5 5
Hence, ๐ค maps ๐ท onto the first quadrant in the ๐ค-plane.
The mapping ๐ = ๐ค 2 maps the first quadrant in the ๐ค-plane onto the upper half ๐-plane.
The mapping
๐ โ ๐ โ1(๐ โ ๐)
=
๐+๐
๐+๐
maps the the upper half ๐-plane onto the unit disk in the ๐-plane.
๐=
Forming the composition of these mappings gives
Potential Theory - Problems
2
๐=
๐โ๐ ๐โ๐ค
=
=
๐ + ๐ ๐ + ๐ค2
1+๐ง 2
๐ โ (1 โ ๐ง)
1+๐ง 2
๐ + (1 โ ๐ง)
=
๐(1 โ ๐ง)2 โ (1 + ๐ง)2
1 โ 2๐๐ง + ๐ง 2
=
๐
.
๐(1 โ ๐ง)2 + (1 + ๐ง)2
1 + 2๐๐ง + ๐ง 2
P8 Let ๐(๐ง) be univalent in {|๐ง| < 1} with ๐(0) = 0. Show that one can find ๐(๐ง) univalent
in the unit disk and satisfying
2
(๐(๐ง)) = ๐(๐ง 2 )
for |๐ง| < 1.
The solution is too complex to be worth bothering with.
P9 Let ๐(๐ง) be univalent and map the annulus {1 < |๐ง| < ๐1 } univalently onto the annulus
{1 < |๐ค| < ๐2 } so that the inner and outer boundaries correspond. Show that ๐(๐ง) = ๐ด๐ง for
some constant ๐ด with |๐ด| = 1, and so ๐1 = ๐2 . What happens when {|๐ง| = ๐1 } maps to
{|๐ค| = 1} and {|๐ง| = 1}?
The solution is too complex to be worth bothering with.
P10 Find a function which maps the strip ๐ = {๐ง: 0 < Im(๐ง) < ๐} into the interior of the
unit circle. Write down a solution of the Dirichlet problem for this strip.
The mapping
๐ = ๐๐ง
maps ๐ onto the upper half ๐-plane, and the mapping
๐โ๐
๐+๐
maps the upper half ๐-plane onto the unit disk in the ๐ค-plane.
๐ค=
Therefore, the mapping
๐ โ ๐ ๐ โ ๐๐ง
๐ค=
=
= ๐(๐ง)
๐ + ๐ ๐ + ๐๐ง
maps ๐ onto the interior of the unit circle.
The Poisson integral formula gives a solution for the unit disk, from which a solution in ๐ can
be obtained.
P11 Find a function which maps the strip ๐ = {๐ง: 0 โค Re(๐ง) โค ๐, Im(๐ง) โฅ 0} onto the upper
half plane so that the origin maps into the origin, and ๐ maps into 1.
Potential Theory - Problems
The strip ๐ is a polygonal region with interior angles ๐ผ1 = ๐ผ2 = ๐โ2 and vertices ๐ง1 = 0 and
๐ง2 = ๐.
Applying the Schwarz-Christoffel transformation gives
๐๐ง
๐ด
= ๐ด(๐ค โ 0)(๐โ2โ๐)โ1 × (๐ค โ 1)(๐โ2โ๐)โ1 = ๐ด๐ค โ1โ2 (๐ค โ 1)โ1โ2 =
.
1โ2
๐๐ค
(๐ค(๐ค โ 1))
Hence
1
๐ง = ๐ดโซ
(๐ค(๐ค โ 1))
= ๐ดโซ
1โ2
๐๐ค = ๐ด โซ
1
1 2
1 2
((๐ค โ 2) โ (2) )
= ๐ด log (๐ค โ
1โ2
(๐ค 2
1
๐๐ค
โ ๐ค)1โ2
๐๐ค
1
+ (๐ค 2 โ ๐ค)1โ2 ).
2
Further manipulation gives
๐๐ง
๐ด
๐ด
1
=
=
.
โ
1
2
๐๐ค (โ1๐ค(1 โ ๐ค))
๐ (๐ค(1 โ ๐ค))1โ2
Let ๐ต = ๐ดโ๐ , then
๐ง = ๐ตโซ
1
(๐ค(1 โ ๐ค))
= ๐ตโซ
1โ2
๐๐ค = ๐ต โซ
1
1 2
1 2
((2) โ (๐ค โ 2) )
1โ2
1
๐๐ค
(๐ค โ ๐ค 2 )1โ2
๐๐ค
1
๐คโ2
= ๐ต sinโ1 (
) + ๐ถ.
1
2
Now ๐ง1 = 0 when ๐ค1 = 0, hence
๐
0 = ๐ต sinโ1(โ1) + ๐ถ = โ ๐ต + ๐ถ.
2
And ๐ง2 = ๐ when ๐ค2 = 1, hence
๐
๐ = ๐ต sinโ1(1) + ๐ถ = ๐ต + ๐ถ.
2
Subtracting gives
๐ต=1
๐.
๐ = ๐๐ต โ {
๐ถ=
2
Hence
Potential Theory - Problems
๐ง = sin
โ1
(
1
2 ) + ๐.
1
2
2
๐คโ
Rearranging gives
1
๐
1
1
1 1
๐ง
๐ค = sin (๐ง โ ) + = โ cos ๐ง + = (1 โ cos ๐ง) = sin2 ( ).
2
2
2
2
2 2
2
```