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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œ“(๐‘ฅ, ๐‘ฆ) be a harmonic conjugate of a potential function ๐œ™(๐‘ฅ, ๐‘ฆ).
Then the analytic function
ฮฉ(๐‘ง) = ๐œ™(๐‘ฅ, ๐‘ฆ) + ๐‘–๐œ“(๐‘ฅ, ๐‘ฆ)
is called the complex potential corresponding to the real potential ๐œ™. The function ๐œ“ is called
the stream function.
The level curves
๐œ™(๐‘ฅ, ๐‘ฆ) = ๐‘1
and ๐œ“(๐‘ฅ, ๐‘ฆ) = ๐‘2
of ๐œ™ and ๐œ“, 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 {โˆฌ (๐‘ข๐‘ฅ2 + ๐‘ข๐‘ฆ2 ) ๐‘‘๐ด: ๐‘ข โˆˆ ๐น}
๐ท
is attained by ๐‘ˆ โˆˆ ๐น, then ๐‘ˆ 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 โˆ’ ๐‘ง0 | + ๐ป(๐‘ฅ, ๐‘ฆ, ๐‘ฅ0 , ๐‘ฆ0 )
2๐œ‹
where ๐ป is harmonic for |๐‘ง| < ๐‘… and
๐บ(๐‘ฅ, ๐‘ฆ, ๐‘ฅ0 , ๐‘ฆ0 ) = โˆ’
๐ป(๐‘… cos ๐œƒ , ๐‘… sin ๐œƒ , ๐‘ฅ0 , ๐‘ฆ0 ) =
1
ln|(๐‘… cos ๐œƒ โˆ’ ๐‘ฅ0 ) + ๐‘–(๐‘… sin ๐œƒ โˆ’ ๐‘ฆ0 )|
2๐œ‹
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 โˆ’ ๐‘ง0 |.
2๐œ‹
The behaviour on ๐ถ is specified so that an application of Greenโ€™s theorem leads to the desired
result.
โˆ’
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‘ƒ(๐‘ฅ, ๐‘ฆ)i + ๐‘„(๐‘ฅ, ๐‘ฆ)j is a smooth vector field on ๐‘…, then
โˆฌ(
๐‘…
๐œ•๐‘„ ๐œ•๐‘ƒ
โˆ’ ) ๐‘‘๐‘ฅ ๐‘‘๐‘ฆ = โˆฎ๐‘ƒ(๐‘ฅ, ๐‘ฆ)๐‘‘๐‘ฅ + ๐‘„(๐‘ฅ, ๐‘ฆ)๐‘‘๐‘ฆ.
๐œ•๐‘ฅ ๐œ•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œ™(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
โˆ‡2 ๐œ™ =
๐œ• 2๐œ™ ๐œ• 2๐œ™
+
=0
๐œ•๐‘ฅ 2 ๐œ•๐‘ฆ 2
is said to be harmonic in ๐ท. The function ๐œ™ 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œ• 2ฮฆ ๐œ• 2ฮฆ
+
=0
๐œ•๐‘ฅ 2 ๐œ•๐‘ฆ 2
is known as Laplace's equation, or the potential equation. The polar form of Laplace's
equation is given by
๐‘Ÿ 2 ฮฆ๐‘Ÿ๐‘Ÿ (๐‘Ÿ, ๐œƒ) + ๐‘Ÿฮฆ๐‘Ÿ (๐‘Ÿ, ๐œƒ) + ฮฆ๐œƒ๐œƒ (๐‘Ÿ, ๐œƒ) = 0
or alternatively
๐œ• 2ฮฆ
๐œ•ฮฆ ๐œ• 2 ฮฆ
+
๐‘Ÿ
+
= 0.
๐œ•๐‘Ÿ 2
๐œ•๐‘Ÿ ๐œ•๐œƒ 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‘Ÿ2
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 โˆ‡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 โˆ’โˆž < ๐‘ฅ < โˆž. Then the function
defined by
๐œ™(๐‘ฅ, ๐‘ฆ) =
๐‘ฆ โˆž
๐‘“(๐‘ )
โˆซ
๐‘‘๐‘ 
๐œ‹ โˆ’โˆž (๐‘ฅ โˆ’ ๐‘ )2 + ๐‘ฆ 2
is a solution of the Dirichlet problem in the upper half plane ๐‘ฆ > 0 with boundary condition
๐œ™(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œƒ ) on the unit circle
๐‘ง = ๐‘’ ๐‘–๐œƒ give a piecewise continuous and bounded function for โˆ’๐œ‹ โ‰ค ๐œƒ โ‰ค ๐œ‹. Then the function
defined by
๐œ™(๐‘ฅ, ๐‘ฆ) =
1 ๐œ‹ ๐‘“(๐‘’ ๐‘–๐‘  )(1 โˆ’ |๐‘ง|2 )
โˆซ
๐‘‘๐‘ 
|๐‘’ ๐‘–๐‘  โˆ’ ๐‘ง|2
2๐œ‹ โˆ’๐œ‹
is a solution of the Dirichlet problem in the open unit disk |๐‘ง| < 1 with boundary condition
๐œ™(cos ๐œƒ , sin ๐œƒ) = ๐‘“(๐‘’ ๐‘–๐œƒ ) at all points of continuity of ๐‘“.
Polar form. Let ๐‘“(๐œƒ) be a piecewise continuous and bounded function on โˆ’๐œ‹ < ๐œƒ < ๐œ‹. Then
the function defined by
๐œ™(๐‘Ÿ, ๐œƒ) =
1 ๐œ‹
๐‘“(๐‘ )(1 โˆ’ ๐‘Ÿ 2 )
โˆซ
๐‘‘๐‘ 
2๐œ‹ โˆ’๐œ‹ 1 โˆ’ 2๐‘Ÿ cos(๐œƒ โˆ’ ๐‘ ) + ๐‘Ÿ 2
is a solution of the Dirichlet problem in the open unit disk |๐‘ง| < 1 with boundary condition
๐œ™(๐‘Ÿ = 1, ๐œƒ) = ๐‘“(๐œƒ) at all points of continuity of ๐‘“.
Poisson kernel
The function ๐‘“: โ„‚ โ†’ โ„ defined by
๐‘…2 โˆ’ ๐‘Ÿ 2
๐‘“(๐‘Ÿ๐‘’ ) = 2
๐‘… โˆ’ 2๐‘Ÿ๐‘… cos ๐œƒ + ๐‘Ÿ 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‘Ÿ < ๐‘….
๐‘–๐œƒ
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 โˆž 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œ€-neighbourhood of the
point (๐‘ฅ0 , ๐‘ฆ0 ). Then there exists a conjugate harmonic function
๐‘ฃ(๐‘ฅ, ๐‘ฆ) = โˆซ ๐‘ข๐‘ฅ (๐‘ฅ, ๐‘ฆ) ๐‘‘๐‘ฆ โˆ’ โˆซ ๐‘ข๐‘ฆ (๐‘ฅ, ๐‘ฆ) ๐‘‘๐‘ฅ โˆ’ โˆฌ 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) = โˆซ ๐‘ฃ๐‘ฆ (๐‘ฅ, ๐‘ฆ) ๐‘‘๐‘ฆ + ๐ถ(๐‘ฅ) = โˆซ 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) = โˆ’๐‘ข๐‘ฆ (๐‘ฅ, ๐‘ฆ) gives
โˆ’๐‘ข๐‘ฆ (๐‘ฅ, ๐‘ฆ) =
๐œ•
(โˆซ ๐‘ข๐‘ฅ (๐‘ฅ, ๐‘ฆ) ๐‘‘๐‘ฆ) + ๐ถ โ€ฒ (๐‘ฅ).
๐œ•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) = โˆซ 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
E = grad ๐œ™ =
๐œ•๐œ™
๐œ•๐œ™
๐œ•๐œ™
i+
j+
k,
๐œ•๐‘ฅ
๐œ•๐‘ฆ
๐œ•๐‘ง
where ๐œ™ is an arbitrary function of position known as the electrostatic potential.
The electrostatic potential function ๐œ™ 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œ“(๐‘ฅ, ๐‘ฆ) defined in some domain ๐ธ in โ„2 , so that the complex
potential function ฮฉ(๐‘ง) = ๐œ™(๐‘ฅ, ๐‘ฆ) + ๐‘–๐œ“(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œ™(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œ“(๐‘ฅ, ๐‘ฆ) = ๐‘2 , curves that are
orthogonal to the family ๐œ™(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
v = grad ๐œ™ =
๐œ•๐œ™
๐œ•๐œ™
๐œ•๐œ™
i+
j+
k,
๐œ•๐‘ฅ
๐œ•๐‘ฆ
๐œ•๐‘ง
where ๐œ™ is an arbitrary function of position known as the velocity potential.
The velocity potential function ๐œ™ 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œ“(๐‘ฅ, ๐‘ฆ) defined in some domain ๐ธ in โ„2 , so that the complex
potential function ฮฉ(๐‘ง) = ๐œ™(๐‘ฅ, ๐‘ฆ) + ๐‘–๐œ“(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œ™(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œ“(๐‘ฅ, ๐‘ฆ) = ๐‘2 , curves that are
orthogonal to the family ๐œ™(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œ™ 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œƒ) be a piecewise continuous function on 0 โ‰ค ๐œƒ โ‰ค 2๐œ‹. Then the
function defined by
1 2๐œ‹
๐‘”(๐‘ )(๐‘… 2 โˆ’ ๐‘Ÿ 2 )
๐œ™(๐‘Ÿ, ๐œƒ) =
โˆซ
๐‘‘๐‘ 
2๐œ‹ 0 ๐‘… 2 โˆ’ 2๐‘…๐‘Ÿ cos(๐œƒ โˆ’ ๐‘ ) + ๐‘Ÿ 2
is harmonic in the open disk |๐‘ง| < ๐‘…, and lim ๐œ™(๐‘Ÿ, ๐œƒ) = ๐‘”(๐œƒ) for all but a finite number of
values of ๐œƒ.
๐‘Ÿโ†’๐‘…
Theorem 6.2.3 Let ๐ท be a simply connected domain with boundary ๐ถ. Let ๐‘ค = ๐œŒ๐‘’ ๐‘–๐œ— = ๐‘“(๐‘ง)
be analytic in ๐ท, and map ๐ท โˆช 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œ™ โ€ฒ (๐‘ข, ๐‘ฃ) =
1 2๐œ‹
๐‘”(๐‘ )(1 โˆ’ ๐œŒ2 )
โˆซ
๐‘‘๐‘ 
2๐œ‹ 0 1 โˆ’ 2๐œŒ cos(๐œ— โˆ’ ๐‘ ) + ๐œŒ2
where ๐‘ข = Re(๐‘ค) = ๐œŒ cos ๐œ— and ๐‘ฃ = Im(๐‘ค) = ๐œŒ sin ๐œ—, solves the Dirichlet problem in the
sense that ๐œ™(๐‘ฅ, ๐‘ฆ) = ๐œ™ 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œ™(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
โˆฌ (๐‘ขโˆ‡2 ๐‘ฃ + โˆ‡๐‘ข โ‹… โˆ‡๐‘ฃ) ๐‘‘๐‘ฅ ๐‘‘๐‘ฆ = โˆซ๐‘ขโˆ‡๐‘ฃ โ‹… N ๐‘‘๐‘ 
๐‘…
๐ถ
and
โˆฌ (๐‘ขโˆ‡2 ๐‘ฃ โˆ’ ๐‘ฃโˆ‡2 ๐‘ข) ๐‘‘๐‘ฅ ๐‘‘๐‘ฆ = โˆซ(๐‘ขโˆ‡๐‘ฃ โ‹… N โˆ’ ๐‘ฃโˆ‡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 ) = โˆ’ 2๐œ‹ ln|๐‘ง โˆ’ 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 โˆช 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 โˆช 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 โˆช 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 โˆช 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 โˆช ๐ถ, then for ๐‘”(๐‘ ) continuous on ๐ถ
Potential Theory - Notes
๐‘ข(๐‘ฅ0 , ๐‘ฆ0 ) = โˆ’ โˆซ(๐‘”(๐‘ )โˆ‡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 โˆ‡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 ) = โˆ’ 2๐œ‹ 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 โˆ‡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 โˆซ๐ถ ๐‘ข ๐‘‘๐‘  = 0 is unique.
Theorem 6.3.7
โˆซ๐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 ) = โˆ’
(ii)
๐บ and its first partial derivatives are continuous in ๐ท โˆช ๐ถ, except at (๐‘ฅ0 , ๐‘ฆ0 ) in ๐ท;
2๐œ‹
ln|๐‘ง โˆ’ ๐‘ง0 | + ๐ป(๐‘ฅ, ๐‘ฆ, ๐‘ฅ0 , ๐‘ฆ0 ) where ๐ป is harmonic in ๐ท;
(iii) โˆ‡๐บ โ‹… N = โˆ’ 1โ„๐ฟ where ๐ฟ is the length of ๐ถ, when (๐‘ฅ, ๐‘ฆ) is on ๐ถ;
(iv) โˆซ๐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 โˆช 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 โˆช ๐ถ, then for ๐‘”(๐‘ ) continuous on ๐ถ
๐‘ข(๐‘ฅ0 , ๐‘ฆ0 ) = โˆซ๐‘”(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 โˆ‡2 ๐‘ข = 0 in ๐ท with โˆ‡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‘˜โˆ’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 โˆ˜ 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‘”โˆ’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‘”โˆ’1 โˆ˜ 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 โˆช 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 โˆ’ ๐‘ฅ1 )(๐›ผ1 โ„๐œ‹)โˆ’1 × (๐‘ง โˆ’ ๐‘ฅ2 )(๐›ผ2 โ„๐œ‹)โˆ’1 × โ‹ฏ × (๐‘ง โˆ’ ๐‘ฅ๐‘› )(๐›ผ๐‘›โ„๐œ‹)โˆ’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œ‹ 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 โˆซ(๐‘ง โˆ’ ๐‘ฅ1 )(๐›ผ1 โ„๐œ‹)โˆ’1 × (๐‘ง โˆ’ ๐‘ฅ2 )(๐›ผ2 โ„๐œ‹)โˆ’1 × โ‹ฏ × (๐‘ง โˆ’ ๐‘ฅ๐‘› )(๐›ผ๐‘›โ„๐œ‹)โˆ’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) = โˆซ(๐‘ง โˆ’ ๐‘ฅ1 )(๐›ผ1 โ„๐œ‹)โˆ’1 × (๐‘ง โˆ’ ๐‘ฅ2 )(๐›ผ2 โ„๐œ‹)โˆ’1 × โ‹ฏ × (๐‘ง โˆ’ ๐‘ฅ๐‘› )(๐›ผ๐‘›โ„๐œ‹)โˆ’1 ๐‘‘๐‘ง
and the linear mapping โ„Ž(๐‘ง) = ๐ด๐‘ง + ๐ต. The linear mapping โ„Ž 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‘› โˆ’ 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œ™(๐‘ฅ, ๐‘ฆ) + ๐‘–๐œ“(๐‘ฅ, ๐‘ฆ) is analytic in a domain
๐ท and that ๐œ“ 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 โˆ’ 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) = โˆ’๐ด log(๐‘ง โˆ’ 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‘Ÿ๐‘’ ๐‘–๐œƒ and a sink of the same
strength at the point ๐‘ง0 . Then the complex velocity potential of the dipole in the direction ๐œƒ at
๐‘ง0 is
ฮฉ(๐‘ง) = โˆ’๐ด๐‘Ÿ
๐‘’ ๐‘–๐œƒ
๐‘ง โˆ’ ๐‘ง0
where ๐‘€ = ๐ด๐‘Ÿ is strength of the dipole.
The velocity potential is
๐‘‘
[log|๐‘ง โˆ’ ๐‘ง0 |]
๐‘‘๐‘›
where ๐‘‘โ„๐‘‘๐‘› is the directional derivative in the direction ๐œƒ.
๐œ™(๐‘ฅ, ๐‘ฆ) = โˆ’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œ™(๐‘ฅ, ๐‘ฆ) = ๐‘ฅ +
๐‘ฅ2
2๐‘ฅ
?
+ ๐‘ฆ2
For very large ๐‘ฅ 2 + ๐‘ฆ 2 , ๐œ™ is approximately ๐‘ฅ, and
๐‘ฃ๐‘ฅ =
๐œ•๐œ™
=1
๐œ•๐‘ฅ
and ๐‘ฃ๐‘ฆ =
๐œ•๐œ™
= 0.
๐œ•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 โˆ’ ๐‘–๐‘ฆ)
2(๐‘ฅ โˆ’ 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 โˆ’ ๐‘–๐‘ฆ)
๐‘ง
๐‘ฅ + ๐‘–๐‘ฆ
๐‘ฅ + ๐‘ฆ2
2๐‘ฅ
2๐‘ฆ
=๐‘ฅ+ 2
+ ๐‘– (๐‘ฆ โˆ’ 2
).
2
๐‘ฅ +๐‘ฆ
๐‘ฅ + ๐‘ฆ2
Let
๐œ“(๐‘ฅ, ๐‘ฆ) = ๐‘ฆ โˆ’
๐‘ฅ2
2๐‘ฆ
.
+ ๐‘ฆ2
Now
๐œ™๐‘ฅ = 1 +
2(๐‘ฅ 2 + ๐‘ฆ 2 ) โˆ’ 4๐‘ฅ 2
2(๐‘ฅ 2 + ๐‘ฆ 2 ) โˆ’ 4๐‘ฆ 2
=
1
โˆ’
= ๐œ“๐‘ฆ
(๐‘ฅ 2 + ๐‘ฆ 2 )2
(๐‘ฅ 2 + ๐‘ฆ 2 )2
and
๐œ™๐‘ฆ =
โˆ’4๐‘ฅ๐‘ฆ
= โˆ’๐œ“๐‘ฅ .
+ ๐‘ฆ 2 )2
(๐‘ฅ 2
Hence, the Cauchy-Riemann equations are satisfied, and ๐œ“ is a harmonic conjugate of ๐œ™. It
follows that
ฮฉ(๐‘ง) = ๐‘ง +
2
2๐‘ฅ
2๐‘ฆ
=๐‘ฅ+ 2
+
๐‘–
(๐‘ฆ
โˆ’
)
๐‘ง
๐‘ฅ + ๐‘ฆ2
๐‘ฅ2 + ๐‘ฆ2
is the complex potential corresponding to ๐œ™, and the streamlines of the flow are given by the
level curves of ๐œ“.
Now
๐‘ฆโˆ’
๐‘ฅ2
2๐‘ฆ
= 0 โ‡’ ๐‘ฆ 3 + ๐‘ฅ 2 ๐‘ฆ โˆ’ 2๐‘ฆ = 0 โ‡’ ๐‘ฅ 2 + ๐‘ฆ 2 = 2,
+ ๐‘ฆ2
which is the equation of a circle of radius โˆš2 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œ™ is that of a uniform stream past a circular
cylinder of radius โˆš2.
SAQ 2 Express โˆ‡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‘Ÿ, ๐œƒ) + ๐‘–๐‘ฃ(๐‘Ÿ, ๐œƒ) be an analytic function, then the polar form of the CauchyRiemann equations give
๐œ•๐‘ข ๐œ•๐‘ฃ
๐œ•๐‘ฃ
๐œ•๐‘ข
=
and ๐‘Ÿ
=โˆ’ .
๐œ•๐‘Ÿ ๐œ•๐œƒ
๐œ•๐‘Ÿ
๐œ•๐œƒ
Since ๐‘ข and ๐‘ฃ are analytic,
๐‘Ÿ
๐œ•
๐œ•๐‘ข
๐œ• ๐œ•๐‘ฃ
๐œ• ๐œ•๐‘ฃ
๐œ• 1 ๐œ•๐‘ข
(๐‘Ÿ ) = ( ) =
( )=โˆ’ (
).
๐œ•๐‘Ÿ ๐œ•๐‘Ÿ
๐œ•๐‘Ÿ ๐œ•๐œƒ
๐œ•๐œƒ ๐œ•๐‘Ÿ
๐œ•๐œƒ ๐‘Ÿ ๐œ•๐œƒ
Hence
๐œ•
๐œ•๐‘ข
๐œ• 1 ๐œ•๐‘ข
(๐‘Ÿ ) +
(
) = 0.
๐œ•๐‘Ÿ ๐œ•๐‘Ÿ
๐œ•๐œƒ ๐‘Ÿ ๐œ•๐œƒ
Using the chain rule gives
๐œ•
๐œ•๐‘ข
๐œ• 1 ๐œ•๐‘ข
๐œ• 2 ๐‘ข ๐œ•๐‘ข 1 ๐œ• 2 ๐‘ข
(๐‘Ÿ ) +
(
)=๐‘Ÿ 2+
+
= 0.
๐œ•๐‘Ÿ ๐œ•๐‘Ÿ
๐œ•๐œƒ ๐‘Ÿ ๐œ•๐œƒ
๐œ•๐‘Ÿ
๐œ•๐‘Ÿ ๐‘Ÿ ๐œ•๐œƒ 2
Which gives
๐‘Ÿ2
๐œ• 2๐‘ข
๐œ•๐‘ข ๐œ• 2 ๐‘ข
+
๐‘Ÿ
+
= 0.
๐œ•๐‘Ÿ 2
๐œ•๐‘Ÿ ๐œ•๐œƒ 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‘Ž along the
axis of the cylinder at a potential of zero.
Hint: by symmetry, the solution must be independent of the polar angle ๐œƒ.
The electrostatic potential ๐œ™(๐‘Ÿ) inside the capacitor is independent of the polar angle ๐œƒ and so
satisfies the polar form of Laplace's equation
๐‘‘2 ๐œ™
๐‘‘๐œ™
+๐‘Ÿ
= 0, ๐‘Ž < ๐‘Ÿ < ๐‘
2
๐‘‘๐‘Ÿ
๐‘‘๐‘Ÿ
with boundary conditions
๐‘Ÿ2
๐œ™(๐‘Ž) = 0, ๐œ™(๐‘) = 10.
The differential equation above is a Cauchy-Euler equation.
Indicial equation:
๐œ†2 = 0 โ‡’ ๐œ† = 0.
General solution:
๐œ™(๐‘Ÿ) = ๐ด + ๐ต ln ๐‘Ÿ.
Applying the boundary conditions gives
๐œ™(๐‘Ž) = ๐ด + ๐ต ln ๐‘Ž = 0 โ‡’ ๐ด = โˆ’๐ต ln ๐‘Ž
and
๐œ™(๐‘) = ๐ด + ๐ต ln ๐‘ = 10 โ‡’ โˆ’๐ต ln ๐‘Ž + ๐ต ln ๐‘ = 10 โ‡’ ๐ต =
Hence, the required potential equation is
10
.
ln ๐‘ โˆ’ ln ๐‘Ž
Potential Theory - SAQโ€™s
๐œ™(๐‘Ÿ) = โˆ’๐ต ln ๐‘Ž +
10
10(ln ๐‘Ÿ โˆ’ ln ๐‘Ž)
ln ๐‘Ÿ =
.
ln ๐‘ โˆ’ ln ๐‘Ž
ln ๐‘ โˆ’ ln ๐‘Ž
SAQ 4 Find the potential in an annulus bounded by circles of radii ๐‘Ž < ๐‘ if the inner circle
has potential ๐›ผ and the outer circle has potential ๐›ฝ.
The electrostatic potential ๐œ™(๐‘Ÿ) inside the capacitor is independent of the polar angle ๐œƒ and so
satisfies the polar form of Laplace's equation
๐‘‘2 ๐œ™
๐‘‘๐œ™
+๐‘Ÿ
= 0, ๐‘Ž < ๐‘Ÿ < ๐‘
2
๐‘‘๐‘Ÿ
๐‘‘๐‘Ÿ
with boundary conditions
๐‘Ÿ2
๐œ™(๐‘Ž) = ๐›ผ,
๐œ™(๐‘) = ๐›ฝ.
The differential equation above is a Cauchy-Euler equation.
Indicial equation:
๐œ†2 = 0 โ‡’ ๐œ† = 0.
General solution:
๐œ™(๐‘Ÿ) = ๐ด + ๐ต ln ๐‘Ÿ.
Applying the boundary conditions gives
๐œ™(๐‘Ž) = ๐ด + ๐ต ln ๐‘Ž = ๐›ผ โ‡’ ๐ด = ๐›ผ โˆ’ ๐ต ln ๐‘Ž
and
๐œ™(๐‘) = ๐ด + ๐ต ln ๐‘ = ๐›ฝ โ‡’ ๐›ผ โˆ’ ๐ต ln ๐‘Ž + ๐ต ln ๐‘ = ๐›ฝ โ‡’ ๐ต =
๐›ฝโˆ’๐›ผ
.
ln ๐‘ โˆ’ ln ๐‘Ž
Hence, the required potential equation is
๐œ™(๐‘Ÿ) = ๐›ผ โˆ’ ๐ต ln ๐‘Ž +
SAQ 5
๐›ฝโˆ’๐›ผ
๐›ผ ln ๐‘ โˆ’ ๐›ฝ ln ๐‘Ž + (๐›ฝ โˆ’ ๐›ผ) ln ๐‘Ÿ
ln ๐‘Ÿ =
.
ln ๐‘ โˆ’ ln ๐‘Ž
ln ๐‘ โˆ’ ln ๐‘Ž
What conditions on the real numbers ๐‘Ž, ๐‘, ๐‘, ๐‘‘ ensure that
๐‘Ž๐‘ฅ 3 + ๐‘๐‘ฅ 2 ๐‘ฆ + ๐‘๐‘ฅ๐‘ฆ 2 + ๐‘‘๐‘ฆ 3
is harmonic.
Let
๐œ™(๐‘ฅ, ๐‘ฆ) = ๐‘Ž๐‘ฅ 3 + ๐‘๐‘ฅ 2 ๐‘ฆ + ๐‘๐‘ฅ๐‘ฆ 2 + ๐‘‘๐‘ฆ 3 .
Then ๐œ™ is harmonic if it has continuous first and second-order partial derivatives in a domain
๐ท, and satisfies Laplaceโ€™s equation
โˆ‡2 ๐œ™ =
Now
๐œ• 2๐œ™ ๐œ• 2๐œ™
+
= 0.
๐œ•๐‘ฅ 2 ๐œ•๐‘ฆ 2
Potential Theory - SAQโ€™s
๐œ•๐œ™
= 3๐‘Ž๐‘ฅ 2 + 2๐‘๐‘ฅ๐‘ฆ + ๐‘๐‘ฆ 2
๐œ•๐‘ฅ
and
๐œ• 2๐œ™
= 6๐‘Ž๐‘ฅ + 2๐‘๐‘ฆ
๐œ•๐‘ฅ 2
๐œ•๐œ™
= 2๐‘๐‘ฅ + 2๐‘๐‘ฅ๐‘ฆ + 3๐‘‘๐‘ฆ 2
๐œ•๐‘ฆ
and
๐œ• 2๐œ™
= 2๐‘๐‘ฅ + 6๐‘‘๐‘ฆ.
๐œ•๐‘ฆ 2
and
So ๐œ™ has continuous first and second-order partial derivatives in the complex plane.
Substituting the second-order partial derivatives into Laplace's equation gives
6๐‘Ž๐‘ฅ + 2๐‘๐‘ฆ + 2๐‘๐‘ฅ + 6๐‘‘๐‘ฆ = 2(3๐‘Ž + ๐‘)๐‘ฅ + 2(๐‘ + 3๐‘‘)๐‘ฆ = 0
which is satisfied when
๐‘ = โˆ’3๐‘Ž
and ๐‘ = โˆ’3๐‘‘ .
Hence
๐œ™(๐‘ฅ, ๐‘ฆ) = ๐‘Ž๐‘ฅ 3 โˆ’ 3๐‘‘๐‘ฅ 2 ๐‘ฆ โˆ’ 3๐‘Ž๐‘ฅ๐‘ฆ 2 + ๐‘‘๐‘ฆ 3 = ๐‘Ž(๐‘ฅ 3 โˆ’ 3๐‘ฅ๐‘ฆ 2 ) + ๐‘‘(๐‘ฆ 3 โˆ’ 3๐‘ฅ 2 ๐‘ฆ).
Now
๐‘ง 3 = (๐‘ฅ + ๐‘–๐‘ฆ)3 = ๐‘ฅ 3 โˆ’ 3๐‘ฅ๐‘ฆ 2 + ๐‘–(3๐‘ฅ 2 ๐‘ฆ โˆ’ ๐‘ฆ 3 ).
Which gives
๐œ™(๐‘ฅ, ๐‘ฆ) = ๐‘Ž Re(๐‘ง 3 ) โˆ’ ๐‘‘ Im(๐‘ง 3 )
for all complex ๐‘ง.
SAQ 6
๐‘ง+1
By considering arg (๐‘งโˆ’1), find a function ๐‘” harmonic in the unit disk such that
๐‘”(๐‘’ ๐‘–๐œƒ ) = {
1
0<๐œƒ<๐œ‹
โˆ’1 โˆ’๐œ‹ < ๐œƒ < 0.
Let ๐‘ง = ๐‘’ ๐‘–๐œƒ , then |๐‘ง| < 1 and
(cos ๐œƒ + 1) + ๐‘– sin ๐œƒ
๐‘’ ๐‘–๐œƒ + 1
arg ( ๐‘–๐œƒ
) = arg (
)
(cos ๐œƒ โˆ’ 1) + ๐‘– sin ๐œƒ
๐‘’ โˆ’1
[(cos ๐œƒ + 1) + ๐‘– sin ๐œƒ][(cos ๐œƒ โˆ’ 1) โˆ’ ๐‘– sin ๐œƒ]
= arg (
)
(cos ๐œƒ โˆ’ 1)2 + sin2 ๐œƒ
โˆ’ sin ๐œƒ
= arg (
๐‘–)
1 โˆ’ cos ๐œƒ
๐œ‹
โˆ’
0<๐œƒ<๐œ‹
2
={ ๐œ‹
โˆ’๐œ‹ < ๐œƒ < 0.
2
So, we define
2
๐‘ง+1
๐‘”(๐‘ง) = โˆ’ arg (
).
๐œ‹
๐‘งโˆ’1
Let
๐‘ง+1
๐‘ง+1
๐‘“(๐‘ง) = log (
)=|
| + ๐‘–๐‘”(๐‘ง).
๐‘งโˆ’1
๐‘งโˆ’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 โˆš(๐‘ฅ โˆ’ ๐‘ฅ0 )2 + (๐‘ฆ โˆ’ ๐‘ฆ0 )2 +
ln โˆš(๐‘ฅ โˆ’ ๐‘ฅ0 )2 + (๐‘ฆ + ๐‘ฆ0 )2
2๐œ‹
2๐œ‹
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 ) = โˆ’
โˆž
๐‘ฆ0 โˆž
๐‘”(๐‘ฅ)
๐‘ข(๐‘ฅ0 , ๐‘ฆ0 ) = โˆซ ๐‘”(๐‘ฅ)๐บ๐‘ฆ (๐‘ฅ, 0, ๐‘ฅ0 , ๐‘ฆ0 ) ๐‘‘๐‘ฅ = โˆซ
๐‘‘๐‘ฅ.
๐œ‹ โˆ’โˆž (๐‘ฅ โˆ’ ๐‘ฅ0 )2 + ๐‘ฆ02
โˆ’โˆž
Let let ๐‘ง0 = ๐‘ฅ0 + ๐‘–๐‘ฆ0 be a point in the open upper half-plane๐ท, and let ๐‘… = ๐ท โˆช ๐ถ where ๐ถ is
the ๐‘ฅ-axis. Then
1
1
ln|๐‘ง โˆ’ ๐‘ง0 | +
ln|๐‘ง โˆ’ ๐‘ง0 |
2๐œ‹
2๐œ‹
1
1
=โˆ’
Re(log(๐‘ง โˆ’ ๐‘ง0 )) +
Re(log(๐‘ง โˆ’ ๐‘ง0 )).
2๐œ‹
2๐œ‹
๐บ(๐‘ฅ, ๐‘ฆ, ๐‘ฅ0 , ๐‘ฆ0 ) = โˆ’
Hence ๐บ is analytic in ๐ท. It follows that
1
1
Re(log(๐‘ง โˆ’ ๐‘ง0 )) =
ln โˆš(๐‘ฅ โˆ’ ๐‘ฅ0 )2 + (๐‘ฆ + ๐‘ฆ0 )2
2๐œ‹
2๐œ‹
is harmonic in ๐ท.
Also, ๐บ is continuous in ๐ท โˆช 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 ) = โˆ’ โˆซ(๐‘”(๐‘ )โˆ‡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 โˆ‡2 ๐‘ข = 0 in ๐ท with ๐‘ข = ๐‘” on ๐ถ.
In this case
N = โˆ’j and ๐‘‘๐‘  = ๐‘‘๐‘ฅ.
Hence, integrating along the ๐‘ฅ-axis gives
โˆž
โˆž
๐‘ข(๐‘ฅ0 , ๐‘ฆ0 ) = โˆซ (๐‘”(๐‘ฅ)โˆ‡๐บ โ‹… j) ๐‘‘๐‘ฅ = โˆซ (๐‘”(๐‘ฅ)๐บ๐‘ฆ (๐‘ฅ, ๐‘ฆ, ๐‘ฅ0 , ๐‘ฆ0 )|๐‘ฆ=0 ) ๐‘‘๐‘ฅ
โˆ’โˆž
โˆ’โˆž
Potential Theory - SAQโ€™s
โˆž
= โˆซ (๐‘”(๐‘ฅ)
โˆ’โˆž
+
1
ln โˆš(๐‘ฅ โˆ’ ๐‘ฅ0 )2 + (๐‘ฆ + ๐‘ฆ0 )2 ] ) ๐‘‘๐‘ฅ
2๐œ‹
๐‘ฆ=0
โˆž
= โˆซ (๐‘”(๐‘ฅ)
โˆ’โˆž
+
๐œ•
1
[โˆ’ ln โˆš(๐‘ฅ โˆ’ ๐‘ฅ0 )2 + (๐‘ฆ โˆ’ ๐‘ฆ0 )2
๐œ•๐‘ฆ 2๐œ‹
๐œ•
1
[โˆ’ ln((๐‘ฅ โˆ’ ๐‘ฅ0 )2 + (๐‘ฆ โˆ’ ๐‘ฆ0 )2 )
๐œ•๐‘ฆ 4๐œ‹
1
ln((๐‘ฅ โˆ’ ๐‘ฅ0 )2 + (๐‘ฆ + ๐‘ฆ0 )2 )] ) ๐‘‘๐‘ฅ
4๐œ‹
๐‘ฆ=0
=โˆ’
1 โˆž
2(๐‘ฆ โˆ’ ๐‘ฆ0 )
2(๐‘ฆ + ๐‘ฆ0 )
โˆซ (๐‘”(๐‘ฅ) [
โˆ’
] ) ๐‘‘๐‘ฅ
2
2
(๐‘ฅ โˆ’ ๐‘ฅ0 ) + (๐‘ฆ โˆ’ ๐‘ฆ0 )
(๐‘ฅ โˆ’ ๐‘ฅ0 )2 + (๐‘ฆ + ๐‘ฆ0 )2
4๐œ‹ โˆ’โˆž
๐‘ฆ=0
=โˆ’
1 โˆž
โˆ’2๐‘ฆ0
2๐‘ฆ0
โˆซ (๐‘”(๐‘ฅ) [
โˆ’
]) ๐‘‘๐‘ฅ
2
2
(๐‘ฅ โˆ’ ๐‘ฅ0 ) + ๐‘ฆ0 (๐‘ฅ โˆ’ ๐‘ฅ0 )2 + ๐‘ฆ02
4๐œ‹ โˆ’โˆž
๐‘ฆ0 โˆž
๐‘”(๐‘ฅ)
= โˆซ
๐‘‘๐‘ฅ.
๐œ‹ โˆ’โˆž (๐‘ฅ โˆ’ ๐‘ฅ0 )2 + ๐‘ฆ02
SAQ 8 Prove that the function ๐‘“(๐‘ง) = ๐‘ง + ๐‘Ž2 ๐‘ง 2 + ๐‘Ž3 ๐‘ง 3 + โ‹ฏ is simple for |๐‘ง| < 1 if
โˆ‘โˆž
๐‘˜=2 ๐‘˜|๐‘Ž๐‘˜ | โ‰ค 1.
Since
โˆž
โˆž
๐ด โˆ‘ ๐‘˜ โ‰ค โˆ‘ ๐‘˜|๐‘Ž๐‘˜ | โ‰ค 1
๐‘˜=2
๐‘˜=2
where ๐ด = min{|๐‘Ž2 |, |๐‘Ž3 |, โ€ฆ }, we have
๐ดโ‰ค
1
โˆ‘โˆž
๐‘˜=2 ๐‘˜
which implies that |๐‘Ž๐‘˜ | โ†’ 0 as ๐‘˜ โ†’ โˆž.
๐‘˜
Hence, by the root test, ๐‘“(๐‘ง) = โˆ‘โˆž
๐‘˜=2 ๐‘Ž๐‘˜ ๐‘ง defines an analytic function on |๐‘ง| < โˆž.
Suppose that ๐‘ง1 โ‰  ๐‘ง2 , then
๐‘“(๐‘ง1 ) โˆ’ ๐‘“(๐‘ง2 ) = (๐‘ง1 + ๐‘Ž2 ๐‘ง12 + ๐‘Ž3 ๐‘ง13 + โ‹ฏ ) โˆ’ (๐‘ง2 + ๐‘Ž2 ๐‘ง22 + ๐‘Ž3 ๐‘ง23 + โ‹ฏ )
= (๐‘ง1 โˆ’ ๐‘ง2 + ๐‘Ž2 (๐‘ง12 โˆ’ ๐‘ง22 ) + ๐‘Ž3 (๐‘ง13 โˆ’ ๐‘ง23 ) + โ‹ฏ )
= (๐‘ง1 โˆ’ ๐‘ง2 )(1 + ๐‘Ž2 (๐‘ง1 + ๐‘ง2 ) + ๐‘Ž3 (๐‘ง12 + ๐‘ง1 ๐‘ง2 + ๐‘ง22 ) + โ‹ฏ ).
Hence, by the reverse triangle inequality
|๐‘“(๐‘ง1 ) โˆ’ ๐‘“(๐‘ง2 )| โ‰ฅ ||๐‘ง1 + ๐‘Ž2 ๐‘ง12 + ๐‘Ž3 ๐‘ง13 + โ‹ฏ | โˆ’ |๐‘ง2 + ๐‘Ž2 ๐‘ง22 + ๐‘Ž3 ๐‘ง23 + โ‹ฏ ||
= |๐‘ง1 โˆ’ ๐‘ง2 |(1 + |๐‘Ž2 |(|๐‘ง1 | + |๐‘ง2 |) + |๐‘Ž3 |(|๐‘ง12 | + |๐‘ง1 ๐‘ง2 | + |๐‘ง22 |) + โ‹ฏ )
โ‰ฅ |๐‘ง1 โˆ’ ๐‘ง2 |(1 + 2|๐‘Ž2 | + 3|๐‘Ž3 | + 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 โˆซ๐ถ|๐‘“ โ€ฒ (๐‘ง)| |๐‘‘๐‘ง|;
(ii)
Prove that the area of ๐ทโ€ฒ is โˆฌ๐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œ‹|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 = โˆซ |๐‘‘๐‘ค|.
๐ถโ€ฒ
Now ๐‘‘๐‘ค = ๐‘“ โ€ฒ (๐‘ง)๐‘‘๐‘ง, which gives
๐ฟ = โˆซ |๐‘“ โ€ฒ (๐‘ง)๐‘‘๐‘ง| = โˆซ|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) โˆˆ ๐ทโ€ฒ . Hence, ๐ทโ€ฒ has an area ๐ด given by
๐ด = โˆฌ |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 ) โˆ’ ๐‘“(๐‘ง2 ) = (3๐‘ง1 + ๐‘ง12 ) โˆ’ (3๐‘ง2 + ๐‘ง22 ) = (3(๐‘ง1 โˆ’ ๐‘ง2 ) + (๐‘ง12 โˆ’ ๐‘ง22 ))
= (๐‘ง1 โˆ’ ๐‘ง2 )(3 + ๐‘ง1 + ๐‘ง2 ).
Hence, by the reverse triangle inequality
|๐‘“(๐‘ง1 ) โˆ’ ๐‘“(๐‘ง2 )| โ‰ฅ ||3๐‘ง1 + ๐‘ง12 | โˆ’ |3๐‘ง2 + ๐‘ง22 || = |๐‘ง1 โˆ’ ๐‘ง2 |(3 + |๐‘ง1 | + |๐‘ง2 |)
โ‰ฅ 5|๐‘ง1 โˆ’ 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 โˆ’ ๐‘–๐‘ฃ๐‘ฆ is analytic outside ๐ถ, i.e. ๐‘ฃ๐‘ฅ โˆ’ 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
way to answer this question.
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 โˆ‘โˆž
โˆ’โˆž ๐‘Ž๐‘˜ ๐‘ง with Im(๐‘Žโˆ’1 ) = 0.
Since ๐‘“ is analytic on the open annulus ๐ด = {1 < |๐‘ง| < โˆž}, it has the Laurent series
representation
โˆž
๐‘“(๐‘ง) = โˆ‘ ๐‘Ž๐‘› ๐‘ง ๐‘›
๐‘›=โˆ’โˆž
for all ๐‘ง โˆˆ 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 โˆ’ ๐‘–๐‘ฃ๐‘ฆ = ๐‘“(๐‘ง) โ‡’
๐‘‘๐‘ง
ฮฉ(๐‘ง) = โˆซ๐‘“(๐‘ง) ๐‘‘๐‘ง.
๐ถ
Which gives, for the circle |๐‘ง| = 1
โˆž
๐œ“(๐‘ฅ, ๐‘ฆ) = Re (โˆซ
โˆ‘ ๐‘Ž๐‘› ๐‘ง ๐‘› ๐‘‘๐‘ง).
๐‘“(๐‘ง) ๐‘‘๐‘ง) = Re (โˆซ
|๐‘ง|=1
|๐‘ง|=1
๐‘›=โˆ’โˆž
On the unit circle ๐‘ง = ๐‘’ , which gives
๐‘–๐œƒ
โˆž
2๐œ‹
๐œ“(๐‘ฅ, ๐‘ฆ) = Re (โˆซ
(๐‘–๐‘’
๐‘–๐œƒ
0
โˆ‘ ๐‘Ž๐‘› ๐‘’ ๐‘–๐‘›๐œƒ ) ๐‘‘๐œƒ)
๐‘›=โˆ’โˆž
โˆž
2๐œ‹
= Re (๐‘– โˆ‘ ๐‘Ž๐‘› โˆซ ๐‘’ ๐‘–(๐‘›+1)๐œƒ ๐‘‘๐œƒ)
0
๐‘›=โˆ’โˆž
2๐œ‹
= Re (๐‘–๐‘Žโˆ’1 โˆซ ๐‘‘๐œƒ) = Re(๐‘–2๐œ‹๐‘Žโˆ’1 )
0
since
2๐œ‹
โˆซ ๐‘’ ๐‘–๐‘˜๐œƒ ๐‘‘๐œƒ = 0
0
for any non-zero integer ๐‘˜.
Consequently, ๐œ“(๐‘ฅ, ๐‘ฆ) is non-zero if Im(๐‘Žโˆ’1 ) = 0.
Potential Theory - Problems
P3 Find a steady state temperature for a thin metal infinite strip {(๐‘ฅ, ๐‘ฆ): โˆ’โˆž < ๐‘ฅ < โˆž, 0 โ‰ค
๐‘ฆ โ‰ค 1} when the upper edge is perfectly insulated and ๐‘ข(โˆ’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
โˆ’โˆž < ๐‘ฅ < โˆž, 0 โ‰ค ๐‘ฆ โ‰ค 1, which satisfies the boundary conditions
๐‘ข(โˆ’1,0) = 3;
๐‘ข(1,0) = 2;
โˆ‡๐‘ข โ‹… N = 0
for
โˆ’โˆž < ๐‘ฅ < โˆž, ๐‘ฆ = 1.
In this case, the unit outward facing normal is N = j, hence
(โˆ‡๐‘ข โ‹… N)|๐‘ฆ=1 =
๐œ•๐‘ข
|
= 0.
๐œ•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 (โˆ‡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(โˆ’1,0) = ๐‘”(โˆ’1) = โˆ’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=โˆ’
5 ๐‘ฅ
โˆ’ .
2 2
Another possibility is ๐‘ข(๐‘ฅ, ๐‘ฆ) = ๐‘”(๐‘ฅ)โ„Ž(๐‘ฆ) where โ„Žโ€ฒ (1) = 0.
๐‘ข(๐‘ฅ, ๐‘ฆ) =
Consider the function ๐‘“(๐‘ง) = ๐‘’ ๐‘๐‘งโˆ’๐‘–๐‘ for real ๐‘. Since ๐‘“ is analytic in ๐ท, it follows that
Re(๐‘’ ๐‘๐‘งโˆ’๐‘–๐‘ ) = Re(๐‘’ ๐‘๐‘ฅ ๐‘’ ๐‘(๐‘ฆโˆ’1)๐‘– ) = ๐‘’ ๐‘๐‘ฅ cos(๐‘(๐‘ฆ โˆ’ 1))
Potential Theory - Problems
is harmonic in ๐ท.
Now
๐œ• ๐‘๐‘ฅ
[๐‘’ cos(๐‘(๐‘ฆ โˆ’ 1))]|
= โˆ’๐‘’ ๐‘๐‘ฅ ๐‘ sin(0) = 0.
๐œ•๐‘ฆ
๐‘ฆ=1
Choosing ๐‘ = ๐œ‹โ„2 gives
๐œ‹
๐‘’ ๐œ‹๐‘ฅโ„2 cos ( (๐‘ฆ โˆ’ 1))
2
which is equal to zero when ๐‘ฆ = 0.
Therefore
๐‘ข(๐‘ฅ, ๐‘ฆ) =
5 ๐‘ฅ
๐œ‹
โˆ’ + ๐ถ๐‘’ ๐œ‹๐‘ฅโ„2 cos ( (๐‘ฆ โˆ’ 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
โˆซ 3 (๐‘ฅ๐‘‘๐‘ฅ + ๐‘ฆ๐‘‘๐‘ฆ + ๐‘ง๐‘‘๐‘ง)
๐ถ ๐‘Ÿ
is independent of the path assuming that it is smooth.
Note: the quantity represented by ๐‘Ÿ 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‘Žโ‰ค๐‘กโ‰ค๐‘
where ๐‘Ž 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‘’
โˆซ
๐‘‘๐‘ก
3
3โ„2
๐ถ ๐‘Ÿ
๐‘Ž
(๐‘ฅ 2 (๐‘ก) + ๐‘ฆ 2 (๐‘ก) + ๐‘ง 2 (๐‘ก))
โˆซ
= โˆ’๐พ๐‘’
2
๐‘
1
(๐‘ฅ 2 (๐‘ก) + ๐‘ฆ 2 (๐‘ก) + ๐‘ง 2 (๐‘ก))
1โ„2
|
๐‘Ž
since
๐‘‘ 2
2
2
๐‘‘
1
1 ๐‘‘๐‘ก [๐‘ฅ (๐‘ก) + ๐‘ฆ (๐‘ก) + ๐‘ง (๐‘ก)]
[
]=โˆ’
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
=โˆ’
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
1โˆ’๐‘Ÿ
1+๐‘Ÿ
โ‰ค ๐‘ข(๐‘Ÿ๐‘’ ๐‘–๐œƒ ) โ‰ค ๐‘ข(0)
.
1+๐‘Ÿ
1โˆ’๐‘Ÿ
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 โˆ’ ๐‘Ÿ 2 )๐‘ข(๐‘’ ๐‘–๐œ™ )
1 2๐œ‹
๐‘ข(๐‘Ÿ๐‘’ ) =
โˆซ
๐‘‘๐œ™
2๐œ‹ 0 1 โˆ’ 2๐‘Ÿ cos(๐œƒ โˆ’ ๐œ™) + ๐‘Ÿ 2
๐‘–๐œƒ
1 2๐œ‹
1 โˆ’ ๐‘Ÿ2
=
โˆซ (
) ๐‘ข(๐‘’ ๐‘–๐œ™ ) ๐‘‘๐œ™
2
2๐œ‹ 0
1 โˆ’ 2๐‘Ÿ cos(๐‘ก) + ๐‘Ÿ
with ๐‘ก = ๐œƒ โˆ’ ๐œ™.
Hence
๐‘ข(0) =
1 2๐œ‹
โˆซ ๐‘ข(๐‘’ ๐‘–๐œ™ ) ๐‘‘๐œ™.
2๐œ‹ 0
Since both the Poisson kernel and ๐‘ข(๐‘’ ๐‘–๐œ™ ) are positive
1 2๐œ‹
1 โˆ’ ๐‘Ÿ2
โˆซ min
๐‘ข(๐‘’ ๐‘–๐œ™ ) ๐‘‘๐œ™ โ‰ค ๐‘ข(๐‘Ÿ๐‘’ ๐‘–๐œƒ )
2
0โ‰ค๐‘กโ‰ค2๐œ‹
2๐œ‹ 0
1 โˆ’ 2๐‘Ÿ cos(๐‘ก) + ๐‘Ÿ
โ‰ค
1 2๐œ‹
1 โˆ’ ๐‘Ÿ2
โˆซ max
๐‘ข(๐‘’ ๐‘–๐œ™ ) ๐‘‘๐œ™.
2๐œ‹ 0 0โ‰ค๐‘กโ‰ค2๐œ‹ 1 โˆ’ 2๐‘Ÿ cos(๐‘ก) + ๐‘Ÿ 2
Since
1 โˆ’ ๐‘Ÿ2
1 โˆ’ ๐‘Ÿ2
1 โˆ’ ๐‘Ÿ2
=
=
0โ‰ค๐‘กโ‰ค2๐œ‹ 1 โˆ’ 2๐‘Ÿ cos(๐‘ก) + ๐‘Ÿ 2
1 + 2๐‘Ÿ + ๐‘Ÿ 2 (1 + ๐‘Ÿ)2
min
and
1 โˆ’ ๐‘Ÿ2
1 โˆ’ ๐‘Ÿ2
1 โˆ’ ๐‘Ÿ2
=
=
.
0โ‰ค๐‘กโ‰ค2๐œ‹ 1 โˆ’ 2๐‘Ÿ cos(๐‘ก) + ๐‘Ÿ 2
1 โˆ’ 2๐‘Ÿ + ๐‘Ÿ 2 (1 โˆ’ ๐‘Ÿ)2
max
We have
1 โˆ’ ๐‘Ÿ 2 1 2๐œ‹
1 โˆ’ ๐‘Ÿ 2 1 2๐œ‹
๐‘–๐œ™
๐‘–๐œƒ
โˆซ ๐‘ข(๐‘’ ) ๐‘‘๐œ™ โ‰ค ๐‘ข(๐‘Ÿ๐‘’ ) โ‰ค
โˆซ ๐‘ข(๐‘’ ๐‘–๐œ™ ) ๐‘‘๐œ™.
(1 + ๐‘Ÿ)2 2๐œ‹ 0
(1 โˆ’ ๐‘Ÿ)2 2๐œ‹ 0
Which gives, since ๐‘Ÿ < 1
1โˆ’๐‘Ÿ
1+๐‘Ÿ
๐‘ข(0) โ‰ค ๐‘ข(๐‘Ÿ๐‘’ ๐‘–๐œƒ ) โ‰ค
๐‘ข(0).
1+๐‘Ÿ
1โˆ’๐‘Ÿ
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 {โˆฌ (๐‘ข๐‘ฅ2 + ๐‘ข๐‘ฆ2 ) ๐‘‘๐ด: ๐‘ข โˆˆ ๐น}
๐ท
is attained by ๐‘ˆ โˆˆ ๐น, show that ๐‘ˆ 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 = โˆž.
Consider the linear fractional transformation
1+๐‘ง
.
1โˆ’๐‘ง
The images of the points โˆ’1, ๐‘–, and 1 under this transformation are
๐‘ค=
๐‘ค(โˆ’1) = 0, ๐‘ค(๐‘–) = ๐‘–,
and ๐‘ค(1) = โˆž.
Therefore, the upper semicircle is mapped to the positive imaginary axis.
The images of the points โˆ’1,0, and 1 under this transformation are
๐‘ค(โˆ’1) = 0, ๐‘ค(0) = 1,
and ๐‘ค(1) = โˆž.
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 โˆ’ ๐‘– โ„2 5 5
Hence, ๐‘ค maps ๐ท onto the first quadrant in the ๐‘ค-plane.
The mapping ๐‘Š = ๐‘ค 2 maps the first quadrant in the ๐‘ค-plane onto the upper half ๐‘Š-plane.
The mapping
๐‘– โˆ’ ๐‘Š โˆ’1(๐‘Š โˆ’ ๐‘–)
=
๐‘–+๐‘Š
๐‘Š+๐‘–
maps the the upper half ๐‘Š-plane onto the unit disk in the ๐œ‰-plane.
๐œ‰=
Forming the composition of these mappings gives
Potential Theory - Problems
2
๐œ‰=
๐‘–โˆ’๐‘Š ๐‘–โˆ’๐‘ค
=
=
๐‘– + ๐‘Š ๐‘– + ๐‘ค2
1+๐‘ง 2
๐‘– โˆ’ (1 โˆ’ ๐‘ง)
1+๐‘ง 2
๐‘– + (1 โˆ’ ๐‘ง)
=
๐‘–(1 โˆ’ ๐‘ง)2 โˆ’ (1 + ๐‘ง)2
1 โˆ’ 2๐‘–๐‘ง + ๐‘ง 2
=
๐‘–
.
๐‘–(1 โˆ’ 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œ‹} into the interior of the
unit circle. Write down a solution of the Dirichlet problem for this strip.
The mapping
๐‘Š = ๐‘’๐‘ง
maps ๐‘† onto the upper half ๐‘Š-plane, and the mapping
๐‘–โˆ’๐‘Š
๐‘–+๐‘Š
maps the upper half ๐‘Š-plane onto the unit disk in the ๐‘ค-plane.
๐‘ค=
Therefore, the mapping
๐‘– โˆ’ ๐‘Š ๐‘– โˆ’ ๐‘’๐‘ง
๐‘ค=
=
= ๐‘“(๐‘ง)
๐‘– + ๐‘Š 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œ‹, Im(๐‘ง) โ‰ฅ 0} onto the upper
half plane so that the origin maps into the origin, and ๐œ‹ maps into 1.
Potential Theory - Problems
The strip ๐‘† is a polygonal region with interior angles ๐›ผ1 = ๐›ผ2 = ๐œ‹โ„2 and vertices ๐‘ง1 = 0 and
๐‘ง2 = ๐œ‹.
Applying the Schwarz-Christoffel transformation gives
๐‘‘๐‘ง
๐ด
= ๐ด(๐‘ค โˆ’ 0)(๐œ‹โ„2โ„๐œ‹)โˆ’1 × (๐‘ค โˆ’ 1)(๐œ‹โ„2โ„๐œ‹)โˆ’1 = ๐ด๐‘ค โˆ’1โ„2 (๐‘ค โˆ’ 1)โˆ’1โ„2 =
.
1โ„2
๐‘‘๐‘ค
(๐‘ค(๐‘ค โˆ’ 1))
Hence
1
๐‘ง = ๐ดโˆซ
(๐‘ค(๐‘ค โˆ’ 1))
= ๐ดโˆซ
1โ„2
๐‘‘๐‘ค = ๐ด โˆซ
1
1 2
1 2
((๐‘ค โˆ’ 2) โˆ’ (2) )
= ๐ด log (๐‘ค โˆ’
1โ„2
(๐‘ค 2
1
๐‘‘๐‘ค
โˆ’ ๐‘ค)1โ„2
๐‘‘๐‘ค
1
+ (๐‘ค 2 โˆ’ ๐‘ค)1โ„2 ).
2
This isn't very helpful.
Further manipulation gives
๐‘‘๐‘ง
๐ด
๐ด
1
=
=
.
โ„
1
2
๐‘‘๐‘ค (โˆ’1๐‘ค(1 โˆ’ ๐‘ค))
๐‘– (๐‘ค(1 โˆ’ ๐‘ค))1โ„2
Let ๐ต = ๐ดโ„๐‘– , then
๐‘ง = ๐ตโˆซ
1
(๐‘ค(1 โˆ’ ๐‘ค))
= ๐ตโˆซ
1โ„2
๐‘‘๐‘ค = ๐ต โˆซ
1
1 2
1 2
((2) โˆ’ (๐‘ค โˆ’ 2) )
1โ„2
1
๐‘‘๐‘ค
(๐‘ค โˆ’ ๐‘ค 2 )1โ„2
๐‘‘๐‘ค
1
๐‘คโˆ’2
= ๐ต sinโˆ’1 (
) + ๐ถ.
1
2
Now ๐‘ง1 = 0 when ๐‘ค1 = 0, hence
๐œ‹
0 = ๐ต sinโˆ’1(โˆ’1) + ๐ถ = โˆ’ ๐ต + ๐ถ.
2
And ๐‘ง2 = ๐œ‹ when ๐‘ค2 = 1, hence
๐œ‹
๐œ‹ = ๐ต sinโˆ’1(1) + ๐ถ = ๐ต + ๐ถ.
2
Subtracting gives
๐ต=1
๐œ‹.
๐œ‹ = ๐œ‹๐ต โ‡’ {
๐ถ=
2
Hence
Potential Theory - Problems
๐‘ง = sin
โˆ’1
(
1
2 ) + ๐œ‹.
1
2
2
๐‘คโˆ’
Rearranging gives
1
๐œ‹
1
1
1 1
๐‘ง
๐‘ค = sin (๐‘ง โˆ’ ) + = โˆ’ cos ๐‘ง + = (1 โˆ’ cos ๐‘ง) = sin2 ( ).
2
2
2
2
2 2
2
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