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# Potential Theory - notes

advertisement ```Potential Theory - Glossary
Analytic function A complex function 𝑤 = 𝑓(𝑧) is said to be analytic at a point 𝑧0 if 𝑓 is
differentiable at 𝑧0 and at every point in some neighbourhood of 𝑧0 . A function is analytic in a
domain 𝐷 if it is analytic at every point in 𝐷.
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 𝜓(𝑥, 𝑦) be a harmonic conjugate of a potential function 𝜙(𝑥, 𝑦).
Then the analytic function
Ω(𝑧) = 𝜙(𝑥, 𝑦) + 𝑖𝜓(𝑥, 𝑦)
is called the complex potential corresponding to the real potential 𝜙. The function 𝜓 is called
the stream function.
The level curves
𝜙(𝑥, 𝑦) = 𝑐1
and 𝜓(𝑥, 𝑦) = 𝑐2
of 𝜙 and 𝜓, respectively, are orthogonal families.
Conformal Mapping Let 𝑤 = 𝑓(𝑧) be a complex mapping defined in a domain 𝐷, and let 𝑧0
be a point in 𝐷. Then 𝑤 is a conformal mapping (or conformal) at 𝑧0 if for every pair of smooth
oriented curves 𝐶1 and 𝐶2 in 𝐷 intersecting at 𝑧0 , the angle between 𝐶1 and 𝐶2 is equal to the
angle between the image curves 𝐶1′ and 𝐶2′ at 𝑓(𝑧0 ) in both magnitude and sense (direction).
If 𝑤 = 𝑓(𝑧) maps a domain 𝐷 onto a domain 𝐷′ and if 𝑤 is conformal at all points in 𝐷, then 𝑤
is a conformal mapping of 𝐷 onto 𝐷′ .
Critical Point Let the complex function 𝑓(𝑧) be analytic in a domain 𝐷 containing the point
𝑧0 . If 𝑓 ′ (𝑧0 ) = 0, then the mapping 𝑤 = 𝑓(𝑧) is not conformal at 𝑧0 , and 𝑧0 is called a critical
point of 𝑓.
Dipole
A pair consisting of a source and a sink is called a dipole (or doublet).
Dirichlet's principle Let 𝐷 be a domain in the plane bounded by a simple closed curve 𝐶.
Let 𝑔 be a real-valued smooth function defined on 𝐶, and let 𝐹 be the family of real-valued
functions with smooth derivatives on 𝐷 and 𝐶 and boundary values 𝑔. If
inf {∬ (𝑢𝑥2 + 𝑢𝑦2 ) 𝑑𝐴: 𝑢 ∈ 𝐹}
𝐷
is attained by 𝑈 ∈ 𝐹, then 𝑈 is the solution of the Dirichlet problem for 𝐷 with boundary
values 𝑔.
Dirichlet problem Let 𝐷 be a domain in the plane bounded by a finite number of nonintersecting simple closed contours 𝐶𝑖 , and let 𝑔𝑖 be 𝑛 continuous functions defined on the 𝑛
contours. The problem of finding a function 𝑢(𝑥, 𝑦), which satisfies Laplace's equation in 𝐷
and which takes on the values 𝑔𝑖 on 𝐶𝑖 is called a Dirichlet problem.
Green’s function Let 𝑧0 = 𝑥0 + 𝑖𝑦0 be a point in a region of the plane bounded by a circle 𝐶
of radius 𝑅. Then a function of the form
1
ln|𝑧 − 𝑧0 | + 𝐻(𝑥, 𝑦, 𝑥0 , 𝑦0 )
2𝜋
where 𝐻 is harmonic for |𝑧| < 𝑅 and
𝐺(𝑥, 𝑦, 𝑥0 , 𝑦0 ) = −
𝐻(𝑅 cos 𝜃 , 𝑅 sin 𝜃 , 𝑥0 , 𝑦0 ) =
1
ln|(𝑅 cos 𝜃 − 𝑥0 ) + 𝑖(𝑅 sin 𝜃 − 𝑦0 )|
2𝜋
Potential Theory - Glossary
is called a Green’s function.
The essential feature of a Green’s function 𝐺 is the singular behaviour specified by
1
ln|𝑧 − 𝑧0 |.
2𝜋
The behaviour on 𝐶 is specified so that an application of Green’s theorem leads to the desired
result.
−
Green's Theorem in the Plane Let 𝑅 be a region in the plane whose boundary 𝐶 consists of
one, or more smooth, non-self-intersecting, closed curves that are positively oriented with
respect to 𝑅. If 𝑃(𝑥, 𝑦)i + 𝑄(𝑥, 𝑦)j is a smooth vector field on 𝑅, then
∬(
𝑅
𝜕𝑄 𝜕𝑃
− ) 𝑑𝑥 𝑑𝑦 = ∮𝑃(𝑥, 𝑦)𝑑𝑥 + 𝑄(𝑥, 𝑦)𝑑𝑦.
𝜕𝑥 𝜕𝑦
𝐶
Harmonic conjugate Suppose that the real function 𝑢(𝑥, 𝑦) is harmonic on a domain 𝐷, and
there exists a real function 𝑣(𝑥, 𝑦) such that the partial derivatives of 𝑢 and 𝑣 satisfy the
Cauchy-Riemann equations throughout 𝐷. Then 𝑣 is said to be a harmonic conjugate of 𝑢.
Furthermore, it follows that 𝑓(𝑧) = 𝑢(𝑥, 𝑦) + 𝑖𝑣(𝑥, 𝑦) is analytic on 𝐷.
Harmonic function A real valued function 𝜙(𝑥, 𝑦) that has continuous first and secondorder partial derivatives in a domain 𝐷, and satisfies Laplace’s equation in two dimensions
∇2 𝜙 =
𝜕 2𝜙 𝜕 2𝜙
+
=0
𝜕𝑥 2 𝜕𝑦 2
is said to be harmonic in 𝐷. The function 𝜙 is called a harmonic (or potential) function.
Laplace's equation (in two dimensions) Let Φ(𝑥, 𝑦) be a real valued function of two real
variables 𝑥 and 𝑦. The partial differential equation
Φ𝑥𝑥 (𝑥, 𝑦) + Φ𝑦𝑦 (𝑥, 𝑦) = 0
or alternatively
𝜕 2Φ 𝜕 2Φ
+
=0
𝜕𝑥 2 𝜕𝑦 2
is known as Laplace's equation, or the potential equation. The polar form of Laplace's
equation is given by
𝑟 2 Φ𝑟𝑟 (𝑟, 𝜃) + 𝑟Φ𝑟 (𝑟, 𝜃) + Φ𝜃𝜃 (𝑟, 𝜃) = 0
or alternatively
𝜕 2Φ
𝜕Φ 𝜕 2 Φ
+
𝑟
+
= 0.
𝜕𝑟 2
𝜕𝑟 𝜕𝜃 2
Level curves Suppose the function 𝑓(𝑧) = 𝑢(𝑥, 𝑦) + 𝑖𝑣(𝑥, 𝑦) is analytic in a domain 𝐷. Then
the real and imaginary parts of 𝑓 can be used to define two families of curves in 𝐷. The
equations
𝑟2
𝑢(𝑥, 𝑦) = 𝑐1
and 𝑣(𝑥, 𝑦) = 𝑐2
where 𝑐1 and 𝑐2 are arbitrary real constants, are called level curves of 𝑢 and 𝑣, respectively.
The level curves are orthogonal families.
Neumann problem Let 𝐷 be a simply connected domain in the plane bounded by a simple
closed smooth contour 𝐶, and let 𝑔 be a function defined on 𝐶. The problem of finding a
function 𝑢(𝑥, 𝑦), which satisfies Laplace's equation in 𝐷 and such that ∇𝑢 ⋅ N = 𝑔 where N is
the unit outward normal on 𝐶 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 𝑦 > 0, and (ii) the unit open disk |𝑧| < 1.
(i) Poisson integral formula for the upper half plane.
Let 𝑓(𝑥) be a piecewise continuous and bounded function on −∞ < 𝑥 < ∞. Then the function
defined by
𝜙(𝑥, 𝑦) =
𝑦 ∞
𝑓(𝑠)
∫
𝑑𝑠
𝜋 −∞ (𝑥 − 𝑠)2 + 𝑦 2
is a solution of the Dirichlet problem in the upper half plane 𝑦 > 0 with boundary condition
𝜙(𝑥, 0) = 𝑓(𝑥) at all points of continuity of 𝑓.
(ii) Poisson integral formula for the unit disk.
Cartesian form. Let 𝑓(𝑧) be a complex function for which the values 𝑓(𝑒 𝑖𝜃 ) on the unit circle
𝑧 = 𝑒 𝑖𝜃 give a piecewise continuous and bounded function for −𝜋 ≤ 𝜃 ≤ 𝜋. Then the function
defined by
𝜙(𝑥, 𝑦) =
1 𝜋 𝑓(𝑒 𝑖𝑠 )(1 − |𝑧|2 )
∫
𝑑𝑠
|𝑒 𝑖𝑠 − 𝑧|2
2𝜋 −𝜋
is a solution of the Dirichlet problem in the open unit disk |𝑧| < 1 with boundary condition
𝜙(cos 𝜃 , sin 𝜃) = 𝑓(𝑒 𝑖𝜃 ) at all points of continuity of 𝑓.
Polar form. Let 𝑓(𝜃) be a piecewise continuous and bounded function on −𝜋 < 𝜃 < 𝜋. Then
the function defined by
𝜙(𝑟, 𝜃) =
1 𝜋
𝑓(𝑠)(1 − 𝑟 2 )
∫
𝑑𝑠
2𝜋 −𝜋 1 − 2𝑟 cos(𝜃 − 𝑠) + 𝑟 2
is a solution of the Dirichlet problem in the open unit disk |𝑧| < 1 with boundary condition
𝜙(𝑟 = 1, 𝜃) = 𝑓(𝜃) at all points of continuity of 𝑓.
Poisson kernel
The function 𝑓: ℂ → ℝ defined by
𝑅2 − 𝑟 2
𝑓(𝑟𝑒 ) = 2
𝑅 − 2𝑟𝑅 cos 𝜃 + 𝑟 2
is known as the Poisson kernel for the disk {𝑧: |𝑧| < 𝑅}. It vanishes at every point of
{𝑧: |𝑧| = 𝑅} except at 𝑧 = 𝑅, where it has a singularity, and has the property of being positive
for 0 < 𝑟 < 𝑅.
𝑖𝜃
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 ∞ is also called a vertex of the polygon.
Simple function A function 𝑤 = 𝑓(𝑧) is said to be simple (or univalent) in a domain 𝐷 if it
is analytic and one-to-one in 𝐷.
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 𝐷 is called streamlining.
Potential Theory - Notes
Harmonic Functions and Conjugates
Theorem Suppose the complex function 𝑓(𝑧) = 𝑢(𝑥, 𝑦) + 𝑖𝑣(𝑥, 𝑦) is analytic in a domain 𝐷.
Then the functions 𝑢(𝑥, 𝑦) and 𝑣(𝑥, 𝑦) are harmonic in 𝐷.
Theorem Suppose that the real function 𝑢(𝑥, 𝑦) is harmonic in an 𝜀-neighbourhood of the
point (𝑥0 , 𝑦0 ). Then there exists a conjugate harmonic function
𝑣(𝑥, 𝑦) = ∫ 𝑢𝑥 (𝑥, 𝑦) 𝑑𝑦 − ∫ 𝑢𝑦 (𝑥, 𝑦) 𝑑𝑥 − ∬ 𝑢𝑥𝑥 (𝑥, 𝑦) 𝑑𝑦𝑑𝑥
defined in this neighbourhood such that 𝑓(𝑧) = 𝑢(𝑥, 𝑦) + 𝑖𝑣(𝑥, 𝑦) is an analytic function.
A harmonic conjugate can be constructed by using the following method.
First confirm that 𝑢(𝑥, 𝑦) is harmonic by verifying that
𝑢𝑥𝑥 (𝑥, 𝑦) + 𝑢𝑦𝑦 (𝑥, 𝑦) = 0.
Then, using the Cauchy-Riemann equation 𝑣𝑦 (𝑥, 𝑦) = 𝑢𝑥 (𝑥, 𝑦) gives
𝑣(𝑥, 𝑦) = ∫ 𝑣𝑦 (𝑥, 𝑦) 𝑑𝑦 + 𝐶(𝑥) = ∫ 𝑢𝑥 (𝑥, 𝑦) 𝑑𝑦 + 𝐶(𝑥)
where 𝐶(𝑥) is a function of 𝑥 alone.
Differentiating both sides with respect to 𝑥 and using the Cauchy-Riemann equation
𝑣𝑥 (𝑥, 𝑦) = −𝑢𝑦 (𝑥, 𝑦) gives
−𝑢𝑦 (𝑥, 𝑦) =
𝜕
(∫ 𝑢𝑥 (𝑥, 𝑦) 𝑑𝑦) + 𝐶 ′ (𝑥).
𝜕𝑥
All terms except those involving 𝑥 will cancel leaving a formula for 𝐶 ′ (𝑥) in terms of 𝑥 alone,
say 𝐶 ′ (𝑥) = 𝑓(𝑥).
Integrating both sides with respect to 𝑥 gives
𝐶(𝑥) = 𝐹(𝑥) + 𝑐
where 𝐹 is an antiderivative of 𝑓 and 𝑐 is an arbitrary constant.
Finally, letting 𝑐 = 0 gives
𝑣(𝑥, 𝑦) = ∫ 𝑢𝑥 (𝑥, 𝑦) 𝑑𝑦 + 𝐹(𝑥).
A similar method can be used if we are given 𝑣(𝑥, 𝑦) and wish to find 𝑢(𝑥, 𝑦).
Laplace’s Equation in Physics
Electromagnetism
For an electrostatic field, the electric field E = (𝐸𝑥 , 𝐸𝑦 , 𝐸𝑧 ), where 𝐸𝑥 , 𝐸𝑦 , 𝐸𝑧 are the Cartesian
components of E, is given by
E = grad 𝜙 =
𝜕𝜙
𝜕𝜙
𝜕𝜙
i+
j+
k,
𝜕𝑥
𝜕𝑦
𝜕𝑧
where 𝜙 is an arbitrary function of position known as the electrostatic potential.
The electrostatic potential function 𝜙 satisfies Laplace's equation and is therefore harmonic in
some domain 𝐷. If we restrict our attention to two dimensions, it follows that there exists a
Potential Theory - Notes
harmonic conjugate function 𝜓(𝑥, 𝑦) defined in some domain 𝐸 in ℝ2 , so that the complex
potential function Ω(𝑧) = 𝜙(𝑥, 𝑦) + 𝑖𝜓(𝑥, 𝑦) is analytic in 𝐸.
The electric field E (i.e. the electron flow) is given by
𝑑Ω
= 𝐸𝑥 + 𝑖𝐸𝑦 .
𝑑𝑧
The level curves 𝜙(𝑥, 𝑦) = 𝑐1 are called equipotential curves, that is, curves along which the
electrostatic potential is constant. Whereas, the level curves 𝜓(𝑥, 𝑦) = 𝑐2 , curves that are
orthogonal to the family 𝜙(𝑥, 𝑦) = 𝑐1, 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 = (𝑣𝑥 , 𝑣𝑦 , 𝑣𝑧 ), where 𝑣𝑥 , 𝑣𝑦 , 𝑣𝑧 are the
Cartesian components of v, is given by
v = grad 𝜙 =
𝜕𝜙
𝜕𝜙
𝜕𝜙
i+
j+
k,
𝜕𝑥
𝜕𝑦
𝜕𝑧
where 𝜙 is an arbitrary function of position known as the velocity potential.
The velocity potential function 𝜙 satisfies Laplace's equation and is therefore harmonic in
some domain 𝐷. If we restrict our attention to two dimensions, it follows that there exists a
harmonic conjugate function 𝜓(𝑥, 𝑦) defined in some domain 𝐸 in ℝ2 , so that the complex
potential function Ω(𝑧) = 𝜙(𝑥, 𝑦) + 𝑖𝜓(𝑥, 𝑦) is analytic in 𝐸. In this case, the complex potential
function Ω(𝑧) is called the complex velocity potential of the flow.
The velocity field v (i.e. the fluid flow) is given by
𝑑Ω
= 𝑣𝑥 + 𝑖𝑣𝑦 .
𝑑𝑧
The level curves 𝜙(𝑥, 𝑦) = 𝑐1 are called equipotential curves, that is, curves along which the
velocity potential is constant. Whereas, the level curves 𝜓(𝑥, 𝑦) = 𝑐2 , curves that are
orthogonal to the family 𝜙(𝑥, 𝑦) = 𝑐1, 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 𝜙 for a given domain 𝐷 and set of
functions 𝑔𝑖 , then the solution is unique.
Theorem 6.2.2
The Dirichlet problem is solvable for the disk |𝑧| < 𝑅.
Corollary 6.2.1 Let 𝑔(𝜃) be a piecewise continuous function on 0 ≤ 𝜃 ≤ 2𝜋. Then the
function defined by
1 2𝜋
𝑔(𝑠)(𝑅 2 − 𝑟 2 )
𝜙(𝑟, 𝜃) =
∫
𝑑𝑠
2𝜋 0 𝑅 2 − 2𝑅𝑟 cos(𝜃 − 𝑠) + 𝑟 2
is harmonic in the open disk |𝑧| < 𝑅, and lim 𝜙(𝑟, 𝜃) = 𝑔(𝜃) for all but a finite number of
values of 𝜃.
𝑟→𝑅
Theorem 6.2.3 Let 𝐷 be a simply connected domain with boundary 𝐶. Let 𝑤 = 𝜌𝑒 𝑖𝜗 = 𝑓(𝑧)
be analytic in 𝐷, and map 𝐷 ∪ 𝐶 in a one to one fashion onto the closed disk |𝑤| ≤ 1 so that 𝐷
maps onto the interior of the disk |𝑤| < 1 and 𝐶 maps onto its boundary |𝑤| = 1. Then
Potential Theory - Notes
𝜙 ′ (𝑢, 𝑣) =
1 2𝜋
𝑔(𝑠)(1 − 𝜌2 )
∫
𝑑𝑠
2𝜋 0 1 − 2𝜌 cos(𝜗 − 𝑠) + 𝜌2
where 𝑢 = Re(𝑤) = 𝜌 cos 𝜗 and 𝑣 = Im(𝑤) = 𝜌 sin 𝜗, solves the Dirichlet problem in the
sense that 𝜙(𝑥, 𝑦) = 𝜙 ′ (𝑢, 𝑣) at corresponding points under the mapping.
Steps for Solving a Dirichlet Problem
(i)
Find an analytic function 𝑓(𝑧) = 𝑢(𝑥, 𝑦) + 𝑖𝑣(𝑥, 𝑦) that maps the domain 𝐷 in the 𝑧plane onto a simpler domain 𝐷′ in the 𝑤-plane, and that maps the boundary curves
𝐶1 , 𝐶2 , … , 𝐶𝑛 onto the curves 𝐶1′ , 𝐶2′ , … , 𝐶𝑛′ , respectively.
(ii)
Transform the boundary conditions on 𝐶1 , 𝐶2 , … , 𝐶𝑛 to boundary conditions on
𝐶1′ , 𝐶2′ , … , 𝐶𝑛′ .
(iii) Solve this new (and easier ) Dirichlet problem in 𝐷′ to obtain a harmonic function
Φ(𝑢, 𝑣).
(iv) Substitute the real and imaginary parts 𝑢(𝑥, 𝑦) and 𝑣(𝑥, 𝑦) of 𝑓 for the variables 𝑢 and 𝑣
in Φ(𝑢, 𝑣). Then the function 𝜙(𝑥, 𝑦) = Φ(𝑢(𝑥, 𝑦), 𝑣(𝑥, 𝑦)) is a solution to the Dirichlet
problem in 𝐷.
Green’s Functions
Theorem
Let 𝐶 be a simple closed smooth contour with interior 𝑅, then
∬ (𝑢∇2 𝑣 + ∇𝑢 ⋅ ∇𝑣) 𝑑𝑥 𝑑𝑦 = ∫𝑢∇𝑣 ⋅ N 𝑑𝑠
𝑅
𝐶
and
∬ (𝑢∇2 𝑣 − 𝑣∇2 𝑢) 𝑑𝑥 𝑑𝑦 = ∫(𝑢∇𝑣 ⋅ N − 𝑣∇𝑢 ⋅ N) 𝑑𝑠.
𝑅
𝐶
where 𝑢 and 𝑣 are functions with continuous first and second order partial derivatives within,
and on 𝐶, N is the unit outward normal on 𝐶, and 𝑑𝑠 is the element of arc length on 𝐶.
Green’s function for the Dirichlet problem Let 𝐷 be a bounded simply connected domain
bounded by a simple closed smooth contour 𝐶, and let 𝑧0 = 𝑥0 + 𝑖𝑦0 be a point in 𝐷. If
𝐺(𝑥, 𝑦, 𝑥0 , 𝑦0 ) is a function satisfying the following properties:
1
(i)
𝐺(𝑥, 𝑦, 𝑥0 , 𝑦0 ) = − 2𝜋 ln|𝑧 − 𝑧0 | + 𝐻(𝑥, 𝑦, 𝑥0 , 𝑦0 ) where 𝐻 is harmonic inside 𝐶;
(ii)
𝐺 is continuous in 𝐷 ∪ 𝐶, except at (𝑥0 , 𝑦0 ) in 𝐷;
(iii) 𝐺(𝑥, 𝑦, 𝑥0 , 𝑦0 ) = 0 when (𝑥, 𝑦) is on 𝐶.
Then 𝐺 is a Green’s function associated with the Dirichlet problem for the region 𝑅 = 𝐷 ∪ 𝐶.
Theorem 6.3.1
If a Green’s function (for the Dirichlet problem) exists, then it is unique.
Theorem 6.3.2
positive in 𝐷.
The Green’s function (for the Dirichlet problem) for the region 𝑅 = 𝐷 ∪ 𝐶 is
Theorem 6.3.3 If 𝐺 is the Green’s function (for the Dirichlet problem) for the region 𝑅 =
𝐷 ∪ 𝐶, then 𝐺(𝑥, 𝑦, 𝑥1 , 𝑦1 ) = 𝐺(𝑥, 𝑦, 𝑥2 , 𝑦2 ) for any pair of distinct points (𝑥1 , 𝑦1 ) and (𝑥2 , 𝑦2 ) in
𝐷.
Theorem 6.3.4 If the Green’s function (for the Dirichlet problem) exists for the region 𝑅 =
𝐷 ∪ 𝐶, then for 𝑔(𝑠) continuous on 𝐶
Potential Theory - Notes
𝑢(𝑥0 , 𝑦0 ) = − ∫(𝑔(𝑠)∇𝐺 ⋅ N) 𝑑𝑠
𝐶
where N is the unit outward normal on 𝐶 and 𝑑𝑠 is the element of arc length on 𝐶, solves the
Dirichlet problem ∇2 𝑢 = 0 in 𝐷 with 𝑢 = 𝑔 on 𝐶.
Theorem 6.3.5 Let 𝑤 = 𝑓(𝑧) map the region 𝑅 onto the unit closed disk |𝑧| ≤ 1 so that 𝐷,
the interior of 𝑅, maps onto the open disk |𝑧| < 1 and 𝐶, the boundary of 𝑅, maps onto the
unit circle |𝑧| = 1. Furthermore, let 𝑓 ′ (𝑧) exist and never vanish in 𝐷, and let 𝑧0 map into the
1
origin. Then 𝐺(𝑥, 𝑦, 𝑥1 , 𝑦1 ) = − 2𝜋 ln|𝑓(𝑧)| is the Green’s function (for the Dirichlet problem)
for 𝑅.
The Neumann Problem
An insulated boundary curve in a heat flow problem corresponds to a boundary condition of
the form ∇𝑢 ⋅ N = 0, and so is an example of a Neumann problem.
The solution 𝑢 of a Neumann problem is not unique since 𝑢 + 𝑐 is also a solution for any
constant 𝑐.
Theorem 6.3.6
The solution 𝑢 of the Neumann problem such that ∫𝐶 𝑢 𝑑𝑠 = 0 is unique.
Theorem 6.3.7
∫𝐶 𝑔(𝑠) 𝑑𝑠 = 0.
A necessary condition for the Neumann problem to have a solution is that
Green’s function for the Neumann problem Let 𝐷 be a bounded simply connected domain
bounded by a simple closed smooth contour 𝐶, and let 𝑧0 = 𝑥0 + 𝑖𝑦0 be a point in 𝐷. If
𝐺(𝑥, 𝑦, 𝑥0 , 𝑦0 ) is a function satisfying the following properties:
1
(i)
𝐺(𝑥, 𝑦, 𝑥0 , 𝑦0 ) = −
(ii)
𝐺 and its first partial derivatives are continuous in 𝐷 ∪ 𝐶, except at (𝑥0 , 𝑦0 ) in 𝐷;
2𝜋
ln|𝑧 − 𝑧0 | + 𝐻(𝑥, 𝑦, 𝑥0 , 𝑦0 ) where 𝐻 is harmonic in 𝐷;
(iii) ∇𝐺 ⋅ N = − 1⁄𝐿 where 𝐿 is the length of 𝐶, when (𝑥, 𝑦) is on 𝐶;
(iv) ∫𝐶 𝐺 𝑑𝑠 = 0.
Then 𝐺 is a Green’s function associated with the Neumann problem for the region 𝑅 = 𝐷 ∪ 𝐶.
Theorem
If a Green’s function (for the Neumann problem) exists, then it is unique.
Theorem 6.3.9 If the Green’s function (for the Neumann problem) exists for the region 𝑅 =
𝐷 ∪ 𝐶, then for 𝑔(𝑠) continuous on 𝐶
𝑢(𝑥0 , 𝑦0 ) = ∫𝑔(𝑠)𝐺 𝑑𝑠
𝐶
where 𝑢 satisfies Theorems 6.3.6 and 6.3.7 and 𝑑𝑠 is the element of arc length on 𝐶, solves the
Neumann problem ∇2 𝑢 = 0 in 𝐷 with ∇𝑢 ⋅ N = 𝑔 on 𝐶.
Conformal Mappings
Theorem Let the complex function 𝑓(𝑧) be analytic in a domain 𝐷 containing a point 𝑧0 . If
𝑓 ′ (𝑧0 ) ≠ 0, then 𝑤 = 𝑓(𝑧) is conformal at 𝑧0 .
Theorem Let the complex function 𝑓(𝑧) be analytic at a critical point 𝑧0 . If 𝑘 > 1 is an
integer such that 𝑓 ′ (𝑧0 ) = 𝑓 ′′ (𝑧0 ) = ⋯ = 𝑓 (𝑘−1) (𝑧0 ) = 0, and 𝑓 (𝑘) (𝑧0 ) ≠ 0, then the angle
between any two smooth curves intersecting at 𝑧0 is increased by a factor 𝑘 by the complex
mapping 𝑤 = 𝑓(𝑧). In particular, 𝑤 = 𝑓(𝑧) is not a conformal mapping at 𝑧0 .
Potential Theory - Notes
Theorem Laplace's equation is preserved under a mapping 𝑤 = 𝑓(𝑧) provided that 𝑓 ′ (𝑧0 ) ≠
0.
Theorem 6.4.1
If 𝑓(𝑧) is simple in a domain 𝐷, then 𝑓 ′ (𝑧0 ) ≠ 0 in 𝐷.
Corollary 6.4.1
If 𝑓(𝑧) is simple in a domain 𝐷, then it is conformal in 𝐷.
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 𝑤 = 𝑓(𝑧) has range 𝐷′ and is simple in a domain 𝐷, and 𝐹(𝑤) is simple in
𝐷′ , then the composite (𝐹 ∘ 𝑓)(𝑧) = 𝐹[𝑓(𝑧)] is simple in 𝐷.
Theorem 6.4.3 Let 𝐶 be a simple closed contour with interior 𝐷. Furthermore, let 𝑓(𝑧) be
analytic within and on 𝐶, and take no value more than once on 𝐶. Then 𝑓(𝑧) is simple in 𝐷.
Theorem 6.4.4 Let 𝑓𝑛 (𝑧) be simple in a domain 𝐷 for all 𝑛 = 1,2,3, …. Furthermore, let 𝑓𝑛 (𝑧)
converge uniformly in compact subsets of 𝐷. Then 𝑓𝑛 (𝑧) is either constant, or simple in 𝐷.
A fundamental problem in the theory of conformal mappings is to find a simple function 𝑤 =
𝑓(𝑧) which maps a given simply connected domain onto the open unit disk |𝑤| < 1.
Theorem 6.4.7 (Riemann Mapping Theorem) Let 𝐷 be a simply connected domain in the
complex plane ℂ with at least two boundary points. Then there exists a simple function 𝑤 =
𝑓(𝑧) which maps 𝐷 onto the open unit disk |𝑤| < 1 in the 𝑤-plane. If we specify that a given
point 𝑧0 in 𝐷 maps into the origin 𝑤 = 0, and a given direction at 𝑧0 is mapped into a given
direction at the origin, then the mapping is unique.
Let 𝐷 and 𝐷′ be simply connected domains in ℂ. We can use the Riemann mapping theorem to
establish the existence of a simple mapping from 𝐷 onto 𝐷′ . According to the theorem there is
a simple mapping 𝑓 from 𝐷 onto the open unit disk |𝑤| < 1. Similarly, there is a simple
mapping 𝑔 from 𝐷′ onto the open unit disk |𝑤| < 1. Since 𝑔 is simple, it has a well defined
inverse function 𝑔−1 that maps the open unit disk |𝑤| < 1 onto 𝐷′ . The desired mapping from
𝐷 onto 𝐷′ is then given by the composition 𝑤 = (𝑔−1 ∘ 𝑓)(𝑧).
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 𝐷 correspond to boundary points of 𝐷′ .
Theorem 6.4.8 Let 𝐷 and 𝐷′ be domains bounded by simple closed contours 𝐶 and 𝐶 ′ ,
respectively. Then the conformal map 𝑓: 𝐷 → 𝐷′ is continuous in 𝐷 ∪ 𝐶 and establishes a oneto-one correspondence between 𝐶 and 𝐶 ′ .
The Schwarz-Christoffel Transformation
The Schwarz-Christoffel transformation provides a recipe for a conformal mapping of the
upper half of the 𝑧-plane to a polygonal region in the 𝑤-plane.
Theorem (Schwarz-Christoffel formula)
domain 𝑦 > 0 and has the derivative
Let 𝑓 be a complex function that is analytic in the
𝑓 ′ (𝑧) = 𝐴(𝑧 − 𝑥1 )(𝛼1 ⁄𝜋)−1 × (𝑧 − 𝑥2 )(𝛼2 ⁄𝜋)−1 × ⋯ × (𝑧 − 𝑥𝑛 )(𝛼𝑛⁄𝜋)−1
Potential Theory - Notes
where 𝑥1 < 𝑥2 < ⋯ < 𝑥𝑛 are arbitrary points on the 𝑥-axis, 0 < 𝛼𝑖 < 2𝜋 for 1 ≤ 𝑖 ≤ 𝑛, and 𝐴
is a complex constant. Then the upper half plane 𝑦 ≥ 0 is mapped by 𝑤 = 𝑓(𝑧) onto an
unbounded polygonal region with interior angles 𝛼1 , 𝛼2 , … , 𝛼𝑛 .
Since 𝑓 ′ (𝑥 + 𝑖𝑦) ≠ 0 when 𝑦 > 0, it follows that the function given by the Schwarz-Christoffel
formula is a conformal mapping in the domain 𝑦 > 0. Furthermore, although the mapping is
defined on the upper half plane 𝑦 ≥ 0, it is only conformal in the domain 𝑦 > 0.
The above theorem provides a formula for the derivative of 𝑓, which may in principle be
integrated to obtain 𝑤 as follows
𝑤 = 𝑓(𝑧) = 𝐴 ∫(𝑧 − 𝑥1 )(𝛼1 ⁄𝜋)−1 × (𝑧 − 𝑥2 )(𝛼2 ⁄𝜋)−1 × ⋯ × (𝑧 − 𝑥𝑛 )(𝛼𝑛⁄𝜋)−1 𝑑𝑧 + 𝐵
where 𝐴 and 𝐵 are complex constants. Thus 𝑓 is the composition of the function
𝑔(𝑧) = ∫(𝑧 − 𝑥1 )(𝛼1 ⁄𝜋)−1 × (𝑧 − 𝑥2 )(𝛼2 ⁄𝜋)−1 × ⋯ × (𝑧 − 𝑥𝑛 )(𝛼𝑛⁄𝜋)−1 𝑑𝑧
and the linear mapping ℎ(𝑧) = 𝐴𝑧 + 𝐵. The linear mapping ℎ allows us to rotate, scale, and
translate the polygonal region produced by 𝑔.
In practice we usually have a certain amount of freedom in the selection of the points 𝑥𝑘 on
the 𝑥-axis. A judicious choice can simplify the computation of 𝑓(𝑧).
The Schwarz-Christoffel formula may be used to construct a mapping of the upper half plane
𝑦 ≥ 0 onto a bounded polygonal region. To do so, we apply the formula using only 𝑛 − 1 on
the 𝑛 interior angles of the bounded polygonal region. Essentially, we are choosing the point
𝑥𝑛 on the 𝑥-axis to lie at infinity.
Flows with Sources and Sinks
Streamlining
Suppose that the complex velocity potential Ω(𝑧) = 𝜙(𝑥, 𝑦) + 𝑖𝜓(𝑥, 𝑦) is analytic in a domain
𝐷 and that 𝜓 is constant on the boundary of 𝐷. Then Ω′ (𝑧) is a complex representation of the
velocity field of a flow of an ideal fluid in 𝐷. Moreover, if a particle is placed in 𝐷 and allowed
to flow with the fluid, then its path 𝑧(𝑡) remains in 𝐷.
Solving a streamlining problem If 𝑤 = 𝑓(𝑧) is a simple mapping of the domain 𝐷 onto a
domain 𝐷′ in the 𝑤-plane such that the image of the boundary 𝐶 of 𝐷 is a horizontal line in the
𝑤-plane, then 𝑤 = Ω(𝑧) is a complex velocity potential of the flow of an ideal fluid in 𝐷 and
Ω′ (𝑧) is a complex representation of its velocity field.
An alternative method is to apply the mapping 𝑤 = 𝑓(𝑧) directly to the complex velocity
potential of a uniform flow in the 𝑤-plane
Ω𝐷′ = 𝑣0 𝑤
⇒
Ω𝐷 = 𝑣0 𝑓(𝑧).
Sources, Sinks and Dipioles
The complex velocity potential of a (simple) source of strength 𝐴 at the point 𝑧0 is
Ω(𝑧) = 𝐴 log(𝑧 − 𝑧0 )
where 𝐴 is a positive real constant.
The complex velocity potential of a (simple) sink of strength 𝐴 at the point 𝑧0 is
Ω(𝑧) = −𝐴 log(𝑧 − 𝑧0 )
Potential Theory - Notes
where 𝐴 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 𝐴 at the point 𝑧1 = 𝑟𝑒 𝑖𝜃 and a sink of the same
strength at the point 𝑧0 . Then the complex velocity potential of the dipole in the direction 𝜃 at
𝑧0 is
Ω(𝑧) = −𝐴𝑟
𝑒 𝑖𝜃
𝑧 − 𝑧0
where 𝑀 = 𝐴𝑟 is strength of the dipole.
The velocity potential is
𝑑
[log|𝑧 − 𝑧0 |]
𝑑𝑛
where 𝑑⁄𝑑𝑛 is the directional derivative in the direction 𝜃.
𝜙(𝑥, 𝑦) = −𝑀
Determining fluid flows produced by sources and sinks If 𝑤 = 𝑓(𝑧) is a simple mapping of
the domain 𝐷 onto a domain 𝐷′ in the 𝑤-plane such that the image of boundary curves,
sources, and sinks for the flow in 𝐷are mapped onto boundary curves, sources, and sinks in 𝐷′
where the complex potential is known to be Ω𝐷′ (𝑤). Then the complex potential in 𝐷 is given
by Ω𝐷 (𝑧) = Ω𝐷′ (𝑓(𝑧)).
Sources, sinks and dipoles, and their potentials, have their counterparts in electrostatics,
magnetostatics, heat transfer, and so on.
Potential Theory - SAQ’s
SAQ 1
Which flow is given by the velocity potential
𝜙(𝑥, 𝑦) = 𝑥 +
𝑥2
2𝑥
?
+ 𝑦2
For very large 𝑥 2 + 𝑦 2 , 𝜙 is approximately 𝑥, and
𝑣𝑥 =
𝜕𝜙
=1
𝜕𝑥
and 𝑣𝑦 =
𝜕𝜙
= 0.
𝜕𝑦
Hence, at large distances from the origin, the flow is that of a uniform stream with velocity 1
to the right.
Consider
𝑧+
2
2
2(𝑥 − 𝑖𝑦)
2(𝑥 − 𝑖𝑦)
= 𝑥 + 𝑖𝑦 +
= 𝑥 + 𝑖𝑦 +
= 𝑥 + 𝑖𝑦 + 2
(𝑥 + 𝑖𝑦)(𝑥 − 𝑖𝑦)
𝑧
𝑥 + 𝑖𝑦
𝑥 + 𝑦2
2𝑥
2𝑦
=𝑥+ 2
+ 𝑖 (𝑦 − 2
).
2
𝑥 +𝑦
𝑥 + 𝑦2
Let
𝜓(𝑥, 𝑦) = 𝑦 −
𝑥2
2𝑦
.
+ 𝑦2
Now
𝜙𝑥 = 1 +
2(𝑥 2 + 𝑦 2 ) − 4𝑥 2
2(𝑥 2 + 𝑦 2 ) − 4𝑦 2
=
1
−
= 𝜓𝑦
(𝑥 2 + 𝑦 2 )2
(𝑥 2 + 𝑦 2 )2
and
𝜙𝑦 =
−4𝑥𝑦
= −𝜓𝑥 .
+ 𝑦 2 )2
(𝑥 2
Hence, the Cauchy-Riemann equations are satisfied, and 𝜓 is a harmonic conjugate of 𝜙. It
follows that
Ω(𝑧) = 𝑧 +
2
2𝑥
2𝑦
=𝑥+ 2
+
𝑖
(𝑦
−
)
𝑧
𝑥 + 𝑦2
𝑥2 + 𝑦2
is the complex potential corresponding to 𝜙, and the streamlines of the flow are given by the
level curves of 𝜓.
Now
𝑦−
𝑥2
2𝑦
= 0 ⇒ 𝑦 3 + 𝑥 2 𝑦 − 2𝑦 = 0 ⇒ 𝑥 2 + 𝑦 2 = 2,
+ 𝑦2
which is the equation of a circle of radius √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 𝜙 is that of a uniform stream past a circular
cylinder of radius √2.
SAQ 2 Express ∇2 𝑢 = 0 in two dimensions in polar coordinates by considering 𝑢 as the real
part of an analytic function and using the Cauchy-Riemann equations in polar form.
Potential Theory - SAQ’s
Let 𝑓(𝑧) = 𝑢(𝑟, 𝜃) + 𝑖𝑣(𝑟, 𝜃) be an analytic function, then the polar form of the CauchyRiemann equations give
𝜕𝑢 𝜕𝑣
𝜕𝑣
𝜕𝑢
=
and 𝑟
=− .
𝜕𝑟 𝜕𝜃
𝜕𝑟
𝜕𝜃
Since 𝑢 and 𝑣 are analytic,
𝑟
𝜕
𝜕𝑢
𝜕 𝜕𝑣
𝜕 𝜕𝑣
𝜕 1 𝜕𝑢
(𝑟 ) = ( ) =
( )=− (
).
𝜕𝑟 𝜕𝑟
𝜕𝑟 𝜕𝜃
𝜕𝜃 𝜕𝑟
𝜕𝜃 𝑟 𝜕𝜃
Hence
𝜕
𝜕𝑢
𝜕 1 𝜕𝑢
(𝑟 ) +
(
) = 0.
𝜕𝑟 𝜕𝑟
𝜕𝜃 𝑟 𝜕𝜃
Using the chain rule gives
𝜕
𝜕𝑢
𝜕 1 𝜕𝑢
𝜕 2 𝑢 𝜕𝑢 1 𝜕 2 𝑢
(𝑟 ) +
(
)=𝑟 2+
+
= 0.
𝜕𝑟 𝜕𝑟
𝜕𝜃 𝑟 𝜕𝜃
𝜕𝑟
𝜕𝑟 𝑟 𝜕𝜃 2
Which gives
𝑟2
𝜕 2𝑢
𝜕𝑢 𝜕 2 𝑢
+
𝑟
+
= 0.
𝜕𝑟 2
𝜕𝑟 𝜕𝜃 2
SAQ 3 Find the potential inside a capacitor consisting of an infinitely long circular cylinder
of radius 𝑏 charged to a potential of 10 volts, and an infinitely long wire of radius 𝑎 along the
axis of the cylinder at a potential of zero.
Hint: by symmetry, the solution must be independent of the polar angle 𝜃.
The electrostatic potential 𝜙(𝑟) inside the capacitor is independent of the polar angle 𝜃 and so
satisfies the polar form of Laplace's equation
𝑑2 𝜙
𝑑𝜙
+𝑟
= 0, 𝑎 < 𝑟 < 𝑏
2
𝑑𝑟
𝑑𝑟
with boundary conditions
𝑟2
𝜙(𝑎) = 0, 𝜙(𝑏) = 10.
The differential equation above is a Cauchy-Euler equation.
Indicial equation:
𝜆2 = 0 ⇒ 𝜆 = 0.
General solution:
𝜙(𝑟) = 𝐴 + 𝐵 ln 𝑟.
Applying the boundary conditions gives
𝜙(𝑎) = 𝐴 + 𝐵 ln 𝑎 = 0 ⇒ 𝐴 = −𝐵 ln 𝑎
and
𝜙(𝑏) = 𝐴 + 𝐵 ln 𝑏 = 10 ⇒ −𝐵 ln 𝑎 + 𝐵 ln 𝑏 = 10 ⇒ 𝐵 =
Hence, the required potential equation is
10
.
ln 𝑏 − ln 𝑎
Potential Theory - SAQ’s
𝜙(𝑟) = −𝐵 ln 𝑎 +
10
10(ln 𝑟 − ln 𝑎)
ln 𝑟 =
.
ln 𝑏 − ln 𝑎
ln 𝑏 − ln 𝑎
SAQ 4 Find the potential in an annulus bounded by circles of radii 𝑎 < 𝑏 if the inner circle
has potential 𝛼 and the outer circle has potential 𝛽.
The electrostatic potential 𝜙(𝑟) inside the capacitor is independent of the polar angle 𝜃 and so
satisfies the polar form of Laplace's equation
𝑑2 𝜙
𝑑𝜙
+𝑟
= 0, 𝑎 < 𝑟 < 𝑏
2
𝑑𝑟
𝑑𝑟
with boundary conditions
𝑟2
𝜙(𝑎) = 𝛼,
𝜙(𝑏) = 𝛽.
The differential equation above is a Cauchy-Euler equation.
Indicial equation:
𝜆2 = 0 ⇒ 𝜆 = 0.
General solution:
𝜙(𝑟) = 𝐴 + 𝐵 ln 𝑟.
Applying the boundary conditions gives
𝜙(𝑎) = 𝐴 + 𝐵 ln 𝑎 = 𝛼 ⇒ 𝐴 = 𝛼 − 𝐵 ln 𝑎
and
𝜙(𝑏) = 𝐴 + 𝐵 ln 𝑏 = 𝛽 ⇒ 𝛼 − 𝐵 ln 𝑎 + 𝐵 ln 𝑏 = 𝛽 ⇒ 𝐵 =
𝛽−𝛼
.
ln 𝑏 − ln 𝑎
Hence, the required potential equation is
𝜙(𝑟) = 𝛼 − 𝐵 ln 𝑎 +
SAQ 5
𝛽−𝛼
𝛼 ln 𝑏 − 𝛽 ln 𝑎 + (𝛽 − 𝛼) ln 𝑟
ln 𝑟 =
.
ln 𝑏 − ln 𝑎
ln 𝑏 − ln 𝑎
What conditions on the real numbers 𝑎, 𝑏, 𝑐, 𝑑 ensure that
𝑎𝑥 3 + 𝑏𝑥 2 𝑦 + 𝑐𝑥𝑦 2 + 𝑑𝑦 3
is harmonic.
Let
𝜙(𝑥, 𝑦) = 𝑎𝑥 3 + 𝑏𝑥 2 𝑦 + 𝑐𝑥𝑦 2 + 𝑑𝑦 3 .
Then 𝜙 is harmonic if it has continuous first and second-order partial derivatives in a domain
𝐷, and satisfies Laplace’s equation
∇2 𝜙 =
Now
𝜕 2𝜙 𝜕 2𝜙
+
= 0.
𝜕𝑥 2 𝜕𝑦 2
Potential Theory - SAQ’s
𝜕𝜙
= 3𝑎𝑥 2 + 2𝑏𝑥𝑦 + 𝑐𝑦 2
𝜕𝑥
and
𝜕 2𝜙
= 6𝑎𝑥 + 2𝑏𝑦
𝜕𝑥 2
𝜕𝜙
= 2𝑏𝑥 + 2𝑐𝑥𝑦 + 3𝑑𝑦 2
𝜕𝑦
and
𝜕 2𝜙
= 2𝑐𝑥 + 6𝑑𝑦.
𝜕𝑦 2
and
So 𝜙 has continuous first and second-order partial derivatives in the complex plane.
Substituting the second-order partial derivatives into Laplace's equation gives
6𝑎𝑥 + 2𝑏𝑦 + 2𝑐𝑥 + 6𝑑𝑦 = 2(3𝑎 + 𝑐)𝑥 + 2(𝑏 + 3𝑑)𝑦 = 0
which is satisfied when
𝑐 = −3𝑎
and 𝑏 = −3𝑑 .
Hence
𝜙(𝑥, 𝑦) = 𝑎𝑥 3 − 3𝑑𝑥 2 𝑦 − 3𝑎𝑥𝑦 2 + 𝑑𝑦 3 = 𝑎(𝑥 3 − 3𝑥𝑦 2 ) + 𝑑(𝑦 3 − 3𝑥 2 𝑦).
Now
𝑧 3 = (𝑥 + 𝑖𝑦)3 = 𝑥 3 − 3𝑥𝑦 2 + 𝑖(3𝑥 2 𝑦 − 𝑦 3 ).
Which gives
𝜙(𝑥, 𝑦) = 𝑎 Re(𝑧 3 ) − 𝑑 Im(𝑧 3 )
for all complex 𝑧.
SAQ 6
𝑧+1
By considering arg (𝑧−1), find a function 𝑔 harmonic in the unit disk such that
𝑔(𝑒 𝑖𝜃 ) = {
1
0<𝜃<𝜋
−1 −𝜋 < 𝜃 < 0.
Let 𝑧 = 𝑒 𝑖𝜃 , then |𝑧| < 1 and
(cos 𝜃 + 1) + 𝑖 sin 𝜃
𝑒 𝑖𝜃 + 1
arg ( 𝑖𝜃
) = arg (
)
(cos 𝜃 − 1) + 𝑖 sin 𝜃
𝑒 −1
[(cos 𝜃 + 1) + 𝑖 sin 𝜃][(cos 𝜃 − 1) − 𝑖 sin 𝜃]
= arg (
)
(cos 𝜃 − 1)2 + sin2 𝜃
− sin 𝜃
= arg (
𝑖)
1 − cos 𝜃
𝜋
−
0<𝜃<𝜋
2
={ 𝜋
−𝜋 < 𝜃 < 0.
2
So, we define
2
𝑧+1
𝑔(𝑧) = − arg (
).
𝜋
𝑧−1
Let
𝑧+1
𝑧+1
𝑓(𝑧) = log (
)=|
| + 𝑖𝑔(𝑧).
𝑧−1
𝑧−1
Potential Theory - SAQ’s
Since 𝑓 is analytic in the unit open disk |𝑧| < 1, it follows that 𝑔(𝑧) = Im(𝑓(𝑧)) 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 √(𝑥 − 𝑥0 )2 + (𝑦 − 𝑦0 )2 +
ln √(𝑥 − 𝑥0 )2 + (𝑦 + 𝑦0 )2
2𝜋
2𝜋
is a Green's function for the upper half-plane problem in the sense that the solution, if it exists,
can be expressed as
𝐺(𝑥, 𝑦, 𝑥0 , 𝑦0 ) = −
∞
𝑦0 ∞
𝑔(𝑥)
𝑢(𝑥0 , 𝑦0 ) = ∫ 𝑔(𝑥)𝐺𝑦 (𝑥, 0, 𝑥0 , 𝑦0 ) 𝑑𝑥 = ∫
𝑑𝑥.
𝜋 −∞ (𝑥 − 𝑥0 )2 + 𝑦02
−∞
Let let 𝑧0 = 𝑥0 + 𝑖𝑦0 be a point in the open upper half-plane𝐷, and let 𝑅 = 𝐷 ∪ 𝐶 where 𝐶 is
the 𝑥-axis. Then
1
1
ln|𝑧 − 𝑧0 | +
ln|𝑧 − 𝑧0 |
2𝜋
2𝜋
1
1
=−
Re(log(𝑧 − 𝑧0 )) +
Re(log(𝑧 − 𝑧0 )).
2𝜋
2𝜋
𝐺(𝑥, 𝑦, 𝑥0 , 𝑦0 ) = −
Hence 𝐺 is analytic in 𝐷. It follows that
1
1
Re(log(𝑧 − 𝑧0 )) =
ln √(𝑥 − 𝑥0 )2 + (𝑦 + 𝑦0 )2
2𝜋
2𝜋
is harmonic in 𝐷.
Also, 𝐺 is continuous in 𝐷 ∪ 𝐶, except at (𝑥0 , 𝑦0 ) in 𝐷.
Finally, since 𝐺(𝑥, 0, 𝑥0 , 𝑦0 ) = 0, we have 𝐺(𝑥, 𝑦, 𝑥0 , 𝑦0 ) = 0 when (𝑥, 𝑦) is on 𝐶.
Hence, 𝐺 is a Green’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
𝑢(𝑥0 , 𝑦0 ) = − ∫(𝑔(𝑠)∇𝐺 ⋅ N) 𝑑𝑠
𝐶
where N is the unit outward normal on 𝐶 and 𝑑𝑠 is the element of arc length on 𝐶, solves the
Dirichlet problem ∇2 𝑢 = 0 in 𝐷 with 𝑢 = 𝑔 on 𝐶.
In this case
N = −j and 𝑑𝑠 = 𝑑𝑥.
Hence, integrating along the 𝑥-axis gives
∞
∞
𝑢(𝑥0 , 𝑦0 ) = ∫ (𝑔(𝑥)∇𝐺 ⋅ j) 𝑑𝑥 = ∫ (𝑔(𝑥)𝐺𝑦 (𝑥, 𝑦, 𝑥0 , 𝑦0 )|𝑦=0 ) 𝑑𝑥
−∞
−∞
Potential Theory - SAQ’s
∞
= ∫ (𝑔(𝑥)
−∞
+
1
ln √(𝑥 − 𝑥0 )2 + (𝑦 + 𝑦0 )2 ] ) 𝑑𝑥
2𝜋
𝑦=0
∞
= ∫ (𝑔(𝑥)
−∞
+
𝜕
1
[− ln √(𝑥 − 𝑥0 )2 + (𝑦 − 𝑦0 )2
𝜕𝑦 2𝜋
𝜕
1
[− ln((𝑥 − 𝑥0 )2 + (𝑦 − 𝑦0 )2 )
𝜕𝑦 4𝜋
1
ln((𝑥 − 𝑥0 )2 + (𝑦 + 𝑦0 )2 )] ) 𝑑𝑥
4𝜋
𝑦=0
=−
1 ∞
2(𝑦 − 𝑦0 )
2(𝑦 + 𝑦0 )
∫ (𝑔(𝑥) [
−
] ) 𝑑𝑥
2
2
(𝑥 − 𝑥0 ) + (𝑦 − 𝑦0 )
(𝑥 − 𝑥0 )2 + (𝑦 + 𝑦0 )2
4𝜋 −∞
𝑦=0
=−
1 ∞
−2𝑦0
2𝑦0
∫ (𝑔(𝑥) [
−
]) 𝑑𝑥
2
2
(𝑥 − 𝑥0 ) + 𝑦0 (𝑥 − 𝑥0 )2 + 𝑦02
4𝜋 −∞
𝑦0 ∞
𝑔(𝑥)
= ∫
𝑑𝑥.
𝜋 −∞ (𝑥 − 𝑥0 )2 + 𝑦02
SAQ 8 Prove that the function 𝑓(𝑧) = 𝑧 + 𝑎2 𝑧 2 + 𝑎3 𝑧 3 + ⋯ is simple for |𝑧| < 1 if
∑∞
𝑘=2 𝑘|𝑎𝑘 | ≤ 1.
Since
∞
∞
𝐴 ∑ 𝑘 ≤ ∑ 𝑘|𝑎𝑘 | ≤ 1
𝑘=2
𝑘=2
where 𝐴 = min{|𝑎2 |, |𝑎3 |, … }, we have
𝐴≤
1
∑∞
𝑘=2 𝑘
which implies that |𝑎𝑘 | → 0 as 𝑘 → ∞.
𝑘
Hence, by the root test, 𝑓(𝑧) = ∑∞
𝑘=2 𝑎𝑘 𝑧 defines an analytic function on |𝑧| < ∞.
Suppose that 𝑧1 ≠ 𝑧2 , then
𝑓(𝑧1 ) − 𝑓(𝑧2 ) = (𝑧1 + 𝑎2 𝑧12 + 𝑎3 𝑧13 + ⋯ ) − (𝑧2 + 𝑎2 𝑧22 + 𝑎3 𝑧23 + ⋯ )
= (𝑧1 − 𝑧2 + 𝑎2 (𝑧12 − 𝑧22 ) + 𝑎3 (𝑧13 − 𝑧23 ) + ⋯ )
= (𝑧1 − 𝑧2 )(1 + 𝑎2 (𝑧1 + 𝑧2 ) + 𝑎3 (𝑧12 + 𝑧1 𝑧2 + 𝑧22 ) + ⋯ ).
Hence, by the reverse triangle inequality
|𝑓(𝑧1 ) − 𝑓(𝑧2 )| ≥ ||𝑧1 + 𝑎2 𝑧12 + 𝑎3 𝑧13 + ⋯ | − |𝑧2 + 𝑎2 𝑧22 + 𝑎3 𝑧23 + ⋯ ||
= |𝑧1 − 𝑧2 |(1 + |𝑎2 |(|𝑧1 | + |𝑧2 |) + |𝑎3 |(|𝑧12 | + |𝑧1 𝑧2 | + |𝑧22 |) + ⋯ )
≥ |𝑧1 − 𝑧2 |(1 + 2|𝑎2 | + 3|𝑎3 | + ⋯ ) > 0.
Potential Theory - SAQ’s
So, 𝑧1 ≠ 𝑧2 implies that 𝑓(𝑧1 ) ≠ 𝑓(𝑧2 ), which is the contrapositive of 𝑓(𝑧1 ) = 𝑓(𝑧2 ) ⇒ 𝑧1 = 𝑧2 ;
the necessary condition for a one-to-one function.
It follows that 𝑓(𝑧) is analytic and one-to-one on |𝑧| < 1, and so is simple there.
SAQ 9 Let 𝑓(𝑧) be simple within and on a simple closed contour 𝐶. Let 𝐷 the interior of 𝐶,
map onto 𝐷′ , and 𝐶 map onto 𝐶 ′ , the boundary of 𝐷′ .
(i)
Prove that the length of 𝐶 ′ is ∫𝐶|𝑓 ′ (𝑧)| |𝑑𝑧|;
(ii)
Prove that the area of 𝐷′ is ∬𝐷|𝑓 ′ (𝑧)|2 𝑑𝑥𝑑𝑦;
(iii) If 𝐶 is a circle of radius 𝑅 centred at the origin, prove that the area of 𝐷 ′ is greater than,
or equal to 𝜋|𝑓 ′ (0)|2 𝑅 2.
(i) Since 𝑓 is simple, the curve 𝐶 is mapped one-to-one onto the curve 𝐶 ′ . Hence, 𝐶 ′ is smooth
and has a length 𝐿 given by
𝐿 = ∫ |𝑑𝑤|.
𝐶′
Now 𝑑𝑤 = 𝑓 ′ (𝑧)𝑑𝑧, which gives
𝐿 = ∫ |𝑓 ′ (𝑧)𝑑𝑧| = ∫|𝑓 ′ (𝑧)||𝑑𝑧|.
𝐶′
𝐶
(ii) Let 𝑧 = 𝑥 + 𝑖𝑦, then the area element 𝑑𝐴 at 𝑧 is 𝑑𝑥𝑑𝑦. The mapping 𝑤 = 𝑓(𝑧) changes
small distances near 𝑧0 by the scale factor |𝑓 ′ (𝑧0 )|. Therefore, 𝑑𝐴 is transformed by 𝑓 into an
element of area of 𝐷′ given by |𝑓 ′ (𝑧)|2 𝑑𝑥𝑑𝑦 at 𝑓(𝑧) ∈ 𝐷′ . Hence, 𝐷′ has an area 𝐴 given by
𝐴 = ∬ |𝑓 ′ (𝑧)|2 𝑑𝑥𝑑𝑦.
𝐷
(iii) Too difficult to waste time on.
SAQ 10
Prove that 𝑤 = 3𝑧 + 𝑧 2 is simple on the disk |𝑧| < 1.
Let 𝑤 = 𝑓(𝑧), then 𝑓(𝑧) is analytic on |𝑧| < 1.
Suppose that 𝑧1 ≠ 𝑧2 , then
𝑓(𝑧1 ) − 𝑓(𝑧2 ) = (3𝑧1 + 𝑧12 ) − (3𝑧2 + 𝑧22 ) = (3(𝑧1 − 𝑧2 ) + (𝑧12 − 𝑧22 ))
= (𝑧1 − 𝑧2 )(3 + 𝑧1 + 𝑧2 ).
Hence, by the reverse triangle inequality
|𝑓(𝑧1 ) − 𝑓(𝑧2 )| ≥ ||3𝑧1 + 𝑧12 | − |3𝑧2 + 𝑧22 || = |𝑧1 − 𝑧2 |(3 + |𝑧1 | + |𝑧2 |)
≥ 5|𝑧1 − 𝑧2 | > 0.
So, 𝑧1 ≠ 𝑧2 implies that 𝑓(𝑧1 ) ≠ 𝑓(𝑧2 ), which is the contrapositive of 𝑓(𝑧1 ) = 𝑓(𝑧2 ) ⇒ 𝑧1 = 𝑧2 ;
the necessary condition for a one-to-one function.
It follows that 𝑓(𝑧) is analytic and one-to-one on |𝑧| < 1, and so is simple there.
Potential Theory - Problems
P1 Consider a 2-dimensional flow outside a simple curve 𝐶 in the (𝑥, 𝑦)-plane. If the
velocity v = (𝑣𝑥 , 𝑣𝑦 ) is such that 𝑣𝑥 − 𝑖𝑣𝑦 is analytic outside 𝐶, i.e. 𝑣𝑥 − 𝑖𝑣𝑦 = 𝑓(𝑧) with 𝑧 =
𝑥 + 𝑖𝑦 and 𝑓(𝑧) analytic, show that the circulation is the same round any simple curve
surrounding 𝐶.
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 𝐶 the unit circle {|𝑧| = 1}. Prove
𝑘
that if the circulation is non-zero, then 𝑓(𝑧) is of the form ∑∞
−∞ 𝑎𝑘 𝑧 with Im(𝑎−1 ) = 0.
Since 𝑓 is analytic on the open annulus 𝐴 = {1 < |𝑧| < ∞}, it has the Laurent series
representation
∞
𝑓(𝑧) = ∑ 𝑎𝑛 𝑧 𝑛
𝑛=−∞
for all 𝑧 ∈ 𝐴.
The velocity field v is given by
𝑑Ω
= 𝑣𝑥 + 𝑖𝑣𝑦 .
𝑑𝑧
Hence
𝑑Ω
= 𝑣𝑥 − 𝑖𝑣𝑦 = 𝑓(𝑧) ⇒
𝑑𝑧
Ω(𝑧) = ∫𝑓(𝑧) 𝑑𝑧.
𝐶
Which gives, for the circle |𝑧| = 1
∞
𝜓(𝑥, 𝑦) = Re (∫
∑ 𝑎𝑛 𝑧 𝑛 𝑑𝑧).
𝑓(𝑧) 𝑑𝑧) = Re (∫
|𝑧|=1
|𝑧|=1
𝑛=−∞
On the unit circle 𝑧 = 𝑒 , which gives
𝑖𝜃
∞
2𝜋
𝜓(𝑥, 𝑦) = Re (∫
(𝑖𝑒
𝑖𝜃
0
∑ 𝑎𝑛 𝑒 𝑖𝑛𝜃 ) 𝑑𝜃)
𝑛=−∞
∞
2𝜋
= Re (𝑖 ∑ 𝑎𝑛 ∫ 𝑒 𝑖(𝑛+1)𝜃 𝑑𝜃)
0
𝑛=−∞
2𝜋
= Re (𝑖𝑎−1 ∫ 𝑑𝜃) = Re(𝑖2𝜋𝑎−1 )
0
since
2𝜋
∫ 𝑒 𝑖𝑘𝜃 𝑑𝜃 = 0
0
for any non-zero integer 𝑘.
Consequently, 𝜓(𝑥, 𝑦) is non-zero if Im(𝑎−1 ) = 0.
Potential Theory - Problems
P3 Find a steady state temperature for a thin metal infinite strip {(𝑥, 𝑦): −∞ < 𝑥 < ∞, 0 ≤
𝑦 ≤ 1} when the upper edge is perfectly insulated and 𝑢(−1,0) = 3 and 𝑢(1,0) = 2. How
many solutions are there for 𝑢(𝑥, 𝑦)?
The steady-state temperature 𝑢 is a solution of Laplace's equation in the domain 𝐷 defined by
−∞ < 𝑥 < ∞, 0 ≤ 𝑦 ≤ 1, which satisfies the boundary conditions
𝑢(−1,0) = 3;
𝑢(1,0) = 2;
∇𝑢 ⋅ N = 0
for
−∞ < 𝑥 < ∞, 𝑦 = 1.
In this case, the unit outward facing normal is N = j, hence
(∇𝑢 ⋅ N)|𝑦=1 =
𝜕𝑢
|
= 0.
𝜕𝑦 𝑦=1
We seek a harmonic function 𝑢, which behaves as required on the boundary, and in particular
satisfies the condition (∇𝑢 ⋅ N)|𝑦=1 = 0.
One possibility is that 𝑢(𝑥, 𝑦) is a function of 𝑥 alone, i.e.
𝑢(𝑥, 𝑦) = 𝑔(𝑥)
for some function 𝑔 of 𝑥.
In order to satisfy Laplace's equation we must have
𝑑2 𝑢 𝑑2𝑢 𝑑2𝑔
+
=
= 0.
𝑑𝑥 2 𝑑𝑦 2 𝑑𝑥 2
Integrating twice gives
𝑔(𝑥) = 𝐴𝑥 + 𝐵
for arbitrary constants 𝐴, 𝐵.
Applying the remaining boundary conditions gives
𝑢(−1,0) = 𝑔(−1) = −𝐴 + 𝐵 = 3
and
𝑢(1,0) = 𝑔(1) = 𝐴 + 𝐵 = 2.
Hence
1
5
and 𝐵 =
2
2
which gives a solution
𝐴=−
5 𝑥
− .
2 2
Another possibility is 𝑢(𝑥, 𝑦) = 𝑔(𝑥)ℎ(𝑦) where ℎ′ (1) = 0.
𝑢(𝑥, 𝑦) =
Consider the function 𝑓(𝑧) = 𝑒 𝑐𝑧−𝑖𝑐 for real 𝑐. Since 𝑓 is analytic in 𝐷, it follows that
Re(𝑒 𝑐𝑧−𝑖𝑐 ) = Re(𝑒 𝑐𝑥 𝑒 𝑐(𝑦−1)𝑖 ) = 𝑒 𝑐𝑥 cos(𝑐(𝑦 − 1))
Potential Theory - Problems
is harmonic in 𝐷.
Now
𝜕 𝑐𝑥
[𝑒 cos(𝑐(𝑦 − 1))]|
= −𝑒 𝑐𝑥 𝑐 sin(0) = 0.
𝜕𝑦
𝑦=1
Choosing 𝑐 = 𝜋⁄2 gives
𝜋
𝑒 𝜋𝑥⁄2 cos ( (𝑦 − 1))
2
which is equal to zero when 𝑦 = 0.
Therefore
𝑢(𝑥, 𝑦) =
5 𝑥
𝜋
− + 𝐶𝑒 𝜋𝑥⁄2 cos ( (𝑦 − 1))
2 2
2
solves the problem for any constant 𝐶, which means there are infinitely many solutions.
P4
Prove the assertion on p.235 that
𝐾𝑒 2
∫ 3 (𝑥𝑑𝑥 + 𝑦𝑑𝑦 + 𝑧𝑑𝑧)
𝐶 𝑟
is independent of the path assuming that it is smooth.
Note: the quantity represented by 𝑟 3 is the distance from the origin.
Let the path 𝐶 be parameterised by
(𝑥(𝑡), 𝑦(𝑡), 𝑧(𝑡))
for
𝑎≤𝑡≤𝑏
where 𝑎 and 𝑏 are the left and right endpoints of 𝐶, respectively.
Since 𝐶 is smooth, the derivatives 𝑥 ′ (𝑡), 𝑦 ′ (𝑡), 𝑧 ′ (𝑡) exist. Which gives
𝑥𝑑𝑥 + 𝑦𝑑𝑦 + 𝑧𝑑𝑧 = 𝑥(𝑡)
𝑑𝑥
𝑑𝑦
𝑑𝑧
𝑑𝑡 + 𝑦(𝑡) 𝑑𝑡 + 𝑧(𝑡) 𝑑𝑡
𝑑𝑡
𝑑𝑡
𝑑𝑡
on 𝐶.
Hence
𝑏
𝐾𝑒 2
𝑥(𝑡)𝑥 ′ (𝑡) + 𝑦(𝑡)𝑦 ′ (𝑡) + 𝑧(𝑡)𝑧 ′ (𝑡)
2
(𝑥𝑑𝑥
+
𝑦𝑑𝑦
+
𝑧𝑑𝑧)
=
𝐾𝑒
∫
𝑑𝑡
3
3⁄2
𝐶 𝑟
𝑎
(𝑥 2 (𝑡) + 𝑦 2 (𝑡) + 𝑧 2 (𝑡))
∫
= −𝐾𝑒
2
𝑏
1
(𝑥 2 (𝑡) + 𝑦 2 (𝑡) + 𝑧 2 (𝑡))
1⁄2
|
𝑎
since
𝑑 2
2
2
𝑑
1
1 𝑑𝑡 [𝑥 (𝑡) + 𝑦 (𝑡) + 𝑧 (𝑡)]
[
]=−
𝑑𝑡 (𝑥 2 (𝑡) + 𝑦 2 (𝑡) + 𝑧 2 (𝑡))1⁄2
2 (𝑥 2 (𝑡) + 𝑦 2 (𝑡) + 𝑧 2 (𝑡))3⁄2
Potential Theory - Problems
=−
𝑥(𝑡)𝑥 ′ (𝑡) + 𝑦(𝑡)𝑦 ′ (𝑡) + 𝑧(𝑡)𝑧 ′ (𝑡)
(𝑥 2 (𝑡) + 𝑦 2 (𝑡) + 𝑧 2 (𝑡))
3⁄2
.
So the line integral depends only on the endpoints of 𝐶, which are independent of its actual
path.
P5 If 𝑢(𝑧) is harmonic and positive in the unit disk and continuous in the closed unit disk
show that
1−𝑟
1+𝑟
≤ 𝑢(𝑟𝑒 𝑖𝜃 ) ≤ 𝑢(0)
.
1+𝑟
1−𝑟
Hint: first find upper and lower bounds for the Poisson kernel of the unit disk.
𝑢(0)
As shown in the proof of Theorem 6.2.2, for the unit disk {𝑧: |𝑧| < 𝑅 = 1}
(1 − 𝑟 2 )𝑢(𝑒 𝑖𝜙 )
1 2𝜋
𝑢(𝑟𝑒 ) =
∫
𝑑𝜙
2𝜋 0 1 − 2𝑟 cos(𝜃 − 𝜙) + 𝑟 2
𝑖𝜃
1 2𝜋
1 − 𝑟2
=
∫ (
) 𝑢(𝑒 𝑖𝜙 ) 𝑑𝜙
2
2𝜋 0
1 − 2𝑟 cos(𝑡) + 𝑟
with 𝑡 = 𝜃 − 𝜙.
Hence
𝑢(0) =
1 2𝜋
∫ 𝑢(𝑒 𝑖𝜙 ) 𝑑𝜙.
2𝜋 0
Since both the Poisson kernel and 𝑢(𝑒 𝑖𝜙 ) are positive
1 2𝜋
1 − 𝑟2
∫ min
𝑢(𝑒 𝑖𝜙 ) 𝑑𝜙 ≤ 𝑢(𝑟𝑒 𝑖𝜃 )
2
0≤𝑡≤2𝜋
2𝜋 0
1 − 2𝑟 cos(𝑡) + 𝑟
≤
1 2𝜋
1 − 𝑟2
∫ max
𝑢(𝑒 𝑖𝜙 ) 𝑑𝜙.
2𝜋 0 0≤𝑡≤2𝜋 1 − 2𝑟 cos(𝑡) + 𝑟 2
Since
1 − 𝑟2
1 − 𝑟2
1 − 𝑟2
=
=
0≤𝑡≤2𝜋 1 − 2𝑟 cos(𝑡) + 𝑟 2
1 + 2𝑟 + 𝑟 2 (1 + 𝑟)2
min
and
1 − 𝑟2
1 − 𝑟2
1 − 𝑟2
=
=
.
0≤𝑡≤2𝜋 1 − 2𝑟 cos(𝑡) + 𝑟 2
1 − 2𝑟 + 𝑟 2 (1 − 𝑟)2
max
We have
1 − 𝑟 2 1 2𝜋
1 − 𝑟 2 1 2𝜋
𝑖𝜙
𝑖𝜃
∫ 𝑢(𝑒 ) 𝑑𝜙 ≤ 𝑢(𝑟𝑒 ) ≤
∫ 𝑢(𝑒 𝑖𝜙 ) 𝑑𝜙.
(1 + 𝑟)2 2𝜋 0
(1 − 𝑟)2 2𝜋 0
Which gives, since 𝑟 < 1
1−𝑟
1+𝑟
𝑢(0) ≤ 𝑢(𝑟𝑒 𝑖𝜃 ) ≤
𝑢(0).
1+𝑟
1−𝑟
Potential Theory - Problems
P6 Let 𝐷 be a domain in the plane bounded by a simple closed curve 𝐶. Let 𝑔 be a realvalued smooth function defined on 𝐶, and let 𝐹 be the family of real-valued functions with
smooth derivatives on 𝐷 and 𝐶 and boundary values 𝑔. If
inf {∬ (𝑢𝑥2 + 𝑢𝑦2 ) 𝑑𝐴: 𝑢 ∈ 𝐹}
𝐷
is attained by 𝑈 ∈ 𝐹, show that 𝑈 is the solution of the Dirichlet problem for 𝐷 with boundary
values 𝑔.
A long and complicated proof is the required answer. Not worth wasting time on.
P7
Find a univalent mapping of the half disk {𝑧: |𝑧| < 1, Im(𝑧) > 0} onto the unit disk.
Let 𝐷 = {𝑧: |𝑧| < 1, Im(𝑧) > 0}. According to the Riemann mapping theorem, there exists a
simple function 𝑤 = 𝑓(𝑧) which maps 𝐷 onto the open unit disk |𝑤| < 1 in the 𝑤-plane.
The boundary of the domain 𝐷 consists of two circular arcs
(i)
the upper semicircle {𝑧: |𝑧| = 1, Im(𝑧) > 0};
(ii)
the 𝑥-axis as a circle through 𝑧 = 0 and 𝑧 = ∞.
Consider the linear fractional transformation
1+𝑧
.
1−𝑧
The images of the points −1, 𝑖, and 1 under this transformation are
𝑤=
𝑤(−1) = 0, 𝑤(𝑖) = 𝑖,
and 𝑤(1) = ∞.
Therefore, the upper semicircle is mapped to the positive imaginary axis.
The images of the points −1,0, and 1 under this transformation are
𝑤(−1) = 0, 𝑤(0) = 1,
and 𝑤(1) = ∞.
Therefore, the 𝑥-axis is mapped to the positive real axis.
The test point 𝑧 = 1⁄2 is mapped to
𝑤(1 + 𝑖) =
1 + 𝑖 ⁄2 3 4
= + 𝑖.
1 − 𝑖 ⁄2 5 5
Hence, 𝑤 maps 𝐷 onto the first quadrant in the 𝑤-plane.
The mapping 𝑊 = 𝑤 2 maps the first quadrant in the 𝑤-plane onto the upper half 𝑊-plane.
The mapping
𝑖 − 𝑊 −1(𝑊 − 𝑖)
=
𝑖+𝑊
𝑊+𝑖
maps the the upper half 𝑊-plane onto the unit disk in the 𝜉-plane.
𝜉=
Forming the composition of these mappings gives
Potential Theory - Problems
2
𝜉=
𝑖−𝑊 𝑖−𝑤
=
=
𝑖 + 𝑊 𝑖 + 𝑤2
1+𝑧 2
𝑖 − (1 − 𝑧)
1+𝑧 2
𝑖 + (1 − 𝑧)
=
𝑖(1 − 𝑧)2 − (1 + 𝑧)2
1 − 2𝑖𝑧 + 𝑧 2
=
𝑖
.
𝑖(1 − 𝑧)2 + (1 + 𝑧)2
1 + 2𝑖𝑧 + 𝑧 2
P8 Let 𝑓(𝑧) be univalent in {|𝑧| < 1} with 𝑓(0) = 0. Show that one can find 𝑔(𝑧) univalent
in the unit disk and satisfying
2
(𝑔(𝑧)) = 𝑓(𝑧 2 )
for |𝑧| < 1.
The solution is too complex to be worth bothering with.
P9 Let 𝑓(𝑧) be univalent and map the annulus {1 < |𝑧| < 𝑅1 } univalently onto the annulus
{1 < |𝑤| < 𝑅2 } so that the inner and outer boundaries correspond. Show that 𝑓(𝑧) = 𝐴𝑧 for
some constant 𝐴 with |𝐴| = 1, and so 𝑅1 = 𝑅2 . What happens when {|𝑧| = 𝑅1 } maps to
{|𝑤| = 1} and {|𝑧| = 1}?
The solution is too complex to be worth bothering with.
P10 Find a function which maps the strip 𝑆 = {𝑧: 0 < Im(𝑧) < 𝜋} into the interior of the
unit circle. Write down a solution of the Dirichlet problem for this strip.
The mapping
𝑊 = 𝑒𝑧
maps 𝑆 onto the upper half 𝑊-plane, and the mapping
𝑖−𝑊
𝑖+𝑊
maps the upper half 𝑊-plane onto the unit disk in the 𝑤-plane.
𝑤=
Therefore, the mapping
𝑖 − 𝑊 𝑖 − 𝑒𝑧
𝑤=
=
= 𝑓(𝑧)
𝑖 + 𝑊 𝑖 + 𝑒𝑧
maps 𝑆 onto the interior of the unit circle.
The Poisson integral formula gives a solution for the unit disk, from which a solution in 𝑆 can
be obtained.
P11 Find a function which maps the strip 𝑆 = {𝑧: 0 ≤ Re(𝑧) ≤ 𝜋, Im(𝑧) ≥ 0} onto the upper
half plane so that the origin maps into the origin, and 𝜋 maps into 1.
Potential Theory - Problems
The strip 𝑆 is a polygonal region with interior angles 𝛼1 = 𝛼2 = 𝜋⁄2 and vertices 𝑧1 = 0 and
𝑧2 = 𝜋.
Applying the Schwarz-Christoffel transformation gives
𝑑𝑧
𝐴
= 𝐴(𝑤 − 0)(𝜋⁄2⁄𝜋)−1 × (𝑤 − 1)(𝜋⁄2⁄𝜋)−1 = 𝐴𝑤 −1⁄2 (𝑤 − 1)−1⁄2 =
.
1⁄2
𝑑𝑤
(𝑤(𝑤 − 1))
Hence
1
𝑧 = 𝐴∫
(𝑤(𝑤 − 1))
= 𝐴∫
1⁄2
𝑑𝑤 = 𝐴 ∫
1
1 2
1 2
((𝑤 − 2) − (2) )
= 𝐴 log (𝑤 −
1⁄2
(𝑤 2
1
𝑑𝑤
− 𝑤)1⁄2
𝑑𝑤
1
+ (𝑤 2 − 𝑤)1⁄2 ).
2
This isn't very helpful.
Further manipulation gives
𝑑𝑧
𝐴
𝐴
1
=
=
.
⁄
1
2
𝑑𝑤 (−1𝑤(1 − 𝑤))
𝑖 (𝑤(1 − 𝑤))1⁄2
Let 𝐵 = 𝐴⁄𝑖 , then
𝑧 = 𝐵∫
1
(𝑤(1 − 𝑤))
= 𝐵∫
1⁄2
𝑑𝑤 = 𝐵 ∫
1
1 2
1 2
((2) − (𝑤 − 2) )
1⁄2
1
𝑑𝑤
(𝑤 − 𝑤 2 )1⁄2
𝑑𝑤
1
𝑤−2
= 𝐵 sin−1 (
) + 𝐶.
1
2
Now 𝑧1 = 0 when 𝑤1 = 0, hence
𝜋
0 = 𝐵 sin−1(−1) + 𝐶 = − 𝐵 + 𝐶.
2
And 𝑧2 = 𝜋 when 𝑤2 = 1, hence
𝜋
𝜋 = 𝐵 sin−1(1) + 𝐶 = 𝐵 + 𝐶.
2
Subtracting gives
𝐵=1
𝜋.
𝜋 = 𝜋𝐵 ⇒ {
𝐶=
2
Hence
Potential Theory - Problems
𝑧 = sin
−1
(
1
2 ) + 𝜋.
1
2
2
𝑤−
Rearranging gives
1
𝜋
1
1
1 1
𝑧
𝑤 = sin (𝑧 − ) + = − cos 𝑧 + = (1 − cos 𝑧) = sin2 ( ).
2
2
2
2
2 2
2
```