Digital Text Book :POTENTIAL THEORY AND ELLIPTIC PARTIAL DIFFERENTIAL EQUATIONS

POTENTIAL THEORY AND
ELLIPTIC PARTIAL
DIFFERENTIAL EQUATIONS
Study Results
BASIL ROY IDICULLA
M O U N T T A B O R T R A I N I N G C O L L E G E , P A T H A N A P U R A M

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Chapter Title Page Number
0 INTRODUCTION 2
1 PRELIMINARIES 3
2 POTENTIAL THEORY AND ELLIPTIC
PARTIAL DIFFERENTIAL EQUATIONS
11
3 APPLICATIONS 31
CONTENTS

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INTRODUCTION
Differential equations serve as an effective tool to study the changes
in the mathematical as well as the physical world. Partial differential
equation of second order arises frequently in mathematical physics. It is of
this reason that the study of such equations are of great value. This project
work is targeting at ”Potential theory and Elliptic partial differential
equations”. Laplace’s equation is a second order partial differential
equation which is elliptic in nature.
This project includes three chapters. The first chapter includes some
important definitions and necessary theorems which are needed for the
following chapters. Second chapter consist potential theory of Laplace’s
equation, solutions of Laplace’s equation, fundamental solution of Green’s
function for Laplace’s equation and examples. Third chapter highlights
some important applications of Laplace’s equation in mathematics as well
as in Physics such as Dirichlet problem for the upper half plane, the
velocity potential for an irrotational flow of an incompressible fluid etc.

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CHAPTER-1
PRELIMINARIES
Definition 1:1
The Laplace’s equation in two dimension is, ∇2
𝑈 =
𝜕2 𝑢
𝜕𝑥2
+
𝜕2 𝑢
𝜕𝑦2
=0
That is, ∇2
𝑢 = 𝑢 𝑥𝑥 + 𝑢 𝑦𝑦 = 0
A solution of two dimensional Laplace’s equation is called two
dimensional harmonic function.
Definition 1:2
The m- dimensional case of Laplace’s equation is called potential
equation. It is denoted by ∆ 𝑚 𝑢 = 𝑢 𝑥
𝛼
𝑥
𝛼
= 0
That is, 𝑢 𝑥
1
𝑥
1
+ 𝑢 𝑥
2
𝑥
2
… 𝑢 𝑥
𝑚
𝑥
𝑚
= 0
Definition 1:3
Any problem in which we are required to find a function u is called a
boundary value problem for Laplace’s equation. Let D be the interior of a

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simple, closed, smooth curve B and f be a conscious function defined on
the boundary B
The problem of finding a harmonic function 𝑢 ( 𝑥, 𝑦) in D such that
it coincides with f on the boundary B is called the first boundary value
problem or Dirichlet problem.
The problem of finding a function 𝑢 ( 𝑥, 𝑦) such that it is harmonic in
D and satisfies
𝜕𝑢
𝜕𝑛
=f( 𝑠) on B,(where
𝜕
𝜕𝑛
is the directional derivative along
the outward normal) with the condition ∫ 𝑓( 𝑠) 𝑑𝑠 = 0𝐵
is called the
second boundary value problem or Neumann problem.
The problem of finding a function 𝑢 ( 𝑥, 𝑦) which is harmonic in D
and satisfies the condition
𝜕𝑢
𝜕𝑛
+ℎ( 𝑠) 𝑢( 𝑠)=0 on B where ℎ( 𝑠) ≥ 0 ,
And ℎ( 𝑠) ≢0is called the third boundary value problem or Robin
problem.
The fourth boundary value problem involves finding a function
𝑢 ( 𝑥, 𝑦) which is harmonic in D and satisfies the boundary conditions of
different types on different portions of the boundary B.
For example,

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𝑢 = f1( 𝑠)on B1 ,
𝜕𝑢
𝜕𝑛
= f2( 𝑠) on B2,where B1UB2 = B
Definition 1:4
Let A 𝑢 𝑥 𝑥 + 𝐵 𝑢 𝑦𝑦 + 𝐶 𝑢 𝑦 𝑦 = 𝐹(𝑥, 𝑦, 𝑢, 𝑢 𝑥, 𝑢 𝑦 ) ……………….(1)
Where A( 𝑥, 𝑦) , B( 𝑥, 𝑦) and C( 𝑥, 𝑦) are functions of 𝑥 and 𝑦
and⎾0 be a curve in the 𝑥, 𝑦 – plane. The problem of finding the solution
𝑢 ( 𝑥, 𝑦) of the partial differential equation (1) in the neighbourhood of ⎾0
satisfying the following conditions.
𝑢 = 𝑓( 𝑠) ……………….(2)
𝜕𝑢
𝜕𝑛
= 𝑔( 𝑠) ……………….(3)
On ⎾0( 𝑠) is called a Cauchy problem. The conditions (2) and (3)
are called the Cauchy conditions.
Example: The problem of vibrations of an infinite string is a Cauchy
problem.
Definition 1:5
If U ( 𝑥, 𝑦) and V( 𝑥, 𝑦) are functions defined inside and on the
boundary B of the closed region D, then

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∫ [
𝜕𝑈
𝜕𝑥
+
𝜕𝑉
𝜕𝑦
] 𝑑𝑆 = ∫(𝑈𝑑𝑦 − 𝑉𝑑𝑥)
𝐵𝐷
Let
U=Ψ
𝜕∅
𝜕𝑥
, V=Ψ
𝜕∅
𝜕𝑦
Then
∫ (Ψ𝑥∅ 𝑥 + Ψ∅ 𝑥𝑥 + Ψ𝑥∅ 𝑥 + Ψ∅ 𝑦𝑦)𝐷
𝑑𝑆 =
∫ Ψ
𝜕∅
𝜕𝑛𝐵
𝑑𝑠…….(4)
On interchanging 𝜙 and Ψ and subtracting one from the other ,we
get
∫ (𝚿𝛁 𝟐
∅ − ∅𝛁 𝟐
𝚿)𝑫
𝒅𝑺 = ∫ (𝚿
𝝏∅
𝝏𝒏
− ∅
𝝏𝚿
𝝏𝒏
)𝑩
𝒅𝒔
……….(5)
The identities (4) and (5) are called Green,s identities
Theorem 1:1 . Fourier integral theorem
Let f ( 𝑥) be a continuous and absolutely integrable function in
(−∞, ∞). Then 𝔉( 𝑓) = F( 𝛼) called the Fourier transform of f ( 𝑥) is
defined as

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F( 𝛼) = 𝔉( 𝑓) =
1
√2𝜋
∫ 𝑓( 𝑥) 𝑒 𝑖𝛼𝑎
𝑑𝑥
∞
−∞
Then
f( 𝑥)=𝔉−1( 𝐹) =
1
√2𝜋
∫ 𝐹( 𝛼) 𝑒−𝑖𝛼𝑥
𝑑𝛼
∞
−∞
Is called the inverse Fourier transform of F( 𝛼)
Example 1 : Find the Fourier transform of
f( 𝑥)=𝑒−| 𝑥|
We have
F( 𝛼) =
1
√2𝜋
∫ 𝑒−| 𝑥|
𝑒 𝑖𝛼𝑥
𝑑𝑥
∞
−∞
=
1
√2𝜋
(∫ 𝑒 𝑥 𝑒 𝑖𝛼𝑥 𝑑𝑥 + ∫ 𝑒−𝑥 𝑒 𝑖𝛼𝑥 𝑑𝑥
∞
0
0
−∞
)
=
1
√2𝜋
[
1
1+𝑖𝛼
+
1
1−𝑖𝛼
]
=√
2
𝜋
[
1
1+𝛼2
]
Theorem 1:2 . Convolution Theorem
Let f( 𝑥) and g( 𝑥) possess Fourier transforms. Define
( 𝑓 ∗ 𝑔)( 𝑥) =
1
√2𝜋
∫ 𝑓( 𝑥 − 𝜉) 𝑔( 𝜉) 𝑑𝜉
∞
−∞
,

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Called the convolution of the function f and g over the interval (−∞,∞) .
𝔉( 𝑓 ∗ 𝑔) = 𝔉( 𝑓) 𝔉( 𝑔)
Theorem 1:3 . Maximum Principle
Suppose that 𝑢( 𝑥, 𝑦) is harmonic in a bounded domain D and
continuous in
𝐷̅ = 𝐷 ∪ 𝐵. Then u attain its maximum on the boundary B of D.
Theorem 1:4 . Minimum Principles
Suppose that 𝑢( 𝑥 , 𝑦) is harmonic in a bounded domain D and
continuous on 𝐷̅ = 𝐷 ∪ 𝐵. Then u attains its minimum on the boundary B
of D
Theorem 1:5 . Uniqueness Theorem
The solution of the Dirichlet problem, if it exists, is unique
Proof
Suppose 𝑢1( 𝑥 , 𝑦) and 𝑢2( 𝑥 , 𝑦) are two solutions of the Dirichlet
problem. That is ∇2
𝑢1 = 0 in D and 𝑢1 = 𝑓( 𝑠) on B
∇2
𝑢2 = 0in D and 𝑢2 = 𝑓( 𝑠) on B
As 𝑢1 and 𝑢2 are harmonic in D, ( 𝑢1 − 𝑢2) is also harmonic in D.

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However ( 𝑢1 − 𝑢2) =0 on B
By the maximum and minimum principle, ( 𝑢1 − 𝑢2) ≡0 in D.Hence the
theorem.
Theorem 1:6 .
The necessary condition for the Neumann problem
Let 𝑢 be a solution of the Neumann problem
∇2
𝑢 = 0in D,And
𝜕𝑢
𝜕𝑛
= 𝑓( 𝑠) on B
Then
∫ 𝑓( 𝑠) 𝑑𝑠
𝐵
= 0
Proof
Put Ψ=1 and 𝜙 =𝑢 in. Then we get
∫ 𝑓( 𝑠) 𝑑𝑠
𝐵
= 0
Thus f( 𝑠) cannot be arbitrarily prescribed as it has to satisfy the above
condition.

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Theorem 1:7 . Gauss divergence theorem
If 𝐹⃗ is a vector point function, finite and differentiable in a reagion
V of space bounded by a closed surfaces S, then the surface integral of the
normal component of𝐹⃗taken over S is equal to the volume integral of the
divergence of 𝐹⃗ taken over V
That is, ∫ 𝐹⃗ 𝑛⃗⃗ 𝑑𝑠 = ∭ (∇.𝐹⃗)𝑣𝑠
𝑑𝑣
Where n is the unit vector in the direction positive (outward drawn)
normal to S.

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CHAPTER 2
POTENTIAL THEORY AND ELLIPTIC
PARTIAL DIFFERENTIAL EQUATIONS
Definition 2:1
The general linear homogeneous second order partial differential
equation in m-space variable𝑥1 𝑥2 …… .. 𝑥 𝑚 is
Lu =𝑎 𝛼𝛽 𝑢 𝑥𝛼𝑥𝛽 + 𝑏 𝛼 𝑢 𝑥𝛼 + 𝑐𝑢 = 0, 𝛼, 𝛽 = 1,2 …… 𝑚……………….(1)
Where the coefficients 𝑎 𝛼𝛽,𝑏 𝛼 and c are continuous functions of
the independent variables 𝑥1, 𝑥2, …… .. 𝑥 𝑚and 𝑎 𝛼𝛽. This equation is said to
be elliptic in a domain D of m-dimensional space, when the quadratic
form.
Q( 𝜆)=𝑎 𝛼𝛽 𝜆 𝛼 𝜆 𝛽 ……………….(2)
Can be expressed as the sum of squares with coefficients of the same
sign, or equivalently Q( 𝜆) is either positive or negative definite in D. The
simplest case is that of the Laplaces equation or potential equation.
∆ 𝑚 𝑢=𝑢 𝑥
𝛼
𝑥
𝛼
=0 ……………….(3)

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Definition 2:2
A function 𝑢( 𝑥) is called harmonic in D, if 𝑢( 𝑥) ∈ 𝐶°
In D+𝜕𝐷 ∈ 𝐶2
in D and ∆ 𝑚 𝑢 = 0 in D
General Solution of the potential equation
In the case of two or three variables, the general solution of the potential
equation can be easily obtained. For m=2( 𝑥1 = 𝑥, 𝑥2 = 𝑦), this is the real
or imaginary part of any analytic function of the complex variable +𝑖𝑦 .
For m=3 ( 𝑥1 = 𝑥, 𝑥2 = 𝑦, 𝑥3 = 𝑧), consider an arbitrary function p( 𝑤, 𝑡)
analytic in the complex variable w for fixed real t. Then, for arbitrary
values of t, both the real and imaginary parts of the function.
𝑢 = 𝑝( 𝑍 + 𝑖𝑥 𝑐𝑜𝑠𝑡 + 𝑖𝑦 sin 𝑡, 𝑡) ……………….(4)
Of the real variables 𝑥, 𝑦, 𝑧 are solutions of the equation ∆ 𝑢= 0. Further
solutions may now be obtained by superposition.
𝑢 = ∫ 𝑝( 𝑧 + 𝑖𝑥 cos 𝑡 + 𝑖𝑦 sin 𝑡, 𝑡)
𝑏
𝑎
dt……………….(5)
If 𝑢( 𝑥, 𝑦) is a solution of Laplace’s equation in a domain D of the ( 𝑥, 𝑦)
plane, the function
𝜐( 𝑥, 𝑦) = 𝑢 [
𝑥
𝑟−2
,
𝑦
𝑟2
] , 𝑟2
= 𝑥2
+ 𝑦2
……………….(6)

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Also satisfies the potential equation and is regular in the domain 𝐷1
obtained from D by inversion with respect to the unit circle.
In general in m-dimensions, if u (𝑥1, 𝑥2 …… . 𝑥 𝑚) satisfies the potential
equation in a bounded domain D, then.
𝜐 =
1
𝑟 𝑚−2
𝑢 [
𝑥1
𝑟2
,
𝑥2
𝑟2
, ……
𝑥 𝑚
𝑟2
], 𝑟2 = 𝑥 𝛼 𝑥 𝛼 ……………….(7)
Also satisfies the potential equation and is regular in the region 𝐷1
obtained from D by inversion with respect to the m-dimensional unit
sphere. Therefore, except for the factor 𝑟2−𝑚
, the harmonic character of a
function is invariant under inversions with respect to spheres.
Examples 2:1
Dirichlet problem for a circle in the 𝒙, 𝒚 plane.
Let the circle C be given by | 𝜉|=R, where 𝜉 = 𝑥 + 𝑖𝑦. The problem is to
find 𝑢( 𝑥, 𝑦) such that
Δ2 𝑢 = 𝑢 𝑥𝑥 + 𝑢 𝑦𝑦 = 0 ……………….(8)
Where 𝑢 = 𝑓( 𝜃) on C
Where 𝜃 is the angular coordinate on C;
That is ,𝜉 = 𝑅𝑒 𝑖𝜃
on C

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Fig 2.1.𝑤and
𝑅2
𝑤̅
are inverse points with respect to C:| 𝑍| = 𝑅
We shall present here a constructional proof for the existence of the
solution.
That is, we shall derive an expression for the solution. Let F ( 𝜉) be an
analytic function in the region enclosed by C, such that the real part of
F( 𝜉) on | 𝜉| =f( 𝜃). Let w be a complex number in the region. The inverse
point of w with respect to C is 𝑅2/𝑤̅, which lies out side C. Here 𝑤̅
denotes the complex conjugate of w. According to the Cauchy integral
formula.
𝐹( 𝑤) =
1
2𝜋𝑖
∫
𝐹( 𝜉)
𝜉 − 𝑤
𝑑𝜉
𝑐
0
W
A
𝑅2/𝑊̅
B

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0 =
1
2𝜋𝑖
∫
𝐹( 𝜉)
𝜉 − 𝑅2/𝑤̅
𝑑𝜉
𝐶
Subtracting
𝐹( 𝑤) =
1
2𝜋𝑖
∫
𝐹( 𝜉)( 𝑅2
− 𝑤𝑤̅)
𝜉( 𝑅2 + 𝑤𝑤̅) − 𝑤𝜉2 − 𝑤𝑅2̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅
𝑑𝜉
𝐶
As 𝜉 lies on C and w inside C, set
𝜉 = 𝑅𝑒 𝑖𝜃,
𝑤 = 𝑟𝑒 𝑖𝜙,
, 𝑟 < 𝑅
Then,
𝐹(𝑟𝑒 𝑖𝜙
) =
1
2𝜋
∫ 𝐹(𝑟𝑒 𝑖𝜃
)
𝑅2
− 𝑟2
𝑅2 + 𝑟2 − 2𝑟𝑅 cos(𝜃 − 𝜙)
𝑑𝜃
2𝜋
0
Taking the real part on both sides, we get
𝑢( 𝑥, 𝑦) =
𝑅2−𝑟2
2𝜋
∫
𝑓(𝜃)𝑑𝜃
𝑅2 +𝑟2 −2𝑟 𝑅 cos(𝜃−𝜙)
2𝜋
0
……………….(9)
Where 𝑟2
= 𝑥2
+ 𝑦2
, tan𝜙 =𝑦/𝑥
Equation (9) is called Poisson’s integral formula in two dimensions. This
completes the proof of the existence of the solution.

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Definition 2:3
In 𝑅 𝑚
the solutions 𝑣 (𝑟) of the potential equation Δ 𝑚 𝑢 = 0, which
depend only on the distance (𝑟 ≠ 0) of a point x from a fixed point a, say,
are given by the equation.
𝑑2 𝑣
𝑑𝑟2
+
𝑚−1
𝑟
= 0
𝑟 = |𝑥 − 𝑎| = √(𝑥 𝛼 − 𝑎 𝛼)(𝑥 𝛼 − 𝑎 𝛼)
} ……….(10)
This equation has solutions
𝑣( 𝑟) = 𝑐1 + 𝑐2 𝑟2−𝑚
, 𝑚 > 2
= 𝑐1 + 𝑐2 log 𝑟, 𝑚 = 2
} ……………….(11)
Where 𝑐1 and 𝑐2 are arbitrary constants. These solutions exhibit the so
called characteristic singularity at r=0. We call
𝑠( 𝑎, 𝑋)
=
𝐿
( 𝑚−2) 𝑤 𝑚
| 𝑎 − 𝑋|2−𝑚, 𝑚 > 2
=
−1
2𝜋
𝑙𝑜𝑔 | 𝑎 − 𝑋 | , 𝑚 = 2
}……………….(12)
The singularity function for Δ 𝑚 𝑢 = 0, where 𝑤 𝑚 is the surface area of the
unit sphere in m dimensions given by

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𝑤 𝑚 = 2(√ 𝜋) 𝑚
∣−
Γ(𝑚 ∕ 2) ……………….(13)
Note 2.1
s(𝑎, 𝑥) has the property that s∈ 𝐶∞ and ∆ 𝑚 𝑠 = 0 for 𝑥 ≠ 𝑎, with a
singularity at x = 𝑎. For m=3, s (𝑎, 𝑥) corresponds physically to the
gravitational potential at the point x of a unit mass concentrated at the
point a.
Definition 2:4
Every solution of the potential equation ∆ 𝑚 𝑢 = 0 in D of the form
𝛾( 𝑎, 𝑥) = 𝑠( 𝑎, 𝑥) + 𝜙( 𝑋), 𝑎 ∊ 𝐷 ……………….(14)
Where 𝜙( 𝑋) ∊ 𝐶2
in D and 𝜙( 𝑋) ∊ 𝐶1
in the closed region D+ 𝜕 D and
∆ 𝑚 𝜙 = 0 in D, is called a fundamental solution relative to D with a
Singularity at a.
Definition 2:5
Green’s formulae
Let 𝑢 and 𝑣 be two functions defined in a domain D bounded by a
piecewise smooth surface 𝜕D. Then
∫ ∇ 𝑚 𝑢∇ 𝑚 𝜐𝑑𝑥 + ∫ 𝜐Δ 𝑚 𝑢𝑑𝑥 = ∫ 𝜐
𝜕𝑢
𝜕𝑣𝜕𝐷𝐷𝐷
𝑑𝑆 ……………….(15)

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And
∫ ( 𝑢∆ 𝑚 𝜐 − 𝜐Δ 𝑚 𝑢) 𝑑𝑥 = ∫ 〈 𝑢
𝜕𝜐
𝜕𝑉
− 𝜐
𝜕𝜐
𝜕𝑉
〉 𝑑𝑆𝜕𝑑𝐷
..…………….(16)
Where
𝜕
𝜕𝑉
represents differentiation in the direction of the outward drawn
normal to S.
Note 2.2
If △m u=0 and v=1 we obtained from (15)
𝜕D⨜uvds=0 …………(17)
where the suffix v denotes normal derivative to the surface 𝜕D. That
means, if a function satisfies ∆𝑚 𝑢 = 0 in a bounded region D and is
continuously differentiable in D+𝜕𝐷 thenthe surface integral of its normal
derivative is zero. Then we get a Neumann problem
That is,
Green’s formulas undergo an important modification if for v we substitute
a function having the characteristic singularity of potential function at an
interior point. Let us choose v to be a fundamental solution
V=ɣ(a,x)= s(a,x) +ɸ(x)

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Theorem 2.1
If u(x)єC1 in D + 𝜕D and єC2 in D and u(x ) is a solution of the potential
equation then for an arbitrarypoint aє D
u(a)=∫ 𝛾( 𝑎, 𝑋) 𝑢( 𝑥) − 𝑢(𝑥)𝛾𝑣(𝑎, 𝑥)𝑑𝑆𝜕𝐷
Proof
Since D is open ,the ball V; |a-x|⩽𝛒 about any point a in D is contained in
D for sufficiently small 𝛒. Since ɣ(a,x) becomes singular at x= a, We
remove V from D and apply Green’s formula (16) for ɣ and u on D-V.
We obtain
∫( 𝛾∆ 𝑚 𝑢 − 𝑢∆ 𝑚 𝛾𝛾) 𝑑𝑋 = ∫( 𝛾𝑢 𝑣 − 𝑢𝛾𝑣) 𝑑𝑆 + ∫ (𝛾𝑢 𝑣 − 𝑢𝛾𝑣)𝑑𝑆
| 𝑋−𝑎|=𝜌𝜕𝐷𝐷−𝑣
Since △m ɣ=0, △mu=0,in D-V in the left hand side is zero
Consider the integral∫ (𝛾𝑢 𝑣 − 𝑢𝛾𝑣)𝑑𝑆| 𝑋−𝑎|=𝜌
and using the fact ɣ(a,x)=s(a,x)+ɸ(x)
Then equation (20) expressed as a sum of two integrals I1 and I2
I1 =⨜(ɸuv-uɸv)ds

20 | P a g e
At any point on the surface of V, x=a +𝛒v and dS=𝛒m-1dw
Therefore
|X-a|= 𝛒⨜suv ds =𝛒m-1
|v|=1⨜s(a.a+𝛒v) uv (a+𝛒v)dw
=
𝜌 𝑚−1
𝑚−2𝑤 𝑚
∫ | 𝑎 − 𝑋|| 𝑣|=1
2-muv(a+𝛒v)dw
=
ρ 𝑚−1ρ2−𝑚
𝑚−2𝑤 𝑚
∫ 𝑢| 𝑣|=1 v(a+ ρv)dw
=
ρ
(𝑚−2)𝑤 𝑚
∫ 𝑢|𝑣| v(a+ ρv)dw
This tends to zero as ρ 0
Further,
-∫ 𝑠𝑣 𝑢 𝑑𝑆 = −ρ| 𝑋−𝑎|=ρ
m-1
∫ 𝑢(𝑎 + 𝑣ρ)| 𝑣|=1
ρv(a,a+ρv)dw
=-
−ρ 𝑚−1
(𝑚−2)𝑤 𝑚
∫ 𝑢(𝑎 + 𝑣ρ)
𝜕
𝜕𝑣| 𝑣|=1
ρ2-mdw
Here
𝜕
𝜕𝑣
= −
𝜕
𝜕ρ
. Therefore
-∫ 𝑠𝑣 𝑢 𝑑𝑆| 𝑋−𝑎|=ρ
=
ρ 𝑚−1
(𝑚−2)𝑤 𝑚
∫ 𝑢(𝑎 + 𝑣ρ)
𝜕
𝜕𝑠
(| 𝑣|
ρ2−𝑚
) dw
=
ρ 𝑚−1
(𝑚−2)𝑤 𝑚
∫ 𝑢(𝑎 + 𝑣ρ)(2 − 𝑚)ρ1−𝑚
| 𝑣|=1
dw

21 | P a g e
=
−1
𝑤 𝑚
∫ 𝑢( 𝑎 + 𝑣ρ) 𝑑𝑤| 𝑣|=1
As ρ 0, This is tends to –u(a). This proves the results (19)
That is, u(a)= ∫ {𝛾( 𝑎, 𝑋) 𝑢 𝑣( 𝑋) − 𝑢( 𝑋) 𝛾𝑣( 𝑎, 𝑋) 𝑑𝑆𝜕𝐷
This formula holds for any a𝞊 D
Note 2.3
If instead a line on 𝜕D, then we get
u(a) = 2∫ {𝛾(𝑎, 𝑋)𝑢𝜕𝐷 v(X) – u(X)𝛾𝑣(𝑎, 𝑋)} dS
Where as if a lies outside D, then
∫ {𝛾( 𝑎, 𝑋) 𝑢 𝑣( 𝑋) − 𝑢(𝑋)𝛾𝑣(𝑎, 𝑋)}𝜕𝐷
dS =0
In particular in 3-dimensions m=3, if ∅( 𝑥) = 0, hen we have for a∈ 𝐷
u(a)=
1
4
∫ {
𝜕𝑢
𝜕𝑣
− 𝑢
𝜕
𝜕𝑣
(
1
𝑟
)} 𝑑𝑆𝜕𝐷
u can be considered as the potential of adistribution consisting of a
single layer surface distribution of density (
1
4𝜋
)
𝜕𝑢
𝜕𝑣
𝑎𝑛𝑑 double layer
dipole distribution of density
−𝑢
4𝜋
on the boundary surface 𝜕𝐷. We had
taken u be an arbitrary function, then for a∈ 𝐷 the formula .

22 | P a g e
u(a)=∫ { 𝛾( 𝑎, 𝑥) 𝑢 𝑣( 𝑋) − 𝑢(𝑋)𝛾𝑣(𝑎, 𝑋)} 𝑑𝑆𝜕𝐷
is modified as
∫ 𝛾∆ 𝑚𝐷
𝑢𝑑𝑋 + ∫ (𝛾
𝜕𝑢
𝜕𝑣
− 𝑢
𝜕𝛾
𝜕𝑣
) 𝑑𝑆𝜕𝐷
Formally, using the Greens formula of the equation.
∫( 𝑢∆ 𝑚 𝑣 − 𝑣∆ 𝑚 𝑢) 𝑑𝑋 + ∫ (𝛾
𝜕𝜐
𝜕𝑣
− 𝑣
𝜕𝑢
𝜕𝜐
) 𝑑𝑆
𝜕𝐷𝐷
For the right hand side, we have
u(a)0= − ∫ 𝑢∆ 𝑚 𝛾 𝑑𝑋 ≡ −∆ 𝑚 𝛾(𝑢( 𝑋))𝐷
Definition 2.6
We define a grees function G(a,X) of the differential expression ∆ 𝑚u for
the region D as a specific fundamental solution of ∆ 𝑚u=0, depending on
the parameter a, of the form
G(a, X) =G(a1, a 2, ….,a n,x1,x2,xn)
=s(a,X)+ 𝜙

23 | P a g e
Which vanishes at all points x on 𝟃D and for which the component 𝟇 is
continuous in D + 𝟃D and harmonic in D. Assuming the existence of a
Green’s functions G, we replace 𝛾 in equation
u(a)= ∫ { 𝛾( 𝑎, 𝑋) 𝑢 𝑣 ( 𝑋) − 𝑢( 𝑋) 𝛾𝑣( 𝑎, 𝑋)} 𝑑𝑆𝜕𝐷
by G and we get the solution of a Dirichlet boundary value problem at
appoint a𝞊D as
u(a)=-∫ 𝑓(𝑥)
𝜕𝐺
𝜕𝑣
(𝑎, 𝑋)𝜕𝐷
𝑑𝑆 --------------->(21)
Example 2.2
Greens function for asphers of radius R,withcentre at origin in m-
dimensions.
Here G(a,X) = 𝛹( 𝑟) =𝛹[
|𝑎|
𝑅
𝑟1]
Where
𝛹( 𝑟) =
1
(𝑚−2)𝑤 𝑚
|𝑋 − 𝑎|2-m , m>2
=
1
2𝜋
log
1
|𝑋−𝑎|
m=2
|a2| =a 𝛼a 𝛼, r1
2=(x𝛼-
𝑅2
|𝑎|2
a 𝛼)( x𝛼 -
𝑅2
|𝑎|2
a 𝛼)

24 | P a g e
=|𝑋 −
−𝑅2
|𝑎|2
𝑎|
2
r1denotes the distance of the point x from the reflected image of the
point in the sphere. This function satisfies all the requirement of the
Green’s function since(i) it is of the form s(𝑎, 𝑥) +𝜙 (𝑥), where 𝜙(𝑥),
where 𝜙(𝑥)is regular in D and continuous inD+𝜕𝐷 and (ii) it vanishes on
𝜕𝐷, since
r=
|𝑎|
𝑅
𝑟1 on 𝜕𝐷
Example 2.3
Green’s function for a positive half plane bounded by x1 =0, m > 2
𝐺(𝑎, 𝑥)=
1
(𝑚−2)𝑤 𝑚
|𝑋 − 𝑎|2−𝑚 − 𝜙( 𝑥),m >2
Where
𝜙(𝑥)=
1
(𝑚−2)𝑤 𝑚
[(𝑥1 + 𝑎1)2
+ ∑ (𝑥 𝛼 − 𝑎 𝛼)2𝑚
𝛼=2 ](2-m)/2
𝜙(𝑥)is obtained by taking the image of ‘a’ in the boundary by x1=0 .We
can find the solution at x = aby
u(a) =∫ 𝑓( 𝑋) {
𝜕
𝜕𝑥1
𝐺( 𝑎, 𝑋)}𝑥1
𝑑𝑆

25 | P a g e
Example 2.4
Green’s function for a circle and poisson’s formula,m=2. This is a
special case of example 2.2
G(a,X) =
1
2𝜋
𝑙𝑜𝑔
1
|𝑋−𝑎|
−
1
2𝜋
log |𝑋 −
𝑅2
|𝑎|2
𝑎|
Of special interest is the case of the Green’s function in two dimension ,for
a domine D which can be mapped conformlly onto the unit circle.
Theorem 2.2
Let F(x+iy)= u+ iv represent a mapping of D+𝜕𝐷 on the unit circle in the
u-v plane, where F(x+iy) is a simple analytic function of the complex
variable x+ iy .Then the Greens function for D is given by
G(a1,a2 ;x,y)= -
1
2𝜋
𝑅𝑒𝑙𝑜𝑔[
𝐹( 𝑎1+ 𝑖𝑎2)−𝐹(𝑥+𝑖𝑦)
𝐹( 𝑎1+𝑖𝑎2) 𝐹( 𝑥+𝑖𝑦)−1
]
Where Re denotes the real part of a complex quantity.
Proof
To show that G=0 on the boundary 𝜕𝐷, we denote that 𝜕𝐷 is
mapped by F onto boundary of the unit circle in the z-plane.
Therefore
F(x+iy)=𝑒 𝑖𝜃

26 | P a g e
When (x,y) lies on 𝜕𝐷 . Set z= x+iy, 𝛼 =a1+ia2
On 𝜕𝐷, G(a1 ,a2 ;x,y) =
−1
2𝜋
Relog[
𝐹( 𝛼)−𝑒 𝑖𝜃
𝐹(𝛼)𝑒 𝑖𝜃−1
]
=
−1
2𝜋
Relog[
𝐹( 𝛼)−𝑒 𝑖𝜃
(𝐹( 𝛼)−𝑒 𝑖𝜃)𝑒 𝑖𝜃
]
=
−1
2𝜋
log 1
=0
To show that G is a fundamental solution of the Laplace equation,
We have to show that it satisfies ∆2u =0 except at (x,y)=(a1,a2)and in the
neighbourhood of this point .
2𝜋𝐺 = − log|( 𝑥, 𝑦) − (𝑎1, 𝑎2)| + 𝑎 ℎ𝑎𝑟𝑚𝑜𝑛𝑖𝑐 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
Since F is a one-one mapping
For any two points (x,y), (a1,a2), the function
log[
𝐹( 𝛼)−𝐹(𝑧)
𝐹(𝛼)̅̅̅̅̅̅̅−𝐹( 𝑧)−1
]
is analytic in z and its real part satisfies Laplace’s euation. In the
neighbourhood of 𝛼
F(z)-F(𝛼)= (z-𝛼)F’(𝛼)-
(𝑧−𝛼)2
2!
F”(𝛼)+…….

27 | P a g e
= (z-𝛼)[𝐹′( 𝛼) −
( 𝑧−𝛼)
2!
𝐹"(𝛼)] +…….
= (z-𝛼)H(z)
Where H(z) is analytic . Due to the conformal nature of the
mapping, H(z) is non zero in a suitable neighbourhood of 𝛼. Therefore
2𝜋G(𝛼,z) = -Re log(
−𝐻(𝑧)
𝐹( 𝛼)̅̅̅̅̅̅̅ 𝐹( 𝑧)−1
)
= -log|z-( 𝛼)| +harmonic function
Therefore, G is the desired Green’s function
Theorem 2.3: Poisson’s Theorem
This theorem gives the solution of Dirichlet’s problem in sphere of
radious R about the origin.
If f(x) C0 on |x|=R, then for m≥2
u(x)= {
𝑅2−| 𝑥|2
𝑅𝑤 𝑚
∫
𝑓( 𝑦)
|𝑥−𝑦| 𝑚
𝑑𝑆| 𝑦|=𝑅
𝑓𝑜𝑟 | 𝑥| < 𝑅
𝑓( 𝑥) 𝑓𝑜𝑟| 𝑥| = 𝑅
Belongs to C0 in |x|< R and u(x) if a solution of the problem
∆ 𝑚 𝑢 = 0 𝑓𝑜𝑟 | 𝑥| < 𝑅, 𝑢 = 𝑓( 𝑥) 𝑓𝑜𝑟 | 𝑥| = 𝑅
(22
)
(23
)

28 | P a g e
We can further show that in |x| <R, u∈ 𝐶∞
Proof
Step 1: We shall first show that u given by(22) satisfies ∆mu=0.
If |x|<R, then |x-y|≠0 in the integrand and (23) can be differentiated under
the integral sign arbitrarilyoften
∆ 𝑚 𝑢( 𝑋) =
1
𝑅𝑤 𝑚
∫ 𝑓(𝑦)∆ 𝑚
| 𝑦|=𝑅
{
𝑅2
− |𝑋|2
|𝑋 − 𝑦| 𝑚
} 𝑑𝑆 = 0 𝑓𝑜𝑟 | 𝑥| < 𝑅
Since∆ 𝑚 {
𝑅2 −|𝑋|2
|𝑋−𝑦| 𝑚
}
In particular u(x)≡1 is a solution of ∆mu=0 in D and u ∈ C2 in D+𝜕D.
Applying (21) in this case,we get
I=
𝑅2−|𝑋|2
𝑅𝑤 𝑚
∫
𝑑𝑆
|𝑋−𝑦| 𝑚| 𝑦|=𝑅
………………………………….(25)
Step 2: To show that on approaching the boundary 𝜕D,u (x) us given by
(22) tends to the prescribedboundary value f(x).
Let 0<p<a, x0 is an arbitrary point on 𝜕D and x is such that |x-x0|<0/2. Then
form (22) and (25) we get
u(x)-f(x0)=
𝑅2−|𝑋|2
𝑅𝑤 𝑚
∫
𝑓( 𝑦)−𝑓(𝑥0)𝑑𝑆
|𝑋−𝑦| 𝑚| 𝑦|=𝑅

29 | P a g e
about x0 construct a sphere with radious 𝜌 and the part of 𝜕D which lies in
this sphere we denote by S1
That is S1 ; |y|=R,|y-x0| ≤ 𝜌,S2 : |y|=R|, |y-x0| >𝜌, 𝜕D=S1 +S2
∫
𝑓( 𝑦) − 𝑓(𝑥0)𝑑𝑆
|𝑋 − 𝑦| 𝑚
| 𝑦|=𝑅
= ( ∫ +
𝑆1
∫
𝑆2
)
𝑓( 𝑦) − 𝑓(𝑥0)𝑑𝑆
|𝑋 − 𝑦| 𝑚
Fig 2.2. Dividing the boundary into two parts S1 and S2 for an arbitrary
point x0on the boundary.
Consider each of these integrals in turn
|
𝑅2
− |𝑋|2
𝑅𝑤 𝑚
∫
𝑓( 𝑦) − 𝑓(𝑋0
)
|𝑋 − 𝑦| 𝑚
𝑆1
𝑑𝑆|
R
𝑆2
𝑆1
𝑋 0
𝜌

30 | P a g e
≤
𝑅2 –| 𝑋|
𝑅𝑤 𝑚
∫
𝑓( 𝑦)−𝑓( 𝑋0 )
| 𝑋−𝑦| 𝑚
𝑑𝑆𝑆1
≤𝑚𝑎𝑥
𝑦∈𝑆1
∣ 𝑓( 𝑦) − 𝑓( 𝑋0) ∣
𝑅2−∣𝑋∣2
𝑅𝑤 𝑚
∫
𝑑𝑆
∣𝑋−𝑦∣ 𝑚∣𝑣∣=𝑅
= 𝑚𝑎𝑥
𝑦∈𝑆1
|f(y)-f(𝑋0)| from(25)
Next 𝑚𝑎𝑥
𝑦∈𝑆1
𝑅2−|𝑋|2
∣𝑋−𝑦∣ 𝑚
= 𝑚𝑎𝑥
𝑦∈𝑆1
(𝑅−|𝑋|)(𝑅+|𝑋|)
∣𝑋−𝑦∣ 𝑚
≤
2𝑅(𝑅−|𝑋|)
(
𝜌
2⁄ ) 𝑚
Since |x-y| = |y-𝑥0
+ 𝑥0
− 𝑥|
= |y-𝑥0
|-|𝑥0
− 𝑥|
=𝜌 − 𝜌
2⁄ = 𝜌
2⁄
If we set 𝑚𝑎𝑥
𝑦∈𝜕𝐷
| 𝑓( 𝑦)| = 𝑀, we obtain the estimate for the second term.

31 | P a g e
APPLICATIONS
Laplace’s Equation
• Separation of variables – two examples
• Laplace’s Equation in Polar Coordinates
– Derivation of the explicit form
– An example from electrostatics
• A surprising applicationof Laplace’s eqn
– Image analysis
– This bit is NOT examined
Laplace’s Equation
𝝏 𝟐
∅
𝝏𝒙 𝟐 +
𝝏 𝟐
∅
𝝏𝒚 𝟐 = 𝟎
In the vector calculus course, this appears as∇2
∅ = 0 where

32 | P a g e
∇= [
𝜕
𝜕𝑥
𝜕
𝜕𝑥
]
Note that the equation has no dependence on time, just on the spatial
variablesx,y. This means that Laplace’s Equation describes steady state
situations such as:
• steady state temperature distributions
• steady state stress distributions
• steady state potential distributions (it is also called the potential equation
• steady state flows, for example in a cylinder, around a corner, …

33 | P a g e
Stress analysis example: Dirichlet conditions
Steady state stress analysis problem, which satisfies Laplace’s
equation; that is, a stretched elastic membrane on a rectangular former
that has prescribedout-of-plane displacements along the boundaries
𝑤 = 𝑤0 𝑠𝑖𝑛
𝜋𝑥
𝑎
To Solve:
𝜕2 𝑤
𝜕𝑥2
+
𝜕2 𝑤
𝜕𝑦2
=0
Solution by separation of variables
w(x,y)=X(x)Y(y)
w = 0
w = 0 a
b
w = 0
Y
X

34 | P a g e
from which X”Y +XY”=0
and so
𝑋"
𝑋
+
𝑌"
𝑌
= 0
as usual
𝑋"
𝑋
=
−𝑌"
𝑌
=k
wherek is a constant that is either equal to, >, or < 0.
Case k=0
X (x)= (Ax + B),Y( y) = (Cy + D)
W(x,0)=0 =>ADx= 0
Either A=0 or D=0
Continue with w(x,y) = ACxy
W(a,y) = 0 => ACay = 0 =>A =0 or C= 0
=>w(x,y)≡ 0
That is, the case k=0 is not possible
Case k>0
Suppose that k2=α2 , so that
W(x, y)= (A cosh αx +Bsinh αx )(C cos αy+ sin αy )
Recall that cosh 0=0,sinh0=0

35 | P a g e
w(o,y)=0 => A(Ccos αy +Dsin αy)=0
C = D= 0 => w(x,y)≡0
Continue with A=0 =>w(x,y)= Bsinfαx(Ccos αy+Dsin αy)
W(x,0)=0 => BCsinh αx =0
B= 0 => w(x,y)≡0
Continue with C=0 =>w(x,y)=BD sinh𝛼xsin𝛼𝑦
w(a,y)= => BDsinh𝛼a sin𝛼𝑦=0
So either B =0 or D =0 => w(x,y)≡0
Again, we find that the case k>0 is not possible
Final case k<0
Suppose that k = -α2
W(x,y) = (A cos αx + B sin αx) (C cosh αy + D sinh αy)
W(0,y)= 0 A(C cosh αy + D sinh αy) = 0
As usual, C = D = 0 w ≡ 0
Continue with A = 0 w(x,y) = B sin αx(C cosh αy + D sinh αy)

36 | P a g e
W(x,0) = 0 BC sin αx = 0
B = 0 w ≡ 0
Continue with C = 0 w(x,y)= BD sin αx sinh αy
W(a,y)=0α BD sinαa sinh αy = 0
B = 0 or D = 0 w ≡ 0
sin αa = 0 α = 𝑛
𝜋
𝑎
w (x,y) = BD sin 𝑛
𝜋
𝑎
x sinh 𝑛
𝜋
2𝑎
𝑦 y
Solution
Applying the first three boundary conditions,
we have 𝑤( 𝑥, 𝑦) = ∑ (𝑘 𝑛 sin
(2𝑛−1)∏
2𝑎
x sinh
(2𝑛−1)𝜋𝑦
2𝑎
)
∞
𝑛=1
The final boundary condition is : w(x,b) = w0 sin
∏𝑥
2𝑎
Which gives: w0 sin
∏𝑥
2𝑎
= ∑ (𝑘 𝑛 sin
(2𝑛−1)∏
2𝑎
x sinh
(2𝑛−1) 𝜋𝑏
2𝑎
)
∞
𝑛=1
We can see from this that n must take only one value, namely 1, so that
K1=
𝑊0
sinh
∏𝑏
2𝑎
And the final solution to the stress distribution is

37 | P a g e
PDEs in other coordinates…
In the vector algebra course, we find that it is often easier to express
problems in coordinates other than (x,y), for example in polar coordinates
(r,Θ)
Recall that in practice, for example for finite element techniques, it is
usual to use curvilinear coordinates… but we won’t go that far
We illustrate the solution of Laplace’s Equation using polar coordinates*
A problem in electrostatics
This is a cross section of a charged
cylindrical rod.
w(x,y) =
𝑤0
sinh
∏𝑏
2𝑎
sin
∏𝑥
𝑎
sinh
∏𝑦
𝑎
Thinstripof
insulating
material
V(rѲ,z) = U
on the upper half
Radius a

38 | P a g e
I could simply Tell you that Laplace’s Equation in cylindrical polars is:
∇2V =
𝜕2 𝑣
𝜕𝑟2
+
1
𝑟
𝜕𝑉
𝜕𝑟
+
1
𝑟2
𝜕2 𝑣
𝜕𝑟ѳ2
+
𝜕2 𝑣
𝜕𝑧2
= 0 ….. brief time out while I derive this
2D Laplace’s Equation in Polar Coordinates
x Ѳ r y
∇2u =
𝜕2 𝑢
𝜕𝑥2
+
𝜕2 𝑢
𝜕𝑦2
=0 where x= x(r,ѳ) , y = y(r,ѳ)
U(x,y) = u(r,ѳ)
So, Laplace’s Equation is ∇2u(r,ѳ) = 0
We next derive the explicit polar form of Laplace’s Equation in 2D
Recall the chain rule :
𝜕𝑢
𝜕𝑥
=
𝜕𝑢
𝜕𝑟
𝜕𝑟
𝜕𝑥
+
𝜕𝑢
𝜕ѳ
𝜕ѳ
𝜕𝑥
Use the product rule to differentiate again
X = r cos Ѳ
Y = r sin Ѳ
r = √ 𝑥2 + 𝑦2
Ѳ = tan−1
(
𝑦
𝑥
)

39 | P a g e
𝜕2 𝑢
𝜕𝑥2
=
𝜕𝑢
𝜕𝑟
𝜕2 𝑢
𝜕𝑥2
+
𝜕
𝜕𝑥
(
𝜕𝑢
𝜕𝑟
)
𝜕𝑟
𝜕𝑥
+
𝜕𝑢
𝜕ѳ
𝜕2ѳ
𝜕𝑥2
+
𝜕
𝜕𝑥
(
𝜕𝑢
𝜕𝑟
)
𝜕ѳ
𝜕𝑥
(*)
and the chain rule again to get these derivatives
𝜕
𝜕𝑥
(
𝜕𝑢
𝜕𝑟
)=
𝜕
𝜕𝑟
(
𝜕𝑢
𝜕𝑟
)
𝜕𝑟
𝜕𝑥
+
𝜕
𝜕ѳ
(
𝜕𝑢
𝜕𝑟
)
𝜕ѳ
𝜕𝑥
=
𝜕2 𝑢
𝜕𝑥2
𝜕𝑟
𝜕𝑥
+
𝜕2 𝑢
𝜕ѳ𝜕𝑟
𝜕ѳ
𝜕𝑥
𝜕
𝜕𝑥
(
𝜕𝑢
𝜕ѳ
) =
𝜕
𝜕𝑟
(
𝜕𝑢
𝜕ѳ
)
𝜕𝑟
𝜕𝑥
+
𝜕
𝜕ѳ
(
𝜕𝑢
𝜕ѳ
)
𝜕ѳ
𝜕𝑥
=
𝜕2 𝑢
𝜕𝑟𝜕ѳ
𝜕𝑟
𝜕𝑥
+
𝜕2 𝑢
𝜕ѳ2
𝜕ѳ
𝜕𝑥
THE REQUIRED PARTIAL DERIVATIVES
X= r cosѲ y = r sin Ѳ r =√𝑥2 + 𝑦2 Ѳ = tan−1
(y/x)
r2 = x2 + y2 2r
𝜕𝑟
𝜕𝑥
= 2x
𝜕𝑟
𝜕𝑥
=
𝑥
𝑟
similarly,
𝜕𝑟
𝜕𝑦
=
𝑦
𝑟
,
𝜕2 𝑟
𝜕𝑥2
= 𝑦2
/𝑟3
,
𝜕2 𝑟
𝜕𝑦2
= 𝑥2
/𝑟3
in like manner…..
𝜕ѳ
𝜕𝑥
= -
𝑦
𝑟2
,
𝜕ѳ
𝜕𝑦
=
𝑥
𝑟2
𝜕2ѳ
𝜕𝑥2
=
2𝑥𝑦
𝑟4
,
𝜕2ѳ
𝜕𝑦2
=
2𝑥𝑦
𝑟4

40 | P a g e
Back to Laplace’s Equation in polar coordinates
Plugging in the formula for the partials on the previous page to the
formulae on the one before that we get:
𝜕2 𝑢
𝜕𝑥2
=
𝜕2 𝑢
𝜕𝑟2
𝑥2
/𝑟2
+
𝜕𝑢
𝜕𝑟
𝑦2
/𝑟3
+
𝜕2 𝑢−
𝜕𝑟𝜕ѳ
2𝑥𝑦
𝑟3
+
𝜕𝑢
𝜕ѳ
2𝑥𝑦
𝑟4
+
𝜕2 𝑢
𝜕ѳ2
𝑦2
/𝑟4
Similarly,
𝜕2 𝑢
𝜕𝑦2
=
𝜕2 𝑢
𝜕𝑟2
𝑦2
/𝑟2
+
𝜕𝑢
𝜕𝑟
𝑥2
/𝑟3
+
𝜕2 𝑢
𝜕𝑟𝜕ѳ
2𝑥𝑦
𝑟3
-
𝜕𝑢
𝜕ѳ
2𝑥𝑦
𝑟4
+
𝜕2 𝑢
𝜕ѳ2
𝑥2
/𝑟4
So Laplace’s Equation in polars is
𝜕2 𝑢
𝜕𝑥2
+
𝜕2 𝑢
𝜕𝑦2
=
𝜕2 𝑢
𝜕𝑟2
+
1
𝑟
𝜕𝑢
𝜕𝑟
+
1
𝑟2
𝜕2 𝑢
𝜕ѳ2
=0
𝜕2 𝑢
𝜕𝑥2
+
𝜕2 𝑢
𝜕𝑦2
=0
Is equivalent to
𝜕2 𝑢
𝜕𝑟2
+
1
𝑟
𝜕𝑢
𝜕𝑟
+
1
𝑟2
𝜕2 𝑢
𝜕ѳ2
=0

41 | P a g e
Example of Laplace in cylindrical Polar Coordinates (r, Ѳ, z)
Consider a cylindrical capacitor z
θ y
0v
Laplace’s Equation in cylindrical polars is: Boundary
conditions
∇2v =
𝜕2 𝑣
𝜕𝑟2
+
1
𝑟
𝜕𝑣
𝜕𝑟
+
1
𝑟2
𝜕2 𝑣
𝜕ѳ2
+
𝜕2 𝑣
𝜕𝑧2
=0 v (a, ѳ) =
U∀ ѳ: 0 ≤ ѳ ≤ 𝜋
v (a, ѳ) = 0 ∀ ѳ: 𝜋 ≤ ѳ ≤ 2𝜋
In the polar system, note that the
solution must repeat itself every
Ѳ = 2π
V should remain finite at r = 0
There is no variation in V in the z- direction, so
𝜕𝑣
𝜕𝑧
= 0
This means we can treat it as a 2D problem
𝜕2 𝑣
𝜕𝑟2
+
1
𝑟
𝜕𝑣
𝜕𝑟
+
1
𝑟2
𝜕2 𝑣
𝜕ѳ2
= 0
Using separation of variables
V = R(r) Θ (Ѳ)
R″Θ + 1/r R′Θ + 1/r2 RΘ″ = 0
- Θ″ = R″+ R′/r
Θ R/r2
R
θ
Thinstripof insulating
mmaterial
V(rѲ,z) = U
on the upper half
Radius a

42 | P a g e
As before, this means
- Θ″ = R″+ R′/r = k, a constant
Θ R/r2
THE CASE K=0
Θ″ = 0 Θ(Ѳ) = a Ѳ + b
R″ + R′/r = 0 R( r )= (C lnr +D)
and so V(r, Ѳ) = (a Ѳ+ b) (C ln r+ D)
The solution has to be periodic in 2π.a = 0
The solution has to remain finite as r 0: c=0
THE CASE K< 0
Suppose that k = -m2
Ѳ″ + m2 Ѳ = 0 Θ (Ѳ)= (Amcoshm Ѳ+ Bmsinhm Ѳ)
R″ + R′/r + m2/r2 R=0 R(r) = (Cmr-m + Dmrm)
The solution has to be peiodic in Ѳ, with period 2π. This implies that
Am=Bm= 0 V(r, Ѳ) ≡ 0
THE CASE K>0
Suppose that k = n2
Ѳ″ + m2 Ѳ = 0 Θ (Ѳ)= (Ancosn Ѳ + Bnsin n Ѳ)
R″ + R′/r – n2 /r2 R = 0 R(r)= (Cnrn + Dnr –n)
V(r, Ѳ) = (Ancos n Ѳ + Bn sin n Ѳ) (Cnrn + Dnr –n)
Evidently, this is periodic with period 2π
To remain finite as r 0 Dn= 0
V(r,Ѳ) = bd = g, a constant

43 | P a g e
THE SOLUTION
V(r, Ѳ) = g + ∑nrn (Ancosn Ѳ + Bnsinn Ѳ)
Notice that we have not yet applied the voltage boundary condition!!
Now is the time to do so
g + ∑n{an (Ancosn Ѳ + Bnsinn Ѳ) = V (a, Ѳ) = U 0≤ Ѳ≤π
0 π<Ѳ≤2π
integrating V from 0 to 2π: ∫ 𝑣
2π
0
(a, Ө) d Ө = π U
Left hand side :∫ 𝑔
2π
0
+ ∑n an (Ancosn Ѳ + Bnsinn Ѳ)d Ѳ = 2πg and so g =
U/2
SOLVING FOR Am and Bm
so far, the solution is V (a, Ө) = U/2 + ∑n an (Ancosn Ѳ + Bnsinn Ѳ)
we apply the orthogonality relationships:
∫ 𝑉(𝑎,
2π
0
Ѳ)cosmѲdѲ =
𝑢
2
∫ 𝑐𝑜𝑠𝑚
2π
0
ѲdѲ +∫
2π
0
∑nan(An
cosnѲ+BnsinnѲ)cosmѲdѲ
U∫ cos mѲdѲ
𝑥
0
=0+ Am am π
0 = Amam π, and so Am = 0, for all m
∫ 𝑉(𝑎,
2π
0
Ѳ)sinmѲdѲ=
𝑢
2
∫ 𝑠𝑖𝑛𝑚
2π
0
ѲdѲ+∫
2π
0
∑ 𝑎 𝑛
𝑛 (An
cosnѲ+BnsinnѲ)sin nѲdѲ
U∫ sin mѲd Ѳ
𝑥
0
=
𝑢
2
[
𝑐𝑜𝑠𝑚Ѳ
𝑚
] + Bnam π
2𝑢
𝑚
= Bnam π, for odd m = (2n-1)
𝑉 (𝑟, Ѳ) =
𝑢
2
+ ∑ 𝑛
2𝑢
(2𝑛−1)πa(2n−1) r (2n-1)sin(2n-1) Ѳ
𝑉 (𝑟,Ѳ) =
𝑢
2
+
2𝑢
π
∑
𝑟(2𝑛−1)
(2𝑛−1) 𝑎(2𝑛−1)
∝
𝑛−1 sin(2n-1) Ѳ
Check for r = a, Ѳ=
π
2
:

44 | P a g e
𝑉 (𝑟,Ѳ) =
𝑢
2
+
2𝑢
π
∑
𝑎(2𝑛−1)
(2𝑛−1) 𝑎(2𝑛−1)
∝
𝑛−1 sin(2n-1)
π
2
=
𝑢
2
+
2𝑢
π
[𝑠𝑖𝑛
π
2
+
1
3
𝑠𝑖𝑛
3π
2
+
1
5
𝑠𝑖𝑛
5π
2
+ ⋯]
=
𝑢
2
+
2𝑢
π
[1 −
1
3
+
1
5
−
1
7
+ ⋯]
=
𝑢
2
+
2𝑢
π
[
π
4
]= U
An application in image analysis
 We saw that the Gaussian is a solution to the heat/diffusion equation
 We have studied Laplace’s equation
 The next few slides hint at the application of what we have done
sofar in image analysis
 This is aimed at engaging your interest in PDEs … it is not
examined
Laplace’s Equation in image analysis
How do we compute the edges?
Image fragment Edge map
 Remove the noiseby smoothing
 Find places where the second derivative
of the image is zero

45 | P a g e
Gaussian smoothing
Blurring with a Gaussian filter is one way to tame noise
Zero crossings of a second derivative, isotropic operator, after Gaussian
smoothing
∇2Ismooth = 0
An application of Laplace’s Equation!

46 | P a g e
Limits of isotropic Gaussian blurring
A noisy image Gaussian blurring
Guassian is isotropic – takes no account of orientation of image features –
so it gives crap edge features
∇2
(G𝝈 * I)= 0
As the blurring is increased, by increasing the standard
deviation of the Gaussian, the structure of the image is quickly lost.

47 | P a g e
Can we do better? Can we make blurring respect edges?
Anisotropic diffusion
∂tI = ∇T (g (x,t) ∇ I)
g(x;t) = 𝑒
|∇I𝜎|2
𝑘2
, for some constant k, or
g(x;t)=
1
1+|∇I𝜎|2/𝑘2
This is non-linear version of Laplace’s Equation, in which the blurring is
small across an edge fearture (low gradient) and large along an edge.

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