A Study of the Use of the Fractional Laplacian in
Extension Problems
Christian K. Akumaglo
10178135
THIS THESIS IS SUBMITTED TO THE UNIVERSITY OF GHANA,
LEGON IN PARTIAL FULFILMENT OF THE REQUIREMENTS
FOR THE AWARD OF MPHIL MATHEMATICS DEGREE
July, 2015
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DECLARATION
This thesis was written in the Department of Mathematics, University of Ghana, Legon from
September 2014 to July 2015 in partial fulfilment of the requirement for the award of Master
of Philosophy degree in Mathematics under the supervision of Dr. Benoit F. Sehba and Dr.
Douglas Adu-Gyamfi of the University of Ghana
I hereby declare that except where due acknowledgement is made, this work has never been
presented wholly or in part for the award of a degree at the University of Ghana or any other
University.
Signature: ...................................................
Student: Christian K. Akumaglo
Signature: ...................................................
Supervisor: Dr. Benoit F. Sehba
Signature: ...................................................
Supervisor: Dr. Douglas Adu-Gyamfi
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Abstract
We obtain the operator square root (−∆)1/2 of the Laplacian, called the fractional Laplacian,
from the harmonic extension problem to the upper half space. It turns out that this operator
maps the Dirichlet boundary condition to the Neumann condition. In this thesis, we extend
the work of [2] by establishing the fractional Laplacian using semi-group methods and also
providing proofs to certain claims and propositions in [2]. We also study some properties of
the fractional Laplacian and relate it to an extension problem.
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DEDICATION
I dedicate this work to my parents, Mr. Emmanuel Akumaglo and Mrs. Diana M. Akumaglo,
my sister Fafali Akumaglo, my dear friend Vera Amo-Asamoah and all my loved ones.
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ACKNOWLEDGEMENTS
My outmost thanks goes to God Almighty from whom flows wisdom, knowledge and under-
standing and for bringing me this far.
My sincere gratitude to my supervisors Dr. Benoit F. Sehba, Dr. Douglas Adu-Gyamfi and
Dr. Margaret McIntyre for their keen interest, support and guidance thereby aiding me in the
successful completion of this work.
Special Thanks to Prof. Dr. Peter Stollman for introducing me this topic.
I am also grateful to Prof. Francis W. K. Allotey, UG-Carnegie “Next Generation of Aca-
demics in Africa Project” for the financial support I received.
I am also grateful to my sister, Fafali Akumaglo and Mr. Nathan Obeng Boafo for their
immense financial support during this program and Mrs Faustina Dzifa Buabeng for helping
me in printing my work.
Many thanks to my parents and course-mates for their support and prayers.
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Contents
Declaration i
Abstract ii
Dedication iii
Acknowledgements iv
1 Introduction 1
1.1 Statement of the Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Notations and Some Concepts . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.2.1 Mean Value Properties . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.2.3 Maximum Principles . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.3 The Poisson Kernel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.4 The Fourier Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
1.5 Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
2 The Fractional Laplacian 11
2.1 The Extension Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
3 Some Properties of PDEs 19
3.1 Harmonic Functions in (n+1+a) Dimensions . . . . . . . . . . . . . . . . 19
3.2 Fundamental Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
3.3 Conjugate Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
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3.4 Poisson Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
4 The Relation of the extension problem to the fractional Laplacian 34
4.1 Proof Using The Poisson Formula . . . . . . . . . . . . . . . . . . . . . . . 34
4.2 Proof Using Fourier Transform . . . . . . . . . . . . . . . . . . . . . . . . 36
5 Reflection Extensions 44
5.1 The Divergent Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
5.2 The Nondivergence Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
5.2.2 Uniqueness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
5.2.3 Existence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
5.3 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
Bibliography 55
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Chapter 1
Introduction
1.1 Statement of the Problem
Given a smooth bounded function f : Rn 7→ R, we want a smooth extension u satisfying
∆u(x,y) = 0, for x ∈ Rn and y > 0 .
u(x,0) = f (x), for x ∈ Rn . (1.1)
The solutions of the Laplace equation are called harmonic functions and so we expect the
solution u of (1.1) to be harmonic. We think of Rn as the boundary hyperplane of (n+1) di-
mensional upper-half space Rn+1+ = {(x,y) : x ∈ R
n, y > 0}, and find a harmonic function
u(x,y) on Rn+1+ , whose boundary values on R
n are f (x). Also, from the definition of f , we
expect u to be bounded and smooth.
Our main reference paper for this thesis is [2]. This thesis is about understanding the role of
the fractional Laplacian in the solution of (1.1) and also we provide proofs to some claims
and propositions as we work through the said paper. This thesis is organized as follows:
Chapter 1 deals with the problem statement, some definitions and basic concepts necessary
for understanding what follows in the subsequent chapters.
In Chapter 2, we will derive the fractional Laplacian (−∆)1/2 for the given problem us-
ing semigroup methods. We extend the given problem to a harmonic function u defined on
Rn× [0,+∞) and later study some properties of this extension problem.
In Chapter 3, we shall look at some properties of harmonic functions since they are the natu-
ral solutions of the given problem in (1.1). We shall derive the fundamental solution for the
Laplace equation and show that all solutions of the differential equations in (2.3) and (2.5)
can be obtained from this fundamental solution. We will derive a Poisson formula to solve
the differential equations in (2.3) and (2.5) among others.
In Chapter 4, we will use the Poisson formula we will derive in chapter 3 to relate the differ-
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ential equations in (2.3) and (2.5) to the fractional Laplacian. We will also show that this can
be done using the Fourier transform.
Chapter 5 deals with the application of the interior Harnack estimates to the differential equa-
tion in (2.3). Here, we will make use of the Harnack inequality obtained by Caffarelli and
Gutierrez in [1].
1.2 Notations and Some Concepts
In the sequel, the variable x is always in Rn, y and z are nonnegative real numbers. The func-
tion u(x,y) is an extension of the real valued function f to the upper half space. We shall use
capital letters, such as X , to refer to a point in Rn+1. So for example, X = (x,y) will mean
x ∈ Rn and y ∈ [0,+∞) .
1.2.1 Mean Value Properties
Let Ω⊂ Rn be a domain and u ∈C(Ω). Then;
Definition 1.2.2. We say that u satisfies ;
i. the first mean value property if
u(x) =
1
αnrn−1
∫
∂Br(x)
u(y) dSy , for any Br(x) ⊂Ω ,
ii. the second mean value property if
u(x) =
n
αnrn
∫
Br(x)
u(y) dSy , for any Br(x)⊂Ω ,
where αn is the surface area of the unit sphere in Rn and dSy is the area element.
1.2.3 Maximum Principles
Assume Ω is a bounded domain in Rn and consider the operator L in Ω of the form
Lu = ai j(x) Di ju+bi(x) Diu+ c(x)u , (1.2)
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where u ∈C2(Ω)∩C(Ω). We assume also that ai j, bi and c are continuous and bounded in
Ω and that L is uniformly elliptic in the sense that it satisfies the ellipticity condition
0 < α(x) |ξ |2 ≤ ai j(x) ξiξ j ≤ β (x) |ξ |2 , (1.3)
for all ξ ∈Rn\{0} and for any x ∈Ω with α(x) and β (x) being the minimum and maximum
eigenvalues of the matrix ai j(x) respectively.
By the ellipticity condition we notice that L is elliptic in Ω if the coefficient matrix ai j(x) is
positive definite for all x ∈Ω.
Definition 1.2.4. Following are the types of ellipticity L may assume in Ω.
i. We say that L is elliptic in Ω if the minimum eigenvalue β > 0 in Ω.
ii. L is strictly elliptic in Ω if 0 < β0 ≤ β , where β0 is any positive real number.
iii. If β/α is bounded in Ω, then we say that L is uniformly elliptic in Ω.
We shall illustrate this with an example.
Example 1.2.5. Let the operator L be given as L = D11 +x1D22. Then from (1.2), we realize
that ai j(x) is a 2× 2 matrix of the form a =
(
1 0
0 x1
)
having eigenvalues λ = 1,x1. If
x1 > 0, then L is elliptic. Also, L is strictly elliptic since for x1 > 1, we can always find a
positive real number c such that 0 < c < 1. But L is not uniformly elliptic since β/α = x1
is infinite as x1→ ∞ and so is not bounded.
Theorem 1.2.6. (The Weak Maximum Principle)[6]
Let the operator L be elliptic in the bounded domain Ω. Let u ∈ C2(Ω)∩C(Ω) satisfying
Lu ≥ 0 in Ω with c = 0 in Ω. Then the nonnegative maximum of u in Ω is attained on ∂Ω.
That is,
sup
Ω
u = sup
∂Ω
u
Proof. We follow the order of proof used in [6] and [7] . Let α(x) and β (x) be the minimum
and maximum eigenvalues of the matrix ai j(x) respectively and suppose that |bi(x)|/α(x) is
bounded. Now, if Lu > 0 in Ω, we realize that u cannot attain a maximum in the interior of
Ω. Let us define for some ε > 0, the function v(x) as
v(x) = u(x)+ εeβx1 , where β is to be determined.
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Then we have
Lv = L
(
u(x)+ εeβx1
)
,
= Lu+β 2 ·a11(x) · εeβx1 +β ·b1(x) · εeβx1 + c(x) · εeβx1 ,
= Lu+ εeβx1
(
β 2 a11(x)+β b1(x)+ c(x)
)
.
But we know that |b1(x)| ≤ kα(x), c(x) = 0 and the matrix a11(x) is positive since L is
elliptic. So a11(x) ≥ α(x) > 0 for any x ∈ Ω. Hence by choosing β large enough, we have
that β 2 a11(x)+ β b1(x)+ c(x) > 0 for all x ∈ Ω. Hence, we have Lv > 0 in Ω. That is,
L(u(x)+ εeβx1)> 0 in Ω and so we have that
sup
Ω
(
u(x)+ εeβx1
)
= sup
∂Ω
(
u(x)+ εeβx1
)
.
If we let ε → 0, we obtain the desired result.
An immediate consequence of the weak maximum principle is given in the Corollary below.
Corollary 1.2.7. (This is due to [6]) Let L be elliptic in the bounded domain Ω⊂Rn. Suppose
that in Ω , we have Lu≥ 0 with c≤ 0 where u ∈C(Ω). Then we have
sup
Ω
u≤ sup
∂Ω
u+ ,
where u+ is the positive part of u.
Proof. Let us consider a subset Ω+ ⊂ Ω such that u > 0 in Ω+. Then if Lu ≥ 0 in Ω, we
notice that Lu = ai j Di ju+bi Diu ≥ −cu ≥ 0 in Ω+ since c ≤ 0. Hence, the maximum of u
on Ω
+
must be attained on ∂Ω+ and also on ∂Ω. Also defining v as in the proof of 1.2.6, we
see that v attains its maximum only on ∂Ω and so we have
sup
Ω
v≤ sup
∂Ω
v+ .
Hence we have that
sup
Ω
u≤ sup
Ω
v≤ sup
∂Ω
v+ ≤ sup
∂Ω
u++ ε sup
∂Ω
eβx1 .
By letting ε → 0, we obtain
sup
Ω
u≤ sup
Ω
v≤ sup
∂Ω
v+ ≤ sup
∂Ω
u+ .
Hence, the result is established.
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Remark 1.2.8. i. The boundedness of the domain Ω guarantees the existence and unique-
ness of a maximum and minimum of u in Ω and also the continuous dependence of
solutions on their boundary values.
ii. If Lu = 0 in Ω in the above corollary, then we have that
sup
Ω
|u|= sup
∂Ω
|u| .
Example 1.2.9. Let Ω be the set Ω = {(x,y) ∈ R2 : 0 < x < pi, 0 < y < pi} and u(x,y) =
sinxsiny. Then we claim that u is a nontrivial solution for the problem
∆u+2u = 0 in Ω,
u = 0 on ∂Ω.
Proof.
∆u =
∂ 2u
∂x2 +
∂ 2u
∂y2 ,
= −sinxsiny− sinxsiny ,
= −2sinxsiny .
So for 0 < x < pi , 0 < y < pi , we have ∆u+2u = −2sinxsiny+2sinxsiny = 0. Also the
boundary ∂Ω = {(0,y) : 0 ≤ y ≤ pi}∪{(pi,y) : 0 ≤ y ≤ pi}∪{(x,0) : 0 ≤ x ≤ pi}∪{(x,pi) :
0≤ x≤ pi}. Consequently, for any (x,y) ∈ ∂Ω, we see that either x = 0 or pi and y = 0 or pi .
Hence sinxsiny = 0 on ∂Ω.
Definition 1.2.10. Let Ω⊂ Rn and u ∈C2(Ω). Then
i. A function u is said to be harmonic if ∆u = 0 in Ω. Note that L = ∆, the Laplacian.
ii. u is said to be a supersolution in Ω if ∆u≤ 0 in Ω.
iii. u is a subsolution if ∆u≥ 0 in Ω.
Now, if u ∈C2(Ω) and for any φ ∈C2(Ω) such that ∆φ > 0 in Ω, then we have ∆(u−φ)< 0
in Ω. Hence, by the maximum principle, u−φ cannot have a local minimum in Ω. This is
so because if u−φ has a local minimum at a point x0 ∈Ω, then ∆φ ≤ 0. From a geometrical
point of view, u− φ has a local minimum at a point x0 ∈ Ω means that φ touches u from
below at the point x0.
Definition 1.2.11. Let Ω⊂ Rn, u ∈C(Ω) and φ ∈C2(Ω). Then
i. u is called a viscosity supersolution of ∆u = 0 in Ω if for any point x0 ∈ Ω such that
u−φ has a local minimum at x0, we have ∆φ(x0)≤ 0.
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ii. u is called a viscosity subsolution of ∆u = 0 in Ω if for any point y0 ∈ Ω with u− φ
having a local minimum at y0, we have ∆φ(y0)≥ 0.
iii. We say u is a viscosity solution if it is a viscosity supersolution and subsolution.
Definition 1.2.12. The difference quotient uεz (x,z) in the direction of z is given by
uεz (x,z) = lim
h→0
u(x,z+h)−u(x,z)
h
, for |h|< ε .
1.3 The Poisson Kernel
The Poisson kernel for the upper half-space is given as
Py(x) =Cn.
y
(|x|2 + y2)n+12
, where Cn =
Γ(n+12 )
pi( n+12 )
(1.4)
Proof. This proof is found in [10]. The Poisson kernel for the upper half-space occurs natu-
rally as the Fourier transform of the Abel kernel K(ξ , t) = e−2pit|ξ |. That is,
P(x,y) =F (K(·,y))(x) =
∫
Rn
e−2pi y|ξ | e−2pi iξ ·x dξ (1.5)
We shall employ the following identities in the proof of (1.4).
(i)
∫
Rn e
−pi α|t|2 e−2pi it.x dt = (α)− n2 e−pi
|x|2
α , α > 0.
(ii)
∫ ∞
−∞ e
iax e−bx
2
dx =
(pi
b
) 1
2 e
−a2
4b .
(iii) e−λ = 1√pi
∫ ∞
0
e−u√
u e
− λ
2
4u du, λ > 0.
(iv)
∫ ∞
0 e
−v e−
|x|2 v
t2 . t−n. v
n−1
2 dv = t
(|x|2+t2)
n+1
2
∫ ∞
0 e
−v v
n−1
2 dv .
By putting λ = 2piy|ξ | in (1.5), we have
Py(x) =
∫
Rn
e−λ e−2pi iξ ·x dξ ,
=
1
√
pi
∫
Rn
(∫ ∞
0
e−u
u
1
2
e
pi2|t|2y2
u du
)
e−2pi it.x dt , from (iii) ,
=
1
√
pi
∫ ∞
0
(∫
Rn
e
pi2|t|2y2
u e−2pi it.x dt
)
e−u
u
1
2
du .
Taking α = pi y
2
u , we obtain
Py(x) =
1
√
pi
∫ ∞
0
(∫
Rn
epi α|t|
2
e−2pi it.x dt
)
e−u
u
1
2
du .
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But
∫
Rn
epi α|t|
2
e−2pi it.x dt = (α)−
n
2 e−
pi |x|2
α , follows from identity (i),
=
(
pi y2
u
)− n2
e
−
(
pi |x|2 x u
pi y2
)
,
=
( u
pi
) n
2
. y−n e
− |x|
2 u
y2 .
Hence, we have by substitution
Py(x) =
1
pi n+12
∫ ∞
0
e−u e
− |x|
2 u
y2 . y−n. u
n−1
2 du ,
=
1
pi( n+12 )
.
y
(|x|2 + y2)
n+1
2
∫ ∞
0
e−u u
n−1
2 du , by (iv) ,
=
Γ(n+12 )
pi n+12
.
y
(y2 + |x|2)
n+1
2
, since Γ(t) =
∫ ∞
0
ut−1 e−u du .
We shall see later in our work that the function u is given as
u(x,y) =
∫
Rn
Py(x−ξ ) f (ξ ) dξ (1.6)
= Cn
∫
Rn
y f (ξ )
(|x−ξ |2 + y2)
n+1
2
dξ ,
where Py(x−ξ ) is the Poisson kernel and Cn is a constant, is a solution of (1.1). We want to
show that the integral in (1.6) converges absolutely.
Now, we know that
∫
Py(x− ξ ) dξ = 1 and also, f is bounded by definition, that is, there
exists β > 0 such that | f | ≤ β , so we have
∣
∣
∣
∣
∫
Rn
Py(x−ξ ) f (ξ ) dξ
∣
∣
∣
∣ ≤
∫
Rn
∣
∣Py(x−ξ ) f (ξ )
∣
∣ dξ
=
∫
Rn
∣
∣Py(x−ξ )
∣
∣ | f (ξ )| dξ
≤ β
∫
Rn
∣
∣Py(x−ξ )
∣
∣ dξ
= β
< +∞ .
By a hyperplane in an n dimensional Euclidean space, we mean a flat, n− 1 dimensional
subset of the n dimensional space that divides the space into two disconnected parts. So the
real line for example, has a point on the line as its hyperplane. For a two-dimensional plane,
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it is a line and for a three-dimensional space, we have a two-dimensional plane being its
hyperplane.
1.4 The Fourier Transform
Let f be an integrable function on Rn. The Fourier transform of f on Rn is the function fˆ
defined by
fˆ (ξ ) =
∫
Rn
e−iξx f (x) dx ,
if the integral exists and x,ξ ∈ Rn. In a similar manner, we define the the inverse Fourier
transform as
f (x) =
∫
Rn
eiξx f (ξ ) dξ ,
provided the integral exists and f (ξ ) is the Fourier transform of f .
We notice that |e−ixξ |= 1, so the Fourier transform converges absolutely for all ξ . That is,
| fˆ (ξ )| =
∣
∣
∣
∣
∫
Rn
e−iξx f (x) dx
∣
∣
∣
∣
≤
∫
Rn
|e−ixξ | | f (x)| dx
=
∫
Rn
| f (x)| dx .
Thus the Fourier transform is a bounded function of ξ for f ∈ L1. We shall state the next two
theorems without proof.
Theorem 1.4.1. (Parseval’s Theorem) Let f (x) and g(x) be integrable and let fˆ (ξ ) and gˆ(ξ )
be their Fourier transform respectively. If f (x), g(x) ∈ L2(Rn), then
∫
Rn
f (x) g(x) dx =
∫
Rn
fˆ (ξ ) gˆ(ξ ) dξ ,
where x,ξ ∈ Rn and the bar denotes complex conjugation.
Theorem 1.4.2. (Plancherel’s Theorem) Let f (x) and g(x) and their Fourier transforms be
defined as in (1.4.1), then
∫
Rn
| f (x)|2 dx =
∫
Rn
| fˆ (ξ )|2 dξ .
This theorem has the physical interpretation that the Fourier transform preserves the energy
of the original quantity.
Theorem 1.4.3. The Leibniz Integral Rule(Measure Theory Version)
Let Y be an open subset of R and X a measure space. Suppose u : X ×Y → R satisfies the
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following conditions:
1) u(x,y) is a Lebesgue integrable function of x for each y ∈ Y .
2) For almost all x ∈ X, the derivative uy exists for all y ∈ Y .
3) There is an integrable function f : X → R such that |uy(x,y)| ≤ f (x) for all y ∈ Y and
almost every x ∈ X.
Then for all y ∈ Y , we have
d
dy
∫
X
u(x,y) dx =
∫
X
uy(x,y) dx .
We show that the function u defined by
u(x,y) =Cn
∫
Rn
y( f (ξ )− f (x))
(|x−ξ |2 + y2)
n+1
2
dξ + f (x)
satisfies the conditions of the Leibniz integral rule.
This is the case since if we let
F =
y( f (ξ )− f (x))
(|x−ξ |2 + y2)
n+1
2
We see that F is an integrable function of x ∈ Rn. Also, the derivative Fy exists for all
y ∈ [0,+∞). So we show that F satisfies condition (iii). We differentiate F partially with
respect to y to get
Fy(x,y) =
1
(|x−ξ |2 + y2)
n+1
2
−
(n+1)y2
(|x−ξ |2 + y2)
n+3
2
Therefore
|Fy(x,y)| ≤
∣
∣
∣
∣
∣
1
(|x−ξ |2 + y2)
n+1
2
−
(n+1)y2
(|x−ξ |2 + y2)
n+3
2
∣
∣
∣
∣
∣
≤
∣
∣
∣
∣
∣
1
(|x−ξ |2 + y2)
n+1
2
∣
∣
∣
∣
∣
≤
1
|x−ξ |n+1 .
Hence, condition (iii) is satisfied.
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1.5 Distributions
Definition 1.5.1. The Support of a Function
Let f be a function on Rn. The support of f , written supp f , is the closure of the set of all
points x ∈ Rn for which f (x) 6= 0. In other words, we say it is the smallest closed set outside
of which f vanishes.
We denote by Cα(Rn), the space of functions on Rn whose partial derivatives of order α
exist and are continuous on Rn. C∞0 (R
n) refers to the space of functions on Rn whose partial
derivatives of all orders exist and are continuous on Rn and whose support is compact,that is,
a closed and bounded subset of Rn.
Definition 1.5.2. Test Functions
Let φ be a function on Rn. We call φ a test function if φ ∈C∞0 (Rn).
Definition 1.5.3. Distributions
A continuous linear functional defined on the space of test functions is called a distribution
or a generalized function.
Definition 1.5.4. Distributional Derivative
Let φ ∈C∞0 (Rn) and α be a multi-index. Then for a smooth function u, we have
(Dαu,φ) =
∫
Rn
Dαu(x)φ(x) dx
= (−1)|α|
∫
Rn
u(x)Dαφ(x) .
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Chapter 2
The Fractional Laplacian
We now introduce the fractional Laplacian as follows. Suppose we solve the differential
equation in (1.1) subject to the given boundary condition to obtain a smooth bounded function
u, then we claim that (−∆)1/2 f (x) = −uy(x,0) where we realize (−∆)1/2 as the operator
T : f 7→ −uy(x,0). We demonstrate the above claim shortly.
Definition 2.0.5. Let X be a Banach space. A one parameter family T (t), 0 ≤ t ≤ ∞, of
bounded operators from X into X is a semigroup of bounded linear operators on X if;
(i) T (0) = I, where I is the identity operator on X .
(ii) T (s+ t) = T (s)T (t), for s, t ≥ 0 (this is known as the semigroup property)
(iii) For t ∈ [0,+∞) and each x ∈ X , the map
t 7→ (T (t))(x) ∈ X is continuous.
That is,
lim
t→0+
(T (t))x = x ∀ x ∈ X .
We say that a semigroup of bounded linear operators T (t), is uniformly continuous if
lim
t→0+
‖(T (t))x− x‖= 0 ∀ x ∈ X .
The one parameter family {e−t
√
−∆ : t ≥ 0} is a semigroup, since if we take the operator T as
T (t) = e−t
√
−∆,
then T (0) = e0 = I , the identity operator.
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Also for any t,s≥ 0, we have
T (s+ t) = e−(t+s)
√
−∆
= e−t
√
−∆ · e−s
√
−∆
= T (t)T (s)
Moreover, the map t 7→ (T (t)) f , where f ∈ X is continuous since
lim
t→0
(T (t)) f (x) = lim
t→0
(e−t
√
−∆) f (x)
= 1 · f (x)
= f (x) .
So {e−t
√
−∆ : t ≥ 0} satisfies the semigroup property and is known as the Poisson semigroup.
Now starting with the Poisson semigroup {e−t
√
−∆ : t ≥ 0} and taking the Fourier transform,
we have
F
[(
e−t
√
−∆
)
f
]
(p) = e−t|p| fˆ (p) , where fˆ is the Fourier transform of f .
Definition 2.0.6. Let u and v be functions on R. The convolution of u and v is the function
u∗ v defined by
u∗ v (x) =
∫
u(x− y) v(y) dy , x,y ∈ R
provided that the integral exists.
Definition 2.0.7. For f ∈ L1(Rn), we have
(
e−t
√
−∆
)
f (x) = (Pt ∗ f )(x)
=
∫
Rn
Pt(x− y) f (y) dy ,
where Pt(x) is the Poisson Kernel defined as
Pt(x) =
1
(2pi)n
∫
Rn
eix.p e−t|p|dp
= Cn.
t
(t2 + |x|2)
n+1
2
, where Cn =
Γ(n+12 )
pi n+12
and
∫
Pt(x) dx = 1 .
Any function that solves the partial differential equation in (1.1) can be represented as the
convolution of the Poisson kernel and some bounded function. We state this as a definition.
Definition 2.0.8. Let u be the solution of (1.1) and Py(x) be the Poisson kernel as defined
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above. Then u can be written as the convolution Py ∗ f , that is
u(x,y) = (Py ∗ f )(x)
=
∫
Rn
Py(x−ξ ) f (ξ ) dξ ,
where f (ξ ) is the boundary condition.
This definition says that u is the Poisson integral of f and so u converges to f .
Let us now evaluate the limit
lim
y→0
f (x)−
(
e−y
√
−∆
)
f (x)
y
.
In evaluating the above limit, we make use of the L’Hopital’s rule where we let
g(y) = f (x)− (e−y
√
−∆) f (x) and h(y) = y
Then g
′
(y) =
√
−∆ (e−y
√
−∆) f (x) and h
′
(y) = 1
Thus
lim
y→0
g(y)
h(y)
= lim
y→0
g
′
(y)
h′(y)
exists and is (
√
−∆) f (x). Hence
lim
y→0
f (x)−
(
e−y
√
−∆
)
f (x)
y
= (
√
−∆) f (x) .
Also, from
Py(x−ξ ) =Cn.
y
(y2 + |x−ξ |2) n+12
and
∫
Py(x) dx = 1
we have
f (x) =Cn
∫
Rn
y f (x)
(y2 + |x−ξ |2) n+12
dξ .
So,
f (x)−
(
e−y
√
−∆
)
f (x) = Cn
∫
Rn
y( f (x)− f (ξ ))
(y2 + |x−ξ |2) n+12
dξ
that is,
f (x)−
(
e−y
√
−∆
)
f (x)
y
= Cn
∫
Rn
f (x)− f (ξ )
(y2 + |x−ξ |2) n+12
dξ .
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Taking the limit as y→ 0, we have
lim
y→0
f (x)−
(
e−y
√
−∆
)
f (x)
y
= Cn
∫
Rn
lim
y→0
{
f (x)− f (ξ )
(y2 + |x−ξ |2) n+12
}
dξ
√
−∆ f (x) = Cn
∫
Rn
f (x)− f (ξ )
|x−ξ |n+1 dξ
that is, (−∆)
1
2 f (x) = Cn
∫
Rn
f (x)− f (ξ )
|x−ξ |n+1 dξ , where Cn =
Γ(n+12 )
pi n+12
.
Remark 2.0.9. We were able to interchange the limit and the integral in our calculations
above using the Monotone Convergence Theorem( MCT). This is because we first see that
the function Ky(x−ξ ) = f (x)− f (ξ )
(y2+|x−ξ |2)
n+1
2
is an increasing sequence as y→ 0+. Also,
lim
y→0+
Ky(x−ξ ) = ( f (x)− f (ξ )) lim
y→0+
1
(y2 + |x−ξ |2)
n+1
2
=
f (x)− f (ξ )
|x−ξ |n+1
= K(x−ξ ) .
And so the conditions necessary to apply the MCT are satisfied.
The above calculations immediately lead us to our next definition.
Definition 2.0.10. For any s ∈ (0,1), we define
(−∆)s f (x) =Cn
∫
Rn
f (x)− f (ξ )
|x−ξ |n+2s dξ , where Cn =
Γ(n+12 )
pi n+12
. (2.1)
The operator (−∆)s is referred to as the fractional Laplacian.
We can now establish the claim that (−∆)1/2 f (x) =−uy(x,0). By definition
u(x,y) =
∫
Rn
Py(x−ξ ) f (ξ ) dξ
= Cn
∫
Rn
y
(y2 + |x−ξ |2) n+12
f (ξ ) dξ
and since
∫
Py(x−ξ ) dξ = 1, we have
f (x) =Cn
∫
Rn
y f (x)
(y2 + |x−ξ |2) n+12
dξ .
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So that
u(x,y) = Cn
∫
Rn
y f (ξ )
(y2 + |x−ξ |2) n+12
dξ + f (x)− f (x)
= Cn
∫
Rn
y f (ξ )
(y2 + |x−ξ |2) n+12
dξ −Cn
∫
Rn
y f (x)
(y2 + |x−ξ |2) n+12
dξ + f (x)
= Cn
∫
Rn
y( f (ξ )− f (x))
(y2 + |x−ξ |2) n+12
dξ + f (x)
And differentiating partially with respect to y and applying the Leibniz integral rule, we get
uy(x,y) = Cn
∫
Rn
∂
∂y
{
y( f (ξ )− f (x))
(y2 + |x−ξ |2) n+12
}
dξ + ∂∂y f (x)
= Cn
∫
Rn
[
f (ξ )− f (x)
(|x−ξ |2 + y2)
n+1
2
−
(n+1)y2( f (ξ )− f (x))
(|x−ξ |2 + y2)
n+3
2
]
dξ .
So uy(x,0) = Cn
∫
Rn
f (ξ )− f (x)
(|x−ξ |2) n+12
dξ .
Thus −uy(x,0) = −Cn
∫
Rn
f (ξ )− f (x)
(|x−ξ |2) n+12
dξ
= Cn
∫
Rn
f (x)− f (ξ )
|x−ξ |n+1 dξ . (2.2)
Hence from (2.1) and (2.2), we have
(−∆)1/2 f (x) =−uy(x,0) .
We can thus define (−∆)1/2 as the operator T : f 7→−uy(· ,0) and we notice that if we replace
the Dirichlet condition f (x) in (1.1) with −uy(x,0) , we obtain −uy(x,y) as the solution of
(1.1) instead of u(x,y). This is because from definition 2.1.4, we expect the solution u of (1.1)
to converge to the boundary condition. This means that the operator T maps the Dirichlet
condition to the Neumann condition. If we apply T twice, we obtain
T (−uy(x,0))(x) = uyy(x,0) = T (T ( f ))(x) =−∆x f (x) ,
which means that −uy(x,y) is the solution in this case.
2.1 The Extension Problem
We shall in this section introduce an extension u of f . We shall study some properties of its
Laplace equation analogous to (1.1).
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Now let f : Rn→ R and consider the extension u : Rn× [0,∞)→ R that satisfies the differ-
ential equation
∆x u+
a
y
uy +uyy = 0
u(x,0) = f (x) , (2.3)
where a is a non-negative integer. But we notice that equation (2.3) can be written as
div (ya ∇u) = 0 (2.4)
since
div (ya ∇u) = ∇ · (ya ∇u) and
∇u =
∂u
∂x1
i1 +
∂u
∂x2
i2 + · · ·+
∂u
∂xn
in +
∂u
∂y in+1 ,
ya ∇u = ya
∂u
∂x1
i1 + · · ·+ ya
∂u
∂xn
in + ya
∂u
∂y in+1 ,
so ∇. (ya ∇u) = ya
∂ 2u
∂x21
+ · · ·+ ya
∂ 2u
∂x2n
+aya−1
∂u
∂y + y
a∂ 2u
∂y2
= ya∆x u+ay
a−1 uy + y
a uyy .
Thus (2.4) says
ya∆x u+ay
a−1 uy + y
a uyy = 0
that is, ∆x u+
a
y
uy +uyy = 0 .
If we make the change of variables z =
( y
1−a
)1−a
, we have y = (1−a) · z(1−a)
−1
,
so ∂ z∂y =
1
(1−a)1−a · (1− a) · y
−a = (1− a)a y−a . We note that, for the change of variables
to be valid, a must be in the interval (−1,1).
Thus
∂u
∂y =
∂u
∂ z ·
∂ z
∂y
= uz · (1−a)
a y−a = k uz · y
−a, where k = (1−a)a
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and
∂ 2u
∂y2 =
∂
∂y(k uz · y
−a)
= k
{
∂
∂y(uz) · y
−a +uz
∂
∂y(y
−a)
}
= k
{
y−a
(
∂uz
∂ z ·
∂ z
∂y
)
+uz
∂
∂y(y
−a)
}
= k
{
y−a(uzz · (1−a)
a y−a)−a y−(1+a) uz
}
= k(1−a)a y−2a uzz− k a y
−(1+a) uz .
Also ay uy = (1−a)
a y−(1+a) ·a uz = k a y−(1+a) uz. Hence
∆x u+
a
y
uy +uyy = ∆x u+(1−a)
2a y−2a uzz
= ∆x u+
(
1−a
y
)2a
uzz .
The transformed equation is then
∆x u+
(
1−a
y
)2a
uzz = 0
∆xu+ z
α uzz = 0 , (2.5)
where α = −2a
1−a
Proposition 2.1.1.
lim
y→0+
yauy(x,y) = (1−a) lim
y→0+
u(x,y)−u(x,0)
y1−a
Proof. The Poisson kernel associated with equation (2.3) is
P(x,y) =Cn,a
y1−a
(|x|2 + y2)
n+1−a
2
,
where Cn,a is a constant which we will calculate later.
Hence we have
u(x,y) =Cn,a
∫
Rn
y1−a
(|x|2 + y2)
n+1−a
2
f (ξ ) dξ .
And so from the first section in Chapter 2, we can write u as
u(x,y) =Cn,a
∫
Rn
y1−a ( f (ξ )− f (x))
(|x|2 + y2)
n+1−a
2
dξ + f (x) .
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By the Leibniz integral rule, we can differentiate under the integral to get
uy(x,y) =Cn,a
∫
Rn
{
(1−a)y−a
(|x−ξ |2 + y2)
n+1−a
2
−
(n+1−a)y2−a
(|x−ξ |2 + y2)
n+3−a
2
}
( f (ξ )− f (x)) dξ
and
−yauy(x,y) =Cn,a
∫
Rn
{
1−a
(|x−ξ |2 + y2)
n+1−a
2
−
(n+1−a)y2
(|x−ξ |2 + y2)
n+3−a
2
}
( f (x)− f (ξ )) dξ .
We multiply the right hand side by y1−a/y1−a and take the limit as y→ 0, to get
−yauy = (1−a)Cn,a ·
y1−a
y1−a
∫
Rn
f (x)− f (ξ )
(|x−ξ |2 + y2)
n+1−a
2
dξ
lim
y→0
−yauy = (1−a) lim
y→0
1
y1−a
Cn,a
∫
Rn
y1−a( f (x)− f (ξ ))
(|x−ξ |2 + y2)
n+1−a
2
dξ
= (1−a) lim
y→0
1
y1−a
(u(x,y)−u(x,0)) .
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Chapter 3
Some Properties of PDEs
We shall in this chapter, study some properties of the partial differential equations (2.3) and
(2.5). That is
∆xu+
a
y
uy +uyy = 0 and
∆xu+ z
αuzz = 0 .
Among these properties are fundamental solutions and conjugate equations. We shall then
vary Poisson formulae to explicitly solve the differential equations in (2.3) and (2.5). Our
starting point is harmonic functions since they are much suited to these types of problems as
they are naturally the solutions of (2.3) and (2.5). Some properties of harmonic functions are
that:
i. The average value of a harmonic function over a spherical surface is equal to its value
at the center of the sphere. This is known as Gauss’s harmonic function theorem.
ii. A harmonic function has no local maxima or minima and since the Laplace’s equation
is linear, the superposition of any two solutions is also a solution.
iii. A harmonic function has a conjugate harmonic function.
iv. A harmonic function is radially symmetric.
We shall study the properties of (2.3) via some of these properties of harmonic functions.
3.1 Harmonic Functions in (n+1+a) Dimensions
We shall obtain various properties of equation (2.3) by studying some properties of its solu-
tions. That is, we want to know among others if increasing the dimension of the domain of u
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will change the form of (2.3) or not.
Let a be a nonnegative integer and suppose u : Rn×R1+a→ R is radially symmetric in the
variable y. That is, if r = |y|= |yˆ|, then u(x,y) = u(x, yˆ).
If we think of u as a function of x and r, then
uxi =
∂u
∂xi , uxixi =
∂ 2u
∂x2i
, ∂u∂y j =
∂u
∂ r ·
∂ r
∂y j =
y j
r ·
∂u
∂ r and
∂ 2u
∂y2j
=
∂u
∂ r ·
∂
∂y j
(y j
r
)
+
y j
r
·
∂
∂y j
(
∂u
∂ r
)
=
(
1
r
−
y2j
r3
)
∂u
∂ r +
y2j
r2
·
∂ 2u
∂ r2 .
So we write an expression for the Laplacian of u as
∆u =
n
∑
i=1
∂ 2u
∂x2i
+
1+a
∑
j=1
[
(
1
r
−
y2j
r3
)
∂u
∂ r +
y2j
r2
∂ 2u
∂y2j
]
= ∆xu+
(
1+a
r
−
1
r
)
∂u
∂ r +
∂ 2u
∂ r2
= ∆xu+
a
r
ur +urr .
Thus (1.1) becomes (2.3) with y replaced by r.
3.2 Fundamental Solution
In looking for explicit solutions to a specific PDE such as the one we have, we often restrict
ourselves to functions in which we can identify some symmetry properties which will afford
us some ease when solving such PDEs.
From section (3.1) above, we realize that the Laplace equation we want to solve is invariant
under rotations and so it will be best to search for radial solutions. A function f is said to be
radial if its value at a point x depends only on |x|.
Let x ∈ Rn and suppose u(x) = v(r) is a solution to the Laplace equation, where r = |x| =
(
x21 + x
2
2 + . . .+ x
2
n
)1/2
. Also, we choose v(r) such that ∆u = 0.
Now ∂ r∂xi =
xi
r , and for each 1≤ i≤ n , we have
uxi =
xi
r
·
dv
dr
and uxixi =
x2i
r2
v
′′
+
(
1
r
−
x2i
r3
)
v
′
.
So ∆u = 0 implies that
x2i
r2
v
′′
+
(
1
r
−
x2i
r3
)
v
′
= 0 , for each i.
20
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Summing over i, we obtain
v
′′
+
(
n−1
r
)
v
′
= 0 .
Separating the variables, we have
v
′′
v′
=
1−n
r
that is,
d
dr
(lnv
′
) =
1−n
r
lnv
′
= (1−n) lnr+ lnc
= lnr1−n + lnc
v
′
= exp
(
lncr1−n
)
.
So if n = 2, we have
v(r) = c lnr+a , where a is some constant.
When n≥ 3, we obtain
v(r) = c
1
rn−2
+a , for some constant a.
Following the strategy in [5] these results thus lead us to define the fundamental solution of
Laplace’s equations.
Definition 3.2.1. Let x ∈ Rn \{0}. Then the function Φ defined as
Φ(x) :=
{
− 12pi log |x| , n = 2
1
n(n−2)α(n) ·
1
|x|n−2 , n≥ 3
is the fundamental solution of the Laplace equation, where α(n) is the volume of the unit
ball in Rn given by α(n) = nΓ(n/2)
2pin/2 .
From Φ(x), we can obtain the fundamental solution of the differential equations in (2.3) and
(2.5) since we know that Φ(x) generates all solutions of the Laplace equation. We achieve
this by considering the fundamental solution of the Laplacian in n + 1 + a dimensions as
captured in the following propositions.
Proposition 3.2.2. If n+a−1 > 1, then the solution Γ(x,y) to differential equation in (2.3)
is
Γ(x,y) =C(n+1+a) ·
1
(|x|2 + y2)
n−1+a
2
,
where Cn+1+a is some constant that depends on n and a.
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From the definition of Φ in 3.2.1, we see that Cn = 1n(n−2)α(n) .
Proof. From Γ(x,y) =Cn+1+a 1
(y2+|x|2)
n−1+a
2
, we have
Γxi(x,y) =−Cn+1+a · (n−1+a) ·
xi
(y2+|x|2)
n+1+a
2
and
Γxixi(x,y)=−Cn+1+a ·(n−1+a)·
1
(y2+|x|2)
n+1+a
2
+Cn+1+a ·(n−1+a)(n+1+a)·
x2i
(y2+|x|2)
n+3+a
2
.
Hence,
∆xΓ=−
n(n−1+a)Cn+1+a
(y2+|x|2)
n+1+a
2
+ (n−1+a)(n+1+a)Cn+1+a|x|
2
(y2+|x|2)
n+3+a
2
.
Also Γy =−(n−1+a)Cn+1+a ·
y
(y2+|x|2)
n+1+a
2
and
Γyy =−
(n−1+a)Cn+1+a
(y2 + |x|2)
n+1+a
2
+
(n−1+a)(n+1+a)Cn+1+a · y2
(y2 + |x|2)
n+3+a
2
.
So
∆xΓ+
a
y
Γy +Γyy = −
(n−1+a)(n+1+a)Cn+1+a
(y2 + |x|2)
n+1+a
2
+
(n−1+a)(n+1+a)(y2 + |x|2)Cn+1+a
(y2 + |x|2)
n+3+a
2
= −
(n−1+a)(n+1+a)Cn+1+a
(y2 + |x|2)
n+1+a
2
+
(n−1+a)(n+1+a)Cn+1+a
(y2 + |x|2)
n+1+a
2
= 0 .
We also compute the value of Cn+1+a as follows.
Let k = n+1+a, then we have
Ck =
1
(n−1+a)(n+1+a)
·
(n+1+a)Γ(n+1+a2 )
2pi n+1+a2
=
1
n−1+a
·
1
2
(n−1+a) ·
Γ(n−1+a2 )
2pi n+1+a2
=
Γ( k2 −1)
4pi k2
.
Proposition 3.2.3. Let u be a solution of (2.3) , then we have
lim
y→0+
yauy =−Cδ0 ,
where δ0 is the Dirac measure on Rn which assigns a unit mass to the point zero.
22
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Proof. We know that we can write the solution u(x,y) of (2.3) as
u(x,y) =Cn
∫
Rn
y1−a( f (ξ )− f (x))
(|x−ξ |2 + y2) n+1−a2
dξ + f (x) .
As in the previous case, the function under the integral satisfies the conditions of the
Leibniz integral rule and so we can differentiate under the integral to get
uy(x,y) =Cn
∫
Rn
[
(1−a)y−a
(|x−ξ |2 + y2) n+1−a2
−
(n+1−a)y2−a
(|x−ξ |2 + y2) n+3−a2
]
( f (ξ )− f (x))dξ
and
yauy(x,y) =Cn
∫
Rn
[
1−a
(|x−ξ |2 + y2) n+1−a2
−
(n+1−a)y2
(|x−ξ |2 + y2) n+3−a2
]
( f (ξ )− f (x))dξ
So, we have
limy→0+ y
auy(x,y) =
lim
y→0+
Cn
∫
Rn
[
1−a
(|x−ξ |2 + y2) n+1−a2
−
(n+1−a) y2
(|x−ξ |2 + y2) n+3−a2
]
( f (ξ )− f (x)) dξ
But if this limit exists, then it must also exist at the point x = 0 and so we have
lim
y→0+
yauy(0,y) = lim
y→0+
Cn
∫
Rn
[
(1−a)ya−n−1− (n+1−a)ya−n−1
]
( f (ξ )− f (0)) dξ ,
lim
y→0+
yauy(0,y) = lim
y→0+
Cn
∫
Rn
−nya−n−1( f (ξ )− f (0)) dξ ,
= −n Cn lim
y→0+
ya−n−1
∫
Rn
( f (ξ )− f (0)) dξ .
Now if n > a, then
ya−n−1 = y−k , for some k ∈ Z+ .
This gives
lim
y→0+
ya−n−1 = lim
y→0+
y−k
= δ0 ,
that is, the limit becomes uncontrollably large as y→ 0+. Therefore, we have
lim
y→0+
yauy(0,y) =−Cδ0 , where C = nCn
∫
Rn
( f (ξ )− f (0)) dξ ,
as required.
23
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We saw in section 2.1 that with an appropriate change of variables, equation (2.5) can be de-
rived from the differential equation (2.3). So with the same change of variables in (3.2.2)we
obtain
Γ˜(x,z) =Cn+1+a
1
(|x|2 +(1−a)2z
2
1−a )
n−1+a
2
as a solution to the differential equation (2.5). We demonstrate this explicitly in the following
claim.
Claim 3.2.4.
Γ˜(x,z) =Cn+1+a
1
(|x|2 +(1−a)2z
2
1−a )
n−1+a
2
is the solution of the differential equation in (2.5).
Proof.
From
Γ˜(x,z) = Cn+1+a
1
(|x|2 +(1−a)2|z|
2
1−a )
n−1+a
2
we have,
Γ˜xi(x,z) = −Cn+1+a(n−1+a)
xi
(|x|2 +(1−a)2|z|
2
1−a )
n+1+a
2
and
Γ˜xixi(x,z) = −
Cn+1+a(n−1+a)
(|x|2 +(1−a)2|z|
2
1−a )
n+1+a
2
+
Cn+1+a(n−1+a)(n+1+a)x2i
(|x|2 +(1−a)2|z|
2
1−a )
n+3+a
2
therefore,
∆xΓ˜ =
Cn+1+a(n−1+a)(n+1+a)|x|2
(|x|2 +(1−a)2|z|
2
1−a )
n+3+a
2
−
Cn+1+an(n−1+a)
(|x|2 +(1−a)2|z|
2
1−a )
n+1+a
2
,
Also,
Γ˜z = −
Cn+1+a(n−1+a)(1−a)z
1+a
1−a
(|x|2 +(1−a)2z
2
1−a )
n+1+a
2
, |z|> 0
and
Γ˜zz =
Cn+1+a(n−1+a)(n+1+a)(1−a)2z
2(1+a)
1−a
(|x|2 +(1−a)2z
2
1−a )
n+3+a
2
−
Cn+1+a(n−1+a)(1+a)z
2a
1−a
(|x|2 +(1−a)2z
2
1−a )
n+1+a
2
.
Multiply Γ˜zz by zα , where α =− 2a1−a to get
zα Γ˜zz =
Cn+1+a(n−1+a)(n+1+a)(1−a)2z
2
1−a
(|x|2 +(1−a)2z
2
1−a )
n+3+a
2
−
Cn+1+a(n−1+a)(1+a)
(|x|2 +(1−a)2z
2
1−a )
n+1+a
2
.
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Therefore
∆xΓ˜+ z
α Γ˜zz =
Cn+1+a(n−1+a)(n+1+a)
[
|x|2 +(1−a)2z
2
1−a
]
(|x|2 +(1−a)2z
2
1−a )
n+3+a
2
−
Cn+1+a(n−1+a)(n+1+a)
(|x|2 +(1−a)2z
2
1−a )
n+1+a
2
=
Cn+1+a(n−1+a)(n+1+a)
(|x|2 +(1−a)2z
2
1−a )
n+1+a
2
−
Cn+1+a(n−1+a)(n+1+a)
(|x|2 +(1−a)2z
2
1−a )
n+1+a
2
= 0 .
Hence, Γ˜ satisfies the differential equation in (2.5).
If we take u(x,z) = Γ˜(x,z), then
uz(x,z) =−
Cn+1+a(n−1+a)(1−a)z
1+a
1−a
(|x|2 +(1−a)2z
2
1−a )
n+1+a
2
.
And taking limits as z→ 0, we obtain
lim
z→0
uz(x,z) = lim
z→0
−
Cn+1+a(n−1+a)(1−a)z
1+a
1−a
(|x|2 +(1−a)2z
2
1−a )
n+1+a
2
.
Again, if this limit exists, then it must be equal to
lim
z→0
uz(0,z) = −(n−1+a)Cn+1+a lim
z→0
z−
n
1−a
= −(n−1+a)Cn+1+a lim
z→0
z−k , where k =
n
1−a
,k ∈ R+
= −γδ0 , where γ = (n−1+a)Cn+1+a .
It follows that uz(x,z)→−γδ0 as z→ 0.
3.3 Conjugate Equation
The solutions of (1.1) and for that matter (2.3) are known as harmonic functions. The re-
sulting pairs of such solutions are called conjugate harmonic functions. This suggests that
associated with (2.3) is a conjugate equation so that if u satisfies (2.3), then its conjugate so-
lution w must also satisfy the conjugate equation associated with (2.3). If u(x,y) is a solution
to (2.3), then we know that any constant times u is also a solution. From the foregoing dis-
cussion, we propose that (3.3.1) below is the conjugate equation to the differential equation
in (2.3).
25
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Proposition 3.3.1. Suppose w(x,y) is the conjugate solution of u(x,y), then w satisfies the
conjugate equation
∆xw−
a
y
wy +wyy = 0 . (3.1)
Proof. Consider w(x,y) := yauy(x,y), then we have
wy = aya−1uy + yauyy, wyy = a(a−1)ya−2uy +2aya−1uyy + yauyyy, wx = ya(ux)y
and ∆xw = ya(∆xu)y = ya∂y∆xu
Hence,we have
∆xw−
a
y
wy +wyy = y
a∂y∆xu−a2ya−2uy−ayauyy +a(a−1)ya−2uy
+2aya−1uyy + y
auyyy
= ya∂y∆xu−aya−2uy +aya−1uyy + yauyyy
= ya
(
∂y∆xu−
a
y2
uy +
a
y
uyy + y
auyyy
)
= ya∂y
(
∆xu+
a
y
uy +uyy
)
= ya∂y ·0 since u satisfies (2.3)
= 0 .
Thus w satisfies (3.1).
Let n = 1, that is x ∈ R and y ∈ [0,+∞). If we set v = ux, we have
vx = ∆xu , vxx = ∂x(∆xu) = ∆xv , vy = (ux)y = (uy)x = ∂xuy
and vyy = (uyy)x = ∂xuyy
Thus
∆xv+
a
y
vy + vyy = ∂x∆xu+
a
y
∂xuy +∂xuyy
= ∂x
(
∆xu+
a
y
uy +uyy
)
= ∂x ·0 since u satisfies (2.3)
= 0 .
This says that v is also a solution of (2.3). Since equation (2.3) is invariant under translations
in the variable x, then we must also have div(ya∇v) = 0.
The above calculations lead us to our next result which we state as a proposition. We shall
start with the following definitions.
Definition 3.3.2. (i). A vector field with the property that the line integral is path indepen-
dent is called a Conservative Field.
26
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(ii). A vector field with a zero curl is said to be Irrotational. An irrotational vector field is
conservative if the domain is simply connected.
Proposition 3.3.3. Let v = ux be a solution of (3.1), then
(i) the vector field (yavy,−yavx) is irrotational.
(ii) there exists a function w such that ∇w = (−yavy,yavx) and
(iii) div(y−a∇w) = 0
Proof. (i) All we need do is to show that the vector field (yavy,−yavx) has a zero curl. We
proceed as follows;
Let F = (yavy,−yavx,0), then
curl F =
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
i j k
∂
∂x
∂
∂y
∂
∂ z
yavy −yavx 0
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
=
(
∂
∂x(−y
avx)−
∂
∂y(y
avy)
)
k
=
(
−yavxx− y
avyy−ay
a−1vy
)
k
= −ya(vxx +
a
y
vy + vyy)k.
= −ya ·0 k
= ~0 , the zero vector.
Hence, the vector field (yavy,−yavx) is irrotational.
(ii) We choose w = yau, then
∇w = ∇(yauy)
=
(
∂
∂x(y
auy),
∂
∂y(y
auy)
)
=
(
ya(ux)y,ay
a−1uy + y
auyy
)
=
(
ya(ux)y,y
a(
a
y
uy +uyy)
)
= (yavy,−y
avx) , since vx =−(
a
y
uy +uyy) .
27
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(iii) From ∇w = (yavy,−yavx), we have
div (y−a∇w) = ∇ · (vy,−vx)
= (vxy,−vxy)
= 0
3.4 Poisson Formula
In this section, we shall develop a Poisson formula to explicitly solve the extension problem
∆xu+
a
y
uy +uyy = 0
u(x,0) = f (x) .
In the proof of (2.3.1), we stated that the Poisson kernel associated with the differential
equation in (2.3) is P(x,y) = Cn,a
y1−a
(|x|2+y2)
n+1−a
2
. We shall now demonstrate how P(x,y) was
obtained. From the previous section, we see that Γ(x,y) is a solution to the differential
equation in (2.3). This means that Γ−a(x,y) is also a solution to the conjugate equation of
(2.3). Note that Γ−a(x,y) is just Γ(x,y) where a is replaced with −a. We differentiate Γ(x,y)
partially with respect to y to obtain
∂y Γa(x,y) =−Cn−1+a(n−1+a)
y
(|x|2 + y2)
n+1+a
2
Replace a by −a and multiply through by −y−a to get
−y−a∂y Γ−a(x,y) =Cn−1−a(n−1−a)
y1−a
(|x|2 + y2)
n+1−a
2
.
But we know that if u is a solution to equation (2.3), then we can write u as
u(x,y) =
∫
Rn
P(x−ξ ,y) f (ξ )dξ , where P is the Poisson kernel given as
P(x,y) =Cn,a
y1−a
(|x|2+|y|2)
n+1−a
2
. From this, the next proposition follows.
Proposition 3.4.1. The Poisson kernel
P(x,y) =Cn,a
y1−a
(|x|2 + |y|2)
n+1−a
2
28
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is a solution to the differential equation in (2.3).
Proof. From
P(x,y) =Cn,a
y1−a
(|x|2 + |y|2)
n+1−a
2
, we have Pxi =−Cn,a(n+1−a)
xiy1−a
(|x|2 + |y|2)
n+3−a
2
and
Pxixi =−Cn,a(n+1−a)
y1−a
(|x|2 + |y|2)
n+3−a
2
+Cn,a(n+1−a)(n+3−a)
x2i y
1−a
(|x|2 + |y|2)
n+5−a
2
.
Hence, we have that
∆xP =−Cn,an(n+1−a)
y1−a
(|x|2 + |y|2)
n+3−a
2
+Cn,a(n+1−a)(n+3−a)
|x|2y1−a
(|x|2 + |y|2)
n+5−a
2
Also
Py =Cn,a(1−a)
y1−a
(|x|2 + |y|2)
n+1−a
2
−Cn,a(n+1−a)
y2−a
(|x|2 + |y|2)
n+3−a
2
and
Pyy = −
Cn,aa(1−a)y−(1+a)
(|x|2 + |y|2)
n+1−a
2
−
Cn,a(n+1−a)(3−2a)y1−a
(|x|2 + |y|2)
n+3−a
2
+
Cn,a(n+1−a)(n+3−a)y3−a
(|x|2 + |y|2)
n+5−a
2
.
Therefore, we have
∆xP+
a
y
Py +Pyy = −
Cn,a(n+1−a)(n+3−a)y1−a
(|x|2 + |y|2)
n+3−a
2
+
Cn,a(n+1−a)(n+3−a)y1−a(|x|2 + y2)
(|x|2 + |y|2)
n+5−a
2
= −
Cn,a(n+1−a)(n+3−a)y1−a
(|x|2 + |y|2)
n+3−a
2
+
Cn,a(n+1−a)(n+3−a)y1−a
(|x|2 + |y|2)
n+3−a
2
= 0 .
Proposition 3.4.2. The Poisson kernel P converges to a multiple of the Dirac delta function
as y→ 0, that is
lim
y→0
P(x,y) =Cn,aδ0 .
29
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Proof. From the definition of P, we have
lim
y→0
P(x,y) =Cn,a lim
y→0
y1−a
(|x|2 + |y|2)
n+1−a
2
.
But if this limit exists, then it must be equal to
lim
y→0
P(0,y) = Cn,a lim
y→0
y1−a
yn+1−a
= Cn,a lim
y→0
y−n
= Cn,aδ0 .
Note that we can also obtain the same result by taking the limit of P(x,y) = y−nP(x/y,1)
as y→ 0. Thus the P(x,y) we obtained is actually the Poisson kernel associated with the
differential equation in (2.3).
We saw earlier that (2.5) was obtained from (2.3) by a suitable change of variables and so
with the same change of variables we have that
P˜(x,z) =Cn,a
(1−a)1−az
(
|x|2 +(1−a)2|z|
2
1−a
) n+1−a
2
is the Poisson kernel associated with (2.5) . We show this explicitly in the next proposition.
Proposition 3.4.3. The corresponding Poisson kernel
P˜(x,z) = C˜n,a
z
(
|x|2 +(1−a)2|z|
2
1−a
) n+1−a
2
solves the differential equation in (2.5), where C˜n,a is a constant that depends on n and a.
Proof. From
P˜(x,z) = C˜n,a
z
(
|x|2 +(1−a)2|z|
2
1−a
) n+1−a
2
we have
P˜xi(x,z) =−C˜n,a
(n+1−a)xiz
(
|x|2 +(1−a)2|z|
2
1−a
) n+3−a
2
and
P˜xixi(x,z) =−C˜n,a
(n+1−a)z
(
|x|2 +(1−a)2|z|
2
1−a
) n+3−a
2
+C˜n,a
(n+1−a)(n+3−a)x2i z
(
|x|2 +(1−a)2|z|
2
1−a
) n+5−a
2
.
30
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Therefore, we have
∆xP˜ =−C˜n,a
n(n+1−a)z
(
|x|2 +(1−a)2|z|
2
1−a
) n+3−a
2
+C˜n,a
(n+1−a)(n+3−a)|x|2z
(
|x|2 +(1−a)2|z|
2
1−a
) n+5−a
2
.
Also
P˜z = C˜n,a
1
(
|x|2 +(1−a)2|z|
2
1−a
) n+1−a
2
−C˜n,a
(n+1−a)(1−a)z
2
1−a
(
|x|2 +(1−a)2|z|
2
1−a
) n+3−a
2
,
and
P˜zz = −
C˜n,a(n+1−a)(1−a)z
1+a
1−a
(
|x|2 +(1−a)2|z|
2
1−a
) n+3−a
2
−
2C˜n,a(n+1−a)z
1+a
1−a
(
|x|2 +(1−a)2|z|
2
1−a
) n+3−a
2
+
C˜n,a(n+1−a)(n+3−a)(1−a)2z
3+a
1−a
(
|x|2 +(1−a)2|z|
2
1−a
) n+5−a
2
,
so
zα P˜zz =−
C˜n,a(n+1−a)(3−a)z
(
|x|2 +(1−a)2|z|
2
1−a
) n+3−a
2
+
C˜n,a(n+1−a)(n+3−a)(1−a)2z
3−a
1−a
(
|x|2 +(1−a)2|z|
2
1−a
) n+5−a
2
.
Therefore
∆xP˜+ z
α P˜zz = −
C˜n,a(n+1−a)(n+3−a)z
(
|x|2 +(1−a)2|z|
2
1−a
) n+3−a
2
+
C˜n,a(n+1−a)(n+3−a)
(
|x|2 +(1−a)2z
2
1−a
)
z
(
|x|2 +(1−a)2|z|
2
1−a
) n+5−a
2
,
= −
C˜n,a(n+1−a)(n+3−a)z
(
|x|2 +(1−a)2|z|
2
1−a
) n+3−a
2
+
C˜n,a(n+1−a)(n+3−a)z
(
|x|2 +(1−a)2|z|
2
1−a
) n+3−a
2
,
= 0 .
We now estimate the value of the constants Cn,a and C˜n,a in the Poisson kernels we have just
considered.
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(i) To find Cn,a, consider
P(x,y) =Cn,a
y1−a
(|x|2 + |y|2)
n+1−a
2
.
Since
∫
P(x,y)dx = 1, we have
Cn,a
∫
Rn
y1−a
(|x|2 + |y|2)
n+1−a
2
dX = 1 , where dX = dx1dx2 . . .dxn .
Let x j = y tan t j. Then dx j = y(1+ tan2 t j)dt j and
dX = ynΠnj=1(1+ tan
2 t j)dt j , |x|
2 = y2
n
∑
j=1
tan2 t j
Hence, we have
Cn,a
yn+1−a
yn+1−a
∫ pi
2
− pi2
· · ·
∫ pi
2
− pi2
Πnj=1(1+ tan
2 t j)
(
1+∑nj=1 tan
2 t j
) n+1−a
2
dt1dt2 . . .dtn = 1 .
That is
Cn,a ·K = 1
∴ Cn,a =
1
K
where K =
∫ pi
2
− pi2
· · ·
∫ pi
2
− pi2
Πnj=1(1+ tan
2 t j)
(
1+∑nj=1 tan
2 t j
) n+1−a
2
dt1dt2 . . .dtn .
(ii) To evaluate C˜n,a , consider
P˜(x,z) = C˜n,a
z
(
|x|2 +(1−a)2|z|
2
1−a
) n+1−a
2
.
Let x j = (1− a)z
1
1−a tan t j . Then dx j = (1− a)z
1
1−a
(
1+ tan2 t j
)
dt j and dX =
(1−a)nz
n
1−aΠnj=1
(
1+ tan2 t j
)
dt j . Hence,
C˜n,a
(1−a)nz
n+1−a
1−a
(1−a)n+1−az
n+1−a
1−a
∫ pi
2
− pi2
· · ·
∫ pi
2
− pi2
Πnj=1(1+ tan
2 t j)
(
1+∑nj=1 tan
2 t j
) n+1−a
2
dt1dt2 . . .dtn = 1 .
32
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That is
C˜n,a ·
(
1
1−a
)1−a
·K = 1
∴ C˜n,a = (1−a)
1−a ·
1
K
C˜n,a = (1−a)
1−aCn,a .
33
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Chapter 4
The Relation of the extension problem to
the fractional Laplacian
We shall in this chapter see the relationship between the differential equations
∆xu+
a
y
uy +uyy = 0 and ∆xu+ z
αuzz = 0
and the operator (−∆)s, where s ∈ (0,1). We shall achieve this by showing that
Proposition 4.0.4.
lim
y→0
yauy(x,y) = uz(x,0) =−(−∆)
s f (x) =
∫
Rn
f (ξ )− f (x)
|ξ − x|n+2s dξ (4.1)
holds up to a constant factor.
The Poisson formula developed in Chapter 3 and the Fourier transform will be used to estab-
lish 4.1. We shall break up Proposition (4.0.4) into two propositions.
4.1 Proof Using The Poisson Formula
First we compute uz(x,0) using the Poisson formula obtained in the previous section. A
Proof is found in [2] , but we shall give an alternate proof as well.
Our first result for this section is stated as follows.
Proposition 4.1.1. Let u(x,z) be the solution of (2.5), then
uz(x,0) =−k(−∆)
1−a
2 f (x) , for some constant k .
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Proof. (See [2]) From the definition of the partial derivative, we have
uz(x,0) = lim
h→0
u(x,h)−u(x,0)
h
In terms of the Poisson kernels, this becomes
uz(x,0) = lim
h→0
1
h
∫
Rn
P˜(x−ξ ,h)( f (ξ )− f (x))dξ
= lim
h→0
1
h
∫
Rn
C˜n,a h
(
|x−ξ |2 +(1−a)2h
2
1−a
) n+1−a
2
( f (ξ )− f (x)) dξ
= lim
h→0
∫
Rn
C˜n,a
(
|x−ξ |2 +(1−a)2h
2
1−a
) n+1−a
2
( f (ξ )− f (x))dξ
=
∫
Rn
lim
h→0
C˜n,a
(
|x−ξ |2 +(1−a)2h
2
1−a
) n+1−a
2
( f (ξ )− f (x)) dξ
(since the integral converges absolutely)
= kCn
∫
Rn
f (ξ )− f (x)
|x−ξ |n+1−a dξ , where C˜n,a = kCn for some constant k
= −k(−∆)
1−a
2 f (x) (See[2]).
Proof. Alternative Proof
We can write u(x,z) as
u(x,z) =Cn,a
∫
Rn
z( f (ξ )− f (x))
(
|x−ξ |2 +(1−a)2z
2
1−a
) n+1−a
2
dξ + f (x) .
Differentiating under the integral, we obtain
uz(x,z) = C˜n,a
∫
Rn




1
(
|x−ξ |2 +(1−a)2z
2
1−a
) n+1−a
2
−
(n+1−a)(1−a)z
2
1−a
(
|x−ξ |2 +(1−a)2z
2
1−a
) n+3−a
2



( f (ξ )− f (x)) dξ
therefore uz(x,0) = C˜n,a
∫
Rn
f (ξ )− f (x)
|x−ξ |n+1−a dξ
= −k(−∆)
1−a
2 f (x) as required.
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4.2 Proof Using Fourier Transform
We can also prove (4.1) using the Fourier transform via the energy of u. One way to do this
is to show that the integral of the energy functional |∇u|2ya coincides with the integral of its
Fourier transform since the Fourier transform preserves the energy of a functional. We state
this in the next proposition.
Proposition 4.2.1. (See [2])
∫
y>0
|∇u|2yadX =
∫
Rn
|ξ |2s | fˆ (ξ )|2 dξ .
Proof. We first reduce the partial differential equation in (2.3) to an ordinary differential
equation by taking the Fourier transform in the variable x. We thus obtain
−|ξ |2uˆ(ξ ,y)+ a
y
uˆy(ξ ,y)+ uˆyy(ξ ,y) = 0
which is an ODE for each fixed value of ξ .
Also, consider the one-dimensional functional for each φ : [0,+∞)→ R , given by
J(φ) :=
∫
y>0
(|φ ′|2 + |φ |2)ya dy
and suppose φ minimizes the functional J(φ) for φ(0) = 1.
Then φ solves the Euler-Lagrange equation
−φˆ(y)+ a
y
φˆy(y)+ φˆyy(y) = 0
subject to the conditions
φ(0) = 1
and
lim
y→∞
φ(y) = 0 .
This is because the Euler-Lagrange equation for J is given by
Fφ −
d
dy
Fφ ′ = 0
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where
F =
(
|φ ′|2 + |φ |2
)
ya
= ya(φ ′)2 + yaφ2 ,
so, Fφ = 2yaφ , Fφ ′ = 2y
aφ ′ and ddyFφ ′ = 2ay
a−1φ ′+2yaφ ′′ ,
hence, Fφ − ddyFφ ′ = 0 becomes
2yaφ −2aya−1φ ′−2yaφ ′′ = 0
φ − a
y
φ ′−φ ′′ = 0 , as required.
We shall show shortly that with a suitable change of variables, the Euler-Lagrange equation
obtained above is in fact the modified Bessel differential equation. But in the mean time, we
know that
|∇u|2 =
∣
∣
∣
∣
∂u
∂x1
∣
∣
∣
∣
2
+ · · ·+
∣
∣
∣
∣
∂u
∂xn
∣
∣
∣
∣
2
+
∣
∣
∣
∣
∂u
∂y
∣
∣
∣
∣
2
=
n
∑
j=1
∣
∣
∣
∣
∂u
∂x j
∣
∣
∣
∣
2
+
∣
∣
∣
∣
∂u
∂y
∣
∣
∣
∣
2
. (4.2)
Also, from
̂( ∂
∂x
)α
f (x) = (iξ )α fˆ (ξ )
and the Plancherel’s theorem, if we take the Fourier transform in the variable x of (4.2),we
have
∫
Rn
∫
y>0
|∇u(x,y)|2 ya dydx =
∫
Rn
∫
y>0
|F (∇u(ξ ,y))|2 ya dydξ
=
∫
Rn
∫
y>0
(
n
∑
j=1
∣
∣
∣
∣F
(
∂u
∂x j
)∣
∣
∣
∣
2
+
∣
∣
∣
∣F
(
∂u
∂y
)∣
∣
∣
∣
2
)
ya dydξ
=
∫
Rn
∫
y>0
(
|ξ |2 |uˆ(ξ ,y)|2 +
∣
∣uˆy(ξ ,y)
∣
∣2
)
ya dydξ ,
where dξ = dξ1dξ2 · · ·dξn .
We now show that the equation
−φˆ(y)+ a
y
φˆy(y)+ φˆyy(y) = 0
is the modified Bessel differential equation. If we make the change of variable
φˆ(y) = ysϕˆ(y),
where y > 0, s = 1−a2 , and 0 < a < 1
37
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then
φˆy = sys−1ϕ(y)+ ysϕˆy and φˆyy = s(s−1)ys−2ϕˆ+2sys−1ϕˆy + ysϕˆyy .
Hence the equation −φˆ(y)+ ay φˆy(y)+ φˆyy(y) = 0, becomes
−ysϕˆ+ a
y
(
sys−1ϕˆ+ ysϕˆy
)
+ s(s−1)ys−2ϕˆ+2sys−1ϕˆy + ysϕˆyy = 0
ysϕˆyy +(a+2s)ys−1ϕˆy−
(
y2−as− s(s−1)
)
ys−2ϕˆ = 0
ys−2{y2ϕˆyy +(a+2s)yϕˆy−
(
y2−as− s(s−1)
)
ϕˆ}= 0 .
But since ys−2 6= 0, we obtain
y2ϕˆyy +(a+2s)yϕˆy−
(
y2−as− s(s−1)
)
ϕˆ = 0 .
And the fact that a+2s = 1, gives
as+ s(s−1) = s(1−2s)+ s(s−1)
= s−2s2 + s2− s
= −s2
Upon substitution, our differential equation then becomes
y2ϕˆyy + yϕˆy− (y2 + s2)ϕˆ = 0
which is the modified Bessel differential equation. At this point in our work, the well known
solution of the modified Bessel equation will help us understand the energy functional asso-
ciated with the fractional Laplacian. We state the result in the form of a lemma.
Lemma 4.2.2. The ordinary differential equation
φyy(y)+
a
y
φy(y)−φ(y) = 0
defined for y ∈ R+, has two linearly independent solutions. In particular
φ(y) = ys (c1Is(y)+ c2Ks(y))
for constants c1,c2 ∈R. Is(y) is the modified Bessel function of the first kind and Ks(y) is the
modified Bessel function of the second kind (See [4]).
Proof. By the change φ(y) = ysϕ(y), the Euler- Lagrange equation for the energy functional
is transformed into the modified Bessel equation
y2ϕyy + yϕy− (y2 + s2)ϕ = 0 , (4.3)
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We shall first solve the Bessel equation
y2ϕyy + yϕy +(y2− s2)ϕ = 0 (4.4)
and then do some substitutions to obtain the solutions of (4.3). We notice that (4.3) has a
regular singular point at y = 0, so we expect solutions of the form
ϕ(y) =
∞
∑
k=0
aky
k+b , where a0 6= 0 (4.5)
and the b and ak’s are to be determined. Differentiate (4.5) with respect to y and substitute
into (4.4) to get
ϕ ′(y) =
∞
∑
k=0
(k+b)ak y
k+b−1, ϕ ′′(y) =
∞
∑
k=0
ak(k+b)(k+b−1) y
k+b−2
and
∞
∑
k=0
ak
(
(k+b)(k+b−1) yk+b +(k+b)yk+b + yk+b+2− s2 yk+b
)
= 0 . (4.6)
Expanding and relabelling the index of summation of the terms yk+b+2, we have
∞
∑
k=0
ak y
k+b+2 = a0y
b+2 +a1y
b+3 +a2y
b+4 + · · ·+any
b+(n+2)+ · · ·
=
∞
∑
k=0
ak−2 y
k+b .
Hence, equation (4.6) becomes
∞
∑
k=0
ak[(k+b)
2− s2] yk+b +
∞
∑
k=2
ak−2 y
k+b = 0 . (4.7)
The power series in (4.7) will vanish identically only when its coefficients are all zero. This
leads us to the following recurrence relations.
(b2− s2)a0 = 0 , for k = 0 . (4.8)
(
(1+b)2− s2
)
a1 = 0 , for k = 1 . (4.9)
(
(k+b)2− s2
)
ak +ak−2 = 0 , for k ≥ 2 . (4.10)
Since a0 6= 0, then (4.8) gives b = ±s. We substitute b = s into (4.9) to get (1+ 2s)a1 = 0.
We notice that choosing b = −s will lead us to the same result. This implies that a1 = 0
39
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except when s =−1/2. From (4.10), we have the following recursion formula.
ak = −
ak−2
(k+ s)2− s2
,
= −
ak−2
k(k+2s)
. (4.11)
From (4.11), we obtain the even-numbered coefficients in terms of a0 as
a2 =−
a0
2(2+2s)
, a4 =−
a2
4(4+2s)
=
a0
2 ·4(2+2s)(4+2s)
a6 =−
a4
6(6+2s)
=−
a0
2 ·4 ·6(2+2s)(4+2s)(6+2s)
In general, we have
a2 j =
(−1) ja0
2 ·4 ·6 · · ·2 j(2+2s)(4+2s)(6+2s) · · ·(2 j+2s)
,
=
(−1) ja0
22 j j!(1+ s)(2+ s) · · ·( j+ s)
.
We also obtain all odd-numbered coefficients a2 j+1 in terms of a1. But we already established
that a1 = 0. Hence, we have that a2 j+1 = 0.
A careful examination of the a2 j’s reveals that we shall obtain a zero in the denominator
when s is a negative integer or a half-integer. That is, if s =−n/2 where n is odd, then (4.11)
has a zero in the denominator for k = n. This leads to the conclusion that except when s is a
negative integer
Js(y) = a0
∞
∑
j=0
(−1) j y2 j+s
22 j j!(1+ s)(2+ s) · · ·( j+ s)
is a solution to (4.4). We have to choose the constant a0, bearing in mind that a0 6= 0. A
standard choice is a0 = 12s Γ(s+1) (See [13]). So we have by substitution that
Js(y) =
∞
∑
j=0
(−1) j
j! (1+ s)(2+ s) · · ·( j+ s)Γ(s+1)
·
(y
2
)2 j+s
.
Since ( j+ s)( j−1+ s) · · ·( j− ( j−1)+ s)Γ(s+1) = Γ( j+ s+1), we have
Js(y) =
∞
∑
j=0
(−1) j
j! Γ( j+ s+1)
·
(y
2
)2 j+s
. (4.12)
The function Js(y) is called the Bessel function of the first kind of order s. When we set
40
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s =−s in (4.12), we obtain
J−s(y) =
∞
∑
j=s
(−1) j
j! ( j− s)!
·
(y
2
)2 j−s
, since 1/Γ(x) = 0 for x = 0,−1,−2,−3, . . . ,
=
∞
∑
m=0
(−1)m+s
m! (m+ s)!
(y
2
)2m+s
.
Comparing Js(y) and J−s(y), we have J−s(y) = (−1)sJs(y). This means that Js(y) and J−s(y)
are two linearly dependent solutions when s is an integer and so we look for second linearly
independent solution .The Weber function, Ys(y) which is also known as Bessel function of
the second kind and given by
Ys(y) =
cos (pis) Js(y)− J−s(y)
sin pis
is taken as the second linearly independent solution.
Now, we realize that if we make the change y→ iy in (4.4), we obtain (4.3). Hence, one
of the solutions of (4.3) is Js(iy). But a common choice is i−sJs(iy) (See [13]), so the first
solution is
Is(y) = i
−s Js(iy) =
∞
∑
m=0
1
m!Γ(m+ s+1)
(y
2
)2m+s
.
A second independent solution is obtained using Ys(iy) for arbitrary s. But the standard
choice according to [13] is
Ks(y) =
I−s(y)− Is(y)
sin pis .
We are interested in the asymptotic behaviour of Is and Ks. This is given by the following;
Is(y)∼
1
Γ(s+1)
(y
2
)s
and Ks(y)∼
Γ(s)
2
(
2
y
)s
, as y→ 0+ .
And as y→+∞, we have
Is(y)∼
1
√
2piy
ey and Ks(y)∼
√
pi
2y
e−y .
Since φ minimizes J, then using the values of Is and Ks as y→∞ we have c1 = 0 and c2 =
Cs, some constant that depends on s.
Hence, the minimizer we seek must be of the form
uˆ(ξ ,y) = fˆ (ξ )φ(|ξ |y) . (4.13)
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We can now return to our computation of the energy functional
∫
y>0 |∇u|
2yadX . We had
∫
y>0
|∇u|2ya dX =
∫
Rn
∫
y>0
(
|ξ |2|uˆ(ξ ,y)|2 + |uˆy(ξ ,y)|2
)
ya dy dξ ,
=
∫
Rn
∫
y>0
(
|ξ |2
∣
∣ fˆ (ξ )φ(|ξ |y)
∣
∣2
+
∣
∣
∣ fˆ (ξ )φ
′
(|ξ |y) · |ξ |
∣
∣
∣
2
)
ya dy dξ , by (4.13)
=
∫
Rn
∫
y>0
|ξ |2| fˆ (ξ )|2
(
|φ(|ξ |y)|2 + |φ ′(|ξ |y)|2
)
ya dy dξ .
Let t = |ξ |y
Then dy = |ξ |−1dt, and ya = ta|ξ |−a .
Therefore,
∫
y>0
|∇u|2yadX =
∫
Rn
∫
y>0
|ξ |2| fˆ (ξ )|2
(
|φ(t)|2 + |φ ′(t)|2
)
|ξ |−a−1ta dt dξ
=
∫
Rn
|ξ |1−a| fˆ (ξ )|2
∫
y>0
(
|φ(t)|2 + |φ ′(t)|2
)
ta dt dξ ,
=
∫
Rn
| fˆ (ξ )|2|ξ |1−aJ(φ) dξ ,
= J(φ)
∫
Rn
| fˆ (ξ )|2|ξ |2s dξ , as required.
This means that the Euler-Lagrange equation for each energy must coincide up to a constant
factor.
Hence, we obtain
− lim
y→+0
yauy(x,y) =C(−∆)
1−a
2 f (x) ,
where the constant C is given by J(φ) (See [2]) .
We shall tackle the proof for the partial differential equation in (2.5). As in the previous case,
we take the Fourier transform in the variable x to obtain
−|ξ |2uˆ(ξ ,z)+ zα uˆzz(ξ ,z) = 0 ,
which is an ordinary differential equation for each value of ξ .
Suppose that φ : [0,+∞)→ R, solves the differential equation
−zαφ ′′(z)+φ(z) = 0
subject to the conditions
φ(0) = 1 and limz→∞φ(z) = 0 .
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Then we notice in this also that the solution we are looking for must be of the form
uˆ(ξ ,z) = uˆ(ξ ,0)φ(|ξ |
2
2−α z) .
So
uˆz(ξ ,z) = uˆz(ξ ,0) · |ξ |
2
2−α ·φ ′(|ξ |
2
2−α z) ,
therefore uˆz(ξ ,0) = uˆz(ξ ,0)|ξ |
2
2−α φ ′(0) ,
= uˆz(ξ ,0)|ξ |1−aφ
′
(0) , since α =− 2a
1−a
.
From the boundary conditions, uˆ(ξ ,0) = fˆ (ξ ) .
Hence
uˆz(ξ ,0) =Ca|ξ |1−a fˆ (ξ ) , where Ca = φ
′
(0).
43
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Chapter 5
Reflection Extensions
In this chapter, we apply the interior Harnack estimates to the differential equation in (2.3).
In order to achieve this, we must show that if (−∆)s f = 0 holds in an open ball, then we can
reflect the solution u and make it a solution of the differential equation (2.3) across y = 0 or
a solution of (2.5) across z = 0 in a suitable sense. Our main tool is the well known Harnack
inequality.
Theorem 5.0.3. (The Harnack Inequality) Suppose Ω⊂ Rn and u≥ 0 is a C2 solution of
Lu = 0 in Ω
where L is the Laplace operator. Suppose Ω0 is a bounded domain such that Ω0 ⊂ Ω is
connected. Then there exists a constant C such that
sup
x∈Ω0
u(x)≤C inf
x∈Ω0
u(x) .
The constant C depends only on the domains Ω and Ω0.
Remark 5.0.4. Literally speaking, Harnack’s inequality asserts that the values of a non-
negative solution of an elliptic PDE are comparable. That is, we can compare the values of
a non-negative solution of an elliptic partial differential equation at least in any subregion
away from the boundary of the domain.
Proof. (The Harnack Inequality)
Let Ω ⊂ Rn and suppose that x ∈ Ω. Let u be a non-negative function satisfying the mean
value property in Ω. Then for y ∈ Br(x) where r > 0, we see that Br(y)⊂ B2r(x) and since u
is non-negative we have
u(y) =
n
αnrn
∫
Br(y)
u dx = −
∫
Br(y)
u dx≤ 2n −
∫
B2r(y)
u(x) dx . (5.1)
44
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In a similar manner, if we take another point z ∈ Br(x), then we see that B2r(x) ⊂ B3r(z).
From the mean value property, we shall obtain
u(z) = −
∫
B3r(z)
u dx≥
(
2
3
)n
−
∫
B2r(x)
u dx . (5.2)
From (5.1) and (5.2) we have
sup
B2r(x)
u≤ 2n −
∫
B2r(x)
u dx and inf
B2r(x)
u≥
2n
3n
−
∫
B2r(x)
u dx .
This implies that
sup
B2r(x)
u≤ 3n inf
B2r(x)
u
From section (2.2), we saw that the differential equation in (2.3) can be written in a diver-
gence form and (2.5) in a nondivergence form. So we shall address both the divergent and
the nondivergent cases.
5.1 The Divergent Case
Lemma 5.1.1 (Found in [2]). Suppose that u : Rn× [0,∞)→ R is a solution of (2.3) such
that for |x| ≤ r
lim
y→0
yauy(x,y) = 0 (5.3)
then the extension to the whole space
u˜(x,y) :=
{
u(x,y) , y≥ 0
u(x,−y) , y < 0
is a solution to
div(|y|a ∇u) = 0
in the weak sense in the (n+ 1) dimensional ball of radius R. That is, BR = {(x,y) : |x|2 +
|y|2 ≤ R2}.
Proof. We first note that since u˜ is a solution in the weak sense, we will need a test function
in our computation.
Let v ∈C∞0 (BR) be a test function.
We want to show that ∫
BR
∇u˜ ·∇v |y|a dX = 0 . (5.4)
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In evaluating the integral above, we realize that the integral vanishes when y = 0. And to
overcome this difficulty, we separate a strip of width say, ε > 0 around y = 0 in the domain
of integration. So we can then write the integral as
∫
BR
∇u˜ ·∇v |y|a dX =
∫
BR\{|y|<ε}
∇u˜ ·∇v |y|a dX +
∫
BR∩{|y|<ε}
∇u˜ ·∇v |y|a dX . (5.5)
And since
div(|y|av∇u˜) = ∇ ·
[
v
(
|y|a
n
∑
j=1
∂ u˜
∂x j
+ |y|au˜y
)
]
,
= ∇v · |y|a∇u˜+ v
(
|y|a∆xu˜+a|y|
a−1u˜y + |y|
au˜yy
)
,
= ∇v ·∇u˜|y|a + v|y|a
(
∆xu˜+
a
|y|
u˜y + u˜yy
)
,
= ∇v ·∇u˜|y|a ,
we have
∫
BR
∇u˜ ·∇v |y|a dX =
∫
BR\{|y|<ε}
div(|y|a v ·∇u˜) dX +
∫
BR∩{|y|<ε}
∇u˜ ·∇v |y|a dX .
Applying Stoke’s theorem to the first integral on the right hand side, we obtain
∫
BR
∇u˜ ·∇v |y|a dX =
∫
BR∩{|y|<ε}
v · u˜y(x,ε) |ε|a dx+
∫
BR∩{|y|<ε}
∇u˜ ·∇v |y|a dX . (5.6)
Now, we see that as ε → 0 , εau˜y(x,ε)→ 0 and so the first integral on the right hand
side of (5.5) converges to zero. Also as ε → 0 , the second integral goes to zero since
|y|a |∇u˜|2 is locally integrable.
Hence, ∫
BR
∇u˜ ·∇v |y|a dX → 0
thus div(|y|a ∇u˜) = 0 , in the whole ball BR .
5.2 The Nondivergence Case
Lemma 5.2.1. Let g be a continuous function on ∂BR such that g(x,z) = g(x,−z). Then
there exists a unique function u ∈C(B¯R) which coincides with g on ∂BR such that
i. u solves uxx + |z|αuzz = 0 in BR∪{z 6= 0}
ii. u ∈C1(BR) and
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iii. uz(x,0) = 0 .
Proof. As pointed out by [2], we see that any viscosity solution of (2.5) is C2 away from
z = 0. It turns out that this solution is a solution in the classical sense in BR∪{z = 0}. The
proof of the lemma is in two parts - uniqueness and existence of such a solution.
5.2.2 Uniqueness
For this case, we want to show that any two solutions u and v satisfying (2.5) must coincide
in the ball BR. We shall follow the proof in [2] .
Let u be a solution satisfying (i), (ii) and (iii). Suppose v is another solution satisfying (i),
(ii) and (iii).
Then for any ε > 0, we set w = u− v+ ε|z|. Then we see that w≤ εR on ∂BR,
since u = v = g on ∂BR and |z| ≤ R.
Also w is harmonic by definition since from
w = u− v+ ε|z| , we have
∆w = ∆u−∆v
= 0 , in BR .
Next, suppose that w has a maximum value at a point x ∈ BR. Then this point cannot be in
BR∩{z 6= 0} since we know that a harmonic function attains its maximum on the boundary
of its domain. So w does not have an interior maximum point in BR∩{z 6= 0}.
This means that any interior maximum point has to be on BR∩{z = 0}
wz+ = uz− vz + ε
wz− = uz− vz− ε
wz+−wz− = 2ε
> 0
therefore wz+ > wz− .
This means that there cannot be a maximum. Therefore we have w < εR in the whole ball
BR. And since ε > 0 is arbitrary, we can let ε → 0. This gives
u− v≤ 0 in BR
u≤ v in BR .
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Similarly, if we replace u by−u and define w = v−u+ε|z| and following the same argument
we obtain
v≤ u in BR .
Hence u = v in BR .
5.2.3 Existence
To prove that there exists a C1 solution to (2.5), we prove that there exists a uniform C1,α
estimate for (2.5). We achieve this by proving a uniform C1,α estimate for solutions to the
problem
∆xu
ε +(|z|+ ε)αuεzz = 0 in BR ,
with uε = g on ∂BR .
For any ε > 0, by the Schauder theory, the given problem has a classical solution (See [2]).
The Hölder estimate of Caffarelli and Gutierrez( 1997) is given as
|u(x)−u(y)| ≤ oscsφ (x,ρ0)u≤
(ρ0
R
)α
M(R)≤C|x− y|α/ε2M(R) , (5.7)
where u is a nonnegative solution of Lφu = 0 in the section Sφ (z, t) defined by
Sφ (z, t) = {y ∈ Rn : φ(y)< `(y)+ t},
where `(y) is the supporting hyperplane of φ at the point (z,φ(z)), φ(y) is a convex function
on Rn, ` is an arbitrary affine function, t > 0 and oscSφ (x,ρ0)u = maxSφ (x,ρ0) u−minSφ (x,ρ0) u.
We will now give the proof of (5.7) which is due to [1], but we will make a definition before
the proof.
Definition 5.2.4 (The Engulfing Property of Sections (See [1])). The engulfing property
of sections states that for any section Sφ (z, t) with y ∈ Sφ (z, t), there exists a constant θ > 0
such that Sφ (z, t)⊂ Sφ (y,θ t)
Proof. (See [1]) Let Sφ (x0,R) be a section and let x,y ∈ Sφ (x0,R/2). Also, let Sφ (x,ρ0) be
the smallest section containing y. That is,
ρ0 = inf {ρ : y ∈ Sφ (x,ρ)} (5.8)
Then by the engulfing property of the sections we have Sφ (x0,R) ⊂ Sφ (x,θR). In particu-
lar, y ∈ Sφ (x,θR) which gives ρ0 ≤ θR. Let T be an affine transformation that normalizes
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Sφ (x0,θR). That is,
B(0,1/n)⊂ T(Sφ (x0,θR))⊂ B(0,1) ,n > 1. (5.9)
So that Sφ (x0,θR)∩Sφ (x0,ρ0) 6= /0 and ρ0 ≤ θR. Then by the properties of sections, we have
B
(
z,K2
( ρ0
Rθ
)ε2
)
⊂ T(Sφ (x,ρ0))⊂ B
(
z,K1
( ρ0
Rθ
)ε1
)
, (5.10)
for some z∈B(0,K3) depending on the sections Sφ (x0,θR) and Sφ (x,ρ0) and Tx∈B
(
z, 12K2
( ρ0
Rθ
)ε2
)
,
where K1, K2 and K3 are some positive constants. We estimate ρ0 from below as follows.
From (5.8) we have
Ty ∈ T
(
Sφ (x,ρ0)
)
⊂ B
(
z,K1
( ρ0
Rθ
)ε1
)
. (5.11)
Then
|Tx−Ty| = |Tx−Tz+Tz−Ty|
≤ |Tx−Tz|+ |Tz−Ty|
≤
1
2
K2
( ρ0
Rθ
)ε2
+K1
( ρ0
Rθ
)ε1
≤ 2K1
( ρ0
Rθ
)ε1
.
Also
|x− y| =
∣
∣T−1Tx−T−1Ty
∣
∣
≤ ‖T−1‖ |Tx−Ty|
≤ ‖T−1‖ 2K1
( ρ0
Rθ
)ε1
.
Hence, we have
Rθ
(2K1‖T−1‖)
1/ε1
· |x− y|1/ε1 ≤ ρ0 . (5.12)
Next, we estimate ρ0 from above as follows. From (∗), we see that if y /∈ Sφ (x,ρ0/2), then
Ty /∈ T(Sφ (x,ρ0/2)). This implies that Ty /∈ B
(
z,K2
( ρ0
2Rθ
)ε2
)
, that is,
|Ty−Tz|> K2
( ρ0
2Rθ
)ε2 . Also, Tx ∈ B
(
z, 12K2
( ρ0
2Rθ
)ε2
)
, and so we have
K2
( ρ0
2Rθ
)ε2
< |Ty−Tz|
≤ |Ty−Tx|+ |Tx−Tz|
≤ |Ty−Tx|+
1
2
K2
( ρ0
2Rθ
)ε2
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That is
1
2
K2
( ρ0
2Rθ
)ε2
< |Ty−Tx|
≤ ‖T‖ |x− y|
ρ0 ≤ 2Rθ ·
(
2‖T‖
K2
)1/ε2
· |x− y|1/ε2
(ρ0
R
)α
≤ (2θ)α
(
2‖T‖
K2
)α/ε2
|x− y|α/ε2
(ρ0
R
)α
≤ C |x− y|α/ε2 (5.13)
The right hand inequality then follows from (5.12) and (5.13).
We have
|u(x)−u(y)| ≤C|x− y|α/ε2M(R) .
By definition of the partial derivative, we have
uεz (x,z) = lim
h→0
u(x,z+h)−u(x,z)
h
, where |h|< ε
and
|uεz (x,z)| ≤ lim
h→0
|u(x,z+h)−u(x,z)|
|h|
= lim
h→0
|u(x,z+h)−u(x,z)|
|h|α/ε2
· |h|α/ε2−1
that is |uεz (x,z)| ≤ CM(R) · |h|
α
ε2
−1
≤ R
α
ε2
−1
CM(R) , where
α
ε2
−1 > 0
This means that uεz is bounded independently of ε in BR.
Now from
∆xu
ε +(|z|+ ε)αuεzz = 0 ,
we have
∆xu
ε =−(|z|+ ε)αuεzz
and
|∆xu
ε | = |(|z|+ ε)α ||uεzz|
≤ |z|α |uεzz| .
Hence, we see that ∆xuε is bounded independently of ε in BR since uεzz is bounded and |z| ≤ R
in BR. Also, the remark by [2] that ∆xuε is bounded in any smaller ball Br where r =(1−δ/2)
holds.
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Furthermore, since uε ∈ C2(BR) and u is symmetric with respect to the hyperplane z = 0,
then uεz (x,0) = 0 .
This is so because
uεz (x,0) = lim
h→0
u(x,h)−u(x,0)
h
, where |h|< ε
|uεz (x,0)|= lim
h→0
|u(x,h)−u(x,0)|
|h|α/ε2
· |h|α/ε2−1
Then from the estimate above, we have
|uεz (x,0)| ≤CM(R) · lim
h→0
|h|α/ε2−1 = 0 .
Again from
∆xu
ε +(|z|+ ε)αuεzz = 0 ,
we have
uεzz =−
∆xuε
(|z|+ ε)α
|uεzz| =
|∆xuε |
|(|z|+ ε)α |
≤
C
|z|α
,
where α =− 2a1−a and since a ∈ (−1,1), we have α < 1 .
Integrating uεzz for any value of x,z such that |x|< 1−2δ and 0 < z < 1−δ , we have
uεz (x,z) =
∫ z
0
uεzz(x,z) ds
=−
∫ z
0
∆xuε(x,s)
(|z|+ ε)α ds
therefore
|uεz (x,z)|=
∣
∣
∣
∣
∫ z
0
∆xuε(x,s)
(|z|+ ε)α ds
∣
∣
∣
∣
≤
∫ z
0
|∆xuε(x,s)|
(|z|+ ε)α ds
≤
∫ z
0
C
|z|α
ds
=
C
zα
· s
∣
∣
∣
z
0
=Cz1−α .
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Hence, we conclude that uεz is C
η in B(1−δ )R independently of ε for any δ and where η =
min(1,1−α).
So, as pointed out in [2], we can take ε → 0 and extract a subsequence that converges to the
solution u we are interested in finding.
Also, uε is smooth for any ε > 0 and satisfies (5.6) and for which the Harnack inequality
from [1] holds. Therefore the same estimate passes to the limit as ε → 0 and the solution of
the original problem we have satisfies the Harnack inequality.
In conclusion, we would like to say that the basic definition of using C2 test functions would
not be enough to show uniqueness when looking for viscosity solutions to (2.5). We shall
use the example given by [2] to illustrate this fact.
Definition 5.2.5. Let Ω be a bounded domain in Rn. Let u ∈ C(Ω) and assume that ϕ ∈
C2(Ω). Then we say;
i. u is a viscosity supersolution of ∆u = 0 in Ω if for any point x0 ∈Ω such that u−ϕ has
a local minimum at x0, we have ∆ϕ(x0)≤ 0.
ii. u is a viscosity subsolution of ∆u = 0 in Ω if for any point y0 ∈Ω with u−ϕ having a
local maximum at y0, we have ∆ϕ(y0)≥ 0.
We say that u is a viscosity solution if u is a viscosity supersolution and a viscosity subsolu-
tion.
Example 5.2.6. Let u(x,z) = |z| and φ be any C2 function touching u from below at a point
(x,0).Then we see that u and φ are equal at the point (x,0). So φ must satisfy (2.5) at the
point (x,0).
Therefore
∆xφ(x,0)+0αφzz(x,0) = ∆xφ(x,0) , α > 0
≤ 0 .
Hence u(x,z) = |z| would be a (non-differentiable) viscosity solution.
Note that u is a viscosity supersolution in this case. If however, we test against functions of
the form φ(x,z) = z2−α , then u would not be a supersolution.
5.3 Conclusion
In summary, we have shown that the fractional Laplacian (−∆)s for s ∈ (0,1) maps the
Dirichlet boundary condition to the Neumann condition. We also verified that the function Φ
defined in (3.2.1) generates all solutions of the differential equations in (2.3) and (2.5). We
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developed Poisson formulae for the differential equations in (2.3) and (2.5) and also related
the said differential equations to the fractional Laplacian. We applied the interior Harnack
estimates to the differential equations in (2.3) and (2.5) using the Harnack inequality obtained
by Caffarelli and Gutierrez (1997). Although the change of variables in section (2.2) used
to obtain (2.5) was valid for a ∈ (−1,1), yet we were unable to give meaning to R1+a when
a ∈ (−1,1). This problem of giving meaning to the space Rn+1+a when a is not an integer
may be the focus of further research in the future.
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References
[1] Caffarelli, L. and Gutierrez, C. (1997). Properties of the Solutions of the Linearized
Monge-Ampere Equation. Amer. J. Math, 119(2):423–465.
[2] Caffarelli, L. and Silvestre, L. (2007). An Extension Problem Related to the Fractional
Laplacian. Communications in Partial Differential Equations, 32:8:1245 – 1260.
[3] Dahlberg, B. E. J. and Kenig, C. E. (1985). Harmonic Analysis and Partial Differential
Equations. Department, Chalmers TU, Univ., Gothenburg, Sweden.
[4] del Mar, G. M. (2009). Gamma convergence of an energy functional related to the
fractional Laplacian. arXiv:0805.3869.
[5] Evans, L. C. (1998). Partial Differential Equations. Graduate Studies in Mathematics
19. American Mathematical Society.
[6] Gilbarg, D. and Trudinger, N. S. (1998). Elliptic Partial Differential Equations of Second
Order. Springer-Verlag Berlin Heidelberg, New York.
[7] Han, Q. and Lin, F. (1997). Elliptic Partial Differential Equations. Courant lecture notes
in mathematics. AMS, Courant Institute, Providence, Rhode Island.
[8] Murray H. Protter, H. F. W. (1984). Maximum principles in differential equations.
Springer.
[9] Pazy, A. (1992). Semigroups of Linear Operators and Applications to Partial Differential
Equations. Springer-Verlag, New York Inc.
[10] Stein, E. M. (1970). Singular Integrals and Differentiability Properties of Functions.
Princeton Mathematical Series 30. Princeton University Press, Princeton, N.J.
[11] Stein, E. M. and Shakarchi, R. (2005). Real Analysis: Measure Theory, Integration
and Hilbert Spaces. Princeton Lectures in Analysis Bk. 3. Princeton University Press,
Princeton and Oxford.
[12] Stein, E. M. and Weiss, G. (1971). Introduction to Fourier Analysis on Euclidean
Spaces. Princeton University Press, Princeton, New Jersey.
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[13] Tenenbaum, M. and Pollard, H. (2005). Ordinary Differential Equations. Harper’s
Mathematics Series. Harper and Row, New York, N.Y.
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