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A symmetry problem of elliptic differential operators in potential theory

Yu-Ping Wang

Sacred Hearts High School

Abstract

This paper is a study of the equation(−E)α2u(x) = f(x), where(−E)α2 is an (elliptic pseudo-differential) operator defined by

(−E)α2f = 1 Γ(α2)

Z

0 tα2−1(Ht∗f)(x)dt, Ht(x) ≡H(x, t) = p 1

(4πt)nη1η2· · ·ηn

exp −

i

x2i it

! ,

where η1, η2,· · ·, ηnare a set of non-negative numbers that specify the operator. Note that it is an extension of the fractional Laplacian operator(−)α2.

In this paper, we construct a solution, noted as Jαf , by Jαf(x) ≡ 1

β(α) Z

Rn

f(y)

|η−1· (x−y)|n−ady, where|η−1· (x−y)|isqni η−1

i (xi−yi), and β(α)−1equals β(α)−1= √ 1

η1η2· · ·ηn

· Γ(n−α2 ) π

n 22αΓ(α2).

Then if we set f =χwhere χis the indicator function andΩ is some bounded domain inRn, then for all bounded domainΩ that is invariant under reflection trans- formation Pm, namely PmΩ=Ω for all m=1, . . . , n, Jαf ≡Jα(x)satisfies

Jα(x) = Jα(Pmx). The reflection transformation is defined as

Pmx=Pm(x1,· · ·, xm,· · ·, xn) = (x1,· · ·,−xm,· · ·, xn), where m=1, 2, . . . , n.

摘摘要要要: 在這篇報告中, 我們要探討一個方程式(−E)α2u=f ,其中(−E)α2 是一個分數 次的橢圓形微分算子, 其定義為

(−E)α2f = 1 Γ(α2)

Z

0 tα2−1(Ht∗f)(x)dt,

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Ht(x) ≡H(x, t) = p 1

(4πt)nη1η2· · ·ηn

exp −

i

x2i it

! ,

其中 η1, η2,· · ·, ηn是一群決定其算子特性的參數. 而它是從一般的分數次拉普拉斯算 子延伸而得到的.

在報告中, 我們也將找出其一個解, 記為 Jαf ,為 Jαf(x) ≡ 1

β(α) Z

Rn

f(y)

|η· (x−y)|n−ady, 其中|η−1· (x−y)|代表qni η−1

i (xi−yi), 而 β(α)−1等於 β(α)−1= √ 1

η1η2· · ·ηn

· Γ(n−α2 ) π

n 22αΓ(α2).

如果在 Jαf中令 f =χ,其中 χ指示函數, 而 Ω 是一個在 Rn中的有界區域, 則對於所有滿足鏡射變換 Pm的 Ω, 更精確的說, 對於 m =1, . . . , n, 都有 PmΩ =Ω, Jαf ≡J(x)滿足

Jα(x) = Jα(Pmx). 鏡射變換定義為

Pmx=Pm(x1,· · ·, xm,· · ·, xn) = (x1,· · ·,−xm,· · ·, xn) 其中 m=1, 2, . . . , n.

1 Introdution

The basic idea of this paper is derived from an important concept in potential theory, the Riesz potential Iαf . It is known that Riesz potential is closely related to the fractional Laplacian operator. It is actually the inverse operator of (−)α2, namely, u(x) = Iαf if (−)α2u = f [1]. Now we let f ≡ χ, where χ is the indicator function. Then this function denoted as Iα(x)in some bounded domainΩ has an interesting property. Iα(x) is radially symmetric to a center of a ball. In other words, u(x)|∂Ω=const. if and only ifΩ is a ball [4].

In this paper, we will extend the fractional Laplacian to an elliptic operator

(−E)α2u= −

n j

ηj 2

∂x2j

!α2 u,

where η1, η2,· · ·, ηn > 0 and they are independent of the variables. The fractional ex- ponent will be defined in the article. We hope to achieve the following things in the paper:

1. Find the solution of(−∆E)α2u = f , which is denoted by Jαf(x). Then u(x) = Jαf if(−E)α2u(x) = f(x).

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2. Discuss the integrability of Jαf .

3. Discuss the symmetry property of the solution of (−E)α2u = χ where Ω is an n-dimensional ellipsoid centered at origin point and axis parallel to the axis (x1, x2,· · ·, xn)of some cartesian coordinate system.

4. Consider symmetry property of the solution of another equation (−E)α2u = χxi, where i=1, 2,· · ·, n. (The antisymmetric property)

But before doing all this, we will first define some concepts.

1.1 Fractional Laplacian

Now we turn to an important concept of this paper: the fractional Laplacian operator (−)α2. Only the fractional exponent of a positive definite operator can be defined, so we need to take a minus sign in front of the ordinary Laplacian∆.

One way to define(−∆)α2 is to use the Gamma functionΓ(α). We can start from the fact that for any number A [1, 3]:

A−s= 1 Γ(s)

Z

0 ts−1e−tAdt. (1)

If we exchange A to a Laplacian, A7→ −∆, sα2 , then we get the definition.

Definition 1. The fractional Laplacian(−)α2 is defined by (−)α2 f = 1

Γ(α2) Z

0 tα2−1et∆f dt, (2)

where

e∆tf(x) =Gt∗f(x) = Z

RnGt(x−y)f(y)dy (3) and

G(x, t) =Gt(x) = (4πt)−nexp(−|x|2

4t ) ≥0. (4)

Gt(x) is called the Gauss-Weierstrass kernel [1]. It is the fundamental solution of heat equation, and it is not difficult to see why we use it to define et∆

∂Gt(x)

∂t =∆Gt(x) ⇐⇒Gt(x) =e∆t, t>0. (5) However, there is a problem in this definition. When α = −2n, where n is a positive integer, then the Γ(1α

2) = Γ(−n)1 part will be zero, and the integral part diverges. We fix this problem by taking the limit

α→2nlim 1 Γ(α2)

Z

0 tα2−1etAf dt (6)

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where A could be any number, and we find this limit to be Anby using the equation Γ(s+1)

A−s+1 = Z

0 tse−Atdt.

So it is reasonable to redefine the fractional Laplacian by taking limits in the definition of it. Now we can define the fractional Laplacian with a positive integer exponent by

(−)n= lim

α→2n(−)α2. (7)

1.2 Riesz potential

Riesz potential is closely related to the fractional Laplacian, for it can be seen as an inverse of the fractional Laplacian [1].

Definition 2. For any n≥2, 0<α<n, and x∈Rnthe Riesz potential is Iαf(x) = (Kα∗f)(x) = 1

γ(α) Z

Rn

f(y)

|x−y|n−ady, (8) where γ(α)is

γ(α) = π

n22αΓ(α2) Γ(n−α2 ) and

Kα= 1

γ(α)|x|α−n (9)

is called the Risz kernel.

We are going to focus on Riesz potential in a compact domainΩ or 1

γ(α) Z

f(y)

|x−y|n−ady= 1 γ(α)

Z Rn

f(y)

|x−y|n−aχdy, (10) where χ is the indicator function. The Riesz potential is a singular integral operator, so the concept of integrability is important. In other words, the question will be for f ∈ Lp(Ω), and Iαf ∈ Lq(Ω), that p, q satisfy some condition which makes Iα : Lp(Ω) → Lq()a bounded operator.

This property can be seen by the Hardy-Littlewood-Sobolev inequality [2]:

Theorem 1. For 0<α<n, 1≤p, q≤∞, Iα : Lp(Ω) →Lq(Ω)is a bounded operator:

kIαfkq ≤Ckfkp, if n p ≤ n

q +α. (11)

Proof. See [2]. This theorem says that if f ∈ Lp(), then for x ∈ Ω, Iαf(x)converges absolutely.

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We are going to see the relationship between fractional Laplacian and the Riesz po- tential.

Theorem 2. For any 0<α<n, if u(x)satisfies the equation

(−)α2u(x) =ρ(x), x∈Rn, (12) then u(x)can be written as the convolution of Kαand f :

u(x) =Iαρ(x) = (Kαρ)(x). (13) Proof. The proof is standard [1]. For convenience, we will recall it in the appendix.

2 Derive J

α

2.1 Extending the fractional Laplacian

Before extending the fractional Laplacian, we will start by looking at the normal Lalp- cian first:

∆≡

i

2

∂x2i. (14)

We will extend this to

E = −

i

ηi 2

∂ξ2i, where(η1, η2,· · ·ηn>0), (15) because it is positive definite, η1, η2, η3· · · > 0. For the specified case η1 =η2 = · · · = ηn=1, it reduces to the ordinary Laplacian.

The question is how to define this operator with a fractional exponent(−∆E)α2. We can do the same as the original fractional Laplacian:

Definition 3. The fractional exponent for the elliptical operator can be written as (−Ef)α2 = 1

Γ(α2) Z

0 tα2−1eEtf dt, (16) where eEtf = (H∗f)(ξ), eEtδ(ξ) ≡H(t, ξ)is the fundamental solution for ∂tu=Eu, and

H(ξ, t) = p 1

(4πt)nη1η2· · ·ηn

exp −

i

ξ2i it

!

≥0. (17)

(17) can be easily calculated,

∂H(ξ, t)

∂tEH(ξ, t) =0(t>0, lim

t→0=δ(x)) (18)

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and we apply (18) to the Fourier transformation

∂ bH(ξ, t)

∂t +

i

ηiξ2i

!

Hb(ξ, t) =0, (19)

Hb(ξ, t) =exp −

i

ηiξ2i

!

t=0, (20)

H(ξ, t) =

i

1 2√

πtηiexp − ξ

2i

4tηi

!

= p 1

(4πt)nη1η2· · ·ηn

exp −

i

ξ2i it

!

. (21)

2.2 The solution for

E

With all these definitions, we can start to derive the solution for fractional elliptic oper- ator associated to∆E.

Theorem 3. The solution for fractional elliptic operator

(−∆E)α2u(x) =ρ(x) (22) can be taken as u(x) = Jαρ(x), where

Jαu(x) ≡ 1 β(α)

Z Rn

f(y)

|η−1· (x−y)|n−ady, (23) and|η−1(x−y)|stands for

q∑ni η−1i (xi−yi), and β(α)−1equals

β(α)−1= √ 1

η1η2· · ·ηn · Γ(n−α2 ) π

n

22αΓ(α2). (24)

Proof. This theorem can be proved by some simple transformation of the variables.

For the equation

(−E)α2u(x) =ρ(x), (25) consider a transformation:

xi 7→ √ξi

ηi. (26)

Then (25) is transformed to

(−∆)α2u˜(ξ) = ˜ρ(ξ). (27) This is just the ordinary fractional Laplacian, so its solution is just the Riesz potential:

˜

u(ξ) = 1 γ(α)

Z Rn

˜ρ(ξ)

|ξζ|n−adζ, (28)

and we can transform it back to xivariable, so the solution will be u(x) = √ 1

η1η2· · ·ηn

· Γ(n−α2 ) πn22αΓ(α2)

Z Rn

f(y)

|η−1· (x−y)|n−ady. (29)

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It is also easy to define the solution for some compact domainΩ; simply set Jαf ≡ 1

β(α) Z

f(y)

|η−1(x−y)|n−αdy= 1 β(α)

Z Rn

f(y)χ

|η−1(x−y)|n−αdy (30) where χis the indicator function.

3 Integrability

We have to discuss the integrability of Jαf . Because Jαf can be turned to Iαf by changing variables, they should satisfy the same inequality. This has been proven to be true, so we can apply everything in the same way.

Theorem 4. Let 0 ≤ q ≤ ∞, 0 < α < n. Then Jα : Lp(Ω) → Lq(Ω)is an continuous operator

kJαfkLq(Ω) ≤CkfkLp(Ω), for any 1 p ≤ 1

q+α

n. (31)

Proof. Before proofing this theorem we need some lemmas.

Lemma 1. If a function f(x)depends only on|η−1x| ≡ r (where the norm stands for (ni η−1i xi)1/2), then we have the integral equality

Z

Rn f(x)dx=ωn Z

0 f(r)rn−1dr, (32)

where ωnis

ωn =√

η1· · ·ηn

πn2

Γ(n2+1). (33)

Proof. We can start from the fact that [3]

Z Rn+

f(x1b1+xb22+ · · · +xbnn)x1a1−1x1a2−1· · ·xann−1dx

= Γ(ab1

1)Γ(ab2

2) · · ·Γ(abn

n) b1· · ·bnΓ(ab1

1 +ba2

2· · · + abn

n) Z

0 f(t)ta1b1+a2b2···+anbn−1dt.

(34)

Rn+is defined as

Rn+= {x∈Rn |x1,· · ·, xn>0}. (35) By setting b1=b2= · · · =bn=2, and a1=a2= · · · =an=1, and a transformation,

xi 7→√

ηixi, i=1, 2,· · ·n, (36) we get

Z Rn+

f(|η−1x|2)dx=√

η1· · ·ηn

πn/2 2nΓ(n2)

Z

0 f(t)tn2−1dt. (37)

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Last, consider a change of variable t=r2: Z

Rn+

f(|η−1x|)dx=√

η1· · ·ηn πn/2 2nΓ(n2 +1)

Z

0 f(r)rn−1dr. (38) By the symmetry of f(|η−1x|), it is easy to check

2n Z

Rn+

f(|η−1x|)dx= Z

Rn f(|η−1x|)dx, (39) then the lemma is proven.

Lemma 2. For some 1≤p, q, r≤∞, if they satisfy 1

r +1= 1 p +1

q, (40)

then

β(α)kJαfkr≤ kfkpkhkq, (41) where

h(x, y) ≡h(η−1(x−y)) = 1

|η−1(x−y)|n−α. (42) Proof. First, we set

|Jαf| = 1 β(α)

Z

Rn f(y)h(x, y)dy

1

β(α) Z

Rn|h(x, y)f(y)|dy

= 1

β(α) Z

Rn|f(y)|pr|f(y)|1−pr|h(x, y)|qr|h(x, y)|1−qrdy. (43) We can see that,

1 r + 1

p −1 r

 + 1

q−1 r



= 1

r + 1

pr/(p−r)+ 1

qr/(q−r) =1. (44) Then we can apply the H ¨older inequality to it:

β(α)|Jαf| ≤

Z

Rn|f(y)|p|h(x, y)|qdy

1r

·

Z

Rn|f(y)|pdy

1p1r

·

Z

Rn|h(x, y)|qdy

1q1r

. (45)

Take both sides to an exponent r, then integrate it by x, and we get

β(α)rkJαfkrr

Z

Rn|f|pdy

 Z

R2n|h|qdxdy



kfkr−pp khkr−qq = kfkrpkhkrq (46) and the lemma is proven.

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Lemma 3. For n≥2, 0<α<n, one has Z

1

|η−1(x−y)|αdyn|En| n−α

||

|En|

1−αn

, (47)

where Enis

En=√

η1· · ·ηn πn/2

Γ(n2+1). (48)

It is the volume of a n-dimensional ellipsoid with axes√ η1,√

η2,· · ·,√ ηn.

Proof. First, we set S ∈ Rn an n-dimensional ellipsoid centered at x with axes√ η1R,

η2R,· · ·,√

ηnR and each parallel to the axis x1, x2,· · ·, xn of coordinate, and |S| = EnRnis the volume of S. Then we set|Ω| = |S|, so that R= (||/|S|)1/n.

Z S

dy

|η−1(x−y)|α = Z

S∩

dy

|η−1(x−y)|α+ Z

S−(S∩Ω)

dy

|η−1(x−y)|α (49)

and Z

dy

|η−1(x−y)|α = Z

S∩Ω

dy

|η−1(x−y)|α+ Z

Ω−(S∩Ω)

dy

|η−1(x−y)|α. (50) Because S− (S∩Ω)is inside S,1(x−y)|1 α ≤R−α; therefore

Z

Ω−(S∩Ω)

dy

|η−1(x−y)|α ≤R−α(|| − |S∩|). (51) Similarly,Ω− (S∩Ω)is outside S, so that

Z

S−(S∩Ω)

dy

|η−1(x−y)|α ≥R−α(|S| − |S∩Ω|) =R−α(|| − |S∩Ω|), (52) thus we get

Z

S−(S∩Ω)

dy

|η−1(x−y)|α ≥ Z

Ω−(S∩Ω)

dy

|η−1(x−y)|α, (53) or

Z

dy

|η−1(x−y)|α ≤ Z

S

dy

|η−1(x−y)|α = Z R

0 r−αrn−1|nEn|dr

= R

n−α

n−αnEn= n|En| n−α

||

|En|

1−αn

. (54)

Replace χf(x)by f(x)in lemma one in (54), and the lemma is proven.

The rest of the proof is obvious. For some r we can let 1

r +1

p =1+1

q (55)

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and if 1≤r≤1/(1−n/α)then n/p≤n/q+n is satisfied. By Lemma 2

kJαfk ≤β(α)−1khkrkfkp. (56) Note that we have replaced f(x)χby f(x), and h(x, y)χby h(x, y)because (57) only integrates over a bounded domain.

Then by Lemma 3

khkr

 nEn

n− (n−α)r

1r ||

|En|

1r+αn−1

. (57)

So the theorem is proven.

4 The symmetry problem

We know that for Iαthe solution of(−∆)α/2u=χhas some very interesting property, such as its the volume on ∂Ω is a constant if any only if Ω is a ball [4].

(−∆E)α2u = χ is invariant under some “elliptical rotation” that preserves|η−1x|, just like the(−)α/2u=χis invariant under rotations that preserve|x|, but the same property cannot carry over; that is, Jα(x)|will not be a constant whereΩ is the ellipsoid with axis parallel to the coordinate. It is because not all the transformation that preserves

|η−1x| preserves an infinitesimal volume dV in Rn, (if we see this transformation as a coordinate transformation, then it means the Jacobian does not equal one) [3], and therefore Jα(x)|∂Ωdoes not satisfy this property.

There is a transformation that preserves|η−1x| and infinitesimal volume. It is the reflection transformation Pm(See Definition 4). It is a discrete transformation, so instead of Jα(x)|∂Ω =const, we will get Jα(x)|∂Ω =Jα(Pmx)|∂Ω. (See Theorem 5.)

Definition 4. We are going to introduce the reflection transformation Pm:RnRn. Pmx=Pm(x1,· · ·, xm,· · ·, xn) = (x1,· · ·,−xm,· · ·xn), (58) where m=1, 2, . . . , n.

For an n-dimensional ellipsoid with axis√ η1,√

η2,· · ·,√

ηn, and each parallel to the axis of the coordinate(x1, x2,· · ·, xn), which will be noted asΩ is symmetric under reflection transformation. That is, for a x∈Ω, then Pmx∈Ω, and for some x∈∂Ω, then Pmx∈∂Ω.

Theorem 5. Let

u(x) ≡ 1 β(α)

Z

dy

|η−1(x−y)|n−α = Jα, (59) then there is a property of u(x)|∂Ω.

u(x)|∂Ω=u(Pmx)|∂Ω, for all n=1, 2,· · ·, n. (60)

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For convenience, we need to use a different kind of coordinate instead of the ordinary Cartesian coordinate.

Definition 5. We are going to define an elliptical coordinate (ρ, φ1· · ·φn−1)with the center at some point p.

x1−p1 = √

η1ρcos φ1

x2−p2 = √

η2ρsin φ1cos φ2

...

xn−1−pn−1 = √

ηn−1ρsin φ1· · ·sin φn−2cos φn−1 xn−pn = √

ηnρsin φ1· · ·sin φn−2sin φn−1 We can set that p∈∂Ω.

Then another coordinate(r, θ1,· · ·θn−1)at a point p0. x1−p01 = √

η1r cos θ1 x2−p02 = pη2r sin θ1cos θ2

...

xn−1−p0n−1 = √

ηn−1r sin θ1· · ·sin θn−2cos θn−1 xn−p0n = √

ηnr sin θ1· · ·sin θn−2sin θn−1 Then we can set that p0 =Pmp∈∂Ω.

With this coordinate, we shall define a subset inRnby

τl(k)= {x ∈|2l−1kρ<2kl}, 1≤l ≤k. (61) It is easy to check out that

[ 1≤l≤k

τl(k)=Ω. (62)

We will do the same to coordinate(r, θ1· · ·θn−1).

τl0(k) = {x∈Ω|2l−1k ≤r<2lk}, 1≤l≤k (63) and

[ 1≤l≤k

τl0(k) =Ω. (64)

Lemma 4. For any k and any 1≤l≤k, it satisfies

|τl(k)| = |τ

0(k)

l |. (65)

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Proof. For any x∈τl(k)it satisfies the condition

 2l−1

k

2

≤ (x1−p1)2

η1 +(x2−p2)2

η2 + · · ·(xm−pm)2 ηm

+ · · · + (xn−1−pn−1)2 ηn−1

+(xn−pn)2 ηn

 2l

k

2 (66)

and

x∈Ω. (67)

If we transform any x ∈ τl(k) with the reflection transformation Pm , then Pmx ≡ x0 satisfies

 2l−1

k

2

≤ (Pmx1−p1)2

η1 +(Pmx2−p2)2 η2

+ · · ·(Pmxm−pm)2 ηm

+ · · · + (Pmxn−pn)2 ηn

 2l

k

2 (68)

or

 2l−1

k

2

≤ (x1−p1)2

η1 +(x2−p2)2 η2

+ · · ·(xm+pm)2 ηm

+ · · · + (xn−1−pn−1)2

ηn−1 +(xn−pn)2 ηn

 2l

k

2

,

(69)

and x0∈Ω. This is exactly the condition that satisfies for any x0τl0(k). Thus,

Pmτl(k) =τl0(k). (70)

Since the reflection transformation preserves the volume, so that

|τl(k)| = |τl0(k)|. (71)

We know that τl(k) and τl0(k) approach to zero as k approaches to infinity. But how exactly and how rapidly it approaches to zero, we can see it by Lemma 5.

Lemma 5. For any integer k, and some 1≤l≤k the volume of τl(k)and τ0(k)satisfy

|τl(k)| ≤C(l)k−n, (72)

|τ

0(k)

l | ≤C(l)k−n, (73)

where C is a constant independent of k but dependent to l.

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Proof. For convenience, we define

σl(k) = {x∈Rn|2l−1kρ<2lk} (74) and

σ

0(k)

l = {x∈Rn |2l−1k ≤r<2kl}. (75) where 1 ≤ l ≤ k. By the definition of (74) and (75), we can see that τl(k)σl(k), and τ

0(k)σ

0(k)

l ; therefore,|τl(k)| ≤ |σl(k)|, and|τ

0(k) l | ≤ |σ

0(k)

l |. Since the volume of σl(k)and σ

0(k)

l can be computed

|σl(k)| = Z

Rnχ

σl(k)dx=ωn Z l

k l1

k

rn−1dr= ωn n−1

 l k

n

 l−1 k

n

. (76)

We have used Lemma 1 in Equation (76). The ωn has been defined in (33). Therefore, we can see that

|τl(k)| ≤ |σl(k)| = ωn n−1

 l k

n

 l−1 k

n

≡C(l)k−n, (77) where C(l)equals

C(l) = ωn

n−1[ln− (l−1)n]. (78) The case for τl0(k)can be proven in the same way.

Now, we can divide the function Jα(x)into Jα(x) = 1

β(α) Z

dy

|η−1(x−y)|n−α (79)

=

1≤l≤k Z

τl(k)

dy

|η−1(x−y)|n−α

1≤l≤k

j(l)α

=

1≤l≤k Z

τl0 (k)

dy

|η−1(x−y)|n−α

1≤l≤k

jα0(l), where

jα(l)(x) ≡ Z

τl(k)

dy

|η−1(x−y)|n−α, (80) and

j0α(l)(x) ≡ Z

τl0 (k)

dy

|η−1(x−y)|n−α. (81) We are going to define an approximation of j(l)

j(l)appx.α(x) = |τl(k)| 1

|η−1(x−y)|n−α, (82) j0appx.α(l) (x) = |τ

0(k)

l | 1

|η−1(x−y0)|n−α, (83) for some y∈τl(k)and y0τ

0(k) l .

(14)

Lemma 6. For any integer k and 1≤l≤k it satisfies

|j(l)α (p) −j(k)appx.α(p)| ≤C0(l)k−α (84) and

|j0α(l)(p0) −j0appx.α(k) (p0)| ≤C0(l)k−α, (85) where C0(l)is independent of k.

Proof. First, it is obvious that

|τl(k)| min

x∈τl(k)

1

|η−1(p−y)|n−α ≤ j(l)α (p), j(l)appx.α(p) ≤ |τl(k)| max

x∈τl(k)

1

|η−1(p−y)|n−α, (86) where

max

y∈τl(k)

1

|η−1(p−y)|n−α = 1

(2l−1k )n−α, (87)

and

min

y∈τl(k)

1

|η−1(p−y)|n−α = 1

(2kl)n−α. (88)

So for a sufficiently large k, it can satisfy

max

y∈τl(k)

|τl(k)|

|η−1(p−y)|n−α − min

x∈τl(k)

|τl(k)|

|η−1(p−y)|n−α = |τl(k)|

(2l−1k )n−α − |τl(k)|

(2lk)n−α, (89) so that

|j(l)α (p) −j(k)appx.α(p)| ≤ |τl(k)| 1

(2l−1k )n−α1 (2lk)n−α

!

≡C0(l)k−α. (90)

We have used Lemma 5 in this equation and C0(l)equals

C0(l) =C(l)2α−n·(l−1)α−n−lα−n . (91) Then the theorem is proven.

Of course, we can basically do the same with|τl0(k)|. Then we can get

|j0α(l)(p0) −j0appx.α(k) (p0)| ≤C0(l)k−α. (92) Note that C0(l)increases as l increases, and by (91) and (78), we can see that C0(l)in- creases in the order of kn−1·kα−n−1=kα−2

Now, back to the main theorem, we can see that jappx.α(k) (p) = j0appx.α(k) (p0) because by Lemma 4|τl(k)| = |τ

0(k)

l |, and 1(p−y)|1 nα = 1

1(p0−y0)|nα for some y ∈ τl(k) and Pmy=y0τ

0(k)

l , therefore

|j0α(l)(p0) −jα(l)(p)| ≤ |jα0(l)(p0) −j0appx.α(k) (p0)| + |jα(l)(p) −j(k)appx.α(p)| ≤2C0(l)k−α, (93)

(15)

thus, for sufficiently large k

|Jα(p) −Jα(p0)| =

k l

j(l)α (p) −j0α(l)(p0)

k l

|jα0(l)(p0) −jα(l)(p)|

k l

C0(l)k−α ≤kC0(k) ·k−α.

(94)

Since kC0(k)increases in the order of kα−1, so (94) will decrease in the order of k−1as k

approaches to infinity. So the theorem is proven. 

4.1 Generalization

In Theorem 5, we have assumedΩ to be an n-dimensional ellipsoid centered at the ori- gin point and it has axis of√

η1,√

η2· · ·√

ηn each parallel to the coordinate(x1· · ·xn). But this assumption is superfluous, for all we need is the restriction forΩ is PmΩ = and Ω is bounded. From (66) to (70) we can see that Lemma 4 still holds under this restriction, and therefore, so does in Theorem 5.

Another assumption that is superfluous is that we only consider p∈ ∂Ω and Pmp∈

∂Ω. That is, we only consider Jα(x)under the restriction Jα(x)|∂Ω. We will extend it to any point p∈Rnand p0 =Pmp∈Rn.

We will redefine the coordinate(ρ, φ1· · ·φn−1) and (r, θ1,· · ·θn−1)in Definition 5 basically in the same way but this time the coordinate will be centered at any point p and Pmp which is not necessary on ∂Ω.|τl(k)|and|τ

0(k)

l |are now written as

τl(k)= {x∈Ω|2l−1kρ<2kl}, kmin≤l≤kmax (95) and

τl0(k) = {x∈|2l−1k ≤r<2kl}, kmin≤l≤kmax, (96) where kmax is defined as∀l > kmax, τl(k) = ∅. Since Pmτl(k) = τ

0(k)

l , so that∀l > kmax, τ

0(k)

l =∅. Such kmaxexists because of the boundedness ofΩ. Similarly, kminis defined as∀l<kmin, τl(k)=∅. If such kmindoes not exist, then set kmin=1.

By this definition, we can get

Ω= [

kmin≤l≤kmax

τl(k)= [

kmin≤l≤kmax

τ

0(k)

l , (97)

(16)

and therefore,

Jα(x) = 1 β(α)

Z

dy

|η−1(x−y)|n−α (98)

=

kmin≤l≤kmax

Z τl(k)

dy

|η−1(x−y)|n−α

kmin≤l≤kmax

j(l)α

=

kmin≤l≤kmax Z

τl0 (k)

dy

|η−1(x−y)|n−α

kmin≤l≤kmax

jα0(l).

The definition of j(l)α (x), jα0(l)(x), jappx.α(l) (x), and j‘(l)appx.α(x) are still the same, so that we can see Theorem 5 still holds.

The generalization of Theorem 5 is:

Theorem 6(generlization). Let u(x) ≡ 1

β(α) Z

dy

|η−1(x−y)|n−α = Jα. (99) For all bounded domainΩ that satisfies Ω=Pm

u(x) =u(Pmx). (100)

4.2 The antisymmetric property

In the equation of (−E)α2u(x) = f(x), we have set f(x) = χ and found out that the solution, noted as Jα(x), has the symmetric property. Now, if we replace the func- tion f(x) by another function g(x) = χxi, where 1 ≤ i ≤ n, then the solution for (−E)α2u(x) =g(x), noted as Jαg(x) =Jα(x)will satisfy another property.

Theorem 7. Let

u(x) ≡ 1 β(α)

Z

xidy

|η−1(x−y)|n−α =Jα(x), (101) then there is a property of u(x)|.

u(x)|∂Ω=u(Pmx)|∂Ω ,for m6=i, (102) u(x)|∂Ω= −u(Pmx)|∂Ω ,for m=i. (103) WhereΩ is an n-dimensional ellipsoid centered at origin point and axis parallel to the coordinate(x1, x2· · ·xn).

Proof. First, we will redefine, j(l)α , j0α(l)as jα(l)(x) ≡

Z τl(k)

xidy

|η−1(x−y)|n−α (104)

(17)

and

j0α(x)(l) ≡ Z

τl0 (k)

xidy

|η−1(x−y)|n−α. (105)

Since τl(k)and τl0(k)are the same as (61) and (63). so that

Jα(x) = 1 β(α)

Z

xidy

|η−1(x−y)|n−α (106)

=

1≤l≤k Z

τl(k)

xidy

|η−1(x−y)|n−α

1≤l≤k

j(l)α

=

1≤l≤k Z

τl0 (k)

xidy

|η−1(x−y)|n−α

1≤l≤k

j0α(l).

Now we are going to do something different. Define a subset in τl(k),

π(k)ml = {x ∈τl(k)|√

ηim−1k ≤xi <√

ηimk}. (107)

Where m is an integer ranges from−k+1 to k. Similarly, we define π

0(k)

ml = {x ∈τ

0(k) l |√

ηim−1k ≤xi<√

ηimk}. (108)

By (107) and (108), we can see that Pmπml(k) = π

0(k)

ml for m 6= i and Pmπ(k)ml = π

0(k)

−m+1l for m=i, therefore,|πml(k)| = |π

0(k)

ml |for m6=i, and|π(k)ml| = |π

0(k)

−m+1l|for m=i.

We can see that|π(k)ml|decay to zero as k approach to infinity, and Lemma 7 told that who rapidly does it approaches to zero.

Lemma 7. For any−k+1≤m≤k,|π(k)ml|and|π

0(k)

ml |satisfy

|πml(k)| ≤Ck−n−1 (109)

and

|π

0(k)

ml | ≤Ck−n−1, (110)

where C is a constant independent of k.

Proof. Set

|π|max≡max{|π(k)−k+1l|,|π−k+2l(k) | · · · |π(k)k−1l|,|πkl(k)|} (111) and

|π|min≡min{|π(k)−k+1l|,|π(k)−k+2l| · · · |πk−1l(k) |,|πkl(k)|}. (112) Notice that,

2k|π|min

k m=−k+1

|π(k)ml| = |τl(k)|. (113)

參考文獻

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