A symmetry problem of elliptic differential operators in potential theory

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 t^α2−1(Ht∗f)(x)_dt, Ht(x) ≡H(_{x, t}) = _p ¹

(4πt)ⁿη₁η₂· · ·ηn

exp −

∑

i

x²_i 4η_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)|_is^q_∑ⁿ_{i η}⁻¹

i (x_i−y_i)_{, and β}(_α)⁻¹_equals β(α)⁻¹= √ ¹

η₁η₂· · ·ηn

· ^Γ(^n−α₂ ) π

n 22^αΓ(^α₂)^.

Then if we set f =χ_Ωwhere χ_Ωis 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, J_αf ≡J_α(x)satisfies

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

P_mx=P_m(x₁,· · ·, x_m,· · ·, x_n) = (x₁,· · ·,−x_m,· · ·, x_n), where m=1, 2, . . . , n.

摘

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 t^α2−1(Ht∗f)(x)_dt,

Ht(x) ≡H(_{x, t}) = _p ¹

(4πt)ⁿη₁η₂· · ·ηn

exp −

∑

i

x²_i 4η_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 J_αf(x) ≡ ¹

β(α) Z

Rⁿ

f(y)

|η· (x−y)|^n−ady, 其中|η⁻¹· (x−y)|_代表^q_∑ⁿ_{i η}⁻¹

i (x_i−y_i)_{, 而 β}(_α)⁻¹_等於 β(α)⁻¹= √ ¹

η₁η₂· · ·ηn

· ^Γ(^n−α₂ ) π

n 22^αΓ(^α₂)^.

如果在 Jαf中令 f =χ_Ω,其中 χ_Ω是指示函數, 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) = I_αf if (−_∆)^α2u = f [1]. Now we let f ≡ χ_Ω, where χ_Ω 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

η_j ∂²

∂x²_j

!^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)_.

2. Discuss the integrability of J_αf .

3. Discuss the symmetry property of the solution of (−_∆E)^α2u = χ_Ω 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 = χ_Ωxi, 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 t^s−1e^−tAdt. (1)

If we exchange A to a Laplacian, A7→ −_{∆, s}→ ^α₂ , then we get the definition.

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

Γ(^α₂) Z _∞

0 t^α2−1e^t∆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 e^t∆

∂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 t^α2−1e^tAf dt (6)

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 t^se^−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 I_αf(x) = (K_α∗f)(x) = ¹

γ(α) Z

Rⁿ

f(y)

|x−y|^{n−ady, (8)} where γ(α)_is

γ(α) = ^π

n22^αΓ(^α₂) Γ(^n−α₂ ) and

K_α= 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

Ω

f(y)

|x−y|^n−ady= ¹ γ(α)

Z Rⁿ

f(y)

|x−y|^{n−aχΩdy, (10)} where χ_Ω 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 ∈ L^p(Ω), and Iαf ∈ L^q(Ω), that p, q satisfy some condition which makes Iα : L^p(Ω) → L^q(_Ω)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 : L^p(Ω) →L^q(Ω)is a bounded operator:

kIαfkq ≤Ckfkp, if n p ≤ ⁿ

q +α. (11)

Proof. See [2]. This theorem says that if f ∈ L^p(_Ω), then for x ∈ _{Ω, Iα}f(x)converges absolutely.

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) =I_αρ(x) = (K_α∗ρ)(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

∂x²_i. (14)

We will extend this to

−_∆E = −

∑

i

η_i ∂²

∂ξ²_i, 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 t^α2−1e^∆Etf dt, (16) where e^∆Etf = (H∗f)(ξ)_{, e}^∆Etδ(ξ) ≡H(t, ξ)is the fundamental solution for ∂tu=_∆Eu, and

H(ξ, t) = _p ¹

(4πt)ⁿη₁η₂· · ·ηn

exp −

∑

i

ξ²_i 4η_it

!

≥0. (17)

(17) can be easily calculated,

∂H(ξ, t)

∂t −_∆EH(ξ, t) =0(t>0, lim

t→0=δ(x)) (18)

and we apply (18) to the Fourier transformation

∂ bH(ξ, t)

∂t +

∑

i

ηiξ²_i

!

Hb(ξ, t) =0, (19)

Hb(ξ, t) =exp −

∑

i

η_iξ²_i

!

t=0, (20)

H(ξ, t) =

∏

i

1 2√

πtη_iexp − ^ξ

2i

4tη_i

!

= _p ¹

(4πt)ⁿη₁η2· · ·ηn

exp −

∑

i

ξ²_i 4η_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) = J_αρ(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∑ⁿ_i η⁻¹_i (xi−yi), and β(α)⁻¹equals

β(α)⁻¹= √ ¹

η₁η₂· · ·ηn · ^Γ(^n−α₂ ) π

n

22^αΓ(^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

η_i. (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)

|ξ−ζ|^{n−adζ, (28)}

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

η₁η2· · ·ηn

· ^Γ(^n−α₂ ) πⁿ²2^αΓ(^α₂)

Z Rⁿ

f(y)

|η⁻¹· (x−y)|^{n−ady. (29)}

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

β(α) Z

Ω

f(y)

|η⁻¹(x−y)|^n−αdy= ¹ β(α)

Z Rⁿ

f(y)χ_Ω

|η⁻¹(x−y)|^{n−αdy (30)} where χ_Ωis the indicator function.

3 Integrability

We have to discuss the integrability of Jαf . Because J_αf can be turned to I_αf 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 : L^p(Ω) → L^q(Ω)is an continuous operator

kJαfk_Lq(Ω) ≤Ckfk_Lp(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 (_∑ⁿ_i η⁻¹_i x_i)^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(x₁^b1+x^b₂²+ · · · +x^b_nⁿ)x₁^a1−1x₁^a2−1· · ·x^a_nⁿ⁻¹dx

= ^Γ(^a_b¹

1)Γ(^a_b²

2) · · ·Γ(^a_bⁿ

n) b₁· · ·bnΓ(^a_b¹

1 +_b^a2

2· · · + ^a_bⁿ

n) Z _∞

0 f(t)t^{a1b1+a2b2···+anbn−1}dt.

(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→√

η_ixi, i=1, 2,· · ·n, (36) we get

Z Rⁿ+

f(|η⁻¹x|²)dx=√

η1· · ·ηn

π^n/2 2ⁿΓ(ⁿ₂)

Z _∞

0 f(t)tⁿ²⁻¹dt. (37)

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

Rⁿ+

f(|η⁻¹x|)dx=√

η₁· · ·η_n π^n/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)kJ_αfk_r≤ kfk_pkhk_q, (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)|J_αf| ≤

_Z

Rⁿ|f(y)|^p|h(x, y)|^qdy

¹_r

·

_Z

Rⁿ|f(y)|^pdy

¹_p−¹_r

·

_Z

Rⁿ|h(x, y)|^qdy

¹_q−¹_r

. (45)

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

β(α)^rkJ_αfk^r_r ≤

Z

Rⁿ|f|^pdy

 Z

R²ⁿ|h|^qdxdy



kfk^r−p_p khk^r−q_q = kfk^r_pkhk^r_q (46) and the lemma is proven.

Lemma 3. For n≥2, 0<α<n, one has Z

Ω

1

|η⁻¹(x−y)|^αdy≤ ⁿ|En| n−α

|_Ω|

|En|

1−^α_n

, (47)

where Enis

En=√

η₁· · ·η_n π^n/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,· · ·,√

η_nR 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= (|_Ω|/|S|)^1/n.

Z S

dy

|η⁻¹(x−y)|^α = Z

S∩Ω

dy

|η⁻¹(x−y)|^α+ Z

S−(S∩Ω)

dy

|η⁻¹(x−y)|^{α (49)}

and Z

Ω

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,_|η−1(x−y)|^{1 α} ≤R^−α; therefore

Z

Ω−(S∩Ω)

dy

|η⁻¹(x−y)|^α ≤R^−α(|_Ω| − |S∩_Ω|). (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(|_Ω| − |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 Ω

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

|_Ω|

|En|

1−^α_n

. (54)

Replace χ_Ωf(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)

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

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

Then by Lemma 3

khkr ≤

 nEn

n− (n−α)r

¹_r |_Ω|

|En|

¹_r+^α_n−1

. (57)

So the theorem is proven.

4 The symmetry problem

We know that for Iαthe solution of(−∆)^α/2u=χ_Ωhas 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_Ω is invariant under some “elliptical rotation” that preserves|η⁻¹x|, just like the(−_∆)^α/2u=χ_Ωis invariant under rotations that preserve|x|, but the same property cannot carry over; that is, J_α(x)|_Ωwill 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 =J_α(Pmx)|_∂Ω. (See Theorem 5.)

Definition 4. We are going to introduce the reflection transformation Pm:Rⁿ →Rⁿ. 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 Pm}x∈Ω, and for some x∈_{∂Ω, then} Pmx∈∂Ω.

Theorem 5. Let

u(x) ≡ ¹ β(α)

Z Ω

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)

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· · ·φ_n−1)with the center at some point p.

x1−p1 = √

η₁ρcos φ1

x2−p2 = √

η₂ρsin φ1cos φ2

...

x_n−1−p_n−1 = √

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

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

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

η₁r cos θ₁ x2−p⁰₂ = ^pη2r sin θ₁cos θ2

...

x_n−1−p⁰_n−1 = √

η_n−1r sin θ₁· · ·sin θn−2cos θ_n−1 xn−p⁰_n = √

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

With this coordinate, we shall define a subset inRⁿby

τ_l^(k)= {x ∈_Ω|2^l−1_k ≤ρ<2_k^l}, 1≤l ≤k. (61) It is easy to check out that

[ 1≤l≤k

τ_l^(k)=_{Ω. (62)}

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

τ_l^0(k) = {x∈Ω|2^l−1_k ≤r<2^l_k}, 1≤l≤k (63) and

[ 1≤l≤k

τ_l^0(k) =_Ω. (64)

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

|τ_l^(k)| = |τ

0(k)

l |. (65)

Proof. For any x∈τ_l^(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 ∈ τ_l^(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

+ · · · + (x_n−1−p_n−1)²

η_n−1 +(xn−pn)² ηn

≤

 2l

k

2

,

(69)

and x⁰∈Ω. This is exactly the condition that satisfies for any x⁰ ∈τ_l^0(k). Thus,

Pmτ_l^(k) =τ_l^0(k). (70)

Since the reflection transformation preserves the volume, so that

|τ_l^(k)| = |τ_l^0(k)|. (71)

We know that τ_l^(k) and τ_l^0(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 τ_l^(k)and τ^0(k)satisfy

|τ_l^(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.

Proof. For convenience, we define

σ_l^(k) = {x∈Rⁿ|2^l−1_k ≤ρ<2^l_k} (74) and

σ

0(k)

l = {x∈_Rⁿ |2^l−1_k ≤r<2_k^l}. (75) where 1 ≤ l ≤ k. By the definition of (74) and (75), we can see that τ_l^(k) ⊆ σ_l^(k), and τ

0(k)⊆σ

0(k)

l ; therefore,|τ_l^(k)| ≤ |σ_l^(k)|, and|τ

0(k) l | ≤ |σ

0(k)

l |. Since the volume of σ_l^(k)and σ

0(k)

l can be computed

|σ_l^(k)| = Z

Rⁿχ

σ_l^(k)dx=ω_n Z ^l

k l−1

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

|τ_l^(k)| ≤ |σ_l^(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 τ_l^0(k)can be proven in the same way.

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

β(α) Z

Ω

dy

|η⁻¹(x−y)|^{n−α (79)}

=

∑

1≤l≤k Z

τ_l^(k)

dy

|η⁻¹(x−y)|^n−α ≡

∑

1≤l≤k

j^(l)_α

=

∑

1≤l≤k Z

τ_l^{0 (k)}

dy

|η⁻¹(x−y)|^n−α ≡

∑

1≤l≤k

j_α^0(l), where

j_α^(l)(x) ≡ Z

τ_l^(k)

dy

|η⁻¹(x−y)|^{n−α, (80)} and

j⁰_α^(l)(x) ≡ Z

τ_l^{0 (k)}

dy

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

j^(l)appx.α(x) = |τ_l^(k)| ¹

|η⁻¹(x−y)|^{n−α, (82)} j⁰_appx.α^(l) (x) = |τ

0(k)

l | ¹

|η⁻¹(x−y⁰)|^{n−α, (83)} for some y∈τ_l^(k)and y⁰∈τ

0(k) l .

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

|j⁰_α^(l)(p⁰) −j⁰_appx.α^(k) (p⁰)| ≤C⁰(l)k^−α, (85) where C⁰(l)is independent of k.

Proof. First, it is obvious that

|τ_l^(k)| min

x∈τ_l^(k)

1

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

x∈τ_l^(k)

1

|η⁻¹(p−y)|^{n−α, (86)} where

max

y∈τ_l^(k)

1

|η⁻¹(p−y)|^n−α = ¹

(2^l−1_k )^{n−α, (87)}

and

min

y∈τ_l^(k)

1

|η⁻¹(p−y)|^n−α = ¹

(2_k^l)^{n−α. (88)}

So for a sufficiently large k, it can satisfy

max

y∈τ_l^(k)

|τ_l^(k)|

|η⁻¹(p−y)|^n−α − min

x∈τ_l^(k)

|τ_l^(k)|

|η⁻¹(p−y)|^n−α = |τ_l^(k)|

(2^l−1_k )^n−α − |τ_l^(k)|

(2^l_k)^{n−α, (89)} so that

|j^(l)_α (p) −j^(k)appx.α(p)| ≤ |τ_l^(k)| ¹

(2^l−1_k )^n−α − ¹ (2^l_k)^n−α

!

≡C⁰(l)k^−α. (90)

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

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

Of course, we can basically do the same with|τ_l^0(k)|. Then we can get

|j⁰_α^(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·k^α−n−1=k^α−2

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

0(k)

l |, and _|η−1(p−y)|^{1 n−α} = ¹

|η⁻¹(p⁰−y⁰)|^n−α for some y ∈ τ_l^(k) and Pmy=y⁰ ∈τ

0(k)

l , therefore

|j⁰_α^(l)(p⁰) −j_α^(l)(p)| ≤ |j_α^0(l)(p⁰) −j⁰_appx.α^(k) (p⁰)| + |j_α^(l)(p) −j^(k)_appx.α(p)| ≤_2C⁰(l)k^−α, (93)

thus, for sufficiently large k

|J_α(p) −J_α(p⁰)| =     

∑

k l

j^(l)_α (p) −j⁰_α^(l)(p⁰)     

≤

∑

k l

|j_α^0(l)(p⁰) −j_α^(l)(p)|

≤

∑

k l

C⁰(l)k^−α ≤kC⁰(k) ·k^−α.

(94)

Since kC⁰(k)increases in the order of k^α−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 =_Ω 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 Pm}p∈

∂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.|τ_l^(k)|and|τ

0(k)

l |are now written as

τ_l^(k)= {x∈Ω|2^l−1_k ≤ρ<2_k^l}, kmin≤l≤kmax (95) and

τ_l^0(k) = {x∈_Ω|2^l−1_k ≤r<2_k^l}, kmin≤l≤kmax, (96) where kmax is defined as∀l > kmax, τ_l^(k) = _{∅. Since Pm}τ_l^(k) = τ

0(k)

l , so that∀l > kmax, τ

0(k)

l =_{∅. Such kmax}exists because of the boundedness ofΩ. Similarly, kminis defined as∀l<k_min, τ_l^(k)=∅. If such kmindoes not exist, then set k_min=1.

By this definition, we can get

Ω= ^[

k_min≤l≤k_max

τ_l^(k)= ^[

k_min≤l≤k_max

τ

0(k)

l , (97)

and therefore,

Jα(x) = ¹ β(α)

Z Ω

dy

|η⁻¹(x−y)|^{n−α (98)}

=

∑

k_min≤l≤kmax

Z τ_l^(k)

dy

|η⁻¹(x−y)|^n−α ≡

∑

k_min≤l≤kmax

j^(l)_α

=

∑

k_min≤l≤k_max Z

τ_l^{0 (k)}

dy

|η⁻¹(x−y)|^n−α ≡

∑

k_min≤l≤k_max

j_α^0(l).

The definition of j^(l)α (x)_{, jα}^0(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

Ω

dy

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

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_Ω and found out that the solution, noted as J_α(x), has the symmetric property. Now, if we replace the func- tion f(x) by another function g(x) = χ_Ωxi, where 1 ≤ i ≤ n, then the solution for (−_∆E)^α2u(x) =g(x), noted as J_αg(x) =J_α(x)will satisfy another property.

Theorem 7. Let

u(x) ≡ ¹ β(α)

Z Ω

xidy

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

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 , j⁰_α^(l)as j_α^(l)(x) ≡

Z τ_l^(k)

x_idy

|η⁻¹(x−y)|^{n−α (104)}

and

j⁰_α(x)^(l) ≡ Z

τ_l^{0 (k)}

xidy

|η⁻¹(x−y)|^{n−α. (105)}

Since τ_l^(k)and τ_l^0(k)are the same as (61) and (63). so that

J_α(x) = ¹ β(α)

Z Ω

xidy

|η⁻¹(x−y)|^{n−α (106)}

=

∑

1≤l≤k Z

τ_l^(k)

xidy

|η⁻¹(x−y)|^n−α ≡

∑

1≤l≤k

j^(l)_α

=

∑

1≤l≤k Z

τ_l^{0 (k)}

x_idy

|η⁻¹(x−y)|^n−α ≡

∑

1≤l≤k

j⁰_α^(l).

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

π^(k)_ml = {x ∈τ_l^(k)|√

η_i^m−1_k ≤x_i <√

η_i^m_k}. (107)

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

0(k)

ml = {x ∈τ

0(k) l |√

η_i^m−1_k ≤x_i<√

η_i^m_k}. (108)

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

0(k)

ml for m 6= i and Pmπ^(k)_ml = π

0(k)

−m+1l for m=i, therefore,|π_ml^(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

|π_ml^(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|,|π_kl^(k)|} (111) and

|π|_min≡min{|π^(k)_−k+1l|,|π^(k)_−k+2l| · · · |π_k−1l^(k) |,|π_kl^(k)|}. (112) Notice that,

2k|π|_min≤

∑

k m=−k+1

|π^(k)_ml| = |τ_l^(k)|. (113)