# Boundary Value Problems Governed by Superdiffusion in the Right Angle: Existence and Regularity

Authors: Ramzet Dzhafarov; Nataliya Vasylyeva
Journal: Journal of Mathematics
 (2018)

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## Abstract

Forα∈(1,2), we analyze a stationary superdiffusion equation in the right angle in the unknown u=u(x1,x2): Dx1αu+Dx2αu=f(x1,x2), where Dxα is the Caputo fractional derivative. The classical solvability in the weighted fractional Hölder classes of the associated boundary problems is addressed.


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## Body

## 1. Introduction

Fractional partial differential equations (FPDE) play a key role in the description of the so-called anomalous phenomena in nature and in the theory of complex systems (see, e.g., [1]). In particular, these equations provide a more faithful representation of the long-memory and nonlocal dependence of many anomalous processes. The signature of an anomalous diffusion species 〈(Δx)2〉 scales as a nonlinear power law in time; i.e., 〈(Δx)2〉~tα, α>0. When α>1, this is referred to as superdiffusion.Superdiffusion is used in modelling turbulent flow [2, 3], chaotic dynamics of classical conservative systems [4], model solute transport in underground aquifers [5–7], and rivers [8–10], biophysics [11], and physical and chemical models described by the Lévy processes [12, 13]. In the present paper, we focus on the boundary value problems to the stationary superdiffusion equation.LetΩ={(x1,x2):x1∈(0,+∞),x2∈(0,+∞)} be the first quarter with a boundary ∂Ω=Γ1∪Γ2, Γ1={(x1,x2):x2=0,x1≥0}, Γ2={(x1,x2):x1=0,x2≥0}. For a fixed α∈(1,2), we consider the linear equation in the unknown function u=u(x1,x2):Ω→R,(1)Dx1αu+Dx2αu=fx1,x2,subject either to the Dirichlet boundary condition (DBC),(2)u=ψ1x1onΓ1andu=ψ2x2onΓ2,or to the Neumann boundary condition (NBC),(3)∂u∂x2=φ1x1onΓ1and∂u∂x1=φ2x2onΓ2,where the functions f, ψ1, ψ2, φ1, φ2 are prescribed.Here, the symbolsDx1α and Dx2α stand for the Caputo fractional derivative of order α with respect to x1 and x2.Introducing the function,(4)ωθy=yθ-1Γθ,θ>0,y≥0,we define the Caputo derivative Dyθv(·,y) as (see, e.g., (2.4.1) in [1]) (5)Dyθv·,y=∂θ∂yθ∫0yωθ-θy-z×v·,z-∑k=0θ-1∂kv·,0∂zkzkk!dz,where ⌈θ⌉ is the ceiling function of θ (i.e., the smallest integer greater than or equal to θ), Γ being the Euler Gamma-function. An equivalent definition in the case of u∈AC⌈θ⌉ reads (see (2.4.15) in [1]) (6)Dyθv·,y=∫0yωθ-θy-z∂θ∂zθv·,zdz.In the limit case θ=⌈θ⌉, the Caputo fractional derivatives of v(·,y) boils down (∂⌈θ⌉/∂y⌈θ⌉)v(·,y).Elliptic boundary value problems,α=2, in domain with conical and dihedral singularities have been extensively studied, starting with Kondratiev’s famous paper [14]. In this field of researches, it should be also noted the works of Kondariev and Oleinik [15], Borsuk and Kondratiev [16], Grisvard [17], and Maz’ya and Plamenevskii [18]. The results of those investigations are formulated in both Wl,p and Cl+β spaces and corresponding weighted classes, where the desired functions together with their derivatives are bounded with some power weights in the Lp or Cl+β norms.Recently, a great attention in the literature has been devoted to the study of boundary value problems to the fractional Laplacian operator-Δsu with order s∈(0,1) (see, e.g., [19–24] and references therein). Boundary value problems for the equation with the main part,(7)∂2∂x2ux,y+Dy2αux,y,α∈0,1,are studied with various approaches, such as spectral technique, method of potential theory, and Fourier method (see [25–28] and references therein). We also quote the works [29, 30], where the authors presented a Galerkin finite element approximation for variational solution to the steady state fractional advection dispersion equation:(8)-Dx1ap0Dx-ν+qxD1-νDx1u+bxDx1u+cxu=f,ν∈0,1,where 0Dx-ν, xD1-ν are left and right fractional integral operators, 0≤p,q,≤1, p+q=1.The goal of the present paper is the proof of the well-posedness and the regularity of solutions to boundary value problems (1)-(3) in weighted Hölder classes. It is worth noting that these classes allow one to control the behavior of the solution near the boundary including the corner point and at the infinity.Outline of the Paper. In the next section, we introduce necessary functional spaces and state the main results (Theorems 3 and 4) along with the general assumptions. In Section 3, we construct the integral representation to u(x1,x2) in the case of homogenous boundary conditions. To this end, we apply Mellin transform and reduce problem (1)-(3) to the linear nonhomogenous difference equation of the first-order with variables coefficients in the two-dimensional case. Section 4 is devoted to some auxiliary results which will play a key role in the investigation. In Section 5, we estimate the seminorms of the minor derivatives Dx1α1Dx2α2u, 0≤α1+α2≤1, and in Section 6 we evaluate the Hölder coefficients of the major derivatives Dx1α1Dx2α2u, 1<α1+α2≤α. In Section 7, using these estimates, we provide the proofs of Theorems 3 and 4. Moreover, in Remark 27 we show how results of Theorems 3 and 4 can be extend to the more general equation compared to (1).


## 2. Functional Setting and Main Results

Throughout this work, the symbolC will denote a generic positive constant, depending only on the structural quantities of the problem. We will carry out our analysis in the framework of the weighted Hölder spaces. Let β∈(0,1) be arbitrary fixed. We denote by r(x) and d(x) the distance from a point x(x1,x2)∈Ω to the origin (0,0) and to the boundary ∂Ω, correspondingly. Then for every x and x¯ from Ω we define r(x,x¯)=min⁡{r(x),r(x¯)} and d(x,x¯)=min{d(x),d(x¯)}. Note that if x,x¯∈∂Ω, then r(x)=d(x) and r(x,x¯)=d(x,x¯).For fixeds∈R, we introduce the Banach spaces Esβ(Ω¯) and Csβ(Ω¯) of the functions v with the norms(9)vEsβΩ¯=supΩ¯⁡r-sxv+vx1,s-β,Ωβ+vx2,s-β,Ωβ,vCsβΩ¯=supΩ¯⁡d-sxv+vx1,s-β,Ωβ+vx2,s-β,Ωβ,where(10)vy,s-β,Ωβ=supy,y¯∈Ω¯⁡r-s+βy,y¯vy-vy¯y-y¯β,vy,s-β,Ωβ=supy,y¯∈Ω¯⁡d-s+βy,y¯vy-vy¯y-y¯β,for y≠y¯.Definition 1.
A functionv=v(x1,x2) belongs to the class Msl+β(Ω¯), for l>0, if v∈Esβ(Ω¯), Dxilv∈Es-lβ(Ω¯), i=1,2, and the norms below are finite, (11)vMsl+βΩ¯=vEsβΩ¯+∑j=1l-1DxjvEs-jβΩ¯+∑i=12DxilvEs-lβΩ¯,if l≥1, and (12)vMsl+βΩ¯=vEsβΩ¯+∑i=12DxilvEs-lβΩ¯in the case of l∈(0,1).Definition 2.
A functionv=v(x1,x2) belongs to the class Nsl+β(Ω¯), for l>0, if v∈Csβ(Ω¯), Dxilv∈Cs-lβ(Ω¯), i=1,2, and the norms below are finite, (13)vNsl+βΩ¯=vCsβΩ¯+∑j=1l-1DxjvCs-jβΩ¯+∑i=12DxilvCs-lβΩ¯,if l≥1, and (14)vNsl+βΩ¯=vCsβΩ¯+∑i=12DxilvCs-lβΩ¯in the case of l∈(0,1).In a similar way we introduce the spacesEsβ(∂Ω) and Msl+β(∂Ω). As for Csβ(∂Ω) and Nsl+β(∂Ω), these spaces concave with Esβ(∂Ω) and Msl+β(∂Ω).M s l + β and Nsl+β defined above are Banach spaces. Indeed, the fact that they are normed spaces is easily seen, whereas the completeness follows from Theorem 2.7 [31] together with standard arguments (see, e.g., Remark 3.1.3 in [32]).We begin to stipulate the general assumptions.h1 (condition on the parameters): we require(15)α-1<β<1,s=α+β,0<s1<2-α.h2 (conditions on the right-hand sides of boundary conditions): we demand(16)ψ1∈Msα+βΓ1∩M-s1α+βΓ1,ψ2∈Msα+βΓ2∩M-s1α+βΓ2,φ1∈Ms-11+βΓ1∩M-s11+βΓ1,φ2∈Ms-11+βΓ2∩M-s1-11+βΓ2.h3 (conditions on the right-hand side of the equation): we suppose(17)eitherf∈Es-αβΩ¯∩E-s1-αβΩ¯(18)orf∈Cs-αβΩ¯∩C-s1-αβΩ¯.We are now in the position to state our main results.Theorem 3.
Let assumptions (h1), (h2), and (17) hold and moreover for every points x1∈Γ1 and x2∈Γ2(19)fx1,0-Dx1αψ1x1=0,f0,x2-Dx2αψ2x2=0,iftheDBC(2)holds;and(20)fx1,0-Dx1α-1φ1x1=0,f0,x2-Dx2α-1φ2x2=0,iftheNBC(3)holds.Then, (1) subject either to the DBC (2) or to the NBC (3), admits a unique classical solution u=u(x1,x2) on Ω, satisfying the regularity(21)u∈Msα+βΩ¯∩M-s1α+βΩ¯andDx1α1Dx2α2u=Dx2α2Dx1α1ufor 0<α1+α2<α. Besides, the following estimates hold:(22)uMsα+βΩ¯+uM-s1α+βΩ¯+uNs1+βΩ¯≤CfEs-αβΩ¯+fE-s1-αβΩ¯+∑i=12ψiMsα+βΓi+ψiM-s1α+βΓi≡FMf,ψ1,ψ2,(23)Dx1α1Dx2α2uEs-α1-α2βΩ¯+Dx1α1Dx2α2uE-s1-α1-α2βΩ¯≤FMf,ψ1,ψ2in the case of DBC (2), and(24)uMsα+βΩ¯+uM-s1α+βΩ¯+uCsβΩ¯+Dx1α1Dx2α2uEs-α1-α2βΩ¯+Dx1α1Dx2α2uE-s1-α1-α2βΩ¯≤CfEs-αβΩ¯+fE-s1-αβΩ¯+∑i=12φiMs-1α-1+βΓi+φiM-s1-1α-1+βΓiif NBC (3) hold.Theorem 4.
Let assumptions (h1), (h2), and (18) hold. Then, boundary value problems (1)-(3) admit a unique classical solution u=u(x1,x2) on Ω, satisfying the regularity(25)u∈Nsα+βΩ¯∩N-s1α+βΩ¯andDx1α1Dx2α2u=Dx2α2Dx1α1ufor 0<α1+α2<α. Besides, the following estimates hold:(26)uNsα+βΩ¯+uN-s1α+βΩ¯≤CfCs-αβΩ¯+fC-s1-αβΩ¯+∑i=12ψiMsα+βΓi+ψiM-s1α+βΓi≡FNf,ψ1,ψ2,(27)Dx1α1Dx2α2uCs-α1-α2βΩ¯+Dx1α1Dx2α2uC-s1-α1-α2βΩ¯≤FNf,ψ1,ψ2,in the case of DBC (2), and(28)uNsα+βΩ¯+uN-s1α+βΩ¯+Dx1α1Dx2α2uCs-α1-α2βΩ¯+Dx1α1Dx2α2uC-s1-α1-α2βΩ¯≤CfCs-αβΩ¯+fC-s1-αβΩ¯+∑i=12φiMs-1α-1+βΓi+φiM-s1-1α-1+βΓiif NBC (3) hold.Remark 5.
It is easily apparent that the functions,(29)f=x12+x22β/2x1x21+x12+x22s1+α/2x12+x22,ψi,φi≡0,i=1,2,satisfy conditions(h2) and (17), (19), and (20).The remainder of the paper is devoted to the proof of Theorems3 and 4 in the DBC case. The proof of these theorems for NBC is almost identical and is left to the interested reader.


## 3. Integral Representation foru(x1,x2) in the Special Case

We first dwell on the special case whereψi≡0 and f is a finite function. Namely (2), (17), and (19) are replaced by the simpler conditions(30)u∂Ω=0,(31)f≡0ifx12+x221/2>R0for some given positive R0 and(32)eitherf∈Es-αβΩ¯,fx1,0=f0,x2=0,orf∈Cs-αβΩ¯.We denote byu⋆(p1,p2) the Mellin transform of the function u(x1,x2). Due to conditions (30)-(32) and assumptions of Theorem 3, we can apply, at least formally, the Mellin transformation to problem (1) and (2) (see for details § 1.4 and § 2.5 in [1]). Then simple calculations lead to the equation (33)u∗p1-α,p2+α+Γ1-p1Γ1+α-p1Γ1-p2Γ1-p2-αu∗p1,p2=Γ1-p1Γ1+α-p1f∗p1,p2+α.Introducing new variables(34)p1=-αq1andp2=αq2,and new functions (35)Vq1,q2=u∗-αq1,αq2,Fq1,q2=Γ1+αq1Γ1+α+αq1f∗-αq1,αq2+α,Pq1,q2=Γ1+αq1Γ1+α+αq1Γ1-αq2Γ1-αq2-α,we rewrite the equation in the more compact form(36)Vq1+1,q2+1+Pq1,q2Vq1,q2=Fq1,q2.Thus, we transform problem (1) and (2) to the linear nonhomogeneous difference equation of the first-order with variable coefficients. In order to solve this equation, we adapt the technique from Section 3 in [33] to our case.Proposition 6.
Letγ¯ denote Euler-Mascheroni constant (see, e.g., Definition in [34]),(37)Re⁡q1≠-m+1α,Re⁡q2≠m+1α,m=0,1,2,…,and let E1(q1) and E2(q2) be arbitrary analytic functions such that(38)E1q1=E1q1+1,E2q2=E2q2+1.Then the function(39)V0q1,q2=πeγ¯αeiπq1Γ1+αq1Γ1-αq2E1q1E2q2solves homogenous equation (36) (i.e., F≡0) and does not have any poles.
If, in addition, the functionsE1(q1) and E2(q2) have no zeros, then the function 1/V0(q1,q2) does not have any poles and the following estimate holds:(40)E1q1E2q2V0q1,q2≤Cq1αRe⁡q1+1/2q21/2-αRe⁡q2×exp⁡-πIm⁡q1-πα2Im⁡q1-πα2Im⁡q2for Imq1,Imq2→+∞ and for every fixed Re⁡q1 and Re⁡q2 satisfying inequalities above.Proof.
In order to verify that the functionV0(q1,q2) solves homogenous equation (36), it is enough to substitute V0 in (36) and take into account the properties of the Gamma-function: Γ(q+1)=qΓ(q). Besides, the properties of the function V0 are simple consequences of the well-known properties of Gamma function:(i)
Γ(z) has simple poles in the points(41)Re⁡z=-m,m=0,1,2,…(ii)
The Stirling asymptotic formula holds:(42)Γz=exp⁡z-12ln⁡z-z+ln⁡2π2+112z+Oz-3asImz→+∞ while Rez remains boundedWe are now in the position to construct the solution of nonhomogeneous equation (36).Till the end of the paper, we assume thatγ∈[0,1] is an arbitrary fixed quantity and define the contour Lγ in the complex plane ζ as(i)
(43)Lγ=Re⁡ζ=-γ,Im⁡ζ∈-∞,+∞,ifγ∈0,1.(ii)
Ifγ=1, then L1 consists of three parts:the half-circleζ+1=γ0,Re⁡ζ>-1,the intervalsReζ=-1,Imζ∈(-∞,-γ0) and Reζ=-1,Imζ∈(γ0,+∞)with the small positive numberγ0, γ0<γ/2.(iii)
The contourL0 (γ=0) is obtained from L1 after its shifting to the right on Reζ=1.Introducing the periodic functionK(ζ) with period 1,(44)Kζ=12icot⁡πζ+isin2πasin2πζ+a,0<a-γ<1,we assert the following results.Proposition 7.
Letζ=-γ+iz, z∈R; conditions (32) and (31) hold, and let(45)Re⁡q1≠-m+1α+γ,Re⁡q2≠m+1α+γ-1,m=0,1,2,….Then the solution of inhomogeneous equation (36) is given by(46)Vq1,q2=V0q1,q22i∫L-γKζFq1+ζ,q2+ζV0q1+ζ+1,q2+ζ+1dζwhere V0(q1,q2) is defined in (39) with E1(q1), E2(q2)≡1.Proof.
First, we prove this statement in the caseγ=1; i.e., Lγ=L1. To this end, we will seek the solution of (36) as a product (47)Vq1,q2=V0q1,q2Yq1,q2,where the unknown function Y(q1,q2) solves the first-order difference inhomogeneous equation with the constant coefficients(48)Yq1+1,q2+1-Yq1,q2=Fq1,q2V0q1+1,q2+1.
Then adapting the technique from [35] to our case, we deduce that(49)Yq1,q2=12i∫L-1Fq1+ζ,q2+ζV0q1+1+ζ,q2+1+ζKζdζ,if condition (45) holds.
Indeed, substitutingY(q1,q2) to (48) and applying Cauchy’s residue theorem arrive at (50)Yq1+1,q2+1-Yq1,q2=12i∫L-1Fq1+1+ζ,q2+1+ζV0q1+2+ζ,q2+2+ζKζdζ-∫L-1Fq1+ζ,q2+ζV0q1+1+ζ,q2+1+ζKζdζ=πresζ=0Fq1+ζ,q2+ζV0q1+1+ζ,q2+1+ζKζ=Fq1,q2V0q1+1,q2+1.To reach these equalities, we essentially used the following properties of the functions F, V0, and K:(i)
F(q1+ζ,q2+ζ)/V0(q1+1+ζ,q2+1+ζ) has no poles if Reζ∈(-1,γ0) and Re⁡q1, Re⁡q2 satisfy (45).(ii)
K(ζ) is periodic of period 1 with simple poles at ζ=0,±1 and multiple poles at ζ=-a.Besides, the following inequality holds:(51)K-γ+iz≤Ce-4πz,z→+∞;e2πz,z→-∞,forγ∈[0,1](iii)
There is the asymptotic representation(52)KζFq1+ζ,q2+ζV0q1+1+ζ,q2+1+ζ→0asIm⁡ζ→0,for the boundedImq1 and Imq2 and if Req1, Req2 meet requirement (45).
Note that statements (i) and (ii) are simple consequences of Proposition6 and (32). As for assessment (iii), it can be easily drawn from (i) and (ii).
After that, we return to the representation ofV(q1,q2) and obtain solution (36) as(53)Vq1,q2=V0q1,q22i∫L-1KζFq1+ζ,q2+ζV0q1+1+ζ,q2+1+ζdζ.This proves Proposition 7 in the case Lγ≔L-1.
Recasting the arguments above in the caseγ∈[0,1) and applying Cauchy theorem allow us to prove Proposition 7 in the case of arbitrary contour Lγ, γ∈[0,1). It completes the proof of Proposition 7.We are now in the position to obtain integral representation of the solutionu(x1,x2). Indeed, Propositions 6 and 7 provide(54)
u
⋆
p
1
,
p
2
=
V
0
-
p
1
/
α
,
p
2
/
α
2
i
∫
L
-
γ
K
ζ
×
F
-
p
1
/
α
+
ζ
,
p
2
/
α
+
ζ
V
0
-
p
1
/
α
+
1
+
ζ
,
p
2
/
α
+
1
+
ζ
d
ζ
=
-
1
2
i
×
∫
L
-
γ
K
ζ
e
-
i
π
ζ
×
Γ
1
-
p
1
+
α
ζ
Γ
1
-
p
2
-
α
-
α
ζ
Γ
1
-
p
1
Γ
1
-
p
2
×
f
⋆
p
1
-
α
ζ
,
p
2
+
α
ζ
+
α
d
ζ
,where(55)Re⁡p1≠m+1-γα,Re⁡p2≠m+1-α+αγ,m=0,1,2,…,γ∈0,1.Note that, conditions on Repi hold, for example, in the domain(56)Re⁡p1<1-γαandRe⁡p2<1-α+γα.At last, we carry out the inverse Mellin transform to derive an explicit integral representation foru(x1,x2),(57)ux1,x2=18πi∫0+∞dt1t1×∫0+∞dt2t2fx1t1,x2t2Lγt1,t2,x1,x2,with (58)Lγt1,t2,x1,x2=∫L-γKζx1t1-αζx2t2α+αζ×e-iπζ∫γ1-i∞γ1+i∞t1-p1Γ1-p1+αζΓ1-p1dp1×∫γ2-i∞γ2+i∞t2-p2Γ1-p2-α-αζΓ1-p2dp2dζ,for Reγ1∈(A1,B1) and Reγ2∈(A2,B2), where Ai and Bi, i=1,2, are chosen that(59)m+1-γα∉A1,B1andm+1-α+γα∉A2,B2,m=0,1,2,….In the light of (56), we can pick up (A1,B1) and (A2,B2) as(60)A1,B1⊆Re⁡γ1<1-γαandA2,B2⊆Re⁡γ2<1-α+γα.


## 4. Some Technical Results

First we introduce some equivalent norms.Proposition 8.
Lets and s1 be any nonnegative numbers. Then for any functions v and w, v∈Esβ(Ω¯)∩E-s1β(Ω¯), and w∈Csβ(Ω¯)∩C-s1β(Ω¯), we have the following norm equivalence: (61)C1vEsβΩ¯+vE-s1βΩ¯≤supΩ¯⁡r-sx×1+r2xs1+s/2vx+supx,x¯∈Ω¯⁡r-s+βx,x¯×1+r2x,x¯s1+s/2vx-vx¯x-x¯β≤C2vEsβΩ¯+vE-s1βΩ¯,C1wCsβΩ¯+wC-s1βΩ¯≤supΩ¯⁡d-sx×1+dxs1+swx+supx,x¯∈Ω¯⁡d-s+βx,x¯×1+dx,x¯s1+swx-wx¯x-x¯β≤C2wCsβΩ¯+wC-s1βΩ¯.The proof of this statement follows with direct calculations. Let(62)a≥1,b∈0,2,z,r∈R.Next we represent certain estimates for the functions(63)I±1b,z,r=∫-∞+∞e±iprΓa±ip-zΓa+b±ipdp,which will be frequently used to evaluate the functions u(x1,x2) and Dxiαu, i=1,2.Proposition 9.
Letε be positive number, ε∈(0,1). Then there are estimates.(64)I±1b,z,r≤C1+z2cosh⁡πz2×e-εr,ifb∈1,2;e-εr1+r-1,ifb∈0,1;1+r-1,ifb=0.Proof.
For simplicity consideration, we puta=1. Here we prove this proposition for the function I+1(b,z,r). The case of I-1(b,z,r) is considered in the similar way. We start our consideration with the case of positive b (i.e., b≠0).
The Stirling asymptotic formula for the Gamma-function,(65)Γim+d≈Ce-π/2mim+dd-1/2×exp⁡imln⁡im+d-im-d×1+Oim+d3,asm→+∞,provides that the integrand Γ(1+i(p-z))/Γ(1+b+ip), b>0, has the order p-b at the infinity. Moreover, the function Γ(1+p¯-iz)/Γ(1+b+p¯) with p¯=Rep¯+ip has no poles in the domain Rep¯>-1. Thus, by the residue theorem, we get (66)I+1b,z,r=e-εr∫-∞+∞Γ1-ε+ip-zΓ1+b-ε+ipeiprdp,r≥0,eεr∫-∞+∞Γ1+ε+ip-zΓ1+b+ε+ipeiprdp,r<0=e-εr∫-∞+∞Γ1-εsgnr+ip-zΓ1+b-εsgnr+ipeiprdp.Then, following [33], we decompose the plane (p,z) into 15 subdomains, as shown in Figure 1(67)p,z=Ω0∪⋃i=17Ωi+∪⋃i=17Ωi-.In this decomposition M is a sufficiently large number such that the terms O((im+d)-3) in (65) can be neglected in all regions Ωi±.
Taking into account this decomposition, we represent the functionI+1(b,z,r) as(68)I+1b,z,r=e-εr×∑i=27∫Ωi+Γ1-εsgnr+ip-zΓ1+b-εsgnr+ipeiprdp,z>2M,∑i=27∫Ωi-Γ1-εsgnr+ip-zΓ1+b-εsgnr+ipeiprdp,z<-2M,∫Ω1+∪Ω1-Γ1-εsgnr+ip-zΓ1+b-εsgnr+ipeiprdp+∫Ω0Γ1-εsgnr+ip-zΓ1+b-εsgnr+ipeiprdp,z<2M.It is apparent that, if z<-2M the function I+1(b,z,r) is estimated analogous to the function I+1(b,z,r) if z>2M. Thus, we consider here just case z>-2M. To this end, we apply Stirling formula (65) and the well-known properties of the Gamma-function. Introducing the functions (69)d1p,z=1-εsgnr+ip-z1+b-εsgnr+ip-εsgnr+1/2,d2p,z=exp⁡ip-zln⁡1-εsgnr+ip-z-ipln⁡1+b-εsgnr+ip,we arrive at the representation to the function Γ(1-εsgnr+i(p-z))/Γ(1+b-εsgnr+ip) in every domain Ωi+, Ω1-, and Ω0,(70)
Γ
1
-
ε
s
g
n
r
+
i
p
-
z
Γ
1
+
b
-
ε
s
g
n
r
+
i
p
=
Φ
0
p
,
z
,
s
g
n
r
,
in
Ω
0
,
d
1
p
,
z
d
2
p
,
z
1
+
b
-
ε
s
g
n
r
+
i
p
b
Φ
1
p
,
z
,
s
g
n
r
,
in
Ω
1
+
,
e
π
z
/
2
d
1
p
,
z
d
2
p
,
z
1
+
b
-
ε
s
g
n
r
+
i
p
b
Φ
2
p
,
z
,
s
g
n
r
,
in
Ω
2
+
,
e
π
p
/
2
1
+
b
-
ε
s
g
n
r
+
i
p
b
Φ
3
p
,
z
,
s
g
n
r
,
in
Ω
3
+
,
e
π
p
-
z
+
π
z
/
2
d
1
p
,
z
d
2
p
,
z
1
+
b
-
ε
s
g
n
r
+
i
p
b
Φ
4
p
,
z
,
s
g
n
r
,
in
Ω
4
+
,
e
π
/
2
p
-
z
1
-
ε
s
g
n
r
+
i
p
-
ε
s
g
n
r
+
1
/
2
Φ
5
p
,
z
,
s
g
n
r
,
in
Ω
5
+
,
e
-
π
z
/
2
d
1
p
,
z
d
2
p
,
z
1
+
b
-
ε
s
g
n
r
+
i
p
b
Φ
6
p
,
z
,
s
g
n
r
,
in
Ω
6
+
∪
Ω
7
+
,
d
1
p
,
z
d
2
p
,
z
1
+
b
-
ε
s
g
n
r
+
i
p
b
Φ
7
p
,
z
,
s
g
n
r
,
in
Ω
1
-
,where the functions Φi(p,z,sgnr), i=0,1,2,…,7, are uniformly bounded with respect to p and z together with (∂Φi/∂z)(p,z,sgnr).
Moreover,(71)∂Φ1∂p,∂Φ2∂p,∂Φ6∂p,∂Φ7∂p≤Cp3,for (p,z) belongs to one of the corresponding domains Ω1+, Ω2+, Ω7+, Ω1-. It is worth mentioning that representation (70) holds for each b∈[0,2].
Ifb∈(1,2], representation (70) with aid (65) and (71) arrives at the inequality(72)I+1b,z,r≤Ce-εr1+zcosh⁡πz2,ifz>2M,e-εr1+z,ifz≤2M.These relations guarantee the first estimate in Proposition 9 for b∈(1,2].
Further, we verify the estimateI+1(b,z,r) if b∈(0,1]. Representation (70) and straightforward calculations lead to the inequality(73)∫Ω0Γ1-εsgnr+ip-zΓ1+b-εsgnr+ipeiprdp+∑i=36∫Ωi+Γ1-εsgnr+ip-zΓ1+b-εsgnr+ipeiprdp≤C1+zcosh⁡πz2.To get the same estimate in the domains Ωi+, i=1,2,7, and Ω1-, it is enough to integrate by parts. Namely, let us consider the case Ω1+. Based on representation (70), we can write(74)∫3M+∞Γ1-εsgnr+ip-zΓ1+b-εsgnr+ipeiprdp=∫3M+∞d1p,zd2p,z1+b-εsgnr+ipbΦ1p,z,sgnr×eiprdp=iei3Mrrd13M,zd23M,z1+b-εsgnr+i3Mb×Φ13M,z,sgnr+ir×∫3M+∞eipr∂∂pd1p,zd2p,z1+b-εsgnr+ipb×Φ1p,z,sgnrdp.It is apparent that (75)d13M,z+d23M,z+Φ13M,z,sgnr≤C∂d1p,z∂p+∂d2p,z∂p≤C1+z2p.Thus, we can enhance estimate (74), taking into account (71), so as to get(76)∫3M+∞eiprΓ1-εsgnr+ip-zΓ1+b-εsgnr+ipdp≤C1+z2r.Recasting the arguments above in the case of domains Ω2+, Ω6+, and Ω7+ yields the same estimate to ∫Ω2+∪Ω7+∪Ω1-eiprΓ1-εsgnr+ip-z/Γ1+b-εsgnr+ipdp.
After that, we combine inequalities (73) and (76) and obtain the required estimate to I+1b,z,r if b∈(0,1].
At last, we are left to produce suitable estimate ofI+10,z,r. Standard direct calculations lead to(77)d1p,z+d2p,z≤C1+zforp,z∈⋃i=26Ωi+∪Ω0andb∈0,2.In conclusion, we derive that(78)∫⋃i=36Ωi+∪Ω0Γ1+ip-zΓ1+ipeiprdp≤Ccosh⁡πz21+z.
After that, we remark that the termsd1(p,z), d2(p,z), and Φi share the same properties in Ω1- and Ωi+, i=1,2,6,7. Thus, the estimate of ∫Ωi+(Γ(1+i(p-z))/Γ(1+ip))eiprdp is very similar and we will confine ourselves the consideration of the case Ω1+.
Simple conclusions draw to representations inΩ1+(79)d1p,z=1+g1z,p,p-zln⁡1+p-z2-pln⁡1+p2=-zln⁡p-z+g2p,z,where(80)g1≤C1+z2p2andg2≤Cp.Therefore, the main term in the integral ∫Ω1+(Γ(1+i(p-z))/Γ(1+ip))eiprdp is(81)∫3M+∞eirpe-iz-izlnpdp=e-iz∫0+∞eirpe-izlnpdp-∫03Meirpe-izlnpdp.Note that the second term here is a regular function of r and z. In order to evaluate the first term in this representation, we apply Proposition 7.1 from [33] and obtain (82)∫0+∞eirpe-izlnpdp=eizlnrzr-1g3z+1+zr-1g4z,where g3 and g4 are twice continuously differentiated and bounded functions for z<2M.
In summary, we obtain the estimate(83)∫Ω1+Γ1+ip-zΓ1+ipeiprdp≤C1+r-1.Finally, this inequality together with (78) completes the proof of Proposition 9.Figure 1Recasting the proof of the previous proposition, we state the results, which will be used to evaluateDxiαiu, i=1,2, 0<αi<α, in Sections 5 and 6.Proposition 10.
Letz∈(-2M,2M), a≥1, b∈[0,2], r∈R; then the inequalities are fulfilled (84)∂∂zI±1b,z,r≤Ccosh⁡πz2×e-εr1,ifb∈1,2,1+r-1,ifb∈0,1,∂∂z∫Ω0Γa±ip-zΓa±ipe±iprdp+∂∂ze∓izlnr∫Ω1+∪Ω1-Γa±ip-zΓa±ipe±iprdp≤C1+r-1.The following results are related to the properties of the function(85)
I
±
2
ε
1
,
ε
2
,
ε
,
b
,
z
,
ρ
≔
∫
-
ε
1
ρ
ε
2
ρ
d
r
e
-
ε
r
×
∫
-
∞
+
∞
Γ
1
±
ε
s
g
n
r
±
i
p
-
z
Γ
1
+
b
±
ε
s
g
n
r
±
i
p
e
±
i
r
p
-
z
d
p
,where ε1, ε2 are some positive constant, ε∈[0,1), b∈[0,2].Proposition 11.
Letz∈R, ρ∈(0,+∞). Then there are the following estimates for every fixed ρ:(i)
(86)I±2ε1,ε2,ε,b,z,ρ≤C1+zcosh⁡πz21+ρε1+ε2ifb∈0,2.(ii)
(87)I±2ε1,ε2,ε,0,z,ρ≤C1+zcosh⁡πz2×1+ρε1+ε2+1ρε1+1ρε2≤C1+z×cosh⁡πz21+ρε1+ε2+ε1ε2+ε2ε1.(iii)
Besides, ifρ=+∞, then(88)I±2ε1,ε2,0,b,z,+∞=∫-∞+∞dr∫-∞+∞Γ1±ip-zΓ1+b±ipe±irp-zdp=πΓ1+b±iz,ifb∈0,2,I±2ε1,ε2,ε,0,z,+∞≔I±2ε,z≤C1+zcosh⁡πz2.Proof.
We will carry out the detailed proof inI+2(ε1,ε2,ε,b,z,ρ) case. The arguments for I-2(ε1,ε2,b,z,ρ) are almost identical and left to the interested reader.
Straightforward calculations lead to the representation(89)
I
+
2
ε
1
,
ε
2
,
ε
,
b
,
z
,
ρ
=
2
∫
-
∞
+
∞
Γ
1
+
i
p
-
z
Γ
1
+
b
+
i
p
×
s
i
n
ε
1
+
ε
2
p
-
z
ρ
p
-
z
e
i
ε
2
-
ε
1
p
-
z
ρ
/
2
d
p
≡
2
∫
-
∞
+
∞
i
2
d
p
.First, we consider case b∈(0,2]. Asymptotic representation (70) with the decomposition in Figure 1 provides estimate in R2∖(Ω0∪Ω3+∪Ω3-∪Ω5+∪Ω5-)(90)i2≤C1+zcoshπz/2p-zp2+1+b-ε2b/2with ε∈(0,1) and(91)i2≤Csin⁡ε1+ε2p-zρε1+ε2p-zρ1+coshπz2ε1+ε2ρin Ω0∪Ω3+∪Ω3-∪Ω5+∪Ω5-. Then, this inequality together with (89) leads to statement (ii).
Let us verify point (ii). One can easily check that(92)sin⁡ε1+ε2p-zρp-z≤Cε1+ε2ρ,ifp-z≤M,p-z-1,ifp-z>M.After that, recasting the proof of Proposition 9 with aid the last inequality arrives at the estimate(93)∑j=36∫Ωj±i2dp+∫Ω0i2dp≤C1+zcosh⁡πz21+ρε1+ε2.
Then we are left to evaluate the terms∫Ωj±i2dp, j=1,2,7. We restrict ourselves the estimate of ∫Ω2+i2dp, the remaining terms are evaluated the same way. Asymptotic (70) with b=0 and the change of variables, p-z=k, provide the representation(94)e-πz/2∫Ω2+i2dp=∫2M+∞d1k+z,zd2k+z,z2ikΦ2k+z,z×eiε2ρk-e-iε1ρkdk.After that, following the proof of Proposition 9, we integrate by parts and deduce (95)
e
-
π
z
/
2
∫
Ω
2
+
i
2
d
p
≤
C
1
ρ
ε
1
+
1
ρ
ε
2
×
d
1
2
M
+
z
,
z
d
2
2
M
+
z
,
z
2
M
×
Φ
2
2
M
+
z
,
z
+
∫
2
M
+
∞
e
i
ε
2
ρ
k
+
e
-
i
ε
1
ρ
k
×
∂
∂
k
d
1
k
+
z
,
z
d
2
k
+
z
,
z
Φ
2
k
+
z
,
z
k
d
k
.The standard calculations lead to(96)∂∂kd1k+z,zd2k+z,zΦ2k+z,zk≤1+zk2.In conclusion, we reach the estimate(97)∫Ω2+i2dp≤Ce-πz/21+z1ρε1+1ρε2.Collecting this inequality with (93), we deduce the first estimate in statement (ii) of Proposition 11.
Further, to improve the estimate forI+2(ε1,ε2,ε,0,z,ρ), we consider two different cases:
(i)(98)2Mρε1+ε2>π2,
(ii)(99)2Mρε1+ε2≤π2.
It is apparent that, in the first case there is(100)1ε1ρ+1ε2ρ<4Mπε1-1+ε2-1ε1+ε2,and, therefore, keeping in mind the previous estimate of I±2(ε1,ε2,ε,0,z,ρ), we deduce that(101)I+2ε1,ε2,ε,0,z,ρ≤C1+zcosh⁡πz2×1+ρε1+ε2+ε1ε2+ε2ε1.
Coming to the second case (i.e.,ρ-1>4M(ε1-1+ε2-1)π-1), the change of the variable, k(ε1+ε2)ρ=v, leads to(102)
e
-
π
z
/
2
∫
Ω
2
+
i
2
d
p
=
∫
2
M
ρ
ε
1
+
ε
2
+
∞
d
1
z
+
v
ρ
ε
1
+
ε
2
,
z
×
d
2
z
+
v
ρ
ε
1
+
ε
2
,
z
sin
⁡
v
v
e
i
v
2
ε
1
+
2
ε
2
-
1
ε
2
-
ε
1
d
v
=
∫
2
M
ρ
ε
1
+
ε
2
π
/
2
d
1
z
+
v
ρ
ε
1
+
ε
2
,
z
×
d
2
z
+
v
ρ
ε
1
+
ε
2
,
z
sin
⁡
v
v
e
i
v
2
ε
1
+
2
ε
2
-
1
ε
2
-
ε
1
d
v
+
∫
π
/
2
+
∞
d
1
z
+
v
ρ
ε
1
+
ε
2
,
z
×
d
2
z
+
v
ρ
ε
1
+
ε
2
,
z
sin
⁡
v
v
e
i
v
2
ε
1
+
2
ε
2
-
1
ε
2
-
ε
1
d
v
.The straightforward calculations ensure the uniformly boundedness of the first term in (102); moreover, this estimate is independent of ε1, ε2, and ρ.
Further, we treat the second term in (102). Integrating by parts leads to (103)
∫
π
/
2
+
∞
d
1
z
+
v
ρ
ε
1
+
ε
2
,
z
d
2
z
+
v
ρ
ε
1
+
ε
2
,
z
×
Φ
2
z
+
v
ρ
ε
1
+
ε
2
,
z
s
i
n
v
v
e
i
v
2
ε
1
+
2
ε
2
-
1
ε
2
-
ε
1
d
v
=
-
i
8
e
i
π
ε
2
-
ε
1
ε
1
+
ε
2
-
1
/
4
4
π
-
π
ε
2
-
ε
1
2
ε
2
+
ε
1
-
2
×
d
1
π
2
ρ
ε
1
+
ε
2
+
z
,
z
d
2
π
2
ρ
ε
1
+
ε
2
+
z
,
z
×
Φ
2
π
2
ρ
ε
1
+
ε
2
+
z
,
z
-
4
4
-
ε
2
-
ε
1
2
ε
2
+
ε
1
-
2
×
∫
π
/
2
+
∞
e
-
i
v
+
i
v
ε
2
-
ε
1
/
2
ε
2
+
ε
1
×
∂
∂
v
d
1
v
ρ
-
1
ε
1
+
ε
2
-
1
+
z
,
z
d
2
v
ρ
-
1
ε
1
+
ε
2
-
1
+
z
,
z
×
Φ
2
v
ρ
-
1
ε
1
+
ε
2
-
1
+
z
,
z
v
-
1
d
v
.After that, the properties of functions d1, d2, and Φ2 allow us to extend this estimate and conclude (104)
∫
π
/
2
+
∞
d
1
z
+
v
ρ
ε
1
+
ε
2
,
z
d
2
z
+
v
ρ
ε
1
+
ε
2
,
z
×
sin
⁡
v
v
e
i
v
2
ε
1
+
2
ε
2
-
1
ε
2
-
ε
1
Φ
2
z
+
v
ρ
ε
1
+
ε
2
,
z
d
v
≤
C
4
4
-
ε
2
-
ε
1
2
ε
2
+
ε
1
-
2
≤
C
1
+
ε
2
ε
1
+
ε
1
ε
2
.Hence, this estimate completes the proof of statement (ii).
Finally, we are left to verify statement (iii). As for the first equality in (iii), it is a simple consequence of Lemma 5.3 [36]. Let us check the second inequality in (iii). To this end, we rewrite I+2(ε,z) as (105)I+2ε,z=∫-∞+∞Γ1+ε+ip-zΓ1+ε+ipdpε-ip-z+∫-∞+∞Γ1-ε+ip-zΓ1-ε+ipdpε+ip-z≡i1+z+i2+z,where we set (106)i1+z=∫-∞+∞Γ1+ε+ip-zΓ1+ε+ip2εdpε2+p-z2,i2+z=-∫-11dq×∫-∞+∞ε2ε+ip-z∂∂1-qεΓ1+qε+ip-zΓ1+qε+ipdp.Then representation (70) with b=0 and direct calculations provide(107)i1+z+i2+z≤C1+zcosh⁡πz2,which implies(108)I+2ε,z≤C1+zcosh⁡πz2.Thus, the claim is proven.Finally, we complete this preliminary section with two estimates that will be frequently used in the following sections.Proposition 12.
Letz∈[-2M,2M], x1, x2∈[0,+∞), r,ρ∈R, ε∈(0,1). Then the following inequalities are fulfilled: (i)
(109)∂∂zI±2ε1,ε2,ε,b,z,ρ≤C1+ρε1+ε2,b∈0,2,1+ρε1+ε2+ε1ε2+ε2ε1,b=0,for every fixedρ. Besides,(110)∂∂zI±2ε1,ε2,ε,0,z,+∞≤C.(ii)
(111)J1x1,x2,t,ρ,M≔∫01dqq∫-2M2M∂∂zexp⁡izqln⁡x2rx1ρdz≤C,where the positive constant C is independent of ρ, ε1, ε2, ε, x1, and x2.Proof.
To show the validity of statement (i), it is enough to recast the arguments in Propositions11 and 10 if z∈[-2M,2M].
Next, we obtain the second inequality from this proposition. For the sake of clarity, we consider casex2r/x1ρ≥1. In the opposite case, we exchange the function eizqln(x2r/x1ρ) by e-izqln(x1ρ/x2r) and repeat the arguments below.
The direct calculations together with the change of variables,v=2Mqln(x2r/x1ρ), reduce the function J1 to the form(112)J1=∫02Mlnx2r/x1ρsin⁡vvdv.At this point, we consider the two different cases for J1:
(i)(113)2Mln⁡x2rx1ρ>π2,
(ii)(114)0≤b2Mln⁡x2rx1ρ≤π2.
In the first case, we have(115)J1≤∫0π/2sin⁡vvdv+∫π/22Mln⁡x2r/x1ρsin⁡vvdv≤π2+-cos⁡2Mln⁡x2r/x1ρ2Mln⁡x2r/x1ρ+∫π/22Mlnx2r/x1ρcos⁡vv2dv≤π2+2π+∫π/2+∞dvv2≤C.Note that we apply integrating by parts in order to reach the last estimate.
Further, we obtain the same estimate in the second case. To this end, we use the easily verified inequality,(116)sin⁡vv≤10<v<π2,and deduce the bound(117)J1≤∫02Mlnx2r/x1ρdv≤∫0π/2dv≤C.In summary, we complete the proof of Proposition 12.


## 5. Estimates on Minor Derivatives ofu(x1,x2)

To evaluate the functionu(x1,x2), we use representation (57) with γ∈(0,1) and γ1=-αγ, γ2=-α(1-γ). The change of variables(118)t1=er1andt2=er2,in the integrals in (57) yields(119)ux1,x2=eiπγx1αγx2α1-δi8πα∫-∞+∞dr1×∫-∞+∞dr2fx1e-r1,x2e-r2Lγr1,r2,x1,x2,where (120)Lγr1,r2,x1,x2=∫-∞+∞K-γ-izα×eπz/αx1-izx2izeizr1-r2∫-∞+∞e-ip1r1×Γ1-ip1-zΓ1+αγ-ip1dp1∫-∞+∞eip2r2×Γ1+ip2-zΓ1+α-αγ-ip2dp2dz,γ∈0,1.


### 5.1. Estimate on Maximum ofux1,x2

Before evaluating the functionsu and Dxiαiu, i=1,2, 0<αi≤1, we describe the suitable properties of the kernel Lγ.Lemma 13.
Letε∈(0,1) and let r1, r2∈R. Then there are estimates. (i)
(121)Lγr1,r2,x1,x2≤Cexp⁡-εr1-εr2×1+r2-1ifγ∈1α,1;1+r1-1ifγ∈0,1-1α;1+r2-11+r1-1ifγ∈1-1α,1α.(ii)
Ifγ=0 or γ=1, the following representations hold:(122)L0r1,r2,x1,x2=exp⁡-εr1-εr2×δr1+δ-r1L01r1,r2,x1,x2+L02r1,r2,x1,x2,L1r1,r2,x1,x2=exp⁡-εr1-εr2×δr2+δ-r2L11r1,r2,x1,x2+L12r1,r2,x1,x2,whereδ(r) is a Dirac delta function and (123)L01r1,r2,x1,x2+L11r1,r2,x1,x2≤C,L02r1,r2,x1,x2≤C1+r1-11+r2,L12r1,r2,x1,x2≤C1+r2-11+r1.(iii)
(124)∫0+∞dr1∫0+∞r2βLγr1,r2,x1,x2dr2≤Cifγ∈1α,1;∫0+∞dr1r1β∫0+∞Lγr1,r2,x1,x2dr2≤Cifγ∈0,1-1α;∫0+∞dr1∫0+∞r2βe-εr1-εr2L12r1,r2,x1,x2dr2+∫0+∞dr1r1β∫0+∞e-εr1-εr2L12r1,r2,x1,x2dr2≤Cifγ=1;∫0+∞dr1∫0+∞r2βe-εr1-εr2L02r1,r2,x1,x2dr2+∫0+∞dr1r1β∫0+∞e-εr1-εr2L02r1,r2,x1,x2dr2≤Cifγ=0.Here the constant C is independent on x1, x2 and γ.Proof.
First of all, we verify statement (i). It is easy to see that inequality (51) guarantees for γ∈[0,1](125)K-γ-izαeπz/α≤Ce-3πz/α,z→+∞,zK1-izα=zK-izα≤C,z≤2M,∂∂zzK1-izα=∂∂zzK-izα≤C,z≤2M. Then the results of Proposition 9 with aid of (125) provide statement (i).
As for statement (ii), we confirm ourselves the case ofγ=0, due to the arguments in the case δ=1 are similar.
First, we representLγ, γ∈(0,1), as (126)Lγr1,r2,x1,x2=e-εr1-εr2∫-∞+∞dzK-γ-izα×eπz/αeizlnx2/x1eizr1-r2∫-∞+∞e-ip1r1×Γ1-εsgnr1-ip1-zΓ1+αγ-εsgnr1-ip1dp1∫-∞+∞eip2r2×Γ1+εsgnr2+ip2-zΓ1+α-αγ+εsgnr2+ip2dp2.Here we use the same reasons as in the proof of Proposition 9.
Then, as mentioned in the proof of Proposition7 (see (ii) there) the function K(-γ-iz/α) has the simple poles if γ=0,1. Thus, following Cauchy’s residue theorem, we rewrite Lγ as(127)Lγr1,r2,x1,x2=e-εr1-εr2L¯01r1,r2+L¯02r1,r2,x1,x2,γ=0,where we set (128)L¯01r1,r2=C∫-∞+∞e-ip1r1∫-∞+∞eip2r2×Γ1+εsgnr2+ip2Γ1+α+εsgnr2+ip2dp2,L¯02r1,r2,x1,x2=∫-∞+∞K-izα×eπz/αeizlnx2r1/r2x1eizr1-r2×∫-∞+∞e-ip1r1e-izlnr1×Γ1-εsgnr1-ip1-zΓ1-εsgnr1-ip1dp1×∫-∞+∞eip2r2eizlnr2×Γ1+εsgnr2+ip2-zΓ1+α+εsgnr2+ip2dp2dz.Concerning the estimate of L¯01(r1,r2), we apply Lemma 5.3 from [36] and deduce(129)
L
¯
0
1
r
1
,
r
2
=
C
δ
-
r
1
+
δ
r
1
×
∫
-
∞
+
∞
e
i
p
2
r
2
Γ
1
+
ε
s
g
n
r
2
+
i
p
2
-
z
Γ
1
+
α
+
ε
s
g
n
r
2
+
i
p
2
d
p
2
.Then the asymptotic representation of Gamma-function (see, e.g., (1.5.15) in [1]),(130)Γp+aΓp+b=pa-b1+Op-1,arg⁡p+a<π,p→+∞,provides the estimate(131)∫-∞+∞eip2r2Γ1+εsgnr2+ip2-zΓ1+α+εsgnr2+ip2dp2≤C.Hence, we are left to evaluate the function L¯02. To this end, we rewrite the term L02 in the form(132)L¯02=∑i=13li,where we put (133)Bz,r1,r2=zK-izαeizr2-r1∫-∞+∞e-ip1r1×Γ1-εsgnr1-ip1-zΓ1-εsgnr1-ip1eizlnr1dp1∫-∞+∞eip2r2×Γ1+εsgnr2+ip2-zΓ1+α+εsgnr2+ip2e-izlnr2dp2,l1=∫z>2MK-izαeizlnx2/x1eizr2-r1∫-∞+∞e-ip1r1×Γ1-εsgnr1-ip1-zΓ1-εsgnr1-ip1dp1∫-∞+∞eip2r2×Γ1+εsgnr2+ip2-zΓ1+α+εsgnr2+ip2dp2,l2=B0,r1,r2∫-2M2Meizlnx2r1/x1r2-1zdz,l3=∫-2M2Meizlnx2r1/x1r2Bz,r1,r2-B0,r1,r2zdz.Treating the first term in (132) via Proposition 9 and inequalities (125), we arrive at(134)l1≤C1+r1-1.Concerning l2, we get(135)B0,r1,r2=CL¯01r1,r2,and therefore(136)l2=CL¯01∫01dqq∫-2M2M∂∂zeizqlnx2r1/x1r2dz.After that, statement (ii) of Proposition 12 (with r=r1,ρ=r2) provides(137)∫01dqq∫-2M2M∂∂zeizqlnx2r1/x1r2dz≤C.In summary, we can conclude that(138)l2+L¯01=δ-r1+δr1×C1+∫01dqq∫-2M2M∂∂zeizqlnx2r1/x1r2dz×∫-∞+∞eip2r2Γ1+εsgnr2+ip2Γ1+εsgnr2+α+ip2dp2≡δ-r1+δr1L01r1,r2,x1,x2,L01r1,r2,x1,x2<C.Finally, coming to l3, we first rewrite it as(139)l3=∫01dq∫-2M2Meizlnx2r1/x1r2∂∂qzBqz,r1,r2dz.Then, the tedious calculations with Propositions 9 and 10 and estimates (125) entail(140)l3≤C1+1+r2r1.Hence, representations (127) and (132) together with estimates (134), (138), and (140) provide statement (ii) for the function L0 where(141)L02r1,r2,x1,x2=l1+l3.At last, the proof of statement (iii) is simple consequences of statements (i) and (ii) from this proposition. Thus, the claim is proven.We are ready now to state estimates of the functionu(x1,x2).Lemma 14.
Let assumptions (h1), (31), and (32) hold. Then the function u(x1,x2) represented with (119) satisfies inequalities (142)supΩ¯⁡r-α-βx1+r2xs1+α/2u+supΩ¯⁡x1-α+x2-αr-βxu≤CR0fEββΩ¯,supΩ¯⁡d-α-βx1+dxs1+αu≤CR0fCββΩ¯.Proof.
At the beginning, we verify the first inequality in this lemma. To this end, we evaluate the termr-α-βxu. Putting γ∈(1/α,1) in (119), we conclude(143)r-α-βxu≤r-βx∫-∞+∞dr1×∫-∞+∞dr2fx1e-r1,x2e-r2-fx1e-r1,x2×Lγr1,r2,x1,x2+r-βx×∫-∞+∞dr1fx1e-r1,x2×∫-∞+∞dr2Lγr1,r2,x1,x2≡r-βxJ1+J2;here we use the simple inequality(144)x1αγx2α1-γ<rαγxrα1-γx=rαx.ConcerningJ1, we apply statement (i) from Lemma 13 with ε>β to deduce(145)J1≤Cfx2,0,Ωβ∫0+∞dr1e-εr1×∫0+∞x2βe-r2-1βe-εr21+r2-1dr2≤Cx2βfx2,0,Ωβ≤Crβxfx2,0,Ωβwith the constant is independent ofx1, x2, and γ.Coming to the termJ2, applying Proposition 11 to the function(146)∫-∞+∞dr2∫-∞+∞e-izr2eip2r2Γ1+ip2-zΓ1+α-αγ+ip2dp2,and keeping in mind Proposition 9 with inequality (125), we get(147)∫-∞+∞Lγr1,r2,x1,x2dr2≤Ce-εr1∫-∞+∞K-γ-izαeπz/α1+z×coshπz/2Γ1+α-αγ+izdz≤Ce-εr1.
In conclusion, we have(148)r-βxJ2≤r-βxfEββΩ¯∫0+∞x22e2r1+x12β/2e-εr1dr1≤CfEββΩ¯,if ε>β.
Then, collecting this inequality with (143) and (145) yields(149)supΩ¯⁡r-α-βxu≤CfEββΩ¯.
Further, we estimater-α-βx1+r2xs1+α/2u. Assumption (31) provides that(150)fx1,x2=F0x1,x21+x12+x22s1+α/2where the function F0 satisfies conditions (31) and (32). Thus, substituting this representation of f in (143) and recasting arguments above, we deduce(151)supΩ¯⁡r-α-βx1+r2xs1+α/2u≤CR0fEββΩ¯.It is easy to see that the second estimate of Lemma 14 is proved with the similar arguments.
Finally, we are left to estimatexi-αr-βxu, i=1,2. For simplicity consideration, we evaluate x1-αr-βxu. Another case is studied with the same arguments. Using statement (ii) of Lemma 13 with γ=1, we rewrite u(x1,x2) as (152)
x
1
-
α
u
x
1
,
x
2
=
C
∫
-
∞
+
∞
d
r
1
∫
-
∞
+
∞
d
r
2
f
x
1
e
-
r
1
,
x
2
e
-
r
2
×
e
-
ε
r
1
+
r
2
L
1
1
r
1
,
r
2
,
x
1
,
x
2
δ
r
2
+
δ
-
r
2
+
C
∫
-
∞
+
∞
d
r
1
×
∫
-
∞
+
∞
d
r
2
f
x
1
e
-
r
1
,
x
2
e
-
r
2
-
f
x
1
e
-
r
1
,
x
2
×
L
1
2
r
1
,
r
2
,
x
1
,
x
2
e
-
ε
r
1
+
r
2
+
C
∫
-
∞
+
∞
d
r
1
f
x
1
e
-
r
1
,
x
2
∫
-
∞
+
∞
L
1
2
r
1
,
r
2
,
x
1
,
x
2
×
e
-
ε
r
1
+
r
2
d
r
2
.After that, statements (ii) and (iii) of Lemma 13 and Proposition 10 yield inequality (153)supΩ¯⁡x1-αr-βxu≤CfEββΩ¯.This completes the proof of Lemma 14.


### 5.2. Estimate onDx1α1Dx2α2ux1,x2, 0<α1+α2≤1

First, based on representation (57) and formula (2.1.17) [1], we obtain(154)
D
x
1
α
1
D
x
2
α
2
u
=
D
x
2
α
2
D
x
1
α
1
u
=
e
i
π
γ
8
π
α
×
x
1
α
γ
-
α
1
x
2
α
1
-
γ
-
α
2
∫
-
∞
+
∞
d
r
1
×
∫
-
∞
+
∞
d
r
2
f
x
1
e
-
r
1
,
x
2
e
-
r
2
-
f
x
1
,
x
2
×
M
γ
r
1
,
r
2
,
x
1
,
x
2
+
e
i
π
γ
f
x
1
,
x
2
8
π
α
×
x
1
α
γ
-
α
1
x
2
α
1
-
γ
-
α
2
∫
-
∞
+
∞
d
r
1
×
∫
-
∞
+
∞
d
r
2
M
γ
r
1
,
r
2
,
x
1
,
x
2
,where we set (155)Mγr1,r2,x1,x2=∫-∞+∞dzK-γ+izαeπzαeizr2-r1e-izlnx1x2×∫-∞+∞e-ip1r1Γ1-ip1-zΓ1+αγ-α1-ip1dp1×∫-∞+∞eip2r2Γ1+ip2-zΓ1+α1-γ-α2+ip2dp2,γ∈α1α,1-α2α.After that, recasting the arguments of Lemma14 with aid of Propositions 9–11, we can state the following results.Lemma 15.
Letε∈(β,1) and γ∈[α1/α,1-α2/α], and let the nonnegative numbers α1 and α2 satisfy 0<α1+α2≤1. Then the kernel Mγ(r1,r2,x1,x2) possesses the following properties: (i)
(156)∫-∞+∞dr1∫-∞+∞dr2Mγr1,r2,x1,x2≤C;(ii)
(157)∫0+∞dr1∫0r1e±r2-1βMγr1,r2,x1,x2dr2+∫0+∞dr2∫0r2e±r1-1βMδr1,r2,x1,x2dr1≤C;(iii)
(158)∫0+∞dr1e±r1-1β∫-r1r1Mγr1,r2,x1,x2dr2+∫0+∞dr2e±r2-1β∫-r2r2Mγr1,r2,x1,x2dr1≤C;(iv)
If eitherγ=1 and α1=1, α2=0 or γ=0 and α2=1, α1=0, then (159)M1r1,r2,x1,x2=δr2+δ-r2M11r1+M12r1,r2,x1,x2,M0r1,r2,x1,x2=δr1+δ-r1M01r2+M02r1,r2,x1,x2,where the functionsM12 and M02 satisfy statements (i)-(iii) of this lemma and(160)M11r1≤Ce-εr1r1,M01r1≤Ce-εr2r2,∫-∞+∞M11r1dr1+∫-∞+∞M01r2dr2≤C.Now we are ready to describe properties of low derivatives to the functionu.Lemma 16.
Let assumptions of Lemma14 hold, and 0<α1+α2≤1. Then there are the estimates.(i)
(161)supΩ¯⁡r-α-β+α1+α2x1+r2xs1+α/2Dx1α1Dx2α2u+supΩ¯⁡r-α-β+α1+α2x1+r2xs1+α/2Dx2α2Dx1α1u+supΩ¯⁡x1-α+1r-βxDx1u+supΩ¯⁡x2-α+1r-βxDx2u≤CR0fEββΩ¯;(ii)
(162)supΩ¯⁡d-α-β+α1+α2x1+dxs1+αDx1α1Dx2α2u+supΩ¯⁡d-α-β+α1+α2x1+dxs1+αDx2α2Dx1α1u≤CR0fCββΩ¯.Proof.
Taking into account (154), we will carry out the detailed proof of the inequality(163)supΩ¯⁡r-α-β+α1+α2x1+r2xs1+α/2Dx1α1Dx2α2u+supΩ¯⁡x1-α+1Dx1u≤CR0fEββΩ¯. The remaining terms in (i) and (ii) are estimated in the same way and with the recasting of the corresponding arguments of Lemma 14.
We begin to evaluate the first term in (163). To this end, we represent Dx1α1Dx2α2u as(164)Dx1α1Dx2α2u=eiπγ8παx1αγ-α1x2α1-γ-α2∑j=19Qj,where we put (165)Q1=∫0+∞dr1∫-r1r1fx1er1,x2er2-fx1er1,x2×Mγr1,r2,x1,x2dr2,Q2=∫0+∞dr1fx1er1,x2-fx1,x2×∫-r1r1Mγr1,r2,x1,x2dr2,Q3=∫-∞0dr1∫r1-r1fx1er1,x2er2-fx1er1,x2×Mγr1,r2,x1,x2dr2,Q4=∫-∞0dr1fx1er1,x2-fx1,x2×∫r1-r1Mγr1,r2,x1,x2dr2,Q5=∫0+∞dr2∫-r2r2fx1er1,x2er2-fx1,x2er2×Mγr1,r2,x1,x2dr1,Q6=∫0+∞dr2fx1,x2er2-fx1,x2×∫-r2r2Mγr1,r2,x1,x2dr1,Q7=∫-∞0dr2∫r2-r2fx1er1,x2er2-fx1,x2er2×Mγr1,r2,x1,x2dr1,Q8=∫-∞0dr2fx1,x2er2-fx1,x2×∫r2-r2Mγr1,r2,x1,x2dr1,Q9=fx1,x2∫-∞+∞dr1∫-∞+∞Mγr1,r2,x1,x2dr2.Statement (ii) of Lemma 15 provides(166)Q1+Q3+Q5+Q7≤Cx2βfx2,0,Ωβ∫0+∞dr1×∫0r1er2-1βMγr1,r2,x1,x2dr2+x1βfx2,0,Ωβ∫0+∞dr2×∫0r2er1-1βMγr1,r2,x1,x2dr1≤Cx1β+x2βfx1,0,Ωβ+fx2,0,Ωβ.As for terms Q2, Q4, Q6, and Q8, we apply statement (iii) of Lemma 15 and deduce (167)Q2+Q4+Q6+Q8≤Cx1β+x2βfx1,0,Ωβ+fx2,0,Ωβ.
Finally, statement (i) of Lemma16 guarantees the estimate to Q9(168)Q9≤CrβxsupΩ¯⁡r-βxf. Hence, representation (164) and inequalities for Qj entail(169)supΩ¯⁡r-α-β+α1+α2xDx1α1Dx2α2≤CfEββΩ¯.In order to complete the proof, we need a similar estimate to the term x1-α+1r-βxDx1u. We exchange Mγ in (154) by the function M1 from (iv) in Lemma 15. After that, recasting the arguments above together with statement (iv) in Lemma 15 derives(170)supΩ¯⁡x1-α+1r-βxDx1u≤CfEββΩ¯.This finishes the proof of the lemma.The following result is a simple consequence of Lemmas14 and 16, interpolation inequalities, and Proposition 8.Proposition 17.
Let assumptions of Lemma14 hold. Then (171)uEsβΩ¯+uE-s1βΩ¯≤CR0fEββΩ¯,uCsβΩ¯+uC-s1βΩ¯≤CR0fCββΩ¯.


## 5.1. Estimate on Maximum ofux1,x2

Before evaluating the functionsu and Dxiαiu, i=1,2, 0<αi≤1, we describe the suitable properties of the kernel Lγ.Lemma 13.
Letε∈(0,1) and let r1, r2∈R. Then there are estimates. (i)
(121)Lγr1,r2,x1,x2≤Cexp⁡-εr1-εr2×1+r2-1ifγ∈1α,1;1+r1-1ifγ∈0,1-1α;1+r2-11+r1-1ifγ∈1-1α,1α.(ii)
Ifγ=0 or γ=1, the following representations hold:(122)L0r1,r2,x1,x2=exp⁡-εr1-εr2×δr1+δ-r1L01r1,r2,x1,x2+L02r1,r2,x1,x2,L1r1,r2,x1,x2=exp⁡-εr1-εr2×δr2+δ-r2L11r1,r2,x1,x2+L12r1,r2,x1,x2,whereδ(r) is a Dirac delta function and (123)L01r1,r2,x1,x2+L11r1,r2,x1,x2≤C,L02r1,r2,x1,x2≤C1+r1-11+r2,L12r1,r2,x1,x2≤C1+r2-11+r1.(iii)
(124)∫0+∞dr1∫0+∞r2βLγr1,r2,x1,x2dr2≤Cifγ∈1α,1;∫0+∞dr1r1β∫0+∞Lγr1,r2,x1,x2dr2≤Cifγ∈0,1-1α;∫0+∞dr1∫0+∞r2βe-εr1-εr2L12r1,r2,x1,x2dr2+∫0+∞dr1r1β∫0+∞e-εr1-εr2L12r1,r2,x1,x2dr2≤Cifγ=1;∫0+∞dr1∫0+∞r2βe-εr1-εr2L02r1,r2,x1,x2dr2+∫0+∞dr1r1β∫0+∞e-εr1-εr2L02r1,r2,x1,x2dr2≤Cifγ=0.Here the constant C is independent on x1, x2 and γ.Proof.
First of all, we verify statement (i). It is easy to see that inequality (51) guarantees for γ∈[0,1](125)K-γ-izαeπz/α≤Ce-3πz/α,z→+∞,zK1-izα=zK-izα≤C,z≤2M,∂∂zzK1-izα=∂∂zzK-izα≤C,z≤2M. Then the results of Proposition 9 with aid of (125) provide statement (i).
As for statement (ii), we confirm ourselves the case ofγ=0, due to the arguments in the case δ=1 are similar.
First, we representLγ, γ∈(0,1), as (126)Lγr1,r2,x1,x2=e-εr1-εr2∫-∞+∞dzK-γ-izα×eπz/αeizlnx2/x1eizr1-r2∫-∞+∞e-ip1r1×Γ1-εsgnr1-ip1-zΓ1+αγ-εsgnr1-ip1dp1∫-∞+∞eip2r2×Γ1+εsgnr2+ip2-zΓ1+α-αγ+εsgnr2+ip2dp2.Here we use the same reasons as in the proof of Proposition 9.
Then, as mentioned in the proof of Proposition7 (see (ii) there) the function K(-γ-iz/α) has the simple poles if γ=0,1. Thus, following Cauchy’s residue theorem, we rewrite Lγ as(127)Lγr1,r2,x1,x2=e-εr1-εr2L¯01r1,r2+L¯02r1,r2,x1,x2,γ=0,where we set (128)L¯01r1,r2=C∫-∞+∞e-ip1r1∫-∞+∞eip2r2×Γ1+εsgnr2+ip2Γ1+α+εsgnr2+ip2dp2,L¯02r1,r2,x1,x2=∫-∞+∞K-izα×eπz/αeizlnx2r1/r2x1eizr1-r2×∫-∞+∞e-ip1r1e-izlnr1×Γ1-εsgnr1-ip1-zΓ1-εsgnr1-ip1dp1×∫-∞+∞eip2r2eizlnr2×Γ1+εsgnr2+ip2-zΓ1+α+εsgnr2+ip2dp2dz.Concerning the estimate of L¯01(r1,r2), we apply Lemma 5.3 from [36] and deduce(129)
L
¯
0
1
r
1
,
r
2
=
C
δ
-
r
1
+
δ
r
1
×
∫
-
∞
+
∞
e
i
p
2
r
2
Γ
1
+
ε
s
g
n
r
2
+
i
p
2
-
z
Γ
1
+
α
+
ε
s
g
n
r
2
+
i
p
2
d
p
2
.Then the asymptotic representation of Gamma-function (see, e.g., (1.5.15) in [1]),(130)Γp+aΓp+b=pa-b1+Op-1,arg⁡p+a<π,p→+∞,provides the estimate(131)∫-∞+∞eip2r2Γ1+εsgnr2+ip2-zΓ1+α+εsgnr2+ip2dp2≤C.Hence, we are left to evaluate the function L¯02. To this end, we rewrite the term L02 in the form(132)L¯02=∑i=13li,where we put (133)Bz,r1,r2=zK-izαeizr2-r1∫-∞+∞e-ip1r1×Γ1-εsgnr1-ip1-zΓ1-εsgnr1-ip1eizlnr1dp1∫-∞+∞eip2r2×Γ1+εsgnr2+ip2-zΓ1+α+εsgnr2+ip2e-izlnr2dp2,l1=∫z>2MK-izαeizlnx2/x1eizr2-r1∫-∞+∞e-ip1r1×Γ1-εsgnr1-ip1-zΓ1-εsgnr1-ip1dp1∫-∞+∞eip2r2×Γ1+εsgnr2+ip2-zΓ1+α+εsgnr2+ip2dp2,l2=B0,r1,r2∫-2M2Meizlnx2r1/x1r2-1zdz,l3=∫-2M2Meizlnx2r1/x1r2Bz,r1,r2-B0,r1,r2zdz.Treating the first term in (132) via Proposition 9 and inequalities (125), we arrive at(134)l1≤C1+r1-1.Concerning l2, we get(135)B0,r1,r2=CL¯01r1,r2,and therefore(136)l2=CL¯01∫01dqq∫-2M2M∂∂zeizqlnx2r1/x1r2dz.After that, statement (ii) of Proposition 12 (with r=r1,ρ=r2) provides(137)∫01dqq∫-2M2M∂∂zeizqlnx2r1/x1r2dz≤C.In summary, we can conclude that(138)l2+L¯01=δ-r1+δr1×C1+∫01dqq∫-2M2M∂∂zeizqlnx2r1/x1r2dz×∫-∞+∞eip2r2Γ1+εsgnr2+ip2Γ1+εsgnr2+α+ip2dp2≡δ-r1+δr1L01r1,r2,x1,x2,L01r1,r2,x1,x2<C.Finally, coming to l3, we first rewrite it as(139)l3=∫01dq∫-2M2Meizlnx2r1/x1r2∂∂qzBqz,r1,r2dz.Then, the tedious calculations with Propositions 9 and 10 and estimates (125) entail(140)l3≤C1+1+r2r1.Hence, representations (127) and (132) together with estimates (134), (138), and (140) provide statement (ii) for the function L0 where(141)L02r1,r2,x1,x2=l1+l3.At last, the proof of statement (iii) is simple consequences of statements (i) and (ii) from this proposition. Thus, the claim is proven.We are ready now to state estimates of the functionu(x1,x2).Lemma 14.
Let assumptions (h1), (31), and (32) hold. Then the function u(x1,x2) represented with (119) satisfies inequalities (142)supΩ¯⁡r-α-βx1+r2xs1+α/2u+supΩ¯⁡x1-α+x2-αr-βxu≤CR0fEββΩ¯,supΩ¯⁡d-α-βx1+dxs1+αu≤CR0fCββΩ¯.Proof.
At the beginning, we verify the first inequality in this lemma. To this end, we evaluate the termr-α-βxu. Putting γ∈(1/α,1) in (119), we conclude(143)r-α-βxu≤r-βx∫-∞+∞dr1×∫-∞+∞dr2fx1e-r1,x2e-r2-fx1e-r1,x2×Lγr1,r2,x1,x2+r-βx×∫-∞+∞dr1fx1e-r1,x2×∫-∞+∞dr2Lγr1,r2,x1,x2≡r-βxJ1+J2;here we use the simple inequality(144)x1αγx2α1-γ<rαγxrα1-γx=rαx.ConcerningJ1, we apply statement (i) from Lemma 13 with ε>β to deduce(145)J1≤Cfx2,0,Ωβ∫0+∞dr1e-εr1×∫0+∞x2βe-r2-1βe-εr21+r2-1dr2≤Cx2βfx2,0,Ωβ≤Crβxfx2,0,Ωβwith the constant is independent ofx1, x2, and γ.Coming to the termJ2, applying Proposition 11 to the function(146)∫-∞+∞dr2∫-∞+∞e-izr2eip2r2Γ1+ip2-zΓ1+α-αγ+ip2dp2,and keeping in mind Proposition 9 with inequality (125), we get(147)∫-∞+∞Lγr1,r2,x1,x2dr2≤Ce-εr1∫-∞+∞K-γ-izαeπz/α1+z×coshπz/2Γ1+α-αγ+izdz≤Ce-εr1.
In conclusion, we have(148)r-βxJ2≤r-βxfEββΩ¯∫0+∞x22e2r1+x12β/2e-εr1dr1≤CfEββΩ¯,if ε>β.
Then, collecting this inequality with (143) and (145) yields(149)supΩ¯⁡r-α-βxu≤CfEββΩ¯.
Further, we estimater-α-βx1+r2xs1+α/2u. Assumption (31) provides that(150)fx1,x2=F0x1,x21+x12+x22s1+α/2where the function F0 satisfies conditions (31) and (32). Thus, substituting this representation of f in (143) and recasting arguments above, we deduce(151)supΩ¯⁡r-α-βx1+r2xs1+α/2u≤CR0fEββΩ¯.It is easy to see that the second estimate of Lemma 14 is proved with the similar arguments.
Finally, we are left to estimatexi-αr-βxu, i=1,2. For simplicity consideration, we evaluate x1-αr-βxu. Another case is studied with the same arguments. Using statement (ii) of Lemma 13 with γ=1, we rewrite u(x1,x2) as (152)
x
1
-
α
u
x
1
,
x
2
=
C
∫
-
∞
+
∞
d
r
1
∫
-
∞
+
∞
d
r
2
f
x
1
e
-
r
1
,
x
2
e
-
r
2
×
e
-
ε
r
1
+
r
2
L
1
1
r
1
,
r
2
,
x
1
,
x
2
δ
r
2
+
δ
-
r
2
+
C
∫
-
∞
+
∞
d
r
1
×
∫
-
∞
+
∞
d
r
2
f
x
1
e
-
r
1
,
x
2
e
-
r
2
-
f
x
1
e
-
r
1
,
x
2
×
L
1
2
r
1
,
r
2
,
x
1
,
x
2
e
-
ε
r
1
+
r
2
+
C
∫
-
∞
+
∞
d
r
1
f
x
1
e
-
r
1
,
x
2
∫
-
∞
+
∞
L
1
2
r
1
,
r
2
,
x
1
,
x
2
×
e
-
ε
r
1
+
r
2
d
r
2
.After that, statements (ii) and (iii) of Lemma 13 and Proposition 10 yield inequality (153)supΩ¯⁡x1-αr-βxu≤CfEββΩ¯.This completes the proof of Lemma 14.


## 5.2. Estimate onDx1α1Dx2α2ux1,x2, 0<α1+α2≤1

First, based on representation (57) and formula (2.1.17) [1], we obtain(154)
D
x
1
α
1
D
x
2
α
2
u
=
D
x
2
α
2
D
x
1
α
1
u
=
e
i
π
γ
8
π
α
×
x
1
α
γ
-
α
1
x
2
α
1
-
γ
-
α
2
∫
-
∞
+
∞
d
r
1
×
∫
-
∞
+
∞
d
r
2
f
x
1
e
-
r
1
,
x
2
e
-
r
2
-
f
x
1
,
x
2
×
M
γ
r
1
,
r
2
,
x
1
,
x
2
+
e
i
π
γ
f
x
1
,
x
2
8
π
α
×
x
1
α
γ
-
α
1
x
2
α
1
-
γ
-
α
2
∫
-
∞
+
∞
d
r
1
×
∫
-
∞
+
∞
d
r
2
M
γ
r
1
,
r
2
,
x
1
,
x
2
,where we set (155)Mγr1,r2,x1,x2=∫-∞+∞dzK-γ+izαeπzαeizr2-r1e-izlnx1x2×∫-∞+∞e-ip1r1Γ1-ip1-zΓ1+αγ-α1-ip1dp1×∫-∞+∞eip2r2Γ1+ip2-zΓ1+α1-γ-α2+ip2dp2,γ∈α1α,1-α2α.After that, recasting the arguments of Lemma14 with aid of Propositions 9–11, we can state the following results.Lemma 15.
Letε∈(β,1) and γ∈[α1/α,1-α2/α], and let the nonnegative numbers α1 and α2 satisfy 0<α1+α2≤1. Then the kernel Mγ(r1,r2,x1,x2) possesses the following properties: (i)
(156)∫-∞+∞dr1∫-∞+∞dr2Mγr1,r2,x1,x2≤C;(ii)
(157)∫0+∞dr1∫0r1e±r2-1βMγr1,r2,x1,x2dr2+∫0+∞dr2∫0r2e±r1-1βMδr1,r2,x1,x2dr1≤C;(iii)
(158)∫0+∞dr1e±r1-1β∫-r1r1Mγr1,r2,x1,x2dr2+∫0+∞dr2e±r2-1β∫-r2r2Mγr1,r2,x1,x2dr1≤C;(iv)
If eitherγ=1 and α1=1, α2=0 or γ=0 and α2=1, α1=0, then (159)M1r1,r2,x1,x2=δr2+δ-r2M11r1+M12r1,r2,x1,x2,M0r1,r2,x1,x2=δr1+δ-r1M01r2+M02r1,r2,x1,x2,where the functionsM12 and M02 satisfy statements (i)-(iii) of this lemma and(160)M11r1≤Ce-εr1r1,M01r1≤Ce-εr2r2,∫-∞+∞M11r1dr1+∫-∞+∞M01r2dr2≤C.Now we are ready to describe properties of low derivatives to the functionu.Lemma 16.
Let assumptions of Lemma14 hold, and 0<α1+α2≤1. Then there are the estimates.(i)
(161)supΩ¯⁡r-α-β+α1+α2x1+r2xs1+α/2Dx1α1Dx2α2u+supΩ¯⁡r-α-β+α1+α2x1+r2xs1+α/2Dx2α2Dx1α1u+supΩ¯⁡x1-α+1r-βxDx1u+supΩ¯⁡x2-α+1r-βxDx2u≤CR0fEββΩ¯;(ii)
(162)supΩ¯⁡d-α-β+α1+α2x1+dxs1+αDx1α1Dx2α2u+supΩ¯⁡d-α-β+α1+α2x1+dxs1+αDx2α2Dx1α1u≤CR0fCββΩ¯.Proof.
Taking into account (154), we will carry out the detailed proof of the inequality(163)supΩ¯⁡r-α-β+α1+α2x1+r2xs1+α/2Dx1α1Dx2α2u+supΩ¯⁡x1-α+1Dx1u≤CR0fEββΩ¯. The remaining terms in (i) and (ii) are estimated in the same way and with the recasting of the corresponding arguments of Lemma 14.
We begin to evaluate the first term in (163). To this end, we represent Dx1α1Dx2α2u as(164)Dx1α1Dx2α2u=eiπγ8παx1αγ-α1x2α1-γ-α2∑j=19Qj,where we put (165)Q1=∫0+∞dr1∫-r1r1fx1er1,x2er2-fx1er1,x2×Mγr1,r2,x1,x2dr2,Q2=∫0+∞dr1fx1er1,x2-fx1,x2×∫-r1r1Mγr1,r2,x1,x2dr2,Q3=∫-∞0dr1∫r1-r1fx1er1,x2er2-fx1er1,x2×Mγr1,r2,x1,x2dr2,Q4=∫-∞0dr1fx1er1,x2-fx1,x2×∫r1-r1Mγr1,r2,x1,x2dr2,Q5=∫0+∞dr2∫-r2r2fx1er1,x2er2-fx1,x2er2×Mγr1,r2,x1,x2dr1,Q6=∫0+∞dr2fx1,x2er2-fx1,x2×∫-r2r2Mγr1,r2,x1,x2dr1,Q7=∫-∞0dr2∫r2-r2fx1er1,x2er2-fx1,x2er2×Mγr1,r2,x1,x2dr1,Q8=∫-∞0dr2fx1,x2er2-fx1,x2×∫r2-r2Mγr1,r2,x1,x2dr1,Q9=fx1,x2∫-∞+∞dr1∫-∞+∞Mγr1,r2,x1,x2dr2.Statement (ii) of Lemma 15 provides(166)Q1+Q3+Q5+Q7≤Cx2βfx2,0,Ωβ∫0+∞dr1×∫0r1er2-1βMγr1,r2,x1,x2dr2+x1βfx2,0,Ωβ∫0+∞dr2×∫0r2er1-1βMγr1,r2,x1,x2dr1≤Cx1β+x2βfx1,0,Ωβ+fx2,0,Ωβ.As for terms Q2, Q4, Q6, and Q8, we apply statement (iii) of Lemma 15 and deduce (167)Q2+Q4+Q6+Q8≤Cx1β+x2βfx1,0,Ωβ+fx2,0,Ωβ.
Finally, statement (i) of Lemma16 guarantees the estimate to Q9(168)Q9≤CrβxsupΩ¯⁡r-βxf. Hence, representation (164) and inequalities for Qj entail(169)supΩ¯⁡r-α-β+α1+α2xDx1α1Dx2α2≤CfEββΩ¯.In order to complete the proof, we need a similar estimate to the term x1-α+1r-βxDx1u. We exchange Mγ in (154) by the function M1 from (iv) in Lemma 15. After that, recasting the arguments above together with statement (iv) in Lemma 15 derives(170)supΩ¯⁡x1-α+1r-βxDx1u≤CfEββΩ¯.This finishes the proof of the lemma.The following result is a simple consequence of Lemmas14 and 16, interpolation inequalities, and Proposition 8.Proposition 17.
Let assumptions of Lemma14 hold. Then (171)uEsβΩ¯+uE-s1βΩ¯≤CR0fEββΩ¯,uCsβΩ¯+uC-s1βΩ¯≤CR0fCββΩ¯.


## 6. Estimates onDx1α1Dx2α2u(x1,x2),1<α1+α2≤α

We begin our consideration with the estimates of the functionDx1αu. Here, we use representation (119) to the function u(x1,x2) with γ=1; i.e., Lγ=L1.Recasting the arguments leading to (127) arrives at (172)ux1,x2=∫-∞+∞∫-∞+∞fey1,ey2×x1αL¯11y1-ln⁡x1,y2-ln⁡x2e-εy1-ln⁡x1dy1dy2+∫-∞+∞∫-∞+∞fey1,ey2×x1αL¯12y1-ln⁡x1,y2-ln⁡x2,x1,x2×e-εy1-lnx1-εy2-ln⁡x2dy1dy2where we put(173)
L
¯
1
1
y
1
-
ln
⁡
x
1
,
y
2
-
ln
⁡
x
2
=
-
1
4
α
×
r
e
s
z
=
0
K
-
1
+
i
z
α
∫
-
∞
+
∞
Γ
1
+
i
p
2
-
z
Γ
1
+
i
p
2
e
-
i
p
2
y
2
-
ln
⁡
x
2
d
p
2
×
∫
-
∞
+
∞
Γ
1
-
i
p
1
-
z
Γ
1
+
α
-
i
p
1
e
i
p
1
y
1
-
ln
⁡
x
1
d
p
1
e
π
z
/
α
e
i
z
y
2
-
y
1
,
L
¯
1
2
y
1
-
ln
⁡
x
1
,
y
2
-
ln
⁡
x
2
,
x
1
,
x
2
=
-
1
8
π
i
×
∫
-
∞
+
∞
d
z
e
π
z
/
α
K
-
1
+
i
z
α
e
i
z
y
2
-
y
1
×
∫
-
∞
+
∞
Γ
1
+
ε
s
g
n
y
1
-
ln
⁡
x
1
-
i
p
1
-
z
Γ
1
+
ε
s
g
n
y
1
-
ln
⁡
x
1
-
i
p
1
e
i
p
1
y
1
-
ln
⁡
x
1
d
p
1
×
∫
-
∞
+
∞
Γ
1
-
ε
s
g
n
y
2
-
ln
⁡
x
2
+
i
p
2
-
z
Γ
1
-
ε
s
g
n
y
2
-
ln
⁡
x
2
+
i
p
2
e
-
i
p
2
y
2
-
ln
⁡
x
2
d
p
2
.After that, straightforward calculations arrive at(174)L¯11y1-ln⁡x1,y2-ln⁡x2=Cδy2-ln⁡x2+δ-y2+ln⁡x2×∫-∞+∞Γ1-ip1eip1y1-ln⁡x1Γ1+α-ip1dp1.Thus, we can rewriteu(x1,x2) in the form(175)ux1,x2=C∫-∞+∞dy1fey1,x2×∫-∞+∞Γ1-ip1Γ1+α-ip1x1αeip1y1-ln⁡x1dp1+C∫-∞+∞fey1,ey2×x1αL¯12y1-ln⁡x1,y2-ln⁡x2,x1,x2×e-εy1-ln⁡x1-εy2-ln⁡x2dy1dy2.We are now in the position to calculateDx1α1u(x1,x2). Indeed, formula (2.1.17) [1] and direct calculations provide the representation(176)Dx1αux1,x2=∑j=13Ujx1,x2 where we put (177)U1=∫-∞+∞fey1,x2R1x1,y1-ln⁡x1dy1,U2=fx1,x2∫-∞+∞∫-∞+∞R2x1,x2,y1-ln⁡x1,y2-ln⁡x2dy1dy2,U3=∫-∞+∞∫-∞+∞fey1,ey2-fx1,x2×R2x1,x2,y1-ln⁡x1,y2-ln⁡x2dy1dy2,with (178)R1x1,y1-ln⁡x1=C∫-∞+∞eip1y1-ln⁡x1dp1,R2x1,x2,y1-ln⁡x1,y2-ln⁡x2=e-εy2-ln⁡x2Dx1α1x1αe-εy1-ln⁡x1L¯12y1-ln⁡x1,y2-ln⁡x2,x1,x2.After that, introducing new variables(179)r1=ln⁡x1-y1,r2=ln⁡x2-y2,and new functions (180)B1z,r1,r2=eπz/αzK-1+izα×eizr2-r1eizlnr2/r1∫-∞+∞e-ip1r1×Γ1-εsgnr1-ip1-zΓ1-εsgnr1-ip1dp1∫-∞+∞eip2r2×Γ1+εsgnr2+ip2-zΓ1+εsgnr2+ip2dp2,R3x1,x2,r1,r2=R2x1,x2,-r1,-r2-e-εr1-εr2B10,r1,r2∫z<2Mdzz×∫01∂∂qe-iqzlnx1r2/x2r1dq,we rewrite the functions R1, R2, and Uj in the more comfortable forms(181)R1x1,-r1=Cδr1+δ-r1,R2x1,x2,-r1,-r2=e-εr1-εr2∫-∞+∞B1z,r1,r2e-izlnx1r2/x2r1dzz,U1x1,x2=Cfx1,x2,U2x1,x2=fx1,x2×∫-∞+∞∫-∞+∞R2x1,x2,-r1,-r2dr1dr2,U3x1,x2=∫-∞+∞∫-∞+∞fx1e-r1,x2e-r2-fx1,x2×R3x1,x2,r1,r2dr1dr2.In order to reach the last equality in (181), we recast the arguments of Lemma 13 which lead to representations (132), (134), and (138). Indeed,(182)B10,r1,r2=Cδ-r1+δr1δ-r2+δr2,and(183)∫z<2Mdzz∫01∂∂qe-iqzlnx1r2/x2r1dq=∫01dqq∫-2M2M∂∂ze-iqzlnx1r2/x2r1dz≤C,where the positive constant C is independent of xi and ri.Hence, we obtain(184)∫-∞+∞∫-∞+∞fx1e-r1,x2e-r2-fx1,x2×B10,r1,r2∫z<2Mdzz×∫01∂∂qe-iqzlnx1r2/x2r1dqdr1dr2=0.


### 6.1. Estimates on MaximumDx1αux1,x2

Further, in virtue of representations (176) and (181), estimates of Dx1αux1,x2 and Dx1αuxi,0,Ωβ follow from the corresponding estimates of the functions Uj(x1,x2).First, we describe the properties of the kernelsR2 and R3.Lemma 18.
Letε∈(β,1), x1, x2∈[0,+∞). Then there are the following estimates. (i)
(185)∫-∞+∞∫-∞+∞R2x1,x2,-r1,-r2dr1dr2≤C,(ii)
(186)∫0+∞dr1∫0r1R3x1,x2,r1,r2e±r2-1βdr2+∫0+∞e±r1-1β∫-r1r1R3x1,x2,r1,r2dr2dr1≤C,(iii)
(187)∫0+∞dr2∫0r2R3x1,x2,r1,r2e±r1-1βdr1+∫0+∞e±r2-1β∫-r2r2R3x1,x2,r1,r2dr1dr2≤C,where the positive constant C is independent of ε, x1, and x2.Proof.
First, applying Proposition11 with b=0 and ρ=+∞ and using definition (85), we have(188)
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,where(189)B0=eπz/αzK-1+izα×e-izlnx1/x2eiqzlnr2/r1I+2ε,zI-2ε,zz=0.Then, applying statement (iii) in Propositions 11 and 12 with aid (125) arrives at the first inequality of this proposition.
Further, we proceed with a detailed proof of statement (ii). The proof of statement (iii) is almost identical and is left to the interested reader.
At the beginning, we evaluate the first term in the left-hand side of the inequality in (ii)(190)J1≔∫0+∞dr1∫0r1e±r2-1βR3x1,x2,r1,r2dr2.Simple calculations lead to(191)R3x1,x2,r1,r2eεr1+εr2=∑j=13R3j, where we set (192)R31=∫z>2MB1z,r1,r2e-izlnx1/x2dzz,R32=∫01dq×∫z<2Me-izlnx1r2/x2r1∂∂qzeizqlnr2/r1×B1qz,r1,r2dz.
At this point, we estimate each termR3j separately.
By Proposition9 and estimates (125),(193)R31≤C1+r1-11+r2-1×∫2M+∞1+z2e-3πz/αcosh⁡πz2dz<C1+r1-11+r2-1where the positive constant C is independent of ri and xi.
ConcerningR32, keeping in mind Propositions 9-10 and estimates (125), we deduce(194)R32≤C1+r1-11+r2-1∫01dq∫z<2Mdz<C1+r1-11+r2-1where the positive constant C is independent of ri and xi.
In summary, we can conclude(195)J1≤C∫0+∞dr11+r1-1e-εr1×∫0r1e±r2-1βe-εr21+r2-1dr2<C.In order to obtain the last inequality, we use the condition ε>β.
Then, we are left to evaluate the second term in inequality (ii)(196)J2≔∫0+∞dr1e±r1-1β∫-r1r1R3x1,x2,r1,r2dr2.
To this end, coming to representation (191), we deduce(197)J2≤J21+J22,with(198)J21=∫0+∞dr1e±r1-1β∫-r1r1R31e-εr1-εr2dr2,J22=∫0+∞dr1e±r1-1β∫-r1r1R32e-εr1-εr2dr2.Proposition 9, statement (ii) in Proposition 11 (with ε1=ε2=1, ρ=r1), and estimate (125) lead to(199)∫-r1r1R31e-εr1-εr2dr2≤C1+r1+r1-1×∫z>2MK-1+izαeπz/α1+z22×cosh2⁡πz2dz<C1+r1+r1-1.Thus, we have(200)J21≤C∫0+∞e-εr1e±r1-1β1+r1+r1-1<C.
Finally, Propositions9 and 10 provide inequality(201)∫-r1r1R32e-εr1-εr2dr2≤C1+r1+r1-1∫01dq∫02Mdz<C1+r1+r1-1,which guarantees that(202)J22≤C∫0+∞e-εr1e±r1-1β1+r1+r1-1dr1<C.Then, collecting this inequality with the estimate for J21 yields(203)J2≤Cwith the positive constant C independent of x1 and x2.
The last inequality together with the analogous estimate forJ1 completes the proof of statement (ii). This finishes the proof of Lemma 18.Now we estimate the functionDx1αu.Lemma 19.
Let assumptions of Lemma14 hold. Then(204)supΩ¯⁡r-βx1+r2xs1+α/2Dx1αu≤CR0fEββΩ¯,supΩ¯⁡d-βx1+dxs1+αDx1αu≤CR0fCββΩ¯.Proof.
Here we prove in detail the estimate(205)supΩ¯⁡r-βxDx1αu≤CfEββΩ¯.The rest inequalities are obtained with the same techniques and with the recasting the arguments in Lemma 14.
By representations (176) and (181), we have(206)Dx1αu≤C∑j=13Uj≤Cfx1,x2+U2+U3.
Further, we estimateUj separately.
By statement (i) in Lemma18,(207)U2≤CrβxsupΩ¯⁡r-βxf.Concerning U3, we rewrite this function in the form analogous to (164) (208)U3≤∑j=18U3j,where we put (209)U31=∫0+∞dr1×∫-r1r1fx1e-r1,x2e-r2-fx1e-r1,x2×R3x1,x2,r1,r2dr2,U32=∫0+∞dr1fx1e-r1,x2-fx1,x2×∫-r1r1R3x1,x2,r1,r2dr2,U33=∫-∞0dr1×∫r1-r1fx1e-r1,x2e-r2-fx1e-r1,x2×R3x1,x2,r1,r2dr2,U34=∫-∞0dr1fx1e-r1,x2-fx1,x2×∫r1-r1R3x1,x2,r1,r2dr2,U35=∫0+∞dr2×∫-r2r2fx1e-r1,x2e-r2-fx1,x2e-r2×R3x1,x2,r1,r2dr1,U36=∫0+∞dr2fx1,x2e-r2-fx1,x2×∫-r2r2R3x1,x2,r1,r2dr1,U37=∫-∞0dr2×∫r2-r2fx1e-r1,x2e-r2-fx1,x2e-r2×R3x1,x2,r1,r2dr1,U38=∫-∞0dr2fx1,x2e-r2-fx1,x2×∫r2-r2R3x1,x2,r1,r2dr1.
It is apparent that the estimates of eachU3j are simple consequence of Lemma 18. Thus, we deduce(210)U3≤Cfx1,0,Ωβ+fx2,0,Ωβx1β+x2β≤CrβxfEββΩ¯.
The last inequality with (206) and (207) provides(211)Dx1αu≤CrβxfEββΩ¯.This finishes the proof of Lemma 19.


### 6.2. Estimates on Hölder Seminorms ofDx1αu(x1,x2)

First of all, we return to representations (176) and (181) and evaluate each function Ui(x1,x2) separately.Proposition 20.
Let assumptions of Lemma14 hold. Then(212)∑i=12U1xi,0,Ωβ+U2xi,0,Ωβx1β+x2β≤CfEββΩ¯.Proof.
It is apparent that (181) and properties of the function f ensure estimates of U1xi,0,Ω(β).
Next we evaluate the termU2x1,0,Ω(β). For simplicity consideration, we assume(213)x1,x¯1∈0,+∞andx1<x¯1,and put(214)Δ1x=x1-x¯1,Δ1U2=U2x1,x2-U2x¯1,x2.In virtue of representation (181) and Lemma 18, we have (215)Δ1U2≤fx1,x2-fx¯1,x2∫-∞+∞dr1×∫-∞+∞dr2R2x1,x2,-r1,-r2+fx¯1,x2×∫-∞+∞dr1∫-∞+∞dr2R2x1,x2,-r1,-r2-R2x¯1,x2,-r1,-r2≤Cfx1,0,ΩβΔx1β+fx¯1,x2∫-∞+∞dr1×∫-∞+∞dr2R2x1,x2,-r1,-r2-R2x¯1,x2,-r1,-r2.Then we are left to tackle the second term in the right-hand side of the last inequality. Simple calculations lead to (216)R2x1,x2,-r1,-r2-R2x¯1,x2,-r1,-r2=Δ1x∫01∂∂qΔ1x+x¯1R2qΔ1x+x¯1,x2,-r1,-r2dq=Δ1x∫01dqe-εr1-εr2×∫-∞+∞e-izlnqΔ1x+x¯1r2/x2r1B1z,r1,r2qΔ1x+x¯1dz.At this point, recasting the proof of statement (i) in Lemma 18 arrives at (217)∫-∞+∞dr1∫-∞+∞dr2R2x1,x2,-r1,-r2-R2x¯1,x2,-r1,-r2≤CΔ1x∫01dqqΔ1x+x¯1<CΔ1xx¯1β∫01dqqΔ1x1-β=CΔ1xβx¯1β.
In conclusion, we obtain the estimate(218)fx¯1,x2∫-∞+∞dr1∫-∞+∞dr2R2x1,x2,-r1,-r2-R2x¯1,x2,-r1,-r2≤Cfx¯1,x2-f0,x2Δ1xβx¯1β.In order to reach the last inequality we use property (32) to the function f.
Summarizing, we have inequality(219)Δ1U2≤Cfx1,0,ΩβΔ1xββ.The same arguments in the case of the difference Δ2U2=U2(x1,x2)-U2(x1,x¯2) guarantee estimate(220)Δ2U2≤Cfx2,0,Ωβx2-x¯2β.
Thus, collecting estimates forΔiU2 we reach the conclusion(221)∑i=12U2xi,0,Ωβ≤CfEββΩ.This completes the proof of Proposition 20.Next we obtain the same results to the functionU3(x1). For each point x1, x¯1, x2, x¯2∈[0,+∞): x1>x¯1 and x2>x¯2, we put(222)Δ1x=x1-x¯1,Δ2x=x2-x¯2.Further, we describe the properties of the kernel R3(x1,x2,r1,r2).Lemma 21.
Letε∈(β,1) and εj, j=1,2,3,4, be some positive numbers, εj∈[0,2], and let(223)Δixx¯j<13,i,j=1,2.Then the kernel R3 possesses the following:(i)
(224)∫0ε3dr1∫-ε1r1ε2r1e±r2-1βR3x1,x2,r1,r2dr2+∫0ε3dr2∫-ε1r2ε2r2e±r1-1βR3x1,x2,r1,r2dr1≤Cε2β+ε1βε3β,(ii)
(225)∫0ε3dr1e±r1-1β∫-ε1r1ε2r1R3x1,x2,r1,r2dr2+∫0ε3dr2e±r2-1β∫-ε1r2ε2r2R3x1,x2,r1,r2dr1≤Cε2ε1+ε1ε2+ε1+ε2ε3ε3β,(iii)
(226)∬R2R3x1,x2,r1,r2dr1dr2+∫-ε4ε3∫-ε1ε2R3x1,x2,r1,r2dr1dr2≤C1+1+ε2ε1+ε1ε21+ε3ε4+ε4ε3,(iv)
(227)∫ε3+∞dr1∫0ε2r1e±r2-1βR3x1,x2,r1+ln⁡1+Δ1xx1,r2-R3x¯1,x2,r1,r2dr2+∫ε3+∞dr1e±r1-1β∫-ε1r1ε2r1R3x1,x2,r1+ln⁡1+Δ1xx1,r2-R3x¯1,x2,r1,r2dr2≤C1+ε2β+ε2ε1+ε1ε2ε3β-1ln⁡1+Δ1xx1,∫ε3+∞dr2∫0ε2r2e±r1-1βR3x1,x2,r1+ln⁡1+Δ1xx1,r2-R3x¯1,x2,r1,r2dr1+∫ε3+∞dr2e±r2-1β∫-ε1r2ε2r2R3x1,x2,r1+ln⁡1+Δ1xx1,r2-R3x¯1,x2,r1,r2dr1≤C1+ε2β+ε2ε1+ε1ε2ε3β-1ln⁡1+Δ1xx1,where the positive constant C is independent of xi, x¯i, εj.Since the proof of this result is technically tedious and repeating certain steps in the proof of Lemma18, we provide one in the Appendix.Then, based on Lemma21, we can conclude the following.Lemma 22.
Let assumptions of Lemma14 hold. Then there is the estimate(228)∑i=12U3xi,0,Ωβ≤CfEββΩ.Proof.
We provide here the estimate ofU3x1,0,Ω(β). The case of U3x2,0,Ω(β) is treated in similar arguments.
It should be noted that it is enough to evaluateU3x1,0,Ω(β) in case of (223). Indeed, in the opposite case, its estimate follows from Lemma 19. Thus, we assume that (223) hold. Introducing the domain(229)D¯1x1,x2=y1,y2∈R2:e-y1-x1≤2Δ1x,e-y2-x2≤2Δ2x,,we rewrite the function U3 in a more suitable form: (230)U3=∬D¯1x1,x2fey1,ey2-fx1,x2×R3x1,x2,ln⁡x1-y1,ln⁡x2-y2dy1dy2+∬R2∖D¯1x1,x2fey1,ey2-fx1,x2×R3x1,x2,ln⁡x1-y1,ln⁡x2-y2dy1dy2.Then we consider the difference(231)Δ1U3≔U3x1,x2-U3x¯1,x2=∑j=14Vj,where we set (232)V1=∬D¯1x1,x2fey1,ey2-fx1,x2×R3x1,x2,ln⁡x1-y1,ln⁡x2-y2dy1dy2,V2=-∬D¯1x1,x2fey1,ey2-fx¯1,x2×R3x¯1,x2,ln⁡x¯1-y1,ln⁡x2-y2dy1dy2,V3=∬R2∖D¯1x1,x2fx¯1,x2-fx1,x2×R3x1,x2,ln⁡x1-y1,ln⁡x2-y2dy1dy2,V4=∬R2∖D¯1x1,x2fey1,ey2-fx¯1,x2×R3x1,x2,ln⁡x1-y1,ln⁡x2-y2-R3x¯1,x2,ln⁡x¯1-y1,ln⁡x2-y2dy1dy2.Then putting(233)a1x1=ln⁡1+2Δ1xx1,a1x¯1=ln⁡1+Δ1xx-1,a3x1=-ln⁡1-2Δ1xx1,a3x¯1=-ln⁡1-3Δ1xx-1,a2=ln⁡1+Δ1xx2,a4=-ln⁡1-2Δ1xx2,and making change of variables (179), we reduce the domain D¯1(x1,x2) to either D1(x1,x2) in the case of y1=-r1+ln⁡x1 or D1(x¯1,x2) if y1=-r1+ln⁡x¯1: (234)D1x1,x2=r1,r2∈R2:-a3x1<r1<a1x1,-a4<r2<a2,D1x¯1,x2=r1,r2∈R2:-a3x¯1<r1<a1x¯1,-a4<r2<a2.At this point, we estimate each term Vj separately.
Change of variables (179) leads to (235)V1=∫0a1x1dr1×∫-a4/a1x1r1a2/a1x1r1fx1e-r1,x2e-r2-fx1,x2×R3x1,x2,r1,r2dr2+∫-a3x10dr1×∫a4/a3x1r1-a2/a3x1r1fx1e-r1,x2e-r2-fx1,x2×R3x1,x2,r1,r2dr2+∫0a2dr2×∫-a3x1/a2r2a1x1/a2r2fx1e-r1,x2e-r2-fx1,x2×R3x1,x2,r1,r2dr1+∫-a40dr2×∫a3x1/a4r2-a1x1/a2r2fx1e-r1,x2e-r2-fx1,x2×R3x1,x2,r1,r2dr1.Applying decomposition like (164) to each term in the representation above, then statements (i) and (ii) in Lemma 21 provide estimate(236)V1≤CΔ1xβfx1,0,Ωβ+fx2,0,Ωβ.
Indeed, let us verify this with the example of the first term in the representation toV1. Standard calculations lead to (237)∫0a1x1dr1∫-a4/a1x1r1a2/a1x1r1fx1e-r1,x2e-r2-fx1,x2R3x1,x2,r1,r2dr2=∫0a1x1dr1∫-a4/a1x1r1a2/a1x1r1fx1e-r1,x2e-r2-fx1e-r1,x2R3x1,x2,r1,r2dr2+∫0a1x1dr1fx1e-r1,x2-fx1,x2×∫-a4/a1x1r1a2/a1x1r1R3x1,x2,r1,r2dr2Then, properties of the function f and Lemma 21 with ε3=a1(x1), ε2=a2/a1(x1), ε1=a4/a1(x1) yield (238)∫0a1x1dr1×∫-a4/a1x1r1a2/a1x1r1fx1e-r1,x2e-r2-fx1,x2×R3x1,x2,r1,r2dr2≤Cfx1,0,Ωβ+fx2,0,Ωβx2βa2β+x2βa4β+x1βa1βx1a2a4+a4a2≤CΔ1xβfx1,0,Ωβ+fx2,0,Ωβ.To reach the last inequality we apply (223) to (233). The remaining terms in the representation of V1 are evaluated in the same way.
ConcerningV2, we repeat the arguments above with changing D(x1,x2) by D(x¯1,x2) and obtain the estimate like (236) to the function V2.
Coming toV3, change of variables (179), Lemma 21, and condition (223) arrive at(239)V3=fx¯1,x2-fx1,x2×∬R2R3x1,x2,r1,r2dr1dr2+∬D1x1,x2R3x1,x2,r1,r2dr1dr2≤CΔ1xβfx1,0,Ωβ.Finally, we are left to estimate V4. Change of variables (179) and statement (iv) in Lemma 21 lead to(240)V4≤CΔ1xβfx1,0,Ωβ+fx2,0,Ωβ.Collecting this inequality with (236) and (239), we get the bound(241)Δ1U3≤CΔ1xβfEββΩ¯.This finishes the proof of Lemma 22.At this point, we come back toDx1αux1,0,Ω(β). Keeping in mind Proposition 20, Lemma 22, and representations (176) and (181), we deduce the inequality(242)Dx1αux1,0,Ωβ+Dx1αux2,0,Ωβ≤CfEββΩ¯.Further, recasting arguments in Proposition20 and Lemmas 22 and 14 and applying Proposition 8, we assert the result.Lemma 23.
Let assumption of Lemma14 hold. Then (243)Dx1αuEββΩ¯+Dx1αuE-s1-αβΩ¯≤CR0fEββΩ¯,Dx1αuCββΩ¯+Dx1αuC-s1-αβΩ¯≤CR0fCββΩ¯.In order to obtain the same results to the functionsDx1α1Dx2α2u and Dx2α2Dx1α1u, 1<α1+α2≤α, it is enough to repeat the arguments from this section. Thus, we have.Remark 24.
Under conditions of Lemma23, the functions Dx1α1Dx2α2u, Dx2α2Dx1α1u, 1<α1+α2≤α, satisfy inequalities (23) and (27) and(244)Dx1α1Dx2α2u=Dx2α2Dx1α1u.


## 6.1. Estimates on MaximumDx1αux1,x2

Further, in virtue of representations (176) and (181), estimates of Dx1αux1,x2 and Dx1αuxi,0,Ωβ follow from the corresponding estimates of the functions Uj(x1,x2).First, we describe the properties of the kernelsR2 and R3.Lemma 18.
Letε∈(β,1), x1, x2∈[0,+∞). Then there are the following estimates. (i)
(185)∫-∞+∞∫-∞+∞R2x1,x2,-r1,-r2dr1dr2≤C,(ii)
(186)∫0+∞dr1∫0r1R3x1,x2,r1,r2e±r2-1βdr2+∫0+∞e±r1-1β∫-r1r1R3x1,x2,r1,r2dr2dr1≤C,(iii)
(187)∫0+∞dr2∫0r2R3x1,x2,r1,r2e±r1-1βdr1+∫0+∞e±r2-1β∫-r2r2R3x1,x2,r1,r2dr1dr2≤C,where the positive constant C is independent of ε, x1, and x2.Proof.
First, applying Proposition11 with b=0 and ρ=+∞ and using definition (85), we have(188)
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,where(189)B0=eπz/αzK-1+izα×e-izlnx1/x2eiqzlnr2/r1I+2ε,zI-2ε,zz=0.Then, applying statement (iii) in Propositions 11 and 12 with aid (125) arrives at the first inequality of this proposition.
Further, we proceed with a detailed proof of statement (ii). The proof of statement (iii) is almost identical and is left to the interested reader.
At the beginning, we evaluate the first term in the left-hand side of the inequality in (ii)(190)J1≔∫0+∞dr1∫0r1e±r2-1βR3x1,x2,r1,r2dr2.Simple calculations lead to(191)R3x1,x2,r1,r2eεr1+εr2=∑j=13R3j, where we set (192)R31=∫z>2MB1z,r1,r2e-izlnx1/x2dzz,R32=∫01dq×∫z<2Me-izlnx1r2/x2r1∂∂qzeizqlnr2/r1×B1qz,r1,r2dz.
At this point, we estimate each termR3j separately.
By Proposition9 and estimates (125),(193)R31≤C1+r1-11+r2-1×∫2M+∞1+z2e-3πz/αcosh⁡πz2dz<C1+r1-11+r2-1where the positive constant C is independent of ri and xi.
ConcerningR32, keeping in mind Propositions 9-10 and estimates (125), we deduce(194)R32≤C1+r1-11+r2-1∫01dq∫z<2Mdz<C1+r1-11+r2-1where the positive constant C is independent of ri and xi.
In summary, we can conclude(195)J1≤C∫0+∞dr11+r1-1e-εr1×∫0r1e±r2-1βe-εr21+r2-1dr2<C.In order to obtain the last inequality, we use the condition ε>β.
Then, we are left to evaluate the second term in inequality (ii)(196)J2≔∫0+∞dr1e±r1-1β∫-r1r1R3x1,x2,r1,r2dr2.
To this end, coming to representation (191), we deduce(197)J2≤J21+J22,with(198)J21=∫0+∞dr1e±r1-1β∫-r1r1R31e-εr1-εr2dr2,J22=∫0+∞dr1e±r1-1β∫-r1r1R32e-εr1-εr2dr2.Proposition 9, statement (ii) in Proposition 11 (with ε1=ε2=1, ρ=r1), and estimate (125) lead to(199)∫-r1r1R31e-εr1-εr2dr2≤C1+r1+r1-1×∫z>2MK-1+izαeπz/α1+z22×cosh2⁡πz2dz<C1+r1+r1-1.Thus, we have(200)J21≤C∫0+∞e-εr1e±r1-1β1+r1+r1-1<C.
Finally, Propositions9 and 10 provide inequality(201)∫-r1r1R32e-εr1-εr2dr2≤C1+r1+r1-1∫01dq∫02Mdz<C1+r1+r1-1,which guarantees that(202)J22≤C∫0+∞e-εr1e±r1-1β1+r1+r1-1dr1<C.Then, collecting this inequality with the estimate for J21 yields(203)J2≤Cwith the positive constant C independent of x1 and x2.
The last inequality together with the analogous estimate forJ1 completes the proof of statement (ii). This finishes the proof of Lemma 18.Now we estimate the functionDx1αu.Lemma 19.
Let assumptions of Lemma14 hold. Then(204)supΩ¯⁡r-βx1+r2xs1+α/2Dx1αu≤CR0fEββΩ¯,supΩ¯⁡d-βx1+dxs1+αDx1αu≤CR0fCββΩ¯.Proof.
Here we prove in detail the estimate(205)supΩ¯⁡r-βxDx1αu≤CfEββΩ¯.The rest inequalities are obtained with the same techniques and with the recasting the arguments in Lemma 14.
By representations (176) and (181), we have(206)Dx1αu≤C∑j=13Uj≤Cfx1,x2+U2+U3.
Further, we estimateUj separately.
By statement (i) in Lemma18,(207)U2≤CrβxsupΩ¯⁡r-βxf.Concerning U3, we rewrite this function in the form analogous to (164) (208)U3≤∑j=18U3j,where we put (209)U31=∫0+∞dr1×∫-r1r1fx1e-r1,x2e-r2-fx1e-r1,x2×R3x1,x2,r1,r2dr2,U32=∫0+∞dr1fx1e-r1,x2-fx1,x2×∫-r1r1R3x1,x2,r1,r2dr2,U33=∫-∞0dr1×∫r1-r1fx1e-r1,x2e-r2-fx1e-r1,x2×R3x1,x2,r1,r2dr2,U34=∫-∞0dr1fx1e-r1,x2-fx1,x2×∫r1-r1R3x1,x2,r1,r2dr2,U35=∫0+∞dr2×∫-r2r2fx1e-r1,x2e-r2-fx1,x2e-r2×R3x1,x2,r1,r2dr1,U36=∫0+∞dr2fx1,x2e-r2-fx1,x2×∫-r2r2R3x1,x2,r1,r2dr1,U37=∫-∞0dr2×∫r2-r2fx1e-r1,x2e-r2-fx1,x2e-r2×R3x1,x2,r1,r2dr1,U38=∫-∞0dr2fx1,x2e-r2-fx1,x2×∫r2-r2R3x1,x2,r1,r2dr1.
It is apparent that the estimates of eachU3j are simple consequence of Lemma 18. Thus, we deduce(210)U3≤Cfx1,0,Ωβ+fx2,0,Ωβx1β+x2β≤CrβxfEββΩ¯.
The last inequality with (206) and (207) provides(211)Dx1αu≤CrβxfEββΩ¯.This finishes the proof of Lemma 19.


## 6.2. Estimates on Hölder Seminorms ofDx1αu(x1,x2)

First of all, we return to representations (176) and (181) and evaluate each function Ui(x1,x2) separately.Proposition 20.
Let assumptions of Lemma14 hold. Then(212)∑i=12U1xi,0,Ωβ+U2xi,0,Ωβx1β+x2β≤CfEββΩ¯.Proof.
It is apparent that (181) and properties of the function f ensure estimates of U1xi,0,Ω(β).
Next we evaluate the termU2x1,0,Ω(β). For simplicity consideration, we assume(213)x1,x¯1∈0,+∞andx1<x¯1,and put(214)Δ1x=x1-x¯1,Δ1U2=U2x1,x2-U2x¯1,x2.In virtue of representation (181) and Lemma 18, we have (215)Δ1U2≤fx1,x2-fx¯1,x2∫-∞+∞dr1×∫-∞+∞dr2R2x1,x2,-r1,-r2+fx¯1,x2×∫-∞+∞dr1∫-∞+∞dr2R2x1,x2,-r1,-r2-R2x¯1,x2,-r1,-r2≤Cfx1,0,ΩβΔx1β+fx¯1,x2∫-∞+∞dr1×∫-∞+∞dr2R2x1,x2,-r1,-r2-R2x¯1,x2,-r1,-r2.Then we are left to tackle the second term in the right-hand side of the last inequality. Simple calculations lead to (216)R2x1,x2,-r1,-r2-R2x¯1,x2,-r1,-r2=Δ1x∫01∂∂qΔ1x+x¯1R2qΔ1x+x¯1,x2,-r1,-r2dq=Δ1x∫01dqe-εr1-εr2×∫-∞+∞e-izlnqΔ1x+x¯1r2/x2r1B1z,r1,r2qΔ1x+x¯1dz.At this point, recasting the proof of statement (i) in Lemma 18 arrives at (217)∫-∞+∞dr1∫-∞+∞dr2R2x1,x2,-r1,-r2-R2x¯1,x2,-r1,-r2≤CΔ1x∫01dqqΔ1x+x¯1<CΔ1xx¯1β∫01dqqΔ1x1-β=CΔ1xβx¯1β.
In conclusion, we obtain the estimate(218)fx¯1,x2∫-∞+∞dr1∫-∞+∞dr2R2x1,x2,-r1,-r2-R2x¯1,x2,-r1,-r2≤Cfx¯1,x2-f0,x2Δ1xβx¯1β.In order to reach the last inequality we use property (32) to the function f.
Summarizing, we have inequality(219)Δ1U2≤Cfx1,0,ΩβΔ1xββ.The same arguments in the case of the difference Δ2U2=U2(x1,x2)-U2(x1,x¯2) guarantee estimate(220)Δ2U2≤Cfx2,0,Ωβx2-x¯2β.
Thus, collecting estimates forΔiU2 we reach the conclusion(221)∑i=12U2xi,0,Ωβ≤CfEββΩ.This completes the proof of Proposition 20.Next we obtain the same results to the functionU3(x1). For each point x1, x¯1, x2, x¯2∈[0,+∞): x1>x¯1 and x2>x¯2, we put(222)Δ1x=x1-x¯1,Δ2x=x2-x¯2.Further, we describe the properties of the kernel R3(x1,x2,r1,r2).Lemma 21.
Letε∈(β,1) and εj, j=1,2,3,4, be some positive numbers, εj∈[0,2], and let(223)Δixx¯j<13,i,j=1,2.Then the kernel R3 possesses the following:(i)
(224)∫0ε3dr1∫-ε1r1ε2r1e±r2-1βR3x1,x2,r1,r2dr2+∫0ε3dr2∫-ε1r2ε2r2e±r1-1βR3x1,x2,r1,r2dr1≤Cε2β+ε1βε3β,(ii)
(225)∫0ε3dr1e±r1-1β∫-ε1r1ε2r1R3x1,x2,r1,r2dr2+∫0ε3dr2e±r2-1β∫-ε1r2ε2r2R3x1,x2,r1,r2dr1≤Cε2ε1+ε1ε2+ε1+ε2ε3ε3β,(iii)
(226)∬R2R3x1,x2,r1,r2dr1dr2+∫-ε4ε3∫-ε1ε2R3x1,x2,r1,r2dr1dr2≤C1+1+ε2ε1+ε1ε21+ε3ε4+ε4ε3,(iv)
(227)∫ε3+∞dr1∫0ε2r1e±r2-1βR3x1,x2,r1+ln⁡1+Δ1xx1,r2-R3x¯1,x2,r1,r2dr2+∫ε3+∞dr1e±r1-1β∫-ε1r1ε2r1R3x1,x2,r1+ln⁡1+Δ1xx1,r2-R3x¯1,x2,r1,r2dr2≤C1+ε2β+ε2ε1+ε1ε2ε3β-1ln⁡1+Δ1xx1,∫ε3+∞dr2∫0ε2r2e±r1-1βR3x1,x2,r1+ln⁡1+Δ1xx1,r2-R3x¯1,x2,r1,r2dr1+∫ε3+∞dr2e±r2-1β∫-ε1r2ε2r2R3x1,x2,r1+ln⁡1+Δ1xx1,r2-R3x¯1,x2,r1,r2dr1≤C1+ε2β+ε2ε1+ε1ε2ε3β-1ln⁡1+Δ1xx1,where the positive constant C is independent of xi, x¯i, εj.Since the proof of this result is technically tedious and repeating certain steps in the proof of Lemma18, we provide one in the Appendix.Then, based on Lemma21, we can conclude the following.Lemma 22.
Let assumptions of Lemma14 hold. Then there is the estimate(228)∑i=12U3xi,0,Ωβ≤CfEββΩ.Proof.
We provide here the estimate ofU3x1,0,Ω(β). The case of U3x2,0,Ω(β) is treated in similar arguments.
It should be noted that it is enough to evaluateU3x1,0,Ω(β) in case of (223). Indeed, in the opposite case, its estimate follows from Lemma 19. Thus, we assume that (223) hold. Introducing the domain(229)D¯1x1,x2=y1,y2∈R2:e-y1-x1≤2Δ1x,e-y2-x2≤2Δ2x,,we rewrite the function U3 in a more suitable form: (230)U3=∬D¯1x1,x2fey1,ey2-fx1,x2×R3x1,x2,ln⁡x1-y1,ln⁡x2-y2dy1dy2+∬R2∖D¯1x1,x2fey1,ey2-fx1,x2×R3x1,x2,ln⁡x1-y1,ln⁡x2-y2dy1dy2.Then we consider the difference(231)Δ1U3≔U3x1,x2-U3x¯1,x2=∑j=14Vj,where we set (232)V1=∬D¯1x1,x2fey1,ey2-fx1,x2×R3x1,x2,ln⁡x1-y1,ln⁡x2-y2dy1dy2,V2=-∬D¯1x1,x2fey1,ey2-fx¯1,x2×R3x¯1,x2,ln⁡x¯1-y1,ln⁡x2-y2dy1dy2,V3=∬R2∖D¯1x1,x2fx¯1,x2-fx1,x2×R3x1,x2,ln⁡x1-y1,ln⁡x2-y2dy1dy2,V4=∬R2∖D¯1x1,x2fey1,ey2-fx¯1,x2×R3x1,x2,ln⁡x1-y1,ln⁡x2-y2-R3x¯1,x2,ln⁡x¯1-y1,ln⁡x2-y2dy1dy2.Then putting(233)a1x1=ln⁡1+2Δ1xx1,a1x¯1=ln⁡1+Δ1xx-1,a3x1=-ln⁡1-2Δ1xx1,a3x¯1=-ln⁡1-3Δ1xx-1,a2=ln⁡1+Δ1xx2,a4=-ln⁡1-2Δ1xx2,and making change of variables (179), we reduce the domain D¯1(x1,x2) to either D1(x1,x2) in the case of y1=-r1+ln⁡x1 or D1(x¯1,x2) if y1=-r1+ln⁡x¯1: (234)D1x1,x2=r1,r2∈R2:-a3x1<r1<a1x1,-a4<r2<a2,D1x¯1,x2=r1,r2∈R2:-a3x¯1<r1<a1x¯1,-a4<r2<a2.At this point, we estimate each term Vj separately.
Change of variables (179) leads to (235)V1=∫0a1x1dr1×∫-a4/a1x1r1a2/a1x1r1fx1e-r1,x2e-r2-fx1,x2×R3x1,x2,r1,r2dr2+∫-a3x10dr1×∫a4/a3x1r1-a2/a3x1r1fx1e-r1,x2e-r2-fx1,x2×R3x1,x2,r1,r2dr2+∫0a2dr2×∫-a3x1/a2r2a1x1/a2r2fx1e-r1,x2e-r2-fx1,x2×R3x1,x2,r1,r2dr1+∫-a40dr2×∫a3x1/a4r2-a1x1/a2r2fx1e-r1,x2e-r2-fx1,x2×R3x1,x2,r1,r2dr1.Applying decomposition like (164) to each term in the representation above, then statements (i) and (ii) in Lemma 21 provide estimate(236)V1≤CΔ1xβfx1,0,Ωβ+fx2,0,Ωβ.
Indeed, let us verify this with the example of the first term in the representation toV1. Standard calculations lead to (237)∫0a1x1dr1∫-a4/a1x1r1a2/a1x1r1fx1e-r1,x2e-r2-fx1,x2R3x1,x2,r1,r2dr2=∫0a1x1dr1∫-a4/a1x1r1a2/a1x1r1fx1e-r1,x2e-r2-fx1e-r1,x2R3x1,x2,r1,r2dr2+∫0a1x1dr1fx1e-r1,x2-fx1,x2×∫-a4/a1x1r1a2/a1x1r1R3x1,x2,r1,r2dr2Then, properties of the function f and Lemma 21 with ε3=a1(x1), ε2=a2/a1(x1), ε1=a4/a1(x1) yield (238)∫0a1x1dr1×∫-a4/a1x1r1a2/a1x1r1fx1e-r1,x2e-r2-fx1,x2×R3x1,x2,r1,r2dr2≤Cfx1,0,Ωβ+fx2,0,Ωβx2βa2β+x2βa4β+x1βa1βx1a2a4+a4a2≤CΔ1xβfx1,0,Ωβ+fx2,0,Ωβ.To reach the last inequality we apply (223) to (233). The remaining terms in the representation of V1 are evaluated in the same way.
ConcerningV2, we repeat the arguments above with changing D(x1,x2) by D(x¯1,x2) and obtain the estimate like (236) to the function V2.
Coming toV3, change of variables (179), Lemma 21, and condition (223) arrive at(239)V3=fx¯1,x2-fx1,x2×∬R2R3x1,x2,r1,r2dr1dr2+∬D1x1,x2R3x1,x2,r1,r2dr1dr2≤CΔ1xβfx1,0,Ωβ.Finally, we are left to estimate V4. Change of variables (179) and statement (iv) in Lemma 21 lead to(240)V4≤CΔ1xβfx1,0,Ωβ+fx2,0,Ωβ.Collecting this inequality with (236) and (239), we get the bound(241)Δ1U3≤CΔ1xβfEββΩ¯.This finishes the proof of Lemma 22.At this point, we come back toDx1αux1,0,Ω(β). Keeping in mind Proposition 20, Lemma 22, and representations (176) and (181), we deduce the inequality(242)Dx1αux1,0,Ωβ+Dx1αux2,0,Ωβ≤CfEββΩ¯.Further, recasting arguments in Proposition20 and Lemmas 22 and 14 and applying Proposition 8, we assert the result.Lemma 23.
Let assumption of Lemma14 hold. Then (243)Dx1αuEββΩ¯+Dx1αuE-s1-αβΩ¯≤CR0fEββΩ¯,Dx1αuCββΩ¯+Dx1αuC-s1-αβΩ¯≤CR0fCββΩ¯.In order to obtain the same results to the functionsDx1α1Dx2α2u and Dx2α2Dx1α1u, 1<α1+α2≤α, it is enough to repeat the arguments from this section. Thus, we have.Remark 24.
Under conditions of Lemma23, the functions Dx1α1Dx2α2u, Dx2α2Dx1α1u, 1<α1+α2≤α, satisfy inequalities (23) and (27) and(244)Dx1α1Dx2α2u=Dx2α2Dx1α1u.


## 7. Proofs of Theorems3 and 4

Under restrictions (30) and (31), the proof of Theorems 3 and 4 follows from the arguments of Sections 3–6. Indeed, Proposition 17 and Lemmas 16, 19, and 23 with Remark 24 provide existence of a solution u which satisfies estimates (22), (23) and (26), (27). Next, direct calculations with aid of representation (176) allow one to verify that the constructed solution u with (57) satisfies (1) and the corresponding boundary conditions. Further, the uniqueness of the solution follows from the coercive estimates. This completes the proof of Theorems 3 and 4 if (30) and (31) hold.At this point, we remove restriction (30). To this end, we introduce the new function(245)W1=x1x12+x22ψ1x1+x2x12+x22ψ2x2.The following properties of the function W1 could be easily checked with aid (2.2) in [37] and properties of the functions ψi and f (see (h2), (19)): if β>α-1, then(246)eitherW1∈Msα+βΩ¯∩M-s1α+βΩ¯,orW1∈Nsα+βΩ¯∩N-s1α+βΩ¯,fx1,0-Dx1αW1x1,x2x2=0-Dx2αW1x1,x2x2=0-Dx1αψ1x1=0,f0,x2-Dx1αW1x1,x2x1=0-Dx2αW1x1,x2x1=0-Dx2αψ2x2=0.After that, we look for the solution to (1), (2) in the form(247)ux,t=W+W1,where the unknown function W solves the boundary value problem with homogenous boundary condition(248)Dx1αW-Dx2αW=f¯x1,x2inΩ,W=0onΓ1∪Γ2,with(249)f¯x1,x2=fx1,x2-Dx1αW1x1,x2-Dx2αW1x1,x2.It is apparent that the function f¯ satisfies conditions (h3), (19). Thus, arguments of Sections 3–6 guarantee the one-valued solvability to (248) and, besides, the function W satisfies properties (22), (23), and (27) and (26). Then, we return the function u and obtain the one-to-one solvability of the original problem, where the solution u possesses the same properties as W. Thus, Theorems 3 and 4 are proven in the absence of (30).Finally, we are left to remove restriction (31). To this end, it is worth to remark that a function f∈Es-αβ(Ω¯) which satisfies condition (31) belongs to f∈E-s¯-αβ(Ω¯) with any positive s¯ and hence also with s¯=s1, s1∈(0,2-α). Further, it is enough to repeat the arguments from Sections 3–7 in order to prove Theorem 3, if f satisfies (17). If f meets requirement (18), the same reasons hold. This finishes the proof of Theorems 3 and 4.Remark 25.
Requirement (30) could be removed without introducing the function W1. To this end, it is enough to recast the arguments of Sections 3–6 to the problems with the homogenous equation and nonhomogeneous boundary conditions (250)Dx1αu-Dx2αu=0inΩ,subject either to the Dirichlet boundary condition (DBC) (251)u=ψ1x1onΓ1andu=ψ2x2onΓ2,or to the Neumann boundary condition (NBC) (252)∂u∂x2=φ1x1onΓ1and∂u∂x1=φ2x2onΓ2.Remark 26.
Arguments from Sections3–6 guarantee that the requirement on the weight s in the assumption(h1) can be changed by(253)s≥α+β.Remark 27.
Actually, with inessential modifications in the proofs, the very same results hold for the more general equation(254)Dx1α1u+Dx2α2u=fx1,x2inΩ,where 1<α1,α2<2. The details are left to the interested reader.



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*Source: 5395124-2018-12-02.xml*