We obtain the pointwise boundary differentiability of strong solu- tions for elliptic equations with the lower order coefficients, the boundary, and the right-hand side term satisfying a Dini type condition

ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu

BOUNDARY REGULARITY FOR NONDIVERGENCE ELLIPTIC EQUATION WITH UNBOUNDED DRIFT

YONGPAN HUANG, QIAOZHU ZHAI, SHULIN ZHOU Communicated by Hongjie Dong

Abstract. We obtain the pointwise boundary differentiability of strong solu- tions for elliptic equations with the lower order coefficients, the boundary, and the right-hand side term satisfying a Dini type condition. Furthermore, we establish a pointwise estimate of strong solutions and show that the gradients of the strong solutions are continuous along the boundary if the drift term, the boundary, and the right-hand side term satisfy a uniform Dini type condition on the boundary.

1. Introduction

In this article, we will study the boundary regularity of strong solutions of elliptic equation with unbounded lower order coefficients. Suppose thatu∈W_loc^2,n(Ω)∩C(Ω) satisfies

Lu:=−aij(x)D_iju+b_i(x)D_iu=f(x) in Ω;

u(x) = 0 on∂Ω; (1.1)

where Ω is a bounded domain in Rⁿ (n ≥2). We use the summation convention over repeated indices and the notationsDi := _∂x^∂

i; Dij :=DiDj. We assume that a_ij, b_i andf are measurable functions on Ω, the matrix (a_ij(x))_n×n is symmetric and satisfies the uniformly elliptic condition

λ|ξ|²≤a_ij(x)ξ_iξ_j ≤λ⁻¹|ξ|², for allξ∈Rⁿ, a.e. x∈Ω, (1.2) with a constantλ∈(0,1], andbi, f ∈Lⁿ(Ω). Throughout this article, the operator Lin (1.1) is applied to functionsuin the classW(Ω) :=W_loc^2,n(Ω)∩C(Ω).

In the following, we extend the results in [15] to elliptic equations with un- bounded lower order term. The boundary differentiability is shown for strong so- lution of nondivergence elliptic equation onC^1,Dini domain with unbounded drift satisfies Dini type condition. Furthermore, we prove that boundary first order derivative is continuous along the boundary.

As for the boundary regularity of nondivergence elliptic equations: If the drift term|b|is bounded, Krylov [8, 9] showed that the solution isC^1,αalong the bound- ary if∂Ω isC^1,1. Lieberman [13] gave a more general estimates. Wang [19] proved

2010Mathematics Subject Classification. 35J25, 35B65.

Key words and phrases. Elliptic equations; strong Solutions; unbounded drift;

boundary regularity.

c

2019 Texas State University.

Submitted March 27, 2018. Published March 12, 2019.

1

a similar pointwise result as in [8, 9] by an iteration method that will be adopted in this paper. Ma and Wang [15] proved a boundaryC^1,ψ estimate for fully nonlinear elliptic equations onC^1,Dini domain. Li and Wang [11, 12] showed the boundary differentiability of solutions of elliptic equations on convex domains. If |b| is un- bounded, Ladyzhenskaya and Ural’tseva in [10] proved boundaryC^1,α estimate of elliptic and parabolic inequalities on W^2,q domain with b ∈ L^q, Φ ∈ L^q, q > n and nonlinear termµ₁|Du|². Apushkinskaya and Nazarov [1] proved the boundary C^1,α estimate for nondivergence parabolic equation with composite righthand side and lower order coefficients, and in [2] they gave a counterexample of Hopf-Oleinik lemma in the elliptic case. Safonov [18] obtained the Hopf-Oleinik lemma on a flat domain for elliptic equations and gave the counterexample which indicated that the Dini condition on|b|can not be removed for our theorem. Nazarov [16] proved the Hopf-Oleinik Lemma and boundary gradient estimate under minimal restrictions on lower-order coefficients. Braga, Moeira and Wang [3] generalized the elliptic case in [10] to Lⁿ viscosity solutions with µ1 = 0 and C^1,Dini boundary value. Some related results concerning Dini continuity can be found in [4, 6, 7, 17, 20, 21].

The following Alexandroff-Bakelman-Pucci maximum principle and Harnack in- equality are our main tools.

Theorem 1.1([5, 18]). LetΩbe a bounded domain inRⁿ, and letube a function inW(Ω)such thatLu≤f inΩ. Suppose that the matrix(a_ij(x))_n×n is symmetric and satisfies the uniformly elliptic condition (1.2), andb_i, f ∈Lⁿ(Ω). Then

sup

Ω

u≤sup

∂Ω

u+Ndiam Ω·e^NkbknLn(Ω)kf⁺k_Ln(Ω), (1.3) where

kbkLⁿ(Ω)=Z

Ω

|b|ⁿdx^1/n

, b= (b₁, b₂, . . . , b_n), (1.4) andN is a positive constant depending only on nandλ.

Theorem 1.2 (Harnack Inequality). Let u be a nonnegative function in W(B8), Lu= f in B8 and bi, f ∈Lⁿ(B8). There exists a positive constant 0 depending only onλandn, such that ifkbk_Ln(B8)≤0, then

sup

B1

u≤C(inf

B1

u+kfk_Ln(B8)), (1.5) whereC is constant depending only onλandn.

Theorem 1.2 follows from the the proof in [18] clearly. The most important thing is that the quantitykbkLⁿ is scaling invariant(see [18, Remark 1.4]) and the Harnack constant is invariant in the iteration procedure. Before we state out our main theorem, for convenience, we give the following notation and definitions.

{ei}ⁿ_i=1 is the standard basis ofRⁿ.

|x|:=Xⁿ

i=1

x²_i^1/2

is the Euclidean norm of x = (x1, x2, . . . , xn) ∈ Rⁿ. a⁺ := max{0, a}. Br :=

{x ∈ Rⁿ : |x| < r}. B_r(x) := x+B_r. Ω_r := Ω∩B_r. Ω_r(x) := Ω∩B_r(x).

diam(Ω) := sup_x,y∈Ω|x−y|.

Qr:={x∈Rⁿ :|xi|< r, i= 1,2, . . . , n}.

kfkLⁿ(Ω):= R

Ω|f(x)|ⁿdx1/n

. W(Ω) :=W_loc^2,n(Ω)∩C(Ω).

Definition 1.3. We say that∂Ω isC^1,Dini atx∈∂Ω, if there exist a unit vector

~

nand a positive constantr0such that 1

r sup

y∈∂Ω,|y−x|≤r

|(y−x)·~n| ≤ω(r), for 0< r≤r₀, where ω(r) is a nonnegative nondecreasing function and satisfiesRr₀

0 ω(r)

r dr <∞.

We say that∂Ω isC^1,Dini if for anyx∈∂Ω,∂Ω isC^1,Dini atx∈∂Ω.

If∂Ω satisfies the pointwiseC^1,Dinicondition at anyx∈∂Ω with the samer0, it follows that∂Ω isC^1,Dini in the classical sense, i.e.,∂Ω can be locally represented as aC¹graph with the gradient being Dini continuous.

Definition 1.4. We say that the function g ∈ Lⁿ(Ω) is C_n^−1,Dini at x ∈ ∂Ω, if there exists a positive constantr0 such that

( 1

|Br(x)∩Ω|

Z

B_r(x)∩Ω

|g(y)|ⁿdy)^1/n≤r⁻¹ω(r)

for each 0< r≤r₀, whereω(r) is a nonnegative nondecreasing function and satisfies Rr₀

0 ω(r)

r dr <∞. Obviously, we havekgkLⁿ(Ω∩Br(x)) ≤ |B1(0)|^1/nω(r)≤2ω(r). We say thatg isC_n^−1,Dini on∂Ω if for anyx∈∂Ω,gisC_n^−1,Dini atx∈∂Ω.

Generally, for any function in L^p(Ω)(1≤p≤ ∞), we can define the pointwise C_p^k,Dini (k∈Z). We say that the functiong ∈L^p(Ω) isC_p^k,Dini atx∈∂Ω, if there exists a positive constant r0 and a k−th order polynomial P_k^x(y) (P_k^x(y) ≡ 0 if k <0) such that

1

|B_r(x)∩Ω|

Z

Br(x)∩Ω

|g(y)−P_k^x(y)|^pdy1/p

≤r^kω(r)

for each 0 < r ≤ r0, where ω(r) is a nonnegative nondecreasing function and satisfiesRr0

0 ω(r)

r dr <∞.

The main results of this paper are Theorems 1.5, 1.9, and Corollary 1.7 below.

Theorem 1.5. Assume that

(1) 0 ∈ ∂Ω, r0 > 0, u ∈ W(Ωr0), u|∂Ω∩Br0 = 0, Lu = f in Ωr0, |b|, f ∈ Lⁿ(Ω_r0)andRr₀

0

kfk_Ln(Ωr)

r dr <∞;

(2) ∂ΩisC^1,Dini at0and |b| isC_n^−1,Dini at0 with the modulus of continuity ω(r)satisfies

ω(r₀)≤minδ 6,1

2,₀

2 and

Z r0

0

ω(r)

r dr≤min

1, δln¹_δ 72M√

nA2

, (1.6) whereδ,M and A2 are constants depending only on λandn (see Lemma 2.2), and0 is the constant in Theorem 1.2.

Then u is differentiable at 0, furthermore, there exist a linear function L(x) and constants α >ˆ 0,Λ>1,C >0 such that

|u(x)−L(x)| ≤Cn

r^αˆ+ω(Λr) +kfkLⁿ(Ω_Λr)+r^αˆ Z r0

r

ω(s) +kfk_Ln(Ω_s)

s^{1+ ˆα} ds +

Z Λr

0

ω(s) +kfkLⁿ(Ω_s)

s dso

r,

(1.7)

for any x ∈ Ωr and 0 < r ≤ ^r_Λ⁰, where C depends on kukL^∞(Ω_r0), kfkLⁿ(Ω_r0), Rr0

0

kfk_Ln(Ωs)

s ds,r0,λandn.

Remark 1.6. (1) The condition (1.6) will always be satisfied for small r0 if the modulus of continuityω(r) satisfies the Dini condition, which will guarantee that the slopes of hyperplanes in the iteration procedure are uniformly bounded (see (2.22)).

(2) We can also deduce pointwise boundary differentiability with nonhomoge- neous pointwise C^1,Dini boundary value as in [15]. Here we only consider the homogeneous boundary value just for convenience.

(3) The modulus of continuity ω(r) is nondecreasing can be replaced by ω(r) satisfies the doubling condition(see [14, Definition 2.3]).

The following corollary is a direct consequence of Theorems 1.5 and 2.1.

Corollary 1.7. Assume that

(1) 0 ∈ ∂Ω, r₀ > 0, u ∈ W(Ω_r0), Lu = f in Ω_r0, u|∂Ω∩Br0 = 0 and |b|, f ∈Lⁿ(Ω_r0);

(2) ∂Ω is C^1,Dini at 0, |b| is C_n^−1,Dini at 0 and f is C_n^−1,Dini at 0 with the modulus of continuityω(r)satisfies

ω(r₀)≤minδ 6,1

2,₀ 2 ,

Z r0

0

ω(r)

r dr≤min

1, δln¹_δ 72M√

nA2

,

whereδ,M andA₂ are the constants in Lemma 2.2, and₀ is the constant in Theorem 1.2.

Then u is differentiable at 0, furthermore, there exist a linear function L(x) and constants α >ˆ 0,Λ>1,C >0 such that for anyx∈Ω_r and0< r≤r₀/Λ,

|u(x)−L(x)| ≤C

r^αˆ+ω(Λr) +r^αˆ Z r₀

r

ω(s) s^{1+ ˆα}ds+

Z Λr

0

ω(s) s ds

r, (1.8) whereC depends onkukL^∞(Ω_r0),r0,λandn.

Remark 1.8. If ∂Ω is C^1,α at 0, |b| is C_n^−1,α at 0 and f is C_n^−1,α at 0 with ω(r) = r^α(0 < α < 1), then u is C^1,βˆ at 0 with ˆβ = min{α,α}ˆ if α 6= ˆα and 0<β <ˆ min{α,α}ˆ ifα= ˆα.

Theorem 1.9. Assume that

(1) 0 ∈ ∂Ω, r0 > 0, u∈ W(Ω3r₀), Lu = f in Ω3r₀, u|_∂Ω∩B3r

0 = 0 and |b|, f ∈Lⁿ(Ω3r₀);

(2) ∂Ω is C^1,Dini, |b| is C_n^−1,Dini and f is C_n^−1,Dini on ∂Ω∩Br₀ uniformly with the modulus of continuityω(r)satisfies

ω(r0)≤minδ 6,1

2,0

2 , Z r₀

0

ω(r)

r dr≤min

1, δln¹_δ 72M√

nA2

,

whereδ,M andA2 are constants in Lemma 2.2, and0 is the constant in Theorem 1.2.

Then there exist constantsα >ˆ 0,Λ>1,C >0 such that for any y, z ∈∂Ω∩Br₀

and0<|y−z|=r≤^r_Λ⁰,

|∇u(y)− ∇u(z)| ≤C

r^αˆ+ω(Λr) +r^αˆ Z r₀

r

ω(s) s^{1+ ˆα}ds+

Z Λr

0

ω(s) s ds

,

where αˆ and Λ are the constants in Corollary 1.7, andC is a constant depending onkukL^∞(Ω_3r0),r0,λandn.

Remark 1.10. If∂Ω isC^1,αon∂Ω∩Br₀,|b|isC_n^−1,α on∂Ω∩Br₀ andf isC_n^−1,α on ∂Ω∩Br0 with ω(r) = r^α(0 < α < 1), then ∇u is C^βˆ along ∂Ω∩Br0 with βˆ= min{α,α}ˆ ifα6= ˆαand 0<β <ˆ min{α,α}ˆ ifα= ˆα.

We shall prove Theorems 1.5 and 1.9 in the next section.

2. Boundary estimates

By standard normalization, it is enough to prove Theorem 2.1, below, instead of proving Theorem 1.5. Since∂Ω isC^1,Dini at 0∈∂Ω, without loss of generality, we assume~n =e_n as the inward normal direction in the following Theorem 2.1.

Consider the normalization of solution,

˜

u(x) = u(r0x)

kukL^∞(Ω_r0)++r₀kfkLⁿ(Ω_r0)+r₀Rr₀ 0

kfkLn(Ωr)

r dr ,

for >0 andx∈Ω˜∩B1, with the normalized domain ˜Ω :={x∈Rⁿ: r0x∈Ω}.

Obviously, ˜u(x) satisfies

k˜uk_L∞( ˜Ω1)≤1 and −˜aij(x)Diju˜(x) + ˜bi(x)Diu˜(x) = ˜f(x) forx∈Ω˜ ∩B₁, where

˜

a_ij(x) =a_ij(r₀x), ˜b_i(x) =r₀b_i(r₀x), f˜(x) = r²₀f(r0x)

kukL^∞(Ω_r0)++r₀kfkLⁿ(Ω_r0)+r₀Rr₀ 0

kfk_Ln(Ωr) r dr

.

Let ˜ω(r) =ω(r₀r). Obviously, 1

r sup

y∈∂Ω,|y|≤r˜

|y·e_n| ≤ω(r),˜ kbk˜ _Ln( ˜Ωr)=kbk_Ln(Ω_r0r)≤2˜ω(r) for 0< r≤1, and

Z 1

0

˜ ω(r)

r dr= Z r₀

0

ω(r) r dr.

Theorem 2.1. Assume that

(1) 0∈∂Ω,u∈W(Ω1),u|_∂Ω∩B1= 0,Lu=f in Ω1, andkuk_L^∞_(Ω1)≤1;

(2) f ∈Lⁿ(Ω₁)withkfk_Ln(Ω₁)≤1 andR1 0

kfk_Ln(Ωr) r dr≤1;

(3) ∂ΩisC^1,Dini at0and |b| isC_n^−1,Dini at0 with the modulus of continuity ω(r)satisfies the normalized conditions

ω(1)≤minδ 6,1

2,0

2 , Z 1

0

ω(r)

r dr≤min

1, δln¹_δ

72M A₂ , (2.1) where ₀ is the constant in Theorem 1.2, and δ, M, A₂ are constants in Lemma 2.2.

Then there exist the three positive constants C, αˆ andΛ(≥324n) depending only onλandn, and there exists a constant θ such that

|u(x)−θxn| ≤Cn

r^αˆ+ω(Λr) +kfk_Ln(ΩΛr)+r^αˆ Z 1

r

ω(s) +kfk_Ln(Ωs)

s^{1+ ˆα} ds +

Z Λr

0

ω(s) +kfkLⁿ(Ω_s)

s dso

r,

(2.2)

for any x∈Ωr andr≤ _Λ¹.

We shall establish Theorem 2.1 by an iteration method which is based on Lemmas 2.2 and 2.3 below. For convenience, we define

γ(r) = 1 r sup

y∈∂Ω,|y|≤r

|y· en| for 0< r≤1.

Obviously,

γ(r)≤ω(r), kbk_Ln(Ω_r)≤2ω(r) for 0< r≤1.

Lemma 2.2. Suppose that 0 ∈ ∂Ω, u ∈ W(Ω₁), u|∂Ω∩B1 = 0, Lu = f in Ω₁, f ∈ Lⁿ(Ω₁), γ(1)≤δ/6 and kbkLⁿ(Ω₁) ≤min{0,1}, where ₀ is the constant in Theorem 1.2 andδ(<1)will be chosen in (2.3). Then there exist positive constants µ <1,M,A₁ andA₂ depending only onλandn. If

kxn−l≤u(x)≤Kxn+B in Ω1, (2.3) for some constants l≥0,B(≥0),kandK with k≤K, then there exist constants

˜k andK˜ such that

˜kxn−A1kfk_Ln(Ω1)−A2(|K|+|k|+l)(γ(1) +kbk_Ln(Ω1))

≤u(x)≤Kx˜ n+A1kfk_Ln(Ω1)+A2(|K|+|k|+B)(γ(1) +kbk_Ln(Ω1)) (2.4) inΩδ, where either

˜k=k−3M√

nl+µ(K−k) and K˜ =K+ 3M√

nB, (2.5)

or

˜k=k−3M√

nl and K˜ =K+ 3M√

nB−µ(K−k). (2.6) Obviously, we have˜k≤K.˜

Proof of Lemma 2.2. First we proof the following.

Claim. There exist positive constants M, ˜δ and C₁ depending only on λand n, such that

(k−3M√

nl)xn−C1kfk_Ln(Ω1)−3M√

n(|k|+l)γ(1)−C1(|k|+l)kbk_Ln(Ω1)

≤u(x)

≤(K+ 3M√

nB)xn+C1kfk_Ln(Ω1)+ 3M√

n(|K|+B)γ(1) +C₁(|K|+B)kbk_Ln(Ω₁) in Ω∩Q˜δ.

Proof. LetM = 1 +²

√n−1

λ (≥3) and(>0) be small enough, such that

3−(1 +)(2 +)(M−1)≥0. (2.7) Let

˜δ= 1 M√

n(≤ 1 3√

n), δ=

˜δ

2M = 1

2√ n(1 + ²

√n−1 λ )²

(2.8)

and

ψ(x) =˜ 4 3

2(xn+γ(1))

δ˜ −(xn+γ(1))² δ˜²

+ λ² 2(n−1)

n−1

X

i=1

|xi|

δ˜ −1+2+

. The barrier function ˜ψ(x) isC² and satisfies the following conditions (observe that 1≤^δ+γ(1)˜ _˜

δ ≤3/2):

ψ(x)˜ ≥1 onQ_1/^√_n∩ {x∈Rⁿ:xn = ˜δ};

ψ(x)˜ ≥0 onQ_1/^√_n∩ {x∈Rⁿ:xn =−γ(1)};

ψ(x)˜ ≥1 on∂Q√1

n ∩ {x∈Rⁿ: −γ(1)< xn <δ};˜

−aij(x)Dijψ(x)˜ ≥0 a.e. inQ_1/^√_n∩ {x∈Rⁿ :−γ(1)< xn<˜δ} ∩Ω;

ψ(x)˜ ≤3(xn+γ(1))

˜δ inQδ˜∩ {x:xn≥ −γ(1)}.

(2.9)

Combining (??) and (2.4), we have L kxn−lψ(x)˜ −u(x)

≤bi(x)Di(kxn−lψ(x))˜ −f(x) in ˜Q∩Ω;

kxn−lψ(x)˜ −u(x)≤ |k|γ(1) on∂( ˜Q∩Ω); (2.10) where ˜Q=Q_1/^√_n∩ {x∈Rⁿ :−γ(1)< x_n<δ}.˜

By the Alexandroff-Bakelman-Pucci maximum principle,

kx_n−lψ(x)˜ −u(x)≤ |k|γ(1) +C₁(|k|+l)kbkLⁿ(Ω₁)+C₁kfkLⁿ(Ω₁) (2.11) in ˜Q∩Ω, whereC1is a constant depending only onλandn.

By (2.4) (fifth inequality), we have u(x)≥(k−3M√

nl)x_n−C₁kfkLⁿ(Ω₁)−3M√

n(|k|+l)γ(1)

−C1(|k|+l)kbk_Ln(Ω1)

(2.12) in Ω∩Q˜δ. As in (2.5), we have

L u(x)−Kx_n−Bψ(x)˜

≤f(x)−b_i(x)D_i(Kx_n+Bψ(x))˜ in ˜Q∩Ω;

u(x)−Kx_n−Bψ(x)˜ ≤ |K|γ(1) on∂( ˜Q∩Ω).

According to the Alexandroff-Bakelman-Pucci maximum principle,

u(x)−Kx_n−Bψ(x)˜ ≤ |K|γ(1) +C₁kfk_Ln(Ω₁)+C₁(|K|+B)kbk_Ln(Ω₁)

in ˜Q∩Ω, where C1 is a constant depending only on λ and n. By (2.4) (fifth inequality), we have

u(x)≤(K+ 3M√

nB)x_n+C₁kfkLⁿ(Ω₁)+ 3M√

n(|K|+B)γ(1) +C1(|K|+B)kbkLⁿ(Ω₁)

(2.13) in Ω∩Q˜δ. By (2.7) and (2.8), the claim follows.

Let Γ =Q_{M δ}∩ {x∈Rⁿ: x_n=δ}. By γ(1)≤δ/6, we have Γ⊂Ω and dist(Γ, ∂Ω)≥δ

2. (2.14)

Next, we show (??) for the two cases: u(δe_n) ≥ ¹₂(K +k)δ and u(δe_n) <

1

2(K+k)δ, corresponding to (??) and (??).

Case 1: u(δen)≥¹₂(K+k)δ. Let v(x) =u(x)−(k−3M√

nl)xn+C1kfk_Ln(Ω1)+ 3M√

n(|k|+l)γ(1) +C₁(|k|+l)kbk_Ln(Ω₁).

Then

v(δen)≥K−k

2 + 3M√ nl

δ+C1kfk_Ln(Ω1)+ 3M√

n(|k|+l)γ(1) +C₁(|k|+l)kbkLⁿ(Ω₁).

(2.15) Since v(x)≥0 for x∈Ω∩Qδ˜, from (2.9) and the interior Harnack inequality, it follows that

sup

Γ

v(x)≤C2 inf

Γ v(x) +kfk_Ln(Ω1)+ (|k|+l)kbk_Ln(Ω1)

, (2.16) where C2(≥1) is a constant depending only on λandn. Combining (2.10),(2.11) andv(x)≥0, we have

inf

Γ v(x)≥n 1 C2

(K−k

2 + 3M√

nl)δ+ 3M√

n(|k|+l)γ(1) + (C1

C₂ −1)

(|k|+l)kbk_Ln(Ω1)+kfk_Ln(Ω1)

o+

:=a.

Let ψ(x) = 3

8

x_n+γ(1) δ

+x_n+γ(1) δ

²

− λ² 4(n−1)

n−1

X

i=1

|xi|

δ −1+²⁺

, (2.17) wheresatisfies (2.2).

The barrier function ψ(x) is C² and satisfies the following conditions (observe that 1≤ ^δ+γ(1)_δ ≤7/6):

ψ(x)≤1 onQ_{M δ}∩ {x∈Rⁿ:x_n=δ};

ψ(x)≤0 onQM δ∩ {x∈Rⁿ:xn=−γ(1)};

ψ(x)≤0 on∂QM δ∩ {x∈Rⁿ:−γ(1)≤xn≤δ};

−a_ij(x)D_ijψ(x)≤0 a.e. inQ_{M δ}∩ {x∈Rⁿ:−γ(1)< x_n < δ} ∩Ω;

ψ(x)≥ xn+γ(1)

3δ inQδ∩ {x: xn ≥ −γ(1)};

ψ(x)≤ xn+γ(1)

δ inQ_{M δ}∩ {x∈Rⁿ:−γ(1)≤x_n ≤δ}.

(2.18)

We claim that L aψ(x)−v(x)

≤bi(x)Di(aψ(x) + (k−3M√

nl)xn)−f inQ˜˜∩Ω;

aψ(x)−v(x)≤2 + 9M√ n

C₂ (|K|+|k|+l)γ(1) on∂(Q˜˜∩Ω);

(2.19)

whereQ˜˜=Q_{M δ}∩ {x∈Rⁿ:−γ(1)< x_n< δ}.

In fact, the first inequality is clear. For the second inequality, we separate the boundary∂(Q˜˜∩Ω) into three parts:

∂Q˜˜∩ {x∈Rⁿ :x_n=δ}, ∂Q˜˜∩ {x∈Rⁿ:−δ < xn< δ} ∩Ω, ∂Ω∩Q.˜˜

The first part is just Γ where v(x)≥a and ψ(x)≤1, then aψ(x)−v(x)≤0 on it. On the second part, sincev(x)≥0 andψ(x)≤0, we haveaψ(x)−v(x)≤0 on them. On the last part, sinceψ(x)≤^xn+γ(1)_δ ≤1 on it by (2.13)(6), we have

aψ(x)−v(x)≤ 1 C2

(K−k

2 + 3M√

nl)δ+ 3M√

n(|k|+l)γ(1)xn+γ(1) δ

≤ 1 C2

(|K|+|k|

2 + 3M√

nl)(xn+γ(1)) + 3M√

n(|k|+l)γ(1)

≤2 + 9M√ n

C₂ (|K|+|k|+l)γ(1),

where we have used −γ(1) ≤ x_n ≤ γ(1) for x ∈ ∂Ω∩Q. By the Alexandroff-˜˜ Bakelman-Pucci maximum principle,

aψ(x)−v(x)≤C3(|K|+|k|+l)(γ(1) +kbk_Ln(Ω1)) +C3kfk_Ln(Ω1) inQ˜˜∩Ω, where we have usedkbk_Ln(Ω1)≤1 andC3 is a constant depending only onλand n.

From (2.13) (fifth inequality), it follows that for allx∈Ω∩Qδ, aψ(x)≥ a

3δ(x_n+γ(1))

≥

(K−k)δ

2C₂ − kfk_Ln(Ω1)−(|k|+l)kbk_Ln(Ω1)

3δ (x_n+γ(1))

≥K−k 6C2

xn− kfk_Ln(Ω1)−(|k|+l)kbk_Ln(Ω1), where we have usedK−k≥0.

Therefore, for allx∈Ωδ, u(x)≥aψ(x) + (k−3M√

nl)xn−(C1+C3)kfk_Ln(Ω1)

−(C3+ 3M√

n+C1)(|K|+|k|+l)(γ(1) +kbk_Ln(Ω1))

≥

k−3M√ nl+ 1

6C2

(K−k)

xn−(C1+C3+ 1)kfk_Ln(Ω1)

−(C3+ 3M√

n+C1+ 1)(|K|+|k|+l)(γ(1) +kbk_Ln(Ω1)).

(2.20)

Let

µ= 1 6C2

, A1=C1+C3+ 1, A2=C1+C3+ 3M√

n+ 1. (2.21) Combining (2.8),(2.15) and (2.16), we have (??) and (??).

Case 2: u(δen)<¹₂(K+k)δ. The proof is similar to that of Case 1. Let v(x) = (K+ 3M√

nB)xn+C1kfk_Ln(Ω1)+ 3M√

n(|K|+B)γ(1) +C₁(|K|+B)kbkLⁿ(Ω₁)−u(x)

forx∈Ω∩Q˜δ. Then v(δen)> K−k

2 + 3M√ nB

δ+C1kfk_Ln(Ω1)+ 3M√

n(|K|+B)γ(1) +C₁(|K|+B)kbk_Ln(Ω₁)).

By the interior Harnack inequality, we have sup

Γ

v≤C2 inf

Γ v+kfkLⁿ(Ω₁)+ (|K|+B)kbkLⁿ(Ω₁)

, whereC₂(≥1) is a constant depending only on λandn. Then

inf

Γ v≥n 1 C2

K−k

2 + 3M√ nB

δ+ 3M√

n(|K|+B)γ(1) + C1

C₂ −1

(kfk_Ln(Ω1)+ (|K|+B)kbk_Ln(Ω1)))o+

:=a.

(2.22)

Letψ(x) be defined by (2.12). As in (2.14), we have L aψ(x)−v(x)

≤biDi(aψ(x)−(|K|+ 3M√

nB)xn) +f(x) in Q˜˜∩Ω;

aψ(x)−v(x)≤ (2 + 9M) C2

(|K|+|k|+B)γ(1) on∂(Q˜˜∩Ω);

(2.23)

whereQ˜˜=QM δ∩ {x∈Rⁿ:−γ(1)< xn< δ}.

Therefore, by the Alexandroff-Bakelman-Pucci maximum principle,

aψ(x)−v(x)≤C3(|K|+|k|+B)(γ(1) +kbk_Ln(Ω1)) +C3kfk_Ln(Ω1), (2.24) inQ˜˜∩Ω, where we have used kbk_Ln(Ω1)≤1, andC3 is a constant depending only onλand n.

By (2.13) (fifth inequality), we have that for anyx∈Ω∩Qδ, a

3δ(xn+γ(1))−v(x)≤C3(|K|+|k|+B)(γ(1) +kbkLⁿ(Ω₁)) +C3kfkLⁿ(Ω₁). Combining (2.17) with (2.19), we have that for allx∈Ωδ,

u(x)≤(K+ 3M√

nB)x_n− a

3δ(x_n+γ(1)) + (C₁+C₃)kfkLⁿ(Ω₁)

+ (C₁+C₃+ 3M√

n)(|K|+|k|+B)(γ(1) +kbkLⁿ(Ω₁))

≤ K+ 3M√

nB− 1 6C2

(K−k)

x_n+ (C₁+C₃+ 1)kfk_Ln(Ω₁)

+ (C₁+C₃+ 3M√

n+ 1)(|K|+|k|+B)(γ(1) +kbkLⁿ(Ω₁)).

(2.25)

Let µ= _6C¹

2, A1 =C1+C3+ 1 andA2 =C1+C3+ 3M√

n+ 1. Combining

(2.7) and (2.20), we have that (??) and (??) hold.

Using induction, the following lemma is a direct consequence of Lemma 2.2.

Lemma 2.3. Suppose that 0 ∈ ∂Ω, u ∈ W(Ω₁), u|∂Ω∩B1 = 0, Lu = f in Ω₁, kukL^∞(Ω₁) ≤ 1, f ∈ Lⁿ(Ω₁) and ω(1) ≤ min{0/2,1/2, δ/6}. Then there exist nonnegative sequences {lm}^∞_m=0, {Bm}^∞_m=0, and sequences {km}^∞_m=0, {Km}^∞_m=0 withk₀=K₀= 0,l₀=B₀= 1, and for m= 0,1,2, . . .,

l_m+1=A₁δ^mkfkLⁿ(Ω_δm)+A₂δ^m(|Km|+|km|+ l_m

δ^m)(γ(δ^m) +kbkLⁿ(Ω_δm)), Bm+1=A1δ^mkfk_Ln(Ω_δm)+A2δ^m(|Km|+|km|+Bm

δ^m)(γ(δ^m) +kbk_Ln(Ω_δm)), and

k_m+1=k_m−3M√ nlm

δ^m+µ(K_m−k_m) and K_m+1=K_m+ 3M√ nBm

δ^m,

or

km+1=km−3M√ nlm

δ^m and Km+1=Km+ 3M√ nBm

δ^m −µ(Km−km), such that

k_mx_n−l_m≤u(x)≤K_mx_n+B_m inΩ_δm, (2.26) whereδ,µ,M,A₁ andA₂ are positive constants given by Lemma 2.2.

Proof of Theorem 2.1. Let {lm}^∞_m=0, {Bm}^∞_m=0, {km}^∞_m=0 and {Km}^∞_m=0 be de- fined by Lemma 2.3. We prove the following claim first.

Claim. There exists a constant C1 depending only λ and n such that for all m= 0,1,2, . . .,

|Km|,|km|,B_m δ^m,l_m

δ^m ≤C₁. (2.27)

Proof. Firstly, notice that we takeK0=k0= 0 andl0=B0= 1, then by induction, we haveK_m≥k_m for allm≥0. Form≥0, we defineS_m=Pm

i=0

B_i

δⁱ +_δ^li_i). For anym≥0, since

Km+1 ≤Km+ 3M√ nBm

δ^m andK0= 0, we have

Km+1≤3M√

nSm form≥0.

Similarly, we have

k_m+1≥ −3M√

nS_m form≥0.

It follows that

|km+1|+|Km+1| ≤6M√

nSm form≥0. (2.28)

Since

B_m+1+l_m+1 δ^m+1 =A₂

δ (γ(δ^m) +kbkLⁿ(Ω_δm))(2|Km|+ 2|km|+B_m+l_m δ^m ) +2A1

δ kfk_Ln(Ω_δm),

form≥1, combining the above identity with (2.23), we obtain Bm+1+lm+1

δ^m+1 ≤ A2

δ γ(δ^m) +kbk_Ln(Ω_δm)

12M√

nS_m−1+Bm+lm

δ^m

+2A₁

δ kfkLⁿ(Ω_δm)

≤ 12M√ nA2

δ (γ(δ^m) +kbkLⁿ(Ω_δm))Sm+2A1

δ kfkLⁿ(Ω_δm).

(2.29)

By the normalized condition, we have

∞

X

i=1

12M√ nA2

δ γ(δⁱ) +kbk_Ln(Ω_δi)

≤

∞

X

i=1

36M√ nA2

δ ω(δⁱ)

≤ 36M√ nA₂ δln¹_δ

Z 1

0

ω(r) r dr≤ 1

2,

(2.30)

and

2A₁ δ

∞

X

i=1

kfk_Ln(Ω_δi)≤ 2A₁ δln¹_δ

Z 1

0

kfk_Ln(Ω_r)

r dr≤ 2A₁

δln¹_δ. (2.31)

From (2.24)-(2.26), it follows that for anym≥1, S_m+1−S₁=

m

X

i=1

Bi+1+li+1

δⁱ⁺¹

≤Sm+1 m

X

i=1

12M√ nA2

δ γ(δⁱ) +kbk_Ln(Ω_δi)

+2A1

δ

m

X

i=1

kfk_Ln(Ω_δi)

≤ 1

2S_m+1+ 2A₁ δln¹_δ. Therefore, for allm≥1,

S_m+1≤ 4A₁

δln(1/δ)+ 2S₁. SinceS0= 2, 0≤S1≤A1+A2, we have

0≤S_m≤2A₁+ 2A₂+ 2 + 4A₁

δln¹_δ for allm≥0.

LetC1= 3M√

n(2A1+ 2A2+ 2 +_δ^4A_ln¹1 δ

). This completes the proof of the claim.

Next we show estimate (??). By Lemma 2.3, we have that for allm≥1, 0≤Km+1−km+1≤(1−µ)(Km−km) + 3M√

nlm+Bm

δ^m or

|Km+1−km+1| ≤(1−µ)|Km−km|+C2(kfk_Ln(Ω_δm−1)+ω(δ^m−1)), whereC2= 3M√

n(A1+ 6A2C1) /δ.

Let 1−µ=δ^αˆ( ˆα >0) . By iteration, we have that for allm≥1,

|Km+1−km+1| ≤C3δ^αmˆ 1 +

Z 1

δ^m

ω(r) +kfkLⁿ(Ω_r)

r^{1+ ˆα} dr , whereC3 is a constant depending only onλandn.

For anym≥1,

Km+1+km+1≤Km+km+µ(Km−km) + 3M√ nBm

δ^m, Km+1+km+1≥Km+km−µ(Km−km)−3M√

nlm

δ^m. Hence,

|(Km+1+k_m+1)−(K_m+k_m)|

≤µ|Km−km|+ 3M√

nlm+Bm

δ^m

≤µ|Km−km|+C4(ω(δ^m−1) +kfk_Ln(Ω_δm−1)),

(2.32)

whereC4 is a constant depending only onλandn. It follows that

∞

X

j=m

|(Kj+1+kj+1)−(Kj+kj)|

≤C₃µ

∞

X

j=m

(δ^j−1)^αˆ 1 +

Z 1

δ^j−1

ω(r) +kfk_Ln(Ω_r)

r^{1+ ˆα} dr +C₄

∞

X

j=m

(ω(δ^j−1) +kfk_Ln(Ω_δjˆ−1)).

(2.33)

Let

F_r:=

Z 1

r

ω(s) +kfk_Ln(Ωs)

s^{1+ ˆα} ds.

By

∞

X

j=m

(δ^j−1)^αˆ Z 1

δ^j−1

ω(r) +kfk_Ln(Ωr)

r^{1+ ˆα} dr

=

∞

X

j=m−1

(δ^αˆ)^jF_δj

= 1

δ^αˆ(1−δ)

∞

X

j=m−1

(δ^j+1)^αˆF_δj ·δ^j−δ^j+1 δ^j

≤ 1

δ^αˆ(1−δ)

∞

X

j=m−1

Z δ^j

δ^j+1

r^α−1ˆ Frdr

= 1

δ^αˆ(1−δ) Z δ^m−1

0

r^α−1ˆ F_rdr

= 1

δ^αˆ(1−δ) ˆα

Z δ^m−1

0

ω(r) +kfk_Ln(Ω_r)

r dr

+ (δ^m−1)^αˆ Z 1

δ^m−1

ω(r) +kfkLⁿ(Ω_r)

r^{1+ ˆα} dr and

∞

X

j=m

(ω(δ^j−1) +kfk_Ln(Ω_δj−1))≤ 1 1−δ

Z δ^m−2

0

ω(r) +kfk_Ln(Ωr)

r dr,

it follows that

∞

X

j=m

|(Kj+1+kj+1)−(Kj+kj)|

≤C₅n

(δ^m−1)^αˆ+ (δ^m−1)^αˆ Z 1

δ^m−1

ω(r) +kfk_Ln(Ω_r)

r^{1+ ˆα} dr +

Z δ^m−2

0

ω(r) +kfk_Ln(Ωr)

r dro

.

(2.34)

whereC₅ is a constant depending only onλandn.

Whilem→ ∞, by lim_r→0+ω(r) = 0 and L’Hospital rule, we have the righthand side of (2.29) tends to 0. Hence{K_m+k_m}^∞_m=0is convergent. Let lim_m→∞^Km+k₂ ^m = θ. Then for all m≥2,

θ−K_m+k_m 2

≤

∞

X

j=m

|K_j+1+k_j+1

2 −K_j+k_j

2 |

≤C5

2 n

(δ^m−1)^αˆ+ (δ^m−1)^αˆ Z 1

δ^m−1

ω(r) +kfk_Ln(Ωr)

r^{1+ ˆα} dr +

Z δ^m−2

0

ω(r) +kfkLⁿ(Ω_r)

r dro

.

(2.35)

For anym≥0 and anyx∈Ωδ^m, we have

|u(x)−θxn| ≤ |u(x)−Km+km

2 xn|+|(Km+km

2 −θ)xn|. (2.36) From (2.21), it follows that

−|Km−km|

2 |xn| −lm≤u(x)−Km+km

2 xn≤|Km−km|

2 |xn|+Bm. Then for anym≥0 and anyx∈Ω_δm,

|u(x)−K_m+k_m

2 x_n| ≤(|Km−k_m|+l_m+B_m

δ^m )δ^m. (2.37) By (2.30) and the inequality above, for allx∈Ωδ^m,m= 2,3, . . .,

|u(x)−θx_n|

≤ |u(x)−Km+km

2 xn|+|(Km+km

2 −θ)xn|

≤

|Km−km|+Bm+lm

δ^m +|Km+km

2 −θ|

δ^m

≤C₆n

(δ^m−1)^αˆ+ (δ^m−1)^αˆ Z 1

δ^m−1

ω(r) +kfk_Ln(Ω_r)

r^{1+ ˆα} dr +ω(δ^m−1) +kfk_Ln(Ω_δm−1)+

Z δ^m−2

0

ω(r) +kfk_Ln(Ωr)

r dro

δ^m,

(2.38)

whereC6 is a constant depending only onλandn.

Let Λ = 1/δ² (≥324n). By (2.33), we have that for allx∈Ωr andr≤1/Λ,

|u(x)−θxn| ≤C7

n

r^αˆ+ω(Λr) +r^αˆ Z 1

r

ω(s) +kfkLⁿ(Ω_s)

s^{1+ ˆα} ds +kfk_Ln(ΩΛr)+

Z Λr

0

ω(s) +kfk_Ln(Ωs)

s dso

r.

This completes the proof of Theorem 2.1.

Proof of Theorem 1.9. Consider |∇u(y)− ∇u(z)|, wherey, z ∈∂Ω∩Br₀ and 0<

|y−z|=r≤ ^r_Λ⁰. By Corollary 1.7, we have ku(x)−L_y(x)kL^∞(Ω_r(y))≤Cn

r^αˆ+ω(Λr) +r^αˆ Z r0

r

ω(s) s^{1+ ˆα}ds+

Z Λr

0

ω(s) s dso

r, ku(x)−Lz(x)k_L^∞_(Ωr(z)) ≤Cn

r^αˆ+ω(Λr) +r^αˆ Z r₀

r

ω(s) s^{1+ ˆα}ds+

Z Λr

0

ω(s) s dso

r.

Noticing that∂Ω isC^1,Dini and the normalization makesω small enough, then there exist a pointp∈Ω and a small positive constantη(<_Λ¹) such thatB_ηr(p)⊂ Ω_r(y)∩Ω_r(z). Then by the triangle inequality, we have

kLy(x)−Lz(x)k_L∞(B_ηr(p)) ≤2Cn

r^αˆ+ω(Λr) +r^αˆ Z r₀

r

ω(s) s^{1+ ˆα}ds+

Z Λr

0

ω(s) s dso

r.

SinceLy(x)−Lz(x) is an affine function, we obtain

|∇Ly(x)− ∇Lz(x)| ≤ 1

ηrkLy(x)−Lz(x)k_L∞(B_ηr(p)). It follows that

|∇L_y(x)− ∇L_z(x)| ≤ 2C η

r^αˆ+ω(Λr) +r^αˆ Z r0

r

ω(s) s^{1+ ˆα}ds+

Z Λr

0

ω(s) s ds

. Hence, fory, z ∈∂Ω∩B_r0, 0<|y−z|=r≤ ^r_Λ⁰, we have

|∇u(y)− ∇u(z)| ≤ 2C η

r^αˆ+ω(Λr) +r^αˆ Z r0

r

ω(s) s^{1+ ˆα}ds+

Z Λr

0

ω(s) s ds

.

This completes the proof.

Acknowledgements. The authors would like to thank the anonymous referee for the carefully reading and for the useful comments and suggestions. This work was supported by NSFC 11401460. This work was done while Y. Huang was visiting School of Mathematical Sciences, Peking University. He would like to thank Peking University for the hospitality.

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Yongpan Huang

Department of Mathematics, Xi’an Polytechnic University, Xi’an 710048, China.

School of Mathematics and Statistics, Xi’an Jiaotong University, Xi’an 710049, China Email address:[email protected], [email protected]

Qiaozhu Zhai

Systems Engineering Institute, Xi’an Jiaotong University, Xi’an 710049, China Email address:[email protected]

Shulin Zhou

School of Mathematical Sciences, Peking University, Beijing, 100871, China Email address:[email protected]