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Slope estimate and boundary differentiability for inhomogeneous infinity Laplace equation on convex domains
Nonlinear Analysis 176 (2018) 36–47
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Nonlinear Analysis www.elsevier.com/locate/na
Slope estimate and boundary differentiability for inhomogeneous infinity Laplace equation on convex domains Xiaomeng Feng, Guanghao Hong* School of Mathematics and Statistics, Xi’an Jiaotong University, Xi’an, 710049, PR China
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Article history: Received 25 January 2018 Accepted 4 June 2018 Communicated by Enzo Mitidieri MSC: primary 35J25 35J70 secondary 49N60
abstract We study the boundary differentiability for inhomogeneous infinity Laplace equations on convex domains Ω with the inhomogeneous term f ∈ C(Ω ) ∩ L∞ (Ω ) and differentiable boundary data g. At a flat point (the boundary point where the blow up of the domain is the half-space), u is differentiable due to a previous result of the second author in Hong (2014). At a corner point (the boundary point where the blow up of the domain is not the half-space), we establish a slope estimate for u, and provide a necessary and sufficient condition for the boundary differentiability of u in this paper. © 2018 Elsevier Ltd. All rights reserved.
Keywords: Boundary differentiability Infinity Laplacian Convex domain Blow up
1. Introduction The infinity Laplace equation △∞ u(x) :=
∑
uxi uxj uxi xj = 0
1≤i,j≤n
was introduced by G. Aronsson in the 1960s [1] as the Euler–Lagrange equation of the sup-norm variational problem of |▽u| or the equivalent absolutely minimizing Lipschitz extension (AMLe) problem. It is a highly degenerate elliptic partial differential equation. In 1993, R. Jensen [13] proved that the Dirichlet problem: △∞ u = 0 in Ω , u = g on ∂Ω has a unique viscosity solution for any bounded domain Ω ⊂ Rn and g ∈ C(∂Ω ). Such a solution u is called an infinity harmonic function. He also proved that a function u ∈ C(Ω ) is an AMLe if and only if u is a
*
Corresponding author. E-mail addresses: [email protected] (X. Feng), [email protected] (G. Hong).
https://doi.org/10.1016/j.na.2018.06.003 0362-546X/© 2018 Elsevier Ltd. All rights reserved.
X. Feng, G. Hong / Nonlinear Analysis 176 (2018) 36–47
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viscosity solution to △∞ u = 0. In 2001, Crandall–Evans–Gariepy [3] provided that u ∈ C(Ω ) is an infinity harmonic function if and only if u enjoys the comparison with cones property. The regularity problems are always the core issues of the elliptic pdes. Evans and Smart [7] proved the interior differentiability of u. This is the best result that has been achieved so far. For dimension 2, Savin [16] and Evans–Savin [6] proved the C 1 and C 1,α regularity earlier. The boundary regularity of u was initially studied by Wang–Yu [17]. They proved the boundary differentiability of u in general dimensions with the assumption that ∂Ω and g are C 1 . They also proved the C 1 boundary regularity of u for dimension 2 if ∂Ω and g are C 2 . The second author of this paper improved Wang–Yu’s first result to that if both ∂Ω and g are differentiable at a boundary point x0 ∈ ∂Ω then u is differentiable at x0 [8]. In another paper [10], Hong constructed a two dimensional counterexample to show that |Du| may not be continuous along the boundary if ∂Ω is merely assumed to be C 1 . In [12], Hong–Liu investigated the boundary regularity of u on convex domains. The boundary points of a convex domain are divided into flat points (where the blow up of the domain is the half-space) and corner points (where the blow up of the domain is not the half-space). The boundary of a convex domain is differentiable at a flat point, so u is differentiable at a flat point if g is differentiable at this point by the former result in [8]. The interesting case is the behavior of u at a corner point. Assume that g is differentiable at a corner point x0 . In general, the solution u need not be differentiable at x0 (see Example 1 in [12]). They proved that the slope function (defined later) S(x0 ) ≤ |Dg(x0 )| and provided a sufficient and necessary condition for the establishment of differentiability of u. Our work in this article is to prove that these two conclusions are also true for the solutions of the inhomogeneous infinity Laplace equation with continuous and bounded right hand side. Let Ω ⊂ Rn be a bounded and connected open set. The Dirichlet problem for inhomogeneous infinity Laplace equation { △∞ u = f, in Ω (1) u = g, on ∂Ω was introduced by Lu–Wang [15]. They achieved the existence and uniqueness of the viscosity solution of (1) under the conditions that f ∈ C(Ω ) with inf Ω f > 0 or supΩ f < 0 and g ∈ C(∂Ω ). They also gave the comparison principle and the comparison with standard functions property. These properties were applied to prove the stability of the inhomogeneous infinity Laplace equation with nonvanishing f . E. Lindgren [14] proved that at an interior point the blow ups of u are linear if f ∈ C(Ω ) and u is differentiable if f ∈ C 1 (Ω ). If f ≡ −1, g ≡ 0, Ω is convex and satisfies an interior sphere condition, Crasta– Fragal` a [4] achieved the solution u is power-concave and of class C 1 , and proved the expected optimal regularity of u is C 1,1/3 . For Serrin-type overdetermined boundary value problems in [5], C 1 regularity continues to hold without the interior sphere condition. In the normalized case with the operator △N ∞ [5], n they also showed that stadium-like domains are precisely the unique convex sets in R where the solution to a Dirichlet problem is of class C 1,1 (Ω ). Now we assume that u ∈ C(Ω ) is the viscosity solution of (1). Define the slope functions as in [17], for ¯, x∈Ω Sr+ (x) =
Sr− (x) =
sup y∈∂(Brn (x)∩Ω)\{x}
u(y) − u(x) , |y − x|
u(x) − u(y) , |y − x| y∈∂(Brn (x)∩Ω)\{x} sup
and Sr (x) := max{Sr+ (x), Sr− (x)}, with Brn (x) := {y ∈ Rn : |y − x| < r} for x ∈ Rn , Brn := Brn (0).
X. Feng, G. Hong / Nonlinear Analysis 176 (2018) 36–47
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From now on, we assume that Ω is convex. Without loss of generality, we assume that 0 is a boundary point of Ω under study. The blow up of Ω at 0 always uniquely exists and is a convex cone. We denote ⋃ Ω0t . Ω0t := {ty : y ∈ Ω } and Ω0∞ := t>0
If Ω0∞ = {xn > 0} under some coordinate system, 0 is called a flat point. Otherwise, 0 is called a corner point and in this case one can prove that there exists δ > 0, such that Ω0∞ ⊂ {xn > δ|xn−1 |} under some coordinate system. If 0 ∈ ∂Ω is a flat point, it is easy to verify that ∂Ω is differentiable at 0. In this case, Hong [9] gave a conclusion. Proposition 1. Let u ∈ C(Ω ) be a viscosity solution of the problem (1) under the conditions that f ∈ C(Ω ) ∩ L∞ (Ω ) and g is differentiable at 0. Assume 0 is a flat point. Then u is differentiable at 0. If 0 is a corner point, u is not necessarily differentiable at 0. We denote C := Ω0∞ and Γ := ∂C. For the inhomogeneous case, the slope function Sr± (x) is not necessarily monotone. So we define S ± (x) := lim Sr± (x) r→0
and S(x) := max{S + (x), S − (x)}. The conclusions of homogeneous case can be extended to the inhomogeneous case. Both slope estimate and the analysis of the differentiability are valid in the inhomogeneous case. In this paper, we prove the following conclusions. In this paper, we prove the following conclusions. Theorem 1. Let u ∈ C(Ω ) be a viscosity solution of the problem (1) under the conditions that f ∈ C(Ω ) ∩ L∞ (Ω ) and g is differentiable at 0. Assume 0 is a corner point. Then S(0) ≤ |Dg(0)|. Theorem 2. Under the conditions of Theorem 1, (i) if Dg(0) ̸∈ C ∪ −C, then u is differentiable at 0; (ii) if Dg(0) ∈ C ∪ −C, then u is differentiable at 0 if and only if S(0) = |Dg(0)|. The strategy that we prove Theorems 1 and 2 is that we first prove the conclusions with two additional assumptions and then we drop the additional assumptions to give the proof of Theorems 1 and 2. These two steps are completed in Sections 2 and 3 separately. 2. Analysis with additional assumptions In this section, we prove the conclusions with two additional assumptions. Theorem 3. Under the conditions of Theorem 1, if the problem (1) also satisfies the assumptions: (A1) Sr± (x) ≥ 1 for all x and r; (A2) − 43 ≤ f ≤ − 14 . Then S(0) ≤ |Dg(0)|. Theorem 4. Under the conditions of Theorem 1, if the problem (1) also satisfies the assumptions (A1) and (A2), then (i) if Dg(0) ̸∈ C ∪ −C, then u is differentiable at 0; (ii) if Dg(0) ∈ C ∪ −C, then u is differentiable at 0 if and only if S(0) = |Dg(0)|.
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Without loss of generality, we assume u(0) = g(0) = 0. We also assume that the conditions in Theorems 3 and 4 are satisfied throughout this section. In this situation, E. Lindgren [14] proved that Sr± (x) + r is monotone in r for x ∈ Ω . After that, Hong [9] proved that Sr± (x) + r is monotone in r for x ∈ Ω . ¯ ∩ Brn , Sr± (x) + r is non-decreasing for 0 < r < r0 . So the limit Proposition 2 ([9]). For a fixed x ∈ Ω 0 ± ± S (x) := limr→0 (Sr (x) + r) exists, and we denote S(x) := max{S + (x), S − (x)}. In order to make the proof clearer, we suppose g is C 1 near 0 and drop it in the end of proof. g is C 1 near 0, i.e. there exists g˜ ∈ C(Rn ) ∩ C 1 (Brn0 ) such that g = g˜|∂Ω . Lemma 1. Suppose g is C 1 near 0. Then for any ε > 0, there exists 0 < r < r0 such that sup
S(y) ≤ max{S(0), |Dg(0)|} + ϵ.
n ¯ y∈Ω∩B r
Proof . For ϵ > 0, since g˜ ∈ C 1 (Brn0 ), there exists 0 < r1 < r0 such that ϵ sup |D˜ g (x)| ≤ |Dg(0)| + 3 x∈Brn 1
and sup x̸=y∈∂Ω∩Brn 1
|u(x) − u(y)| ≤ |x − y|
sup x̸=y∈∂Ω∩Brn 1
|˜ g (x) − g˜(y)| |x − y|
ϵ ≤ sup |D˜ g (x)| ≤ |Dg(0)| + . n 3 x∈Br
(2)
1
Since limr→0 Sr (0) + r = S(0), there exists 0 < r2 ≤ min( r21 , 3ϵ ), such that ϵ Sr2 (0) + r2 ≤ S(0) + . 3 From the continuity of u, there exists 0 < r3 ≪ r2 , such that sup x∈∂Brn (y)∩Ω 2
|u(x) − u(y)| ϵ ≤ Sr2 (0) + r2 ≤ S(0) + r2 3
for
y ∈ ∂Ω ∩ Brn3 .
(3)
Combining (2) and (3), we have |u(x) − u(y)| |x − y| x∈∂(Brn (y)∩Ω)\{y} 2 ϵ ≤ max{S(0), |Dg(0)|} + for y ∈ ∂Ω ∩ Brn3 . 3
Sr2 (y) ≤
sup
By Proposition 2, we have |u(x) − u(y)| 2ϵ ≤ Sr2 (y) + r2 ≤ max{S(0), |Dg(0)|} + |x − y| 3 for x ∈ Ω ∩ Brn3 and y ∈ ∂Ω ∩ Brn3 . From the continuity of u again, there exists 0 < r4 ≤ |u(x) − u(y)|
sup y∈∂B n (x)∩Ω r3 /2
r3 2
≤ max{S(0), |Dg(0)|} +
2ϵ 3
(4) r3 2 ,
such that
¯ ∩ Brn . for x ∈ Ω 4
Combining (4), (5) and Proposition 2, we have S(x) ≤ S r3 (x) + 2
r3 |u(x) − u(y)| r3 = sup + ≤ max{S(0), |Dg(0)|} + ϵ n 2 |x − y| 2 y∈∂(B (x)∩Ω)\{x} r3 /2
¯∩ for x ∈ Ω
Brn4 .
Finally we choose r(ϵ, u) = r4 . □
(5)
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X. Feng, G. Hong / Nonlinear Analysis 176 (2018) 36–47
Now we employ the standard blow up argument at 0 as in [14]. We say that w is a blow-up of u at 0, if for some subsequence λm → 0 such that w(x) = lim um (x) λm →0
where um (x) =
u(λm x) . λm
In order to prove u is differential at 0, it suffices to prove that there is a unique linear blow up at 0 as in [8] and [14], i.e. to prove w(x) = a · x = Dg(0) · x at the corner point 0 where · is the inner product. According to the conclusions in [2], to prove the blow up at 0 is linear, we shall prove a subsequence of {um } converges locally uniformly firstly. In the following Lemmas 2 and 3, we also suppose g is C 1 near 0. Lemma 2. For any sequence of λm → 0, set Ω0m = λ−1 m Ω and um (x) =
u(λm x) , x ∈ Ω0m . λm
Then a subsequence of {um } converges to an infinity harmonic function w ∈ C(C) locally uniformly. Proof . Since Lemma 1, lim S(y) ≤ max{S(0), |Dg(0)|},
y→0
then Lip(u(x), BR ∩ Ω ) ≤ max{S(0), |Dg(0)|} + 1 for some R > 0. The function um (x) satisfies the equation △∞ um (x) = λm f (λm x), x ∈ Ω0m . Fix λ > 0. For 0 < λm < λ, the Lipschitz constant for um (x) on B Lip(um (x), B
R λm
R λm
∩ Ω0m :
∩ Ω0m ) = Lip(u(x), BR ∩ Ω ),
then Lip(um (x), B R ∩ Ω0m ) ≤ max{S(0), |Dg(0)|} + 1 λ
and |um (x)| ≤
R (max{S(0), |Dg(0)|} + 1). λ
By using Arzela–Ascoli Theorem, there exists a subsequence of {um } that converges to a function w ∈ C(C) locally uniformly and w satisfies △∞ w = 0 in C in the viscosity sense. □
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Lemma 3. Let w be a blow up of u at 0 as above. Then w has the following properties: ¯ (1) Lip(w, C) ≤ max{S(0), |Dg(0)|}, w ∈ C(C); (2) w(y) = Dg(0) · y for all y ∈ Γ ; (3) If S(0) = S + (0), then there exists e ∈ C¯ ∩ S n−1 , such that w(te) = S(0)t for t ∈ [0, +∞); if S(0) = S − (0), then there exists e ∈ C¯ ∩ S n−1 , such that w(te) = −S(0)t for t ∈ [0, +∞). Proof . The proof is totally the same as the proof of Lemma 4 in [12]. From Lemma 2, (1) is easy to get. k k For any y ∈ Γ \ {0}, let yk → y, yk ∈ C. Then choose zm ∈ ∂Ω satisfying zm = τm λm yk + (1 − τm )λm y with τm ∈ [0, 1). Then using Lemma 1, i.e. the Lipschitz continuity of u, and the continuity of Dg near 0, we can get (2). Suppose S(0) = S + (0). Using the definition of Sr+ (0), we can get the direction er ∈ C ∩ S n−1 which is maximum slope direction from 0 to ∂(Brn ∩ Ω ). By the compactness of C ∩ S n−1 , we have a subsequence erj → e as rj → 0. Then we can check that w(te) = S(0)t for t ∈ [0, +∞) by using the comparison with cones property. □ Proof of Theorem 3. Suppose g is C 1 near 0. The proof of the slope estimate is the same as Theorem 2 in [12]. We only describe it briefly and refer the reader to [12] for the details. We prove S(0) ≤ |Dg(0)| by contradiction. Assume |Dg(0)| < S(0). Without loss of generality, we suppose S(0) = S + (0). Using (2) and (3) of Lemma 3 and the key fact that 0 is a corner point, we can manage to find a point P ∈ Γ such that |w(e) − w(P )| > S(0). |e − P | It is a contradiction to (1) of Lemma 3. Next, we drop the C 1 assumption on g. Let u± be constructed as in the proof of Theorem 3 in [12]. Find two functions g ± ∈ C 1 near 0 satisfying g − ≤ g ≤ g + and Dg ± (0) = Dg(0). Let u± solve the problem { △∞ u± = f, in Br ∩ C u± = g ± , on ∂(Br ∩ C). From the above proof, S[u± ](0) ≤ |Dg ± (0)| = |Dg(0)|. By comparison principle, u− ≤ u ≤ u+ in Br ∩ C. So S + (0) ≤ S + [u+ ](0) and S − (0) ≤ S − [u− ](0). Therefore, S(0) ≤ max{S + [u+ ](0), S − [u− ](0)} ≤ max{S[u+ ](0), S[u− ](0)} ≤ |Dg(0)|.
□
Lemma 4. Suppose g is C 1 near 0. ¯ then (1) If Dg(0) ∈ C¯ ∪ −C, lim sup S(y) ≤ S(0); y→0
(2) Dg(0) ∈ C implies S(0) = S + (0); Dg(0) ∈ −C implies S(0) = S − (0). Proof . The proof of (1) is essentially the same as the proof of Lemma 3 in [12]. If |Dg(0)| ≤ S(0), applying Dg(0) Dg(0) (or − |Dg(0)| ) ∈ C and the Lemma 1, it is clear to get. Suppose |Dg(0)| > S(0). Let the direction ν = |Dg(0)| angle θ = dS n−1 (ν, Γ ∩ ∂B1 ). We consider the boundary value g at the corner point 0, then we have cos θ|Dg(0)| ≤ S(0)
and θ > 0.
Denote C τ = {x ∈ C : dS n−1 (
x , Γ ∩ ∂B1 ) > τ }. |x|
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Then we can prove that for any τ > 0, there exists r(τ ) > 0 such that C τ ∩ Br(τ ) ⊂ Ω ⊂ C. Fix ϵ > 0, and let τ > 0 be such that cos(θ − 3τ )|Dg(0)| ≤ cos θ|Dg(0)| + 2ϵ . Applying the differentiability ) g (x) g (x)| ≤ |Dg(0)| + 2ϵ and dS n−1 ( D˜ of g near 0, let 0 < r1 < τ r(τ be such that supx∈Br |D˜ , Dg(0) ) < τ . 4 1 |D˜ g (x)| |Dg(0)| Then for any x ∈ ∂Ω ∩ B r1 , we have 2
sup y∈∂Ω∩B n
r1 /2
ϵ) |g(x) − g(y)| ( ≤ |Dg(0)| + cos(θ − 3τ ) ≤ cos θ|Dg(0)| + ϵ. |x − y| 2 (x)
The rest of the proof of (1) is similar to the proof of Lemma 1 in this paper. The proof of (2) is the same as Lemma 5 in [12]. We prove it by contradiction. If Dg(0) ∈ C and ¯ ≤ S(0) applying (1) in S(0) > S + (0). Then S(0) = S − (0) > 0. Let w be as in Lemma 2, then Lip(w, C) this lemma. From Lemma 3(3), there exists e ∈ C¯ ∩ S n−1 , such that w(te) = −S(0)t for all t ∈ [0, +∞). As in the proof of Theorem 3, we will make contradiction by finding a point P ∈ Γ such that |w(e) − w(P )| > S(0). |e − P |
□
Proof of Theorem 4. Suppose g is C 1 near 0. As S(0) ≥ 1, |Dg(0)| > 0. Let w be as in Lemma 2. Applying Lemma 3, we have w|Γ (y) = Dg(0) · y, and ¯ ≤ max{S(0), |Dg(0)|} = |Dg(0)|. Lip(w, C)
(6)
The rest can be proved in the same way as Theorem 4 in [12]. Without loss of generality, Dg(0) = en . If Dg(0) ̸∈ C ∪ −C, there exist two points in {x + ten } ∩ Γ with different sign of t for any x ∈ C. Using (6) on the points x and x + ten , we get w(x) ≤ xn
and w(x) ≥ xn .
Then w(x) = Dg(0) · x = xn
for all
x ∈ C.
If Dg(0) ∈ C ∪ −C, the necessity is evident. Suppose S(0) = |Dg(0)|. Note that e = en (or e = −en ) and there exists only one point in {x + ten } ∩ Γ . We use the points x and x + ten ∈ Γ to prove only one inequality between w(x) and xn as the above case. The other one can be achieved by using (6) on the points x and te with S(0) = |Dg(0)| = 1. Now we drop the C 1 assumption on g. Let u± be constructed as in the proof of Theorem 3. By comparison principle, u− ≤ u ≤ u+ in Br ∩ C. If u± are differentiable at 0, then u is also differentiable at 0. Theorem 4 is established. □
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3. Proof of Theorems 1 and 2 Now we drop the assumptions (A1) and (A2) in this section. These assumptions are not restrictions since we can extend u as in [14] and [9]. Define 4
U (X) = αu(x) + xn+1 − β|xn+2 | 3 1
with X = (x, xn+1 , xn+2 ) ∈ Rn+2 , α =
1/3 41/3 ∥f ∥ ∞ L (Ω)
and β =
34/3 . 21/3 ·4
By the result in [11], the function
U (X) is a viscosity solution of the problem △∞ U (X) = F (X) in Σ , U (X) = G(X) on ∂Σ , where Σ = Ω × R2 , F (X) = C
f (x) 4∥f ∥L∞ (Ω)
−
1 2
(7) 4 3
and G(X) = αg(x) + xn+1 − β|xn+2 | . The regularity up to
1 1, 3
on U in general dimension also holds for u. The origin 0 ∈ ∂Σ is also a corner point of Σ , F (X) ∈ C(Σ ) ∩ L∞ (Σ ) and G(X) is differentiable at 0 ¯ ∩ B n+2 and 0 < r < r0 , we define with G(0) = U (0) = 0. For X ∈ Σ r0 Sr+ [U ](X) =
sup Y ∈∂(Brn+2 (X)∩Σ )\{X}
U (Y ) − U (X) |Y − X|
and Sr− [U ](X) =
U (X) − U (Y ) . |Y − X| Y ∈∂(B n+2 (X)∩Σ )\{X} sup
r
It is easy to check that the problem (7) satisfies the assumptions (A1) and (A2): (A1) Sr± [U ](X) ≥ 1 for all X and r; (A2) − 43 ≤ F ≤ − 14 . Thus, we define S ± [U ](X) = lim Sr± [U ](X), r→0
and S[U ](X) = max{S + [U ](X), S − [U ](X)}. Then we have the conclusions of Theorems 3 and 4 on U with K := C × R2 : S[U ](0) ≤ |DG(0)|; and (i) If DG(0) ̸∈ K ∪ −K, then U is differentiable at 0; (ii) If DG(0) ∈ K ∪ −K, then U is differentiable at 0 if and only if S[U ](0) = |D(0)|. Since u does not satisfy the conditions (A1) and (A2), we only have S ± (x) = lim Sr± (x) r→0
and S(x) = max{S + (x), S − (x)}. We compare S(0) with S[U ](0).
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Lemma 5. S[U ](0) =
√ α2 S(0)2 + 1.
Proof . Firstly, we prove S[U ](0) ≤
√
α2 S(0)2 + 1. Without loss of generality, we suppose that
S[U ](0) = S + [U ](0) = lim Sr+ [U ](0), S + [U ](0) ≥ S − [U ](0). r→0
For any r > 0, by the definition of Sr+ [U ](0), we consider our argument in the following two cases. Case a: There exists Xr ∈ ∂(Brn+2 ∩ Σ ), such that Sr+ [U ](0) =
U (Xr ) ; |Xr |
Case b: There exists a sequence {Xrk } ⊂ ∂(Brn+2 ∩ Σ ), Xrk → 0, as k → ∞, such that U (Xrk ) . k→∞ |Xrk |
Sr+ [U ](0) = lim For Case a, we have
U (Xr ) |Xr | xr xr,n+1 xr,n+2 u(x)r xr −2 =( , , ) · (α , 1, −βxr,n+2 |xr,n+2 | 3 ) |Xr | |Xr | |Xr | |xr | |xr | √ 2 ≤ α2 Sρ+r (0)2 + 1 + β 2 |xr,n+2 | 3
Sr+ [U ](0) =
where Xr = (xr , xr,n+1 , xr,n+2 ) and ρr = |xr |. For Case b, we have U (Xrk ) k→∞ |Xrk | xk xkr,n+1 xkr,n+2 u(xkr ) xkr −2 = lim ( rk , , ) · (α , 1, −βxkr,n+2 |xkr,n+2 | 3 ) k k k k k→∞ |Xr | |Xr | |Xr | |xr | |xr | √
Sr+ [U ](0) = lim
2
≤ lim α2 Sρ+k (0)2 + 1 + β 2 |xkr,n+2 | 3 r k→∞ √ ≤ α2 S + (0)2 + 1 where Xr = (xkr , xkr,n+1 , xkr,n+2 ) and ρkr = |xkr |. Then, we have S + [U ](0) = lim Sr+ [U ](0) ≤ r→0
√ α2 S + (0)2 + 1.
√ Next, we give the proof of S[U ](0) ≥ α2 S(0)2 + 1. If S(0) = 0, it is easy to verify that S[U ](0) ≥ 1. Assume S(0) > 0 and S(0) = S + (0) ≥ S − (0). Note that S + (0) = lim Sr+ (0). r→0
Hence, for any 0 < ε <
S(0) 2 ,
there exists a sequence {ρj } with ρj → 0, such that S + (0) −
ε ε ⩽ Sρ+j (0) ⩽ S + (0) + 2 2
for small enough {ρj }. For any ρj > 0, we consider the argument in two cases.
X. Feng, G. Hong / Nonlinear Analysis 176 (2018) 36–47
Case a’: There exists xρj ∈ ∂(Bρnj ∩ Ω ), such that Sρ+j (0) =
u(xρj ) , |xρj |
then S + (0) − ε ⩽
u(xρj ) ⩽ S + (0) + ε; |xρj |
Case b’: There exists a sequence {xlρj } ⊂ ∂(Bρnj ∩ Ω ), xlρj → 0, as l → ∞, such that Sρ+j (0) = lim
l→∞
u(xlρj ) |xlρj |
i.e. for large enough l, Sρ+j (0) −
u(xlρj ) ε ε ⩽ ⩽ Sρ+j (0) + , l 2 |xρj | 2
then S + (0) − ε ⩽
u(xlρj ) |xlρj |
⩽ S + (0) + ε.
Since S(0) > 0 and u(0) = 0, then u(xρj ), u(xlρj ) ̸= 0. For Case a’, we choose ( Xρj =
) 2 |xρj | ,0 . xρj , αu(xρj )
Take Rρj = |Xρj |, then + SR [U ](0) ρj
U (Xρj ) ≥ = |Xρj |
(
) ( ) 2 xρj |xρj | u(xρj ) xρj , ,0 · α , 1, 0 . |Xρj | αu(xρj )|Xρj | |xρj | |xρj |
Note that ⏐( )⏐ 2 ⏐ xρj ⏐ |xρj | ⏐ ⏐ ⏐ |Xρ | , αu(xρ )|Xρ | , 0 ⏐ = 1 j j j and (
where t = |(α
) ( ) 2 xρ j |xρj | u(xρj ) xρj , ,0 = t α , 1, 0 |Xρj | αu(xρj )|Xρj | |xρj | |xρj |
−1 u(xρj ) xρj . |xρj | |xρj | , 1, 0)|
Then
⏐( )⏐ √ √ ⏐ ⏐ u(x ) x u(xρj ) ρ ρ j j + 2+1≥ ⏐ α ⏐ = α2 ( SR , 1, 0 ) α2 (S + (0) − ε)2 + 1. [U ](0) ≥ ⏐ ⏐ ρj |xρ | |xρ | |xρ | j
j
j
45
X. Feng, G. Hong / Nonlinear Analysis 176 (2018) 36–47
46
Taking ρj → 0, Rρj → 0. It follows that S + [U ](0) ≥
√
α2 (S + (0) − ε)2 + 1.
In Case b’, using the same argument in Case a’ to analysis xlρj , we obtain √ + SR [U ](0) ≥ α2 (S + (0) − ε)2 + 1. l ρ j
Taking l → ∞, xlρj → 0 and Rρl → 0. It follows that j
S + [U ](0) ≥
√
α2 (S + (0) − ε)2 + 1.
Therefore, take ε → 0, we obtain S + [U ](0) ≥
√ α2 S + (0)2 + 1.
If S(0) = S − (0) ≥ S + (0), using the similar argument, we have √ S − [U ](0) ≥ α2 S − (0)2 + 1. The lemma follows. □ Remark 1. From the proof of Lemma 5, one can easily see that if S[U ](0) = S + [U ](0), then √ S(0) = S + (0) and S + [U ](0) = α2 S + (0)2 + 1; if S[U ](0) = S − [U ](0), then S(0) = S − (0) and S − [U ](0) =
√ α2 S − (0)2 + 1.
Proof of Theorem 1. Note that S[U ](0) ≤ |DG(0)|, S[U ](0) =
√
α2 S(0)2
√ 2 + 1 and |DG(0)| = α2 |Dg(0)| + 1.
Since S(0) is nonnegative, then we get the slope estimate S(0) ≤ |Dg(0)|.
□
Proof of Theorem 2. Case (i) Assume that Dg(0) ̸∈ C ∪ −C. Note that DG(0) = (αDg(0), 1, 0) and K = C × R2 . Then DG(0) ̸∈ K ∪ −K. Applying (i) in Theorem 4, we get that U is differentiable at 0. Since 4 U (X) = αu(x) + xn+1 − β|xn+2 | 3 , we obtain that u is differentiable at 0. Case (ii) Assume that Dg(0) ∈ C ∪ −C. If u is differentiable at 0, it is clear that S(0) = |Dg(0)|. Suppose S(0) = |Dg(0)|. Since √ √ 2 2 2 S[U ](0) = α S(0) + 1 and |DG(0)| = α2 |Dg(0)| + 1, then S[U ](0) = |DG(0)|. Applying Theorem 4 (ii), U is differentiable at 0. It follows that u is also differentiable at 0. □
X. Feng, G. Hong / Nonlinear Analysis 176 (2018) 36–47
47
Acknowledgments This work is supported by the National Nature Science Foundation of China: NSFC 11301411 and 11671316. We would like to thank the referees for their careful reading and helpful comments and suggestions which improved the presentation of the paper. References
[1] G. Aronsson, Extension of functions satisfying Lipschitz conditions, Ark. Mat. 6 (1967) 551–561. [2] M.G. Crandall, L.C. Evans, A remark on infinity harmonic functions, USA-Chile Workshop on Nonlinear Analysis, Electron. J. Differ. Equ. Conf. 06 (2001) 123–129. [3] M.G. Crandall, L.C. Evans, R.F. Gariepy, Optimal Lipschitz extensions and the infinity Laplacian, Calc. Var. Partial Differential Equations 13 (2) (2001) 123–139. [4] G. Crasta, I. Fragal` a, On the Dirichlet and Serrin problems for the inhomogeneous infinity Laplacian in convex domains: Regularity and geometric results, Arch. Ration. Mech. Anal. 218 (2015) 1577–1607. [5] G. Crasta, I. Fragal` a, Characterization of stasiums-like domains via boundary value problems for the infinity Laplacian, Nonlinear Anal. TMA 133 (2016) 228–249. [6] L.C. Evans, O. Savin, C 1,α regularity for infinity harmonic functions in two dimensions, Calc. Var. Partial Differential Equations 32 (3) (2008) 325–347. [7] L.C. Evans, C.K. Smart, Everywhere differentiability of infinity harmonic functions, Calc. Var. Partial Differential Equations 42 (1–2) (2011) 289–299. [8] G. Hong, Boundary differentiability of infinity harmonic functions, Nonlinear Anal. TMA 93 (2013) 15–20. [9] G. Hong, Boundary differentiability for inhomogeneous infinity Laplace equations, Electron. J. Differential Equations 72 (2014) 1–6. [10] G. Hong, Counterexample to C 1 boundary regularity of infinity harmonic functions, Nonlinear Anal. TMA 104 (2014) 120–123. [11] G. Hong, X. Feng, Superposition principle on the viscosity solutions of infinity Laplace equations, Nonlinear Anal. TMA 171 (2018) 32–40. [12] G. Hong, D. Liu, Slope estimate and boundary differentiability of infinity harmonic functions on convex domains, Nonlinear Anal. TMA 139 (2016) 158–168. [13] R. Jensen, Uniqueness of Lipschitz extensions minimizing the sup-norm of the gradient, Arch. Ration. Mech. Anal. 123 (1993) 51–74. [14] E. Lindgren, On the regularity of solutions of the inhomogeneous infinity Laplace equation, Proc. Amer. Math. Soc. 142 (1) (2014) 277–288. [15] G. Lu, P. Wang, Inhomogeneous infinity Laplace equation, Adv. Math. 217 (4) (2008) 1838–1868. [16] O. Savin, C 1 regularity for infinity harmonic functions in two dimensions, Arch. Ration. Mech. Anal. 176 (3) (2005) 351–361. [17] C.Y. Wang, Y.F. Yu, C 1 -boundary regularity of planar infinity harmonic functions, Math. Res. Lett. 19 (4) (2012) 823–835.
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Nonlinear Analysis www.elsevier.com/locate/na
Slope estimate and boundary differentiability for inhomogeneous infinity Laplace equation on convex domains Xiaomeng Feng, Guanghao Hong* School of Mathematics and Statistics, Xi’an Jiaotong University, Xi’an, 710049, PR China
article
info
Article history: Received 25 January 2018 Accepted 4 June 2018 Communicated by Enzo Mitidieri MSC: primary 35J25 35J70 secondary 49N60
abstract We study the boundary differentiability for inhomogeneous infinity Laplace equations on convex domains Ω with the inhomogeneous term f ∈ C(Ω ) ∩ L∞ (Ω ) and differentiable boundary data g. At a flat point (the boundary point where the blow up of the domain is the half-space), u is differentiable due to a previous result of the second author in Hong (2014). At a corner point (the boundary point where the blow up of the domain is not the half-space), we establish a slope estimate for u, and provide a necessary and sufficient condition for the boundary differentiability of u in this paper. © 2018 Elsevier Ltd. All rights reserved.
Keywords: Boundary differentiability Infinity Laplacian Convex domain Blow up
1. Introduction The infinity Laplace equation △∞ u(x) :=
∑
uxi uxj uxi xj = 0
1≤i,j≤n
was introduced by G. Aronsson in the 1960s [1] as the Euler–Lagrange equation of the sup-norm variational problem of |▽u| or the equivalent absolutely minimizing Lipschitz extension (AMLe) problem. It is a highly degenerate elliptic partial differential equation. In 1993, R. Jensen [13] proved that the Dirichlet problem: △∞ u = 0 in Ω , u = g on ∂Ω has a unique viscosity solution for any bounded domain Ω ⊂ Rn and g ∈ C(∂Ω ). Such a solution u is called an infinity harmonic function. He also proved that a function u ∈ C(Ω ) is an AMLe if and only if u is a
*
Corresponding author. E-mail addresses: [email protected] (X. Feng), [email protected] (G. Hong).
https://doi.org/10.1016/j.na.2018.06.003 0362-546X/© 2018 Elsevier Ltd. All rights reserved.
X. Feng, G. Hong / Nonlinear Analysis 176 (2018) 36–47
37
viscosity solution to △∞ u = 0. In 2001, Crandall–Evans–Gariepy [3] provided that u ∈ C(Ω ) is an infinity harmonic function if and only if u enjoys the comparison with cones property. The regularity problems are always the core issues of the elliptic pdes. Evans and Smart [7] proved the interior differentiability of u. This is the best result that has been achieved so far. For dimension 2, Savin [16] and Evans–Savin [6] proved the C 1 and C 1,α regularity earlier. The boundary regularity of u was initially studied by Wang–Yu [17]. They proved the boundary differentiability of u in general dimensions with the assumption that ∂Ω and g are C 1 . They also proved the C 1 boundary regularity of u for dimension 2 if ∂Ω and g are C 2 . The second author of this paper improved Wang–Yu’s first result to that if both ∂Ω and g are differentiable at a boundary point x0 ∈ ∂Ω then u is differentiable at x0 [8]. In another paper [10], Hong constructed a two dimensional counterexample to show that |Du| may not be continuous along the boundary if ∂Ω is merely assumed to be C 1 . In [12], Hong–Liu investigated the boundary regularity of u on convex domains. The boundary points of a convex domain are divided into flat points (where the blow up of the domain is the half-space) and corner points (where the blow up of the domain is not the half-space). The boundary of a convex domain is differentiable at a flat point, so u is differentiable at a flat point if g is differentiable at this point by the former result in [8]. The interesting case is the behavior of u at a corner point. Assume that g is differentiable at a corner point x0 . In general, the solution u need not be differentiable at x0 (see Example 1 in [12]). They proved that the slope function (defined later) S(x0 ) ≤ |Dg(x0 )| and provided a sufficient and necessary condition for the establishment of differentiability of u. Our work in this article is to prove that these two conclusions are also true for the solutions of the inhomogeneous infinity Laplace equation with continuous and bounded right hand side. Let Ω ⊂ Rn be a bounded and connected open set. The Dirichlet problem for inhomogeneous infinity Laplace equation { △∞ u = f, in Ω (1) u = g, on ∂Ω was introduced by Lu–Wang [15]. They achieved the existence and uniqueness of the viscosity solution of (1) under the conditions that f ∈ C(Ω ) with inf Ω f > 0 or supΩ f < 0 and g ∈ C(∂Ω ). They also gave the comparison principle and the comparison with standard functions property. These properties were applied to prove the stability of the inhomogeneous infinity Laplace equation with nonvanishing f . E. Lindgren [14] proved that at an interior point the blow ups of u are linear if f ∈ C(Ω ) and u is differentiable if f ∈ C 1 (Ω ). If f ≡ −1, g ≡ 0, Ω is convex and satisfies an interior sphere condition, Crasta– Fragal` a [4] achieved the solution u is power-concave and of class C 1 , and proved the expected optimal regularity of u is C 1,1/3 . For Serrin-type overdetermined boundary value problems in [5], C 1 regularity continues to hold without the interior sphere condition. In the normalized case with the operator △N ∞ [5], n they also showed that stadium-like domains are precisely the unique convex sets in R where the solution to a Dirichlet problem is of class C 1,1 (Ω ). Now we assume that u ∈ C(Ω ) is the viscosity solution of (1). Define the slope functions as in [17], for ¯, x∈Ω Sr+ (x) =
Sr− (x) =
sup y∈∂(Brn (x)∩Ω)\{x}
u(y) − u(x) , |y − x|
u(x) − u(y) , |y − x| y∈∂(Brn (x)∩Ω)\{x} sup
and Sr (x) := max{Sr+ (x), Sr− (x)}, with Brn (x) := {y ∈ Rn : |y − x| < r} for x ∈ Rn , Brn := Brn (0).
X. Feng, G. Hong / Nonlinear Analysis 176 (2018) 36–47
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From now on, we assume that Ω is convex. Without loss of generality, we assume that 0 is a boundary point of Ω under study. The blow up of Ω at 0 always uniquely exists and is a convex cone. We denote ⋃ Ω0t . Ω0t := {ty : y ∈ Ω } and Ω0∞ := t>0
If Ω0∞ = {xn > 0} under some coordinate system, 0 is called a flat point. Otherwise, 0 is called a corner point and in this case one can prove that there exists δ > 0, such that Ω0∞ ⊂ {xn > δ|xn−1 |} under some coordinate system. If 0 ∈ ∂Ω is a flat point, it is easy to verify that ∂Ω is differentiable at 0. In this case, Hong [9] gave a conclusion. Proposition 1. Let u ∈ C(Ω ) be a viscosity solution of the problem (1) under the conditions that f ∈ C(Ω ) ∩ L∞ (Ω ) and g is differentiable at 0. Assume 0 is a flat point. Then u is differentiable at 0. If 0 is a corner point, u is not necessarily differentiable at 0. We denote C := Ω0∞ and Γ := ∂C. For the inhomogeneous case, the slope function Sr± (x) is not necessarily monotone. So we define S ± (x) := lim Sr± (x) r→0
and S(x) := max{S + (x), S − (x)}. The conclusions of homogeneous case can be extended to the inhomogeneous case. Both slope estimate and the analysis of the differentiability are valid in the inhomogeneous case. In this paper, we prove the following conclusions. In this paper, we prove the following conclusions. Theorem 1. Let u ∈ C(Ω ) be a viscosity solution of the problem (1) under the conditions that f ∈ C(Ω ) ∩ L∞ (Ω ) and g is differentiable at 0. Assume 0 is a corner point. Then S(0) ≤ |Dg(0)|. Theorem 2. Under the conditions of Theorem 1, (i) if Dg(0) ̸∈ C ∪ −C, then u is differentiable at 0; (ii) if Dg(0) ∈ C ∪ −C, then u is differentiable at 0 if and only if S(0) = |Dg(0)|. The strategy that we prove Theorems 1 and 2 is that we first prove the conclusions with two additional assumptions and then we drop the additional assumptions to give the proof of Theorems 1 and 2. These two steps are completed in Sections 2 and 3 separately. 2. Analysis with additional assumptions In this section, we prove the conclusions with two additional assumptions. Theorem 3. Under the conditions of Theorem 1, if the problem (1) also satisfies the assumptions: (A1) Sr± (x) ≥ 1 for all x and r; (A2) − 43 ≤ f ≤ − 14 . Then S(0) ≤ |Dg(0)|. Theorem 4. Under the conditions of Theorem 1, if the problem (1) also satisfies the assumptions (A1) and (A2), then (i) if Dg(0) ̸∈ C ∪ −C, then u is differentiable at 0; (ii) if Dg(0) ∈ C ∪ −C, then u is differentiable at 0 if and only if S(0) = |Dg(0)|.
X. Feng, G. Hong / Nonlinear Analysis 176 (2018) 36–47
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Without loss of generality, we assume u(0) = g(0) = 0. We also assume that the conditions in Theorems 3 and 4 are satisfied throughout this section. In this situation, E. Lindgren [14] proved that Sr± (x) + r is monotone in r for x ∈ Ω . After that, Hong [9] proved that Sr± (x) + r is monotone in r for x ∈ Ω . ¯ ∩ Brn , Sr± (x) + r is non-decreasing for 0 < r < r0 . So the limit Proposition 2 ([9]). For a fixed x ∈ Ω 0 ± ± S (x) := limr→0 (Sr (x) + r) exists, and we denote S(x) := max{S + (x), S − (x)}. In order to make the proof clearer, we suppose g is C 1 near 0 and drop it in the end of proof. g is C 1 near 0, i.e. there exists g˜ ∈ C(Rn ) ∩ C 1 (Brn0 ) such that g = g˜|∂Ω . Lemma 1. Suppose g is C 1 near 0. Then for any ε > 0, there exists 0 < r < r0 such that sup
S(y) ≤ max{S(0), |Dg(0)|} + ϵ.
n ¯ y∈Ω∩B r
Proof . For ϵ > 0, since g˜ ∈ C 1 (Brn0 ), there exists 0 < r1 < r0 such that ϵ sup |D˜ g (x)| ≤ |Dg(0)| + 3 x∈Brn 1
and sup x̸=y∈∂Ω∩Brn 1
|u(x) − u(y)| ≤ |x − y|
sup x̸=y∈∂Ω∩Brn 1
|˜ g (x) − g˜(y)| |x − y|
ϵ ≤ sup |D˜ g (x)| ≤ |Dg(0)| + . n 3 x∈Br
(2)
1
Since limr→0 Sr (0) + r = S(0), there exists 0 < r2 ≤ min( r21 , 3ϵ ), such that ϵ Sr2 (0) + r2 ≤ S(0) + . 3 From the continuity of u, there exists 0 < r3 ≪ r2 , such that sup x∈∂Brn (y)∩Ω 2
|u(x) − u(y)| ϵ ≤ Sr2 (0) + r2 ≤ S(0) + r2 3
for
y ∈ ∂Ω ∩ Brn3 .
(3)
Combining (2) and (3), we have |u(x) − u(y)| |x − y| x∈∂(Brn (y)∩Ω)\{y} 2 ϵ ≤ max{S(0), |Dg(0)|} + for y ∈ ∂Ω ∩ Brn3 . 3
Sr2 (y) ≤
sup
By Proposition 2, we have |u(x) − u(y)| 2ϵ ≤ Sr2 (y) + r2 ≤ max{S(0), |Dg(0)|} + |x − y| 3 for x ∈ Ω ∩ Brn3 and y ∈ ∂Ω ∩ Brn3 . From the continuity of u again, there exists 0 < r4 ≤ |u(x) − u(y)|
sup y∈∂B n (x)∩Ω r3 /2
r3 2
≤ max{S(0), |Dg(0)|} +
2ϵ 3
(4) r3 2 ,
such that
¯ ∩ Brn . for x ∈ Ω 4
Combining (4), (5) and Proposition 2, we have S(x) ≤ S r3 (x) + 2
r3 |u(x) − u(y)| r3 = sup + ≤ max{S(0), |Dg(0)|} + ϵ n 2 |x − y| 2 y∈∂(B (x)∩Ω)\{x} r3 /2
¯∩ for x ∈ Ω
Brn4 .
Finally we choose r(ϵ, u) = r4 . □
(5)
40
X. Feng, G. Hong / Nonlinear Analysis 176 (2018) 36–47
Now we employ the standard blow up argument at 0 as in [14]. We say that w is a blow-up of u at 0, if for some subsequence λm → 0 such that w(x) = lim um (x) λm →0
where um (x) =
u(λm x) . λm
In order to prove u is differential at 0, it suffices to prove that there is a unique linear blow up at 0 as in [8] and [14], i.e. to prove w(x) = a · x = Dg(0) · x at the corner point 0 where · is the inner product. According to the conclusions in [2], to prove the blow up at 0 is linear, we shall prove a subsequence of {um } converges locally uniformly firstly. In the following Lemmas 2 and 3, we also suppose g is C 1 near 0. Lemma 2. For any sequence of λm → 0, set Ω0m = λ−1 m Ω and um (x) =
u(λm x) , x ∈ Ω0m . λm
Then a subsequence of {um } converges to an infinity harmonic function w ∈ C(C) locally uniformly. Proof . Since Lemma 1, lim S(y) ≤ max{S(0), |Dg(0)|},
y→0
then Lip(u(x), BR ∩ Ω ) ≤ max{S(0), |Dg(0)|} + 1 for some R > 0. The function um (x) satisfies the equation △∞ um (x) = λm f (λm x), x ∈ Ω0m . Fix λ > 0. For 0 < λm < λ, the Lipschitz constant for um (x) on B Lip(um (x), B
R λm
R λm
∩ Ω0m :
∩ Ω0m ) = Lip(u(x), BR ∩ Ω ),
then Lip(um (x), B R ∩ Ω0m ) ≤ max{S(0), |Dg(0)|} + 1 λ
and |um (x)| ≤
R (max{S(0), |Dg(0)|} + 1). λ
By using Arzela–Ascoli Theorem, there exists a subsequence of {um } that converges to a function w ∈ C(C) locally uniformly and w satisfies △∞ w = 0 in C in the viscosity sense. □
X. Feng, G. Hong / Nonlinear Analysis 176 (2018) 36–47
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Lemma 3. Let w be a blow up of u at 0 as above. Then w has the following properties: ¯ (1) Lip(w, C) ≤ max{S(0), |Dg(0)|}, w ∈ C(C); (2) w(y) = Dg(0) · y for all y ∈ Γ ; (3) If S(0) = S + (0), then there exists e ∈ C¯ ∩ S n−1 , such that w(te) = S(0)t for t ∈ [0, +∞); if S(0) = S − (0), then there exists e ∈ C¯ ∩ S n−1 , such that w(te) = −S(0)t for t ∈ [0, +∞). Proof . The proof is totally the same as the proof of Lemma 4 in [12]. From Lemma 2, (1) is easy to get. k k For any y ∈ Γ \ {0}, let yk → y, yk ∈ C. Then choose zm ∈ ∂Ω satisfying zm = τm λm yk + (1 − τm )λm y with τm ∈ [0, 1). Then using Lemma 1, i.e. the Lipschitz continuity of u, and the continuity of Dg near 0, we can get (2). Suppose S(0) = S + (0). Using the definition of Sr+ (0), we can get the direction er ∈ C ∩ S n−1 which is maximum slope direction from 0 to ∂(Brn ∩ Ω ). By the compactness of C ∩ S n−1 , we have a subsequence erj → e as rj → 0. Then we can check that w(te) = S(0)t for t ∈ [0, +∞) by using the comparison with cones property. □ Proof of Theorem 3. Suppose g is C 1 near 0. The proof of the slope estimate is the same as Theorem 2 in [12]. We only describe it briefly and refer the reader to [12] for the details. We prove S(0) ≤ |Dg(0)| by contradiction. Assume |Dg(0)| < S(0). Without loss of generality, we suppose S(0) = S + (0). Using (2) and (3) of Lemma 3 and the key fact that 0 is a corner point, we can manage to find a point P ∈ Γ such that |w(e) − w(P )| > S(0). |e − P | It is a contradiction to (1) of Lemma 3. Next, we drop the C 1 assumption on g. Let u± be constructed as in the proof of Theorem 3 in [12]. Find two functions g ± ∈ C 1 near 0 satisfying g − ≤ g ≤ g + and Dg ± (0) = Dg(0). Let u± solve the problem { △∞ u± = f, in Br ∩ C u± = g ± , on ∂(Br ∩ C). From the above proof, S[u± ](0) ≤ |Dg ± (0)| = |Dg(0)|. By comparison principle, u− ≤ u ≤ u+ in Br ∩ C. So S + (0) ≤ S + [u+ ](0) and S − (0) ≤ S − [u− ](0). Therefore, S(0) ≤ max{S + [u+ ](0), S − [u− ](0)} ≤ max{S[u+ ](0), S[u− ](0)} ≤ |Dg(0)|.
□
Lemma 4. Suppose g is C 1 near 0. ¯ then (1) If Dg(0) ∈ C¯ ∪ −C, lim sup S(y) ≤ S(0); y→0
(2) Dg(0) ∈ C implies S(0) = S + (0); Dg(0) ∈ −C implies S(0) = S − (0). Proof . The proof of (1) is essentially the same as the proof of Lemma 3 in [12]. If |Dg(0)| ≤ S(0), applying Dg(0) Dg(0) (or − |Dg(0)| ) ∈ C and the Lemma 1, it is clear to get. Suppose |Dg(0)| > S(0). Let the direction ν = |Dg(0)| angle θ = dS n−1 (ν, Γ ∩ ∂B1 ). We consider the boundary value g at the corner point 0, then we have cos θ|Dg(0)| ≤ S(0)
and θ > 0.
Denote C τ = {x ∈ C : dS n−1 (
x , Γ ∩ ∂B1 ) > τ }. |x|
X. Feng, G. Hong / Nonlinear Analysis 176 (2018) 36–47
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Then we can prove that for any τ > 0, there exists r(τ ) > 0 such that C τ ∩ Br(τ ) ⊂ Ω ⊂ C. Fix ϵ > 0, and let τ > 0 be such that cos(θ − 3τ )|Dg(0)| ≤ cos θ|Dg(0)| + 2ϵ . Applying the differentiability ) g (x) g (x)| ≤ |Dg(0)| + 2ϵ and dS n−1 ( D˜ of g near 0, let 0 < r1 < τ r(τ be such that supx∈Br |D˜ , Dg(0) ) < τ . 4 1 |D˜ g (x)| |Dg(0)| Then for any x ∈ ∂Ω ∩ B r1 , we have 2
sup y∈∂Ω∩B n
r1 /2
ϵ) |g(x) − g(y)| ( ≤ |Dg(0)| + cos(θ − 3τ ) ≤ cos θ|Dg(0)| + ϵ. |x − y| 2 (x)
The rest of the proof of (1) is similar to the proof of Lemma 1 in this paper. The proof of (2) is the same as Lemma 5 in [12]. We prove it by contradiction. If Dg(0) ∈ C and ¯ ≤ S(0) applying (1) in S(0) > S + (0). Then S(0) = S − (0) > 0. Let w be as in Lemma 2, then Lip(w, C) this lemma. From Lemma 3(3), there exists e ∈ C¯ ∩ S n−1 , such that w(te) = −S(0)t for all t ∈ [0, +∞). As in the proof of Theorem 3, we will make contradiction by finding a point P ∈ Γ such that |w(e) − w(P )| > S(0). |e − P |
□
Proof of Theorem 4. Suppose g is C 1 near 0. As S(0) ≥ 1, |Dg(0)| > 0. Let w be as in Lemma 2. Applying Lemma 3, we have w|Γ (y) = Dg(0) · y, and ¯ ≤ max{S(0), |Dg(0)|} = |Dg(0)|. Lip(w, C)
(6)
The rest can be proved in the same way as Theorem 4 in [12]. Without loss of generality, Dg(0) = en . If Dg(0) ̸∈ C ∪ −C, there exist two points in {x + ten } ∩ Γ with different sign of t for any x ∈ C. Using (6) on the points x and x + ten , we get w(x) ≤ xn
and w(x) ≥ xn .
Then w(x) = Dg(0) · x = xn
for all
x ∈ C.
If Dg(0) ∈ C ∪ −C, the necessity is evident. Suppose S(0) = |Dg(0)|. Note that e = en (or e = −en ) and there exists only one point in {x + ten } ∩ Γ . We use the points x and x + ten ∈ Γ to prove only one inequality between w(x) and xn as the above case. The other one can be achieved by using (6) on the points x and te with S(0) = |Dg(0)| = 1. Now we drop the C 1 assumption on g. Let u± be constructed as in the proof of Theorem 3. By comparison principle, u− ≤ u ≤ u+ in Br ∩ C. If u± are differentiable at 0, then u is also differentiable at 0. Theorem 4 is established. □
X. Feng, G. Hong / Nonlinear Analysis 176 (2018) 36–47
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3. Proof of Theorems 1 and 2 Now we drop the assumptions (A1) and (A2) in this section. These assumptions are not restrictions since we can extend u as in [14] and [9]. Define 4
U (X) = αu(x) + xn+1 − β|xn+2 | 3 1
with X = (x, xn+1 , xn+2 ) ∈ Rn+2 , α =
1/3 41/3 ∥f ∥ ∞ L (Ω)
and β =
34/3 . 21/3 ·4
By the result in [11], the function
U (X) is a viscosity solution of the problem △∞ U (X) = F (X) in Σ , U (X) = G(X) on ∂Σ , where Σ = Ω × R2 , F (X) = C
f (x) 4∥f ∥L∞ (Ω)
−
1 2
(7) 4 3
and G(X) = αg(x) + xn+1 − β|xn+2 | . The regularity up to
1 1, 3
on U in general dimension also holds for u. The origin 0 ∈ ∂Σ is also a corner point of Σ , F (X) ∈ C(Σ ) ∩ L∞ (Σ ) and G(X) is differentiable at 0 ¯ ∩ B n+2 and 0 < r < r0 , we define with G(0) = U (0) = 0. For X ∈ Σ r0 Sr+ [U ](X) =
sup Y ∈∂(Brn+2 (X)∩Σ )\{X}
U (Y ) − U (X) |Y − X|
and Sr− [U ](X) =
U (X) − U (Y ) . |Y − X| Y ∈∂(B n+2 (X)∩Σ )\{X} sup
r
It is easy to check that the problem (7) satisfies the assumptions (A1) and (A2): (A1) Sr± [U ](X) ≥ 1 for all X and r; (A2) − 43 ≤ F ≤ − 14 . Thus, we define S ± [U ](X) = lim Sr± [U ](X), r→0
and S[U ](X) = max{S + [U ](X), S − [U ](X)}. Then we have the conclusions of Theorems 3 and 4 on U with K := C × R2 : S[U ](0) ≤ |DG(0)|; and (i) If DG(0) ̸∈ K ∪ −K, then U is differentiable at 0; (ii) If DG(0) ∈ K ∪ −K, then U is differentiable at 0 if and only if S[U ](0) = |D(0)|. Since u does not satisfy the conditions (A1) and (A2), we only have S ± (x) = lim Sr± (x) r→0
and S(x) = max{S + (x), S − (x)}. We compare S(0) with S[U ](0).
X. Feng, G. Hong / Nonlinear Analysis 176 (2018) 36–47
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Lemma 5. S[U ](0) =
√ α2 S(0)2 + 1.
Proof . Firstly, we prove S[U ](0) ≤
√
α2 S(0)2 + 1. Without loss of generality, we suppose that
S[U ](0) = S + [U ](0) = lim Sr+ [U ](0), S + [U ](0) ≥ S − [U ](0). r→0
For any r > 0, by the definition of Sr+ [U ](0), we consider our argument in the following two cases. Case a: There exists Xr ∈ ∂(Brn+2 ∩ Σ ), such that Sr+ [U ](0) =
U (Xr ) ; |Xr |
Case b: There exists a sequence {Xrk } ⊂ ∂(Brn+2 ∩ Σ ), Xrk → 0, as k → ∞, such that U (Xrk ) . k→∞ |Xrk |
Sr+ [U ](0) = lim For Case a, we have
U (Xr ) |Xr | xr xr,n+1 xr,n+2 u(x)r xr −2 =( , , ) · (α , 1, −βxr,n+2 |xr,n+2 | 3 ) |Xr | |Xr | |Xr | |xr | |xr | √ 2 ≤ α2 Sρ+r (0)2 + 1 + β 2 |xr,n+2 | 3
Sr+ [U ](0) =
where Xr = (xr , xr,n+1 , xr,n+2 ) and ρr = |xr |. For Case b, we have U (Xrk ) k→∞ |Xrk | xk xkr,n+1 xkr,n+2 u(xkr ) xkr −2 = lim ( rk , , ) · (α , 1, −βxkr,n+2 |xkr,n+2 | 3 ) k k k k k→∞ |Xr | |Xr | |Xr | |xr | |xr | √
Sr+ [U ](0) = lim
2
≤ lim α2 Sρ+k (0)2 + 1 + β 2 |xkr,n+2 | 3 r k→∞ √ ≤ α2 S + (0)2 + 1 where Xr = (xkr , xkr,n+1 , xkr,n+2 ) and ρkr = |xkr |. Then, we have S + [U ](0) = lim Sr+ [U ](0) ≤ r→0
√ α2 S + (0)2 + 1.
√ Next, we give the proof of S[U ](0) ≥ α2 S(0)2 + 1. If S(0) = 0, it is easy to verify that S[U ](0) ≥ 1. Assume S(0) > 0 and S(0) = S + (0) ≥ S − (0). Note that S + (0) = lim Sr+ (0). r→0
Hence, for any 0 < ε <
S(0) 2 ,
there exists a sequence {ρj } with ρj → 0, such that S + (0) −
ε ε ⩽ Sρ+j (0) ⩽ S + (0) + 2 2
for small enough {ρj }. For any ρj > 0, we consider the argument in two cases.
X. Feng, G. Hong / Nonlinear Analysis 176 (2018) 36–47
Case a’: There exists xρj ∈ ∂(Bρnj ∩ Ω ), such that Sρ+j (0) =
u(xρj ) , |xρj |
then S + (0) − ε ⩽
u(xρj ) ⩽ S + (0) + ε; |xρj |
Case b’: There exists a sequence {xlρj } ⊂ ∂(Bρnj ∩ Ω ), xlρj → 0, as l → ∞, such that Sρ+j (0) = lim
l→∞
u(xlρj ) |xlρj |
i.e. for large enough l, Sρ+j (0) −
u(xlρj ) ε ε ⩽ ⩽ Sρ+j (0) + , l 2 |xρj | 2
then S + (0) − ε ⩽
u(xlρj ) |xlρj |
⩽ S + (0) + ε.
Since S(0) > 0 and u(0) = 0, then u(xρj ), u(xlρj ) ̸= 0. For Case a’, we choose ( Xρj =
) 2 |xρj | ,0 . xρj , αu(xρj )
Take Rρj = |Xρj |, then + SR [U ](0) ρj
U (Xρj ) ≥ = |Xρj |
(
) ( ) 2 xρj |xρj | u(xρj ) xρj , ,0 · α , 1, 0 . |Xρj | αu(xρj )|Xρj | |xρj | |xρj |
Note that ⏐( )⏐ 2 ⏐ xρj ⏐ |xρj | ⏐ ⏐ ⏐ |Xρ | , αu(xρ )|Xρ | , 0 ⏐ = 1 j j j and (
where t = |(α
) ( ) 2 xρ j |xρj | u(xρj ) xρj , ,0 = t α , 1, 0 |Xρj | αu(xρj )|Xρj | |xρj | |xρj |
−1 u(xρj ) xρj . |xρj | |xρj | , 1, 0)|
Then
⏐( )⏐ √ √ ⏐ ⏐ u(x ) x u(xρj ) ρ ρ j j + 2+1≥ ⏐ α ⏐ = α2 ( SR , 1, 0 ) α2 (S + (0) − ε)2 + 1. [U ](0) ≥ ⏐ ⏐ ρj |xρ | |xρ | |xρ | j
j
j
45
X. Feng, G. Hong / Nonlinear Analysis 176 (2018) 36–47
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Taking ρj → 0, Rρj → 0. It follows that S + [U ](0) ≥
√
α2 (S + (0) − ε)2 + 1.
In Case b’, using the same argument in Case a’ to analysis xlρj , we obtain √ + SR [U ](0) ≥ α2 (S + (0) − ε)2 + 1. l ρ j
Taking l → ∞, xlρj → 0 and Rρl → 0. It follows that j
S + [U ](0) ≥
√
α2 (S + (0) − ε)2 + 1.
Therefore, take ε → 0, we obtain S + [U ](0) ≥
√ α2 S + (0)2 + 1.
If S(0) = S − (0) ≥ S + (0), using the similar argument, we have √ S − [U ](0) ≥ α2 S − (0)2 + 1. The lemma follows. □ Remark 1. From the proof of Lemma 5, one can easily see that if S[U ](0) = S + [U ](0), then √ S(0) = S + (0) and S + [U ](0) = α2 S + (0)2 + 1; if S[U ](0) = S − [U ](0), then S(0) = S − (0) and S − [U ](0) =
√ α2 S − (0)2 + 1.
Proof of Theorem 1. Note that S[U ](0) ≤ |DG(0)|, S[U ](0) =
√
α2 S(0)2
√ 2 + 1 and |DG(0)| = α2 |Dg(0)| + 1.
Since S(0) is nonnegative, then we get the slope estimate S(0) ≤ |Dg(0)|.
□
Proof of Theorem 2. Case (i) Assume that Dg(0) ̸∈ C ∪ −C. Note that DG(0) = (αDg(0), 1, 0) and K = C × R2 . Then DG(0) ̸∈ K ∪ −K. Applying (i) in Theorem 4, we get that U is differentiable at 0. Since 4 U (X) = αu(x) + xn+1 − β|xn+2 | 3 , we obtain that u is differentiable at 0. Case (ii) Assume that Dg(0) ∈ C ∪ −C. If u is differentiable at 0, it is clear that S(0) = |Dg(0)|. Suppose S(0) = |Dg(0)|. Since √ √ 2 2 2 S[U ](0) = α S(0) + 1 and |DG(0)| = α2 |Dg(0)| + 1, then S[U ](0) = |DG(0)|. Applying Theorem 4 (ii), U is differentiable at 0. It follows that u is also differentiable at 0. □
X. Feng, G. Hong / Nonlinear Analysis 176 (2018) 36–47
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Acknowledgments This work is supported by the National Nature Science Foundation of China: NSFC 11301411 and 11671316. We would like to thank the referees for their careful reading and helpful comments and suggestions which improved the presentation of the paper. References
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