Partial regularity for an exponential PDE in crystal surface models
aa r X i v : . [ m a t h . A P ] J a n EXPONENTIAL NON-LINEARITY IN CRYSTAL SURFACE MODELS
XIANGSHENG XU
Abstract.
We consider the existence of a solution to the boundary value problem for the equation − div (cid:16) D ( ∇ u ) ∇ e − div( |∇ u | p − ∇ u + β |∇ u | − ∇ u ) (cid:17) + au = f . This problem is derived from the mathe-matical modeling of crystal surfaces. The analytical difficulty is due to the fact that the smallesteigenvalue of the mobility matrix D ( ∇ u ) is not bounded away from 0 below and the inside operatoris an exponential function composed with a linear combination of the p-Laplace operator and the1-Laplace operator. Known existence results on problems related to ours either have to allow thepossibility that the exponent in the equation be a measure or assume that data are suitably smallin order to eliminate the possibility. In this paper we show the existence of a non-measure-valuedweak solution without any smallness assumption on the data. We achieve this by employing apower series expansion technique. Introduction
Let Ω be a bounded domain in R N with smooth boundary ∂ Ω and ν the unit outward normalto ∂ Ω. In this paper we consider the boundary value problem − div (cid:16) D ( ∇ u ) ∇ e − div( ∂ z E ( ∇ u )) (cid:17) + au ∋ f in Ω,(1.1) D ( ∇ u ) ∇ e − div( ∂ z E ( ∇ u )) · ν ∋ ∂ Ω,(1.2) ∇ u · ν = 0 on ∂ Ω(1.3)for given data D ( ∇ u ) , E ( ∇ u ) , a , and f with properties:(H1) The matrix D ( ∇ u ) has the expression D ( ∇ u ) = I − q |∇ u | (1 + q |∇ u | ) ∇ u ⊗ ∇ u, where I is the N × N identity matrix and q is a positive number;(H2) The function E = E ( z ) is given by E ( z ) = 1 p | z | p + β | z | , z ∈ R N , p > , β > , and hence its subgradient ∂ z E ( z ) is a multi-valued function(1.4) ∂ z E ( z ) = (cid:26) | z | p − z + β | z | − z if z = 0, β [ − , N if z = 0,which explains the inclusion sign “ ∈ ” in (1.1)-(1.2);(H3) a ∈ (0 , ∞ ) , f ∈ W ,p (Ω) ∩ L ∞ (Ω). Mathematics Subject Classification.
Key words and phrases.
Crystal surface models, existence, exponential nonlinearity, the 1-Laplace operator, non-linear fourth order equations.Department of Mathematics and Statistics, Mississippi State University, Mississippi State, MS 39762.
Email :[email protected].
Our interest in this problem originated in the mathematical modeling of crystal surface growth.In this case, u is the surface height, D ( ∇ u ) is the so-called mobility [15], and R Ω E ( ∇ u ) dx representsthe surface energy. Currently, it is well accepted [1, 5, 6, 7, 8, 19, 20] that the evolution of a crystalsurface below the roughing temperature can be accurately described by the following continuumequation(1.5) ∂ t u ∈ div (cid:16) D ( ∇ u ) ∇ e − div ( ∂ z E ( ∇ u )) (cid:17) . Our equation (1.1) is obtained by discretizing the time derivative in the above equation.Crystal surfaces are known to develop facets, where ∇ u = 0. To define D ( ∇ u ) there, we observethat (cid:12)(cid:12)(cid:12)(cid:12) ∇ u ⊗ ∇ u |∇ u | (cid:12)(cid:12)(cid:12)(cid:12) ≤ |∇ u | . Thus it is natural for us to set D ( ∇ u ) = I on the set where ∇ u = 0.Observe that each entry of D ( ∇ u ) is bounded by 2 and(1.6) D ( ∇ u ) ξ · ξ = | ξ | − q ( ∇ u · ξ ) |∇ u | (1 + q |∇ u | ) ≥
11 + q |∇ u | | ξ | for each ξ ∈ R N . Hence equation (1.5) degenerates on the set {|∇ u | = ∞} .Continuum models of this type are phenomenological in nature. That is, they are derived fromempirical data and observed phenomenons, not first principles. Hence their mathematical validationis important. Unfortunately, current analytical results are still far-lacking. For example, theexistence assertion for (1.5) coupled with initial boundary conditions is still open. The mainmathematical challenge is the exponential non-linearity involved. The function e s decays rapidlyto 0 as s → −∞ . Thus it is extremely difficult to derive any estimates for the exponent term near −∞ . In a sense the authors in [14, 7, 8, 19] circumvented this issue by allowing the possibility thatthe exponent term be a measure. In fact, an explicit solution was obtained in [14] which showedthat this possibility did occur. Our investigations here reveal that if we design our approximatescheme right we can eliminate the singularity in the exponent. To describe our method, we let τ = i , i = 1 , , · · · . We approximate E ( z ) by(1.7) E τ ( z ) = 1 p ( | z | + τ ) p + β ( | z | + τ ) and D ( ∇ u ) by(1.8) D τ ( ∇ u ) = (1 + τ ) I − q ( |∇ u | + τ ) (1 + q |∇ u | ) ∇ u ⊗ ∇ u, respectively. Then formulate our approximating problems as follows: − div ( D τ ( ∇ u ) ∇ ρ ) + τ ln( ρ + L ) + au = f in Ω , (1.9) − div ( ∇ E τ ( ∇ u ) + τ ∇ u ) + τ u = ln( ρ + L ) in Ω,(1.10) ∇ u · ν = ∇ ρ · ν = 0 on ∂ Ω , (1.11)where L >
0. This is similar to what the author did in [19] except that here we have introduced apositive L . Surprisingly, the number L makes all the difference. It turns out that if we choose L suitably large then ln( ρ + L ) ∈ L s (Ω) for each s ≥
1. Thus no singularity occurs in the exponentand we can take τ → ρ ∈ − L + e − div( ∂ z E ( ∇ u )) . Substitute this into (1.9) in the limit to obtain the original equation (1.1).
XPONENTIAL NON-LINEARITY IN CRYSTAL SURFACE MODELS 3
Our starting point is the following three a priori estimates (cid:12)(cid:12)(cid:12)(cid:12)Z Ω ln( ρ + L ) dx (cid:12)(cid:12)(cid:12)(cid:12) ≤ c, (1.12) k u k W ,p (Ω) ≤ c, (1.13) Z Ω |∇√ ρ + L | q |∇ u | dx ≤ c. (1.14)Here and in what follows the letter c denotes a generic positive constant. In theory, its value can becomputed from various given data. We must extract enough information from these three estimatesand the three equations (1.9)-(1.11) to justify passing to the limit. The first issue is that to be ableto apply Poincar´e’s inequality (see Lemma 2.2 below) we need to know the average of ρ + L overa set of positive measure is finite and (1.12) is far from doing that. We must bridge this gap toprevent the ρ -component of our approximate solutions from converging to infinity a.e. on Ω [19].The second issue is how to estimate the function ln( ρ + L ) near ρ = − L . If we compare this withthe singularity of the function at infinity, we can bound the function by ( ρ + L ) ε , ε >
0, as ρ goes to ∞ , while as ρ → − L + the function is dominated by ( ρ + L ) − ε . Since ρ satisfies a non-homogeneousequation, it does not seem possible that one can obtain any integral estimates for the latter. Ourinvestigations reveal that the two issues are interconnected and they can be addressed via thepower series expansion for ln( ρ τ + L ). In this respect, we would like to mention [11, 13], where thepower series expansion for e − div( ∂ z E ( ∇ u )) was employed. However, the subsequent application of theFourier transform required the authors there to assume that D ( ∇ u ) = I and the exponent term belinear, i.e., div( ∂ z E ( ∇ u )) = ∆ u . They also needed the given data to be suitably small. We havemanaged to remove these restrictions. Even though the problems in [11, 13] are time-dependent,we believe that the technique developed here is still applicable, and we will carry out this study ina future paper.In view of our analysis, we can give the following definition of a weak solution. Definition 1.1.
We say that a triplet ( u, ρ, ϕ ) is a weak solution to (1.1) - (1.3) if the followingconditions hold: (D1) ρ ∈ W , (Ω) with ρ ≥ − L for some L > , u ∈ W , ∞ (Ω) , ϕ ∈ ( L ∞ (Ω)) N , and div( |∇ u | p − ∇ u + β ϕ ) ∈ L s (Ω) for each s ≥ ; (D2) ϕ ( x ) ∈ ∂ z H ( ∇ u ( x )) for a.e. x ∈ Ω , where (1.15) H ( z ) = | z | , and ρ + L = e − div( |∇ u | p − ∇ u + β ϕ ) ; (D3) There hold Z Ω D ( ∇ u ) ∇ ρ · ∇ ξ dx + a Z Ω uξ dx = Z Ω f ξ dx, Z Ω ( |∇ u | p − ∇ u + β ϕ ) · ∇ ξ dx = Z Ω ln( ρ + L ) ξ dx for all ( ξ , ξ ) ∈ W , (Ω) × W ,p (Ω) . Our main result is the following
Theorem 1.2 (Main theorem) . Assume that (H1) -(H3) hold and Ω is a bounded domain in R N with C , boundary. Then there is a weak solution to (1.1) - (1.3) . Throughout the remainder of the paper we shall assume(1.16) 1 < p ≤ , N > . XIANGSHENG XU
This is done mainly for the convenience in applying the Sobolev inequality and also avoiding non-essential complications. Cases where p > N = 2 [20] are simpler, and we leave them tothe interested reader.This paper is organized as follows: In Section 2 we collect a few known results. Three keypreparatory lemmas are established in Section 3. The proof of the main theorem is given in Section4. Finally, we make some remarks about our convention. If a, b ∈ [0 , ∞ ) and β >
0, we have( a + b ) β ≤ (cid:26) a β + b β if β ≤ β − (cid:0) a β + b β (cid:1) if β > . That is, we always have ( a + b ) β ≤ c (cid:0) a β + b β (cid:1) . When an occasion arises for this inequality, it willbe used without acknowledgment. Other frequently used inequalities include Young’s inequality(1.17) ab ≤ εa p + 1 ε q/p b q , ε > , p, q > p + q = 1and the interpretation inequality(1.18) k f k q ≤ ε k f k r + ε − σ k f k p , ε > , p ≤ q ≤ r , and σ = (cid:16) p − q (cid:17) / (cid:16) q − r (cid:17) , where k · k p denotes the norm in the space L p (Ω). In the applications of the Sobolev inequality(1.19) k u k p ∗ ≤ c ( k∇ u k p + k u k ) , p ∗ = N pN − p , it is understood that 1 ≤ p < N because the case where p = N can always be handled separately.2. Preliminaries
In this section we collect a few known results that are useful to us.Our existence theorem is based upon the following fixed point theorem, which is often called theLeray-Schauder Theorem ([9], p.280).
Lemma 2.1.
Let B be a map from a Banach space B into itself. Assume: (LS1) B is continuous; (LS2) the images of bounded sets of B are precompact; (LS3) there exists a constant c such that k z k B ≤ c for all z ∈ B and σ ∈ [0 , satisfying z = σB ( z ) .Then B has a fixed point. Lemma 2.2.
Let Ω be a bounded domain in R N with Lipschitz boundary and ≤ p < N . Thenthere is a positive number c = c ( N ) such that (2.1) k u − u S k p ∗ ≤ cd N +1 − pN | S | p k∇ u k p for each u ∈ W ,p (Ω) ,where S is any measurable subset of Ω with | S | > , u S = | S | R S udx , and d is the diameter of Ω . This lemma can be inferred from Lemma 7.16 in [9]. Also see [10, 16]. It is a version of Poincar´e’sinequality.
Lemma 2.3.
Let { y n } , n = 0 , , , · · · , be a sequence of positive numbers satisfying the recursiveinequalities (2.2) y n +1 ≤ cb n y αn for some b > , c, α ∈ (0 , ∞ ) . XPONENTIAL NON-LINEARITY IN CRYSTAL SURFACE MODELS 5 If y ≤ c − α b − α , then lim n →∞ y n = 0 . This lemma can be found in ([4], p.12).3.
Three key lemmas
In this section we prove three key lemmas. They lay the foundation for our existence theorem.The first lemma deals with the exponent in our problem.Let E τ be given as in (1.7). Define F τ ( s ) = ( s + τ ) p − + β ( s + τ ) − on [0 , ∞ ) . Then we can easily verify ∇ E τ ( z ) = F τ ( | z | ) z. Remember p ∈ (1 , (cid:16) ( | z | + τ ) p − z − ( | y | + τ ) p − y (cid:17) · ( z − y ) ≥ ( p − (cid:0) | y | + | z | (cid:1) p − | z − y | , (cid:16) ( | z | + τ ) − z − ( | y | + τ ) − y (cid:17) · ( z − y ) ≥ , z, y ∈ R N . Subsequently,(3.1) ( F τ ( | z | ) z − F τ ( | y | ) y ) · ( z − y ) ≥ ( p − (cid:0) | y | + | z | (cid:1) p − | z − y | for all z, y ∈ R N . Lemma 3.1.
Let Ω be a bounded domain in R N with Lipschitz boundary ∂ Ω . Consider the problem − div (cid:0) F τ ( |∇ u | ) ∇ u (cid:1) + τ u = f in Ω,(3.2) ∇ u · ν = 0 on ∂ Ω,(3.3) where p > , f ∈ L pp − (Ω) . Then there is a unique weak solution u to the above problem in thespace W ,p (Ω) . Furthermore, if f also lies in the space L s (Ω) with (3.4) s > Np ,u is bounded and we have the estimate (3.5) k u k ∞ ≤ c k u k + c ( k f k s ) p − + √ τ , where c depends only on N, p, q, Ω . If, in addition, ∂ Ω is C , , then for each ℓ > N there is apositive number c such that (3.6) k∇ u k ∞ ≤ c k∇ u k + c ( k f k ℓ ) p − + c √ τ + c. Once again, c here is independent of τ . This lemma is more or less known. A local version of (3.6) can be found in [18]. We will offer asimpler proof here.
Proof.
The existence of a solution can be established by showing the functional Z Ω E τ ( ∇ u ) dx − Z Ω f udx has a minimizer in W ,p (Ω), while the uniqueness can be inferred from (3.1). We shall omit thedetails.Without loss of generality, assume max Ω u = k u k ∞ . XIANGSHENG XU
Select(3.7) k ≥ ( k f k s ) p − + √ τ . as below. Let k n = k − k n +1 ,y n = k ( u − k n ) + k , n = 0 , , , · · · . Using ( u − k n +1 ) + as a test function in (3.2), we derive, with the aid of H¨older’s inequality and theSobolev inequality (1.19), that Z Ω (cid:0) |∇ ( u − k n +1 ) + | + τ (cid:1) p − |∇ ( u − k n +1 ) + | dx ≤ Z Ω f ( u − k n +1 ) + dx ≤ Z { u ≥ k n +1 } | f | NpNp − N + p dx ! Np − N + pNp (cid:18)Z Ω (cid:2) ( u − k n +1 ) + (cid:3) NpN − p dx (cid:19) N − pNp ≤ c k f k s |{ u ≥ k n +1 }| Np − N + pNp − s (cid:0) k∇ ( u − k n +1 ) + k p + k ( u − k n +1 ) + k (cid:1) ≤ ck p − |{ u ≥ k n +1 }| Np − N + pNp − s (cid:0) k∇ ( u − k n +1 ) + k p + y n +1 (cid:1) . Here the last step is due to (3.7). On the other hand, we can deduce from (1.16) that Z Ω |∇ ( u − k n +1 ) + | p dx ≤ Z { u ≥ k n +1 } (cid:0) |∇ ( u − k n +1 ) + | + τ (cid:1) p − (cid:0) |∇ ( u − k n +1 ) + | + τ (cid:1) dx ≤ ck p − |{ u ≥ k n +1 }| Np − N + pNp − s (cid:0) k∇ ( u − k n +1 ) + k p + y n +1 (cid:1) + τ p |{ u ≥ k n +1 }|≤ (cid:0) k∇ ( u − k n +1 ) + k pp + y pn +1 (cid:1) + ck p |{ u ≥ k n +1 }| Np − N + pN ( p − − ps ( p − + k p |{ u ≥ k n +1 }| . The last step is due to Young’s inequality (1.17) and (3.7). Subsequently, Z Ω |∇ ( u − k n +1 ) + | p dx ≤ ck p |{ u ≥ k n +1 }| Np − N + pN ( p − − ps ( p − + cy pn +1 + k p |{ u ≥ k n +1 }| . Apply the Sobolev inequality again to deduce y n +1 ≤ k ( u − k n +1 ) + k NpN − p |{ u ≥ k n +1 }| − N − pNp ≤ c (cid:0) k∇ ( u − k n +1 ) + k p + y n +1 (cid:1) |{ u ≥ k n +1 }| − N − pNp ≤ ck |{ u ≥ k n +1 }| Np − N + pN ( p − − s ( p − + ck |{ u ≥ k n +1 }| N +1 N + cy n +1 |{ u ≥ k n +1 }| − N − pNp . (3.8)Note that y n ≥ Z { u ≥ k n +1 } ( u − k n ) + dx ≥ k n +1 |{ u ≥ k n +1 }| . Moreover, α ≡ ps − N ( p − N s >
XPONENTIAL NON-LINEARITY IN CRYSTAL SURFACE MODELS 7
Subsequently, k |{ u ≥ k n +1 }| Np − N + pN ( p − − s ( p − = k |{ u ≥ k n +1 }| α ≤ ( n +1)(1+ α ) k α y αn . Without loss of generality, we may assume s ≤ N . Then1 − N − pN p − α = p − p + N − s ( p − N s > , N − α = N − s ( p − N s ≥ . It follows that y n +1 |{ u ≥ k n +1 }| − N − pNp = y n +1 |{ u ≥ k n +1 }| α +1 − N − pNp − α ≤ c ( n +1) α k α y αn ,k |{ u ≥ k n +1 }| N +1 N = k |{ u ≥ k n +1 }| α + N − α ≤ c ( n +1)(1+ α ) k α y αn . Collect the preceding inequalities in (3.8) to get y n +1 ≤ cb n k α y αn , b > . By Lemma 2.3, if we choose k so large that y ≤ (cid:16) ck α (cid:17) − α b − α , then u ≤ k on Ω.In view of (3.7), it is enough for us to take k = c Z Ω | u | dx + ( k f k s ) p − + √ τ . This implies the desired result.To obtain (3.6), we first derive a differential inequality satisfied by w = |∇ u | . To this end, we first observe that our solution u actually lies in W ,r (Ω) for each r ≥ x j , j ∈ { , · · · , N } , to derive − div (cid:0) ∇ z E τ ( ∇ u ) ∇ u x j (cid:1) + τ u x j = f x j . It is easy to verify(3.9) ∇ E τ ( z ) = ( | z | + τ ) p − (cid:18) I + ( p − z ⊗ z | z | + τ (cid:19) + β ( | z | + τ ) − (cid:18) I − z ⊗ z | z | + τ (cid:19) . XIANGSHENG XU
It is easy to verify that for each z ∈ R N ∇ E τ ( ∇ u ) z · z = ( w + τ ) p − (cid:18) | z | + ( p −
2) ( ∇ u · z ) w + τ (cid:19) + β ( w + τ ) − (cid:18) | z | − ( ∇ u · z ) w + τ (cid:19) ≥ ( w + τ ) p − (cid:18) | z | − (2 − p ) + w | z | w + τ (cid:19) + β ( w + τ ) − (cid:18) | z | − w | z | w + τ (cid:19) ≥ (1 − (2 − p ) + )( w + τ ) p − | z | . (3.10)We also have |∇ E τ ( ∇ u ) | ≤ c h ( w + τ ) p − + β ( w + τ ) − i . (3.11)Multiply through (3.9) by u x j to obtain −
12 div (cid:16) ∂ z E τ ( ∇ u ) ∇ u x j (cid:17) + ∂ z E τ ( ∇ u ) ∇ u x j · ∇ u x j + τ u x j = f x j u x j . (3.12)By (3.10),(3.13) ∇ z E τ ( ∇ u ) ∇ u x j · ∇ u x j ≥ . Use this in (3.12), sum up the resulting inequality over j , and thereby obtain(3.14) − div (cid:0) ∇ z E τ ( ∇ u ) ∇ w (cid:1) ≤ ∇ f ∇ u. The estimate (3.6) will be established in two steps. First, we obtain a local interior estimate,while the boundary estimate will be achieved by flattening the relevant portion of the boundary.To do the local estimate, we fix a point z ∈ Ω. Then pick a number R from (0 , dist( z , ∂ Ω)).Define a sequence of concentric balls B R n ( z ) in Ω as follows: B R n ( z ) = { z : | z − z | < R n } , where R n = R R n +1 , n = 0 , , , · · · . Choose a sequence of smooth functions θ n so that θ n ( z ) = 1 in B R n ( z ) ,θ n ( z ) = 0 outside B R n − ( z ) , |∇ θ n ( z ) | ≤ c n R for each z ∈ R N , and0 ≤ θ n ( z ) ≤ R N .Select(3.15) K ≥ ( R − Nℓ k f k ℓ,B R ( z ) ) pp − + τ pp − ( R − Nℓ k u k ℓ,B R ( z ) ) pp − + 1as below. Set K n = K − K n +1 , n = 0 , , , · · · ,v = ( w + τ ) p . XPONENTIAL NON-LINEARITY IN CRYSTAL SURFACE MODELS 9
We use θ n +1 ( v − K n +1 ) + as a test function in (3.14) to derive Z Ω θ n +1 ∇ E τ ( ∇ u ) ∇ w · ∇ ( v − K n +1 ) + dx ≤ − Z Ω θ n +1 ∇ θ n +1 · ∇ E τ ( ∇ u ) ∇ w ( v − K n +1 ) + dx +2 Z Ω ∇ f · ∇ uθ n +1 ( v − K n +1 ) + dx. (3.16)Now we proceed to analyze each term in the above inequality. In view of (3.10), we have Z Ω θ n +1 ∇ E τ ( ∇ u ) ∇ w · ∇ ( v − K n +1 ) + dx = 2 p Z Ω θ n +1 ( w + τ ) − p ∇ E τ ( ∇ u ) ∇ ( v − K n +1 ) + · ∇ ( v − K n +1 ) + dx ≥ p − p Z Ω θ n +1 |∇ ( v − K n +1 ) + | dx. (3.17)With (3.11) and (3.15) in mind, we can estimate the second term in (3.16) as follows: − Z Ω θ n +1 ∇ θ n +1 · ∇ E τ ( ∇ u ) ∇ w ( v − K n +1 ) + dx ≤ c n R Z Ω θ n +1 h β ( w + τ ) − p − i |∇ ( v − K n +1 ) + | ( v − K n +1 ) + dx ≤ ε Z Ω θ n +1 |∇ ( v − K n +1 ) + | dx + c ( ε )4 n R β K p − p n +1 Z B Rn ( z ) (cid:2) ( v − K n +1 ) + (cid:3) dx ≤ ε Z Ω θ n +1 |∇ ( v − K n +1 ) + | dx + c ( ε )4 n R Z B Rn ( z ) (cid:2) ( v − K n +1 ) + (cid:3) dx. As for the last integral in (3.16), we recall from (3.2) thatdiv ( F τ ( w ) ∇ u ) = τ u − f. Consequently, the last term in (3.16) = 2 Z Ω ∇ f · ∇ uθ n +1 ( v − K n +1 ) + dx = 2 Z Ω ∇ f · F τ ( w ) ∇ uθ n +1 ( v − K n +1 ) + F τ ( w ) dx = − Z Ω f ( τ u − f ) θ n +1 ( v − K n +1 ) + F τ ( w ) dx − Z Ω f ∇ u · θ n +1 ∇ θ n +1 ( v − K n +1 ) + dx − Z Ω θ n +1 f ∇ u · F τ ( w ) ∇ ( v − K n +1 ) + F τ ( w ) dx ≡ I + I + I . (3.18)We easily see that 1 F τ ( w ) ≤ ( w + τ ) − p = v − pp Hence, I ≤ c Z Ω | ( τ u − f ) f | θ n +1 v − pp ( v − K n +1 ) + dx ≤ c Z Q Rn ( z ) ( f + τ u ) v p dx, (3.19)where Q R n ( z ) = { z ∈ B R n ( z ) : v ( z ) ≥ K n +1 } . Similarly, I ≤ c n R Z Ω | f | w θ n +1 ( v − K n +1 ) + dx ≤ c n R Z Ω | f | v p θ n +1 ( v − K n +1 ) + dx ≤ c n R Z Q Rn ( z ) (cid:2) ( v − K n +1 ) + (cid:3) dx + c Z Q Rn ( z ) f v p dx. To estimate I , we observe that( v − K n +1 ) + F τ ( w ) = s ( s + K n +1 ) − − pp + β ( s + K n +1 ) − p (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) s =( v − K n +1 ) + Then we can easily check (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) dds s ( s + K n +1 ) − − pp + β ( s + K n +1 ) − p !(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ c ( s + K n +1 ) − − pp + β ( s + K n +1 ) − p . This immediately implies that (cid:12)(cid:12)(cid:12)(cid:12) F τ ( w ) ∇ ( v − K n +1 ) + F τ ( w ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ c (cid:12)(cid:12) ∇ ( v − K n +1 ) + (cid:12)(cid:12) . Subsequently, I = − Z Ω θ n +1 f ∇ u · F τ ( w ) ∇ ( v − K n +1 ) + F τ ( w ) dx ≤ ε Z Ω θ n +1 |∇ ( v − K n +1 ) + | dx + c ( ε ) Z Q Rn ( z ) f v p dx. (3.20)With the aid of (3.17)-(3.20), we can deduce from (3.16) that Z Ω θ n +1 |∇ ( v − K n +1 ) + | dx ≤ c n R Z Q Rn ( z ) (cid:2) ( v − K n +1 ) + (cid:3) dx + c Z Q Rn ( z ) ( f + τ u ) v p dx. (3.21)Now set y n = Z B Rn ( z ) (cid:2) ( v − K n ) + (cid:3) dx. We wish to show that the sequence { y n } satisfies (2.2). To this end, we estimate y n ≥ Z Q Rn ( z ) v (cid:18) − K n K n +1 (cid:19) dx ≥ n +2 Z Q Rn ( z ) v dx. XPONENTIAL NON-LINEARITY IN CRYSTAL SURFACE MODELS 11
Consequently, Z Q Rn ( z ) f v p dx ≤ Z Q Rn ( z ) v dx ! p Z Q Rn ( z ) | f | pp − dx ! p − p ≤ n +2 p y p n k f k ℓ,B R ( z ) | Q R n ( z ) | p − p − ℓ ≤ c n +2 p R ( − Nℓ ) K p − p y p n | Q R n ( z ) | p − p − ℓ . The last step is due to (3.15). By the same token, Z Q Rn ( z ) ( τ u ) v p dx ≤ c n +2 p R ( − Nℓ ) K p − p y p n | Q R n ( z ) | p − p − ℓ . Substituting the two preceding inequalities into (3.21) yields Z Ω θ n +1 |∇ ( v − K n +1 ) + | dx ≤ c n R (cid:18) y n + R Nℓ K p − p y p n | Q R n ( z ) | p − p − ℓ (cid:19) . (3.22)By Poincar´e’s inequality, y n +1 ≤ Z Ω (cid:2) θ n +1 ( v − K n +1 ) + (cid:3) dx ≤ (cid:18)Z Ω (cid:2) θ n +1 ( v − K n +1 ) + (cid:3) NN − dx (cid:19) N − N | Q R n ( z ) | N ≤ c Z Ω (cid:12)(cid:12) ∇ (cid:0) θ n +1 ( v − K n +1 ) + (cid:1)(cid:12)(cid:12) dx | Q R n ( z ) | N ≤ c Z Ω θ n +1 (cid:12)(cid:12) ∇ ( v − K n +1 ) + (cid:12)(cid:12) dx | Q R n ( z ) | N + c n R y n | Q R n ( z ) | N . This combined with (3.22) yields(3.23) y n +1 ≤ c n R (cid:18) y n | Q R n ( z ) | N + R Nℓ K p − p y p n | Q R n ( z ) | p − p − ℓ + N (cid:19) . Note that y n ≥ Z Q Rn ( z ) (cid:2) K n +1 − K n ) + (cid:3) dx ≥ K n +2 | Q R n ( z ) | ,α ≡ − ℓ + 2 N = 2( ℓ − N ) N ℓ > . It immediately follows that y n | Q R n ( z ) | N = y n | Q R n ( z ) | ℓ | Q R n ( z ) | α ≤ cR Nℓ ( n +2) α K α y αn ,K p − p y p n | Q R n ( z ) | p − p − ℓ + N ≤ c ( n +2) (cid:16) p − p + α (cid:17) K α y αn . Use these in (3.23) to derive y n +1 ≤ c (cid:16) p − p + α (cid:17) n R ℓ − N ) ℓ K α y αn = c (cid:16) p − p + α (cid:17) n R Nα K α y αn . By Proposition 2.3, if we choose K so large that y ≤ cK R N , then sup B R ( z ) v ≤ K. In view of (3.15), it is enough for us to take K = (cid:16) cy R N (cid:17) + ( R − Nℓ k f k ℓ,B R ( z ) ) pp − + τ pp − ( R − Nℓ k u k ℓ,B R ( z ) ) pp − + 1 . Recall that y = Z B R ( z ) "(cid:18) v − K (cid:19) + dx ≤ Z B R ( z ) ( w + τ ) p dx ≤ c Z B R ( z ) |∇ u | p dx + cτ p R N . Hence, sup B R ( z ) |∇ u | ≤ sup B R ( z ) v p ≤ c (cid:18) R − B R ( z ) |∇ u | p dx (cid:19) p + c √ τ + ( R − Nℓ k f k ℓ,B R ( z ) ) p − + τ p − ( R − Nℓ k u k ℓ,B R ( z ) ) p − + 1 . This is the so-called local interior estimate. Now we proceed to derive the boundary estimate.Suppose z ∈ ∂ Ω. Our assumption on the boundary implies that there exist a neighborhood U ( z )of z and a C , diffeomorphism T defined on U ( z ) such that the image of U ( z ) ∩ Ω under T isthe half ball B + δ ( y ) = { y : | y − y | < δ, y > } , where δ > y = T ( z ). This implies that wehave flatten U ( z ) ∩ ∂ Ω into a region in the plane y = 0 in the y space [3]. Set˜ u = u ◦ T − , ˜ w = w ◦ T − . We can choose T so that ˜ u satisfies the boundary condition(3.24) ˜ u y | y =0 = ∂ ˜ u∂ n (cid:12)(cid:12)(cid:12)(cid:12) y =0 = 0 . One way of doing this is to pick T = f ( z )... f N ( z ) so that the graph of f ( z ) = 0 is U ( z ) ∩ ∂ Ω andthe set of vectors {∇ f , · · · , ∇ f N } is orthogonal. By a result in [21], ˜ w satisfies the equation − div (cid:2) ( J T T ∇ z E τ ( ∇ u ) J T ) ◦ T − ∇ ˜ w (cid:3) ≤ h ( ∇ z E τ ( ∇ u ) J T ) ◦ T − ∇ ˜ w + 2( ∇ f · ∇ u ) ◦ T − in B + δ ( y ) , (3.25)where J T is the Jacobian matrix of T , i.e., J T = ∇ T , (cid:0) J T ◦ T − ∇ ˜ u (cid:1) i is the i-th component of the vector J T ◦ T − ∇ ˜ u , and the row vector h is roughlydiv( J T T J T ) and is, therefore, bounded by our assumption on T . In view of (3.24), we can extend ˜ u across the line y = 0 by setting˜ u ( − y , y , · · · , y N ) = ˜ u ( y , y , · · · , y N ) . XPONENTIAL NON-LINEARITY IN CRYSTAL SURFACE MODELS 13
Now equation (3.25) is satisfied in the whole ball B δ ( y ). That is, you have turned y into aninterior point, and thus the method employed to prove (3.24) becomes applicable. This means thatwe have the estimate k∇ u k ∞ ≤ c k∇ u k p + c ( k f k ℓ ) p − + + c ( τ k u k ℓ ) p − + c √ τ + c. By the interpretation inequality ([9], p.146), k∇ u k p ≤ ε k∇ u k ∞ + c ( ε ) k∇ u k . To complete the proof, we claim(3.26) k τ u k λ ≤ k f k λ for each λ ≥ . To see this, we introduce the function(3.27) h ε ( s ) = s > ε , s if | s | ≤ ε , − s < − ε , ε > . Obviously, lim ε → = sgn ( s ) ≡ s > s = 0, − s < λ ∈ [1 , ∞ ) be given. Then the function ( | u | + ε ) λ − h ε ( u ) is an increasing function of u . We easilycheck that it lies in W , (Ω) and ∇ (cid:0) | u | λ − u (cid:1) = (cid:2) ( λ − | u | + ε ) λ − | h ε ( u ) | + ( | u | + ε ) λ − h ′ ε ( u ) (cid:3) ∇ u .Multiply through (3.2) by this function and integrate the resulting equation over Ω to obtain Z Ω h ( λ − | u | + ε ) λ − | h ε ( u ) | + ( | u | + ε ) λ − h ′ ε ( u ) i F τ ( |∇ u | ) |∇ u | dx + τ Z Ω ( | u | + ε ) λ − h ε ( u ) u dx = Z Ω f ( | u | + ε ) λ − h ε ( u ) dx ≤ Z Ω | f | ( | u | + ε ) λ − dx. (3.28)Dropping the first integral in the above inequality and then let ε → τ Z Ω | u | λ ≤ Z Ω | f || u | λ − dx ≤ k f k λ k u k λ − λ , from which (3.26) follows. This completes the proof. (cid:3) Further regularity results for solutions to equations of the p-Laplace type can be found in [2, 17]and the references therein.Let D τ ( ∇ u ) be given as in (1.8). It is easy for us to verify that D τ ( ∇ u ) ξ · ξ = (1 + τ ) | ξ | − q ( ∇ u · ξ ) ( |∇ u | + τ ) (1 + q |∇ u | ) ≥ (cid:18)
11 + q |∇ u | + τ (cid:19) | ξ | for each ξ ∈ R N .(3.30)Furthermore, each entry in D τ ( ∇ u ) is bounded by 2 + τ .Let L >
0. Consider the boundary value problem − div ( D τ ( ∇ u ) ∇ ρ ) + τ ln( ρ + L ) = f in Ω , (3.31) ∇ ρ · ν = 0 on ∂ Ω.(3.32)A solution to this problem is a function ρ ∈ W , (Ω) such thatln( ρ + L ) ∈ L (Ω) and(3.33) Z Ω D τ ( ∇ u ) ∇ ρ ∇ ϕdx + τ Z Ω ln( ρ + L ) ϕdx = Z Ω f ϕdx for each ϕ ∈ W , (Ω) . (3.34) Of course, (3.33) implies ρ > − L a.e. on Ω. Lemma 3.2.
For each f ∈ L (Ω) there is a unique solution to (3.31) - (3.32) .Proof. For the existence part, we consider the approximate problem − div( D τ ( ∇ u ) ∇ ρ δ ) + δρ δ + τ ψ L,δ ( ρ δ ) = f in Ω , (3.35) ∇ ρ δ · ν = 0 on ∂ Ω,(3.36)where δ ∈ (0 ,
1) and(3.37) ψ L,δ ( s ) = (cid:26) ln ( s + L + δ ) if s > − L ,ln δ if s ≤ − L .Existence of a solution to this problem can be established via the Leray-Schauder Theorem. Tosee this, we define an operator B from L (Ω) into itself as follows: We say w = B ( v ) if w solvesproblem − div( D τ ( ∇ u ) ∇ w ) + δw = f − τ ψ L,δ ( v ) in Ω , (3.38) ∇ w · ν = 0 on ∂ Ω.(3.39)Note that ψ L,δ ( s ) ≥ ln δ . Thus for v ∈ L (Ω) we have ψ L,δ ( v ) ∈ L q (Ω) for each ≥
1. Problem(3.38)-(3.39) has a unique weak solution w in the space W , (Ω). That is, B is well-defined. It isalso easy for us to see that B is continuous and maps bounded sets into compact ones. Now weverify (LS3) in Lemma 2.1. Suppose that σ ∈ (0 , , w ∈ L (Ω) are such that w = σB ( w ), i.e., − div( D τ ( ∇ u ) ∇ w ) + δw = σ ( f − τ ψ L,δ ( w )) in Ω , (3.40) ∇ w · ν = 0 on ∂ Ω.(3.41)Use w as a test function in (3.40) to get τ Z Ω |∇ w | dx + δ Z Ω w dx ≤ σ Z Ω f wdx − στ Z Ω ( ψ L,δ ( w ) − ψ L,δ (0)) wdx − στ ψ L,δ (0) Z Ω wdx ≤ δ Z Ω w dx + 12 Z Ω ( f − ψ L,δ (0)) dx. This implies Z Ω w dx ≤ c ( δ ) . Thus (LS3) in Lemma 2.1 holds. As a result, problem (3.35)-(3.36) has a solution.Next, we proceed to show that we can take δ → s z ≡ − L − δ is the zero point of ψ L,δ , i.e., ψ L,δ ( s z ) = 0. Use ρ δ − s z as a test function in (3.31) to get τ Z Ω |∇ ρ δ | dx + δ Z Ω ( ρ δ − s z ) dx + τ Z Ω ψ L,δ ( ρ δ )( ρ δ − s z ) dx ≤ Z Ω ( f − δs z )( ρ δ − s z ) dx ≤ k f − δs z k NN +2 k ( ρ δ − s z ) k NN − ≤ c k f − δs z k NN +2 ( k∇ ρ δ k + k ( ρ δ − s z ) k ) ≤ ε k∇ ρ δ k + c k f − δs z k NN +2 + c k f − δs z k NN +2 k ( ρ δ − s z ) k . XPONENTIAL NON-LINEARITY IN CRYSTAL SURFACE MODELS 15
Consequently, Z Ω |∇ ρ δ | dx + Z Ω ψ L,δ ( ρ δ )( ρ δ − s z ) dx ≤ c k f − δs z k NN +2 k ( ρ δ − s z ) k + c k f − δs z k NN +2 ≤ c k ( ρ δ − s z ) k + c. (3.42)This together with the definition of ψ L,δ implies Z Ω | ρ δ − s z | dx = Z { ρ δ − s z ≤− (1 − δ ) } | ρ δ − s z | dx + Z {− (1 − δ ) <ρ δ − s z < δ } | ρ δ − s z | dx + Z { ρ δ − s z ≥ δ } | ρ δ − s z | dx ≤ | ln δ | Z Ω ψ L,δ ( ρ δ )( ρ δ − s z ) dx + 1 δ | Ω | + 1ln(1 + δ ) Z Ω ψ L,δ ( ρ δ )( ρ δ − s z ) dx. (3.43)Combining this with (3.42), we conclude that there exists a δ ∈ (0 ,
1) such that(3.44) k ρ δ k ≤ c for all δ ≤ δ . In light of the Sobolev inequality (1.19), we see that { ρ δ } is bounded in W , (Ω). We may assumethat(3.45) ρ δ → ρ weakly in W , (Ω), strongly in L (Ω), and a.e. on Ω.By suitably modifying the test function in (3.28) (i.e., use ( | ψ L,δ ( ρ δ ) | + ε ) λ − h ε ( ρ δ − s z ) instead),we can derive that k τ ψ L,δ ( ρ δ ) k ≤ k f k . (3.46)By Fatou’s lemma, we have Z Ω | ln( ρ + L ) | dx ≤ lim δ → Z Ω ψ L,δ ( ρ δ ) dx ≤ c. We are ready to pass to the limit in (3.35).The uniqueness of a solution to (3.31)-(3.32) is trivial because ln( ρ τ + L ) is strictly monotone.The proof is complete. (cid:3) Lemma 3.3.
For each positive integer j the function ln j (1 + s ) can be represented as a power series (3.47) ln j (1 + s ) = ∞ X n = j a ( j ) n s n on ( − , .Furthermore, (3.48) lim sup n →∞ | a ( j ) n | n ≤ . Proof.
It is well known that ln(1 + s ) = ∞ X n =1 ( − n − n s n on ( − , Thus the fact that the function ln j (1 + s ) does have a power series representation (3.47) is simplythe repeated application of the theorem for the Cauchy product of power series. We just need tofocus on (3.48). If j = 1, (3.48) is trivially true. If j = 2, we invoke the just-mentioned Cauchyproduct theorem , thereby obtainingln (1 + s ) = ∞ X n =1 ( − n − n s n ∞ X n =1 ( − n − n s n = ∞ X n =2 a (2) n s n on (-1, 1) , where a (2) n = ( − n − n − X k =1 k ( n − k ) . It is easy to check n ≥ k ( n − k ) ≥ n − k = 1 , · · · , n − n − n ≤ | a (2) n | ≤ j = m ≥
2. Then for each ε > J such that(3.49) | a ( m ) n | ≤ (1 + ε ) n whenever n ≥ J .Using the Cauchy product again, we haveln m +1 (1 + s ) = ∞ X n = m a ( m ) n s n ∞ X n =1 ( − n − n s n = ∞ X n = m +1 a ( m +1) n s n on (-1, 1) , where a ( m +1) n = n − X k = m ( − n − k − a ( m ) k n − k . Set L J = max {| a ( m ) m | , · · · , | a ( m ) J − |} . By (3.49), we obtain (cid:12)(cid:12)(cid:12) a ( m +1) n (cid:12)(cid:12)(cid:12) ≤ max { L J , (1 + ε ) n } (cid:18) · · · + 1 n − m (cid:19) ≤ max { L J , (1 + ε ) n } (1 + ln( n − m )) for n ≥ m + 1 . Here we have used the estimate for the partial sums of the harmonic series, i.e., 1 + + · · · + n < n for n >
1. We are ready to estimatelim sup n →∞ (cid:12)(cid:12)(cid:12) a ( m +1) n (cid:12)(cid:12)(cid:12) n ≤ ε. Since ε is arbitrary, we yield (3.48). (cid:3) Proof of the main theorem
The proof of the main theorem will be divided into several claims. We begin by showing theexistence of a solution to our approximate problems.
Claim 4.1.
Let the assumptions of the main theorem hold. Then there is a weak solution ( ρ, u ) to (1.9) - (1.11) in the space W , (Ω) × W , (Ω) . XPONENTIAL NON-LINEARITY IN CRYSTAL SURFACE MODELS 17
Proof.
The existence assertion will be established via the Leray-Schauder Theorem. To do this, wedefine an operator B from W , (Ω) into itself as follows: For each v ∈ W , (Ω) we first solve theproblem − div ( D τ ( ∇ v ) ∇ ρ ) + τ ln( ρ + L ) = f − av in Ω , (4.1) ∇ ρ · ν = 0 on ∂ Ω . (4.2)By Lemma 3.2, there is a unique weak solution ρ ∈ W , (Ω) with ln( ρ + L ) ∈ L (Ω) to the aboveproblem. We use the function ρ so obtained to form the problem − div (cid:0) ( F τ ( |∇ u | ) + τ ) ∇ u (cid:1) + τ u = ln( ρ + L ) in Ω , (4.3) ∇ u · ν = 0 on ∂ Ω . (4.4)Note that the difference between (4.3) and (3.2) is that here we have added a τ to F τ . This is toensure that we can obtain a solution u in W , (Ω). Obviously, the conclusions of Lemma 3.1 stillhold for (4.3)-(4.4). Thus there is a unique weak solution u ∈ W , (Ω) to (4.3)-(4.4). We define B ( v ) = u. We can easily conclude that B is well-defined. Claim 4.2. B is continuous and maps bounded sets into precompact ones.Proof. We first show that { v n } is bounded in W , (Ω) ⇒ { B ( v n ) } is precompact in W , (Ω) . To see this, set u n = B ( v n ) . Then we have − div ( D τ ( ∇ v n ) ∇ ρ n ) + τ ln( ρ n + L ) = f − av n in Ω , (4.5) − div (cid:0) ( F τ ( |∇ u n | ) + τ ) ∇ u n (cid:1) + τ u n = ln( ρ n + L ) in Ω , (4.6) ∇ u n · ν = 0 on ∂ Ω , (4.7) ∇ ρ n · ν = 0 on ∂ Ω . (4.8)Use ρ n − (1 − L ) as a test function in (4.5), then employ a calculation similar to (3.42) and (3.43),and thereby obtain Z Ω |∇ ρ n | dx + Z Ω | ρ n | dx ≤ c. Moreover, the proof of (3.26) implies that(4.9) k τ ln( ρ n + L ) k ≤ k f − av n k ≤ c. Next, we use u n as a test function in (4.6) to get Z Ω |∇ u n | dx + Z Ω u n dx ≤ c ( τ ) . We may assume u n → u weakly in W ,p (Ω), strongly in L p (Ω), and a.e. on Ω ,ρ n → ρ weakly in W , (Ω), strongly in L (Ω), and a.e. on Ω(pass to subsequences if necessary.) This combined with (4.9) impliesln( ρ n + L ) → ln( ρ + L ) weakly in L (Ω). With this in mind, we derive from (3.1) and (4.6) that τ Z Ω |∇ ( u n − u ) | dx ≤ Z Ω (cid:2) ( F τ ( |∇ u n | ) + τ ) ∇ u n − ( F τ ( |∇ u | ) + τ ) ∇ u (cid:3) · ∇ ( u n − u ) dx = Z Ω (ln( ρ n + L ) − τ u n )( u n − u ) dx − Z Ω ( F τ ( |∇ u | ) + τ ) ∇ u · ∇ ( u n − u ) dx → n → ∞ .(4.10)If v n → v strongly in W , (Ω), then we can infer from (1.8) that D τ ( ∇ v n ) → D τ ( ∇ v ) strongly in ( L s (Ω)) N × N for each s ≥ (cid:3) We still need to show that there is a positive number c such that(4.11) k u k W , (Ω) ≤ c for all u ∈ W , (Ω) and σ ∈ (0 ,
1] satisfying u = σB ( u ) . This equation is equivalent to the boundary value problem − div ( D τ ( ∇ u ) ∇ ρ ) + τ ln( ρ + L ) = f − au in Ω , (4.12) − div (cid:18) ( F τ ( (cid:12)(cid:12)(cid:12) ∇ uσ (cid:12)(cid:12)(cid:12) ) + τ ) ∇ u (cid:19) + τ u = σ ln( ρ + L ) in Ω , (4.13) ∇ u · ν = ∇ ρ · ν = 0 on ∂ Ω . (4.14)Use ln( ρ + L ) as a test function in (4.12) to get(4.15) τ Z Ω |∇ ρ | ρ + L dx + τ Z Ω ln ( ρ + L ) ≤ Z Ω f ln( ρ + L ) dx − a Z Ω u ln( ρ + L ) dx. We can show that the last integral in the preceding inequality is non-negative by using u as a testfunction in (4.13) as follows:(4.16) σ Z Ω u ln( ρ + L ) dx = τ Z Ω u dx + Z Ω ( F τ ( (cid:12)(cid:12)(cid:12) ∇ uσ (cid:12)(cid:12)(cid:12) ) + τ ) |∇ u | dx ≥ . Substituting this into (4.15), we obtain Z Ω ln ( ρ + L ) ≤ c ( τ ) . This combined with (4.16) yields the desired result. (cid:3)
We indicate the dependence of our approximate solutions on τ by writing ρ = ρ τ , u = u τ . Then problem (1.9)-(1.11) becomes − div ( D τ ( ∇ u τ ) ∇ ρ τ ) + τ ln( ρ τ + L ) = f − au τ in Ω , (4.17) − div (cid:0) ( F τ ( |∇ u τ | ) + τ ) ∇ u τ (cid:1) + τ u τ = ln( ρ τ + L ) in Ω , (4.18) ∇ u τ · ν = ∇ ρ τ · ν = 0 on ∂ Ω . (4.19)We proceed to derive estimates for { u τ , ρ τ } that are uniform in τ . XPONENTIAL NON-LINEARITY IN CRYSTAL SURFACE MODELS 19
Claim 4.3.
We have Z Ω (cid:12)(cid:12)(cid:12) ∇ p ρ τ + L (cid:12)(cid:12)(cid:12) q dx ≤ c, q = 2 pp + 1 , (4.20) k u τ k W .p (Ω) ≤ c. (4.21) Proof.
Multiply through (4.18) by τ and add the resulting equation to (4.17) to get(4.22) − div ( D τ ( ∇ u τ ) ∇ ρ τ ) − τ div (cid:0) ( F τ ( |∇ u τ | ) + τ ) ∇ u τ (cid:1) + ( a + τ ) u τ = f. Integrate the above equation over Ω to obtain(4.23) (cid:12)(cid:12)(cid:12)(cid:12)Z Ω u τ dx (cid:12)(cid:12)(cid:12)(cid:12) = 1 a + τ (cid:12)(cid:12)(cid:12)(cid:12)Z Ω f dx (cid:12)(cid:12)(cid:12)(cid:12) ≤ c. Use ln( ρ τ + L ) as a test function in (4.17) to get Z Ω (cid:18)
11 + q |∇ u τ | + τ (cid:19) |∇ ρ τ | ρ τ + L dx + τ Z Ω ln ( ρ τ + L ) dx ≤ Z Ω ( f − au τ ) ln( ρ τ + L ) dx. (4.24)Use u τ , f as test functions in (4.18) successively to get Z Ω (cid:0) F τ ( |∇ u τ | ) + τ (cid:1) |∇ u τ | dx + τ Z Ω u τ dx = Z Ω u τ ln( ρ τ + L ) dx, (4.25) Z Ω (cid:0) F τ ( |∇ u τ | ) + τ (cid:1) ∇ u τ · ∇ f dx + τ Z Ω u τ f dx = Z Ω f ln( ρ τ + L ) dx. Use the above two equations in (4.24) to deduce Z Ω q |∇ u τ | )( ρ τ + L ) |∇ ρ τ | dx + τ Z Ω ln ( ρ τ + L )( ρ τ ) dx + a Z Ω (cid:0) F τ ( |∇ u τ | ) + τ (cid:1) |∇ u τ | dx + aτ Z Ω u τ ≤ Z Ω (cid:0) F τ ( |∇ u τ | ) + τ (cid:1) ∇ u τ · ∇ f dx + τ Z Ω u τ f dx. (4.26)Note that a Z Ω |∇ u τ | p dx + aβ Z Ω |∇ u τ | dx = a Z Ω (cid:0) |∇ u τ | + τ (cid:1) p − ( |∇ u τ | + τ ) dx + aβ Z Ω (cid:0) |∇ u τ | + τ (cid:1) − ( |∇ u τ | + τ ) dx ≤ a Z Ω F τ ( |∇ u τ | ) |∇ u τ | dx + aτ p | Ω | + aβ τ | Ω | . (4.27)Here we have used the fact that p ≤
2. Use this again to get Z Ω (cid:12)(cid:12) F τ ( |∇ u τ | ) ∇ u τ · ∇ f (cid:12)(cid:12) dx ≤ Z Ω |∇ u τ | p − |∇ f | dx + β Z Ω |∇ f | dx ≤ k∇ u τ k p − p k∇ f k p + c k∇ f k . (4.28) Plug this and (4.27) into (4.27) and apply Young’s inequality (1.17) in the resulting inequalityappropriately to derive Z Ω q |∇ u τ | )( ρ τ + L ) |∇ ρ τ | dx + τ Z Ω ln ( ρ τ + L ) dx + Z Ω |∇ u τ | p dx + τ Z Ω |∇ u τ | dx + τ u τ dx ≤ c Z Ω |∇ f | p dx + cτ Z Ω |∇ f | dx + cτ Z Ω | f | dx + c ≤ c. (4.29)By virtue of the Sobolev inequality and (4.23), we have Z Ω | u τ | p dx ≤ p − (cid:18)Z Ω (cid:12)(cid:12)(cid:12)(cid:12) u τ − | Ω | Z Ω u τ dx (cid:12)(cid:12)(cid:12)(cid:12) p dx + 1 | Ω | p (cid:18)Z Ω u τ dx (cid:19) p (cid:19) ≤ c Z Ω |∇ u τ | p dx + c ≤ c. (4.30)This gives (4.21). Let q be give as in (4.20). Then q − q = p . We calculate from (4.29) that Z Ω (cid:12)(cid:12)(cid:12) ∇ p ρ τ + L (cid:12)(cid:12)(cid:12) q dx = Z Ω (1 + q |∇ u τ | ) q (cid:12)(cid:12) ∇√ ρ τ + L (cid:12)(cid:12) q (1 + q |∇ u τ | ) q dx ≤ Z Ω (cid:12)(cid:12) ∇√ ρ τ + L (cid:12)(cid:12) q |∇ u τ | dx ! q (cid:18)Z Ω (1 + q |∇ u τ | ) q − q dx (cid:19) − q ≤ c. (4.31)The proof is complete. (cid:3) The following claim constitutes the core of our development.
Claim 4.4.
For each
L > there is a positive number c = c ( L ) such that (4.32) Z Ω | ln( ρ τ + L ) | dx ≤ c. Proof.
Suppose that there is a subsequence of { ρ τ } , still denoted by { ρ τ } , such that(4.33) lim τ → |{ ρ τ < − L }| = δ > . Since (cid:16)p ρ τ + L − (cid:17) + (cid:12)(cid:12)(cid:12)(cid:12) { ρ τ < − L } = 0 , we can conclude from Lemma 2.2 and (4.31) that Z Ω (cid:16)p ρ τ + L − (cid:17) + dx ≤ c Z Ω (cid:12)(cid:12)(cid:12)(cid:12) ∇ (cid:16)p ρ τ + L − (cid:17) + (cid:12)(cid:12)(cid:12)(cid:12) dx ≤ c. Note that ln + ( ρ τ + L ) = 2 ln + (1 + p ρ τ + L − ≤ (cid:16)p ρ τ + L − (cid:17) + . Integrate (4.18) over Ω to get Z Ω ln( ρ τ + L ) dx = τ Z Ω u τ dx. This gives(4.34) Z Ω ln − ( ρ τ + L ) dx = Z Ω ln + ( ρ τ + L ) dx − τ Z Ω u τ dx. XPONENTIAL NON-LINEARITY IN CRYSTAL SURFACE MODELS 21
Consequently, Z Ω | ln( ρ τ + L ) | dx = Z Ω ln − ( ρ τ + L ) + Z Ω ln + ( ρ τ + L ) dx = 2 Z Ω ln + ( ρ τ + L ) dx − τ Z Ω u τ dx ≤ Z Ω (cid:16)p ρ τ + L − (cid:17) + dx + c ≤ c. If our assumption (4.33) is not true, then we have(4.35) lim τ → |{ ρ τ ≤ − L }| = 0 . Remember that |{ ρ τ < − L }| + |{ ρ τ ≥ − L }| = | Ω | . Hence(4.36) lim τ → |{ ρ τ ≥ − L }| = | Ω | . From here on we assume that
L > . (4.37)Then pick a number β ≥
1. We use (cid:2) ( − ρ τ ) β − ( L − β (cid:3) + as a test function in (4.17) to get β Z { ρ τ ≤− ( L − } ( − ρ τ ) β − |∇ ρ τ | q |∇ u τ | dx ≤ Z Ω ( au τ − f ) h ( − ρ τ ) β − ( L − β i + dx. (4.38)Note that ( − ρ τ ) β − |∇ ρ τ | χ { ρ τ ≤− ( L − } = 4( β + 1) (cid:12)(cid:12)(cid:12)(cid:12) ∇ h ( − ρ τ ) β +12 − ( L − β +12 i + (cid:12)(cid:12)(cid:12)(cid:12) . Plug this into (4.38) to get Z Ω
11 + q |∇ u τ | (cid:12)(cid:12)(cid:12)(cid:12) ∇ h ( − ρ τ ) β +12 − ( L − β +12 i + (cid:12)(cid:12)(cid:12)(cid:12) dx ≤ ( β + 1) β Z Ω ( au τ − f ) h ( − ρ τ ) β − ( L − β i + dx ≤ ( β + 1) Z { ρ τ ≤− ( L − } | au τ − f | ( − ρ τ ) β − dx. (4.39)Let q be given as in (4.20). Using a calculation similar to (4.31), we arrive at(4.40) (cid:13)(cid:13)(cid:13)(cid:13) ∇ h ( − ρ τ ) β +12 − ( L − β +12 i + (cid:13)(cid:13)(cid:13)(cid:13) q ≤ c ( β + 1) Z { ρ τ ≤− ( L − } | au τ − f | ( − ρ τ ) β − dx ! . Obviously, h ( − ρ τ ) β +12 − ( L − β +12 i + (cid:12)(cid:12)(cid:12)(cid:12) { ρ τ ≥ − L } = 0 . This together with (4.36) enables us to invoke Lemma 2.2. Upon doing so, we obtain (cid:13)(cid:13)(cid:13)(cid:13)h ( − ρ τ ) β +12 − ( L − β +12 i + (cid:13)(cid:13)(cid:13)(cid:13) q ∗ ≤ c (cid:13)(cid:13)(cid:13)(cid:13) ∇ h ( − ρ τ ) β +12 − ( L − β +12 i + (cid:13)(cid:13)(cid:13)(cid:13) q ≤ c ( β + 1) Z { ρ τ ≤− ( L − } | au τ − f | ( − ρ τ ) β − dx ! ≤ c ( β + 1) Z { ρ τ ≤− ( L − } ( − ρ τ ) ( β − p ∗ p ∗− dx ! p ∗− p ∗ . (4.41)Remember that − L ≤ ρ τ ≤ − ( L −
1) on the set { ρ τ ≤ − ( L − } . With this in mind, we estimate Z { ρ τ ≤− ( L − } ( − ρ τ ) β +12 dx ≤ (cid:13)(cid:13)(cid:13)(cid:13)h ( − ρ τ ) β +12 − ( L − β +12 i + (cid:13)(cid:13)(cid:13)(cid:13) + c ( L − β +12 ≤ c ( β + 1) Z { ρ τ ≤− ( L − } ( − ρ τ ) ( β − p ∗ p ∗− dx ! p ∗− p ∗ + c ( L − β +12 ≤ c ( β + 1) L β (cid:16) − p ∗− p ∗ (cid:17) ( L − + p ∗− p ∗ Z { ρ τ ≤− ( L − } ( − ρ τ ) β +12 dx ! p ∗− p ∗ + c ( L − β +12 . (4.42)Set Y τ = Z { ρ τ ≤− ( L − } ( − ρ τ ) β +12 dx. If lim sup τ → Y τ >
1, then there is a subsequence of { Y τ } , still denoted by { Y τ } , such that(4.43) Y τ > . As a result, we can conclude from (4.42) that Y τ ≤ c ( β + 1) L β (cid:16) − p ∗− p ∗ (cid:17) ( L − + p ∗− p ∗ ( Y τ ) p ∗− p ∗ + c ( L − β +12 ≤ c ( β + 1) L β (cid:16) − p ∗− p ∗ (cid:17) ( L − + p ∗− p ∗ ( Y τ ) + c ( L − β +12 . (4.44)This is a quadratic inequality in ( Y τ ) . Solving it yields Y τ ≤ c ( β + 1) L β (cid:16) − p ∗− p ∗ (cid:17) ( L − + p ∗− p ∗ + vuuut c ( β + 1) L β (cid:16) − p ∗− p ∗ (cid:17) ( L − + p ∗− p ∗ + 4 c ( L − β +12 ≤ c ( β + 1) L β (cid:16) − p ∗− p ∗ (cid:17) ( L − + p ∗− p ∗ + c ( L − β +14 . (4.45)In view of (4.43), we always have(4.46) Y τ ≤ c ( β + 1) L β (cid:16) − p ∗− p ∗ (cid:17) ( L − + p ∗− p ∗ + c ( L − β +14 + 1 at least for a subsequence. XPONENTIAL NON-LINEARITY IN CRYSTAL SURFACE MODELS 23
We easily see that ln( ρ τ + L ) = ln L + ln (cid:16) ρ τ L (cid:17) = ln L − ∞ X n =1 nL n ( − ρ τ ) n for (cid:12)(cid:12) ρ τ L (cid:12)(cid:12) < β = 2 n − Z { ρ τ ≤− ( L − } ( − ρ τ ) n dx ≤ cnL n (cid:16) − p ∗− p ∗ (cid:17) L (cid:16) − p ∗− p ∗ (cid:17) ( L − + p ∗− p ∗ + c ( L − n + 1 ≤ cnL n (cid:16) − p ∗− p ∗ (cid:17) L − c ( L − n + 1 . (4.48)Equipped with this, we estimate Z Ω ln − ( ρ τ + L ) dx ≤ c ln L + ∞ X n =1 nL n Z { ρ τ ≤− ( L − } ( − ρ τ ) n dx ≤ c ln L + cL − ∞ X n =1 L ( p ∗− n p ∗ + c ∞ X n =1 n (cid:18) √ L − L (cid:19) n + ∞ X n =1 nL n ≤ c ( L ) . (4.49)We can now conclude the lemma by appealing to (4.34). (cid:3) Claim 4.5.
There exist a subsequence of { ρ τ } , still denoted by { ρ τ } , and a finite a.e. function ρ such that (4.50) ρ τ → ρ a.e. on Ω as τ → .Proof. Let q be given as (4.20). We easily obtain that { arctan( √ ρ τ + L ) } is bounded in W ,q (Ω).Hence we can extract an a.e. convergent subsequence, which we still denote by (cid:8) arctan( √ ρ τ + L ) (cid:9) .Since the function arctan( √ s + L ) is a strictly increasing function of s , { ρ τ } also converges a.e..We call the limit ρ . To see that ρ is finite a.e. on Ω, we appeal to Fatou’s lemma and (4.32) to get(4.51) Z Ω | ln( ρ + L ) | dx = Z Ω lim τ → | ln( ρ τ + L ) | dx ≤ lim sup τ → Z Ω | ln( ρ τ + L ) | dx ≤ c. The proof is complete. (cid:3)
Claim 4.6.
The sequence { ln( ρ τ + L ) } is bounded in L s (Ω) for each s ≥ .Proof. Since ρ is finite a.e., there must exist a positive number L such that(4.52) |{ ρ ≤ L }| > . According to Egoroff’s theorem, for each ε > K ⊂ { ρ ≤ L } such that |{ ρ ≤ L } \ K | < ε and ρ τ → ρ uniformly on K . We take ε = |{ ρ ≤ L }| . Then the measure ofthe corresponding K is bigger than |{ ρ ≤ L }| . We easily conclude from the uniform convergencethat there is a number τ ∈ (0 ,
1) such that | ρ τ − ρ | ≤ K for each τ ≤ τ .Consequently,(4.53) ρ τ ≤ L + 1 on K for τ ≤ τ . We deduce from Lemma 2.2 and (4.20) that Z Ω (cid:20)(cid:16)p ρ τ + L − p L + 1 + L (cid:17) + (cid:21) NqN − q dx ! N − qN ≤ c Z Ω (cid:12)(cid:12)(cid:12)(cid:12) ∇ (cid:16)p ρ τ + L − p L + 1 + L (cid:17) + (cid:12)(cid:12)(cid:12)(cid:12) q dx ≤ c, (4.54)from whence follows Z Ω | ρ τ | NpN ( p +1) − p dx ≤ c. Recall that for each α > ρ τ →∞ ln α ( ρ τ + L ) ρ NpN ( p +1) − p τ = 0 . This immediately implies that { ln + ( ρ τ + L ) } is bounded in L s (Ω) for each s ≥ R Ω | ln( ρ + L ) | dx < ∞ . We have |{ ρ = − L }| = 0 . There must exist an ε > |{ ρ + L ≥ ε }| > . We can infer from Egorroff’s theorem that there is a subset K ⊂ { ρ + L ≥ ε } with positive measuresuch that ρ τ + L ≥ ε K at least for τ sufficiently small.If ε ≥
1, then |{ ρ τ + L ≥ }| ≥ |{ ρ τ + L ≥ ε }| ≥ | K | > . Thus (4.48) remains valid. For each positive integer j we have from the Binomial Theorem thatln j ( ρ τ + L ) = (cid:16) ln L + ln (cid:16) ρ τ L (cid:17)(cid:17) j = j X m =0 (cid:18) jm (cid:19) ln j − m L ln m (cid:16) ρ τ L (cid:17) . We estimate from (3.47) and (4.48) that Z { ρ τ ≤− L +1 } (cid:12)(cid:12)(cid:12) ln m (cid:16) ρ τ L (cid:17)(cid:12)(cid:12)(cid:12) dx ≤ ∞ X n = m (cid:12)(cid:12)(cid:12) a ( m ) n (cid:12)(cid:12)(cid:12) L n Z { ρ τ + L ≤ } ( − ρ τ ) n dx ≤ ∞ X n = m (cid:12)(cid:12)(cid:12) a ( m ) n (cid:12)(cid:12)(cid:12) L n cnL n (cid:16) − p ∗− p ∗ (cid:17) L − c ( L − n + 1 ≤ cL − ∞ X n = m n (cid:12)(cid:12)(cid:12) a ( m ) n (cid:12)(cid:12)(cid:12) L ( p ∗− n p ∗ + c ∞ X n = m (cid:12)(cid:12)(cid:12) a ( m ) n (cid:12)(cid:12)(cid:12) (cid:18) √ L − L (cid:19) n + ∞ X n = m (cid:12)(cid:12)(cid:12) a ( m ) n (cid:12)(cid:12)(cid:12) L n . (4.55)Remember L >
1. The root test and (3.48) asserts that each series on the right-hand side in thepreceding inequality is convergent. This gives the desired result.
XPONENTIAL NON-LINEARITY IN CRYSTAL SURFACE MODELS 25 If ε <
1, we change the test function to (cid:2) ( − ρ τ ) β − ( L − ε ) β (cid:3) + in the proof of (4.38). All thesubsequent calculations remain valid with L − L − ε . We are eventually led to(4.56) Z { ρ τ ≤− L + ε } ( − ρ τ ) n dx ≤ cnL n (cid:16) − p ∗− p ∗ (cid:17) L − ε + c ( L − ε ) n + 1 . In view of (4.55), we can conclude the claim. (cid:3)
Now we are in a position to invoke Lemma 3.1. Upon doing so, we arrive at(4.57) k u τ k ∞ ≤ c, k∇ u τ k ∞ ≤ c. Claim 4.7.
The sequence { u τ } is precompact in W ,s (Ω) for each s < ∞ . Therefore, we mayassume that {∇ u τ } converges a.e. on Ω . The essence of the proof has already been demonstrated in Claim 4.2. The only difference hereis that in (4.10) we use (3.1) instead. We shall omit the details.
Claim 4.8.
The sequence { ρ τ } is bounded in W , (Ω) .Proof. Let L be given as in (4.53). We use ( ρ τ − L − + as a test function in (4.17) and keep(4.57) in mind to get Z Ω |∇ ( ρ τ − L − + | dx ≤ c Z Ω ( f − au τ )( ρ τ − L − + dx ≤ c k f − au τ k NN +2 k ( ρ τ − L − + k NN − ≤ c k∇ ( ρ τ − L − + k . The last step is due to Lemma 2.2. Hence k∇ ( ρ τ − L − + k ≤ c . Apply Lemma 2.2 again to get Z Ω | ρ τ | NN − dx ≤ c. Use ρ τ − L as a test function in (4.17) to obtain the desired result. (cid:3) We infer from (1.8) and Claim 4.7 that for a.e. z ∈ Ω D τ ( ∇ u τ ( z )) → (cid:26) D ( ∇ u ( z )) if ∇ u ( z ) = 0, I if ∇ u ( z ) = 0 . That is, each entry of D τ ( ∇ u τ ) converges a.e on Ω. It is also bounded. Therefore, D τ ( ∇ u τ ) ∇ ρ τ → D ( ∇ u ( z )) ∇ ρ weakly in (cid:0) L (Ω) (cid:1) N . We may assume that ∇ u τ ( |∇ u τ | + τ ) → ϕ weak ∗ in ( L ∞ (Ω)) N . We claim(4.58) ϕ ( x ) ∈ ∂ z H ( ∇ u ( x )) for a.e. x ∈ Ω,where H is given as in (1.15). To see this, we derive Claim 4.7 that ∇ u τ ( z )( |∇ u τ ( z ) | + τ ) → ∇ u ( z ) |∇ u ( z ) | = ϕ ( z ) for a.e. z ∈ {|∇ u | > } . We always have (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∇ u τ ( |∇ u τ | + τ ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ . Consequently, | ϕ | ≤
1. This gives (4.58).We are ready to pass to the limit in (4.17)-(4.19) to conclude the proof of the main theorem.
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