Uniqueness of solution of an inverse source problem for ultrahyperbolic equations
aa r X i v : . [ m a t h . A P ] S e p Uniqueness of solution of an inverse source problemfor ultrahyperbolic equations
Fikret G¨olgeleyen ∗ and Masahiro Yamamoto † , ‡ ∗ Department of Mathematics, Faculty of Arts and Sciences, Zonguldak B¨ulentEcevit University, Zonguldak 67100 Turkey † Department of Mathematical Sciences, The University of Tokyo, 3-8-1 Komaba,Meguro, Tokyo 153-8914 Japan ‡ Honorary Member of Academy of Romanian Scientists,Splaiul Independentei Street, no 54, 050094 Bucharest Romania
E-mail: [email protected], [email protected]
ABSTRACT
The aim of this article is to investigate the uniqueness of solution of an inverse problemfor ultrahyperbolic equations. We first reduce the inverse problem to a Cauchy problemfor an integro-differential equation and then by using a pointwise Carleman type inequalitywe prove the uniqueness.
1. INTRODUCTION AND THE MAIN RESULT
In this article, we consider an inverse problem for an ultrahyperbolic equation.One of our motivations to deal with this equation is its interesting structure fromthe point of view of the theory of partial differential equations. For instance, de-pending on the specific form of initial conditions, solutions possess both hyperbolicand elliptic properties (see [11]). Another motivation is recent discussions on thepossibility of physics in multiple time dimensions, (e.g., [3,16,17]). Namely, in somesuperstring theories which attempt to unify the general theory of relativity and thequantum mechanics, extra dimensions are required for the consistency of theory.When the presence of more than one temporal dimension is considered, the mathe-matical model occurs as an ultrahyperbolic equation (e.g., [7]). More precisely, the1aper [7] asserts that the equation in a form of ∂ xx u ( x, y , ..., y m ) − m X j =1 ∂ y j y j u ( x, y , ..., y m ) = F ( x, y , ..., y m )is of central physical importance, which describes the dynamical evolution of manyphysical quantities of classical and quantum field theories including the compo-nents of the electromagnetic fields in the case of a single time dimension, while theequation in a form of n X i =1 ∂ x i x i u ( x , ..., x n , y , ..., y m ) − m X j =1 ∂ y j y j u ( x , ..., x n , y , ..., y m ) = F ( x , ..., x n , y , ..., y m )is fundamental where x ∈ R n and y ∈ R m are respectively time-like variables andspace-like variables.Let n, m ≥
2. Inspired by [7,16,17], we here consider an ultrahyperbolic equationin u ( x, y ) := u ( x , ..., x n , y , ..., y m ), which is associated with general geometry inthe space variables y : Lu ( x, y ) ≡ n X i =1 ∂ x i x i u ( x, y ) − m X i,j =1 a ij ( x, y , ..., y m − ) u y i y j ( x, y )+ n X i =1 a i ( x, y ) u x i ( x, y ) + m X j =1 b j ( x, y ) u y j ( x, y ) + a ( x, y ) u ( x, y )= f ( x, y ) g ( x, y , ..., y m − ) (1)in the domain Ω = D × G . Here D ⊂ R n and G ⊂ R m are bounded domains,and we assume that Ω ⊂ R n + m is supported by the plane x = 0, G = G ′ × I with an open interval I and G ′ ⊂ R m − , and the coefficients are assumed to sat-isfy a ij ( x, y , ..., y m − ) ∈ C ( D × G ′ ) , a i ( x, y ) , b j ( x, y ) ∈ C (cid:0) Ω (cid:1) ( i = 0 , , ..., n ; j =1 , ..., m ) , f ∈ C (cid:0) Ω (cid:1) .The purpose of this article is to investigate the uniqueness of solution of the2ollowing problem: Problem.
For given u ( x, y , ..., y m − ) , find a pair of functions ( u ( x, y ) , g ( x, y , ..., y m − )) in Ω satisfying equation (1), Cauchy data u (0 , x , ..., x n , y ) = u x (0 , x , ..., x n , y ) = 0 (2) and the additional information u ( x, y , ..., y m − ,
0) = u ( x, y , ..., y m − ) . (3)This is an inverse problem of determining a factor g which is independent of thecomponent y m of the source in (1) which causes the action under consideration.This inverse problem is called an inverse source problem.Our main result is stated in Theorem 1: Theorem 1.
Let f ( x, y ′ , = 0 and the functions a ij satisfy − m X i,j =1 ∂a ij ∂x ξ i ξ j ≥ α | ξ | , α > for any ξ ∈ R m and ( x, y ′ ) ∈ D × G ′ . Then Problem has at most one solution ( u, g ) such that ( u, g ) ∈ C (Ω) × C ( D × G ′ ) . Inverse problems for ultrahyperbolic equations were studied in [1, 2, 4, 9], wherethe key method is based on weighted a priori estimates and was firstly developed byBukhgeim and Klibanov [4]. A uniqueness theorem for ultrahyperbolic equations,is given by [4] for a bounded domain with Dirichlet and Numann type condition ona part of the boundary. In [1] and [2], uniqueness is invesigated in an unboundeddomain with an additional information for the solution of direct problem at y = 0 . In [9], H¨older stability estimates were obtained in a bounded domain by some lateralboundary data. A major difference of our work from the existing results is that, in3roblem, additional information is given only at y m = 0 . As for the direct problem (1) - (2) with given f g , it is known that the prob-lem of determination of the function u from relations (1) and (2) is ill-posed in theHadamard sense (see [12], Chapter 4). By using the mean-value theorem of Asgeirs-son, it was shown by [6] that the existence of solutions fails if the initial conditionsare not properly prescribed. We refer to [5, 8, 13–15], as for the uniqueness resultsfor various Cauchy, Dirichlet and Neumann problems for ultrahyperbolic equations.Finally, in [7] it is proved that under a nonlocal constraint, the initial value problemis well-posed for initial data given on a codimension-one hypersurface.
2. KEY CARLEMAN ESTIMATE
We set x = ( x , ′ x ) ∈ R n , ′ x = ( x , ..., x n ) ∈ R n − ,y = ( y ′ , y m ) ∈ R m , y ′ = ( y , ..., y m − ) ∈ R m − , and ∂ x i = ∂∂x i , ∂ y j = ∂∂y j , ∇ ′ x = ( ∂ x , ∂ x , · · · , ∂ x n ) , ∇ y = ( ∂ y , ∂ y , · · · , ∂ y m ) , ∆ x = n X i =1 ∂ x i x i , ∆ ′ x = n X i =2 ∂ x i x i . In order to prove Theorem 1, the key tool is an Carleman type inequality whichwill be presented in Lemma 1 below. First of all, we reduce equation (1) to amore suitable form by introducing a new variable e x = √ x − η , that is, x = ( e x + η ) , where 2 η = min { α , γ } , the parameters α , γ will be specified later, η > . Without loss of generality, we assume that √ x − η > , i.e., x > η ,and so we have e x > e u ( e x , ′ x, y ) ≡ u (cid:16) ( e x + η ) , ′ x, y (cid:17) , by using the4elations u x = e u e x d e x dx = e u e x e x + η ; u x x = e u e x e x ( e x + η ) − − e u e x ( e x + η ) − , we have( e x + η ) − e u e x e x + ∆ ′ x e u − m X i,j =1 e a ij e u y i y j + n X i =2 e a i e u x i + m X j =1 e b j e u y j + e a e u e x + e a e u = e f e g, where e a ij = a ij ( ( e x + η ) , ′ x, y ′ ) , i, j = 1 , ..., m ; e a i = a i ( ( e x + η ) , ′ x, y ) , i =0 , , , ..., n ; e b j = b j ( ( e x + η ) , ′ x, y ) , j = 1 , , ..., m ; e a = a ( ( e x + η ) , ′ x, y ) − ( e x + η ) − ; e f = f ( ( e x + η ) , ′ x, y ) , e g = g ( ( e x + η ) , ′ x, y ′ ) . For the sake of simplicity, let us denote e u, e a ij , e x , e a k , e b s , e f , e g by u, a ij , x , a s ,b j , f, g respectively, where i, j = 2 , ..., m ; k = 0 , , , ..., n ; s = 1 , , ..., m. Then wecan write ( x + η ) − u x x + ( x + η ) (∆ ′ x u − m X i,j =1 a ij u y i y j )+ ( x + η ) ( n X i =1 a i u x i + m X j =1 b j u y j + a u )= ( x + η ) f g. (5)We set L u ≡ ( x + η ) − u x x + ( x + η ) ∆ ′ x u − m X i,j =1 a ij u y i y j . We introduceΩ γ = ( ( x, y ); x > , < δx + 12 n X i =2 ( x i − x i ) + 12 m − X i =1 ( y i − y i ) + 12 y m < γ ) ,where 0 < γ < , δ > , (cid:0) x , y (cid:1) ∈ Ω , x = (cid:0) , x , ..., x n (cid:1) , y = (cid:0) y , y , ..., y m − , (cid:1) .
5n Ω γ we define the weight function χ = exp (cid:0) λψ − ν (cid:1) , ψ ( x ) = δx + 12 n X i =2 ( x i − x i ) + 12 m − X i =1 ( y i − y i ) + 12 y m + α ,where α > , γ + α = ρ < , α < ψ ( x ) < ρ, and λ, ν, δ are positive parameterssatisfying some additional conditions which are specified later.The following Carleman estimate is the key for the proof of Theorem 1. Lemma 1.
Let condition (4) be satisfied. With arbitrarily fixed constant
M > ,we assume that || a ij || C (Ω) ≤ M and the number γ is ”small”, that is < γ < min { ,
43 ( M (cid:0) M m ε − + m + m ( m + 1) (cid:1) ) − / } , (6) where < ε < α m . Then there exists a constant δ ∗ = δ ∗ ( α , M, n, m, ν ) > suchthat for any δ > δ ∗ there exists λ ∗ = λ ∗ ( δ ) such that the following estimate holds: ψ ν +1 ( L ϕ ) χ − λνβ ϕ ( x + η ) ( L ϕ ) χ ≥ λνδ ( x + η ) − ϕ x χ + 2 λν ( x + η ) |∇ ′ x ϕ | χ +2 λν ( x + η ) |∇ y ϕ | χ + λ ν δ ψ − ν − ϕ χ + D ( ϕ ) (7)for all ϕ ( x, y ) ∈ C (cid:0) Ω γ (cid:1) and λ > λ ∗ . In (7), β ≡ β ( n, m ) = n + 2 + M m ((1 +3 √ γ ) m +1) and D ( ϕ ) is described by a divergence form which includes the function ϕ and is given explicitly in the proofs of the lemmata below.
3. THE PROOF OF LEMMA 1.
In the proof of Lemma 1, we shall use two Lemmata 2 and 3.
Lemma 2.
Under the hypothesis of Lemma 1, there exists a constant δ =6 ( α , M, n, m, ν ) > such that for any δ > δ there exists λ = λ ( δ ) such that ψ ν +1 ( L ϕ ) χ ≥ λνδ ( x + η ) − ϕ x χ − λν ( x + η ) ( β − |∇ ′ x ϕ | χ + λνδα ( x + η ) |∇ y ϕ | χ + 2 δ λ ν ( x + η ) − ψ − ν − ϕ χ + σ ( λ, δ ) ϕ χ + D ( χϕ ) + D ( χϕ ) , (8)for all λ > λ and ϕ ∈ C (Ω γ ) . Here σ ( λ, δ ) = λ ν σ + λ ν σ ,σ = ψ − ν − ( − δ ( x + η ) − + 2 ( x + η ) ( β − |∇ ′ x ψ | − δα ( x + η ) |∇ y ψ | ) ,σ = ( ν + 1) ψ − ν − ( − δ ( x + η ) − + 2 ( x + η ) ( β − |∇ ′ x ψ | − δα ( x + η ) ( ν + 1) |∇ y ψ | ) + ψ − ν − (6 δ ( x + η ) − − x + η ) ( β − n −
1) + δα ( x + η ) m ) , and the terms D ( χϕ ) , D ( χϕ ) are given by divergence forms which include thefunction ϕ and are given in the proof explicitly. The proof of Lemma 2 is technical and lenghty, and we postpone it to Appendix.In (8), the signs of the terms of |∇ ′ x ϕ | and |∇ y ϕ | are different. Thus we needto perform another estimation: Lemma 3.
The following equality holds: − ( x + η ) ϕ ( L ϕ ) χ = ϕ x χ + χ ( x + η ) |∇ ′ x ϕ | − m X i,j =1 a ij ϕ y i ϕ y j + σ ( λ, δ ) ϕ χ + D ( ϕ ) (9)for any function ϕ ∈ C (Ω) . Here σ ( λ, δ ) = λ ν σ + λνσ + σ , = − ψ − ν − δ + ( x + η ) |∇ ′ x ψ | − m X i,j =1 a ij ψ y i ψ y j ,σ = − ( ν + 1) ψ − ν − δ + ( x + η ) |∇ ′ x ψ | − m X i,j =1 a ij ψ y i ψ y j + ψ − ν − ( x + η ) n − − m X i,j =1 ∂a ij ∂y j ψ y i − m X i =1 a ij ,σ = 12 m X i,j =1 ∂ a ij ∂y i ∂y j ( x + η ) and D ( ϕ ) = − (( ϕϕ x + λνδψ − ν − ϕ ) χ ) x − n X i =2 ∂∂x i (( ϕϕ x i + λνψ x i ψ − ν − ϕ ) χ ( x + η ) )+ m X i =1 ∂∂y i m X j =1 (cid:18) a ij (cid:16) ϕ y j ϕ + λνψ y j ψ − ν − ϕ (cid:17) − ∂a ij ∂y j ϕ (cid:19) χ ( x + η ) . Lemma 3 can be proved easily by direct calculations and we omit the proof here.
Now we will proceed to the completion of the proof of Lemma 1. We multiplyequality (9) by 2 λνβ and add to inequality (8) to have ψ ν +1 ( L ϕ ) χ − λνβ ( x + η ) ϕ ( L ϕ ) χ ≥ λνδ ( x + η ) − ϕ x χ + 2 λν ( x + η ) |∇ ′ x ϕ | χ + λνδα ( x + η ) |∇ y ϕ | χ − λνβ ( x + η ) χ m X i,j =1 a ij ϕ y i ϕ y j +2 λ ν δ ( x + η ) − ψ − ν − ϕ χ + σ ( λ, δ ) ϕ χ + D ( χϕ ) + D ( χϕ ) + 2 λνβ D ( ϕ ) , (10)for δ > δ , λ > λ , where σ ( λ, δ ) = σ ( λ, δ ) + 2 λνβ σ ( λ, δ ) . We set δ = α (cid:0) mβ γM (cid:1) .Since − m X i,j =1 a ij ϕ y i ϕ y j ≥ − M m |∇ y ϕ | ,
8e can estimate the coefficient of |∇ y ϕ | : λνδα ( x + η ) |∇ y ϕ | χ − λνβ ( x + η ) m X i,j =1 a ij ϕ y i ϕ y j χ ≥ λν ( x + η ) ( δα − β mM ( x + η )) |∇ y ϕ | χ ≥ λν ( x + η ) |∇ y ϕ | χ (11)for δ ≥ δ . As for the coefficient of ϕ , we can write σ ( λ, δ ) in the form σ ( λ, δ ) = λ ν σ + λ ν σ + λνσ , where σ = σ + 2 β σ , σ = σ + 2 β σ , σ = 2 β σ . Since the functions a ij , ψ, ψ x i , ψ y j are bounded in the space C (cid:0) Ω γ (cid:1) , it is clearthat the function ˜ σ = σ δ νψ − ν − is bounded uniformly with respect to ( x, y ) ∈ Ω γ : | ˜ σ | ≤ M , M > . Then we see that λ ν δ ( x + η ) − ψ − ν − + λ ν σ > λ ν δ ψ − ν − (cid:18) δ ˜ σ (cid:19) ≥ λ ν δ ψ − ν − (cid:18) − δ M (cid:19) = λ ν δ ψ − ν − (cid:18) − δ δ (cid:19) ≥ λ ν δ ψ − ν − for δ ≥ δ = 2 M . Here we note that ( x + η ) − ≥ ( γ ) − > . On the other hand, it is obvious that, for fixed δ ≥ δ , ν >
1, the functions σ and σ are also bounded on Ω γ , that is, there exist constants M , M > | σ | ≤ M , | σ | ≤ M . Thus we have λ ν δ ( x + η ) − ψ − ν − + λ ν σ + λ ν σ + λνσ ≥ λ ν δ ψ − ν − − λ ν M − λνM ≥ λ ≥ λ = max (cid:8) M , √ M (cid:9) , which yields2 λ ν δ ( x + η ) − ψ − ν − − σ ( λ, δ ) ≥ λ ν δ ψ − ν − . (13)Consequently, inequalities (10), (11) and (13) imply that ψ ν +1 ( L ϕ ) χ − λνβ ϕ ( L ϕ ) χ ≥ λνδ ( x + η ) − ϕ χ + 2 λν ( x + η ) χ |∇ ′ x ϕ | χ +2 λν ( x + η ) |∇ y ϕ | χ + λ ν δ ψ − ν − ϕ χ + D ( ϕ ) , for δ ≥ δ ∗ = max { δ , δ , δ } and λ ≥ λ ∗ = max { λ , λ } , where D ( ϕ ) = D ( χϕ ) + D ( χϕ ) + 2 λνβ D ( ϕ ). Thus the proof of Lemma 1 is complete.
4. THE PROOF OF THEOREM 1
Let ( u, g ) be a solution to (1) - (3) with u ≡ γ . Since f ( x, y ′ , = 0 and f ∈ C (cid:0) Ω (cid:1) , there exists a number 0 < γ < f ( x, y ) = 0 also in Ω γ . Weassume that γ, which was introduced before, satisfies this condition. We define anew unknown function w = uf in Ω γ . Then dividing equation (5) by f ( x, y ) andtaking into account relations (2)-(3), we obtain( x + η ) − w x x + ( x + η ) (∆ ′ x w − m X i,j =1 a ij w y i y j )+ ( x + η ) ( n X i =1 ¯ a i w x i + m X j =1 ¯ b j w y j + ¯ a w )= ( x + η ) g, (14) w (0 , ′ x, y ) = w x (0 , ′ x, y ) = 0 , (15) w ( x, y ′ ,
0) = 0 , (16)10here ¯ a = (( x + η ) − f x x + ∆ ′ x f − m X i,j =1 a ij f y i y j + n X i =1 a i f x i + m X j =1 b j f y j + a f ) f − , ¯ a = (2 ( x + η ) − f x + a f ) f − , ¯ a i = (2 f x i + a i f ) f − , i = 2 , ..., n, ¯ b j = ( − m X i =1 ( a ji f y j + a ij f y i ) + b j f ) f − , j = 1 , ..., m. Differentiating equation (14) with respect to y m , setting z = w y m and using (16),we obtain the integro-differential equation( x + η ) − z x x + ( x + η ) ∆ ′ x z − m X i,j =1 a ij z y i y j +( x + η ) n X i =1 ¯ a i z x i + m X j =1 ¯ b j z y j + ¯ a z +( x + η ) n X i =1 ∂ ¯ a i ∂y m I x i z + m X j =1 ∂ ¯ b j ∂y m I y j z + ∂ ¯ a ∂y m Iz = 0 (17)with the Cauchy data z (0 , ′ x, y ) = z x (0 , ′ x, y ) = 0 , (18)where Iz = R y m z ( x, y ′ , τ ) dτ , y m ≥ R y m z ( x, y ′ , τ ) dτ , y m < ,I x i z = ∂∂x i R y m z ( x, y ′ , τ ) dτ , y m ≥ ∂∂x i R y m z ( x, y ′ , τ ) dτ , y m < , i = 1 , .., n ;11 y j z = ∂∂y j R y m z ( x, y ′ , τ ) dτ , y m ≥ ∂∂y j R y m z ( x, y ′ , τ ) dτ , y m < , j = 1 , ..., m. We now prove that, if z ( x, y ) satisfies (17) and (18), then z ( x, y ) = 0 in Ω γ .From (17), we obtain( L z ) = ( x + η ) n X i =1 (cid:18) ¯ a i z x i + ∂ ¯ a i ∂y m I x i z (cid:19) + m X j =1 (cid:18) ¯ b j z y j + ∂ ¯ b j ∂y m I y j z (cid:19) + ¯ a z + ∂ ¯ a ∂y m Iz (cid:19) ≤ M max { n, m } ( x + η ) |∇ x z | + n X i =1 ( I x i z ) + |∇ y z | + m X j =1 ( I y j z ) + z + ( Iz ) , (19)where M > M and k f k C (Ω γ ) .On the other hand, by Lemma 1, we can write( L ϕ ) χ + ( x + η ) ( L ϕ ) χ + λ ν β ϕ χ ≥ ψ ν +1 ( L ϕ ) χ − λνβ ϕ ( x + η )( L ϕ ) χ ≥ λνδ ( x + η ) − ϕ x χ + 2 λν ( x + η ) |∇ ′ x ϕ | χ +2 λν ( x + η ) |∇ y ϕ | χ + λ ν δ ( x + η ) − ψ − ν − ϕ χ + D ( ϕ ) (20)for δ > δ ∗ , λ > λ ∗ . Taking ϕ ≡ z in (20) and using (19), we obtain6 M max { n, m } ( x + η ) ( |∇ x z | + n X i =1 ( I x i z ) + |∇ y z | + m X j =1 ( I y j z ) + ( Iz ) ) χ (1 + ( x + η ) )+(6 M max { n, m } ( x + η ) (1 + ( x + η ) ) + λ ν β ) z χ ≥ λνδ ( x + η ) − z x χ + 2 λν ( x + η ) |∇ ′ x z | χ +2 λν ( x + η ) |∇ y z | χ + λ ν δ ( x + η ) − ψ − ν − z χ + D ( z ) (21)12ere we shall use the following lemma, whose proof is given in Appendix. Lemma 4.
The following relations hold: Z Ω γ ( Iz ) χ d Ω γ ≤ γ Z Ω γ z χ d Ω γ , Z Ω γ ( I x i z ) χ d Ω γ ≤ γ Z Ω γ z x i χ d Ω γ , Z Ω γ (cid:0) I y j z (cid:1) χ d Ω γ ≤ γ Z Ω γ z y j χ d Ω γ , where i = 1 , ..., n ; j = 1 , ..., m. Integrating inequality (21) on Ω γ and using Lemma 4, we have6 M max { n, m } (1 + γ ) Z Ω γ (1 + ( x + η ) ) ( x + η ) × ( |∇ x z | + |∇ y z | + z ) χ d Ω γ + λ ν β Z Ω γ z χ d Ω γ ≥ λνδ ( x + η ) − Z Ω γ z x χ d Ω γ + 2 λν Z Ω γ ( x + η ) |∇ ′ x z | χ d Ω γ +2 λν Z Ω γ ( x + η ) |∇ y z | χ d Ω γ + λ ν δ Z Ω γ ψ − ν − z χ d Ω γ + D ( z ) . Hence, if λ ≥ λ ∗ = 12 M max { n, m } (1 + γ ) > ν ≥ δ − (1 + β + ( γ ) ), thenwe obtain Z Ω γ ( λ ν z + λν ( z x + ( x + η ) |∇ ′ x z | + ( x + η ) |∇ y z | )) χ d Ω γ ≤ − Z Ω γ D ( z ) d Ω γ . (22)Passing to the limit as λ → ∞ in (22), we conclude that Z Ω γ z d Ω γ ≤ , which means that z = 0 in Ω γ .Varying the point x = (cid:0) , x , x , ..., x n + m (cid:1) of the plane x = 0, we estab-13ish that z = 0 on ˜Ω γ = { ( x, y ) ∈ Ω; 0 ≤ δx ≤ γ } , that is ∂w∂y m = 0 on ˜Ω γ .Then from equation (14) by condition (16) we conclude that g ( x, y ′ ) = 0 on˜Ω ′ γ = { ( x, y ′ ) ∈ D × G ′ ; 0 ≤ δx ≤ γ } , where ˜Ω ′ γ the is projection of Ω onto R n + m − . Repeating the same argument, we see that z = 0 in ˜Ω γ and g ( x, y ′ ) = 0on ˜Ω ′ γ . Thus, continuing the argument, we complete the proof.
5. APPENDIX5.1. Proof of Lemma 2
We introduce a new function ϑ = χϕ. Using the relations ϕ x x = χ − (cid:0) ϑ x x + 2 λνψ − ν − ψ x ϑ x + δ φ ϑ (cid:1) ∆ ′ x ϕ = χ − (cid:0) ∆ ′ x ϑ + 2 λνψ − ν − ( ∇ ′ x ψ, ∇ ′ x ϑ ) + φ ϑ (cid:1) , m X i,j =1 a ij ϕ y i y j = χ − m X i,j =1 a ij (cid:16) ϑ y i y j + λνψ − ν − ψ y i ϑ y j + λνψ − ν − ψ y j ϑ y i (cid:17) + φ ϑ ,
14e obtain ψ ν +1 ( L ϕ ) χ = ψ ν +1 (cid:26) ( x + η ) − ϑ x x + ( x + η ) ∆ ′ x ϑ − m X i,j =1 a ij ϑ y i y j + ϑ (cid:16) ( x + η ) − δ φ + ( x + η ) ( φ − φ ))+2 λνψ − ν − δ ( x + η ) − ϑ x + ( x + η ) ( ∇ ′ x ψ, ∇ ′ x ϑ ) − m X i,j =1 a ij ψ y i ϑ y j (cid:27) ≥ λν (cid:26) ( x + η ) − ϑ x x + ( x + η ) ∆ ′ x ϑ − m X i,j =1 a ij ϑ y i y j + ϑ (( x + η ) − δ φ + ( x + η ) ( φ − φ )) (cid:27) × δ ( x + η ) − ϑ x + ( x + η ) ( ∇ ′ x ψ, ∇ ′ x ϑ ) − m X i,j =1 a ij ψ y i ϑ y j = : X k =1 T k , (23)where we set φ : = φ ( λ, ν, ψ ) = λ ν ψ − ν − − λν ( ν + 1) ψ − ν − ,φ : = φ ( λ, ν, ψ ) = λ ν φ ( ν, ψ ) − λνφ ( ν, ψ ) ,φ : = φ ( λ, ν, ψ ) = λ ν φ ( ν, ψ ) − λνφ ( ν, ψ )and φ ( ν, ψ ) : = ψ − ν − |∇ ′ x ψ | , φ ( ν, ψ ) = ( ν + 1) ψ − ν − |∇ ′ x ψ | − ( n − ψ − ν − φ ( ν, ψ ) : = ψ − ν − m X i,j =1 a ij ψ y i ψ y j ,φ ( ν, ψ ) : = m X i,j =1 a ij (cid:16) (( ν + 1) ψ − ν − − ψ − ν − ) ψ y i ψ y j (cid:17) k a ij k C (Ω) ≤ M , (cid:12)(cid:12) ψ x i (cid:12)(cid:12) ≤ √ γ, (2 ≤ i ≤ n ) and | ψ y k | ≤ √ γ (1 ≤ k ≤ m )in Ω γ , we estimate the terms T i , ≤ i ≤
14 as follows: T = 4 λνδ ( x + η ) − ϑ x ϑ x x = d ( ϑ ) + 4 λνδ ( x + η ) − ϑ x , (24)where d ( ϑ ) = 2 λνδ (( x + η ) − ϑ x ) x ; T = 4 λνδϑ x ∆ ′ x ϑ = 4 λνδ n X i =2 (( ϑ x ϑ x i ) x i − λνδ (cid:0) ϑ x i (cid:1) x )= : d ( ϑ ) ; (25) T = − λνδϑ x m X i,j =1 a ij ϑ y i y j = d ( ϑ ) − λνδ m X i,j =1 ( − ∂a ij ∂y j ϑ y i ϑ x + ∂a ij ∂x ϑ y i ϑ y j ) ≥ d ( ϑ ) − λνδ m X i,j =1 (cid:12)(cid:12)(cid:12)(cid:12) ∂a ij ∂y j (cid:12)(cid:12)(cid:12)(cid:12) | ϑ y i ϑ x | + 2 λνδα |∇ y ϑ | ≥ d ( ϑ ) + 2 λνδ ( α − m ( x + η ) ε ) |∇ y ϑ | − λνδ m M ε ( x + η ) ϑ x , (26)where d ( ϑ ) = − λνδ m P i,j =1 (2 ( a ij ϑ y i ϑ x ) y j − (cid:0) a ij ϑ y i ϑ y j (cid:1) x ) . Next T = 4 λνδ ( x + η ) − ϑϑ x φ = d ( ϑ ) + 4 λνδ ϑ ( x + η ) − φ +4 λνδ ϑ ( ν + 1) ( x + η ) − φ , (27)where d ( ϑ ) = 2 λνδ (( x + η ) − ϑ φ ) x and we set φ := φ ( λ, ν, ψ ) = λ ν ψ − ν − − λν ( ν + 2) ψ − ν − .
16e have T = 4 λνδϑ x ϑφ x = d ( ϑ ) + 4 λνδ ( ν + 1) ϑ φ n X i =2 ψ i + 12 ( n − ν + 1) λνψ − ν − ! , (28)where d ( ϑ ) = 2 λνδ (cid:0) ϑ φ x (cid:1) x ; T = − λνδϑ x ϑφ = d ( ϑ ) + 2 λνδϑ ( φ ) x , (29)where d ( ϑ ) = − λνδ (cid:0) ϑ φ (cid:1) x ; T = 4 λν n X i =2 ψ x i ϑ x i ϑ x x = d ( ϑ ) + 2 λνϑ x , (30)where d ( ϑ ) = 4 λν n P i =2 ( (cid:0) ψ x i ϑ x i ϑ x (cid:1) x − λν (cid:0) ψ x i ϑ x (cid:1) x i ); T = 4 λν ( x + η ) n X i,j =2 ψ x i ϑ x i ϑ x j x j = d ( ϑ ) − λν ( x + η ) n X i,j =2 δ ij ϑ x i ϑ x j + 2 λν ( x + η ) n X i,j =2 ϑ x j = d ( ϑ ) − λν ( x + η ) n X i =2 ϑ x i (2 − ( n − , (31)where d ( ϑ ) = 2 λν n P i,j =2 (cid:18) ψ x i ϑ x i ϑ x j ( x + η ) ) x j − ( x + η ) (cid:16) ψ x i ϑ x j (cid:17) x i (cid:19) .T = − λν ( x + η ) n X i =2 ψ x i ϑ x i m X k,s =1 a ks ϑ y k y s = d ( ϑ ) + 2 λν ( x + η ) n X i =2 m X k,s =1 (2 (cid:0) ψ x i a ks (cid:1) y s ϑ y k ϑ x i − (cid:0) ψ x i a ks (cid:1) x i ϑ y k ϑ y s ) ≥ − λν ( x + η ) n X i =2 m X k,s =1 (cid:16) (cid:12)(cid:12)(cid:12)(cid:0) ψ x i a ks (cid:1) y s ϑ y k ϑ x i (cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:0) ψ x i a ks (cid:1) x i ϑ y k ϑ y s (cid:12)(cid:12)(cid:12)(cid:17) ≥ − λν ( x + η ) M (1 + p γ ) (cid:16) mn |∇ y ϑ | + m |∇ ′ x ϑ | (cid:17) , (32)17here d ( ϑ ) = − λν ( x + η ) m P k,s =1 m P k,s =1 (2 (cid:0) ψ x i a ks ϑ x i ϑ y k (cid:1) s − (cid:0) ψ x i a ks ϑ y k ϑ y s (cid:1) x i ); T = 4 λν n X i =2 ψ x i ϑ x i ϑ ( x + η ) (( x + η ) − δ φ + ( x + η ) ( φ − φ ))= d ( ϑ ) − λνϑ n X i =2 ( ψ x i δ φ + ( x + η ) ψ x i ( φ − φ )), (33)where d ( ϑ ) = 2 λν n P i =2 ( ϑ ψ x i ( x + η ) (( x + η ) − δ φ + ( x + η ) ( φ − φ ))) x i ; T = − λν m X i,j =1 a ij ψ y i ϑ y j ϑ x x = d ( ϑ )+4 λν m X i,j =1 ∂a ij ∂x ψ y i ϑ y j ϑ x − λν m X i,j =1 (cid:18) ∂a ij ∂y j ψ y i + a ij δ ij (cid:19) ϑ x ≥ − λν m X i,j =1 (cid:12)(cid:12)(cid:12)(cid:12) ∂a ij ∂x ψ y i ϑ y j ϑ x (cid:12)(cid:12)(cid:12)(cid:12) − λν m X i,j =1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:18) ∂a ij ∂y j ψ y i + a ij δ ij (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) ϑ x ≥ − λν ( x + η ) p γm |∇ y ϑ | − λνM ( x + η ) − m υ x − λνM p γm ϑ x − λνM mϑ x , (34)where d ( ϑ ) = − λν m P i,j =1 ( (cid:0) a ij ψ y i ϑ y j ϑ x (cid:1) x + 2 λν (cid:0) a ij ψ y i ϑ x (cid:1) y j ); T = − λν ( x + η ) m X i,j =1 a ij ψ y i ϑ y j n X s =2 ϑ x s x s = d ( ϑ ) + 4 λν ( x + η ) m X i,j =1 n X s =2 ∂a ij ∂x s ψ y i ϑ y j ϑ x s − λν ( x + η ) m X i,j =1 n X s =2 (cid:18) ∂a ij ∂y j ψ y i + a ij δ ij (cid:19) ϑ x s ≥ − λν ( x + η ) M p γ (cid:16) m |∇ ′ x ϑ | + ( n − m |∇ y ϑ | (cid:17) − λν ( x + η ) M m ( p γm + 1) |∇ ′ x ϑ | , (35)18here d ( ϑ ) = − λν (( x + η ) m P i,j =1 n P s =2 (cid:0) a ij ψ y i ϑ y j ϑ x s (cid:1) x s +2 λν ( x + η ) (cid:0) a ij ψ y i ϑ x s (cid:1) y j ); T = 4 λν ( x + η ) m X i,j,k,s =1 a ij ψ y i ϑ y j a ks ϑ y k y s = d ( ϑ ) − λν ( x + η ) m X i,j,k,s =1 (2 (cid:0) a ij a ks ψ y i (cid:1) y s ϑ y j ϑ y k − (cid:0) a ij a ks ψ y i (cid:1) y j ϑ y k ϑ y s ) ≥ − λν ( x + η ) m X i,j,k,s =1 (2 (cid:12)(cid:12)(cid:12)(cid:0) a ij a κs ψ y i (cid:1) y s (cid:12)(cid:12)(cid:12) (cid:12)(cid:12) ϑ y j ϑ y k (cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:0) a ij a κs ψ y i (cid:1) y j (cid:12)(cid:12)(cid:12) | ϑ y k ϑ y s | ) ≥ − λν ( x + η ) M (cid:16) p γ + 1 (cid:17) m |∇ y ϑ | , (36)where d ( ϑ ) = 2 λν ( x + η ) m P i,j,k,s =1 (2( a ij ψ y i ϑ y j ϑ y k a ks ) y s − (cid:0) a ij a ks ψ y i ϑ y k ϑ y s (cid:1) y j ); T = − λν ( x + η ) m X i,j =1 a ij ψ y i ϑ y j ϑ (( x + η ) − δ φ + ( x + η ) ( φ − φ ))= d ( ϑ ) + 2 λνϑ m X i,j =1 ( a ij ψ y i ( δ φ + ( x + η ) ( φ − φ ))) y j , (37)where d ( ϑ ) = − λν m P i,j =1 ( a ij ψ y i ϑ ( x + η ) (( x + η ) − δ φ + ( x + η ) ( φ − φ ))) y j . Then by relations (24)-(37), we see that ψ ν +1 ( L ϕ ) χ ≥ λνδ ( x + η ) − β ϑ − λν ( x + η ) ( β − |∇ ′ x ϑ | + 2 λν ( x + η ) β |∇ y ϑ | + (cid:0) λ ν β + λ ν β (cid:1) ϑ + D ( ϑ ) , (38)where β = 2 − M m ( x + η ) ( M mε − − mδ − − δ − ( p γm + 1)( x + η )) ,β = δ ( α − mε ) − β ,β = p γm + (2 n (1 + p γ ) + ( n − p γ + 3 m M (2 p γ + 1)) ( x + η ) mM = 4 δ ( x + η ) − ψ − ν − + 4 δ ( ν + 1) ( x + η ) − ψ − ν − +4 δ ( ν + 1) ψ − ν − |∇ ′ x ψ | + β ,β = 2 δ ( φ ) x + 2 |∇ ′ x ψ | (( x + η ) ( φ − φ ) − δ ψ − ν − φ )+2 m X i,j =1 ( a ij ψ y i (( x + η ) ( φ − φ ) + δ ψ − ν − )) y j ,β = − δ ( x + η ) − ( ν + 1) ψ − ν − + 4 δ ( ν + 1)( 12 ( n − ν + 1) ψ − ν − −
12 ( ν + 2) ψ − ν − |∇ ′ x ψ | − δ ( x + η ) −
12 ( ν + 2) ψ − ν − ) − δ ( φ ) x +2 |∇ ′ x ψ | (( x + η ) ( φ − φ ) + δ ( ν + 1) ψ − ν − φ )+2 m X i,j =1 ( a ij ψ y i (( x + η ) ( φ − φ ) − δ ( ν + 1) ψ − ν − )) y j and D ( ϑ ) = P k =1 d k ( ϑ ) . Now we shall evaluate the expressions β , β , β , β in (38), respectively.Since δ ≥ , x + η < γ, and M m (cid:0) γ (cid:1) ( M mε − + m + ( √ γm + 1) γ ) < β > − M m ( x + η ) ( M mε − + m + ( p γm + 1)( x + η )) > − M m (cid:18) γ (cid:19) (cid:18) M mε − + m + ( p γm + 1) 34 γ (cid:19) > , which implies2 λνδ ( x + η ) − ϑ x (2 − M m ε − ( x + η ) − M m δ − ( x + η ) − M δ − m ( p γm + 1) ( x + η ) ) ≥ λνδ ( x + η ) − ϑ x . (39)Next, by choosing 0 < ε < α m , and setting δ = mα √ γl , we see that β > δ ( α − mε ) − m p γl ≥ δα (40)20or δ ≥ δ , where l = 1 + ( n − M γ + (2 nM + 3 M m )(1 + 2 p γ ) 34 (cid:16) γ (cid:17) / . It is clear that β > δ ( ν + 1) ( x + η ) − ψ − ν − − | β | = 4 δ ( ν + 1) ( x + η ) − ψ − ν − (1 − δ (cid:12)(cid:12)(cid:12) ˜ β (cid:12)(cid:12)(cid:12) ) (41)where ˜ β = δ ( x + η ) ψ ν +3 1 ν +1 β . Since the functions a ij and ψ bounded in the space C (cid:0) Ω γ (cid:1) , the function ˜ β is bounded uniformly with respect to x ∈ Ω γ , that is there exist a number M > (cid:12)(cid:12)(cid:12) ˜ β (cid:12)(cid:12)(cid:12) ≤ M . Then β > δ ( ν + 1) ( x + η ) − ψ − ν − (42)for δ ≥ δ = √ M . By the same reasons, β is also bounded on Ω γ : that is | β | ≤ M , M > . Moreover, since 2 λ ν − λ ν M ≥ λ ≥ λ = M ( δ, ν, γ ) , ν > , we can write λ ν β + λ ν β ≥ λ ν δ ( ν + 1) ( x + η ) − ψ − ν − − λ ν M = 2 λ ν δ ( x + η ) − ψ − ν − + 2 λ ν δ ( x + η ) − ψ − ν − − λ ν M ≥ λ ν δ ( x + η ) − ψ − ν − (43)for δ ≥ δ = max { , δ , δ } , ν > , λ ≥ λ . ψ ν +1 ( L ϕ ) χ ≥ λνδ ( x + η ) − ϑ x − λν ( x + η ) ( β − |∇ ′ x ϑ | + λνδα ( x + η ) |∇ y ϑ | + 2 λ ν δ ( x + η ) − ψ − ν − ϑ + d ( ϑ ) (44)for δ ≥ δ , λ ≥ λ , ν > . Finally, taking into account the equality ϑ = χϕ and the relations ϑ x = ϕ x χ − λ ν δ ψ − ν − ϕ χ − λν ( ν + 1) δ ψ − ν − ϕ χ − λν (cid:0) δψ − ν − ϕ χ (cid:1) x , |∇ ′ x ϑ | = |∇ ′ x ϕ | χ − λ ν |∇ ′ x ψ | ψ − ν − ϕ χ − λν ( ν + 1) |∇ ′ x ψ | ψ − ν − ϕ χ + λν ( n − ψ − ν − ϕ χ − λν n X i =2 (cid:0) ψ x i ψ − ν − ϕ χ (cid:1) x i , |∇ y ϑ | = |∇ y ϕ | χ − λ ν |∇ y ψ | ψ − ν − ϕ χ − λν ( ν + 1) |∇ y ψ | ψ − ν − ϕ χ + λνmψ − ν − ϕ χ − λν m X k =1 (cid:0) ψ y k ψ − ν − ϕ χ (cid:1) y k , from (44) we obtain (8), where D ( χϕ ) = − λ ν δ (cid:16) ( x + η ) − ϕ ψ − ν − χ (cid:17) x − λ ν δα m X i =1 (cid:0) ( x + η ) ϕ ψ y i ψ − ν − χ (cid:1) y i +2 λ ν ( x + η ) ( β − n X i =2 (cid:0) ϕ ψ i ψ − ν − χ (cid:1) x i . (45)Thus the proof of Lemma 2 is complete. Let K be the part of the domain Ω γ produced by the plane y m = 0 and σ ( x, y ′ )is the distance between the boundary point ( x, y ) of Ω γ and the point ( x, y ′ , ∈ K. χ is monotonic with respect to y m , we have Z Ω γ χ ( Iz ) d Ω γ = Z K Z σ ( x,y ′ ) χ ( Iz ) dy m + Z − σ ( x,y ′ ) χ ( Iz ) dy m ! dxdy ′ ≤ Z K (cid:18) √ γ Z a Z y m z ( x, y ′ , τ ) χ dτ dy m + √ γ Z − a Z y m z ( x, y ′ , τ ) χ dτ dy m (cid:19) dxdy ′ ≤ γ Z K ( Z a z ( x, y ′ , τ ) χ ( x, y ′ , τ ) dτ + Z − a z ( x, y ′ , τ ) χ ( x, y ′ , τ ) dτ ) dxdy ′ = γ Z Ω γ z χ d Ω γ for a = σ ( x, y ′ ). Here we used the relations Z a χ ( x, y ) Z y m z ( x, y ′ , τ ) dτ dy m = Z a Z y m χ ( x, y ) z ( x, y ′ , τ ) dτ dy m = Z a z ( x, y ′ , τ ) Z aτ χ dy m dτ ≤ Z a z ( x, y ′ , τ ) χ ( x, y ′ , τ )( a − τ ) dτ ≤ √ γ Z a z ( x, y ′ , τ ) χ ( x, y ′ , τ ) dτ and Z − a χ Z y m z ( x, y ′ , τ ) dτ dy m = Z − a z ( x, y ′ , τ ) Z τ − a χ dy m dτ ≤ √ γ Z − a z ( x, y ′ , τ ) χ ( x, y ′ , τ ) dτ . Similarly we can prove the rest part.
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