Hindawi Publishing Corporation Abstract and Applied Analysis Volume 2015, Article ID 276810, 11 pages http://dx.doi.org/10.1155/2015/276810
Research Article The Dirichlet Problem for Second-Order Divergence Form Elliptic Operators with Variable Coefficients: The Simple Layer Potential Ansatz Alberto Cialdea, Vita Leonessa, and Angelica Malaspina Department of Mathematics, Computer Science and Economics, University of Basilicata, Viale dell’Ateneo Lucano 10, 85100 Potenza, Italy Correspondence should be addressed to Alberto Cialdea;
[email protected] Received 9 June 2015; Accepted 18 October 2015 Academic Editor: Giovanni P. Galdi Copyright © 2015 Alberto Cialdea et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. We investigate the Dirichlet problem related to linear elliptic second-order partial differential operators with smooth coefficients in divergence form in bounded connected domains of R𝑚 (𝑚 ≥ 3) with Lyapunov boundary. In particular, we show how to represent the solution in terms of a simple layer potential. We use an indirect boundary integral method hinging on the theory of reducible operators and the theory of differential forms.
1. Introduction As remarked in [1, p. 121], elliptic operators with variable coefficients naturally arise in several areas of physics and engineering. In this paper, we study the Dirichlet problem related to a scalar elliptic second-order differential operator with smooth coefficients in divergence form in a bounded simply connected domain of R𝑚 (𝑚 ≥ 3) with Lyapunov boundary. This is a classical problem which nowadays can be treated in several ways. In particular, different potential methods have been developed for such operators (see, e.g., [1–6]). In the present paper, we obtain the solution of the Dirichlet problem by means of a simple layer potential instead of the classical double layer potential (see, e.g., [6, pp. 73–75]). We use an indirect boundary integral method introduced for the first time in [7] for the 𝑚-dimensional Laplacian. It requires neither the knowledge of pseudodifferential operators nor the use of hypersingular integrals, but it hinges on the theory of singular integral operators and the theory of differential forms (for details of the method, see, e.g., [8, Section 2]). The method has been also used to treat different boundary value problems in simply connected domains: the Neumann problem for Laplace equation (via a double layer potential), the Dirichlet problem for the Lam´e and Stokes systems, the four
boundary value problems of the theory of thermoelastic pseudooscillations, the traction problem for Lam´e and Stokes systems, the four basic boundary value problems arising in couple-stress elasticity, and the two boundary value problems of the linear theory of viscoelastic materials with voids (see [9, 10] and the references therein). The method can be applied also in multiply connected domains, as shown for the Laplacian, the linearized elastostatics, and the Stokes system (see [11] and the references therein). The present paper is organized as follows. In Section 2, after giving preliminary results, we make use of Fichera’s construction of a principal fundamental solution [12] and we prove some identities for the related nuclear double form. Section 3 is devoted to the study of the Dirichlet problem. It contains the main results concerning the reduction of a certain singular integral operator acting in spaces of differential forms and the integral representation of the solution of the Dirichlet problem by means of a simple layer potential.
2. Preliminary Results Let Ω be a bounded domain (open connected set) of R𝑚 (𝑚 ≥ 3).
2
Abstract and Applied Analysis In this paper, we deal with the Dirichlet problem: 𝐸𝑢 = 0 in Ω,
We remark that (see, e.g., [13, p. 285])
𝑢 = 𝑓 on Σ,
where 𝐸 is a scalar second-order differential operator (throughout this paper, we use the Einstein summation convention): 𝐸𝑢 (𝑥) =
𝜕𝑢 (𝑥) 𝜕 (𝑎𝑖𝑗 (𝑥) ). 𝑖 𝜕𝑥 𝜕𝑥𝑗
(2)
We suppose that the coefficients 𝑎𝑖𝑗 are defined on 𝑇, 𝑇 being an open ball containing Ω, and we assume that they belong to 𝐶2,𝜆 (𝑇), 0 < 𝜆 ≤ 1. Moreover, assume that 𝐴 = (𝑎𝑖𝑗 )𝑖,𝑗=1,...,𝑚 is a symmetric contravariant positive-definite tensor. Then, 𝐸 is a uniform elliptic operator; that is, there exists 𝑐 > 0 such that 𝑎𝑖𝑗 (𝑥)𝜉𝑖 𝜉𝑗 ≥ 𝑐|𝜉|2 , for every (𝜉1 , . . . , 𝜉𝑚 ) ∈ R𝑚 and for any 𝑥 ∈ 𝑇. For the sake of simplicity, we suppose that the determinant |𝐴| of 𝐴 is equal to 1. It is known that to the contravariant tensor 𝐴 there corresponds a covariant tensor 𝐴−1 = (𝑎𝑖𝑗 )𝑖,𝑗=1,...,𝑚 such that 𝑖𝑗
𝑎 𝑎𝑗ℎ =
𝛿ℎ𝑖 ,
for every 𝑖, ℎ = 1, . . . , 𝑚,
(3)
𝛿ℎ𝑖 being the Kronecker delta. A differential form of degree 𝑘 (in short a 𝑘-form) on 𝑇 is a function defined on 𝑇 whose values are in the 𝑘-covectors space of R𝑚 . A 𝑘-form 𝑢 can be represented as 𝑢=
1 d𝑥𝑠1 ⋅ ⋅ ⋅ d𝑥𝑠𝑘 𝑢 𝑘! 𝑠1 ⋅⋅⋅𝑠𝑘
(4)
with respect to an admissible coordinate system (𝑥1 , . . . , 𝑥𝑚 ), where 𝑢𝑠1 ⋅⋅⋅𝑠𝑘 are the components of a skew-symmetric covariant tensor (for details about differential forms, we refer to [13, 14]). The symbol 𝐶𝑘ℎ (𝑇) means the space of all 𝑘-forms whose components are continuously differentiable up to the order ℎ in a coordinate system of class 𝐶ℎ+1 (and then in every coordinate system of class 𝐶ℎ+1 ). If 𝑢 ∈ 𝐶𝑘1 (𝑇), the differential of 𝑢 is a (𝑘 + 1)-form defined as 1 𝜕𝑢𝑠1 ⋅⋅⋅𝑠𝑘 𝑗 𝑠1 d𝑥 d𝑥 ⋅ ⋅ ⋅ d𝑥𝑠𝑘 . d𝑢 = 𝑘! 𝜕𝑥𝑗
(5)
Further, if 𝑢 ∈ 𝐶𝑘0 (𝑇), the adjoint of 𝑢 is the following (𝑚 − 𝑘)form: ∗𝑢 =
1 𝛿1⋅⋅⋅𝑚 𝑘! (𝑚 − 𝑘)! 𝑗1 ⋅⋅⋅𝑗𝑘 𝑖𝑘+1 ⋅⋅⋅𝑖𝑚
(6)
⋅ 𝑎𝑠1 𝑗1 ⋅ ⋅ ⋅ 𝑎𝑠𝑘 𝑗𝑘 𝑢𝑠1 ⋅⋅⋅𝑠𝑘 d𝑥𝑖𝑘+1 ⋅ ⋅ ⋅ d𝑥𝑖𝑚 , ⋅⋅⋅𝑝𝑟 where 𝛿𝑞𝑝11⋅⋅⋅𝑞 is the generalized Kronecker delta (𝑟 ≤ 𝑚). We 𝑟 recall that (see, e.g., [15, p. 127]) 𝑗 ⋅⋅⋅𝑗 𝑗
⋅⋅⋅𝑗
ℎ
⋅⋅⋅ℎ
𝑗 ⋅⋅⋅𝑗 𝑗
⋅⋅⋅𝑗
𝛿ℎ1 ⋅⋅⋅ℎ𝑠 ℎ𝑠+1 ⋅⋅⋅ℎ𝑚 𝛿𝑘 𝑠+1⋅⋅⋅𝑘 𝑚 = (𝑚 − 𝑠)!𝛿ℎ1 ⋅⋅⋅ℎ𝑠 𝑘𝑠+1 ⋅⋅⋅𝑘𝑚 . 1
𝑠 𝑠+1
𝑚
𝑠+1
𝑚
∗ ∗ 𝑢 = (−1)𝑘(𝑚+1) 𝑢.
(1)
1
𝑠 𝑠+1
𝑚
(7)
(8)
If 𝑢 ∈ 𝐶𝑘1 (𝑇), we define the codifferential of 𝑢 as the following (𝑘 − 1)-form: 𝛿𝑢 = (−1)𝑚(𝑘+1)+1 ∗ d ∗ 𝑢.
(9)
A differential double form 𝑢ℎ,𝑘 (𝑥, 𝑦) of degree ℎ with respect to 𝑥 and of degree 𝑘 with respect to 𝑦 (in short a double (ℎ, 𝑘)form) is represented as 𝑢ℎ,𝑘 (𝑥, 𝑦) =
1 (𝑥, 𝑦) d𝑥𝑠1 ⋅ ⋅ ⋅ d𝑥𝑠ℎ d𝑦𝑗1 ⋅ ⋅ ⋅ d𝑦𝑗𝑘 . 𝑢 ℎ!𝑘! 𝑠1 ⋅⋅⋅𝑠ℎ 𝑗1 ⋅⋅⋅𝑗𝑘
(10)
If ℎ = 𝑘, we denote it briefly by 𝑢𝑘 (𝑥, 𝑦). We proceed to introduce the following double 𝑘-form (see [13, p. 204]) defined, for every 𝑥, 𝑦 ∈ 𝑇, 𝑥 ≠ 𝑦, as 𝜆 𝑘 (𝑥, 𝑦) =
1 𝐿 (𝑥, 𝑦) (𝑘!)2
(11) 𝑠1
𝑠𝑘
𝑖1
𝑖𝑘
⋅ 𝑎𝑠1 ⋅⋅⋅𝑠𝑘 𝑖1 ⋅⋅⋅𝑖𝑘 (𝑦) d𝑥 ⋅ ⋅ ⋅ d𝑥 d𝑦 ⋅ ⋅ ⋅ d𝑦 , where, for 𝑘 ≤ 𝑚, 𝑎 𝑠1 𝑖1 ⋅ ⋅ ⋅ 𝑎𝑠1 𝑖𝑘 . 𝑙 ⋅⋅⋅𝑙 .. . 𝑎𝑠1 ⋅⋅⋅𝑠𝑘 𝑖1 ⋅⋅⋅𝑖𝑘 = . d . = 𝛿𝑖11 ⋅⋅⋅𝑖𝑘𝑘 𝑎𝑠1 𝑙1 ⋅ ⋅ ⋅ 𝑎𝑠𝑘 𝑙𝑘 , 𝑎𝑠 𝑖 ⋅ ⋅ ⋅ 𝑎𝑠 𝑖 𝑘1 𝑘 𝑘
(12)
𝐿 (𝑥, 𝑦) =
(2−𝑚)/2 1 [𝑎𝑖𝑗 (𝑦) (𝑥𝑖 − 𝑦𝑖 ) (𝑥𝑗 − 𝑦𝑗 )] (𝑚 − 2) 𝜔𝑚
(13)
(𝜔𝑚 being the hypersurface measure of the unit sphere in R𝑚 ) is a parametrix in the sense of Hilbert and E.E. Levi for the operator 𝐸. We recall that (if we write 𝑢ℎ,𝑘 (𝑥, 𝑦) = O(|𝑥 − 𝑦|𝛼 ), 𝑢ℎ,𝑘 (𝑥, 𝑦) being a double (ℎ, 𝑘)-form, we mean that all its components are O(|𝑥 − 𝑦|𝛼 )) 2−𝑚 𝐿 (𝑥, 𝑦) = O (𝑥 − 𝑦 ) , 1−𝑚 d𝑥 𝐿 (𝑥, 𝑦) = O (𝑥 − 𝑦 ) , 1−𝑚 d𝑦 𝐿 (𝑥, 𝑦) = O (𝑥 − 𝑦 )
(14)
(15)
(see [13, Section 9]). The next results provide other properties of 𝐿 and 𝜆 𝑘 . Lemma 1. For every 𝑝 = 1, . . . , 𝑚, 𝜕𝐿 (𝑥, 𝑦) 𝜕𝐿 (𝑥, 𝑦) =− + 𝑀 (𝑥, 𝑦) , 𝑝 𝜕𝑦 𝜕𝑥𝑝
(16) 𝑥, 𝑦 ∈ 𝑇, 𝑥 ≠ 𝑦,
where 𝑀(𝑥, 𝑦) = O(|𝑥 − 𝑦|2−𝑚 ).
Abstract and Applied Analysis
3
Proof. Taking definition (13) into account, we have
where
𝜕𝐿 (𝑥, 𝑦) −𝑚/2 1 =− [𝑎 (𝑦) (𝑥𝑖 − 𝑦𝑖 ) (𝑥𝑗 − 𝑦𝑗 )] 𝜕𝑥𝑝 2𝜔𝑚 𝑖𝑗 ⋅ {𝑎𝑖𝑗 (𝑦) [𝛿𝑝𝑖 (𝑥𝑗 − 𝑦𝑗 ) + 𝛿𝑝𝑗 (𝑥𝑖 − 𝑦𝑖 )]} =−
−𝑚/2 1 [𝑎𝑖𝑗 (𝑦) (𝑥𝑖 − 𝑦𝑖 ) (𝑥𝑗 − 𝑦𝑗 )] 𝜔𝑚
2−𝑚 𝜖𝑘,𝑘+1 (𝑥, 𝑦) = O (𝑥 − 𝑦 ) .
(17)
⋅ [𝑎𝑝𝑗 (𝑦) (𝑥𝑗 − 𝑦𝑗 )] .
⋅
𝜕𝐿 (𝑥, 𝑦) −𝑚/2 1 =− [𝑎 (𝑦) (𝑥𝑖 − 𝑦𝑖 ) (𝑥𝑗 − 𝑦𝑗 )] 𝜕𝑦𝑝 2𝜔𝑚 𝑖𝑗 ⋅[ − ⋅
𝜕𝑎𝑖𝑗 (𝑦) 𝜕𝑦𝑝
= 𝑀 (𝑥, 𝑦) +
𝑗
𝑗
(𝑥 − 𝑦 ) −
𝛿𝑝𝑗
𝛿𝑝1⋅⋅⋅𝑚 𝐿 (𝑥, 𝑦) d𝑥𝑞1 1 ⋅⋅⋅𝑝𝑘 𝑞1 ⋅⋅⋅𝑞𝑚−𝑘
∗𝑥 𝜆 𝑘 (𝑥, 𝑦) =
−𝑚/2 1 [𝑎 (𝑦) (𝑥𝑖 − 𝑦𝑖 ) (𝑥𝑗 − 𝑦𝑗 )] 2𝜔𝑚 𝑖𝑗 𝑖
(24) ⋅ ⋅ ⋅ d𝑥
𝑞𝑚−𝑘
𝑝1
𝑝𝑘
d𝑦 ⋅ ⋅ ⋅ d𝑦 .
1 (𝑚 − 𝑘)! (𝑘!)2
⋅ 𝛿𝑗1⋅⋅⋅𝑚 [(𝑎𝑠1 𝑗1 ⋅ ⋅ ⋅ 𝑎𝑠𝑘 𝑗𝑘 ) (𝑥) 1 ⋅⋅⋅𝑗𝑘 𝑞1 ⋅⋅⋅𝑞𝑚−𝑘
(18)
𝑖
(𝑥 − 𝑦 )]}
− (𝑎𝑠1 𝑗1 ⋅ ⋅ ⋅ 𝑎𝑠𝑘 𝑗𝑘 ) (𝑦)] ⋅ 𝐿 (𝑥, 𝑦) 𝑎𝑠1 ⋅⋅⋅𝑠𝑘 𝑝1 ⋅⋅⋅𝑝𝑘 (𝑦) d𝑥𝑞1 ⋅ ⋅ ⋅ d𝑥𝑞𝑚−𝑘 d𝑦𝑝1 ⋅ ⋅ ⋅ d𝑦𝑝𝑘
−𝑚/2 1 [𝑎𝑖𝑗 (𝑦) (𝑥𝑖 − 𝑦𝑖 ) (𝑥𝑗 − 𝑦𝑗 )] 𝜔𝑚
⋅ [𝑎𝑝𝑗 (𝑦) (𝑥𝑗 − 𝑦𝑗 )] = 𝑀 (𝑥, 𝑦) −
(−1)𝑘(𝑚−𝑘) 𝑘! (𝑚 − 𝑘)!
On the other hand,
(𝑥𝑖 − 𝑦𝑖 ) (𝑥𝑗 − 𝑦𝑗 )]
{𝑎𝑖𝑗 (𝑦) [−𝛿𝑝𝑖
Proof. First, we prove (19). It follows from (12), (3), and (7) that
∗𝑦 𝜆 𝑚−𝑘 (𝑥, 𝑦) =
On the other hand,
(23)
+
𝜕𝐿 (𝑥, 𝑦) 𝜕𝑥𝑝
(25)
1 𝛿𝑗1⋅⋅⋅𝑚 (𝑎𝑠1 𝑗1 ⋅ ⋅ ⋅ 𝑎𝑠𝑘 𝑗𝑘 ) (𝑦) ⋅⋅⋅𝑗 𝑞 ⋅⋅⋅𝑞 (𝑚 − 𝑘)! (𝑘!)2 1 𝑘 1 𝑚−𝑘
⋅ 𝐿 (𝑥, 𝑦) 𝑎𝑠1 ⋅⋅⋅𝑠𝑘 𝑝1 ⋅⋅⋅𝑝𝑘 (𝑦) d𝑥𝑞1 ⋅ ⋅ ⋅ d𝑥𝑞𝑚−𝑘 d𝑦𝑝1 ⋅ ⋅ ⋅ d𝑦𝑝𝑘 .
and this yields the claim.
From (12), (3), and (7), we have that
The identities proved in the next proposition generalize the ones obtained by Colautti [16, p. 309] for the Laplacian. Proposition 2. Let 𝜆 𝑘 be the double 𝑘-form defined by (11). Then, for every 𝑥 ≠ 𝑦, the following properties hold: ∗𝑥 𝜆 𝑘 (𝑥, 𝑦) = (−1)𝑘(𝑚−𝑘) ∗𝑦 𝜆 𝑚−𝑘 (𝑥, 𝑦) + 𝜏𝑚−𝑘,𝑘 (𝑥, 𝑦) , 𝑘 ≤ 𝑚,
(19)
∗𝑥 𝜆 𝑘 (𝑥, 𝑦) =
⋅ 𝛿𝑗1⋅⋅⋅𝑚 [(𝑎𝑠1 𝑗1 ⋅ ⋅ ⋅ 𝑎𝑠𝑘 𝑗𝑘 ) (𝑥) 1 ⋅⋅⋅𝑗𝑘 𝑞1 ⋅⋅⋅𝑞𝑚−𝑘 − (𝑎𝑠1 𝑗1 ⋅ ⋅ ⋅ 𝑎𝑠𝑘 𝑗𝑘 ) (𝑦)] 𝐿 (𝑥, 𝑦) ⋅ 𝑎𝑠1 ⋅⋅⋅𝑠𝑘 𝑝1 ⋅⋅⋅𝑝𝑘 (𝑦) d𝑥𝑞1 ⋅ ⋅ ⋅ d𝑥𝑞𝑚−𝑘 d𝑦𝑝1 ⋅ ⋅ ⋅ d𝑦𝑝𝑘 +
where
1 (𝑚 − 𝑘)! (𝑘!)2
(26)
1 (𝑚 − 𝑘)!𝑘!
𝐿 (𝑥, 𝑦) d𝑥𝑞1 ⋅ ⋅ ⋅ d𝑥𝑞𝑚−𝑘 d𝑦𝑝1 ⋅ ⋅ ⋅ d𝑦𝑝𝑘 ⋅ 𝛿𝑝1⋅⋅⋅𝑚 1 ⋅⋅⋅𝑝𝑘 𝑞1 ⋅⋅⋅𝑞𝑚−𝑘 3−𝑚 𝜏𝑚−𝑘,𝑘 (𝑥, 𝑦) = O (𝑥 − 𝑦 ) , ∗𝑥 d𝑥 𝜆 𝑘 (𝑥, 𝑦) = (−1)𝑚𝑘+1 ∗𝑦 d𝑦 𝜆 𝑚−𝑘−1 (𝑥, 𝑦) + 𝛾𝑚−𝑘−1,𝑘 (𝑥, 𝑦) , 𝑘 < 𝑚,
(20) (21)
where 𝛾𝑚−𝑘−1,𝑘 (𝑥, 𝑦) = O(|𝑥 − 𝑦|2−𝑚 ); and 𝛿𝑥 𝜆 𝑘+1 (𝑥, 𝑦) = d𝑦 𝜆 𝑘 (𝑥, 𝑦) + 𝜖𝑘,𝑘+1 (𝑥, 𝑦) , 𝑘 < 𝑚, (22)
= 𝜏𝑚−𝑘,𝑘 (𝑥, 𝑦) + (−1)𝑘(𝑚−𝑘) ∗𝑦 𝜆 𝑚−𝑘 (𝑥, 𝑦) , where 𝜏𝑚−𝑘,𝑘 (𝑥, 𝑦) satisfies (20) on account of (𝑎𝑠1 𝑗1 ⋅ ⋅ ⋅ 𝑎𝑠𝑘 𝑗𝑘 ) (𝑥) − (𝑎𝑠1 𝑗1 ⋅ ⋅ ⋅ 𝑎𝑠𝑘 𝑗𝑘 ) (𝑦) = O (𝑥 − 𝑦) (27) and (8).
4
Abstract and Applied Analysis
Now we pass to show (21). With calculations analogue to (26), we have that ∗𝑥 d𝑥 𝜆 𝑘 (𝑥, 𝑦) =
1 1⋅⋅⋅𝑚 𝛿𝑗𝑗 ⋅⋅⋅𝑗 𝑖 ⋅⋅⋅𝑖 (𝑚 − 𝑘 − 1)! (𝑘!)2 1 𝑘 𝑘+2 𝑚
+ 𝛾𝑚−𝑘−1,𝑘 (𝑥, 𝑦) = (−1)𝑚𝑘+1 ∗𝑥 d𝑥 𝜆 𝑘 (𝑥, 𝑦)
𝜕𝐿 (𝑥, 𝑦) 𝑎𝑠1 ⋅⋅⋅𝑠𝑘 𝑖1 ⋅⋅⋅𝑖𝑘 (𝑦) 𝜕𝑥𝑠
1⋅⋅⋅𝑚 ⋅ 𝛿𝑗𝑗 (𝑎𝑠𝑗 𝑎𝑠1 𝑗1 ⋅ ⋅ ⋅ 𝑎𝑠𝑘 𝑗𝑘 ) (𝑦) 1 ⋅⋅⋅𝑗𝑘 𝑖𝑘+2 ⋅⋅⋅𝑖𝑚
(𝑥, 𝑦) + 𝛾𝑚−𝑘−1,𝑘 (𝑥, 𝑦) , + 𝛾𝑚−𝑘−1,𝑘
𝜕𝐿 (𝑥, 𝑦) 𝜕𝑥𝑠
(30) (28)
⋅ 𝑎𝑠1 ⋅⋅⋅𝑠𝑘 𝑖1 ⋅⋅⋅𝑖𝑘 (𝑦) d𝑥𝑖𝑘+2 ⋅ ⋅ ⋅ d𝑥𝑖𝑚 d𝑦𝑖1 ⋅ ⋅ ⋅ d𝑦𝑖𝑘
(𝑥, 𝑦) + 𝛾𝑚−𝑘−1,𝑘 (𝑥, 𝑦)) . = (−1)𝑚𝑘 (𝛾𝑚−𝑘−1,𝑘
𝜕𝐿 (𝑥, 𝑦) 𝜕𝑥𝑠
𝛿𝑥 𝜆 𝑘+1 (𝑥, 𝑦) = (−1)𝑚−𝑘 ∗𝑥 d𝑥 ∗𝑦 𝜆 𝑚−𝑘−1 (𝑥, 𝑦)
(𝑥, 𝑦) = O(|𝑥 − 𝑦|1−𝑚 ) thanks to (15) and (27). where 𝛾𝑚−𝑘−1,𝑘 Moreover,
1 1⋅⋅⋅𝑚 𝛿𝑗𝑞 ⋅⋅⋅𝑞 𝑖 ⋅⋅⋅𝑖 𝑘! [(𝑚 − 𝑘 − 1)!]2 1 𝑚−𝑘−1 1 𝑘 (29)
𝜕 [𝐿 (𝑥, 𝑦) 𝑎𝑠1 ⋅⋅⋅𝑠𝑚−𝑘−1 𝑝1 ⋅⋅⋅𝑝𝑚−𝑘−1 (𝑦)] 𝜕𝑦𝑠
⋅
𝜖𝑘,𝑘+1 (𝑥, 𝑦) = (−1)𝑚−𝑘 ∗𝑦 𝛾𝑘,𝑚−𝑘−1 (𝑥, 𝑦) 1 𝑘! [(𝑚 − 𝑘 − 1)!]2
𝜕 ⋅ 𝐿 (𝑥, 𝑦) 𝑠 𝑎𝑠1 ⋅⋅⋅𝑠𝑚−𝑘−1 𝑝1 ⋅⋅⋅𝑝𝑚−𝑘−1 (𝑦) 𝜕𝑦 ⋅ d𝑥 ⋅ ⋅ ⋅ d𝑥
𝑠𝑚−𝑘−1
𝑚𝑘
(−1) d𝑦 ⋅ ⋅ ⋅ d𝑦 = − 𝑘! (𝑚 − 𝑘 − 1)! 𝑖1
𝑎𝑠𝑗 (𝑦) ⋅ 𝛿𝑗𝑖1⋅⋅⋅𝑚 1 ⋅⋅⋅𝑖𝑘 𝑠1 ⋅⋅⋅𝑠𝑚−𝑘−1
𝑖𝑘
(33)
and hence the claim with
𝜕𝐿 (𝑥, 𝑦) (𝑦) 𝜕𝑦𝑠
1⋅⋅⋅𝑚 ⋅ 𝛿𝑗𝑞 (𝑎𝑠𝑗 𝑎𝑝1 𝑞1 ⋅ ⋅ ⋅ 𝑎𝑝𝑚−𝑘−1 𝑞𝑚−𝑘−1 ) (𝑦) 1 ⋅⋅⋅𝑞𝑚−𝑘−1 𝑖1 ⋅⋅⋅𝑖𝑘
𝑠1
+ 𝜖𝑘,𝑘+1 (𝑥, 𝑦)
= d𝑦 𝜆 𝑘 (𝑥, 𝑦) + 𝜖𝑘,𝑘+1 (𝑥, 𝑦) ,
(−1)𝑘(𝑚−𝑘−1) 𝑘! (𝑚 − 𝑘 − 1)!
⋅ d𝑥𝑠1 ⋅ ⋅ ⋅ d𝑥𝑠𝑚−𝑘−1 d𝑦𝑖1 ⋅ ⋅ ⋅ d𝑦𝑖𝑘 +
(32)
where 𝜖𝑘,𝑘+1 (𝑥, 𝑦) = O(|𝑥−𝑦|2−𝑚 ). Now, by using (21) and (8), we get
+ (−1)𝑚−𝑘 ∗𝑦 𝛾𝑘,𝑚−𝑘−1 (𝑥, 𝑦)
Arguing again as in (26) and taking Lemma 1 into account, we get
𝑎𝑠𝑗 𝛿𝑗𝑖1⋅⋅⋅𝑚 1 ⋅⋅⋅𝑖𝑘 𝑠1 ⋅⋅⋅𝑠𝑚−𝑘−1
= (−1)𝑚−𝑘 ∗𝑦 ∗𝑥 d𝑥 𝜆 𝑚−𝑘−1 (𝑥, 𝑦)
𝛿𝑥 𝜆 𝑘+1 (𝑥, 𝑦) = (−1)𝑚−𝑘−𝑘𝑚+1 ∗𝑦 ∗𝑦 d𝑦 𝜆 𝑘 (𝑥, 𝑦)
⋅ d𝑥𝑠1 ⋅ ⋅ ⋅ d𝑥𝑠𝑚−𝑘−1 d𝑦𝑖1 ⋅ ⋅ ⋅ d𝑦𝑖𝑘 .
∗𝑦 d𝑦 𝜆 𝑚−𝑘−1 (𝑥, 𝑦) =
+ (−1)𝑚𝑘+1 ∗𝑥 d𝑥 𝜏𝑚−𝑘−1,𝑘+1 (𝑥, 𝑦)
+ 𝜖𝑘,𝑘+1 (𝑥, 𝑦) ,
⋅ (𝑎𝑠𝑗 𝑎𝑝1 𝑞1 ⋅ ⋅ ⋅ 𝑎𝑝𝑚−𝑘−1 𝑞𝑚−𝑘−1 ) (𝑦) ⋅
(31)
Finally, we prove (22). Thanks to (9) and (19), we have
⋅ d𝑥𝑖𝑘+2 ⋅ ⋅ ⋅ d𝑥𝑖𝑚 d𝑦𝑖1 ⋅ ⋅ ⋅ d𝑦𝑖𝑘 ,
∗𝑦 d𝑦 𝜆 𝑚−𝑘−1 (𝑥, 𝑦) =
where both 𝛾𝑚−𝑘−1,𝑘 (𝑥, 𝑦) and 𝛾𝑚−𝑘−1,𝑘 (𝑥, 𝑦) are O(|𝑥−𝑦|2−𝑚 ). Then, we obtain the claim by setting
𝛾𝑚−𝑘−1,𝑘 (𝑥, 𝑦)
1 (𝑚 − 𝑘 − 1)!𝑘!
𝑎𝑠𝑗 (𝑦) ⋅ ⋅ 𝛿𝑗𝑖1⋅⋅⋅𝑚 𝑘+2 ⋅⋅⋅𝑖𝑚 𝑖1 ⋅⋅⋅𝑖𝑘
(−1)𝑚𝑘 𝑎𝑠𝑗 (𝑦) 𝛿1⋅⋅⋅𝑚 𝑘! (𝑚 − 𝑘 − 1)! 𝑗𝑖1 ⋅⋅⋅𝑖𝑘 𝑠1 ⋅⋅⋅𝑠𝑚−𝑘−1
⋅ 𝑀 (𝑥, 𝑦) d𝑥𝑠1 ⋅ ⋅ ⋅ d𝑥𝑠𝑚−𝑘−1 d𝑦𝑖1 ⋅ ⋅ ⋅ d𝑦𝑖𝑘
1 ⋅ d𝑥𝑖𝑘+2 ⋅ ⋅ ⋅ d𝑥𝑖𝑚 d𝑦𝑖1 ⋅ ⋅ ⋅ d𝑦𝑖𝑘 + (𝑚 − 𝑘 − 1)! (𝑘!)2
= 𝛾𝑚−𝑘−1,𝑘 (𝑥, 𝑦) +
𝜕𝐿 (𝑥, 𝑦) 𝑠1 d𝑥 ⋅ ⋅ ⋅ d𝑥𝑠𝑚−𝑘−1 d𝑦𝑖1 ⋅ ⋅ ⋅ d𝑦𝑖𝑘 𝜕𝑥𝑠
+
⋅ [(𝑎𝑠𝑗 𝑎𝑠1 𝑗1 ⋅ ⋅ ⋅ 𝑎𝑠𝑘 𝑗𝑘 ) (𝑥) − (𝑎𝑠𝑗 𝑎𝑠1 𝑗1 ⋅ ⋅ ⋅ 𝑎𝑠𝑘 𝑗𝑘 ) (𝑦)]
⋅
+ 𝜖𝑘,𝑘+1 (𝑥, 𝑦) .
(34)
Proposition 3. If 𝑢 ∈ 𝐶𝑘2 (𝑇), then (𝛿d + d𝛿) 𝑢 = −𝐸𝑢 + 𝐹𝑢,
(35)
where 𝐹 is a linear first-order differential operator whose coefficients depend only on first- and second-order derivatives of entries of the tensor 𝐴.
Abstract and Applied Analysis
5
In particular,
Proof. We begin by observing that
(𝛿𝑥 d𝑥 + d𝑥 𝛿𝑥 ) 𝜆 𝑘 (𝑥, 𝑦) = 𝐹𝑥 [𝜆 𝑘 (𝑥, 𝑦)] , 𝑥 ≠ 𝑦.
d𝛿𝑢 = (−1)𝑚(𝑘+1)+1 d
(36)
𝜕 1 𝛿1⋅⋅⋅𝑚 𝑎𝑗ℎ 𝑎𝑖𝑘+1 ℎ𝑘+1 ⋅ ⋅ ⋅ 𝑎𝑖𝑚 ℎ𝑚 𝛿𝑗1⋅⋅⋅𝑚 𝑎𝑠1 𝑗1 ⋅ ⋅ ⋅ 𝑎𝑠𝑘 𝑗𝑘 𝑗 𝑢𝑠1 ⋅⋅⋅𝑠𝑘 d𝑥𝑖2 ⋅ ⋅ ⋅ d𝑥𝑖𝑘 1 ⋅⋅⋅𝑗𝑘 𝑖𝑘+1 ⋅⋅⋅𝑖𝑚 𝜕𝑥 (𝑘 − 1)!𝑘! (𝑚 − 𝑘)! ℎℎ𝑘+1 ⋅⋅⋅ℎ𝑚 𝑖𝑖2 ⋅⋅⋅𝑖𝑘
+ (−1)𝑚(𝑘+1)+1 d
𝜕 1 𝑎𝑗ℎ 𝑎𝑖𝑘+1 ℎ𝑘+1 ⋅ ⋅ ⋅ 𝑎𝑖𝑚 ℎ𝑚 ⋅ 𝛿𝑗1⋅⋅⋅𝑚 (𝑎𝑠1 𝑗1 ⋅ ⋅ ⋅ 𝑎𝑠𝑘 𝑗𝑘 ) 𝛿1⋅⋅⋅𝑚 1 ⋅⋅⋅𝑗𝑘 𝑖𝑘+1 ⋅⋅⋅𝑖𝑚 𝜕𝑥𝑗 (𝑘 − 1)!𝑘! (𝑚 − 𝑘)! ℎℎ𝑘+1 ⋅⋅⋅ℎ𝑚 𝑖𝑖2 ⋅⋅⋅𝑖𝑘
(37)
⋅ 𝑢𝑠1 ⋅⋅⋅𝑠𝑘 d𝑥𝑖2 ⋅ ⋅ ⋅ d𝑥𝑖𝑘 .
Since 𝐴 is symmetric and |𝐴| = 1, we get
⋅
𝛿𝑗1⋅⋅⋅𝑚 𝑎𝑠1 𝑗1 ⋅ ⋅ ⋅ 𝑎𝑠𝑘 𝑗𝑘 𝑎𝑖𝑘+1 ℎ𝑘+1 ⋅ ⋅ ⋅ 𝑎𝑖𝑚 ℎ𝑚 1 ⋅⋅⋅𝑗𝑘 𝑖𝑘+1 ⋅⋅⋅𝑖𝑚 =
𝑠1 ⋅⋅⋅𝑠𝑘 ℎ𝑘+1 ⋅⋅⋅ℎ𝑚 𝛿1⋅⋅⋅𝑚
(39) On the other hand,
1 𝜕 𝑠 ⋅⋅⋅𝑠 d𝛿𝑢 = − 𝛿ℎ𝑖1 ⋅⋅⋅𝑖𝑘 𝑖 (𝑎𝑗ℎ 2 𝑘 𝜕𝑥 1 (𝑘 − 1)!𝑘! 𝜕𝑢𝑠1 ⋅⋅⋅𝑠𝑘 𝜕𝑥𝑗
𝛿d𝑢 = (−1)𝑚(𝑘+2)+1
) d𝑥𝑖1 d𝑥𝑖2 ⋅ ⋅ ⋅ d𝑥𝑖𝑘
+ (−1)𝑚(𝑘+1)+1 1⋅⋅⋅𝑚 ⋅ 𝛿ℎℎ 𝑘+1 ⋅⋅⋅ℎ𝑚 𝑖𝑖
2
d𝛿𝑢 + 𝛿d𝑢 = −
1 (𝑘 − 1)!𝑘! (𝑚 − 𝑘)!
⋅
1⋅⋅⋅𝑚 ⋅⋅⋅𝑖𝑘 𝛿𝑗1 ⋅⋅⋅𝑗𝑘 𝑖𝑘+1 ⋅⋅⋅𝑖𝑚
𝜕𝑢𝑠1 ⋅⋅⋅𝑠𝑘 𝜕 (𝑎𝑗ℎ 𝑎𝑠1 ℎ1 ⋅ ⋅ ⋅ 𝑎𝑠𝑘 ℎ𝑘 ) d𝑥𝑖1 ⋅ ⋅ ⋅ d𝑥𝑖𝑘 . 𝑝 𝜕𝑥 𝜕𝑥𝑗
Then,
𝜕2 𝑢𝑠 ⋅⋅⋅𝑠 1 1 𝑠 ⋅⋅⋅𝑠 1⋅⋅⋅𝑚 𝛿ℎ𝑖1 ⋅⋅⋅𝑖𝑘 𝑎𝑗ℎ 𝑖 1 𝑗𝑘 d𝑥𝑖1 d𝑥𝑖2 ⋅ ⋅ ⋅ d𝑥𝑖𝑘 + (−1)𝑚(𝑘+2)+1 𝛿𝑞𝑗 ⋅⋅⋅𝑗 𝑖 ⋅⋅⋅𝑖 2 𝑘 1 𝜕𝑥 𝜕𝑥 (𝑘 − 1)!𝑘! (𝑚 − 𝑘 − 1)! (𝑘!)2 𝑘+2 𝑚 1 𝑘
⋅ d𝑥𝑖1 d𝑥𝑖2 ⋅ ⋅ ⋅ d𝑥𝑖𝑘 + (−1)𝑚(𝑘+1)+1
𝜕2 𝑢𝑠1 ⋅⋅⋅𝑠𝑘 𝜕𝑥𝑝 𝜕𝑥
d𝑥𝑖1 ⋅ ⋅ ⋅ d𝑥𝑖𝑘 − 𝑗
𝑗ℎ 1 𝑠 ⋅⋅⋅𝑠 𝜕𝑎 𝜕𝑢𝑠1 ⋅⋅⋅𝑠𝑘 𝛿ℎ𝑖1 ⋅⋅⋅𝑖𝑘 (𝑘 − 1)!𝑘! 2 𝑘 𝜕𝑥𝑖1 𝜕𝑥𝑗
1 𝛿1⋅⋅⋅𝑚 𝑎𝑗ℎ 𝑎𝑖𝑘+1 ℎ𝑘+1 ⋅ ⋅ ⋅ 𝑎𝑖𝑚 ℎ𝑚 𝛿1⋅⋅⋅𝑚 (𝑘 − 1)!𝑘! (𝑚 − 𝑘)! ℎℎ𝑘+1 ⋅⋅⋅ℎ𝑚 𝑖𝑖2 ⋅⋅⋅𝑖𝑘 𝑗1 ⋅⋅⋅𝑗𝑘 𝑖𝑘+1 ⋅⋅⋅𝑖𝑚
𝜕𝑢𝑠1 ⋅⋅⋅𝑠𝑘 𝑖 𝑖 1 𝜕 (𝑎𝑠1 𝑗1 ⋅ ⋅ ⋅ 𝑎𝑠𝑘 𝑗𝑘 ) d𝑥 1 d𝑥 2 ⋅ ⋅ ⋅ d𝑥𝑖𝑘 + (−1)𝑚(𝑘+2)+1 1 𝑗 𝜕𝑥 𝜕𝑥𝑖 (𝑚 − 𝑘 − 1)! (𝑘!)2
1⋅⋅⋅𝑚 1⋅⋅⋅𝑚 ⋅ 𝛿𝑞𝑗 𝑎𝑝𝑞 𝑎𝑖𝑘+2 𝑗𝑘+2 ⋅ ⋅ ⋅ 𝑎𝑖𝑚 𝑗𝑚 𝛿ℎℎ 𝑘+2 ⋅⋅⋅𝑗𝑚 𝑖1 ⋅⋅⋅𝑖𝑘 1 ⋅⋅⋅ℎ𝑘 𝑖𝑘+2 ⋅⋅⋅𝑖𝑚
⋅
1 (𝑚 − 𝑘 − 1)! (𝑘!)2
1⋅⋅⋅𝑚 1⋅⋅⋅𝑚 ⋅ 𝛿𝑞𝑗 𝑎𝑝𝑞 𝑎𝑖𝑘+2 𝑗𝑘+2 ⋅ ⋅ ⋅ 𝑎𝑖𝑚 𝑗𝑚 𝛿ℎℎ 𝑘+2 ⋅⋅⋅𝑗𝑚 𝑖1 ⋅⋅⋅𝑖𝑘 1 ⋅⋅⋅ℎ𝑘 𝑖𝑘+2 ⋅⋅⋅𝑖𝑚
1⋅⋅⋅𝑚 ⋅ 𝑎𝑝𝑞 𝑎𝑖𝑘+2 𝑗𝑘+2 ⋅ ⋅ ⋅ 𝑎𝑖𝑚 𝑗𝑚 ⋅ 𝛿ℎℎ 𝑎𝑗ℎ 𝑎𝑠1 ℎ1 ⋅ ⋅ ⋅ 𝑎𝑠𝑘 ℎ𝑘 1 ⋅⋅⋅ℎ𝑘 𝑖𝑘+2 ⋅⋅⋅𝑖𝑚
⋅
⋅ 𝑢𝑠1 ⋅⋅⋅𝑠𝑘 ] d𝑥𝑖1 d𝑥𝑖2 ⋅ ⋅ ⋅ d𝑥𝑖𝑘 .
(38)
and, keeping in mind (7), we get
⋅
𝜕 𝜕 [𝑎𝑗ℎ 𝑎𝑖𝑘+1 ℎ𝑘+1 ⋅ ⋅ ⋅ 𝑎𝑖𝑚 ℎ𝑚 𝑗 (𝑎𝑠1 𝑗1 ⋅ ⋅ ⋅ 𝑎𝑠𝑘 𝑗𝑘 ) 𝜕𝑥𝑖1 𝜕𝑥
𝜕𝑢𝑠1 ⋅⋅⋅𝑠𝑘 𝑖 𝜕 (𝑎𝑗ℎ 𝑎𝑠1 ℎ1 ⋅ ⋅ ⋅ 𝑎𝑠𝑘 ℎ𝑘 ) d𝑥 1 ⋅ ⋅ ⋅ d𝑥𝑖𝑘 + (−1)𝑚(𝑘+1)+1 𝑝 𝜕𝑥 𝜕𝑥𝑗
𝜕 1 𝜕 𝛿1⋅⋅⋅𝑚 𝛿1⋅⋅⋅𝑚 (𝑎𝑗ℎ 𝑎𝑖𝑘+1 ℎ𝑘+1 ⋅ ⋅ ⋅ 𝑎𝑖𝑚 ℎ𝑚 ) 𝑗 (𝑎𝑠1 𝑗1 ⋅ ⋅ ⋅ 𝑎𝑠𝑘 𝑗𝑘 ) 𝑢𝑠1 ⋅⋅⋅𝑠𝑘 d𝑥𝑖1 d𝑥𝑖2 ⋅ ⋅ ⋅ d𝑥𝑖𝑘 𝜕𝑥 (𝑘 − 1)!𝑘! (𝑚 − 𝑘)! ℎℎ𝑘+1 ⋅⋅⋅ℎ𝑚 𝑖𝑖2 ⋅⋅⋅𝑖𝑘 𝑗1 ⋅⋅⋅𝑗𝑘 𝑖𝑘+1 ⋅⋅⋅𝑖𝑚 𝜕𝑥𝑖1
+ (−1)𝑚(𝑘+1)+1
1 𝜕2 1⋅⋅⋅𝑚 1⋅⋅⋅𝑚 𝑗ℎ 𝑖𝑘+1 ℎ𝑘+1 𝑖𝑚 ℎ𝑚 𝛿ℎℎ 𝛿 𝑎 𝑎 ⋅ ⋅ ⋅ 𝑎 (𝑎𝑠1 𝑗1 ⋅ ⋅ ⋅ 𝑎𝑠𝑘 𝑗𝑘 ) 𝑗 ⋅⋅⋅ℎ 𝑖 ⋅⋅⋅𝑖 ⋅⋅⋅𝑗 𝑖 ⋅⋅⋅𝑖 𝜕𝑥𝑖1 𝜕𝑥𝑗 (𝑘 − 1)!𝑘! (𝑚 − 𝑘)! 𝑘+1 𝑚 𝑖2 𝑘 1 𝑘 𝑘+1 𝑚
⋅ 𝑢𝑠1 ⋅⋅⋅𝑠𝑘 d𝑥𝑖1 d𝑥𝑖2 ⋅ ⋅ ⋅ d𝑥𝑖𝑘 = − −
𝜕𝑢𝑠1 ⋅⋅⋅𝑠𝑘 1 𝜕 𝑚! − (𝑚 − 𝑘)! 𝜕𝑎𝑗𝑝 𝜕𝑢𝑠1 ⋅⋅⋅𝑠𝑘 𝑠1 𝑗𝑝 𝑠1 𝑠𝑘 (𝑎 ) d𝑥 ⋅ ⋅ ⋅ d𝑥 + d𝑥 ⋅ ⋅ ⋅ d𝑥𝑠𝑘 𝑘! 𝜕𝑥𝑝 𝜕𝑥𝑗 𝑚!𝑘! 𝜕𝑥𝑝 𝜕𝑥𝑗
𝑚! − (𝑚 − 𝑘)! 𝑠1 𝑠2 ⋅⋅⋅𝑠𝑘 𝜕𝑎𝑗ℎ 𝜕𝑢𝑠1 ⋅⋅⋅𝑠𝑘 𝑖1 (𝑚 − 𝑘 + 1)! 𝑗𝑠2 ⋅⋅⋅𝑠𝑘 𝜕𝑎𝑠1 ℎ 𝜕𝑢𝑠1 ⋅⋅⋅𝑠𝑘 𝑖1 (𝑚 − 𝑘)! 𝑗𝑠2 ⋅⋅⋅𝑠𝑘 d𝑥 ⋅ ⋅ ⋅ d𝑥𝑖𝑘 − 𝛿 d𝑥 ⋅ ⋅ ⋅ d𝑥𝑖𝑘 + 𝛿ℎ𝑖 ⋅⋅⋅𝑖 𝛿 𝑖 𝑗 𝑚! (𝑘 − 1)!𝑘! 2 𝑘 𝜕𝑥 1 𝜕𝑥 𝑚! (𝑘 − 1)!𝑘! 𝑖1 𝑖2 ⋅⋅⋅𝑖𝑘 𝑚! (𝑘 − 1) !2 ℎ𝑖2 ⋅⋅⋅𝑖𝑘 𝜕𝑥𝑗 𝜕𝑥𝑖1
(40)
6
Abstract and Applied Analysis 𝜕𝑎𝑠1 ℎ 𝜕𝑢𝑠1 ⋅⋅⋅𝑠𝑘 𝑖1 (𝑚 − 𝑘)! 𝑗𝑠2 ⋅⋅⋅𝑠𝑘 𝜕𝑎𝑠1 ℎ 𝜕𝑢𝑠1 ⋅⋅⋅𝑠𝑘 𝑖1 (𝑚 − 𝑘 + 1)! 𝑗𝑠2 ⋅⋅⋅𝑠𝑘 𝜕2 𝑎𝑠1 ℎ 𝑖𝑘 𝑖𝑘 d𝑥 ⋅ ⋅ ⋅ d𝑥 − 𝛿 d𝑥 ⋅ ⋅ ⋅ d𝑥 − 𝑢 d𝑥𝑖1 ⋅ ⋅ ⋅ d𝑥𝑖𝑘 𝛿 𝑚! (𝑘 − 1)! ℎ𝑖2 ⋅⋅⋅𝑖𝑘 𝜕𝑥𝑖1 𝜕𝑥𝑗 𝑠1 ⋅⋅⋅𝑠𝑘 𝜕𝑥ℎ 𝜕𝑥𝑗 𝑚! (𝑘 − 1) !2 ℎ𝑖2 ⋅⋅⋅𝑖𝑘 𝜕𝑥𝑖1 𝜕𝑥𝑗 1 1 = − 𝐸𝑢𝑠1 ⋅⋅⋅𝑠𝑘 d𝑥𝑠1 ⋅ ⋅ ⋅ d𝑥𝑠𝑘 + 𝐹𝑢𝑠1 ⋅⋅⋅𝑠𝑘 d𝑥𝑠1 ⋅ ⋅ ⋅ d𝑥𝑠𝑘 𝑘! 𝑘!
⋅
(41)
and this proves (35). Finally, (36) follows from (35). Finally, following Fichera we employ the parametrix 𝐿 to construct a principal fundamental solution of the differential operator 𝐸 (see [12]). Lemma 4. There exists 𝜁(𝑥, 𝑦) such that the function 𝑆 (𝑥, 𝑦) = 𝐿 (𝑥, 𝑦) + 𝜁 (𝑥, 𝑦) , 𝑥 ∈ 𝑇, 𝑦 ∈ 𝑇,
(42)
is a principal fundamental solution of 𝐸. In particular, we have 𝐸𝑥 𝑆 (𝑥, 𝑦) = 0, 𝑆 (𝑥, 𝑦) = 0,
𝑥 ∈ 𝑇, 𝑦 ∈ 𝑇, 𝑥 ≠ 𝑦, 𝑥 ∈ 𝜕𝑇, 𝑦 ∈ 𝑇, 2−𝑚
𝑆 (𝑥, 𝑦) = O (𝑥 − 𝑦
(43)
).
Moreover, for every 𝑥 ≠ 𝑦, 2−𝑚−𝛾 d𝑥 𝜁 (𝑥, 𝑦) = O (𝑥 − 𝑦 )
(44)
for some 0 < 𝛾 ≤ 1. Proof. The existence of 𝜁(𝑥, 𝑦) can be obtained as the solution of a certain integral equation (see [12, §2]). In [12] properties (43) are proved and 𝜁(𝑥, 𝑦) is written as 𝜁 (𝑥, 𝑦) = 𝐺 (𝑥, 𝑦) + ∫ 𝐿 (𝑥, 𝑤) 𝑅 (𝑤, 𝑦) d𝑤, 𝑇
(45)
where 𝐺 is a smooth function on 𝑇 and, for some 0 < 𝛾 ≤ 1, 1−𝑚−𝛾 𝑅 (𝑤, 𝑦) = O (𝑤 − 𝑦 ).
(46)
Then, by (15),
𝑦 ∈ Σ associated with the operator 𝐸 and defined as ]𝑖 (𝑦) = 𝑎𝑖𝑗 (𝑦)𝑛𝑗 (𝑦) (𝑖 = 1, . . . , 𝑚). By 𝜕𝑢/𝜕] we denote the conormal derivative 𝜕𝑢 𝜕𝑢 = 𝑎𝑖𝑗 𝑛𝑗 𝑖 . 𝜕] 𝜕𝑦
(48)
As usual, the symbols 𝐿𝑝 (Σ) and 𝑊1,𝑝 (Σ) (1 < 𝑝 < +∞) stand for the classical Lebesgue and Sobolev spaces, respectively. 𝑝 By 𝐿 𝑘 (Σ), we denote the space of all 𝑘-forms whose components are 𝐿𝑝 real-valued functions in a coordinate system of class 𝐶1 (and then in every coordinate system of class 𝐶1 ). We will look for the solution of the Dirichlet problem for the operator 𝐸 in the domain Ω in the form of a simple layer potential. To this end, we introduce the space S𝑝 . Definition 5. The function 𝑢 belongs to S𝑝 if and only if there exists 𝜑 ∈ 𝐿𝑝 (Σ) such that it can be represented by means of a simple layer potential; that is, 𝑢 (𝑥) = ∫ 𝜑 (𝑦) 𝑆 (𝑥, 𝑦) d𝜎𝑦 , 𝑥 ∈ Ω. Σ
(49)
Specifically our aim is to give an existence and uniqueness theorem for the Dirichlet problem 𝑢 ∈ S𝑝 , 𝐸𝑢 = 0 𝑢=𝑓
in Ω,
(50)
on Σ, 𝑓 ∈ 𝑊1,𝑝 (Σ) .
First, we prove the following formula. Proposition 6. For any 𝑢 ∈ 𝑊1,𝑝 (Σ),
d𝑥 𝜁 (𝑥, 𝑦) = d𝑥 𝐺 (𝑥, 𝑦) + ∫ d𝑥 𝐿 (𝑥, 𝑤) 𝑅 (𝑤, 𝑦) d𝑤 𝑇
2−𝑚−𝛾 = O (𝑥 − 𝑦 ).
(47)
𝜕 𝜕 (∫ 𝑢 (𝑥) 𝐿 (𝑧, 𝑥) d𝜎𝑥 ) d𝜎𝑧 𝜕]𝑧 Σ 𝜕]𝑥 = d𝑧 ∫ d𝑢 (𝑥) ∧ 𝜆 𝑚−2 (𝑧, 𝑥) Σ
3. The Dirichlet Problem In this section, we suppose that the domain Ω is such that R𝑚 \ Ω is connected and such that its boundary Σ = 𝜕Ω is a Lyapunov surface (i.e., Σ ∈ 𝐶1,𝜆 , 0 < 𝜆 ≤ 1). By 𝑛(𝑦) = (𝑛1 (𝑦), . . . , 𝑛𝑚 (𝑦)), we denote the outwards unit normal vector at the point 𝑦 ∈ Σ and by ](𝑦) = (]1 (𝑦), . . . , ]𝑚 (𝑦)) we denote the conormal vector at the point
(51)
+ ∫ 𝑢 (𝑥) ∧ 𝐹𝑧 [𝜆 𝑚−1 (𝑧, 𝑥)] Σ
− ∫ 𝑢 (𝑥) ∧ 𝜂𝑚−1 (𝑧, 𝑥) , 𝑧 ∈ Σ, Σ
where 𝐹 is the linear first-order differential operator considered in Proposition 3 and 𝜂𝑚−1 (𝑧, 𝑥) = O (|𝑧 − 𝑥|1−𝑚 ) .
(52)
Abstract and Applied Analysis
7
Proof. Set, for every 𝑧 ∉ Σ,
Therefore, 𝑉 (𝑧) = − ∫ 𝑢 (𝑥) Σ
𝑈 (𝑧) = ∫ d𝑢 (𝑥) ∧ 𝜆 𝑚−2 (𝑧, 𝑥) , Σ
= 𝑉0 (𝑧) d𝑧1 ⋅ ⋅ ⋅ d𝑧𝑚 ,
(53)
𝛿𝑉 (𝑧) = (−1)𝑚(𝑚+1)+1 ∗ d ∗ 𝑉 (𝑧) = − ∗ d𝑉0 (𝑧)
𝑉 (𝑧) = ∫ 𝑢 (𝑥) ∧ d𝑧 [𝜆 𝑚−1 (𝑧, 𝑥)] . Σ
=−∗ On account of (22) and (36), we get
=−
𝜕𝑉0 (𝑧) 𝑗 d𝑧 𝜕𝑧𝑗
Σ
Σ
Σ
lim 𝛿𝑉 (𝑧 ) = −
𝑧 → 𝑧
= ∫ 𝑢 (𝑥) ∧ 𝛿𝑧 d𝑧 [𝜆 𝑚−1 (𝑧, 𝑥)] (54)
− ∫ 𝑢 (𝑥) ∧ 𝐹𝑧 [𝜆 𝑚−1 (𝑧, 𝑥)] + ∫ 𝑢 (𝑥) ∧ d𝑧 [𝜖𝑚−2,𝑚−1 (𝑧, 𝑥)] = 𝛿𝑉 (𝑧) − ∫ 𝑢 (𝑥) ∧ 𝐹𝑧 [𝜆 𝑚−1 (𝑧, 𝑥)]
and this concludes the proof.
Σ
= d𝑧 ∫ d𝑢 (𝑥) ∧ 𝑠𝑚−2 (𝑧, 𝑥) , 𝑢 ∈ 𝑊 Σ
Σ
and (52) follows from (23). ̂𝚥 On the other hand, if 𝐴 ̂𝚤 is the minor of 𝐴−1 obtained deleting the 𝑖th row and the 𝑗th column, for 𝑧 ∈ Ω, 𝑥 ∈ Σ we get
d𝑧 [𝜆 𝑚−1 (𝑧, 𝑥)] = d𝑧 𝐿 (𝑧, 𝑥) 𝑗 ⋅ 𝐴 𝑖 (𝑥) d𝑧1 ⋅ ⋅ ⋅ ̂𝚤⋅ ⋅ ⋅ d𝑧𝑚 d𝑥1 ⋅ ⋅ ⋅ ̂𝚥 ⋅ ⋅ ⋅ d𝑥𝑚
𝜕𝐿 (𝑧, 𝑥) d𝜎𝑥 d𝑧1 ⋅ ⋅ ⋅ d𝑧𝑚 . 𝜕]𝑥
(Σ) ,
where 𝑠(𝑧, 𝑥) and 𝑠𝑘 (𝑧, 𝑥) denote the fundamental solution for Laplace equation and the double 𝑘-form associated with 𝑠(𝑧, 𝑥), respectively. We recall that if 𝐵 and 𝐵̃ are two Banach spaces and 𝐶 : 𝐵 → 𝐵̃ is a continuous linear operator, we say that 𝐶 can be reduced on the left if there exists a continuous linear operator 𝐶 : 𝐵̃ → 𝐵 such that 𝐶 𝐶 = 𝐼 + 𝐾, where 𝐼 stands for the identity operator on 𝐵 and 𝐾 : 𝐵 → 𝐵 is compact. One of the main properties of such operators is that equation 𝐶𝛼 = 𝛽 has a solution if and only if ⟨𝛾, 𝛽⟩ = 0 for any 𝛾 such that 𝐶∗ 𝛾 = 0, 𝐶∗ being the adjoint of 𝐶 (see [17, 18]). 𝑝
Theorem 8. Let 𝐽̃ : 𝐿𝑝 (Σ) → 𝐿 1 (Σ) be the singular integral operator defined as
=
=−
(58) 1,𝑝
+ ∫ 𝑢 (𝑥) ∧ 𝜂𝑚−1 (𝑧, 𝑥)
𝜕𝐿 (𝑧, 𝑥) d𝜎𝑥 d𝑧1 ⋅ ⋅ ⋅ d𝑧𝑚 𝜕𝑥𝑖
(57)
𝜕 𝜕 (∫ 𝑢 (𝑥) 𝑠 (𝑧, 𝑥) d𝜎𝑥 ) d𝜎𝑧 𝜕𝑛𝑧 Σ 𝜕𝑛𝑥
Σ
⋅ 𝑛𝑗 (𝑥)
𝜕𝑉0 (𝑧) d𝜎𝑧 𝜕]𝑧
Remark 7. We note that (51) generalizes the following identity (see [7] [8, Proposition 2.2]):
Σ
𝜕𝐿 (𝑧, 𝑥) 𝑖𝑗 = 𝑎 (𝑥) 𝑛𝑗 (𝑥) d𝜎𝑥 d𝑧1 ⋅ ⋅ ⋅ d𝑧𝑚 = −𝑎𝑖𝑗 (𝑥) 𝜕𝑧𝑖
𝜕𝑉0 (𝑧) 𝜕𝑉 (𝑧) d𝜎𝑧 = − 0 d𝜎𝑧 . 𝑗 𝜕𝑧 𝜕]𝑧
Then, if 𝑧 ∈ Σ,
+ ∫ 𝑢 (𝑥) ∧ d𝑧 [𝜖𝑚−2,𝑚−1 (𝑧, 𝑥)]
𝜕𝐿 (𝑧, 𝑥) (−1)𝑖−𝑗 𝜕𝑧𝑖 𝑗 ⋅ 𝐴 𝑖 (𝑥) d𝑧1 ⋅ ⋅ ⋅ d𝑧𝑚 (−1)𝑗−1 d𝑥1 ⋅ ⋅ ⋅ ̂𝚥 ⋅ ⋅ ⋅ d𝑥𝑚
𝜕𝑉0 (𝑧) (−1)ℎ−1 d𝑧1 ⋅ ⋅ ⋅ ̂ℎ ⋅ ⋅ ⋅ d𝑧𝑚 𝜕𝑧𝑗
= −𝑎𝑗ℎ (𝑧) 𝑛ℎ (𝑧)
= − ∫ 𝑢 (𝑥) ∧ d𝑧 𝛿𝑧 [𝜆 𝑚−1 (𝑧, 𝑥)]
(56)
𝜕𝑉 (𝑧) 1 𝛿1⋅⋅⋅𝑚 𝑎𝑗ℎ (𝑧) 0 𝑗 d𝑧𝑘2 ⋅ ⋅ ⋅ d𝑧𝑘𝑚 𝜕𝑧 (𝑚 − 1)! ℎ𝑘2 ⋅⋅⋅𝑘𝑚
= −𝑎𝑗ℎ (𝑧)
d𝑈 (𝑧) = − ∫ 𝑢 (𝑥) ∧ d𝑧 d𝑥 [𝜆 𝑚−2 (𝑧, 𝑥)]
Σ
𝜕𝐿 (𝑧, 𝑥) d𝜎𝑥 d𝑧1 ⋅ ⋅ ⋅ d𝑧𝑚 𝜕]𝑥
(55)
̃ (𝑥) = ∫ 𝜑 (𝑦) d𝑥 [𝑆 (𝑥, 𝑦)] d𝜎𝑦 , 𝐽𝜑 Σ
(59) 𝑝
𝜑 ∈ 𝐿 (Σ) , 𝑥 ∈ Σ. Then, 𝐽̃ can be reduced on the left by the operator 𝐽 : → 𝐿𝑝 (Σ):
𝑝 𝐿 1 (Σ)
𝐽 𝜓 (𝑧) = ∗ ∫ 𝜓 (𝑥) ∧ d𝑧 [𝜆 𝑚−2 (𝑧, 𝑥)] , 𝑧 ∈ Σ, Σ
Σ
(60)
8
Abstract and Applied Analysis
where the symbol ∗ means that if 𝑤 = 𝑤0 d𝜎 is an (𝑚 − 1)-form Σ
on Σ, then ∗𝑤 = 𝑤0 .
Then, 𝐽 𝐽𝜑 (𝑧)
Σ
1 = − 𝜑 (𝑧) 4
Proof. We start with the observation that ̃ (𝑥) = ∫ 𝜑 (𝑦) d𝑥 [𝐿 (𝑥, 𝑦)] d𝜎𝑦 𝐽𝜑
+ ∫ 𝜑 (𝑦) d𝜎𝑦 ∫
Σ
Σ
+ ∫ 𝜑 (𝑦) d𝑥 [𝜁 (𝑥, 𝑦)] d𝜎𝑦
(61)
Since 𝜕/𝜕]𝑥 𝐿(𝑥, 𝑦) = O(|𝑥 − 𝑦|1−𝑚+𝜆 ), 𝐾 is a compact operator. Thus,
and then ̃ = 𝐽 𝐽𝜑 + 𝐽 𝑍𝜑. 𝐽 𝐽𝜑
(62)
̃ = 𝐽 𝐽𝜑 + 𝐽 𝑍𝜑 = − 1 𝜑 + (𝐾2 + 𝑄 + 𝐽 𝑍) 𝜑 𝐽 𝐽𝜑 4
The operator 𝐽 𝑍 is compact because of (44). Concerning 𝐽 𝐽, keeping in mind Proposition 6 and setting 𝑢(𝑥) = ∫Σ 𝜑(𝑦)𝐿(𝑥, 𝑦)d𝜎𝑦 , we get 𝐽 𝐽𝜑 (𝑧) = ∗ ∫ ∫ 𝜑 (𝑦) d𝑥 [𝐿 (𝑥, 𝑦)] d𝜎𝑦 ∧ d𝑧 [𝜆 𝑚−2 (𝑧, 𝑥)] Σ
Σ Σ
=
𝑝
Theorem 9. Given 𝜔 ∈ 𝐿 1 (Σ), there exists a solution of the singular integral equation ̃ (𝑥) = 𝜔 (𝑥) , 𝜑 ∈ 𝐿𝑝 (Σ) , 𝑥 ∈ Σ 𝐽𝜑
(67)
∫ 𝛾∧𝜔=0
(68)
if and only if
Σ
𝜕 𝜕 𝐿 (𝑧, 𝑥) d𝜎𝑥 ∫ 𝑢 (𝑥) 𝜕]𝑧 Σ 𝜕]𝑥
Σ
(63)
𝑞
for every weakly closed form 𝛾 ∈ 𝐿 𝑚−2 (Σ)(1/𝑝 + 1/𝑞 = 1). 𝑞
̃ that Proof. Denote by 𝐽̃∗ : 𝐿 𝑚−2 (Σ) → 𝐿𝑞 (Σ) the adjoint of 𝐽; is,
− ∗ ∫ 𝑢 (𝑥) ∧ 𝐹𝑧 [𝜆 𝑚−1 (𝑧, 𝑥)] Σ
Σ
𝐽̃∗ 𝛾 (𝑥) = ∫ 𝛾 (𝑦) ∧ d𝑦 [𝑆 (𝑥, 𝑦)] ,
+ ∗ ∫ 𝑢 (𝑥) ∧ 𝜂𝑚−1 (𝑧, 𝑥) Σ
=
Σ
Σ
𝜕 𝜕 𝐿 (𝑧, 𝑥) d𝜎𝑥 + 𝑄𝜑 (𝑧) . ∫ 𝑢 (𝑥) 𝜕]𝑧 Σ 𝜕]𝑥
Since 𝐹𝑧 [𝜆 𝑚−1 (𝑧, 𝑥)] = O(|𝑧 − 𝑥|1−𝑚 ) and in view of (52), 𝑄 is a compact operator from 𝐿𝑝 (Σ) into itself. In view of the Stokes formula for 𝑢 and on account of known properties of potentials (see, e.g., [6, p. 35]), we get 𝜕 𝜕 𝐿 (𝑧, 𝑥) d𝜎𝑥 ∫ 𝑢 (𝑥) 𝜕]𝑧 Σ 𝜕]𝑥
1 𝜕 1 = (− 𝜑 (𝑧) + ∫ 𝜑 (𝑦) 𝐿 (𝑧, 𝑦) d𝜎𝑦 ) 2 2 𝜕]𝑧 Σ 1 𝜕 𝜕 + ∫ [− 𝜑 (𝑥) + ∫ 𝜑 (𝑦) 𝐿 (𝑥, 𝑦) d𝜎𝑦 ] 𝐿 (𝑧, 𝑥) d𝜎𝑥 2 𝜕]𝑥 𝜕]𝑧 Σ Σ 𝜕 1 𝜕 = − 𝜑 (𝑧) + ∫ 𝜑 (𝑦) d𝜎𝑦 ∫ 𝐿 (𝑥, 𝑦) 𝐿 (𝑧, 𝑥) d𝜎𝑥 . 4 𝜕]𝑧 Σ Σ 𝜕]𝑥
𝑥 ∈ Σ.
(69)
From Theorem 8, it follows that operator 𝐽̃ can be reduced on the left; therefore, (67) admits a solution 𝜑 ∈ 𝐿𝑝 (Σ) if and only if ∫ 𝛾 ∧ 𝜔 = 0, Σ
𝑞 ∀𝛾 ∈ 𝐿 𝑚−2 (Σ) , 𝐽̃∗ 𝛾 = 0.
(70)
On the other hand, 𝐽̃∗ 𝛾 = 0 if and only if 𝛾 is a weakly closed form; that is, ∫ 𝛾 ∧ d𝑔 = 0, ∀𝑔 ∈ 𝐶∞ (R𝑚 ) .
𝜕 𝜕 = [𝑢 (𝑧) + ∫ 𝑢 (𝑥) 𝐿 (𝑧, 𝑥) d𝜎𝑥 ] 𝜕]𝑧 𝜕] Σ 𝑥 𝜕 1 𝜕 𝜕 = (1 − ) 𝑢 (𝑧) + ∫ 𝑢 (𝑥) 𝐿 (𝑧, 𝑥) d𝜎𝑥 2 𝜕]𝑧 𝜕]𝑧 Σ 𝜕]𝑥
(66)
is a Fredholm operator and the assertion is proved.
= ∗ ∫ d𝑢 (𝑥) ∧ d𝑧 [𝜆 𝑚−2 (𝑧, 𝑥)] Σ
(65)
1 + 𝑄𝜑 (𝑧) = − 𝜑 (𝑧) + 𝐾2 𝜑 (𝑧) + 𝑄𝜑 (𝑧) . 4
Σ
= 𝐽𝜑 (𝑥) + 𝑍𝜑 (𝑥)
Σ
𝜕 𝜕 𝐿 (𝑥, 𝑦) 𝐿 (𝑧, 𝑥) d𝜎𝑥 𝜕]𝑥 𝜕]𝑧
(71)
Σ
In fact, if ∫ 𝛾 (𝑦) ∧ d𝑦 [𝑆 (𝑥, 𝑦)] = 0,
(64)
Σ
a.e. 𝑥 ∈ Σ,
(72)
we have ∫ 𝑝 (𝑥) d𝜎𝑥 ∫ 𝛾 (𝑦) ∧ d𝑦 [𝑆 (𝑥, 𝑦)] = 0, Σ
Σ
(73) 𝜆
∀𝑝 ∈ 𝐶 (Σ)
Abstract and Applied Analysis
9 Lemma 10. For every 𝑓 ∈ 𝑊1,𝑝 (Σ), there exists a solution of the boundary value problem
and then 0 = ∫ 𝛾 (𝑦) ∧ d𝑦 ∫ 𝑝 (𝑥) 𝑆 (𝑥, 𝑦) d𝜎𝑥 = ∫ 𝛾 ∧ d𝑢 Σ
Σ
Σ
𝑤 ∈ S𝑝 ,
(74)
𝐸𝑤 = 0
for any smooth solution 𝑢 of 𝐸𝑢 = 0 in Ω. Therefore, we have ∫ 𝛾 (𝑦) ∧ d𝑦 [𝑆 (𝑥, 𝑦)] = 0, Σ
∀𝑥 ∈ 𝑇 \ Ω.
(75)
Let us consider
d𝑤 = d𝑓
in Ω,
(81)
on Σ.
Its solution 𝑤 is a simple layer potential (49) whose density 𝜑 ̃ = d𝑓 (see (59)). solves 𝐽𝜑 Proof. Consider the following singular integral equation:
𝑧 (𝑥) = ∫ 𝛾 (𝑦) ∧ d𝑦 [𝑆 (𝑥, 𝑦)] , 𝑥 ∈ 𝑇. Σ
(76)
If V ∈ 𝐶∞ (𝑇) and 𝜂 ∈ 𝐶1 (Ω) ∩ 𝐶2 (Ω) are such that 𝐸𝜂 = 𝐸V in Ω and 𝜂 = 0 on Σ, we have
Ω
= ∫ 𝐸𝜂 (𝑥) d𝑥 ∫ 𝛾 (𝑦) ∧ d𝑦 [𝑆 (𝑥, 𝑦)] Ω
Σ
(77)
Σ
Ω
(78) 𝑦 ∈ Σ.
∫ 𝑧𝐸V d𝑥 = ∫ 𝛾 (𝑦) ∧ d𝑦 ∫ 𝐸𝜂 (𝑥) 𝑆 (𝑥, 𝑦) d𝑥 = ∫ 𝛾 (𝑦) ∧ d𝑦 ∫
Σ
𝜕𝜂 (𝑥) 𝑆 (𝑥, 𝑦) d𝜎𝑥 𝜕]
Σ
(79)
= 0. We have proved that 𝑧 = 0 on Σ, 𝑧 = 0 in Ω, and then 𝑧 = 0 in 𝑇. Therefore,
(85)
(86) 𝜕𝑢 = 0 on Σ, + 𝜕] we get 𝑢𝑖 = 𝛼𝑖 in Ω. Choose (𝑐1 , 𝑐2 ) ≠ (0, 0) such that 𝑐1 𝛼1 + 𝑐2 𝛼2 = 0 and set 𝑢 (𝑥) = ∫ (𝑐1 𝛾1 (𝑦) + 𝑐2 𝛾2 (𝑦)) 𝑆 (𝑥, 𝑦) d𝜎𝑦 .
(87)
Since 𝑢 = 𝑐1 𝛼1 + 𝑐2 𝛼2 = 0 in Ω, 𝑢 satisfies the following boundary value problem:
Σ
= ∫ 𝛾 (𝑦) ∧ d𝑦 ∫ 𝐸𝜑 (𝑥) 𝑆 (𝑥, 𝑦) d𝑥
𝑖 = 1, 2.
We note that 𝛾1 and 𝛾2 are H¨older continuous functions. With potentials 𝑢𝑖 being smooth solutions of the problem
Σ
0 = ∫ 𝑧𝐸𝜑 d𝑥 = ∫ 𝐸𝜑 d𝑥 ∫ 𝛾 (𝑦) ∧ d𝑦 [𝑆 (𝑥, 𝑦)]
𝑇
a.e. 𝑥 ∈ Σ. (84)
𝐸𝑢 = 0 in Ω,
𝜕𝜂 =∫ (𝑥) d𝜎𝑥 ∫ 𝛾 (𝑦) ∧ d𝑦 [𝑆 (𝑥, 𝑦)] Σ 𝜕] Σ
Σ
1 𝜕 − 𝛾 (𝑥) + ∫ 𝛾 (𝑦) 𝑆 (𝑥, 𝑦) d𝜎𝑦 = 0, 2 𝜕]𝑦 Σ
𝑢𝑖 (𝑥) = ∫ 𝛾𝑖 (𝑦) 𝑆 (𝑥, 𝑦) d𝜎𝑦 ,
Ω
𝑇
(83)
Obviously, the constant functions are eigensolutions of (84). We want to show that they are the only ones. Let 𝛾1 and 𝛾2 be two linearly independent eigensolutions of (84) and set
In view of (72), we find
𝑇
1 𝜕 − 𝜓 (𝑥) + ∫ 𝜓 (𝑦) 𝑆 (𝑥, 𝑦) d𝜎𝑦 = 0, 2 𝜕]𝑥 Σ
Proof. The Fredholm equation (83) has the same number of linearly independent solutions of the following equation:
𝜕𝜂 ∫ 𝑆 (𝑥, 𝑦) 𝐸𝜂 (𝑥) d𝑥 = ∫ 𝑆 (𝑥, 𝑦) (𝑥) d𝜎𝑥 , 𝜕] Ω Σ
Σ
in which the unknown is 𝜑 ∈ 𝐿𝑝 (Σ) and the datum is 𝑝 d𝑓 ∈ 𝐿 1 (Σ). With conditions (68) being satisfied, in view of Theorem 9 there exists a solution 𝜑 of (82).
The dimension of A is 1.
From the Green formulas we have
Σ
(82)
a.e. 𝑥 ∈ Σ.
= ∫ 𝛾 (𝑦) ∧ d𝑦 ∫ 𝐸𝜂 (𝑥) 𝑆 (𝑥, 𝑦) d𝑥.
Ω
𝑥 ∈ Σ,
Σ
Lemma 11. Let A be the eigenspace of the Fredholm integral equation
∫ 𝑧𝐸V d𝑥 = ∫ 𝑧𝐸𝜂 d𝑥 Ω
∫ 𝜑 (𝑦) d𝑥 [𝑆 (𝑥, 𝑦)] d𝜎𝑦 = d𝑓 (𝑥) ,
(80)
= ∫ 𝛾 (𝑦) ∧ d𝜑 (𝑦) , Σ
𝐸𝑢 = 0 in 𝑇 \ Ω, 𝑢 = 0 on Σ,
(88)
𝑢 = 0 on 𝜕𝑇. ∘
∞
for any 𝜑 ∈ 𝐶 (𝑇). This implies (71) and the theorem is proved.
By Green’s formula, 𝑢 = 0 in 𝑇 \ Ω and therefore 𝑢 = 0 in 𝑇. This implies 𝑐1 𝛾1 + 𝑐2 𝛾2 = 0, which is a contradiction.
10
Abstract and Applied Analysis
Lemma 12. Given 𝑐 ∈ R, there exists a solution of the following boundary value problem: V ∈ S𝑝 , 𝐸V = 0
in Ω,
V=𝑐
on Σ.
(89)
It is given by V (𝑥) = 𝑐 ∫ 𝜓0 (𝑦) 𝑆 (𝑥, 𝑦) d𝜎𝑦 , 𝑥 ∈ Ω, Σ
(90)
where 𝜓0 is the unique element of A such that ∫ 𝜓0 (𝑦) 𝑆 (𝑥, 𝑦) d𝜎𝑦 = 1, Σ
∀𝑥 ∈ Ω.
Σ
(92)
we have that 𝑃𝜓 = 𝑐 in Ω. As in Lemma 11, this implies that if 𝑐 = 0, we have that 𝜓 = 0. Then, 𝑐 ≠ 0. Function 𝜓0 = (1/𝑐)𝜓 satisfies (91) and V given by (90) is solution of (89). Theorem 13. The Dirichlet problem (50) has a unique solution for every 𝑓 ∈ 𝑊1,𝑝 (Σ). In particular, the density 𝜑 of 𝑢 can be written as 𝜑 = 𝜑0 + 𝜓, where 𝜑0 solves the singular integral system a.e. 𝑥 ∈ Σ
(93)
and 𝜓 ∈ A. Proof. Let 𝑤 be a solution of the boundary value problem (81). Since d𝑤 = d𝑓 on Σ, 𝑤 = 𝑓−𝑐 on Σ for some 𝑐 ∈ R. Function 𝑢 = 𝑤 + V, where V is given by (90), solves problem (50). Consider now two solutions of the same problem (50): 𝑢 (𝑥) = ∫ 𝜑 (𝑦) 𝑆 (𝑥, 𝑦) d𝜎𝑦 , Σ
(94)
𝑢 (𝑥) = ∫ 𝜑 (𝑦) 𝑆 (𝑥, 𝑦) d𝜎𝑦 . Σ
Therefore, the potential V (𝑥) = ∫ 𝜓 (𝑦) 𝑆 (𝑥, 𝑦) d𝜎𝑦 , Σ
(95)
where 𝜓 = 𝜑 − 𝜑 , solves the problem V ∈ S𝑝 , 𝐸V = 0
in Ω,
V=0
on Σ.
(96)
Since ∫ 𝜓 (𝑦) d𝑥 [𝑆 (𝑥, 𝑦)] d𝜎𝑦 = 0 on Σ, Σ
̃ = N (𝐽) ̃ N (𝐽 𝐽)
(98)
is true, 𝐽 still provides equivalence in a certain sense. In fact, we have the following lemma.
𝑃𝜓 (𝑥) = ∫ 𝜓 (𝑦) 𝑆 (𝑥, 𝑦) d𝜎𝑦
Σ
We end this section by observing that when we study the Dirichlet problem (50), we need to solve the singular integral ̃ = d𝑓, 𝜑 ∈ 𝐿𝑝 (Σ). We have proved that this equation 𝐽𝜑 equation can be reduced to a Fredholm one by means of the operator 𝐽 . This reduction is not an equivalent reduction in the usual sense (see, e.g., [18, pp. 19-20]); that is, it is not true that N(𝐽 ) = {0}, N(𝐽 ) being the kernel of the operator 𝐽 . However, if the condition
(91)
Proof. Let 𝜓 ∈ A, 𝜓 ≠ 0. Setting
∫ 𝜑0 (𝑦) d𝑥 [𝑆 (𝑥, 𝑦)] d𝜎𝑦 = d𝑓 (𝑥) ,
̃ = 0 (see (66)). By standard arguments, 𝜓 is we have 𝐽 𝐽𝜓 H¨older continuous and then V ∈ 𝐶0 (Ω) ∩ 𝐶2 (Ω). The weak maximum principle (see, e.g., [19, p. 32]) shows that V = 0 in Ω; that is, 𝑢 = 𝑢 .
(97)
Lemma 14. If condition (98) holds, the singular integral ̃ = equation (82) is equivalent to the Fredholm equation 𝐽 𝐽𝜑 𝐽 d𝑓. Proof. Condition (98) implies that if 𝑔 is such that there exists ̃ = 𝑔, then this equation is a solution 𝜑 of the equation 𝐽𝜑 ̃ = 𝐽 𝑔. Since the equation 𝐽𝜑 ̃ = d𝑓 satisfied if and only if 𝐽 𝐽𝜑 ̃ = d𝑓 if and only is solvable (see Lemma 10), we have that 𝐽𝜑 ̃ = 𝐽 d𝑓. if 𝐽 𝐽𝜑 Condition (98) is satisfied, for example, if the differential operator 𝐸 has constant coefficients. This can be proved as in [20, Remark 1, p. 1045], replacing the Laplacian and the normal derivative by 𝐸 and the conormal derivative, respectively.
Conflict of Interests The authors declare that there is no conflict of interests regarding the publication of this paper.
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