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ISRN Mathematical Analysis
Volume 2011, Article ID 582097, 16 pages
Research Article

On Removable Sets of Solutions of Elliptic Equations

Department of Nonlinear Analyses, Institute of Mathematics and Mechanics of NAS of Azerbaijan, 9, F. Agaev street, Baku AZ1141, Azerbaijan

Received 8 March 2011; Accepted 20 April 2011

Academic Editors: G. L. Karakostas, G. Mantica, X. B. Pan, and C. Zhu

Copyright © 2011 Tair S. Gadjiev 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 consider a nondivergent elliptic equation of second order whose leading coefficients are from some weight space. The sufficient condition of removability of a compact with respect to this equation in the weight space of Hölder functions was found.

Let 𝐷 be a bounded domain situated in 𝑛-dimensional Euclidean space 𝐸𝑛 of the points 𝑥=(𝑥1,,𝑥𝑛),𝑛3,andlet𝜕𝐷 be its boundary. Consider in 𝐷 the following elliptic equation:𝑢=𝑛𝑖,𝑗=1𝑎𝑖𝑗(𝑥)𝑢𝑖𝑗+𝑛𝑖=1𝑏𝑖(𝑥)𝑢𝑖+𝑐(𝑥)𝑢=0,(1) in supposition that 𝑎𝑖𝑗(𝑥) is a real symmetric matrix, moreover 𝜔(𝑥) is a positive measurable function satisfying the doubling condition: for concentric balls 𝐵𝑥𝑅 or 𝑅 and 2𝑅 radius, there exists such a constant 𝛾𝜔𝐵𝑥𝑅𝛾𝜔𝐵𝑥2𝑅,(2) where for the measurable sets 𝐸. 𝜔(𝐸) means 𝐸𝜔(𝑦)𝑑𝑦𝛾||𝜉||2𝜔(𝑥)𝑛𝑖,𝑗=1𝑎𝑖𝑗(𝑥)𝜉𝑖𝜉𝑗𝛾1𝜔(𝑥)||𝜉||2;𝜉𝐸𝑛,𝑥𝐷,(3)𝑎𝑖𝑗(𝑥)𝐶1𝜔𝐷;𝑖,𝑗,1,,𝑛,(4)||𝑏𝑖(𝑥)||𝑏0;𝑏0𝑐(𝑥)0;𝑖=1,,𝑛;𝑥𝐷.(5) Here 𝑢𝑖=𝜕𝑢/𝜕𝑥𝑖,𝑢𝑖𝑗=𝜕2𝑢/𝜕𝑥𝑖𝜕𝑥𝑗;𝑖,𝑗=1,,𝑛;𝛾(0,1] and 𝑏00 are constants. Besides we will suppose that the lower coefficients of the operator are measurable functions in 𝐷. Let 𝜆(0,1) be a number. Denote by 𝐶0,𝜆(𝐷) a Banach space of the functions 𝑢(𝑥) defined in 𝐷 with the finite norm;𝑢𝐶𝜆𝜔(𝐷)=sup𝑥𝐷𝜔(𝑥)||𝑢(𝑥)||+sup𝑥,𝑦𝐷𝑥𝑦||𝑢(𝑥)𝑢(𝑦)||𝜔||𝑥𝑦||𝜆.(6)

The compact 𝐸𝐷 is called removable with respect to (1) in the space 𝐶𝜆𝜔(𝐷) if from𝑢=0,𝑥𝐷𝐸,𝑢𝜕𝐷𝐸=0,𝑢(𝑥)𝐶𝜆𝜔(𝐷),(7) it follows that 𝑢(𝑥)0 in 𝐷.

The aim of the given paper is finding sufficient condition of removability of a compact with respect to (1) in the space 𝐶𝜆𝜔(𝐷). This problem have been investigated by many researchers. For the Laplace equation the corresponding result was found by Carleson [1]. Concerning the second-order elliptic equations of divergent structure, we show in this direction the papers [2, 3]. For a class of nondivergent elliptic equations of the second order with discontinuous coefficients the removability condition for a compact in the space C𝜆(𝐷) was found in [4]. mention also papers [57] in which the conditions of removability for a compact in the space of continuous functions have been obtained. The removable sets of solutions of the second-order elliptic and parabolic equations in nondivergent form were considered in [810]. In [11], Kilpelӓinen and Zhong have studied the divergent quasilinear equation without minor members and proved the removability of a compact. Removable sets for pointwise solutions of elliptic partial differential equations were found by Diederich [12]. Removable singularities of solutions of linear partial differential equations were considered in [13]. Removable sets at the boundary for subharmonic functions have been investigated by Dahlberg [14]. Denote by 𝐵𝑅(𝑧) and 𝑆𝑅(𝑧) the ball {𝑥|𝑥𝑧|<𝑅} and the sphere

{𝑥|𝑥𝑧|=𝑅} of radius 𝑅 with the center at the point 𝑧𝐸𝑛 respectively. We’ll need the following generalization of mean value theorem belonging to Gerver and Landis [15] in weight case.

Lemma 1. Let the domain 𝐺 be situated between the spheres 𝑆𝑅(0)and 𝑆2𝑅(0), moreover let the intersection 𝜕𝐺{𝑥𝑅<|𝑥|<2𝑅}be a smooth surface. Further, let in 𝐺 the uniformly positive definite matrix 𝑎𝑖𝑗(𝑥);𝑖,𝑗=1,,𝑛 and the function 𝑢(𝑥)𝐶2(𝐺)𝐶1𝜔(𝐺) be given. Then there exists the piecewise smooth surface Σ  dividing in 𝐺 the spheres 𝑆𝑅(0) and 𝑆2𝑅(0)such that Σ𝜔|||𝜕𝑢𝜕𝜈|||𝑑𝑠𝐾oscu𝐺𝜔(𝐺)𝑅2.(8) Here 𝐾>0 is a constant depending only on the matrix 𝑎𝑖𝑗(𝑥)and 𝑛, and 𝜕𝑢/𝜕𝜈 is a derivative by a conormal determined by the equality 𝜕𝑢(𝑥)𝜕𝜈=𝑛𝑖,𝑗=1𝑎𝑖𝑗(𝑥)𝜕𝑢(𝑥)𝜕𝑥𝑖cos𝑛,𝑥𝑗1/2,(9) where cos(𝑛,𝑥𝑗);𝑗=1,,𝑛 are direction cosines of a unit external normal vector to Σ.

Proof. Let 𝐺𝑅𝑛 be a bounded domain 𝑓(𝑥)𝐶2(𝐺). Then there exists a finite number of balls {𝐵x𝜈𝑟𝜈},𝜈=1,2,,𝑁 which cover 𝑄𝑓 and such that if we denote by 𝑆𝜈, the surface of 𝜈th ball, then 𝑁𝜈=1𝑆𝜈𝜔(𝑥)||𝑓||𝑑𝑥<𝜀.(10)
Decompose 𝑂𝑓 into two parts: 𝑂𝑓=𝑂𝑓𝑂𝑓, where 𝑂𝑓 is a set of points 𝑂𝑓 for which 2𝑓0,𝑂𝑓 is a set of points for which 2𝑓=0.
The set 𝑂𝑓 has 𝑛-dimensional Lebesgue measure equal zero, as on the known implicit function theorem, the 𝑂𝑓 lies on a denumerable number of surfaces of dimension 𝑛1. If we use the absolute continuity of integral 𝜔(𝐷)=𝐷𝜔(𝑥)𝑑𝑥(11) with respect to Lebesque measure 𝐷 and the above said, we get that the set 𝑂𝑓 may be included into the set 𝐷 for which 𝜔(𝐷)<𝜂,𝜂>0 will be chosen later. Let for each point 𝑥𝑂𝑓, there exist such 𝑟𝑥 that 𝐵𝑥𝑟𝑥 and 𝐵𝑥6𝑟𝑥 are contained in 𝐷𝐺. Then 6𝑟𝑥5𝑟𝑥𝑑𝑟𝑆𝑥𝑟𝜔(𝜎)𝑑𝜎𝜔𝐵𝑥6𝑟𝑥,(12) therefore there exists such 5𝑟𝑥𝑡6𝑟𝑥 that 𝑟𝑥𝑆𝑥𝑟𝜔(𝜎)𝑑𝜎𝜔𝐵𝑥6𝑟𝑥.(13) Then 𝑆𝑥𝑟𝜔(𝜎)||𝑓||𝑑𝜎𝐶𝑡𝑆𝑥𝑟𝜔(𝜎)𝑑𝜎(6𝐶)𝑟𝑟𝑥𝑆𝑥𝑟𝜔(𝜎)𝑑𝜎(6𝐶)𝜔𝐵𝑥6𝑟𝑥(6𝐶)𝛾3𝜔𝐵𝑥𝑟𝑥𝑐0𝜔𝐵𝑥𝑡/5,(14) where 𝐶=sup𝐷|2𝑓|,𝛼=diam𝐺,𝑐0=(6𝐶)𝛾3.
Now by a Banach process [4, page 126] from the ball system {𝐵𝑥𝑡/5} we choose such a denumerable number of notintersecting balls {𝐵𝑥𝜈𝑡𝜈/5},𝜈=1,2,,𝑁 that the ball of five-times greater radius {𝐵𝑥𝜈𝑡𝜈} cover the whole 𝑂𝑓 set. We again denote these balls by {𝐵𝑥𝜈𝑡𝜈/5},𝜈=1,2,,𝑁 and their surface by 𝑆𝜈. Then by virtue of (5) 𝜈=1𝑆𝜈𝜔(𝜎)||𝑓||𝑑𝜎𝐶0𝜔(𝐺)<𝐶0𝜂.(15) Now let 𝑥𝑂𝑓. Then 6𝑟𝑥5𝑟𝑥𝑑𝑟𝑆𝑥𝑟𝜔(𝜎)𝑑𝜎𝜔𝐵𝑥6𝑟𝑥.(16) Therefore there exists such 5𝑟𝑥𝑡6𝑟𝑥 that 𝑟𝑥𝑆𝑥𝑟𝜔(𝜎)𝑑𝜎𝜔𝐵𝑥6𝑟𝑥.(17)
Assign arbitrary 𝜂>0. By virtue of that |𝑓|𝑆𝑥𝑡𝜂𝑡, for sufficiently small 𝑡 we have 𝑆𝑥𝑡𝜔(𝜎)||𝑓||𝑑𝜎𝜂𝑡𝑆𝑥𝑡𝜔(𝜎)𝑑𝜎(2𝜂)𝑟𝑥𝑆𝑥𝑡𝜔(𝜎)𝑑𝜎(2𝜂)𝜔𝐵𝑥2𝑟𝑥(6𝐶)(2𝜂)𝛾1𝜔𝐵𝑥𝑡/5𝜂𝐶1𝜔𝐵𝑥𝑡/5.(18) Again by means of Banach process and by virtue of (43) we get 𝑁𝜈=1𝑆𝑛𝜈𝜔(𝜎)||𝑓||𝑑𝜎𝜂𝐶1𝜔(𝐷),(19) where 𝑆𝑛𝜈 is the surface of balls in the second covering.
Combining the spherical surfaces 𝑆𝜈 and 𝑆𝜈 we get that the open balls system covers the closed set 𝑂𝑓. Then a finite subcovering may be choosing from it. Let them be the balls 𝐵1,𝐵2,,𝐵𝑥 and their surfaces are 𝑆1,𝑆2,,𝑆𝑁. We get from inequalities (4) and (7) 𝑁𝜈=1𝑆𝜈||𝑓||𝜔(𝜎)𝑑𝜎𝐶1𝜔𝐷+𝐶0𝜂.(20)
Put now 𝜀=[𝐶1𝜔(𝐷)+𝐶0]𝜂.
Following [2], assume 𝜀=𝜔(𝐷)oscu𝐺𝑅2,(21) and according to Lemma 1 for a given 𝜀 we will find the balls 𝐵1,𝐵2,,𝐵𝑥 and exclude them from the domain 𝐺. Put 𝐷=𝐷𝑁𝜈=1𝐵𝜈 intersect with 𝐺 a closed spherical layer 𝑅1+14|𝑥|𝑅1+14.(22) We denote the intersection by 𝐺. We can assume that the function 𝑢(𝑥) is defined in some 𝛿 vicinity 𝐺𝛿 of set 𝐺. Take 𝛿<𝑅/4 so that oscu𝐺𝛿2osc𝐺u.(23)
On a closed set 𝐺 we have 𝑓0. Consider on 𝐺𝛿 the equation system 𝑑𝑥𝑑𝑡=𝑢𝑥.(24)
Let some surface 𝑆 touches the direction of the field at each its point, then 𝑆|||𝜕𝑢𝜕𝑛|||𝑑𝜎=0,(25) since 𝜕𝑢/𝜕𝑛 is identically equal to zero at 𝑆.
We will use it in constructing the needed surface of Σ. Tubular surfaces whose generators will be the trajectories of the system (50) constitute the basis of Σ.
They will add nothing to the integral we are interested in. These surfaces will have the form of thin tubes that cover 𝐺. Then we shall put partitions to some of these tubes. Lets construct tubes. Denote by 𝐸 the intersection of 𝐺 with sphere |𝑥|=𝑅(1+3/4).
Let 𝑁 be a set of points 𝐸. Where field direction of system (50) touches the sphere |𝑥|=𝑅(1+3/4). Cover 𝑁 with such an open on the sphere |𝑥|=𝑅(1+3/4) set 𝐹 that 𝐹𝜔(𝑥)|||𝜕𝑢𝜕𝑛|||𝑑𝜎𝜔(𝐺)oscu𝐷𝑅2.(26) It will be possible if on 𝑁(𝜕𝑢/𝜕𝑅)0.
Put 𝐸=𝐸𝐹. Cover 𝐸 on the sphere by a finite number of open domains with piece-wise smooth boundaries. We shall call them cells. We shall control their diameters in estimation of integrals that we need. The surface remarked by the trajectories lying in the ball |𝑥|(7/4)𝑅 and passing through the bounds of cells we shall call tube.
So, we obtained a finite number of tubes. The tube is called open if not interesting, this tube one can join by a broken line the point of its corresponding cell with a spherical layer (5/4)𝑅𝛿<|𝑥|<(7/4)𝑅. Choose the diameters of cells so small that the trajectory beams passing through each cell could differ no more than 𝛿/2𝑛.
By choose of cells diameters the tubes will be contained in 54𝑅𝛿<|𝑥|<54𝑅.(27)
Let also the cell diameter be chosen so small that the surface that is orthogonal to one trajectory of the tube intersects the other trajectories of the tube at an angle more than 𝜋/4.
Cut off the open tube by the hypersurface in the place where it has been imbedded into the layer 54𝑅𝛿2<|𝑥|<54𝑅(28) at first so that the edges of this tube be embedded into this layer.
Denote these cutoff tubes by 𝑇1,𝑇2,,𝑇𝑆. If each open tube is divided with a partition, then a set-theoretical sum of closed tubes, tubes 𝑇1,𝑇2,,𝑇𝑆 their partitions spheres 𝑆1,𝑆2,,𝑆𝑁, and the set 𝐹 on the sphere |𝑥|=(7/4)𝑅 divides the spheres |𝑥|=𝑅 and |𝑥|=2𝑅. Note that 𝑆𝜔|𝜕𝑢/𝜕𝑛|𝑑𝜎 along the surface of each tube equals to zero, since 𝜕𝑢/𝜕𝑛 identically equals to zero.
Now we have to choose partitions so that the integral 𝑆𝜔|𝜕𝑢/𝜕𝑛|𝑑𝜎 was of the desired value. Denote by 𝑈𝑖 the domain bounded by 𝑇𝑖 with corresponding cell and hypersurface cutting off this tube. We have 𝑈𝑖𝑈𝑗= and therefore 𝑚𝑖=1𝜔𝑈𝑖<2𝜔(𝐷).(29)
Consider a tube 𝑇𝑖 and corresponding domain 𝑈𝑖. Choose any trajectory on this tube. Denote it by 𝐿𝑖. The length 𝜇𝑖𝐿𝑖 of the curve 𝐿𝑖 satisfies the inequality 𝜇𝑖𝐿𝑖𝑅2.(30)
Let introduce on 𝐿𝑖 a parameter 𝑙 (length of the arc), counted from the cell. By 𝜎𝑖(𝑙) denote the cross-section by 𝑈𝑖 hypersurface passing thought the point, corresponding to 𝑙 and orthogonal to the trajectory 𝐿𝑖 at this point. Let the diameter of cells be so small 𝐿𝑖𝑑𝑙𝜎𝑖(𝑙)𝜔(𝑥)𝑑𝜎<2𝜔𝑈𝑖.(31)
Then by Chebyshev inequality a set 𝐻 points 𝑙𝐿𝑖 where 𝜎𝑖(𝑙)𝜔(𝑥)𝑑𝜎>8𝑅𝜔𝑈𝑖(32) satisfies the inequality 𝜇𝑖𝐻<𝑅/4 and hence by virtue of (55) for 𝐸=𝐿𝑖𝐻 it is valid and 𝜇1𝐸>𝑅/4.(33)
At the points of the curve 𝐿𝑖 the derivative 𝜕𝑢/𝜕𝑙 preserves its sign, and therefore 𝐸|||𝜕𝑢𝜕𝑙|||𝑑𝑙𝐿𝑖|||𝜕𝑢𝜕𝑙|||𝑑𝑙oscu𝐷𝛿.(34)
Hence, by using (65) and a mean value theorem for one variable function we find that there exists 𝑙0𝐸𝜕𝑢𝜕𝑙𝑙=𝑙04𝑅oscu𝐷𝛿.(35)
But on the other hand 𝜕𝑢𝜕𝑙𝑙=𝑙0=||𝑢||𝑙=𝑙0.(36)
Together with (67) it gives ||𝑢||𝑙=𝑙0𝜎𝑖𝑙0𝜔(𝑥)𝑑𝜎8𝑅4𝑅𝜔𝑈𝑖oscu𝐷.(37)
Now, let the diameter of cells be still so small that 𝜎𝑖𝑙0𝜔(𝑥)||𝑢||𝑑𝜎164𝑅𝜔𝑈𝑖oscu𝐷,(38) (we can do it, since the derivatives 𝜕𝑢/𝜕𝑥𝑖are uniformly continuous). Therefore according to (53) 𝑆𝑖=1𝜎𝑖𝑙0𝜔(𝑥)||𝑢||𝑑𝜎164𝑅𝜔𝑈𝑖oscu𝐷.(39)
Define by Σ a set-theoretical sum of all closed tubes, all open tubes 𝑇𝑖, all 𝜎𝑖(𝑙0), all spheres 𝑆𝑖 and sets 𝐹 on the sphere |𝑥|=(7/4)𝑅. Then, we get by (4), (49), (51), and (73) Σ𝜔(𝑥)|||𝜕𝑢𝜕𝑛|||𝑑𝜎𝐾𝜔(𝐷)oscu𝐷𝑅𝑝.(40)
Then, we get by (4), (49), (51), (73) Σ𝜔(𝑥)|||𝜕𝑢𝜕𝑛|||𝑑𝜎𝐾𝜔(𝐷)oscu𝐷𝑅𝑝.(41)
The lemma is proved.

Denote by 𝑊12,𝜔(𝐷) the Banach space of the functions 𝑢(𝑥) defined in 𝐷 with the finite norm𝑢𝑊12,𝜔(𝐷)=𝐷𝜔𝑢2+𝑛𝑖=1𝑢2𝑖𝑑𝑥1/2,(42) and let 𝑜𝑊12,𝜔(𝐷) be a completion of 𝐶0(𝐷) by the norm of the space 𝑊12,𝜔(𝐷).

By 𝑚𝑠𝐻(𝐴) we will denote the Hausdorff measure of the set 𝐴 of order 𝑠>0. Further, everywhere the notation 𝐶() means that the positive constant 𝐶 depends only on the content of brackets.

Theorem 2. Let D be a bounded domain in 𝐸𝑛 and let 𝐸𝐷 be a compact. If with respect to the coefficients of the operator the conditions (3)–(5) are fulfilled, then for removability of the compact E with respect to the (1) in the space 𝐶𝜆𝜔(𝐷) it sufficies that 𝑚𝑛2+𝜆𝐻(𝐸)=0.(43)

Proof. At first we show that without loss of generality we can suppose the condition 𝜕𝐷𝐶1 is fulfilled. Suppose that the condition (43) provides the removability of the compact 𝐸 for the domains, whose boundary is the surface of the class 𝐶1, but 𝜕𝐷𝐶1, and by fulfilling (43) the compact 𝐸 is not removable. Then the problem (7) has a nontrivial solution 𝑢(𝑥), moreover 𝑢|𝐸=𝑓(𝑥) and 𝑓(𝑥)0. We always can suppose the lowest coefficients of the operator is infinitely differentiable in 𝐷. Moreover, without loss of generality, we'll suppose that the coefficients of the operator are extended to a ball 𝐵𝐷 with saving the conditions (3)–(5). Let 𝑓+(𝑥)=max{𝑓(𝑥),0},𝑓(𝑥)=min{𝑓(𝑥),0}, and 𝑢±(𝑥) be generalized by Wiener (see [15]) solutions of the boundary value problems 𝑢±=0,𝑥𝐷𝐸,u±𝜕𝐷𝐸=0,u±𝐸=𝑓±.(44)
Evidently, 𝑢(𝑥)=𝑢+(𝑥)+𝑢(𝑥). Further, let 𝐷 be such a domain that 𝜕𝐷𝐶1,𝐷𝐷,𝐷𝐵,and 𝜗±(𝑥) be solutions of the problems 𝜗±=0,𝑥𝐷𝐸,𝜗±𝜕𝐷=0,𝜗±𝐸=𝑓±,𝜗±(𝑥)C𝜆𝜔𝐷.(45) By the maximum principle for 𝑥𝐷, 0𝑢+(𝑥)𝜗+(𝑥),𝜗(𝑥)𝑢(𝑥)0.(46) But according to our supposition, 𝜗+(𝑥)𝜗(𝑥)0. Hence, it follows that 𝑢(𝑥)0. So, we'll suppose that 𝜕𝐷𝐶1. Now, let 𝑢(𝑥) be a solution of the problem (7), and the condition (43) be fulfilled. Give an arbitrary 𝜀>0. Then there exists a sufficiently small positive number 𝛿 and a system of the balls {𝐵𝑟𝑘(𝑥𝑘)},𝑘=1,2,, such that 𝑟𝑘<𝛿,𝐸𝑘=1𝐵𝑟𝑘(𝑥𝑘) and 𝑘=1𝑟𝑛2+𝜆𝑘<𝜀.(47)
Consider a system of the spheres {𝐵2𝑟𝑘(𝑥𝑘)}, and let 𝐷𝑘=𝐷𝐵2𝑟𝑘(𝑥𝑘),𝑘=1,2,. Without loss of generality we can suppose that the cover {𝐵2𝑟𝑘(𝑥𝑘)} has a finite multiplicity 𝑎0(𝑛). By the Landis-Gerver theorem, for every 𝑘, there exists a piece-wise smooth surface Σ𝑘 dividing in 𝐷𝑘 the spheres 𝑆𝑟𝑘(𝑥𝑘) and 𝑆2𝑟𝑘(𝑥𝑘), such that Σ𝑘𝜔|||𝜕𝑢𝜕𝜈|||𝑑𝑠𝐾oscu𝐷𝑘𝜔𝐷𝑘𝑟2𝑘.(48) Since 𝑢(𝑥)𝐶𝜆𝜔(𝐷), there exists a constant 𝐻1>0 depending only on the function 𝑢(𝑥) such that oscu𝐷𝑘𝜔𝐻12𝑟𝑘𝜆.(49) Besides, 𝜔𝐷𝑘mesn𝐵2𝑟𝑘𝑥𝑘=Ω𝑛2𝑛𝑟𝑛𝑘;𝑘=1,2,,(50) where Ω𝑛=mes𝑛𝐵1(0). Using (49) and (50) in (48), we get Σ𝑘𝜔|||𝜕𝑢𝜕𝜈|||𝑑𝑠𝐶1𝑟n2+𝜆𝑘;𝑘=1,2,,(51) where 𝐶1=𝐾𝐻12𝑛+𝜆.
Let 𝐷Σ be an open set situated in 𝐷𝐸 whose boundary consists of unification of Σ and Γ, where Σ=𝑘=1Σ𝑘,Γ=𝜕𝐷𝑘=1𝐷+𝑘,𝐷+𝑘 is a part of 𝐷𝑘 remaining after the removing of points situated between Σ and 𝑆2𝑟𝑘(𝑥𝑘);𝑘=1,2,. Denote by 𝐷Σ the arbitrary connected component 𝐷Σ, and by we denote the elliptic operator of divergent structure =𝑛𝑖,𝑗=1𝜕𝜕𝑥𝑖𝑎𝑖𝑗(𝑥)𝜕𝜕𝑥𝑗.(52)
According to Green formula for any functions 𝑧(𝑥) and 𝑊(𝑥) belonging to the intersection 𝐶2(𝐷Σ)𝐶1(𝐷Σ), we have 𝐷Σ(𝑧𝛽𝛽𝑧)𝑑𝑥=𝜕𝐷Σ𝑧𝜕𝛽𝜕𝜈𝛽𝜕𝑧𝜕𝜈𝑑𝑠.(53)
Since 𝜕𝐷𝐶1, then 𝑢(𝑥)𝐶1(𝐷Σ)𝐶1(𝐷Σ)(𝑥)𝐶1(DΣ) (see [16]). From (53) choosing the functions 𝑧=1,𝛽=𝜔𝑢2, we have 𝐷Σ𝜔𝑢2𝑑𝑥=2𝜕𝐷Σ𝜔𝑢𝜕𝑢𝜕𝜈𝑑𝑠+𝜕𝐷Σ𝜔𝑥𝑖𝑢2𝑑𝑠.(54)
But |𝑢(𝑥)|𝑀< for 𝑥𝐷. Let us put the condition 𝜔𝑥𝑖<𝑐𝜔.()
By virtue of condition (52) and 𝜕𝐷Σ𝜔𝑢2𝑑𝑠<𝐶3𝑀𝜀,subject to (51) and (47), we conclude 𝐷Σ𝜔𝑢2𝑑𝑥2𝑀𝑎0𝑘=1Σ𝑘𝜔|||𝜕𝑢𝜕𝜈|||𝑑𝑠+𝐷Σ𝜔𝑢2𝑑𝑥2𝑀𝑎0𝐶1𝑘=1𝑟𝑛2+𝛼𝑘+𝜀𝑀𝑐2<𝐶3𝜀,(55) where 𝐶3=2𝑀𝑎0𝐶1.
On the other hand 𝜔𝑢2=6𝑢𝜔(𝑢)+2𝑛𝑖,𝑗=1𝜔𝑎𝑖𝑗𝑢𝑖𝑢𝑗+(2𝑢+1)𝑛𝑖,𝑗=1𝑎𝑖𝑗𝑢𝑥𝑗𝜔𝑥𝑖+𝑛𝑖,𝑗=1𝜕𝑎𝑖𝑗𝜕𝑥𝑖𝑢𝜔𝑥𝑗+𝑛𝑖,𝑗=1𝑎𝑖𝑗𝑢𝜔𝑥𝑖𝑥𝑗(56) and besides, 𝑢=𝑢+𝑛𝑖=1𝑑𝑖(𝑥)𝑢𝑖𝑐(𝑥)𝑢,(57) where 𝑑𝑖(𝑥)=𝑛𝑗=1𝜕𝑎𝑖𝑗(𝑥)𝜕𝑥𝑗𝑏𝑖(𝑥),𝑖=1,,𝑛.(58) It is evident that by virtue of conditions (4) and (5) |𝑑𝑖(𝑥)|𝑑0<;𝑖=1,,𝑛. Thus, from (55) we obtain 6𝐷Σ𝑢𝜔𝑛𝑖=1𝑑𝑖(𝑥)𝑢𝑖𝑑𝑥6𝐷Σ𝑢2𝑐(𝑥)𝑑𝑥+2𝐷Σ𝑛𝑖,𝑗=1𝜔(𝑥)𝑎𝑖𝑗𝑢𝑖𝑢𝑗𝑑𝑥+(2𝑢+1)𝐷Σ𝑛𝑖,𝑗=1𝑎𝑖𝑗𝑢𝑗𝜔𝑥𝑖𝑑𝑥+𝐷Σ𝑛𝑖,𝑗=1𝜕𝑎𝑖𝑗𝜕𝑥𝑗𝑢𝜔𝑥𝑖𝑑𝑥+||𝑢||2𝑑𝑥+𝐷Σ𝑛𝑖,𝑗=1𝑎𝑖𝑗𝑢𝜔𝑥𝑖𝑥𝑗𝑑𝑥<𝐶3𝜀.(59)
Hence, for any 𝛼>0 it follows that 2𝛾𝐷Σ𝜔||𝑢||2𝑑𝑥<6𝑑0𝐷Σ𝜔|𝑢|||𝑢𝑖||𝑑𝑥+6𝐷Σ𝑢2𝜔(𝑥)+(2𝑢+1)𝐷Σ𝑎𝑖𝑗𝑢𝑗𝜔𝑥𝑖𝑑𝑥+𝑑0𝐷Σ𝑢𝜔2𝑥𝑖𝑑𝑥+𝐷Σ𝑎𝑖𝑗𝑢𝜔𝑥𝑖𝑥𝑗+𝐶3𝜀6𝑑0𝜀𝐷Σ|𝑢|2𝑑𝑥+6𝑑0𝜀2𝐷Σ𝜔2||𝑢||2𝑑𝑥+(2𝑛+1)𝐷Σ𝑢𝑗𝜔𝑑𝑥+𝑑0𝐷Σ𝑢𝜔𝑑𝑥+𝛾𝐶4𝜀6𝑑0𝜀𝑀mes𝑛𝐷+(2𝑀+1)𝛾𝜀mes𝑛𝐷+𝑑0𝑀𝜔(𝐷)+𝛾𝐶4𝑀𝜔(𝐷)+𝐶3𝜀.(60)
If we take into account that |||𝜔𝑥𝑖𝑥𝑗|||<𝐶4𝜔(𝑥),(61) then from here we have that 𝐷Σ𝜔2||𝑢||2𝑑𝑥𝐶5,(62) where 𝐶5=(6𝑑0+(2𝑀+1))𝑀mes𝑛𝐷+(𝑑0𝑀+𝛾𝐶4𝑀)𝜔(𝐷)+𝐶3/𝛾. Without loss of generality we assume that 𝜀1. Hence we have𝐷𝜔2|𝑢|2𝑑𝑥𝐶6.
Thus 𝑢(𝑥)𝑊12,𝜔(𝐷). From the boundary condition and mes𝑛1(𝜕𝐷𝐸)=0 we get 𝑢(𝑥)𝑊12,𝜔(𝐷). Now, let 𝜎2 be a number which will be chosen later, 𝐷+Σ={𝑥𝑥𝐷Σ,𝑢(𝑥)>0}. Without loss of generality, we suppose that the set 𝐷+Σ is not empty. Supposing in (53) 𝑧=1,𝛽=𝜔𝑢𝜎, we get 𝐷+Σ(𝜔𝑢𝜎)𝑑𝑥=𝜎𝜕𝐷+Σ𝜔𝜈𝑢𝜎+𝜎𝑢𝜎1𝜕𝑢𝜕𝜈𝑑𝑠𝑀𝜎𝜕𝐷+Σ𝜔𝑑𝑠+𝜎𝑀𝜎1𝜕𝐷+Σ|||𝜕𝑢𝜕𝜈|||𝑑𝑠𝐶5𝑎0,𝑀,𝜎,𝐶1𝜀.(63)
But, on the other hand, (𝑢𝜎)=𝑛𝑖,𝑗=1𝜕𝜕𝑥𝑖𝑎𝑖𝑗𝜕𝜔𝑢𝜎𝜕𝑥𝑗=𝑛𝑖,𝑗=1𝜕𝜕𝑥𝑖𝑎𝑖𝑗𝜔𝜎𝑢𝜎1𝜕𝑢𝜕𝑥𝑗+𝑛𝑖,𝑗=1𝜕𝜕𝑥𝑖𝑎𝑖𝑗𝜔𝑥𝑖𝜕𝑢𝜎𝜕𝑥𝑗=𝑛𝑖,𝑗=1𝜕𝜕𝑥𝑖𝑎𝑖𝑗𝜔𝜎𝑢𝜎1𝜕𝑢𝜕𝑥𝑗+𝑛𝑖,𝑗=1𝜕𝜕𝑥𝑖𝑎𝑖𝑗𝜎𝑢𝜎1𝜔𝑥𝜕𝑢𝜕𝑥𝑗=𝜎𝜔𝑢𝜎1(𝑢)+𝜎𝜔𝜕𝜕𝑥𝑖𝑎𝑖𝑗𝑢𝜎1𝜕𝑢𝜕𝑥𝑗+𝜎𝑢𝜎1𝜕𝜕𝑥𝑖𝑎𝑖𝑗𝜔𝜕𝑢𝜕𝑥𝑗+𝛽=𝜎𝜔𝑢𝜎1(𝑢)+𝜎𝜔𝑢𝜎1𝜕𝜕𝑥𝑖𝑎𝑖𝑗𝜕𝑢𝜕𝑥𝑗+𝜎𝜔𝑎𝑖𝑗𝑢𝑥𝑗(𝜎1)𝑢𝜎2𝑢𝑥𝑖+𝜎𝑢𝜎1𝜔𝑥𝑖𝑎𝑖𝑗𝜕𝑢𝜕𝑥𝑗+𝜎𝑢𝜎1𝜔𝜕𝜕𝑥𝑖𝑎𝑖𝑗𝜕𝑢𝜕𝑥𝑗+𝛽=3𝜎𝜔𝑢𝜎1𝑀(𝑢)+𝜎(𝜎1)𝑎𝑖𝑗𝑢𝑥𝑖𝑢𝑥𝑗𝑢𝜎2𝜔+𝜎𝑢𝜎1𝜔𝑥𝑖𝑎𝑖𝑗𝑢𝑥𝑗+𝛽=𝜎𝐷+Σ𝑑𝑖(𝑥)𝑢𝑥𝑖𝑢𝜔𝑑𝑥𝜎(𝜎1)𝐷+Σ𝑢𝜎𝜔(𝑥)𝑐(𝑥)𝑑𝑥+𝜎(𝜎1)𝐷+Σ𝑛𝑖,𝑗=1𝑢𝜎2𝜔(𝑥)𝑎𝑖𝑗𝑢𝑖𝑢𝑗𝑑𝑥+(2𝑢+1)𝐷+Σ𝑛𝑖,𝑗=1𝑎𝑖𝑗𝑢𝑗𝜔𝑥𝑗𝑢𝜎1.(64)
Hence, we conclude 𝜎(𝜎1)𝐷+Σ𝜔2𝑢𝜎2||𝑢||2𝑑𝑥𝑑0𝐷+Σ𝑢𝜎1𝜔𝑢𝑖𝑑𝑥𝑑0𝐷+Σ𝑢𝜎1𝜔𝑢𝑖𝑑𝑥𝑑0𝜀2𝐷+Σ𝑢𝜎𝑑𝑥.(65)
Let 𝐷+={𝑥𝑥𝐷,𝑢(𝑥)>0},𝐷+1 an arbitrary connected component of 𝐷+. Subject to the arbitrariness of 𝜀 from (65) we get (𝜎1)𝛾𝐷+1𝜔𝑢𝜎2||𝑢||2𝑑𝑥𝑑0𝐷+1𝜔𝑢𝜎1𝑛𝑖=1||𝑢𝑖||𝑑𝑥.(66)
Thus, for any𝜇>0(𝜎1)𝛾𝐷+1𝜔𝑢𝜎2||𝑢||2𝑑𝑥𝑑0𝜇2𝐷+1𝜔𝑢𝜎2𝑛𝑖=1||𝑢𝑖||2𝑑𝑥+𝑑02𝜇𝐷+1𝜔𝑢𝜎𝑑𝑥𝑑0𝜇𝑛2𝐷+1𝜔𝑢𝜎2||𝑢||2𝑑𝑥+𝑑02𝜇𝐷+1𝜔𝑢𝜎𝑑𝑥.(67).
But, on the other hand, 𝐼=𝜎𝑛𝑖=1𝐷+1𝑥𝑖𝜔𝑢𝜎1𝑢𝑖𝑑𝑥=𝑛𝑖=1𝐷+1𝑥𝑖𝜔(𝑢𝜎)𝑖𝑑𝑥=𝑛𝐷+1𝜔𝑢𝜎𝑑𝑥.(68) and besides, for any𝛽>0𝐼=𝜎𝛽2𝐷+1𝑟2𝜔𝑢𝜎𝑑𝑥+𝜎2𝛽𝐷+1𝑢𝜎2𝜔2||𝑢||2𝑑𝑥.(69)
Then 𝐼𝜎𝛽2𝐷+1𝑟2𝜔𝑢𝜎𝑑𝑥+𝜎2𝛽𝐷+1𝜔2||𝑢||2𝑢𝜎2𝑑𝑥,(70) where 𝑟=|𝑥|. Denote by 𝑘(𝐷) the quantity sup𝑥𝐷|𝑥|. Without loss of generality we’ll suppose that 𝑘(𝐷)=1. Then 𝐼𝜎2𝛽𝐷+1𝜔𝑢𝜎𝑑𝑥+𝜎2𝛽𝐷+1𝜔2𝑢𝜎2||𝑢||2𝑑𝑥.(71)
Thus, 𝑛𝜎𝛽2𝐷+1𝜔𝑢𝜎𝑑𝑥+𝜎2𝛽𝐷+1𝜔2𝑢𝜎2||𝑢||2𝑑𝑥.(72)
Now, choosing 𝛽=𝑛/𝜎, we finally obtain 𝐷+1𝜔𝑢𝜎𝑑𝑥𝜎2𝑛2𝐷+1𝜔2𝑢𝜎2||𝑢||2𝑑𝑥.(73)
Subject to (73) in (67), we conclude (𝜎1)𝛾𝐷+1𝜔2𝑢𝜎2||𝑢||2𝑑𝑥𝑑0𝜀𝑛2+𝑑0𝜎22𝜀𝑛2𝐷+1𝜔2𝑢𝜎2||𝑢||2𝑑𝑥.(74)
Now choose 𝜇 such that (𝜎1)𝛾>𝑑0𝜇𝑛2+𝑑0𝜎22𝜇𝑛2.(75)
Then from (73)–(75) it will follow that 𝑢(𝑥)0 in 𝐷+1, and thus 𝑢(𝑥)0 in 𝐷. Suppose that 𝜇=(𝜎1)𝛾/𝑑0𝑛. Then (75) is equivalent to the condition 𝑛>𝜎𝜎12𝑑0𝛾2.(76)
At first, suppose that 𝑛>𝑑0𝛾2.(77)
Let’s choose and fix such a big 𝜎2 that by fulfilling (77) the inequality (76) is true. Thus, the theorem is proved, if with respect to 𝑛 the condition (77) is fulfilled. Show that it is true for any 𝑛3. For that, at first, note that if 𝑘(𝐷)1, then condition (77) will take the form 𝑛>𝑑0𝑘(𝐷)𝛾2.(78)
Now, let the condition (77) be not fulfilled. Denote by 𝑘 the least natural number for which 𝑛+𝑘>𝑑0𝛾2.(79)
Consider (𝑛+𝑘)-dimensional semicylinder 𝐷=𝐷×(𝛿0,𝛿0)××(𝛿0,𝛿0), where the number 𝛿0>0 will be chosen later. Since 𝜔(𝐷)=1, then 𝜔(𝐷)1+𝛿0𝑘. Let’s choose and fix 𝛿0 so small that along with the condition (79), the condition 𝑛+𝑘>𝑑0𝜔𝐷𝛾2(80) was fulfilled too.
Let 𝑦=𝑥1,,𝑥𝑛,𝑥𝑛+1,,𝑥𝑛+𝑘,𝐸=𝐸×𝛿0,𝛿0××𝛿0,𝛿0𝑘times.(81)
Consider on the domain 𝐷 the equation 𝜗=𝑛𝑖,𝑗=1𝑎𝑖𝑗(𝑥)𝜗𝑖𝑗+𝑘𝑖=1𝜕2𝜗𝜕𝑥2𝑛+𝑖+𝑛𝑖=1𝑏𝑖(𝑥)𝜗𝑖+𝑐(𝑥)𝜗=0.(82)
It is easy to see that the function 𝜗(𝑦)=𝑢(𝑥) is a solution of (82) in 𝐷𝐸. Besides, 𝑚𝑛+𝑘2+𝜆𝐻(𝐸)=(2𝛿0)𝑘𝑚𝑛2+𝜆𝐻(𝐸)=0, the function 𝜗(𝑦) vanishes on (𝜕𝐷×[𝛿0,𝛿0]××[𝛿0,𝛿0]𝑘times)𝐸 and 𝜕𝜗/𝜕𝜈=0 at 𝑥𝑛+𝑖=±𝛿0,𝑖=1,,𝑘, where 𝜕/𝜕𝜈 is a derivative by the conormal generated by the operator . Noting that 𝛾()=𝛾(),𝑑0()=𝑑0() and subject to the condition (80), from the proved above we conclude that 𝜗(𝑦)0, that is, 𝐷. The theorem is proved.

Remark 3. As is seen from the proof, the assertion of the theorem remains valid if instead of the condition (4) it is required that the coefficients 𝑎𝑖𝑗(𝑥)(𝑖,𝑗=1,,𝑛) have to satisfy in domain 𝐷 the uniform Lipschitz condition with weight.


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