On Removable Sets of Solutions of Elliptic Equations Tair

Tair S. Gadjiev, Mahammad-Rza I. Arazm, and Vafa A. Mamedova Department of Nonlinear Analyses, Institute of Mathematics and Mechanics of NAS of Azerbaijan, 9, F. Agaev street, Baku AZ1141, Azerbaijan Correspondence should be addressed to Mahammad-Rza I. Arazm, shafa mamedova@mail.ru Received 8 March 2011; Accepted 20 April 2011 Academic Editors: G. L. Karakostas, G. Mantica, X. B. Pan, and C. Zhu Copyright q 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 E n of the points x x 1 , . . ., x n , n ≥ 3, and let ∂D be its boundary.Consider in D the following elliptic equation: in supposition that a ij x is a real symmetric matrix, moreover ω x is a positive measurable function satisfying the doubling condition: for concentric balls B x R or R and 2R radius, there exists such a constant γ ω B x R ≥ γω B x 2R , 2 where for the measurable sets E. ω E means E ω y dy γ |ξ| 2 ω x ≤ n i,j 1 a ij x ξ i ξ j ≤ γ −1 ω x |ξ| 2 ; ξ ∈ E n , x ∈ D, 3 Here u i ∂u/∂x i , u ij ∂ 2 u/∂x i ∂x j ; i, j 1, . . ., n; γ ∈ 0, 1 and b 0 ≥ 0 are constants.Besides we will suppose that the lower coefficients of the operator L are measurable functions in D.
Let λ ∈ 0, 1 be a number.Denote by C 0,λ D a Banach space of the functions u x defined in D with the finite norm; The compact E ⊂ D is called removable with respect to 1 in the space 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 C λ ω 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 C λ D was found in 4 .mention also papers 5-7 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 8-10 .In 11 , Kilpelinen 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 B R z and S R z the ball {x : |x − z| < R} and the sphere {x : |x − z| R} of radius R with the center at the point z ∈ E n respectively.We will 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 S R 0 and S 2R 0 , moreover let the intersection ∂G ∩ {x : R < |x| < 2R} be a smooth surface.Further, let in G the uniformly positive definite matrix a ij x ; i, j 1, . . ., n and the function u x ∈ C 2 G ∩ C 1 ω G be given.Then there exists the piecewise smooth surface Σ dividing in G the spheres S R 0 and S 2R 0 such that Here K > 0 is a constant depending only on the matrix a ij x and n, and ∂u/∂ν is a derivative by a conormal determined by the equality where cos n, x j ; j 1, . . ., n are direction cosines of a unit external normal vector to Σ.

14
where Now by a Banach process 4, page 126 from the ball system {B x t/5 } we choose such a denumerable number of notintersecting balls {B x ν t ν /5 }, ν 1, 2, . . ., N that the ball of fivetimes greater radius {B x ν t ν } cover the whole O f set.We again denote these balls by {B x ν t ν /5 }, ν 1, 2, . . ., N and their surface by S ν .Then by virtue of 5 Therefore there exists such 5r x ≤ t ≤ 6r x that r x S x r ω σ dσ ≤ ω B x 6r x .17 Assign arbitrary η > 0. By virtue of that |∇f| S x t ≤ η • t, for sufficiently small t we have

18
Again by means of Banach process and by virtue of 43 we get where S n ν is the surface of balls in the second covering.
Combining the spherical surfaces S ν and S ν we get that the open balls system covers the closed set O f .Then a finite subcovering may be choosing from it.Let them be the balls B 1 , B 2 , . . ., B x and their surfaces are S 1 , S 2 , . . ., S N .We get from inequalities 4 and 7 and according to Lemma 1 for a given ε we will find the balls B 1 , B 2 , . . ., B x and exclude them from the domain G.
We denote the intersection by G .We can assume that the function u x is defined in some δ vicinity G δ of set G .Take δ < R/4 so that oscu On a closed set G we have ∇f / 0. Consider on G δ the equation system dx dt u x .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 It will be possible if on N ∂u/∂R ≡ 0.
Put E E \ F. 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 R − δ < |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

R. 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 at first so that the edges of this tube be embedded into this layer.Denote these cutoff tubes by T 1 , T 2 , . . ., T S .If each open tube is divided with a partition, then a set-theoretical sum of closed tubes, tubes T 1 , T 2 , . . ., T S their partitions spheres S 1 , S 2 , . . ., S N, and the set F on the sphere |x| 7/4 R divides the spheres |x| R and |x| 2R.Note that S ω|∂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 S ω|∂u/∂n|dσ was of the desired value.Denote by U i the domain bounded by T i with corresponding cell and hypersurface cutting off this tube.We have U i ∩ U j ∅ and therefore Consider a tube T i and corresponding domain U i .Choose any trajectory on this tube.Denote it by L i .The length μ i L i of the curve L i satisfies the inequality Let introduce on L i a parameter l length of the arc , counted from the cell.By σ i l denote the cross-section by U i hypersurface passing thought the point, corresponding to l and orthogonal to the trajectory L i at this point.Let the diameter of cells be so small 31 Then by Chebyshev inequality a set H points l ∈ L i where satisfies the inequality μ i H < R/4 and hence by virtue of 55 for E L i \ H it is valid and At the points of the curve L i the derivative ∂u/∂l preserves its sign, and therefore Hence, by using 65 and a mean value theorem for one variable function we find that there exists But on the other hand ∂u ∂l l l 0 |∇u| l l 0 .36 Together with 67 it gives Now, let the diameter of cells be still so small that The lemma is proved.
Denote by W 1 2,ω D the Banach space of the functions u x defined in D with the finite norm Proof.At first we show that without loss of generality we can suppose the condition ∂D ∈ C 1 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 C 1 , but ∂D ∈ C 1 , 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 L is infinitely differentiable in D.Moreover, without loss of generality, we'll suppose that the coefficients of the operator L are extended to a ball B ⊃ D 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 Evidently, u x u x u − x .Further, let D be such a domain that ∂D ∈ C 1 , D ⊂ D , D ⊂ B, and ϑ ± x be solutions of the problems By the maximum principle for x ∈ D, But according to our supposition, ϑ x ≡ ϑ − x ≡ 0. Hence, it follows that u x ≡ 0. So, we'll suppose that ∂D ∈ C 1 .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 Consider a system of the spheres {B 2r k x k }, and let D k D ∩ B 2r k x k , k 1, 2, . . . .Without loss of generality we can suppose that the cover {B 2r k x k } has a finite multiplicity a 0 n .By the Landis-Gerver theorem, for every k, there exists a piece-wise smooth surface Σ k dividing in D k the spheres S r k x k and S 2r k x k , such that Since u x ∈ C λ ω D , there exists a constant H 1 > 0 depending only on the function u x such that oscu where Ω n mes n B 1 0 .Using 49 and 50 in 48 , we get where C 1 KH 1 2 n λ .Let D Σ be an open set situated in D \ E whose boundary consists of unification of Σ and Γ, where According to Green formula for any functions z x and W x belonging to the intersection From 53 choosing the functions z 1, β ωu 2 , we have Let us put the condition By virtue of condition 52 and ∂D Σ ωu 2 ds < C 3 Mε, subject to 51 and 47 , we conclude

55
where On the other hand and besides, where It is evident that by virtue of conditions 4 and 5 |d i x | ≤ d 0 < ∞; i 1, . . ., n.Thus, from 55 we obtain 6

ISRN Mathematical Analysis
If we take into account that then from here we have that Thus u x ∈ W 1 2,ω D .From the boundary condition and mes n−1 ∂D ∩ E 0 we get u x ∈ W 1  2,ω D .Now, let σ ≥ 2 be a number which will be chosen later, D Σ {x : x ∈ D Σ , u x > 0}.Without loss of generality, we suppose that the set D Σ is not empty.Supposing in 53 z 1, β ωu σ , we get

63
But, on the other hand, Hence, we conclude Thus, for any μ > 0

67
. But, on the other hand, ωu σ dx 68 and besides, for any β > 0 Let's choose and fix such a big σ ≥ 2 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 n ≥ 3.For that, at first, note that if k D / 1, then condition 77 will take the form 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 a ij x i, j 1, . . ., n have to satisfy in domain D the uniform Lipschitz condition with weight.

Theorem 2 .
D be a completion of C ∞ 0 D by the norm of the space W1  2,ω D .By m s H 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.Let D be a bounded domain in E n and let E ⊂ D be a compact.If with respect to the coefficients of the operator L the conditions 3 -5 are fulfilled, then for removability of the compact E with respect to the 1 in the space C λ ω D it sufficies that remaining after the removing of points situated between Σ and S 2r k x k ; k 1, 2, . . . .Denote by D Σ the arbitrary connected component D Σ , and by M we denote the elliptic operator of divergent structure
{x : x ∈ D, u x > 0}, D 1 an arbitrary connected component of D .Subject to the arbitrariness of ε from 65 we get
38we can do it, since the derivatives ∂u/∂x i are uniformly continuous .Therefore according to 53