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In this paper 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 Holder functions was found.

Let D be a bounded domain situated in -dimensional Euclidean space of the points be its boundary. Consider in the following elliptic equation

in supposition that is a real symmetric matrix, moreover

Here is non-negative function from

and are constants. Besides we’ll assume that the minor coefficients of the operator are measurable in. Let be some number.

The compact is called removable with respect to the Equation (1) in the space if from

it follows that in.

The paper is organized as follows. In Section 2, we present some definitions and auxiliary results. In Section 3 we give the main results of the sufficient condition of removability of compact.

The aim of the given paper is finding sufficient condition of removability of a compact with respect to the Equation (1) in the space. This problem have been investigated by many researchers. For the Laplace equation the corresponding result was found by L. Carleson [

Diederich [

In previous work, authors considered Direchlet problems for linear equations in some space of functions. In this work we consider Newman problem for quasilinear equations and sufficient conditions of removability of a compact in the weight space of Holder functions is obtained. The application value of the research in many physic problems.

Denote by and the ball and the sphere of radius with the center at the point respectively. We’ll need the following generalization of mean value theorem belonging to E.M. Landis and M.L. Gerver [

Lemma. Let the domain be situated between the spheres and, moreover the intersection be a smooth surface. Further, let in the uniformly positive definite matrix

and the function be given. Then there exists the piece-wise smooth surface dividing in the spheres and such that

Here is a constant depending only on the matrix and, is a derivative by a conormal determined by the equality

where are direction cosines of a unit external normal vector to.

Proof. Let be a bounded domain

. Then for any there exists a finite number of balls which cover and such that if we denote by, the surface of -th ball, then

Decompose into two parts:, where

is a set of points for which, is a set of points for which.

The set has -dimensional Lebesque measure equal zero, as on the known implicit function theorem, the lies on a denumerable number of surfaces of dimension. If we use the absolute continuity of integral

with respect to Lebesque measure and above said we get that the set may be included into the set for which will be choosen later. Let for each point there exist such that and are contained in. Then

therefore there exists such that

Then

where

.

Now by a Banach process ([

and their surface by. Then by virtue of (4)

Now let. Then

Therefore there exists such that

Assign arbitrary. By virtue of that, for sufficiently small we have

Again by means of Banach process and by virtue of (6) we get

where is the surface of balls in the second covering.

Combining the spherical surfaces and we get that the open balls system cover the closed set. Then a finite subcovering may be choosing from it. Let they be the balls and their surfaces is.

We get from inequalities (3) and (5)

Put now.

Following [

and according to lemma 1 well find the balls for given and exclude then from the domain. Put

intersect with a closed spherical layer

We denote the intersection by. We can assume that the function is defined in some vicinity

of set. Take so that

On a closed set we have. Consider on the equation system

Let a such from surface that it touches to field direction at any his point, then

since is identically equal to zero at.

We shall use it in constructing the needed surface of. Tubular surfaces whose generators will be the trajectories of the system (10) constitute the basis of.

They will add nothing to the integral we are interested in. These surfaces will have the form of thin tubes that cover. Then we shall put partitions to some of these tubes. Lets construct tubes. Denote by the intersection of with sphere.

Let be a set of points. Where field direction of system (10) touches the sphere. Cover with such an open on the sphere set that

It will be possible if on.

Put. Cover 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

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. Choose the diameters of cells so small that the trajectory beams passing through each cell, could differ no more than.

By choose of cells diameters the tubes will be contained in

Let also the cell diameter be chosen so small that the surface that is orthogonal to one trajectory of the tube intersect the other trajectories of the tube at an angle more than.

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 cut off tubes by. If each open tube is divided with a partition, then a set-theoretical sum of closed tubes, tubes their partitions spheres and the set on the sphere divides the spheres and. Note that along the surface of each tube equals to zero, since identically equals to zero.

Now we have to choose partitions so that the integral

was of the desired value. Denote by the domain bounded by with corresponding cell and hypersurface cutting off this tube. We have and therefore

Consider a tube and corresponding domain. Choose any trajectory on this tube. Denote it by. The length of the curve satisfies the inequality

On introduce a parameter in -length of the are counted from cell. By denote the cross-section by hypersurface passing thought the point, corresponding to and orthogonal to the trajectory at this point. Let the diameter of cells be so small

Then by Chebyshev inequality a set points where

satisfies the inequality and hence by virtue of (13) for it is valid and

At the points of the curve the derivative preserves its sign, and therefore

Hence, by using (15) and a mean value theorem for one variable function we find that there exists

But on the other hand

Together with (16) it gives

Now, let the diameter of cells be still so small that

Now by we denote a set-theoretic sum of all open tubes all thought tubes all all spheres

and sets on the sphere.

Then, we get by Equations (3), (9), (11) and (17)

The lemma is proved.

Denote by the Banach space of the functions defined in with the finite norm

and let be a completion of by the norm of the space.

By we’ll denote the Hausdorff measure of the set of order. Further everywhere the notation means, that the positive constant depends only on the content of brackets.

Theorem 1. Let be a bounded domain in, be a compact. If with respect to the coefficients of the operator the conditions (2)-(5) are fulfilled, then for removability of the compact with respect to the Equation (1) in the space it sufficies that

Proof. At first we show that without loss of generality we can suppose the condition is fulfilled. Suppose, that the condition (7) provides the removability of the compact for the domains, whose boundary is the surface of the class, but and by fulfilling (7) the compact is not removable. Then the problem (6) has non-trivial solution, moreover and. We always can suppose the lowest coefficients of the operator are infinitely differentiable in. Moreover, without loss of generality, we’ll suppose that the coefficients of the operator are extended to a ball with saving the conditions (2)- (5). Let, and be generalized by Wiener (see [

Evidently, by. Further, let

be such a domain, that and be solutions of the problems

By the maximum principle for

But according to our supposition. Hence, it follows, that. So, we’ll suppose that. Now, let be a solution of the problem (6), and the condition (7) be fulfilled. Give an arbitrary. Then there exists a sufficiently small positive number and a system of the balls

such that and

Consider a system of the spheres, and let. Without loss of generality we can suppose that the cover has a finite multiplicity. By lemma for every there exists a piece-wise smooth surface dividing in the spheres and, such that

Since, there exists a constant depending only on the function such that

Besides,

where. Using (10) and (11) in (9), we get

where.

Let be an open set situated in whose boundary consists of unification of and, where

is a part of remaining after the removing of points situated between and. Denote by the arbitrary connected component, and by we denote the elliptic operator of divergent structure

According to Green formula for any functions and

belonging to the intersection

we have

Since then

(see [

But for. Let’s assume that the condition

is fulfilled. By virtue of condition (*) and

subject to (12) and (8) we conclude

where

On the other hand

and besides,

where

It is evident that by virtue of conditions (3)-(4) Thus, from (13) we obtain

Let’s estimate the nonlinear member on the right part applying the inequality

Hence, for any applying Cauchy inequality we have

If we’ll take into account that

then from here we have that

where

.

Without loss of generality we assume that. Hence we have

Thus. From the boundary condition and we get. Now, let be a number which will be chosen later,

. Without loss of generalitywe suppose that the set isn’t empty. Supposing in (13) we get

But, on the other hand

Hence, we conclude

Let, be an arbitrary connected component of. Subject to the arbitrariness of from (16) we get

Thus, for any

But, on the other hand

and besides, for any

Then

where. Denote by the quantity.

Without loss of generality we’ll suppose, that. Then

Thus,

Now, choosing we finnaly obtain

Subject to Equation (18) in Equation (17) ,we conclude

Now choose such that

Then from Equations (18)-(20) it will follow that in, and thus in. Suppose that

.

Then Equation (20) is equivalent to the condition

At first, suppose that

Let’s choose and fix such a big that by fulfilling (22) the inequality (21) was true. Thus, the theorem is proved, if with respect to the condition (22) is fulfilled. Show that it is true for any. For that, at first, note that if, then condition (22) will take the form

Now, let the condition (22) be not fulfilled. Denote by the least natural number for which

Consider -dimensional semi-cylinder

where the number will be chosen later. Since

, then. Let’s choose and fix

so small that along with the condition (23) the condition

was fulfilled too.

Let

Consider on the domain the equation

It is easy to see that the function is a solution of the Equation (25) in. Besides,

the function vanishes on

and

at, where is a derivative by the conormal generated by the operator. Noting that and subject to the condition (24), from the proved above we conclude that, i.e.. The theorem is proved.

Remark. As is seen from the proof, the assertion of the theorem remains valid if instead of the condition (3) it is required that the coefficients have to satisfy in domain the uniform Lipschitz condition with weight.

Thus in this paper the sufficient condition for removability of the compact respect Newman problem for quasilinear equation in classes in the weight space of Holder functions is obtained.