On Removable Sets of Solutions of Neuman Problem for Quasilinear Elliptic Equations of Divergent Form ()
1. Introduction
Let D be a bounded domain situated in 
-dimensional Euclidean space 
of the points 
be its boundary. Consider in 
the following elliptic equation
(1)
in supposition that 
is a real symmetric matrix, moreover
(2)
(3)
(4)
(5)
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
(6)
it follows that 
in
.
2. Auxiliary Results
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 [1]. Concerning the second order elliptic equations of divergent structure, we show in this direction the papers [2,3]. For a class of non-divergent elliptic equations of the second order with discontinuous coefficients the removability condition for a compact in the space 
was found in [4]. Mention also papers [5-9] 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 [10-12]. In [13], T. Kilpelainen and X. Zhong have studied the divergent quasilinear equation without minor members, proved the removability of a compact. Removable sets for pointwise solutions of elliptic partial differential equations were found by J.
Diederich [14]. Removable singularities of solutions of linear partial differential equations were considered in R. Harvey, J. Polking paper [15]. Removable sets at the boundary for subharmonic functions have been investigated by B. Dahlberg [16]. Also we mentioned the papers of A.V.Pokrovskii [17,18].
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 [8] in weight case.
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 ([4], p.126) from the ball system 
we choose such a denumerable number of not-intersecting balls 
that the ball of five times greater radius 
cover the whole 
set. We again denote these balls by
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 [2], assume
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
(we can do it, since the derivatives 
are uniformly continuous). Therefore according to (12)
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.
3. Main Results
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
(7)
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 [8]) solutions of the boundary value problems
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
(8)
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
(9)
Since
, there exists a constant 
depending only on the function 
such that
(10)
Besides,
(11)
where
. Using (10) and (11) in (9), we get
(12)
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
(13)
Since 
then
(see [9]). From (13) choosing the functions 
we have
But 
for
. Let’s assume that the condition
(*)
is fulfilled. By virtue of condition (*) and
subject to (12) and (8) we conclude
(14)
where 
On the other hand
and besides,
where 
It is evident that by virtue of conditions (3)-(4) 
Thus, from (13) we obtain
(15)
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
(16)
Let
, 
be an arbitrary connected component of
. Subject to the arbitrariness of 
from (16) we get
Thus, for any 
(17)
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
(18)
Subject to Equation (18) in Equation (17) ,we conclude
(19)
Now choose 
such that
(20)
Then from Equations (18)-(20) it will follow that 
in
, and thus 
in
. Suppose that
.
Then Equation (20) is equivalent to the condition
(21)
At first, suppose that
(22)
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
(23)
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
(24)
was fulfilled too.
Let
Consider on the domain 
the equation
(25)
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.