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.