On Removable Sets of Solutions of Elliptic Equations
Tair S. Gadjiev,1Mahammad-Rza I. Arazm,1and Vafa A. Mamedova1
Academic Editor: G. L. Karakostas, C. Zhu, X. B. Pan, G. Mantica
Received08 Mar 2011
Accepted20 Apr 2011
Published29 Jun 2011
Abstract
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 HΓΆlder functions was found.
Let 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 is a positive measurable function satisfying the doubling condition: for concentric balls or and radius, there exists such a constant
where for the measurable sets . means
Here and are constants. Besides we will suppose that the lower coefficients of the operator are measurable functions in . Let be a number. Denote by a Banach space of the functions defined in with the finite norm;
The compact is called removable with respect to (1) in the space if from
it follows that in .
The aim of the given paper is finding sufficient condition of removability of a compact with respect to (1) in the space . 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 was found in [4]. mention also papers [5β7] 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 [8β10]. In [11], KilpelΣinen 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 and the ball and the sphere
of radius with the center at the point respectively. Weβll need the following generalization of mean value theorem belonging to Gerver and Landis [15] in weight case.
Lemma 1. Let the domain be situated between the spheres and , moreover let the intersection be a smooth surface. Further, let in the uniformly positive definite matrix and the function be given. Then there exists the piecewise smooth surface ββdividing in the spheres and such that
Here is a constant depending only on the matrix and , 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 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 Lebesgue 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 the above said, we get that the set may be included into the set for which will be chosen 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, page 126] from the ball system we choose such a denumerable number of notintersecting 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 (5)
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 (43) we get
where is the surface of balls in the second covering. Combining the spherical surfaces and we get that the open balls system covers the closed set . Then a finite subcovering may be choosing from it. Let them be the balls and their surfaces are . We get from inequalities (4) and (7)
Put now . Following [2], assume
and according to Lemma 1 for a given we will find the balls and exclude them 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 some surface touches the direction of the field at each its point, then
since is identically equal to zero at . We will use it in constructing the needed surface of . Tubular surfaces whose generators will be the trajectories of the system (50) 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 (50) 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 intersects 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 cutoff 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
Let introduce on a parameter (length of the arc), counted from the 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 (55) for it is valid and
At the points of the curve the derivative 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
Together with (67) 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 (53)
Define by a set-theoretical sum of all closed tubes, all open tubes , all , all spheres and sets on the sphere . Then, we get by (4), (49), (51), and (73)
Then, we get by (4), (49), (51), (73)
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 will 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 2. Let be a bounded domain in and let be a compact. If with respect to the coefficients of the operator the conditions (3)β(5) are fulfilled, then for removability of the compact with respect to the (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 (43) provides the removability of the compact for the domains, whose boundary is the surface of the class , but , and by fulfilling (43) the compact is not removable. Then the problem (7) has a nontrivial solution , moreover and . We always can suppose the lowest coefficients of the operator is 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 (3)β(5). Let , and be generalized by Wiener (see [15]) solutions of the boundary value problems
Evidently, . 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 (7), and the condition (43) 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 the Landis-Gerver theorem, 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 (49) and (50) in (48), 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 [16]). From (53) choosing the functions , we have
But for . Let us put the condition
By virtue of condition (52) and subject to (51) and (47), we conclude
where . On the other hand
and besides,
where
It is evident that by virtue of conditions (4) and (5) . Thus, from (55) we obtain
Hence, for any it follows that
If we 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 generality, we suppose that the set is not empty. Supposing in (53) , we get
But, on the other hand,
Hence, we conclude
Let an arbitrary connected component of . Subject to the arbitrariness of from (65) we get
Thus, for any. But, on the other hand,
and besides, for any Then
where . Denote by the quantity . Without loss of generality weβll suppose that . Then
Thus,
Now, choosing , we finally obtain
Subject to (73) in (67), we conclude
Now choose such that
Then from (73)β(75) it will follow that in , and thus in . Suppose that . Then (75) is equivalent to the condition
At first, suppose that
Letβs choose and fix such a big that by fulfilling (77) the inequality (76) is true. Thus, the theorem is proved, if with respect to the condition (77) is fulfilled. Show that it is true for any . For that, at first, note that if , then condition (77) will take the form
Now, let the condition (77) be not fulfilled. Denote by the least natural number for which
Consider -dimensional semicylinder , where the number will be chosen later. Since , then . Letβs choose and fix so small that along with the condition (79), the condition
was fulfilled too. Let
Consider on the domain the equation
It is easy to see that the function is a solution of (82) 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 (80), from the proved above we conclude that , that is, . 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 have to satisfy in domain the uniform Lipschitz condition with weight.
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