Consider the principal part of the problem (1)
Condition 1 Assume the following conditions are satisfied:
-
(1)
E is a UMD space, ;
-
(2)
, is a VMO modulus of a, , ;
-
(3)
and for , , . a.e. ;
-
(4)
is a uniformly R-positive operator in E and
-
(5)
, and is a VMO modulus of .
First, we obtain an integral representation formula for solutions.
Lemma 1 Let Condition 1 hold and . Then, for all solutions u of the problem (5) belonging to , we have
where
here , are uniformly bounded operators,
and the expression is a scalar multiple of .
Proof Consider the problem (5) for and , i.e.,
Let be a solution of the problem (7). Taking into account the equality and Theorem A7, we get
Setting in above, we get (6) for . Then the density argument, using Theorem A3, gives the conclusion for
Consider the problem (5) on , i.e.,
□
Theorem 1 Suppose Condition 1 is satisfied. Then there exists a number such that the uniform coercive estimate
(9)
holds for large enough and , , , where C is a positive constant depending only on p, , , .
Proof By Lemma 1, for any solution of the problem (8), we have
(10)
where
(11)
here , are uniformly bounded operators,
and
Moreover, from (10) and (11), clearly, we get
(12)
where the expression differs from only by a constant.
Consider the operators
Since the operators and are regular on , by using the positivity properties of A and the analyticity of semigroups in a similar way as in [[30], Theorem 3.1], we get
(13)
Since the Hilbert operator is bounded in for a UMD space E, we have
(14)
Thus, by virtue of Theorems A4, A6 and in view of (10)-(12) for any , there exists a positive number such that
Hence, the estimates (13)-(15) imply (9). □
Theorem 2 Assume Condition 1 holds. Let be a solution of (4). Then, for sufficiently large , , the following coercive uniform estimate holds:
(16)
Proof This fact is shown by a covering and flattening argument, in a similar way as in Theorem A8. Particularly, by partition of unity, the problem is localized. Choosing diameters of supports for corresponding finite functions, by using Theorem 1, Theorems A4, A6, A7 and embedding Theorem A1 (see the same technique for DOEs with continuous coefficients [19, 20]), we obtain the assertion.
Let denote the operator in generated by the problem (4) for , i.e.,
□
Theorem 3 Assume Condition 1 holds. Then, for all , and for large enough , the problem (5) has a unique solution . Moreover, the following coercive uniform estimate holds:
(17)
Proof First, let us show that the operator has a left inverse. Really, it is clear that
By Theorem A1 for , we have
Then, by virtue of (16) and in view of the above relations, we infer, for all and sufficiently large , that there is a small ε and such that
From the estimate (18) for , we get
(19)
The estimate (19) implies that (4) has a unique solution and the operator has a bounded inverse in its rank space. We need to show the rank space coincides with all the space . It suffices to prove that there is a solution for all . This fact can be derived in a standard way, approximating the equation with a similar one with smooth coefficients [19, 20]. More precisely, by virtue of [[23], Theorem 3.4], UMD spaces satisfy the multiplier condition. Moreover, by the part (2) of Lemma A1, given with VMO modules , we can find a sequence of mollifiers functions converging to a in BMO as with a VMO modulus such that . In a similar way, it can be derived for the operator function . □
Result 1 Theorem 3 implies that the resolvent satisfies the following sharp uniform estimate:
(20)
for , and .
The estimate (20) particularly implies that the operator Q is uniformly positive in and generates analytic semigroups for (see, e.g., [[22], §1.14.5]).
Remark 1 Conditions , arise due to nonlocality of the boundary conditions (4). If the boundary conditions are local, then the conditions mentioned above are not required any more.
Consider the problem (1), where is the same boundary condition as in (4). Let denote a differential operator generated by the problem (1). We will show the separability and Fredholmness of (1).
Theorem 4
Assume
-
(1)
Condition 1 holds;
-
(2)
for any , there is such that for a.e. and for , ,
Then, for all and for large enough , , there is a unique solution of the problem (1) and the following coercive uniform estimate holds:
(21)
Proof It is sufficient to show that the operator has a bounded inverse from to . Put , where
By the second assumption and Theorem A1, there is a small ε and such that
(22)
In view of estimates (17) and (22), we have
(23)
for and . By Theorem 3, the operator has a bounded inverse from to for sufficiently large . So, (23) implies the following uniform estimate:
It is clear that
Then, by above relation and by virtue of Theorem 3, we get the assertion. □
Theorem 4 implies the following result.
Result 2 Suppose all the conditions of Theorem 4 are satisfied. Then the resolvent of the operator satisfies the following sharp uniform estimate:
for , and .
Consider the problem (1) for , i.e.,
Theorem 5 Assume all the conditions of Theorem 4 hold and . Then the problem (24) is Fredholm from into .
Proof Theorem 4 implies that the operator has a bounded inverse from to for large enough ; that is, the operator is Fredholm from into . Then, by virtue of Theorem A2 and by perturbation theory of linear operators, we obtain the assertion. □