We consider the Thomas-Fermi-Dirac-von Weizsäcker equation = , , where is a parameter, , is 1-periodic in , for , and 0 is in a spectral gap of the operator . Using a new infinite-dimensional linking theorem, we prove that, for sufficiently small , this equation has a nontrivial solution.

1. Introduction and Statement of Results

In this paper, we consider the following equation: where ,   is a parameter, and is the standard Sobolev space with norm .

The potential function is -periodic in for . Under this assumption, , the spectrum of the operator is a purely continuous spectrum that is bounded below and consists of closed disjoint intervals ([1, Theorem  XIII.100]). Thus, the complement consists of open intervals called spectral gaps. We assume the following:(v) is 1-periodic in for and 0 is in a spectral gap of , where .

A solution of (1) is called nontrivial if . Our main result is as follows.

Theorem 1. Suppose that and is satisfied. Then, there exists such that, for any , the problem (1) has a nontrivial solution.

Equation (1) arises in the study of the Thomas-Fermi-Dirac-von Weizsäcker (TFDW) model for atoms and molecules with no external potential, where is the electron density. In [2, section VIII], Lieb studied the existence and symmetric and asymptotic properties of solutions to for various choices of . Here, , and are constants and . The term is called the Dirac term with strength . In [3], Le Bris considered the minimizing problem with where and are positive constant, and is the Coulomb potential created by the atomic nuclei. Le Bris proved that there exists a , such that, for , the problem admits a minimizer. Moreover, if is small enough, then the minimizer is unique. Using the Lagrange multiplier, it is easy to see that the minimizer of (5) is a solution to for some . However, in a recent paper [4] by Lu and Otto, the authors proved that for sufficiently large , the variational problem (5) with does not have a minimizer. Equation (1) with a periodic function is used to describe a Hartree model for crystals (see [5, 6]). Moreover, when , (1) is often referred to the Thomas-Fermi-von Weizsäcker model in the literature. One can see [7] and references therein. Finally, we should mention that the so-called Schrödinger-Poisson-Slater equation is also related to (1) and has attracted much attention in recent years (see [5, 722]).

The variational functional for (1) is given by In other words, the critical points of are solutions to (1). Hence, it is natural to use critical point theory to obtain solutions to (1). Under assumption (v), the quadratic form has infinite-dimensional positive and negative spaces. It can be shown that for sufficiently small , has a global infinite-dimensional linking geometry (for its definition, see (6.4) of Willem [23]). However, the difficulty rises when the classical infinite-dimensional linking theorem (see [23]) is used to obtain a critical point to . This theorem requires the functional to satisfy some upper semicontinuous assumption (see (6.3) of [23]). However, because the nonlinearity is neither positive definite nor negative definite in whenever , does not satisfy the upper semicontinuous assumption. To overcome this difficulty, we use a new infinite-dimensional linking theorem in [24] to obtain a sequence (see Definition 6) for . We can prove that a sequence for is bounded in if is sufficiently small. This result is new and original. Finally, through the concentration-compactness principle and the sequence, a nontrivial solution to (1) is obtained. Our method can be used to study more general equation like

Notation. denotes an open ball of radius and center . For a Banach space , we denote the dual space of by and denote strong and weak convergence in by and , respectively. For , we denote the Fréchet derivative of at by . The Gateaux derivative of is denoted by , for all .   and denote the standard space and the locally -integrable function space, respectively . Let be a domain in (). is the space of infinitely differentiable functions with compact support in . We use and to mean and , respectively.

2. Variational Setting for (1)

Let be the Hilbert space with inner product For , by the Lax-Milgram theorem, the equation has a unique solution (see Proposition  2.2 of [10]). And, by Theorem   of [25], can be expressed by The function has the following properties.

Lemma 2. (i) There exists a positive constant such that, for any , (ii) For any , in .

Proof. The inequality can be derived from the the Hardy-Littlewood-Sobolev inequality (see (1.1) of [26]) by choosing ,  ,  , and there. For a proof of the other properties of in this lemma, one can see [8, 10, 12].

By this lemma, the functional in (9) is well defined in . And a direct computation shows that the derivative of is It is easy to verify that is a functional in . Moreover, we have the following.

Lemma 3. The following statements are equivalent:(i) is a solution of (1);(ii) is a critical point of .

Under the assumption , there is a standard variational setting for the quadratic form . One can see section 6.4 of [23]. But for the convenience of the reader, we state it here.

Let be the operator defined by (3). We denote by the square root of the absolute value of . The domain of is the space On , we choose the inner product and the corresponding norm . Since lies in a gap of the essential spectrum of , there exists an orthogonal decomposition such that and are the positive and negative spaces corresponding to the spectral decomposition of . Since is -periodic for all variables, they are invariant under the action of , that is, for any or and for any , is also in or . Furthermore, Let , be the orthogonal projections. By (21), Moreover, by , we have

By (22) and (9), Moreover, by (19),

3. A Global Linking Geometry for

Let be a total orthonormal sequence in and For and with , set

Definition 4. Let . is called weakly sequentially continuous if and are such that ; then, for any ,  .

Lemma 5. The functional satisfies the following.(a) is weakly sequentially continuous and, for every , (b)There exist ,  , with and such that, for ,

Proof. (a) Let and be such that as . It follows that
Since implies in for any , we get that, for any , as , By Lemma 2, we have that is bounded in . Therefore, up to a subsequence, as . It follows that in for any . This yields that, for any , as , From (25), (30), (31), and (32), we deduce that, as , Hence, is weakly sequentially continuous. Moreover, it is easy to see that maps bounded sets into bounded sets; hence .(b)If , then and . Using (17), we get from (24) that, for any , where comes from the Sobolev inequality , for all . It follows that we can choose sufficiently small such that, for ,
Since , there exists such that Let be such that From , we deduce that there exists such that, for any , From (36) and (38), we deduce that, for any and , From Theorem  1.1 of [27], we deduce that there exists such that, for any ,
Let be such that . And let . Since there exists a continuous projection we deduce that there exists a constant such that, for any , Combining (39)–(42), we deduce that there exists a constant such that, for any , It follows that Moreover, for and , (39) implies Together with (44), this implies that there exists such that
From (39) and the definition of (see (26)), we have, for , Choosing , (47) and (35) give that Together with (46), this yields (29).

4. Boundedness of Sequence of

Definition 6. Let . A sequence is called a sequence for , if

In this section, we will prove that if is sufficiently small, then a sequence for is bounded in .

Since , there exists such that where , .

Lemma 7. Suppose that . If is a sequence for , then

Proof. Let . It is easy to verify that and , for all . Together with the fact that is a sequence of , this implies , where denotes the infinitesimal depending only on ; that is, as . Choosing in (19), by (50) and the fact that ,  , and are nonnegative in , we have where in and is the Sobolev constant. Note that on . Together with (52), this implies Similarly, we can prove The result of this lemma follows from these two limits.

Lemma 8. There exists such that if and is a sequence for , then is bounded in .

Proof. By , we have Choosing and in (25), (55) implies that By these two equalities and (see (23)), we obtain
Let and , where comes from (50). Then, and Since is a solution of the equation by in and the standard elliptic estimates (see, [28, Theorem  8.17]), we get that there exists a positive constant that is independent of and , such that, for any , Let . By (61), Together with in (see Lemma 2(ii)), this implies By (57), (59), and (63), we have Since ( is the constant coming from (37)), , and , where is a positive constant independent of and , we have By Lemma 7, we deduce that and if . Together with (63), this implies that, for , Using the Hardy-Littlewood-Sobolev inequality (see, (1.1) of [26]), we get that where and are positive constants independent of and . By Lemma 7, we have Combining (68) and (67) with (66) yields that, for , Let . Then, for , Combining (70) with (69) yields that, for , Let be such that . Then, (71) implies that
From and , we obtain It follows that where is a positive constant depending only on <