Research Article | Open Access
Ground State for the Schrödinger Operator with the Weighted Hardy Potential
We establish the existence of ground states on for the Laplace operator involving the Hardy-type potential. This gives rise to the existence of the principal eigenfunctions for the Laplace operator involving weighted Hardy potentials. We also obtain a higher integrability property for the principal eigenfunction. This is used to examine the behaviour of the principal eigenfunction around 0.
In this paper, we investigate the existence of ground states of the Schrödinger operator associated with the quadratic form where belongs to the Lorentz space and is the largest constant (whenever exists) for which the form is nonnegative. This assumption implies, when , that the potential term is continuous in , where is the Sobolev space obtained as the completion of with respect to the norm We are mainly interested in the case of the Hardy-type potential with . Assuming that is positive on a set of positive measure, the constant is given by the variational problem and the continuity of implies that . If problem (1.3) has a minimizer , then it satisfies A solution of (1.4) is understood in the weak sense for every .
Since is also a minimizer for , we may assume that a.e. on . In particular, when with , then on by the Harnack inequality . If the potential term is weakly continuous in , for example, when with and , then there exists a minimizer for . We will call the minimizer of (1.3) a ground state of finite energy. In general, (1.3) may not have a minimizer. This is the case for the Hardy potential with the corresponding optimal constant . In fact, the ground state of finite energy is a particular case of the generalized ground state, defined as follows (see [2–4]).
Definition 1.1. Let be an open set, and let be as in (1.1). A sequence of nonnegative functions is said to be a null sequence for the functional if , as , and there exists a nonnegative function such that for each .
Let us recall that the capacity of a compact set relative to an open set , with , is given by In the case , we use notation (see ).
Theorem 1.2. Let be a measurable function bounded on every compact subset of , where is a closed set of capacity zero, and assume that for all . Then, if admits a null sequence , then the sequence converges weakly in to a unique (up to a multiplicative constant) positive solution of (1.4).
This theorem gives rise to the definition of the generalized ground state.
Definition 1.3. A unique positive solution of (1.4) is called a generalized ground state of the functional , if the functional admits a null sequence weakly convergent to .
If , then the functional has a ground state of infinite norm, while (1.3) has no minimizer in .
It is important to note that the functional with the optimal constant does not necessarily have a ground state. We quote the following statement from .
Theorem 1.4. Let be a measurable function bounded on every compact subset of , where is a closed set of capacity zero, and assume that for all . Then either admits a null sequence, or there exists a function , positive and continuous on , such that
For example, let be a continuous function on such that for , for and for . Then, and the functional does not admit a null sequence. From Theorem 1.4 follows that satisfies (1.7) with some function positive on .
Obviously, ground states of finite norm are principal eigenfunctions of (1.4). There is a quite extensive literature on principal eigenfunctions with indefinite weight functions for elliptic operators on or on unbounded domains of , with the Dirichlet boundary conditions. We mention papers [2, 6–13], where the existence of principal eigenfunctions has been established under various assumptions on weight functions. These conditions require that a potential belongs to some Lebesgue space, for example with . These results have been recently greatly improved in papers [14, 15], where potentials from the Lorentz spaces have been considered. To describe the results from [14, 15] we recall the definition of the Lorentz space [16–18].
Let be a measurable function. We define the distribution function and a nonincreasing rearrangement of in the following way We now set The Lorentz space is defined by The functional is only a quasinorm. To obtain a norm we replace by in the definition of , that is, the norm is given by equipped with the norm is a Banach space.
However, these conditions do not cover the singular weight functions considered in this paper. By contrast, in our approach, we give an exact upper bound for the principal eigenvalue which allows us to prove the existence of the principal eigenfunction. We point out that if , then the functional is continuous on , but not necessarily weakly continuous.
The paper is organized as follows. In Section 2, we prove the existence of minimizers with finite norm and also with infinite norm . In Section 3 we discuss perturbation of a given quadratic form with . We show that if has ground state, then this property is stable under small perturbations of . This is not true if does not have a ground state; rather it is stable under larger perturbation of . The final Section is devoted to a higher integrability property of minimizers of in the case where with . We also examine the behaviour of the principal eigenfunction around 0.
Throughout this paper, in a given Banach space, we denote strong convergence by “” and weak convergence by “”. The norms in the Lebesgue space , , are denoted by .
2. Existence of Minimizers
We consider the Hardy-type potential with . In Theorem 2.2 we formulate conditions on guaranteeing the existence of a principal eigenfunction. Let and . In our approach to problem (1.3), the following two limits play an important role: it is assumed that the following limits exist a.e. Both functions satisfy . We now define the following infima: (we use the notation instead of ) and
Lemma 2.1. The following holds true
Proof. Let . Testing with gives Letting and using the Lebesgue dominated convergence theorem, we obtain The inequality follows. The proof of the inequality is similar.
Proof. Let be a minimizing sequence for , that is, We can assume, up to a subsequence, that in , , and in for some . Let . We then have We define a radial function such that , for and for . Let . In what follows, we use to denote a quantity such that for each , as . Thus, where We now estimate the integrals involving and . We have By the uniform convergence of to , we see that for sufficiently large uniformly in . For we have It is clear that is a quantity of type . Therefore, we have In a similar way, we obtain for sufficiently large. We now fix so that (2.15) and (2.16) hold. Consequently, we have We now estimate in the following way Since in , we obtain the following estimate This, combined with (2.9), gives the following estimate Let . We deduce from (2.17) and (2.20) that Letting , we obtain It then follows from (2.17) that Since is arbitrary, we get , and the result follows.
In what follows, we denote , assuming that this limit exists. As a direct consequence of Theorem 2.2, we obtain the following result.
Theorem 2.3. Let , and assume that is continuous at 0. Further, suppose that and . If , then there exists a minimizer for .
Remark 2.4. has a minimizer also in the following cases, corresponding formally to or taking the value .(i)Let and . If , then a minimizer for exists.(ii)Let and . If , then a minimizer for exists.(iii)If , and ≢ 0 on , then has a minimizer.
We point out that Theorem 2.3 and the results described in Remark 2.4 can be deduced from [19, Theorem 1.2]. Unlike in paper , to obtain Theorem 2.3 we avoided the use of the concentration-compactness principle.
Example 2.5. Let where is a constant to be chosen later and , , and are continuous bounded functions satisfying the following conditions: for , for , for , and for for . Further we assume that for , where , and constants. A function for small is given by for . We have Both limits are uniform. Since and are bounded, and are positive and finite. We have for large. By Theorem 2.2, with has a minimizer.
Example 2.6. Consider a sequence of functions of the form , where , are constants and and are continuous functions satisfying the following conditions: (a), on , , for , (b) on for , (c) on , and . Then and for . We show that for sufficiently large satisfies the conditions of Theorem 2.3. Let (one can take any other function from which is on ). Thus as . So we can find so that
Proposition 2.7. Let . Suppose that , , and . Assume that there exists a ball such that for and . If Then . (Hence, there exists a minimizer for .)
Proof. Let . Then Hence, Since , we deduce from the above inequality that where denotes the first eigenvalue for “” in with the Dirichlet boundary conditions. We now estimate . We test with for . We have Hence Combining this with (2.34), we derive Therefore if (2.31) holds.
The estimate (2.31) has terms that are easy to compute but are of course not optimal. In particular, the factor can be replaced by the first eigenvalue of the Laplacian on a unit ball with Dirichlet boundary conditions.
If is a continuous bounded and nonnegative function such that on and (or on , ), then does not have a minimizer. Indeed, suppose that on and that has a minimizer . Then, by the Hardy inequality, we obtain So is a minimizer for , which is impossible.
We now construct a ground state with infinite norm.
Theorem 2.8. Let , and assume that the function satisfies Then the form with and (see (2.41) below) admits a ground state satisfying The function is uniquely defined by its values on , and, moreover, the function is a minimizer for the problem
Proof. The problem (2.41) is a compact variational problem that has a minimizer which satisfies with the Neumann boundary conditions. Since the test functions satisfy for , one has Note that is also a minimizer, so we may assume that is nonnegative. We now extend the function from to by using (2.40) and denote the extended function again by . Since satisfies (2.41), the extended function is of class and satisfies in a weak sense. From this and the Harnack inequality on bounded subsets of it follows that is positive on and subsequently there exists a constant such that We can now explain the choice of the exponent in the constraint from (2.41): with any other choice, the resulting Neumann condition would not yield the continuity of the derivatives of the extended function on the spheres . Finally, we show that is a ground state for the corresponding quadratic form with . Using the ground state formula (2.10), from  and (2.45), we have for and for , as . Since uniformly on compact sets, this implies that is a ground state for . By (2.45) and the Sobolev inequality, .
3. Perturbations from Virtual Ground States
In this section, we show that if a potential term admits a (generalized or large or virtual) ground state, then its weakly continuous perturbations in the suitable direction will admit a ground state with the finite norm. Then, we investigate potentials that do not give rise to a ground state with finite norm.
We need the following existence result.
Proposition 3.1. Let be positive on a set of positive measure, and let Assume that is positive on a set of positive measure and that the functional is weakly continuous in , and let If , then there exists a minimizer for .
Proof. Let be a minimizing sequence for (3.2), that is, and . We may assume that, up to a subsequence, in and . Let . Then,
Let . Then . Assuming that we get
From this, we deduce that which is impossible. Hence, From this and the lower semicontinuity of the norm with respect to weak convergence, we derive that is a minimizer and in .
Proposition 3.1 is related to [19, Theorem 1.7] which asserts that a potential of the form , with a subcritical potential (for the definition of a subcritical potential see ), has a principal eigenfunction. This follows from the fact that is weakly continuous in (see ) and the potential admits a principal eigenfunction.
Remark 3.2. (i) If , then , but not necessarily .
(ii) If, in Proposition 3.1, assumption is replaced by , then is attained.
Example 3.3. Let be a continuous function such that , on and . Define , where and are constants. Let and . The functional is weakly continuous in . It is easy to show that for every there exists such that for . By Proposition 3.1 has a minimizer for .
We now give a sufficient condition for the inequality .
Theorem 3.4. Suppose that and satisfy assumptions of Proposition 3.1. Moreover, assume that the quadratic form has a positive ground state , possibly with infinite norm, and that if is a null sequence corresponding to , then Then and has a minimizer.
Proof. It suffices to show that the inequality fails for some . We have for sufficiently large , which completes the proof of the theorem.
Note that the conditions of Theorem 3.4 are satisfied if, in particular, on , with the strict inequality on a set of positive measure. Indeed, the sequence converges weakly in to , and the condition follows from the Fatou lemma.
The situation becomes different if does not have a ground state. The absence of the ground state is stable property under small (in some sense) compact perturbation, but not under compact perturbations that are not small.
Theorem 3.5. Assume that satisfies the conditions of Proposition 3.1 and that (1.7) holds. (This occurs under conditions of Theorem 1.4 if has no ground state.) Let be as in (1.7). Then, for every , the functional has no ground state and . Furthermore, if the functional is weakly continuous in , the same conclusion holds for .
Proof. First, we observe that the constants and corresponding to and , respectively, are equal. Indeed, since , one has by monotonicity. On the other hand, it follows from (1.7) that for which implies . Let satisfy . Then which implies that, up to subsequence, a.e. If were a null sequence, it would converge in and it would have a limit zero. Therefore, admits no null sequence and consequently no ground state. Assume now that the functional is weakly continuous in . Let be a minimizing sequence for . If has a subsequence weakly convergent in to some , then it is easy to see that would be a minimizer for and thus a ground state for . Therefore, . By the weak continuity of , we get and thus This yields . Then, Since , this implies that has no ground state.
Theorem 3.5 concerns with small perturbations of a potential that does not change the constant or the absence of a ground state. The next theorem shows that a large compact perturbation of the potential term yields a ground state of finite norm.
Theorem 3.6. Assume that satisfies conditions of Proposition 3.1 and that is such that the functional is weakly continuous in . Then, for every there exists such that has a ground state of finite norm corresponding to the energy constant (3.2).
Proof. Assume without loss of generality that is positive on a set of positive measure. Let and consider Since defines an equivalent norm on , it is easy to show that there exists a minimizer for . It is clear that this minimizer is also a ground state of corresponding to the optimal constant .
If we assume additionally that is positive on a set of positive measure, then it is easy to show that is a continuous decreasing function of with and . In particular, if (1.7) holds with a weight satisfying , then . In other words, given and as in Theorem 3.6, the potential admits a ground state whenever .
For further results of that nature, we refer to paper .
4. Behaviour of a Ground State Around 0
In what follows we consider the potential of the Hardy-type , where is continuous and and . The corresponding ground state, if it exists, is denoted by , which is chosen to be positive on . Obviously the ground state satisfies in a weak sense.
We need the following extension of the Hardy inequality: let be a bounded domain and , then for every , there exists a constant such that for every (see ).
Proposition 4.1. Let Then for some and .
Proof. Let be such that on , , on and . For simplicity, we set , . We define , where and . Testing (4.1) with , we get Applying the Young inequality to the third term on the left side, we get where is a small number to be suitably chosen. Since the second integral on the left side is nonnegative, this inequality can be rewritten in the following form: Multiplying this inequality by ( and noting that , we get We now observe that Hence, (4.7) takes the form Since , we can choose so that . By the continuity of , there exists such that for . This is now used to estimate the first integral on the right side of (4.9): Applying the Hardy inequality (4.2), we get