Abstract
Keywords:
Schrödinger operator; critical potential; asymptotic expansion1 Introduction
In this paper, we consider a family of Schrödinger operators which are the perturbation of in the form
on , . Here . are the polar coordinates on , and is a real continuous function. is a nonzero continuous function satisfying
Here . Let denote the Laplace operator on the sphere . Assume that
If (2) holds, then in (see [1]).
Under the assumption on V, we know that has discrete eigenvalues when λ is large enough, and each discrete eigenvalue tends to zero when λ tends to some (see Section 2). We study the asymptotic behaviors of the discrete eigenvalues of in this paper. The asymptotic behaviors for Schrödinger operators with fast decaying potentials were studied by Klaus and Simon [2]. In [2], they studied the convergence rate of discrete eigenvalues of when . is the value at which some discrete eigenvalue tends to zero. The main method they used in their paper is the BirmanSchwinger technique.
In order to use the BirmanSchwinger technique to , we need to get the asymptotic expansion of for α near zero, , which was studied by Wang [1]. In this paper, we first show that there exists some such that when , has discrete eigenvalues. Then, we define the BirmanSchwinger kernel for and find that there is onetoone correspondence between the discrete eigenvalues of and the discrete eigenvalues of . Hence, the asymptotic expansion of the discrete eigenvalue of can be got through the asymptotic expansion of the discrete eigenvalue of . In our main results, we need to use that is a bounded operator from to . To get that, we add a strong condition on V (i.e., in (1)). We show that is a family of compact operators converging to and obtain the asymptotic expansions of the discrete eigenvalues of by functional calculus. After that, the convergence rate of the smallest discrete eigenvalue of is obtained.
Here is the plan of our work. In Section 2, we recall some results of and define the BirmanSchwinger kernel for . The relationship between the eigenvalues of these two kinds of operators is studied. In Section 3, we first study the asymptotic behavior of the discrete eigenvalues of . Then the convergence rate of the smallest discrete eigenvalue of is obtained. We get the leading term and the estimate of the remainder term of the smallest discrete eigenvalue.
Let us introduce some notations first.
Notation The scalar product on and is denoted by and that on by . , , , denotes the weighted Sobolev space of order r with volume element . The duality between and is identified with the product. Denote . Notation stands for the space of continuous linear operators from to . The complex plane ℂ is slit along positive real axis so that and with are holomorphic there.
2 Some results for
Assume that are the polar coordinates on . Then the condition
implies
in (see [1]).
Now, we recall some results on the resolvent and the Schrödinger group for the unperturbed operator . Let
Denote
For , let denote the multiplicity of as the eigenvalue of . Let , , denote an orthogonal basis of consisting of eigenfunctions of :
Let denote the orthogonal projection in onto the subspace spanned by the eigenfunctions of associated with the eigenvalue , and let denote the orthogonal projection in onto the eigenfunction :
Here , and is the largest integer which is not larger than ν. For , let be the largest integer strictly less than ν. When , set . Define by , if , , otherwise. One has .
The following is the asymptotic expansion for the resolvent .
Theorem 2.1 (Theorem 2.2 [1])
The following asymptotic expansion holds forznear zero with:
Here
witha polynomial inρof degree j:
First, we show that has discrete eigenvalues when λ is large enough. In fact, we need only to show that there exists a function such that .
From the assumption on V, we know that there exists a point such that . Choose small enough such that for all , . For , , , one has
when λ is large enough, one has . This means that has discrete eigenvalues when λ is large enough.
has a continuous spectrum for because exists and equals zero (see [3]). We know that . Hence, from the continuity of a discrete spectrum of , we know that there exists some such that when , has eigenvalues less than zero, and when , . So, has an eigenvalue at the bottom of its spectrum for . In Section 3 (Proposition 3.1), we prove that is simple and the corresponding eigenfunction can be chosen to be positive everywhere. (There are many results about the simplicity of the smallest eigenvalue of the Schrödinger operator without singularity, but there is no result which can be used directly, because the potential we use in this paper has singularity at zero. Theorem XIII.48 [4] can treat the Schrödinger operator with the potential which has singularity at zero, but the positivity of potential is needed. Hence, we give this result.) From the discussion above and the continuity of a discrete spectrum, one has that tends to zero at some λ. The asymptotic behavior of is studied in this paper.
To study the eigenvalues of , we first define a family of BirmanSchwinger kernel operators. Let
Then we have the following result.
(a) Let
Thenis injective fromAtoB, andis injective fromBtoA.
(b) The multiplicity ofαas the eigenvalue ofis exactly the multiplicity ofas the eigenvalue of.
Proof (a) First, we prove that is injective from A to B. Note that if , then
It follows that is injective from A to B.
Next, we show that is injective from B to A. If , then
It follows that is injective from B to A.
(b) From (a), one has . This means that the multiplicity of α as the eigenvalue of is exactly the multiplicity of as the eigenvalue of . □
From the last proposition, we know that there exists onetoone correspondence between the discrete eigenvalues of and the discrete eigenvalues of . Hence, we can study the eigenvalues of first.
3 Asymptotic expansion of the eigenvalues
If and V are defined as above, we show that if has the eigenvalue less than zero, then the smallest eigenvalue of is simple. We use Theorems XIII.44, XIII.45 [4] to prove it.
Proposition 3.1Supposehas an eigenvalue at the bottom of its spectrum. Then this eigenvalue is simple and the corresponding eigenfunction can be chosen to be a positive function.
Proof Let be a smooth nonincreasing function such that if and if . Let . Set , , , . From the proof of Theorem XIII.47 [4], we know that is positivity preserving and acts irreducibly on . Hence, by Theorem XIII.45 [4], if converges to H and converges to in the strong resolvent sense, then is positivity preserving and acts irreducibly on . By Theorems XIII.43 and XIII.44 [4], we can get the result. Since is the core for all and P, and for any , in , then we have the necessary strong resolvent convergence by Theorem VIII.25(a) [5]. This ends the proof. □
Proposition 3.2Assume that. infor any, .
Proof If , then . For any test function , we have . If , then we have in for . It follows because ϕ and belong to . Hence, in . □
Proposition 3.3Assume that. is a compact operator for. Andconverges toin operator norm sense.
Proof For , . Since is a bounded operator from to , and V is a compact operator from to , then is a compact operator on . Using a similar method to that in Proposition 2.2, we can show that and have the same nonzero eigenvalues, and for the same eigenvalue , the multiplicity of as the eigenvalue of and the multiplicity of as the eigenvalue of are the same. Hence, is a compact operator. Because
and if , then in . Hence, in operator norm sense as . This means that is a compact operator. □
Lemma 3.4Suppose, are two bounded selfadjoint operators on a Hilbert spaceH. Set
Proof By the definition of , one has
This ends the proof. □
Lemma 3.5Supposeis a family of compact selfadjoint operators on a separable Hilbert spaceH, andforαnear zero. Set
Then:
(a) is an eigenvalue of, andconverges when. Moreover, if, thenis an eigenvalue of.
(b) Suppose thatis an eigenvalue ofof the multiplicity ofm. Then there aremeigenvalues (counting multiplicity), (), ofnear. Moreover, we can choosesuch that (), is the eigenvector ofcorresponding to (), andconverges as. Ifconverges to, thenis the eigenvector ofcorresponding to.
Proof (a) By the minmax principle, we know that is an eigenvalue of . By Lemma 3.4, one has
It follows that converges to the eigenvalue of .
(b) Because is a compact operator and is an eigenvalue of , then is a discrete spectrum of . Then there exists a constant small enough such that has only one eigenvalue in (). For α small enough, has exactly m eigenvalues (counting multiplicity) in because the eigenvalues of converge to the eigenvalues of by part (a) of lemma. Suppose the m eigenvalues, near , of are , and the corresponding eigenvectors are such that . Let
Then . Let , then . For α near zero, one has
Let . Let be an element in A such that acquires the minimum value. Then we have
and
In the last equality, we use the fact
and
It follows that . Let . Then for because if , and . This ends the proof. □
Here, or 1, , (), are compact operators, and for β near zero. Set
Then, by the minmax principle, is an eigenvalue of . Moreover, if , then is a discrete eigenvalue of because is a compact operator. If is an eigenvalue of of multiplicity m, without loss, we can suppose that . Then there exist exactly m eigenvalues (counting multiplicity), , of near . By Lemma 3.5, we know that there exists a family of normalized eigenvectors of such that , (), and () converge as . Suppose that converge to for all j such that . Then . can be extended to a standard orthogonal basis. Set
Then we have the following.
Lemma 3.6, are given as before. Then the eigenvalue of, () has the following form:
Here
Proof If , then is the discrete eigenvalue of . Suppose that the multiplicity of is m, and suppose that as before. Hence, we can choose small enough such that there is only one eigenvalue in . We know that () converge to . It follows that if δ is small enough, there are exactly m eigenvalues (counting multiplicity) of in for β small. Set
Then
Since
then
Then
In particular,
In the last step, we use that
Similarly, we can get
and
□
First, we study the asymptotic expansion of the smallest eigenvalue of . By Proposition 3.1, we know that is a simple eigenvalue of , and the corresponding eigenfunction can be chosen to be positive. We suppose is a positive eigenfunction corresponding to . Then . Without loss of generality, we can suppose that . Then we can get the following result.
Lemma 3.7Assume that. Set. is defined as above. Thenconverges inwhen. Ifandconverges toϕ, thenϕis the eigenfunction of, and.
Proof By the assumption of , one has . One can check that converges in as by Lemma 3.5. And also, by Lemma 3.5, we know that ϕ is the normalized eigenfunction of corresponding to . ϕ is a positive function since is a positive function. Let , then and u is a positive function because . Then
In the last equality, we use the fact that
with , and . This ends the proof. □
Theorem 3.8Assume that. ϕis defined in Lemma 3.7. If, one of three exclusive situations holds:
Proof
(a) By Theorem 2.1, one has
Then if , we can get in . Because is the simple eigenvalue of , then is the simple eigenvalue of . Since , one has that is monotonous with respect to λ and so is the . Hence, and the eigenvalues of are monotonous with respect to λ. Therefore, we have that is the biggest eigenvalue of . If not, suppose that is an eigenvalue of , then by the continuity and monotony of the eigenvalue of with respect to λ, we know that there exists a constant such that . It follows that is an eigenvalue of . This is contradictory to that is the smallest eigenvalue. By Lemma 3.7, we know the normalized eigenfunction of converges to ϕ. It follows with . Then
Here is the eigenvalue of corresponding to the eigenfunction . By Lemma 3.6, we should compute . Let . From the definition of ϕ, one has . Hence,
In the last equality, we use the fact , which can be obtained by Proposition 3.2. Since , we have by Theorem 3.1 [1]. So, ψ is the ground state of . We also have
Hence,
It follows
So, with . By the Proposition 2.2, one has . It follows
Since , we can get the leading term of is with .
By Lemma 3.7, one has
Then we have
one has . To get the leading term of , we can suppose that . Then, by comparing the leading term, we can get . It follows
By Lemma 3.7, we know that . Using the same argument as before, we can conclude
with . As above, we can get that
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
Both authors contributed equally to the manuscript and read and approved the final manuscript.
Acknowledgements
This research is supported by the Natural Science Foundation of China (11101127,11271110) and the Natural Science Foundation of Educational Department of Henan Province (2011B110014).
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