^{1} Mathematics and Statistics School, Henan University of Science and Technology, No. 263, Luo-Long District, Kai-Yuan Road, Luoyang City, Henan Province, 471023, China
^{2} College of Civil Engineering and Architecture, Zhejiang University, B505 Anzhong Building, 866 Yuhangtang Road, Hangzhou, Zhejiang Province, 310058, China
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Abstract
A family of Schrödinger operators, P(λ)=P0+λV, is studied in this paper. Here P0=−Δ+f(x) with f(x)∼1|x|2 when |x| is large enough and V(x)=O(|x|−2−ϵ) for some ϵ>0. We show that each discrete eigenvalue of P(λ) tends to 0 when λ tends to some λ0. We get asymptotic behavior of the smallest discrete eigenvalue when λ tends to λ0.
In this paper, we consider a family of Schrödinger operators P(λ) which are the perturbation of P0 in the form
P(λ)=P0+λVfor λ≥0
on L2(Rd), d≥2. Here P0=−Δ+q(θ)r2. (r,θ) are the polar coordinates on Rd, and q(θ) is a real continuous function. V≤0 is a non-zero continuous function satisfying
|V(x)|≤C〈x〉−ρ0for some ρ0>2.(1)
Here 〈x〉=(1+|x|2)1/2. Let Δs denote the Laplace operator on the sphere Sd−1. Assume that
Under the assumption on V, we know that P(λ) has discrete eigenvalues when λ is large enough, and each discrete eigenvalue tends to zero when λ tends to some λ0 (see Section 2). We study the asymptotic behaviors of the discrete eigenvalues of
P(λ) 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 H(λ)=−Δ+λV when λ→λ0. λ0 is the value at which some discrete eigenvalue ei(λ) tends to zero. The main method they used in their paper is the Birman-Schwinger technique.
In order to use the Birman-Schwinger technique to P(λ), we need to get the asymptotic expansion of (P0−α)−1 for α near zero, α<0, which was studied by Wang [1]. In this paper, we first show that there exists some λ0 such that when λ>λ0, P(λ) has discrete eigenvalues. Then, we define the Birman-Schwinger kernel K(α) for P(λ) and find that there is one-to-one correspondence between the discrete eigenvalues
of K(α) and the discrete eigenvalues of P(λ). Hence, the asymptotic expansion of the discrete eigenvalue of P(λ) can be got through the asymptotic expansion of the discrete eigenvalue of K(α). In our main results, we need to use that K(α) is a bounded operator from L2 to L2. To get that, we add a strong condition on V (i.e., ρ0>6 in (1)). We show that K(α) is a family of compact operators converging to K(0) and obtain the asymptotic expansions of the discrete eigenvalues of K(α) by functional calculus. After that, the convergence rate of the smallest discrete
eigenvalue of P(λ) is obtained.
Here is the plan of our work. In Section 2, we recall some results of P0 and define the Birman-Schwinger kernel K(α) for P(λ). 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
K(α). Then the convergence rate of the smallest discrete eigenvalue of P(λ) 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 L2(R+;rd−1dr) and L2(Rd) is denoted by 〈⋅,⋅〉 and that on L2(Sd−1) by (⋅,⋅). Hr,s(Rd), r∈Z, s∈R, denotes the weighted Sobolev space of order r with volume element 〈x〉2sdx. The duality between H1,s and H−1,−s is identified with the L2 product. Denote H0,s=L2,s. Notation L(Hr,s,Hr′,s′) stands for the space of continuous linear operators from Hr,s to Hr′,s′. The complex plane ℂ is slit along positive real axis so that zν=eνlnz and lnz=ln|z|+iargz with 0<argz<2π are holomorphic there.
2 Some results for P0
Assume that (r,θ) are the polar coordinates on Rd. Then the condition
Now, we recall some results on the resolvent and the Schrödinger group for the unperturbed
operator P0. Let
σ∞={ν;ν=λ+(d−2)24,λ∈σ(−Δs+q(θ))}.
Denote
σk=σ∞∩[0,k],k∈N.
For ν∈σ∞, let nν denote the multiplicity of λν=ν2−(d−2)24 as the eigenvalue of −Δs+q(θ). Let φν(j), ν∈σ∞, 1≤j≤nν denote an orthogonal basis of L2(Sd−1) consisting of eigenfunctions of −Δs+q(θ):
(−Δs+q(θ))φν(j)=λνφν(j),(φν(i),φν(j))=δij.
Let πν denote the orthogonal projection in L2(Sd−1) onto the subspace spanned by the eigenfunctions of −Δs+q(θ) associated with the eigenvalue λν, and let πν(i) denote the orthogonal projection in L2(Sd−1) onto the eigenfunction φν(i):
Here ν′=ν−[ν], and [ν] is the largest integer which is not larger than ν. For ν>0, let [ν]− be the largest integer strictly less than ν. When ν=0, set [ν]−=0. Define δν by δν=1, if ν∈σ∞∩N, δν=0, otherwise. One has [ν]=[ν]−+δν.
The following is the asymptotic expansion for the resolvent R0(z)=(P0−z)−1.
First, we show that P(λ) has discrete eigenvalues when λ is large enough. In fact, we need only to show that there exists a function ψ∈L2(Rd) such that 〈ψ,P(λ)ψ〉<0.
From the assumption on V, we know that there exists a point x0∈Rd such that V(x0)=infx∈RdV(x). Choose δ>0 small enough such that for all x∈B(x0,δ), V(x)<12V(x0). For ψ∈C0∞(Rd), ∥ψ(x)∥=1, suppψ⊂B(x0,δ), one has
〈ψ,P(λ)ψ〉=〈ψ,P0ψ〉+λ〈ψ,Vψ〉<〈ψ,P0ψ〉+λ2V(x0),
when λ is large enough, one has 〈ψ,P(λ)ψ〉<0. This means that P(λ) has discrete eigenvalues when λ is large enough.
P(λ) has a continuous spectrum [0,∞) for λ≥0 because lim|x|→∞V(x) exists and equals zero (see [3]). We know that σ(P(0))=σ(P0)=[0,∞). Hence, from the continuity of a discrete spectrum of P(λ), we know that there exists some λ0 such that when λ>λ0, P(λ) has eigenvalues less than zero, and when λ≤λ0, σ(P(λ))=[0,∞). So, P(λ) has an eigenvalue e1(λ)<0 at the bottom of its spectrum for λ>λ0. In Section 3 (Proposition 3.1), we prove that e1(λ) 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 e1(λ) tends to zero at some λ. The asymptotic behavior of e1(λ) is studied in this paper.
To study the eigenvalues of P(λ), we first define a family of Birman-Schwinger kernel operators. Let
(b) The multiplicity ofαas the eigenvalue ofP(λ)is exactly the multiplicity ofλ−1as the eigenvalue ofK(α).
Proof (a) First, we prove that |V|1/2 is injective from A to B. Note that if ψ∈A, then
K(α)ϕ=λ−1ϕ
with ϕ=|V|1/2ψ. And if ϕ=0, then
ψ=−λ(P0−α)−1Vψ=λ(P0−α)−1|V|1/2ϕ=0.
It follows that |V|1/2 is injective from A to B.
Next, we show that (P0−α)−1|V|1/2 is injective from B to A. If ϕ∈B, then
(P(λ)−α)ψ=0,with ψ=(P0−α)−1|V|1/2ϕ.
And if ψ=0, then
0=|V|1/2ψ=K(α)ϕ=λ−1ϕ.
It follows that (P0−α)−1|V|1/2 is injective from B to A.
(b) From (a), one has dimA=dimB. This means that the multiplicity of α as the eigenvalue of P(λ) is exactly the multiplicity of λ−1 as the eigenvalue of K(α). □
From the last proposition, we know that there exists one-to-one correspondence between
the discrete eigenvalues of P(λ) and the discrete eigenvalues of K(α). Hence, we can study the eigenvalues of K(α) first.
3 Asymptotic expansion of the eigenvalues
If P0 and V are defined as above, we show that if P0+V has the eigenvalue less than zero, then the smallest eigenvalue of P0+V is simple. We use Theorems XIII.44, XIII.45 [4] to prove it.
Proposition 3.1SupposeP0+Vhas 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 0≤χ(t)≤1 be a smooth nonincreasing function such that χ(t)=1 if |t|<1 and χ(t)=0 if t>2. Let χn(t)=χ(t/n). Set Vn=χn(r)q(θ)r2+V−〈x〉, H0=−Δ+〈x〉, Hn=H0+Vn, H=P=P0+V. From the proof of Theorem XIII.47 [4], we know that e−tH0 is positivity preserving and {e−tH0}∪L∞ acts irreducibly on L2. Hence, by Theorem XIII.45 [4], if Hn converges to H and H−Vn converges to H0 in the strong resolvent sense, then e−tH is positivity preserving and {e−tH}∪L∞ acts irreducibly on L2. By Theorems XIII.43 and XIII.44 [4], we can get the result. Since C0∞(Rd) is the core for all Pn and P, and for any ψ∈C0∞(Rd), Vnψ→(q(θ)r2+V)ψ in L2, then we have the necessary strong resolvent convergence by Theorem VIII.25(a) [5]. This ends the proof. □
Proof If u∈H−1,s, then F0u∈H1,−s. For any test function ϕ∈C0∞(Rn), we have 〈P0F0u,ϕ〉=〈u,F0P0ϕ〉. If 0∉σ∞, then we have limz→0(P0−z)−1=F0 in H−1,s for ℑz>0. It follows 〈u,F0P0ϕ〉=limz→0〈u,(P0−z)−1P0ϕ〉=limz→0〈u,ϕ−z(P0−z)−1ϕ〉=〈u,ϕ〉 because ϕ and P0ϕ belong to H−1,s. Hence, P0F0u=u in H−1,s. □
Proof For α<0, K(α)=|V|1/2(P0−α)−1|V|1/2. Since (P0−α)−1 is a bounded operator from L2(Rd) to H1(Rd), and V is a compact operator from H1(Rd) to L2(Rd), then V(P0−α)−1 is a compact operator on L2(Rd). Using a similar method to that in Proposition 2.2, we can show that V(P0−α)−1 and K(α) have the same non-zero eigenvalues, and for the same eigenvalue e(α), the multiplicity of e(α) as the eigenvalue of V(P0−α)−1 and the multiplicity of e(α) as the eigenvalue of K(α) are the same. Hence, K(α) is a compact operator. Because
Lemma 3.5SupposeT(α)is a family of compact self-adjoint operators on a separable Hilbert spaceH, andT(α)=T0+o(|α|ϵ)forαnear zero. Set
μk(α)=infϕ1,…,ϕksup∥ψ∥=1,ψ∈[ϕ1,…,ϕk]⊥〈ψ,T(α)ψ〉.
Then:
(a) μk(α)is an eigenvalue ofT(α), andμk(α)converges whenα→0. Moreover, ifμk(α)→μk, thenμkis an eigenvalue ofT0.
(b) Suppose thatE0≠0is an eigenvalue ofT0of the multiplicity ofm. Then there aremeigenvalues (counting multiplicity), Ek(α) (1≤k≤m), ofT(α)nearE0. Moreover, we can choose{ϕk(α);1≤k≤m}such that(ϕk(α),ϕj(α))=δkj (1≤k,j≤m), ϕk(α)is the eigenvector ofT(α)corresponding toEk(α) (Ek(α)→E0), andϕk(α)converges asα→0. Ifϕk(α)converges toϕk, thenϕkis the eigenvector ofT0corresponding toE0.
Proof (a) By the min-max principle, we know that μk(α) is an eigenvalue of T(α). By Lemma 3.4, one has
|μk(α)−μk(0)|≤∥T(α)−T0∥=O(|α|ϵ).
It follows that μk(α) converges to the eigenvalue of T0.
(b) Because T0 is a compact operator and E0≠0 is an eigenvalue of T0, then E0 is a discrete spectrum of T0. Then there exists a constant δ>0 small enough such that T0 has only one eigenvalue E0 in B(E0,δ) (={z∈C;|z−E0|<δ}). For α small enough, T(α) has exactly m eigenvalues (counting multiplicity) in B(E0,δ) because the eigenvalues of T(α) converge to the eigenvalues of T0 by part (a) of lemma. Suppose the m eigenvalues, near E0, of T(α) are E1(α),E2(α),…,Em(α), and the corresponding eigenvectors are ψ1(α),ψ2(α),…,ψm(α) such that 〈ψk(α),ψj(α)〉=δkj. Let
Pα=−12πi∮|E−E0|=δ(T(α)−E)−1dE.
Then Pα=∑k=1m〈⋅,ψk(α)〉ψk(α). Let Pα(k)=〈⋅,ψk(α)〉ψk(α), then Pα=∑k=1mPα(k). For α near zero, one has
It follows that ∥Pα(k)ϕk−ϕk∥=O(|α|ϵ). Let ϕk(α)=Pα(k)ϕk∥Pα(k)ϕk∥. Then 〈ϕk(α),ϕj(α)〉=0 for k≠j because Pα(k)Pα(j)=0 if k≠j, and ∥ϕk(α)−ϕk∥≤∥Pα(k)ϕk−ϕk∥+∥(1−1∥Pα(k)ϕk∥)Pα(k)ϕk∥=O(|α|ϵ). This ends the proof. □
Let 0<α1<α2<⋯<αi<⋯<αm and
T(β)=T0+∑i=1mβαi(lnβ)δiTi+Tr(β).
Here, δi=0 or 1, T0≥0, Ti (1≤i≤m), Tr(β) are compact operators, and Tr(β)=O(|β|αm+ϵ) for β near zero. Set
es=infϕ1,…,ϕssup∥ψ∥=1,ψ∈[ϕ1,…,ϕs]⊥〈ψ,T0ψ〉.
Then, by the min-max principle, es is an eigenvalue of T0. Moreover, if es≠0, then es is a discrete eigenvalue of T0 because T0 is a compact operator. If es≠0 is an eigenvalue of T0 of multiplicity m, without loss, we can suppose that es=es+1=⋯=es+m−1. Then there exist exactly m eigenvalues (counting multiplicity), es(β),es+1(β),…,es+m−1(β), of T(β) near es. By Lemma 3.5, we know that there exists a family of normalized eigenvectors {ϕj(β);j=s,s+1,…,s+m−1} of T(β) such that T(β)ϕj(β)=ej(β)ϕj(β), 〈ϕj(β),ϕk(β)〉=δjk (j,k=s,s+1,…,s+m−1), and ϕj(β) (j=s,s+1,…,s+m−1) converge as β→0. Suppose that ϕj(β) converge to ϕj for all j such that ej≠0. Then 〈ϕs,ϕj〉=δsj. {ϕs} can be extended to a standard orthogonal basis. Set
Proof If es≠0, then es is the discrete eigenvalue of T0. Suppose that the multiplicity of es is m, and suppose that es=es+1=⋯=es+m−1 as before. Hence, we can choose δ>0 small enough such that there is only one eigenvalue es in B(es,δ)={z∈C;|z−es|<δ}. We know that ej(β) (j=s,s+1,…,s+m−1) converge to es. It follows that if δ is small enough, there are exactly m eigenvalues (counting multiplicity) of T(β) in B(es,δ) for β small. Set
First, we study the asymptotic expansion of the smallest eigenvalue e1(λ) of P(λ). By Proposition 3.1, we know that e1(λ) is a simple eigenvalue of P(λ), and the corresponding eigenfunction can be chosen to be positive. We suppose u(λ) is a positive eigenfunction corresponding to e1(λ). Then u˜(λ)=|V|1/2u(λ)∈L2(Rd). Without loss of generality, we can suppose that ∥u˜(λ)∥L2(Rd)=1. Then we can get the following result.
Lemma 3.7Assume that0∉σ∞. Setν0=min{ν;ν∈σ∞}. u˜is defined as above. Thenu˜(λ)converges inL2(Rd)whenλ→λ0. Ifν0<1andu˜(λ)converges toϕ, thenϕis the eigenfunction ofK(0), and〈ϕ,|V|1/2Gν0,0πν0|V|1/2ϕ〉≠0.
Proof By the assumption of u˜(λ), one has K(e1(λ))u˜(λ)=λ−1u˜(λ). One can check that u˜(λ) converges in L2(Rd) as λ→λ0 by Lemma 3.5. And also, by Lemma 3.5, we know that ϕ is the normalized eigenfunction of K(0) corresponding to E0. ϕ is a positive function since u˜(λ) is a positive function. Let u=F0|V|1/2ϕ, then P(λ0)u=0 and u is a positive function because |V|1/2u=|V|1/2F0|V|1/2ϕ=K(0)ϕ=λ0−1ϕ. Then
with dν0=−e−12iπν022ν0+1Γ(ν0+1), and Cν0=dν0bν0,0. This ends the proof. □
Theorem 3.8Assume that0∉σ∞. ϕis defined in Lemma 3.7. Ifρ0>6, one of three exclusive situations holds:
(a) Ifσ1=∅, then
e1(λ)=−c(λ−λ0)+o(λ−λ0)
withc=(λ0∥F0|V|12ϕ∥)−2≠0.
(b) Ifν0=1, then
e1(λ)=−cλ−λ0ln(λ−λ0)+o(λ−λ0ln(λ−λ0))
withc=λ0−2〈ϕ,|V|1/2G1,0π1|V|1/2ϕ〉−1≠0.
(c) Ifν0<1, then
e1(λ)=c((λ−λ0)1ν0)+o((λ−λ0)1ν0)
withc=λ0−2〈ϕ,|V|1/2Gν0,0πν0|V|1/2ϕ〉−1≠0.
Proof
(a) By Theorem 2.1, one has
R0(α)=F0+αF1+R0(1)(α),in L(−1,s;1,−s),s>3.
Then if ρ0>6, we can get K(α)=K(0)+|V|1/2(αF1+R0(1)(α))|V|1/2 in L(0,0;0,0). Because e1(λ) is the simple eigenvalue of P(λ), then λ−1 is the simple eigenvalue of K(e1(λ)). Since V≤0, one has that P(λ) is monotonous with respect to λ and so is the e1(λ). Hence, K(e1(λ)) and the eigenvalues of K(e1(λ)) are monotonous with respect to λ. Therefore, we have that λ−1 is the biggest eigenvalue of K(e1(λ)). If not, suppose that a>λ−1 is an eigenvalue of K(e1(λ)), then by the continuity and monotony of the eigenvalue of K(e1(λ)) with respect to λ, we know that there exists a constant λ′<λ such that λ∈σ(K(e1(λ′))). It follows that e1(λ′)<e1(λ) is an eigenvalue of P0+λV. This is contradictory to that e1(λ) is the smallest eigenvalue. By Lemma 3.7, we know the normalized eigenfunction u˜(λ) of K(e1(λ)) converges to ϕ. It follows u˜(λ)=Pλϕ∥Pλϕ∥ with Pλ=−12πi∮|E−E0|=δ(K(e1(λ))−E)−1dE. Then
Here μ(e1(λ)) is the eigenvalue of K(e1(λ)) corresponding to the eigenfunction u˜(λ). By Lemma 3.6, we should compute 〈ϕ,|V|1/2F1|V|1/2ϕ〉. Let ψ=F0|V|1/2ϕ. From the definition of ϕ, one has K(0)ϕ=λ0−1ϕ. Hence,
(P0+λ0V)ψ=P0F0|V|1/2ϕ+λ0VF0|V|1/2ϕ=0.
In the last equality, we use the fact P0F0|V|1/2ϕ=|V|1/2ϕ, which can be obtained by Proposition 3.2. Since ν0>1, we have ψ∈L2(Rd) by Theorem 3.1 [1]. So, ψ is the ground state of P(λ0). We also have
So, μ(α)=λ0−1+c1α+o(|α|1+ϵ) with c1=∥F0|V|12ϕ∥2. By the Proposition 2.2, one has μ(e1(λ))=λ−1. It follows
λ−1=λ0−1+c1e1(λ)+O(|e1(λ)|1+ϵ).
Since λ−1=λ0−1−λ0−2(λ−λ0)+O(|λ−λ0|2), we can get the leading term of e1(λ) is −c(λ−λ0) with c=(λ0∥F0|V|12ϕ∥)−2.
(b) If ν0=1, then
K(α)=K(0)+αlnα|V|1/2G1,0π1|V|1/2+O(α).
By Lemma 3.7, one has
〈ϕ,|V|1/2G1,0π1|V|1/2ϕ〉≠0.
Then we have
μ(α)=λ0−1+c1αlnα+o(α)
with c1=〈ϕ,|V|1/2G1,0π1|V|1/2ϕ〉. As in (a), using μ(e1(λ))=λ−1 and
λ−1=λ0−1−λ0−2(λ−λ0)+O(|λ−λ0|2),
one has −λ−2(λ−λ0)+O(|λ−λ0|2)=ce1(λ)lne1(λ)+o(e1(λ)). To get the leading term of e1(λ), we can suppose that e1(λ)=(λ−λ0)f(λ). Then, by comparing the leading term, we can get f(λ)=1/ln(λ−λ0). It follows
e1(λ)=−cλ−λ0ln(λ−λ0)+o(λ−λ0ln(λ−λ0))
with c=λ0−2〈ϕ,|V|1/2G1,0π1|V|1/2ϕ〉−1.
(c) If ν0<1, one has
K(α)=K(0)+∑0<ν≤1αν|V|1/2Gν,0πν|V|1/2+O(|α|).
By Lemma 3.7, we know that 〈ϕ,|V|1/2Gν0,0πν0|V|1/2ϕ〉−1≠0. Using the same argument as before, we can conclude
μ(α)=λ0−1+c1αν0+o(|α|ν0)
with c1=〈ϕ,|V|1/2Gν0,0πν0|V|1/2ϕ〉. As above, we can get that
e1(λ)=c(λ−λ0)1ν0+o((λ−λ0)1ν0)
with c=λ0−2〈ϕ,|V|1/2Gν0,0πν0|V|1/2ϕ〉−1. □
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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