On the Thomas Fermi ground state in a harmonic potential

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1 Asymptotic Analysis DOI 0.333/ASY IOS Press On the Thomas Fermi ground state in a harmonic potential Clément Gallo a, and Dmitry Pelinovsky b, a Institut de Mathématiques et de Modélisation de Montpellier, Université Montpellier II, Montpellier, France b Department of Mathematics, McMaster University, Hamilton, ON L8S 4K, Canada Abstract. We study non-linear ground states of the Gross Pitaevskii equation in the space of one, two and three dimensions with a radially symmetric harmonic potential. The Thomas Fermi approximation of ground states on various spatial scales was recently justified using variational methods. We justify here the Thomas Fermi approximation on an uniform spatial scale using the Painlevé-II equation. In the space of one dimension, these results allow us to characterize the distribution of eigenvalues in the point spectrum of the Schrödinger operator associated with the non-linear ground state. Keywords: Gross Pitaevskii equation, Thomas Fermi, Bose Einstein, hydrodynamics limit, ground state, Painlevé II. Introduction Recent experiments with Bose Einstein condensates [5] have stimulated new interest in the Gross Pitaevskii equation with a harmonic potential. We take this equation in the form iεu t + ε Δu + x u u u = 0, x R d, t R +,. where the space dimension d is one, two or three, ux, t C is the wave function of the repulsive Bose gas in the mean-field approximation and ε is a small parameter that corresponds to the Thomas Fermi approximation of a nearly compact atomic cloud [7,8]. A ground state of the Bose Einstein condensate is a positive, time-independent solution ux, t = η ε x of the Gross Pitaevskii equation.. More precisely, η ε : R d R satisfies the stationary Gross Pitaevskii equation ε Δη ε x + x η ε x η 3 εx = 0, x R d,. η ε x > 0forallx R d,andη ε has a finite energy E ε η ε, where E ε is given by E ε u = ε u + x u + u4 dx. R d * Supported in part by the project ANR-07-BLAN-050 of the Agence Nationale de la Recherche. ** Supported in part by the NSERC //$ IOS Press and the authors. All rights reserved

2 54 C. Gallo and D. Pelinovsky / On the Thomas Fermi ground state in a harmonic potential For d =, existence and uniqueness of a radially symmetric ground state η ε for a fixed, sufficiently small ε>0 is proven in [3], Theorem., similarly to earlier works of Brezis and Oswald [4] and Aftalion, Alama and Bronsard [] in bounded domains. It is also shown in [3] that η ε x convergesto η 0 x asε 0forallx R,whereη 0 is the Thomas Fermi s compactly supported function { η 0 x = x / for x <, 0 for x >..3 To be precise, [3], Proposition., states that for d =, ε>0sufficiently small, x 0 η ε x Cε /3 exp for x,.4 and 4ε /3 0 x / η ε x Cε /3 x / for x ε /3.5 η ε η 0 C K C K ε,.6 where K is any compact subset of {x R : x < } and C and C K are ε-independent positive constants. The method used by Ignat and Millot in the case d = to prove the existence of a radially symmetric ground state η ε can be extended to the cases d =, 3. The uniqueness of the positive, radially symmetric ground state does not follow from [3] for d = 3, but we can establish it by a different method, according to the following proposition. This proposition is proved in Section 5. Proposition.. The stationary Gross Pitaevskii equation. has at most one positive radial solution in L R d, for any d. Our main goal in this paper is to prove a uniform asymptotic approximation of the ground state η ε on R d, in the limit ε 0, for d =,, 3. At least two attempts have been made in physics literature [3,4] to establish connection between the non-linear ground state η ε for d = and solutions of the Painlevé-II equation 4ν y + yνy ν 3 y = 0, y R..7 This equation arises as the formal limit as ε 0 of the differential equation satisfied by ν ε : 4 ε /3 y ν ε y ε /3 dν εy + yν ε y ν 3 εy = 0, y, ε /3, where ν ε is defined by η ε x = ε /3 ν ε y, y = x..8 ε /3 The convergence of η ε to η 0 as ε 0 suggests that we should consider the Hasting McLeod solution ν 0 of the Painlevé-II equation [], which is the unique solution of.7 such that ν 0 y y / as y + and ν 0 y 0 as y.

3 C. Gallo and D. Pelinovsky / On the Thomas Fermi ground state in a harmonic potential 55 In both papers [3,4], the asymptotic solution η ε is constructed at three spatial scales I: x ε /3, II: x ε /3,+ ε /3 and III: x + ε /3. Solutions of the Painlevé-II equation.7 are used at the intermediate scale II for matching conditions and connection formulas between the WKB solutions at the inner scale I and the Airy function solutions at the outer scale III. The same formal approach is also developed in [9] for approximations of excited states of the stationary Gross Pitaevskii equation in the case d =. We address the problem of uniform asymptotic approximations of the ground state η ε of the stationary Gross Pitaevskii equation. using the Hasting McLeod solution of the Painlevé-II equation.7. Our main result Theorem in Section establishes this approximation on a rigorous level. In the case when d =, we also study eigenvalues of the Schrödinger operator L ε + = ε x + V ε x, V ε x = 3η εx + x, that arises in the linearization of the stationary Gross Pitaevskii equation. at the ground state η ε. We prove in Section 3 that the spectrum of L ε + in L R consists of an infinite sequence of positive eigenvalues {λ ε n} n such that for any fixed integer k, λ ε k, λ ε k μ k ε /3 as ε 0,.9 where μ k is the kth eigenvalue of the Schrödinger operator M 0 = 4 y + W 0 y, W 0 y = 3ν 0y y. We note that M 0 arises in the linearization of the Painlevé-II equation.7 at the Hasting McLeod solution ν 0. Therefore, the scaling transformation.8 leading to the Painlevé-II equation.7 becomes useful for analysis of eigenvalues of the Schrödinger operator L ε +. It is clear from the shape of η ε that the operator L ε + has a double-well potential V ε x with two symmetric minima converging to ±asε 0, while the operator M 0 has a single-well potential W 0 y. These facts explain both the asymptotic correspondence between eigenvalues of L ε + and M 0 and the double degeneracy of each pair of eigenvalues in the asymptotic limit.9. Formal results of the semiclassical theory for the operator L ε + are collected in Section 4. While a different technique is exploited in our previous work [9], the result.9 provides the same kind of asymptotic behavior for the smallest eigenvalue of L ε + as the one we obtained for the lowest eigenvalue of the simplified operator L ε + = ε x + V 0 x, V 0 x = 3η 0x + x. The spectral stability of the ground state in the Gross Pitaevskii equation. is deducted from the analysis of the symplectically coupled eigenvalue problem for Schrödinger operators L ε + and L ε,where L ε = ε x + Ṽεx, Ṽ ε x = η εx + x = ε η ε x η ε x.

4 56 C. Gallo and D. Pelinovsky / On the Thomas Fermi ground state in a harmonic potential Unfortunately, the asymptotic scaling.8 leading to the Painlevé-II equation.7 does not give a correct scaling of the eigenvalues of L ε nor the eigenvalues of the spectral stability problem because the potential Ṽεx is a single well with a nearly flat bottom on the interval [, ], which is mapped to [0, ε /3 ] by the change of variable y = x /ε /3. Analysis of the eigenvalues of the spectral stability problem and construction of excited states of the stationary Gross Pitaevskii equation are two open problems beyond the scope of this article. Notations. If A and B are two quantities depending on a parameter ε belonging to a neighborhood E of 0, Aε Bε indicates that there exists a positive constant C such that Aε CBε for every ε E. Aε ε 0 BεifAε/Bε asε 0. Aε = OBε as ε 0ifAε/Bε remains bounded as ε 0. Let F x be a function defined in a neighborhood of.givenα R,{f m } m N R,andγ>0, the notation F x x α x f m x γm m=0 means that for every M N, F x x α M f m x γm = O x α γm+ m=0 as x and, moreover, that the asymptotic series can be differentiated term by term. We use the following spaces: H R = s 0 H s R, where H s R is the standard Sobolev space. L rr d is the subspace of radially symmetric functions in L R d. Note that if f L rr d, then f L R d = S d r d fr dr, 0 where S d is the surface of the unit sphere in R d. Similarly, B d is the volume of the unit ball in R d.. Uniform asymptotic expansion of η ε In what follows, d =, or 3 and ε > 0 is sufficiently small such that, as it is proved in [3], Theorem., there exists a positive classical solution η ε of ε Δη ε x + x η ε x η 3 εx = 0, x R d..

5 C. Gallo and D. Pelinovsky / On the Thomas Fermi ground state in a harmonic potential 57 Moreover, this ground state η ε is radially symmetric, so that we can define a function ν ε on J ε :=, ε /3 ]by x η ε x = ε /3 ν ε, x R d.. ε /3 Let y = x /ε /3 be a new variable. Notice that y covers once J ε as x covers R +. It is equivalent for η ε to solve. and for ν ε to solve the differential equation 4 ε /3 y ν ε y ε /3 dν εy + yν ε y ν 3 εy = 0, y J ε..3 Let N 0 be an integer. We look for ν ε using the form ν ε y = N ε n/3 ν n y + ε N+/3 R N,ε y, y J ε..4 n=0 Expansion.4 provides a solution of Eq..3 if {ν n } 0 n N and R N,ε satisfy Eqs.5.7. ν 0 solves the Painlevé-II equation 4ν 0 y + yν 0 y ν 3 0y = 0, y R,.5 for n N, ν n solves where 4ν ny + W 0 yν n y = F n y, y R,.6 W 0 y = 3ν 0y y and F n y = n,n,n 3 <n n +n +n 3 =n ν n yν n yν n3 y dν n y 4yν n y, R N,ε solves where 4 ε /3 y R N,ε + ε /3 dr N,ε + W 0 R N,ε = F N,ε y, R N,ε, y J ε,.7 F N,ε y, R = 4yν N + dν N N 3 n= ε n/3 ε 4N+/3 R 3. N n=0 n +n =n 0 n,n N ε n/3 n +n +n 3 =n+n+ 0 n,n,n 3 N N+ ν n ν n R 3 n=n+ ν n ν n ν n3 ε n/3 ν n N+ R

6 58 C. Gallo and D. Pelinovsky / On the Thomas Fermi ground state in a harmonic potential Notice that for 0 n N, ν n y is defined for all y R and does not depend on ε, whereas R N,ε yis a priori only defined for y J ε. Appropriate solutions of system.5.7 enable us to prove the following theorem. Theorem. Let ν 0 be the unique solution of the Painlevé-II equation.5 such that ν 0 y y / as y + and ν 0 y 0 as y. For n, there exists a unique solution ν n of Eq..6 in H R. For every N 0, there exists ε N > 0 and C N > 0 such that for every 0 <ε<ε N, there is a solution R N,ε C J ε of Eq..7 with x R N,ε L J ε C N ε d /3 and x R N,ε H R d, such that the unique radially symmetric ground state of Eq.. in L R d is given by η ε x = ε /3 N n=0 ε n/3 ν n x ε /3 ε /3 + ε N/3+ R N,ε x ε /3, x R d..8 Remark.. For d = 3, the remainder term in.8 may have the same order as the last term in the sum, because of the growth of the upper bound on R N,ε L J ε as ε 0. Remark.. The uniqueness of the ground state is proved in [3] for d = without assuming the radial symmetry of the ground state. The method of Ignat and Millot can be extended to the case d =, but apparently not to d = 3. Proposition. states the uniqueness of a positive, radially symmetric ground state in L R d ford =,, 3. This proposition is proved in Section 5. The proof of Theorem is described in the following three subsections. Notice first that it is sufficient to prove the theorem for an arbitrarily large value of N. Indeed, for every integer N 0 > 0, the result of the theorem for N<N 0 is a direct consequence of the result for N = N 0. Also, for convenience, we shall assume in the sequel that N... Construction of ν n for 0 n N We are looking for a solution ν ε y of Eq..3 that satisfies the following limit as ε 0: ε /3 ν ε ε /3 x ε 0 { x / for x [, ], 0 for x. Therefore, we choose ν 0 y to be the unique solution of the Painlevé-II equation.5 that satisfies the asymptotic behavior ν 0 y y / as y + and converges to zero as y. Existence and uniqueness of this solution are proved by Hastings and McLeod []. Asymptotic behavior of ν 0 y as y ± is described in more details in Theorem.7 of [8]. These results are combined together in the following proposition.

7 C. Gallo and D. Pelinovsky / On the Thomas Fermi ground state in a harmonic potential 59 Proposition. [8,]. The Painlevé-II equation 4ν y + yνy ν 3 y = 0, y R, admits a unique solution ν 0 C R such that ν 0 y y / as y + and ν 0 y 0 as y. Moreover, ν 0 is strictly increasing on R, ν 0 has exactly one zero on R, which is an inflection point of ν 0. The behavior of ν 0 as y is described by ν 0 y = π y /4 e /3 y3/ + O y 3/4 0,.9 y whereas as y +, it is described by ν 0 y y + y/ n=0 where b 0 =, b = 0 and for n 0, b n+ = 4 9n b n b n,.0 y 3n/ n+ m= b m b n+ m n+ n+ l l= m= b l b m b n+ l m. Next, we construct ν n H R forn by induction on n. Forn 0, we consider the following property: where k {,..., n} β = { 5/ if d =, / if d =, 3. ν k H R solves.6 with n replaced by k, ν k y y + yβ k g k,m y 3m/ for some {g k,m } m N, m=0 ν k y 0, y H 0 is empty and, therefore, true by convention. Fix n and assume that H n is true. We are going to construct ν n such that H n is satisfied. We will make use of the following two lemmas, which are proved in Sections 6 and 7. Lemma.. Let W C R such that W L R + and such that there exists C 0, C +, A + > 0 with W x C + x for x A +, W x C 0 for x R and W x 0 for x A +. H n

8 60 C. Gallo and D. Pelinovsky / On the Thomas Fermi ground state in a harmonic potential Let f L R such that x α f L A +, + for some α>0. Let ϕ = x + W f H R. Then, as x +, ϕx = O x α+.. Moreover, if f and W admit asymptotic series fx x + x α + m=0 + c m x γm, W x x v m x γm. x + m=0 for some coefficients {c m } m N,{v m } m N and γ > 0 such that 3/γ is an integer, then ϕ admits an asymptotic series ϕx x + x α+ + m=0 for some coefficients {d m } m N. In particular, as x +, d m x γm.3 ϕ x = O x α+, ϕ x = O x α+3..4 Lemma.. Let W 0 y := 3ν 0 y y, where ν 0y is the solution of the Painlevé-II equation.5 given in Proposition.. Then, W min := inf y R W 0y > 0. From the asymptotic behaviors of ν 0 y asy ±, we infer that W 0 y y as y + and W 0 y y as y..5 Let us consider the operator M 0 := 4 y + W 0 y on L R with the domain, DomM 0 = { u L R: 4u + W 0 u L R }. The Schrödinger operator M 0 arises in the linearization of the Painlevé-II equation at ν = ν 0.The spectrum of M 0 is purely discrete and, thanks to Lemma., it consists of a sequence of strictly positive

9 C. Gallo and D. Pelinovsky / On the Thomas Fermi ground state in a harmonic potential 6 eigenvalues which goes to infinity. If n =, it follows from the choice of ν 0 and from properties.9.0 that and F y y + dy / + y 7/ + m=0 3m + b m+ 3m+/ 3m + y 3m/.6 F y 0. y.7 Thanks to Lemma., we can look for ν solution of.6 with n = intheform ν y = dφy W 0 yy / + ν y, y R, where Φ C R is such that Φy 0ify /, Φy ify. Then, ν has to solve 4 ν y + W 0 y ν y = F y := F y dy / Φy + 4 d dφy, y R..8 dy W 0 yy / From the asymptotic expansions.9.0 of ν 0, we infer that W 0 also admits asymptotic expansions as y ±.Sinceν 0 C R, it follows from.5.8 that F H R. Then, property H follows from Lemma. applied on the one side to ν := M0 F with α = 7/, so that ν y = Oy 9/ asy +, and on the other side to y ν y with α arbitrarily large. Furthermore, if n, we have F n y = 0<n,n,n 3 <n n +n +n 3 =n 3 0<n,n <n n +n =n ν n yν n yν n3 y ν 0 yν n yν n y dν n y 4yν n y. Thanks to H n, all the terms in the right-hand side admit an asymptotic expansion at ±. More precisely, F n y y + yβ+ n + m=0 f n,m y 3m/ for some coefficients {f n,m } m N, whereas F n y 0. y

10 6 C. Gallo and D. Pelinovsky / On the Thomas Fermi ground state in a harmonic potential Since F n C R andn, we deduce that F n H R, and we can define ν n = M 0 F n H R. By Lemma. with γ = 3/ andα = n β, we then have ν n y y + yβ n + m=0 g n,m y 3m/ for some coefficients {g n,m } m N,and ν n y 0 y where we have applied Lemma. with γ = 3/ to the function ν n y. Therefore, H n istrue,which completes the construction by induction of the sequence of solutions {ν n y} n of the inhomogeneous equations.6... Construction of R N,ε In this subsection, we construct a solution R N,ε to Eq..7, such that given the ν n s constructed in Section., expansion.4 provides a solution of Eq..3. The solution R N,ε of Eq..7 is obtained by a fixed point argument. In order to explain the functional framework in which the fixed point theorem will be applied, let us first introduce the functional spaces and L ε = { u L locj ε : ε /3 y d/4 / u L J ε } H ε = { u L ε: ε /3 y d/4 u L J ε and ε /3 y d/4 / W / 0 u L J ε }, endowed with their respective squared norms and ε /3 u ε := u L = ε ε /3 y d/ u dy ε /3 u H := [ ε 4 ε /3 y d/ u + ε /3 y d/ W 0 u ] dy. We are looking for a solution R N,ε y ofeq..7onj ε such that the function R N,ε ε /3 x is regular on R d. As a result, it is convenient for the sequel to introduce the map T ε : L ε L rr d defined for u L ε by T ε u z := u ε /3 ε /3 z,

11 C. Gallo and D. Pelinovsky / On the Thomas Fermi ground state in a harmonic potential 63 which makes the link between functions defined on J ε and radial functions defined on R d,intermsof the variable z = ε /3 x R d. An easy calculation shows that T ε is a bijection from L ε into L rr d, and that for every u L ε, T ε u L R d = Sd ε d /3 u ε..9 Moreover, T ε induces a bijection from H ε into Q ε := { u L R d : and for every u H ε, [ u + W 0 ε /3 ε /3 z u ] } dz< R d T ε u [ Q ε = T ε u + W 0 ε /3 ε /3 z T ε u ] S d dz = R d ε d /3 u H..0 ε Let us rewrite Eq..7 for the remainder term R N,ε y in the operator form M ε R N,ε y = F N,ε y, R N,ε, y J ε,. where M ε is the self-adjoint operator on L ε defined by { M ε := 4 ε /3 y d/+ y ε /3 y d/ y + W 0 y = T ε K ε T ε, Dom M ε = { u L ε: K ε T ε u L R d}. and K ε denotes the Schrödinger operator on L R d, K ε := Δ + W 0 ε /3 ε /3 z. The solution R N,ε of the non-linear equation. will be obtained from the fixed point theorem applied to the map Φ N,ε : R M ε F N,ε, R, which will be shown to be continuous from H ε into itself. First, we shall prove the following lemma. Lemma.3. The operator M ε is invertible, and for every f L ε, M ε f H ε W / min f ε. Proof. Let us consider the continuous, bilinear, coercive form on Q ε defined by [ au, v = u v + W0 ε /3 ε /3 z uv ] dz. R d

12 64 C. Gallo and D. Pelinovsky / On the Thomas Fermi ground state in a harmonic potential By the Cauchy Schwarz inequality and Lemma., for every f L ε, v T ε fvdz R d defines a continuous linear form on Q ε. Thus, by the Lax Milgram theorem [0], there exists a unique ψ Q ε such that for every v Q ε, aψ, v = T ε fvdz. R d Moreover, ψ Q ε is radial and satisfies K ε ψ = T ε f in D R d. Thus, ϕ := T ε ψ H ε DomM ε satisfies M ε ϕ = f. From.0 and a calculation similar to.9, we also check that ϕ H = εd /3 aψ, ψ = εd /3 T ε fψdz = ε S d S d R d f ε ϕ ε W / min f ε ϕ H ε ε /3 ε /3 y d/ fϕdy from which the upper bound on ϕ = M ε f in H ε follows. Next, we prove that R F N,ε, R continuously maps H ε into L ε. We write F N,ε y, R = F N,0 y + G N,ε y, R, where F N,0 = 4yν N + dν N n +n +n 3 =N+ 0 n,n,n 3 N ν n ν n ν n3.3 and N G N,ε = n= 3 ε n/3 N+ n=n+ n +n +n 3 =n+n+ 0 n,n,n 3 N N ν n ν n ν n3 3 n= ε n/3 n +n =n 0 n,n N ν n ν n R ε n/3 ν n N+ R ε 4N+/3 R 3..4

13 C. Gallo and D. Pelinovsky / On the Thomas Fermi ground state in a harmonic potential 65 We first show that F N,0 L ε. Indeed, from the properties of the ν n s, we infer that F N,0 y 0 y and F N,0 also admits an asymptotic expansion as y +, with 4yν Ny + dν Ny = O y β N and if n + n + n 3 = N +, O y 9/ N if d = andn, n, n 3 > 0, ν n ν n ν n3 y = O y 3/ N if d = andn or n or n 3 = 0, O y / N if d notice that n + n + n 3 = N + with 0 n, n, n 3 N implies that at most one of the numbers n, n, n 3 is equal to 0. Since N, we deduce that in any case, F N,0 y = O y 9/ as y +, while F N,0 y y 0. Therefore, for α>0 sufficiently large and ε<, ε /3 ε /3 y d/ F N,0 dy ε /3 α + y dy + y 9 ε /3 y d/ dy. In the case d =, the second integral in the right-hand side is estimated by ε /3 whereas for d, ε /3 y 9 ε /3 y / dy ε 6/3 y 9 ε /3 y d/ dy Therefore in both cases F N,0 L ε and ε /3 z 9 z / dz ε 6/3 / z 9 dz + ε6/3 ε /3 9 ε /3 y 9 dy. / dz, z / F N,0 ε..5 Similarly, the term which does not depend on R in the right-hand side of.4 is O L ε ε /3. Let R H ε. To estimate the linear term in R in the definition of G N,ε, notice that if n + n = n, then n or n is not equal to 0, thus ν n ν n y = Oy asy +. In particular, ν n ν n L Rand ν n ν n R ε ν n ν n L R R ε ν n ν n L RW / min R H ε R H ε..6

14 66 C. Gallo and D. Pelinovsky / On the Thomas Fermi ground state in a harmonic potential In order to estimate the quadratic and cubic terms in the right-hand side of.4, the following lemma will be useful. Lemma.4. Let p =, or 3. There exists a ε-independent constant C>0 such that for every ε>0, if u H ε then u p L ε and u p ε Cε p d /3 u p H ε. Proof. Let u H ε. We have checked in.0 that T ε u Q ε H R d and T ε u H R d T ε u Qε ε d /3 u H ε..7 By Sobolev embeddings, it follows that T ε u L p R d, and u p ε ε d /3 T ε u p L R d = εd /3 T ε u p L p R d ε d /3 T ε u p H R d ε p d /3 u p H ε,.8 where we have also made use of.9 with u replaced by u p. Remark.3. The statement of Lemma.4 can be extended for all values of p for which H R d is continuously embedded into L p R d, that is p for d =, p< for d = and p 3ford = 3. Thanks to Lemma.4, for any integer k, ν k R ε ν k L Rε d /3 R H ε ε d /3 R H ε,.9 whereas for k = 0, ν 0 R ε ν 0 L J εε d /3 R H ε ε d/3 R H ε..30 On the other side, R 3 ε ε d /3 R 3 H,.3 ε thanks to Lemma.4 again. By Lemma.3 as well as bounds.5,.6,.9.3, Φ N,ε R RN,ε 0 H ε /3 + ε /3 R ε H ε + ε N+ d/3 R H + ε ε4n+6 d/3 R 3 H, ε where R 0 N,ε := M ε F N,0.

15 C. Gallo and D. Pelinovsky / On the Thomas Fermi ground state in a harmonic potential 67 In particular, for ε>0 sufficiently small and for some ε-independent constant C>0, Φ N,ε maps the ball B ε := B H ε R 0 N,ε, Cε /3 into itself, where we have used the assumption N. Similarly, there exists an ε-independent constant C >0 such that for every R, R in B ε, Φ N,ε R Φ N,ε R H ε Cε /3 R R H ε. As a result, provided ε is sufficiently small, Φ N,ε is a contraction on B ε. The Fixed Point Theorem ensures that Φ N,ε has a unique fixed point R N,ε B ε. In particular, R N,ε RN,ε 0 H ε /3. ε.3 We next prove that R N,ε satisfies the regularity properties stated in Theorem. The fixed point R N,ε H ε of Φ N,ε has been constructed in such a way that T ε R N,ε H R d solves the equation K ε T ε R N,ε = T ε F N,ε, R N,ε L R d..33 Thanks to Lemma.3 and.5, we obtain R N,ε 0 H F N,0 ε. ε.34 Thus,.3 yields R N,ε ε R N,ε H ε. As a result, from.5,.6,.9.3, we infer F N,ε, R N,ε ε From.33,.36 and.9 we deduce K ε T ε R L N,ε R d = T ε F N,ε, R N,ε L R d ε d /3..37 Next, we use the following lemma, which is proved in Section 8. Lemma.5. K ε LL R d, H R d is uniformly bounded in ε. As a result, we infer from the Sobolev embedding of H R d intol R d thatt ε R N,ε H R d and R N,ε L J ε = T ε R N,ε L R d ε d /3..38 Moreover, a bootstrapping argument shows that T ε R N,ε C R d. As a result, R N,ε C J ε.

16 68 C. Gallo and D. Pelinovsky / On the Thomas Fermi ground state in a harmonic potential.3. ν ε y > 0 for all y J ε We have constructed above {ν n } n 0 and R N,ε in such a way that η ε x:= ε /3 N n=0 = ε /3 ν ε x ε /3 x ε n/3 ν n + ε N/3+ T ε R ε /3 N,ε ε /3 x, x R d,.39 is a classical, radially symmetric solution of Eq... In order to claim that η ε is a ground state, it is sufficient to check that η ε x > 0, for every x R d, which is equivalent to ν ε y > 0, for every y J ε. For every n, ν n L R. Therefore, from.38,.39, since N, we deduce the existence of a constant C>0such that for every y J ε, ν ε y ν 0 y Cε /3. Since ν 0 y increases from 0 to + as y goes from to +, we deduce that for ε, ν ε y ν 0 Cε /3 > 0, y [, ε /3]. Coming back to the variable x, it follows that η ε x > 0, x + ε /3 /..40 It remains to prove that η ε x > 0forall x > + ε /3 /. Assume by contradiction that η ε is not strictly positive on R d. Then, let r ε = inf { r>0, η ε r = 0 } + ε /3 /,, where for convenience, since η ε is radial, we denote η ε x = η ε x. By construction, η ε r ε = 0and η εr ε 0. If η εr ε = 0, then η ε 0, because η ε r satisfies the differential equation r d d r d dr d dr η ε r + ε r + η ε r η ε r = 0. This is a contradiction with.40. Thus, η εr ε < 0. Let r ε := sup { r>r ε, η ε r < 0forr r ε, r } r ε, + ]. Then, for every r r ε, r ε, d r d dr d dr η ε r = rd ε r + η ε r η ε r 0

17 C. Gallo and D. Pelinovsky / On the Thomas Fermi ground state in a harmonic potential 69 and we deduce by integration that for every r r ε, r ε, and r d η εr r d ε η εr ε < 0 r η ε r rε d η εr ε s d ds. r ε.4 The right-hand side in.4 is a negative, decreasing function of r, which implies r ε =+,aswellas a contradiction with the fact that η ε r 0asr +. Therefore η ε r > 0forallr R Spectrum of the Schrödinger operator L ε + in the case d = Consider the Schrödinger operator L ε + = ε x + V ε x, V ε x = 3η εx + x associated with the stationary Gross Pitaevskii equation. linearized at the ground state η ε.itisa self-adjoint operator on L R. Since the potential V ε x is confining in the sense of V ε x + as x, L ε + has compact resolvent and a purely discrete spectrum. By Sturm Liouville theory, the eigenvalues of L ε +, denoted {λ ε n} n sorted in increasing order are simple. Moreover, thanks to the even symmetry of V ε on R, the eigenfunctions of L ε + corresponding to λ ε n are even resp. odd in x if n is odd resp. even. If λ is an eigenvalue of L ε + and ϕ L R is a corresponding eigenfunction, we define a function v L ε by x ϕx = v, x R +. ε /3 Let us denote W ε y = 3ν εy y. Then, ϕ L R is an even eigenfunction of L ε + corresponding to the eigenvalue λ if and only if v L ε satisfies the differential equation 4 ε /3 y / y ε /3 y / y + W ε y vy = ε /3 λvy, y J ε, 3. and the Neumann boundary condition ϕ 0 = ε /3 ε /3 y / v y y=ε /3 = 0. NC Similarly, ϕ L R is an odd eigenfunction of L ε + corresponding to the eigenvalue λ if and only if v L ε satisfies 3. and the Dirichlet boundary condition ϕ0 = v ε /3 = 0. DC

18 70 C. Gallo and D. Pelinovsky / On the Thomas Fermi ground state in a harmonic potential As a result, the eigenvalues of L ε + are directly related to the eigenvalues of the two self-adjoint operators on L ε, ˇM ε and M ε,where { ˇM ε := 4 ε /3 y / y ε /3 y / y + W ε y, Dom ˇM ε = { v L ε: ˇM ε v L ε and v satisfies NC } and M ε is defined similarly by replacing NC by DC in the definition of the domain. Namely, if we denote {ˇμ ε n} n resp. { μ ε n} n the eigenvalues of ˇM ε resp. M ε sorted in increasing order, then for every n, ˇμ ε n = ε /3 λ ε n and μ ε n = ε /3 λ ε n. As ε 0, the eigenvalue problems 3. for the operators ˇM ε and M ε formally converge to the eigenvalue problem for the Schrödinger operator M 0 defined after Lemma., 4 y + W 0 y vy = μvy, y R, whereμ = ε /3 λ. By the discussion below Lemma., the purely discrete spectrum of M 0 in L R consists of an increasing sequence of positive eigenvalues {μ n } n. We shall prove that the eigenvalues of L ε + converge to the eigenvalues of M 0 as ε 0, according to the following result. Theorem. The spectrum of L ε + consists of an increasing sequence of positive eigenvalues {λ ε n} n such that for each n, λ ε n lim ε 0 ε /3 = lim ε 0 λ ε n ε /3 = μ n. 3. Proof. We prove only the convergence of μ ε n = λ ε n /ε/3 to μ n,foreveryn. The proof of the convergence of ˇμ ε n = λ ε n /ε/3 to μ n is identical. Denote by, and the scalar product and the norm in L R, and by, ε and ε the scalar product and the norm in L ε.ifu, v L R, u v means that u, v = 0, whereas if u, v L ε, u ε v means that u, v ε = 0. We denote by R ε v = Qε v, v v ε the Rayleigh quotient for the operator M ε,whereq ε denotes the corresponding bilinear form Q ε u, v = 4 ε /3 y / W ε y y u y v + uyvy dy, ε /3 y / J ε defined for u, v H ε. Similarly, Rv = Qv, v v

19 C. Gallo and D. Pelinovsky / On the Thomas Fermi ground state in a harmonic potential 7 denotes the Rayleigh quotient for M 0,whereQ is the corresponding bilinear form Qu, v = 4 y u y v + W 0 yuyvy dy R defined for u, v {u H R: W / 0 u L R}. Let ũ ε n resp. u n denote an eigenfunction of M ε resp. M 0 corresponding to the eigenvalue μ ε n resp. μ n, normalized by ũ ε n ε = resp. u n =. The eigenvalues of M 0 are given by the Max Min principle: μ n = inf Rv, 3.3 v DomM 0 v u,...,u n whereas the eigenvalues of M ε are similarly given by μ ε n = inf R ε v. v Dom M ε v εũ ε,...,ũε n 3.4 Let us fix δ 0, /3. Let Φ C R be an non-decreasing function such that Φ 0onR and Φ on [, +. For ε>0sufficiently small, we also define χ ε Cc Rby x + ε /3 ε /3 x χ ε x = Φ ε /3 ε δ Φ ε /3 ε δ, such that χ ε is even, χ ε on[ ε δ, ε δ ] and Suppχ ε [ ε /3 the following properties: i n μ ε n = μ n + O ε /3 δ, ii n for every k n +, χ ε u k, ũ ε n ε = O ε /3 δ/, iii n for every k n, χ ε ũ ε k, u n = O ε /3 δ/, iv n inf χ ε ũ ε n cu n = O ε /3 δ/, c R v n inf χ ε u n cũ ε n ε = O ε /3 δ/, c R, ε /3 ]. We shall prove recursively where for n =, iii and v have to be understood as empty properties. Let us fix n and assume that G k istrueforeveryk {,..., n } for n =, this condition is empty, therefore true by convention. The proof of G n is then divided in five steps. Step Upper bound on μ ε n. First, we shall prove that n R ε vn ε = μn + O ε /3 δ, where vn ε = χ ε u n k= G n χε u n, ũ ε k εũε k. 3.5 Then, thanks to 3.4, since v ε n Spanũ ε,..., ũε n ε L ε by construction, 3.5 yields μ ε n μ n + O ε /3 δ. 3.6

20 7 C. Gallo and D. Pelinovsky / On the Thomas Fermi ground state in a harmonic potential From i k and ii k, which are satisfied for k n thanks to the recursion assumption, we have R ε vn ε Q ε χ ε u n, χ ε u n n = k= με k χ εu n, ũ ε k ε χ ε u n ε n k= χ εu n, ũ ε k ε Next, ε /3 χ ε u n / χ ε = εu n dy ε /3 / ε /3 y / = + O ε /3 δ ε δ u ε δ n dy + = Qε χ ε u n, χ ε u n + Oε /3 δ χ ε u n ε + Oε /3 δ. 3.7 ε δ y ε /3 / χ εu n dy. 3.8 ε /3 y / The last term in the right-hand side of 3.8 is estimated as follows χ εu n ε δ y ε /3 / ε /3 y dy u / y ε δ n dy exp ε δ ε /3, 3.9 where we have used the following lemma. Lemma 3.. For every m, there exists a constant C m > 0 such that for every y R, u m y C m exp y 3.0 and u my C m y + exp y. 3. Proof. Since W 0 y + as y, we can fix b n > 0 such that inf{w 0 y: y b n } > 4 + μ n. Then, 4 y + W 0 y μ n e y = W 0 y μ n 4 e y 0, y >b n. Since u n solves the eigenvalue problem 4 y + W 0 y μ n un = 0, y R, thanks to [], Corollary.8, there exists C>0 such that u n y Ce y, y b n +. Bound 3.0 follows, since u n DomM 0 H R L R. Then, from the differential equation M 0 u n = μ n u n and thanks to the asymptotic behavior of W 0,weinfer u ny = μ n u n y W 0 yu n y y + e y, y R. 3. 4

21 C. Gallo and D. Pelinovsky / On the Thomas Fermi ground state in a harmonic potential 73 By integration of 3. between and y, we deduce, for y<0, u ny = y u nsds y + e y. The same kind of estimate is obtained for y>0byintegration of 3. between y and +, which provides 3. and completes the proof of Lemma 3.. Using Lemma 3. again, as well as the normalization of u n, we infer that ε δ ε δ u n dy = + O ε / From 3.8, 3.9 and 3.3, we deduce that χ ε u n ε = + O ε /3 δ. 3.4 On the other side, Q ε χ ε u n, χ ε u n = [4 ε /3 y / y χ ε u n + W ε χ ε u n ] dy J ε ε /3 y / = 4 ε /3 y / χ ε u n dy + 8 ε /3 y / χ εχ ε u nu n dy J ε J ε ε δ + 4 ε /3 y / u ε δ n dy + 4 ε /3 y / χ εu ε δ y ε /3 n dy / ε δ/ W ε u n + ε δ/ ε /3 y dy + W ε χ ε u n ε dy. 3.5 / δ/ y ε /3 / ε /3 y / The first two integrals in the right-hand side of 3.5 are Oε /3, because u n H R, and χ ε L J ε ε /3 max { ε /3 y / : y Supp χ ε } 3/. The fourth and last integrals in the right-hand side of 3.5 are also Oε /3, thanks to Lemma 3.. From Lemma 3., we also infer that ε δ ε /3 y / u ε δ n dy = + O ε /3 δ ε δ + u ε δ n dy = u n dy + O ε /3 δ. 3.6

22 74 C. Gallo and D. Pelinovsky / On the Thomas Fermi ground state in a harmonic potential From Theorem and from the decay properties of the function ν n for n provided in H n, we deduce that ν ε = ν 0 + ε /3 r ε,wherer ε = O L R and ν 0 r ε = O L R as ε 0. As a result, W ε W 0 = 3ν ε ν 0 L R, and W ε W 0 L R ε / Then, since W 0 y = Oy asy ±, W ε ε /3 y W / 0 ε /3 δ. 3.8 L ε δ/,ε δ/ As a result, using once more Lemma 3., ε δ/ ε δ/ W ε u + n ε /3 y dy = W / 0 u n dy + O ε /3 δ. 3.9 Finally, we get from 3.7, , 3.9 and the estimates on the other term in the right-hand side of 3.5: R ε v ε n = Run + O ε /3 δ = μ n + O ε /3 δ, 3.0 which completes the proof of 3.5 and of its corollary 3.6. Step Asymptotic behavior of the eigenfunction ũ ε n. Property i n will be obtained as a consequence of 3.6 and of the converse inequality μ n μ ε n + O ε /3 δ. The proof of the latter inequality is delivered in Step 3. The proof uses the following properties of the eigenfunction ũ ε n corresponding to the nth eigenvalue μ ε n of M ε. Lemma 3.. There exists a constant C n > 0 such that for every y J ε and ε>0 sufficiently small, ũ ε ny C n e y, 3. whereas ũ ε n y C n y + e y if y 0, C n y + e y + exp ε /3 if 0 <y ε /3 4, C n exp ε /3 /4 if ε /3 <y ε /3. ε /3 y / 3.

23 C. Gallo and D. Pelinovsky / On the Thomas Fermi ground state in a harmonic potential 75 Proof. In order to prove 3., we come back to the eigenfunction x ϕ ε nx =ũ ε n ε /3 of L ε + corresponding to the eigenvalue λ ε n = με nε /3.Since ũ ε n H = ε Qε ũ ε n, ũ ε n = μ ε n ũ ε n ε = με n, it follows from 3.6 and Lemma.4 that for ε sufficiently small, ϕ ε n L R = ũ ε n L J C ε μ ε n C μ n + O ε /3 δ c n, 3.3 where c n > 0isanε-independent constant. Since W 0 y y as y ±, we can fix a n large enough such that inf{w 0 y : y a n } > 4 + μ n. Then, using 3.7 and 3.6, we obtain, for x < a n ε /3 and for ε small enough, ε x + x + 3ηε ε /3 μ ε n exp x x = ε /3 ε /3 4x + W ε ε /3 ε /3 μ ε n exp x ε /3 4 + inf { } W 0 y: y a n μn + O ε /3 δ exp x On the other side, ϕ ε n solves the differential equation ε x + x + 3ηε ε /3 μ ε n ϕ ε n = ε /3 ε /3 Thus, ε x + x + 3ηε ε /3 μ ε n ψ ε n± 0, x < a n ε /3 /, 3.6 where ψn±x ε = c n exp a n x ± ϕ ε nx. Moreover, from 3.3, we get ε /3 ψ ε n± ± an ε /3 / 0. As a result, since for ε small enough, we also have like in 3.4 x + 3ηε ε /3 μ x ε n = ε /3 W ε μ ε n ε /3 ε /3 inf { W 0 y: y a n } μn + O ε /3 δ > 0, 3.7

24 76 C. Gallo and D. Pelinovsky / On the Thomas Fermi ground state in a harmonic potential the maximum principle ensures that ψ ε n±x 0, x < a n ε /3 /, which is equivalent to ϕ ε nx c n exp In terms of ũ ε n, it means that a n x ε /3, x < a n ε /3 /. ũ ε ny c n e an e y, a n y ε / On the other side, for x + a n ε /3 / and for ε sufficiently small, we obtain like in 3.4 ε x + x + 3ηε ε /3 μ ε n exp x x = ε /3 ε /3 4x + W ε ε /3 μ ε n ε /3 exp x ε /3 x ε /3 4 + W 0 4ε /3 x /ε /3 ε /3 W 0 x /ε /3 + W ε x /ε /3 W 0 x /ε /3 μ n + O ε /3 exp x ε / Thus, exp x ε /3 is a positive, continuous supersolution of ε x + x + 3ηε ε /3 μ ε n ϕ = 0 in {x: x > + a n ε /3 / }. From a slightly modified version of [], Corollary.8, we deduce that ϕ ε nx c n exp + a n x ε /3, x + a n + ε /3 /. More precisely, the constant c n e an+ above has been chosen in such a way that the inequality holds for x = + a n + ε /3 /, and the result in [] ensures that then the inequality holds for any x such that x + a n + ε /3 /.Intermsofũ ε n, it means that ũ ε ny c n e an+ e y, y a n Then, 3. follows from 3.3, 3.8 and We next prove 3.. From 3. and the differential equation M ε ũ ε n = μ ε nũ ε n, we infer that for every y J ε, y ε /3 y / y ũ ε n y C n e y 4 ε /3 y / μ n + sup y R W 0 y y + O ε /3 δ, 3.3 y

25 C. Gallo and D. Pelinovsky / On the Thomas Fermi ground state in a harmonic potential 77 where we have also used 3.6 and 3.7. The estimate 3. in the case y<0directly follows by integration of 3.3 between and y: ũ ε y n ε /3 y /ũε y y n = y ε /3 y / y ũ ε n sds y + e y. 3.3 As for the case 0 <y< ε /3, integration of 3.3 between y and ε /3 gives ε /3 y /ũε y ũε ε /3 n n y + e y, 3.33 which provides thanks to the triangular inequality ũ ε y n y + e y + ũε ε / n Using basic integration, we also have ε ũ ε /3 ε n ũ ε /3 n 4 ε /3 / ũε s ũ ε = n n ε /3 / ε /3 s / ε /3 /4 ds + ũε n ε /3 / ε /3 / ε /3 /4 ds. ε /3 s / 3.35 Since the last integral in the right-hand side of 3.35 is bounded from below by ε /3 4, we deduce from 3.35, 3.33 and 3. that ũε ε /3 n exp ε / Combining 3.36 and 3.34, we get 3. in the case when 0 <y< ε /3. Finally, we consider the case when ε /3 <y<ε /3. Integration of 3.3 between ε /3 and y yields ũ ε y n ũε ε /3 y ε /3 y / + n exp ε /3 y / ε /3 4 + ε /3 exp ε /3 / ε /3 s + e s ds ε /3 s / ε /3 ε /3 / ds ε /3 s / exp ε /3 /4, 3.37 ε /3 y /

26 78 C. Gallo and D. Pelinovsky / On the Thomas Fermi ground state in a harmonic potential where we have also used This completes the proof of 3. and the proof of Lemma 3.. Step 3 Lower bound on μ ε n and proof of i n. In order to show that i n holds, we next prove the converse inequality μ n μ ε n + O ε /3 δ, 3.38 which will be deduced from 3.3 and R ṽn ε = μ ε n + O ε /3 δ, where ṽn ε = χ ε ũ ε n χε ũ ε n, u k uk In order to prove 3.39, we proceed similarly as for the proof of 3.5. First, since iii k is assumed to be satisfied for k n, R ṽn ε Qχ ε ũ ε = n, χ ε ũ ε n n k= μ k χ ε ũ ε n, u k χ ε ũ ε n n k= χ εũ ε n, u k n k= = Qχ εũ ε n, χ ε ũ ε n μ n χ ε ũ ε n, u n + Oε /3 δ χ ε ũ ε n χ ε ũ ε n, u n + Oε /3 δ Then, thanks to Lemma 3. and the normalization of ũ ε n, χ ε ũ ε n = ε /3 / ε /3 / χ ε ũ ε n dy = + O ε /3 δ ε δ = + O ε /3 δ J ε ũ ε n = + O ε /3 δ. ε δ ũ ε n ε /3 y / dy + ε δ y ε /3 / ε /3 y / dy + O ε /3 χ ε ũ ε n dy 3.4 Similarly, using Lemma 3. and 3.8 and proceeding as in 3.5, we get Q χ ε ũ ε n, χ ε ũ ε + n = 4 y χε ũ ε n = 4 + χ ε ũ ε n + W 0 χ ε ũ ε n dy dy χ εχ ε ũε n ũε n dy O ε /3 δ ε δ ε /3 y / ũ ε ε δ n dy + 4 χ ε δ y ε /3 ε ũ ε n dy /

27 C. Gallo and D. Pelinovsky / On the Thomas Fermi ground state in a harmonic potential 79 ε δ/ + W 0 ũ ε ε δ/ n dy + ε δ/ y ε /3 / = 4 + O ε /3 δ J ε ε /3 y / ũ ε n dy ε δ/ W ε ũ ε + n ε δ/ ε /3 y dy + / O ε /3 δ W 0 χ ε ũ ε n dy = Q ε ũ ε n, ũ ε n + O ε /3 δ = μ ε n + O ε /3 δ. 3.4 In order to deduce 3.39 from 3.40, it remains to estimate the scalar product χ ε ũ ε n, u n. Notice that in the case when n =, this term does not exists, and there is then nothing to do. From iv n,there exists c n R such that χ ε ũ ε n c n u n ε /3 δ/ Then, by triangular inequality, and thanks to 3.4 for n replaced by n, whereas c n c n u n χ ε ũ ε n + χ ε ũ ε n ε /3 δ/, 3.44 c n χ ε ũ ε n, u n χ ε ũ ε n, c n u n χ ε ũ ε n + χ ε ũ ε ε /3 y n, ũ ε / n + ũ ε n, ũ ε n ε ε /3 δ/, 3.45 where the first term in the right-hand side of 3.45 has been estimated thanks to the Cauchy Schwarz inequality, 3.43 and 3.4. The second term has been estimated thanks to Lemma 3. for ũ ε n and for ũ ε n, and the last one is equal to 0. We deduce from 3.44 and 3.45 that χε ũ ε n, u n = O ε /3 δ/ Then, 3.39 and 3.38 follow from and Property i n is a direct consequence of 3.38 and 3.6. Step 4 Proof of ii n and iv n. From the definition of ṽ ε n in 3.39, it is clear that ṽ ε n Spanu,..., u n. Thus, ṽ ε n can be decomposed as ṽ ε n = c ε nu n + w ε n, where c ε n R and w ε n Spanu,..., u n. 3.47

28 80 C. Gallo and D. Pelinovsky / On the Thomas Fermi ground state in a harmonic potential From 3.39 and i n,wehave μ n + O ε /3 δ = μ ε n + O ε /3 δ = R ṽn ε c ε = n μ n + Qwn, ε wn ε c ε n + wn ε cε n μ n + wn ε μ n+ c ε n + wn ε. It follows that μ n+ μ n wn ε ε /3 δ ṽn ε Thanks to the definition of ṽn ε in 3.39, property iii k for k n as well as 3.46, ṽ n ε χ ε ũ ε n ε /3 δ/ On the other side, 3.4 ensures that χ ε ũ ε n asε 0. As a result, ṽn asε ε 0, and 3.48 implies ε /3 δ/ w ε n Moreover, from Lemmas 3. and 3., we infer that for any k, ε δ χ ε ũ ε n, u k = χ ε ũ ε nu k dy + χ ε ũ ε nu k dy ε δ ε δ y ε /3 / = + O ε /3 δ J ε χ ε u k ũ ε n ε /3 y / dy + O ε / From 3.50 and 3.5 we deduce in particular that for every k n +, O ε /3 + χ ε u k, ũ ε n ε + O ε /3 δ = χ ε ũ ε ṽε n, u k = n, u k = w ε n, u k = O ε /3 δ/, 3.5 which proves ii n. Then, iv n is a consequence of the triangular inequality, 3.49 and 3.50: χ ε ũ ε n c ε nu n χ ε ũ ε n ṽn ε + wn ε ε /3 δ/. Step 5 Proof of iii n and v n. Like in 3.47, we decompose v ε n as v ε n = c ε n ũ ε n + w ε n, where c ε n R and w ε n Span ũ ε,..., ũ ε n ε. From 3.5 for n replaced by n andi n,wehave μ ε n + O ε /3 δ = μ n + O ε /3 δ = R ε v ε c ε n = n μ ε n + Q ε w n, ε w n ε c ε n + w n ε ε cε n μ ε n + w n ε ε μ ε n c ε n + w n ε. ε

29 C. Gallo and D. Pelinovsky / On the Thomas Fermi ground state in a harmonic potential 8 Using i n and i n, it follows that μn μ n + O ε /3 δ w n ε ε = μ ε n μ ε n w n ε ε ε/3 δ v ε n ε Thanks to the definition of v ε n given by 3.5 and property ii k for k n, v ε n χ ε u n ε ε /3 δ/ Thanks to 3.4 for n replaced by n, χ ε u n ε asε 0, thus v ε n ε asε 0. As a result, we deduce from 3.53 that w ε n ε ε /3 δ/. Then, for every k n, weget χε ũ ε k, u n + O ε /3 δ = χ ε u n, ũ ε k ε = vn, ε ũ ε k ε = w n, ε ũ ε k ε using similar arguments as in the derivation of 3.5. Moreover, χε u n, ũ ε k ε = + O ε /3 δ χε ũ ε k, u n 3.55 = O ε /3 δ/, 3.56 ε δ y ε /3 / χ ε u n ũ ε k dy χ ε u n ũ ε k + dy ε δ y ε /3 / ε /3 y / = + O ε /3 δ χ ε ũ ε k, u n + O ε /3, 3.57 where the two integrals in the right-hand side of 3.57 have been estimated thanks to the Cauchy Schwarz inequality, Lemma 3. and the normalization condition ũ ε k ε =. The combination of 3.56 and 3.57 completes the proof of iii n. Then, v n follows from the triangular inequality, 3.54 and 3.55: χ ε u n c ε n ũ ε n ε χ ε u n v ε n ε + w ε n ε ε /3 δ/. It completes the proof of G n, and therefore the proof of Theorem. 4. Semi-classical limit for eigenvalues of L ε + We list here formal results of the semi-classical theory that describe the distribution of eigenvalues of L ε +. We will show that the standard Bohr Sommerfeld quantization rule does not give the correct asymptotic behavior of the eigenvalues of L ε + as ε 0 because the potential V ε x depends on ε. Nevertheless, the Bohr Sommerfeld quantization rule gives the correct scaling Oε /3 in agreement with the asymptotic limit 3. in Theorem.

30 8 C. Gallo and D. Pelinovsky / On the Thomas Fermi ground state in a harmonic potential Eigenvalue problem for operator L ε + can be rewritten in the form x + ε V ε x ux = ε λux, x R. 4. By properties of η ε following from Theorem, the potential V ε x has the properties: V ε x C R foranysmallε>0, lim ε 0 V ε x = V 0 x, where V 0 CRisgivenby V 0 x = { x, x, x, x, V ε x takes its absolute minimum at ±a ε for any small ε 0anda ε asε 0, V ε x + as x for any small ε 0. If V ε x is replaced by V 0 x, the eigenvalue problem 4. takes a simplified form x + ε V 0 x ux = ε λux, x R, 4. which describes the eigenvalues of the operator L ε + mentioned in the Introduction. As it is well known see a recent review in [5], the eigenvalues of the Schrödinger operator x + ε V x, with a smooth, ε-independent double well potential V x, are twice degenerate in the semi-classical limit ε 0. Namely, the eigenvalues are grouped by pairs. In each pair, the two eigenvalues are exponentially close one from another as ε 0. The asymptotic distribution of these pairs of eigenvalues is determined by the Bohr Sommerfeld quantization rule. Let us try to apply the Bohr Sommerfeld quantization rule to the eigenvalue problems 4. and 4. for the operators L ε + and L ε +, in spite of the fact that this rule was proved rigorously by Fedoryuk [6] only for a class of ε-independent, analytic potentials. Since neither 4. nor 4. satisfies assumptions of the main theorem in [6], this application is purely formal. According to the standard Bohr Sommerfeld rule, the consequent eigenvalues λ ε n and λ ε n of the Schrödinger equation 4. with the double-well potential V ε x would be given asymptotically by x ε + λ x ε λ λ Vε xdx επ n, as ε 0, for fixed n, 4.3 where x ε ±λ are the roots of V ε x = λ on R +, such that 0 <x ε λ < <x ε +λ <. Let us use the scaling y = x ε /3, V ε x = ε /3 W ε y, λ = ε /3 μ, 4.4 where W ε y = 3ν εy y and μ is a new eigenvalue. The Bohr Sommerfeld rule is rewritten in an equivalent form by y ε + μ y ε μ μ Wε y ε /3 y dy πn, as ε 0, for fixed n,

31 C. Gallo and D. Pelinovsky / On the Thomas Fermi ground state in a harmonic potential 83 where y ε ±μ are the roots of W ε y = μ on R, such that <y ε μ < 0 <y ε +μ <. Takingthe limit ε 0forafixedn, we obtain y+ μ y μ μ W0 ydy πn for fixed n, 4.5 where W 0 y = 3ν0 y y and y ±μ are the roots of W 0 y = μ on R. The new expression is the Bohr Sommerfeld quantization rule for the Schrödinger operator M 0 = 4 y + W 0 anditisonlyvalid for large n. Therefore, the Bohr Sommerfeld quantization rule 4.3 does not recover the statement of Theorem correctly. Meantime, it still implies that the eigenvalues λ ε n and λε n for a fixed n are scaled as Oε /3 asε 0. The discrepancy of the Bohr Sommerfeld rule is explained by the fact that the smooth potential V ε x in the eigenvalue problem 4. depends on ε. Note that the limit ε 0 can be computed exactly for the simplified eigenvalue problem 4. thanks to the scaling transformation 4.4. In this case, the limiting formula 4.5 holds with W 0 y replaced by y for y 0and y for y 0, so that y μ = μ and y + μ = μ/. In other words, 0 μ μ/ μ + y dy + μ y dy πn for fixed n 0 and the computations of integrals gives μ n πn /3, in agreement with the behavior On /3 of eigenvalues of the Schrödinger operator with a linearly growing potential as y [7]. Therefore, the Bohr Sommerfeld quantization rule suggests that the eigenvalues { λ ε n} n of the simplified operator L ε + considered in our previous work [9] satisfy the asymptotic limit λε lim n ε 0 ε /3 = lim ε 0 λε n ε /3 = πn /3 for fixed n. 4.6 However, the justification of the asymptotic limit 4.6 cannot rely on the work of Fedoryuk [6] because the ε-independent potential V 0 x in the simplified eigenvalue problem 4. is continuous but not C on R. 5. Proof of Proposition. For a radial function ux = u x solution to.,. can be rewritten as ε r d d r d u + r u u = 0, r 0, u 0 = dr Let u, v L R d be two radial positive solutions of 5.. Up to a change of u and v, we assume u0 v0. Let ρ = u/v. Then, a straightforward calculation shows that ε d dr v r d ρ = r d v 4 ρ ρ.

32 84 C. Gallo and D. Pelinovsky / On the Thomas Fermi ground state in a harmonic potential If u0 = v0, since u 0 = v 0 = 0, we have u v. Let us assume by contradiction u0 <v0, that is, ρ0 <. Then r vr r d ρ r is strictly decreasing in a neighborhood of 0, and therefore, since ρ 0 = 0, we infer ρ r < 0forr>0 sufficiently small. Let r 0 = inf { r>0, ρ r = 0 }. ρ is decreasing on 0, r 0, and since for every r, 0<ρr <ρ0 <, we infer that v 4 ρρ < 0 on that interval. As a result, r vr r d ρ r is decreasing on 0, r 0. Moreover, if r 0 was finite, we would have vr 0 r0 d ρ r 0 < 0, which would be a contradiction with the definition of r 0. Thus r 0 =+, such that both ρ and vr r d ρ r are decreasing on R +. We deduce ρ ρ = In particular, the integral dr <+ vr rd converges. On the other side, since v L R d, 0 vr r d dr<+. ρ rdr v ρ dr <0. vr rd Thanks to the Cauchy Schwarz inequality, it turns out that = dr / vr r d dr which gives contradiction. Thus, ρ0 = andu v. / vr r dr <, d 6. Proof of Lemma. Let α> be like in the assumption of the lemma, and A = x α f L R + <. We first prove. by contradiction. We proceed as follows. We suppose that. is not true. Namely, we make the assumption ϕx O x α+, x α f L A +, +. G α If α>, we prove that G α implies G α, such that after a finite number of steps, G α implies G α for some α 0, ]. On the other side, we show that for 0 <α, G α yields a contradiction. If. is not true, then, up to a change of f and ϕ into f and ϕ, there exists a sequence x n n n0 where n 0 >A, such that x n, x n A + and x α nw x n ϕx n >n.

33 C. Gallo and D. Pelinovsky / On the Thomas Fermi ground state in a harmonic potential 85 Then, x α nϕ x n = x α nw x n ϕx n x α nfx x α nw x n ϕx n A>n A. For n n 0 >A,wedefine y n = sup { y>x n, x x n, y, x α W xϕx A>n A/ }. By continuity of W and ϕ, foreveryn n 0, either y n =+ or ϕy n = n + A y α nw y n. 6. We distinguish the two following cases: A There exists n n 0 such that y n =+. B For every n n 0, y n < +. In case B, extracting a subsequence of x n n n0 if necessary, one can assume that x n0 <y n0 <x n0 + <y n0 + <x n0 + <. For n n 0 +, we define x n = inf { y<x n, x y, x n, x α W xϕx A>3n A/4 }. Since y n <x n and y α n W y n ϕy n A = n A/ < 3n A/4 we deduce x n >y n >. Moreover, by continuity, ϕ x n = 3n + A/4 x α nw x n, and ϕx > 3n + A/4x α W x for x> x n, x close to x n. Therefore ϕ x n 3n + A 4 d dx x α W x C n x α+ n W x n 3n + A x= x n 4 α + W L C + for some C > 0. By definition of y n and x n,foreveryx x n, y n, x α ϕ x n A. Thus, ϕ x ϕ x n + n A C n x α+ n x x n y α dy W x n + n A x α+ n W x n x x n y α dy =: G nx. 6.

34 86 C. Gallo and D. Pelinovsky / On the Thomas Fermi ground state in a harmonic potential Notice that G n x n < 0, whereas { + if α, G n + = if α>, where for α>, g n g n n A α x α n > 0 asn +. As a result, for n sufficiently large, since G n is increasing on x n, +, G n vanishes exactly once on that interval. Moreover, this unique zero z n of G n is defined by zn x n y α dy = C n n A x α+ n W x n, thus z n = x n + O. x n By integration of 6., we infer that for x x n, y n, x ϕx ϕ x n + G n ydy x n zn ϕ x n + G n ydy x n C n ϕ x n x α+ n W x n z n x n 3n + A 4 x α nw x n C n x α+3 n W x n for some constant C > 0. Therefore, for n large enough, for every x x n, y n, since W is increasing on A +, +, ϕx 5n 8 x α nw x n 5n 8 x α W x. For n sufficiently large, 5n/8 > n + A/, and it provides a contradiction with 6., which means that case B can not happen. In case A, for every x x n, x α ϕ x n A/ > Therefore ϕ x 0asx,otherwiseϕ would not be in L R. Thus, for every x x n, ϕ x 0, and therefore ϕx 0asx.If0 <α, 6.3 provides a contradiction with the fact that

35 C. Gallo and D. Pelinovsky / On the Thomas Fermi ground state in a harmonic potential 87 ϕ x 0asx.Ifα>, integration of 6.3 between x and + yields ϕ x n A α x α. 6.4 This is a contradiction with ϕx 0, if <α. Finally, if α>, by integration of 6.4, Thus, ϕx n A + y α dy = α x ϕx x O x α. n A α α x α. Since the assumption x α f L A +, + implies x α f L A +, +, we have proved that G α implies G α ifα>. The proof of. is completed by induction. Then, since ϕ = Wϕ f, we deduce ϕ x = O x α. 6.5 We next prove that ϕ x = { O x α if α>, o if 0 <α. 6.6 By integration of 6.5, if α>, ϕ x has a limit as x +. This limit can only be 0, because ϕ L. 6.6 is then obtained by integration of 6.5 between x an +.Ifα, 6.6 is a consequence of the fact that ϕx 0andϕ x 0asx +. Let χ C R be such that { 0 if x, χx = if x. For m N,letϕ m, f m C R be the functions defined by and ϕ m x = χxx α+γm+ f m x = ϕ mx + W xϕ m x. From now on, we assume that f and W have asymptotic series. as x +,sothat f m x x + x + + x α+γm v k x γk + α + γm + α + γm + x α+γm+3/γ k=0 + x α+γm ṽ k x γk, k=0