existence and stability of standing waves - Semantic Scholar

Schrödinger equations with a spatially decaying nonlinearity: existence and stability of standing waves Fran¸cois Genoud and Charles A. Stuart IACS-FSB, Section de Mathématiques, ´ Ecole Polytechnique Fédérale de Lausanne CH-1015 Lausanne, Switzerland [email protected], [email protected]

June 5, 2007 Abstract For N ≥ 3 and p > 1, we consider the nonlinear Schrödinger equation i∂t w + ∆x w + V (x) |w|p−1 w = 0 where w = w(t, x) : R × RN → C with a potential V that decays at infinity like |x|−b for some b ∈ (0, 2). A standing wave is a solution of the form w(t, x) = eiλt u(x) where λ > 0 and u : RN → R. For 1 < p < 1 + (4 − 2b)/(N − 2), we establish the existence of a C 1 -branch of standing waves parametrized by frequencies λ in a right neighbourhood of 0. We also prove that these standing waves are orbitally stable if 1 < p < 1 + (4 − 2b)/N and unstable if 1 + (4 − 2b)/N < p < 1 + (4 − 2b)/(N − 2).

1

Introduction

This paper is concerned with the existence and stability of standing waves for the nonlinear Schrödinger equation i∂t w + ∆x w + V (x) |w|p−1 w = 0 where w = w(t, x) : R × RN → C

(1.1)

with N ≥ 3 and p > 1. A standing wave is a solution of the form w(t, x) = eiλt u(x) where λ > 0 and u : RN → R

(1.2)

and for functions of this type (1.1) is satisfied if and only if u is a solution of the stationary equation ∆u + V (x) |u|p−1 − λu = 0 for x ∈ RN .

1

(1.3)

This issue has been the subject of intensive study for a long time (see [7], [3], [12] and the references therein) and it is well-known (see [4]) that it is appropriate to seek solutions of (1.1) with w ∈ C [0, T ), H 1 (RN ) ∩ C 1 (0, T ), H −1 (RN ) for some T > 0 and correspondingly, solutions of (1.3) with u ∈ H 1 (RN ). Our work on this problem was stimulated by two recent contributions, [3] and [12]. In [3], De Bouard and Fukuizumi assume that the potential V has the following properties: (D1) V ∈ C(RN \{0}, R) with V ≥ 0, but V 6≡ 0 and V ∈ Lθ (|x| < 1) where θ = 2N/{N + 2 − (N − 2)p}. (D2) There exist b ∈ (0, 2), C > 0 and α > {N + 2 − (N − 2)p}/2 > b such that V (x) − |x|−b ≤ C |x|−α for all x ∈ RN with |x| ≥ 1. Under these assumptions it is known (see [8], [17]) that, (A) for all λ > 0, the equation (1.3) has a solution uλ ∈ H 1 (RN )\{0} that is characterized as a minimum of the action with a constraint, and it is shown in [3] that, (B) if p < 1 + (4 − 2b)/N, there exists λ0 > 0 such that, for λ ∈ (0, λ0 ), the standing wave associated with uλ is an orbitally stable solution of (1.1). These results were extended and improved by Jeanjean and Le Coz [12] who replace (D1) and (D2) by the weaker assumptions: (J1) V ∈ Lθloc (RN ) for some θ > 2N/{N + 2 − (N − 2)p}. (J2) There exists b ∈ (0, 2) such that lim |x|b V (x) = 1. |x|→∞

They prove that, (C) for 1 0 such that, for λ ∈ (0, λ0 ), there exists a positive solution uλ ∈ H 1 (RN )\{0} of (1.3) with kuλ kH 1 → 0 and |uλ |L∞ → 0 as λ → 0 and the standing wave associated with uλ is an orbitally stable solution of (1.1). In fact, it should be emphasized that Jeanjean and Le Coz also generalize the treatment to more general nonlinearities g ∈ C 1 (R) such that |g 0 (s)| g 0 (s) = 1 and lim < ∞ for some β ∈ [0, N 4−2 ), s→∞ s→0+ psp−1 sβ

g(0) = 0, lim

rather than the special case g(s) = |s|p−1 s discussed in [3]. Although we deal here with (1.1), we claim that our method can easily be modified to cover this more general situation. The approaches used in [3] and [12] are similar. First the existence of standing waves for all λ in an interval (0, λ1 ) is established using a variational argument. De Bouard and Fukuizumi use minimization on the Nehari manifold whereas Jeanjean and Le Coz use a version of the mountain pass theorem. Then the stability of these 2

solutions is established by using a variational criterion (see Proposition 2 of [3]) that is a by-product of the rather complete analysis in [7] of various criteria for the orbital stability of special solutions of Hamiltonian systems. The method used in [3] and [12] to show that the stability criterion is satisfied when λ is close to 0 goes as follows. First of all it is shown that a rescaled version of the solution uλ converges in H 1 (RN ) as λ → 0 to the unique positive solution ψ ∈ H 1 (RN ) of the equation ∆u + |x|−b |u|p−1 u − u = 0 for x ∈ RN . (1.4) Then it is shown that this solution is non-degenerate in the sense that v = 0 is the only solution of ∆v + p |x|−b |ψ|p−1 v − v = 0 (1.5) in H 1 (RN ). From this it is deduced that the rescaled version of the stability criterion holds in the limit λ → 0 and consequently also for λ sufficiently close to 0. The most difficult part of this programme consists of proving the non-degeneracy of ψ and for that an auxiliary equation is introduced, p ∆v + (δ + |x|−b )v+ − (1 + δψ p−1 )v = 0 in [3]

(1.6)

p ∆v + (δe−|x| + |x|−b )v+ − (1 + δe−|x| ψ p−1 )v = 0 in [12],

(1.7)

and where δ > 0 is small. Clearly, ψ is a positive solution of (1.6), respectively (1.7), for all δ. Let M (δ) denote the Morse index of ψ as a solution of (1.6), respectively (1.7). It is shown that, if (1.5) has a non-trivial solution, then M (δ) ≥ 2. However, for δ > 0 and small, it is also proved that (1.6), respectively (1.7), has a positive ground state ϕ(δ) ∈ H 1 (RN ) and the Morse index of ϕ(δ) is ≤ 1. Finally, in both cases, Yanagida’s uniqueness theorem [22] is used to show that ψ = ϕ(δ), from which it follows that (1.5) cannot have a non-trivial solution. Our approach differs from [3] and [12] in the following aspects. (1) For 1 < p < 1 + (4 − 2b)/(N − 2), we give a direct proof that ψ is a nondegenerate solution of (1.5) without introducing an auxiliary equation like (1.6) or (1.7) and hence we also avoid recourse to Yanagida’s difficult uniqueness result. In fact, we prove that ∆ + p |x|−b |ψ|p−1 − 1 : H 1 (RN ) → H −1 (RN ) is an isomorphism.

(1.8)

(2) We exploit (1.8), through an implicit function theorem, to establish the existence of a smooth branch λ 7−→ uλ of positive solutions of (1.3) mapping an interval (0, λ0 ) into H 1 (RN ). Furthermore, as λ → 0, we have that kuλ kH 1 → 0 and |uλ |L∞ → 0 if 1 < p < 1 + (4 − 2b)/N, whereas |uλ |L2 → ∞ if 1 + (4 − 2b)/N 0 if 1 0 where γ =

4 − 2b − (N − 2)(p − 1) > 0, 2(p − 1)

limλ→0 λ−γ+1 |uλ |L2 (RN ) = L2 > 0, limλ→0 λ−α |uλ |L∞ (RN )

= 0 for any α
1 for 1 < p < 1+ 4−2b , we see that the the branch (λ, uλ ) N bifurcates from (0, 0) in R × H in this case, whereas for 1 + 4−2b < p < 1 + 4−2b , N N −2 we have asymptotic bifurcation as λ → 0. We also have that |uλ |L∞ (RN ) → 0 as λ → 0 for all p ∈ (1, 1 + 4−2b ). N −2 The definition of orbital stability for solutions of (1.1) that are standing waves is recalled in Section 4 where we also show that the assumptions required for the general results in [7] are all satisfied, thus reducing the proof of stability to an analysis of the monotonicity of the L2 -norm on the branch {uλ : λ ∈ (0, λ0 )}. The perturbation method that we use to obtain a smooth branch of solutions of 4

(1.3) has its origins in [19]. (See also [16], [18], [1] and [2] for other work in this direction.) It is set out in Section 3 and relies on the existence of a non-trivial, non-degenerate solution ψ of the limit equation (1.4) that arises as λ → 0 after an appropriate rescaling of (1.3). A self-contained presentation of the existence and non-degeneracy of ψ is given in Section 2. The proof of part (a) of Theorem 1.1 appears at the end of Section 3 and that of part (b) at the end of Section 4.

1.1

A weaker notion of stability

Theorem 1.1(b) establishes the orbital stability in the usual sense of the timeperiodic solution eiλt uλ (x) of (1.1) for positive frequencies λ near 0. There is a (potentially) weaker notion of stability which can be proved much more easily and which holds globally (i.e. for all λ > 0 ) under similar assumptions. Without aiming for the most general hypotheses on V, we formulate a result of this kind for comparison. (A1) V ∈ L∞ (RN ) and V ≥ 0 on RN . (A2) V (x) → 0 as |x| → ∞ and there exist z ∈ RN , B > 0 and b ∈ (0, 2) such that V (x) ≥ B |x|−b for all x ∈ Ω = {ty : t ≥ 1 and |z − y| ≤ 1}. Note that (V2) implies (A2). For 1 0, we set 1

N

Z

S(c) = {u ∈ H (R ) :

u2 dx = c2 } and m(c) = inf E(u). u∈S(c)

RN

If u ∈ S(c) and E(u) = m(c), there is a Lagrange multiplier λ ∈ R such that (λ, u) satisfies (1.3). It follows from (A1) that u ∈ C 1 (RN ). Let W (c) = {u ∈ C 1 (RN ) ∩ S(c) : E(u) = m(c) and u > 0 on RN }. Theorem 1.2 Suppose that (A1) and (A2) are satisfied and that 1 0, there exist uc ∈ W (c) and an associated Lagrange multiplier λc such that λc > 0, lim λc = 0 and lim λc = ∞. c→0

c→∞

Furthermore, eiλc t uc (x) is a stable solution of (1.1) in the following sense. For every ε > 0, there exists δ > 0 such that, for any w0 ∈ H 1 (RN , C) such that kw0 − uc kH 1 < δ, the initial value problem for (1.1) with w(0, ·) = w0 has a unique solution, defined for all t ≥ 0, and it satisfies

sup inf{ w(t, ·) − eiθ v H 1 : θ ∈ R and v ∈ W (c)} < ε. (1.11) t≥0

5

Proof. This follows from Theorems 3.1 and 4.1 together with Propositions 2.3 and 5.1 of [10]. Remarks. The condition (1.11) corresponds to orbital stability in the usual sense (c.f. (4.3)) whenever W (c) = {uc }. This amounts to showing that the minimizer of E on S(c) is unique, but uniqueness results of this kind for non-autonomous equations do not seem to be known. Stability in the sense of (1.11) was first established in [5] for the autonomous (i.e. V constant) version of (1.1) in which case it is known to be optimal (see Section 8.3 of [4]). One easily deduces that kuc kH 1 → 0 as c → 0 (see Remark 1 on page 175 of [17]). Hence the solutions (λc , uc ) of (1.3) bifurcate from (0, 0) in R×H as does the smooth local branch (λ, uλ ) given by Theorem 1.1(a) when 1 < p < 1 + (4 − 2b)/N .

1.2

The homogeneous case

Our hypotheses (V1) to (V3) are satisfied by the homogeneous radial potential V (x) = |x|−b where b ∈ (0, 2).

(1.12)

In this case the branch of standing waves given by Theorem 1.1 does indeed have a global extension to the whole interval (0, ∞). It can be obtained directly from the change of variables used in Section 3 and the ground state ψ of (1.4) that is established in Section 2. One simply checks that the function 2−b

1

uλ (x) = λ 2(p−1) ψ(λ 2 x) for x ∈ RN and λ > 0 satisfies (1.3) for all λ > 0. Then Z Z 2−b −N 2 p−1 2 uλ (x) dx = λ RN

(1.13)

ψ(x)2 dx

RN

and we see immediately that (1.9) and (1.10) are satisfied for all λ > 0.

2

Limit problem

In this section we study the limit problem −∆u + u − |x|−b |u|p−1 u = 0,

x ∈ RN ,

(2.1)

where N ≥ 3, 0 < b < 2, 1 < p < 1 + (4 − 2b)/(N − 2). A variational approach is used to prove the existence of a positive radial solution ψ ∈ H 1 (RN ) of (2.1). Such a solution is obtained by constrained minimization and called a ground state. Then we establish some properties of ψ, in particular that ψ ∈ C 2 (RN \{0})∩L∞ (RN ), is decreasing and vanishes exponentially as |x| → ∞. Finally, the non-degeneracy of ψ is proven. This is the key ingredient needed to make use of the implicit function theorem in Section 3.

6

2.1

Existence of a ground state

It is well-known that one can seek non-trivial solutions of (2.1) as minimizers of the action Z Z 1 1 2 2 S(u) = |∇u| + u dx − |x|−b |u|p+1 dx (2.2) 2 RN p + 1 RN under the natural constraint Z Z 2 2 |∇u| + u dx − RN

|x|−b |u|p+1 dx = 0.

(2.3)

RN

Let us make this statement more precise by introducing the basic assumptions of the variational method. We follow closely the arguments presented in [20]. We denote by H the Sobolev space H 1 (RN , R), by h·, ·i its inner product and k · k its norm. It follows from (A.2) in Appendix A that there exists K > 0 such that Z |x|−b |u|p+1 dx ≤ Kkukp+1 for all u ∈ H. (2.4) RN

We can now define the action functional S : H → R by Z 1 1 2 S(u) = kuk − Φ(u) where Φ(u) = |x|−b |u|p+1 dx. 2 p + 1 RN

(2.5)

The auxiliary functional 1 1 J(u) = kuk2 − φ(u) where φ(u) = 2 2

Z

|x|−b |u|p+1 dx,

(2.6)

RN

is used to set the constraint (2.3). We shall prove in Lemma 2.1 below that S, J ∈ C 2 (H, R). We say that u ∈ H \ {0} is a solution of (2.1) if u = ∇Φ(u),

(2.7)

where ∇Φ is the gradient of Φ, that is, h∇Φ(u), vi = Φ0 (u)v

for all u, v ∈ H.

Obviously u is a solution of (2.1) if S 0 (u) = 0 and u 6= 0. On the other hand, since 2J(u) = S 0 (u)u, J(u) = 0 if u is a solution of (2.1). We shall get solutions of (2.1) by minimizing S under the constraint J(u) = 0. Let us therefore introduce the so-called Nehari manifold V = {u ∈ H \ {0} : J(u) = 0} and set m = inf{S(u) : u ∈ V }.

(2.8)

The homogeneity of the nonlinearity provides us with the following useful properties of Φ and φ. For u ∈ H \ {0}, φ(u) = Φ0 (u)u = (p + 1)Φ(u) > 0 and 0

Z

φ (u)u = (p + 1)

|x|−b |u|p+1 dx = (p + 1)φ(u).

(2.9) (2.10)

RN

The following lemma concerns the regularity of S and J. Its proof is part (iii) of Lemma B.1 in Appendix B. 7

Lemma 2.1 Φ ∈ C 2 (H, R) with Z

0

Φ (u)v =

|x|−b |u|p−1 uv dx

(2.11)

RN

and 00

Z

Φ (u)[v, w] = p

|x|−b |u|p−1 vw dx

for all u, v, w ∈ H.

(2.12)

RN

We are now in a position to collect some important properties of V . Lemma 2.2 (i) There exists δ > 0 such that φ(u) = kuk2 ≥ δ for all u ∈ V . (ii) V ⊂ H is a C 2 -manifold of codimension 1. (iii) For all u ∈ V , S(u) = A(p)kuk2 , where A(p) = [(p − 1)/2(p + 1)] > 0. (iv) If u ∈ V and S(u) = m, then u is a solution of (2.1). Proof. (i) Thanks to (2.4) we have, for all u ∈ V , kuk2 = φ(u) ≤ Kkukp+1 . Since p > 1, it follows that there exists δ > 0 such that kuk2 ≥ δ for all u ∈ V . (ii) For u ∈ V , (2.10) implies that p+1 1−p 1 φ(u) = φ(u) < 0. J 0 (u)u = kuk2 − φ0 (u)u = φ(u) − 2 2 2

(2.13)

Therefore, V is a C 2 -manifold of codimension 1, by the definition (2.8). (iii) For all u ∈ V , 1 1 1 (p − 1) S(u) = kuk2 − Φ(u) = kuk2 − φ(u) = kuk2 2 2 p+1 2(p + 1)

by (2.9).

(iv) If u minimizes S under the constraint J(u) = 0, there exists a real number λ such that S 0 (u)v = λJ 0 (u)v for all v ∈ H. On the other hand, S 0 (u)u = 2J(u) = 0 since u ∈ V . Hence 0 = S 0 (u)u = λJ 0 (u)u with J 0 (u)u < 0 by (2.13). Thus λ = 0 and S 0 (u)v = 0 for all v ∈ H. Remark. (i) and (iii) of Lemma 2.2 imply that m > 0 and that any minimizing sequence {un } ⊂ V is bounded. Lemma 2.3 (i) There exists a unique function t : H \ {0} → ((0, ∞)) such that if u ∈ H \ {0} and s > 0, then su ∈ V if and only if s = t(u). Moreover t ∈ C 2 (H \ {0}, R). (ii) For all u ∈ H \ {0}, we have t(u) ≤ 1 if J(u) ≤ 0 and 8

t(u) ≥ 1 if

J(u) ≥ 0.

(2.14)

1

Proof. Set t(u) = {kuk2 /φ(u)} p−1 for all u ∈ H \ {0}.

To prove the existence of solutions, we still need the following lemma, which is proven in Appendix C. Lemma 2.4 Let N ≥ 3, 0 < b < 2, and 1 < p < 1 + (4 − 2b)/(N − 2). Then the functional φ defined by (2.6) is weakly sequentially continuous on H. We can now state the main result of this section. Theorem 2.1 There exists a function ψ ∈ H such that ψ ∈ V and S(ψ) = m. Moreover, ψ is non-negative, spherically symmetric and radially non-increasing. Proof. Let {vn } ⊂ V be a minimizing sequence: S(vn ) → m as n → ∞. Since φ, k · k and S do not change when replacing u by |u|, we can assume vn ≥ 0. Let vn∗ denote the Schwarz symmetrization of vn (see e.g. [15]). Then Z Z Z Z 2 ∗ 2 2 |∇vn | dx ≥ |∇vn∗ |2 dx (vn ) dx, vn dx = RN

RN

RN

RN

and Z RN

|x|−b vnp+1 dx

Z ≤

−b ∗

(|x| )

p+1 ∗

|vn |

Z dx =

RN

RN

|x|−b |vn∗ |p+1 dx

since the mapping s 7→ |s|p+1 is non-decreasing on [0, ∞). Hence kvn k2 ≥ kvn∗ k2 , φ(vn ) ≤ φ(vn∗ ) and Φ(vn ) ≤ Φ(vn∗ ). Therefore, 1 1 J(vn∗ ) = kvn∗ k2 − φ(vn∗ ) ≤ J(vn ) = 0 2 2 and so by (2.14), t(vn∗ ) ≤ 1. On setting un = t(vn∗ )vn∗ , we have that un ∈ V and, by Lemma 2.2(iii), m ≤ S(un ) = A(p)kun k2 = A(p)kt(vn∗ )vn∗ k2 ≤ A(p)kvn∗ k2 ≤ A(p)kvn k2 = S(vn ) → m as n → ∞. Thus, {un } is a minimizing sequence and hence is bounded. We can assume that un * u weakly in H as n → ∞. Then Lemma 2.4 implies that φ(u) = lim φ(un ) = lim kun k2 ≥ δ > 0, n→∞

n→∞

since un ∈ V . Thus, u 6= 0. Furthermore, 1 1 1 1 J(u) = kuk2 − φ(u) ≤ lim inf kun k2 − φ(u) 2 2 2 n→∞ 2 1 1 = lim φ(un ) − φ(u) = 0. 2 n→∞ 2 Hence t(u) ≤ 1. Then m ≤ S(t(u)u) = A(p)kt(u)uk2 ≤ A(p)kuk2 ≤ A(p) lim inf kun k2 = lim S(un ) = m. n→∞

9

n→∞

(2.15)

We conclude that t(u) = 1, u ∈ V and S(u) = m. Indeed t(u) < 1 in (2.15) would imply m < m, which is a contradiction. Setting ψ = u, we have ψ = ψ ∗ and so ψ is non-negative, spherically symmetric and radially non-increasing. This completes the proof of Theorem 2.1. We conclude this section by the following lemma, which is central in the proof that the ground state ψ is non-degenerate. Lemma 2.5 Consider the bilinear form S 00 (ψ) : H × H → R. We have that S 00 (ψ)[ψ, ψ] < 0 and

S 00 (ψ)[ξ, ξ] ≥ 0 for all ξ ∈ H such that J 0 (ψ)ξ = 0.

Proof. Set R(u) = S(t(u)u) for all u ∈ H \ {0}, where t is the function given by Lemma 2.3. Then R ∈ C 2 (H \ {0}) and ψ is a local minimum of R by Theorem 2.1. Hence R0 (ψ)ξ = 0 and R00 (ψ)[ξ, ξ] ≥ 0 for all ξ ∈ H. Using the fact that S 0 (ψ) = 0, a short computation gives, for all ξ ∈ H, R00 (ψ)[ξ, ξ] = (t0 (ψ)ξ)2 S 00 (ψ)[ψ, ψ] + 2(t0 (ψ)ξ)S 00 (ψ)[ψ, ξ] + S 00 (ψ)[ξ, ξ].

(2.16)

To discuss the sign of S 00 (ψ)[ξ, ξ], it is convenient to rewrite the two first terms in the right hand side of (2.16) in terms of J. Since S 0 (u)u = 2J(u), we have S 00 (u)[u, v] + S 0 (u)v = 2J 0 (u)v. But S 0 (ψ) = 0 and so S 00 (ψ)[ψ, ξ] = 2J 0 (ψ)ξ for all ξ ∈ H. Furthermore, since J(t(u)u) = 0 for all u ∈ H \ {0}, derivation at u = ψ gives (t0 (ψ)ξ)J 0 (ψ)ψ + J 0 (ψ)ξ = 0 for all ξ ∈ H. Since J 0 (ψ)ψ < 0 by 2.13, this implies that t0 (ψ)ξ = −(J 0 (ψ)ξ)/(J 0 (ψ)ψ) for all ξ ∈ H. Therefore, (2.16) reads (J 0 (ψ)ξ) 0 (J 0 (ψ)ξ)2 0 J (ψ)ψ − 4 J (ψ)ξ + S 00 (ψ)[ξ, ξ] (J 0 (ψ)ψ)2 (J 0 (ψ)ψ) (J 0 (ψ)ξ)2 = −2 0 + S 00 (ψ)[ξ, ξ] for all ξ ∈ H. J (ψ)ψ

R00 (ψ)[ξ, ξ] = 2

(2.17) (2.18)

Hence S 00 (ψ)[ξ, ξ] = R00 (ψ)[ξ, ξ] ≥ 0 for all ξ ∈ H such that J 0 (ψ)ξ = 0. On the other hand, S 00 (ψ)[ψ, ψ] = 2J 0 (ψ)ψ < 0. The proof is complete. Remarks. Since ψ is a solution of (2.1), we have, for all ξ ∈ H, 1 1 p+1 0 1−p J 0 (ψ)ξ = S 00 (ψ)[ψ, ξ] = hψ, ξi − φ0 (ψ)ξ = hψ, ξi − Φ (ψ)ξ = hψ, ξi. 2 2 2 2 Therefore, the condition J 0 (ψ)ξ = 0 is equivalent to hψ, ξi = 0 and the tangent space to V at ψ is orthogonal to ψ. This is due to homogeneity. Lemma 2.5 expresses the fact that the Morse index of S 00 at ψ is equal to one.

10

2.2

Further properties of the ground state

In this section we establish some important properties of the ground state ψ. We start with the following proposition concerning spherically symmetric functions of H. Proposition 2.1 Let N ≥ 3 and u : RN → R be a spherically symmetric function. Let v : (0, ∞) → R be such that u(x) = v(r) for r = |x|, x ∈ RN \ {0}. Define N −1 ξ : (0, ∞) → R by ξ(r) = r 2 v(r) for all r > 0. Then u ∈ H 1 (RN ) implies ξ ∈ H01 ((0, ∞)). Proof. First remark that if u : RN → R has generalized derivatives, so has v. (Use test functions with compact support in RN \ {0}.) Then the following identities hold: Z ∞ Z ∞ Z N −1 2 2 r v(r) dr = ωN ξ(r)2 dr, (2.19) u(x) dx = ωN RN

Z

0

Z

2

∞

|∇u(x)| dx = ωN RN

and

0

r

N −1 0

Z

2

v (r) dr = ωN

0

∞

0

2 ( 1−N )ξ(r)/r + ξ 0 (r) dr 2 (2.20)

Z

∞ 0

2

Z

ξ (r) dx = 0

0

∞

( N 2−1 )r

N −3 2

v(r) + r

N −1 2

2 v 0 (r) dr,

(2.21)

where ωN > 0 is the area of the unit sphere in RN . Now suppose that u ∈ H 1 (RN ). Then ξ ∈ L2 ((0, ∞)) by (2.19). Moreover, (2.21) gives Z ∞ Z ∞ Z ∞ 2 0 2 N −1 v(r) dr + rN −1 v 0 (r)2 dr}. ξ (r) dx ≤ C{ r r2 0 0 0 On one hand, Z ∞ Z Z 2 u(x)2 N −1 v(r) −1 r dr = ωN dx ≤ C |∇u(x)|2 dx < ∞ 2 2 r |x| N N 0 R R

(2.22)

by inequality in dimension N ≥ 3. On the other hand we have that R ∞ Hardy’s N −1 0 r v (r)2 dr < ∞ by (2.20). Thus ξ ∈ H 1 ((0, ∞)). 0 We still have to prove that ξ(r) → 0 as r → 0. Since ξ ∈ H 1 ((0, ∞)), there exists ξ(0) = limr→0 ξ(r). Then (2.22) shows that ξ(r)/r ∈ L2 ((0, ∞)) and hence ξ(r)/r ∈ L1 (0, 1). This implies ξ(0) = 0, which completes the proof. Remark. The identities (2.19) and (2.20) show that if u ∈ H 1 (RN ) is radial and N −1 v : (0, ∞) → R is such that u(x) = v(r) for r = |x|, x ∈ RN \ {0}, then r 2 v(r) ∈ N −1 L2 ((0, ∞)) and r 2 v 0 (r) ∈ L2 ((0, ∞)). We introduce two Hilbert spaces which will be useful later: R∞ L2r ≡ {v : (0, ∞) → R : 0 rN −1 v(r)2 dr < ∞} with inner product hu, vi2 = R ∞ N −1 1/2 r uv dr and norm |u|2 = hu, ui2 0 11

L2r ⊃ Hr1 ≡ {v ∈ L2r : v 0 ∈ L2r } with inner product hu, vi1 = u0 v 0 }dr

R∞ 0

rN −1 {uv +

Proposition 2.2 Let u ∈ H be a spherically symmetric solution of (2.1) and let v(r) = u(x) for r = |x|, x ∈ RN \ {0}. Then v ∈ Hr1 and v is a weak solution of v 00 +

N −1 0 v − v + r−b |v|p−1 v = 0. r

(2.23)

Proof. It follows from Proposition 2.1 that v ∈ Hr1 . For ϕ ∈ C0∞ ((0, ∞)), let ψ(x) = ϕ(r). Then ψ ∈ C0∞ (RN ) and Z Z ∞ −b p−1 0= ∇u∇ψ + [1− |x| |u| ]uψ dx = ωN rN −1 {v 0 ϕ0 +[1−r−b |v|p−1 ]vϕ}dx. RN

0

Thus v ∈ Hr1 and rN −1 v 0 admits a generalized derivative on (0, ∞) with (rN −1 v 0 )0 = rN −1 [1 − r−b |v|p−1 ]v. Since r1−N ∈ C ∞ ((0, ∞)), it follows that v 0 admits a generalized derivative on (0, ∞) with (v 0 )0 = (r1−N [rN −1 v 0 ])0 = (1 − N )r−N [rN −1 v 0 ] + r1−N [rN −1 v 0 ]0 (N − 1) 0 =− v + [1 − r−b |v|p−1 ]v, r showing that v is a weak solution of (2.23).

The next two lemmas are concerned with regularity and asymptotic properties of H 1 -solutions of linear ordinary differential equations of the type (2.23). We state these results in a sufficiently general form to use them in another context later. Let f : (0, ∞) → R. We say that f (r) → 0 exponentially as r → ∞ if there exists > 0 such that er f (r) → 0 as r → ∞. Let γ > 0, µ ≥ 0 and Q : (0, ∞) → R be a continuous function such that rb Q(r) is bounded on (0, ∞) and Q(r) → 0 exponentially as r → ∞. Consider the ordinary differential equation v 00 +

µ N −1 0 v − γv − 2 v + Q(r)v = 0 for r > 0. r r

(2.24)

Lemma 2.6 Let v ∈ Hr1 be a weak solution of (2.24). Then v ∈ C 3 ((0, ∞)) and v is a classical solution of (2.24). Proof. Since v ∈ Hr1 , v ∈ C((0, ∞)) and v 0 ∈ L2loc ((0, ∞)). Hence, from (2.24), we 2 deduce that v ∈ Hloc ((0, ∞)) from which it follows that v ∈ C 1 ((0, ∞)). Returning to (2.24), we see that v ∈ C 2 ((0, ∞)) and then v ∈ C 3 ((0, ∞)). Lemma 2.7 Let v ∈ Hr1 be a solution of (2.24). (i) limr→0 v(r) and limr→0 rv 0 (r) exist and are finite. 12

(ii) v(r), v 0 (r) → 0 exponentially as r → ∞. Lemma 2.7 is proven in Appendix D. We can now prove the following properties of the ground state. Theorem 2.2 Let ψ be the solution of (2.1) given by Theorem 2.1. (i) ψ ∈ C(RN ) ∩ C 2 (RN \ {0}). In particular ψ is a classical solution of (2.1). (ii) ψ is radially decreasing on RN . (iii) ψ(x) > 0 for all x ∈ RN . (iv) ψ(x) → 0 exponentially as |x| → ∞. Proof. (i) Let ϕ ∈ Hr1 be such that ψ(x) = ϕ(r) for r = |x|, x ∈ RN \ {0}. By Proposition 2.2, ϕ ∈ Hr1 and ϕ is a weak solution of (2.23). Then ϕ ∈ C 3 ((0, ∞)) by Lemma 2.6. A fortiori ϕ ∈ C 2 ((0, ∞)). Furthermore, by Lemma 2.7, there exists ϕ(0) = limr→0 ϕ(r) > 0. Hence ψ ∈ C(RN ) ∩ C 2 (RN \ {0}) and ψ is a classical solution of (2.1). (ii) It follows from Theorem 2.1 that ϕ is radially non-increasing. Since ϕ is a non-trivial classical solution of (2.23), ϕ cannot be constant on an interval. (iii) By construction, ψ ≥ 0. Thus, (ii) implies that ϕ(r) > 0 for all r > 0. (iv) The exponential decay follows from Lemma 2.7. Remarks. One can show, using the general results of [22], that there exists at most one solution of (2.1) that is spherically symmetric and positive. Therefore, ψ is the unique positive spherically symmetric solution of (2.1). It is worth noting that, in general, the ground state ψ 6∈ C 1 (RN ). Indeed, for 1 < b < 2, the following argument shows that ϕ0 (r) → −∞ as r → 0. As we have seen, ϕ is a solution of (2.23), which reads (rN −1 ϕ0 )0 = rN −1 [ϕ − r−b ϕp ]. Hence, for 0 < r < 1, 0

ϕ (1) − r

N −1

0

Z

1

tN −1 [ϕ(t) − t−b ϕ(t)p ]dt,

ϕ (r) = r

that is, 0

ϕ (r) = r

1−N

0

Z

{ϕ (1) −

1

tN −1 [ϕ(t) − t−b ϕ(t)p ]dt}.

r

Suppose that 0

Z

ϕ (1) −

1

tN −1 [ϕ(t) − t−b ϕ(t)p ]dt → 0 as r → 0,

r 0

otherwise ϕ is not bounded as r → 0. Then, by De L’Hospital’s rule, we get R1 ϕ0 (1) − r tN −1 [ϕ(t) − t−b ϕ(t)p ]dt rN −1 [ϕ(r) − r−b ϕ(r)p ] = lim lim r→0 r→0 rN −1 (N − 1)rN −2 rϕ(r) − r1−b ϕ(r)p = lim = −∞ r→0 N −1 if b > 1, since ϕ(0) > 0. Hence ϕ0 (r) → −∞ as r → 0. 13

2.3

Nondegeneracy of the ground state

We denote by H −1 ≡ H ∗ the topological dual space of H and by h·, ·iH −1 ×H : H −1 × H → R the duality map. As usual, H is identified with a subspace of H −1 by setting Z hu, viH −1 ×H =

u(x)v(x)dx for u, v ∈ H. RN

Next, we define T : H → H −1 by hT (u), viH −1 ×H = S 0 (u)(v) for u, v ∈ H so that T (u) = −∆u + u − |x|−b |u|p−1 u. Then T ∈ C 1 (H, H −1 ) by Lemma 2.1 and hT 0 (u)w, viH −1 ×H = S 00 (u)[w, v] for u, v, w ∈ H. Thus, T 0 (u)w = −∆w + w − p |x|−b |u|p−1 w

for u, w ∈ H.

The ground state ψ satisfies T (ψ) = 0. We shall prove in this section that T 0 (ψ) : H → H −1 is an isomorphism. Note that T 0 (ψ) = −∆ + 1 + K, where Ku = −p|x|−b ψ p−1 u for all u ∈ H. It is well-known that −∆ + 1 : H → H −1 is an isomorphism. Furthermore, since ψ(x) → 0 as |x| → ∞, K is a compact linear operator, as proven in Lemma 2.8 below. Therefore, it suffices to show that T 0 (ψ) is injective. For this, we suppose that v ∈ H \ {0} and T 0 (ψ)v = 0. That is, −∆v + v − p|x|−b ψ p−1 v = 0.

(2.25)

Then, we proceed as follows. We first prove that v is spherically symmetric, which turns (2.25) into a linear ordinary differential equation. As in [12], this makes use of a decomposition of v on a basis of spherical harmonics. Our information about the Morse index of ψ and the spectral theory of linear ordinary differential operators then provides us with oscillation properties of v. Namely, v has exactly one zero on the half real line. Finally we get a contradiction by comparison between the radial equations satisfied by v and ψ respectively. We first show that K is compact. This is a special case of the following lemma which is proved in Appendix E. Lemma 2.8 Let w ∈ L∞ (RN ) and consider the operator C : H → H −1 defined by Cu = |x|−b wu for u ∈ H. Then C is a compact linear operator. Before we prove the spherical symmetry of v, we need some extra regularity of v given by the following result which is proved in Appendix F. Lemma 2.9 Let v ∈ H be a solution of (2.25). Then v ∈ L∞ (RN ) ∩ C(RN ).

14

Remark. The proof of Lemma 2.9 shows that Kv ∈ L2 (RN ) and hence v ∈ H 2 (RN ) if 2 < N/b. This is true for any 0 < b < 2 when N ≥ 4 and for 0 < b < 3/2 when N = 3. We can now prove the spherical symmetry of v. See [12] for similar results under slightly different hypotheses. Lemma 2.10 If v ∈ H is a solution of (2.25), then v is spherically symmetric. Proof. Since v ∈ L2 (RN ), v admits a decomposition of the form (see Appendix G) v(x) =

ak ∞ X X

vkm (r)Ykm (ϑ),

(2.26)

k=0 m=1

where r = |x|, ϑ = x/r for x ∈ RN \ {0}. In this representation, k k−2 ak = − for k ≥ 2, a1 = N and a0 = 1. N +k−1 N +k−3 Ykm , 1 ≤ m ≤ ak , is a spherical harmonic function that satisfies ∆Ykm = −

µk m Y r2 k

for m = 1, . . . , ak ,

(2.27)

with µk = k(N + k − 2) for all k ∈ N. {Ykm } is an orthonormal basis of L2 (S N −1 ) and so the function vkm (r) is given by the projection of v on the spherical harmonic Ykm : Z m vk (r) = v(rϑ)Ykm (ϑ)dϑ for m = 1, . . . , ak , for all k ∈ N. (2.28) S N −1

Note that vkm ∈ C(0, ∞) since v ∈ L∞ (RN ) ∩ C(RN ) by Lemma 2.9. The series in (2.26) is convergent in the sense of L2 (RN ). We shall prove that vkm ≡ 0 for m = 1, . . . , ak , for all k ≥ 1. Since v satisfies (2.25), we have Z v{−∆ϕ + ϕ − p|x|−b ψ p−1 ϕ}dx = 0 for all ϕ ∈ C0∞ (RN ). (2.29) RN

On setting ϕ(x) = ξ(|x|)Ykm (x/|x|) for ξ ∈ C0∞ (0, ∞), we have that ϕ ∈ C0∞ (RN ) and, using (2.27) and (2.29), Z ∞Z N − 1 0 µk ξ + 2 ξ + ξ − pr−b ψ p−1 ξ Ykm (ϑ)dϑ dr = 0. rN −1 v −ξ 00 − r r 0 S N −1 This with (2.28) and Fubini’s theorem yields Z ∞ N − 1 0 µk rN −1 vkm −ξ 00 − ξ + 2 ξ + ξ − pr−b ψ p−1 ξ dr = 0 for all ξ ∈ C0∞ (0, ∞). r r 0

15

By Lemma 2.6, this implies that vkm ∈ C 3 (0, ∞) and satisfies the ordinary differential equation (vkm )00 +

N −1 m 0 µk (vk ) − vkm − 2 vkm + pr−b ψ p−1 vkm = 0. r r

To simplify the notations, we set w ≡ vkm from now on. Thus, w ∈ C 3 (0, ∞) and w00 +

N −1 0 µk w − w − 2 w + pr−b ψ p−1 w = 0. r r

(2.30)

As before, let us write ψ(x) = ϕ(r) for r = |x|, x ∈ RN , where ψ is the ground state. Then ϕ satisfies the equation ϕ00 +

N −1 0 ϕ − ϕ + r−b ϕp = 0. r

(2.31)

Since ϕ ∈ C 3 (0, ∞) by Lemma 2.6, one can differentiate (2.31) and get ϕ000 +

N − 1 00 N − 1 0 ϕ − ϕ − ϕ0 − br−b−1 ϕp + pr−b ϕp−1 ϕ0 = 0. r r2

(2.32)

On setting ζ = ϕ0 , (2.32) reads ζ 00 +

N −1 0 N −1 ζ −ζ − ζ − br−b−1 ϕp + pr−b ϕp−1 ζ = 0. r r2

(2.33)

Let us write Su = u00 + N r−1 u0 − u. For 0 ≤ α < β ≤ ∞, equations (2.30) and (2.33) give the Lagrange identity for w and ζ: r

N −1

0

0

(ζw − ζ w)

|βα

Z

β

rN −1 {ζSw − wSζ}dr α Z Z β N −3 r wζ dr − b = (µk − N + 1)

=

β

rN −b−2 ϕp w dr. (2.34)

α

α

Now we suppose that w 6≡ 0 and without loss of generality, we assume that (α, β) is a maximal interval where w > 0. If α > 0, then w(α) = 0 and w0 (α) > 0. If β < ∞, then w(β) = 0 and w0 (β) < 0. Furthermore, the functions w and ϕ satisfy the hypotheses of Lemma 2.7 and ζ 0 = −( N r−1 ϕ0 − ϕ + r−b ϕp ) by (2.31). Therefore, lim rN −1 (ζw0 − ζ 0 w) = 0 and

r→0

lim rN −1 (ζw0 − ζ 0 w) = 0.

r→∞

Hence, since ζ = ϕ0 < 0, we have in any case rN −1 (ζw0 − ζ 0 w) |βα ≥ 0. Thus (2.34) implies Z β Z β N −3 (µk − N + 1) r wζ dr ≥ b rN −b−2 ϕp w dr > 0. α

α

But ζ < 0 and so we must have N − 1 − µk > 0. Since µk = k(N + k − 2), this implies k = 0. The proof is complete. 16

From now on, we write v(x) = ve(r) and ψ(x) = ϕ(r) for r = |x| > 0. Equation (2.25) then reads N −1 0 ve00 + ve − ve + pr−b ϕp−1 ve = 0. (2.35) r Before we prove that ve has exactly one zero in (0, ∞), we need a technical lemma. On setting f (r, s) = −s + r−b sp for r, s > 0, (2.36) equations (2.31) and (2.35) take the form ϕ00 +

N −1 0 ϕ + f (r, ϕ) = 0 r

(2.37)

and

N −1 0 ve + ∂2 f (r, ϕ)e v = 0. (2.38) r The following integral identities are crucial consequences of (2.37) and (2.38). ve00 +

Lemma 2.11 Let ve ∈ Hr1 be a solution of (2.38). Then we have Z ∞ rN −1 {f (r, ϕ) − ∂2 f (r, ϕ)ϕ}e v dr = 0

(2.39)

0

and

Z

∞

rN −1 {2f (r, ϕ) + r∂1 f (r, ϕ)}e v dr = 0.

(2.40)

0

Proof. The proof needs several integration by parts. We let the reader use Lemma 2.7 to check that all boundary terms vanish. For the first identity, multiplying (2.37) by rN −1 ve, (2.38) by rN −1 ϕ and subtracting gives Z ∞ {(rN −1 ϕ0 )0 ve − (rN −1 ve0 )0 ϕ + rN −1 [f (r, ϕ)e v − ∂2 f (r, ϕ)ϕe v ]}dr = 0. 0

Integrating by parts, one gets Z ∞ rN −1 {f (r, ϕ)e v − ∂2 f (r, ϕ)ϕe v }dr = 0. 0

The second identity needs a bit more work. Start on multiplying (2.38) by rN −1 rϕ0 and integrate: Z ∞ {(rN −1 ve0 )0 rϕ0 + rN ∂2 f (r, ϕ)e v ϕ0 }dr = 0. 0

Integration by parts on the first term yields Z ∞ {−rN −1 ve0 (rϕ0 )0 + rN ∂2 f (r, ϕ)e v ϕ0 }dr = 0. 0

By (2.37), (rϕ0 )0 = (2 − N )ϕ0 − rf (r, ϕ) and so Z ∞ {−rN −1 ve0 [(2 − N )ϕ0 − rf (r, ϕ)] + rN ∂2 f (rϕ)e v ϕ0 }dr = 0. 0

17

Integrating by parts the term in ve0 ϕ0 and using (2.37) again, one gets Z ∞ Z ∞ N −1 r f (r, ϕ)e v dr + rN {f (r, ϕ)e v 0 + ∂2 f (r, ϕ)e v ϕ0 }dr = 0. (N − 2) 0

But Z ∞

Z

0

N

(2.41)

0

r f (r, ϕ)e v dr = − 0

∞

{N rN −1 f (r, ϕ) + rN ∂1 f (r, ϕ) + rN ∂2 f (r, ϕ)ϕ0 }e v dr.

0

Hence (2.41) reads Z ∞ (N − 2) rN −1 f (r, ϕ)e v dr Z ∞ 0 {−N rN −1 f (r, ϕ)e v − rN ∂1 f (r, ϕ)e v − rN ∂2 f (r, ϕ)ϕ0 ve + rN ∂2 f (r, ϕ)e v ϕ0 = 0, + 0

so that

∞

Z

rN −1 {2f (r, ϕ) + r∂1 f (r, ϕ)}e v dr = 0.

0

The next step in our proof is to show that ve has exactly one zero in (0, ∞). This result is obtained via oscillation theory for ordinary differential equations. We shall work in the Hilbert spaces L2r and Hr1 introduced in the remark following Proposition 2.1. We can define the bilinear form β : Hr1 × Hr1 → R by Z ∞ β(u, v) = rN −1 {u0 v 0 + uv − pr−b ϕp−1 uv}dr for all u, v ∈ Hr1 . (2.42) 0

It is proven in Appendix H that there exists a unique selfadjoint operator A : L2r ⊃ D(A) → L2r with the following properties: D(A) = {u ∈ Hr1 : ∃w ∈ L2r , β(u, v) = hw, vi2 ∀v ∈ Hr1 } and Au = w. (2.43) It turns out that A is a generalized Sturm-Liouville operator with separated boundary conditions for (2.35). Moreover the essential spectrum σess (A) ⊂ [1, ∞). The following property of β will be useful. Lemma 2.12 The bilinear form β : Hr1 × Hr1 → R defined by (2.42) is such that β(ϕ, ϕ) < 0 and

β(ξ, ξ) ≥ 0 for all ξ ∈ Hr1 such that hϕ, ξi1 = 0.

Proof. We have that 00

Z

S (ψ)[u, v] =

∇u∇v + uv − p|x|−b ψ p−1 uv dx for all u, v ∈ H.

RN

18

Therefore, the result follows from Lemma 2.5 and the first remark following Lemma 2.5. Now we turn our attention to the spectral properties of the Sturm-Liouville operator A. Lemma 2.13 Let ve ∈ Hr1 be a nontrivial solution of (2.35). Then ve ∈ ker(A) and there exists a unique r0 ∈ (0, ∞) such that ve(r0 ) = 0 and ve changes sign at r0 . Proof. Since ve is a solution of (2.35), we have β(e v , w) = 0 for all w ∈ Hr1 . Hence ve ∈ D(A) and Ae v = 0. By hypothesis ve 6= 0 and so 0 is an eigenvalue of A. Moreover, it is an isolated eigenvalue multiplicity since σess (A) ⊂ [1, ∞). R ∞ofNfinite −1−b p The integral identity (2.39) gives 0 r ϕ ve dr = 0. Hence ve changes sign on (0, ∞), at least once. Therefore (see [21], Theorem 14.10) ve cannot be an eigenfunction for the smallest eigenvalue of A. Thus, A has at least one negative eigenvalue, say λ1 < 0. In fact, A has only one negative eigenvalue. As we shall see, this is a consequence of the behavior of the bilinear form β. Let ξ1 ∈ D(A) be such that Aξ1 = λ1 ξ1 . Suppose that there exists λ2 < 0, λ2 6= λ1 and ξ2 ∈ D(A) such that Aξ2 = λ2 ξ2 . We have β(ξ1 , ξ2 ) = hAξ1 , ξ2 i2 = λ1 hξ1 , ξ2 i2

and β(ξ2 , ξ1 ) = hAξ2 , ξ1 i2 = λ2 hξ2 , ξ1 i2 .

The symmetry of β then implies that (λ1 − λ2 )hξ1 , ξ2 i2 = 0. Hence hξ1 , ξ2 i2 = 0 and β(ξ1 , ξ2 ) = 0. It is possible to choose α1 , α2 ∈ R such that α1 ξ1 + α2 ξ2 ∈ ϕ⊥ ≡ {ξ ∈ Hr1 : hϕ, ξi1 = 0}. Then Lemma 2.12 implies 0 ≤ β(α1 ξ1 + α2 ξ2 , α1 ξ1 + α2 ξ2 ) = α12 β(ξ1 , ξ1 ) + α22 β(ξ2 , ξ2 ) = α12 hAξ1 , ξ1 i2 + α22 hAξ2 , ξ2 i2 = α12 λ1 |ξ1 |22 + α22 λ2 |ξ2 |22 < 0. This is a contradiction. Thus, A has exactly one negative eigenvalue. Hence ve is an eigenfunction of A for its second eigenvalue and Theorem 14.10 of [21] implies that ve has exactly one zero r0 ∈ (0, ∞). We are now in a position to prove that the ground state ψ is nondegenerate. Proposition 2.3 The operator T 0 (ψ) : H → H −1 , defined by T 0 (ψ)u = −∆u + u − p|x|−b ψ p−1 u

for all u ∈ H,

is an isomorphism. Proof. According to the discussion at the beginning of this section, we only need to prove that T 0 (ψ) is injective. By reductio ad absurdum, we have seen sofar (Lemmas 2.10 and 2.13) that if v ∈ ker(T 0 (ψ)) \ {0}, then v is a spherically symmetric function such that ve satisfies (2.35), has exactly one zero r0 ∈ (0, ∞)

19

and changes sign at r0 . Without loss of generality, suppose that ve(r) < 0 for all r < r0 and ve(r) > 0 for all r > r0 . The integral identities (2.39) and (2.40) imply Z ∞ rN −1 {C[f (r, ϕ)−∂2 f (r, ϕ)ϕ]+[2f (r, ϕ)+r∂1 f (r, ϕ)]}e v dr = 0 for any C ∈ R. 0

Replacing f by its definition (2.36), we get Z ∞ rN −1 {C(1 − p)r−b ϕp − 2ϕ + (2 − b)r−b ϕp }e v dr = 0. 0

Since (1 − p)r−b ϕp > 0, this yields Z ∞ rN −1 (1 − p)r−b ϕp {C − g(r)}e v dr = 0,

(2.44)

0

where g(r) = [2rb ϕ(r)1−p + (b − 2)]/(1 − p). Since ϕ is decreasing, b > 0 and 1 − p < 0, g is decreasing on (0, ∞). Therefore, on choosing C = g(r0 ), the integrand in (2.44) becomes positive on (0, ∞) \ {r0 }. This is a contradiction and so ve ≡ 0.

3

A branch of nontrivial solutions

Under the hypotheses (V1) to (V3), we consider the nonlinear problem −∆u + λu − V (x)|u|p−1 u = 0,

u ∈ H = H 1 (RN ),

(3.1)

where λ ∈ R and N ≥ 3. A solution of (3.1) is a couple (λ, u) ∈ R × H such that (3.1) holds in H −1 . Under these hypotheses, we prove part (a) of Theorem 1.1. Our proof is based on an implicit function theorem. However, since (0, 0) ∈ R × H can be a bifurcation point for (3.1), we must first rescale the variables so as to eliminate this.

3.1

Scaling and bifurcation

For λ > 0, we set λ = k 2 , k > 0, and make a change of variables via the linear operator S(k) : H → H defined by 2−b

S(k)v(x) = k p−1 v(kx) for k > 0.

(3.2)

Clearly S ∈ C((0, ∞), B(H, H)) and S(k) : H → H is invertible for all k > 0. Furthermore, if u, v ∈ H are such that u = S(k)v, then we have that |u|2L2 (RN ) = k β |v|2L2 (RN )

(3.3)

|∇u|2L2 (RN ) = k β+2 |∇v|2L2 (RN ) ,

(3.4)

and where β = [4 − 2b − N (p − 1)]/(p − 1).

20

On setting u = S(k)v and y = kx ∈ RN , the problem (3.1) for λ > 0 is transformed into −∆v + v − k −b V (y/k)|v|p−1 v = 0,

v ∈ H.

(3.5)

By assumptions (V1) and (V2), k −b V (y/k) → |y|−b as k → 0, for all y 6= 0. We are thus led to define the limit problem which was studied in Section 2. −∆v + v − |y|−b |v|p−1 v = 0,

v ∈ H.

(3.6)

The following lemma establishes the existence of a branch of non-trivial solutions of (3.5)-(3.6). Proposition 3.1 Let the hypotheses (V1)-(V3) be satisfied and let ψ be the solution of (3.6) given by Theorem 2.1. There exist k0 > 0 and v ∈ C([0, k0 ), H) ∩ C 1 ((0, k0 ), H) such that v(0) = ψ and (k, v(k)) is a non-trivial solution of (3.5) for all k ∈ (0, k0 ). Furthermore, v(k) ∈ C(RN ) ∩ L∞ (RN ) for all k ∈ [0, k0 ) and |v(k)|L∞ (RN ) remains bounded as k → 0. The first part of Proposition 3.1 is proved by applying an implicit function theorem to the function F : R × H → H −1 defined by −∆v + v − k −b V (y/k)|v|p−1 v if k > 0, F (k, v) = −∆v + v − |y|−b |v|p−1 v if k = 0, and F (k, v) ≡ F (−k, v) for all k < 0. We need the following properties of F . Remark. Observe that in the case of the homogeneous potential (1.12), F (k, v) = F (0, v) for all k ∈ R and that v(k) = ψ for all k. Lemma 3.1 Let the hypotheses (V1)-(V3) be satisfied and F : R × H → H −1 be defined as above. (i) F ∈ C(R × H, H −1 ). (ii) For all k ∈ R, F (k, ·) : H → H −1 is Fréchet differentiable and −∆w + w − pk −b V (y/k)|v|p−1 w if k > 0, Dv F (k, v)w = −∆w + w − p|y|−b |v|p−1 w if k = 0, for all v, w ∈ H. Moreover Dv F ∈ C(R × H, B(H, H −1 )). (iii) F ∈ C 1 ((0, ∞) × H, H −1 ). The proof of Lemma 3.1 is given in Appendix I. Proof of Proposition 3.1. We have F (0, ψ) = 0 in H −1 . Furthermore, by Lemma 3.1(iii), Dv F is continuous at (0, ψ) ∈ R × H and Proposition 2.3 implies that Dv F (0, ψ) is an isomorphism. Hence, by a version of the implicit function theorem (see [13], Chapter XVII, Section 4), there exists an open set U ⊂ R × H, > 0 and a unique function η : (−, ) → H such that (0, ψ) ∈ U and {(k, v) ∈ U : F (k, v) = 0} = {(k, η(k)) : k ∈ (−, )}. 21

Since F (k, v) = F (−k, v) for all (k, v) ∈ R × H, it follows that η(k) = η(−k) for all k ∈ (−, ). Moreover, η ∈ C((−, ), H) since F is continuous. The uniqueness of η and the fact that ψ 6= 0 together imply that the solutions (k, η(k)) are non-trivial for all k ∈ (−, ). Now the continuity of Dv F and the fact that the set of all isomorphisms from H to H −1 is an open subset of B(H, H −1 ) together imply that there exists a number δ ∈ (0, ) such that Dv F (k, η(k)) is an isomorphism for all k ∈ (−δ, δ). Let us fix an h ∈ (0, δ). We have F (h, η(h)) = 0. By applying the implicit function theorem to F at the point (h, η(h)) ∈ R × H, we get an 1 > 0 and a function η˜ : (h − 1 , h+ 1 ) → H such that F (k, η˜(k)) = 0 for all k ∈ (h − 1 , h+ 1 ). For 1 small enough, h − 1 > 0 and now Lemma 3.1 (ii) implies that η˜ ∈ C 1 ((h − 1 , h + 1 ), H). Thus, choosing 1 such that h − 1 > 0 and h + 1 < , we conclude by uniqueness that η˜ = η |(h−1 ,h+1 ) ∈ C 1 ((h − 1 , h + 1 ), H). Since this reasoning can be applied to any h ∈ (0, δ), there exists an 2 ∈ (0, ) such that η |(0,2 ) ∈ C 1 ((0, 2 ), H). Setting k0 = 2 and v = η |[0,2 ) completes the proof of the first part of Proposition 3.1. The additional information about v(k) follows from the next lemma, completing the proof. Lemma 3.2 (i) If (k, v) ∈ (0, ∞) × H satisfies (3.5), then v ∈ C(RN ) ∩ L∞ (RN ). (ii) For any M > 0, there exists a constant C(M ) such that |v|L∞ ≤ C(M ) for all solutions (k, v) ∈ (0, ∞) × H of (3.5) with kvk ≤ M. This result is proved in Appendix J. Proof of Theorem 1.1(a). We return to the original variables on setting λ0 = k02 and then u(λ) = S(λ1/2 )v(λ1/2 ) for all λ ∈ (0, λ0 ), 1/2

where v : [0, λ0 ) → H is given by Proposition 3.1. Then u ∈ C 1 ((0, λ0 ), H) and clearly (λ, u(λ)) is a nontrivial solution of (3.1) for all λ ∈ (0, λ0 ). The information about the limits follows immediately from (3.3) and (3.4) since |v(k)| L2 (RN ) → |ψ|L2 (RN ) > 0 and |∇v(k)| L2 (RN ) → |∇ψ|L2 (RN ) > 0 as k → 0. In particular, kv(k)k remains bounded as k → 0 since v(k) → ψ in H. By Lemma 3.2, this implies that v(k) ∈ C(RN ) ∩ L∞ (RN ) and that |v(k)| L∞ (RN ) remains bounded as k → 0. From 2−b

(3.2), we have that |uλ | L∞ (RN ) = k p−1 |v(k)| L∞ (RN ) and so λ−α |uλ | L∞ (RN ) → 0 as 2−b λ → 0 for α < 2(p−1) . The only statement in part (a) of Theorem 1.1 that still has to be proved is the positivity of uλ , equivalently v(k). This is done in part (iii) of Lemma 4.2.

4

Stability

Theorem 1.1(a) establishes the existence of standing waves for i∂t w + ∆x w + V (x) |w|p−1 w = 0 where w = w(t, x) : R × RN → C 22

(4.1)

for sufficiently small positive frequencies λ. We set ψλ (t, x) = eiλt u(λ)(x)

(4.2)

where u : (0, λ0 ) → H is the function given by Theorem 1.1(a). Thus ψλ is a periodic function of time and we define its orbit Θ(λ) = {eiθ u(λ) : θ ∈ R}. A standing wave ψλ is said to be orbitally stable in H 1 (RN , C) if, for any ε > 0, there exists δ > 0 such that, whenever kw(0, ·) − u(λ)kH 1 (RN ,C) < δ, sup inf kw(t, ·) − eiθ u(λ)kH 1 (RN ,C) < ε. t≥0 θ∈R

(4.3)

Roughly speaking, orbital stability means that if the initial data w(0, ·) is close to Θ(λ), then the solution w(t, ·) exists for all t ≥ 0 and stays close to Θ(λ) for all t. Both necessary and sufficient conditions for stability in this sense have been established in [7] for a general class of Hamiltonian systems. Most of this section is devoted to checking that, under the hypotheses (V1) to (V3), the basic assumptions 1 to 3 of [7] are satisfied. Once this is done, Theorem 3 of [7] shows that orbital stability/instability follows from d00 (λ) > 0/ < 0 where, in our context d00 (λ) =

1 d |uλ |2L2 . 2 dλ

(4.4)

d We end this section by determining the sign of dλ |uλ |2L2 for λ close to zero, thus proving part (b) of Theorem 1.1. Note that, for 1 < p < 1 + 4−2b , global in time N existence of solutions with initial conditions close to the standing wave comes as a by-product of this result.

4.1

Hamiltonian formalism

We first express (4.1) in the formalism of [7]. In that paper, the authors deal with real infinite-dimensional Hamiltonian systems of the form d ϕ(t) = JE 0 (ϕ(t)), dt

(4.5)

where ϕ : R → X, with X a real Hilbert space, J : X ∗ → X a skew-symmetric linear operator and E : X → R the Hamiltonian. In our context, we set 0 R−1 1 N H = H (R , R), X = H × H, J = , −R−1 0 where R : H → H ∗ is the Riesz isomorphism. In the context of (4.1), we consider the dynamics of ϕ(t) = (Rew(t, ·), Imw(t, ·)), t ∈ R, where w is a solution of (4.1). The relevant conserved ”physical” quantities are the energy and the charge defined respectively by Z Z Z 1 1 1 2 p+1 |∇ϕ| dx − V (x)|ϕ| dx and Q(ϕ) = |ϕ|2 dx. E(ϕ) = 2 RN p + 1 RN 2 RN 23

Clearly Q ∈ C 2 (X, R). By Proposition 2.2 of [10], |ϕ| ∈ H whenever ϕ ∈ X and the map ϕ 7→ |ϕ| is continuous from Lq × Lq into Lq for all q ≥ 1. Thus Lemma B.1(iii) shows that E ∈ C 2 (X, R). The equation (4.5) is supposed to be invariant under some one-parameter group of unitary operators from X into X. In our case, this group is given by (cos t)I −(sin t)I T (t) = , t ∈ R. (sin t)I (cos t)I Standing waves of (4.1) are solutions of (4.5) of the form ϕ(t) = T (λt)ϕ for some λ ∈ R and ϕ ∈ X. For such solutions, (4.5) reduces to E 0 (ϕ) + λQ0 (ϕ) = 0.

(4.6)

When we make the usual identification H = H 1 (RN , R) ⊂ H −1 (RN , R) = H ∗ ,

R = −∆ + 1,

this is equivalent to −∆w − V (x)|w|p−1 w + λw = 0,

w ∈ H 1 (RN , C),

(4.7)

for ϕ = (Rew, Imw) ∈ X. Theorem 3 of [7] requires three assumptions to be satisfied. Assumption 1 of [7] concerns time local well-posedness of the initial-value problem for (4.1), including conservation of charge and energy. As we show in Appendix K, this is ensured by our hypotheses (V1) and (V2) through Theorem 4.3.1 of [4]. Assumption 2 of [7] concerns the existence of a smooth branch λ 7−→ ϕλ of nontrivial solutions of (4.5). This is ensured by Theorem 1.1(a) on setting ϕλ = (u(λ), 0) for λ ∈ (0, λ0 ). Remark. Although Theorem 1.1(a) asserts the positivity of u(λ), we do not make use of this property in this section since we deduce the positivity of u(λ) from Lemma 4.2(iii) below. The stability criteria for the standing wave T (λt)ϕλ are formulated in [7] in terms of the function d : (0, λ0 ) → R defined by d(λ) = E(ϕλ ) + λQ(ϕλ ). Before stating Assumption 3 of [7] we need a little more notation. A bounded linear operator Hλ : X → X ∗ is defined by Hλ = E 00 (ϕλ ) + λQ00 (ϕλ ) and the spectrum of Hλ is defined by

∗

e : X → X is not an isomorphism , R e= S(Hλ ) = µ ∈ R : Hλ − µR 24

R 0 0 R

.

e−1 Hλ : X → X is a bounded self-adjoint In fact, as is seen from Lemma 4.2 below, R e−1 Hλ ) operator and its spectrum coincides with S(Hλ ). That is, S(Hλ ) = σ(R where, as usual, we denote by σ and σess the spectrum and essential spectrum of a self-adjoint operator mapping a space into itself. Assumption 3 of [7] requires that, for all λ in some open interval (a, b) ⊂ (0, λ0 ), the following holds: (H1) There exists aλ < 0 such that S(Hλ ) ∩ (−∞, 0) = {aλ } and e = 1. dim ker(Hλ − aλ R) (H2) ker Hλ = span{T 0 (0)ϕλ } = span{(0, u(λ))}. (H3) S(Hλ ) \ {aλ , 0} is bounded away from 0 in R. Under these assumptions, it is proven in [7] that, for all λ ∈ (a, b), the standing wave T (λt)ϕλ is orbitally stable if and only if the function d : (0, λ0 ) → R defined by d(λ) = E(ϕλ ) + λQ(ϕλ ). is convex in a neighborhood of λ. But we have d0 (λ) = E 0 (λ)

d d ϕλ + Q(ϕλ ) + λQ0 (ϕλ ) ϕλ = Q(ϕλ ) dλ dλ

since E 0 (λ) + λQ0 (ϕλ ) = 0 by (4.6). Thus, (4.4) holds and the stability/instability of T (λt)ϕλ is established if we prove that (H1)-(H3) are satisfied and that d |u(λ)|2L2 (RN ) > 0/ < 0. dλ

(4.8)

We begin with the conditions (H1) to (H3). For λ ∈ (0, λ0 ), we must study the e : X → X ∗ . It is straightforward to check that operator Hλ − µR A(λ) − µR 0 e= H λ − µR , 0 B(λ) − µR where A(λ), B(λ) : H → H −1 are defined by A(λ) = −∆ + λ − pV (x) |u(λ)|p−1

and B(λ) = −∆ + λ − V (x) |u(λ)|p−1 .

With (H3) in mind, we begin by showing that for ξ > 0 but small enough, S(Hλ ) ∩ (−∞, ξ) is a discrete set. Lemma 4.1 Let the hypotheses (V1) to (V3) be satisfied and λ ∈ (0, λ0 ). Then e : X → X ∗ is a Fredholm operator for all µ < min{1, λ}. Furthermore, H λ − µR for λ > 0 and ξ < min{1, λ}, S(Hλ ) ∩ (−∞, ξ) contains only a finite number of points. Proof. Clearly it is enough to prove these properties separately for the operators A(λ) − µR : H → H −1 and B(λ) − µR : H → H −1 . For µ 6= 1, A(λ) − µR = (1 − µ)[−∆ +

λ−µ p − V (x) |u(λ)|p−1 ]. 1−µ 1−µ 25

Recalling that u(λ) ∈ L∞ (RN ) and that |x|b V (x) is also bounded on RN , it follows from Lemma 2.8 that Cv = pV (x) |u(λ)|p−1 v defines a compact linear operator C : H → H −1 . We also have that −∆ + ζ : H → H −1 is an isomorphism for all ζ > 0. Combining these properties we see that A(λ) − µR : H → H −1 is a Fredholm operator for µ < min{1, λ}. This implies that R−1 A(λ) − µI : H → H is a Fredholm operator for µ < min{1, λ}. But R−1 A(λ) : H → H is a bounded selfadjoint operator (see part (i) of Lemma 4.2), so this means that σess (R−1 A(λ)) ⊂ [min{1, λ}, ∞). Thus, if ξ < min{1, λ}, R−1 A(λ) − µI : H → H is an isomorphism for all but a finite number of points µ ∈ (−∞, ξ). The operator B(λ) − µR can be discussed in exactly the same way, proving the lemma. By the preceding lemma, for 0 < λ < 1, the problem of studying S(Hλ ) ∩ e for µ < λ. The results in Section (−∞, λ) is reduced to the study of ker(Hλ − µR) 3 enable us to do this for λ close to 0 using perturbation theory. For this we set k = λ1/2 and use the change of variables (3.2). Then we find that S(k)u e ⇐⇒ ∈ ker(Hk2 − µR) S(k)v 2 ˜ u k A(k) − µ(−k 2 ∆ + 1) 0 ∈ ker , (4.9) ˜ v − µ(−k 2 ∆ + 1) 0 k 2 B(k) where ˜ A(k) = −∆ + 1 − pk −b V (y/k) |vk |p−1 and ˜ B(k) = −∆ + 1 − k −b V (y/k) |vk |p−1 , for all k ∈ (0, k0 ). We also define ˜ A(0) = −∆ + 1 − p|y|−b |v0 |p−1

˜ and B(0) = −∆ + 1 − |y|−b |v0 |p−1 . 1/2

Here vk ≡ v(k) denotes the function given by Proposition 3.1 for 0 ≤ k < k0 = λ0 . ˜ ˜ Note that A(k) = Du F (k, vk ) ∈ B(H, H −1 ) and that B(k) ∈ B(H, H −1 ), too. It is convenient to formulate some of our discussion in terms of operators from H to H. Therefore we define, for all k ∈ [0, k0 ), ˜ ˜ and M (k) = R−1 B(k). L(k) = R−1 A(k) We denote the associated quadratic forms by D E ˜ u = hL(k)u, uiH×H for u ∈ H, ak (u) = A(k)u, H −1 ×H D E ˜ bk (u) = B(k)u, u = hM (k)u, uiH×H for u ∈ H, H −1 ×H R ck (u) = h(−k 2 ∆ + 1)u, uiH −1 ×H = RN k 2 |∇u|2 + u2 dx for u ∈ H. 1/2

Observe that for k > 0, ck k·k.

is a norm on H that is equivalent to the usual one,

26

Lemma 4.2 (i) L(0) : H → H is a self-adjoint isomorphism, inf σess (L(0)) ≥ 1 and γ ≡ inf σ(L(0)) is a simple eigenvalue of L(0) with ker{L(0) − γI} = {η} and γ < 0. There exists δ > 0 such that ⊥

a0 (u) ≥ δ kuk2 for all u ∈ η = {u ∈ H : hη, uiH×H = 0}. (ii) M (0) : H → H is a bounded self-adjoint operator, inf σess (M (0)) ≥ 1 and 0 = inf σ(M (0)) is a simple eigenvalue of M (0) with ker M (0) = {ψ}. There exists δ > 0 such that ⊥

b0 (u) ≥ δ kuk2 for all u ∈ ψ = {u ∈ H : hψ, uiH×H = 0}. (iii) There exist k1 ∈ (0, k0 ) and γk ∈ (−∞, 0) such that, for all k ∈ (0, k1 ], e − µ(−k 2 ∆ + 1)] 6= {0}} ∩ (−∞, 0] = {γk } {µ ∈ R : ker[A(k) and e {µ ∈ R : ker[B(k) − µ(−k 2 ∆ + 1)] 6= {0}} ∩ (−∞, 0] = {0}. e − γk (−k 2 ∆ + 1)] = span{wk }, Furthermore, there exists wk ∈ H such that ker[A(k) e whereas ker B(k) = span{vk } where vk is given by Proposition 3.1. We also have ∞ N that vk ∈ L (R ) ∩ C(RN ) with vk (x) > 0 for all x ∈ RN \{0}. (iv) For all k ∈ (0, k1 ], S(Hk2 ) ∩ (−∞, 0] = {k 2 γk , 0} and we have that S(k)wk 0 2 e ker(Hk2 − k γk R) = span and ker Hk2 = span . 0 S(k)vk Proof. (i) For u, v ∈ H, hL(0)u, viH×H

D

e = A(0)u, v

Z

E H −1 ×H

=

∇u · ∇v + [1 − p |x|−b |v0 |p−1 ]uv dx

RN 00

= S (ψ)[u, v] since v0 = ψ, showing that L(0) : H → H is bounded and self-adjoint. Defining C : H → H −1 by Cu = p|y|−b |ψ|p−1 u, it follows from Lemma 2.8 that C is ˜ compact. But then L(0) = (−∆ + 1)−1 A(0) = I − R−1 C where R−1 C : H → H is compact. It follows that L(0) − µI : H → H is a Fredholm operator for all µ < 1 and so σess (L(0)) ⊂ [1, ∞). e By Proposition 2.3, L(0) : H → H is an isomorphism since A(0) = T (ψ). Hence H can be split as an orthogonal direct sum E ⊕ F where there exists > 0 such that L(0)E = E, hL(0)u, uiH×H ≤ − kuk2 for u ∈ E, L(0)F = F, hL(0)u, uiH×H ≥ kuk2 for u ∈ F. However Lemma 2.5 implies that dim E = 1. Setting E = span{η}, part (i) is proved. 27

(ii) As in part (i), ˜ M (0) = R−1 B(0) = I − R−1 C : H → H and M (0) − µI = (1 − µ)I − R−1 C where now C : H → H −1 is defined by Cv = |y|−b |v0 |p−1 v. Thus M (0) − µI : H → H is a bounded self-adjoint Fredholm operator for µ < 1 and consequently, inf σess (M (0)) ≥ 1. Next we observe that by definition, v0 = v(0) = ψ > 0 e and we also have that M (0)v0 = R−1 B(0)ψ = 0. Hence 0 ∈ σ(M (0)). But M (0) ∈ B(H, H) is self-adjoint and so R 0 ≥ Λ ≡ inf σ(M (0)) = inf{hM (k)w, wiH×H : w ∈ H and RN w2 dx = 1} D E R = inf{ [−∆ + 1 − |x|−b ψ p−1 ]w, w : w ∈ H and RN w2 dx = 1} H −1 ×H R R 2 −b p−1 = inf{ RN |∇w| + [1 − |x| ψ ]w2 dx : w ∈ H and RN w2 dx = 1}. Since Λ < 1 = inf σess (M (0)), it is an eigenvalue of M (0) and all minimizers of the above quadratic form are eigenfunctions associated with Λ. Thus, if M (0)φ = Λφ with φ ∈ H \ {0}, we also have that M (0) |φ| = Λ |φ|. Hence 0 = hM (0)ψ, |φ|iH×H = hψ, M (0) |φ|iH×H = Λ hψ, |φ|iH×H R R = Λ RN [(−∆ + 1)ψ] |φ| dx = Λ RN [|x|−b ψ p ] |φ| dx. Since ψ > 0 on RN , this implies that Λ = 0. Furthermore this means that v = |φ| satisfies Lv = 0 in H −1 where L = ∆ − [1 − |x|−b ψ p−1 ]. By a trivial modification of Lemma 2.9 we see that |φ| ∈ C(RN ) ∩ L∞ (RN ). Since [1 − |x|−b ψ p−1 (x)] is bounded for x in compact subsets of RN \ {0}, it now follows easily from the Harnack inequality (Theorem 8.20 of [9]) that |φ(x)| > 0 for Rall x ∈ RN \ {0}. Hence φ does not change sign on RN \ {0} and consequently RN φv0 dx 6= 0 for any eigenfunction φ associated with the eigenvalue Λ = 0. It follows that 0 = inf σ(M (0)) is a simple eigenvalue of M (0). Thus there exists δ > 0 such that b0 (u) = hM (0)u, uiH×H ≥ δ kuk2 for all u ∈ ψ ⊥ . (iii) For any k ∈ (0, k0 ) and u ∈ H \ {0}, we have that h[Du F (k, vk ) − Du F (0, v0 )]u, ui ak (u) − a0 (u) = . 2 kuk kuk2 Using Lemma 3.1(ii), it follows that there exists k1 ∈ (0, k0 ) such that, for all k ∈ (0, k1 ], ak (u) − a0 (u) δ ≤ for all u ∈ H \ {0}, 2 kuk2 where δ > 0 is given by part (i). It follows that ak (u) ≥

δ kuk2 for all u ∈ F = η ⊥ . 2

Since L(0)η = γη and γ < 0, k1 can be chosen so that ak (η) ≤ k ∈ (0, k1 ]. 28

γ 2

kηk2 < 0 for all

Let us now fix k ∈ (0, k1 ] and consider the minimisation problem ak (u) mk = inf{ak (u) : u ∈ H and ck (u) = 1} = inf : u ∈ H \ {0} . ck (u) We have that mk ≤ ak (η)/ck (η) < 0 and that ak (u) = kuk2 − hCu, uiH −1 ×H for all u ∈ H, where C : H → H −1 is defined by Cu = pzk |x|−b u with zk (x) = (|x|/k)b V (x/k) |vk (x)|p−1 . Recalling that vk ∈ L∞ (RN ), it follows that zk ∈ L∞ (RN ) and hence, from Lemma 2.8, that C is compact. This implies that ak : H → R is weakly sequentially lower semi-continuous. Now consider a minimizing sequence {un } ⊂ H: ak (un ) → mk and ck (un ) = 1. The sequence {un } is bounded in H and so we may suppose that un * u∗ weakly in H. Then ak (u∗ ) ≤ lim inf n→∞ ak (un ) = mk < 0 and so u∗ 6= 0. Similarly, t = ck (u∗ ) ≤ 1 and since u∗ 6= 0, t > 0. Thus ck (t−1/2 u∗ ) = 1 and mk ≤ ak (t−1/2 u∗ ) = t−1 ak (u∗ ) ≤ t−1 mk ≤ mk . Hence, t = ck (u∗ ) = 1 and ak (u∗ ) = mk . Therefore there is a Lagrange multiplier 2 e γk such that A(k)u ∗ = γk (−k ∆ + 1)u∗ and, in fact, γk = mk < 0. e Suppose now that µ ≤ 0 and that A(k)z = µ(−k 2 ∆ + 1)z for some z ∈ H \ {0}. Then D E D E e e µ h(−k 2 ∆ + 1)z, u∗ iH −1 ×H = A(k)z, u∗ = A(k)u ∗, z H −1 ×H 2

H −1 ×H

= γk h(−k 2 ∆ + 1)u∗ , ziH −1 ×H = γk h(−k ∆ + 1)z, u∗ iH −1 ×H . E D e = 0 if µ 6= γk . Now, since Thus h(−k 2 ∆ + 1)z, u∗ iH −1 ×H = A(k)z, u∗ H −1 ×H

2

h(−k ∆ + 1)z, u∗ iH −1 ×H = sck (u∗ ) = s if z = su∗ for some s ∈ R, it follows that z and u∗ are linearly independent. Furthermore, for all s, t ∈ R, D E e ak (su∗ + tz) = s2 ak (u∗ ) + t2 ak (z) + 2st A(k)z, u∗ = s2 γk + t2 µck (z) ≤ 0, H −1 ×H

showing that ak (u) ≤ 0 for all u ∈ S = span{u∗ , z}. Since dim S = 2, S ∩ F 6= {0} and we have a contradiction because ak (u) > 0 for all u ∈ F \ {0}. Therefore we must have µ = γk and it again follows that ak (u) ≤ 0 for all u ∈ S. The positivity of ak on F now implies that dim S = 1. Setting wk = u∗ , we have proved that e e ker[A(k) − γk (−k 2 ∆ + 1)] = span{wk } and that ker[A(k) − µ(−k 2 ∆ + 1)] = {0} for µ ∈ (−∞, 0] \ {γk }. e We now turn to the properties of B(k). The constant k1 can be adjusted so that, for all k ∈ (0, k1 ], bk (u) − b0 (u) 1 ak (u) − a0 (u) δ = ≤ for all u ∈ H \ {0}, p 2 kuk2 kuk2 29

where δ > 0 is given by part (ii). It follows that bk (u) ≥

δ kuk2 for all u ∈ G = ψ ⊥ . 2

e e Noting that B(k)v k = F (k, vk ) we have that vk ∈ ker B(k). Suppose now that e µ ≤ 0 and that B(k)z = µ(−k 2 ∆ + 1)z for some z ∈ H\{0}. Then D E D E

e e µ (−k 2 ∆ + 1)z, vk H −1 ×H = B(k)z, vk = B(k)v , z = 0, k H −1 ×H

H −1 ×H

showing that vk and z are linearly independent if µ < 0. But then bk (u) ≤ 0 for all u ∈ T = span{vk , z}, contradicting the fact that bk (u) > 0 for all u ∈ G \ {0}. Hence µ = 0 and we now have that bk (u) = 0 for all u ∈ T . Hence T ∩G = {0} and e this implies that z ∈ span{vk }. Thus we have proved that ker B(k) = span{vk } 2 e and that ker[B(k) − µ(−k ∆ + 1)] = {0} for µ < 0. Setting Mk = inf{bk (u) : u ∈ H and ck (u) = 1}, it follows as for mk , that if Mk < 0 then Mk is attained and there exists w∗ ∈ H 2 e such that ck (w∗ ) = 1 and B(k)w ∗ = Mk (−k ∆ + 1)w∗ . Since we have shown that e − µ(−k 2 ∆ + 1)] = {0} if µ < 0, this means that Mk ≥ 0 and consequently ker[B(k) Mk = 0 because bk (vk ) = 0. But bk (|vk |) = bk (vk ), from which it follows that e B(k) |vk | = 0 and hence that |vk | ∈ span{vk }. Recalling that vk → v0 = ψ > 0 in H as k → 0 , it follows that vk ≥ 0 a.e. on RN . By Lemma 3.2, vk ∈ C(RN )∩L∞ (RN ) and so V (x/k)vkp−1 (x) is bounded on any compact subset of RN \{0}. Harnack’s inequality (Theorem 8.20 of [9]) for the operator Lk = ∆ − [1 − k −b V (x/k)vkp−1 (x)] now implies that vk > 0 on RN \ {0}. This completes the proof of part (iii). (iv) This follows from (4.9) and part (iii). Lemma 4.3 Under the assumptions (V1) to (V3), there exists λ1 > 0 such that, for all λ ∈ (0, λ1 ), the conditions (H1), (H2) and (H3) are satisfied. Proof. Let λ1 = k1 and then set aλ = λγ(λ1/2 ) for λ ∈ (0, λ1 ]. Recalling that u(λ) = S(λ1/2 )v(λ1/2 ) we see that Lemmas 4.1 and 4.2 yield the result.

4.2

Verification of the stability criterion

For all k ∈ (0, k1 ), we have by (3.2) that |u(k 2 )|2L2 = |S(k)vk |2L2 = k β |vk |2L2 where β = [4 − 2b − N (p − 1)]/(p − 1). Hence

d d |u(k 2 )|2L2 = βk β−1 |vk |2L2 + k β 2 vk , vk L2 dk dk

d = k β−1 {β|vk |2L2 + 2k vk , vk L2 }. dk

(4.10)

Since |vk |L2 → |v0 |L2 = |ψ|L2 > 0 as k → 0, the right hand

side of (4.10) will have d the same sign as β for k small enough if we prove that k vk , dk vk L2 → 0 as k → 0. 30

Lemma 4.4 Let v ∈ C([0, k0 ), H) ∩ C 1 ((0, k0 ), H) be the function given by Proposition 3.1. Then

d k vk , vk L2 → 0 as k → 0. (4.11) dk Proof. Since F (k, vk ) = 0, we have Dk F (k, vk ) + Dv F (k, vk )wk = 0 for all k ∈ d vk . According to the proof of Proposition 3.1, there exists (0, k0 ) where wk = dk k ∗ ∈ (0, k0 ) such that Dv F (k, vk ) is an isomorphism for all k ∈ (0, k ∗ ). Hence, wk = −[Dv F (k, vk )]−1 Dk F (k, vk ) for all k ∈ (0, k ∗ ). We find that kwk = −[Dv F (k, vk )]−1 k −b W (y/k)vkp , where W is the function in (V3). (Look at the expression for Dk G(k, u) = − Dk F (k, u) in the proof of Lemma 3.1 and recall that vk ≥ 0.) Since

[Dv F (k, vk )]−1 −1 → [Dv F (0, ψ)]−1 −1 as k → 0, B(H

,H)

B(H

,H)

it suffices to show that ξk ≡ k −b W (y/k)vkp → 0 in H −1 as k → 0.

(4.12)

Then we will have kwk → 0 in H and (4.11) will hold. But ξk ≡ zk (x) |x|−b vkp for x 6= 0 and k ∈ (0, k0 ) where zk (x) = (|x|/k)b W (x/k). It follows from (V3) that |zk |L∞ ≤ Q = sup |x|b |W (x)| < ∞ x6=0

and that zk (k, x) → 0 as k → 0 for all x 6= 0. Hence R −b p hξk , φi −1 vk |φ| dx N |zk | |x| H ×H ≤ sup R . kξk kH −1 = sup kφk kφk φ∈H\{0} φ∈H\{0} Now

R sup

RN

φ∈H\{0}

|zk | |x|−b v0p |φ| dx → 0 as k → 0 kφk

by Lemma A.1(ii) and R sup φ∈H\{0}

R sup φ∈H\{0}

RN

RN

|zk | |x|−b |vkp − v0p | |φ| dx ≤ kφk

|zk | |x|−b p |vk + v0 |p−1 |vk − v0 | |φ| dx ≤ pQD kvk + v0 kp−1 kvk − v0 k kφk

by (A.2) below. This shows that kξk kH −1 → 0 as k → 0, completing the proof of the lemma. Proof of Theorem 1.1(b). Note that β 6= 0 if and only if p 6= 1 + 4−2b . The N ∗ ¯ ¯ discussion above shows that there exists k ∈ (0, k ) such that, for all k ∈ (0, k), d dk d 1 d 2d00 (λ) = |u(λ)|2L2 = |u(k 2 )|2L2 = |u(k 2 )|2L2 dλ dλ dk 2k dk has the same sign as β for p 6= 1 + 4−2b . By Lemma 4.3, we may choose k so that N ¯ where λ = k 2 . the spectral hypotheses (H1)-(H3) are satisfied on the interval (0, λ) Hence part (b) of Theorem 1.1 follows from Theorem 3 of [7]. 31

A

Estimates

Lemma A.1 Suppose that b ∈ (0, 2) and 1 0 such that Z |z| |x|−b |v|p−1 − |w|p−1 |ϕ| |ξ| dx RN ≤ C{ z[ |v|p−1 − |w|p−1 ] Lβ |ϕ|Lγ |ξ|Lγ + z[ |v|p−1 − |w|p−1 ] Lσ |ϕ|Lτ |ξ|Lτ }

(A.1)

for all z ∈ L∞ (RN ) and v, w, ϕ, ξ ∈ H, where (p − 1)β = γ = 2∗ and (p − 1)σ = τ = p + 1. In particular, there exists D > 0 such that Z |z| |x|−b |v|p−1 |ϕ| |ξ| dx ≤ D |z|L∞ kvkp−1 kϕk kξk (A.2) RN

for all z ∈ L∞ (RN ) and v, w, ϕ, ξ ∈ H. (ii) If zh ∈ L∞ (RN ) with |zh |L∞ ≤ M < ∞ for all h ≥ 0 and zh (x) → 0 for almost all x ∈ RN as h → 0, then, for all v ∈ H, R |zh | |x|−b |v|p−1 |ϕ| |ξ| dx RN → 0 as h → 0. (A.3) sup kϕk kξk ϕ,ξ∈H\{0} Proof. Using Hölder’s inequality with four exponents, we have Z |z| |x|−b |v|p−1 − |w|p−1 |ϕ| |ξ| dx B(0,1) Z ≤{ |x|−bα dx}1/α z[ |v|p−1 − |w|p−1 ] Lβ |ϕ|Lγ |ξ|Lγ B(0,1)

where α1 + β1 + γ2 = 1 with α, β, γ ≥ 1. With our choice of β and γ, these conditions are satisfied with α = 1/( N2 − p−1 ). Since bα < N by the restrictions on p, we have 2∗ that Z |x|−bα dx < ∞. B(0,1)

Hence, Z

|z| |x|−b |v|p−1 − |w|p−1 |ϕ| |ξ| dx

B(0,1)

≤ K z[ |v|p−1 − |w|p−1 ]

Z Lβ

|ϕ|Lγ |ξ|Lγ where K = {

|x|−bα dx}1/α . (A.4)

B(0,1)

Since |x|−b ≤ 1 on RN \B(0, 1), we also have that Z |z| |x|−b |v|p−1 − |w|p−1 |ϕ| |ξ| dx RN \B(0,1) ≤ z[ |v|p−1 − |w|p−1 ] Lσ |ϕ|Lτ |ξ|Lτ 32

(A.5)

where σ1 + τ2 = 1 with σ, τ ≥ 1. Setting C = max{K, 1}, we obtain (A.1). Putting w = 0 in (A.1) we get Z |z| |x|−b |v|p−1 |ϕ| |ξ| dx RN

≤ C |z|L∞ {|v|p−1 |ϕ|Lγ |ξ|Lγ + |v|p−1 |ϕ|Lτ |ξ|Lτ } L(p−1)β L(p−1)σ = C |z|L∞ {|v|p−1 |ϕ|L2∗ |ξ|L2∗ + |v|Lp−1 p+1 |ϕ|Lp+1 |ξ|Lp+1 L2∗ ≤ 2CC1 |z|L∞ kvkp−1 kϕk kξk , by the Sobolev embedding since 2 < p + 1 < 2∗ , proving (A.2). (ii) Since γ, τ ∈ [2, 2∗ ], it follows from (A.1), that there is a constant C2 such that Z |zh | |x|−b |v|p−1 |ϕ| |ξ| dx ≤ C2 { zh |v|p−1 Lβ + zh |v|p−1 Lσ } kϕk kξk RN

for ϕ, ξ ∈ H where zh |v|p−1 β ≤ M β |v|2∗ and zh |v|p−1 σ ≤ M σ |v|p+1 a.e. on RN . Dominated convergence now shows that zh |v|p−1 β → 0 and zh |v|p−1 σ → 0 as h → 0, L L as required.

B

Some useful continuity and differentiability properties

The following applies to the function Φ defined in Section 2 when we put z = 1 and also to the function F (k, ·) defined in Section 3 whith z(x) = (|x|/|k|)b V (x/|k|). Lemma B.1 Let b ∈ (0, 2) and 1 < p < 1 + (4 − 2b)/(N − 2). Consider a function z ∈ L∞ (RN ) and f (s) = |s|p−1 s for s ∈ R. (i) For u ∈ H, we set Ψ(u) = z |x|−b f (u).

(B.1)

Then Ψ ∈ C(H, H −1 ) and there is a constant C such that kΨ(u)kH −1 ≤ C |z|L∞ kukp for all u ∈ H. (ii) For u, ϕ, ξ ∈ H, we set Z Θ(u)[ϕ, ξ] =

z |x|−b |u|p−1 ϕξ dx

(B.2)

RN

Then Θ(u) : H × H → R is a bounded, symmetric bilinear form. There is a unique operator B(u) ∈ B(H, H −1 ) such that hB(u)ϕ, ξiH −1 ×H = Θ(u)[ϕ, ξ] for all ϕ, ξ ∈ H. Furthermore, B ∈ C(H, B(H, H −1 )) and there is a constant C such that kB(u)kB(H,H −1 ) ≤ C |z|L∞ kukp−1 for all u ∈ H. 33

(B.3)

(iii) For the function Φ defined by 1 Φ(u) = p+1

Z

z|x|−b |u|p+1 dx,

RN

we have that Φ ∈ C 2 (H, R) with Φ0 (u)ϕ = hΨ(u), ϕiH −1 ×H and Φ00 (u)[ϕ, ξ] = p hB(u)ϕ, ξiH −1 ×H for all u, ϕ, ξ ∈ H. Proof. (i) Observing that |f (u) − f (v)| ≤ |u|p−1 |u − v| + |u|p−1 − |v|p−1 |v| , we have that, for u, v, ϕ ∈ H, Z |z| |x|−b |f (u) − f (v)| |ϕ| dx RN Z Z −b p−1 ≤ |z| |x| |u| |u − v| |ϕ| dx + RN

|z| |x|−b |u|p−1 − |v|p−1 |v| |ϕ| dx.

RN

Putting v = 0, we see that Z Z −b |z| |x| |f (u)| |ϕ| dx ≤ RN

RN

|z| |x|−b |u|p |ϕ| dx ≤ D |z|L∞ kukp kϕk

by (A.2), proving that Ψ(u) ∈ H −1 with kΨ(u)kH −1 ≤ D |z|L∞ kukp for all u ∈ H. Then, using (A.1) and (A.2), we find that there exists a constant K such that Z |z| |x|−b |f (u) − f (v)| |ϕ| dx ≤ D |z|L∞ kukp−1 ku − vk kϕk RN + K |z|L∞ { |u|p−1 − |v|p−1 Lβ + |u|p−1 − |v|p−1 Lσ } kvk kϕk where (p − 1)β = 2∗ and (p − 1)σ = p + 1. This implies that kΨ(u) − Ψ(v)kH −1 ≤ D |z|L∞ kukp−1 ku − vk + K |z| ∞ { |u|p−1 − |v|p−1 L

Lβ

+ |u|p−1 − |v|p−1 Lσ } kvk .

Now the Nemytskii operator u 7→ |u|p−1 is continuous from L(p−1)β (RN ) into Lβ (RN ) and from L(p−1)σ (RN ) into Lσ (RN ). Since H is continuously embedded in both L(p−1)β (RN ) and L(p−1)σ (RN ), this proves that Ψ ∈ C(H, H −1 ). (ii) For u, ϕ, ξ ∈ H, Z |Θ(u)[φ, ξ]| ≤ |z| |x|−b |u|p−1 |ϕ| |ξ| dx ≤ D |z|L∞ kukp−1 kϕk kξk , RN

by (A.2), showing that Θ(u) : H × H → R is a bounded form and that kB(u)kB(H,H −1 ) ≤ D |z|L∞ kukp−1 for all u ∈ H. The continuity of B follows from (A.1) since it implies that 34

h[B(u) − B(v)]ϕ, ξi −1 H ×H ≤ K |z|L∞ { |u|p−1 − |v|p−1 Lβ + |u|p−1 − |v|p−1 Lσ } kϕk kξk and we have already seen that u 7→ |u|p−1 maps H continuously into both Lβ (RN ) and Lσ (RN ). (iii) It is sufficient to prove that Φ is Gâteaux differentiable with Φ0 = Ψ and then that Ψ : H → H −1 is Fréchet differentiable with Ψ0 = pB. For t ∈ R\{0} and u, v ∈ H, we have that Φ(u + tv) − Φ(u) = t

Z

−b

z |x| RN

|u + tv|p+1 − |u|p+1 { }dx (p + 1)t

where |u + tv|p+1 − |u|p+1 ≤ (p + 1)f (u + stv)tv for some s = s(t, x) ∈ [0, 1] and so

|u + tv|p+1 − |u|p+1 p ≤ |u + stv| |v| ≤ {|u| + |v|}p |v| (p + 1)t

for 0 < |t| ≤ 1. Noting that |u| + |v| ∈ H, (A.2) shows that z |x|−b {|u| + |v|}p |v| ∈ L1 (RN ) and so it follows from dominated convergence that Z Φ(u + tv) − Φ(u) lim = z |x|−b f (u)vdx, t→0 t RN showing that Φ is Gâteaux differentiable at u with Φ0 (u)v = hΨ(u), viH −1 ×H . Next, for ϕ ∈ H, we consider Z f (u + tv) − f (u) Ψ(u + tv) − Ψ(u) z |x|−b { = ,ϕ }ϕdx t t RN H −1 ×H where |f (u + tv) − f (u)| = p |u + stv|p−1 |tv| for some s = s(t, x) ∈ [0, 1] ≤ p{|u| + |v|}p−1 |tv| for 0 < |t| ≤ 1. Thus f (u + tv) − f (u) −b z |x| { }ϕ ≤ |z| |x|−b p{|u| + |v|}p−1 |v| |ϕ| . t

(B.4)

Since |u|+|v| ∈ H, it follows from (A.2) that |z| |x|−b {|u|+|v|}p−1 |v| |ϕ| ∈ L1 (RN ) and hence, by dominated convergence, that Ψ(u + tv) − Ψ(u) lim ,ϕ = p hB(u)v, ϕiH −1 ×H . t→∞ t H −1 ×H

35

This means that hΨ(u + sv), ϕiH −1 ×H is differentiable with respect to s ∈ R and d hΨ(u + sv), ϕiH −1 ×H = p hB(u + sv)v, ϕiH −1 ×H , which is a continuous function ds of s by (ii). Hence Z 1 d hΨ(u + v) − Ψ(u), ϕiH −1 ×H = hΨ(u + sv), ϕiH −1 ×H ds 0 ds Z 1 p hB(u + sv)v, ϕiH −1 ×H ds = 0

and so Z

1

h[B(u + sv) − B(u)]v, ϕiH −1 ×H ds.

hΨ(u + v) − Ψ(u) − pB(u)v, ϕiH −1 ×H = p 0

Finally hΨ(u + v) − Ψ(u) − pB(u)v, ϕi −1 H ×H Z 1 kB(u + sv) − B(u)kB(H,H −1 ) kvk kϕk ds, ≤p 0

showing that Z

1

kΨ(u + v) − Ψ(u) − pB(u)vkH −1 ≤ p

kB(u + sv) − B(u)kB(H,H −1 ) ds kvk 0

and hence that Ψ : H → H −1 is differentiable at u with Ψ0 (u) = pB(u).

C

Proof of Lemma 2.4

Let {un } ⊂ H be such that un * u weakly in H. We have Z |φ(un ) − φ(u)| ≤ |x|−b |un |p+1 − |u|p+1 dx. RN

For any R > 0, we have, by Hölder’s inequality, Z |x|−b |un |p+1 − |u|p+1 dx B(0,R)

Z ≤{ B(0,R)

−br

|x|

1/r

dx}

Z

|un |p+1 − |u|p+1 s dx}1/s

{ B(0,R)

for all r ≥ 1, 1/r + 1/s = 1. The first integral of the right hand side converges if N − br > 0, which is equivalent to s > N/(N − b). On the other hand, by the compactness of the Sobolev embeddings on bounded domains and the continuity of the mapping u 7→ |u|p+1 from L(p+1)s (B(0, R)) to Ls (B(0, R)), s ≥ 1, one shall get |un |p+1 − |u|p+1 s → 0 as n → ∞, L (B(0,R))

36

if one can choose s ≥ 1 such that (p + 1)s ∈ [1, N2N ). It is possible to find s ≥ 1 −2 satisfying both conditions if 2N N < , N −b (p + 1)(N − 2) which is equivalent to p < 1 + (4 − 2b)/(N − 2). Hence Z |x|−b |un |p+1 − |u|p+1 dx ≤ C |un |p+1 − |u|p+1 Ls (B(0,R)) → 0 as n → ∞. B(0,R)

Let ε > 0. For R ≥ ε−1/b , Z Z −b p+1 p+1 |x| |un | − |u| dx ≤ ε RN \B(0,R)

|un |p+1 + |u|p+1 dx ≤ Cε

RN \B(0,R)

by the Sobolev embedding theorem and the boundedness of {un } in H. Since ε > 0 is arbitrary, this concludes the proof.

D

Proof of Lemma 2.7

We shall derive the limits (i) and (ii) of Lemma 2.7 from the asymptotic behavior of Hr1 -solutions of the equation (D.1). Let f : (0, ∞) → R. We say that f (r) → ∞ exponentially as r → ∞ if there exists > 0 such that e−r f (r) → ∞ as r → ∞. Lemma D.1 Let γ > 0, µ ≥ 0 and Q : (0, ∞) → R be a continuous function such that rb Q(r) is bounded on (0, ∞) and Q(r) → 0 exponentially as r → ∞. Consider the ordinary differential equation v 00 +

N −1 0 µ v − γv − 2 v + Q(r)v = 0 for r > 0. r r

(D.1)

(A) There exist two linearly p independent solutions η0 , η1 of (D.1) on (0, ∞) such that, for α± = (2 − N ± (N − 2)2 + 4µ)/2, we have: (Ai) η0 (r) = [1 + δ0 (r)]rα+ and η00 (r) = [−α+ + %0 (r)]rα+ −1 as r → 0, where δ0 (r), %0 (r) → 0 as r → 0. (Aii) η1 (r) = [1 + δ1 (r)]rα− and η10 (r) = [α− + %1 (r)]rα− −1 as r → 0, where δ1 (r), %1 (r) → 0 as r → 0. (B) There exist two linearly independent solutions ξ0 , ξ1 of (D.1) on (0, ∞) having the following properties: √

(Bi) ξ0 (r) = ε0 (r)e− γr and ξ00 (r) = δ0 (r)e− where ε0 (r), δ0 (r) → 0 as r → ∞.

√

(Bii) |ξ1 (r)|, |ξ10 (r)| → ∞ exponentially as r → ∞.

37

γr

as r → ∞,

Proof of Lemma 2.7. Since v ∈ Hr1 and α− − 1 ≤ 1 − N ≤ −2 for all µ ≥ 0, v has the asymptotic behavior (Ai) of Lemma D.1 as r → 0, respectively (Bi) as r → ∞. Thus (ii) is satisfied. For (i), note that α+ ≥ 0 for all µ ≥ 0. Proof of Lemma D.1. We use Theorem 9.1, Chap. XI of [11]. Let us first consider the behavior as r → 0. We write (D.1) in the form (rN −1 v 0 )0 + rN −1 [−γ − µ/r2 + Q(r)]v = 0.

(D.2)

Then we perform the change of variables r = 1 − t,

u(t) = v(1 − t) and h(t) = Q(1 − t).

Equation (D.2) becomes [(1 − t)N −1 u0 ]0 + (1 − t)N −1 [−γ − µ/(1 − t)2 + h(t)]u = 0.

(D.3)

We will deduce the behavior of v as r → 0+ from the behavior of u as t → 1− . In the notations of Theorem 9.1, Chap. XI of [11], we have ω = 1,

p(t) = (1 − t)N −1

and q(t) = (1 − t)N −1 [−γ − µ/(1 − t)2 + h(t)].

As t → 1, the leading term in (D.3) is −µ/(1 − t)N −3 . We shall thus compare (D.3) with [(1 − t)N −1 x0 ]0 − µ(1 − t)3−N x = 0. (D.4) That is, we choose q0 (t) = −µ(1 − t)3−N . Equation (D.4) is nonoscillatory at t = 1 and we have principal and non-principal solutions respectively given by x0 (t) = (1 − t)α+

and x1 (t) = (1 − t)α− ,

where

p (N − 2)2 + 4µ . α± = 2 Note that α− < 0 for all µ ≥ 0 and α+ ≥ 0 for all µ ≥ 0, with equality if and only if µ = 0. Let us now check the hypothesis (9.22) of Theorem 9.1. We have that Z 1 Z ω |x0 (t)x1 (t)| · |q(t) − q0 (t)|dt = (1 − t)2−N (1 − t)N −1 | − γ + h(t)|dt Z 1 Z 1 ≤ γ(1 − t)dt + (1 − t)|h(t)|dt < ∞ 2−N ±

b 1−b b since R ω (1−t)|h(t)| = (1−t) |h(t)|(1−t) with (1−t) |h(t)| bounded and 1−b > −1. ( means that the integral is taken over a left neighborhood of ω.) Thus, the hypotheses of Theorem 9.1 are verified and we conclude that (D.3) has a pair of local solutions u0 , u1 satisfying, as t → 1,

ui ∼ xi and

p u0i p x0i = +o ui xi 38

(D.5) 1 |x0 x1 |

,

(D.6)

for i = 0, 1. (D.5) immediately gives u0 (t) = [1 + δ˜0 (t)](1 − t)α+

and u1 (t) = [1 + δ˜1 (t)](1 − t)α−

as t → 1,

where δ˜0 (t), δ˜1 (t) → 0 as t → 1. In the original variables, this implies the existence of two local solutions η0 , η1 of (D.1) (ηi (r) = ui (1 − r)) such that η0 (r) = [1 + δ0 (r)]rα+

and η1 (r) = [1 + δ1 (r)]rα−

as r → 0,

where δ0 (r), δ1 (r) → 0 as r → 0. For the derivatives, (D.6) implies (1 − t)N −1 u0i (1 − t)N −1 x0i = + o((1 − t)N −2 ), ui xi

i = 0, 1.

After some algebra, one finds that, as t → 1, and u01 (t) = [−α− + %˜1 (t)](1 − t)α− −1 ,

u00 (t) = [α+ + %˜0 (t)](1 − t)α+ −1

where %˜0 (t), %˜1 (t) → 0 as t → 1. In the original variables, this reads η00 (r) = [−α+ + %0 (r)]rα+ −1

and η10 (r) = [α− + %1 (r)]rα− −1

as r → 0,

whith %0 (r), %1 (r) → 0 as r → 0. This proves (A). Let us now study the behavior of solutions as r → ∞. We make the change of 1−N variables v(r) = r 2 ψ(r). (D.1) then becomes (N − 1)(N − 3) 1 00 ψ − γψ − + µ 2 ψ + Q(r)ψ = 0. (D.7) 4 r We compare this equation with x00 − γx = 0, which has principal and non-principal solutions respectively given by x0 (r) = e−

√

√

γr

and x1 (r) = e

γr

.

The hypotheses of Theorem 9.1, Chap. XI of [11] are satisfied since Z ∞ Z ∞ (N − 1)(N − 3) 1 Q(r) − |x0 (r)x1 (r)|·|q(r)−q0 (r)|dr = + µ dr < ∞ 4 r2 by the exponential decay of Q as r → ∞. Then (D.7) possesses two solutions ψ0 , ψ1 such that ψi ∼ xi (D.8) and

x0 ψi0 = i +o ψi xi

1 |x0 x1 |

as r → ∞,

(D.9)

for i = 0, 1. Hence ψ0 (r) = [1 + θ0 (r)]e−

√

γr

√

,

and

ψ1 (r) = [1 + θ1 (r)]e

γr

ψi0 √ = ± γ + λi (r), ψi where θi (r), λi (r) → 0 as r → ∞ for i = 0, 1. On returning to the original variables, it is not difficult to see that the exponential behavior (B) holds. 39

E

Proof of Lemma 2.8

Let {un } ⊂ H, un * u weakly in H. We prove that Cun → u in H −1 . Let ϕ ∈ H, kϕk ≤ 1. We have Z |hCun − Cu, ϕi| ≤ |x|−b |w| |un − u||ϕ|dx. RN

Let R > 0. Since w is bounded, the generalized Hölder’s inequality gives Z |x|−b |w| |un − u||ϕ|dx B(0,R) Z Z −bα 1/α ≤ K{ |x| dx} { |un − u|β dx}1/β |ϕ|L2∗ (B(0,R)) B(0,R) ZB(0,R) Z |un − u|β dx}1/β , ≤ K{ |x|−bα dx}1/α { B(0,R)

(E.1)

B(0,R)

where α ≥ 1 and 1/α + 1/β + 1/2∗ = 1, i.e. 1/α + 1/β = (N + 2)/2N . We shall use the compactness of the Sobolev embedding on bounded domains. The first integral in the right hand side of (E.1) is finite if N − bα > 0. This is equivalent to β > 2N/(N + 2 − 2b). But 2N/(N + 2 − 2b) < 2∗ since b < 2 and so we can 2N choose β ∈ ( N +2−2b , 2∗ ]. Hence Z |x|−b |w| |un − u||ϕ|dx → 0 as n → ∞. (E.2) B(0,R)

We estimate the integral outside the ball B(0, R) as follows. For ε > 0 given, there exists Rε > 0 such that |x|−b |w(x)| ≤ ε for all x ∈ RN such that |x| ≥ Rε . Then, by the Cauchy-Schwarz inequality, Z Z −b |x| |w| |un − u||ϕ|dx ≤ ε RN \B(0,Rε )

|un − u||ϕ|dx

RN \B(0,Rε )

≤ ε|un − u|L2 (RN \B(0,Rε )) |ϕ|L2 (RN \B(0,Rε )) ≤ ε|un − u|L2 (RN \B(0,Rε )) ≤ 2M ε, (E.3) where M > 0 is such that kun k ≤ M for all n and kuk ≤ M . Since ε > 0 is arbitrary, (E.3) and (E.2) with R = Rε yield the result.

F

Proof of Lemma 2.9

First we refer to [20], Proposition 4.3, for the proof of the following result. Proposition F.1 Let h be a tempered distribution on RN and consider the equation ∆u + λu + h = 0 (F.1) in the sense of distributions, for λ < 0. 40

(i) There is a unique tempered distribution satisfying (F.1). (ii) If h ∈ Lq (RN ) for some 1 ≤ q ≤ ∞, then the solution u ∈ Lq (RN ). Furthermore, if 1 < q < ∞, then u ∈ W 2,q (RN ). Proof of Lemma 2.9. We proceed by a boot-strap argument. For v ∈ H, set Kv = p|x|−b ψ p−1 v. Then (2.25) reads −∆v + v = Kv. We shall prove that Kv ∈ Lr (RN ) for all r ∈ [1, N/b). Then Proposition F.1 will imply that v ∈ W 2,r (RN ) for all r ∈ (1, N/b). The boundedness and the continuity of v then follow by the Sobolev embedding theorem, since b < 2. First remark that for any 1 ≤ r ≤ 2∗ , Z Z r |Kv| dx ≤ C ψ (p−1)r |v|r dx < ∞ |x|≥1

|x|≥1

by Hölder’s inequality and the exponential decay of ψ. Now assume that v ∈ Lq (RN ) for some q ≥ 2∗ and let 1 ≤ r ≤ 2∗ . Since ψ is bounded, we have, by Hölder’s inequality, Z Z r |Kv| dx = |x|−br ψ (p−1)r |v|r dx |x|≤1 |x|≤1 Z Z 0 0 −brs 1/s ≤ C{ |x| dx} { |v|rs dx}1/s , |x|≤1

|x|≤1

for all s ≥ 1, 1/s + 1/s0 = 1. The first integral of the right R hand side converges if N − brs > 0 and the second does if 1 ≤ rs0 ≤ q. Thus, RN |Kv|r dx < ∞ if one can choose s ≥ 1 such that r < min{N/bs, q/s0 } and 1 ≤ r ≤ 2∗ . Setting rq = maxs≥1 min{N/bs, q/s0 }, where 1/s0 = 1 − 1/s, we get rq =

1 q

1 +

b N

.

Observe that rq is increasing in q ∈ [2∗ , ∞) and r2∗ > 1. Therefore, for all q ≥ 2∗ , Kv ∈ Lr (RN ) for all r ∈ [1, rq ). Hence, by Proposition F.1, v ∈ W 2,r (RN ) for all r ∈ (1, rq ). Then, by the Sobolev embedding theorem, v ∈ L∞ (RN ) ∩ C(RN ) if rq > 41

N , 2

i.e. if q >

N . 2−b

Hence we may assume from now on that N/(2 − b) ≥ 2∗ , otherwise the proof is complete. This condition is equivalent to N ≥ 2(3 − b). Then, for q ∈ [2∗ , N/(2 − b)], the Sobolev embedding theorem gives v ∈ Lτ (RN ) for all τ ∈ [1, T (q)), where N rq T (q) = = N − 2rq

1 q

1 − 2−b N

N for q < 2−b

and T

N 2−b

= ∞.

N Remark that T (q) is increasing in q ∈ [2∗ , 2−b ]. We discuss separately two cases. N ] is degenerate and we have Case (a): If N = 2(3 − b), the interval [2∗ , 2−b ∗ T (2 ) = ∞. Hence v ∈ Lτ (RN ) for all τ ∈ [1, ∞)

and so Kv ∈ Lr (RN ) for all r ∈ [1, r∞ ), v ∈ W 2,r (RN ) for all r ∈ (1, r∞ ), where r∞ = N/b > N/2. Hence v ∈ L∞ (RN ) ∩ C(RN ). N ] is non-degenerate. Set Case (b): If N > 2(3 − b), then the interval [2∗ , 2−b q0 =

2N N −2

and

qn+1 = T (qn ) if qn ≤

N . 2−b

Let us suppose that qn < N/(2 − b) for all n. Then {qn } is a bounded increasing sequence and N 2∗ ≤ Q ≡ lim qn ≤ . n→∞ 2−b Then from the definition of T , we obtain Q = T (Q) =

1 Q

1 . − 2−b N

This is equivalent to 1 = 1 + [(2 − b)/N ]Q. Since b 6= 2, this implies Q = 0, which N is a contradiction. Therefore, there exists n0 ≥ 0 such that qn0 ∈ [2∗ , 2−b ) and T (qn0 ) ≥ N/(2 − b). Thus v ∈ Lτ (RN ) for all τ ∈ [1, T (qn0 )). N In particular, for all q ∈ [2∗ , 2−b ), v ∈ Lτ (RN ) for all τ ∈ [1, T (q)). Since limq→N/(2−b) T (q) = ∞, this shows that v ∈ Lτ (RN ) for all τ ∈ [1, ∞). Hence, as in part (a), v ∈ L∞ (RN ) ∩ C(RN ).

42

G

Spherical harmonics

In this appendix we prove that any v ∈ L2 (RN ) admits a decomposition of the form (2.26). It is usually proven in textbooks (see e.g. [6], Sections 8.6.7 and 8.6.8) that any function ϕ ∈ L2 (S N −1 ) admits a decomposition on spherical harmonics of the form ak ∞ X X m N −1 ϕ(ϑ) = ϕm , (G.1) k Yk (ϑ) for ϑ ∈ S k=0 m=1

with the same notations as in the proof of Lemma 2.10. Here, S N −1 denotes the unit sphere in RN . We shall use this result to prove that (2.26) holds for all v ∈ L2 (RN ). For v ∈ L2 (RN ), there exists a function ve : (0, ∞) × S N −1 → RN such that Z ∞Z Z 2 v dx = rN −1 ve(r, ϑ)2 dϑ dr. (G.2) ∞> RN

S N −1

0

R

Hence, by Fubini’s theorem, S N −1 ve(r, ϑ)2 dϑ < ∞ for almost every r > 0. That is, ve(r, ·) ∈ L2 (S N −1 ) and, therefore, for almost every fixed r > 0, there exist coefficients v(r)m k such that ve(r, ϑ) =

ak ∞ X X

m v(r)m k Yk (ϑ),

(G.3)

k=0 m=1

the series in (G.3) being convergent in the sense of L2 (S N −1 ). Moreover, the coefficients v(r)m k are given by the formula Z m v(r)k = ve(r, ϑ)Ykm (ϑ)dϑ. S N −1

The Cauchy-Schwarz inequality in L2 (S N −1 ) then implies Z Z Z m m 2 2 2 Yk (ϑ) dϑ = (v(r)k ) ≤ ve(r, ϑ) dϑ S N −1

so that

∞

Z

r

N −1

2 (v(r)m k ) dr

0

showing that

S N −1

∞

≤

r

N −1

∈

L2r .

ak n X X

Z S N −1

0

v(r)m k

Sn (x) =

Z

S N −1

ve(r, ϑ)2 dϑ

ve(r, ϑ)2 dϑ dr < ∞,

Now we set

m N v(r)m k Yk (ϑ) for r = |x|, ϑ = x/r, x ∈ R \ {0},

k=0 m=1

and prove that kSn − vkL2 (RN ) → 0 as n → ∞. Indeed we have that 2 Z ∞Z ak n X X 2 N −1 m m kSn − vkL2 (RN ) = r v(r)k Yk (ϑ) − ve(r, ϑ) dϑ dr 0 S N −1 k=0 m=1

2 Z ∞ ak ∞ X

X

m = rN −1 v(r)m Y (ϑ) dr → 0

k k

0 m=1 k=n+1

L2 (S N −1 )

43

as n → ∞ by the dominated convergence theorem, since

2

∞ a ak ∞ X k

X X X

2 m m N −1 N −1 (v(r)m v(r)k Yk (ϑ) =r r

k )

k=n+1 m=1

k=n+1 m=1

L2 (S N −1 )

≤r

N −1

ak ∞ X X

2 (v(r)m k )

k=0 m=1 N −1

ke v (r, ·)k2L2 (S N −1 ) Z N −1 ve(r, ϑ)2 dϑ ∈ L1 ((0, ∞)). =r

=r

S N −1

H

Existence and properties of the operator defined by (2.43)

We begin by showing that the bilinear form β defined by (2.42) satisfies the hypotheses of the First Representation Theorem in Kato [14], Theorems 2.1 and 2.6 in Chapter VI. This ensures that a self-adjoint operator is defined by (2.43). Then we establish the additional properties of A that are required by the hypotheses of Theorem 14.10 of [21], which is used in Section 2.3. First of all we observe that the form β defined by (2.42) is simply the restriction of S 00 (ψ) to the radially symmetric elements of H = H 1 (RN ). More precisely, for all u, v ∈ Hr1 , ωN β(u, v) = S 00 (ψ)[e u, ve] where u e(x) = u(r), ve(x) = v(r).

(H.1)

But S 00 (ψ) is a bounded symmetric bilinear form on H by Lemma 2.1. It follows that β : Hr1 → R is a bounded, symmetric bilinear form. (H.2) Next we consider β as a densely defined form on L2r . Let B : D(B) ⊂ L2r → R denote the quadratic form defined by D(B) = Hr1

and B(u) = β(u, u) for u ∈ D(B).

(H.3)

Lemma H.1 For 0 < b < 2 and 1 < p < 1 + (4 − 2b)/(N − 2), the form B : D(B) ⊂ L2r → R is bounded below and closed. Proof. Referring to (H.1), we have that, for all u ∈ D(B), Z ∞ Z N −1 2 e p−1 u ωN r Q(r)u(r) dr = p |x|−b |ψ| e(x)2 dx, RN

0

where Q(r) = pr−b ψ(r)p−1 . Since ψe ∈ L∞ (RN ), we have

44

Z ωN

0

∞

r

N −1

b Z |e u (x)| Q(r)u(r) dr ≤ C |e u(x)|2−b dx |x| RN Z Z 2−b b |e u(x)|2 2 u e(x)2 dx} 2 ≤ C{ 2 dx} { N N |x| ZR ZR 2−b b |∇e u|2 dx} 2 { ≤ C1 { u e(x)2 dx} 2 RN RN Z ∞ Z ∞ 2−b b N −1 0 2 r u (r) dr} 2 { rN −1 u(r)2 dr} 2 = C1 ωN { 0 0 Z bε2/b ∞ N −1 0 2 ≤ C1 ωN r u (r) dr 2 0 Z (2 − b)ε−(2−b)/2 ∞ N −1 r u(r)2 dr + 2 0 2

for any ε > 0. The third inequality follows by Hardy’s inequality and the last one by Young’s inequality. Hence Z ∞ B(u) = rN −1 {(u0 )2 + u2 − Qu2 }dr 0 Z bε2/b ∞ N −1 0 2 r u (r) dr ≥ 1 − C1 2 0 Z (2 − b)ε−(2−b)/2 ∞ N −1 + 1 − C1 r u(r)2 dr. 2 0 −(2−b)/2

Choosing ε = ( C21 b )b/2 and the setting γ = 1 − C1 (2−b)ε 2 , it follows that Z ∞ rN −1 u(r)2 dr for all u ∈ D(B), (H.4) B(u) ≥ γ 0

showing that B : D(B) ⊂ L2r → L2r is bounded below. To prove that B is closed, we consider a sequence {un } ⊂ D(B) such that un → u in L2r

and B(un − um ) → 0 as n, m → ∞.

(H.5)

We must deduce from this that u ∈ D(B) and B(un − u) → 0 as n → ∞. From (H.4), we have that Z ∞ B(un − um ) = rN −1 {(u0n − u0m )2 + (un − um )2 − Q(un − um )2 }dr 0 Z bε2/b ∞ N −1 0 ≥ 1 − C1 r (un − u0m )2 dr 2 0 −(2−b)/2 Z ∞ (2 − b)ε + 1 − C1 rN −1 (un − um )2 }dr 2 0 Z 1 ∞ N −1 0 0 2 ≥ r (un − um ) dr 2 0 Z (2 − b)ε−(2−b)/2 ∞ N −1 + 1 − C1 r (un − um )2 }dr 2 0 45

if we choose ε = ( C21 b )b/2 . It follows from (H.5) that {un } is a Cauchy sequence in Hr1 . This implies that u ∈ Hr1 = D(B) and then the continuity of B : Hr1 → R shows that B(un − u) → 0. Thus we have shown that B : D(B) ⊂ L2r → L2r is a closed quadratic form. Lemma H.2 For 0 < b < 2 and 1 < p < 1 + (4 − 2b)/(N − 2), the operator A : D(A) ⊂ L2r → L2r is self-adjoint and bounded below. Proof. This follows from Theorem 2.6 on page 323 of [14]. In fact A is an extension of the differential operator defined formally by τ (u) = r1−N {−(rN −1 u0 ) + rN −1 [1 − Q(r)]u} for u ∈ C0∞ ((0, ∞)), where as above Q(r) = pr−b ψ(r)p−1 . We still have to show that A : D(A) ⊂ L2r → L2r is a self-adjoint extension of τ with separated boundary conditions and that inf σess (A) > 0, where σess (A) denotes the essential spectrum of A. In order to discuss the self-adjoint extensions of τ , we introduce the following notation. Let AC((0, ∞)) denote the space of all absolutely continuous functions on (0, ∞) and set D(T ) = {u ∈ L2r : u, u0 ∈ AC((0, ∞)) and τ (u) ∈ L2r }. For all u, v ∈ D(T ), it follows that lim rN −1 {u(r)v 0 (r) − u0 (r)v(r)} = [u, v]θ ,

r→θ

θ = 0, ∞,

exist and are finite, see [21] Theorem 3.10. Thus we can define a subspace of D(T0 ) of D(T ) by D(T0 ) = {u ∈ D(T ) : [u, v]0 = [u, v]∞ = 0 for all u ∈ D(T )}. We now have three operators associated with τ acting in L2r , namely Tc : D(Tc ) = C0∞ ((0, ∞)) ⊂ L2r → L2r , T0 : D(T0 ) ⊂ L2r → L2r

and T : D(T ) ⊂ L2r → L2r ,

where T c ⊂ T0 ⊂ T. From the results in Chapter 3 of [21], we have that T and T0 = T ∗ are closed operators and T0 is the closure of Tc . Lemma H.3 With the above notation, T0 ⊂ A. Proof. Let u ∈ D(T0 ). Since T0 is the closure of Tc , there is a sequence {un } ⊂ D(Tc ) = C0∞ ((0, ∞)) such that un → u and Tc un → T0 u in L2r . Then, Z

∞

B(un − um ) = Z0 ∞ =

rN −1 {(u0n − u0m )2 + (un − um )2 − Q(un − um )2 }dr rN −1 [Tc (un − um )](un − um )dr → 0

0

46

and, since B is closed by Lemma H.1, this implies that u ∈ D(B) = Hr1 and B(un − u) → 0. In fact, the proof of Lemma H.1 shows that un → u in Hr1 . Now, for any v ∈ Hr1 , this implies that Z ∞ Z ∞ N −1 rN −1 [T0 (u)]v dr, r [Tc un ]v dr = lim β(u, v) = lim β(un , v) = lim n→∞

n→∞

n→∞

0

0

showing that u ∈ D(A) and A(u) = T0 u.

Referring to Section 2.3 in Chapter VI of [14], it is easy to see that A is the Friedrichs’ extension of T0 . However, a more explicit description of D(A) can be obtained by calculating the deficiency indices of T0 using the asymptotic formulae in Lemma D.1. Recall that τ is said to be in the limit circle case, l.c.c., at r = 0 (respectively, r = ∞) if, for all λ ∈ C, all solutions of τ (u) = λu are in L2r (0, 1) (respectively, L2r (1, ∞)). When τ is not in the l.c.c. it is said to be in the limit point case, l.p.c.. By Theorem 5.3 of [21], the l.c.c. occurs at r = 0/r = ∞ if all solutions of τ (u) = 0 are in L2r (0, 1)/L2r (1, ∞). Thus, using our Lemma D.1, we see that for N ≥ 4, τ is in the l.p.c. at r = 0 and the l.p.c. at r = ∞; for N = 3, τ is in the l.c.c. at r = 0 and the l.p.c. at r = ∞. From Theorem 5.7 of [21] this implies that, for N ≥ 4, T0 has deficiency indices (0, 0) whereas, for N = 3, they are (1, 1). Furthermore, by Theorem 5.8 of [21], for N ≥ 4, T0 = A = T and, for N = 3, T0 is not self-adjoint and there exists a non-trivial real solution, w, of τ (u) = 0 such that D(A) = {u ∈ D(T ) : [u, w]0 = 0}. This means that for all N ≥ 3, the extension A of T0 is a self-adjoint restriction of T that has separated boundary in the sense defined on page 61 of [21]. To complete our discussion of A we must show that its essential spectrum is bounded away from zero. By Lemma D.1, we see that for λ < 1, the differential operator τ − λ is non-oscillatory on (0, ∞). By Theorem 14.9 of [21] this implies that inf σess (A) ≥ 1.

I

Proof of Lemma 3.1

Under the hypotheses (V1) to (V3), we define a function G : R × H → H −1 by G(k, u) = z(k, x) |x|−b f (u) where f (s) = |s|p−1 s and z : R × (RN \{0}) is defined by (|x|/|k|)b V (x/|k|) for k 6= 0 and x 6= 0 z(k, x) = 1 for k = 0 and x 6= 0. We observe that |z(k, ·)|L∞ ≤ M < ∞ for all k ∈ R and that limk→0 z(k, x) = 1 for all x 6= 0. Also for x 6= 0, z(·, x) ∈ C 1 ((0, ∞)) with ∂k z(k, x) = −k −1 (|x|/|k|)b W (x/|k|) 47

where W is the function in (V3). It follows immediately from Lemma B.1 that for all k ∈ R, G(k, ·) ∈ C 1 (H, H −1 ) with Du G(k, u)v = pz(k, x) |x|−b |u|p−1 v for all u, v ∈ H. We need to prove that (i) G ∈ C(R × H, H −1 ), (ii) Du G ∈ C(R × H, B(H, H −1 )), (iii) G ∈ C 1 ((0, ∞) × H, H −1 ). Proof of (i): Fix (k, u) ∈ R × H and consider (h, v) ∈ R × H. Then kG(k, u) − G(h, v)kH −1 ≤ kG(k, u) − G(h, u)kH −1 + kG(h, u) − G(h, v)kH −1 where Z G(h, u) − G(h, v) = 0

1

d G(h, tu + (1 − t)v)dt = dt

Z

1

Du G(h, tu + (1 − t)v)(u − v) dt 0

and so Z

1

kDu G(h, tu + (1 − t)v)kB(H,H −1 ) dt ku − vk Z 1 ktu + (1 − t)vkp−1 dt ku − vk ≤ C |z(h, ·)|L∞

kG(h, u) − G(h, v)kH −1 ≤

0

0

by Lemma B.1(ii). Note that the constant C is independent of h, u and v and that |z(h, ·)|L∞ ≤ M < ∞ for all h ∈ R. If ku − vk ≤ 1, we have that ktu + (1 − t)v)k ≤ kuk + 1 for all t ∈ [0, 1] and it follows that kG(h, u) − G(h, v)kH −1 ≤ CM (kuk + 1)p−1 ku − vk for all h ∈ R and ku − vk ≤ 1. Given ε > 0, there exists δ > 0 such that kG(h, u) − G(h, v)kH −1 ≤ ε for all h ∈ R and ku − vk ≤ δ. For all ϕ ∈ H, we have that Z hG(k, u) − G(h, u), ϕiH −1 ×H =

{z(k, x) − z(h, x)} |x|−b f (u)ϕ dx

RN

and so, setting h = k + s and ws = z(k, ·) − z(k + s, ·), we have that, for all u ∈ H\{0}, R |ws | |x|−b |u|p |ϕ| dx RN kG(k, u) − G(k + s, u)kH −1 ≤ sup . kϕk ϕ∈H\{0} Given ε > 0, it follows from Lemma A.1(ii) that there exists δ1 > 0 such that kG(k, u) − G(k + s, u)kH −1 ≤ ε for all |s| ≤ δ1 . These inequalities prove the continuity of G at (k, u).

48

Proof of (ii): Following the notation of Lemma B.1, let Bk (u)v = z(k, x) |x|−b |u|p−1 v for u, v ∈ H. We again fix (k, u) ∈ R × H and consider (h, v) ∈ R × H. Then kDu G(k, u) − Du G(h, v)kB(H,H −1 ) ≤ kDu G(k, u) − Du G(h, u)kB(H,H −1 ) + kDu G(h, u) − Du G(h, v)kB(H,H −1 ) = p{kBk (u) − Bh (u)kB(H,H −1 ) + kBh (u) − Bh (v)kB(H,H −1 ) }. Since |z(h, ·)|L∞ ≤ M < ∞ for all h ∈ R, it follows from Lemma B.1(ii) that the functions {Bh : H → B(H, H −1 )}h∈R are equicontinuous at u and hence, given any ε > 0, there exists δ > 0 such that kBh (u) − Bh (v)kB(H,H −1 ) ≤ ε for all h ∈ R if ku − vk ≤ δ. Next we note that R kBk (u) − Bh (u)kB(H,H −1 ) ≤

sup ϕ,ξ∈H\{0}

RN

|ws | |x|−b |u|p−1 |ϕ| |ξ| ds kϕk kξk

where h = k + s and ws is defined above. It follows from Lemma A.1(ii) that lim kBk (u) − Bk+s (u)kB(H,H −1 ) ,

s→0

proving the continuity of Du G at (k, u). Proof of (iii): Fixing k > 0 and u ∈ H, we have that, for s 6= 0, Z z(k + s, x) − z(k, x) −b G(k + s, u) − G(k, u) = ,ϕ |x| f (u)ϕ dx s s RN H −1 ×H for all ϕ ∈ H. First of all we observe that ∂k z(k, ·) ∈ L∞ (RN ) with |∂k z(k, ·)|L∞ ≤ Q/k where Q = supx6=0 |x|b |W (x)| < ∞. Thus, it follows from Lemma B.1(i) that ∂k z(k, x) |x|−b f (u) ∈ H −1 . For all s ≥ −k/2 and x 6= 0, we also have that |z(k + s, x) − z(k, x)| ≤ |∂k z(k + ts, x)| |s| for some t ∈ [0, 1], and so z(k + s, x) − z(k, x) ≤ 2Q/k for all s ≥ −k/2 and all x 6= 0. s Thus, for all ϕ ∈ H, G(k + s, u) − G(k, u) −b − ∂k z(k, x) |x| f (u), ϕ s H −1 ×H Z z(k + s, x) − z(k, x) = − ∂k z(k, x) |x|−b f (u)ϕ dx s RN 49

so that

G(k + s, u) − G(k, u)

−b

− ∂k z(k, x) |x| f (u)

−1 s H R −b p |u| |ϕ| dx N |q(s, x)| |x| ≤ sup R kϕk ϕ∈H\{0} where q(s, x) = {z(k + s, x) − z(k, x)}/s − ∂k z(k, x). Since |q(s, ·)|L∞ ≤ 3Q/k for all s ≥ −k/2 and q(s, x) → 0 as s → 0 for all x 6= 0, it follows from Lemma A.1(ii) that

G(k + s, u) − G(k, u)

−b

− ∂ z(k, x) |x| f (u) k

−1 → 0 as s → 0. s H Thus G : R × H → H −1 is differentiable with respect to k at (k, u) and Dk G(k, u) = ∂k z(k, x) |x|−b f (u). Finally kDk G(k, u) − Dk G(h, v)kH −1 ≤ kDk G(k, u) − Dk G(h, u)kH −1 + kDk G(h, u) − Dk G(h, v)kH −1 where R kDk G(k, u) − Dk G(h, u)kH −1 ≤

sup

RN

ϕ∈H\{0}

|∂k z(k, x) − ∂k z(h, x)| |x|−b |u|p |ϕ| dx kϕk

and it follows from Lemma A.1(ii) that kDk G(k, u) − Dk G(h, u)kH −1 → 0 as h → k > 0. On the other hand, since ∂k z(h, ·) ∈ L∞ (RN ) for h ≥ k/2, it follows from Lemma B.1(iii) that Dk G(h, ·) ∈ C 1 (H, H −1 ) with Du Dk G(h, w)ϕ = p∂k z(h, x) |x|−b |w|p−1 ϕ for all w, ϕ ∈ H. Furthermore, there exists a constant C such that kDu Dk G(h, w)kB(H,H −1 ) ≤ C kwkp−1 for all w ∈ H where C is independent of h ≥ k/2 since |∂k z(h, ·)|L∞ ≤ 2Q/k for h ≥ k/2. It follows that kDk G(h, u) − Dk G(h, v)kH −1 ≤ C(kuk + 1)p−1 ku − vk for all h ≥ k/2 and for all v ∈ H with ku − vk ≤ 1. Combining these inequalities we have that Dk G : (0, ∞) × H → H −1 is continuous at (k, u). Since we already have that Du G ∈ C((0, ∞) × H, B(H, H −1 )), we now have that G ∈ C 1 ((0, ∞) × H, H −1 ).

50

J

Proof of Lemma 3.2

This follows by a boot-strap argument similar to the one used to prove Lemma 2.9, so we need only indicate the main steps. By (V1) and (V2), there exists a constant C such that |x|b |V (x)| ≤ C for all x ∈ RN \{0} and hence, for any (k, v) ∈ (0, ∞) × H, −b k V (x/k) ≤ C |x|−b for all k > 0 and x ∈ RN \{0}. If (k, v) is a weak solution of (3.5), we have that −∆v + v = Nk (v) in H −1 (RN ) where Nk (v) = k −b V (x/k) |v|p−1 v and |Nk (v)| ≤ C |x|−b |v|p . Therefore, in the proof of Lemma 2.9, we replace the expression Kv by Nk (v). Thus, if we have that v ∈ Lq (RN ) for some q ≥ 2∗ , it follows from Z Z Z 0 0 r −brt 1/t |Nk (v)| dx ≤ C{ |x| dx} { |v|prt dx}1/t |x|≥1

with t = 1 +

pN 2b

|x|≥1

and 1/t + 1/t0 = 1 that Nk (v) ∈ Lr (|x| ≥ 1) provided that R 1 since q ≥ 2∗ , p < 1 + 4−2b and N ≥ 2. Thus, if we have N −2 q N ∗ that v ∈ L (R ) for some q ≥ 2 , it follows that Nk (v) ∈ Lr (RN ) provided that r ∈ (max{1, R}, Rq ). Starting with q = 2∗ , the boot-strap argument used to prove Lemma 2.9, can now be repeated using Rq instead of rq . The function T (q) is here replaced by T1 (q) =

1 p q

−

(2−b) N

pN for q ∈ [2∗ , 2−b ).

Note that we cannot have Q = T1 (Q) with Q ≥ 2∗ since p < 1 +

51

4−2b . N −2

K

Time local well-posedness

In this section we check the hypotheses of Theorem 4.3.1 of [4]. This theorem concerns time local existence of solutions of (4.1) and conservation of charge and energy for this equation. In our context, it can be formulated as follows. Theorem K.1 Denote H 1 = H 1 (RN , C). Let g ∈ C(H 1 , H −1 ) be a nonlinearity such that g = g1 + g2 , where g1 , g2 ∈ C(H 1 , H −1 ) satisfy the following hypotheses. For i = 1, 2: (a) gi (0) = 0 and there exists Gi ∈ C 1 (H 1 , R) such that Gi (0) = 0 and gi = G0i . (b) There exist ri , ρi ∈ [2, 2∗ ) such that for every M > 0 there exists Ci (M ) > 0 such that |gi (u) − gi (v)|Lρ0i ≤ Ci (M )|u − v|Lri for all u, v ∈ H 1 satisfying kukH 1 + kvkH 1 ≤ M , where 1/ρi + 1/ρ0i = 1. Consider the initial-value problem i∂t w + ∆w + g(w) = 0,

w(0) = ϕ ∈ H 1 .

(K.1)

Denote G = G1 + G2 and define the charge and energy respectively by Z 1 1 Q(u) = |u|L2 and E(u) = |∇u|2 dx − G(u) for all u ∈ H 1 . 2 2 RN Then, for all ϕ ∈ H 1 , there exist T = T (ϕ) > 0 and a unique solution w ∈ C([0, T ), H 1 ) ∩ C 1 ([0, T ), H −1 ) of (K.1). Furthermore, there is conservation of charge and energy, that is, Q(w(t)) = Q(ϕ) and

E(w(t)) = E(ϕ) for all t ∈ [0, T ).

The notion of solution in Theorem K.1 must be understood in the following sense: w ∈ C([0, T ), H 1 ) ∩ C 1 ([0, T ), H −1 ) is a solution of (K.1) if i∂t w + ∆w + g(w) = 0 in H −1 for all t ∈ [0, T ) and w(0) = ϕ. Lemma K.1 Suppose that V satisfies the hypotheses (V1) and (V2). Set g(u) = V (x)|u|p−1 u. Then there exist g1 and g2 satisfying the hypotheses (a) and (b) of Theorem K.1 such that g = g1 + g2 . Proof. Put g1 (u) = χB(0,1) (x)g(u) and g2 (u) = [1 − χB(0,1) (x)]g(u). By hypotheses (V1) and (V2), V (x)|x|b ∈ L∞ (RN ). It is trivial to check that the results of Appendix B remain true when replacing H by H 1 . Therefore, it follows from Lemma B.1(iii) that, for i = 1, 2, gi satisfies (a) with Z 1 z1 (x)|x|−b |u|p+1 dx G1 (u) = p + 1 RN 52

and

1 G2 (u) = p+1

Z

z2 (x)|x|−b |u|p+1 dx,

RN

where z1 (x) = χB(0,1) (x)V (x)|x|b and z2 (x) = [1 − χB(0,1) (x)]V (x)|x|b . For (b), suppose that u, v ∈ H 1 are such that kukH 1 + kvkH 1 ≤ M for some M > 0, and let ϕ ∈ C0∞ (RN , C) ⊂ H 1 . For g1 , we use Hölder’s inequality with four exponents α, β, r1 , ρ1 ≥ 1, 1/α + 1/β + 1/r1 + 1/ρ1 = 1, and the Sobolev embedding. We have that Z Z p−1 ≤ |u| u − |v|p−1 v |ϕ|dx [g (u) − g (v)]ϕ dx χ (x)|V (x)| 1 1 B(0,1) N R RN Z 1 Z −b p−1 ≤C |x| p|tu + (1 − t)v| (u − v)dt |ϕ|dx 0 ZB(0,1) ≤C |x|−b (|u| + |v|)p−1 |u − v||ϕ|dx ZB(0,1) Z −bα 1/α ≤ C{ |x| dx} { (|u| + |v|)(p−1)β dx}1/β × B(0,1)

B(0,1)

× |u − v|Lr1 |ϕ|Lρ1 p−1 ≤ C1 |u| + |v| (p−1)β |u − v|Lr1 |ϕ|Lρ1 L

≤ C1 (|u|L(p−1)β + |v|L(p−1)β )p−1 |u − v|Lr1 |ϕ|Lρ1 ≤ C2 (kukH 1 + kvkH 1 )p−1 |u − v|Lr1 |ϕ|Lρ1 ≤ C2 M p−1 |u − v|Lr1 |ϕ|Lρ1 if α, β > 1 are such that N − bα > 0 and (p − 1)β ∈ [1, 2∗ ]. It is not difficult to check that these conditions can be fulfilled if one chooses r1 = ρ1 ∈ (r0 , 2∗ ), where r0 =

4N . 2N − 2b − (p − 1)(N − 2)

The non-degeneracy of this interval is equivalent to p < 1 + (4 − 2b)/(N − 2), which is true by hypothesis. Hence, the hypothesis (b) of Theorem K.1 is satisfied by g1 . On the other hand, since |x|−b ≤ 1 on RN \ B(0, 1), it follows by Hölder’s inequality with three exponents and by the Sobolev embedding that Z [g (u) − g (v)]ϕ dx 2 2 N R Z ≤ [1 − χB(0,1) (x)]|V (x)| |u|p−1 u − |v|p−1 v |ϕ|dx RN Z 1 Z −b p−1 ≤C |x| p|tu + (1 − t)v| (u − v)dt |ϕ|dx RN \B(0,1) 0 Z ≤C (|u| + |v|)p−1 |u − v||ϕ|dx RN \B(0,1)

p−1 ≤ C1 |u| + |v| H 1 |u − v|Lp+1 |ϕ|Lp+1 ≤ C1 M p−1 |u − v|Lp+1 |ϕ|Lp+1 , proving hypothesis (b) for g2 with r2 = ρ2 = p + 1 ∈ [2, 2∗ ). 53

References [1] A. Ambrosetti and M. Badiale, Variational perturbative methods and bifurcation of bound states from the essential spectrum, Proc. Roy. Soc. Edinburgh Sect. A 128 (1998), no. 6, 1131-1161. [2] A. Ambrosetti and A. Malchiodi, Perturbation methods and semilinear elliptic problems on Rn , Birkhäuser (2006). [3] A. de Bouard and R. Fukuizumi, Stability of standing waves for nonlinear Schrödinger equations with inhomogeneous nonlinearities, Ann. Henri Poincaré 6 (2005), no. 6, 1157-1177. [4] T. Cazenave, Semilinear Schr¨ odinger equations, Courant Lecture Notes in Mathematics, AMS (2003). [5] T. Cazenave and P.-L. Lions, Orbital stability of standing waves for some nonlinear Schrödinger equations, Comm. Math. Phys. 85 (1982), no. 4, 549561. [6] S. Chatterji, Cours d’analyse, Vol. 3, Presses polytechniques et universitaires romandes (1998). [7] M. Grillakis, J. Shatah and W. Strauss, Stability theory of solitary waves in the presence of symmetry I, J. Funct. Anal. 74 (1987), no. 1, 160197. [8] R. Fukuizumi and M. Ohta, Instability of standing waves for nonlinear Schrödinger equations with inhomogeneous potentials, Preprint. [9] D. Gilbarg and N.S. Trudinger, Elliptic partial differential equations of second order, Second Edition, Springer (1983). [10] H. Hajaiej and C.A. Stuart, On the variational approach to the stability of standing waves for the nonlinear Schr¨ odinger equation, Advanced Nonlinear Studies 4 (2004), 469-501. [11] P. Hartman, Ordinary differential equations, Birkha¨ user (1982). [12] L. Jeanjean and S. Le Coz, An existence and stability result for standing waves of nonlinear Schrödinger equations, Adv. Differential Equations 11 (2006), no. 7, 813-840. [13] L. Kantorovitch and G. Akilov, Analyse fonctionnelle, Vol. II, Editions Mir (1981). [14] T. Kato, Perturbation theory for linear operators, Springer (1995). [15] S. Kesavan, Symmetrization and applications, Series in Analysis, World Scientific (2006).

54

[16] R.J. Magnus, On the asymptotic properties of solutions to a differential equation in a case of bifurcation without eigenvalues, Proc. Roy. Soc. Edinburgh Sect. A 104 (1986), no. 1-2, 137-159. [17] C.A. Stuart, Bifurcation for Dirichlet problems without eigenvalues, Proc. London Math. Soc. (3) 45 (1982), 169-192. [18] C.A. Stuart, Bifurcation of homoclinic orbits and bifurcation from the essential spectrum, SIAM J. Math. Anal. 20 (1989), 1145-1171. [19] C.A. Stuart, A global branch of solutions to a semilinear equation on an unbounded interval, Proc. Roy. Soc. Edinburgh Sect. A 101 (1985), no. 3-4, 273-282. [20] C.A. Stuart, Bifurcation in Lp (RN ) for a semilinear elliptic equation, Proc. London Math. Soc. (3) 57 (1988), no. 3, 511-541. [21] J. Weidmann, Spectral theory of ordinary differential operators, Springer (1987). [22] E. Yanagida, Uniqueness of positive radial solutions of ∆u + g(r)u + h(r)up = 0 in Rn , Arch. Rational Mech. Anal. 115 (1991), no. 3, 257-274.

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