Positive solutions to a class of quasilinear elliptic equations on ℝ

Antonio Ambrosetti ^{∗} and Zhi-Qiang Wang ^{†}

Key words: Fully nonlinear elliptic equations, Perturbation theory.

1 Introduction

This Note is concerned with solutions of perturbations to the following problem on the real line ℝ

\begin{cases}−u′′ + u − k(u^{2})′′u = u^{p}, \\ u ∈ H^{1}(ℝ) u > 0, \end{cases} (1)

where k ∈ ℝ and p > 1. Consider

\begin{cases}−u′′ + (1 + εa(x)) u − k ((1 + εb(x)) (u^{2})′′u = (1 + εc(x)) u^{p}, \\ u ∈ H^{1}(ℝ) u > 0, \end{cases} (2)

where a, b, c are assumed to be real valued functions belonging to the class S,

S = {h(x) = h_{1}(x) + h_{2}(x) : h_{1} ∈ L^{r}(ℝ) ∩ L^{∞}(ℝ), for some r ∈ [1, ∞), h_{2} ∈ L^{∞}(ℝ), lim_{|x|→∞} h_{2}(x) = 0}.

Our main existence result is the following Theorem 1.1 There is k_{0} > 0 such that for k > −k_{0} and a, b, c ∈ S, equation (2) has a solution provided |ε| is sufficiently small. Solutions of (2) will be found as critical points of a functional I_{ε} of the form I_{ε}(u) = I_{0}(u) + εG(u), u ∈ H^{1}(ℝ), (3) where

I_{0}(u) = \frac{1}{2} ∫_{ℝ} (|u′|^{2} + u^{2}) dx + k ∫_{ℝ} u^{2}|u′|^{2}dx − \frac{1}{p + 1} ∫_{ℝ} |u|^{p+1}dx (4)

and G is the perturbation

G(u) = \frac{1}{2} ∫_{ℝ} (a(x)u^{2}) dx + k ∫_{ℝ} b(x)u^{2}|u′|^{2}dx − \frac{1}{p + 1} ∫_{ℝ} c(x)|u|^{p+1}dx. (5)

Critical points of I_{ε} will be found using the abstract perturbation procedure introduced in [1, 2]. Due to the translation invariance, the critical points of I_{0} appear as manifolds. We first show that I_{0} has a unique positive critical point z, up to translations, which is nondegenerate, in the sense that I′′_{0}(z) has a one dimensional kernel (given by the translations). Then, for |ε| small, I_{ε} will be shown to have a critical point which is close to the critical manifold of I_{0}.

∗SISSA, via Beirut 2-4, Trieste 34014 Italy. Supported by MIUR under the national project Variational methods and nonlinear differential equations

†Utah State Univ., Logan, UT 84322 USA. Z-Q.W. thanks the hospitality of SISSA where this work was initiated.

Equations like (1) and (2) have been recently studied in [9] (see also the references therein for physical motivations) where the following problem was considered

\begin{cases}−u′′ + V (x)u − k(u^{2})′′u = λu^{p}, \\ u ∈ H^{1}(ℝ) u > 0. \end{cases}

For k > 0, for several classes of potentials V (x) and certain ranges of p > 1, ground state solutions were constructed as minimizers of a constrained minimization problem, with λ being the Lagrange multiplier. An Orlicz space approach was used in [8] to obtain ground state solutions for prescribed λ > 0. The concentration compactness principle was used in these papers to overcome the lack of compactness.

On the other hand, the perturbative approach from [1, 2] allows us to construct solutions for (2) for a broader range of k, V and p, including cases when the ground state solution may not exist. We also give a different approach for obtaining ground state solutions for prescribed λ > 0, and this approach is technically simpler than that in [8].

In Section 2 we first study the ground state solutions. Then Section 3 is devoted to proving Theorem 1.1.

2 Ground state solutions

We first consider the problem

\begin{cases}−u′′ + V (x)u − k(u^{2})′′u = λu^{p}, \\ u ∈ H^{1}(ℝ) u > 0. \end{cases} (6)

We will consider λ > 0 as a fixed parameter. Since in the following the arguments are independent of such λ > 0, we will take λ ≡ 1 for simplicity of notation.

We assume always that

(V 0) inf_{ℝ} V (x) > 0.

Let

X = {u ∈ H^{1}(ℝ) : ∫_{ℝ} V (x)u^{2}dx < +∞}

endowed with norm

∥u∥^{2} = ∫_{ℝ} (|u′|^{2} + V (x)u^{2}) dx.

The corresponding scalar product in X will be denoted by (·, ·). Then solutions of (6) are positive critical points of

I(u) = \frac{1}{2}∥u∥^{2} + k|uu′|_{2}^{2} − \frac{1}{p + 1}|u|_{p+1}^{p+1},

where we use |u|_{q} to denote the L^{q}(ℝ) norm. Let

γ(u) := (I′(u), u) = ∥u∥^{2} + 4k|uu′|_{2}^{2} − |u|_{p+1}^{p+1}

and define the Nehari manifold by setting

M = {u ∈ X \ {0} : γ(u) = 0}.

In the next two lemmas we collect the main properties of M that we need in the sequel.

Lemma 2.1 Let k > 0 and p ≥ 3. Then M ≠ ∅. Moreover, for any u ∈ X \ {0} with γ(u) < 0 there exists a unique θ_{u} ∈]0, 1[ such that θ_{u}u ∈ M.

Proof. Fix u ∈ X \ {0}. One has γ(θu) = θ^{2}g(θ), where g(θ) = ∥u∥^{2} + 4kθ^{2}|uu′|_{2}^{2} − θ^{p−1}|u|_{p+1}^{p+1}. Then g(0) = ∥u∥^{2} > 0 and, for θ > 0 one has

g′(θ) = 8kθ|uu′|_{2}^{2} − (p − 1)θ^{p−2}|u|_{p+1}^{p+1}. (7)

Let us distinguish between the case p > 3 and p = 3. In the former, from (7) we find that g′(θ) < 0, resp. > 0, for

θ > θ^{∗} := (\frac{8k|uu′|_{2}^{2}}{(p − 1)|u|_{p+1}^{p+1}} )^{1/(p−3)} , resp θ < θ^{∗},

and the lemma follows when p > 3.

Next, let us consider the case p = 3. Recalling that inf{∫_{ℝ} |ϕ′|^{2}dx : ϕ ∈ H^{1}(ℝ), ∫_{ℝ} ϕ^{2}dx = 1} = 0, we infer that for all ε > 0 there exists ϕ_{ε} ∈ H^{1}(ℝ), with ∫_{ℝ} ϕ_{ε}^{2}dx = 1 such that ∫_{ℝ} |(ϕ_{ε})′|^{2}dx = ε. Taking u_{ε}^{2} = ϕ_{ε} we get:

|u_{ε}u′_{ε}|_{2}^{2} = ∫_{ℝ} u_{ε}^{2}|u′_{ε}|^{2}dx = \frac{1}{4} ∫_{ℝ} |(u_{ε}^{2})′|^{2}dx = \frac{1}{4} ∫_{ℝ} |ϕ′_{ε}|^{2}dx = \frac{1}{4}ε,

while |u_{ε}|_{4}^{4} = |ϕ_{ε}|_{2}^{2} = 1. Corresponding to such u_{ε}, we find ( recall that p = 3):

g(θ) = ∥u_{ε}∥^{2} + α_{ε}θ^{2}, where α_{ε} := 4k|u_{ε}u′_{ε}|_{2}^{2} − |u_{ε}|_{4}^{4} = kε − 1 .

Choosing ε < k we get that α_{ε} < 0 and the result follows.

Lemma 2.2 Let p ≥ 3. Then M is a manifold and (if M ≠ ∅) one has:

(i) there exists ρ > 0 such that ∥u∥ ≥ ρ, ∀ u ∈ M.

(ii) If u^{∗} ∈ M is a constrained critical point of I on M, then I′(u^{∗}) = 0 and u^{∗} is a solution of (6).

(iii) m := inf_{u∈M} I(u) ≥ const. > 0.

Proof. From γ(u) = 0 it follows that

4k|uu′|_{2}^{2} = |u|_{p+1}^{p+1} − ∥u∥^{2}. (8)

Then for u ∈ M one has

(γ′(u), u) = 2∥u∥^{2} + 16k|uu′|_{2}^{2} − (p + 1)|u|_{p+1}^{p+1} = −2∥u∥^{2} + (3 − p)|u|_{p+1}^{p+1} < 0. (9)

In particular, M is a manifold. Furthermore, (8) also yields

∥u∥^{2} = −4k|uu′|_{2}^{2} + |u|_{p+1}^{p+1} ≤ C∥u∥^{p+1} (u ∈ M),

proving (i). In addition, for the u^{∗} in the lemma, there exists μ ∈ ℝ such that I′(u^{∗}) = μγ′(u^{∗}). Taking the scalar product, one finds (I′(u^{∗}), u^{∗}) = μ(γ′(u^{∗}), u^{∗}). Since u^{∗} ∈ M then (I′(u^{∗}), u^{∗}) = 0 while (γ′(u^{∗}), u^{∗}) < 0, according to (9). Thus μ = 0, proving (ii). Using again (8), we infer that

I_{|M}(u) = \frac{1}{4}∥u∥^{2} + (\frac{1}{4} − \frac{1}{p + 1}) |u|_{p+1}^{p+1}. (10)

Therefore, if p ≥ 3 and taking into account (i), we deduce that m ≥ const. > 0.

For future references let us recall the following lemma from [9].

Lemma 2.3 Let u_{n} ⇀ u in X and set v_{n} = u_{n} − u. Then

lim inf |u_{n}u′_{n}|_{2}^{2} ≥ lim inf |v_{n}v′_{n}|_{2}^{2} + |uu′|_{2}^{2} .

We are now in position to prove the existence of ground states for (6).

Theorem 2.4 Let (V 0) hold and suppose that either V satisfies

(V 1) lim_{|x|→∞} V (x) = +∞;

or is bounded and satisfies

(V 2) V is periodic in x;

(V 3) lim_{|x|→∞} V (x) exists and denoted by \overline{V} such a limit, one has that \overline{V} = sup_{ℝ} V (x).

Then for k > 0 and p ≥ 3, m is always achieved at some u ∈ M, which is a solution of (6).

Remarks. (i) When V satisfies (V 0), (V 3) and is of the form V (x) = 1 + εa(x) we could also apply Theorem 1.1, with b(x) ≡ c(x) ≡ 0. However, let us point out that (V 3) implies that εa(x) < 0. On the contrary, no such restriction is required in Theorem 1.1 where, in addition, we allow k to be negative and p > 1.

(ii) There are cases not covered by Theorem 2.4 in which M is not empty and I_{|M} is bounded below.

For example, if k < 0 and 1 < p < 3 one has that g(θu) = θ^{2} (∥u∥^{2} + 4k|uu′|_{2}^{2}θ^{2} − |u|_{p+1}^{p+1}θ^{p−1}) and this implies that ∀ u ∈ X \ {0}, ∃ θ_{u} such that θ_{u}u ∈ M. Moreover,

I(u) = \frac{p − 1}{2(p + 1)}∥u∥^{2} + k \frac{p − 3}{p + 1}|uu′|_{2}^{2} > 0, ∀ u ∈ M.

Although we do not know in the present generality if the inf_{M} I(u) is achieved, we will show later on that a ground state exists provided k > −k_{0}, p > 1, V (x) = 1 + εa(x), a ∈ S and εa(x) < 0, see the Remark at the end of Section 3.

(iii) The proof below will make it clear that assumption (V 1) can be replaced by any other condition that assures the compact embedding from X into L^{p+1}(ℝ).

Proof. Let u_{n} ∈ M be a minimizing sequence. Clearly u_{n} is bounded and we can assume that u_{n} ⇀ u in X. First, some preliminary remarks are in order. Since u_{n} ∈ M, then |u_{n}|_{p+1}^{p+1} = ∥u_{n}∥^{2} + 4k|u_{n}u′_{n}|_{2}^{2} ≥ ∥u_{n}∥^{2} and therefore (i) of lemma 2.1 yields

lim |u_{n}|_{p+1}^{p+1} > 0. (11)

Furthermore, setting as in the preceding Lemma v_{n} = u_{n} − u, we claim that

γ(u) + lim inf γ(v_{n}) ≤ 0, (12)

so that either γ(u) ≤ 0 or lim inf γ(v_{n}) ≤ 0 (or both). To prove (12) we argue as follows. Since u_{n} ∈ M, then 0 = γ(u_{n}) = ∥u_{n}∥^{2} + 4k|u_{n}u′_{n}|_{2}^{2} − |u_{n}|_{p+1}^{p+1}. By a well known result by Brezis and Lieb [5],

|u_{n}|_{p+1}^{p+1} = |v_{n}|_{p+1}^{p+1} + |u|_{p+1}^{p+1} + o(1).

Also, ∥u_{n}∥^{2} = ∥v_{n}∥^{2} + ∥u∥^{2} + o(1). Using these equations as well as Lemma 2.3 we infer:

0 = ∥u_{n}∥^{2} + 4k|u_{n}u′_{n}|_{2}^{2} − |u_{n}|_{p+1}^{p+1}
= ∥v_{n}∥^{2} + ∥u∥^{2} + 4k|u_{n}u′_{n}|_{2}^{2} − |v_{n}|_{p+1}^{p+1} − |u|_{p+1}^{p+1} + o(1) ≥ lim inf γ(v_{n}) + γ(u),

proving (12). We now distinguish among the three assumptions made on V.

(a) If (V 1) holds then X is compacly embedded in L^{p+1}, see [4], and hence u_{n} → u strongly in L^{p+1} (up to a subsequence). From (11) it follows that u ≠ 0. Moreover, one has

γ(v_{n}) = ∥v_{n}∥^{2} + 4k|v_{n}v′_{n}|_{2}^{2} − |v_{n}|_{p+1}^{p+1} ≥ ∥v_{n}∥^{2} − |v_{n}|_{p+1}^{p+1} = ∥v_{n}∥^{2} + o(1).

If lim ∥v_{n}∥^{2} ≠ 0 (otherwise we have done) it follows that lim inf γ(v_{n}) > 0. Then from (12) it follows that γ(u) < 0. So, u ≠ 0 and Lemma 2.1 imply that there exists θ = θ_{u} ∈]0, 1[ such that θu ∈ M. Since u_{n} is a minimizing sequence and using (10), we deduce

m + o(1) = I(u_{n}) = \frac{1}{4}∥u_{n}∥^{2} + \frac{p − 3}{4(p + 1)}|u_{n}|_{p+1}^{p+1}
≥ \frac{1}{4}∥u∥^{2} + \frac{p − 3}{4(p + 1)}|u|_{p+1}^{p+1} + o(1)
= \frac{1}{4} θ^{−2}∥θu∥^{2} + \frac{p − 3}{4(p + 1)} θ^{−(p+1)}|θu|_{p+1}^{p+1} + o(1).

Since 0 < θ < 1 it follows that

m > \frac{1}{4}∥θu∥^{2} + \frac{p − 3}{4(p + 1)}|θu|_{p+1}^{p+1} = I(θu),

a contradiction because θu ∈ M. Thus, v_{n} → 0 strongly in X, u ∈ M and u is a minimizer of I_{|M}.

(b) Suppose that (V 2) holds. From (11) and by the Concentration Compactness Principle [6], we infer that there exist r > 0 and y_{n} ∈ ℝ such that

lim inf_{n→∞} ∫_{y_{n}−r}^{y_{n}+r} |u_{n}|^{p+1}dx > 0. (13)

We may assume y_{n} are integer multipliers of the period of V. Then the sequence ũ_{n}(x) = u_{n}(x + y_{n}) is also a minimizing sequence because I(ũ_{n}) = I(u_{n}) in view of the periodicity of V. Without loss of generality, assume that y_{n} = 0 and hence (13) implies that u ≠ 0. If γ(u) = 0 it is easy to see that ∥u_{n} − u∥ → 0 and I(u) = m. Otherwise suppose that γ(u) ≠ 0. If γ(u) < 0 we argue as in the preceding point (a). It remains to consider the case in which, up to a subsequence, lim γ(v_{n}) < 0. Then, for n large, there exist θ_{n} = θ_{v_{n}} ∈]0, 1[ such that θ_{n}v_{n} ∈ M. Furthermore one has that lim sup θ_{n} < 1, otherwise, along a subsequence, θ_{n} → 1 and hence γ(v_{n}) = γ(θ_{n}v_{n}) + o(1) = o(1), a contradiction. As before, we find

m + o(1) = I(u_{n}) ≥ \frac{1}{4} θ_{n}^{−2}∥θ_{n}v_{n}∥^{2} + \frac{p − 3}{4(p + 1)} θ_{n}^{−(p+1)}|θ_{n}v_{n}|_{p+1}^{p+1} + o(1).

Therefore one has again m > I(θ_{n}v_{n}), for n large, a contradiction.

(c) Suppose that (V 3) holds. First consider

\overline{I}(u) = \frac{1}{2} ∫_{ℝ} (|u′|^{2} + \overline{V} u^{2}) dx + k ∫_{ℝ} u^{2}|u′|^{2}dx − \frac{1}{p + 1} ∫_{ℝ} |u|^{p}dx.

By the result proved in point (b), \overline{I} has a critical point \overline{u} which is a minimizer of \overline{I} on \overline{M} := \{u ∈ H^{1}(ℝ) : (\overline{I}′(u), u) = 0\}. It is readly proved that \overline{u} > 0 on ℝ. Also, we use the notation \overline{m}, \overline{γ} or \overline{θ} for this problem, with obvious meaning. If V is not constant (otherwise (b) applies), one easily gets γ(u) < \overline{γ}(u) for all u. In particular, γ(\overline{u}) < \overline{γ}(\overline{u}) = 0. Then θ(\overline{u}) ∈]0, 1[, and

m ≤ I(θ(\overline{u})\overline{u}) < \overline{I}(\overline{u}) = \overline{m}. (14)

Let us take again a minimizing sequence for I and let us show that u ≠ 0. Otherwise, from I(u_{n})−\overline{I}(u_{n}) = ∫_{ℝ}(V (x) − \overline{V})u_{n}^{2}dx and since lim_{|x|→∞} V (x) = \overline{V} we get

lim \overline{I}(u_{n}) = lim I(u_{n}) = m.

For the same reason, we also have 0 = γ(u_{n}) = \overline{γ}(u_{n}) + o(1). Let \overline{θ}_{n} be such that \overline{θ}_{n}u_{n} ∈ \overline{M}. From \overline{γ}(\overline{θ}_{n}u_{n}) = 0 we infer for p > 3:

\overline{θ}_{n}^{2} (\overline{θ}_{n}^{p−3}|u_{n}|_{p+1}^{p+1} − 4k|u_{n}u′_{n}|_{2}^{2}) = ∥u_{n}∥^{2} + o(1).

Since ∥u_{n}∥ and |u_{n}u′_{n}|_{2} are bounded and using (11) we deuce that \overline{θ}_{n} is bounded. The case p = 3 can be handled similarly. Then, using also the property of the Nehari manifold that I(λu) ≤ I(u) for all u ∈ M and all λ > 0, we find

\overline{m} ≤ \overline{I}(\overline{θ}_{n}u_{n}) = I(\overline{θ}_{n}u_{n}) + o(1) ≤ I(u_{n}) + o(1) = m + o(1),

a contradiction with (14). This proves that u ≠ 0. Now we proceed as before: one has that m + o(1) = I(u_{n}) = \overline{I}(v_{n}) + I(u) + o(1) and since 0 = γ(u_{n}) ≥ γ(u) + lim inf \overline{γ}(v_{n}), we get that either γ(u) ≤ 0 or lim inf \overline{γ}(v_{n}) ≤ 0, or both. Then the arguments carried out above yield a contradiction, proving that ∥u_{n} − u∥ → 0, u ∈ M and m is achieved there.

In any case, we have shown that u ∈ M is a minimizer for I_{|M} provided that one among (V 1), (V 2) or (V 3) holds. Using (ii) of Lemma 2.2 it follows that u solves the equation in (6). It remains to show that u > 0 and this follows as in [9]. Since |u_{n}| is also a minimizer, one infers that u ≥ 0. Note that the equation in (6) can be written in the form

−(1 + 2ku^{2})u′′ = (−V (x) + 2k(u′)^{2} + |u|^{p−1}) u. (15)

If u(x_{0}) = 0 for some x_{0} ∈ ℝ, then u′(x_{0}) = 0 and by the uniqueness of the Cauchy problem, (15) implies that u ≡ 0, a contradiction. This shows that u > 0 and completes the proof.

The autonomous problem. In the remaining part of this section, we consider the autonomous problem (1). Setting v = u′, (1) is equivalent to the Hamiltonian system on ℝ^{2}

\begin{cases}u′ & = v \\ v′ & = (u − 2kv^{2}u − |u|^{p−1}u)(1 + 2ku^{2})^{−1}. \end{cases} (16)

The energy

E(u, v) = (1 + 2ku^{2})\frac{v^{2}}{2} − \frac{u^{2}}{2} + \frac{|u|^{p+1}}{p + 1}

is constant along the trajectories of (16) and a standard phase plane analysis shows that there exists k_{0} > 0 such that (0, 0) is a hyperbolic equilibrium provided k > −k_{0} and p > 1. Moreover, for such k, p, the equation E(u, v) = 0 defines an isolated orbit (z(x), z′(x)) homoclinic to (0, 0), with z(x) > 0 for all x ∈ ℝ. By homoclinic to (0, 0) we mean, as usual, that z(x) → 0 and z′(x) → 0 as |x| → ∞. In addition, z is even in x and z, z′, z′′ all tend to zero exponentially as |x| → ∞. This z(x) is the unique even positive solution of (1). Of course, all translations z_{ξ}(x) = z(x + ξ), ξ ∈ ℝ, are also solutions of (1) and form a one dimensional manifold Z = {z_{ξ} : ξ ∈ ℝ}. By the uniqueness of the Cauchy problem, it is easy to see that the dimension of Ker[I′′_{0}(z_{ξ})] equals the dimension of the tangent space to the homoclinic trajectory (z(x), z′(x)) in ℝ^{2}. This means that dim(Ker[I′′_{0}(z_{ξ})]) = 1. Since, of course, the tangent space at z_{ξ} to Z, T_{z_{ξ}}Z, is a subset of Ker[I′′_{0}(z_{ξ})], by dimensionality one infers that Z is nondegenerate in the sense that

T_{z_{ξ}}Z = Ker[I′′_{0}(z_{ξ})], ∀ z_{ξ} ∈ Z. (ND)

In conclusion, we have shown:

Proposition 2.5 There exits k_{0} > 0 such that for all k > −k_{0} and all p > 1, equation (1) has a one dimensional manifold Z of critical points such that (ND) holds.

Remark. Let us explicitely point out that, for any k > −k_{0} there is c_{k} > 0 such that (1 + 2kz^{2}(x)) ≥ c_{k} for all x ∈ ℝ.

3 Proof of theorem 1.1

Solutions of (2) will be found as critical points of the functional I_{ε} defined in (3). This will be done by means of the perturbation method discussed in [1, 2], that we briefly recall for the reader convenience. For more details we refer to the forementioned papers, as well as [3].

Here we merely take X = H^{1}(ℝ), with the standard norm ∥u∥^{2} = ∫_{ℝ} (|u′|^{2} + u^{2}) dx. Moreover it is always understood that k > −k_{0}. We look for critical points of I_{ε} of the form z_{ξ} + w, where z_{ξ} ∈ Z and w ∈ (T_{z_{ξ}}Z)^{⊥}, where Z has been introduced in the last part of Section 2. Let P_{ξ} denote the projection of X to (T_{z_{ξ}}Z)^{⊥}. We first solve the auxiliary equation P_{ξ}I′_{ε}(z_{ξ} + w) = 0, namely

P_{ξ}[I′_{0}(z_{ξ} + w) + εG′(z_{ξ} + w)] = 0. (17)

Since I′_{0}(z_{ξ}) = 0 one finds I′_{0}(z_{ξ} + w) = I′′_{0}(z_{ξ}) + R(w), where R(w) = o(∥w∥) as ∥w∥ → 0. Then (17) is equivalent to P_{ξ}I′′_{0}(z_{ξ}) + εP_{ξ}G′(z_{ξ} + w) + P_{ξ}R(w) = 0. The following Lemma holds

Lemma 3.1 Let L_{ξ} := P_{ξ}I′′_{0}(z_{ξ}). There exists c > 0 such that for all ξ ∈ ℝ

|(L_{ξ}w, w)| ≥ c∥w∥^{2}, ∀ w ∈ (T_{z_{ξ}}Z)^{⊥}. (18)

Proof. It suffices to consider the linearized operator

I′′_{0}(z_{ξ}) : v ↦ − ((1 + 2kz_{ξ}^{2})v′)′ + (1 − 4kz_{ξ}z′′_{ξ} − 2k(z′_{ξ})^{2} − pz_{ξ}^{p−1}) v,

and to remark that, according to the Sturmian theory, it has a sequence of simple eigenvalue λ_{i}. The first one, λ_{1} is negative (this is due to the fact that z_{ξ} is a mounatin-pass critical point of I_{0}); moreover, by the nondegeneracy condition (ND) it follows that λ_{2} = 0 < λ_{3} < . . .. Finally it is easy to check that the inequalities λ_{1} < 0 < λ_{3} are uniform with respect to ξ ∈ ℝ.

From the preceding Lemma it follows that (17) can be written as w = (−L_{ξ})^{−1}[εP_{ξ}G′(z_{ξ} + w) + P_{ξ}R(w)]. Since the right hand side is a contraction, one can see that, for |ε| is sufficiently small, there exists ρ(ε), with ρ(ε) → 0 as |ε| → 0, such that (17) has a unique solution w_{ξ,ε} of class C^{1} with respect to ξ satisfying

∥w_{ξ,ε}∥ ≤ ρ(ε). (19)

Once the auxiliary equation is solved, we consider the one dimensional functional

Φ_{ε}(ξ) = I_{ε}(z_{ξ} + w_{ξ,ε}).

According to the results of [1, 2],

(∗) if ξ_{0} ∈ ℝ is such that Φ′_{ε}(ξ_{0}) = 0 then z_{ξ_{0}} + w_{ξ_{0},ε} is a critical point of I_{ε} and hence a solution of (2).

We shall show that

lim_{|ξ|→∞} Φ_{ε}(ξ) = const. (20)

and this will imply, in view of (∗), that (2) has a solution. Thus the rest of this section is devoted to show that (20) holds. For simplicity, we consider b(x) = c(x) ≡ 0, so that

G(u) = \frac{1}{2} ∫_{ℝ} a(x)u^{2}dx.

However, from the proof it will be clear that the result holds for non zero b, c, too. The procedure we follow is similar to that used in [3], but the presence of the term (u^{2})′′u requires different arguments here. In order to show that (20) holds we shall prove that

Proposition 3.2 For all |ε| small, lim_{|ξ|→∞} ∥w_{ξ,ε}∥ = 0.

The proof of this Proposition is postponed after several preliminary Lemmas. In the sequel, for brevity, we will write w_{ξ} instead of w_{ξ,ε}. First, of all, since w_{ξ} is uniformly bounded in ξ ∈ ℝ, then w_{ξ} ⇀ w_{∞}(= w_{∞,ε}), along a sequence |ξ_{n}| → ∞. Such a w_{ξ} satisfies the auxiliary equation (17). A direct computation yields

I′′_{0}(z_{ξ})[w] = −w′′ + w − 2kwz_{ξ}(z_{ξ})′′ − 2kz_{ξ}^{2}w′′ − kw(z_{ξ}^{2})′′ − 4kz_{ξ}(z_{ξ})′w′ − pz_{ξ}^{p−1}w,
G′(z_{ξ} + w) = a(x)(z_{ξ} + w),
R(w) = z_{ξ}^{p} − (z_{ξ} + w)^{p} − k [(w^{2})′′(z_{ξ} + w) + 2(z_{ξ}w)′′w] + pz_{ξ}^{p−1}w.

Therefore, applying the projection P_{ξ} we find there exists α_{ξ} ∈ ℝ such that w_{ξ} satisfies the following differential equation

\{ \begin{matrix}−w′′_{ξ} + w_{ξ} − 2kw_{ξ}z_{ξ}(z_{ξ})′′ − 2kz_{ξ}^{2}(w_{ξ})′′ − kw_{ξ}(z_{ξ}^{2})′′ − 4kz_{ξ}(z_{ξ})′w′_{ξ} \\ +εa(x)(z_{ξ} + w_{ξ}) + z_{ξ}^{p} − (z_{ξ} + w_{ξ})^{p} − k [(w_{ξ}^{2})′′(z_{ξ} + w_{ξ}) + 2(z_{ξ}w_{ξ})′′w_{ξ}] + α_{ξ}z′_{ξ} = 0. \end{matrix} (21)

Here we have used the fact that T_{z_{ξ}}Z is spanned by z′_{ξ}. Multiplying (21) by w_{∞}, integrating on ℝ and taking into account that w_{∞} ⊥ T_{z_{ξ}}Z, we get

\{ \begin{matrix}∫_{ℝ}(w′_{ξ}w′_{∞} + w_{ξ}w_{∞})dx + k ∫_{ℝ}(w_{ξ}^{2})′(w_{ξ}w_{∞})′dx \\ +ε ∫_{ℝ} a(x)(z_{ξ} + w_{ξ})w_{∞}dx + ∫_{ℝ} (z_{ξ}^{p} − (z_{ξ} + w_{ξ})^{p})) w_{∞}dx − ∫_{ℝ} ℓ_{ξ}w_{∞}dx = 0, \end{matrix} (22)

where

ℓ_{ξ} = 2kw_{ξ}z_{ξ}(z_{ξ})′′ + 2kz_{ξ}^{2}(w_{ξ})′′ + kw_{ξ}(z_{ξ}^{2})′′ + 4kz_{ξ}(z_{ξ})′(w_{ξ})′ + k(w_{ξ}^{2})′′z_{ξ} + 2k(z_{ξ}w_{ξ})′′w_{ξ}.

We need to estimate the above integrals. This will be done in the next three lemmas.

Lemma 3.3

(i) lim_{|ξ|→∞} ∫_{ℝ}(w′_{ξ}w′_{∞} + w_{ξ}w_{∞})dx = ∥w_{∞}∥^{2};

(ii) ∫_{ℝ}(w_{ξ}^{2})′(w_{ξ}w_{∞})′dx = ∫_{ℝ} |(w_{∞}^{2})′|^{2}dx + 2 ∫_{ℝ} |(w_{ξ} − w_{∞})′|^{2}w_{∞}^{2}dx + o(1), as |ξ| → ∞.

Proof. (i) follows immediately, because lim ∫_{ℝ}(w′_{ξ}w′_{∞} + w_{ξ}w_{∞})dx = (w_{ξ}, w_{∞}) → ∥w_{∞}∥^{2}. To prove (ii) we recall that w_{ξ}w_{∞} ⇀ w_{∞}^{2} in X as |ξ| → ∞ yielding:

\{ \begin{matrix}∫_{ℝ}(w_{ξ}^{2})′(w_{ξ}w_{∞})′dx = ∫_{ℝ}(w_{ξ}^{2} − w_{∞}^{2} + w_{∞}^{2})′(w_{ξ}w_{∞})′dx \\ = ∫_{ℝ} [(w_{ξ} − w_{∞})′(w_{ξ} + w_{∞}) + (w_{ξ} − w_{∞})(w_{ξ} + w_{∞})′] (w_{ξ}w_{∞})′dx + ∫_{ℝ} |(w_{∞}^{2})′|^{2}dx + o(1). \end{matrix} (23)

Moreover

∫_{ℝ}(w_{ξ} −w_{∞})′(w_{ξ} +w_{∞})(w_{ξ}w_{∞})′dx = ∫_{ℝ} [(w_{ξ} − w_{∞})′(w_{ξ} − w_{∞})(w_{ξ}w_{∞})′ + 2w_{∞}(w_{ξ} − w_{∞})′(w_{ξ}w_{∞})′] dx.

But

|∫_{ℝ}(w_{ξ} − w_{∞})′(w_{ξ} − w_{∞})(w_{ξ}w_{∞})′dx| =

= |∫_{|x|<R}(w_{ξ} − w_{∞})′(w_{ξ} − w_{∞})(w_{ξ}w_{∞})′dx| + |∫_{|x|>R}(w_{ξ} − w_{∞})′(w_{ξ} − w_{∞})(w_{ξ}w_{∞})′dx|

≤ C∥w_{ξ} − w_{∞}∥_{L^{∞}(−R,R)} + C ∫_{|x|>R} w_{∞}^{2}dx + C ∫_{|x|>R}(w′_{∞})^{2}dx.

Thus

lim_{|ξ|→∞} ∫_{ℝ}(w_{ξ} − w_{∞})′(w_{ξ} − w_{∞})(w_{ξ}w_{∞})′dx = 0.

Furthermore

∫_{ℝ} w_{∞}(w_{ξ} − w_{∞})′(w_{ξ}w_{∞})′dx =
= ∫_{ℝ} w_{∞}(w_{ξ} − w_{∞})′((w_{ξ} − w_{∞})w_{∞})′dx + ∫_{ℝ} w_{∞}(w_{ξ} − w_{∞})′(w_{∞}^{2})′dx
= ∫_{ℝ} |(w_{ξ} − w_{∞})′|^{2}w_{∞}^{2}dx + ∫_{ℝ} w_{∞}(w_{ξ} − w_{∞})′(w_{ξ} − w_{∞})(w_{∞})′dx
= ∫_{ℝ} |(w_{ξ} − w_{∞})′|^{2}w_{∞}^{2}dx + o(1).

The last two equations imply

lim_{|ξ|→∞} ∫_{ℝ}(w_{ξ} − w_{∞})′(w_{ξ} + w_{∞})(w_{ξ}w_{∞})′dx = 0. (24)

Similarly

lim_{|ξ|→∞} ∫_{ℝ}(w_{ξ} − w_{∞})(w_{ξ} + w_{∞})′(w_{ξ}w_{∞})′dx = 0.

Substituting this and (24) into (23), (ii) follows.

Lemma 3.4

(i) ∫_{ℝ} [z_{ξ}^{p} − (z_{ξ} + w_{ξ})^{p})] w_{∞}dx = ∫_{ℝ} |w_{∞}|^{p}dx + o(1), as |ξ| → ∞;

(ii) lim_{|ξ|→∞} ∫_{ℝ} a(x)z_{ξ}w_{∞}dx = 0

Proof. To prove (i) it suffices to remark that, obviously, ∫_{ℝ} z_{ξ}^{p}w_{∞}dx = o(1) as |ξ| → ∞ and that z_{ξ} + w_{ξ} ⇀ w_{∞} as well as (z_{ξ} + w_{ξ})^{p} ⇀ (w_{∞})^{p}.

As for (ii), let a = a_{1} + a_{2} with a_{1} ∈ L^{r}(ℝ) and a_{2} ∈ L^{∞}(ℝ) and set, for i = 1, 2 and R > 0, A_{i}(R) = ∫_{|x|<R} |a_{i}(x)z_{ξ}w_{∞}|dx and B_{i}(R) = ∫_{|x|>R} |a_{i}(x)z_{ξ}w_{∞}|dx. One has (r′ denotes the conjugate exponent of r):

A_{1}(R) ≤ max_{|x|<R} |z_{ξ}(x)| (∫_{|x|<R} |a_{1}(x)|^{r}dx)^{1/r} (∫_{|x|<R} |w_{∞}(x)|^{r′}dx)^{1/r′} ≤ C_{1} max_{|x|<R} |z_{ξ}(x)|,

B_{1}(R) ≤ C_{2} (∫_{|x|>R} |a_{1}(x)|^{r}dx)^{1/r} (∫_{|x|>R} |z_{ξ}(x)|^{r′}dx)^{1/r′} ≤ C_{3} (∫_{|x|>R} |z_{ξ}(x)|^{r}dx)^{1/r} .

Given η > 0, choose R > 0 such that B_{1}(R) < η. Since max_{|x|<R} |z_{ξ}(x)| → 0 as |ξ| → ∞, we deduce that A_{1}(R) + B_{1}(R) ≤ η + o(1) as |ξ| → ∞. Moreover

A_{2}(R) ≤ C_{4} max_{|x|<R} |z_{ξ}(x)|, B_{2}(R) ≤ C_{5} ∫_{|x|>R} |z_{ξ}(x)|dx

and, as before, one finds that A_{2}(R) + B_{2}(R) ≤ η + o(1) as |ξ| → ∞. Then (ii) follows.

Lemma 3.5 There results:

lim_{|ξ|→∞} ∫_{ℝ} ℓ_{ξ}w_{∞}dx = 0.

Proof. Note that every term of ℓ_{ξ}w_{∞} contains z_{ξ} or z′_{ξ} and also w_{∞} or w′_{∞}. Then we write each term on (−R, R) and on ℝ \ (−R, R). In (−R, R) we use that z_{ξ} and z′_{ξ} are small for |ξ| → ∞; on ℝ \ (−R, R) we use that ∫_{|x|>R} w_{∞}^{2} and ∫_{|x|>R}(w′_{∞})^{2} are small if R is large.

We are now in position to show Lemma 3.6 For ε small enough, one has that w_{ξ,ε} ⇀ 0 as |ξ| → ∞. .

Proof. We have to show that w_{∞} = 0. Using Lemmas 3.3, 3.4 and 3.5 and passing to the limit in (22) as |ξ| → ∞ we get

∥w_{∞}∥^{2} + ε ∫_{ℝ} a(x)w_{∞}^{2}dx = ∫_{ℝ} |w_{∞}|^{p+1}dx − k ∫_{ℝ} |(w_{∞}^{2})′|^{2}dx − 2k lim ∫_{ℝ} |(w_{ξ} − w_{∞})′|^{2}w_{∞}^{2}dx.

This implies, for some constants C_{i} > 0 independent of ξ and ε,

(1 − εC_{1})∥w_{∞}∥^{2} ≤ C_{2} ∥w_{∞}∥^{p+1} + |k| C_{3} ∥w_{∞}∥^{4} + 2|k| C_{4} ∥w_{∞}∥^{2} lim ∥w_{ξ} − w_{∞}∥^{2}.

From (19), we have that ∥w_{∞}∥ ≤ ρ(ε) and ∥w_{ξ} − w_{∞}∥ ≤ ρ(ε) with ρ(ε) → 0 as |ε| → 0. Then we may choose |ε| small such that the right hand side of the preceding inequality is smaller or equal to ∥w_{∞}∥^{2}/2, while (1 − εC_{1}) > 1/2 and this implies w_{∞} = 0.

We are now in position to prove Proposition 3.2. Multiplying (21) by w_{ξ} and integrate on ℝ, we get

\begin{cases}(L_{ξ}w_{ξ}, w_{ξ}) = & ε ∫_{ℝ} a(x)(z_{ξ} + w_{ξ})w_{ξ}dx + ∫_{ℝ} [(z_{ξ} + w_{ξ})^{p} − z_{ξ}^{p} − p|z_{ξ}|^{p−1}w_{ξ}] w_{ξ}dx \\ & +k ∫_{ℝ} [(w_{ξ}^{2})′′z_{ξ} + (w_{ξ}^{2})′′w_{ξ} + 2(z_{ξ}w_{ξ})′′w_{ξ}] w_{ξ}dx. \end{cases} (25)

Since w_{ξ} → 0 in Lloc^{q} for any q ≥ 1 and a = a_{1} + a_{2} with a_{1} ∈ L^{r} and a_{2} ∈ L^{∞}, then

lim_{|ξ|→∞} ∫_{|x|<R} a(x)w_{ξ}^{2}dx = 0, ∀ R > 0. (26)

Moreover, using that a_{2}(x) → 0 as |x| → ∞, then for any δ > 0 we choose R > 0 such that |a_{2}(x)| < δ for all |x| > R. Then we get

|∫_{|x|>R} a(x)w_{ξ}^{2}dx| ≤ C_{1} ∫_{|x|>R} |a_{1}(x)|^{r}dx + C_{2} δ.

This and (26) imply

lim_{|ξ|→∞} ∫_{ℝ} a(x)w_{ξ}^{2}dx = 0.

Similarly,

lim_{|ξ|→∞} ∫_{ℝ} a(x)z_{ξ}w_{ξ}dx = 0.

For the second term in (25) we find readily

|∫_{ℝ}[(z_{ξ} + w_{ξ})^{p} − z_{ξ}^{p} − p|z_{ξ}|^{p−1}w_{ξ}]w_{ξ}dx| ≤ C∥w_{ξ}∥^{p+1}.

In the last part of (25) each term contains either three w_{ξ} and w′_{ξ} or four of them. Then

|∫_{ℝ}[(w_{ξ}^{2})′′z_{ξ}w_{ξ} + (w_{ξ}^{2})′′w_{ξ}^{2} + 2(z_{ξ}w_{ξ})′′w_{ξ}^{2}]dx| ≤ C(∥w_{ξ}∥^{3} + ∥w_{ξ}∥^{4}).

Thus from (25) we get

C_{1}∥w_{ξ}∥^{2} ≤ C_{2} (∥w_{ξ}∥^{p+1} + ∥w_{ξ}∥^{3} + ∥w_{ξ}∥^{4}) + o(1).

Since ∥w_{ξ}∥ ≤ ρ(ε), and ρ(ε) → 0, we must have ∥w_{ξ}∥ → 0 as |ξ| → ∞. This completes the proof of proposition 3.2.

We are now ready to show that (20) holds, namely that

lim_{|ξ|→∞} Φ_{ε}(ξ) = const.

Below, of course, we take |ε| small in such a way that all the preceding results hold true. With easy calculation we find:

Φ_{ε}(ξ) = I_{0}(z_{ξ} + w_{ξ}) + \frac{ε}{2} ∫_{ℝ} a(x)(z_{ξ} + w_{ξ})^{2}dx = I_{0}(z) + o(∥w_{ξ}∥) + \frac{ε}{2} ∫_{ℝ} a(x)(z_{ξ} + w_{ξ})^{2}dx.

Here we have used that I_{0}(z_{ξ}) ≡ I_{0}(z). Arguing as before one has that ∫_{ℝ} a(x)(z_{ξ} + w_{ξ})^{2}dx → 0 as |ξ| → ∞. In addition, we know by Proposition 3.2 that ∥w_{ξ}∥ → 0 as |ξ| → ∞. Then we infer that lim_{|ξ|→∞} Φ_{ε}(ξ) = I_{0}(z_{ξ}) ≡ const, proving (20). Since lim_{|ξ|→∞} Φ_{ε}(ξ) = const, then Φ_{ε} has at least a stationary point ξ_{0}. Thus, according to (∗), u_{ε} = z_{ξ_{0}} + w_{ξ_{0},ε} is a solution of (2).

Let us finally show that for any fix k > −k_{0}, the solution u_{ε} is positive on ℝ, provided |ε| is small enough. Given R > 0 let ζ_{R} > 0 be the minimum of z_{ξ_{0}}(x) on the interval [−R, R]. Recalling that ∥w_{ξ_{0},ε}∥ ≤ ρ(ε), and ρ(ε) → 0, we find ε_{R} > 0 such that u_{ε}(x) > ζ_{R}/2 > 0 for all x ∈ [−R, R]. By regularity, u_{ε} ∈ C^{1} ∩ L^{∞} and u_{ε}(x) → 0 as |x| → ∞. Then we can choose R such that |u_{ε}(x)|^{p−1} < 1/2 for all |x| > R and all |ε| < ε_{R}. Now, suppose by contradiction that u_{ε} takes some negative values on |x| > R and let x_{1}, |x_{1}| > R, a point where u_{ε}(x_{1}) = min{u_{ε}(x) : |x| > R} < 0. We claim that, taking |ε| possibly smaller,

−(1 + 2ku_{ε}^{2}(x_{1}))u′′_{ε}(x_{1}) ≤ 0. (27)

This is trivial if k ≥ 0. When −k_{0} < k < 0 we use the fact that u_{ε} = z_{ξ_{0}} + w_{ξ_{0},ε} and that ∥w_{ξ_{0},ε}∥_{L^{∞}} → 0 as |ε| → 0. Using the Remark after Proposition 2.5, it follows that (27) holds for k ∈] − k_{0}, 0[, provided |ε| is sufficiently small. Next, from (15), namely

−(1 + 2ku^{2})u′′ = (−(1 + εa(x)) + 2k(u′)^{2} + |u|^{p−1}) u,

we infer that

(−(1 + εa(x_{1})) + |u_{ε}(x_{1})|^{p−1}) u_{ε}(x_{1}) = −(1 + 2ku_{ε}^{2}(x_{1}))u′′_{ε}(x_{1}) ≤ 0.

On the other side, choose ε_{1} > 0 in such a way that (1 + εa(x)) > 1/2 for |ε| < ε_{1}. Then, taking |ε| < min{ε_{R}, ε_{1}} we get that (1 + εa(x_{1})) > 1/2 > |u_{ε}(x_{1})|^{p−1} and thus

(−(1 + εa(x_{1})) + |u_{ε}(x_{1})|^{p−1}) u_{ε}(x_{1}) > 0,

a contradiction, proving that u_{ε} > 0. This completes the proof of Theorem 1.1.

Remarks. (i) The results of [2] would imply that the solution found above is a mountain-pass solution, provided εa(x) < 0, |ε| small. Therefore, for all k > −k_{0} and all p > 1 Theorem 1.1 yields, for |ε| small, the existence of a ground state when εa(x) < 0. Let us point out that, setting V = 1 + εa(x) with a ∈ S, assumption (V 3) in Theorem 2.4 holds whenever εa(x) < 0. Of course, unlike in Theorem 2.4, here we deal only with perturbed potentials but cover a broader range of k and p.

(ii) According to [1, 2] one finds that

Φ_{ε}(ξ) = I_{0}(z) + εΓ(ξ) + o(|ε|),

where

Γ(ξ) = \frac{1}{2} ∫_{ℝ} a(x + ξ)z^{2}dx + k ∫_{ℝ} b(x + ξ)z^{2}(z′)^{2}dx − \frac{1}{p + 1} ∫_{ℝ} c(x + ξ)z^{p+1}dx.

As in [3] one can show that Γ(ξ) → 0 as |ξ| → ∞ provided a, b, c ∈ L^{r} ∩ L^{∞}. If Γ(ξ) ≢ 0 then Γ has a maximum or minimum which gives rise to a positive solution of (2). In general, we do not know if a, b, c ≢ 0 implies that Γ(ξ) ≢ 0 and this is the reason why we have carried out the preceding arguments leading to show that Φ_{ε} has a stationary point provided a, b, c ∈ S. However, in some circumstances it is possible to work with Γ and this often permits to obtain sharper results. For instance, conditions on a, b, c can be imposed so that one can even obtain multiplicity results. We will shortly illustrate such a fact with a specific example. Let us assume that a, b, c are smooth and even. By a stright calculation one finds

Γ(0) = \underbrace{\frac{1}{2} ∫_{ℝ} a(x)z^{2}dx}_{A} +k \underbrace{∫_{ℝ} b(x)z^{2}(z′)^{2}dx}_{B} − \underbrace{\frac{1}{p + 1} ∫_{ℝ} c(x)z^{p+1}dx}_{C} = A + kB − C,

Γ′(0) = \frac{1}{2} ∫_{ℝ} a′(x)z^{2}dx + k ∫_{ℝ} b′(x)z^{2}(z′)^{2}dx − \frac{1}{p + 1} ∫_{ℝ} c′(x)z^{p+1}dx = 0,
Γ′′(0) = \underbrace{\frac{1}{2} ∫_{ℝ} a′′(x)z^{2}dx}_{A′′} +k \underbrace{∫_{ℝ} b′′(x)z^{2}(z′)^{2}dx}_{B′′} − \underbrace{\frac{1}{p + 1} ∫_{ℝ} c′′(x)z^{p+1}dx}_{C′′} = A′′ + kB′′ − C′′.

Therefore ξ = 0 is a stationary point of Γ and if, e.g. A + kB − C < 0 and A′′ + kB′′ − C′′ ≠ 0, then ξ = 0 gives rise to a solution of (2). Furthermore, if A′′ + kB′′ − C′′ < 0 the function Γ achieves its global minimum at some ξ^{∗} ≠ 0 and such ξ^{∗} gives rise to a second solution of (2) u_{ε}^{∗} ≃ z_{ξ^{∗}} + w_{ξ^{∗}}, which is asymmetric (namely not even). Let us point out that such u_{ε}^{∗} is the ground state of (2), according to the discussion made in the item (i) above. For example, if B′′ > 0 and A′′ − C′′ < 0, then the ground state of (2) is asymmetric provided k < k^{∗} = (C′′ − A′′)/B′′.

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