Ground State Solutions for the Periodic Discrete Nonlinear Schrödinger Equations with Superlinear Nonlinearities

Theorem 1. Suppose that conditions (V₁), (f₁)–(f₄) are satisfied. Then, one has the following conclusions.

• (1)

  If either σ = 1 and β ≠ ∞ or σ = −1 and α ≠ −∞, then (7) has at least a nontrivial ground state solution.

• (2)

  If either σ = 1 and β = ∞ or σ = −1 and α = −∞, then (7) has no nontrivial solution.

Theorem 2. Suppose that conditions (V₁), (f₁)–(f₄) are satisfied and f_n is odd in u. If either σ = 1 and β ≠ ∞ or σ = −1 and α ≠ −∞, then (7) has infinitely many pairs of geometrically distinct solutions.

Remark 3. In [8], the author considered (7) with f_n defined by

which obviously satisfies (f₁)–(f₄); the author also discussed the case where f satisfies the Ambrosetti-Rabinowitz condition; that is, there exists μ > 2 such that Clearly, (23) implies that F_n(u) ≥ c | u|^μ > 0 for |u | ≥ 1. So, it is a stronger condition than (f₃).

Remark 4. In [9], the author assumed that f_n satisfies the following condition: there exists θ ∈ (0,1) such that

Obviously, (24) implies (23) with μ = 1 + (1/θ), so it is a stronger condition than the Ambrosetti-Rabinowitz condition. In our paper, the nonlinearities satisfy more general superlinear assumptions instead of (24) which also implies (f₄). However, we do not assume that f_n is differentiable and satisfies (24), ℳ is not a C¹ manifold of E, and the minimizers on ℳ may not be critical points of J. Hence, the method of [9] does not apply any more. Nevertheless, ℳ is still a topological manifold, naturally homeomorphic to the unit sphere in E⁺ (see in detail in Section 3). We use the generalized Nehari manifold approach developed by Szulkin and Weth which is based on reducing the strongly indefinite variational problem to a definite one and prove that the minimizers of J on ℳ are indeed critical points of J.

Remark 5. In [7], it is shown that (7) has at least a nontrivial solution u ∈ l² if f satisfies (V₁), (f₂), (f₃), and the following conditions:

•  

  (B₁) F_n(u) ≥ 0 for any u ∈ ℝ and H_n(u): = (1/2)f_n(u)u − F_n(u) > 0 if u ≠ 0,

•  

  (B₂) H_n(u) → ∞ as |u | → ∞, and there exist r₀ > 0 and γ > 1 such that |f_n(u)|^γ/|u|^γ ≤ c₀H_n(u) if |u | ≥ r₀, where c₀ is a positive constant,

Lemma 6. F_n(u) > 0 and (1/2)f_n(u)u > F_n(u) for all u ≠ 0.

Proof. By (f₂) and (f₄), it is easy to get that

Set H_n(u) = (1/2)f_n(u)u − F_n(u). It follows from (f₄) that So, (1/2)f_n(u)u > F_n(u) for all u ≠ 0.

Proposition 7 (see [16], [17].)Let u, s, v ∈ ℝ be numbers with s ≥ −1 and w : = su + v ≠ 0. Then,

Lemma 8. If u ∈ ℳ, then

Hence, u is the unique global maximum of $$.

Proof. We rewrite J by

Since u ∈ ℳ, we have

Together with Proposition 7, we know that

The proof is complete.

Lemma 9. (a) There exists α > 0 such that $$, where S_α : = {u ∈ E⁺ : ∥u∥ = α}.

(b)$$ for every u ∈ ℳ.

Proof. (a) By (f₁) and (f₂), it is easy to show that for any ε > 0, there exists c_ε > 0 such that

∥·∥ is equivalent to the E norm on E⁺ and E ⊂ l^q for 2 ≤ q ≤ ∞ with $$. Hence, for any ε ∈ (0, 1/2) and u ∈ E⁺, we have which implies $$ for some α > 0 (small enough), where $$.

The first inequality is a consequence of Lemma 8 since for every u ∈ ℳ, there is s > 0 such that $$.

(b) For u ∈ ℳ, by (25), we have

Hence, $$.

Lemma 10. Let 𝒲 ⊂ E⁺∖{0} be a compact subset. Then, there exists R > 0 such that J ≤ 0 on E(u)∖B_R(0) for every u ∈ 𝒲, where B_R(0) denotes the open ball with radius R and center 0.

Proof. Suppose by contradiction that there exist u^(k) ∈ 𝒲 and w^(k) ∈ E(u^(k)), k ∈ ℕ, such that J(w^(k)) > 0 for all k and ∥w^(k)∥ → ∞ as k → ∞. Without loss of generality, we may assume that ∥u^(k)∥  = 1 for k ∈ ℤ. Then, there exists a subsequence, still denoted by the same notation, such that u^(k) → u ∈ E⁺. Set v^(k) = w^(k)/∥w^(k)∥   = s^(k)u^(k) + v^(k)−. Then,

By (25), we have Consequently, we know that $$ and $$. Passing to a subsequence if necessary, we assume that $$, v^(k)⇀v, $$, and $$ for every n. Hence, $$ and $$. It follows that for n₀ ∈ ℤ with $$, $$, as k → ∞. Then, by (f₃), we have which contradicts with (35).

Lemma 11. For each u ∈ E⁺∖{0}, the set $$ consists of precisely one point which is the unique global maximum of $$.

Proof. By Lemma 8, it suffices to show that $$. Since $$, we may assume that u ∈ S⁺. By Lemma 10, there exists R > 0 such that J ≤ 0 on E(u)∖B_R(0) provided that R is large enough. By Lemma 9 (a), J(tu) > 0 for small t > 0. Moreover, J ≤ 0 on $$. Hence, $$.

Let v^(k)⇀v in $$. Then, $$ as k → ∞ for all n after passing to a subsequence if necessary. Hence, $$. Let φ(v) = ∑_n∈ℤ χ_nF_n(v_n). Then,

that is, φ is a weakly lower semicontinuous. From the weak lower semi-continuity of the norm, it is easy to see that J is weakly upper semicontinuous on $$. Therefore, $$ for some $$. By the proof of Lemma 10, u₀ is a critical point of $$. It follows that (J^′(u₀), u₀) = (J^′(u₀), z) = 0 for all z ∈ E and hence u₀ ∈ ℳ. To summarize, $$.

Lemma 12. J is coercive on ℳ; that is, J(u) → ∞ as ∥u∥ → ∞, u ∈ ℳ.

Proof. Suppose, by contradiction, that there exists a sequence {u^(k)} ⊂ ℳ such that ∥u^(k)∥ → ∞ and J(u^(k)) ≤ d for some d ∈ [c, ∞). Let v^(k) = u^(k)/∥u^(k)∥. Then, there exists a subsequence, still denoted by the same notation, such that v^(k)⇀v and $$ for every n as k → ∞.

First, we know that there exist δ > 0 and n_k ∈ ℤ such that

Indeed, if not, then v^(k)+ → 0 in l^∞ as k → ∞. By Lemma 9(b), $$, which means that $$ is bounded. For q > 2, Then, v^(k)+ → 0 in all l^q, q > 2. By (32), for any s ∈ ℝ, which implies that $$ as k → ∞.

Since $$ for s ≥ 0, Lemma 8 implies that

as k → ∞. This is a contradiction if $$.

Due to the periodicity of coefficients, both J and ℳ are invariant under T-translation. Making such shifts, we can assume that 1 ≤ n_k ≤ T − 1 in (39). Moreover, passing to a subsequence if needed, we can assume that n_k = n₀ is independent of k. Next, we may extract a subsequence, still denoted by {v^(k)}, such that $$ for all n ∈ ℤ. In particular, for n = n₀, inequality (39) shows that $$ and hence v⁺ ≠ 0.

Since $$ as k → ∞, it follows again from (f₃) and Fatou’s lemma that

a contradiction again. The proof is finished.

Lemma 13. (a) The mapping $$ is continuous.

(b) The mapping $$ is a homeomorphism between S⁺ and ℳ, and the inverse of m is given by m⁻¹(u) = u⁺/∥u⁺∥, where S⁺ : = {u ∈ E⁺ : ∥u∥ = 1}.

(c) The mapping m⁻¹ : ℳ ↦ S⁺ is the Lipschitz continuous.

Proof. (a) Let (u^(k)) ⊂ E⁺∖{0} be a sequence with u^(k) → u. Since $$, without loss of generality, we may assume that ∥u^(k)∥ = 1 for all k. Then, $$. By Lemma 10, there exists R > 0 such that

It follows from Lemma 12 that $$ is bounded. Passing to a subsequence if needed, we may assume that where $$ by Lemma 9(b). Moreover, by Lemma 11, Therefore, using the weak lower semicontinuity of the norm and φ (defined in Lemma 11), we get which implies that all inequalities above must be equalities and $$. By Lemma 11, $$ and hence $$.

(b) This is an immediate consequence of (a).

(c) For u, v ∈ ℳ, by (b), we have

Lemma 14. (a)  $$ and

(b)  Ψ ∈ C¹(S⁺, ℝ) and

(c)  {w_n} is a Palais-Smale sequence for Ψ if and only if {m(w_n)} is a Palais-Smale sequence for J.

(d)  w ∈ S⁺ is a critical point of Ψ if and only if m(w) ∈ ℳ is a nontrivial critical point of J. Moreover, the corresponding values of   Ψ and J coincide and $$.

Proof. (a) We put $$, so we have u = (∥u⁺∥/∥w∥)w + u⁻. Let z ∈ E⁺. Choose δ > 0 such that w_t : = w + tz ∈ E⁺∖{0} for |t | < δ and put $$. We may write $$ with s_t > 0. From the proof of Lemma 13, the function t ↦ s_t is continuous. Then, s₀ = ∥u⁺∥/∥w∥. By Lemma 11 and the mean value theorem, we have

with some η_t ∈ (0,1). Similarly, with some τ_t ∈ (0,1). Combining these inequalities and the continuity of function t ↦ s_t, we have Hence, the Gâteaux derivative of $$ is bounded linear in z and continuous in w. It follows that $$ is of class C¹ (see [15]).

(b) It follows from (a) by noting that $$ since w ∈ S⁺.

(c) Let {w_n} be a Palais-Smale sequence for Ψ, and let u_n = m(w_n) ∈ ℳ. Since for every n ∈ ℤ, we have an orthogonal splitting $$; using (b), we have

because J^′(u_n)v = 0 for all v ∈ E(w_n) and E(w_n) is orthogonal to $$. Using (b) again, we have Therefore, According to Lemma 9(b) and Lemma 12, $$. Hence, {w_n} is a Palais-Smale sequence for Ψ if and only if {u_n} is a Palais-Smale sequence for J.

(d) By (57), Ψ^′(w) = 0 if and only if J^′(m(w)) = 0. The other part is clear.

Proof of Theorem 1. (1) We know that c > 0 by Lemma 9(a). If u₀ ∈ ℳ satisfies J(u₀) = c, then m⁻¹(u₀) ∈ S⁺ is a minimizer of Ψ and therefore a critical point of Ψ and also a critical point of J by Lemma 14. We shall show that there exists a minimizer u ∈ ℳ of J|_ℳ. Let {w^(k)} ⊂ S⁺ be a minimizing sequence for Ψ. By Ekeland’s variational principle, we may assume that Ψ(w^(k)) → c and Ψ^′(w^(k)) → 0 as k → ∞. Then, J(u^(k)) → c and J^′(u^(k)) → 0 as k → ∞ by Lemma 14(c), where u^(k) : = m(w^(k)) ∈ ℳ. By Lemma 12, {u^(k)} is bounded, and hence {u^(k)} has a weakly convergent subsequence.

First, we show that there exist δ > 0 and n_k ∈ ℤ such that

Indeed, if not, then u^(k) → 0 in l^∞ as k → ∞. From the simple fact that for q > 2, we have u^(k) → 0 in all l^q, q > 2. By (32), we know that which implies that $$ as k → ∞. Therefore, Then, ∥u^(k)+∥² → 0 as k → ∞, contrary to Lemma 9(b).

From the periodicity of the coefficients, we know that J and J^′ are both invariant under T-translation. Making such shifts, we can assume that 1 ≤ n_k ≤ T − 1 in (58). Moreover, passing to a subsequence, we can assume that n_k = n₀ is independent of k.

Next, we may extract a subsequence, still denoted by {u^(k)}, such that u^(k)⇀u and $$ for all n ∈ ℤ. Particularly, for n = n₀, inequality (58) shows that $$, so u ≠ 0. Moreover, we have

that is, u is a nontrivial critical point of J.

Finally, we show that J(u) = c. By Lemma 6 and Fatou’s lemma, we have

Hence, J(u) = c. That is, u is a nontrivial ground state solution of (7).

(2) If β = ∞, by way of contradiction, we assume that (7) has a nontrivial solution u ∈ E. Then, u is a nonzero critical point of J in E. Thus, J^′(u) = 0. But by Lemma 6,

This is a contradiction, so the conclusion holds.

This completes the proof of Theorem 1.

Proof of Theorem 2. It is easy to see that mappings m, m⁻¹ are equivariant with respect to the ℤ-action by Lemma 13; hence, the orbits 𝒪(u) ⊂ ℳ consisting of critical points of J are in 1-1 correspondence with the orbits 𝒪(w) ⊂ S⁺ consisting of critical points of Ψ by Lemma 14(d). Next, we may choose a subset ℱ ⊂ K such that ℱ = −ℱ and ℱ consists of a unique representative of ℤ-orbits. So, we only need to prove that the set ℱ is infinite. By contradiction, we assume that

Let where γ denotes genus and j ∈ ℕ. We consider the sequence of the Lusternik-Schnirelmann values of Ψ defined by

Now, we claim that

Firstly, we show that

In fact, there exist v^(k), w^(k) ∈ ℱ, and g_k, l_k ∈ ℤ such that v^(k)*g_k ≠ w^(k)*l_k for all k and Let m_k = g_k − l_k. Passing to a subsequence, v^(k) = v ∈ ℱ, w^(k) = w ∈ ℱ, and either m_k = m ∈ ℤ for all k or |m_k | → ∞. In the first case, 0 < ∥v^(k)*g_k − w^(k)*l_k∥ = ∥v − w*m∥ = κ for all k. In the second case, w*m_k⇀0 and therefore κ = lim _k→∞∥v − w*m_k∥ ≥ ∥v∥ = 1. By (70), $$ or 1.

Next, we consider a pseudogradient vector field of Ψ  [18]; that is, there exists a Lipschitz continuous map V: S⁺∖K → T_wS⁺ and for all w ∈ S⁺∖K,

Let η : 𝒟 → S⁺∖K be the corresponding Ψ-decreasing flow defined by where 𝒟 = {(t, w) : w ∈ S⁺∖K, T⁻(w) < t < T⁺(w)} ⊂ ℝ × (S⁺∖K), and T⁻(w) < 0, T⁺(w) > 0 are the maximal existence times of the trajectory t → η(t, w) in negative and positive direction. By the continuity property of the genus, there exists δ > 0 such that $$, where $$ and δ < κ/2. Following the deformation argument (Lemma A.3), we choose ε = ε(δ) > 0 such that Then, for every $$, there exists t ∈ [0, T⁺(w)) such that Ψ(η(t, w)) < c_k − ε. Hence, we may define the entrance time map which satisfies r(w) < T⁺(w) for every $$. Since c_k − ε is not a critical value of Ψ by (74), it is easy to see that r is a continuous and even map. It follows that the map is odd and continuous. Then, $$, and consequently, So, $$. Therefore, $$. Moreover, the definition of c_k and of c_k+1 implies that $$ if c_k < c_k+1 and $$ if c_k = c_k+1. Since $$, c_k < c_k+1. Therefore, there is an infinite sequence {±w_k} of pairs of geometrically distinct critical points of Ψ with Ψ(w_k) = c_k, which contradicts with (66). Therefore, the set ℱ is infinite.

This completes the proof of Theorem 2.