On Standing Wave Solutions for Discrete Nonlinear Schrödinger Equations

Lemma 1. If V satisfies the condition (V₁), then

• (1)

  for any 2 ≤ p ≤ ∞, the embedding map from E into l^p(ℤ^m) is compact,

• (2)

  the spectrum σ(L) is discrete and consists of simple eigenvalues accumulating to +∞.

Theorem 2. Suppose that conditions (V₁) and (f₁)–(f₄) are satisfied. Then one has the following conclusions.

• (1)

  If σ = −1, ω ≤ λ₁, (4) has no nontrivial solution.

• (2)

  If σ = 1, ω < λ₁, (4) has a nontrivial ground state solution.

• (3)

  If σ = 1, ω < λ₁, and f(n, u) is odd in u for each n ∈ ℤ^m, (4) has infinitely many pairs of solutions ±u^(k) in E.

Remark 3. In [20], the author considered the following DNLS equation:

where there exists a positive constant $$, such that for any n ∈ ℤ^m, $$. Clearly, (18) corresponds (4) if we let f(n, u) = γ_nf(u). Therefore, (18) is a special case of (4).

In [20], the nonlinearity f ∈ C¹(ℝ) satisfies the following condition:

which implies (8). So it is a stronger condition than (f₃). Therefore, our results generalize the corresponding ones.

Remark 4. In [16], the authors also considered (18) and assumed that the nonlinearity f ∈ C¹(ℝ) satisfies the classical Ambrosetti-Rabinowitz superlinear condition (8). Clearly, it is a stronger condition than (f₃).

Since ω < λ₁, we may introduce an equivalent norm in E by setting

and then the functional J can be rewritten as

To prove the multiplicity results, we need the following lemma.

Lemma 5 (see [21].)Let S = {w ∈ E : ∥w∥ = 1}. If E is a infinite-dimensional Hilbert space, Φ ∈ C¹(S, ℝ) is even and bounded below and satisfies the Palais-Smale condition. Then Φ has infinitely many pairs of critical points.

Lemma 6. Suppose that conditions (V₁) and (f₁)–(f₄) are satisfied. Then one has

• (1)

  F(n, u) > 0 and (1/2)f(n, u)u > F(n, u) for all u ≠ 0,

• (2)

  J(u) > 0, for all u ∈ 𝒩.

Proof. (1) From (f₂) and (f₄), it is easy to get that

By (f₄), we have So (1/2)f(n, u)u > F(n, u) for all u ≠ 0.

(2) For all u ∈ 𝒩, by (1), we have

Lemma 7. Suppose that conditions (V₁) and (f₁)–(f₄) are satisfied, and let $$ Then one has the following.

• (1)

  I^′(u) = o(∥u∥) as u → 0.

• (2)

  s ↦ I^′(su)u/s is strictly increasing for all u ≠ 0 and s > 0.

• (3)

  I(su)/s² → ∞ uniformly for u on the weakly compact subsets of E∖{0}, as s → ∞.

Proof. (1) and (2) are easy to be shown from (f₂) and (f₄), respectively. Next, we verify (3). Let W ⊂ E∖{0} be weakly compact and let {u^(k)} ⊂ W. It suffices to show that if s^(k) → ∞ as k → ∞, then so does a subsequence of I(s^(k)u^(k))/(s^(k)) ². Passing to a subsequence if necessary, u^(k)⇀u ∈ E∖{0} and $$ for every n, as k → ∞.

Since $$ and u^(k) ≠ 0, by (f₃) and (23), we have

Lemma 8. Under the assumptions (V₁) and (f₁)–(f₄), for each w ∈ E∖{0}, there exists a unique s_w > 0 such that s_ww ∈ 𝒩.

Proof. Let g(s) : = J(sw),s > 0. Note that

and from (2) of Lemma 7, then there exists a unique s_w, such that g^′(s) > 0 whenever 0 < s < s_w, g^′(s) < 0 whenever s > s_w, and g^′(s_w) = J^′(s_ww)w = 0. So s_ww ∈ 𝒩.

Remark 9. By (1) and (3) of Lemma 7, g(s) > 0 for s > 0 small and g(s) < 0 for s > 0 large. Together with Lemma 8, we have that s_w is a unique maximum of g(s) and s_ww is the unique point on the ray s ↦ sw (s > 0) which intersects with 𝒩. That is, u ∈ 𝒩 is the unique maximum of J on the ray. Therefore, we may define the mapping $$ and m : S → 𝒩 by setting

where S = {u ∈ E : ∥u∥ = 1}.

Lemma 10. For each compact subset 𝒱 ⊂ S, there exists a constant C_𝒱 such that s_w ≤ C_𝒱 for all w ∈ 𝒱.

Proof. Suppose that, by contradiction, $$ as k → ∞. By Lemma 6 and (f₃), we have

This is a contradiction.

Lemma 11. (1) The mapping $$ is continuous.

(2) The mapping m is a homeomorphism between S and 𝒩, and the inverse of m is given by m⁻¹(u) = u/∥u∥.

Proof. (1) Suppose that w_n → w ≠ 0. Since $$ for each t > 0, we may assume that w_n ∈ S for all n. Write $$. By Lemmas 8 and 10, $$ is bounded, and hence $$ after passing to a subsequence if needed. Since 𝒩 is closed and $$. Hence $$ by the uniqueness of s_w of Lemma 8. (2) This is an immediate consequence of (1).

Lemma 12. J satisfies the Palais-Smale condition on 𝒩.

Proof. Let {u^(k)} ⊂ 𝒩 be a sequence such that J(u^(k)) ≤ d for some d > 0 and J^′(u^(k)) → 0 as k → ∞.

Firstly, we prove that {u^(k)} is bounded. In fact, if not, we may assume by contradiction that ∥u^(k)∥ → ∞ as k → ∞. Let v^(k) = u^(k)/∥u^(k)∥. Then there exists a subsequence, still denoted by the same notation, such that v^(k)⇀v in  E as k → ∞.

Suppose that v = 0. For every s > 0, from Remark 9, we have

This is a contradiction if $$. Therefore, v ≠ 0.

According to Lemma 7(3), we have

a contradiction again. Thus, {u^(k)} is bounded.

Finally, we show that there exists a convergent subsequence of {u^(k)}. Actually, there exists a subsequence, still denoted by the same notation, such that u^(k)⇀u. By Lemma 1, for any 2 ≤ q ≤ ∞, then

Note that

The first term (J^′(u^(k)) − J^′(u), (u^(k) − u)) → 0 as k → ∞ because of the weak convergence.

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

Then, Combining (32) and the boundedness of {u^(k)}, the above inequality implies It follows from (33) that u^(k) → u in E; that is, J satisfies Palais-Smale condition.

The proof is complete.

Lemma 13. (1) $$, and

(2) Ψ ∈ C¹(S, ℝ), and

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

(4) w is a critical point of Ψ if and only if m(w) is a nontrivial critical point of J. Moreover, the corresponding values of Ψ and J coincide and inf _SΨ = inf _𝒩J.

Proof. (1) Let w ∈ E∖{0} and z ∈ E. By Remark 9 and the mean value theorem, we obtain

where |t| is small enough and τ_t ∈ (0,1). Similarly, where η_t ∈ (0,1). From the proof of Lemma 11, the function w ↦ s_w is continuous, combining these two inequalities that Hence the Gâteaux derivative of $$ is bounded linear in z and continuous in w. It follows that $$ is a class of C¹ (see [19, Proposition 1.3]).

(2) follows from (1). Note only that since w ∈ S,    $$.

(3) Let {w_n} be a Palais-Smale sequence for Ψ, and let u_n = m(w_n) ∈ 𝒩. Since for every w_n ∈ S we have an orthogonal splitting $$, using (2) we have

Using (2) again, then Therefore, According to Lemma 6, for u_n ∈ 𝒩, J(u_n) > 0, so there exists a constant c > 0 such that J(u_n) > c. And since $$, $$. Together with 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.

(4) By (45), Ψ^′(w) = 0 if and only if J^′(m(w)) = 0. The other part is clear.

Proof of Theorem 2. (1) If σ = −1, ω ≤ λ₁, we suppose that (4) has a nontrivial solution u ∈ E. Then u is a nonzero critical point of J in E and J^′(u) = 0. But

This is a contradiction.

(2) If σ = 1, ω < λ₁. We firstly show that Ψ satisfies the Palais-Smale condition.

Let {w^(k)} be a Palais-Smale sequence for Ψ; then {u^(k)} is a Palais-Smale sequence for J by Lemma 13(3), where u^(k): = m(w^(k)) ∈ 𝒩. From Lemma 12, u^(k) → u after passing to a subsequence and w^(k) → m⁻¹(u), so Ψ satisfies the Palais-Smale condition.

Let {w^(k)} ⊂ S be a minimizing sequence for Ψ. By Ekeland′s variational principle, we may assume that Ψ^′(w^(k)) → 0 as k → ∞, so {w^(k)} is a Palais-Smale sequence for Ψ. By Palais-Smale condition, w^(k) → w after passing to a subsequence if needed. Hence w is a minimizer for Ψ and therefore a critical point of Ψ, and then u = m(w) is a critical point of J and is also a minimizer for J by Lemma 13. Therefore, u is a ground state solution of (4).

(3) If σ = 1, ω < λ₁, and f(n, u) is odd in u for each n ∈ ℤ^m, then J is even and so is Ψ. Since inf _SΨ = inf _𝒩J > 0 and Ψ satisfies the Palais-Smale condition, Ψ has infinitely many pairs of critical points by Lemma 5. It follows that (4) has infinitely many pairs of solutions ±u^(k) in E from Lemma 13.

This completes Theorem 2.