Standing Wave Solutions for the Discrete Coupled Nonlinear Schrödinger Equations with Unbounded Potentials

Lemma 1. If V_i,   i = 1,2, satisfy the condition (10), then for any 2 ≤ p ≤ ∞, E₁ and E₂ are compactly embedded into l^p and denote the best embedding constant $$ and $$, respectively. Furthermore, the spectra σ(L₁) and σ(L₂) are discrete, respectively.

Lemma 2. Assume that ω₁ < λ₁,   ω₂ < λ₂, and (10) holds. Then the Nehari manifold N is nonempty in E₁ × E₂. Furthermore, for (ϕ, ψ) ∈ N, J(tϕ, tψ) attains a unique maximum point at t = 1.

Proof. First we show that N ≠ ∅.

From (15) and (16), we rewrite

Let (ϕ, ψ)∈(E₁ − {0})×(E₂ − {0}); then by (19) Notice that $$ and $$; by (20), we see that I(tϕ, tψ) > 0 for t > 0 small enough and I(tϕ, tψ) < 0 for t > 0 large enough. As a consequence, there exists t₀ > 0 such that I(t₀ϕ, t₀ψ) = 0; that is, (t₀ϕ, t₀ψ) ∈ N.

Let F(t) = J(tϕ, tψ), (ϕ, ψ) ∈ N. Computing the derivative of F, we have

This shows that t = 1 is a unique maximum point. The proof is completed.

Lemma 3. Assume that ω₁ < λ₁,   ω₂ < λ₂, and (10) holds. Then there exists η > 0 such that J(ϕ, ψ) ≥ η, for all (ϕ, ψ) ∈ N.

Proof. Since λ₁ is the smallest eigenvalue of E₁ and λ₂ is the smallest eigenvalue of E₂, from the definition of the constant α_p and β_p, we get $$ and $$. For any (ϕ, ψ) ∈ N, we have

where a^* = max  {a₁, a₂, a₃} and $$.

Let

By (22), it is easy to see that

and this implies that Moreover, we have Let $$; then we get J(ϕ, ψ) ≥ η, for all (ϕ, ψ) ∈ N. The proof is completed.

Theorem 4. Assume that ω₁ < λ₁,   ω₂ < λ₂, and (10) holds. Then system (13) has a nontrivial solution in E₁ × E₂; that is, system (1) has a nontrivial standing wave solution.

Proof. Let d be given by (27). By Lemma 2, N is nonempty and there exists a sequence {(ϕ^(k), ψ^(k))} ⊂ N such that

By Lemma 3, d > 0 and d ≤ D = max _k{J(ϕ^(k), ψ^(k))} < ∞. By virtue of (26), we have

Thus, sequences {ϕ^(k)} and {ψ^(k)} are bounded in Hilbert spaces E₁ and E₂, respectively. Therefore, there exist subsequences of {ϕ^(k)} and {ψ^(k)} (denoted by itself) that weakly converge to some ϕ^* ∈ E₁ and ψ^* ∈ E₂, respectively. By Lemma 1, we get, for any 2 ≤ p ≤ ∞, By virtue of (15) and (16), we have First, we claim that

According to (30), it suffices to show that

In fact,

Thus Hölder inequality and (30) imply the (33) holds.

Next, we show that (ϕ^*, ψ^*) ∈ N and J(ϕ^*, ψ^*) = d.

Since E₁ and E₂ are Hilbert spaces, by (32) we have

which implies $$. Through a similar argument to the proof of Lemma 2, we know that I(tϕ^*, tψ^*) is positive as t is small enough. Therefore there exists t^* ∈ (0,1] such that I(t^*ϕ^*, t^*ψ^*) = 0 which implies (t^*ϕ^*, t^*ψ^*) ∈ N. Thus we have J(t^*ϕ^*, t^*ψ^*) = (1/4)W(t^*) and by (32), W(1) = 4d, where Clearly, W(t) is strictly increasing on 0 < t < ∞. Therefore by (27), This implies that t^* = 1 and J(ϕ^*, ψ^*) = d.

Finally, we will prove (ϕ^*, ψ^*) is a nontrivial solution to system (13).

Since (ϕ^*, ψ^*) is an energy minimizer on Nehari manifold N, there exists a Lagrange multiplier Λ such that

for any (ϕ, ψ) ∈ E₁ × E₂. Let (ϕ, ψ) = (ϕ^*, ψ^*) in (38). (J^′(ϕ^*, ψ^*), (ϕ^*, ψ^*)) = I(ϕ^*, ψ^*) = 0 implies that but Thus, Λ = 0 and for any (ϕ, ψ) ∈ E₁ × E₂. Take (ϕ, ψ) = (e^(k), 0) and (ϕ, ψ) = (0, e^(k)) in (41) for k ∈ ℤ, where We see that J^′(ϕ^*, ψ^*) = 0. Thus, (ϕ^*, ψ^*) is a nontrivial solution to system (13). The proof is completed.

Theorem 5. Assume that ω₁ < λ,   ω₂ < λ, a₃ > max  {a₁, a₂, ((λ − ω₂)/(λ − ω₁))a₁, ((λ − ω₁)/(λ − ω₂))a₂}, and (10) holds. Then system (43) has a nontrivial standing wave solution $$ in E × E with $$ and $$.

Proof. By Theorem 4, we know that system (43) has a nontrivial standing wave solution $$ in E × E.

Now we will prove that $$ and $$.

Since $$, we know that $$. If one of the components $$, say $$, then $$. For ϵ small enough, we consider $$; by a similar argument to the proof of Lemma 2, we know that there exists t^* such that $$; that is, $$.

By (20) and $$, we have $$, where

and $$.

We noticed that $$ and

For the sake of simplicity, we let

If ω₁ ≤ ω₂ < λ, then Thus, a₃ > a₁ and (47) yields a₁BD < a₁a₃B².

If ω₂ < ω₁ < λ, then by (45),

Thus, a₃ > ((λ − ω₂)/(λ − ω₁))a₁ and (48) yields a₁BD < a₁a₃B².

From the above arguments, if a₃ > max  {a₁, ((λ − ω₂)/(λ − ω₁))a₁}, then a₁BD < a₁a₃B².

For ϵ small enough, we have

Hence, by (49), we have This is a contradiction. So, $$.

Similarly, if $$ and a₃ > max  {a₂, ((λ − ω₁)/(λ − ω₂))a₂}, then $$. The proof is completed.