Existence of Nontrivial Solutions and High Energy Solutions for a Class of Quasilinear Schrödinger Equations via the Dual-Perturbation Method

Lemma 1. If assumptions (V), (h₁), and (h₂) hold, then the functional I_θ is well defined on E and I_θ ∈ C¹(E, R).

Proof. By conditions (h₁) and (h₂), the properties (f₂), (f₃), (f₇), and (f₁₁) imply that there exists δ > 0 such that

Hence for all v ∈ R. By (26) and the continuity of the embedding E↪L^s(R^N) (s ∈ [2, 2^*]), Hence I_θ is well defined in E.

Now, we prove that I_θ ∈ C¹(E, R). It suffices to prove that

For any v, ϕ ∈ E and 0<|t | < 1, by the mean value theorem, (25) and (f₂)-(f₃), we have

The Hölder inequality implies that Hence, by the Lebesgue theorem, we have for all ϕ ∈ E. Now, we show that $$, i = 1,2, are continuous. Indeed, if v_n → v in E, then v_n → v in L^s(R^N) for all s ∈ [2, 2^*].

On the space $$, we define the norm

Then Moreover, on the space $$, we define the norm By (25), we have where q = p/2 and r = p/(p − 2). Then Theorem A.4 in [18] implies as n → +∞. If h(x, f(v_n))f^′(v_n) − h(x, f(v))f^′(v) = y_n + z_n with y_n ∈ L²(R^N) and z_n ∈ L^r(R^N), one has Hence and hence as n → ∞. Therefore, Ψ₁ ∈ C¹(E, R).

Define

with the norm $$. On the space $$, we define the norm On the space $$, we define the norm From v_n → v in E, one has $$ and as n → ∞. Since |f(v)f^′(v)|≤|v|, by the following Lemma 2, we have If f(v_n)f^′(v_n) − f(v)f^′(v) = y_n + z_n with $$ and $$, one has Hence and hence as n → ∞. Therefore, Ψ₂ ∈ C¹(E, R). This completes the proof.

Lemma 2. Assume that 1 ≤ p, q, r, s < +∞, g ∈ C(R^N × R) and

Then, for every $$, $$, and the operator is continuous.

Proof. Let η(s) be a smooth cut-off function such that η(s) = 1 for |s | ≤ 1 and η(s) = 0 for |s | ≥ 2. Define

We can assume that p/r ≤ q/s. Hence for all (x, v) ∈ R^N × R. Assume v_n → v in $$. Then v_n → v in $$ and g(·, v_n) → g(·, v) in $$. As in the proof of Lemma A.1 in [18], there exists a subsequence {w_n} of {v_n} and $$ such that w_n(x) → v(x) and |v(x)|, |w_n(x)| ≤ α(x) for a.e. x ∈ R^N. Hence, from (51), one has a.e. on R^N. It follows from the Lebesgue theorem that g₁(·, w_n) → g₁(·, v) in $$. Consequently, g₁(·, v_n) → g₁(·, v) in $$. Similarly, we can prove g₂(·, v_n) → g₂(·, v) in $$. Since it follows that g(·, v_n) → g(·, v) in $$. This completes the proof.

Lemma 3. Let (V), (h₁), and (h₂) hold. Then every bounded sequence {v_n} ⊂ E with $$ possesses a convergent subsequence.

Proof. Since {v_n} ⊂ E is bounded, then, by the compactness of the embedding E↪L^s(R^N) (2 ≤ s < 2^*), passing to a subsequence, one has v_n⇀v in E, v_n → v in L^s(R^N) for all 2 ≤ s < 2^*, and v_n(x) → v(x) for a.e. x ∈ R^N. By (25)

as n → ∞. Similarly, $$ as n → ∞. Hence, by the property of (f₈), we have where o_n(1) → 0 as n → ∞. This shows that $$ as n → ∞. This completes the proof.

Lemma 4. If v_n(x) → v(x) a.e. in R^N and $$, then $$ as n → ∞.

Theorem 5. Assume conditions (V), (h₁)–(h₃) hold. Let {θ_n}⊂(0,1] be such that θ_n → 0. Let v_n ∈ E be a critical point of $$ with $$ for some constant c independent of n. Then, up to subsequence, one has v_n → v in E, $$ and v is a critical point of I.

Proof. By (h₂), for 0 < ε₀ < (1/4)(1/2 − 1/μ)a₀, there exists δ₀ > 0 such that

By (h₁), for δ₀≤|s | ≤ r (r is the constant appearing in condition (h₃)), we have where C is the constant appearing in condition (h₁). Hence Since lim _|x|→∞V(x) = +∞, there exists ρ₀ > 0 such that for all |x | ≥ ρ₀. Hence Since v_n is a critical point of $$, for all ϕ ∈ E. Consequently, taking ϕ = f(v_n)/f^′(v_n) ∈ E, by (h₃) and (f₆) we have and hence for some constant C independent of n. By the boundedness of $$, there exists C₂ > 0 such that for all n. Hence, by the Sobolev embedding theorem, one has

Next, we prove that f(v_n) ∈ L^∞(R^N) and $$, where the positive constant C is independent of n. Setting T > 2, r > 0, define $$, where b : R → R is a smooth function satisfying b(s) = s for |s | ≤ T − 1, b(−s) = −b(s); b^′(s) = 0 for s ≥ T, and b^′(s) is decreasing in [T − 1, T].

This means that $$, for |v_n | ≤ T − 1; $$, for T − 1≤|v_n | ≤ T; $$, for |v_n | ≥ T, where T − 1 ≤ C_T ≤ T.

Let $$; then ϕ ∈ E. By (61) 〈I^′(v_n), ϕ〉 = 0. Hence

where For T − 1≤|v_n | ≤ T, $$. By the properties of f and b, the mean value theorem implies Hence Consequently, Combining (67) and (68), we have For any ε > 0, by (h₁) and (h₂), there exists C(ε) > 0 such that Combining (66), (72), and (73), one has By the Hölder inequality and (65), Moreover, Hence Since 4 < p < 2(2^*), d = 2^*/(8N/(4N − (p − 4)(N − 2))) = 2^*/2 − p/4 + 1 > 1. Set q = 8N/(4N − (p − 4)(N − 2)). Then Take r = r₀ such that (2 + r₀)q = 2(2^*). Since $$, $$. Hence, from (65), we have Since $$ as T → +∞, taking T → +∞ in (78) with r = r₀, we have Set 2 + r₁ = (2 + r₀)d. Then Inductively, we have where (2 + r_i) = dⁱ(2 + r₀)  (i = 0,1, …, k), and is convergent as k → ∞. Let $$. Then C_k → C_∞ > 0 as k → ∞. Hence Let k → ∞; by (65), we have Hence, by (f₉) and (85), we have By (63) we know that $$ is bounded, and hence {v_n} is bounded in E. Up to subsequence, one has v_n⇀v in E, v_n → v in L^s(R^N) for s ∈ [2, 2^*), and v_n(x) → v(x) a.e. x ∈ R^N.

Now, we show that v is a critical point of I. For any $$ with ψ ≥ 0, by (85), we know that ϕ = ψexp (−f(v_n)) ∈ E. Take ϕ = ψexp (−f(v_n)) as the test function in (61); we have

By |∇(v_n − v)|²ψexp (−f(v_n))f^′(v_n) ≥ 0, one has Since θ_n → 0, by (63) as n → ∞. Moreover, notice that v_n⇀v in E, v_n → v in L^s(R^N) for s ∈ [2, 2^*), and v_n(x) → v(x) a.e. x ∈ R^N; by Hölder inequality and Lebesgue theorem, we have Hence, from (87), we have For any φ ∈ E with φ ≥ 0, by (85) we know that ζ : = φexp (f(v)) ∈ E. By Theorem 2.8 in [19], there exists a sequence $$ such that ψ_n ≥ 0 and ψ_n → ζ and ψ_n(x) → ζ(x) for a.e. x ∈ R^N. Take ψ = ψ_n in (91), and let n → ∞; we have The opposite inequality can be obtained by taking ϕ = ψexp (f(v_n)) and ζ = φexp (−f(v)). Consequently, This shows that v ∈ E is a critical point of I, and by taking φ = f(v)/f^′(v) ∈ E, one has

Finally, taking ϕ = f(v_n)/f^′(v_n) as the test function in (61), we have

Since by Fatou’s Lemma, (63), (94), (95), up to subsequence, one has Hence $$ as n → ∞. Set w_n : = v_n − v ∈ E. By (f₈), (f₁₂), and (85), one has Consequently, by (f₉), (98), and Lemma 4, one has as n → ∞. Therefore, v_n → v in E. This completes the proof.

Theorem 6. Assume conditions (V), (h₁)–(h₃) hold; then (1) has a weak solution.

Proof. First, we prove that, for each θ ∈ (0,1], I_θ satisfy the Cerami condition. Indeed, let {v_n} ⊂ E be an arbitrary Cerami sequence of I_θ. Set ϕ = f(v_n)/f^′(v_n). Then $$. Similar to the proof of (63), we can prove that {v_n} is bounded in E. Hence, by Lemma 3, the sequence {v_n} possesses a convergent subsequence in E. This shows that I_θ satisfy the Cerami condition.

Next, for any ε > 0, by (h₁), (h₂), (f₃), and (f₇), there exists C(ε) > 0 such that

for all (x, t) ∈ R^N × R. For small 0 < ρ ≪ 1, set Then, from (101), for v ∈ S_ρ, for small ε > 0 and ρ > 0. Moreover, by (h₃), for any (x, z) ∈ R^N × R with |z | ≥ r, one has Since μ > 4, there exists a constant 4 < α < min {μ, 2(2^*)}. Hence, by (f₅), we have uniformly in x ∈ R^N. Consequently, there exist constants τ > 1 such that for all x ∈ R^N. For any finite-dimensional subspace $$, by the equivalency of all norms in the finite-dimensional space, there is a constant a > 0 such that By (h₁), (h₂), and (106), there exists a positive constant C > 0 such that Since 4 < α < 2(2^*), by (f₃), (107), and (108), we have for all $$. Hence there exists a large R > 0 such that I_θ < 0 on $$. Set a fixed $$ with ∥e∥_E = 1. For any fixed T > ρ, define the path $$ by h_T(t) = tTe. Then for large T > 0, by (109), one has Hence by Theorem 2.2 with the Cerami condition in [20], I_θ possesses a critical value where Consequently, by Theorem 5, we know that (1) has a weak solution. This completes the proof of Theorem 6.

Remark 7. Let v⁺ = max {v, 0} and v⁻ = max {−v, 0}. Set

instead of I(u) and I_θ(u), respectively. Then, under the conditions of Theorem 6, we can obtain the existence of a positive solution and a negative solution for (1).

Theorem 8. Assume conditions (V), (h₁)–(h₃) hold. If h(x, s) is odd in s, then (1) has a sequence {v_m} of solutions such that I(v_m)→+∞.

Proof. Consider the eigenvalue of the operate L = −Δ + V. By assumption (V) and the compactness of the embedding E↪L²(R^N), we know that the spectrum σ(L) = {λ₁, λ₂, …, λ_n⋯} of L with

and λ_n → +∞ as n → +∞ (see page 3820 in [21]). Let φ_n be the eigenfunction corresponding to λ_n. By regularity argument we know that φ_n ∈ E. Set E_n = span {φ₁, φ₂, …, φ_n}. Then we can decompose the space E as E = E_n ⊕ W_n for n = 1,2, …, where W_n is orthogonal to E_n in E. For ρ > 0, set By (109) there exists r_n > 0 independent of θ such that Set where γ(·) is the genus. Let We have the following three facts (we refer the reader to [16] for their proofs).

•   

  Fact (1). For each B ∈ Γ_j, if 0 < ρ < r_n for all n ≥ j, then B∩∂Q_ρ∩W_j−1 ≠ ∅.

•   

  Fact (2). There exist constants α_j ≤ β_j such that c_j(θ)∈[α_j, β_j] and α_j → +∞ as j → +∞.

•   

  Fact (3).  c_j(θ), j = 1,2, … are critical values of I_θ.

Consequently, Theorem 8 follows from Theorem 5 and the above Facts (2)-(3). This completes the proof.

Corollary 9. If the following conditions (h₄) and (h₅) are used in place of (h₃); then the conclusions of Theorem 5, Theorem 6, and Theorem 8 hold:

•  

  (h₄) lim _|s|→+∞inf H(x, s) > 0 uniformly in x ∈ ℝ^N,

•  

  (h₅) there exist μ > 4 and τ > 0 such that

  $$ ()

for all x ∈ ℝ^N and |s | ≥ τ.

Proof. By (h₄), there are constants λ > 0 and r₁ > 0 such that whenever |s | ≥ r₁, one has

Set r = max {τ, r₁}. Then, by (h₅), for all x ∈ ℝ^N and |s | ≥ r. Therefore, condition (h₃) holds. This completes the proof.