On some classes of generalized Schrödinger equations

Proof of Theorem 1.2

(i):

By Lemma 2.1(i) and Sobolev embedding

Therefore I_λ,2 is coercive. Since I_λ,2 is weakly lower semicontinuous, we conclude that I_λ,2 is bounded from below. On the other hand, since

for all v ∈ d(m). We conclude from Lemma 2.1(i), L’Hospital and Lebesgue Dominated Convergence Theorem that

for all v ∈ d(m). Since v=∑j=1d(m)vλjej, where d(m) := 1 + ∑i=2m dim V_λi, we get

Since v ∈ 𝓢_d(m),

for all v ∈ 𝓢_d(m), because λ > ϑ(0)λ_m. Therefore, there exist ε, δ > 0 such that

for all 0 < s < δ and v ∈ 𝓢_d(m). Fixing 0 < s_∗ < δ, we have

Since I_λ,2 is coercive, it is standard to prove that it satisfies the (PS)_c condition. Finally, as I_λ,2 is an even C¹-functional, it follows from Theorem 9.1 in [18] (see also [4]) that I_λ,2 has at least d(m) pairs of critical points. □

Lemma 2.2

If ϑ satisfies (ϑ₁) − (ϑ₃) and λ > (α²/4)λ₁, the following claims hold:

1. The set 𝓕 is open and nonempty;

2. ∂ 𝓕 = {v ∈ H01 (Ω) : ∥v∥² = (4λ/α²)∫_Ω v² dx};

3. 𝓕^c = {v ∈ H01 (Ω) : ∥v∥² ≥ (4λ/α²)∫_Ωv² dx};

4. 𝓝 ⊂ 𝓕;

5. 𝓢 ∩ 𝓕 ≠ ∅.

Proof

(i) Since λ > (α²/4)λ₁, any eigenfunction associated to λ₁ belongs to 𝓕. Moreover, 𝓕 = Φ⁻¹(− ∞, 0) where Φ: H01 (Ω) → ℝ is the continuous functional defined by Φ(v) = ∥v∥² − (4λ/α²)∫_Ωv² dx. Items (ii) and (iii) are immediate.

(iv) If v ∈ 𝓝 then, by Lemma 2.1(ii), we obtain

(v) It is sufficient to choose a normalized (in H01 (Ω)) eigenfunction associated to λ₁. □

Lemma 2.3

Suppose that ϑ verifies (ϑ₁) − (ϑ₃) and let h_v : [0, ∞) → R be defined by h_v(s) = I_λ,4(sv).

1. For each v ∈ 𝓕, there exists a unique s_v > 0 such that hv′ (s) > 0 in (0, s_v), hv′ (s_v) = 0 and hv′ (s) < 0 in (s_v, ∞). Moreover, sv ∈ 𝓝 if, and only if, s = s_v;

2. For each v ∈ 𝓕^c ∖ {0}, hv′ (s) > 0 for all s ∈ (0, ∞).

Proof

1. Observe that h_v(0) = 0. Moreover, for each v ∈ 𝓕, we have

   hv(s)s2=12∥v∥2−λ∫[v≠0]G(sv)(sv)2v2dx. (2.1)

   Thus, in view of Lemma 2.1(ii), L’Hôspital rule and Lebesgue’s dominated convergence theorem, it follows that

   lims→∞hv(s)s2=12∥v∥2−(4λ/α2)∫Ωv2dx<0.

   Showing that lim_s→∞h_v(s) = − ∞. Moreover, h_v(s) is positive for s small enough. Indeed, reasoning as in the previous limit, we get

   lims→0+hv(s)s2=12∥v∥2−λ∫[v≠0]G(sv)(sv)2v2dx=12∥v∥2>0.

   Hence, there exists a global maximum point s_v > 0 of h_v which, by Lemma 2.1(ii), is the unique critical point of h_v.

2. If v ∈ 𝓕^c ∖ {0}, then ∥v∥² ≥ (4λ/α²)∫_Ωv²dx. Thus, by Lemma 2.1(ii), it follows that

   hv′(s)s=∥v∥2−λ∫[v≠0]g(sv)svv2dx≥λ∫[v≠0]4α2−g(sv)svv2dx>0,∀s>0.

   Consequently, hv′ (s) > 0 for all s ∈ (0, ∞). □

Lemma 2.4

If ϑ verifies (ϑ₁) − (ϑ₃), the following claims hold:

1. There exists τ > 0 such that s_v ≥ τ, for all v ∈ 𝓢_𝓕;

2. For each compact set 𝓦 ⊂ 𝓢_𝓕 there exists C_𝓦 > 0 such that s_v ≤ C_𝓦, for all v ∈ 𝓦;

3. The map m̂ : 𝓕 → 𝓝 given by m̂(v) = s_vv is continuous and m := m̂_|𝓢𝓕 is a homeomorphism between 𝓢_𝓕 and 𝓝. Moreover, m⁻¹(v) = v/∥v∥.

Proof

1. Suppose that there exists {v_n} ⊂ 𝓢_𝓕 with s_n := s_vn → 0. In this case, we get v ∈ H01 (Ω) with v_n ⇀ v in H01 (Ω). It follows from Lemma 2.1(ii) that

   1=λ∫Ωg(snvn)vndx≤(4/α2)λsn∫Ωvn2dx. (2.2)

   By passing to the limit as n → ∞ in the last inequality, we get a contradiction.

2. Let {v_n} ⊂ 𝓦 be a sequence such that s_n := s_vn → ∞. Since 𝓦 is compact, up to a subsequence, we get v ∈ 𝓦 such that v_n → v in H01 (Ω). Hence, passing to the lower limit as n → ∞ in

   1=∥vn∥2≥λ∫[v≠0]g(snvn)snvnvn2χ[vn≠0]dx,

   it follows from Lemma 2.1(ii) that

   ∥v∥2=1≥(4λ/α2)∫Ωv2dx,

   showing that v ∈ 𝓕^c. Since v ∈ 𝓦 ⊂ 𝓕, we have a contradiction.

3. We are going to prove that m̂ is continuous. Let {v_n} ⊂ 𝓕 and v ∈ 𝓕 be such that v_n → v in H01 (Ω). Since m̂(sw) = m̂(w) for all w ∈ 𝓕 and s > 0, we can assume that {v_n} ⊂ S_𝓕. Hence,

   sn=sn∥vn∥2=λ∫Ωg(snvn)vndx, (2.3)

   where s_n := s_vn. By (A₁) and (A₂), it follows that, passing to a subsequence, s_n → s > 0. Thence, passing to the limit as n → ∞ in (2.3), we have

   s=s∥v∥2=λ∫Ωg(sv)vdx,

   showing that m̂(v_n) = s_nv_n → sv = m̂(v). The second part of (A₃) is immediate. □

Lemma 2.5

Suppose that ϑ satisfies (ϑ₁) − (ϑ₃). Then I_λ,4 is bounded from below in 𝓝.

Proof

By Lemma 2.1(iii), we get

Therefore I_λ,4 is bounded from below in 𝓝. □

Lemma 2.6

Suppose that ϑ verifies (ϑ₁) − (ϑ₃). Then,

1. Ψ̂_λ,4 ∈ C¹(𝓕, ℝ) and

   Ψ^λ,4′(u)v=∥m^(u)∥∥u∥Iλ,4′(m^(u))v,∀u∈Fand∀v∈H01(Ω).

2. Ψ_λ,4 ∈ C¹(𝓢_𝓕, ℝ) and

   Ψλ,4′(u)v=∥m(u)∥Iλ,4′(m(u))v,∀v∈TuSF.

3. If {u_n} is a (PS)_c sequence for Ψ_λ,4 then {m(u_n)} is a (PS)_c sequence for I_λ,4. If {u_n} ⊂ 𝓝 is a bounded (PS)_c sequence for I_λ,4 then {m⁻¹(u_n)} is a (PS)_c sequence for Ψ_λ,4.

4. u is a critical point of Ψ_λ,4 if, and only if, m(u) is a nontrivial critical point of I_λ,4. Moreover, the corresponding critical values coincide and

   infSFΨλ,4=infNIλ,4.

Proposition 2.7

Suppose that (ϑ₁) − (ϑ₃) hold. If {v_n} ⊂ 𝓢_𝓕 is such that dist(v_n, ∂ 𝓢_𝓕) → 0, then there exists v ∈ H01 (Ω) ∖ {0} such that v_n ⇀ v in H01 (Ω), s_vn → ∞ and

Proof

Since {v_n} ⊂ 𝓢_𝓕 is bounded, up to a subsequence, there exists v ∈ H01 (Ω) with v_n ⇀ v in H01 (Ω). Since dist(v_n, ∂ 𝓢_𝓕) → 0, there exists {w_n} ⊂ ∂𝓢_𝓕 such that ∥v_n − w_n∥ → 0 as n → ∞. Thus,

Therefore,

By using compact embedding from H01 (Ω) into L²(Ω), it follows that

Thus v ≠ 0. Suppose by contradiction that, for some subsequence, {s_vn} is bounded. In this case, passing again to a subsequence, there exists s₀ > 0 (see Lemma 2.4(A₁)) such that

It follows from (2.7) and

that

Combining last equality and Lemma 2.1(ii), we obtain

But this contradicts (2.6). Showing that s_vn → ∞. Finally, from s_vn → ∞, Lemma 2.1(iii) and Fatou Lemma, we get

□

Proposition 2.8

Suppose that (ϑ₁) − (ϑ₃) hold and λ > (α²/4)λ₁. Then Ψ_λ,4 satisfies the (PS)_c condition.

Proof

By Lemmas 2.4(A₃) and 2.6(iii), it is sufficient to show that I_λ,4 satisfies the (PS)_c condition. For this, let {w_n} ⊂ 𝓝 be a (PS)_c sequence for I_λ,4. We are going to prove that {w_n} is bounded in H01 (Ω). Indeed, otherwise, up to a subsequence, we have ∥w_n∥ → ∞. Define v_n := w_n/∥w_n∥ = m⁻¹(w_n) ∈ 𝓢_𝓕. Thus {v_n} is bounded in H01 (Ω) and

Consequently, there exists v ∈ H01 (Ω) such that

Suppose by contradiction that v = 0. Since {Ψ_λ,4(v_n)} is bounded, it follows that there exists C > 0 such that

By Lemma 2.1(ii), L’Hôspital rule and compact embedding, passing to the limit as n → ∞ in (2.10), we get

a clear contradiction. Thereby, we conclude that v ≠ 0.

Since {w_n} ⊂ 𝓝 is a (PS)_c sequence for functional I_λ,4, we get

Dividing last equality by ∥w_n∥, we have

Passing to the limit as n → ∞, it follows from Lemma 2.1(ii) that

Now we are going to consider two cases:

1. If (4λ/α²) ≠ λ_j, whatever j > 1, it follows from (2.11) that v = 0. But this is a contradiction. Therefore {w_n} is bounded in H01 (Ω).

2. If (4λ/α²) = λ_j, for some j > 1, then (2.11) implies that v is an eigenfunction associated to λ_j. From (2.11), it follows also that ∥v∥² = (4λ/α²)∫_Ωv²dx, i.e., v ∈ ∂𝓕. On the other hand,

   (4λ/α2)∫Ωv2dx=∥v∥2≤lim infn→∞∥vn∥2=1.

Suppose that

In this case, since

passing to the limit as n → ∞ in

and using Lemma 2.1(ii), L’Hôspital rule and (2.12), we conclude that Ψ_λ,4(v_n) → ∞, a contradiction with (2.8). Consequently,

showing that v = e_j and

By using (2.9) and (2.15), we derive v_n → v in H01 (Ω) with v ∈ ∂𝓢_𝓕 (see (2.14)). Invoking Proposition 2.7, we conclude that

Since (2.16) cannot occurs, we conclude that {v_n} is bounded.

Hence, there exists v ∈ H01 (Ω) such that v_n ⇀ v in H01 (Ω) up to a subsequence. Since v_n ⇀ v, to finish the proof we just have to prove that ∥v_n∥ → ∥v∥. To this end, it is sufficient to note that since {v_n} is a (PS)_c sequence, we have

Passing to the limit as n → ∞ in the previous equality, we get

Then (2.17) and Lebesgue’s convergence theorem imply that

□

Lemma 2.9

Let B and C be sets in 𝓔.

1. If x ≠ 0, then y({x} ∪ {− x}) = 1;

2. If there exists an odd map φ ∈ C(B, C), then y(B) ≤ y(C). In particular, if B ⊂ C then y(B) ≤ y(C).

3. If there exists an odd homeomorphism φ : B → C, then y(B) = y(C). In particular, if B is homeomorphic to the unit sphere in Rⁿ, then y(B) = n.

4. If B is a compact set, then there exists a neighborhood K ∈ 𝓔 of B such that y(B) = y(K).

5. If y(C) < ∞, then y(B∖C) ≥ y(B) − y(C).

6. If y(A) ≥ 2, then A has infinitely many points.

Lemma 2.10

Suppose that (ϑ₁) − (ϑ₃) hold and λ > (α²/4)λ_m. Then,

1. y_d(m) ≠ ∅;

2. y₁ ⊃ y₂ ⊃ … ⊃ y_d(m);

3. If φ ∈ C(𝓢_𝓕, 𝓢_𝓕) and is odd, then φ(y_j) ⊂ y_j, for all 1 ≤ j ≤ d(m);

4. If B ∈ y_j and C ∈ 𝓔 with y(C) ≤ s < j ≤ d(m), then B∖C ∈ y_j−s.

Proof

(i) Let S_d(m) be the unit sphere of V₁ ⊕ V₂ ⊕ … ⊕ V_m. Since λ > (α²/4)λ_m, it is clear that S_d(m) ⊂ 𝓢_𝓕. Moreover, by Lemma 2.9(iii), we have y(S_d(m)) = d(m). Showing that S_d(m) ∈ y_d(m). (ii) It is immediate. (iii) It follows directly from Lemma 2.9(ii). (iv) It is a consequence of Lemma 2.9(v). □

Lemma 2.11

Suppose (ϑ₁) − (ϑ₃) hold. Then,

Proof

First inequality follows from Lemma 2.5. On the other hand, the monotonicity of the levels c_j is a consequence of Lemma 2.10(ii). □

Proposition 2.12

Suppose that ϑ satisfies (ϑ₁) − (ϑ₃) and λ > (α²/4)λ_m. If c_j = … = c_j+p ≡ c, j + p ≤ d(m), then y(K_c) ≥ p + 1, where K_c := {v ∈ 𝓢_𝓕 : Ψ_λ,4(v) = c and Ψλ,4′ (v) = 0}.

Proof

Suppose that y(K_c) ≤ p. By Proposition 2.8 and Lemma 2.11, K_c is a compact set. Thus, by Lemma 2.9(iv), there exists a compact neighborhood K (in H01 (Ω)) of K_c such that y(K) ≤ p. Defining M := K ∩ 𝓢_𝓕, we derive from Lemma 2.9(ii) that y(M) ≤ p. Despite the noncompleteness of 𝓢_𝓕 we can still use Theorem 3.11 in [23] (see also Remark 3.12 in [23]) to ensure the existence of an odd homeomorphisms family η(., t) of 𝓢_𝓕 such that, for each u ∈ 𝓢_𝓕, the map

Observe that, although 𝓢_𝓕 is non-complete, from Proposition 2.7 and (2.20), for all u ∈ 𝓢_𝓕, maps t ↦ η(u, t) are well defined in t ∈ [0, ∞). Consequently, it makes sense the third claim of Theorem 3.11 in [23], namely,

Let us choose B ∈ y_j+p such that sup_B Ψ_λ,4 ≤ c + ε. From Lemma 2.10(iv), B∖M ∈ y_j. It follows again from Lemma 2.10(iii) that η(B∖M, 1) ∈ y_j. Therefore, from (2.21) and the definition of c, we have

that is a contradiction. Then y(K_c) ≥ p + 1. □

Proof of Theorem 1.2

(ii):

Note that 0 ≤ c_j < ∞ are critical levels of Ψ_λ,4. In fact, suppose by contradiction that c_j is regular for some j. Invoking Theorem 3.11 in [23], with β = c_j, ε = 1, N = ∅, there exist ε > 0 and a family of odd homeomorphisms η(., t) satisfying the properties of referred theorem. Choosing B ∈ y_j such that sup_B ψ < c_j + ε and arguing as in the proof of Proposition 2.12 we get a contradiction.

Finally, if the levels c_j, 1 ≤ j ≤ d(m), are different from each other, by Proposition 2.6(iv) the result is proved. On the other hand, if c_j = c_j+1 ≡ c for some 1 ≤ j ≤ d(m), it follows from Proposition 2.12 that y(K_c) ≥ 2. Combining last inequality with Lemma 2.9(vi) and Proposition 2.6(iv), we conclude that (P_λ,q) has infinitely many pairs of nontrivial solutions. □