Infinitely many radial and non-radial sign-changing solutions for Schrödinger equations

Proof

Suppose that {v_n} is a Palais-Smale (Shortly: (PS) ) sequence of ℐ for some c ∈ ℝ, that is,

We claim that the sequence {v_n} is bounded in E, otherwise there should be a subsequence, still denoted v_n, such that ‖v_n‖ → ∞ as n → ∞. Let un=vn∥vn∥ , so that there exists u ∈ E satisfying u_n → u in L^q (ℝ^N) (2 ≤ q < 2^*), u_n(x) → u(x) a.e. x ∈ ℝ^N. Let

If meas(Ω) is not equal to 0, then lim_n→∞ |v_n(x)| = +∞ almost everywhere in Ω. With applying (f₂) and Fatou's lemma, we calculate directly

Then we obtain that

This contradiction indicates that meas(Ω) = 0, so that u(x) = 0 a.e. x ∈ ℝ^N and u_n → 0 in L^s (ℝ^N) (2 ≤ s < 2^*). By (f₁), we have

Combining with (f₃), we deduce

Hence,

This is a contradiction with μ > 2. So {v_n} is bounded in E and then there exists v ∈ E satisfying v_n → v in L^q (ℝ^N) (2 ≤ q < 2^*), v_n(x) → v(x) a.e. x ∈ ℝ^N. Generalized control convergence theorem gives us that (f₁) and (f₂) imply that

Therefore v_n → v in E, and hence ℐ satisfies the Palais-Smale condition.

Proof

(i). From (f₁)–(f₃), for any fixed δ > 0 we can find C_δ > 0 such that

where p is given in (f₂). Then we have

Noting that p > 2, taking δ small enough we conclude that there is 0 < ρ < 1 satisfying

(ii). It follows from (f₁)–(f₃) that there exists C₁, C₂ > 0 such that

So we deduce

For μ > 2 and because of the equivalence of norms in finite dimensional space, we see that ℐ(u) = −∞ as ‖u‖ → ∞, u ∈ E_m. Therefore the result follows.

Proof

(i) Let u ∈ E_r, and define for any v ∈ E_r,

so that ℰ ∈ C¹(E_r, ℝ) and ℰ is weakly lower semicontinuous. By (f₁)–(f₂) and the Sobolev embedding we then obtain

which implies

and this means that ℰ(v) → +∞ as ‖v‖ → ∞, where C = C(u) is a constant depending on u. This shows that ℰ is coercive. Moreover, we know that ℰ is bounded below and it keeps the boundness of sets as a mapping. We now estimate

so that ℰ is also strictly convex. Hence, from [2, Theorems 1.5.6 and 1.5.8], ℰ admits a unique minimizer v = B(u) ∈ E_r, which is the unique solution to problem (3.2). Clearly, B also maps bounded sets into bounded sets by (f₁)–(f₃).

Next, we give the proof of the continuity of B. Assume that {u_n} ⊂ E_r with u_n → u in E_r. Denote v = B(u) and v_n = B (u_n). Since 〈ℰ′(v_n), v_n〉 = 0 we have

Combining with (f₁)–(f₂), we can obtain the boundness of {v_n} in E_r. Assume that v_n ⇀ v^* in E_r, after extracting a subsequence, then for any w ∈ E_r

which implies that v^* = B(u). So v = v^* by the uniqueness.

We show that ‖v_n − v‖ → 0 as n → ∞. From 〈ℰ′(v) − ℰ′(v_n), v − v_n〉 = 0, we can write

So, by the general Lebesgue control convergence theorem, v_n → v in E_r. Hence, B is continuous.

• (ii) By using the fact that B(u) is a solution of problem (3.2), one obtains

  〈ℐr′(u),u−B(u)〉=‖u−B(u)‖2,

  for all u ∈ E_r.

• (iii) Notice that for all ϕ∈C0∞(ℝN)∩Er ,

  〈ℐr′(u),ϕ〉=∫ℝN[∇(u−B(u))⋅∇ϕ+V(x)(u−B(u))ϕ]dx.

  Then we have

  ‖ℐr′(u)‖≤C∥u−B(u)∥    for all u∈Er.

• (iv) Contrarily, suppose that it exists {u_n} ⊂ E_r with

  ℐr(un)∈[c,d] and  ‖ℐr′(un)‖≥α,

  such that

  ‖un−B(un)‖→0 as    n→∞,

  and hence from (iii) we deduce that ‖ℐr′(un)‖→0 . But it is impossible, so the proof is complete.

Proof

Let u ∈ E_r and v = B(u). Conditions (f₁) and (f₂) imply that for any ∊ > 0 there exists a constant C_∊ > 0 depending on ∊ to satisfy

Based on dist (v, Y⁻) ≤ ‖v⁺‖, combined with (3.3) and Sobolev's inequality, it is true that

Let's make ∊ and ε small enough, then

and thus one can find ε₀ > 0 such that

As a result, it gives us that B(u)∈Yε− for any u∈Yε− . Suppose that there exists u∈Yε− with B(u) = u, hence we can say that u ∈ Y⁻. What's more, if u is nontrivial, it follows from the maximum principle that u < 0 in ℝ^N. Similarly, the second conclusion is also true.

Proof

Set Kc1:=Kc∩W and Kc2:=Kc\W . Lemma 2.1 tells us that K_c is compact and then d:=dist(Kc1,Erw)>0 , where Erw=Er\W . Furthermore, choose δ∈(0,d8) so small that ℬ3δ(Kc2)⊂N and ℬ3δ(Kc1)⊂W . In fact, there are a, ε₀ > 0 satisfying

Thanks to Lemma 3.1, one can always find b > 0 satisfying

Taking generality into account, we can still assume ε0≤bδ32 . Take a locally Lipschitz continuous cut-off function χ : E_r → [0, 1] such that

Decreasing ε₀ if necessary, by Lemma 2.1 there is δ′ > 0 such that

Thus, χ(u) = 0 for any u ∈ ℬ_δ′ (K). From Lemma 3.3, we know that χ(·)𝒜(·) is locally Lipschitz continuous on E_r, where 𝒜(u) = u − A(u). We study the following initial value problem

which deduces that η(t, u) is well-defined and continuous on ℝ⁺ × E_r.

Let σ(t,u)=η(16εbt,u) . It is sufficient to easily show (a) and (d). For (b), conversely, we suppose that there exists u∈ℐrc+ε\(N∪W) such that σ(1,u)∉ℐrc−ε . It is a fact that u ∉ ℬ_3δ (K_c), since ℬ3δ(Kc)⊂ℬ3δ(Kc1)∪ℬ3δ(Kc2)⊂W∪N . Observe that for any t ∈ [0, 1]

This says that for any t ∈ [0, 1], σ(t, u) ∉ ℬ_2δ (K_c). Lemma 3.3 tells us that

so that ℐ_r(σ(t, u)) is not increased for t ≥ 0, and thus σ(t,u)∉ℐrc−ε for any t ∈ [0, 1]. Therefore,

This shows that for any t ∈ [0, 1] we have

Therefore, from Lemma 3.3, we have

which is a contradiction suggesting that (b) holds.

If K_c \ W = ∅, we can see that ℬ_3δ (K_c) ⊂ W. According to the proof of (b), we can derive (c). Finally, (e) is the result of Lemma 3.3 (see [5, 21, 22]). This proves the lemma.

Proof

Define

where k ≥ 2, γ(P) denotes the Krasnoselskir's genus of P (see [32]) and

Define

We claim that 0 < a ≤ c_k < +1. In fact, G_m ≠ ∅ since id ∈ G_m, m ∈ ℕ. Thus, c_k < +∞. Now, we seek to prove 0 < a ≤ c_k. From [29, Proposition 9.18], Ψ_k satisfies:

• (1) Ψ_k ≠ ∅ and Ψ_k+1 ⊂ Ψ_k for all k ≥ 2,

• (2) Let ϕ ∈ C(E_r, E_r) be an odd function and ϕ(u) = u, u ∈ ∂B_Rm ∩ H_m, then ϕ( ) ∈ Ψ_k if Ψ ∈ Ψ_k for every k ≥ 2,

• (3) if Ψ ∈ Ψ_k, X = −X is an open set and γ(X) ≤ s < k with k − s ≥ 2, then Ψ\X ∈ Ψ_k−s.

Define the attracting domain of 0 in E_r as

where τ is defined in (3.5). Since 0 is a local minimum of ℐ_r, combining with the continuous dependence of ODE on initial data, we have seen that 𝒪 is open. In fact, Yε+∩Yε−⊂O and ∂𝒪 is an invariant set. As in [4, Lemma 3.4], one obtains

We choose Ψ ∈ Ψ_k such that

Define

So that 𝒟 is a bounded open symmetric set with 0 ∈ 𝒟 and D¯⊂BRm∩Hm . The Borsuk-Ulam theorem gives us that γ(∂𝒟) = m. In addition, ψ(∂𝒟) ⊂ ∂𝒟 from the continuity of ψ. Consequently,

According to [32, Proposition 5.4], one can see that

Since Yε+∩Yε−∩∂O=∅ , we obtain γ(W ∩ ∂𝒪) ≤ 1, and then one can derive

so that

As a result, Ψ∩Erw≠∅ for every Ψ ∈ Ψ_k with k ≥ 2. We can choose ρ given in Lemma 2.2 so small that ∂B_ρ ⊂ 𝒪. Therefore, from Lemma 2.2 and the property of τ, we have

Thus, c_k ≥ α > 0.

Next, we will show that Kck∩Erw≠∅ . By the contradiction, we may assume that Kck∩Erw=∅ . From the definition of c_k, we can find Ψ ∈ Ψ_k satisfying

Then Ψ∩Erw⊂ℐrck+ε . It follows from Lemma 3.4 that there is a map σ ∈ C([0, 1] × E_r, E_r) and ε > 0 such that σ(1, ·) is an odd function, σ(1,u)=u,u∈ℐrck−2ε and

Combining with the definition of W, it follows that σ(1, W) \ W = ∅ from (e) of Lemma 3.4. Then, by (3.8), one has see that

which is contrary to the definition of c_k, since σ(1, Ψ) ∈ Ψ_k. So Kck∩Erw≠∅ .

We also note that c_k+1 ≥ c_k for any k ≥ 2 because of the property (1) of Ψ_k. Finally, we’re arguing to prove that ℐ_r has infinitely many sign-changing critical points by proving that c_k → ∞ as k → ∞. By using reduction to absurdity, we assume that c_k → c < ∞ as k → ∞. As a result of Lemma 2.2, one can see that K_c is compact and nonempty. In addition, we claim that

Let {u^k} be a sequence of sign-changing solutions of problem (3.1) such that ℐ_r (u^k) → c as k → ∞. Since Lemma 2.2, up to a subsequence, there exists u∈Erw such that u^k → u as k → ∞. In fact, 〈ℐr′(uk),(uk)±〉=0 implies that

which yields that ‖(u^k)^±‖ ≥ δ₀ > 0, where δ₀ is a constant independent of k. Consequently, u is still a sign-changing solution and then Kcw≠∅ .

Assume that γ(Kcw)=n . One can always find an open neighborhood 𝒩 in E_r satisfying Kcw∈N and γ(N¯)=n on account of 0∉Kcw , the compactness of Kcw and the continuous property of the genus. Lemma 3.4 then says that there exists ∊ > 0 and a map σ ∈ C([0, 1] × E_r, E_r) such that σ(1, ·) is an odd function, σ(1,u)=u,u∈ℐrc−2ε and

Because c_k → c as k → ∞, one may choose k large enough to get

In fact, c_k+n ≥ c_k for any k ≥ 2. The definition of c^k+n brings us that there exists Ψ ∈ Ψ_k+n, i.e.,

where ψ ∈ G_m, m ≥ k + n, γ(P¯)≤m−(k+n) , so that

Therefore, Ψ∩Erw⊂ℐrc+ε . Set P₁ = P ∪ ψ⁻¹(𝒩). Undoubtedly, P₁ is symmetric and open, and then

Thus

Notice that

where the fact σ(1, W))\W = ∅ also plays a role. Hence, ck≤supu∈Ψ˜∩Erwℐr(u)≤c−ε . This contradicts (3.9) and proves to be complete.

Proof

For k ≥ 2, set

and

Define

where ℐ_o stands for ℐ|_Eo. Certainly, Gmo≠∅ , based on the fact that id∈Gmo for all m ∈ ℕ. Thus, cko<+∞ . Notice that 0<α≤cko<+∞ from Lemma 3.1. Now, we seek to prove that ℐ_o has one nontrivial critical point u_k and ℐo(uk)=cko . From the definition of cko , one can find a sequence {u_n,k} ⊂ E_o such that

Lemma 2.1 gives us that ℐ_o satisfies the (PS) condition. Therefore, u_n,k → u_k in E_o, u_k is one nontrivial critical point of ℐ_o and ℐo(uk)=cko .

As in [29, Proposition 9.33], one can prove that cko→∞ as k → ∞. This brings us that ℐ_o(u) admits infinitely many critical points and hence completes the proof.