On a class of critical elliptic systems in ℝ⁴

Definition 1.1

The pair (u, v) is called a nontrivial solution iff u ≠ 0, v ≠ 0; we call (u, v) is positive iff u > 0, v > 0; the nontrivial solution (u₀, v₀) is called a ground state solution to (1.1) iff I(u₀, v₀) ≤ I(u, v) for all nontrivial critical points (u, v).

Theorem 1.2

Let α₁(Ω) > 0 be the first eigenvalue of –Δ in H01 (Ω).

1. Assume μ₁, μ₂, β, y > 0. If min(λ₁, λ₂) ≤ –α₁(Ω), then system (1.1) has no solution;

2. Assume Ω is star-shaped or Ω = ℝ⁴. If y ≤ 0 and min(λ₁, λ₂) > 0, then (1.9) has no nontrivial solution.

Remark 1.3

Assume that –α₁(Ω) < λ_i < 0. Let 𝓢 be the sharp constant of D^1,2(ℝ⁴) ↪ L⁴(ℝ⁴). Then B_μi mentioned in (1.10) satisfies (See [9]):

Theorem 1.4

Let λ₁, λ₂ > –α₁(Ω), μ₁, μ₂, β, y > 0. Assume that λ₁ satisfies one of the following two conditions:

1. λ₁ < 0 and β>(∫Ω|∇v0|2+λ22v02)−2y(∫Ωuμ12qpv02)∫Ωuμ12v02, for some 0 ≠ v₀ ∈ H01 (Ω);

2. λ₁ > 0 with Ω is a (smooth) starshaped domain,

and that λ₂ satisfies one of the following two conditions:

1. λ₂ < 0;

2. λ₂ > 0 with Ω is a (smooth) starshaped domain.

Then the system (1.1) has a ground state solution provided that one of the following cases holds:

1. min(λ₁, λ₂) < 0 < max(λ₁, λ₂) and β > max(μ₁, μ₂);

2. min(λ₁, λ₂) > 0 and β > max(μ₁, μ₂);

3. max(λ₁, λ₂) < 0 and β > μ1μ2.

Remark 1.5

If, for example, (1₁) and (2₁) hold, then only the case (c) occurs by default.

Lemma 2.1

[Pohozaev identity] Suppose u, v are (smooth) functions satisfying

where Ω ⊂ ℝⁿ(n ≥ 3); g and f are continuous on ℝ². Then we have

where 2∗=2nn−2,G(u,v)=∫0ug(t,v)dt,F(u,v)=∫0vf(u,t)dt and ν⃗ denotes the outward normal to ∂Ω.

Proof

We multiply the two equations of (2.1) by (x⃗ ⋅ ∇u), (x⃗ ⋅ ∇v) respectively and integrate over Ω. Then add them up, we have

By Pohozaev identity (cf. [7]), we have

and

Similarly, we have

Then plug (2.4), (2.5) and (2.6) into (2.3), we obtain (2.2), which completes the proof.□

Proof of Theorem 1.2

1. Recall system (1.1), we know that β > 0, y > 0, μ₁ > 0, μ₂ > 0. If λ₁ = min(λ₁, λ₂) ≤ –α₁ (Ω) and let φ be the corresponding eigenfunction of α₁ (Ω). Suppose that (u, v) is a solution of system (1.1), i.e., u > 0, v > 0. Then we multiply (1.9) with (φ, 0) and integrate by parts, we obtain

   0≥(λ1+α1(Ω))∫Ωuφ=∫Ω∇u∇φ+λuφ=∫Ω(μ1u3+βuv2+2qpyu2qp−1v2)φ>0,

   which is impossible. Similarly, if λ₂ = min (λ₁, λ₂) ≤ –α₁(Ω) and we multiply (1.1) with (0, φ), we have

   0≥(λ2+α1(Ω))∫Ωvφ=∫Ω(μ2v3+βu2v+2yu2qpv)φ>0,

   which leads to a contradiction. So system (1.1) has no solution.

2. Let (u, v) be a solution of the system (1.9). We take

   G(u,v)=−λ12u2+μ14u4+β2u2v2+yu2qpv2,F(u,v)=−λ22v2+μ24v4+β2u2v2+yu2qpv2.

   By the Pohozaev identity in Lemma 2.1, we have

   ∫Ω(|∇u|2+|∇v|2+2(λ1u2+λ2v2)−(μ1u4+μ2v4+2βu2v2+4yu2qpv2))=−12∫∂Ω(|∇u|2+|∇v|2)(x→⋅ν→)ds. (2.7)

   One the other hand,

   ∫Ω(|∇u|2+|∇v|2+λ1u2+λ2v2−(μ1u4+μ2v4+2βu2v2)−(2qp+2)yu2qpv2)=0. (2.8)

   Therefore,

   ∫Ω(λ1u2+λ2v2−(2−2qp)yu2qpv2)=−12∫∂Ω(|∇u|2+|∇v|2)(x→⋅ν→)ds.

   When one of the following two conditions holds: (a) Ω = ℝ⁴ or (b) Ω is star-shaped, i.e., (x⃗⋅ ν⃗) > 0 a.e. ∂Ω, then we have – 12 ∫_∂Ω(|∇u|² + |∇v|²)(x⃗⋅ ν⃗)ds ≤ 0, it follows that if λ₁, λ₂ > 0, y ≤ 0, system (1.9) has no nontrivial solution.□

Lemma 3.1

(see [23]) Let

Then c := d𝓢² is attained by (u, v) if and only if (u, v) = (t₁U_ε,z, t₂U_ε,z), where 𝓢 is the best Sobolev embedding constant from D^1,2(ℝ⁴) to L⁴(ℝ⁴) and

is the Talanti function satisfying –ΔU = U³ in ℝ⁴ and (t₁, t₂) satisfies

Moreover, if β > max{μ₁, μ₂}, then t₁, t₂ > 0.

Lemma 3.2

Assume μ₁, μ₂, β, y > 0.

1. If min(λ₁, λ₂) < 0 < max(λ₁, λ₂), then ∃a₁, a₂ > 0 such that (3.5) has a unique positive solution k and ∃ Q < 0 such that

   λ1a12+λ2a222−∥Vε∥2qp+22qp+2∥Vε∥22ya12qpa22k2qp≤Q<0, (3.7)

   for ρ > 0 and ε > 0 small enough.

2. If min(λ₁, λ₂) > 0, then ∀ a₁, a₂ > 0, (3.5) has a unique positive solution k and ∃ Q < 0 such that (3.7) holds for ∀ ε > 0 small enough.

3. If max(λ₁, λ₂) < 0, then ∀ a₁, a₂ > 0, (3.5) has a unique positive solution k and ∃ Q < 0 such that (3.7) holds for ∀ ρ, ε > 0 small enough.

Proof

Since G_ε(k) ∈ C¹(ℝ) and monotonically increasing on [0, +∞) with

It is sufficient to verify that the unique solution k > 0 exists as long as

1. Without loss of generality, we assume λ₁ < 0 < λ₂ in this case. Let

   Ai:=∥∇Vε∥22+λi∥Vε∥22,i=1,2.

   Then we have A₁ < A₂ and A₂ > 0. If there exists a₁, a₂ > 0 such that

   (a2a1)2>−A1A2, (3.9)

   then Pε=a12(A1+(a2a1)2A2)>0 and (3.8) holds, which means (3.5) has positive solution k. Next we will show that ∀ a > 0, a₁ = a, a₂ = −λ12λ2 a satisfy (3.9). In fact, – A1A2 < 0 for ε > 0 small enough. Recall that λ₁ < 0 ≤ λ₂, A₂ > 0 and

   |Vε∥2≤|B(0,2ρ)|14⋅∥Vε∥4,

   here |B(0, 2ρ)| denotes the measure of the ball B(0, 2ρ), then |B(0, 2ρ)| = 8π²ρ⁴ when N = 4, we have

   −A1A2=−∥∇Vε∥22+λ1∥Vε∥22∥∇Vε∥22+λ2∥Vε∥22≤−S2+o(ε2)+λ1(|B(0,2ρ)|12∥Vε∥42)∥∇Vε∥22+λ2∥Vε∥22=−S2+22πρ2Sλ1+o(ε2)∥∇Vε∥22+λ2∥Vε∥22. (3.10)

   Take ρ > 0 small enough and satisfying

   S+22πρ2λ1>0, (3.11)

   here 𝓢 is the best Sobolev embedding constant from D^1,2(ℝ^N) to L⁴(ℝ^N). Thus we have 𝓢² + 2 2 πρ² λ₁𝓢 + o(ε²) > 12 (𝓢² + 2 2 πρ²λ₁𝓢) > 0 when ε > 0 small enough and

   −S2+22πρ2λ1S+o(ε2)∥∇Vε∥22+λ2∥Vε∥22<−12(S2+22πρ2λ1S)∥∇Vε∥22+λ2∥Vε∥22≤−12(S2+22πρ2λ1S)S2+o(ε2)+λ2(|B(0,2ρ)|12∥Vε∥42)=−12(S2+22πρ2λ1S)S2+22πρ2λ2S+o(ε2)<−12(S2+22πρ2λ1S)32(S2+22πρ2λ2S)<0. (3.12)

   Together with (3.10), when ε > 0 and ρ > 0 small enough, we obtain – A1A2 < 0. Then ∀ a > 0, (a₁, a₂) = (a,−λ12λ2a) satisfies (3.9). Therefore (3.5) has a positive solution, and in particular

   λ1a12+λ2a222<0,a1,a2>0.

   So we may choose Q=λ1a12+λ2a222<0 and then (3.7) holds.

2. For the case λ₁, λ₂ > 0, then obviously P_ε > 0. Therefore, the solution k exists for ∀ a₁, a₂ > 0. In the following proposition, we will show that if the positive solution k exists, then k has positive lower bound. Note that

   ∥Vε∥2qp+22qp+2∥Vε∥22=∫B2ρ(0)(Ψ(x)⋅Uε,0(x))2qp+2dx∫B2ρ(0)(Ψ(x)⋅Uε,0(x))2dx=∫02ρεr3(1+r2)−2qp+2dr+o(1)8−qpε2qp∫02ρεr3(1+r2)−2dr+o(1)→+∞,ε→0+,

   we can take G=λ1a12+λ2a22ya12qpa22k02qp>0, here k₀ will be given by Proposition 3.3. Then we obtain that

   λ1a12+λ2a222−∥Vε∥2qp+22qp+2∥Vε∥22ya12qpa22k2qp<λ1a12+λ2a222−Gya1a22k0=−λ1a12+λ2a222

   when ε > 0 is small enough, so we may let Q=−λ1a12+λ2a222, then Q < 0 and satisfies (3.7).

3. Recall the definition of A_i in the case (a) and λ₁, λ₂ < 0, if ρ satisfies

   0<ρ<(−S22πmin(λ1,λ2))12,

   then

   Ai=∥∇Vε∥22+λi∥Vε∥22≥S2+22πρ2λiS+o(ε2)>12(S2+22πρ2λiS)>0,i=1,2

   for ε > 0 small enough. Similar to the proof of the case (a), we obtain Pε=a12(A1+(a2a1)2A2)>0 for any a₁, a₂ > 0, so (3.5) has a unique positive solution. Further, if we take Q=λ1a12+λ2a222, then (3.7) holds.□

Proposition 3.3

Assume μ₁, μ₂, β, y > 0, then for all the three cases of Lemma 3.2, ∃ a₁, a₂ > 0 such that the positive solution k of (3.5) exists and satisfies:

for ε > 0 small enough, here k₀ is a constant depending on μ_i, β, y, a₁, a₂.

Proof

1. We are going to show that 0 < k < 1.

   1. For the case min(λ₁, λ₂) < 0 < max (λ₁, λ₂). According to the proof of (a) in Lemma 3.2, ∀ a > 0, ∃ (a₁, a₂) = (a,−λ12λ2a) such that k exists and λ1a12+λ2a22 < 0. Then

      Gε(1)=((μ1a14+2βa12a22+μ2a24)∥Vε∥44)+((2qp+2)ya12qpa22∥Vε∥2qp+22qp+2)−((a12+a22)∥∇Vε∥22+(λ1a12+λ2a22)∥Vε∥22)≥(μ1a14+2βa12a22+μ2a24)(S2+o(ε4))−(a12+a22)(S2+o(ε2))−C(λ1a12+λ2a22)(ε2|ln⁡ε|+o(ε2))=(μ1a14+2βa12a22+μ2a24−(a12+a22))S2+o(ε).

      We can take a > 0 large enough, such that a₁ and a₂ satisfy

      μ1a14+2βa12a22+μ2a24−(a12+a22)>0.

      Therefore, G_ε(1) > 0 for ε > 0 small enough, and we observe that ∃ k ∈ (0, 1) such that G_ε(k) = 0, here k depends on a₁, a₂, ε only, therefore it is the unique solution of the algebraic equation (3.5).

   2. For the case λ₁, λ₂ > 0. It follows from λ1a12+λ2a22 > 0 and

      ∥Vε∥2≤|B(0,2ρ)|14⋅∥Vε∥4,|B(0,2ρ)|=12π2(2ρ)4=8π2ρ4,

      that

      Gε(1)=((μ1a14+2βa12a22+μ2a24)∥Vε∥44)+((2qp+2)ya12qpa22∥Vε∥2qp+22qp+2)−((a12+a22)∥∇Vε∥22+(λ1a12+λ2a22)∥Vε∥22)≥(μ1a14+2βa12a22+μ2a24)(S2+o(ε2))−(a12+a22)(S2+o(ε4))−22πρ2(λ1a12+λ2a22)(S+o(ε4))=S[(μ1a14+2βa12a22+μ2a24−(a12+a22))S−22πρ2(λ1a12+λ2a22)]+o(ε2).

      According to the case (b) of Lemma 3.2, ∀ a₁, a₂ > 0, the unique solution k of (3.5) exists. So we can take a₁, a₂ large enough such that

      (μ1a14+2βa12a22+μ2a24−(a12+a22))S−22πρ2(λ1a12+λ2a22)>0,

      and then G_ε(1) > 0 for ε small enough, which means 0 < k < 1 since G_ε(0) < 0.

   3. For the case of max(λ₁, λ₂) < 0, we see that λ1a12+λ2a22 < 0. We can apply the same calculation as that used in i) to show that for a₁, a₂ large enough and ρ, ε > 0 small enough, there holds 0 < k < 1.

2. We show that k has a lower bound, i.e., k > k₀ > 0.

   1. Assume λ₁ < 0 < λ₂. As the Step 1, we consider a₁, a₂ so that λ1a12+λ2a22 < 0. Together with

      ∥Vε∥2qp+2≤|B(0,2ρ)|p−q4(p+q)⋅∥Vε∥4,

      we have

      Gε(k)≤(μ1a14+2βa12a22+μ2a24)(S2+o(ε4))k2−(λ1a12+λ2a22)|B(0,2ρ)|12(S+o(ε4))+(2qp+2)ya12qpa22|B(0,2ρ)|p−q2p(Sqp+1+o(ε4))k2qp−(a12+a22)(S2+o(ε2))=:G∗(k)+o(ε),

      here

      G∗(k)=(μ1a14+2βa12a22+μ2a24)S2k2+(2qp+2)ya12qpa22|B(0,2ρ)|p−q2pSqp+1k2qp−(a12+a22)S2−(λ1a12+λ2a22)|B(0,2ρ)|12S.

      As the proof of Lemma 3.2, we take ρ > 0 such that 𝓢 + 2 2 πρ²λ₁ > 0 and (a₁, a₂) = (a,−λ12λ2a) . Then we obtain that – S+22πρ2λ2S+22πρ2λ1 < 0, it follows that (a1a2)2>−S+22πρ2λ2S+22πρ2λ1 always holds, which means that

      G∗(0)=−(a12+a22)S2−(λ1a12+λ2a22)|B(0,2ρ)|12S<0.

      Together with limk→+∞ G^*(k) = +∞, there exists k₀ > 0 dependent on μ_i, β, y, a₁, a₂ such that G^*(k₀) < 0. Thus

      Gε(k0)≤G∗(k0)+o(ε)<0,

      when ε is small enough and then we obtain k₀ < k.

   2. Assume that λ_i > 0, i = 1, 2. In this case, λ1a12+λ2a22 > 0, it follows that

      Gε(k)≤(μ1a14+2βa12a22+μ2a24)(S2+o(ε4))k2−(λ1a12+λ2a22)(Cε2|ln⁡ε|+o(ε2))−(a12+a22)(S2+o(ε2))+(2qp+2)ya12qpa22|B(0,2ρ)|p−q2p(Sqp+1+o(ε4))k2qp≤(μ1a14+2βa12a22+μ2a24)S2k2+(8π2ρ4)p−q2p(2qp+2)ya12qpa22Sqp+1k2qp−(a12+a22)S2+o(ε)=:G∗(k)+o(ε),

      here

      G∗(k)=(μ1a14+2βa12a22+μ2a24)S2k2+(8π2ρ4)p−q2p(2qp+2)ya12qpa22Sqp+1k2qp−(a12+a22)S2.

      Obviously, G_*(0) < 0 and limk→+∞ G_*(k) = +∞, then ∃ k₀ > 0 dependent on μ_i, β, y, a₁, a₂ such that G_*(k₀) < 0. Thus G_ε(k₀) ≤ G_*(k₀) + o(ε²) < 0 for ε > 0 small enough. Recall that k is the unique positive solution satisfying G_ε(k) = 0, we obtain k₀ < k.

   3. For the case max(λ₁, λ₂) < 0, we have λ1a12+λ2a22 < 0 as that in case 1). According to the proof of Lemma 3.2, we take ρ > 0 satisfying 𝓢 + 2 2 πρ²λ_i > 0, i = 1, 2. Thus we can take a₁, a₂ > 0, note that (a1a2)2>−S+22πρ2λ2S+22πρ2λ1, then we have

      G∗(0)=−(a12+a22)S2−(λ1a12+λ2a22)|B(0,2ρ)|12S<0.

      Therefore, similar to the case 1), we get a constant k₀ > 0 independent of ε such that k₀ < k.□