A system of equations involving the fractional p-Laplacian and doubly critical nonlinearities

Definition 1.1

We say a pair ( u , v ) ∈ X is a positive weak solution of the system ( S ) if u , v > 0 , and for every ( ϕ , ψ ) ∈ X , it holds

Definition 1.2

1. We say a weak solution ( u , v ) of ( S ) is of the synchronized form if u = λ w , v = μ w for some constants λ , μ and a common function w ∈ W ˙ s , p ( R N ) .

2. We say a weak solution ( u , v ) of ( S ) is a ground state solution if ( u , v ) is a minimizer of S α , β (see (1.4)).

Theorem 1.3

Let ( u 0 , v 0 ) be any positive ground state solution of ( S ). If one of the following conditions hold

1. 1 < β < p ,

2. β = p and α < p ,

3. β > p and α < p ,

then, there exists unique τ min > 0 satisfying

where h is defined by (1.6). Moreover,

where U and V are two positive ground state solutions of (1.2). Further, λ p s * − p = p s * p s * + α τ min β .

Remark 1.4

Since for p ≠ 2 , uniqueness of ground state solutions of (1.2) is not yet known, we are not able to conclude whether any ground state solution of ( S ) is of the synchronized form, i.e., of the form of ( λ U , τ min λ U ) or not.

Theorem 1.5

Assume that one of the following conditions hold:

1. If N 2 s < p < N s , α , β > p and

   (1.10) 0 < γ ≤ p s * ( p s * − p ) p min 1 α α − p β − p β − p p , 1 β β − p α − p α − p p .

2. If 2 N N + 2 s < p < N 2 s , α , β < p and

   (1.11) γ ≥ p s * ( p s * − p ) p max 1 α p − β p − α p − β p , 1 β p − α p − β p − α p .

Then the least energy A = s N ( k 0 + ℓ 0 ) S N ∕ s p and A is attained by ( k 0 1 ∕ p U , ℓ 0 1 ∕ p U ) , where U is a minimizer of (1.1), k 0 , ℓ 0 satisfies (1.9) and

Theorem 1.6

Assume that 2 N N + 2 s < p < N 2 s and α , β < p . There exists γ 1 > 0 such that for any γ ∈ ( 0 , γ 1 ) , there exists a solution ( k ( γ ) , ℓ ( γ ) ) of (1.9) such that ( k ( γ ) 1 ∕ p U , ℓ ( γ ) 1 ∕ p U ) is a positive solution of system ( S ˜ γ ) with J ( k ( γ ) 1 ∕ p U , ℓ ( γ ) 1 ∕ p U ) > A ˜ , where U is a minimizer of (1.1),

and

Theorem 1.7

Assume that 2 N N + 2 s < p < N 2 s and α , β < p . Then the following system of equations

admit a radial positive solution ( u 0 , v 0 ) .

Lemma 2.1

Suppose α , β > 1 such that α + β = p s * . Then

1. S α , β = h ( τ min ) S .

2. S α , β has minimizers ( U , τ min U ) , where U is a ground state solution of (1.2) and τ min satisfies

   τ p − 1 ( p s * + α τ β − β τ β − p − p s * τ p s * − p ) = 0 .

Proof

Let { ( u n , v n ) } be a minimizing sequence in X for S α , β . Choose τ n > 0 such that ‖ v n ‖ L p s * ( R N ) = τ n ‖ u n ‖ L p s * ( R N ) . Now set, w n = v n τ n . Therefore, ‖ u n ‖ L p s * ( R N ) = ‖ w n ‖ L p s * ( R N ) and applying Young’s inequality,

Therefore,

Thus, as n → ∞ , we have h ( τ min ) S ≤ S α , β . For the reverse inequality, we choose u = U , v = τ min U to obtain h ( τ min ) S ≥ S α , β . In Lemma 2.2, we will show that point τ min exists. This proves (i).

(ii) Taking ( u , v ) = ( U , τ min U ) , a simple computation yields that

By using (i), we infer that ( U , τ min U ) is a minimizer of S α , β . Further, since τ min is a critical point of h , computing h ′ ( τ min ) = 0 yields that τ min satisfies

Lemma 2.2

Let α , β > 1 , and α + β = p s * . Then (2.1) always has at least one root τ > 0 , and for any root τ > 0 , the problem ( S ) has positive solutions ( λ U , μ U ) , where

Moreover, if one of the following conditions hold

1. 1 < β < p ,

2. β = p and α < p ,

3. β > p and α < p ,

then, τ min > 0 and h ( τ min ) < 1 . In all other cases, τ min = 0 .

Proof

Clearly, if τ > 0 solves

then ( λ U , μ U ) with μ = τ λ solves ( S ). Thus to prove the required result, it is enough to show that (2.1) has positive roots τ , which we discuss in the following cases.

Case 1: If 1 < β < p .

Therefore, lim τ → 0 + g ( τ ) = − ∞ .

Now, if α ≥ p , then g ( 1 ) = α − β > 0 . Thus, there exists τ ∈ ( 0 , 1 ) such that g ( τ ) = 0 .

If 1 < α < p , then we have p s * − p < p s * − α = β , and consequently, lim τ → ∞ g ( τ ) = ∞ . Thus, there exists τ > 0 such that g ( τ ) = 0 .

Also observe that, by direct computation, we obtain

Thus, f ( τ ) ≥ 0 for all τ > 0 and f ( 0 ) = 0 . This together with the fact that lim τ → 0 + g ( τ ) = − ∞ implies h ′ ( τ ) < 0 in τ ∈ ( 0 , ε ) for some ε > 0 . This means h is a decreasing function near 0. Combining this with the fact that h ( 0 ) = 1 and lim τ → ∞ h ( τ ) = 1 , we conclude that there exists a point τ min ∈ ( 0 , ∞ ) such that min τ ≥ 0 h ( τ ) = h ( τ min ) < 1 , and this holds for all α > 1 .

Case 2: If β = p .

In this case, g becomes g ( τ ) = α ( 1 + τ p ) − p s * τ α . Hence, g ( 0 ) = α > 0 and g ( 1 ) = α − p .

(i) If α = p , then N = 2 s p and g ( τ ) = p − p τ p . Thus, there exists a unique root τ 1 = 1 of g . Also note that h is increasing near 0. Hence, τ 1 is the maximum point of h with h ( τ 1 ) > 1 . In this case, min τ ≥ 0 h ( τ ) = h ( τ min ) = h ( 0 ) .

(ii) If 1 < α < p , then we have 2 s p < N < s p ( p + 1 ) . Observe that g ( 0 ) = α > 0 , g ( 1 ) = α − p < 0 , and lim τ → ∞ g ( τ ) = + ∞ . Also note that, g is decreasing in 0 , ( p s * ∕ p ) 1 p − α and increasing in ( p s * ∕ p ) 1 p − α , ∞ . Therefore, g has exactly one critical point ( p s * ∕ p ) 1 p − α and two roots τ i ( i = 1 , 2 ) with τ 1 ∈ ( 0 , ( p s * ∕ p ) 1 p − α ) , τ 2 ∈ ( ( p s * ∕ p ) 1 p − α , ∞ ) .

Further, note that in this case h is increasing near 0, which leads that first positive critical point of h , i.e., τ 1 is the local maximum for h and h ( τ 1 ) > 1 . Further, as lim τ → ∞ h ( τ ) = 1 , the second root of g , i.e., τ 2 , becomes the second and last critical point of h and h ( τ 2 ) < 1 . Therefore, in this case, τ min = τ 2 > 0 is the minimum point of h with h ( τ min ) < 1 .

(iii) If α > p , then s p < N < 2 s p and g ( 0 ) > 0 . We see that g is increasing in 0 , ( p ∕ p s * ) 1 α − p and decreasing in ( p ∕ p s * ) 1 α − p , ∞ . This together with the fact lim τ → ∞ g ( τ ) = − ∞ leads that there exists a unique τ > 0 such that g ( τ ) = 0 .

Since in this case, h is increasing near 0, so at τ , h attains the maximum with h ( τ ) > 1 . Hence, h has no other critical point, and therefore, h ( τ min ) = h ( 0 ) .

Case 3: If β > p .

If 1 < α ≤ p , then g ( 1 ) = α − β ≤ 0 . Since g ( 0 ) > 0 , there is a τ ∈ ( 0 , 1 ] such that g ( τ ) = 0 . If α > p and α > β , then g ( 1 ) > 0 and lim τ → ∞ g ( τ ) = − ∞ . Thus, there exists τ ∈ ( 1 , ∞ ) such that g ( τ ) = 0 . If α > p and α ≤ β , then g ( 1 ) ≤ 0 . As g ( 0 ) > 0 , thus there exists τ ∈ ( 0 , 1 ] such that g ( τ ) = 0 . Next we analyze τ min in case 3 in the following three subcases.

(i) β > p and α > p .

Observe that in this case we have

Hence, without loss of generality, we can assume α ≥ β .

Claim 1: g ( τ ) > 0 for τ ∈ [ 0 , 1 ) . Indeed, by using (2.2), τ ∈ [ 0 , 1 ) implies τ α , τ β > τ p s * − p . Therefore,

where in the last inequality, we have used the fact that τ β − p < 1 ⇒ τ β − p < β ≤ α . This proves claim 1.

Claim 2: g is monotonically decreasing for τ ≥ 1 . Indeed, τ ≥ 1 implies τ α ≥ τ p . Therefore, by using (2.2), we have

This proves claim 2. Also observe that g ( 1 ) ≥ 0 and g ( τ ) → − ∞ as τ → ∞ . Combining these facts along with claim 1 and 2 proves that g has only one root say τ in ( 0 , ∞ ) , which in turn implies h has only one critical point τ in ( 0 , ∞ ) . Since β > p implies h is increasing near 0, so at τ , h attains the maximum with h ( τ ) > 1 . Combining this with lim τ → ∞ h ( τ ) = 1 proves that h ( τ min ) = h ( 0 ) = 1 , i.e, τ min = 0 .

(ii) β > p and α < p .

In this case, g ( 0 ) > 0 , g ( 1 ) < 0 , and we claim g is strictly decreasing in ( 0 , 1 ) . Indeed, α < p ⇒ τ p < τ α for τ ∈ ( 0 , 1 ) . Also β > p ⇒ α < p s * − p . Therefore,

Claim: g has only one critical point in ( 1 , ∞ ) . Indeed,

So to prove that g has only critical point in ( 1 , ∞ ) , it is enough to show that g 1 has only one root in ( 1 , ∞ ) . Observe that, g 1 ( 0 ) < 0 , lim τ → ∞ g 1 ( τ ) = ∞ and a straight forward computation yields that g 1 is a decreasing function in 0 , p s * ( p s * − p ) p β 1 p − α and g 1 is an increasing function in p s * ( p s * − p ) p β 1 p − α , ∞ . Thus, g 1 has only one root. Hence, the claim follows. Next, we observe that α < p ⇒ β > p s * − p , and therefore, lim τ → ∞ g ( τ ) = ∞ .

Combining all the aforementioned observations and claim, it follows that g has only one critical point in ( 0 , ∞ ) and two roots τ 1 , τ 2 with τ 1 ∈ ( 0 , 1 ) and τ 2 ∈ ( 1 , ∞ ) . Hence, h has exactly two critical points τ 1 , τ 2 . Since h is increasing near 0 leads to the conclusion that first positive critical point of h , i.e., τ 1 is the local maximum for h and h ( τ 1 ) > 1 and since lim τ → ∞ h ( τ ) = 1 at the second critical point of h , i.e., at τ 2 , we have h ( τ 2 ) < 1 . Therefore, in this case, τ min = τ 2 > 0 is the minimum point of h with h ( τ min ) < 1 .

(iii) β > p , α = p .

In this case, g ( 0 ) > 0 and α = p ⇒ β = p s * − p . Therefore,

i.e., g is a strictly decreasing function. Also, observe that lim τ → ∞ g ( τ ) = − ∞ . Hence, g has only one root in ( 0 , ∞ ) , i.e., h has only critical point τ in ( 0 , ∞ ) . Since β > p implies h is increasing near 0, so at τ , h attains the maximum with h ( τ ) > 1 . Combining this with lim τ → ∞ h ( τ ) = 1 proves that h ( τ min ) = h ( 0 ) = 1 , i.e, τ min = 0 .□