A uniqueness result for the fractional Schrödinger-Poisson system with strong singularity

Li Wang; Renhua Chen; Qiaocheng Zhong; Jun Wang; Fang Li

doi:10.1515/math-2024-0076

Artikel Open Access

A uniqueness result for the fractional Schrödinger-Poisson system with strong singularity

Li Wang , Renhua Chen , Qiaocheng Zhong , Jun Wang und Fang Li

Veröffentlicht/Copyright: 31. Dezember 2024

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Abstract

This article considers existence of solution for a class of fractional Schrödinger-Poisson system. By using the Nehari method and the variational method, we obtain a uniqueness result for positive solutions.

Keywords: fractional Schrödinger-Poisson system; strong singularity; uniqueness; positive solutions

MSC 2010: 35A15; 35B09; 35B38; 35J75

1 Introduction and main result

This article is devoted to existence result for the following fractional Schrödinger-Poisson system

(1) ( − Δ ) s u + ϕ ( x ) u = h ( x ) u − θ − g ( x ) u q , in Ω , ( − Δ ) t ϕ ( x ) = u 2 , in Ω , u > 0 , in Ω , u = ϕ = 0 , in R 3 \ Ω ,

where s , t ∈ ( 0 , 1 ) with 2 t + 4 s > 3 , Ω ⊂ R 3 is a bounded open domain with smooth boundary ∂ Ω , g ∈ L ∞ ( Ω ) and inf Ω g ≥ 0 , θ > 1 , 0 < q < 1 and h ∈ L 1 ( Ω ) with h > 0 a.e. in Ω . The operator ( − Δ ) s is given by, up to a normalization constant,

( − Δ ) s ψ ( x ) = 2 lim ε → 0 + ∫ R 3 \ B ε ( x ) ψ ( x ) − ψ ( y ) ∣ x − y ∣ 3 + 2 s d y , ∀ x ∈ R 3

along ψ ∈ C 0 ∞ ( R 3 ) . B ε ( x ) ≔ { y ∈ R 3 : ∣ y − x ∣ < ε } with ε > 0 and x ∈ R 3 . We can know the fractional Laplace operator ( − Δ ) τ becomes − Δ as τ → 1 − , see [1, Proposition 4.4]. This operator arises wildly in the applied sciences, for example, plasma physics [2], finance [3], Signorini problems [4], and so on. For more information about operator ( − Δ ) τ , please refer to [1,5–7] and the references therein.

Owing to their wide range of applications to physical models, such as non-Newtonian fluids, chemical heterogenous catalysts, etc. [8,9], nonlinear elliptic equations with singularities have received much attention in the past decade. For more information about the physical context of singularity, see [10,11] and the references therein.

The motivation to study the system like (1) is expressed in following Schrödinger-Poisson equation:

(2) − Δ u + V ( x ) u + ϕ u = g ( x , u ) , in R 3 , − Δ ϕ = u 2 , in R 3 ,

which has been studied by Benci and Fortunato [12]. In this equation, we can find that the potential is decided by wave function. The term ϕ u is related to electric field. More information about system (2) can be found in [13] and the references therein.

In [14], Teng studied the following fractional Schrödinger-Poisson system:

(3) ( − Δ ) t u + V ( x ) u + ϕ u = μ ∣ u ∣ p − 1 u + ∣ u ∣ 2 t * − 2 u , in R 3 , ( − Δ ) s ϕ = u 2 , in R 3 ,

where 1 < p < 2 t * − 1 , μ > 0 and s , t ∈ ( 0 , 1 ) with 2 s + 2 t > 3 . In that article, the author showed the existence of solution by the method of Pohožaev-Nehari manifold, where μ is big enough. However, in [15], Wei obtained the existence of infinitely many solutions for the following fractional Schrödinger-Maxwell equations:

(4) ( − Δ ) s u + V ( x ) u + ϕ u = h ( x , u ) , in R 3 , ( − Δ ) s ϕ = K s u 2 , in R 3 ,

by the Fountain theorem, where K s > 0 is related only to s and nonlinearity term h and potential V satisfy some suitable assumptions. More results about the aforementioned problem can be seen in [16,17].

Lazer and Mckenna [18] considered the singular boundary-value problem

(5) − Δ u + p ( x ) u ( x ) − γ = 0 , x ∈ Ω , u = 0 , x ∈ ∂ Ω .

They showed that there exists a solution, which is smooth on Ω and continuous on Ω ¯ when γ > 1 . But the solution is in H 0 1 ( Ω ) if and only if γ < 3 . The authors in [19] provided an extension of this result and revealed the role of 3, and they have given a local description of the solution set. For the Kirchhoff-type fractional Laplacian problems, we can refer to [20]. Motivated by the aforementioned work, the purpose of the present article is to prove the uniqueness of solution for the strong singular fractional Schrödinger-Poisson problems with θ > 1 , which has not been studied according to my knowledge.

The main difficulties in dealing with this problem are that the fractional Laplacian is a nonlocal operator, u − θ is nonintegrable for u ∈ E 0 and the h ( x ) is indefinite. Here, we point out that the typical sub/supersolution method by variational techniques [21] may be ineffective for h ( x ) in our article. So, we will use the spirit of the variational technique of Sun in [22] to overcome these difficulties and establish some suitable constraints to recover integrability.

First, we give some notations. We define W s , 2 ( Ω ) ( s ∈ ( 0 , 1 ) ) with the norm

(6) ‖ u ‖ W s , 2 ( Ω ) = ∫ Ω ∣ u ( x ) ∣ 2 d x 1 2 + ∫ Ω ∫ Ω ∣ u ( x ) − u ( y ) ∣ 2 ∣ x − y ∣ 3 + 2 s d x d y 1 2 .

Set Q = ( R 3 × R 3 ) \ ( C Ω × C Ω ) with C Ω = R 3 \ Ω . We define

E = u ∣ u : R 3 → R , u is measurable , u ∣ Ω ∈ L 2 ( Ω ) , and ∬ Q ∣ u ( x ) − u ( y ) ∣ 2 ∣ x − y ∣ 3 + 2 s d x d y < ∞ ,

which is endowed with the norm

(7) ‖ u ‖ E = ∫ Ω ∣ u ( x ) ∣ 2 d x 1 2 + ∬ Q ∣ u ( x ) − u ( y ) ∣ 2 ∣ x − y ∣ 3 + 2 s d x d y 1 2 .

The space E 0 denotes the closure of C 0 ∞ ( Ω ) in E . By Theorems 6.5 and 7.1 in [1], we can obtain that space E 0 is a Hilbert space, which is endowed with the scalar product and norm

(8) ⟨ u , w ⟩ ≔ ⟨ u , w ⟩ E 0 = ∬ Q ( u ( x ) − u ( y ) ) ( w ( x ) − w ( y ) ) ∣ x − y ∣ 3 + 2 s d x d y ,

(9) ‖ u ‖ ≔ ‖ u ‖ E 0 = ∬ Q ∣ u ( x ) − u ( y ) ∣ 2 ∣ x − y ∣ 3 + 2 s d x d y 1 2

for all u , w ∈ E 0 . The norm (9) is equipped with the usual one defined in (6). Since u = 0 a.e. in R 3 \ Ω , we can extend the integral in (7), (8), and (9) to R 3 . By results of [1], we can see that E 0 ( Ω ) is continuously embedded in L r ( Ω ) for r ∈ [ 2 , 2 s * ] and the embedding is compact when r ∈ [ 1 , 2 s * ) , where 2 s * = 6 3 − 2 s . Furthermore, according to Theorem 6.5 in [1], we can define S s > 0 which is a best constant as follows:

(10) S s = inf u ∈ E 0 \ { 0 } ‖ u ‖ 2 ‖ u ‖ L 2 s * ( Ω ) 2 .

Form s , t ∈ ( 0 , 1 ) with 2 t + 4 s > 3 , it is easy to see that 12 3 + 2 t < 6 3 − 2 s , and we have E 0 ↪ L 12 3 + 2 t ( Ω ) . Considering for all u ∈ E 0 , we can define the linear functional L u on space D t , 2 ( R 3 ) as

L u ( v ) = ∫ R 3 u 2 v d x , ∀ v ∈ D t , 2 ( R 3 ) .

The function space D t , 2 ( R 3 ) is the completion of C 0 ∞ ( R 3 ) under the norm

‖ u ‖ D t , 2 ( R 3 ) = ∬ R 3 × R 3 ∣ u ( x ) − u ( y ) ∣ 2 ∣ x − y ∣ 3 + 2 t d x d y 1 ⁄ 2 .

Then, by combining with (10) and the Hölder inequality, we can infer there exist some positive numbers C 1 , C 2 such that

(11) ∣ L u ( v ) ∣ ≤ ∫ Ω ∣ u ( x ) ∣ 12 3 + 2 t d x 3 + 2 t 6 ∫ Ω ∣ v ( x ) ∣ 6 3 − 2 t d x 3 − 2 t 6 ≤ C 1 ‖ u ‖ 2 S t − 1 2 ‖ v ‖ = C 2 ‖ u ‖ 2 ‖ v ‖ .

Hence, we can obtain that there is a unique function ϕ u t ∈ E 0 such that

(12) ∫ Ω u 2 w d x = ∫ Ω ( − Δ ) t 2 ϕ u t ⋅ ( − Δ ) t 2 w d x , ∀ u , w ∈ E 0

by the Lax-Milgram theorem, i.e., the equation

( − Δ ) t ϕ u t = u 2 , in x ∈ Ω

has a weak solution ϕ u t . Furthermore, it holds

(13) ϕ u t = c t ∫ Ω u 2 ( y ) ∣ x − y ∣ 3 − 2 t d y = c t 1 ∣ x ∣ 3 − 2 t ∗ u 2 , ∀ x ∈ Ω ,

which is known as the t -Riesz potential and c t is a positive number. Equation (13) ensures that ϕ u t > 0 for all u ≠ 0 . In problem (1), we replace ϕ by ϕ u t to obtain the equation:

(14) ( − Δ ) s u + ϕ u t ( x ) u = h ( x ) u − θ − g ( x ) u q , in Ω , u > 0 , in Ω , u = ϕ = 0 , in R 3 \ Ω .

Note that if problem (14) has a weak solution u ∈ E 0 , which means system (1) has a weak solution ( u , ϕ u ) and u > 0 satisfies

⟨ u , ψ ⟩ + ∫ Ω ϕ u t u ψ d x + ∫ Ω g ( x ) u q ψ d x − ∫ Ω h ( x ) u − θ ψ d x = 0 , ∀ ψ ∈ E 0 .

Associated to problem (14), we give the energy functional I : E 0 → R by

I ( u ) = 1 2 ‖ u ‖ 2 + 1 4 ∫ Ω ϕ u t ( x ) u 2 d x + 1 1 + q ∫ Ω g ( x ) ∣ u ∣ 1 + q d x − 1 1 − θ ∫ Ω h ( x ) ∣ u ∣ 1 − θ d x .

In this problem, the main difficulties become that u − θ is nonintegrable for u ∈ E 0 and any inequality involving u ∈ E 0 will not provide significant assistance. Hence, there are many difficulties when we use standard variational methods to find solutions. Since, in the case 0 < θ < 1 , it is well known that the energy functional is continuous, which is completely different from the case θ > 1 . Another difficulty is that h ( x ) is indefinite. In general, we can apply the subsupersolution method to deal with singularity [21,23]. However, this method does not work when considering such a general measurable function h ( x ) > 0 in our article. So we use variational techniques in [24,25] to overcome these difficulties and restore integrability by establishing the appropriate constraints.

First, we define the operator Φ : E 0 → E 0 as follows:

Φ [ u ] = ϕ u t .

Then, we will give some properties about Φ , which is helpful for our next research. The proof is the same as in [14,26].

Lemma 1.1

∀ u ∈ E 0 ,

Φ is continuous;
Φ maps a bounded set into a bounded set;
if u n ⇀ u in E 0 , we can obtain Φ [ u n ] → Φ [ u ] in E 0 and ∫ Ω ϕ u n t u n v d x → ∫ Ω ϕ u t u v d x for every v ∈ E 0 ;
τ 2 Φ [ u ] = Φ [ τ u ] , ∀ τ ∈ R ;
Set Ψ ( u ) = ∫ Ω ϕ u t u 2 d x , then Ψ : E 0 → E 0 is continuous and
⟨ Ψ ′ ( u ) , v ⟩ = 4 ∫ Ω ϕ u t u v d x , ∀ v ∈ E 0 .

At the end of this section, we give a detailed description of our result.

Theorem 1.1

Assume 0 < h ∈ L 1 ( Ω ) , θ > 1 , and s , t ∈ ( 0 , 1 ) with 2 t + 4 s > 3 . Then, system (1) has a unique positive solution ( u ˆ , ϕ u ˆ t ) ∈ E 0 × E 0 if and only if there exists u * ∈ E 0 such that

(15) ∫ Ω h ( x ) ∣ u * ∣ 1 − θ d x < + ∞ .

2 Proof of Theorem 1.1

Since I is not well defined on E 0 when θ > 1 , then we define two suitable constrained sets

N * ≔ u ∈ E 0 ∣ ‖ u ‖ 2 + ∫ Ω ϕ u t ( x ) u 2 d x + ∫ Ω g ( x ) ∣ u ∣ 1 + q d x − ∫ Ω h ( x ) ∣ u ∣ 1 − θ d x = 0

and

N ≔ u ∈ E 0 ∣ ‖ u ‖ 2 + ∫ Ω ϕ u t ( x ) u 2 d x + ∫ Ω g ( x ) ∣ u ∣ 1 + q d x − ∫ Ω h ( x ) ∣ u ∣ 1 − θ d x ≥ 0 ,

which is useful for us to obtain the solution of problem (14). Next we will present properties of these two sets.

Lemma 2.1

Assume that (15) holds and θ > 1 , then the set N * is nonempty and N is an unbounded nonempty closed set in E 0 .

Proof

By (15), we have that there exists u ∈ E 0 such that ∫ Ω h ( x ) ∣ u ∣ 1 − θ d x < + ∞ . Since α > 0 , by Lemma 1.1 (4), we define

J u ( α ) ≔ I ( α u ) = α 2 2 ‖ u ‖ 2 + α 4 4 ∫ Ω ϕ u t ( x ) u 2 d x + α 1 + q 1 + q ∫ Ω g ( x ) ∣ u ∣ 1 + q d x − α 1 − θ 1 − θ ∫ Ω h ( x ) ∣ u ∣ 1 − θ d x .

Then, we can infer that lim α → 0 + J u ( α ) = + ∞ , lim α → + ∞ J u ( α ) = + ∞ and by a direct computation, we have that

J u ′ ( α ) = α ‖ u ‖ 2 + α 3 ∫ Ω ϕ u t ( x ) u 2 d x + α q ∫ Ω g ( x ) ∣ u ∣ 1 + q d x − α − θ ∫ Ω h ( x ) ∣ u ∣ 1 − θ d x .

Thus, we also infer that lim α → 0 + J u ′ ( α ) = − ∞ , lim α → + ∞ J u ′ ( α ) = + ∞ . Furthermore, we also have

J u ″ ( α ) = ‖ u ‖ 2 + 3 α 2 ∫ Ω ϕ u t ( x ) u 2 d x + q α q − 1 ∫ Ω g ( x ) ∣ u ∣ 1 + q d x + θ α − θ − 1 ∫ Ω h ( x ) ∣ u ∣ 1 − θ d x > 0 ,

which implies J u ′ ( α ) is nondecreasing for α > 0 . So, there is an unique α ( u ) > 0 such that J u ′ ( α ( u ) ) = 0 , and we have J u ′ ( α ) > 0 when α > α ( u ) , which shows t u ∈ N and α ( u ) u ∈ N * for t ≥ α ( u ) . Hence, we can obtain N * and N are nonempty. Moreover, it is clear that N is unbounded. As well as there exists a unique point α 0 ∈ ( 0 , ∞ ) such that J u ( α 0 ) ≤ J u ( α ) for all α > 0 .

Next, we will show that N is a closed set. First, we assume { u n } ⊂ N with u n → u in E 0 and ∫ Ω h ( x ) ∣ u n ∣ 1 − θ d x < + ∞ , which is sufficient to prove u ∈ N . Since

‖ u n ‖ 2 + ∫ Ω ϕ u n t ( x ) u n 2 d x + ∫ Ω g ( x ) ∣ u n ∣ 1 + q d x ≥ ∫ Ω h ( x ) ∣ u n ∣ 1 − θ d x .

It is simple to check J ( ∣ u ∣ ) = J ( u ) . Then, combining with Fatou’s lemma and Lemma 1.1, one has

‖ u ‖ 2 + ∫ Ω ϕ u t ( x ) u 2 d x + ∫ Ω g ( x ) ∣ u ∣ 1 + q d x ≥ liminf n → ∞ ∫ Ω h ( x ) ∣ u n ∣ 1 − θ d x ≥ ∫ Ω h ( x ) ∣ u ∣ 1 − θ d x ,

which ensures u ∈ N . □

Lemma 2.2

Let 0 < φ ∈ E 0 and u ∈ N * , then there exist ε > 0 and a C 1 function α : B ε → R + such that α ( ξ ) ( u + ξ φ ) ∈ N * when ‖ ξ φ ‖ < ε .

Proof

First, set G : R × R → R with

G ( α , ξ ) = α 1 + θ ‖ u + ξ φ ‖ 2 + α 3 + θ ∫ Ω ϕ u + ξ φ t ( x ) ∣ u + ξ φ ∣ 2 d x + α q + θ ∫ Ω g ( x ) ∣ u + ξ φ ∣ 1 + q d x − ∫ Ω h ( x ) ∣ u + ξ φ ∣ 1 − θ d x ,

where u ∈ N * . Consequently, by a simple calculation, one has

G α ( α , ξ ) = ( 1 + θ ) α θ ‖ u + ξ φ ‖ 2 + ( 3 + θ ) α 2 + θ ∫ Ω ϕ u + ξ φ t ( x ) ∣ u + ξ φ ∣ 2 d x + ( q + θ ) α q + θ − 1 ∫ Ω g ( x ) ∣ u + ξ φ ∣ 1 + q d x .

Since u ∈ N * , we can bring point ( 1 , 0 ) into the aforementioned equations to obtain

G ( 1 , 0 ) = ‖ u ‖ 2 + ∫ Ω ϕ u t ( x ) u 2 d x + ∫ Ω g ( x ) ∣ u ∣ 1 + q d x − ∫ Ω h ( x ) ∣ u ∣ 1 − θ d x = 0

and

G α ( 1 , 0 ) = ( 1 + θ ) ‖ u ‖ 2 + ( 3 + θ ) ∫ Ω ϕ u t ( x ) u 2 d x + ( q + θ ) ∫ Ω g ( x ) ∣ u ∣ 1 + q d x > 0 .

By applying the implicit function theorem, we can obtain a positive number ε and a C 1 function α = α ( ξ ) > 0 ( ξ ∈ R ) satisfying α ( 0 ) = 1 , α ( ξ ) ( u + ξ φ ) ∈ N * for ‖ ξ φ ‖ < ε . □

Next, we recall the Ekeland variational principle, which is very useful for us to prove Theorem 2.1.

Lemma 2.3

(Ekeland variational principle [27]) Assume that Z is a Banach space, and let Ψ ∈ C 1 ( Z , R ) be bounded below, v ∈ Z and ε , δ > 0 . If

Ψ ( v ) ≤ inf Z Ψ + ε ,

then there exists u ∈ Z such that

Ψ ( u ) ≤ inf Z Ψ + 2 ε , ∥ Ψ ′ ( u ) ∥ < 8 ε δ , ‖ u − v ‖ ≤ 2 δ .

Theorem 2.1

Assume that h ∈ L 1 ( Ω ) is a positive function and θ > 1 , then problem (14) has a unique positive solution u ˆ ∈ E 0 if and only if h satisfies (15).

Remark 2.1

Set

m = inf u ∈ N I ( u ) .

First, we need to prove that m can be achieved on manifold N * , that is, there is 0 < u ˆ ∈ N * such that I ( u ˆ ) = m . Next, we show that u ˆ is a solution of problem (14). Finally, the solution is unique.

Proof

The necessity is easy to prove. Then, we only prove the sufficiency condition. Since θ > 1 , we can immediately obtain

I ( u ) = 1 2 ‖ u ‖ 2 + 1 4 ∫ Ω ϕ u t ( x ) u 2 d x + 1 1 + q ∫ Ω g ( x ) ∣ u ∣ 1 + q d x − 1 1 − θ ∫ Ω h ( x ) ∣ u ∣ 1 − θ d x ≥ 1 2 ‖ u ‖ 2 ,

which shows that the coercivity and boundedness of I from below on E 0 . Therefore, the real number m = inf u ∈ N I ( u ) is well defined. We know that N is closed by Lemma 2.1. Then, by applying the Ekeland’s variational principle, it is easy to obtain that there is a sequence { u n } ⊂ N satisfying

I ( u n ) < m + 1 n ,
I ( u n ) − 1 n ‖ u n − u ‖ ≤ I ( u ) , ∀ u ∈ N .

It is simple to obtain I ( ∣ u ∣ ) = I ( u ) , which means that we can assume u n ≥ 0 in Ω . Since u n ∈ N , there holds

∫ Ω h ( x ) ∣ u n ∣ 1 − θ d x < + ∞ ,

which ensures that u n ( x ) > 0 a.e. x ∈ Ω . Obviously, it is easy to prove the boundedness of { u n } in E 0 . Then, we conclude that, up to a subsequence, there is 0 < u ˆ ∈ E 0 such that, as n → ∞ ,

(16) u n ⇀ u ˆ , in E 0 , u n → u ˆ , a.e. in Ω , u n → u ˆ , in L r ( Ω ) for 2 ≤ r < 2 s * .

Then, we can deduce ∫ Ω h ( x ) ∣ u ˆ ∣ 1 − θ d x < ∞ by Fatou’s lemma and u ˆ ( x ) > 0 a.e. x ∈ Ω . Now, we need to check that u ˆ ∈ N * and I ( u ˆ ) = m . We prove it in two cases.

Case 1. Let 0 ≤ φ ∈ E 0 and { u n } ⊂ N \ N * for n big enough.

Since θ > 1 , then ∣ ξ ∣ 1 − θ is nonincreasing for any ξ , which is enough to obtain that

h ( x ) ∣ u n ∣ 1 − θ > h ( x ) ∣ u n + ξ φ ∣ 1 − θ ,

where h > 0 and u n ( x ) > 0 a.e. x ∈ Ω , φ ≥ 0 and ξ > 0 , then we immediately have

∫ Ω h ( x ) ∣ u n ∣ 1 − θ d x ≥ ∫ Ω h ( x ) ∣ u n + ξ φ ∣ 1 − θ d x .

Furthermore, we obtain

‖ u n ‖ 2 + ∫ Ω ϕ u n t ( x ) u n 2 d x + ∫ Ω g ( x ) ∣ u n ∣ 1 + q d x > ∫ Ω h ( x ) ∣ u n ∣ 1 − θ d x ≥ ∫ Ω h ( x ) ∣ u n + ξ φ ∣ 1 − θ d x ,

where { u n } ⊂ N \ N * . Then, we choose a sufficiently small number ξ > 0 to make

‖ u n + ξ φ ‖ 2 + ∫ Ω ϕ u n + ξ φ t ( x ) ( u n + ξ φ ) 2 d x + ∫ Ω g ( x ) ∣ u n + ξ φ ∣ 1 + q d x > ∫ Ω h ( x ) ∣ u n + ξ φ ∣ 1 − θ d x ,

which implies that u n + ξ φ ∈ N . Then, according to (ii), we have

‖ u n − ( u n + ξ φ ) ‖ n ≥ I ( u n ) − I ( u n + ξ φ ) ,

that is,

ξ ‖ φ ‖ n ≥ ‖ u n ‖ 2 − ‖ u n + ξ φ ‖ 2 2 + ∫ Ω ϕ u n t ( x ) u n 2 − ϕ u n + ξ φ t ( x ) ( u n + ξ φ ) 2 d x 4 + ∫ Ω g ( x ) ∣ u n ∣ 1 + q − g ( x ) ∣ u n + ξ φ ∣ 1 + q d x 1 + q − ∫ Ω h ( x ) ∣ u n ∣ 1 − θ − h ( x ) ∣ u n + ξ φ ∣ 1 − θ d x 1 − θ .

Therefore, by dividing by ξ > 0 and taking the infimum limit as ξ → 0 , and then combining with Lemma 1.1 (5) and Fatou’s Lemma, we can obtain

‖ φ ‖ n + ⟨ u n , φ ⟩ + ∫ Ω ϕ u n t ( x ) u n φ d x + ∫ Ω g ( x ) ∣ u n ∣ q φ d x ≥ liminf ξ → 0 ∫ Ω h ( x ) ∣ u n − h ( x ) ∣ u n ∣ 1 − θ + ξ φ ∣ 1 − θ ξ ( 1 − θ ) d x ≥ ∫ Ω liminf ξ → 0 h ( x ) ∣ u n − h ( x ) ∣ u n ∣ 1 − θ + ξ φ ∣ 1 − θ ξ ( 1 − θ ) d x = ∫ Ω h ( x ) ∣ u n ∣ − θ φ d x .

Passing the aforementioned inequality to the infimum limit, let n → ∞ , by the property (3) of Lemma 1.1 and Fatou’s Lemma again, we deduce

(17) ⟨ u ˆ , φ ⟩ + ∫ Ω ϕ u ˆ t ( x ) u ˆ φ d x + ∫ Ω g ( x ) ∣ u ˆ ∣ q φ d x ≥ ∫ Ω h ( x ) ∣ u ˆ ∣ − θ φ d x , ∀ φ ∈ E 0 , φ ≥ 0 .

Choosing φ = u ˆ , one has u ˆ ∈ N . Then, according to Lemma 2.1, it is obvious that there is a unique positive constant number α ( u ˆ ) such that I ( α ( u ˆ ) u ˆ ) = min α > 0 I ( α u ) and α ( u ˆ ) u ˆ ∈ N * . Since u n ⇀ u ˆ in E 0 , by the weakly lower semi-continuity, we arrive at

liminf n → ∞ ‖ u n ‖ ≥ ‖ u ˆ ‖ .

While, from θ > 1 , 0 < q < 1 and Fatou’s lemma again, we can immediately infer that

− 1 1 − θ ∫ Ω h ( x ) ∣ u ˆ ∣ 1 − θ d x ≤ liminf n → ∞ − 1 1 − θ ∫ Ω h ( x ) ∣ u n ∣ 1 − θ d x .

By means of calculations, we have that

m = lim n → ∞ I ( u n ) = liminf n → ∞ ‖ u n ‖ 2 2 + 1 4 ∫ Ω ϕ u n t ( x ) u n 2 d x + 1 1 + q ∫ Ω g ( x ) ∣ u n ∣ 1 + q d x − 1 1 − θ ∫ Ω h ( x ) ∣ u n ∣ 1 − θ d x = liminf n → ∞ ‖ u n ‖ 2 2 − 1 1 − θ ∫ Ω h ( x ) ∣ u n ∣ 1 − θ d x + 1 4 ∫ Ω ϕ u ˆ t ( x ) u ˆ 2 d x + 1 1 + q ∫ Ω g ( x ) ∣ u ˆ ∣ 1 + q d x ≥ ‖ u ˆ ‖ 2 2 − 1 1 − θ ∫ Ω h ( x ) ∣ u ˆ ∣ 1 − θ d x + 1 4 ∫ Ω ϕ u ˆ t ( x ) u ˆ 2 d x + 1 1 + q ∫ Ω g ( x ) ∣ u ˆ ∣ 1 + q d x = I ( u ˆ ) ≥ I ( α ( u ˆ ) u ˆ ) ≥ m ,

which provides

(18) I ( u ˆ ) = m , u ˆ ∈ N * ( this is α ( u ˆ ) = 1 ) .

Case 2. Assume that there is a subsequence of { u n } , which is still denoted as { u n } , such that { u n } ⊂ N * . Let 0 < φ ∈ E 0 . By applying Lemma 2.2 with u = u n and choosing ξ > 0 sufficiently small, we can obtain that there is a sequence of C 1 functions α n = α n ( ξ ) such that α n ( 0 ) = 1 and α n ( ξ ) ( u n + ξ φ ) ∈ N * . Then, one has

(19) α n 2 ( ξ ) ‖ u n + ξ φ ‖ 2 + α n 4 ( ξ ) ∫ Ω ϕ u n + ξ φ t ( x ) ( u n + ξ φ ) 2 d x + α n 1 + q ( ξ ) ∫ Ω g ( x ) ∣ u n + ξ φ ∣ 1 + q d x − α n 1 − θ ( ξ ) ∫ Ω h ( x ) ∣ u n + ξ φ ∣ 1 − θ d x = 0

and

(20) ‖ u n ‖ 2 + ∫ Ω ϕ u n t ( x ) u n 2 d x + ∫ Ω g ( x ) ∣ u n ∣ 1 + q d x − ∫ Ω f ( x ) ∣ u n ∣ 1 − θ d x = 0 .

By combining with (19) and (20), we arrive at

( α n 2 ( ξ ) − 1 ) ‖ u n + ξ φ ‖ 2 + ‖ u n + ξ φ ‖ 2 − ‖ u n ‖ 2 + ( α n 4 ( ξ ) − 1 ) ∫ Ω ϕ u n + ξ φ t ( x ) ( u n + ξ φ ) 2 d x + ∫ Ω ϕ u n + ξ φ t ( x ) ( u n + ξ φ ) 2 d x − ∫ Ω ϕ u n t ( x ) u n 2 d x + ( α n 1 + q ( ξ ) − 1 ) ∫ Ω g ( x ) ∣ u n + ξ φ ∣ 1 + q d x + ∫ Ω g ( x ) ∣ u n + ξ φ ∣ 1 + q d x − ∫ Ω g ( x ) ∣ u n ∣ 1 + q d x − ( α n 1 − θ ( ξ ) − 1 ) ∫ Ω h ( x ) ∣ u n + ξ φ ∣ 1 − θ d x + ∫ Ω h ( x ) ∣ u n + ξ φ ∣ 1 − θ − h ( x ) ∣ u n ∣ 1 − θ d x = 0 .

By dividing by ξ > 0 in both sides of the equation, we have that

α n ( ξ ) − 1 ξ ( α n ( ξ ) + 1 ) ‖ u n + ξ φ ‖ 2 + α n 4 ( ξ ) − 1 α n ( ξ ) − 1 ∫ Ω ϕ u n + ξ φ t ( x ) ( u n + ξ φ ) 2 d x + α n 1 + q ( ξ ) − 1 α n ( ξ ) − 1 ∫ Ω g ( x ) ∣ u n + ξ φ ∣ 1 + q d x − α n 1 − θ ( ξ ) − 1 α n ( ξ ) − 1 ∫ Ω h ( x ) ∣ u n + ξ φ ∣ 1 − θ d x + ‖ u n + ξ φ ‖ 2 − ‖ u n ‖ 2 ξ + ∫ Ω ϕ u n + ξ φ t ( x ) ( u n + ξ φ ) 2 − ϕ u n t ( x ) u n 2 ξ d x + ∫ Ω g ( x ) ∣ u n + ξ φ ∣ 1 + q − g ( x ) ∣ u n ∣ 1 + q ξ d x ≤ 0 ,

where θ > 1 . Letting ξ → 0 , we can deduce

ω n , 0 ′ 2 ‖ u n ‖ 2 + 4 ∫ Ω ϕ u n t ( x ) u n 2 d x + ( 1 + q ) ∫ Ω g ( x ) ∣ u n ∣ 1 + q d x − ( 1 − θ ) ∫ Ω h ( x ) ∣ u n ∣ 1 − θ d x + 2 ⟨ ∇ u n , ∇ φ ⟩ + 4 ∫ Ω ϕ u n t ( x ) u n φ d x + ( 1 + q ) ∫ Ω g ( x ) ∣ u n ∣ q φ d x ≤ 0 ,

where we have set ω n , 0 ′ = lim ξ → 0 α n ( ξ ) − 1 ξ ∈ [ − ∞ , + ∞ ] . We can choose ξ k → 0 when the limit does not exist, then lim k → ∞ α n ( ξ k ) − 1 ξ k ∈ [ − ∞ , + ∞ ] . Since { u n } is bounded in E 0 , then we can infer that ω n , 0 ′ ≠ + ∞ by the aforementioned inequality. As mentioned earlier, we also know that ω n , 0 ′ cannot go to + ∞ as n → ∞ . Then ω n , 0 ′ is uniformly bounded from above for all n .

In addition, to prove ω n , 0 ′ ≥ C for n large, we come back condition (ii). By combining with (19) and (20), we can have that

∣ α n ( ξ ) − 1 ∣ ‖ u n ‖ n + ξ α n ( ξ ) ‖ φ ‖ n ≥ ‖ u n − α n ( ξ ) ( u n + ξ φ ) ‖ n ≥ I ( u n ) − I [ α n ( ξ ) ( u n + ξ φ ) ] = ‖ u n ‖ 2 − ‖ α n ( ξ ) ( u n + ξ φ ) ‖ 2 2 + ∫ Ω ϕ u n t u n 2 d x − ∫ Ω ϕ α n ( ξ ) ( u n + ξ φ ) t ( x ) [ α n ( ξ ) ( u n + ξ φ ) ] 2 d x 4 + ∫ Ω g ( x ) ∣ u n ∣ 1 + q d x − ∫ Ω g ( x ) ∣ α n ( ξ ) ( u n + ξ φ ) ∣ 1 + q d x 1 + q − ∫ Ω h ( x ) ∣ u n ∣ 1 − θ d x − ∫ Ω h ( x ) ∣ α n ( ξ ) ( u n + ξ φ ) ∣ 1 − θ d x 1 − θ = ‖ u n ‖ 2 − ‖ α n ( ξ ) ( u n + ξ φ ) ‖ 2 2 + ∫ Ω ϕ u n t u n 2 d x − ∫ Ω ϕ α n ( ξ ) ( u n + ξ φ ) t ( x ) [ α n ( ξ ) ( u n + ξ φ ) ] 2 d x 4 + ∫ Ω g ( x ) ∣ u n ∣ 1 + q d x − ∫ Ω g ( x ) ∣ α n ( ξ ) ( u n + ξ φ ) ∣ 1 + q d x 1 + q − 1 1 − θ ‖ u n ‖ 2 + ∫ Ω ϕ u n t ( x ) u n 2 d x + ∫ Ω g ( x ) ∣ u n ∣ 1 + q d x − [ α n 2 ( ξ ) ‖ u n + ξ φ ‖ 2 + ∫ Ω ϕ α n ( ξ ) ( u n + ξ φ ) t ( x ) [ α n ( ξ ) ( u n + ξ φ ) ] 2 d x + ∫ Ω g ( x ) ∣ α n ( ξ ) ( u n + ξ φ ) ∣ 1 + q d x = 1 2 − 1 1 − θ ( ‖ u n ‖ 2 − ∣ α n ( ξ ) ∣ 2 ‖ ( u n + ξ φ ) ‖ 2 ) + 1 4 − 1 1 − θ ∫ Ω ϕ u n t u n 2 d x − ∫ Ω ϕ α n ( ξ ) ( u n + ξ φ ) t [ α n ( ξ ) ( u n + ξ φ ) ] 2 d x + 1 1 + q − 1 1 − θ ∫ Ω g ( x ) ∣ u n ∣ 1 + q d x − ∫ Ω g ( x ) ∣ α n ( ξ ) ( u n + ξ φ ) ∣ 1 + q d x = 1 + θ 2 ( 1 − θ ) [ ( α n ( ξ ) 2 − 1 ) ‖ ( u n + ξ φ ) ‖ 2 + ( ‖ u n + ξ φ ‖ 2 − ‖ u n ‖ 2 ) ] + 3 + θ 4 ( 1 − θ ) ( α n ( ξ ) 4 − 1 ) ∫ Ω ϕ u n + ξ φ t ( u n + ξ φ ) 2 d x + ∫ Ω ϕ u n + ξ φ t ( u n + ξ φ ) 2 − ϕ u n t u n 2 d x + q + θ ( 1 + q ) ( 1 − θ ) ( α n ( ξ ) 1 + q − 1 ) ∫ Ω g ( x ) ∣ u n + ξ φ ∣ 1 + q d x + ∫ Ω g ( x ) ∣ u n + ξ φ ∣ 1 + q − g ( x ) ∣ u n ∣ 1 + q d x .

By dividing the inequality simultaneously by ξ > 0 again, and then letting ξ → 0 , we arrive at

∣ ω n , 0 ′ ∣ ‖ u n ‖ n + lim ξ → 0 α n ( ξ ) ‖ φ ‖ n ≥ 1 + θ 1 − θ [ ω n , 0 ′ ‖ u n ‖ 2 + ⟨ u n , φ ⟩ ] + 3 + θ 1 − θ ω n , 0 ′ ∫ Ω ϕ u n t u n 2 d x + ∫ Ω ϕ u n t u n φ d x + q + θ 1 − θ ω n , 0 ′ ∫ Ω g ( x ) ∣ u n ∣ 1 + q d x + ∫ Ω g ( x ) ∣ u n ∣ q φ d x .

Then, we have ω n , 0 ′ ≠ − ∞ since θ > 1 . As mentioned earlier, we also know that ω n , 0 ′ cannot go to − ∞ as n → ∞ , this is to say that ω n , 0 ′ > C is uniform for some positive constant C . Therefore, we have ∣ ω n , 0 ′ ∣ ≤ C for n large.

Next, we apply condition (ii) again to obtain

α n ( ξ ) − 1 ξ ‖ u n ‖ n + α n ( ξ ) ‖ φ ‖ n ≥ 1 ξ ⋅ ‖ u n − α n ( ξ ) ( u n + ξ φ ) ‖ n ≥ 1 ξ ( I ( u n ) − I [ α n ( ξ ) ( u n + ξ φ ) ] ) = ‖ u n ‖ 2 − ‖ α n ( ξ ) ( u n + ξ φ ) ‖ 2 2 ξ + ∫ Ω ϕ u n t u n 2 d x − ∫ Ω ϕ α n ( ξ ) ( u n + ξ φ ) t ( x ) [ α n ( ξ ) ( u n + ξ φ ) ] 2 d x 4 ξ + ∫ Ω g ( x ) ∣ u n ∣ 1 + q d x − ∫ Ω g ( x ) ∣ α n ( ξ ) ( u n + ξ φ ) ∣ 1 + q d x ξ ( 1 + q ) − ∫ Ω h ( x ) ∣ u n ∣ 1 − θ d x − ∫ Ω h ( x ) ∣ α n ( ξ ) ( u n + ξ φ ) ∣ 1 − θ d x ξ ( 1 − θ ) = − 1 2 ξ [ ( α n ( ξ ) 2 − 1 ) ‖ ( u n + ξ φ ) ‖ 2 + ( ‖ u n + ξ φ ‖ 2 − ‖ u n ‖ 2 ) ] − 1 4 ξ ( α n ( ξ ) 4 − 1 ) ∫ Ω ϕ u n + ξ φ t ( u n + ξ φ ) 2 d x + ∫ Ω ϕ u n + ξ φ t ( u n + ξ φ ) 2 − ϕ u n t u n 2 d x − 1 ( 1 + q ) ξ ( α n ( ξ ) 1 + q − 1 ) ∫ Ω g ( x ) ∣ u n + ξ φ ∣ 1 + q d x + ∫ Ω g ( x ) ∣ u n + ξ φ ∣ 1 + q − g ( x ) ∣ u n ∣ 1 + q d x + 1 ( 1 − θ ) ξ ( α n ( ξ ) 1 − θ − 1 ) ∫ Ω h ( x ) ∣ u n + ξ φ ∣ 1 − θ d x + ∫ Ω h ( x ) ∣ u n + ξ φ ∣ 1 − θ − h ( x ) ∣ u n ∣ 1 − θ d x = − α n ( ξ ) − 1 ξ α n ( ξ ) + 1 2 ‖ ( u n + ξ φ ) ‖ 2 + α n ( ξ ) 4 − 1 4 ( α n ( ξ ) − 1 ) ∫ Ω ϕ u n + ξ φ t ( u n + ξ φ ) 2 d x + α n ( ξ ) 1 + q − 1 ( 1 + q ) ( α n ( ξ ) − 1 ) ∫ Ω g ( x ) ∣ u n + ξ φ ∣ 1 + q d x − α n ( ξ ) 1 − θ − 1 ( 1 − θ ) ( α n ( ξ ) − 1 ) ∫ Ω h ( x ) ∣ u n + ξ φ ∣ 1 − θ d x − ‖ u n + ξ φ ‖ 2 − ‖ u n ‖ 2 2 ξ + ∫ Ω ϕ u n + ξ φ t ( u n + ξ φ ) 2 − ϕ u n t u n 2 4 ξ d x + ∫ Ω g ( x ) ∣ u n + ξ φ ∣ 1 + q − g ( x ) ∣ u n ∣ 1 + q ( 1 + q ) ξ d x − ∫ Ω h ( x ) ∣ u n − h ( x ) ∣ u n ∣ 1 − θ + ξ φ ∣ 1 − θ ( 1 − θ ) ξ d x .

Letting ξ → 0 in the aforementioned inequality again, by Fatou’s lemma, we have the following estimate:

∣ ω n , 0 ′ ∣ ‖ u n ‖ n + lim ξ → 0 α n ( ξ ) ‖ φ ‖ n ≥ − ω n , 0 ′ ‖ u n ‖ 2 + ∫ Ω ϕ u n t u n 2 d x + ∫ Ω g ( x ) ∣ u n ∣ 1 + q d x − ∫ Ω h ( x ) ∣ u n ∣ 1 − θ d x − ⟨ u n , φ ⟩ + ∫ Ω ϕ u n t u n φ d x + ∫ Ω g ( x ) ∣ u n ∣ q φ d x + ∫ Ω liminf ξ → 0 h ( x ) ∣ u n + ξ φ ∣ 1 − θ − h ( x ) ∣ u n ∣ 1 − θ ( 1 − θ ) ξ d x = − ⟨ u n , φ ⟩ + ∫ Ω ϕ u n t u n φ d x + ∫ Ω g ( x ) ∣ u n ∣ q φ d x + ∫ Ω h ( x ) ∣ u n ∣ − θ φ d x ( since u n ∈ N * ) .

Since ∣ ω n , 0 ′ ∣ ≤ C for n large, then we can apple Fatou’s lemma again in turn to obtain the integrability of h ( x ) u ˆ − θ φ and

(21) ⟨ u ˆ , φ ⟩ + ∫ Ω ϕ u ˆ t u ˆ φ d x + ∫ Ω g ( x ) ∣ u ˆ ∣ q φ d x − ∫ Ω h ( x ) ∣ u ˆ ∣ − θ φ d x ≥ 0 .

Particularly, by choosing φ = u ˆ in (21), we have u ˆ ∈ N . Moreover, following as in Case 1 of the prove, we can derive

(22) I ( u ˆ ) = m , u ˆ ∈ N * ( this is α ( u ˆ ) = 1 ) .

We next need to prove that problem (14) has a solution u ˆ . From (17), (18), (21), and (22), we can obtain the result in all cases that u ˆ ∈ N * and

(23) ⟨ u ˆ , φ ⟩ + ∫ Ω ϕ u ˆ t u ˆ φ d x + ∫ Ω g ( x ) ∣ u ˆ ∣ q φ d x − ∫ Ω h ( x ) ∣ u ˆ ∣ − θ φ d x ≥ 0 , ∀ φ ∈ E 0 , φ ≥ 0 .

For any ψ ∈ E 0 , by defining Ω ε = { x ∈ R 3 : u ˆ + ε ψ ≤ 0 } and choosing Ψ ε = u ˆ + ε ψ in (23), we obtain

0 ≤ 1 ε ⟨ u ˆ , Ψ ε + ⟩ + ∫ Ω [ ϕ u ˆ t u ˆ Ψ ε + + g ( x ) ∣ u ˆ ∣ q Ψ ε + − h ( x ) u ˆ − θ Ψ ε + ] d x = 1 ε ⟨ u ˆ , Ψ ε ⟩ − ⟨ u ˆ , Ψ ε − ⟩ + ∫ Ω − ∫ Ω ε [ ϕ u ˆ t u ˆ Ψ ε + g ( x ) u ˆ q Ψ ε − h ( x ) u ˆ − θ Ψ ε ] d x = 1 ε ‖ u ˆ ‖ 2 + ∫ Ω ϕ u ˆ t u ˆ 2 + ∫ Ω g ( x ) u ˆ 1 + q d x − ∫ Ω h ( x ) u ˆ 1 − θ d x + ⟨ u ˆ , ψ ⟩ + ∫ Ω ϕ u ˆ t u ˆ ψ d x + ∫ Ω g ( x ) u ˆ q ψ d x − ∫ Ω h ( x ) u ˆ − θ ψ d x − 1 ε ⟨ u ˆ , Ψ ε − ⟩ + ∫ Ω ε [ ϕ u ˆ t u ˆ ( u ˆ + ε ψ ) + g ( x ) u ˆ q ( u ˆ + ε ψ ) − h ( x ) u ˆ − θ ( u ˆ + ε ψ ) ] d x ≤ ⟨ u ˆ , ψ ⟩ + ∫ Ω ϕ u ˆ t u ˆ ψ d x + ∫ Ω g ( x ) u ˆ q ψ d x − ∫ Ω h ( x ) u ˆ − θ ψ d x − 1 ε ⟨ u ˆ , Ψ ε − ⟩ + ε ∫ Ω ε [ ϕ u ˆ t u ˆ ψ + g ( x ) u ˆ q ψ ] d x .

Since meas { u ˆ + ε ψ ≤ 0 } → 0 as ε → 0 , we can deduce

∫ Ω ε [ ϕ u ˆ t u ˆ ψ + g ( x ) u ˆ q ψ ] d x → 0 .

Letting

Θ ε ( x , y ) ≔ ( Ψ ε − ( x ) − Ψ ε − ( y ) ) ( u ˆ ( x ) − u ˆ ( y ) ) ∣ x − y ∣ 3 + 2 s .

Since the fractional kernel has the symmetry, then we can infer that

(24) ⟨ u ˆ , Ψ ε − ⟩ = ∬ Ω ε × Ω ε Θ ε ( x , y ) d x d y + 2 ∬ Ω ε × Ω ε c Θ ε ( x , y ) d x d y ≤ − ε ∬ Ω ε × Ω ε Θ ( x , y ) d x d y + 2 ∬ Ω ε × Ω ε c Θ ( x , y ) d x d y ≤ 2 ε ∬ Ω ε × R 3 ∣ Θ ( x , y ) ∣ d x d y ,

where we set

Θ ( x , y ) ≔ ( ψ ( x ) − ψ ( y ) ) ( u ˆ ( x ) − u ˆ ( y ) ) ∣ x − y ∣ 3 + 2 s .

Clearly Θ ∈ L 1 ( R 3 × R 3 ) , then for σ > 0 , there is R σ big enough such that

∫ supp ψ ∫ R 3 \ B R σ ∣ Θ ( x , y ) ∣ d x d y < σ 2 .

Also, according to the definition of Ω ε , we can know that Ω ε ⊂ supp ψ and ∣ Ω ε × B R σ ∣ → 0 as ε → 0 + . Then, there exist positive numbers ε σ and δ σ such that

∣ Ω ε × B R σ ∣ < δ σ and ∬ Ω ε × B R σ ∣ Θ ( x , y ) ∣ d x d y < σ 2 ,

where 0 < ε < ε σ . Therefore, we have

∬ Ω ε × R 3 ∣ Θ ( x , y ) ∣ d x d y < σ .

As ε → 0 + , we can infer that

lim ε → 0 + ∬ Ω ε × R 3 ∣ Θ ( x , y ) ∣ d x d y = 0 .

Hence, from (24), we have

lim ε → 0 + 1 ε ⟨ u ˆ , Ψ ε − ⟩ = 0 .

Then, we conclude that

⟨ u ˆ , ψ ⟩ + ∫ Ω ϕ u ˆ t u ˆ ψ d x + ∫ Ω g ( x ) u ˆ q ψ d x − ∫ Ω h ( x ) u ˆ − θ ψ d x ≥ 0

for all ψ ∈ E 0 . Hence, replacing − ψ by ψ , this inequality also holds. Then, we have

(25) ⟨ u ˆ , ψ ⟩ + ∫ Ω ϕ u ˆ t u ˆ ψ d x + ∫ Ω g ( x ) u ˆ q ψ d x − ∫ Ω h ( x ) u ˆ − θ ψ d x = 0

for all ψ ∈ E 0 . Thus, problem (14) has a positive solution u ˆ .

We next show uniqueness of the solution. First, we assume that problem (14) has another solution v ˆ . By using (25), we have

(26) ⟨ u ˆ , u ˆ − v ˆ ⟩ + ∫ Ω ϕ u ˆ t u ˆ ( u ˆ − v ˆ ) d x + ∫ Ω g ( x ) u ˆ q ( u ˆ − v ˆ ) d x − ∫ Ω h ( x ) u ˆ − θ ( u ˆ − v ˆ ) d x = 0

and

(27) ⟨ v ˆ , u ˆ − v ˆ ⟩ + ∫ Ω ϕ v ˆ t v ˆ ( u ˆ − v ˆ ) d x + ∫ Ω g ( x ) v ˆ q ( u ˆ − v ˆ ) d x − ∫ Ω h ( x ) v ˆ − θ ( u ˆ − v ˆ ) d x = 0 .

Then, we can infer

(28) ‖ u ˆ − v ˆ ‖ + ∫ Ω ( ϕ u ˆ t u ˆ − ϕ v ˆ t v ˆ ) ( u ˆ − v ˆ ) d x + ∫ Ω g ( x ) ( u ˆ q − v ˆ q ) ( u ˆ − v ˆ ) d x − ∫ Ω h ( x ) ( u ˆ − θ − v ˆ − θ ) ( u ˆ − v ˆ ) d x = 0 .

It is simple to obtain the basic inequalities

( m q − n q ) ( m − n ) ≥ 0 , ( m − θ − n − θ ) ( m − n ) ≤ 0 , ∀ m , n > 0 ,

where 0 < q < 1 and θ > 1 . Then, we have

∫ Ω h ( x ) ( u ˆ − θ − v ˆ − θ ) ( u ˆ − v ˆ ) d x ≤ 0 , ∫ Ω g ( x ) ( u ˆ q − v ˆ q ) ( u ˆ − v ˆ ) d x ≥ 0 .

Since

( − Δ ) t ϕ u ˆ t = u ˆ 2 , ( − Δ ) t ϕ v ˆ t = v ˆ 2 ,

we have

( − Δ ) t ( ϕ u ˆ t − ϕ v ˆ t ) = u ˆ 2 − v ˆ 2 ,

that is,

‖ ϕ u ˆ t − ϕ v ˆ t ‖ 2 = ∫ Ω ( ϕ u ˆ t − ϕ v ˆ t ) ( u ˆ 2 − v ˆ 2 ) d x .

We can obtain that

∫ Ω ( ϕ u ˆ t u ˆ − ϕ v ˆ t v ˆ ) ( u ˆ − v ˆ ) d x = ∫ Ω ( ϕ u ˆ t u ˆ 2 + ϕ v ˆ t v ˆ 2 − ϕ v ˆ t v ˆ u ˆ − ϕ u ˆ t u ˆ v ˆ ) d x = ∫ Ω [ ϕ u ˆ t u ˆ 2 + ϕ v ˆ t v ˆ 2 − ( ϕ v ˆ t + ϕ u ˆ t ) u ˆ v ˆ ] d x ≥ ∫ Ω ϕ u ˆ t u ˆ 2 + ϕ v ˆ t v ˆ 2 − ( ϕ v ˆ t + ϕ u ˆ t ) u ˆ 2 + v ˆ 2 2 d x = 1 2 ∫ Ω ( ϕ u ˆ t u ˆ 2 + ϕ v ˆ t v ˆ 2 − ϕ u ˆ t v ˆ 2 − ϕ v ˆ t u ˆ 2 ) d x = 1 2 ∫ Ω ( ϕ u ˆ t − ϕ v ˆ t ) ( u ˆ 2 − v ˆ 2 ) d x = 1 2 ‖ ϕ u ˆ t − ϕ v ˆ t ‖ 2 ≥ 0 .

Consequently, by using (28), we have that ‖ u ˆ − v ˆ ‖ ≤ 0 . Then we can deduce ‖ u ˆ − v ˆ ‖ = 0 , which means u ˆ = v ˆ . Therefore, problem (14) has the unique positive solution u ˆ .□

Proof of Theorem 1.1

By Theorem 2.1, ( u ˆ , ϕ u ˆ ) ∈ E 0 × E 0 is the unique positive solution of system (1) if and only if (15) holds. So, the proof of Theorem 1.1 is easily obtained.□

Acknowledgements

The authors would like to thank Professor Giovanni Molica Bisci for valuable comments and suggestions on improving this article.

Funding information: This work was supported by the National Natural Science Foundation of China (No. 12161038 and 12301584), the Natural Science Foundation of Jiangxi Province (20232BAB201009), and the Science and Technology Project Foundation of the Department of Education (GJJ2400901).
Author contributions: All authors have accepted responsibility for the entire content of this manuscript and consented to its submission to the journal, reviewed all the results, and approved the final version of the manuscript. LW provided the idea for the study and led the implementation review and revision of the manuscript. RC and QZ proposed research ideas and writing original draft preparation. JW checked original draft preparation and FL provided the funding support. All authors have read and agreed to the published version of the manuscript.
Conflict of interest: The authors state no conflict of interest.
Data availability statement: Data sharing is not applicable to this article as no datasets were generated or analyzed during the current study.

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Received: 2022-03-09

Revised: 2023-11-21

Accepted: 2024-09-23

Published Online: 2024-12-31

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fractional Schrödinger-Poisson system; strong singularity; uniqueness; positive solutions

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