On a fractional Schrödinger-Poisson system with strong singularity

Theorem 1.1

Assume λ > 0 , γ > 1 and ( V 1 ) , ( V 2 ) , ( f 1 ) hold. Then system ( SP λ ) has a unique positive solution u λ ∈ E if and only if there exists u 0 ∈ E satisfying

Theorem 1.2

Assume λ > 0 , γ > 1 and ( V 1 ) , ( V 2 ) hold, further suppose f 1 , f 2 ∈ L 2 1 + γ ( R 3 ) are two nonnegative functions satisfying ∫ R 3 f i ∣ u 0 ∣ 1 − γ d x < + ∞ , i = 1 , 2 , then f 1 ≥ f 2 implies u 1 ≥ u 2 , where u 1 , u 2 are two solutions of system ( SP λ ) corresponding to f 1 , f 2 and provided by Theorem 1.1, respectively.

Theorem 1.3

Assume γ > 1 and ( V 1 ) , ( V 2 ) hold, further suppose that f satisfies ( f 1 ) and (1.6). For 0 < λ 1 ≤ λ 2 , we have u λ 1 ≥ u λ 2 , where u λ 1 , u λ 2 are two solutions of system ( SP λ ) corresponding to λ 1 , λ 2 and provided by Theorem 1.1, respectively.

Theorem 1.4

Assume γ > 1 and ( V 1 ) , ( V 2 ) , ( f 1 ) hold. For any sequence { λ n } ⊂ ( 0 , 1 ) with λ n → 0 , the sequence { u λ n } of positive solutions provided by Theorem 1.1  strongly converges to w 0 in E , where w 0 is the unique positive solution of

Lemma 2.1

Assume 0 < s < 1 and ( V 1 ) , ( V 2 ) hold. Then the embedding E ↪ L p ( R 3 ) is continuous for p ∈ [ 2 , 2 s ∗ ] and is compact for p ∈ [ 2 , 2 s ∗ ) . Hence, there are constants C p > 0 such that ‖ u ‖ p ≤ C p ‖ u ‖ E , ∀ u ∈ E , p ∈ [ 2 , 2 s ∗ ] .

Lemma 2.2

For 4 s + 2 t > 3 and any u ∈ H s ( R 3 ) , the following statements hold

1. ϕ u t : H s ( R 3 ) → D t , 2 ( R 3 ) is continuous and maps bounded sets into bounded sets;

2. ϕ u t ≥ 0 . Moreover, ϕ u t > 0 if u ≢ 0 ;

3. For u ∈ E , we have ‖ ϕ u t ‖ D t , 2 ( R 3 ) 2 = ∫ R 3 ϕ u t u 2 d x ≤ C ‖ u ‖ E 4 , where C is positive and independent of u ;

4. ϕ τ u t = τ 2 ϕ u t , ∀ τ ∈ R ;

5. Suppose u n ⇀ u in E , then

   ∫ R 3 ϕ u n t u n v d x → ∫ R 3 ϕ u t u v d x , ∀ v ∈ E

   and

   ∫ R 3 ϕ u n t u n 2 d x → ∫ R 3 ϕ u t u 2 d x ;

6. For u , v ∈ H s ( R 3 ) , ∫ R 3 ( ϕ u t u − ϕ v t v ) ( u − v ) d x ≥ 1 2 ‖ ϕ u t − ϕ v t ‖ D t , 2 ( R 3 ) 2 .

Proof of Theorem 1.1

(Necessity) Assume u is a solution to ( SP λ ) , that is u > 0 and fulfils (1.5). Selecting v = u in (1.5), one then has

(Sufficiency) This portion will be divided into six steps.

Step 1. N i ( λ ) ≠ ∅ , i = 1 , 2 .

Taking u ∈ E satisfying ∫ R 3 f ( x ) ∣ u ∣ 1 − γ d x < + ∞ , we have

Define g ( η ) = η d I λ ( η u ) d η , we have

Obviously, g ( η ) is increasing on ( 0 , + ∞ ) together with lim η → 0 + g ( η ) = − ∞ and lim η → + ∞ g ( η ) = + ∞ . Thus, there exists a unique η ( u ) > 0 such that I λ ( η ( u ) u ) = min η > 0 I λ ( η u ) and g ( η ( u ) ) = 0 , i.e.

which means that η ( u ) u ∈ N 2 ( λ ) . Moreover, (1.6) reveals the existence of η ( u 0 ) > 0 satisfying η ( u 0 ) u 0 ∈ N 2 ( λ ) ⊂ N 1 ( λ ) . Therefore, N i ( λ ) ≠ ∅ , i = 1 , 2 for any λ > 0 .

Step 2. N 1 ( λ ) is unbounded and closed in E with ‖ u ‖ E ≥ C 1 , ∀ u ∈ N 1 ( λ ) for some suitable constant C 1 > 0 .

By Step 1, η u 0 ∈ N 1 ( λ ) for every η ≥ η ( u 0 ) , hence N 1 ( λ ) is an unbounded set in E . Fatou’s lemma and Lemma 2.2 ( v ) show that N 1 ( λ ) is closed directly. Next, we want to prove that ‖ u ‖ E ≥ C 1 , ∀ u ∈ N 1 ( λ ) for some suitable constant C 1 > 0 . If not, there exists a sequence { u n } ⊂ N 1 ( λ ) with u n → 0 in E . Considering u n ∈ N 1 ( λ ) and γ > 1 , it follows from Lemma 2.2 ( v ) and the reverse form of Hölder’s inequality that (together with u n ≢ 0 since γ > 1 ).

This together with ∫ R 3 f 2 1 + γ d x 1 + γ 2 > 0 results in ‖ u n ‖ 2 2 → + ∞ , which is impossible. Hence, ‖ u ‖ E ≥ C 1 , ∀ u ∈ N 1 ( λ ) for some C 1 > 0 .

Step 3. Properties of the minimizing sequence { u n } .

By Step 2, For every u ∈ N 1 ( λ ) , we have ‖ u ‖ E ≥ C 1 . Furthermore, it follows from γ > 1 , λ > 0 , Lemma 2.2 (ii) and (2.2) that

which implies that I λ ( u ) is coercive and bounded from below on N 1 ( λ ) . Therefore, inf N 1 ( λ ) I λ exists. Thus, we apply the Ekeland variational principle to get a minimizing sequence { u n } ⊂ N 1 ( λ ) such that:

1. I λ ( u n ) < inf N 1 ( λ ) I λ + 1 n ;

2. I λ ( z ) ≥ I λ ( u n ) − 1 n ‖ u n − z ‖ E , ∀ z ∈ N 1 ( λ ) .

The coercivity of I λ on N 1 ( λ ) implies the existence of a constant C 2 > 0 such that ‖ u n ‖ E ≤ C 2 . To sum up, C 1 ≤ ‖ u n ‖ E ≤ C 2 . So there exists u λ ∈ E such that, going if necessary to a subsequence, we have

Since I λ ( ∣ u ∣ ) ≤ I λ ( u ) , one can assume u n ≥ 0 , then u λ ≥ 0 . Using Lemma 2.2 (iii), { u n } ⊂ N 1 ( λ ) and Fatou’s lemma, we further get ∫ R 3 f u λ 1 − γ d x < + ∞ which implies u λ > 0 a.e. in R 3 .

Step 4. u λ ∈ N 2 ( λ ) , inf N 1 ( λ ) I λ = I λ ( u λ ) and

The above result will be proved by considering two situations on whether { u n } ⊂ N 1 ( λ ) ⧹ N 2 ( λ ) or { u n } ⊂ N 2 ( λ ) .

Case 1. { u n } ⊂ N 1 ( λ ) ⧹ N 2 ( λ ) for n large enough.

Fix 0 ≤ ψ ∈ E , since f is nonnegative, we can derive from { u n } ⊂ N 1 ( λ ) ⧹ N 2 ( λ ) and γ > 1 that

Thus, there exists η > 0 sufficiently small satisfying

so u n + η ψ ∈ N 1 ( λ ) . In virtue of condition (2) by choosing z = u n + η ψ , one has

Dividing by η > 0 and letting η → 0 + , then we deduce from Fatou’s lemma that

Letting n → ∞ , one can obtain from Fatou’s lemma and Lemma 2.2 ( v ) that

Choosing ψ = u λ in (3.2) leads to u λ ∈ N 1 ( λ ) , ∫ R 3 f u λ 1 − γ d x < + ∞ . Then, by Step 1, there exists a unique η ( u λ ) > 0 satisfying η ( u λ ) u λ ∈ N 2 ( λ ) and I λ ( η ( u λ ) u λ ) = min η > 0 I λ ( η u λ ) . Therefore, by Fatou’s lemma and Lemma 2.2 ( v ) once again, we derive that

This together with the uniqueness of η ( u λ ) leads to η ( u λ ) = 1 . So, we have

Moreover, we can also obtain that lim inf n → ∞ ‖ u n ‖ E 2 = ‖ u λ ‖ E 2 and a subsequence of { u n } (still denoted by { u n } ) fulfils lim n → ∞ ‖ u n ‖ E 2 = ‖ u λ ‖ E 2 , which means u n → u λ in E .

Case 2. There is a subsequence of { u n } (we still call u n ) belonging to N 2 ( λ ) .

Fix 0 ≤ ψ ∈ E , since { u n } is bounded, one can obtain from γ > 1 and Lemma 2.2 (iii) that

By Step 1, there are some functions h n , ψ ( η ) : [ 0 , + ∞ ) → ( 0 , + ∞ ) satisfying

By the dominated convergence theorem, Lemma 2.2 ( v ), ∫ R 3 f ( x ) u n 1 − γ d x < + ∞ and γ > 1 , we know that h n , ψ ( η ) is continuous for all η ≥ 0 . To facilitate the following proof, we define

If the right term is nonexistent, one can select η k → 0 (instead of η → 0 ) with η k > 0 and h n , ψ ′ ( 0 ) = lim k → ∞ h n , ψ ( η k ) − 1 η k ∈ [ − ∞ , + ∞ ] . Using Lemma 2.2 (iv), u n ∈ N 2 ( λ ) and h n , ψ ( η ) ( u n + η ψ ) ∈ N 2 ( λ ) , one can obtain

Consequently, we have

Dividing by η > 0 and letting η → 0 + , then one can get from u n ∈ N 2 ( λ ) , the continuity of h n , ψ ( η ) , Lemma 2.2 ( v ) and γ > 1 that

which implies that h n , ψ ′ ( 0 ) ≠ + ∞ . Next, we want to show the existence of C 3 > 0 satisfying h n , ψ ′ ( 0 ) ≤ C 3 uniformly in n . Fix n , without loss of generality, suppose h n , ψ ′ ( 0 ) ≥ 0 . By Lemma 2.2 (ii) and the above inequality, we have

This together with C 1 ≤ ‖ u n ‖ E ≤ C 2 leads to the existence of C 3 > 0 satisfying

We claim that h n , ψ ′ ( 0 ) is bounded from below uniformly for n large enough satisfying

Suppose h n , ψ ′ ( 0 ) ≤ 0 , otherwise the declaration is obvious. Hence, h n , ψ ( η ) < 1 for η > 0 small. In virtue of condition (2) again, we obtain

By (3.5), Lemma 2.2 (iv) and u n ∈ N 2 ( λ ) , one has

Letting η → 0 + and combing with C 1 ≤ ‖ u n ‖ E ≤ C 2 , the continuity of h n , ψ ( η ) , γ > 1 and Lemma 2.2 ( v ), one has

where we used h n , ψ ′ ( 0 ) ≤ 0 . Thus, the sign of the first coefficient reveals that

for suitable constant C 4 . This together with (3.4) results in

with C = max { C 3 , ∣ C 4 ∣ } . Again, by condition (2), we obtain

Letting η → 0 + , similar to the above discussion, one can get

We can further get from ∣ h n , φ ′ ( 0 ) ∣ ≤ C for large n and C 1 ≤ ‖ u n ‖ E ≤ C 2 that

Letting n → ∞ , it follows from Fatou’s lemma and Lemma 2.2 ( v ) that

for every 0 ≤ ψ ∈ E . Similar to Case 1, we have

in Case 2. Therefore, we can deduce from (3.2), (3.3), (3.7) and (3.8) that in both cases, up to subsequence, u n → u λ in E , inf N 1 ( λ ) I λ = I λ ( u λ ) , u λ ∈ N 2 ( λ ) and

Step 5. u λ > 0 is a solution of system ( SP λ ) .

Since u λ ∈ N 2 ( λ ) , we have

For every ε > 0 and v ∈ E , set ψ ε = u λ + ε v , we have

In virtue of the proof of [16, Theorem 3.2], we have

Setting Ω ε = { x ∈ R 3 : ψ ε ≤ 0 } and inserting ψ = ψ ε + into (3.9) together with (3.10)–(3.11), we obtain

Letting ε → 0 + , thanks to (3.12) and the fact that ∣ Ω ε ∣ → 0 as ε → 0 + , one has

Substituting v for − v , one can get the reserved inequality. So we have

Similar to [12, Theorem 6.3], we get u λ ∈ C l o c α ( R 3 ) for some α ∈ ( 0 , s ) . Since u λ ≥ 0 and u λ ≢ 0 , by the strong maximum principle in [38, Theorem 2.6], we can obtain that u λ > 0 in R 3 and u λ ∈ E is a solution to ( SP λ ) .

Step 6. The solution u λ is unique.

Assume u ∗ ∈ E also solves ( SP λ ) , then one has

Taking v = u λ − u ∗ in (3.13)–(3.14), then subtracting (3.13) from (3.14) and using Lemma 2.2 (vi), we get

Therefore, ‖ u λ − u ∗ ‖ E 2 = 0 , which implies u λ = u ∗ and the solution u λ is unique.□