Groundstate for the Schrödinger-Poisson-Slater equation involving the Coulomb-Sobolev critical exponent

Theorem 1.1

Assume μ = 0 . Then, equation (1.1) has no nontrivial solutions.

Theorem 1.2

Let 3 < p < 6 and μ > 0 . Then, equation (1.1) has a positive ground state solution on ℳ .

Remark 1.1

1. In [19], when r = 3 , in order to overcome the loss of compactness of the Sobolev embedding E ↪ L 3 ( R 3 ) , the authors worked on the radial space E rad to obtain a compactness result, namely, the embedding E rad ↪ L 3 ( R 3 ) is compact. In other words, r = 3 is no longer the Coulomb-Sobolev critical exponent in the radial space E rad .

   We recover Theorem 1.2 in [19] using the Nehari-Pohozaev manifold method instead of using a perturbation method and the mountain pass theorem. However, the most important and interesting point of our approach is that it handles a Schrödinger-Poisson-Slater equation with the Coulomb-Sobolev critical growth.

2. The lack of compactness. A general tool that could be useful to overcome the lack of compactness is the Schwartz rearrangement map u → u ∗ . The following properties are well known (see [11]):

   ∫ R 3 ∣ ∇ u ∗ ∣ 2 d x ⩽ ∫ R 3 ∣ ∇ u ∣ 2 d x ;

   ∫ R 3 ∫ R 3 ∣ u ( x ) ∣ 2 ∣ u ( y ) ∣ 2 ∣ x − y ∣ d x d y ⩽ ∫ R 3 ∫ R 3 ∣ u ∗ ( x ) ∣ 2 ∣ u ∗ ( y ) ∣ 2 ∣ x − y ∣ d x d y ;

   ∫ R 3 ∣ u ∣ p d x = ∫ R 3 ∣ u ∗ ∣ p d x , 1 ⩽ p < + ∞ .

   For (1.1), however, if we use the Schwartz rearrangement technique, it is impossible to overcome the lack of compactness due to the nonlocal term.

3. For (1.1), it is difficult to prove the boundedness of (PS) sequence of J . In the present article, we have opted to use the Nehari-Pohozaev manifold ℳ to avoid the difficulty.

Lemma 3.1

The functional J is unbounded from below.

Proof

Let u ∈ E and u t = t u ( t b x ) , b = 1 / 2 , t > 0 . By the standard scaling, we have

and

Hence,

Based on p > 3 , we see that J ( u t ) → − ∞ as t → + ∞ .□

Lemma 3.2

Let

for t ⩾ 0 and u ∈ ℳ . Then, f has a unique critical point, corresponding to its maximum.

Proof

For u ∈ ℳ , we have

Therefore,

Noting f ( 0 ) = 0 , we see that f has a unique positive critical point. The proof of Lemma 3.2 is completed.□

Lemma 3.3

ℳ is a C 1 -manifold and every critical point of J in ℳ is a critical point of J.

Proof

We proceed in three steps.

Step 1. We claim inf ℳ J > 0 and 0 ∉ ∂ ℳ .

Once u ∈ ℳ , then

and so

Moreover, we see that 0 ∉ ∂ ℳ .

Step 2. We claim that ℳ is a C 1 -manifold.

By the implicit function theorem, it only needs I ′ ( u ) ≠ 0 for any u ∈ ℳ . We prove it by argument of contradiction. In particular, we suppose that I ′ ( u ) = 0 for some u ∈ ℳ . Thus, in a weak sense, there holds

Multiplying (3.2) by u and integrating, we have

The Pohozaev identity corresponding to equation (3.2) is

It follows from p I ( u ) = 0 that

Multiplying (3.4) by 1 3 and applying I ( u ) = 0 , we obtain

Multiplying (3.3) by 1 p and using I ( u ) = 0 again, we have

Applying the relation ∫ R 3 ∣ ∇ u ∣ 2 d x = 1 2 L ( u ) to the above equation, we obtain

Finally, taking L ( u ) = 2 3 ∫ R 3 ∣ u ∣ 3 d x and ∫ R 3 ∣ ∇ u ∣ 2 d x = 1 2 L ( u ) into (3.4), we obtain

This a contradiction. Thus, ℳ is a C 1 -manifold.

Step 3. We claim that every critical point of J on ℳ is a critical point of J in E .

Assume that u is a critical point of J on ℳ , there exists a Lagrange multiplier λ such that J ′ ( u ) = λ I ′ ( u ) . It can be written, in a weak sense, as follows:

That is,

We see that λ ≠ 1 , it remains now to prove that λ = 0 . Denote

We can establish the following equations:

where the second equation follows by multiplying (3.5) by u and integrating, and the third equality is the Pohozaev identity corresponding to equation (3.5).

It follows from the first and the third equations in (3.6) that

Applying the the first and the second equations in (3.6), we have

In particular,

It follows from (3.7) and (3.8) that

That is,

Noting that

we obtain

Since p > 3 , we have

Therefore, we obtain J ′ ( u ) = 0 , i.e., u is a critical point of J . The proof is complete.□

Lemma 3.4

Let { u n } ⊂ ℳ be a minimizing sequence of J. Then, { u n } is a bounded ( PS ) inf ℳ J sequence for J. Then, up to a subsequence { u n } , still denoted itself, a number k ∈ N ∪ { 0 } , and a finite sequence

of critical points problem (1.1) and k sequences { ξ n 1 } , … , { ξ n k } ⊂ R 3 , such that as n → + ∞ ,

Proof

Step 1. Let { u n } ⊂ ℳ be a minimizing sequence of J in ℳ , i.e., J ( u n ) → inf ℳ J as n → ∞ . We claim that { u n } is a (PS) sequence of J . In fact, by the Ekeland variational principle (see Theorem 8.5 in [28]), there exists { λ n } ⊂ R such that

Then,

In particular,

We also have ( λ n ↛ 0 )

It is similar to the Step 3 in Lemma 3.3 that

Thus, we obtain that { u n } ⊂ ℳ is (PS) sequence of J . In particular,

It follows from J ′ ( u n ) → 0 and I ( u n ) = 0 that

By (3.1), there holds

and so we obtain that ∫ R 3 ∣ u n ∣ p d x is bounded.

Step 2. We claim that { u n } is bounded in E .

In fact, by P ( u n ) = o ( 1 ) , u ∈ ℳ , and (3.13), we deduce that when n → ∞ ,

Now, for each u ∈ E \ { 0 } , define

By a simple computation,

It follows from (3.15) that lim t → 0 + ‖ γ ( t ) ‖ 2 = + ∞ and lim t → + ∞ ‖ γ ( t ) ‖ 2 = 0 . Therefore, for any u ∈ E \ { 0 } , there exists a unique T u > 0 such that

This and (3.15) lead to

Next, we prove that ‖ u n ‖ is bounded by contradiction argument. We suppose that there exists a subsequence of u n denoted by itself such that ‖ u n ‖ → + ∞ as n → ∞ . Then, by (3.16) and (3.17), there exists T n ≔ T u n such that

Here, { v n } ⊂ E is defined by v n ( x ) ≔ T n − 1 u n ( T n − 1 2 x ) . By (3.14),