In this section, we obtain the existence of positive ground state solution of (1.6) by using Theorem1.2with λ=1, q= 4. Similarly to Section 4, we first estimate the level of mountain critical of the functional ˆJ corresponding to (1.6) and show that the critical level is below the non-compactness level of ˆJ by using approximation techniques. Then we are devoted to verify that the (PS) sequence of the functional ˆJ is also the one of the approximation functional associated to ˆJ. Finally, by the regularity theory of the elliptic equation, the positive ground state solution of (1.6) is obtained. In order to find the weak solutions to (1.6) and it is natural to consider the energy functional on H:
Jˆ(u) = 1
2kuk2+ b
4kuk4−
Z
B
1
6+|x|α|u|6+|x|α. Then we have from Lemma2.2 that ˆJ is well defined onH and is ofC1, and
(Jˆ0(u),v) = (1+bkuk2)
Z
B
∇u∇v−
Z
B
|u|4+|x|αuv, u,v∈ H.
It is standard to verify that the weak solutions of (1.6) correspond to the critical points of the functional ˆJ.
Lemma 6.1. Letα2,β2,γ2 >0and define f2 :[0,∞)→Ras f2(t) = α2
2 t2+ β2
4 t4− γ2 6 t6. Then
sup
t∈[0,∞)
f2(t) = 6α2β2γ2
+β32+4α2γ2
q
β22+4α2γ2+β22 q
β22+4α2γ2
24γ22 .
Proof. Fort ≥0, we have
f20(t) =α2t+β2t3−γ2t5=t(α2+β2t2−γ2t4). Letα2+β2t2−γ2t4 =0, we write at
t2 = q
β22+4α2γ2+β2 2γ2
.
Substituting it into f2(t), the result is valid. The proof is completed.
Lemma 6.2. Let
g2(t) = t
2
2kuεk2+ bt
4
4 kuεk4−t6 6
Z
R3|uε|6, then we have, asε→0+,
sup
t≥0
g2(t)≤Λ1+O(ε), whereΛ1= b4S3+ b243S6+ 16S√
S4b2+4S+24b2S4√
S4b2+4S.
Proof. It follows from Lemma6.1and the estimate (2.2) that g2(t) = t
2
2kuεk2+ bt
4
4 kuεk4−t6 6
Z
R3|uε|6
= t
2
2(S3/2+O(ε)) + bt
4
4 (S3+O(ε))− t
6
6(S3/2+O(ε3))
≤ b 4S3+ b
3
24S6+ 1 6Sp
S4b2+4S+ b
2
24S4p
S4b2+4S+O(ε), forε >0 sufficiently small. The proof is completed.
From Lemma 3.2, we know that the functional ˆJ possesses the mountain pass geometry.
Then there is a(PS)c2 sequence{un} ⊂Hfor ˆJ with the property that Jˆ(un)→c2, kJˆ0(un)kH−1 →0, n→∞, wherec2is given by
c2= inf
ˆ
γ∈Γmax
t∈[0,1]
Jˆ(γ(t)), and ˆΓ= {γ∈ C([0, 1],H):γ(0) =0, ˆJ(γ(1))<0}.
In the following we give an estimate of the upper bound of the critical level c2 by using above two lemmas.
Lemma 6.3. There holds0< c2 <Λ1.
Proof. Similar to Lemma5.4, there exists R2 >0 sufficiently large, such that ˆJ(R2uε)≤ 0 forε small enough, hence, we can find 0< tε <R2 satisfying
0< η2 ≤c2≤ max
t∈[0,R2]
Jˆ(tuε) =Jˆ(tεuε).
Since dtd Jˆ(tuε)|t=tε =0, we have
tεkuεk2+bt3εkuεk4=
Z
Bt5ε+|x|α|u|6+|x|α. Hence we deduce from (2.2) that
S32 +O(ε) +bt2ε(S3+O(ε)) =t4ε Z
B
|uε|6+t4ε Z
B
t|εx|α|uε|6+|x|α− |uε|6
=t4ε[S32 +O(ε3) +Aε]
=t4ε[S32 +O(ε3) +O(εα|logε|) +O(ε3/2)],
(6.1)
where Aε = O(εα|logε|) +O(ε3/2) is given in [21, page 14]. For convenience, we set A = S32 +O(ε),B = b(S3+O(ε))andC= S32 +O(ε3) +O(εα|logε|) +O(ε3/2). Thus, (6.1) can be rewritten asA+Bt2ε =Ct4ε. It is easy to see that
t2ε = B+√
B2+4AC
2C = bS
3+O(ε) +pb2S6+4S3+O(ε) +O(εα|logε|) 2S3/2+O(ε3/2) +O(εα|logε|) .
Thereby,t2ε >1 forεsmall enough, which implies Combining (6.3), (6.4) with (6.2) and using Lemma6.2, we derive
Jˆ(tεuε) = t
By choosingε>0 small enough, we derive by (6.5),
0<η2≤ c2≤ Jˆ(tεuε)<Λ1. The proof is completed.
Lemma 6.4. If{un}is a(PS)c2 sequence ofJ, then there exists uˆ ∈ H such that, up to a subsequence, un*u and Jˆ0(u) =0.
Proof. By Lemma3.3,{un}is bounded inH and hence, going if necessary to a subsequence, we may assume that un * u in H. Let A > 0 be such that R
where p(x) =6+|x|α. Hence, the Vitali theorem (see [28]) leads to norm and Fatou’s lemma that
c2≤ Jˆ(tuu)− 1 which is impossible. Thus,R
B|∇u|2 = A2and ˆJ0(u) =0. The proof is completed.
In order to obtain the nontrivial solution of (1.6), we need define the functional ˆI : H→R by
By repeating the arguments used in Lemma 3.3, it is easy to show that {un} is bounded in H. Then passing to a subsequence, we can find u ∈ H such that un * u in H. Now, let vn = un−u, we claim that kvnk → 0. In fact, we use an argument of contradiction and suppose that there exists a subsequence still denoted by {vn}such that kvnk → l˜ > 0. It is easy to verify that
kunk2= kvnk2+kuk2+o(1) (6.7) and
kunk4 =kvnk4+kuk4+2kvnk2kuk2+o(1). (6.8) From the Brezis–Lieb lemma in [9], we have
Z
Recall that ˆI0(un)→0 asn→∞, there holds by (6.7),
On the other hand, combining (6.7), (6.8) with (6.9) leads to Iˆ(un)−Iˆ(u) +o(1)
Similarly, by using (6.10) again, we deduce
o(1) =hIˆ0(un),uni −(kuk2+bl˜2kuk2+bkuk4−
It follows from (6.12) and (6.13) that Iˆ(u) = Iˆ(un)− This together with (6.14) ensures that
Iˆ(u) =c2−
which contradicts to (6.11). Therefore vn →0 strongly in H, or equivalently, un → uin Has n→∞. The proof is completed.
Proof of Theorem1.9. By Lemmas 6.4, 6.5, we know that the assumptions in Theorem 1.2 are valid. Hence, (1.6) possesses a nonnegative nontrivial ground state solution u ∈ H, which satisfies the following equation in weak sense
−
1+b
Z
B
|∇u|2
∆u= u5+|x|α inB.
Let us define
ˆ
g(u(x)) = u
5+|x|α
1+bR
B|∇u|2, x∈B.
It follows from Lemma2.2thatR
Bu6+32|x|α ≤C, which implies a= gˆ(u)
1+|u| ∈ L32(B).
Hence, we deduce immediately from Lemma 5.7 that u ∈ Lq(B) for any 1 < q < ∞. Then, there holds ˆg(u) ∈ Lq(B) for any 1 < q < ∞. By the Calderón–Zygmund inequality and Lp estimate given in [16,30], we derive u ∈ W2,q(B), whence also u ∈ C1,α2(B) by Sobolev embedding theorem for any 0 < α2 < 1. Moreover, the Harnack inequality [32] implies u(x)>0 for allx ∈B. The proof is completed.
7 Acknowledgement
This work was partially supported by National Natural Science Foundation of China (Grant Nos. 12071266, 12026217, 12026218) and Shanxi Scholarship Council of China (Grant No.
2020-005). The author would like to express their sincere gratitude to anonymous referees for his/her constructive comments for improving the quality of this paper.
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