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On a nonlocal problem with critical Sobolev growth
Accepted Manuscript On a nonlocal problem with critical Sobolev growth
Jixiu Wang
PII: DOI: Reference:
S0893-9659(19)30274-5 https://doi.org/10.1016/j.aml.2019.06.030 AML 5959
To appear in:
Applied Mathematics Letters
Received date : 14 May 2019 Revised date : 23 June 2019 Accepted date : 23 June 2019 Please cite this article as: J. Wang, On a nonlocal problem with critical Sobolev growth, Applied Mathematics Letters (2019), https://doi.org/10.1016/j.aml.2019.06.030 This is a PDF file of an unedited manuscript that has been accepted for publication. As a service to our customers we are providing this early version of the manuscript. The manuscript will undergo copyediting, typesetting, and review of the resulting proof before it is published in its final form. Please note that during the production process errors may be discovered which could affect the content, and all legal disclaimers that apply to the journal pertain.
*Manuscript Click here to view linked References
ON A NONLOCAL PROBLEM WITH CRITICAL SOBOLEV GROWTH JIXIU, WANG Abstract. We obtain positive solutions for a nonlinear fractional Laplacian problem with critical Sobolev growth via localization principle and variational methods.
Keywords: Fractional Laplacian; Critial Sobolev growth; Positive solutions; Variational methods 2010 Mathematics Subject Classification: 35B09, 35J20
1. Introduction Let Ω ⊂ Rn , n ≥ 2, be a bounded smooth domain and paper, we consider the problem q−1 2♯ −1 A1/2 u = g(x)(u − k)+ + u (1.1) u>0 u = 0
denote 2♯ = 2n/(n − 1). In this in Ω, in Ω, on ∂Ω,
where 2 ≤ q < 2♯ , k ∈ (0, ∞) is an arbitrary number, and A1/2 is the square root of the minus Laplacian operator −∆ on Ω with zero Dirichlet boundary values on ∂Ω, as introduced in Cabré and Tan [3]. More precisely, let {λk , ϕk }k≥1 be the corresponding eigenvalues and eigenfunctions of P −∆ on Ω with ϕk |∂Ω = 0 and normalized by kϕk kL2 (Ω) = 1. Then, for u = k≥1 ck ϕk P P 1/2 2 satisfying k≥1 λk ck ϕk . Therefore, a k≥1 ck λk < ∞, A1/2 u is defined as A1/2 u = natural function space of A1/2 in the sense of functional analysis is given by X X 1/2 V0 (Ω) = u ∈ L2 (Ω) : u = ck ϕk satisfying c2k λk < ∞ . k≥1
k≥1
A positive function u ∈ V0 (Ω) is called a solution of problem (1.1) if for every v = P k≥1 dk ϕk ∈ V0 (Ω), there holds Z X ♯ 1/2 q−1 + u2 −1 vdx. g(x)(u − k)+ ck dk λk = k≥1
Ω
We remark that Cabré and Tan [3] contributed a systematical study on A1/2 in partial differential equations. It turns out that problem (1.1) is difficult to deal with directly due to the nonlocality of A1/2 . To handle problem (1.1), we apply the localization principle developed by Cabré This work is financially supported by NSFC (No. 11501186).
1
2
J. WANG
and Tan [3]. First denote C = Ω × (0, ∞) and ∂L C = ∂Ω × [0, ∞)
with coordinates (x, y) ∈ C for x ∈ Ω and y ≥ 0. Then introduce the function space 1 H0,L (C) = u ∈ H 1 (C) : u = 0 a.e. on ∂L C R equipped with norm kuk2 ≡ C |∇u(x, y)|2 dxdy. According to [3], for every u ∈ V0 (Ω), 1 (C), which is harmonic in C and equals to u on there exists a unique function u ¯ in H0,L Ω × {y = 0} in the sense of trace. Moreover, problem (1.1) turns out to be equivalent to the following problem u(x, y) = 0, u ¯>0 in C, ∆¯ (1.2) u ¯=0 on ∂L C, ♯ −1 q−1 ∂ u 2 ¯ = g(x)(¯ u − k) +u ¯ on Ω × {y = 0}, ν
+
where ∂ν u ¯ = ∂u ¯/∂ν is the unit outward normal derivative of u ¯ on Ω × {y = 0}. For later use, we present an important trace inequality [7]: there exists C > 0 such n+1 ∞ that, for every wR belonging to D 1,2 (Rn+1 + ), the closure of C0 (R+ ) under the seminorm 2 2 kwkD1,2 (Rn+1 ) = Rn+1 |∇w(x, y)| dxdy, there holds +
(1.3) Let
+
C
Z
Rn
2♯
|w(x, 0)| dx
2/2♯
≤
Z
Rn+1 +
|∇w(x, y)|2 dxdy.
R Rn+1 |∇w(x, y)|2 dxdy + n+1 1,2 S = inf ), w ≡ 6 0 : w ∈ D (R + R |w(x, 0)|2♯ dx2/2♯ Rn
be the best constant for inequality (1.3). It is known [4] that S is attained by the functions (1−n)/2 (1.4) Uǫ,x0 (x, y) = cn ǫ(n−1)/2 |x − x0 |2 + (y + ǫ)2
for all ǫ > 0 and x0 ∈ Rn . We recall that Z Z 2 |∇Uǫ,x0 | dxdy = Rn+1 +
Rn
♯
|Uǫ,x0 (x, 0)|2 dx = S n .
Now we can state our assumptions and result. Throughout we assume that 2 ≤ q < 2♯ and g satisfies (G1) g ∈ C(Ω) is a nonnegative nontrivial function, and (G2) g ∈ L2n/(2n−(n−1)q) (Ω). In addition, if q = 2, then kgkLn (Ω) < S. Then, we have Theorem 1.1. Assume that 2 ≤ q < 2♯ and g satisfies (G1) (G2). Then problem (1.1) admits a positive solution for every k ∈ (0, ∞). Problem (1.1) is inspired by the interesting work [5] of Gazzola, where the problem q−1 p−2 p∗ −1 in Ω, −div(|∇u| ∇u) = g(x)(u − k)+ + u (1.5) u>0 in Ω, u = 0 on ∂Ω
FRACTIONAL PROBLEM
3
was studied with 1 < p < n and p∗ = np/(n − p), and positive solutions were obtained among other results. Problem (1.5) can be viewed as a variant of the well known BrézisNirenberg problem of [2]. Li and Xiang [6] also extends the above problem to the setting of fractional Laplacion operators which is differently defined from ours. To prove Theorem 1.1, we turn to the equivalent problem (1.2). Note that problem 1 (C) → R being given by (1.2) is variational with the energy functional J : H0,L Z 1 1 2♯ 1 q 2 uk − g(x)(¯ u − k)+ + ♯ u ¯ dx J (¯ u) = k¯ 2 2 + Ω q
1 (C). So we only need to find a critical point of J . It is direct to verify that for u ¯ ∈ H0,L J satisfies the mountain pass geometry, which inspires us to apply the Mountain Pass Lemma (see e.g. Struwe [8]) in the below. Our notations are standard. We use C to denote positive constants that are different from line to line.
2. Proof of the main result In this section we prove Theorem 1.1 via the well known mountain pass lemma. Lemma 2.1. Under the assumptions (G1) and (G2), there holds and
q−1 q−1 → g(u − k)+ g(um − k)+
g(um − k)q+ → g(u − k)q+
2n
in L n+1 (Ω) in L1 (Ω)
1 (C) that converges weakly to u ∈ H 1 (C). for every sequence (um ) ⊂ H0,L 0,L
Lemma 2.1 is an easy consequence of Vitali’s convergence theorem. We omit the details, see also Gazzola [5]. Lemma 2.2. J satisfies the (P S)c condition provided c < S n /(2n). 1 (C) is such that J (u ) → c and J ′ (u ) → 0 as m → ∞. We Proof. Suppose (um ) ⊂ H0,L m m 1 (C) provided c < S n /(2n). have to prove that (um ) converges up to a subsequence in H0,L 1 (C). Indeed, for any β ≥ 2, Step 1. We claim that (um ) is a bounded sequence in H0,L we have 1 c + o(1) + o(1)kum k = J (um ) − hJ ′ (um ), um i. β In the case q > 2, taking β = q yields Z Z 1 ♯ 1 k 1 1 q−1 2 (um )2+ . − g(um − k)+ + − ♯ kum k + c + o(1) + o(1)kum k = 2 q q Ω q 2 Ω
By the assumption g ≥ 0, we obtain
(2.1)
c + o(1) + o(1)kum k ≥
1 1 − 2 q
kum k2 .
In the case q = 2, taking 2♯ > β > 2 yields kgkLn (Ω) 1 1 (2.2) c + o(1) + o(1)kum k ≥ − 1− kum k2 , 2 β S
4
J. WANG
R where we have used Hölder inequality and the assumption (G2) to get the estimate Ω g(u− k)2+ ≤ (kgkLn (Ω) /S)kuk2 . The claim follows from (2.1) and (2.2). 1 (C). By Step 1, we Step 2. We prove that (um ) has a convergent subsequence in H0,L may assume that, up to a subsequence, 1 um ⇀ u in H0,L (C),
um (·, 0) → u(·, 0) in L2 (Ω),
um (·, 0) → u(·, 0) a.e. in Ω, 1 (C). Write v = u − u. We need to prove that kv k → 0. for some function u ∈ H0,L m m m Easy to derive that J ′ (u) = 0. Moreover,
kum k2 = kuk2 + kvm k2 + o(1), and by Lemma 2.1,
Z
Ω
g(um − k)q+ →
Z
Ω
g(u − k)q+ ,
and by the Lemma of Brézis and Lieb [1], Z Z Z ♯ ♯ ♯ u2+ + (vm )2+ + o(1). (um )2+ = Ω
Ω
Ω
Hence,
1 ♯ 1 J (um ) = J (u) + kvm k2 − ♯ kvm k2 2♯ + o(1). L (Ω) 2 2 Furthermore, J (u) ≥ 0 holds since J ′ (u) = 0. This implies (2.3)
1 1 ♯ J (um ) ≥ kvm k2 − ♯ kvm k2 2♯ + o(1). L (Ω) 2 2
On the other hand, remember that J ′ (um ) → 0. We deduce hJ ′ (um ), um −ui−hJ ′ (u), um − ui = o(1), from which follows Z ♯ 2 kvm k − (vm )2+ = o(1). Ω
R ♯ We may assume kvm k2 → b ≥ 0. Then from the above we derive Ω (vm )2+ → b. Hence, ♯ the trace inequality (1.3) gives b ≥ Sb2/2 . Thus, either b = 0 or b ≥ S n . If b > 0, then (2.3) implies c ≥ S n /(2n). We reach a contradiction since we assume c < S n /(2n). Hence, 1 (C). The proof is complete. b = 0. That is, vm → 0 in H0,L Now let
n o 1 Γ = γ ∈ C([0, ∞), H0,L (C)) : γ(0) = 0 and lim inf J (γ(t)) < 0 t→∞
and (2.4)
Lemma 2.3. c0 < S n /(2n).
c0 = inf max J (γ(t)). γ∈Γ t≥0
FRACTIONAL PROBLEM
5
Proof. With no loss of generality, assume 0 ∈ Ω and g(0) > 0. The assumption (G1) allows to assume that Br (0) ⊂ {x ∈ Ω : g(x) > g0 } for some constant g0 > 0 and some r > 0 sufficiently small. Let Uǫ,x0 be defined as in (1.4) and write Uǫ = Uǫ,0 . Let 0 < ρ < r ¯ such that 0 ≤ η ≤ 1 and η(x, y) ≡ 1 for |x| ≤ ρ/2 and y ≥ 0, and and let η ∈ C ∞ (C) η(x, y) ≡ 0 for |x| ≥ ρ. Let uǫ = ηUǫ and take the particular path γ(t) = tuǫ for t ≥ 0. We show that maxt>0 J (γ(t)) < S n /(2n). Then the desired result follows. For this path γ, we have Z ♯ Z t2 ♯ 1 t2 q 2 u2ǫ . g(tuǫ − k)+ − ♯ J (γ(t)) = kuǫ k − 2 q Ω 2 Ω Let tǫ be such that J (γ(tǫ )) = maxt>0 J (γ(t)). Claim that 0 < c1 ≤ tǫ ≤ c2 holds for some constants c1 , c2 independent of ǫ. First note that ♯ Z ♯ t2 t2 u2 . J (γ(t)) ≤ kuǫ k2 − ♯ 2 2 Ω ǫ R ♯ By letting t¯ǫ = 2♯ kuǫ k2 /2 Ω u2ǫ , we infer from the above inequality that J(γ(t)) ≤ 0 holds for all t ≥ t¯ǫ , which implies tǫ ≤ t¯ǫ ≤ c2 . On the other hand, standard computation shows that there exist b, ρ > 0 sufficiently small such that J (u) ≥ b for all kuk = ρ. This implies that tǫ can not converges to zero. Therefore, there exists c1 > 0 such that tǫ ≥ c1 for ǫ sufficiently small. The claim is proved. By [9, Proposition 4.2], there hold R ♯ kuǫ k2 = S n + O(ǫn−1 ) and Ω u2ǫ = S n + O(ǫn ). Thus, for ǫ sufficiently small, ! n 2♯ Z 2♯ − 1 2−1 ♯ S t t t t2ǫ u2 = kuǫ k2 − ǫ♯ + ǫ − ǫ ♯ S n + O(ǫn−1 ), 2 2 Ω ǫ 2n 2 2 which implies that Sn 1 − J (γ(tǫ )) ≤ 2n q
(2.5)
Z
Ω
g(tǫ uǫ − k)q+ + O(ǫn−1 )
♯
♯ 2 2 due to the fact that R maxt≥0 ((t q − 1)/2 − (t − 1)/2 ) = 0. We estimate Ω g(tǫ uǫ − k)+ as follows. For ǫ sufficiently small, Z Z Z n−1 n−1 q q ǫ− 2 q = Cǫn− 2 q (tǫ Uǫ − k)+ ≥ C g(tǫ uǫ − k)+ ≥ g0 Bǫ (0)
Bρ/2 (0)
Ω
for some C > 0 independent of ǫ. In the case q = n = 2, we note that on the set {ǫ < |x| < ǫ3/4 }, there holds ǫ1/2 ǫ−1/4 Uǫ (x) ≥ √ ≥ √ > 2k 2|x| 2
for ǫ sufficiently small. Hence, Z Z (tǫ Uǫ − k)2+ ≥ C g0 Bρ/2 (0)
{ǫ<|x|<ǫ3/4 }
ǫ 1 = Cǫ log 2 |x| ǫ
for some C > 0 independent of ǫ. Thus, there exists C > 0 such that ( n−1 Z Cǫn− 2 q , if q > 2 or n > 2 q g(tǫ uǫ − k)+ ≥ (2.6) Cǫ log 1ǫ if q = n = 2. Ω
6
J. WANG
Combining (2.5) and (2.6) yields the result. The proof is complete.
Now we are in the position to prove Theorem 1.1. Proof of Theorem 1.1. Since J satisfies the geometry of mountain pass, there exists a 1 (C) satisfying J (u ) → c and J ′ (u ) → 0 as m → ∞, where c sequence (um ) ⊂ H0,L m 0 m 0 is defined as in (2.4). Therefore, Lemmas 2.2 and 2.3 imply that problem (1.1) admits a nonnegative nontrivial solution. Finally, a maximum principle of Cabré and Tan [3] implies that the solution is positive in Ω. The proof is complete. References [1] Brézis, H. and Lieb, E.: A relation between pointwise convergence of functions and convergence of functionals. Proc. Amer. Math. Soc. 88, 486-490 (1983). [2] Brézis, H and Nirenberg, L.: Positive solutions of nonlinear elliptic equations involving critical Sobolev exponents. Comm. Pure Appl. Math. 36, 437-477 (1983). [3] Cabré, X. and Tan, J.: Positive solutions of nonlinear problems involving the square root of the Laplacian. Adv. Math. 224, 2052-2093 (2010). [4] Escobar, J.: Sharp constant in a Sobolev trace inequality. Indiana Univ. Math. J. 37, 687-698 (1988). [5] Gazzola F.: Critical growth quasilinear elliptic problems with shifting subcritical perturbation. Diff. Integral Equ. 14, 513-528 (2001). [6] Li, Q. and Xiang C.-L.: The fractional problem with shifting subcritical perturbation. Submitted. [7] Lions, P.L.: The concentration-compactness principle in the calculus of variations. The limit case II. Rev. Mat. Iberoamericana 1, 45-121 (1985). [8] Struwe M.: Variational methods. Applications to nonlinear partial differential equations and Hamiltonian systems. Second edition. Ergebnisse der Mathematik und ihrer Grenzgebiete (3), 34. SpringerVerlag, Berlin, 1996. [9] Tan J.: The Brezis-Nirenberg type problem involving the square root of the Laplacian. Calc. Var. Partial Differential Equations 42, 21-41 (2011). (Jixiu, Wang) School of Mathematics and Statistics, Hubei University of Arts and Science, Xiangyang, 441053, P.R.China. E-mail address: [email protected]
Jixiu Wang
PII: DOI: Reference:
S0893-9659(19)30274-5 https://doi.org/10.1016/j.aml.2019.06.030 AML 5959
To appear in:
Applied Mathematics Letters
Received date : 14 May 2019 Revised date : 23 June 2019 Accepted date : 23 June 2019 Please cite this article as: J. Wang, On a nonlocal problem with critical Sobolev growth, Applied Mathematics Letters (2019), https://doi.org/10.1016/j.aml.2019.06.030 This is a PDF file of an unedited manuscript that has been accepted for publication. As a service to our customers we are providing this early version of the manuscript. The manuscript will undergo copyediting, typesetting, and review of the resulting proof before it is published in its final form. Please note that during the production process errors may be discovered which could affect the content, and all legal disclaimers that apply to the journal pertain.
*Manuscript Click here to view linked References
ON A NONLOCAL PROBLEM WITH CRITICAL SOBOLEV GROWTH JIXIU, WANG Abstract. We obtain positive solutions for a nonlinear fractional Laplacian problem with critical Sobolev growth via localization principle and variational methods.
Keywords: Fractional Laplacian; Critial Sobolev growth; Positive solutions; Variational methods 2010 Mathematics Subject Classification: 35B09, 35J20
1. Introduction Let Ω ⊂ Rn , n ≥ 2, be a bounded smooth domain and paper, we consider the problem q−1 2♯ −1 A1/2 u = g(x)(u − k)+ + u (1.1) u>0 u = 0
denote 2♯ = 2n/(n − 1). In this in Ω, in Ω, on ∂Ω,
where 2 ≤ q < 2♯ , k ∈ (0, ∞) is an arbitrary number, and A1/2 is the square root of the minus Laplacian operator −∆ on Ω with zero Dirichlet boundary values on ∂Ω, as introduced in Cabré and Tan [3]. More precisely, let {λk , ϕk }k≥1 be the corresponding eigenvalues and eigenfunctions of P −∆ on Ω with ϕk |∂Ω = 0 and normalized by kϕk kL2 (Ω) = 1. Then, for u = k≥1 ck ϕk P P 1/2 2 satisfying k≥1 λk ck ϕk . Therefore, a k≥1 ck λk < ∞, A1/2 u is defined as A1/2 u = natural function space of A1/2 in the sense of functional analysis is given by X X 1/2 V0 (Ω) = u ∈ L2 (Ω) : u = ck ϕk satisfying c2k λk < ∞ . k≥1
k≥1
A positive function u ∈ V0 (Ω) is called a solution of problem (1.1) if for every v = P k≥1 dk ϕk ∈ V0 (Ω), there holds Z X ♯ 1/2 q−1 + u2 −1 vdx. g(x)(u − k)+ ck dk λk = k≥1
Ω
We remark that Cabré and Tan [3] contributed a systematical study on A1/2 in partial differential equations. It turns out that problem (1.1) is difficult to deal with directly due to the nonlocality of A1/2 . To handle problem (1.1), we apply the localization principle developed by Cabré This work is financially supported by NSFC (No. 11501186).
1
2
J. WANG
and Tan [3]. First denote C = Ω × (0, ∞) and ∂L C = ∂Ω × [0, ∞)
with coordinates (x, y) ∈ C for x ∈ Ω and y ≥ 0. Then introduce the function space 1 H0,L (C) = u ∈ H 1 (C) : u = 0 a.e. on ∂L C R equipped with norm kuk2 ≡ C |∇u(x, y)|2 dxdy. According to [3], for every u ∈ V0 (Ω), 1 (C), which is harmonic in C and equals to u on there exists a unique function u ¯ in H0,L Ω × {y = 0} in the sense of trace. Moreover, problem (1.1) turns out to be equivalent to the following problem u(x, y) = 0, u ¯>0 in C, ∆¯ (1.2) u ¯=0 on ∂L C, ♯ −1 q−1 ∂ u 2 ¯ = g(x)(¯ u − k) +u ¯ on Ω × {y = 0}, ν
+
where ∂ν u ¯ = ∂u ¯/∂ν is the unit outward normal derivative of u ¯ on Ω × {y = 0}. For later use, we present an important trace inequality [7]: there exists C > 0 such n+1 ∞ that, for every wR belonging to D 1,2 (Rn+1 + ), the closure of C0 (R+ ) under the seminorm 2 2 kwkD1,2 (Rn+1 ) = Rn+1 |∇w(x, y)| dxdy, there holds +
(1.3) Let
+
C
Z
Rn
2♯
|w(x, 0)| dx
2/2♯
≤
Z
Rn+1 +
|∇w(x, y)|2 dxdy.
R Rn+1 |∇w(x, y)|2 dxdy + n+1 1,2 S = inf ), w ≡ 6 0 : w ∈ D (R + R |w(x, 0)|2♯ dx2/2♯ Rn
be the best constant for inequality (1.3). It is known [4] that S is attained by the functions (1−n)/2 (1.4) Uǫ,x0 (x, y) = cn ǫ(n−1)/2 |x − x0 |2 + (y + ǫ)2
for all ǫ > 0 and x0 ∈ Rn . We recall that Z Z 2 |∇Uǫ,x0 | dxdy = Rn+1 +
Rn
♯
|Uǫ,x0 (x, 0)|2 dx = S n .
Now we can state our assumptions and result. Throughout we assume that 2 ≤ q < 2♯ and g satisfies (G1) g ∈ C(Ω) is a nonnegative nontrivial function, and (G2) g ∈ L2n/(2n−(n−1)q) (Ω). In addition, if q = 2, then kgkLn (Ω) < S. Then, we have Theorem 1.1. Assume that 2 ≤ q < 2♯ and g satisfies (G1) (G2). Then problem (1.1) admits a positive solution for every k ∈ (0, ∞). Problem (1.1) is inspired by the interesting work [5] of Gazzola, where the problem q−1 p−2 p∗ −1 in Ω, −div(|∇u| ∇u) = g(x)(u − k)+ + u (1.5) u>0 in Ω, u = 0 on ∂Ω
FRACTIONAL PROBLEM
3
was studied with 1 < p < n and p∗ = np/(n − p), and positive solutions were obtained among other results. Problem (1.5) can be viewed as a variant of the well known BrézisNirenberg problem of [2]. Li and Xiang [6] also extends the above problem to the setting of fractional Laplacion operators which is differently defined from ours. To prove Theorem 1.1, we turn to the equivalent problem (1.2). Note that problem 1 (C) → R being given by (1.2) is variational with the energy functional J : H0,L Z 1 1 2♯ 1 q 2 uk − g(x)(¯ u − k)+ + ♯ u ¯ dx J (¯ u) = k¯ 2 2 + Ω q
1 (C). So we only need to find a critical point of J . It is direct to verify that for u ¯ ∈ H0,L J satisfies the mountain pass geometry, which inspires us to apply the Mountain Pass Lemma (see e.g. Struwe [8]) in the below. Our notations are standard. We use C to denote positive constants that are different from line to line.
2. Proof of the main result In this section we prove Theorem 1.1 via the well known mountain pass lemma. Lemma 2.1. Under the assumptions (G1) and (G2), there holds and
q−1 q−1 → g(u − k)+ g(um − k)+
g(um − k)q+ → g(u − k)q+
2n
in L n+1 (Ω) in L1 (Ω)
1 (C) that converges weakly to u ∈ H 1 (C). for every sequence (um ) ⊂ H0,L 0,L
Lemma 2.1 is an easy consequence of Vitali’s convergence theorem. We omit the details, see also Gazzola [5]. Lemma 2.2. J satisfies the (P S)c condition provided c < S n /(2n). 1 (C) is such that J (u ) → c and J ′ (u ) → 0 as m → ∞. We Proof. Suppose (um ) ⊂ H0,L m m 1 (C) provided c < S n /(2n). have to prove that (um ) converges up to a subsequence in H0,L 1 (C). Indeed, for any β ≥ 2, Step 1. We claim that (um ) is a bounded sequence in H0,L we have 1 c + o(1) + o(1)kum k = J (um ) − hJ ′ (um ), um i. β In the case q > 2, taking β = q yields Z Z 1 ♯ 1 k 1 1 q−1 2 (um )2+ . − g(um − k)+ + − ♯ kum k + c + o(1) + o(1)kum k = 2 q q Ω q 2 Ω
By the assumption g ≥ 0, we obtain
(2.1)
c + o(1) + o(1)kum k ≥
1 1 − 2 q
kum k2 .
In the case q = 2, taking 2♯ > β > 2 yields kgkLn (Ω) 1 1 (2.2) c + o(1) + o(1)kum k ≥ − 1− kum k2 , 2 β S
4
J. WANG
R where we have used Hölder inequality and the assumption (G2) to get the estimate Ω g(u− k)2+ ≤ (kgkLn (Ω) /S)kuk2 . The claim follows from (2.1) and (2.2). 1 (C). By Step 1, we Step 2. We prove that (um ) has a convergent subsequence in H0,L may assume that, up to a subsequence, 1 um ⇀ u in H0,L (C),
um (·, 0) → u(·, 0) in L2 (Ω),
um (·, 0) → u(·, 0) a.e. in Ω, 1 (C). Write v = u − u. We need to prove that kv k → 0. for some function u ∈ H0,L m m m Easy to derive that J ′ (u) = 0. Moreover,
kum k2 = kuk2 + kvm k2 + o(1), and by Lemma 2.1,
Z
Ω
g(um − k)q+ →
Z
Ω
g(u − k)q+ ,
and by the Lemma of Brézis and Lieb [1], Z Z Z ♯ ♯ ♯ u2+ + (vm )2+ + o(1). (um )2+ = Ω
Ω
Ω
Hence,
1 ♯ 1 J (um ) = J (u) + kvm k2 − ♯ kvm k2 2♯ + o(1). L (Ω) 2 2 Furthermore, J (u) ≥ 0 holds since J ′ (u) = 0. This implies (2.3)
1 1 ♯ J (um ) ≥ kvm k2 − ♯ kvm k2 2♯ + o(1). L (Ω) 2 2
On the other hand, remember that J ′ (um ) → 0. We deduce hJ ′ (um ), um −ui−hJ ′ (u), um − ui = o(1), from which follows Z ♯ 2 kvm k − (vm )2+ = o(1). Ω
R ♯ We may assume kvm k2 → b ≥ 0. Then from the above we derive Ω (vm )2+ → b. Hence, ♯ the trace inequality (1.3) gives b ≥ Sb2/2 . Thus, either b = 0 or b ≥ S n . If b > 0, then (2.3) implies c ≥ S n /(2n). We reach a contradiction since we assume c < S n /(2n). Hence, 1 (C). The proof is complete. b = 0. That is, vm → 0 in H0,L Now let
n o 1 Γ = γ ∈ C([0, ∞), H0,L (C)) : γ(0) = 0 and lim inf J (γ(t)) < 0 t→∞
and (2.4)
Lemma 2.3. c0 < S n /(2n).
c0 = inf max J (γ(t)). γ∈Γ t≥0
FRACTIONAL PROBLEM
5
Proof. With no loss of generality, assume 0 ∈ Ω and g(0) > 0. The assumption (G1) allows to assume that Br (0) ⊂ {x ∈ Ω : g(x) > g0 } for some constant g0 > 0 and some r > 0 sufficiently small. Let Uǫ,x0 be defined as in (1.4) and write Uǫ = Uǫ,0 . Let 0 < ρ < r ¯ such that 0 ≤ η ≤ 1 and η(x, y) ≡ 1 for |x| ≤ ρ/2 and y ≥ 0, and and let η ∈ C ∞ (C) η(x, y) ≡ 0 for |x| ≥ ρ. Let uǫ = ηUǫ and take the particular path γ(t) = tuǫ for t ≥ 0. We show that maxt>0 J (γ(t)) < S n /(2n). Then the desired result follows. For this path γ, we have Z ♯ Z t2 ♯ 1 t2 q 2 u2ǫ . g(tuǫ − k)+ − ♯ J (γ(t)) = kuǫ k − 2 q Ω 2 Ω Let tǫ be such that J (γ(tǫ )) = maxt>0 J (γ(t)). Claim that 0 < c1 ≤ tǫ ≤ c2 holds for some constants c1 , c2 independent of ǫ. First note that ♯ Z ♯ t2 t2 u2 . J (γ(t)) ≤ kuǫ k2 − ♯ 2 2 Ω ǫ R ♯ By letting t¯ǫ = 2♯ kuǫ k2 /2 Ω u2ǫ , we infer from the above inequality that J(γ(t)) ≤ 0 holds for all t ≥ t¯ǫ , which implies tǫ ≤ t¯ǫ ≤ c2 . On the other hand, standard computation shows that there exist b, ρ > 0 sufficiently small such that J (u) ≥ b for all kuk = ρ. This implies that tǫ can not converges to zero. Therefore, there exists c1 > 0 such that tǫ ≥ c1 for ǫ sufficiently small. The claim is proved. By [9, Proposition 4.2], there hold R ♯ kuǫ k2 = S n + O(ǫn−1 ) and Ω u2ǫ = S n + O(ǫn ). Thus, for ǫ sufficiently small, ! n 2♯ Z 2♯ − 1 2−1 ♯ S t t t t2ǫ u2 = kuǫ k2 − ǫ♯ + ǫ − ǫ ♯ S n + O(ǫn−1 ), 2 2 Ω ǫ 2n 2 2 which implies that Sn 1 − J (γ(tǫ )) ≤ 2n q
(2.5)
Z
Ω
g(tǫ uǫ − k)q+ + O(ǫn−1 )
♯
♯ 2 2 due to the fact that R maxt≥0 ((t q − 1)/2 − (t − 1)/2 ) = 0. We estimate Ω g(tǫ uǫ − k)+ as follows. For ǫ sufficiently small, Z Z Z n−1 n−1 q q ǫ− 2 q = Cǫn− 2 q (tǫ Uǫ − k)+ ≥ C g(tǫ uǫ − k)+ ≥ g0 Bǫ (0)
Bρ/2 (0)
Ω
for some C > 0 independent of ǫ. In the case q = n = 2, we note that on the set {ǫ < |x| < ǫ3/4 }, there holds ǫ1/2 ǫ−1/4 Uǫ (x) ≥ √ ≥ √ > 2k 2|x| 2
for ǫ sufficiently small. Hence, Z Z (tǫ Uǫ − k)2+ ≥ C g0 Bρ/2 (0)
{ǫ<|x|<ǫ3/4 }
ǫ 1 = Cǫ log 2 |x| ǫ
for some C > 0 independent of ǫ. Thus, there exists C > 0 such that ( n−1 Z Cǫn− 2 q , if q > 2 or n > 2 q g(tǫ uǫ − k)+ ≥ (2.6) Cǫ log 1ǫ if q = n = 2. Ω
6
J. WANG
Combining (2.5) and (2.6) yields the result. The proof is complete.
Now we are in the position to prove Theorem 1.1. Proof of Theorem 1.1. Since J satisfies the geometry of mountain pass, there exists a 1 (C) satisfying J (u ) → c and J ′ (u ) → 0 as m → ∞, where c sequence (um ) ⊂ H0,L m 0 m 0 is defined as in (2.4). Therefore, Lemmas 2.2 and 2.3 imply that problem (1.1) admits a nonnegative nontrivial solution. Finally, a maximum principle of Cabré and Tan [3] implies that the solution is positive in Ω. The proof is complete. References [1] Brézis, H. and Lieb, E.: A relation between pointwise convergence of functions and convergence of functionals. Proc. Amer. Math. Soc. 88, 486-490 (1983). [2] Brézis, H and Nirenberg, L.: Positive solutions of nonlinear elliptic equations involving critical Sobolev exponents. Comm. Pure Appl. Math. 36, 437-477 (1983). [3] Cabré, X. and Tan, J.: Positive solutions of nonlinear problems involving the square root of the Laplacian. Adv. Math. 224, 2052-2093 (2010). [4] Escobar, J.: Sharp constant in a Sobolev trace inequality. Indiana Univ. Math. J. 37, 687-698 (1988). [5] Gazzola F.: Critical growth quasilinear elliptic problems with shifting subcritical perturbation. Diff. Integral Equ. 14, 513-528 (2001). [6] Li, Q. and Xiang C.-L.: The fractional problem with shifting subcritical perturbation. Submitted. [7] Lions, P.L.: The concentration-compactness principle in the calculus of variations. The limit case II. Rev. Mat. Iberoamericana 1, 45-121 (1985). [8] Struwe M.: Variational methods. Applications to nonlinear partial differential equations and Hamiltonian systems. Second edition. Ergebnisse der Mathematik und ihrer Grenzgebiete (3), 34. SpringerVerlag, Berlin, 1996. [9] Tan J.: The Brezis-Nirenberg type problem involving the square root of the Laplacian. Calc. Var. Partial Differential Equations 42, 21-41 (2011). (Jixiu, Wang) School of Mathematics and Statistics, Hubei University of Arts and Science, Xiangyang, 441053, P.R.China. E-mail address: [email protected]