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Boundary singularities of solutions to semilinear fractional equations
Phuoc-Tai Nguyen, Laurent Veron, Laurent Eron
To cite this version:
Phuoc-Tai Nguyen, Laurent Veron, Laurent Eron. Boundary singularities of solutions to semilin-
ear fractional equations. Advanced Nonlinear Studies, Walter de Gruyter GmbH, In press. �hal-
01516714v3�
FRACTIONAL EQUATIONS
PHUOC-TAI NGUYEN AND LAURENT V ´ERON
Abstract. We prove the existence of a solution of (−∆)su+f(u) = 0 in a smooth bounded domain Ω with a prescribed boundary valueµin the class of Radon measures for a large class of continuous functions f satisfying a weak singularity condition expressed under an integral form. We study the existence of a boundary trace for positive moderate solutions. In the particular case wheref(u) =upandµis a Dirac mass, we show the existence of several critical exponentsp. We also demonstrate the existence of several types of separable solutions of the equation (−∆)su+up= 0 inRN+.
2010 Mathematics Subject Classification. 35J66, 35J67, 35R06, 35R11. Key words: s-harmonic functions, semilinear fractional equations, boundary trace.
Contents
1. Introduction 1
2. Linear problems 5
2.1. s-harmonic functions 5
2.2. Green kernel, Poisson kernel and Martin kernel 5
2.3. Boundary trace 7
2.4. Weak solutions of linear problems 10
3. Nonlinear problems 10
3.1. Subcritical absorption 10
3.2. Power absorption 15
Appendix A. Appendix - Separable solutions 22
A.1. Separable s-harmonic functions 22
A.2. The nonlinear problem 27
References 29
1. Introduction
Let Ω ⊂ R
Nbe a bounded domain with C
2boundary and s ∈ (0, 1). Define the s-fractional Laplacian as
(−∆)
su(x) := lim
ε→0
(−∆)
sεu(x) where
(−∆)
sεu(x) := a
N,sˆ
RN\Bε(x)
u(x) − u(y)
|x − y|
N+2sdy, a
N,s:= Γ(N/2 + s)
π
N/2Γ(2 − s) s(1 − s).
We denote by G
Ωsand M
sΩthe Green kernel and the Martin kernel of (−∆)
sin Ω respectively.
Denote by G
Ωsand M
Ωsthe Green operator and the Martin operator (see section 2 for more details). Further, for φ ≥ 0, denote by M(Ω, φ) the space of Radon measures τ on Ω satisfying
´
Ω
φd|τ | < ∞ and by M(∂Ω) the space of bounded Radon measures on ∂Ω. Let ρ(x) be the distance from x to ∂Ω. For β > 0, set
Ω
β:= {x ∈ Ω : ρ(x) < β}, D
β:= {x ∈ Ω : ρ(x) > β}, Σ
β:= {x ∈ Ω : ρ(x) = β}.
1
Definition 1.1. We say that a function u ∈ L
1loc(Ω) possesses a s-boundary trace on ∂Ω if there exists a measure µ ∈ M(∂Ω) such that
β→0
lim β
1−sˆ
Σβ
|u − M
Ωs[µ]|dS = 0. (1.1)
The s-boundary trace of u is denoted by tr
s(u).
Let τ ∈ M(Ω, ρ
s), µ ∈ M(∂Ω) and f ∈ C( R ) be a nondecreasing function with f (0) = 0.
In this paper, we study boundary singularity problem for semilinear fractional equation of the form
(−∆)
su + f(u) = τ in Ω tr
s(u) = µ
u = 0 in Ω
c.
(1.2) We denote by X
s(Ω) ⊂ C( R
N) the space of test functions ξ satisfying
(i) supp (ξ) ⊂ Ω, ¯
(ii) (−∆)
sξ(x) exists for all x ∈ Ω and |(−∆)
sξ(x)| ≤ C for some C > 0,
(iii) there exists ϕ ∈ L
1(Ω, ρ
s) and
0> 0 such that |(−∆)
sξ| ≤ ϕ a.e. in Ω, for all ∈ (0,
0].
Definition 1.2. Let τ ∈ M(Ω, ρ
s) and µ ∈ M(∂Ω). A function u is called a weak solution of (1.2) if u ∈ L
1(Ω), f (u) ∈ L
1(Ω, ρ
s) and
ˆ
Ω
(u(−∆)
sξ + f (u)ξ) dx = ˆ
Ω
ξdτ + ˆ
Ω
M
Ωs[µ](−∆)
sξ dx, ∀ξ ∈ X
s(Ω). (1.3) The boundary value problem with measure data for semilinear elliptic equations
−∆u + f (u) = 0 in Ω
u = µ on ∂Ω, (1.4)
was first studied by A. Gmira and L. V´ eron in [18] and then the typical model, i.e. problem (1.4) with f (u) = u
p(p > 1), has been intensively investigated by numerous authours (see [22–26]
and references therein). They proved that if f is a continuous, nondecreasing function satisfying ˆ
∞1
[f(t) − f (−t)]s
−1−pcdt < ∞, (1.5) where p
c:=
N+1N−1, then problem (1.4) admits a unique weak solution. In particular, when f (u) = u
pwith 1 < p < p
cand µ = kδ
0with 0 ∈ ∂Ω and k > 0, there exists a unique solution u
kof (1.4). It was showed [22, 26] that the sequence {u
k} is increasing and converges to a function u
∞which is a solution of the equation in (1.4).
To our knowledge, few papers concerning boundary singularity problem for nonlinear frac- tional elliptic equation have been published in the literature. The earliest works in this direction are the papers [10, 17] by P. Felmer et al. which deal with the existence, nonexistence and as- ymptotic behavior of large solutions for equations involving fractional Laplacian. Afterwards, N. Abatangelo [1] presented a suitable setting for the study of fractional Laplacian equations in a measure framework and provided a fairly comprehensive description of large solutions which improve the results in [10, 17]. Recently, H. Chen et al. [9] investigated semilinear elliptic equa- tions involving measures concentrated on the boundary by employing approximate method.
In the present paper, we aim to establish the existence and uniqueness of weak solutions of (1.2). To this end, we develop a theory for linear equations associated to (1.2)
(−∆)
su = τ in Ω tr
s(u) = µ
u = 0 in Ω
c.
(1.6)
Existence and uniqueness result for (1.6) is stated in the following proposition.
Proposition A. Assume s ∈ (
12, 1). Let τ ∈ M(Ω, ρ
s) and µ ∈ M(∂Ω). Then problem (1.6) admits a unique weak solution. The solution is given by
u = G
Ωs[τ ] + M
Ωs[µ]. (1.7)
Moreover, there exists a positive constant c = c(N, s, Ω) such that
kuk
L1(Ω)≤ c(kτ k
M(Ω,ρs)+ kµk
M(∂Ω)). (1.8) This proposition allows to study semilinear equation (1.2). We first deal with the case of L
1data.
Theorem B. Assume s ∈ (
12, 1). Let f ∈ C(R) be a nondecreasing function satisfying tf(t) ≥ 0 for every t ∈ R .
I. Existence and uniqueness. For every τ ∈ L
1(Ω, ρ
s) and µ ∈ L
1(∂Ω), problem (1.2) admits a unique weak solution u. Moreover,
u = G
Ωs[τ − f (u)] + M
Ωs[µ] in Ω, (1.9)
− G
Ωs[τ
−] − M
Ωs[µ
−] ≤ u ≤ G
Ωs[τ
+] + M
Ωs[µ
+] in Ω. (1.10) II. Monotonicity. The mapping (τ, µ) 7→ u is nondecreasing.
Remark. The restriction s ∈ (
12, 1) is due to the fact that in this range of s, tr
s( G [τ ]) = 0 for every τ ∈ M(Ω, ρ
s) (see Proposition 2.11). We conjecture that this still holds if s ∈ (0,
12].
We reveal that, in measures framework, because of the interplay between the nonlocal oper- ator (−∆)
sand the nonlinearity term f(u), the analysis is much more intricate and there are 3 critical exponents
p
∗1:= N + 2s
N , p
∗2:= N + s
N − s , p
∗3:= N N − 2s .
This yields substantial new difficulties and leads to disclose new types of results. The new aspects are both on the technical side and on the one of the new phenomena observed.
Theorem C. Assume s ∈ (
12, 1). Let f ∈ C(R) be a nondecreasing function, tf (t) ≥ 0 for every
t ∈ R and ˆ
∞1
[f(s) − f (−s)]s
−1−p∗2ds < ∞. (1.11) I. Existence and Uniqueness. For every τ ∈ M(Ω, ρ
s) and µ ∈ M(∂Ω) there exists a unique weak solution of (1.2). This solution satisfies (1.9) and (1.10). Moreover, the mapping (τ, µ) 7→ u is nondecreasing.
II. Stability. Assume {τ
n} ⊂ M(Ω, ρ
s) converges weakly to τ ∈ M(Ω, ρ
s) and {µ
n} ⊂ M(∂Ω) converges weakly to µ ∈ M(∂Ω). Let u and u
nbe the unique weak solutions of (1.2) with data (τ, µ) and (τ
n, µ
n) respectively. Then u
n→ u in L
1(Ω) and f (u
n) → f (u) in L
p(Ω, ρ
s).
If µ is a Dirac mass concentrated at a point on ∂Ω, we obtain the behavior of the solution near that boundary point.
Theorem D. Under the assumption of Theorem C, let z ∈ ∂Ω, k > 0 and u
Ωz,kbe the unique weak solution of
(−∆)
su + f (u) = 0 in Ω tr
s(u) = kδ
zu = 0 in Ω
c.
(1.12) Then
Ω3x→z
lim
u
Ωz,k(x)
M
sΩ(x, z) = k. (1.13)
We next assume that 0 ∈ ∂Ω. Let 0 < p < p
∗2and denote by u
Ωkthe unique weak solution of
(−∆)
su + u
p= 0 in Ω tr
s(u) = kδ
0u = 0 in Ω
c.
(1.14) By Theorem C, u
Ωk≤ kM
sΩ(·, 0) and k 7→ u
Ωkis increasing. Therefore, it is natural to investigate lim
k→∞u
Ωk. This is accomplishable thanks to the study of separable solutions of
(−∆)
su + u
p= 0 in R
N+u = 0 in R
N−(1.15) with p > 1. Denote by
S
N−1:=
σ = (cos φ σ
0, sin φ) : σ
0∈ S
N−2, −
π2≤ φ ≤
π2the unit sphere in R
Nand by S
+N−1:= S
N−1∩ R
N+the upper hemisphere. Writing separable solution under the form u(x) = u(r, σ) = r
−p−12sω(σ), with r > 0 and σ ∈ S
+N−1, we obtain that ω satisfies
A
sω − L
s, 2sp−1
ω + ω
p= 0 in S
+N−1ω = 0 in S
−N−1, (1.16)
where A
sis a nonlocal operator naturally associated to the s-fractional Laplace-Beltrami oper- ator and L
s, 2sp−1
is a linear integral operator with kernel. In analyzing the spectral properties of A
swe prove
Theorem E. Let N ≥ 2, s ∈ (0, 1) and p > p
∗1.
I- If p
∗2≤ p < p
∗3there exists no positive solution of (1.16) belonging to W
0s,2(S
+N−1).
II- If p
∗1< p < p
∗2there exists a unique positive solution ω
∗∈ W
0s,2(S
+N−1) of (1.16).
As a consequence of this result we obtain the behavior of u
Ωkwhen k → ∞.
Theorem F Assume s ∈ (
12, 1). Let Ω = R
N+or Ω be a bounded domain with C
2boundary containing 0.
I- If p ∈ (p
∗1, p
∗2) then u
Ω∞:= lim
k→0u
Ωkis a positive solution of (−∆)
su + u
p= 0 in Ω
u = 0 in Ω
c. (1.17)
(i) If Ω = R
N+then u
RN
∞+
(x) = |x|
−p−12sω
∗(σ), with σ = x
|x| ∀x ∈ R
N+. (ii) If Ω is a bounded C
2domain with ∂Ω containing 0 then
lim
Ω3x→0
x
|x| =σ∈S+N−1
|x|
p−12su
Ω∞(x) = ω
∗(σ), (1.18)
locally uniformly on S
+N−1. In particular, there exists a positive constant c depending on N , s, p and the C
2norm of ∂Ω such that
c
−1ρ(x)
s|x|
−(p+1)sp−1≤ u
Ω∞(x) ≤ cρ(x)
s|x|
−(p+1)sp−1∀x ∈ Ω. (1.19) II- Assume p ∈ (0, p
∗1]. Then lim
k→∞u
Ωk= ∞ in Ω.
The main ingredients of the present study: estimates on Green kernel and Martin kernel, the-
ory for linear fractional equations in connection with the notion s−boundary trace as mentioned
above, similarity transformation and the study of equation (1.16).
The paper is organized as follows. In Section 2, we present important properties of s-boundary trace and prove Proposition A. Theorems B,C,D and F are obtained in Section 3. Finally, in Appendix, we discuss separable solutions of (1.15) and demonstrate Theorem E.
2. Linear problems
Throughout the present paper, we denote by c, c
0, c
1, c
2, C, ... positive constants that may vary from line to line. If necessary, the dependence of these constants will be made precise.
2.1. s-harmonic functions. We first recall the definition of s-harmonic functions (see [3, page 46], [4, page 230], [6, page 20]). Denote by (X
t, P
x) the standard rotation invariant 2s-stable L´ evy process in R
N(i.e. homogeneous with independent increments) with characteristic function
E
0e
iξXt= e
−t|ξ|2s, ξ ∈ R
N, t ≥ 0.
Denote by E
xthe expectation with respect to the distribution P
xof the process starting from x ∈ R
N. We assume that sample paths of X
tare right-continuous and have left-hand limits a.s.
The process (X
t) is Markov with transition probabilities given by P
t(x, A) = P
x(X
t∈ A) = µ
t(A − x),
where µ
tis the one-dimensional distribution of X
twith respect to P
0. It is well known that (−∆)
sis the generator of the process (X
t, P
x).
For each Borel set D ⊂ R
N, set t
D:= inf{t ≥ 0 : X
t6∈ D}, i.e. t
Dis the first exit time from D. If D is bounded then t
D< ∞ a.s. Denote
E
xu(X
tD) = E
x{u(X
tD) : t
D< ∞}.
Definition 2.1. Let u be a Borel measurable function in R
N. We say that u is s-harmonic in Ω if for every bounded open set D b Ω,
u(x) = E
xu(X
tD), x ∈ D.
We say that u is singular s-harmonic in Ω if u is s-harmonic and u = 0 in Ω
c. Put
D
s:=
u : R
N7→ R : Borel measurable such that ˆ
RN
|u(x)|
(1 + |x|)
N+2s.
The following result follows from [5, Corollary 3.10 and Theorem 3.12] and [6, page 20] (see also [20]).
Proposition 2.2. Let u ∈ D
s.
(i) u is s-harmonic in Ω if and only if (−∆)
su = 0 in Ω in the sense of distributions.
(ii) u is singular s-harmonic in Ω if and only if u is s-harmonic in Ω and u = 0 in Ω
c. 2.2. Green kernel, Poisson kernel and Martin kernel. In what follows the notation f ∼ g means: there exists a positive constant c such that c
−1f < g < cf in the domain of the two functions or in a specified subset of this domain.
Denote by G
Ωsthe Green kernel of (−∆)
sin Ω. Namely, for every y ∈ Ω, ( (−∆)
sG
Ωs(·, y) = δ
yin Ω
G
Ωs(·, y) = 0 in Ω
c,
where δ
yis the Dirac mass at y. By combining [1, Lemma 3.2] and [14, Corollary 1.3]), we get Proposition 2.3. (i) G
Ωsis in continuous, symmetric, positive in {(x, y) ∈ Ω × Ω : x 6= y} and G
Ωs(x, y) = 0 if x or y belongs to Ω
c.
(ii) (−∆)
sG
Ωs(x, ·) ∈ L
1(Ω
c) for every x ∈ Ω and (−∆)
sG
Ωs(x, y) ≤ 0 for every x ∈ Ω and y ∈ Ω
c.
(iii) There holds G
Ωs(x, y) ∼ min n
|x − y|
2s−N, ρ(x)
sρ(y)
s|x − y|
−No
∀(x, y) ∈ Ω × Ω, x 6= y. (2.1)
The similarity constant in the above estimate depends only on Ω and s.
Denote by G
Ωsthe associated Green operator G
Ωs[τ ] =
ˆ
Ω
G
Ωs(·, y)dτ(y) τ ∈ M(Ω, ρ
s).
Put
k
s,γ:=
p
∗3if γ ∈ [0,
N−2sNs)
N+s
N−2s+γ
if γ ∈ [
NN−2ss, s]. (2.2) H. Chen and L. V´ eron obtained the following estimate for Green operator [12, Proposition 2.3 and Proposition 2.6].
Lemma 2.4. Assume γ ∈ [0, s] and k
s,γbe as in (2.2).
(i) There exists a constant c = c(N, s, γ, Ω) > 0 such that
G
Ωs[τ ]
Mks,γ(Ω,ρs)≤ c kτ k
M(Ω,ργ)∀τ ∈ M(Ω, ρ
γ). (2.3) (ii) Assume {τ
n} ⊂ M(Ω, ρ
γ) converges weakly to τ ∈ M(Ω, ρ
γ). Then G
Ωs[τ
n] → G
Ωs[τ ] in L
p(Ω, ρ
s) for any p ∈ [1, k
s,γ).
Let P
sΩbe the Poisson kernel of (−∆)
sdefined by (see [7]) P
sΩ(x, y) := −a
N,−sˆ
Ω
G
Ωs(x, z)
|z − y|
N+2sdz, ∀x ∈ Ω, y ∈ Ω
c.
The relation between P
sΩand G
Ωsis expressed in [1, Proposition 2] (see also [14, Theorem 1.4], [4, Lemma 2], [14, Theorem 1.5]).
Proposition 2.5. (i) P
sΩ(x, y) = −(−∆)
sG
Ωs(x, y) for every x ∈ Ω and y ∈ Ω
c. Moreover, P
sΩis continuous in Ω × Ω
c.
(ii) There holds
P
sΩ(x, y) ∼ ρ(x)
sρ(y)
s(1 + ρ(y))
s1
|x − y|
N, ∀x ∈ Ω, y ∈ Ω
c. (2.4) The similarity constant in the above estimate depends only on Ω and s.
Denote by P
Ωsthe corresponding operator defined by P
Ωs[ν](x) =
ˆ
Ωc
P
sΩ(x, y)dν(y), ν ∈ M(Ω
c).
Fix a reference point x
0∈ Ω and denote by M
sΩthe Martin kernel of (−∆)
sin Ω, i.e.
M
sΩ(x, z) = lim
Ω3y→z
G
Ωs(x, y)
G
Ωs(x
0, y) , ∀x ∈ R
N, z ∈ ∂Ω.
By [15, Theorem 3.6], the Martin boundary of Ω can be identified with the Euclidean boundary
∂Ω. Denote by M
Ωsthe associated Martin operator M
Ωs[µ] =
ˆ
∂Ω
M
sΩ(·, z)dµ(z), µ ∈ M(∂Ω).
The next result [4, 15] is important in the study of s-harmonic functions, which give a unique presentation of s−harmonic functions in terms of Martin kernel.
Proposition 2.6. (i) The mapping (x, z) 7→ M
sΩ(x, z) is continuous on Ω × ∂Ω. For any z ∈ ∂Ω, the function M
sΩ(., z ) is singular s-harmonic in Ω with M
sΩ(x
0, z) = 1. Moreover, if z, z
0∈ ∂Ω, z 6= z
0then lim
x→z0M
sΩ(x, z) = 0.
(ii) There holds
M
sΩ(x, z) ∼ ρ(x)
s|x − z|
−N∀x ∈ Ω, z ∈ ∂Ω. (2.5) The similarity constant in the above estimate depends only on Ω and s.
(iii) For every µ ∈ M
+(∂Ω) the function M
Ωs[µ] is singular s-harmonic in Ω with u(x
0) =
µ( R
N). Conversely, if u is a nonnegative singular s-harmonic function in Ω then there exists a
unique µ ∈ M
+(∂Ω) such that u = M
Ωs[µ] in R
N.
(iv) If u is a nonnegative s-harmonic function in Ω then there exists a unique µ ∈ M
+(∂Ω) such that
u(x) = M
Ωs[µ](x) + P
Ωs[u](x) ∀x ∈ Ω.
Lemma 2.7. (i) There exists a constant c = c(N, s, γ, Ω) such that
M
Ωs[µ]
M
N+γ N−s(Ω,ργ)
≤ c kµk
M(∂Ω), ∀µ ∈ M(∂Ω), γ > −s. (2.6) (ii) If {µ
n} ⊂ M(∂Ω) converges weakly to µ ∈ M(∂Ω) then M
Ωs[µ
n] → M
Ωs[µ] in L
p(Ω, ρ
γ) for every 1 ≤ p <
NN−s+γ.
Proof. (i) By using (2.5) and a similar argument as in the proof of [2, Theorem 2.5], we obtain (2.6).
(ii) By combining the fact that M
sΩ(x, z) = 0 for every x ∈ Ω
c, z ∈ ∂Ω and Proposition 2.6 (i) we deduce that for every x ∈ R
N, M
sΩ(x, ·) ∈ C(∂Ω). It follows that M
Ωs[µ
n] → M
Ωs[µ]
everywhere in Ω. Due to (i) and the Holder inequality, we deduce that, for any 1 ≤ p ≤
N+γN−s, { M
Ωs[µ
n]} is uniformly integrable with respect to ρ
γdx. By invoking Vitali’s theorem, we obtain
the convergence in L
p(Ω, ρ
γ).
2.3. Boundary trace. We recall that, for β > 0,
Ω
β:= {x ∈ Ω : ρ(x) < β}, D
β:= {x ∈ Ω : ρ(x) > β}, Σ
β:= {x ∈ Ω : ρ(x) = β}.
The following geometric property of C
2domains can be found in [26].
Proposition 2.8. There exists β
0> 0 such that
(i) For every point x ∈ Ω
β0, there exists a unique point z
x∈ ∂Ω such that |x − z
x| = ρ(x).
This implies x = z
x− ρ(x)n
zx.
(ii) The mappings x 7→ ρ(x) and x 7→ z
xbelong to C
2(Ω
β0) and C
1(Ω
β0) respectively. Fur- thermore, lim
x→zx∇ρ(x) = −n
zx.
Proposition 2.9. Assume s ∈ (0, 1). Then there exist positive constants c = c(N, Ω, s) such that, for every β ∈ (0, β
0),
c
−1≤ β
1−sˆ
Σβ
M
sΩ(x, y)dS(x) ≤ c ∀y ∈ ∂Ω. (2.7) Proof. For r
0> 0 fixed, by (2.5),
ˆ
Σβ\Br0(y)
M
sΩ(x, y)dS(x) ≤ c
1β
s, (2.8) which implies
β→0
lim ˆ
Σβ\Br0(y)
M
sΩ(x, y)dS(x) = 0 ∀y ∈ ∂Ω. (2.9) Note that for r
0fixed, the rate of convergence is independent of y.
In order to prove (2.7) we may assume that the coordinates are placed so that y = 0 and the tangent hyperplane to ∂Ω at 0 is x
N= 0 with the x
Naxis pointing into the domain. For x ∈ R
Nput x
0= (x
1, · · · , x
N−1). Pick r
0∈ (0, β
0) sufficiently small (depending only on the C
2characteristic of Ω) so that
1
2 (|x
0|
2+ ρ(x)
2) ≤ |x|
2∀x ∈ Ω ∩ B
r0(0).
Hence if x ∈ Σ
β∩ B
r0(0) then
14(|x
0| + β) ≤ |x|. Combining this inequality and (2.5) leads to ˆ
Σβ∩Br0(0)
M
sΩ(x, 0)dS(x) ≤ c
2β
sˆ
Σβ,0
(|x
0| + β)
−NdS(x)
≤ c
2β
sˆ
|x0|<r0
(|x
0| + β )
−Ndx
0= c
3β
s−1.
Therefore, for β < r
0,
β
1−sˆ
Σβ∩Br0(0)
M
sΩ(x, 0)dS(x) ≤ c
4. (2.10) By combining estimates (2.8) and (2.10), we obtain the second estimate in (2.7). The first
estimate in (2.7) follows from (2.5).
As a consequence, we get the following estimates.
Corollary 2.10. Assume s ∈ (0, 1). For every µ ∈ M
+(∂Ω) and β ∈ (0, β
0), there holds c
−1kµk
M(∂Ω)≤ β
1−sˆ
Σβ
M
Ωs[µ]dS ≤ c kµk
M(∂Ω), (2.11) with c is as in (2.7).
Proposition 2.11. Assume s ∈ (
12, 1). Then there exists a constant c = c(s, N, Ω) such that for any τ ∈ M(Ω, ρ
s) and any 0 < β < β
0,
β
1−sˆ
Σβ
G
Ωs[τ ]dS ≤ c ˆ
Ω
ρ
sd|τ |. (2.12)
Moreover,
β→0
lim β
1−sˆ
Σβ
G
Ωs[τ ]dS = 0. (2.13)
Proof. Without loss of generality, we may assume that τ > 0. Denote v := G
Ωs[τ ]. We first prove (2.12). By Fubini’s theorem and (2.5),
ˆ
Σβ
v(x)dS(x) ≤ c
5ˆ
Ω
ˆ
Σβ∩Bβ 2
(y)
|x − y|
2s−NdS(x) dτ (y)
+ β
sˆ
Ω
ˆ
Σβ\Bβ
2
(y)
|x − y|
−NdS(x) ρ(y)
sdτ (y)
:= I
1,β+ I
2,β. Note that, if x ∈ Σ
β∩ B
β2
(y) then β/2 ≤ ρ(y) ≤ 3β/2. Therefore β
1−sI
1,β≤ c
6β
1−2sˆ
Σβ∩Bβ
2
(y)
|x − y|
2s−NdS(x) ˆ
Ω
ρ(y)
sdτ (y)
≤ c
6β
1−2sˆ
β/20
r
2s−Nr
N−2dr ˆ
Ω
ρ(y)
sdτ (y)
≤ c
7ˆ
Ω
ρ(y)
sdτ(y),
where the last inequality holds since s >
12. On the other hands, we have I
2,β≤ c
7β
sˆ
∞β/2
r
−Nr
N−2dr ˆ
Ω
ρ(y)
sdτ (y) = c
8β
s−1ˆ
Ω
ρ(y)
sdτ (y).
Combining the above estimates, we obtain (2.12).
Next we demonstrate (2.13). Given ∈ (0, kτ k
M(Ω,ρs)) and β
1∈ (0, β
0) put τ
1= τ χ
Dβ¯ 1and τ
2= τ χ
Ωβ1
. We can choose β
1= β
1() such that ˆ
Ωβ1
ρ(y)
sdτ(y) ≤ . (2.14)
Thus the choice of β
1depends on the rate at which ´
Ωβ
ρ
sdτ tends to zero as β → 0.
Put v
i:= G
Ωs[τ
i]. Then, for 0 < β < β
1/2, ˆ
Σβ
v
1(x) dS(x) ≤ c
9β
sβ
−N1ˆ
Ω
ρ(y)
sdτ
1(y),
which yields
β→0
lim β
1−sˆ
Σβ
v
1(x) dS(x) = 0. (2.15)
On the other hand, due to (2.12), β
1−sˆ
Σβ
v
2dS ≤ c
10ˆ
Ω
ρ
sdτ
2≤ c
11∀β < β
0. (2.16)
From (2.15) and (2.16), we obtain (2.13).
Lemma 2.12. Assume s ∈ (
12, 1). Let u, w ∈ D
sbe two nonnegative functions satisfying ( (−∆)
su ≤ 0 ≤ (−∆)
sw in Ω,
u = 0 in Ω
c. (2.17)
If u ≤ w in R
Nthen (−∆)
su ∈ M(Ω, ρ
s) and there exists a measure µ ∈ M
+(∂Ω) such that
β→0
lim β
1−sˆ
Σβ
|u − M
Ωs[µ]|dS = 0. (2.18)
Moreover, if µ = 0 then u = 0.
Proof. By the assumption, there exists a nonnegative Radon measure τ on Ω such that (−∆)
su =
−τ .
We first prove that τ ∈ M
+(Ω, ρ
s). Define M ˜
sΩ(x, z) := lim
Ω3y→z
G
Ωs(x, y)
ρ(y)
s. (2.19)
By [1, page 5547], there holds is a positive constant c = c(Ω, s) such that
M ˜
sΩ(x, z) ∼ ρ(x)
s|x − z|
−N, ∀x ∈ Ω, z ∈ ∂Ω, (2.20) where the similarity constant depends only on Ω and s. This follows
c
−112< c
−113ˆ
∂Ω
ρ(x)|x − z|
−NdS(z)
≤ ρ(x)
1−sˆ
∂Ω
M ˜
sΩ(x, z)dS(z)
≤ c
13ˆ
∂Ω
ρ(x)|x − z|
−NdS(z) < c
12∀x ∈ Ω.
(2.21)
We define
E
Ωs[u](z) := lim
Ω3x→z
´ u(x)
∂Ω
M ˜
sΩ(x, y)dS(y) z ∈ ∂Ω.
For any β ∈ (0, β
0), denote by τ
βthe restriction of τ to D
βand by v
βthe restriction of u on Σ
β. By [1, Theorem 1.4], there exists a unique solution v
βof
(−∆)
sv
β= −τ
βin D
βE
Dsβ[v
β] = 0 on Σ
βv
β= u|
Dcβ
in D
βc. Moreover, the solution can be written as
v
β+ G
Dsβ[τ
β] = P
Dsβ[u|
Dcβ
] in D
β. (2.22)
By the maximum principle [1, Lemma 3.9], v
β= u and P
Dsβ[u|
Dcβ
] ≤ w a.e. in R
N. This,
together with (2.22), implies that G
Dsβ[τ
β] ≤ w in D
β. Letting β → 0 yields G
Ωs[τ ] < ∞. For
fixed x
0∈ Ω, by (2.1), G
Ωs(x
0, y) > cρ(y)
sfor every y ∈ Ω. Hence the finiteness of G
Ωs[τ ] implies
that τ ∈ M
+(Ω, ρ
s).
We next show that there exists a measure µ ∈ M
+(∂Ω) such that (2.18) holds. Put v = u + G
Ωs[τ ] then v is a nonnegative singular s-harmonic in Ω due to the fact that G
Ωs[τ ] = 0 in Ω
c. By Proposition 2.2 and Proposition 2.6 (iii), there exists µ ∈ M
+(∂Ω) such that v = M
Ωs[µ]
in R
N. By Proposition 2.11, we obtain (2.18). If µ = 0 then v = 0 and thus u = 0.
Definition 2.13. A function u possesses a s-boundary trace on ∂Ω if there exists a measure µ ∈ M(∂Ω) such that
β→0
lim β
1−sˆ
Σβ
|u − M
Ωs[µ]|dS = 0. (2.23)
The s-boundary trace of u is denoted noted by tr
s(u).
Remark. (i) The notation of s-boundary trace is well defined. Indeed, suppose that µ and µ
0satisfy (2.23). Put v = ( M
Ωs[µ − µ
0])
+. Clearly v ≤ M
Ωs[|µ| + |µ
0|], v = 0 in Ω
cand lim
β→0β
1−s´
Σβ
|v|dS = 0. By Kato’s inequality [8, Theorem 1.2], (−∆)
sv ≤ 0 in Ω. Therefore, we deduce v ≡ 0 from Lemma 2.12. This implies M
Ωs[µ − µ
0] ≤ 0. By permuting the role of µ and µ
0, we obtain M
Ωs[µ − µ
0] ≥ 0. Thus µ = µ
0.
(ii) It is clear that for every µ ∈ M(∂Ω), tr
s( M
Ωs[µ]) = µ. Moreover, if s >
12, by Proposi- tion 2.11, for every τ ∈ M(Ω, ρ
s), tr
s( G
Ωs[τ ]) = 0.
(iii) This kind of boundary trace was first introduced by P.-T. Nguyen and M. Marcus [21]
in order to investigate semilinear elliptic equations with Hardy potential. In the present paper we prove that it is still an effective tools in the study of nonlocal fractional elliptic equations.
2.4. Weak solutions of linear problems.
Definition 2.14. Let τ ∈ M(Ω, ρ
s) and µ ∈ M(∂Ω). A function u is called a weak solution of (1.6) if u ∈ L
1(Ω) and
ˆ
Ω
u(−∆)
sξ dx = ˆ
Ω
ξdτ + ˆ
Ω
M
Ωs[µ](−∆)
sξ dx, ∀ξ ∈ X
s(Ω). (2.24) Proof of Proposition A. The uniqueness follows from [12, Proposition 2.4]. Let u be as in (1.7). By [12],
ˆ
Ω
(u − M
Ωs[µ])(−∆)
sξ dx = ˆ
Ω
G
Ωs[τ ](−∆)
sξ dx = ˆ
Ω
ξdτ ∀ξ ∈ X
s(Ω).
This implies (2.24) and therefore u is the unique solution of (1.6). Since s ∈ (
12, 1), by Propo- sition 2.11, tr
s(u) = tr
s(M
Ωs[µ]) = µ. Finally, estimate (1.8) follows from Lemma 2.4 and
Lemma 2.7.
3. Nonlinear problems
In this section, we study the nonlinear problem (1.2). The definition of weak solutions of (1.2) is given in Definition 1.2.
3.1. Subcritical absorption. Proof of Theorem B.
Monotonicity. Let τ, τ
0∈ L
1(Ω, ρ
s), µ, µ
0∈ L
1(∂Ω) and u and u
0be the weak solutions of (1.2) with data (τ, µ) and (τ
0, µ
0) respectively. We will show that if τ ≤ τ
0and µ ≤ µ
0then u ≤ u
0in Ω. Indeed, put v := (u − u
0)
+, it is sufficient to prove that v ≡ 0. Since (1.9) holds, it follows
|u| ≤ G
Ωs[|τ | + |f (u)|] + M
Ωs[|µ|] in Ω.
Similarly
|u
0| ≤ G
Ωs[|τ
0| + |f (u
0)|] + M
Ωs[|µ
0|] in Ω.
Therefore
0 ≤ v ≤ |u| + |u
0| ≤ G
Ωs[|τ | + |τ
0| + |f (u)| + |f (u
0)|] + M
Ωs[|µ| + |µ
0|] := w.
By Kato inequality, the assumption τ ≤ τ
0and the monotonicity of f, we obtain
(−∆)
sv ≤ sign
+(u − u
0)(τ − τ
0) − sign
+(u − u
0)(f (u) − f (u
0)) ≤ 0.
Therefore
(−∆)
sv ≤ 0 ≤ (−∆)
sw in Ω.
Since µ ≤ µ
0, it follows that tr
s(v) = 0. By Lemma 2.12, v = 0 and thus u ≤ u
0. Existence.
Step 1: Assume that τ ∈ L
∞(Ω) and µ ∈ L
∞(∂Ω).
Put ˆ f (t) := f (t + M
Ωs[µ]) − f ( M
Ωs[µ]) and ˆ τ := τ − f ( M
Ωs[µ]). Then ˆ f is nondecreasing and t f ˆ (t) ≥ 0 for every t ∈ R and ˆ τ ∈ L
1(Ω, ρ
s). Consider the problem
( (−∆)
sv + ˆ f (v) = ˆ τ in Ω
v = 0 in Ω
c. (3.1)
By [11, Proposition 3.1] there exists a unique weak solution v of (3.1). It means that v ∈ L
1(Ω), f ˆ (v) ∈ L
1(Ω, ρ
s) and
ˆ
Ω
(v(−∆)
sξ + ˆ f (v)ξ) dx = ˆ
Ω
ξ τ dx, ˆ ∀ξ ∈ X
s(Ω). (3.2) Put u := v + M
Ωs[µ] then u ∈ L
1(Ω) and f (u) ∈ L
1(Ω, ρ
s). By (3.2) u satisfies (1.3).
Step 2: Assume that 0 ≤ τ ∈ L
1(Ω, ρ
s) and 0 ≤ µ ∈ L
1(∂Ω).
Let {τ
n} ⊂ C
1(Ω) be a nondecreasing sequence converging to τ in L
1(Ω, ρ
s) and {µ
n} ⊂ C
1(∂Ω) be a nondecreasing sequence converging to µ in L
1(∂Ω). Then { M
Ωs[µ
n]} is nondecreas- ing and by Lemma 2.7 (ii) it converges to M
Ωs[µ] a.e. in Ω and in L
p(Ω, ρ
s) for every 1 ≤ p < p
∗2. Let u
nbe the unique solution of (1.2) with τ and µ replaced by τ
nand µ
nrespectively. By step 1 and the monotonicity of f , we derive that {u
n} and {f(u
n)} are nondecreasing. Moreover
ˆ
Ω
(u
n(−∆)
sξ + f(u
n)ξ) dx = ˆ
Ω
ξdτ
n+ ˆ
Ω