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Asymptotic behavior of positive solutions of a singular nonlinear Dirichlet problem
J. Math. Anal. Appl. 369 (2010) 719–729
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Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa
Asymptotic behavior of positive solutions of a singular nonlinear Dirichlet problem Sabrine Gontara, Habib Mâagli, Syrine Masmoudi, Sameh Turki ∗ Département de Mathématiques, Faculté des Sciences de Tunis, Campus Universitaire, 2092 Tunis, Tunisia
a r t i c l e
i n f o
a b s t r a c t Let Ω be a C 1,1 -bounded domain in Rn for n 2. In this paper, we are concerned with the asymptotic behavior of the unique positive classical solution to the singular boundary-value problem u + a(x)u −σ = 0 in Ω , u |∂Ω = 0, where σ 0, a is a nonnegative function in α (Ω), 0 < α < 1 and there exists c > 0 such that 1 a(x)(δ(x))λ m (Log ( ω ))μk c. C loc k δ(x) k=1 c Here λ 2, μk ∈ R, ω is a positive constant and δ(x) = dist(x, ∂Ω). © 2010 Elsevier Inc. All rights reserved.
Article history: Received 21 December 2009 Available online 3 April 2010 Submitted by V. Radulescu Keywords: Asymptotic behavior Dirichlet problem Singular equation
1. Introduction We are interested in this paper on the estimates of positive solutions for nonlinear problems of the type
−u = a(x) g (u ), x ∈ Ω, u > 0 in Ω, u |∂Ω = 0,
(1.1)
where Ω is a C 1,1 -bounded domain in Rn (n 2). The existence of such solutions and their asymptotic behavior have been investigated by many authors (see [1–9,12,15] and the references therein). For a ≡ 1 on Ω and g (u ) = u −σ for σ > 1, Crandall, Rabinowitz and Tartar proved in [6] that the problem (1.1) has a unique classical solution u in Ω and showed that this solution satisfies the following estimates
1+2σ
c 1 δ(x)
2 u (x) c 2 δ(x) 1+σ ,
near the boundary ∂Ω,
(1.2)
where c 1 , c 2 are positive constants and δ(x) = dist(x, ∂Ω). In [9], Lazer and McKenna established the estimate (1.2) on Ω and instead of a ≡ 1 on Ω , they assumed that 0 < b1 a(x)(δ(x))λ b2 , for all x ∈ Ω , where b1 , b2 are positive constants and λ ∈ (0, 2); then they proved that for σ > 1, there exist two positive constants c 1 and c 2 such that
1+2σ
c 1 δ(x)
2−λ u (x) c 2 δ(x) 1+σ ,
for x ∈ Ω.
(1.3)
Most recently, applying Karamata regular variation theory, many authors studied the exact asymptotic behavior of solutions of the problem (1.1) (see [1–5,15]). For instance, in [15], Zhang has dealt with functions in the Karamata class called regularly varying at zero with index −γ , γ ∈ R which we recall in the following
*
Corresponding author. E-mail addresses: [email protected] (S. Gontara), [email protected] (H. Mâagli), [email protected] (S. Masmoudi), [email protected] (S. Turki). 0022-247X/$ – see front matter doi:10.1016/j.jmaa.2010.04.008
©
2010 Elsevier Inc. All rights reserved.
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Definition 1. (See [11,13].) A positive measurable function k defined on some interval (0, η), varying at zero with index −γ , γ ∈ R, if for each ξ > 0
lim
t →0+
k(ξ t )
η > 0, is called regularly
= ξ −γ .
k(t )
Remark 1. As typical example of function k in this class, we quote
k(t ) = t −γ
m
−μi 1
Log i
,
t
i =1
where Log i (x) = Log ◦ Log ◦ · · · ◦ Log(x) (i times),
μi ∈ R .
In [15], Zhang studied the problem (1.1) under the following assumptions:
(H1 ) g ∈ C 1 ((0, ∞), (0, ∞)), g (s) 0 for all s > 0 and g is regularly varying at zero with index −γ , γ > 0. α (Ω), 0 < α < 1 and there exist η > 0 and h ∈ C ((0, η), (0, ∞)) nonincreasing and (H2 ) a is a nonnegative function in C loc regularly varying at zero with index −λ, 0 < λ < 2, such that a(x)
lim
δ(x)→0 h (δ(x))
= c 0 > 0.
Then he proved the following result: Theorem 1. (See [15].) Suppose (H1 ) and (H2 ) hold. Then the problem (1.1) has a unique solution u ∈ C 2+α (Ω) ∩ C (Ω) such that
u (x)
lim
δ(x)→0 p (δ(x))
1
γ +1
= c0
,
where p is the unique local solution to the problem
− p (s) = h(s) g p (s) , s ∈ (0, η), p (s) > 0, s ∈ (0, η), p (0) = 0.
In [12], the authors extended the above result. In fact, they allowed both terms a(x) and g (u ) to be in more general classes of functions. More precisely, the function a was assumed to satisfy a similar asymptotic behavior and was required to be in the elliptic Kato class K (Ω) studied and introduced in [10] and [14]. For the reader’s convenience we recall the definition of K (Ω) and we give a class of functions belonging to this class. The result proved in [12] will be detailed in Section 2. Definition 2. A Borel measurable function q in Ω belongs to the Kato class K (Ω) if
δ( y ) G (x, y ) q( y ) dy = 0. δ(x)
lim sup
α →0 x∈Ω
B (x,α )∩Ω
Here and always G denotes the Green function for the Laplace operator in Ω , with zero Dirichlet boundary condition. Example 1. (See [10] and [14].) (i) For p > n2 , we have L p (Ω) ⊂ K (Ω).
(ii) The function x → (δ(1x))λ is in K (Ω) if and only if λ < 2. In fact, we have the more general example: Example 2. (See [12].) Let m ∈ N∗ and
ψ(t ) = t −λ
m
Logk
k =1
ω be a sufficiently large positive constant such that the function
−μk
ω t
is defined and positive on (0, 2d) where d = diam(Ω).
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Let p > n2 , then if one of the following conditions is satisfied
• λ<2− • λ=2−
n p n p
and and
μk ∈ R, for 1 k m, μ1 = μ2 = · · · = μl−1 = 1 − 1p , μl > 1 − 1p , μk ∈ R for l + 1 k m,
we have the function x → ψ(δ(x)) f (x) is in K (Ω), provided that f ∈ L p (Ω). Remark 2. Let k be a function regularly varying at zero with index −γ . If 0 < γ < 2, then by [12, Corollary 1] we have k(δ(.)) ∈ K (Ω). To simplify our statements, the letter c will denote a generic positive constant which may vary from line to line and for two nonnegative functions f and g on a set S, we write f (x) g (x), for x ∈ S, if there exists a constant c > 0 not depending on x such that
1 c
g (x) f (x) cg (x),
∀x ∈ S .
In this paper, we study the asymptotic behavior to the solution of the following problem
where
−u = a(x)u −σ , x ∈ Ω, u > 0 in Ω, u |∂Ω = 0,
(1.4)
σ 0 and the function a satisfies the following hypothesis
α (Ω), 0 < α < 1, satisfying for all x ∈ Ω (H) a is a nonnegative function in C loc
−μk m −λ ω a(x) δ(x) Logk . δ(x)
k =1
Here and throughout this paper, m ∈ N∗ and satisfy one of the following conditions
ω is a positive constant large enough. The real numbers λ and μk , 1 k m,
• λ < 2, μ1 , μ2 , . . . , μm ∈ R, and • λ = 2, μ1 = μ2 = · · · = μl−1 = 1, μl > 1, μk ∈ R for l + 1 k m. Remark 3. It is obvious to see from Example 2 that if the function a satisfies hypothesis (H), then a is in the Kato class K (Ω) and which implies by [12, Theorem 2] that problem (1.4) has a unique solution u ∈ C 2+α (Ω) ∩ C (Ω). The asymptotic behavior of such solution depends on the choice of the constants λ and results into the following theorems.
σ . Hence we divide our main
Theorem 2. Assume that a satisfies (H). If 1 − σ < λ < 2 and μk ∈ R for 1 k m, then the solution u of the problem (1.4) satisfies for each x ∈ Ω
u (x)
2−λ
δ(x) σ +1
m k =1
Remark 4. If we consider satisfies 2−λ
u (x) δ(x) σ +1 ,
Logk
ω
−σ +μk1 .
δ(x)
σ > 1, λ ∈ (0, 2) and μk = 0 for all 1 k m in Theorem 2, the solution u of the problem (1.4)
for x ∈ Ω,
which improves the inequalities (1.3) found in [9] by Lazer and McKenna. Theorem 3. Assume that a satisfies (H). If λ = 2, μ1 = μ2 = · · · = μl−1 = 1, μl > 1 and μk ∈ R for l + 1 k m, then the solution u of the problem (1.4) satisfies for each x ∈ Ω
u (x) Logl
ω δ(x)
1σ−+μ1l m k=l+1
Logk
ω δ(x)
−σ +μk1 .
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Theorem 4. Assume that a satisfies (H) and λ 1 − σ . Let u be a solution of the problem (1.4).
• If λ < 1 − σ , then u (x) δ(x),
for x ∈ Ω.
• If λ = 1 − σ , μ1 = μ2 = · · · = μl−1 = 1, μl > 1 and μk ∈ R for l + 1 k m, then u (x) δ(x),
for x ∈ Ω.
• If λ = 1 − σ , μ1 = μ2 = · · · = μl−1 = 1, μl < 1 and μk ∈ R for l + 1 k m, then
1σ−+μ1l −σ +μk1 m ω ω u (x) δ(x) Logl Logk , δ(x) δ(x)
for x ∈ Ω.
k=l+1
• If λ = 1 − σ and μ1 = μ2 = · · · = μm = 1, then
u (x) δ(x)Logm+1
ω
σ +1 1 ,
δ(x)
for x ∈ Ω.
Remark 5. We mention that if σ = 0 and the function a satisfies an exact asymptotic behavior, the results in Theorems 2, 3 and 4 have been stated in [12]. For the convenience of the readers, this result is recalled in Lemma 2 stated below. The remaining part of this paper is organized as follows. In Section 2, we state some technical lemmas which will be useful for the proof of our results. Section 3 is devoted to prove our main results in Theorems 2, 3 and 4. 2. Technical lemmas Lemma 1. Let η > 0, m ∈ N∗ , l ∈ N∗ , βl = 0 and βk ∈ R for k l + 1. Then the function θ defined on (0, η) by
θ(t ) =
m
−βk
ω
Logk
,
t
k=l
satisfies: θ (t )
(i) limt →0+ t θ(t ) = 0. θ (t ) (ii) limt →0+ t θ (t ) = −1. θ (t )
(iii) limt →0+ t 2 θ(t ) = 0. Proof. By an elementary calculus, we have
θ (t ) 1 = h(t ), θ(t ) t
(2.1)
where
h(t ) =
m
βk
k
−1 Log j
j =1
k=l
ω t
.
(2.2)
Since limt →0+ h(t ) = 0, then (i) holds. Now, we shall prove (ii). By differentiation, we have
θ (t ) = −
1 t2
h(t )θ(t ) +
1 1 h (t )θ(t ) + h(t )θ (t ). t t
So, by using (2.1), we obtain
t θ (t )
θ (t )
= −1 +
th (t )
We claim that
lim
t →0+
th (t ) h(t )
= 0.
h(t )
+ h(t ).
(2.3)
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To prove the claim, we write (2.2) of the form
h(t ) =
m
βk S k (t ),
k=l
where
S k (t ) =
k
−1
ω
Log j
.
t
j =1
Using that
S k (t ) S k (t )
1 j
k
=−
t
−1
t
S k (t )
=
1 Log t
+o
,
t
j =1 i =1
it follows that for each k ∈ N∗ ,
S k (t )
ω
Log i
1 Log t
if t → 0+ .
,
This implies that
th (t ) = t
m
βk S k (t ) =
1 Log t
k=l
+o
1
Log t
for t → 0+ .
h(t ),
So, we deduce
th (t )
lim
t →0+
h(t )
= 0.
This, together with (2.3), proves (ii). Finally, (iii) holds immediately from (i) and (ii).
2
The following lemma is stated in [12, Example 6]. To present it we need the following notation. For a nonnegative measurable function a in Ω , we denote by Va the potential of a defined in Ω by Va(x) = Ω G (x, y )a( y ) dy. Lemma 2. (See [12].) Let m ∈ N∗ and ω be a sufficiently large positive real number such that the function ψ is positive and defined on (0, 1) by
ψ(t ) = t −λ
m
−μk
Logk
k =1
ω
,
t
where the constants λ and μk , 1 k m, satisfy one of the following conditions:
• λ < 2, μk ∈ R, for 1 k m, • λ = 2, μ1 = μ2 = · · · = μl−1 = 1, μl > 1 and μk ∈ R for l + 1 k m. α (Ω) ∩ K (Ω), 0 < α < 1, such that Let a be a nonnegative function in C loc
lim
δ(x)→0
a(x) = c 0 > 0. ψ(δ(x))
Then we have the following estimates: (a) If λ = 2, then
Va(x)
∼
δ(x)→0
c Logl
ω
1−μl m
δ(x)
Logk
k=l+1
ω δ(x)
(b) If 1 < λ < 2, then
−μk m 2−λ ω Va(x) ∼ c δ(x) Logk . δ(x)→0 δ(x)
k =1
−μk .
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(c) If λ = 1 and μ1 = μ2 = · · · = μm = 1, then
Va(x)
∼
δ(x)→0
c δ(x) Logm+1
(d) If λ = 1 and l = inf{1 k m, (i) If μl > 1, then
Va(x)
δ(x)
.
μk = 1} exists, then:
∼
c δ(x).
∼
c δ(x) Logl
δ(x)→0
ω
(ii) If μl < 1, then
Va(x)
δ(x)→0
ω
1−μl m
Logk
δ(x)
k=l+1
ω δ(x)
−μk .
(e) If λ < 1, then
Va(x)
∼
δ(x)→0
c δ(x).
Now, let ϕ be a positive eigenfunction corresponding to the first smallest eigenvalue λ1 of the operator in Ω with zero Dirichlet boundary condition and satisfying ϕ ∞ = supx∈Ω |ϕ (x)| = 1. It is well known that for each x ∈ Ω ,
ϕ (x) δ(x).
(2.4)
Lemma 3. The function θ defined in Lemma 1 satisfies
θ ϕ (x) θ δ(x) ,
if x is near ∂Ω.
Proof. We intend to prove that for all k ∈ N∗
Logk
ω ω Logk , if x is near ∂Ω. ϕ (x) δ(x)
(2.5)
Using (2.4), there exists c > 0 such that
ω
ω ωc Log Log . Log c δ(x) ϕ (x) δ(x) 1
(2.6)
Moreover, we have
Log( ct ) Log( 1t )
if t → 0+ .
→ 1,
(2.7)
This, together with (2.6), proves (2.5) for k = 1. By induction, we suppose that (2.5) holds for k ∈ N∗ . Then, there exists c > 0 such that
1 c
Logk
ω
δ(x)
Logk
ω ω c Logk . ϕ (x) δ(x)
Using again (2.7), it follows that
Logk+1
ω ω Logk+1 , if x is near ∂Ω. ϕ (x) δ(x)
This ends the proof.
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3. Proof of theorems Our idea is to compare the solution u of problem (1.4) to the function
v (x) =
α
ϕ (x) θ ϕ (x) , for x ∈ Ω,
m
where θ(t ) = k=l (Logk ( ωt ))−βk , βl = 0, βk ∈ R for k l + 1 and An elementary calculus shows that
α ∈ [0, 1].
v (x) = −θ ϕ (x) ϕ α −2 (x) W (x), where
(3.1)
2 θ (ϕ (x)) θ (ϕ (x)) θ (ϕ (x)) 2 2
− ϕ (x) + λ1 ϕ (x) α + ϕ (x) . W (x) = ∇ ϕ (x) α (1 − α ) − 2αϕ (x) θ(ϕ (x)) θ(ϕ (x)) θ(ϕ (x))
So, it is clear that the asymptotic behavior of v near the boundary ∂Ω depends on the choice of α . Now, we recall that ϕ > 0 in Ω and ∇ ϕ (x) = 0, for each x ∈ ∂Ω , then for each c 1 , c 2 > 0 and β , μ > 0, there exists c > 1 such that for each x ∈ Ω ,
1 c
μ c 1 ϕ β (x) + c 2 ∇ ϕ (x) c .
(3.2)
Proof of Theorem 2. Let 1 − σ < λ < 2 and put (2.4), it follows that
lim
δ(x)→0
ϕ (x)
α=
μk 2−λ σ +1 and βk = σ +1 for 1 k m, then 0 < α < 1. From Lemma 1 and
θ (ϕ (x)) (ϕ (x))2 θ (ϕ (x)) = lim = 0. θ(ϕ (x)) δ(x)→0 θ(ϕ (x))
This, together with (3.1) and (3.2), leads to
− v (x) θ ϕ (x) ϕ α −2 (x),
if δ(x) → 0.
Then
α (σ +1)−2 (σ +1) θ ϕ (x) − v (x) v σ (x) ϕ (x) ,
if δ(x) → 0.
So
−μk m −λ ω − v (x) v σ (x) ϕ (x) Logk , ϕ (x)
if δ(x) → 0.
k =1
Using Lemma 3, (2.4) and (H), we get
− v (x) v σ (x) a(x), This means that there exist
1 c
if δ(x) → 0.
η > 0 and c > 0 such that
a(x) − v (x) v σ (x) ca(x),
for x ∈ Ωη ,
where Ωη = {x ∈ Ω, δ(x) < η}. This implies that for x ∈ Ωη ,
− v 1 (x) v σ1 (x) a(x) − v 2 (x) v σ2 (x), 1
(3.3)
1
where v 1 = c 1 v = c − σ +1 v and v 2 = c 2 v = c σ +1 v. Now, we claim that the solution u of (1.4) satisfies
v 1 (x) u (x) v 2 (x),
for x ∈ Ωη .
(3.4)
Indeed, let A = {x ∈ Ωη , u (x) < v 1 (x)} and put w = u − v 1 , then w = 0 on ∂Ω . Since σ 0 and the function a is nonnegative, it follows from (3.3) that − w 0 on A. The maximum principle asserts that w = 0 on ∂ A and A must be empty. Thus w 0 on Ωη . By the same argument we also have u v 2 on Ωη and this proves (3.4). Finally, since the functions u and v are positive and continuous in the compact Ω \ Ωη , then we have
u (x) v (x),
for x ∈ Ω \ Ωη .
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So, we deduce from (3.4), (2.4) and Lemma 3 that
u (x)
2−λ
δ(x) σ +1
m
Logk
−σ +μk1
ω
for x ∈ Ω.
,
δ(x)
k =1
2
Proof of Theorem 3. Let λ = 2 and suppose that μ1 = μ2 = · · · = μl−1 = 1, βl = μσl+−11 and βk = σμ+k1 for l + 1 k m. Then (3.1) becomes
v (x) = −
μl > 1 and μk ∈ R for l + 1 k m. Put α = 0,
2 θ (ϕ (x)) ϕ (x)θ (ϕ (x))
. λ1 ϕ 2 (x) − ϕ ( x ) ∇ ϕ (x) θ (ϕ (x))
From Lemma 1 and (2.4), it yields
lim
δ(x)→0
ϕ (x)
θ (ϕ (x)) = −1. θ (ϕ (x))
According to (3.2), it follows that
− v (x)
θ (ϕ (x)) , ϕ (x)
if δ(x) → 0.
Then
− v (x) v σ (x)
θ (ϕ (x)) σ θ ϕ (x) , ϕ (x)
if δ(x) → 0.
(3.5)
Now, we have
θ (ϕ (x)) h(ϕ (x)) = 2 , ϕ (x)θ(ϕ (x)) ϕ (x) where
h(t ) =
m
βk
l
−1
ω
Log j
t
j =1
k=l
=
k
−1
Log j
j =1
ω t
βl + o(1) ,
if t → 0+ .
μl > 1 which implies that βl > 0, we obtain −1 l θ (ϕ (x)) 1 ω Log j , if δ(x) → 0. ϕ (x)θ(ϕ (x)) ϕ 2 (x) ϕ (x)
Using the fact that
j =1
This, together with (3.5), leads to
− v (x) v σ (x) Since
1
m
ω
−β j (σ +1)
j =l (Log j ( ϕ (x) ))
l
ϕ 2 (x)
ω
j =1 (Log j ( ϕ (x) ))
if δ(x) → 0.
,
μ1 = μ2 = · · · = μl−1 = 1, μl = 1 + βl (σ + 1) and μk = βk (σ + 1) for l + 1 k m, then −μ j m 1 ω − v (x) v σ (x) 2 Log j , if δ(x) → 0. ϕ (x) ϕ (x) j =1
We deduce from (H), (2.4) and Lemma 3 that
− v (x) v σ (x) a(x),
if δ(x) → 0.
The rest of the proof is similar to the one of Theorem 2.
2
Proof of Theorem 4. Let λ 1 − σ . The solution u of the problem (1.4) verifies the integral equation
u (x) = Ω
G (x, y )a( y )u −σ ( y ) dy .
(3.6)
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It is well known that for each x, y ∈ Ω ,
δ(x)δ( y ) cG (x, y ). Then
u (x) c δ(x).
(3.7)
This, together with (3.6) and (H), implies that
u (x) c
m −λ−σ
G (x, y ) δ( y )
Logk
k =1
Ω
ω
−μk
δ( y )
dy .
(3.8)
μ = λ + σ , we obtain −μk m −μ+ ω u (x) c G (x, y ) δ( y ) Logk dy , δ( y )
Now, putting
k =1
Ω
where
μ+ = max(0, μ).
• Suppose that λ < 1 − σ , then we have 0 μ+ < 1. So, it follows from Lemma 2(e) that u (x) c δ(x). This, together with (3.7), leads to
u (x) δ(x),
for x ∈ Ω.
• Suppose that λ = 1 − σ , we distinguish three cases. Case 1. Suppose that (3.8), we have
μ1 = μ2 = · · · = μl−1 = 1, μl > 1 and μk ∈ R for l + 1 k m. Then, by using Lemma 2(d)(i) and
u (x) c δ(x). According to (3.7), we obtain
u (x) δ(x),
for x ∈ Ω.
Case 2. Suppose that μ1 = μ2 = · · · = μl−1 = 1, μl < 1 and μ −1 μ Put α = 1, βl = σl+1 and βk = σ +k1 for l + 1 k m. From (3.1), we have
v (x) = θ ϕ (x) −λ1 ϕ 2 (x)
μk ∈ R for l + 1 k m.
θ(ϕ (x))
∇ ϕ (x) 2 2 + ϕ (x) θ (ϕ (x)) . + 1 + ϕ (x)θ (ϕ (x)) θ (ϕ (x))
(3.9)
By Lemma 1 and (2.4), it yields
lim
δ(x)→0
ϕ (x)
θ (ϕ (x)) = 0. θ(ϕ (x))
Then, if x is near ∂Ω , we have
θ(ϕ (x)) θ(ϕ (x)) +1 . ϕ (x)θ (ϕ (x)) ϕ (x)θ (ϕ (x)) This implies that
−λ1 ϕ 2 (x)
θ(ϕ (x)) ϕ (x)θ(ϕ (x)) + , 1 −λ1 ϕ (x)θ (ϕ (x)) θ (ϕ (x))
Now, we have
ϕ (x)θ(ϕ (x)) ϕ 2 (x) = , θ (ϕ (x)) h(ϕ (x))
if δ(x) → 0.
(3.10)
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S. Gontara et al. / J. Math. Anal. Appl. 369 (2010) 719–729
where
h(t ) =
m
βk
l
ω t
−1
ω
Log j
t
j =1
Since
−1 Log j
j =1
k=l
=
k
βl + o(1) ,
if t → 0+ .
μl < 1, we have βl < 0 and it follows that −1 l ω −h(t ) Log j , if t → 0+ .
(3.11)
t
j =1
Using the fact that there exists c > 0 such that
1 c
t t
l
Log j
ω
c,
t
j =1
for t small enough,
we have
1 c
ϕ 2 (x) −
ϕ 2 (x) c ϕ (x), if δ(x) → 0, h(ϕ (x))
which implies that there exists c > 0 such that
1 c
ϕ 2 (x) −
ϕ (x)θ(ϕ (x)) c ϕ (x), for x near ∂Ω. θ (ϕ (x))
So, using (3.10), we get
1 c
ϕ 2 (x) −λ1 ϕ 2 (x)
θ(ϕ (x)) + 1 c ϕ (x), ϕ (x)θ (ϕ (x))
for x near ∂Ω.
(3.12)
Now, since from Lemma 1 and (2.4), we have
ϕ (x)θ (ϕ (x)) = −1, δ(x)→0 θ (ϕ (x)) lim
it follows that
∇ ϕ (x) 2 2 + ϕ (x) θ (ϕ (x)) ∇ ϕ (x) 2 , θ (ϕ (x))
if δ(x) → 0.
(3.13)
This, together with (3.12) and (3.2), implies by (3.9) that
v (x) θ ϕ (x) ,
if δ(x) → 0.
Using
h(ϕ (x))θ(ϕ (x)) , θ ϕ (x) = ϕ (x) we get
v (x) v σ (x) ϕ σ −1 (x)h ϕ (x) θ σ +1 ϕ (x) , which implies from (3.11) that
m
− v (x) v σ (x) ϕ σ −1 (x) Finally, since λ = 1 − σ , that
ω
if δ(x) → 0,
−(σ +1)β j
j =l (Log j ( ϕ (x) ))
l
,
ω
j =1 (Log j ( ϕ (x) ))
if δ(x) → 0.
μ1 = μ2 = · · · = μl−1 = 1, μl = 1 + βl (σ + 1) and μk = βk (σ + 1) for 1 + l k m, it follows
− v (x) v σ (x) ϕ −λ (x)
m
Log j
j =1
ω ϕ (x)
−μ j ,
if δ(x) → 0.
S. Gontara et al. / J. Math. Anal. Appl. 369 (2010) 719–729
729
We deduce from (H), (2.4) and Lemma 3 that
− v (x) v σ (x) a(x),
if δ(x) → 0.
The same argument used in Theorem 2 ends the proof. Case 3. Suppose that We consider
v (x) = ϕ (x)θ
μ1 = μ2 = · · · = μm = 1.
ϕ (x) , for x ∈ Ω, 1
where θ(t ) = (Logm+1 ( ωt )) σ +1 . Using (3.1), we notice that v satisfies as in Case 2, the following equality
v (x) = θ ϕ (x) −λ1 ϕ 2 (x)
θ(ϕ (x))
∇ ϕ (x) 2 2 + ϕ (x) θ (ϕ (x)) . + + 1 ϕ (x)θ (ϕ (x)) θ (ϕ (x))
Moreover, from Lemma 1 and (2.4), we find again
lim
δ(x)→0
ϕ (x)
θ (ϕ (x)) = 0 and θ(ϕ (x))
ϕ (x)θ (ϕ (x)) = −1. δ(x)→0 θ (ϕ (x)) lim
Then (3.10) and (3.13) continuous to hold. Now, since
ϕ (x)θ(ϕ (x)) ϕ 2 (x) = , θ (ϕ (x)) h(ϕ (x)) where
h(t ) =
−1 m+1 −1 ω Log j , σ +1 t j =1
then we find again (3.12). Hence to achieve the proof we use the same argument as in Case 2.
2
Acknowledgments We would like to thank the anonymous referees for their careful reading of the paper and their helpful suggestions.
References [1] F.-C. Cîrstea, V. Radulescu, Existence and uniqueness of blow-up solutions for a class of logistic equations, Commun. Contemp. Math. 4 (2002) 559–585. [2] F.-C. Cîrstea, V. Radulescu, Uniqueness of the blow-up boundary solution of logistic equations with absorption, C. R. Acad. Sci. Paris Sér. I 335 (2002) 447–452. [3] F.-C. Cîrstea, V. Radulescu, Asymptotics for the blow-up boundary solution of the logistic equation with absorption, C. R. Acad. Sci. Paris Sér. I 336 (2003) 231–236. [4] F.-C. Cîrstea, V. Radulescu, Extremal singular solutions for degenerate logistic-type equations in anisotropic media, C. R. Acad. Sci. Paris Sér. I 339 (2004) 119–124. [5] F.-C. Cîrstea, An extreme variation phenomenon for some nonlinear elliptic problems with boundary blow-up, C. R. Acad. Sci. Paris Sér. I 339 (2004) 689–694. [6] M.G. Crandall, P.H. Rabinowitz, L. Tartar, On a Dirichlet problem with a singular nonlinearity, Comm. Partial Differential Equations 2 (1977) 193–222. [7] M. Ghergu, V. Radulescu, Singular Elliptic Problems. Bifurcation and Asymptotic Analysis, Oxford Lecture Ser. Math. Appl., vol. 37, Oxford University Press, 2008. [8] A. Kristàly, V. Radulescu, C. Varga, Variational Principles in Mathematical Physics, Geometry and Economics: Qualitative Analysis of Nonlinear Equations and Unilateral Problems, Encyclopedia Math. Appl., vol. 136, Cambridge University Press, Cambridge, 2010. [9] A.C. Lazer, P.J. McKenna, On a singular elliptic boundary value problem, Proc. Amer. Math. Soc. 111 (1991) 721–730. [10] H. Mâagli, M. Zribi, On a new Kato class and singular solutions of a nonlinear elliptic equation in bounded domain of Rn , Positivity 9 (2005) 667–686. [11] V. Maric, Regular Variation and Differential Equations, Lecture Notes in Math., vol. 1726, Springer-Verlag, Berlin, 2000. [12] S.B. Othman, H. Mâagli, S. Masmoudi, M. Zribi, Exact asymptotic behavior near the boundary to the solution for singular nonlinear Dirichlet problems, Nonlinear Anal. 71 (2009) 4137–4150. [13] S.I. Resnick, Extreme Values, Regular Variation and Point Processes, Springer-Verlag, New York, 1987. [14] N. Zeddini, Positive solutions for a singular nonlinear problem on a bounded domain in R2 , Nonlinear Anal. 18 (2003) 97–118. [15] Z. Zhang, The asymptotic behavior of the unique solution for the singular Lane–Emdem–Fowler equation, J. Math. Anal. Appl. 312 (2005) 33–43.
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Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa
Asymptotic behavior of positive solutions of a singular nonlinear Dirichlet problem Sabrine Gontara, Habib Mâagli, Syrine Masmoudi, Sameh Turki ∗ Département de Mathématiques, Faculté des Sciences de Tunis, Campus Universitaire, 2092 Tunis, Tunisia
a r t i c l e
i n f o
a b s t r a c t Let Ω be a C 1,1 -bounded domain in Rn for n 2. In this paper, we are concerned with the asymptotic behavior of the unique positive classical solution to the singular boundary-value problem u + a(x)u −σ = 0 in Ω , u |∂Ω = 0, where σ 0, a is a nonnegative function in α (Ω), 0 < α < 1 and there exists c > 0 such that 1 a(x)(δ(x))λ m (Log ( ω ))μk c. C loc k δ(x) k=1 c Here λ 2, μk ∈ R, ω is a positive constant and δ(x) = dist(x, ∂Ω). © 2010 Elsevier Inc. All rights reserved.
Article history: Received 21 December 2009 Available online 3 April 2010 Submitted by V. Radulescu Keywords: Asymptotic behavior Dirichlet problem Singular equation
1. Introduction We are interested in this paper on the estimates of positive solutions for nonlinear problems of the type
−u = a(x) g (u ), x ∈ Ω, u > 0 in Ω, u |∂Ω = 0,
(1.1)
where Ω is a C 1,1 -bounded domain in Rn (n 2). The existence of such solutions and their asymptotic behavior have been investigated by many authors (see [1–9,12,15] and the references therein). For a ≡ 1 on Ω and g (u ) = u −σ for σ > 1, Crandall, Rabinowitz and Tartar proved in [6] that the problem (1.1) has a unique classical solution u in Ω and showed that this solution satisfies the following estimates
1+2σ
c 1 δ(x)
2 u (x) c 2 δ(x) 1+σ ,
near the boundary ∂Ω,
(1.2)
where c 1 , c 2 are positive constants and δ(x) = dist(x, ∂Ω). In [9], Lazer and McKenna established the estimate (1.2) on Ω and instead of a ≡ 1 on Ω , they assumed that 0 < b1 a(x)(δ(x))λ b2 , for all x ∈ Ω , where b1 , b2 are positive constants and λ ∈ (0, 2); then they proved that for σ > 1, there exist two positive constants c 1 and c 2 such that
1+2σ
c 1 δ(x)
2−λ u (x) c 2 δ(x) 1+σ ,
for x ∈ Ω.
(1.3)
Most recently, applying Karamata regular variation theory, many authors studied the exact asymptotic behavior of solutions of the problem (1.1) (see [1–5,15]). For instance, in [15], Zhang has dealt with functions in the Karamata class called regularly varying at zero with index −γ , γ ∈ R which we recall in the following
*
Corresponding author. E-mail addresses: [email protected] (S. Gontara), [email protected] (H. Mâagli), [email protected] (S. Masmoudi), [email protected] (S. Turki). 0022-247X/$ – see front matter doi:10.1016/j.jmaa.2010.04.008
©
2010 Elsevier Inc. All rights reserved.
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S. Gontara et al. / J. Math. Anal. Appl. 369 (2010) 719–729
Definition 1. (See [11,13].) A positive measurable function k defined on some interval (0, η), varying at zero with index −γ , γ ∈ R, if for each ξ > 0
lim
t →0+
k(ξ t )
η > 0, is called regularly
= ξ −γ .
k(t )
Remark 1. As typical example of function k in this class, we quote
k(t ) = t −γ
m
−μi 1
Log i
,
t
i =1
where Log i (x) = Log ◦ Log ◦ · · · ◦ Log(x) (i times),
μi ∈ R .
In [15], Zhang studied the problem (1.1) under the following assumptions:
(H1 ) g ∈ C 1 ((0, ∞), (0, ∞)), g (s) 0 for all s > 0 and g is regularly varying at zero with index −γ , γ > 0. α (Ω), 0 < α < 1 and there exist η > 0 and h ∈ C ((0, η), (0, ∞)) nonincreasing and (H2 ) a is a nonnegative function in C loc regularly varying at zero with index −λ, 0 < λ < 2, such that a(x)
lim
δ(x)→0 h (δ(x))
= c 0 > 0.
Then he proved the following result: Theorem 1. (See [15].) Suppose (H1 ) and (H2 ) hold. Then the problem (1.1) has a unique solution u ∈ C 2+α (Ω) ∩ C (Ω) such that
u (x)
lim
δ(x)→0 p (δ(x))
1
γ +1
= c0
,
where p is the unique local solution to the problem
− p (s) = h(s) g p (s) , s ∈ (0, η), p (s) > 0, s ∈ (0, η), p (0) = 0.
In [12], the authors extended the above result. In fact, they allowed both terms a(x) and g (u ) to be in more general classes of functions. More precisely, the function a was assumed to satisfy a similar asymptotic behavior and was required to be in the elliptic Kato class K (Ω) studied and introduced in [10] and [14]. For the reader’s convenience we recall the definition of K (Ω) and we give a class of functions belonging to this class. The result proved in [12] will be detailed in Section 2. Definition 2. A Borel measurable function q in Ω belongs to the Kato class K (Ω) if
δ( y ) G (x, y ) q( y ) dy = 0. δ(x)
lim sup
α →0 x∈Ω
B (x,α )∩Ω
Here and always G denotes the Green function for the Laplace operator in Ω , with zero Dirichlet boundary condition. Example 1. (See [10] and [14].) (i) For p > n2 , we have L p (Ω) ⊂ K (Ω).
(ii) The function x → (δ(1x))λ is in K (Ω) if and only if λ < 2. In fact, we have the more general example: Example 2. (See [12].) Let m ∈ N∗ and
ψ(t ) = t −λ
m
Logk
k =1
ω be a sufficiently large positive constant such that the function
−μk
ω t
is defined and positive on (0, 2d) where d = diam(Ω).
S. Gontara et al. / J. Math. Anal. Appl. 369 (2010) 719–729
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Let p > n2 , then if one of the following conditions is satisfied
• λ<2− • λ=2−
n p n p
and and
μk ∈ R, for 1 k m, μ1 = μ2 = · · · = μl−1 = 1 − 1p , μl > 1 − 1p , μk ∈ R for l + 1 k m,
we have the function x → ψ(δ(x)) f (x) is in K (Ω), provided that f ∈ L p (Ω). Remark 2. Let k be a function regularly varying at zero with index −γ . If 0 < γ < 2, then by [12, Corollary 1] we have k(δ(.)) ∈ K (Ω). To simplify our statements, the letter c will denote a generic positive constant which may vary from line to line and for two nonnegative functions f and g on a set S, we write f (x) g (x), for x ∈ S, if there exists a constant c > 0 not depending on x such that
1 c
g (x) f (x) cg (x),
∀x ∈ S .
In this paper, we study the asymptotic behavior to the solution of the following problem
where
−u = a(x)u −σ , x ∈ Ω, u > 0 in Ω, u |∂Ω = 0,
(1.4)
σ 0 and the function a satisfies the following hypothesis
α (Ω), 0 < α < 1, satisfying for all x ∈ Ω (H) a is a nonnegative function in C loc
−μk m −λ ω a(x) δ(x) Logk . δ(x)
k =1
Here and throughout this paper, m ∈ N∗ and satisfy one of the following conditions
ω is a positive constant large enough. The real numbers λ and μk , 1 k m,
• λ < 2, μ1 , μ2 , . . . , μm ∈ R, and • λ = 2, μ1 = μ2 = · · · = μl−1 = 1, μl > 1, μk ∈ R for l + 1 k m. Remark 3. It is obvious to see from Example 2 that if the function a satisfies hypothesis (H), then a is in the Kato class K (Ω) and which implies by [12, Theorem 2] that problem (1.4) has a unique solution u ∈ C 2+α (Ω) ∩ C (Ω). The asymptotic behavior of such solution depends on the choice of the constants λ and results into the following theorems.
σ . Hence we divide our main
Theorem 2. Assume that a satisfies (H). If 1 − σ < λ < 2 and μk ∈ R for 1 k m, then the solution u of the problem (1.4) satisfies for each x ∈ Ω
u (x)
2−λ
δ(x) σ +1
m k =1
Remark 4. If we consider satisfies 2−λ
u (x) δ(x) σ +1 ,
Logk
ω
−σ +μk1 .
δ(x)
σ > 1, λ ∈ (0, 2) and μk = 0 for all 1 k m in Theorem 2, the solution u of the problem (1.4)
for x ∈ Ω,
which improves the inequalities (1.3) found in [9] by Lazer and McKenna. Theorem 3. Assume that a satisfies (H). If λ = 2, μ1 = μ2 = · · · = μl−1 = 1, μl > 1 and μk ∈ R for l + 1 k m, then the solution u of the problem (1.4) satisfies for each x ∈ Ω
u (x) Logl
ω δ(x)
1σ−+μ1l m k=l+1
Logk
ω δ(x)
−σ +μk1 .
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S. Gontara et al. / J. Math. Anal. Appl. 369 (2010) 719–729
Theorem 4. Assume that a satisfies (H) and λ 1 − σ . Let u be a solution of the problem (1.4).
• If λ < 1 − σ , then u (x) δ(x),
for x ∈ Ω.
• If λ = 1 − σ , μ1 = μ2 = · · · = μl−1 = 1, μl > 1 and μk ∈ R for l + 1 k m, then u (x) δ(x),
for x ∈ Ω.
• If λ = 1 − σ , μ1 = μ2 = · · · = μl−1 = 1, μl < 1 and μk ∈ R for l + 1 k m, then
1σ−+μ1l −σ +μk1 m ω ω u (x) δ(x) Logl Logk , δ(x) δ(x)
for x ∈ Ω.
k=l+1
• If λ = 1 − σ and μ1 = μ2 = · · · = μm = 1, then
u (x) δ(x)Logm+1
ω
σ +1 1 ,
δ(x)
for x ∈ Ω.
Remark 5. We mention that if σ = 0 and the function a satisfies an exact asymptotic behavior, the results in Theorems 2, 3 and 4 have been stated in [12]. For the convenience of the readers, this result is recalled in Lemma 2 stated below. The remaining part of this paper is organized as follows. In Section 2, we state some technical lemmas which will be useful for the proof of our results. Section 3 is devoted to prove our main results in Theorems 2, 3 and 4. 2. Technical lemmas Lemma 1. Let η > 0, m ∈ N∗ , l ∈ N∗ , βl = 0 and βk ∈ R for k l + 1. Then the function θ defined on (0, η) by
θ(t ) =
m
−βk
ω
Logk
,
t
k=l
satisfies: θ (t )
(i) limt →0+ t θ(t ) = 0. θ (t ) (ii) limt →0+ t θ (t ) = −1. θ (t )
(iii) limt →0+ t 2 θ(t ) = 0. Proof. By an elementary calculus, we have
θ (t ) 1 = h(t ), θ(t ) t
(2.1)
where
h(t ) =
m
βk
k
−1 Log j
j =1
k=l
ω t
.
(2.2)
Since limt →0+ h(t ) = 0, then (i) holds. Now, we shall prove (ii). By differentiation, we have
θ (t ) = −
1 t2
h(t )θ(t ) +
1 1 h (t )θ(t ) + h(t )θ (t ). t t
So, by using (2.1), we obtain
t θ (t )
θ (t )
= −1 +
th (t )
We claim that
lim
t →0+
th (t ) h(t )
= 0.
h(t )
+ h(t ).
(2.3)
S. Gontara et al. / J. Math. Anal. Appl. 369 (2010) 719–729
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To prove the claim, we write (2.2) of the form
h(t ) =
m
βk S k (t ),
k=l
where
S k (t ) =
k
−1
ω
Log j
.
t
j =1
Using that
S k (t ) S k (t )
1 j
k
=−
t
−1
t
S k (t )
=
1 Log t
+o
,
t
j =1 i =1
it follows that for each k ∈ N∗ ,
S k (t )
ω
Log i
1 Log t
if t → 0+ .
,
This implies that
th (t ) = t
m
βk S k (t ) =
1 Log t
k=l
+o
1
Log t
for t → 0+ .
h(t ),
So, we deduce
th (t )
lim
t →0+
h(t )
= 0.
This, together with (2.3), proves (ii). Finally, (iii) holds immediately from (i) and (ii).
2
The following lemma is stated in [12, Example 6]. To present it we need the following notation. For a nonnegative measurable function a in Ω , we denote by Va the potential of a defined in Ω by Va(x) = Ω G (x, y )a( y ) dy. Lemma 2. (See [12].) Let m ∈ N∗ and ω be a sufficiently large positive real number such that the function ψ is positive and defined on (0, 1) by
ψ(t ) = t −λ
m
−μk
Logk
k =1
ω
,
t
where the constants λ and μk , 1 k m, satisfy one of the following conditions:
• λ < 2, μk ∈ R, for 1 k m, • λ = 2, μ1 = μ2 = · · · = μl−1 = 1, μl > 1 and μk ∈ R for l + 1 k m. α (Ω) ∩ K (Ω), 0 < α < 1, such that Let a be a nonnegative function in C loc
lim
δ(x)→0
a(x) = c 0 > 0. ψ(δ(x))
Then we have the following estimates: (a) If λ = 2, then
Va(x)
∼
δ(x)→0
c Logl
ω
1−μl m
δ(x)
Logk
k=l+1
ω δ(x)
(b) If 1 < λ < 2, then
−μk m 2−λ ω Va(x) ∼ c δ(x) Logk . δ(x)→0 δ(x)
k =1
−μk .
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S. Gontara et al. / J. Math. Anal. Appl. 369 (2010) 719–729
(c) If λ = 1 and μ1 = μ2 = · · · = μm = 1, then
Va(x)
∼
δ(x)→0
c δ(x) Logm+1
(d) If λ = 1 and l = inf{1 k m, (i) If μl > 1, then
Va(x)
δ(x)
.
μk = 1} exists, then:
∼
c δ(x).
∼
c δ(x) Logl
δ(x)→0
ω
(ii) If μl < 1, then
Va(x)
δ(x)→0
ω
1−μl m
Logk
δ(x)
k=l+1
ω δ(x)
−μk .
(e) If λ < 1, then
Va(x)
∼
δ(x)→0
c δ(x).
Now, let ϕ be a positive eigenfunction corresponding to the first smallest eigenvalue λ1 of the operator in Ω with zero Dirichlet boundary condition and satisfying ϕ ∞ = supx∈Ω |ϕ (x)| = 1. It is well known that for each x ∈ Ω ,
ϕ (x) δ(x).
(2.4)
Lemma 3. The function θ defined in Lemma 1 satisfies
θ ϕ (x) θ δ(x) ,
if x is near ∂Ω.
Proof. We intend to prove that for all k ∈ N∗
Logk
ω ω Logk , if x is near ∂Ω. ϕ (x) δ(x)
(2.5)
Using (2.4), there exists c > 0 such that
ω
ω ωc Log Log . Log c δ(x) ϕ (x) δ(x) 1
(2.6)
Moreover, we have
Log( ct ) Log( 1t )
if t → 0+ .
→ 1,
(2.7)
This, together with (2.6), proves (2.5) for k = 1. By induction, we suppose that (2.5) holds for k ∈ N∗ . Then, there exists c > 0 such that
1 c
Logk
ω
δ(x)
Logk
ω ω c Logk . ϕ (x) δ(x)
Using again (2.7), it follows that
Logk+1
ω ω Logk+1 , if x is near ∂Ω. ϕ (x) δ(x)
This ends the proof.
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3. Proof of theorems Our idea is to compare the solution u of problem (1.4) to the function
v (x) =
α
ϕ (x) θ ϕ (x) , for x ∈ Ω,
m
where θ(t ) = k=l (Logk ( ωt ))−βk , βl = 0, βk ∈ R for k l + 1 and An elementary calculus shows that
α ∈ [0, 1].
v (x) = −θ ϕ (x) ϕ α −2 (x) W (x), where
(3.1)
2 θ (ϕ (x)) θ (ϕ (x)) θ (ϕ (x)) 2 2
− ϕ (x) + λ1 ϕ (x) α + ϕ (x) . W (x) = ∇ ϕ (x) α (1 − α ) − 2αϕ (x) θ(ϕ (x)) θ(ϕ (x)) θ(ϕ (x))
So, it is clear that the asymptotic behavior of v near the boundary ∂Ω depends on the choice of α . Now, we recall that ϕ > 0 in Ω and ∇ ϕ (x) = 0, for each x ∈ ∂Ω , then for each c 1 , c 2 > 0 and β , μ > 0, there exists c > 1 such that for each x ∈ Ω ,
1 c
μ c 1 ϕ β (x) + c 2 ∇ ϕ (x) c .
(3.2)
Proof of Theorem 2. Let 1 − σ < λ < 2 and put (2.4), it follows that
lim
δ(x)→0
ϕ (x)
α=
μk 2−λ σ +1 and βk = σ +1 for 1 k m, then 0 < α < 1. From Lemma 1 and
θ (ϕ (x)) (ϕ (x))2 θ (ϕ (x)) = lim = 0. θ(ϕ (x)) δ(x)→0 θ(ϕ (x))
This, together with (3.1) and (3.2), leads to
− v (x) θ ϕ (x) ϕ α −2 (x),
if δ(x) → 0.
Then
α (σ +1)−2 (σ +1) θ ϕ (x) − v (x) v σ (x) ϕ (x) ,
if δ(x) → 0.
So
−μk m −λ ω − v (x) v σ (x) ϕ (x) Logk , ϕ (x)
if δ(x) → 0.
k =1
Using Lemma 3, (2.4) and (H), we get
− v (x) v σ (x) a(x), This means that there exist
1 c
if δ(x) → 0.
η > 0 and c > 0 such that
a(x) − v (x) v σ (x) ca(x),
for x ∈ Ωη ,
where Ωη = {x ∈ Ω, δ(x) < η}. This implies that for x ∈ Ωη ,
− v 1 (x) v σ1 (x) a(x) − v 2 (x) v σ2 (x), 1
(3.3)
1
where v 1 = c 1 v = c − σ +1 v and v 2 = c 2 v = c σ +1 v. Now, we claim that the solution u of (1.4) satisfies
v 1 (x) u (x) v 2 (x),
for x ∈ Ωη .
(3.4)
Indeed, let A = {x ∈ Ωη , u (x) < v 1 (x)} and put w = u − v 1 , then w = 0 on ∂Ω . Since σ 0 and the function a is nonnegative, it follows from (3.3) that − w 0 on A. The maximum principle asserts that w = 0 on ∂ A and A must be empty. Thus w 0 on Ωη . By the same argument we also have u v 2 on Ωη and this proves (3.4). Finally, since the functions u and v are positive and continuous in the compact Ω \ Ωη , then we have
u (x) v (x),
for x ∈ Ω \ Ωη .
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So, we deduce from (3.4), (2.4) and Lemma 3 that
u (x)
2−λ
δ(x) σ +1
m
Logk
−σ +μk1
ω
for x ∈ Ω.
,
δ(x)
k =1
2
Proof of Theorem 3. Let λ = 2 and suppose that μ1 = μ2 = · · · = μl−1 = 1, βl = μσl+−11 and βk = σμ+k1 for l + 1 k m. Then (3.1) becomes
v (x) = −
μl > 1 and μk ∈ R for l + 1 k m. Put α = 0,
2 θ (ϕ (x)) ϕ (x)θ (ϕ (x))
. λ1 ϕ 2 (x) − ϕ ( x ) ∇ ϕ (x) θ (ϕ (x))
From Lemma 1 and (2.4), it yields
lim
δ(x)→0
ϕ (x)
θ (ϕ (x)) = −1. θ (ϕ (x))
According to (3.2), it follows that
− v (x)
θ (ϕ (x)) , ϕ (x)
if δ(x) → 0.
Then
− v (x) v σ (x)
θ (ϕ (x)) σ θ ϕ (x) , ϕ (x)
if δ(x) → 0.
(3.5)
Now, we have
θ (ϕ (x)) h(ϕ (x)) = 2 , ϕ (x)θ(ϕ (x)) ϕ (x) where
h(t ) =
m
βk
l
−1
ω
Log j
t
j =1
k=l
=
k
−1
Log j
j =1
ω t
βl + o(1) ,
if t → 0+ .
μl > 1 which implies that βl > 0, we obtain −1 l θ (ϕ (x)) 1 ω Log j , if δ(x) → 0. ϕ (x)θ(ϕ (x)) ϕ 2 (x) ϕ (x)
Using the fact that
j =1
This, together with (3.5), leads to
− v (x) v σ (x) Since
1
m
ω
−β j (σ +1)
j =l (Log j ( ϕ (x) ))
l
ϕ 2 (x)
ω
j =1 (Log j ( ϕ (x) ))
if δ(x) → 0.
,
μ1 = μ2 = · · · = μl−1 = 1, μl = 1 + βl (σ + 1) and μk = βk (σ + 1) for l + 1 k m, then −μ j m 1 ω − v (x) v σ (x) 2 Log j , if δ(x) → 0. ϕ (x) ϕ (x) j =1
We deduce from (H), (2.4) and Lemma 3 that
− v (x) v σ (x) a(x),
if δ(x) → 0.
The rest of the proof is similar to the one of Theorem 2.
2
Proof of Theorem 4. Let λ 1 − σ . The solution u of the problem (1.4) verifies the integral equation
u (x) = Ω
G (x, y )a( y )u −σ ( y ) dy .
(3.6)
S. Gontara et al. / J. Math. Anal. Appl. 369 (2010) 719–729
727
It is well known that for each x, y ∈ Ω ,
δ(x)δ( y ) cG (x, y ). Then
u (x) c δ(x).
(3.7)
This, together with (3.6) and (H), implies that
u (x) c
m −λ−σ
G (x, y ) δ( y )
Logk
k =1
Ω
ω
−μk
δ( y )
dy .
(3.8)
μ = λ + σ , we obtain −μk m −μ+ ω u (x) c G (x, y ) δ( y ) Logk dy , δ( y )
Now, putting
k =1
Ω
where
μ+ = max(0, μ).
• Suppose that λ < 1 − σ , then we have 0 μ+ < 1. So, it follows from Lemma 2(e) that u (x) c δ(x). This, together with (3.7), leads to
u (x) δ(x),
for x ∈ Ω.
• Suppose that λ = 1 − σ , we distinguish three cases. Case 1. Suppose that (3.8), we have
μ1 = μ2 = · · · = μl−1 = 1, μl > 1 and μk ∈ R for l + 1 k m. Then, by using Lemma 2(d)(i) and
u (x) c δ(x). According to (3.7), we obtain
u (x) δ(x),
for x ∈ Ω.
Case 2. Suppose that μ1 = μ2 = · · · = μl−1 = 1, μl < 1 and μ −1 μ Put α = 1, βl = σl+1 and βk = σ +k1 for l + 1 k m. From (3.1), we have
v (x) = θ ϕ (x) −λ1 ϕ 2 (x)
μk ∈ R for l + 1 k m.
θ(ϕ (x))
∇ ϕ (x) 2 2 + ϕ (x) θ (ϕ (x)) . + 1 + ϕ (x)θ (ϕ (x)) θ (ϕ (x))
(3.9)
By Lemma 1 and (2.4), it yields
lim
δ(x)→0
ϕ (x)
θ (ϕ (x)) = 0. θ(ϕ (x))
Then, if x is near ∂Ω , we have
θ(ϕ (x)) θ(ϕ (x)) +1 . ϕ (x)θ (ϕ (x)) ϕ (x)θ (ϕ (x)) This implies that
−λ1 ϕ 2 (x)
θ(ϕ (x)) ϕ (x)θ(ϕ (x)) + , 1 −λ1 ϕ (x)θ (ϕ (x)) θ (ϕ (x))
Now, we have
ϕ (x)θ(ϕ (x)) ϕ 2 (x) = , θ (ϕ (x)) h(ϕ (x))
if δ(x) → 0.
(3.10)
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S. Gontara et al. / J. Math. Anal. Appl. 369 (2010) 719–729
where
h(t ) =
m
βk
l
ω t
−1
ω
Log j
t
j =1
Since
−1 Log j
j =1
k=l
=
k
βl + o(1) ,
if t → 0+ .
μl < 1, we have βl < 0 and it follows that −1 l ω −h(t ) Log j , if t → 0+ .
(3.11)
t
j =1
Using the fact that there exists c > 0 such that
1 c
t t
l
Log j
ω
c,
t
j =1
for t small enough,
we have
1 c
ϕ 2 (x) −
ϕ 2 (x) c ϕ (x), if δ(x) → 0, h(ϕ (x))
which implies that there exists c > 0 such that
1 c
ϕ 2 (x) −
ϕ (x)θ(ϕ (x)) c ϕ (x), for x near ∂Ω. θ (ϕ (x))
So, using (3.10), we get
1 c
ϕ 2 (x) −λ1 ϕ 2 (x)
θ(ϕ (x)) + 1 c ϕ (x), ϕ (x)θ (ϕ (x))
for x near ∂Ω.
(3.12)
Now, since from Lemma 1 and (2.4), we have
ϕ (x)θ (ϕ (x)) = −1, δ(x)→0 θ (ϕ (x)) lim
it follows that
∇ ϕ (x) 2 2 + ϕ (x) θ (ϕ (x)) ∇ ϕ (x) 2 , θ (ϕ (x))
if δ(x) → 0.
(3.13)
This, together with (3.12) and (3.2), implies by (3.9) that
v (x) θ ϕ (x) ,
if δ(x) → 0.
Using
h(ϕ (x))θ(ϕ (x)) , θ ϕ (x) = ϕ (x) we get
v (x) v σ (x) ϕ σ −1 (x)h ϕ (x) θ σ +1 ϕ (x) , which implies from (3.11) that
m
− v (x) v σ (x) ϕ σ −1 (x) Finally, since λ = 1 − σ , that
ω
if δ(x) → 0,
−(σ +1)β j
j =l (Log j ( ϕ (x) ))
l
,
ω
j =1 (Log j ( ϕ (x) ))
if δ(x) → 0.
μ1 = μ2 = · · · = μl−1 = 1, μl = 1 + βl (σ + 1) and μk = βk (σ + 1) for 1 + l k m, it follows
− v (x) v σ (x) ϕ −λ (x)
m
Log j
j =1
ω ϕ (x)
−μ j ,
if δ(x) → 0.
S. Gontara et al. / J. Math. Anal. Appl. 369 (2010) 719–729
729
We deduce from (H), (2.4) and Lemma 3 that
− v (x) v σ (x) a(x),
if δ(x) → 0.
The same argument used in Theorem 2 ends the proof. Case 3. Suppose that We consider
v (x) = ϕ (x)θ
μ1 = μ2 = · · · = μm = 1.
ϕ (x) , for x ∈ Ω, 1
where θ(t ) = (Logm+1 ( ωt )) σ +1 . Using (3.1), we notice that v satisfies as in Case 2, the following equality
v (x) = θ ϕ (x) −λ1 ϕ 2 (x)
θ(ϕ (x))
∇ ϕ (x) 2 2 + ϕ (x) θ (ϕ (x)) . + + 1 ϕ (x)θ (ϕ (x)) θ (ϕ (x))
Moreover, from Lemma 1 and (2.4), we find again
lim
δ(x)→0
ϕ (x)
θ (ϕ (x)) = 0 and θ(ϕ (x))
ϕ (x)θ (ϕ (x)) = −1. δ(x)→0 θ (ϕ (x)) lim
Then (3.10) and (3.13) continuous to hold. Now, since
ϕ (x)θ(ϕ (x)) ϕ 2 (x) = , θ (ϕ (x)) h(ϕ (x)) where
h(t ) =
−1 m+1 −1 ω Log j , σ +1 t j =1
then we find again (3.12). Hence to achieve the proof we use the same argument as in Case 2.
2
Acknowledgments We would like to thank the anonymous referees for their careful reading of the paper and their helpful suggestions.
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