- Email: [email protected]
Asymptotic behavior of positive solutions of a nonlinear Dirichlet problem
J. Math. Anal. Appl. 409 (2014) 925–933
Contents lists available at ScienceDirect
Journal of Mathematical Analysis and Applications journal homepage: www.elsevier.com/locate/jmaa
Asymptotic behavior of positive solutions of a nonlinear Dirichlet problem Sonia Ben Othman ∗ , Bilel Khamessi Département de Mathématiques, Faculté des Sciences de Tunis, Campus Universitaire, 2092 Tunis, Tunisia
article
info
Article history: Received 12 June 2013 Available online 31 July 2013 Submitted by V. Radulescu Keywords: Dirichlet problem Positive solution Sub and supersolutions Asymptotic behavior Karamata function
abstract We take up the existence and the asymptotic behavior of a classical solution to the following semilinear Dirichlet problem
−∆u = a(x)g (u), x ∈ Ω , u > 0 in Ω , u|∂ Ω = 0, γ
where Ω is a C 1,1 -bounded domain in RN , N ≥ 2 and the function a belongs to Cloc (Ω ), (0 < γ < 1) such that there exist c1 , c2 > 0 satisfying for each x ∈ Ω , c1 δ(x)−λ1 exp
η δ(x)
z1 (s) s
ds
≤ a(x) ≤ c2 δ(x)−λ2 exp
η
z2 (s)
δ(x)
s
ds ,
where η > diam(Ω ), δ(x) = dist (x, ∂ Ω ), λ1 ≤ λ2 ≤ 2 and for i ∈ {1, 2}, zi is a continuous function on [0, η] with zi (0) = 0. Our arguments are based on the sub–supersolution method with Karamata regular variation theory. © 2013 Elsevier Inc. All rights reserved.
1. Introduction Let Ω be a bounded C 1,1 domain in RN , N ≥ 2. We consider the following nonlinear Dirichlet problem
−1u = a(x)g (u), x ∈ Ω , u > 0 in Ω , u|∂ Ω = 0,
(1.1) γ
where g is a nonnegative function in C 1 ((0, ∞)) and a is a positive function in Cloc (Ω ), satisfying some appropriate conditions related to Karamata regular variation theory. Our main purpose is to prove the existence of a classical solution to problem (1.1) and to give estimates on such a solution. Several articles have been devoted to the study of problems of type (1.1) involving singular or sublinear nonlinearities. We refer to [1–6,8–10,12–14,16,17,19,20] and the references therein. Namely, the following problem
−1u = a(x)uσ , x ∈ Ω , σ < 1, u > 0 in Ω , u|∂ Ω = 0
(1.2)
is investigated by many authors under different kinds of assumptions on the weight a(x) (see [1,3–6,9,10,12,16,19,20]).
∗
Corresponding author. E-mail addresses: [email protected], [email protected] (S. Ben Othman), [email protected] (B. Khamessi).
0022-247X/$ – see front matter © 2013 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.jmaa.2013.07.063
926
S. Ben Othman, B. Khamessi / J. Math. Anal. Appl. 409 (2014) 925–933
In [6], Crandall et al. considered problem (1.2) where σ < −1 and a ≡ 1. They proved that there exists a unique positive classical solution u of problem (1.2) satisfying 2
2
c1 δ(x) 1−σ ≤ u(x) ≤ c2 δ(x) 1−σ ,
near the boundary ∂ Ω ,
where c1 , c2 are positive constants and δ(x) = dist(x, ∂ Ω ). Later in [13], Lazer and McKenna extended the result stated in [6] to a large class of functions which are not necessarily bounded in Ω . Indeed they assumed that if b1 δ(x)−λ ≤ a(x) ≤ b2 δ(x)−λ ,
x ∈ Ω,
where b1 , b2 > 0 and λ ∈ (0, 2), then problem (1.2) has a unique positive classical solution u in Ω such that 2−λ
2
c1 δ(x) 1−σ ≤ u(x) ≤ c2 δ(x) 1−σ ,
for x ∈ Ω ,
where c1 and c2 are positive constants. In [3], Brezis and Oswald studied (1.2) where 0 < σ < 1 and the function a is positive and bounded in Ω and they proved that (1.2) has a unique positive solution. Recently, applying Karamata regular variation theory, many authors have studied the asymptotic behavior of solutions of problem (1.2) (see [2–5,12,16]). This paper is motivated by a recent result stated in [16] which will be useful to our study. To describe this result in more detail, we need some notations. We denote by K the set of all Karamata functions L defined on (0, η] by L(t ) := c exp
η
z ( s) s
t
ds ,
for some η > 0, where c > 0 and z is a continuous function on [0, η] with z (0) = 0. It is clear that L is in K if and only if L is a positive function in C 1 ((0, η]), for some η > 0, such that lim
tL′ (t ) L(t )
t →0+
= 0.
Let d := diam(Ω ) and η > d. For λ ≤ 2, σ < 1 and L ∈ K such that (0, d) by
1, 1 1−σ η L(s) ds , t s ΨL,λ,σ (t ) := 1 (L(t )) 1−σ , 1 t L(s) 1−σ , ds 0
η 0
t 1−λ L(t )dt < ∞, we define the function ΨL,λ,σ on
if λ < 1 + σ , if λ = 1 + σ , if 1 + σ < λ < 2, if λ = 2.
s
For two nonnegative functions f and g defined on a set S, the notation f (x) ≈ g (x), x ∈ S means that there exists c > 0 such that 1c f (x) ≤ g (x) ≤ cf (x), for all x ∈ S. In [16], Mâagli studied problem (1.2), where σ < 1 and the function a verifies the following hypothesis. γ
(H1 ) a ∈ Cloc (Ω ) , (0 < γ < 1) satisfying for each x ∈ Ω , a(x) ≈ δ(x)−λ L(δ(x)), where λ ≤ 2 and L ∈ K such that
η 0
t 1−λ L(t )dt < ∞.
Then, by using the sub–supersolution method and some potential theory tools, the author proved in [16] the following result. Theorem 1. Assume (H1 ). Then for each σ < 1, problem (1.2) has a unique classical solution u satisfying for x ∈ Ω , u(x) ≈ δ(x)
−λ ) min(1, 12−σ
ΨL,λ,σ (δ(x)).
(1.3)
These estimates improve and generalize those established previously. In this paper, we take up problem (1.1) and as an extension of the above result, we prove the existence of a classical solution of problem (1.1) where both terms a(x) and g (u) in (1.1) are required to be in a more general class of functions. Moreover, estimates (1.3) continue to hold as it can be seen in our main result. Let us consider the following hypotheses.
S. Ben Othman, B. Khamessi / J. Math. Anal. Appl. 409 (2014) 925–933
927
γ
(A1 ) a is a positive function in Cloc (Ω ) , (0 < γ < 1), satisfying for x ∈ Ω δ(x)−λ1 L1 (δ(x)) ≤ a(x) ≤ δ(x)−λ2 L2 (δ(x)), where λ1 ≤ λ2 ≤ 2 and for i ∈ {1, 2}, Li ∈ K defined on (0, η], η > d such that η
t 1−λi Li (t )dt < ∞.
0
(g1 ) g is a nonnegative function in C 1 ((0, ∞)) satisfying and g (u) ≤ c2 uσ2
c1 uσ1 ≤ g (u) for 0 < u ≤ 1
for u > 0,
where σ2 ≤ σ1 < 1 and 0 < c1 < c2 . Remark 1. (i) If λ1 = λ2 , then hypothesis (A1 ) is satisfied if and only if L1 ≤ L2 . (ii) Let λ ≤ 2 and L ∈ K defined on (0, η], η > d, such that η
t 1−λ L(t )dt < ∞.
0
If we take λ1 = λ2 = λ, L1 = 1c L and L2 = cL for some c > 1, then (A1 ) is equivalent to (H1 ). Otherwise, a function a which satisfies (A1 ), does not necessarily satisfy (H1 ) (see Example 1). (iii) A function g satisfying (g1 ) does not need to be monotonous on (0, ∞) as in [16] (see Example 2). Remark 2. For i ∈ {1, 2}, we need to verify condition Lemma 2.
η 0
t 1−λi Li (t )dt < ∞ in hypothesis (A1 ), only if λi = 2. This is due to
Let us illustrate our hypotheses by some examples. Example 1. Let ψ be a function defined on (0, η], (η > d) by
ψ(t ) =
1 t
3 2
sin2
1 t
1
+ . t
Then there exists c > 0 such that 1 t
≤ ψ(t ) ≤
c 3
t2
,
t ∈ (0, η].
Hence, the function a(x) = ψ(δ(x)) satisfies (A1 ) and not (H1 ). Example 2. Let g be a function defined on (0, ∞) by
2π 1 + 3π t , 3 t g (t ) := 3π + sin(3π t ), 3π , t
if t ∈
0,
if t ∈
1 3
1
3
,
,1 ,
if t ∈ (1, ∞).
Then g satisfies hypothesis (g1 ) with σ1 =
1 2
, σ2 = −1, c1 = 1 and c2 = 3π + 1.
Now, we are ready to present our main result. Theorem 2. Assume (A1 ) and (g1 ). Then problem (1.1) has a positive solution u ∈ C (Ω ) ∩ C 2+γ (Ω ) satisfying for each x ∈ Ω 1 c
2−λ min(1, 1−σ1 )
δ(x)
1
ΨL1 ,λ1 ,σ1 (δ(x)) ≤ u(x) ≤ c δ(x)
2−λ min(1, 1−σ2 ) 2
ΨL2 ,λ2 ,σ2 (δ(x)),
(1.4)
where c > 0. Moreover the uniqueness of such a solution holds if g is nonincreasing on (0, ∞). The main body of the paper is organized as follows. In Section 2, we state some already known results on Karamata functions which will be useful for our study. Section 3 is devoted to the proof of Theorem 2. The last section is reserved for some applications. Throughout, the letter c will denote a generic constant which may vary from line to line.
928
S. Ben Othman, B. Khamessi / J. Math. Anal. Appl. 409 (2014) 925–933
2. Preliminary results In what follows, we recapitulate some properties of functions belonging to the class K . Lemma 1 (See [4,18]). Let L1 , L2 ∈ K be defined on (0, η] for some η > d. Let p ∈ R and ε > 0. Then we have the following. p (i) L1 L2 ∈ K , L1 ∈ K . ε (ii) limt →0+ t L1 (t ) = 0. (iii) The function η
t −→
L1 (s) s
t
(iv) If
η 0
L1 (s) ds s
ds ∈ K .
converges, then the function t
t −→
L1 (s) s
0
ds ∈ K .
Example 3. Let m ∈ N∗ and η > 0. Let (µ1 , µ2 , . . . , µm ) ∈ Rm and w be a sufficiently large positive real number such that the function
L(t ) =
logk
w µk t
1≤k≤m
is defined and positive on (0, η], where logk t = log ◦ · · · ◦ log t (k times). Then we have L ∈ K . Applying Karamata’s theorem, we get the following. Lemma 2 (See [18]). Let γ ∈ R and L be a function in K defined on (0, η]. We have the following. η (i) If γ > −1, then 0 sγ L(s)ds converges and t
1+γ
t →0+
0
(ii) If γ < −1, then η
t 1+γ L(t )
sγ L(s)ds ∼
η 0
sγ L(s)ds diverges and
sγ L(s)ds ∼ −
t 1+γ L(t ) 1+γ
t →0+
t
.
.
In the sequel, let ϕ be the positive normalized eigenfunction corresponding to the first smallest eigenvalue λ of Laplace operator ∆ in Ω with zero Dirichlet boundary condition. It is well known that ϕ ∈ C 2 (Ω ) and satisfies for x ∈ Ω ,
ϕ(x) ≈ δ(x).
(2.1)
See [15]. Lemma 3. Let L ∈ K . Then we have for x ∈ Ω , L(ϕ(x)) ≈ L(δ(x)). Proof. Let L ∈ K defined on (0, η], (η > d). There exist a constant c > 0 and z ∈ C ([0, η]) such that z (0) = 0 and for each t ∈ (0, η] L(t ) := c exp
η
t
z ( s) s
ds .
Using (2.1), there exists c0 > 1 such that for each x ∈ Ω , 1 c0
ϕ(x) ≤ δ(x) ≤ c0 ϕ(x).
Let ξ := sups∈[0,η] |z (s)|. From (2.2), we get that
ϕ(x)
z (s)
δ(x)
s
ds ≤ ξ log(c0 ).
It follows that 1 ξ
c0
ξ
L(ϕ(x)) ≤ L(δ(x)) ≤ c0 L(ϕ(x)).
This ends the proof.
(2.2)
S. Ben Othman, B. Khamessi / J. Math. Anal. Appl. 409 (2014) 925–933
929
3. Proof of Theorem 2 To prove our existence result, we adopt a sub–supersolution method. We consider the more general problem
−1u = f (x, u), x ∈ Ω , u > 0 in Ω , u|∂ Ω = 0.
(3.1)
Definition 1. A function v ∈ C (Ω ) ∩ C 2+γ (Ω ) is called a subsolution of problem (3.1) if
−1v ≤ f (x, v), x ∈ Ω , v > 0 in Ω , v|∂ Ω = 0.
If the above inequality is reversed, v is called a supersolution of problem (3.1). Lemma 4 (See [7]). Let f be a locally Hölder continuous function in Ω ×(0, ∞) and continuously differentiable with respect to the second variable. Suppose that problem (3.1) has a supersolution u¯ and a subsolution u such that u ≤ u¯ on Ω , then problem (3.1) has a solution u ∈ C (Ω ) ∩ C 2+γ (Ω ) satisfying u ≤ u ≤ u¯
in Ω .
Lemma 5 (See [11]). If f is nonincreasing with respect to the second variable in (0, ∞), then problem (3.1) has at most one solution u ∈ C (Ω ) ∩ C 2+γ (Ω ). In what follows, let a and g be two functions satisfying respectively (A1 ) and (g1 ). Applying Lemma 3, there exist m1 , m2 > 0 such that for x ∈ Ω , 1 m1
ϕ(x)−λ1 L1 (ϕ(x)) ≤ δ(x)−λ1 L1 (δ(x)) ≤ m1 ϕ(x)−λ1 L1 (ϕ(x))
(3.2)
ϕ(x)−λ2 L2 (ϕ(x)) ≤ δ(x)−λ2 L2 (δ(x)) ≤ m2 ϕ(x)−λ2 L2 (ϕ(x)).
(3.3)
and 1 m2
Combining hypothesis (A1 ) and the fact that ϕ ∈ C 2 (Ω ), we deduce that the function ϕ −λi Li (ϕ(·)) satisfies (H1 ), for i ∈ {1, 2}. We consider the following nonlinear problems
−1u = 1 c ϕ(x)−λ1 L (ϕ(x))uσ1 , 1 1 m1 u > 0 in Ω , u|∂ Ω = 0
x ∈ Ω,
(3.4)
and
−1u = m2 c2 ϕ(x)−λ2 L2 (ϕ(x))uσ2 , u > 0 in Ω , u|∂ Ω = 0.
x ∈ Ω,
(3.5)
Due to Theorem 1, problem (3.4) (respectively (3.5)) has a unique positive solution u1 (respectively u2 ) in C (Ω ) ∩ C 2 (Ω ) satisfying for x ∈ Ω , u1 (x) ≈ θλ1 (δ(x)) and
u2 (x) ≈ θλ2 (δ(x)),
(3.6)
where for i ∈ {1, 2}, θλi is the function defined on (0, η], (η > d) by
θλi (t ) = t
2−λ min(1, 1−σi ) i
ΨLi ,λi ,σi (t ).
(3.7)
The following proposition plays a key role to determine a subsolution and a supersolution of problem (1.1). Proposition 1. There exists a positive constant k such that for all t ∈ (0, η], we have
θλ1 (t ) ≤ kθλ2 (t ).
(3.8)
Proof. Put δ = min(1, 1−σ1 ) − min(1, 1−σ2 ). Since λ1 ≤ λ2 ≤ 2 and σ2 ≤ σ1 < 1, then 1−σ2 ≤ 1−σ1 and δ ≥ 0. By (3.7), 1 2 2 1 we have for t ∈ (0, η] 2−λ
2−λ
2−λ
2−λ
θλ1 1 (t ) = t δ ΨL1 ,λ1 ,σ1 (t )ΨL−2 ,λ (t ). 2 ,σ2 θλ2 1 Using Lemma 1(i), we deduce that ΨL1 ,λ1 ,σ1 ΨL−2 ,λ ∈ K . So, to reach (3.8), we split the proof into two cases. 2 ,σ2
930
S. Ben Othman, B. Khamessi / J. Math. Anal. Appl. 409 (2014) 925–933
Case 1. δ > 0. Since δ > 0 is equivalent to 1−σ2 < 1−σ1 and 1 + σ2 < λ2 , we conclude by Lemma 1(ii) that 2 1 2−λ
lim
t →0
2−λ
θλ1 1 (t ) = 0. (t ) = lim t δ ΨL1 ,λ1 ,σ1 (t )ΨL−2 ,λ 2 ,σ2 t →0 θλ2
Hence, there exists k > 0 such that for t ∈ (0, η]
θλ1 (t ) ≤ k. θλ2 Case 2. δ = 0. In this case, we have 1−σ2 = 1−σ1 or λ2 ≤ 1 + σ2 . Then, using the fact that λ1 ≤ λ2 and σ2 ≤ σ1 , we deduce 2 1 that for t ∈ (0, η] 2−λ
2−λ
1 1 1−σ t − 1−σ t 1 2 L2 (s) L1 (s) ds ds , s s 0 0 1 L1 (t ) 1−σ2 , L2 ( t ) θλ1 1 1 η − 1−σ (t ) := η L (s) 1−σ 2 2 L2 ( s ) 1 θλ2 ds ds , s s t t 1, 1 η − 1−σ 2 L2 (s) ds ,
t
s
if λ1 = λ2 = 2, if λ1 = λ2 , σ1 = σ2 and 1 + σ2 < λ2 < 2, if λ1 = λ2 , σ1 = σ2 and 1 + σ2 = λ2 , if λ2 < 1 + σ2 , if λ1 < 1 + σ1 and 1 + σ2 = λ2 .
Note that throughout the proof, we use Remark 1(i) and Lemma 1. So, we distinguish the following cases.
• If λ1 = λ2 = 2, then we have L1 ≤ L2 in (0, η], (η > d). This implies that 1 1 1−σ t − 1−σ 1 2 θλ1 L2 (s) . ds (t ) ≤ θλ2 s 0 η L (s) 1 1 Since 1−σ ≤ 1−σ and 0 < 0 2s ds < ∞, we deduce that there exists k > 0 such that 2 1 θλ1 ≤ k, θλ2
on (0, η].
• If λ1 = λ2 , σ1 = σ2 and 1 + σ2 < λ2 < 2, then for t ∈ (0, η] 1 θλ1 L1 (t ) 1−σ2 ≤ 1. (t ) = θλ2 L2 (t ) • If λ1 = λ2 , σ1 = σ2 and λ2 = 1 + σ2 , then for t ∈ (0, η] 1 1 η 1−σ − 1−σ 2 2 θλ1 L2 (s) (t ) ≤ ds = 1. θλ2 s t • If λ2 < 1 + σ2 , then for each t ∈ (0, η] θλ1 (t ) = 1. θλ2 • If λ2 = 1 + σ2 and λ1 < 1 + σ1 , then for t ∈ (0, η] 1 η − 1−σ 2 θλ1 L2 (s) (t ) = ds . θλ2 s t η L (s) Since 0 < 0 2s ds < ∞, we reach (3.8). This completes the proof.
Now, we are ready to state the proof of our main result. Proof of Theorem 2. Let u1 be the unique solution of problem (3.4) and u2 be the unique solution of problem (3.5). It follows from (3.6) that there exist k1 , k2 > 0 such that for each x ∈ Ω , 1 k1
θλ1 (δ(x)) ≤ u1 (x) ≤ k1 θλ1 (δ(x))
(3.9)
S. Ben Othman, B. Khamessi / J. Math. Anal. Appl. 409 (2014) 925–933
931
and 1 k2
θλ2 (δ(x)) ≤ u2 (x) ≤ k2 θλ2 (δ(x)).
(3.10)
Let u and u¯ be the functions defined in Ω respectively by and u¯ := d2 u2 ,
u := d1 u1
(3.11)
where d1 = min(1, ∥u ∥ ) and d2 = max(1, kk1 k2 ) with k the positive constant given by (3.8). Then using hypotheses (A1 ), 1 ∞ (g1 ), (3.2) and (3.3), we obtain that 1
−1u ≤ d1 1 c1 δ(x)−λ1 L1 (δ(x))uσ1 ≤ a(x)g (u), u > 0 in Ω , u|∂ Ω = 0
x ∈ Ω,
−∆u¯ ≥ d2 2 c2 δ(x)−λ2 L2 (δ(x))¯uσ2 ≥ a(x)g (¯u), u¯ > 0 in Ω , u¯ |∂ Ω = 0.
x ∈ Ω,
1−σ
and 1−σ
So u and u¯ are respectively a subsolution and a supersolution of problem (1.1). Moreover by (3.9)–(3.11), we conclude that for each x ∈ Ω , u(x) ≤ u¯ (x). γ
Since a ∈ Cloc (Ω ) and g ∈ C 1 ((0, ∞)), we deduce from Lemma 4 that problem (1.1) has a solution u in C (Ω ) ∩ C 2+γ (Ω ) satisfying u ≤ u ≤ u¯ . This together with (3.6) and (3.11) implies that u satisfies (1.4). Finally, if g is nonincreasing on (0, ∞), then by Lemma 5, the solution u is unique. 4. Applications 4.1. First application Let a be a function satisfying (A1 ) and g be a function satisfying (g1 ) and β < 1. We are interested in the following Dirichlet problem
−1u +
u>0
β
|∇ u|2 = a(x)g (u), u in Ω , u|∂ Ω = 0.
x ∈ Ω,
(4.1)
Put v = u1−β , then by a simple calculus, we prove that v satisfies
−β 1 −1v = (1 − β)a(x)v 1−β g (v 1−β ), v > 0 in Ω , v|∂ Ω = 0.
x ∈ Ω,
(4.2)
Let h be the function defined on (0, ∞) by h(v) = (1 − β)v
β − 1−β
1
g (v 1−β )
and put
µ1 =
σ1 − β 1−β
and µ2 =
σ2 − β . 1−β
Then it is obvious to see that the function h satisfies c1 v µ1 ≤ h(v)
for 0 < v ≤ 1
and
h(v) ≤ c2 v µ2
for v > 0.
Therefore, it follows from Theorem 2 that problem (4.2) has a positive solution v ∈ C (Ω ) ∩ C 2 (Ω ) satisfying 1 c
v1 ≤ v ≤ c v2 ,
932
S. Ben Othman, B. Khamessi / J. Math. Anal. Appl. 409 (2014) 925–933
where c > 0 and for i ∈ {1, 2}, vi is the function defined on Ω by
vi (x) := δ(x)
2−λ min(1, 1−µi ) i
ΨLi ,λi ,µi (δ(x)).
Consequently, we deduce that problem (4.1) has a positive solution u ∈ C (Ω ) ∩ C 2 (Ω ) satisfying 1 c
1
1
v11−β ≤ u ≤ c v21−β ,
where c is a positive constant. 4.2. Second application Let a be a function defined in Ω by a(x) =
1
δ(x)
3 2
sin2
1
δ(x)
+
1
δ(x)
.
Then there exists c > 0 such that for each x ∈ Ω 1
δ(x)
≤ a(x) ≤
c 3
δ(x) 2
. γ
Hence, the function a satisfies (A1 ). Let b be a positive function in Cloc (Ω ), (0 < γ < 1) such that for all x ∈ Ω ,
δ(x)−λ1 L1 (δ(x)) ≤ b(x) ≤ δ(x)−λ2 L2 (δ(x)), where for i ∈ {1, 2}, λi ∈ R and Li ∈ K . Let g be a function defined on (0, ∞) by
2π 1 + 3π t , 3 t g (t ) := 3π + sin(3π t ), 3π , t
if t ∈
if t ∈
0, 1 3
1
3
,
,1 ,
if t ∈ (1, ∞).
Then we have 1
t 2 ≤ g (t ) for 0 < t ≤ 1
and g (t ) ≤ (3π + 1)
1 t
for t > 0.
Let f ∈ C 1 ((0, ∞)) be a function satisfying d1 uα1 ≤ f (u) for 0 < u ≤ 1
and f (u) ≤ d2 uα2
for u > 0,
where α2 ≤ α1 < 1 and 0 < d1 < d2 . Let r ∈ R, and we are interested in the existence of positive continuous solutions of the following system
−1u = a(x)g (u), in Ω , −1v = b(x)ur f (v), in Ω , u, v > 0 in Ω , u|∂ Ω = v|∂ Ω = 0.
(4.3)
By Theorem 2, there exists a classical positive solution u of problem (1.1) satisfying for x ∈ Ω 1 c
1
δ(x) ≤ u(x) ≤ c δ(x) 4 ,
(4.4)
where c is a positive constant. So, we distinguish the following cases. Case 1. r ≥ 0. From estimates (4.4), we have 1 c
r
δ(x)−λ1 +r L1 (δ(x)) ≤ b(x)ur (x) ≤ c δ(x)−λ2 + 4 L2 (δ(x)). η
r
Now suppose that λ1 − r ≤ λ2 − 4r ≤ 2 and 0 t 1−λ2 + 4 L2 (t )dt < ∞. Then applying Theorem 2, we conclude that system (4.3) has a classical solution (u, v) such that 1 c
1
δ(x) ≤ u(x) ≤ c δ(x) 4
S. Ben Othman, B. Khamessi / J. Math. Anal. Appl. 409 (2014) 925–933
933
and 1 m
δ(x)
2−λ1 +r min(1, 1−α ) 1 Ψ
L1 ,λ1 −r ,α1
min(1,
(δ(x)) ≤ v(x) ≤ mδ(x)
2−λ2 + r 4 1−α2
)
ΨL2 ,λ2 − 4r ,α2 (δ(x)),
where c and m are positive constants. Case 2. r < 0. From estimates (4.4), we deduce that 1
r
δ(x)−λ1 + 4 L1 (δ(x)) ≤ b(x)ur (x) ≤ c δ(x)−λ2 +r L2 (δ(x)). c η Now suppose that λ1 − 4r ≤ λ2 − r ≤ 2 and 0 t 1−λ2 +r L2 (t )dt < ∞.
Then applying again Theorem 2, we conclude that system (4.3) has a classical solution (u, v) such that 1 c
1
δ(x) ≤ u(x) ≤ c δ(x) 4
and 1 m
δ(x)
min(1,
2−λ1 + r 4 1−α1
)
ΨL1 ,λ1 − 4r ,α1 (δ(x)) ≤ v(x) ≤ mδ(x)
2−λ2 +r ) min(1, 1−α 2 Ψ
L2 ,λ2 −r ,α2
(δ(x)),
where c and m are positive constants. References [1] I. Bachar, H. Mâagli, N. Zeddini, Estimates on the Green function and existence of positive solutions of nonlinear singular elliptic equations, Commun. Contemp. Math. 3 (2003) 401–434. [2] S. Ben 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. [3] H. Brezis, L. Oswald, Remarks on sublinear elliptic equations, Nonlinear Anal. 10 (1986) 55–64. [4] R. Chemmam, H. Mâagli, S. Masmoudi, M. Zribi, Combined effects in nonlinear singular elliptic problems in a bounded domain, Adv. Nonlinear Anal. 1 (2012) 301–318. [5] F.C. Cîrstea, V.D. Rădulescu, Extermal singular solutions for degenerate logistic-type equations in anisotropic media, CR Acad. Sci. Paris, Ser. Je 339 (2004) 119–124. [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] S. Cui, Existence and nonexistence of positive solutions for singular semilinear elliptic boundary value problems, Nonlinear Anal. 41 (2000) 149–176. [8] M. Ghergu, V.D. Rădulescu, On a class of sublinear singular elliptic problems with convection term, J. Math. Anal. Appl. 311 (2005) 635–646. [9] M. Ghergu, V.D. Rădulescu, Singular Elliptic Problems. Bifurcation and Asymptotic Analysis, in: Oxford Lecture Series in Mathematics and Applications, vol. 37, Oxford University Press, 2008, p. 320. [10] M. Ghergu, V.D. Rădulescu, Nonlinear PDEs Mathematical Models in Biology, Chemistry and Population Genetics, in: Springer Monographs in Mathematics, Springer Verlag, 2012. [11] D. Gilbarg, N.S. Trudinger, Elliptic Partial Differential Equations of Second Order, third ed., Springer Verlag, Berlin, 1983. [12] S. Gontara, H. Mâagli, S. Masmoudi, S. Turki, Asymptotic behavior of positive solutions of a singular nonlinear Dirichlet problem, J. Math. Anal. Appl. 369 (2010) 719–729. [13] A.C. Lazer, P.J. McKenna, On a singular elliptic boundary value problem, Proc. Amer. Math. Soc. 111 (1991) 721–730. [14] H. Mâagli, On the Solutions of a Singular Nonlinear Periodic Boundary Value Problem, Potential Anal. 14 (2001) 437–447. [15] H. Mâagli, Inequalites for the Riesz potentials, Arch. Inequal. Appl. 1 (2003) 295–304. [16] H. Mâagli, Asymptotic behavior of positive solution of a semilinear Dirichlet problem, Nonlinear Anal. 74 (2011) 2941–2947. [17] H. Mâagli, M. Zribi, Existence and estimates of solutions for singular nonlinear elliptic problems, J. Math. Anal. Appl. 263 (2001) 522–542. [18] R. Seneta, Regular Varying Functions, in: Lectures Notes in Math., Springer-Verlag, Berlin, 1976, p. 508. [19] N. Zeddini, Positive solutions for a singular nonlinear problem on a bounded domain in R2 , Potential Anal. 18 (2003) 97–118. [20] Z. Zhang, The asymptotic behavior of the unique solution for the singular Lane–Emden–Fowler equation, J. Math. Anal. Appl. 312 (2005) 33–43.
Contents lists available at ScienceDirect
Journal of Mathematical Analysis and Applications journal homepage: www.elsevier.com/locate/jmaa
Asymptotic behavior of positive solutions of a nonlinear Dirichlet problem Sonia Ben Othman ∗ , Bilel Khamessi Département de Mathématiques, Faculté des Sciences de Tunis, Campus Universitaire, 2092 Tunis, Tunisia
article
info
Article history: Received 12 June 2013 Available online 31 July 2013 Submitted by V. Radulescu Keywords: Dirichlet problem Positive solution Sub and supersolutions Asymptotic behavior Karamata function
abstract We take up the existence and the asymptotic behavior of a classical solution to the following semilinear Dirichlet problem
−∆u = a(x)g (u), x ∈ Ω , u > 0 in Ω , u|∂ Ω = 0, γ
where Ω is a C 1,1 -bounded domain in RN , N ≥ 2 and the function a belongs to Cloc (Ω ), (0 < γ < 1) such that there exist c1 , c2 > 0 satisfying for each x ∈ Ω , c1 δ(x)−λ1 exp
η δ(x)
z1 (s) s
ds
≤ a(x) ≤ c2 δ(x)−λ2 exp
η
z2 (s)
δ(x)
s
ds ,
where η > diam(Ω ), δ(x) = dist (x, ∂ Ω ), λ1 ≤ λ2 ≤ 2 and for i ∈ {1, 2}, zi is a continuous function on [0, η] with zi (0) = 0. Our arguments are based on the sub–supersolution method with Karamata regular variation theory. © 2013 Elsevier Inc. All rights reserved.
1. Introduction Let Ω be a bounded C 1,1 domain in RN , N ≥ 2. We consider the following nonlinear Dirichlet problem
−1u = a(x)g (u), x ∈ Ω , u > 0 in Ω , u|∂ Ω = 0,
(1.1) γ
where g is a nonnegative function in C 1 ((0, ∞)) and a is a positive function in Cloc (Ω ), satisfying some appropriate conditions related to Karamata regular variation theory. Our main purpose is to prove the existence of a classical solution to problem (1.1) and to give estimates on such a solution. Several articles have been devoted to the study of problems of type (1.1) involving singular or sublinear nonlinearities. We refer to [1–6,8–10,12–14,16,17,19,20] and the references therein. Namely, the following problem
−1u = a(x)uσ , x ∈ Ω , σ < 1, u > 0 in Ω , u|∂ Ω = 0
(1.2)
is investigated by many authors under different kinds of assumptions on the weight a(x) (see [1,3–6,9,10,12,16,19,20]).
∗
Corresponding author. E-mail addresses: [email protected], [email protected] (S. Ben Othman), [email protected] (B. Khamessi).
0022-247X/$ – see front matter © 2013 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.jmaa.2013.07.063
926
S. Ben Othman, B. Khamessi / J. Math. Anal. Appl. 409 (2014) 925–933
In [6], Crandall et al. considered problem (1.2) where σ < −1 and a ≡ 1. They proved that there exists a unique positive classical solution u of problem (1.2) satisfying 2
2
c1 δ(x) 1−σ ≤ u(x) ≤ c2 δ(x) 1−σ ,
near the boundary ∂ Ω ,
where c1 , c2 are positive constants and δ(x) = dist(x, ∂ Ω ). Later in [13], Lazer and McKenna extended the result stated in [6] to a large class of functions which are not necessarily bounded in Ω . Indeed they assumed that if b1 δ(x)−λ ≤ a(x) ≤ b2 δ(x)−λ ,
x ∈ Ω,
where b1 , b2 > 0 and λ ∈ (0, 2), then problem (1.2) has a unique positive classical solution u in Ω such that 2−λ
2
c1 δ(x) 1−σ ≤ u(x) ≤ c2 δ(x) 1−σ ,
for x ∈ Ω ,
where c1 and c2 are positive constants. In [3], Brezis and Oswald studied (1.2) where 0 < σ < 1 and the function a is positive and bounded in Ω and they proved that (1.2) has a unique positive solution. Recently, applying Karamata regular variation theory, many authors have studied the asymptotic behavior of solutions of problem (1.2) (see [2–5,12,16]). This paper is motivated by a recent result stated in [16] which will be useful to our study. To describe this result in more detail, we need some notations. We denote by K the set of all Karamata functions L defined on (0, η] by L(t ) := c exp
η
z ( s) s
t
ds ,
for some η > 0, where c > 0 and z is a continuous function on [0, η] with z (0) = 0. It is clear that L is in K if and only if L is a positive function in C 1 ((0, η]), for some η > 0, such that lim
tL′ (t ) L(t )
t →0+
= 0.
Let d := diam(Ω ) and η > d. For λ ≤ 2, σ < 1 and L ∈ K such that (0, d) by
1, 1 1−σ η L(s) ds , t s ΨL,λ,σ (t ) := 1 (L(t )) 1−σ , 1 t L(s) 1−σ , ds 0
η 0
t 1−λ L(t )dt < ∞, we define the function ΨL,λ,σ on
if λ < 1 + σ , if λ = 1 + σ , if 1 + σ < λ < 2, if λ = 2.
s
For two nonnegative functions f and g defined on a set S, the notation f (x) ≈ g (x), x ∈ S means that there exists c > 0 such that 1c f (x) ≤ g (x) ≤ cf (x), for all x ∈ S. In [16], Mâagli studied problem (1.2), where σ < 1 and the function a verifies the following hypothesis. γ
(H1 ) a ∈ Cloc (Ω ) , (0 < γ < 1) satisfying for each x ∈ Ω , a(x) ≈ δ(x)−λ L(δ(x)), where λ ≤ 2 and L ∈ K such that
η 0
t 1−λ L(t )dt < ∞.
Then, by using the sub–supersolution method and some potential theory tools, the author proved in [16] the following result. Theorem 1. Assume (H1 ). Then for each σ < 1, problem (1.2) has a unique classical solution u satisfying for x ∈ Ω , u(x) ≈ δ(x)
−λ ) min(1, 12−σ
ΨL,λ,σ (δ(x)).
(1.3)
These estimates improve and generalize those established previously. In this paper, we take up problem (1.1) and as an extension of the above result, we prove the existence of a classical solution of problem (1.1) where both terms a(x) and g (u) in (1.1) are required to be in a more general class of functions. Moreover, estimates (1.3) continue to hold as it can be seen in our main result. Let us consider the following hypotheses.
S. Ben Othman, B. Khamessi / J. Math. Anal. Appl. 409 (2014) 925–933
927
γ
(A1 ) a is a positive function in Cloc (Ω ) , (0 < γ < 1), satisfying for x ∈ Ω δ(x)−λ1 L1 (δ(x)) ≤ a(x) ≤ δ(x)−λ2 L2 (δ(x)), where λ1 ≤ λ2 ≤ 2 and for i ∈ {1, 2}, Li ∈ K defined on (0, η], η > d such that η
t 1−λi Li (t )dt < ∞.
0
(g1 ) g is a nonnegative function in C 1 ((0, ∞)) satisfying and g (u) ≤ c2 uσ2
c1 uσ1 ≤ g (u) for 0 < u ≤ 1
for u > 0,
where σ2 ≤ σ1 < 1 and 0 < c1 < c2 . Remark 1. (i) If λ1 = λ2 , then hypothesis (A1 ) is satisfied if and only if L1 ≤ L2 . (ii) Let λ ≤ 2 and L ∈ K defined on (0, η], η > d, such that η
t 1−λ L(t )dt < ∞.
0
If we take λ1 = λ2 = λ, L1 = 1c L and L2 = cL for some c > 1, then (A1 ) is equivalent to (H1 ). Otherwise, a function a which satisfies (A1 ), does not necessarily satisfy (H1 ) (see Example 1). (iii) A function g satisfying (g1 ) does not need to be monotonous on (0, ∞) as in [16] (see Example 2). Remark 2. For i ∈ {1, 2}, we need to verify condition Lemma 2.
η 0
t 1−λi Li (t )dt < ∞ in hypothesis (A1 ), only if λi = 2. This is due to
Let us illustrate our hypotheses by some examples. Example 1. Let ψ be a function defined on (0, η], (η > d) by
ψ(t ) =
1 t
3 2
sin2
1 t
1
+ . t
Then there exists c > 0 such that 1 t
≤ ψ(t ) ≤
c 3
t2
,
t ∈ (0, η].
Hence, the function a(x) = ψ(δ(x)) satisfies (A1 ) and not (H1 ). Example 2. Let g be a function defined on (0, ∞) by
2π 1 + 3π t , 3 t g (t ) := 3π + sin(3π t ), 3π , t
if t ∈
0,
if t ∈
1 3
1
3
,
,1 ,
if t ∈ (1, ∞).
Then g satisfies hypothesis (g1 ) with σ1 =
1 2
, σ2 = −1, c1 = 1 and c2 = 3π + 1.
Now, we are ready to present our main result. Theorem 2. Assume (A1 ) and (g1 ). Then problem (1.1) has a positive solution u ∈ C (Ω ) ∩ C 2+γ (Ω ) satisfying for each x ∈ Ω 1 c
2−λ min(1, 1−σ1 )
δ(x)
1
ΨL1 ,λ1 ,σ1 (δ(x)) ≤ u(x) ≤ c δ(x)
2−λ min(1, 1−σ2 ) 2
ΨL2 ,λ2 ,σ2 (δ(x)),
(1.4)
where c > 0. Moreover the uniqueness of such a solution holds if g is nonincreasing on (0, ∞). The main body of the paper is organized as follows. In Section 2, we state some already known results on Karamata functions which will be useful for our study. Section 3 is devoted to the proof of Theorem 2. The last section is reserved for some applications. Throughout, the letter c will denote a generic constant which may vary from line to line.
928
S. Ben Othman, B. Khamessi / J. Math. Anal. Appl. 409 (2014) 925–933
2. Preliminary results In what follows, we recapitulate some properties of functions belonging to the class K . Lemma 1 (See [4,18]). Let L1 , L2 ∈ K be defined on (0, η] for some η > d. Let p ∈ R and ε > 0. Then we have the following. p (i) L1 L2 ∈ K , L1 ∈ K . ε (ii) limt →0+ t L1 (t ) = 0. (iii) The function η
t −→
L1 (s) s
t
(iv) If
η 0
L1 (s) ds s
ds ∈ K .
converges, then the function t
t −→
L1 (s) s
0
ds ∈ K .
Example 3. Let m ∈ N∗ and η > 0. Let (µ1 , µ2 , . . . , µm ) ∈ Rm and w be a sufficiently large positive real number such that the function
L(t ) =
logk
w µk t
1≤k≤m
is defined and positive on (0, η], where logk t = log ◦ · · · ◦ log t (k times). Then we have L ∈ K . Applying Karamata’s theorem, we get the following. Lemma 2 (See [18]). Let γ ∈ R and L be a function in K defined on (0, η]. We have the following. η (i) If γ > −1, then 0 sγ L(s)ds converges and t
1+γ
t →0+
0
(ii) If γ < −1, then η
t 1+γ L(t )
sγ L(s)ds ∼
η 0
sγ L(s)ds diverges and
sγ L(s)ds ∼ −
t 1+γ L(t ) 1+γ
t →0+
t
.
.
In the sequel, let ϕ be the positive normalized eigenfunction corresponding to the first smallest eigenvalue λ of Laplace operator ∆ in Ω with zero Dirichlet boundary condition. It is well known that ϕ ∈ C 2 (Ω ) and satisfies for x ∈ Ω ,
ϕ(x) ≈ δ(x).
(2.1)
See [15]. Lemma 3. Let L ∈ K . Then we have for x ∈ Ω , L(ϕ(x)) ≈ L(δ(x)). Proof. Let L ∈ K defined on (0, η], (η > d). There exist a constant c > 0 and z ∈ C ([0, η]) such that z (0) = 0 and for each t ∈ (0, η] L(t ) := c exp
η
t
z ( s) s
ds .
Using (2.1), there exists c0 > 1 such that for each x ∈ Ω , 1 c0
ϕ(x) ≤ δ(x) ≤ c0 ϕ(x).
Let ξ := sups∈[0,η] |z (s)|. From (2.2), we get that
ϕ(x)
z (s)
δ(x)
s
ds ≤ ξ log(c0 ).
It follows that 1 ξ
c0
ξ
L(ϕ(x)) ≤ L(δ(x)) ≤ c0 L(ϕ(x)).
This ends the proof.
(2.2)
S. Ben Othman, B. Khamessi / J. Math. Anal. Appl. 409 (2014) 925–933
929
3. Proof of Theorem 2 To prove our existence result, we adopt a sub–supersolution method. We consider the more general problem
−1u = f (x, u), x ∈ Ω , u > 0 in Ω , u|∂ Ω = 0.
(3.1)
Definition 1. A function v ∈ C (Ω ) ∩ C 2+γ (Ω ) is called a subsolution of problem (3.1) if
−1v ≤ f (x, v), x ∈ Ω , v > 0 in Ω , v|∂ Ω = 0.
If the above inequality is reversed, v is called a supersolution of problem (3.1). Lemma 4 (See [7]). Let f be a locally Hölder continuous function in Ω ×(0, ∞) and continuously differentiable with respect to the second variable. Suppose that problem (3.1) has a supersolution u¯ and a subsolution u such that u ≤ u¯ on Ω , then problem (3.1) has a solution u ∈ C (Ω ) ∩ C 2+γ (Ω ) satisfying u ≤ u ≤ u¯
in Ω .
Lemma 5 (See [11]). If f is nonincreasing with respect to the second variable in (0, ∞), then problem (3.1) has at most one solution u ∈ C (Ω ) ∩ C 2+γ (Ω ). In what follows, let a and g be two functions satisfying respectively (A1 ) and (g1 ). Applying Lemma 3, there exist m1 , m2 > 0 such that for x ∈ Ω , 1 m1
ϕ(x)−λ1 L1 (ϕ(x)) ≤ δ(x)−λ1 L1 (δ(x)) ≤ m1 ϕ(x)−λ1 L1 (ϕ(x))
(3.2)
ϕ(x)−λ2 L2 (ϕ(x)) ≤ δ(x)−λ2 L2 (δ(x)) ≤ m2 ϕ(x)−λ2 L2 (ϕ(x)).
(3.3)
and 1 m2
Combining hypothesis (A1 ) and the fact that ϕ ∈ C 2 (Ω ), we deduce that the function ϕ −λi Li (ϕ(·)) satisfies (H1 ), for i ∈ {1, 2}. We consider the following nonlinear problems
−1u = 1 c ϕ(x)−λ1 L (ϕ(x))uσ1 , 1 1 m1 u > 0 in Ω , u|∂ Ω = 0
x ∈ Ω,
(3.4)
and
−1u = m2 c2 ϕ(x)−λ2 L2 (ϕ(x))uσ2 , u > 0 in Ω , u|∂ Ω = 0.
x ∈ Ω,
(3.5)
Due to Theorem 1, problem (3.4) (respectively (3.5)) has a unique positive solution u1 (respectively u2 ) in C (Ω ) ∩ C 2 (Ω ) satisfying for x ∈ Ω , u1 (x) ≈ θλ1 (δ(x)) and
u2 (x) ≈ θλ2 (δ(x)),
(3.6)
where for i ∈ {1, 2}, θλi is the function defined on (0, η], (η > d) by
θλi (t ) = t
2−λ min(1, 1−σi ) i
ΨLi ,λi ,σi (t ).
(3.7)
The following proposition plays a key role to determine a subsolution and a supersolution of problem (1.1). Proposition 1. There exists a positive constant k such that for all t ∈ (0, η], we have
θλ1 (t ) ≤ kθλ2 (t ).
(3.8)
Proof. Put δ = min(1, 1−σ1 ) − min(1, 1−σ2 ). Since λ1 ≤ λ2 ≤ 2 and σ2 ≤ σ1 < 1, then 1−σ2 ≤ 1−σ1 and δ ≥ 0. By (3.7), 1 2 2 1 we have for t ∈ (0, η] 2−λ
2−λ
2−λ
2−λ
θλ1 1 (t ) = t δ ΨL1 ,λ1 ,σ1 (t )ΨL−2 ,λ (t ). 2 ,σ2 θλ2 1 Using Lemma 1(i), we deduce that ΨL1 ,λ1 ,σ1 ΨL−2 ,λ ∈ K . So, to reach (3.8), we split the proof into two cases. 2 ,σ2
930
S. Ben Othman, B. Khamessi / J. Math. Anal. Appl. 409 (2014) 925–933
Case 1. δ > 0. Since δ > 0 is equivalent to 1−σ2 < 1−σ1 and 1 + σ2 < λ2 , we conclude by Lemma 1(ii) that 2 1 2−λ
lim
t →0
2−λ
θλ1 1 (t ) = 0. (t ) = lim t δ ΨL1 ,λ1 ,σ1 (t )ΨL−2 ,λ 2 ,σ2 t →0 θλ2
Hence, there exists k > 0 such that for t ∈ (0, η]
θλ1 (t ) ≤ k. θλ2 Case 2. δ = 0. In this case, we have 1−σ2 = 1−σ1 or λ2 ≤ 1 + σ2 . Then, using the fact that λ1 ≤ λ2 and σ2 ≤ σ1 , we deduce 2 1 that for t ∈ (0, η] 2−λ
2−λ
1 1 1−σ t − 1−σ t 1 2 L2 (s) L1 (s) ds ds , s s 0 0 1 L1 (t ) 1−σ2 , L2 ( t ) θλ1 1 1 η − 1−σ (t ) := η L (s) 1−σ 2 2 L2 ( s ) 1 θλ2 ds ds , s s t t 1, 1 η − 1−σ 2 L2 (s) ds ,
t
s
if λ1 = λ2 = 2, if λ1 = λ2 , σ1 = σ2 and 1 + σ2 < λ2 < 2, if λ1 = λ2 , σ1 = σ2 and 1 + σ2 = λ2 , if λ2 < 1 + σ2 , if λ1 < 1 + σ1 and 1 + σ2 = λ2 .
Note that throughout the proof, we use Remark 1(i) and Lemma 1. So, we distinguish the following cases.
• If λ1 = λ2 = 2, then we have L1 ≤ L2 in (0, η], (η > d). This implies that 1 1 1−σ t − 1−σ 1 2 θλ1 L2 (s) . ds (t ) ≤ θλ2 s 0 η L (s) 1 1 Since 1−σ ≤ 1−σ and 0 < 0 2s ds < ∞, we deduce that there exists k > 0 such that 2 1 θλ1 ≤ k, θλ2
on (0, η].
• If λ1 = λ2 , σ1 = σ2 and 1 + σ2 < λ2 < 2, then for t ∈ (0, η] 1 θλ1 L1 (t ) 1−σ2 ≤ 1. (t ) = θλ2 L2 (t ) • If λ1 = λ2 , σ1 = σ2 and λ2 = 1 + σ2 , then for t ∈ (0, η] 1 1 η 1−σ − 1−σ 2 2 θλ1 L2 (s) (t ) ≤ ds = 1. θλ2 s t • If λ2 < 1 + σ2 , then for each t ∈ (0, η] θλ1 (t ) = 1. θλ2 • If λ2 = 1 + σ2 and λ1 < 1 + σ1 , then for t ∈ (0, η] 1 η − 1−σ 2 θλ1 L2 (s) (t ) = ds . θλ2 s t η L (s) Since 0 < 0 2s ds < ∞, we reach (3.8). This completes the proof.
Now, we are ready to state the proof of our main result. Proof of Theorem 2. Let u1 be the unique solution of problem (3.4) and u2 be the unique solution of problem (3.5). It follows from (3.6) that there exist k1 , k2 > 0 such that for each x ∈ Ω , 1 k1
θλ1 (δ(x)) ≤ u1 (x) ≤ k1 θλ1 (δ(x))
(3.9)
S. Ben Othman, B. Khamessi / J. Math. Anal. Appl. 409 (2014) 925–933
931
and 1 k2
θλ2 (δ(x)) ≤ u2 (x) ≤ k2 θλ2 (δ(x)).
(3.10)
Let u and u¯ be the functions defined in Ω respectively by and u¯ := d2 u2 ,
u := d1 u1
(3.11)
where d1 = min(1, ∥u ∥ ) and d2 = max(1, kk1 k2 ) with k the positive constant given by (3.8). Then using hypotheses (A1 ), 1 ∞ (g1 ), (3.2) and (3.3), we obtain that 1
−1u ≤ d1 1 c1 δ(x)−λ1 L1 (δ(x))uσ1 ≤ a(x)g (u), u > 0 in Ω , u|∂ Ω = 0
x ∈ Ω,
−∆u¯ ≥ d2 2 c2 δ(x)−λ2 L2 (δ(x))¯uσ2 ≥ a(x)g (¯u), u¯ > 0 in Ω , u¯ |∂ Ω = 0.
x ∈ Ω,
1−σ
and 1−σ
So u and u¯ are respectively a subsolution and a supersolution of problem (1.1). Moreover by (3.9)–(3.11), we conclude that for each x ∈ Ω , u(x) ≤ u¯ (x). γ
Since a ∈ Cloc (Ω ) and g ∈ C 1 ((0, ∞)), we deduce from Lemma 4 that problem (1.1) has a solution u in C (Ω ) ∩ C 2+γ (Ω ) satisfying u ≤ u ≤ u¯ . This together with (3.6) and (3.11) implies that u satisfies (1.4). Finally, if g is nonincreasing on (0, ∞), then by Lemma 5, the solution u is unique. 4. Applications 4.1. First application Let a be a function satisfying (A1 ) and g be a function satisfying (g1 ) and β < 1. We are interested in the following Dirichlet problem
−1u +
u>0
β
|∇ u|2 = a(x)g (u), u in Ω , u|∂ Ω = 0.
x ∈ Ω,
(4.1)
Put v = u1−β , then by a simple calculus, we prove that v satisfies
−β 1 −1v = (1 − β)a(x)v 1−β g (v 1−β ), v > 0 in Ω , v|∂ Ω = 0.
x ∈ Ω,
(4.2)
Let h be the function defined on (0, ∞) by h(v) = (1 − β)v
β − 1−β
1
g (v 1−β )
and put
µ1 =
σ1 − β 1−β
and µ2 =
σ2 − β . 1−β
Then it is obvious to see that the function h satisfies c1 v µ1 ≤ h(v)
for 0 < v ≤ 1
and
h(v) ≤ c2 v µ2
for v > 0.
Therefore, it follows from Theorem 2 that problem (4.2) has a positive solution v ∈ C (Ω ) ∩ C 2 (Ω ) satisfying 1 c
v1 ≤ v ≤ c v2 ,
932
S. Ben Othman, B. Khamessi / J. Math. Anal. Appl. 409 (2014) 925–933
where c > 0 and for i ∈ {1, 2}, vi is the function defined on Ω by
vi (x) := δ(x)
2−λ min(1, 1−µi ) i
ΨLi ,λi ,µi (δ(x)).
Consequently, we deduce that problem (4.1) has a positive solution u ∈ C (Ω ) ∩ C 2 (Ω ) satisfying 1 c
1
1
v11−β ≤ u ≤ c v21−β ,
where c is a positive constant. 4.2. Second application Let a be a function defined in Ω by a(x) =
1
δ(x)
3 2
sin2
1
δ(x)
+
1
δ(x)
.
Then there exists c > 0 such that for each x ∈ Ω 1
δ(x)
≤ a(x) ≤
c 3
δ(x) 2
. γ
Hence, the function a satisfies (A1 ). Let b be a positive function in Cloc (Ω ), (0 < γ < 1) such that for all x ∈ Ω ,
δ(x)−λ1 L1 (δ(x)) ≤ b(x) ≤ δ(x)−λ2 L2 (δ(x)), where for i ∈ {1, 2}, λi ∈ R and Li ∈ K . Let g be a function defined on (0, ∞) by
2π 1 + 3π t , 3 t g (t ) := 3π + sin(3π t ), 3π , t
if t ∈
if t ∈
0, 1 3
1
3
,
,1 ,
if t ∈ (1, ∞).
Then we have 1
t 2 ≤ g (t ) for 0 < t ≤ 1
and g (t ) ≤ (3π + 1)
1 t
for t > 0.
Let f ∈ C 1 ((0, ∞)) be a function satisfying d1 uα1 ≤ f (u) for 0 < u ≤ 1
and f (u) ≤ d2 uα2
for u > 0,
where α2 ≤ α1 < 1 and 0 < d1 < d2 . Let r ∈ R, and we are interested in the existence of positive continuous solutions of the following system
−1u = a(x)g (u), in Ω , −1v = b(x)ur f (v), in Ω , u, v > 0 in Ω , u|∂ Ω = v|∂ Ω = 0.
(4.3)
By Theorem 2, there exists a classical positive solution u of problem (1.1) satisfying for x ∈ Ω 1 c
1
δ(x) ≤ u(x) ≤ c δ(x) 4 ,
(4.4)
where c is a positive constant. So, we distinguish the following cases. Case 1. r ≥ 0. From estimates (4.4), we have 1 c
r
δ(x)−λ1 +r L1 (δ(x)) ≤ b(x)ur (x) ≤ c δ(x)−λ2 + 4 L2 (δ(x)). η
r
Now suppose that λ1 − r ≤ λ2 − 4r ≤ 2 and 0 t 1−λ2 + 4 L2 (t )dt < ∞. Then applying Theorem 2, we conclude that system (4.3) has a classical solution (u, v) such that 1 c
1
δ(x) ≤ u(x) ≤ c δ(x) 4
S. Ben Othman, B. Khamessi / J. Math. Anal. Appl. 409 (2014) 925–933
933
and 1 m
δ(x)
2−λ1 +r min(1, 1−α ) 1 Ψ
L1 ,λ1 −r ,α1
min(1,
(δ(x)) ≤ v(x) ≤ mδ(x)
2−λ2 + r 4 1−α2
)
ΨL2 ,λ2 − 4r ,α2 (δ(x)),
where c and m are positive constants. Case 2. r < 0. From estimates (4.4), we deduce that 1
r
δ(x)−λ1 + 4 L1 (δ(x)) ≤ b(x)ur (x) ≤ c δ(x)−λ2 +r L2 (δ(x)). c η Now suppose that λ1 − 4r ≤ λ2 − r ≤ 2 and 0 t 1−λ2 +r L2 (t )dt < ∞.
Then applying again Theorem 2, we conclude that system (4.3) has a classical solution (u, v) such that 1 c
1
δ(x) ≤ u(x) ≤ c δ(x) 4
and 1 m
δ(x)
min(1,
2−λ1 + r 4 1−α1
)
ΨL1 ,λ1 − 4r ,α1 (δ(x)) ≤ v(x) ≤ mδ(x)
2−λ2 +r ) min(1, 1−α 2 Ψ
L2 ,λ2 −r ,α2
(δ(x)),
where c and m are positive constants. References [1] I. Bachar, H. Mâagli, N. Zeddini, Estimates on the Green function and existence of positive solutions of nonlinear singular elliptic equations, Commun. Contemp. Math. 3 (2003) 401–434. [2] S. Ben 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. [3] H. Brezis, L. Oswald, Remarks on sublinear elliptic equations, Nonlinear Anal. 10 (1986) 55–64. [4] R. Chemmam, H. Mâagli, S. Masmoudi, M. Zribi, Combined effects in nonlinear singular elliptic problems in a bounded domain, Adv. Nonlinear Anal. 1 (2012) 301–318. [5] F.C. Cîrstea, V.D. Rădulescu, Extermal singular solutions for degenerate logistic-type equations in anisotropic media, CR Acad. Sci. Paris, Ser. Je 339 (2004) 119–124. [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] S. Cui, Existence and nonexistence of positive solutions for singular semilinear elliptic boundary value problems, Nonlinear Anal. 41 (2000) 149–176. [8] M. Ghergu, V.D. Rădulescu, On a class of sublinear singular elliptic problems with convection term, J. Math. Anal. Appl. 311 (2005) 635–646. [9] M. Ghergu, V.D. Rădulescu, Singular Elliptic Problems. Bifurcation and Asymptotic Analysis, in: Oxford Lecture Series in Mathematics and Applications, vol. 37, Oxford University Press, 2008, p. 320. [10] M. Ghergu, V.D. Rădulescu, Nonlinear PDEs Mathematical Models in Biology, Chemistry and Population Genetics, in: Springer Monographs in Mathematics, Springer Verlag, 2012. [11] D. Gilbarg, N.S. Trudinger, Elliptic Partial Differential Equations of Second Order, third ed., Springer Verlag, Berlin, 1983. [12] S. Gontara, H. Mâagli, S. Masmoudi, S. Turki, Asymptotic behavior of positive solutions of a singular nonlinear Dirichlet problem, J. Math. Anal. Appl. 369 (2010) 719–729. [13] A.C. Lazer, P.J. McKenna, On a singular elliptic boundary value problem, Proc. Amer. Math. Soc. 111 (1991) 721–730. [14] H. Mâagli, On the Solutions of a Singular Nonlinear Periodic Boundary Value Problem, Potential Anal. 14 (2001) 437–447. [15] H. Mâagli, Inequalites for the Riesz potentials, Arch. Inequal. Appl. 1 (2003) 295–304. [16] H. Mâagli, Asymptotic behavior of positive solution of a semilinear Dirichlet problem, Nonlinear Anal. 74 (2011) 2941–2947. [17] H. Mâagli, M. Zribi, Existence and estimates of solutions for singular nonlinear elliptic problems, J. Math. Anal. Appl. 263 (2001) 522–542. [18] R. Seneta, Regular Varying Functions, in: Lectures Notes in Math., Springer-Verlag, Berlin, 1976, p. 508. [19] N. Zeddini, Positive solutions for a singular nonlinear problem on a bounded domain in R2 , Potential Anal. 18 (2003) 97–118. [20] Z. Zhang, The asymptotic behavior of the unique solution for the singular Lane–Emden–Fowler equation, J. Math. Anal. Appl. 312 (2005) 33–43.