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On a class of quasilinear problems with sign-changing nonlinearities
Nonlinear Analysis 64 (2006) 1977 – 1983 www.elsevier.com/locate/na
On a class of quasilinear problems with sign-changing nonlinearities D.D. Hai∗ , Xiangsheng Xu Department of Mathematics, Mississippi State University, Mississippi State, MS 39762, USA Received 31 October 2004; accepted 11 July 2005
Abstract We prove existence and nonexistence of positive solutions for the quasilinear problem − p u + mp (u) = a(x)f (u) u=0
in ,
on j,
where p u = div(|∇u|p−2 ∇u), p (u) = |u|p−2 u, p > 1, m 0, is a bounded domain in R N , and the coefficient a(x), is allowed to change sign. © 2005 Published by Elsevier Ltd. Keywords: p-Laplacian; Positive solutions; Sign-changing coefficient
1. Introduction Consider the boundary value problem −p u + mp (u) = a(x)f (u) in , u = 0 on j,
(I)
where is a bounded domain in R N with smooth boundary j, p u = div(|∇u|p−2 ∇u), p (u)=|u|p−2 u, p > 1, m 0, a : → R, f : [0, ∞) → R, and is a positive parameter. We are interested in positive solutions for (I) when the coefficient a(x) is allowed to change sign. ∗ Corresponding author. Tel.: +1 601 325 3414; fax: +1 601 325 0005.
E-mail address: offi[email protected] (D.D. Hai). 0362-546X/$ - see front matter © 2005 Published by Elsevier Ltd. doi:10.1016/j.na.2005.07.026
1978
D.D. Hai, X. Xu / Nonlinear Analysis 64 (2006) 1977 – 1983
When p = 2 and m = 0, the existence and nonexistence of positive solutions to (I) were obtained in [1–3,6,8]. It was established in [1–3,6] that problem (I) has a positive solution for > 0 small when f (0) > 0 and the boundary value problem −z = a(x) in z = 0 on j f has a positive solution z with jz/jn < 0 on j, where n denotes the unit outward normal vector. Nonexistence results for positive solutions of (I) for large and f strictly superlinear at ∞ were obtained by Cac et al. [2] in the case where is a ball, and by Kazdan and Warner [8] for a general domain . In this paper, we shall study existence and nonexistence of positive solutions of the quasilinear problem (I) in certain ranges of . Our results unify and extend corresponding results in [1–3,6,8].
2. Main results Throughout the paper, we assume that a ∈ L∞ () and f (u) = f (0) for u0. We make the following assumptions: (H.1) f : [0, ∞) → R is continuous and f (0) > 0. (H.2) The problem −p z + mp (z) = a(x) in z = 0 on j
(*)
has a positive solution z with jz/jn < 0 on j. Our main results are Theorem 1. Let (H.1) and (H.2) hold. Then there exists a positive number 0 such that problem (I) has a positive solution for 0 < < 0 . Theorem 2. Suppose that (H.3) the solution z of (*) is positive at some point x0 ∈ . and (H.4) f is differentiable on [0, ∞), f (u) > 0, f (u)0 for all u 0, and
0
∞
du <∞ f 1/(p−1) (u)
hold. Then there exists a positive number ∗ > 0 such that (I) has no positive solutions for > ∗ .
D.D. Hai, X. Xu / Nonlinear Analysis 64 (2006) 1977 – 1983
1979
Theorem 3. Suppose that (H.5) there exist , r > 0 and x1 ∈ such that a(x)
f or a.e. x ∈ B(x1 , r),
and (H.6) there exists C > 0 such that f (u)Cup−1
for all u > 0
hold. Then there exists a positive number ∗ > 0 such that (I) has no positive solutions for > ∗ . Remarks. 1. Theorem 1 was established in [1–3,6] for p = 2, m = 0, under an assumption that is equivalent to (H.2). 2. As an example of a function a(x) that satisfies condition (H.2), we take ⎧ 1 ⎪ ⎨ −1 if d(x, j) , n an (x) = 1 ⎪ ⎩1 if d(x, j) > . n By the strong maximum principle [12], the solution z0 of −p z0 + mp (z0 ) = 1 in , z0 = 0 on j is positive in and satisfies jz/jn < 0 on j. Since (an ) converges to 1 in L2 (), we see, in view of Lemma 1 below, that an satisfies (H.2) if n is large enough. 3. Condition (H.6) follows from (H.4) since u ∞ u dx dx < ∞ for u > 0. 1/(p−1) 1/(p−1) 1/(p−1) 2f (u) (x) f (x) u/2 f 0 If a is continuous in then (H.5) is a consequence of (H.3), by the maximum principle. 4. Theorem 2 was established in [8, Theorem 3.16] for p = 2 and m = 0. Nonexistence of positive solutions to (I) was also established in [3, Theorem 7] in the case p=2, m=0, is the unit ball, a(x) = a(r) is nonincreasing with a(x) > in B(0, ) for some > 0, and ∞ du < ∞, g(u) 0 where g(u) = inf u x f (x). Hence Theorem 3 generalizes Theorem 7 in [3] even in the case p = 2. We first establish Lemma 1. Let h , h ∈ L∞ (), ∈ (0, ) for some > 0. Let u and u be the solutions of − p u + mp (u ) = h u=0
on j,
in ,
1980
D.D. Hai, X. Xu / Nonlinear Analysis 64 (2006) 1977 – 1983
and − p u + mp (u) = h in , u=0
on j,
respectively. Suppose that there exists C > 0 such that h ∞ C for ∈ (0, ), and h − hL2 → 0
as → 0.
u − uC 1 → 0
as → 0.
Then
Proof. The proof is similar to that of Lemma 2 in [7], where m = 0 was considered. According to [9], there exists ∈ (0, 1) and k > 0 such that 1/(p−1)
u C 1, kh ∞
kC 1/(p−1) .
Multiplying the equation −p u + p u + m(p (u ) − p (u)) = h − h by u − u and integrating gives p−2 p−2 (|∇u | ∇u − |∇u| ∇u, ∇u − ∇u) dx (h − h)(u − u) dx
h − hL2 u − uL2 .
Using Lemma 30.1 in [10, p.146], we get q (|∇u |p−2 ∇u − |∇u|p−2 ∇u, ∇u − ∇u) dx C1 ∇u − ∇uL2 ,
where q = max(p, 2) and C1 is a positive constant independent of u . Hence ∇u − ∇uL2 → 0
as → 0.
¯ it follows from standard arguments that Since u is uniformly bounded in C 1, (), u − uC 1 → 0 as → 0. We are now ready to give the proof of the main theorems. ¯ define u = A v to be the solution of Proof of Theorem 1. For v ∈ C(), − p u + mp (u) = a(x)f (v) in , u=0
on j.
D.D. Hai, X. Xu / Nonlinear Analysis 64 (2006) 1977 – 1983
1981
Then there exist k > 0 and ∈ (0, 1) such that uC 1, k(a∞ f (v)∞ )1/(p−1) .
(1)
¯ → C() ¯ is completely continuous. We It follows from (1) and Lemma 1 that A : C() shall verify that (I) has a solution for a∞ M < k 1−p ,
(2)
¯ be such that where M = sup0 v 1 |f (v)|. Indeed, let u ∈ C() u = A u
for some ∈ [0, 1].
Using (1) and (2), we conclude that u∞ = 1. By the Leray Schauder fixed point theorem, A has a fixed point u˜ with u˜ ∞ 1. By (1), u˜ ∞ → 0 as → 0. It remains to show that u˜ > 0 in for > 0 small. Let u˜ u = 1/(p−1) , z1 = (f (0))1/(p−1) z. Then we have −p u + mp (u ) = a(x)f (u˜ ) and −p z1 + mp (z1 ) = a(x)f (0). By Lemma 1, u − z1 C 1 → 0
as → 0.
Since z1 > 0 in and jz1 /jn < 0 on j, we deduce that u > 0 in for sufficiently small. This completes the proof of Theorem 1. Proof of Theorem 2. We shall use the ideas in [8]. Suppose that > 1 and u is a positive solution of (I). Let ∞ ∞ dx dx v= , M= . 1/(p−1) 1/(p−1) f (x) f (x) u 0 By (H.4) and Remark 3, there exists C > 0 such that f (u)Cup−1
for all u > 0.
Let v = (v − M)/1/(p−1) . Then −M v 0 in , and a calculation shows that p u |∇u|p f (u) − + mp (v ) f (u) f 2 (u) mup−1 m − a(x) + − a(x) + f (u) C
−p v + mp (v ) =
v 0
on j.
in
1982
D.D. Hai, X. Xu / Nonlinear Analysis 64 (2006) 1977 – 1983
Let z be the solution of − p z + mp (z ) = −a(x) + z = 0
on j.
m C
in ,
By the weak comparison principle [4,5,11], v z . It follows from Lemma 1 that z + zC 1 → 0
as → ∞,
where z is defined in (H.3). Hence z(x0 ) z (x0 ) < − < 0 for large, 2 which implies 1/(p−1) z(x0 ) , 2 a contradiction if is large enough. This completes the proof of Theorem 2. −M < v(x0 ) − M = 1/(p−1) v (x0 ) 1/(p−1) z (x0 ) −
Proof of Theorem 3. Suppose that u is a positive solution of (I). Then we have −p u f (u) + m = a(x) p−1 , up−1 u or
|∇u|p−2 ∇u −div up−1
−
(p − 1)|∇u|p f (u) + m = a(x) p−1 . p u u
(3)
Let B = B(x1 , r/2) and ∈ C0∞ (B) with 0 on B and ≡ / 0. Multiplying (3) by and integrating, we obtain |∇u|p−2 ∇u.∇ f (u) dx a(x) dx − m dx up−1 up−1 B B B (C − m) dx. (4) B
By Holder’s inequality, |∇u|p−2 ∇u.∇ |∇u|p−1 |∇ | |∇u|p p−1 dx dx dx p−1 p up−1 p B B u B u 1 + |∇ |p dx. p B Combining (4) and (5), we get p−1 |∇u|p 1 dx (C − m) dx − |∇ |p dx. p p p B B u B Next, we show that |∇u|p dx K, p B u
(5)
(6)
(7)
D.D. Hai, X. Xu / Nonlinear Analysis 64 (2006) 1977 – 1983
1983
where K is a constant independent of u and . To this ends, let ∈ C0∞ (B1 ) with 0 on B1 ≡ B(x1 , r). Multiplying (3) by p and integrating gives p−1 |∇u|p−2 ∇u ∇ |∇u|p p dx dx − (p − 1) p up up−1 B1 B1 f (u) = a(x) p−1 p dx − m p dx (C − m) p dx 0, u B1 B1 B1 if C > m. From this and Holder’s inequality, we deduce |∇u|p p p−1 |∇u|p−2 ∇u.∇ dx p dx (p − 1) p u up−1 B1 B1 |∇u|p p p−1 p−1 dx + p |∇|p dx, p p u B1 B1 which implies (p − 1)2 p
B1
|∇u|p p dx p p−1 up
B1
|∇|p dx
and (7) follows if we choose so that = 1 on B. In view of (5) and (6), we reach a contradiction if is sufficiently large. This completes the proof of Theorem 3. References [1] G.A. Afrouzi, K.J. Brown, Positive solutions for a semilinear problem with sign-changing nonlinearity, Nonlinear Anal. 36 (1999) 507–510. [2] N.P. Cac, A.M. Fink, J.A. Gatica, Nonnegative solutions to the radial Laplacian with nonlinearity that changes sign, Proc. Am. Math. Soc. 123 (1995) 1393–1398. [3] N.P. Cac, J.A. Gatica, Y. Li, Positive solutions to semilinear problems with coefficient that changes sign, Nonlinear Anal. 37 (1999) 501–510. [4] P. Drabek, J. Hernandez, Existence and uniqueness of positive solutions for some quasilinear elliptic problems, Nonlinear Anal. 44 (2001) 189–204. [5] J. Garcia-Melian, J. Sabina de Lis, Maximum and comparison principles for operators involving the p-Laplacian, J. Math. Anal. Appl. 218 (1998) 49–65. [6] D.D. Hai, Positive solutions to a class of elliptic boundary value problems, J. Math. Anal. Appl. 227 (1998) 195–199. [7] D.D. Hai, On a class of sublinear quasilinear elliptic problems, Proc. Am. Math. Soc. 131 (2003) 2409–2414. [8] J.L. Kazdan, F.W. Warner, Remarks on some quasilinear elliptic equations, Commun. Pure Appl. Math. 28 (1975) 567–597. [9] G.M. Lieberman, Boundary regularity for solutions of degenerate elliptic equations, Nonlinear Anal. 12 (1988) 1203–1219. [10] T. Oden, Qualitative Methods in Nonlinear Mechanics, Prentice-Hall, Englewood Cliffs, New Jersey, 1986. [11] P. Tolksdorf, On the Dirichlet problem for quasilinear equations in domains with conical boundary points, Commun. Partial Differential Equations 8 (1983) 773–817. [12] J.L. Vazquez, A strong maximum principle for some quasilinear elliptic equations, Appl. Math. Optim. 12 (1984) 191–202.
On a class of quasilinear problems with sign-changing nonlinearities D.D. Hai∗ , Xiangsheng Xu Department of Mathematics, Mississippi State University, Mississippi State, MS 39762, USA Received 31 October 2004; accepted 11 July 2005
Abstract We prove existence and nonexistence of positive solutions for the quasilinear problem − p u + mp (u) = a(x)f (u) u=0
in ,
on j,
where p u = div(|∇u|p−2 ∇u), p (u) = |u|p−2 u, p > 1, m 0, is a bounded domain in R N , and the coefficient a(x), is allowed to change sign. © 2005 Published by Elsevier Ltd. Keywords: p-Laplacian; Positive solutions; Sign-changing coefficient
1. Introduction Consider the boundary value problem −p u + mp (u) = a(x)f (u) in , u = 0 on j,
(I)
where is a bounded domain in R N with smooth boundary j, p u = div(|∇u|p−2 ∇u), p (u)=|u|p−2 u, p > 1, m 0, a : → R, f : [0, ∞) → R, and is a positive parameter. We are interested in positive solutions for (I) when the coefficient a(x) is allowed to change sign. ∗ Corresponding author. Tel.: +1 601 325 3414; fax: +1 601 325 0005.
E-mail address: offi[email protected] (D.D. Hai). 0362-546X/$ - see front matter © 2005 Published by Elsevier Ltd. doi:10.1016/j.na.2005.07.026
1978
D.D. Hai, X. Xu / Nonlinear Analysis 64 (2006) 1977 – 1983
When p = 2 and m = 0, the existence and nonexistence of positive solutions to (I) were obtained in [1–3,6,8]. It was established in [1–3,6] that problem (I) has a positive solution for > 0 small when f (0) > 0 and the boundary value problem −z = a(x) in z = 0 on j f has a positive solution z with jz/jn < 0 on j, where n denotes the unit outward normal vector. Nonexistence results for positive solutions of (I) for large and f strictly superlinear at ∞ were obtained by Cac et al. [2] in the case where is a ball, and by Kazdan and Warner [8] for a general domain . In this paper, we shall study existence and nonexistence of positive solutions of the quasilinear problem (I) in certain ranges of . Our results unify and extend corresponding results in [1–3,6,8].
2. Main results Throughout the paper, we assume that a ∈ L∞ () and f (u) = f (0) for u0. We make the following assumptions: (H.1) f : [0, ∞) → R is continuous and f (0) > 0. (H.2) The problem −p z + mp (z) = a(x) in z = 0 on j
(*)
has a positive solution z with jz/jn < 0 on j. Our main results are Theorem 1. Let (H.1) and (H.2) hold. Then there exists a positive number 0 such that problem (I) has a positive solution for 0 < < 0 . Theorem 2. Suppose that (H.3) the solution z of (*) is positive at some point x0 ∈ . and (H.4) f is differentiable on [0, ∞), f (u) > 0, f (u)0 for all u 0, and
0
∞
du <∞ f 1/(p−1) (u)
hold. Then there exists a positive number ∗ > 0 such that (I) has no positive solutions for > ∗ .
D.D. Hai, X. Xu / Nonlinear Analysis 64 (2006) 1977 – 1983
1979
Theorem 3. Suppose that (H.5) there exist , r > 0 and x1 ∈ such that a(x)
f or a.e. x ∈ B(x1 , r),
and (H.6) there exists C > 0 such that f (u)Cup−1
for all u > 0
hold. Then there exists a positive number ∗ > 0 such that (I) has no positive solutions for > ∗ . Remarks. 1. Theorem 1 was established in [1–3,6] for p = 2, m = 0, under an assumption that is equivalent to (H.2). 2. As an example of a function a(x) that satisfies condition (H.2), we take ⎧ 1 ⎪ ⎨ −1 if d(x, j) , n an (x) = 1 ⎪ ⎩1 if d(x, j) > . n By the strong maximum principle [12], the solution z0 of −p z0 + mp (z0 ) = 1 in , z0 = 0 on j is positive in and satisfies jz/jn < 0 on j. Since (an ) converges to 1 in L2 (), we see, in view of Lemma 1 below, that an satisfies (H.2) if n is large enough. 3. Condition (H.6) follows from (H.4) since u ∞ u dx dx < ∞ for u > 0. 1/(p−1) 1/(p−1) 1/(p−1) 2f (u) (x) f (x) u/2 f 0 If a is continuous in then (H.5) is a consequence of (H.3), by the maximum principle. 4. Theorem 2 was established in [8, Theorem 3.16] for p = 2 and m = 0. Nonexistence of positive solutions to (I) was also established in [3, Theorem 7] in the case p=2, m=0, is the unit ball, a(x) = a(r) is nonincreasing with a(x) > in B(0, ) for some > 0, and ∞ du < ∞, g(u) 0 where g(u) = inf u x f (x). Hence Theorem 3 generalizes Theorem 7 in [3] even in the case p = 2. We first establish Lemma 1. Let h , h ∈ L∞ (), ∈ (0, ) for some > 0. Let u and u be the solutions of − p u + mp (u ) = h u=0
on j,
in ,
1980
D.D. Hai, X. Xu / Nonlinear Analysis 64 (2006) 1977 – 1983
and − p u + mp (u) = h in , u=0
on j,
respectively. Suppose that there exists C > 0 such that h ∞ C for ∈ (0, ), and h − hL2 → 0
as → 0.
u − uC 1 → 0
as → 0.
Then
Proof. The proof is similar to that of Lemma 2 in [7], where m = 0 was considered. According to [9], there exists ∈ (0, 1) and k > 0 such that 1/(p−1)
u C 1, kh ∞
kC 1/(p−1) .
Multiplying the equation −p u + p u + m(p (u ) − p (u)) = h − h by u − u and integrating gives p−2 p−2 (|∇u | ∇u − |∇u| ∇u, ∇u − ∇u) dx (h − h)(u − u) dx
h − hL2 u − uL2 .
Using Lemma 30.1 in [10, p.146], we get q (|∇u |p−2 ∇u − |∇u|p−2 ∇u, ∇u − ∇u) dx C1 ∇u − ∇uL2 ,
where q = max(p, 2) and C1 is a positive constant independent of u . Hence ∇u − ∇uL2 → 0
as → 0.
¯ it follows from standard arguments that Since u is uniformly bounded in C 1, (), u − uC 1 → 0 as → 0. We are now ready to give the proof of the main theorems. ¯ define u = A v to be the solution of Proof of Theorem 1. For v ∈ C(), − p u + mp (u) = a(x)f (v) in , u=0
on j.
D.D. Hai, X. Xu / Nonlinear Analysis 64 (2006) 1977 – 1983
1981
Then there exist k > 0 and ∈ (0, 1) such that uC 1, k(a∞ f (v)∞ )1/(p−1) .
(1)
¯ → C() ¯ is completely continuous. We It follows from (1) and Lemma 1 that A : C() shall verify that (I) has a solution for a∞ M < k 1−p ,
(2)
¯ be such that where M = sup0 v 1 |f (v)|. Indeed, let u ∈ C() u = A u
for some ∈ [0, 1].
Using (1) and (2), we conclude that u∞ = 1. By the Leray Schauder fixed point theorem, A has a fixed point u˜ with u˜ ∞ 1. By (1), u˜ ∞ → 0 as → 0. It remains to show that u˜ > 0 in for > 0 small. Let u˜ u = 1/(p−1) , z1 = (f (0))1/(p−1) z. Then we have −p u + mp (u ) = a(x)f (u˜ ) and −p z1 + mp (z1 ) = a(x)f (0). By Lemma 1, u − z1 C 1 → 0
as → 0.
Since z1 > 0 in and jz1 /jn < 0 on j, we deduce that u > 0 in for sufficiently small. This completes the proof of Theorem 1. Proof of Theorem 2. We shall use the ideas in [8]. Suppose that > 1 and u is a positive solution of (I). Let ∞ ∞ dx dx v= , M= . 1/(p−1) 1/(p−1) f (x) f (x) u 0 By (H.4) and Remark 3, there exists C > 0 such that f (u)Cup−1
for all u > 0.
Let v = (v − M)/1/(p−1) . Then −M v 0 in , and a calculation shows that p u |∇u|p f (u) − + mp (v ) f (u) f 2 (u) mup−1 m − a(x) + − a(x) + f (u) C
−p v + mp (v ) =
v 0
on j.
in
1982
D.D. Hai, X. Xu / Nonlinear Analysis 64 (2006) 1977 – 1983
Let z be the solution of − p z + mp (z ) = −a(x) + z = 0
on j.
m C
in ,
By the weak comparison principle [4,5,11], v z . It follows from Lemma 1 that z + zC 1 → 0
as → ∞,
where z is defined in (H.3). Hence z(x0 ) z (x0 ) < − < 0 for large, 2 which implies 1/(p−1) z(x0 ) , 2 a contradiction if is large enough. This completes the proof of Theorem 2. −M < v(x0 ) − M = 1/(p−1) v (x0 ) 1/(p−1) z (x0 ) −
Proof of Theorem 3. Suppose that u is a positive solution of (I). Then we have −p u f (u) + m = a(x) p−1 , up−1 u or
|∇u|p−2 ∇u −div up−1
−
(p − 1)|∇u|p f (u) + m = a(x) p−1 . p u u
(3)
Let B = B(x1 , r/2) and ∈ C0∞ (B) with 0 on B and ≡ / 0. Multiplying (3) by and integrating, we obtain |∇u|p−2 ∇u.∇ f (u) dx a(x) dx − m dx up−1 up−1 B B B (C − m) dx. (4) B
By Holder’s inequality, |∇u|p−2 ∇u.∇ |∇u|p−1 |∇ | |∇u|p p−1 dx dx dx p−1 p up−1 p B B u B u 1 + |∇ |p dx. p B Combining (4) and (5), we get p−1 |∇u|p 1 dx (C − m) dx − |∇ |p dx. p p p B B u B Next, we show that |∇u|p dx K, p B u
(5)
(6)
(7)
D.D. Hai, X. Xu / Nonlinear Analysis 64 (2006) 1977 – 1983
1983
where K is a constant independent of u and . To this ends, let ∈ C0∞ (B1 ) with 0 on B1 ≡ B(x1 , r). Multiplying (3) by p and integrating gives p−1 |∇u|p−2 ∇u ∇ |∇u|p p dx dx − (p − 1) p up up−1 B1 B1 f (u) = a(x) p−1 p dx − m p dx (C − m) p dx 0, u B1 B1 B1 if C > m. From this and Holder’s inequality, we deduce |∇u|p p p−1 |∇u|p−2 ∇u.∇ dx p dx (p − 1) p u up−1 B1 B1 |∇u|p p p−1 p−1 dx + p |∇|p dx, p p u B1 B1 which implies (p − 1)2 p
B1
|∇u|p p dx p p−1 up
B1
|∇|p dx
and (7) follows if we choose so that = 1 on B. In view of (5) and (6), we reach a contradiction if is sufficiently large. This completes the proof of Theorem 3. References [1] G.A. Afrouzi, K.J. Brown, Positive solutions for a semilinear problem with sign-changing nonlinearity, Nonlinear Anal. 36 (1999) 507–510. [2] N.P. Cac, A.M. Fink, J.A. Gatica, Nonnegative solutions to the radial Laplacian with nonlinearity that changes sign, Proc. Am. Math. Soc. 123 (1995) 1393–1398. [3] N.P. Cac, J.A. Gatica, Y. Li, Positive solutions to semilinear problems with coefficient that changes sign, Nonlinear Anal. 37 (1999) 501–510. [4] P. Drabek, J. Hernandez, Existence and uniqueness of positive solutions for some quasilinear elliptic problems, Nonlinear Anal. 44 (2001) 189–204. [5] J. Garcia-Melian, J. Sabina de Lis, Maximum and comparison principles for operators involving the p-Laplacian, J. Math. Anal. Appl. 218 (1998) 49–65. [6] D.D. Hai, Positive solutions to a class of elliptic boundary value problems, J. Math. Anal. Appl. 227 (1998) 195–199. [7] D.D. Hai, On a class of sublinear quasilinear elliptic problems, Proc. Am. Math. Soc. 131 (2003) 2409–2414. [8] J.L. Kazdan, F.W. Warner, Remarks on some quasilinear elliptic equations, Commun. Pure Appl. Math. 28 (1975) 567–597. [9] G.M. Lieberman, Boundary regularity for solutions of degenerate elliptic equations, Nonlinear Anal. 12 (1988) 1203–1219. [10] T. Oden, Qualitative Methods in Nonlinear Mechanics, Prentice-Hall, Englewood Cliffs, New Jersey, 1986. [11] P. Tolksdorf, On the Dirichlet problem for quasilinear equations in domains with conical boundary points, Commun. Partial Differential Equations 8 (1983) 773–817. [12] J.L. Vazquez, A strong maximum principle for some quasilinear elliptic equations, Appl. Math. Optim. 12 (1984) 191–202.