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On the one-dimensional p -Laplacian with a singular nonlinearity
Nonlinear Analysis 75 (2012) 3994–4005
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On the one-dimensional p-Laplacian with a singular nonlinearity✩ Wenshu Zhou a , Xulong Qin b , Guokai Xu c , Xiaodan Wei a,∗ a
Department of Mathematics, Dalian Nationalities University, Dalian 116600, China
b
Department of Mathematics, Sun Yat-sen University, Guangzhou 510275, China
c
College of Electromechanical and Information Engineering, Dalian Nationalities University, Dalian 116600, China
article
info
Article history: Received 2 July 2011 Accepted 17 February 2012 Communicated by S. Carl MSC: 34B18
abstract In this paper, we are concerned with the existence and nonexistence of positive solutions of boundary value problems for the one-dimensional p-Laplacian with a singular nonlinearity. In the case of the model equation, we give the necessary and sufficient conditions of the existence of positive solutions for both the Dirichlet problem and the periodic problem. © 2012 Elsevier Ltd. All rights reserved.
Keywords: p-Laplacian Singularity Positive solution Existence Nonexistence
1. Introduction In this paper, we are concerned with the existence and nonexistence of positive solutions for the singular p-Laplacian equation:
′ φp (u′ ) − h(t , u)|u′ |p + f (t , u, u′ ) = 0,
0 < t < 1,
(1.1)
with either the Dirichlet boundary conditions u(1) = u(0) = 0,
(1.2)
or the periodic boundary conditions u′ (1) = u′ (0) = u(1) = u(0) = 0,
(1.3)
where φp (s) = |s| s with p > 1, h(t , z ) is nonnegative and continuous in [0, 1] × (R \ {0}) and may be singular at z = 0, and f (t , z , r ) is nonnegative and continuous on [0, 1] × R × R. The model equation is p−2
′ |u′ |p φp (u′ ) − λ m + f (t , u, u′ ) = 0, u
0 < t < 1,
(1.4)
where λ, m > 0. ✩ This work is supported by NNSF of China (grant nos. 10901030, 11071100, 11001278), Key Technology in the National Research and Development Pillar Program (no. 2009BAH41B05), China Postdoctoral Science Foundation (no. 20090450191), and Dalian Nationalities University (no. DC110109). ∗ Corresponding author. Tel.: +86 411 87189428; fax: +86 411 87189428. E-mail address: [email protected] (X. Wei).
0362-546X/$ – see front matter © 2012 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2012.02.015
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Besides singularity, another main feature of (1.1) is that the lower term has p-growth with respect to the first order derivative. Eq. (1.4) is strongly connected to the degenerate parabolic equation ut = u[φp (ux )]x − λ|ux |p ,
(1.5)
where λ > 0 is a constant, which appears in the investigation of groundwater flow in a water-absorbing fissurized porous rock and non-Newtonian filtration [1–4]. Indeed, if the solution u of (1.5) is of the form u = t −1/(p−1) w(x) with w > 0, then w must satisfy (in some sense)
[φp (w ′ )]′ − λ
|w ′ |p 1 + = 0. w p−1
The existence of positive solutions for (1.4) with different boundary conditions has recently been studied in [2,5–8]. In [2,5,6], the equations considered may have time singularities. In the present paper, we extend some existence results in [5,7,8] and obtain some sufficient conditions on the nonexistence of positive solutions. As by-products of our main results, we prove, under some auxiliary assumptions on f , that the condition m < p is necessary and sufficient for the existence of positive solutions to problem (1.4)–(1.2), and the condition 1 6 m < p is necessary and sufficient for the existence of positive solutions to problem (1.4)–(1.3). Our arguments on the existence are based on regularization technique, while in the arguments on the nonexistence, Gronwall’s inequality, embedding inequality and a scaling technique will play important roles. In the case where h ≡ 0, (1.1) becomes
′ φp (u′ ) + f (t , u, u′ ) = 0,
0 < t < 1.
(1.6)
Recently, problem (1.6)–(1.2) have been studied extensively, see for example [9–27], in which f (t , z , r ) may change sign and be singular at t = 0, t = 1 and z = 0. Some basic results were obtained in those papers. We point out that because the singularity at u = 0 and the p-growth with respect to u′ as a whole appear in (1.1), it is not in considerations of those papers. By a solution u of problem (1.1)–(1.2) we mean that u ∈ W , where W := {w ∈ C 1 [0, 1]; w > 0 in (0, 1), φp (w ′ ) ∈ C 1 (0, 1)},
and u satisfies (1.1) and (1.2). If u is a solution problem (1.1)–(1.2) and satisfies (1.3), then we call it is a solution for problem (1.1)–(1.3). To study the existence and nonexistence of solutions, we impose some additional conditions on h and f . We first study the existence of solutions. For h, we suppose that 0 6 h(t , z ) 6 g (z ),
∀t ∈ [0, 1], ∀z > 0,
(1.7)
where g : (0, +∞) → [0, +∞) is continuous in (0, +∞) and differentiable in (0, ξ ) for some ξ > 0, such that g ′ 6 0 in (0, ξ ), 1
lim
s→0+
lim g (s) = +∞,
s→0+
[g (z )]1/p dz < +∞.
(1.8)
(1.9)
s
For f , we suppose that there exist constants β > α > 0, γ > 0 such that
α 6 f (t , z , r ) 6 β + γ |r |p−1 ,
∀(t , z , r ) ∈ [0, 1] × [0, +∞) × R.
(1.10)
We obtain Theorem 1.1. Let (1.7)–(1.10) be satisfied. Then there exists a solution for problem (1.1)–(1.2). Theorem 1.2. Let h(t , z ) ≡ g (z ) satisfy (1.8) and (1.9), and suppose (1.10) holds. Then there exists a solution for problem 1 (1.1)–(1.3) if and only if lims→0+ s g (z )dz = +∞. Remark 1.1. Some examples of the functions satisfying (1.8)–(1.9) are: (1) (2) (3) (4)
g (z ) g (z ) g (z ) g (z )
= λ/z m , where λ > 0 and 0 < m < p; = (ln z )2 ; = 1/(ez − 1); = (ln z )2 if 0 < z < ϵ ; g (z ) = (ln ϵ)2 if z > ϵ , where ϵ > 0.
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Next we study the nonexistence of solutions for problem (1.1)–(1.2). We assume that h(t , z ) > g (z ) > 0,
∀t ∈ [0, 1], ∀z > 0,
(1.11)
where g : (0, +∞) → [0, +∞) is a continuous function such that
lim g (s) = +∞,
lim
s→0+
s→0+
1
[g (t )]1/p dt = +∞,
(1.12)
s
s 1/p lim [g (s)]1/p e 1 [g (t )] dt := ϱ0 ∈ [0, +∞).
(1.13)
s→0+
Denote by λ1 the first eigenvalue of the following problem (see [27]):
′ φp (u′ ) + λφp (u) = 0,
0 < t < 1;
u(1) = u(0) = 0.
(1.14)
Theorem 1.3. Let (1.10)–(1.13) be satisfied. If β eγ 6 λ1 , then problem (1.1)–(1.2) has no solution. Remark 1.2. Suppose that (1.10) holds and that there exists a continuous function g : (0, +∞) → [0, +∞) such that h(t , z ) > g (z ) > 0,
∀t ∈ [0, 1], ∀z > 0.
(1.15)
Observe that if u is a solution of problem (1.1)–(1.2), then v = δ u with δ > 0 is a positive solution of problem:
′ φp (v ′ ) − hδ (t , v)|v ′ |p + fδ (t , v, v ′ ) = 0, v(1) = v(0) = 0,
0 < t < 1,
(1.16)
where hδ (t , z ) =
1
δ
h t,
z
,
δ
fδ (t , z , r ) = δ p−1 f (t , z /δ, r /δ).
By (1.10), there holds for fδ fδ (t , z , r ) 6 δ p−1 (β + γ δ −(p−1) |r |p−1 ) 6 δ p−1 β + γ |r |p−1 ,
∀(t , z , r ) ∈ [0, 1] × [0, +∞) × R.
By (1.15), we have hδ (t , z ) > gδ (z ) :=
1
δ
g
z
δ
,
∀(t , z ) ∈ [0, 1] × [0, +∞).
According to Theorem 1.3, if gδ satisfies (1.12) and (1.13) and if δ p−1 β eγ 6 λ1 , then problem (1.16) has no solution, so does not problem (1.1)–(1.2). As a consequence of Theorem 1.3 and Remark 1.2, we obtain the following nonexistence result without additional requirements on f .
Corollary 1.1. Let (1.10) and (1.15) be satisfied. If there exists δ0 > 0 such that gδ (z ) = 1δ g δz all δ ∈ (0, δ0 ), then problem (1.1)–(1.2) has no solution. Proof. The proof can be completed by choosing a small δ ∈ (0, δ0 ) such that δ p−1 β eγ < λ1 .
satisfies (1.12) and (1.13) for
Note that in the model case where g (z ) = λ/z with λ, m > 0, gδ (z ) = λδ /z satisfy (1.12) and (1.13) if and only if m > p or m = p and λδ p−1 > 1. By Corollary 1.1, there does not exist a solution for problem (1.1)–(1.2) when m > p provided that (1.10) holds. For m = p, however, we still can prove the same conclusion by an argument based on Gronwall’s inequality. Precisely, we obtain m
m−1
m
Theorem 1.4. suppose that (1.10) holds and that h satisfies h(t , z ) >
λ zm
,
∀t ∈ [0, 1], ∀z > 0,
(1.17)
where m, λ > 0. If m > p, then problem (1.1)–(1.2) has no solution. Proof. It suffices to prove that problem (1.1)–(1.2) has no solution when m = p. Suppose that this is not so. Then there exists a solution for problem (1.1)–(1.2) when m = p.
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Note that u′ (1) 6 0 6 u′ (0). Integrating (1.1) over (0, 1) and integrating by parts yield 1
1
h(t , u)|u′ |p dt 6
f (t , u, u′ )dt . 0
0
By (1.17) with m = p, we have 1
0
|u′ |p up
dt 6
1
1
λ
f (t , u, u′ )dt := C .
(1.18)
0
By Hölder’s inequality, we arrive at, for t ∈ [0, 1], u( t ) =
t
′
t
u ds 6
up
0
0
|u′ |p
1/p
t
p
(p−1)/p ,
u p−1 ds
ds 0
from which and (1.18) it follows that p
u p−1 6 C 1/(p−1)
t
p
u p−1 ds,
∀t ∈ [0, 1].
0
By Gronwall’s inequality, we have that u ≡ 0, which is a contradiction.
By combining the above theorems, we immediately obtain Corollary 1.2. Let λ, m > 0, and suppose that (1.10) holds. Then problem (1.4)–(1.2) admits at least a solution if and only if m < p, and problem (1.4)–(1.3) admits at least a solution if and only if 1 6 m < p. The plan of the paper is as follows. In Section 2, we are concerned with the existence of solutions and the proofs of Theorems 1.1 and 1.2. Section 3 is devoted to the nonexistence of solutions for problem (1.1)–(1.2) and the proof of Theorem 1.3. 2. Proofs of Theorems 1.1 and 1.2 We begin with the proof of Theorem 1.1. Like in [8], we will use regularization technique to prove it. To do this, a suitable regularized problem must be considered instead of problem (1.1)–(1.2). A key point is to find a lower solution for the regularized problem. Assume that (1.7)–(1.10) hold. We define Ψ : [0, +∞) → [0, +∞) as follows
Ψ (s) =
s
[g (t )]1/p dt ,
0,0
(s > 0) (s = 0).
Since lims→0+ g (s) = ∞, we have that Ψ ′ > 0 in (0, s0 ) for some s0 ∈ (0, 1), hence, Ψ has the inverse Ψ −1 on [0, s0 ] with
Ψ −1 (0) = 0;
′ −1/p Ψ −1 (s) = g Ψ −1 (s) > 0,
s ∈ (0, s0 ).
(2.1)
By mean value theorem for differentials, there exists s1 ∈ (0, s0 ) such that
Ψ −1 (s) 6 s,
∀s ∈ [0, s1 ].
(2.2)
For ε ∈ (0, s1 ), we define aε (t , z , r ) : [0, 1] × R × R → R by aε (t , z , r ) = h(t , iε (z ))|r |p − f (t , iε (z ), r ), where
−1 2 Ψ (ε ), iε (z ) = z , −1 ε ,
(z < Ψ −1 (ε2 )) (Ψ −1 (ε2 ) 6 z 6 ε −1 ) (z > ε−1 ).
By (1.7), 0 6 h(t , iε (z )) 6 g (iε (z )) 6 Cε , where Cε = maxΨ −1 (ε2 )6z 6ε−1 g (z ). By (1.10) and the inequality: xp−1 6 1 + xp (∀x > 0), we have
|aε (t , z , r )| 6 Cε |r |p + β + γ |r |p−1 6 (Cε + β + γ )H (|r |) for all (t , z , r ) ∈ [0, 1] × R × R, where H (s) = 1 + sp for s > 0. Define Lε : W → C (0, 1) by
′ (Lε u)(t ) = − φp (u′ ) (t ) + aε (t , u(t ), u′ (t )),
0 < t < 1.
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W. Zhou et al. / Nonlinear Analysis 75 (2012) 3994–4005
Instead of problem (1.1)–(1.2), we consider the regularized problem with ε ∈ (0, s1 ):
(Lε u)(t ) = 0,
0 < t < 1;
u(1) = u(0) = ε.
(2.3)
A function u ∈ W is called an upper solution (a lower solution) of problem (2.3) if Lε u > (6)0 in (0, 1) and u(t ) > (6)ε at t = 0, 1. We will use the upper and lower solution method (see [16, Theorem 1 and Remark 2.4]) to prove the existence of solutions
+∞
p−1
s for problem (2.3). Since 0 ds = +∞, the condition (2.3) in [16] is satisfied. Then, to obtain a solution of problem H (s) (2.3), it suffices to find a lower solution and an upper solution. By direct calculations, one can verify the following lemma, which will be used to construct the upper and lower solutions of problem (2.3).
Lemma 2.1. Let β and γ be the same as in (1.10). Then the function defined on [0, 1]
1/(p−1) t eγ (1/2−s) − 1 ds, 0 6 t < 1/2, γ 0 1/(p−1) 1 γ (s−1/2) 1/(p−1) e −1 V (t ) = β · ds, 1/2 6 t 6 1, γ t p/(p−1) p/(p−1) 1 (p − 1) 1 − − t , p 2 2
(γ > 0)
(γ = 0)
belongs to W and is a solution of problem
′ − φp (u′ ) = β + γ |u′ |p−1 ,
0 < t < 1;
u(1) = u(0) = 0.
By Lemma 2.1, we obtain Lemma 2.2. Let (1.7)–(1.10) hold. Then there exists a constant ε1 ∈ (0, s1 ), such that for any ε ∈ (0, ε1 ), uε (t ) := V (t ) + ε is an upper solution of problem (2.3) and satisfies Ψ −1 (ε 2 ) < uε < ε −1 on [0, 1]. Proof. Clearly, there exists ε1 ∈ (0, s1 ) such that max[0,1] V + 1 < ε1−1 . By (2.2), Ψ −1 (ε 2 ) < uε (t ) = V (t ) + ε < ε −1 on [0, 1] for all ε ∈ (0, ε1 ). Also, we have
′
Lε uε (t ) = − φp (uε ) (t ) + h(t , uε (t ))|uε (t )|p − f (t , uε (t ), uε (t ))
′
′
′
′ ′ ′ > − φp (uε ) (t ) − f (t , uε (t ), uε (t )) = β + γ |V ′ (t )|p−1 − f (t , V (t ) + ϵ, V ′ (t )) > 0, 0 < t < 1. The proof is complete.
Next, we construct the lower solutions of problem (2.3). Let V be the same as in Lemma 2.1. Choosing a constant ε2 ∈ (0, ε1 ) such that
ε2 (max V + 1) 6 min{s1 , ξ }, [0,1]
ε2 p−1 [g (ξ )]−(p−1)/p (β + γ max |V ′ |p−1 ) + ε2 p max |V ′ |p − α 6 0, [0,1]
[0,1]
(2.4)
where ξ , α, β, γ and s1 are the same as before. By (2.1)–(2.2), we have, for all ε ∈ (0, ε2 ),
Ψ −1 (ε 2 ) < Ψ −1 ε2 (V (t ) + ε) < ξ , ∀t ∈ [0, 1], Ψ −1 ε2 (V (t ) + ε) 6 uε (t ) = V (t ) + ε, ∀t ∈ [0, 1], Ψ
−1
(ε2 ε) 6 ε.
(2.5) (2.6) (2.7)
We have Lemma 2.3. Let (1.7)–(1.10) hold. Then for any ε ∈ (0, ε2 ), uε (t ) := Ψ −1 (ε2 (V (t )+ε)) is a lower solution of problem (2.3) and satisfies uε > uε > Ψ −1 (ε 2 ) on [0, 1].
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Proof. By (2.5) and (2.7), we have that 0 < uε (1) = uε (0) 6 ε and uε > uε > Ψ −1 (ε 2 ) on [0, 1] for all ε ∈ (0, ε2 ). By (1.7), (1.8) and (2.5), we have g ′ (uε ) 6 0, g (uε ) > g (ξ ) and h(t , iε (uε (t ))) = h(t , uε (t )) 6 g (uε (t )) for all ε ∈ (0, ε2 ). Noticing (1.10) and (2.4), we obtain, for ε ∈ (0, ε2 ),
′
Lε uε (t ) = − φp (u′ε ) (t ) + h(t , iε (uε (t )))|u′ε (t )|p − f (t , iε (uε (t )), u′ε (t ))
′
6 − φp (u′ε ) (t ) + g (uε (t ))|u′ε (t )|p − α
′ = −ε2 p−1 [g (uε (t ))]−(p−1)/p |V ′ (t )|p−2 V ′ (t ) + (p − 1)/pε2 p g ′ (uε (t ))[g (uε (t ))]−2 |V ′ (t )|p + ε2 p |V ′ (t )|p − α ′ 6 −ε2 p−1 [g (uε (t ))]−(p−1)/p |V ′ (t )|p−2 V ′ (t ) + ε2 p |V ′ (t )|p − α = ε2 p−1 [g (uε (t ))]−(p−1)/p (β + γ |V ′ (t )|p−1 ) + ε2 p |V ′ (t )|p − α
6 ε2 p−1 [g (ξ )]−(p−1)/p (β + γ max |V ′ |p−1 ) + ε2 p max |V ′ |p − α [0,1]
6 0,
[0,1]
0 < t < 1.
Thus the proof is complete.
By Lemmas 2.2 and 2.3, it follows from [16, Theorem 1 and Remark 2.4] that for any fixed ε ∈ (0, ε2 ), problem (2.3) admits a solution uε ∈ C 1 [0, 1] with φp (u′ε ) ∈ C (0, 1) such that uε (t ) > uε (t ) > uε (t ), and, for all ε ∈ (0, ε2 ),
∀t ∈ [0, 1],
(2.8)
′ φp (u′ε ) (t ) − h(t , uε (t ))|u′ε (t )|p + f (t , uε (t ), u′ε (t )) = 0,
t ∈ (0, 1),
(2.9)
φp (u′ε )
with uε (1) = uε (0) = ε . Since uε ∈ C [0, 1], it follows from (2.9) that ∈ C (0, 1), so uε ∈ W . Next we estimate u′ε . To do this, we first prove a general conclusion, which, in the sequel, will be used repeatedly. 1
1
Lemma 2.4. Suppose that (1.10) holds. If u ∈ W satisfies u(1) = u(0) = 0 and
′ φp (u′ ) (t ) + f (t , u(t ), u′ (t )) > 0,
0 < t < 1,
then
|u′ (t )| 6 Cβ,γ , ∀t ∈ [0, 1], γ 1/(p−1) β(e −1) if γ > 0; Cβ,γ = β 1/(p−1) if γ = 0. where Cβ,γ = γ
(2.10)
Proof. Noticing u(1) = u(0) = 0 and u > 0 on [0, 1], we have u′ (0) > 0 > u′ (1).
(2.11)
By (1.10), we arrive at
′ φp (u′ ) (t ) + β + γ |u′ (t )|p−1 > 0,
t ∈ (0, 1).
(2.12)
Let w = φp (u′ ). Then we derive from (2.12) that w ′ + β + γ |w| > 0 in (0, 1), i.e. w(t )
0
′
1
β + γ |s|
> 0,
ds + t
which and (2.11) imply that 1 > |w(t )|
0
1
β + γs
ds 6 1,
w(t ) 0
∀t ∈ (0, 1), w
ds + t > 0 for t ∈ [0, 1], so that β+γ |s| 1
0
ds 6 1 on [0, 1], therefore β+γ |s| 1
∀t ∈ [0, 1].
From this, one can easily obtain the desired result. The proof is complete.
Lemma 2.5. For any δ ∈ (0, 1/2), there exists a positive constant C (δ) independent of ε such that for all ε ∈ (0, ε2 ),
|u′ε (t )| 6 Cβ,γ ,
∀t ∈ [0, 1],
|u′ε (t2 ) − u′ε (t1 )| 6 C (δ)|t2 − t1 |θ ,
(2.13)
∀t2 , t1 ∈ [δ, 1 − δ],
where θ = 1/(p − 1) if p > 2; θ = 1 if 1 < p < 2.
(2.14)
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W. Zhou et al. / Nonlinear Analysis 75 (2012) 3994–4005
Proof. We first prove that for any ε ∈ (0, ε2 ), uε (t ) > ε,
∀t ∈ [0, 1].
Suppose that this is not so. Noticing uε (1) = uε (0) = ε , we see that there exist some ε ∈ (0, ε2 ) and some t ∗ ∈ (0, 1) such that uε (t ∗ ) = min uε < ε,
u′ε (t ∗ ) = 0.
[0,1]
It follows from (2.9) that
′ φp (u′ε ) (t ∗ ) + f (t ∗ , uε (t ∗ ), 0) = 0. ′ By (1.10), φp (u′ε ) (t ∗ ) 6 −α < 0. By continuity, there exists a sufficiently small δ > 0 such that ′ φp (u′ε ) (t ) < 0, ∀t ∈ (t ∗ , t ∗ + δ). Noticing u′ε (t ∗ ) = 0, we have
φp (u′ε (t )) < 0,
∀t ∈ (t ∗ , t ∗ + δ),
hence u′ε (t ) < 0,
∀t ∈ (t ∗ , t ∗ + δ),
therefore uε (t ) < uε (t ∗ ) = min uε ,
∀t ∈ (t ∗ , t ∗ + δ),
[0,1]
which is a contradiction. Next, we prove (2.13) by Lemma 2.4. Let wε = uε − ε . Then wε satisfies
wε (1) = wε (0) = ε;
wε (t ) > 0,
∀t ∈ [0, 1].
On the other hand, it follows from (2.9) that
′ φp (u′ε ) (t ) + f (t , uε (t ), u′ε (t )) > 0,
t ∈ (0, 1),
by (1.10), we obtain
′ φp (u′ε ) (t ) + β + γ |u′ε (t )|p−1 > 0,
t ∈ (0, 1),
hence
′ φp (wε′ ) (t ) + β + γ |wε′ (t )|p−1 > 0,
t ∈ (0, 1).
Taking f (t , z , r ) ≡ β + γ |r | in Lemma 2.4 yield |wε′ | 6 Cβ,γ on [0, 1], so |u′ε | 6 Cβ,γ on [0, 1]. Thus, (2.13) holds. Below, we prove (2.14). By (2.8) and (2.13), it is easy to derive from (2.9) that for any δ ∈ (0, 1/2) there exists a constant C (δ) > 0 independent of ε , such that for all ε ∈ (0, ε2 ) p−1
′ φp (u′ε ) (t ) 6 C (δ),
δ 6 t 6 1 − δ.
(2.15)
Recalling the inequality (cf. [28])
[φp (η) − φp (η′ )] · [η − η′ ] >
C1 |η − η′ |p , C2 (|η| + |η′ |)p−2 |η − η′ |2 ,
(p > 2) (1 < p < 2)
for all η, η′ ∈ R, where Ci (i = 1, 2) are positive constants, we derive, by (2.15), that if p > 2, then
|u′ε (t2 ) − u′ε (t1 )|p 6 C1−1 [u′ε (t2 ) − u′ε (t1 )] · [|u′ε (t2 )|p−2 u′ε (t2 ) − |u′ε (t1 )|p−2 u′ε (t1 )] 6 C (δ)|u′ε (t2 ) − u′ε (t1 )| |t2 − t1 |, ∀t2 , t1 ∈ [δ, 1 − δ], hence
|u′ε (t2 ) − u′ε (t1 )| 6 C (δ)|t2 − t1 |1/(p−1) ,
∀t2 , t1 ∈ [δ, 1 − δ],
and if p ∈ (1, 2), then
|u′ε (t2 ) − u′ε (t1 )|2 [|u′ε (t2 )| + |u′ε (t1 )|]p−2 6 C2−1 [u′ε (t2 ) − u′ε (t1 )] · [|u′ε (t2 )|p−2 u′ε (t2 ) − |u′ε (t1 )|p−2 u′ε (t1 )] 6 C (δ)|u′ε (t2 ) − u′ε (t1 )| |t2 − t1 |, ∀t2 , t1 ∈ [δ, 1 − δ],
(2.16)
W. Zhou et al. / Nonlinear Analysis 75 (2012) 3994–4005
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so that, using (2.13) yields
|u′ε (t2 ) − u′ε (t1 )| 6 C (δ)|t2 − t1 |[|u′ε (t2 )| + |u′ε (t1 )|]2−p 6 C (δ)|t2 − t1 | for all t2 , t1 ∈ [δ, 1 − δ]. This ends our proof.
Proof of Theorem 1.1. By (2.13) and (2.14) and using the Arzela´ –Ascoli theorem, there exist a subsequence of {uε }, still denoted by {uε }, and a function u ∈ C 1 (0, 1) ∩ C [0, 1] such that, as ε → 0+ , uε → u
uniformly in C [0, 1],
(2.17)
uε → u uniformly in C 1 [δ, 1 − δ],
where δ ∈ (0, 1/2), hence, from uε (1) = uε (0) = ε and (2.8) it follows that u(1) = u(0) = 0 and u > 0 in (0, 1). We now show that u satisfies (1.1). Integrating (2.9) from t0 to t (0 < t0 , t < 1) gives
φp (u′ε (t )) =
t t0
h(s, uε (s))|u′ε (s)|p − f (s, uε (s), u′ε (s)) ds + φp (u′ε (t0 )),
so that, passing to the limit as ε → 0+ and using Lebesgue’s dominated convergence theorem yield
φp (u′ (t )) =
t
h(s, u(s))|u′ (s)|p − f (s, u(s), u′ (s)) ds + φp (u′ (t0 )).
(2.18)
t0
This shows that φp (u′ (t )) ∈ C 1 (0, 1), so (1.1) is satisfied. Next we show that u ∈ C 1 [0, 1]. Since uε (1) = uε (0) = ε and uε > ε , we see that uε (t ) − uε (0)
u′ε (0) = lim
t
t →0+
> 0,
u′ε (1) = lim
uε (t ) − uε (1) t −1
t →1−
6 0.
Integrating (2.9) over (0, 1) and using Lemma 2.4, we derive that 1
h(t , uε (t ))|u′ε (t )|p dt 6
0
1
0
f (t , uε (t ), u′ε (t ))dt 6 C ,
where C > 0 is independent of ε , and passing to the limit as ε → 0+ and using Fatou’s lemma, we obtain 1
h(t , u(t ))|u′ (t )|p dt < +∞. 0
So, h(t , u)|u′ |p ∈ L1 (0, 1). By (2.18), the function χ (t ) = φp (u′ (t )) is absolutely continuous on [0, 1]. Since u′ (t ) = φq (χ(t )) ( 1p + 1q = 1), u′ ∈ C [0, 1]. Hence, u is a solution of problem (1.1)–(1.2). The proof to Theorem 1.1 is complete. Next we prove Theorem 1.2. Proof of Theorem 1.2. Under the hypotheses of Theorem 1.2, it follows from Theorem 1.1 that there exists a solution u for problem (1.1)–(1.2). We begin with the proof of the sufficiency part, and let 1
g (t )dt = +∞.
lim
s→0+
s
Clearly, it suffices to show that u′ (1) = u′ (0) = 0. Assume that u′ (0) ̸= 0. Since u′ (0) > 0, we obtain u′ (0) > 0. By continuity, there exists some t0 ∈ (0, 1) such that
u′ (0) 2u′ (0) > u′ (t ) > 2 > 0 for all t ∈ [0, t0 ]. Thus, u has the inverse u−1 on [0, t0 ]. Moreover, u′ (0)/2 < u′ (u−1 (y)) < 2u′ (0) for all y ∈ [0, u(t0 )]. Then, we have, for 0 < t < t0 , t0
g (u(s))|u (s)| ds > ′
p
p
2
t
=
u′ (0)
u′ (0) 2
t0
g (u(s))ds
t
p
2
>
u′ (0)
u(t0 ) u(t )
p
1 2u′ (0)
g (y) u′ (u−1 (y))
u(t0 ) u(t )
dy
g (y)dy.
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Using lims→0+
s
t0
g (u(s))|u′ (s)|p ds = +∞,
lim
t →0+
g (t )dt = +∞, we obtain
t
which is a contradiction. Indeed, integrating (1.1) with h(t , u) ≡ g (u) over (0, 1) and using u′ (1) 6 0 6 u′ (0), we get 1
g (u(t ))|u′ (t )|p dt 6
1
f (t , u(t ), u′ (t ))dt < +∞.
0
0
Hence, u′ (0) = 0. Similarly, one has that u′ (1) = 0. Next we prove the necessity. Let 1
g (t )dt < +∞.
lim
s→0+
s
It suffices to prove that u′ (1) ̸= u′ (0). Define K (s), G(s) : [0, +∞) → R by K ( s) =
s
g (s)dy,
0,0
(s > 0)
G(s) =
(s = 0),
s
(y) − Kp− 1
e
dy,
0,0
(s > 0) (s = 0).
Clearly, G ∈ C 1 [0, +∞) ∩ C 2 (0, +∞), and G satisfies G′ > 0 and G′′ = −1/(p − 1)gG′ .
(2.19)
By (2.19), we have
′ p−1 p−1 ′ φp ([G(u)]′ ) + G′ (u) f (t , u, u′ ) = G′ (u) φp (u′ ) − g (u)|u′ |p + f (t , u, u′ ) = 0,
t ∈ (0, 1),
hence
′ p−1 φp ([G(u)]′ ) = − G′ (u) f (t , u, u′ ),
t ∈ (0, 1).
Integrating the equation over (0, 1) yields
φp ([G(u)] )(1) − φp ([G(u)] )(0) = − ′
′
1
p−1
G ′ ( u)
f (t , u, u′ )dt < 0.
0
Noticing G′ (u(1)) = G′ (u(0)) = 1, we have
φp (u′ (1)) − φp (u′ (0)) < 0. This shows that u′ (1) ̸= u′ (0). The proof is complete.
3. Proof of Theorem 1.3 In what follows, we assume that (1.10)–(1.13) hold. 1 Since lims→0+ s [g (t )]1/p dt = +∞, we have, by Hölder’s inequality, that 1
g (t )dt = +∞.
lim
s→0+
(3.1)
s
Define ρ(s), ϱ(s) : [0, +∞) → [0, +∞) by
ρ(s) =
s
e 1 g (t )dt , 0,
(s > 0) (s = 0),
ϱ(s) =
s
1/p dt
e 1 [g (t )] 0,
,
(s > 0) (s = 0).
Clearly, ρ ′ , ϱ′ > 0 in (0, +∞). By (1.13), ϱ ∈ C 1 [0, +∞) and ϱ′ (0) = ϱ0 . The following lemma is the key of the proof of Theorem 1.3. Lemma 3.1. Let g satisfy (1.12) and (1.13). Then the function
s 0 ρ(t )[ϱ′ (t )]p dt , Θ (s) = ρ(s) 0,
(s > 0) (s = 0)
W. Zhou et al. / Nonlinear Analysis 75 (2012) 3994–4005
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belongs to C 1 [0, +∞) and satisfies
Θ ′ (s) + g (s)Θ (s) = [ϱ′ (s)]p , Θ (s) 6 [ϱ(s)] ,
∀s > 0,
(3.2)
∀s > 0.
p
(3.3)
Proof. Clearly, Θ ∈ C 1 (0, +∞). By direct calculations, one can verify (3.2). Since ρ ′ > 0 in (0, +∞), we have, by the definition of Θ , s
0 6 Θ (s) 6
[ϱ′ (t )]p dt ,
∀s ∈ (0, 1),
0
hence lims→0+ Θ (s) = 0. This shows that Θ (s) is continuous at s = 0. Using [ϱ′ (s)]p = g (s)[ϱ(s)]p and ρ ′ (s) = g (s)ρ(s), we have
s Θ ( s) =
0
ρ(t )g (t )[ϱ(t )]p dt = ρ(s)
s 0
ρ ′ (t )[ϱ(t )]p dt , ρ(s)
∀s ∈ (0, 1),
integrating by parts yields, for s ∈ (0, 1),
Θ ( s) =
1
[ρ(t )(ϱ(t )) ] | dt − p p
ρ(s)
s 0
s
ρ(t )ϱ (t )[ϱ(t )] ′
p−1
dt
0
= [ϱ(t )]p −
s
p
ρ(s)
ρ(t )ϱ′ (t )[ϱ(t )]p−1 dt 0
6 [ϱ(s)]p .
This proves (3.3). It remains to show that Θ ′ (s) is continuous at s = 0. By (3.3) proved above, we arrive at 06
Θ (s) s
6
[ϱ(s)]p s
,
∀s > 0,
so that, by ϱ′ (0) = ϱ0 , we have that Θ ′ (0) = 0. Using [ϱ′ (s)]p = g (s)[ϱ(s)]p and ρ ′ (s) = g (s)ρ(s), we derive from (3.2) that
lim Θ ′ (s) = lim [ϱ′ (s)]p − g (s)Θ (s)
s→0+
s→0+
p
g (s)
s
ρ(t )[ϱ′ (t )]p dt = lim [ϱ (s)] 1 − ρ(s)[ϱ′ (s)]p s→0+ s ρ(t )[ϱ′ (t )]p dt 0 ′ p = lim [ϱ (s)] 1 − , ρ(s)[ϱ(s)]p s→0+ ′
0
using l’Hospital’s rule and noticing lims→0+ g (s) = +∞, we have
ρ(s)[ϱ′ (s)]p lim Θ (s) = ϱ 1 − lim ′ p−1 + [ϱ(s)]p ρ(s)g (s) s→0+ pρ(s)ϱ (s)[ϱ(s)] s→0+ [ϱ′ (s)]p p = ϱ0 1 − lim ′ p−1 + [ϱ ′ (s)]p s→0+ pϱ (s)[ϱ(s)]
p 0
′
= ϱ0p 1 − lim
s→0+
1
p[g (s)]−(p−1)/p
+1
= 0. This shows that Θ ′ (s) is continuous at s = 0. The proof is complete.
We also need the following embedding theorem (cf. [27, Lemma 3]). Lemma 3.2. We have
λ1
1
|w|p dt 6 0
1
|w ′ |p dt ,
∀w ∈ W01,p [0, 1] \ {0}.
(3.4)
0
Moreover, the equality in (3.4) holds if and only if w is an eigenfunction corresponding to the first eigenvalue λ1 of problem (1.14).
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W. Zhou et al. / Nonlinear Analysis 75 (2012) 3994–4005
Proof of Theorem 1.3. Assume on the contrary that there exists a solution u of problem (1.1) and (1.2). By Lemma 3.1, we have that Θ (u(t )) ∈ C 1 [0, 1] with Θ (u(1)) = Θ (u(0)) = 0. By (3.1) and by a proof similar to that of sufficiency part of Theorem 1.2, we also have that u′ (1) = u′ (0) = 0.
(3.5)
Multiplying (1.1) by Θ (u) and integrating over (0, 1), we obtain
1 |u′ |p Θ ′ (u) + h(t , u)|u′ |p Θ (u) − f (t , u, u′ )Θ (u) dt = 0, 0
by h(t , u) > g (u), we have 1
|u′ |p [Θ ′ (u) + g (u)Θ (u)]dt 6
1
f (t , u, u′ )Θ (u)dt , 0
0
from this and Lemma 3.1 it follows that 1
|u′ |p [ϱ′ (u)]p dt 6
1
f (t , u, u′ )[ϱ(u)]p dt , 0
0
that is 1
|[ϱ(u)]′ |p dt 6 0
1
f (t , u, u′ )[ϱ(u)]p dt . 0
By Lemma 2.4 and (1.10), we derive that f (t , u, u′ ) 6 β eγ , hence 1
|[ϱ(u)]′ |p dt 6 β eγ
0
1
[ϱ(u)]p dt .
(3.6)
0
If β eγ < λ1 , it follows from (3.4) and (3.6) that 1
[ϱ(u)]p dt = 0. 0
This implies that u ≡ 0, which is a contradiction. If β eγ = λ1 , using the second conclusion of Lemma 3.2 and noticing (3.6) yield that ϱ(u) must be an eigenvalue function corresponding to the eigenvalue λ1 of problem (1.14), so that ϱ(u) satisfies
′ φp ([ϱ(u)]′ ) + λ1 [ϱ(u)]p−1 = 0,
t ∈ (0, 1).
Integrating the equation over (0, 1) yields
φp ([ϱ(u)] )(1) − φp ([ϱ(u)] )(0) = −λ1 ′
′
1
ϱ(u)
p−1
dt .
0
Using ϱ′ (0) = ϱ0 and (3.5), we find that the left side in the above equality equals zero, hence, 1
p−1 ϱ(u) dt = 0.
0
This implies that u ≡ 0, which is a contradiction. The proof is complete. Acknowledgments The authors would like to express their deep thanks to the referees for their careful reading of the manuscript and important comments which improve the paper. References [1] G.I. Barenblatt, M. Bertsch, A.E. Chertock, V.M. Prostokishin, Self-similar intermediate asymptotic for a degenerate parabolic filtration–absorption equation, Proc. Natl. Acad. Sci. 18 (2000) 9844–9848. [2] M. Bertsch, M. Ughi, Positivity properties of viscosity solutions of a degenerate parabolic equation, Nonlinear Anal. TMA 14 (1990) 571–592. [3] E. Chasseigne, J.L. Vázquez, The pressure equation in the fast diffusion range, Rev. Mat. Iberoam. 19 (3) (2003) 873–917. [4] Z. Yao, W. Zhou, Nonuniqueness of solutions for a singular diffusion problem, J. Math. Anal. Appl. 325 (2007) 183–204. [5] S. Staněk, Positive solutions of singular Dirichlet boundary value problems with time and space singularities, Nonlinear Anal. TMA 71 (2009) 4893–4905.
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Contents lists available at SciVerse ScienceDirect
Nonlinear Analysis journal homepage: www.elsevier.com/locate/na
On the one-dimensional p-Laplacian with a singular nonlinearity✩ Wenshu Zhou a , Xulong Qin b , Guokai Xu c , Xiaodan Wei a,∗ a
Department of Mathematics, Dalian Nationalities University, Dalian 116600, China
b
Department of Mathematics, Sun Yat-sen University, Guangzhou 510275, China
c
College of Electromechanical and Information Engineering, Dalian Nationalities University, Dalian 116600, China
article
info
Article history: Received 2 July 2011 Accepted 17 February 2012 Communicated by S. Carl MSC: 34B18
abstract In this paper, we are concerned with the existence and nonexistence of positive solutions of boundary value problems for the one-dimensional p-Laplacian with a singular nonlinearity. In the case of the model equation, we give the necessary and sufficient conditions of the existence of positive solutions for both the Dirichlet problem and the periodic problem. © 2012 Elsevier Ltd. All rights reserved.
Keywords: p-Laplacian Singularity Positive solution Existence Nonexistence
1. Introduction In this paper, we are concerned with the existence and nonexistence of positive solutions for the singular p-Laplacian equation:
′ φp (u′ ) − h(t , u)|u′ |p + f (t , u, u′ ) = 0,
0 < t < 1,
(1.1)
with either the Dirichlet boundary conditions u(1) = u(0) = 0,
(1.2)
or the periodic boundary conditions u′ (1) = u′ (0) = u(1) = u(0) = 0,
(1.3)
where φp (s) = |s| s with p > 1, h(t , z ) is nonnegative and continuous in [0, 1] × (R \ {0}) and may be singular at z = 0, and f (t , z , r ) is nonnegative and continuous on [0, 1] × R × R. The model equation is p−2
′ |u′ |p φp (u′ ) − λ m + f (t , u, u′ ) = 0, u
0 < t < 1,
(1.4)
where λ, m > 0. ✩ This work is supported by NNSF of China (grant nos. 10901030, 11071100, 11001278), Key Technology in the National Research and Development Pillar Program (no. 2009BAH41B05), China Postdoctoral Science Foundation (no. 20090450191), and Dalian Nationalities University (no. DC110109). ∗ Corresponding author. Tel.: +86 411 87189428; fax: +86 411 87189428. E-mail address: [email protected] (X. Wei).
0362-546X/$ – see front matter © 2012 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2012.02.015
W. Zhou et al. / Nonlinear Analysis 75 (2012) 3994–4005
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Besides singularity, another main feature of (1.1) is that the lower term has p-growth with respect to the first order derivative. Eq. (1.4) is strongly connected to the degenerate parabolic equation ut = u[φp (ux )]x − λ|ux |p ,
(1.5)
where λ > 0 is a constant, which appears in the investigation of groundwater flow in a water-absorbing fissurized porous rock and non-Newtonian filtration [1–4]. Indeed, if the solution u of (1.5) is of the form u = t −1/(p−1) w(x) with w > 0, then w must satisfy (in some sense)
[φp (w ′ )]′ − λ
|w ′ |p 1 + = 0. w p−1
The existence of positive solutions for (1.4) with different boundary conditions has recently been studied in [2,5–8]. In [2,5,6], the equations considered may have time singularities. In the present paper, we extend some existence results in [5,7,8] and obtain some sufficient conditions on the nonexistence of positive solutions. As by-products of our main results, we prove, under some auxiliary assumptions on f , that the condition m < p is necessary and sufficient for the existence of positive solutions to problem (1.4)–(1.2), and the condition 1 6 m < p is necessary and sufficient for the existence of positive solutions to problem (1.4)–(1.3). Our arguments on the existence are based on regularization technique, while in the arguments on the nonexistence, Gronwall’s inequality, embedding inequality and a scaling technique will play important roles. In the case where h ≡ 0, (1.1) becomes
′ φp (u′ ) + f (t , u, u′ ) = 0,
0 < t < 1.
(1.6)
Recently, problem (1.6)–(1.2) have been studied extensively, see for example [9–27], in which f (t , z , r ) may change sign and be singular at t = 0, t = 1 and z = 0. Some basic results were obtained in those papers. We point out that because the singularity at u = 0 and the p-growth with respect to u′ as a whole appear in (1.1), it is not in considerations of those papers. By a solution u of problem (1.1)–(1.2) we mean that u ∈ W , where W := {w ∈ C 1 [0, 1]; w > 0 in (0, 1), φp (w ′ ) ∈ C 1 (0, 1)},
and u satisfies (1.1) and (1.2). If u is a solution problem (1.1)–(1.2) and satisfies (1.3), then we call it is a solution for problem (1.1)–(1.3). To study the existence and nonexistence of solutions, we impose some additional conditions on h and f . We first study the existence of solutions. For h, we suppose that 0 6 h(t , z ) 6 g (z ),
∀t ∈ [0, 1], ∀z > 0,
(1.7)
where g : (0, +∞) → [0, +∞) is continuous in (0, +∞) and differentiable in (0, ξ ) for some ξ > 0, such that g ′ 6 0 in (0, ξ ), 1
lim
s→0+
lim g (s) = +∞,
s→0+
[g (z )]1/p dz < +∞.
(1.8)
(1.9)
s
For f , we suppose that there exist constants β > α > 0, γ > 0 such that
α 6 f (t , z , r ) 6 β + γ |r |p−1 ,
∀(t , z , r ) ∈ [0, 1] × [0, +∞) × R.
(1.10)
We obtain Theorem 1.1. Let (1.7)–(1.10) be satisfied. Then there exists a solution for problem (1.1)–(1.2). Theorem 1.2. Let h(t , z ) ≡ g (z ) satisfy (1.8) and (1.9), and suppose (1.10) holds. Then there exists a solution for problem 1 (1.1)–(1.3) if and only if lims→0+ s g (z )dz = +∞. Remark 1.1. Some examples of the functions satisfying (1.8)–(1.9) are: (1) (2) (3) (4)
g (z ) g (z ) g (z ) g (z )
= λ/z m , where λ > 0 and 0 < m < p; = (ln z )2 ; = 1/(ez − 1); = (ln z )2 if 0 < z < ϵ ; g (z ) = (ln ϵ)2 if z > ϵ , where ϵ > 0.
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W. Zhou et al. / Nonlinear Analysis 75 (2012) 3994–4005
Next we study the nonexistence of solutions for problem (1.1)–(1.2). We assume that h(t , z ) > g (z ) > 0,
∀t ∈ [0, 1], ∀z > 0,
(1.11)
where g : (0, +∞) → [0, +∞) is a continuous function such that
lim g (s) = +∞,
lim
s→0+
s→0+
1
[g (t )]1/p dt = +∞,
(1.12)
s
s 1/p lim [g (s)]1/p e 1 [g (t )] dt := ϱ0 ∈ [0, +∞).
(1.13)
s→0+
Denote by λ1 the first eigenvalue of the following problem (see [27]):
′ φp (u′ ) + λφp (u) = 0,
0 < t < 1;
u(1) = u(0) = 0.
(1.14)
Theorem 1.3. Let (1.10)–(1.13) be satisfied. If β eγ 6 λ1 , then problem (1.1)–(1.2) has no solution. Remark 1.2. Suppose that (1.10) holds and that there exists a continuous function g : (0, +∞) → [0, +∞) such that h(t , z ) > g (z ) > 0,
∀t ∈ [0, 1], ∀z > 0.
(1.15)
Observe that if u is a solution of problem (1.1)–(1.2), then v = δ u with δ > 0 is a positive solution of problem:
′ φp (v ′ ) − hδ (t , v)|v ′ |p + fδ (t , v, v ′ ) = 0, v(1) = v(0) = 0,
0 < t < 1,
(1.16)
where hδ (t , z ) =
1
δ
h t,
z
,
δ
fδ (t , z , r ) = δ p−1 f (t , z /δ, r /δ).
By (1.10), there holds for fδ fδ (t , z , r ) 6 δ p−1 (β + γ δ −(p−1) |r |p−1 ) 6 δ p−1 β + γ |r |p−1 ,
∀(t , z , r ) ∈ [0, 1] × [0, +∞) × R.
By (1.15), we have hδ (t , z ) > gδ (z ) :=
1
δ
g
z
δ
,
∀(t , z ) ∈ [0, 1] × [0, +∞).
According to Theorem 1.3, if gδ satisfies (1.12) and (1.13) and if δ p−1 β eγ 6 λ1 , then problem (1.16) has no solution, so does not problem (1.1)–(1.2). As a consequence of Theorem 1.3 and Remark 1.2, we obtain the following nonexistence result without additional requirements on f .
Corollary 1.1. Let (1.10) and (1.15) be satisfied. If there exists δ0 > 0 such that gδ (z ) = 1δ g δz all δ ∈ (0, δ0 ), then problem (1.1)–(1.2) has no solution. Proof. The proof can be completed by choosing a small δ ∈ (0, δ0 ) such that δ p−1 β eγ < λ1 .
satisfies (1.12) and (1.13) for
Note that in the model case where g (z ) = λ/z with λ, m > 0, gδ (z ) = λδ /z satisfy (1.12) and (1.13) if and only if m > p or m = p and λδ p−1 > 1. By Corollary 1.1, there does not exist a solution for problem (1.1)–(1.2) when m > p provided that (1.10) holds. For m = p, however, we still can prove the same conclusion by an argument based on Gronwall’s inequality. Precisely, we obtain m
m−1
m
Theorem 1.4. suppose that (1.10) holds and that h satisfies h(t , z ) >
λ zm
,
∀t ∈ [0, 1], ∀z > 0,
(1.17)
where m, λ > 0. If m > p, then problem (1.1)–(1.2) has no solution. Proof. It suffices to prove that problem (1.1)–(1.2) has no solution when m = p. Suppose that this is not so. Then there exists a solution for problem (1.1)–(1.2) when m = p.
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Note that u′ (1) 6 0 6 u′ (0). Integrating (1.1) over (0, 1) and integrating by parts yield 1
1
h(t , u)|u′ |p dt 6
f (t , u, u′ )dt . 0
0
By (1.17) with m = p, we have 1
0
|u′ |p up
dt 6
1
1
λ
f (t , u, u′ )dt := C .
(1.18)
0
By Hölder’s inequality, we arrive at, for t ∈ [0, 1], u( t ) =
t
′
t
u ds 6
up
0
0
|u′ |p
1/p
t
p
(p−1)/p ,
u p−1 ds
ds 0
from which and (1.18) it follows that p
u p−1 6 C 1/(p−1)
t
p
u p−1 ds,
∀t ∈ [0, 1].
0
By Gronwall’s inequality, we have that u ≡ 0, which is a contradiction.
By combining the above theorems, we immediately obtain Corollary 1.2. Let λ, m > 0, and suppose that (1.10) holds. Then problem (1.4)–(1.2) admits at least a solution if and only if m < p, and problem (1.4)–(1.3) admits at least a solution if and only if 1 6 m < p. The plan of the paper is as follows. In Section 2, we are concerned with the existence of solutions and the proofs of Theorems 1.1 and 1.2. Section 3 is devoted to the nonexistence of solutions for problem (1.1)–(1.2) and the proof of Theorem 1.3. 2. Proofs of Theorems 1.1 and 1.2 We begin with the proof of Theorem 1.1. Like in [8], we will use regularization technique to prove it. To do this, a suitable regularized problem must be considered instead of problem (1.1)–(1.2). A key point is to find a lower solution for the regularized problem. Assume that (1.7)–(1.10) hold. We define Ψ : [0, +∞) → [0, +∞) as follows
Ψ (s) =
s
[g (t )]1/p dt ,
0,0
(s > 0) (s = 0).
Since lims→0+ g (s) = ∞, we have that Ψ ′ > 0 in (0, s0 ) for some s0 ∈ (0, 1), hence, Ψ has the inverse Ψ −1 on [0, s0 ] with
Ψ −1 (0) = 0;
′ −1/p Ψ −1 (s) = g Ψ −1 (s) > 0,
s ∈ (0, s0 ).
(2.1)
By mean value theorem for differentials, there exists s1 ∈ (0, s0 ) such that
Ψ −1 (s) 6 s,
∀s ∈ [0, s1 ].
(2.2)
For ε ∈ (0, s1 ), we define aε (t , z , r ) : [0, 1] × R × R → R by aε (t , z , r ) = h(t , iε (z ))|r |p − f (t , iε (z ), r ), where
−1 2 Ψ (ε ), iε (z ) = z , −1 ε ,
(z < Ψ −1 (ε2 )) (Ψ −1 (ε2 ) 6 z 6 ε −1 ) (z > ε−1 ).
By (1.7), 0 6 h(t , iε (z )) 6 g (iε (z )) 6 Cε , where Cε = maxΨ −1 (ε2 )6z 6ε−1 g (z ). By (1.10) and the inequality: xp−1 6 1 + xp (∀x > 0), we have
|aε (t , z , r )| 6 Cε |r |p + β + γ |r |p−1 6 (Cε + β + γ )H (|r |) for all (t , z , r ) ∈ [0, 1] × R × R, where H (s) = 1 + sp for s > 0. Define Lε : W → C (0, 1) by
′ (Lε u)(t ) = − φp (u′ ) (t ) + aε (t , u(t ), u′ (t )),
0 < t < 1.
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W. Zhou et al. / Nonlinear Analysis 75 (2012) 3994–4005
Instead of problem (1.1)–(1.2), we consider the regularized problem with ε ∈ (0, s1 ):
(Lε u)(t ) = 0,
0 < t < 1;
u(1) = u(0) = ε.
(2.3)
A function u ∈ W is called an upper solution (a lower solution) of problem (2.3) if Lε u > (6)0 in (0, 1) and u(t ) > (6)ε at t = 0, 1. We will use the upper and lower solution method (see [16, Theorem 1 and Remark 2.4]) to prove the existence of solutions
+∞
p−1
s for problem (2.3). Since 0 ds = +∞, the condition (2.3) in [16] is satisfied. Then, to obtain a solution of problem H (s) (2.3), it suffices to find a lower solution and an upper solution. By direct calculations, one can verify the following lemma, which will be used to construct the upper and lower solutions of problem (2.3).
Lemma 2.1. Let β and γ be the same as in (1.10). Then the function defined on [0, 1]
1/(p−1) t eγ (1/2−s) − 1 ds, 0 6 t < 1/2, γ 0 1/(p−1) 1 γ (s−1/2) 1/(p−1) e −1 V (t ) = β · ds, 1/2 6 t 6 1, γ t p/(p−1) p/(p−1) 1 (p − 1) 1 − − t , p 2 2
(γ > 0)
(γ = 0)
belongs to W and is a solution of problem
′ − φp (u′ ) = β + γ |u′ |p−1 ,
0 < t < 1;
u(1) = u(0) = 0.
By Lemma 2.1, we obtain Lemma 2.2. Let (1.7)–(1.10) hold. Then there exists a constant ε1 ∈ (0, s1 ), such that for any ε ∈ (0, ε1 ), uε (t ) := V (t ) + ε is an upper solution of problem (2.3) and satisfies Ψ −1 (ε 2 ) < uε < ε −1 on [0, 1]. Proof. Clearly, there exists ε1 ∈ (0, s1 ) such that max[0,1] V + 1 < ε1−1 . By (2.2), Ψ −1 (ε 2 ) < uε (t ) = V (t ) + ε < ε −1 on [0, 1] for all ε ∈ (0, ε1 ). Also, we have
′
Lε uε (t ) = − φp (uε ) (t ) + h(t , uε (t ))|uε (t )|p − f (t , uε (t ), uε (t ))
′
′
′
′ ′ ′ > − φp (uε ) (t ) − f (t , uε (t ), uε (t )) = β + γ |V ′ (t )|p−1 − f (t , V (t ) + ϵ, V ′ (t )) > 0, 0 < t < 1. The proof is complete.
Next, we construct the lower solutions of problem (2.3). Let V be the same as in Lemma 2.1. Choosing a constant ε2 ∈ (0, ε1 ) such that
ε2 (max V + 1) 6 min{s1 , ξ }, [0,1]
ε2 p−1 [g (ξ )]−(p−1)/p (β + γ max |V ′ |p−1 ) + ε2 p max |V ′ |p − α 6 0, [0,1]
[0,1]
(2.4)
where ξ , α, β, γ and s1 are the same as before. By (2.1)–(2.2), we have, for all ε ∈ (0, ε2 ),
Ψ −1 (ε 2 ) < Ψ −1 ε2 (V (t ) + ε) < ξ , ∀t ∈ [0, 1], Ψ −1 ε2 (V (t ) + ε) 6 uε (t ) = V (t ) + ε, ∀t ∈ [0, 1], Ψ
−1
(ε2 ε) 6 ε.
(2.5) (2.6) (2.7)
We have Lemma 2.3. Let (1.7)–(1.10) hold. Then for any ε ∈ (0, ε2 ), uε (t ) := Ψ −1 (ε2 (V (t )+ε)) is a lower solution of problem (2.3) and satisfies uε > uε > Ψ −1 (ε 2 ) on [0, 1].
W. Zhou et al. / Nonlinear Analysis 75 (2012) 3994–4005
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Proof. By (2.5) and (2.7), we have that 0 < uε (1) = uε (0) 6 ε and uε > uε > Ψ −1 (ε 2 ) on [0, 1] for all ε ∈ (0, ε2 ). By (1.7), (1.8) and (2.5), we have g ′ (uε ) 6 0, g (uε ) > g (ξ ) and h(t , iε (uε (t ))) = h(t , uε (t )) 6 g (uε (t )) for all ε ∈ (0, ε2 ). Noticing (1.10) and (2.4), we obtain, for ε ∈ (0, ε2 ),
′
Lε uε (t ) = − φp (u′ε ) (t ) + h(t , iε (uε (t )))|u′ε (t )|p − f (t , iε (uε (t )), u′ε (t ))
′
6 − φp (u′ε ) (t ) + g (uε (t ))|u′ε (t )|p − α
′ = −ε2 p−1 [g (uε (t ))]−(p−1)/p |V ′ (t )|p−2 V ′ (t ) + (p − 1)/pε2 p g ′ (uε (t ))[g (uε (t ))]−2 |V ′ (t )|p + ε2 p |V ′ (t )|p − α ′ 6 −ε2 p−1 [g (uε (t ))]−(p−1)/p |V ′ (t )|p−2 V ′ (t ) + ε2 p |V ′ (t )|p − α = ε2 p−1 [g (uε (t ))]−(p−1)/p (β + γ |V ′ (t )|p−1 ) + ε2 p |V ′ (t )|p − α
6 ε2 p−1 [g (ξ )]−(p−1)/p (β + γ max |V ′ |p−1 ) + ε2 p max |V ′ |p − α [0,1]
6 0,
[0,1]
0 < t < 1.
Thus the proof is complete.
By Lemmas 2.2 and 2.3, it follows from [16, Theorem 1 and Remark 2.4] that for any fixed ε ∈ (0, ε2 ), problem (2.3) admits a solution uε ∈ C 1 [0, 1] with φp (u′ε ) ∈ C (0, 1) such that uε (t ) > uε (t ) > uε (t ), and, for all ε ∈ (0, ε2 ),
∀t ∈ [0, 1],
(2.8)
′ φp (u′ε ) (t ) − h(t , uε (t ))|u′ε (t )|p + f (t , uε (t ), u′ε (t )) = 0,
t ∈ (0, 1),
(2.9)
φp (u′ε )
with uε (1) = uε (0) = ε . Since uε ∈ C [0, 1], it follows from (2.9) that ∈ C (0, 1), so uε ∈ W . Next we estimate u′ε . To do this, we first prove a general conclusion, which, in the sequel, will be used repeatedly. 1
1
Lemma 2.4. Suppose that (1.10) holds. If u ∈ W satisfies u(1) = u(0) = 0 and
′ φp (u′ ) (t ) + f (t , u(t ), u′ (t )) > 0,
0 < t < 1,
then
|u′ (t )| 6 Cβ,γ , ∀t ∈ [0, 1], γ 1/(p−1) β(e −1) if γ > 0; Cβ,γ = β 1/(p−1) if γ = 0. where Cβ,γ = γ
(2.10)
Proof. Noticing u(1) = u(0) = 0 and u > 0 on [0, 1], we have u′ (0) > 0 > u′ (1).
(2.11)
By (1.10), we arrive at
′ φp (u′ ) (t ) + β + γ |u′ (t )|p−1 > 0,
t ∈ (0, 1).
(2.12)
Let w = φp (u′ ). Then we derive from (2.12) that w ′ + β + γ |w| > 0 in (0, 1), i.e. w(t )
0
′
1
β + γ |s|
> 0,
ds + t
which and (2.11) imply that 1 > |w(t )|
0
1
β + γs
ds 6 1,
w(t ) 0
∀t ∈ (0, 1), w
ds + t > 0 for t ∈ [0, 1], so that β+γ |s| 1
0
ds 6 1 on [0, 1], therefore β+γ |s| 1
∀t ∈ [0, 1].
From this, one can easily obtain the desired result. The proof is complete.
Lemma 2.5. For any δ ∈ (0, 1/2), there exists a positive constant C (δ) independent of ε such that for all ε ∈ (0, ε2 ),
|u′ε (t )| 6 Cβ,γ ,
∀t ∈ [0, 1],
|u′ε (t2 ) − u′ε (t1 )| 6 C (δ)|t2 − t1 |θ ,
(2.13)
∀t2 , t1 ∈ [δ, 1 − δ],
where θ = 1/(p − 1) if p > 2; θ = 1 if 1 < p < 2.
(2.14)
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Proof. We first prove that for any ε ∈ (0, ε2 ), uε (t ) > ε,
∀t ∈ [0, 1].
Suppose that this is not so. Noticing uε (1) = uε (0) = ε , we see that there exist some ε ∈ (0, ε2 ) and some t ∗ ∈ (0, 1) such that uε (t ∗ ) = min uε < ε,
u′ε (t ∗ ) = 0.
[0,1]
It follows from (2.9) that
′ φp (u′ε ) (t ∗ ) + f (t ∗ , uε (t ∗ ), 0) = 0. ′ By (1.10), φp (u′ε ) (t ∗ ) 6 −α < 0. By continuity, there exists a sufficiently small δ > 0 such that ′ φp (u′ε ) (t ) < 0, ∀t ∈ (t ∗ , t ∗ + δ). Noticing u′ε (t ∗ ) = 0, we have
φp (u′ε (t )) < 0,
∀t ∈ (t ∗ , t ∗ + δ),
hence u′ε (t ) < 0,
∀t ∈ (t ∗ , t ∗ + δ),
therefore uε (t ) < uε (t ∗ ) = min uε ,
∀t ∈ (t ∗ , t ∗ + δ),
[0,1]
which is a contradiction. Next, we prove (2.13) by Lemma 2.4. Let wε = uε − ε . Then wε satisfies
wε (1) = wε (0) = ε;
wε (t ) > 0,
∀t ∈ [0, 1].
On the other hand, it follows from (2.9) that
′ φp (u′ε ) (t ) + f (t , uε (t ), u′ε (t )) > 0,
t ∈ (0, 1),
by (1.10), we obtain
′ φp (u′ε ) (t ) + β + γ |u′ε (t )|p−1 > 0,
t ∈ (0, 1),
hence
′ φp (wε′ ) (t ) + β + γ |wε′ (t )|p−1 > 0,
t ∈ (0, 1).
Taking f (t , z , r ) ≡ β + γ |r | in Lemma 2.4 yield |wε′ | 6 Cβ,γ on [0, 1], so |u′ε | 6 Cβ,γ on [0, 1]. Thus, (2.13) holds. Below, we prove (2.14). By (2.8) and (2.13), it is easy to derive from (2.9) that for any δ ∈ (0, 1/2) there exists a constant C (δ) > 0 independent of ε , such that for all ε ∈ (0, ε2 ) p−1
′ φp (u′ε ) (t ) 6 C (δ),
δ 6 t 6 1 − δ.
(2.15)
Recalling the inequality (cf. [28])
[φp (η) − φp (η′ )] · [η − η′ ] >
C1 |η − η′ |p , C2 (|η| + |η′ |)p−2 |η − η′ |2 ,
(p > 2) (1 < p < 2)
for all η, η′ ∈ R, where Ci (i = 1, 2) are positive constants, we derive, by (2.15), that if p > 2, then
|u′ε (t2 ) − u′ε (t1 )|p 6 C1−1 [u′ε (t2 ) − u′ε (t1 )] · [|u′ε (t2 )|p−2 u′ε (t2 ) − |u′ε (t1 )|p−2 u′ε (t1 )] 6 C (δ)|u′ε (t2 ) − u′ε (t1 )| |t2 − t1 |, ∀t2 , t1 ∈ [δ, 1 − δ], hence
|u′ε (t2 ) − u′ε (t1 )| 6 C (δ)|t2 − t1 |1/(p−1) ,
∀t2 , t1 ∈ [δ, 1 − δ],
and if p ∈ (1, 2), then
|u′ε (t2 ) − u′ε (t1 )|2 [|u′ε (t2 )| + |u′ε (t1 )|]p−2 6 C2−1 [u′ε (t2 ) − u′ε (t1 )] · [|u′ε (t2 )|p−2 u′ε (t2 ) − |u′ε (t1 )|p−2 u′ε (t1 )] 6 C (δ)|u′ε (t2 ) − u′ε (t1 )| |t2 − t1 |, ∀t2 , t1 ∈ [δ, 1 − δ],
(2.16)
W. Zhou et al. / Nonlinear Analysis 75 (2012) 3994–4005
4001
so that, using (2.13) yields
|u′ε (t2 ) − u′ε (t1 )| 6 C (δ)|t2 − t1 |[|u′ε (t2 )| + |u′ε (t1 )|]2−p 6 C (δ)|t2 − t1 | for all t2 , t1 ∈ [δ, 1 − δ]. This ends our proof.
Proof of Theorem 1.1. By (2.13) and (2.14) and using the Arzela´ –Ascoli theorem, there exist a subsequence of {uε }, still denoted by {uε }, and a function u ∈ C 1 (0, 1) ∩ C [0, 1] such that, as ε → 0+ , uε → u
uniformly in C [0, 1],
(2.17)
uε → u uniformly in C 1 [δ, 1 − δ],
where δ ∈ (0, 1/2), hence, from uε (1) = uε (0) = ε and (2.8) it follows that u(1) = u(0) = 0 and u > 0 in (0, 1). We now show that u satisfies (1.1). Integrating (2.9) from t0 to t (0 < t0 , t < 1) gives
φp (u′ε (t )) =
t t0
h(s, uε (s))|u′ε (s)|p − f (s, uε (s), u′ε (s)) ds + φp (u′ε (t0 )),
so that, passing to the limit as ε → 0+ and using Lebesgue’s dominated convergence theorem yield
φp (u′ (t )) =
t
h(s, u(s))|u′ (s)|p − f (s, u(s), u′ (s)) ds + φp (u′ (t0 )).
(2.18)
t0
This shows that φp (u′ (t )) ∈ C 1 (0, 1), so (1.1) is satisfied. Next we show that u ∈ C 1 [0, 1]. Since uε (1) = uε (0) = ε and uε > ε , we see that uε (t ) − uε (0)
u′ε (0) = lim
t
t →0+
> 0,
u′ε (1) = lim
uε (t ) − uε (1) t −1
t →1−
6 0.
Integrating (2.9) over (0, 1) and using Lemma 2.4, we derive that 1
h(t , uε (t ))|u′ε (t )|p dt 6
0
1
0
f (t , uε (t ), u′ε (t ))dt 6 C ,
where C > 0 is independent of ε , and passing to the limit as ε → 0+ and using Fatou’s lemma, we obtain 1
h(t , u(t ))|u′ (t )|p dt < +∞. 0
So, h(t , u)|u′ |p ∈ L1 (0, 1). By (2.18), the function χ (t ) = φp (u′ (t )) is absolutely continuous on [0, 1]. Since u′ (t ) = φq (χ(t )) ( 1p + 1q = 1), u′ ∈ C [0, 1]. Hence, u is a solution of problem (1.1)–(1.2). The proof to Theorem 1.1 is complete. Next we prove Theorem 1.2. Proof of Theorem 1.2. Under the hypotheses of Theorem 1.2, it follows from Theorem 1.1 that there exists a solution u for problem (1.1)–(1.2). We begin with the proof of the sufficiency part, and let 1
g (t )dt = +∞.
lim
s→0+
s
Clearly, it suffices to show that u′ (1) = u′ (0) = 0. Assume that u′ (0) ̸= 0. Since u′ (0) > 0, we obtain u′ (0) > 0. By continuity, there exists some t0 ∈ (0, 1) such that
u′ (0) 2u′ (0) > u′ (t ) > 2 > 0 for all t ∈ [0, t0 ]. Thus, u has the inverse u−1 on [0, t0 ]. Moreover, u′ (0)/2 < u′ (u−1 (y)) < 2u′ (0) for all y ∈ [0, u(t0 )]. Then, we have, for 0 < t < t0 , t0
g (u(s))|u (s)| ds > ′
p
p
2
t
=
u′ (0)
u′ (0) 2
t0
g (u(s))ds
t
p
2
>
u′ (0)
u(t0 ) u(t )
p
1 2u′ (0)
g (y) u′ (u−1 (y))
u(t0 ) u(t )
dy
g (y)dy.
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W. Zhou et al. / Nonlinear Analysis 75 (2012) 3994–4005
1
Using lims→0+
s
t0
g (u(s))|u′ (s)|p ds = +∞,
lim
t →0+
g (t )dt = +∞, we obtain
t
which is a contradiction. Indeed, integrating (1.1) with h(t , u) ≡ g (u) over (0, 1) and using u′ (1) 6 0 6 u′ (0), we get 1
g (u(t ))|u′ (t )|p dt 6
1
f (t , u(t ), u′ (t ))dt < +∞.
0
0
Hence, u′ (0) = 0. Similarly, one has that u′ (1) = 0. Next we prove the necessity. Let 1
g (t )dt < +∞.
lim
s→0+
s
It suffices to prove that u′ (1) ̸= u′ (0). Define K (s), G(s) : [0, +∞) → R by K ( s) =
s
g (s)dy,
0,0
(s > 0)
G(s) =
(s = 0),
s
(y) − Kp− 1
e
dy,
0,0
(s > 0) (s = 0).
Clearly, G ∈ C 1 [0, +∞) ∩ C 2 (0, +∞), and G satisfies G′ > 0 and G′′ = −1/(p − 1)gG′ .
(2.19)
By (2.19), we have
′ p−1 p−1 ′ φp ([G(u)]′ ) + G′ (u) f (t , u, u′ ) = G′ (u) φp (u′ ) − g (u)|u′ |p + f (t , u, u′ ) = 0,
t ∈ (0, 1),
hence
′ p−1 φp ([G(u)]′ ) = − G′ (u) f (t , u, u′ ),
t ∈ (0, 1).
Integrating the equation over (0, 1) yields
φp ([G(u)] )(1) − φp ([G(u)] )(0) = − ′
′
1
p−1
G ′ ( u)
f (t , u, u′ )dt < 0.
0
Noticing G′ (u(1)) = G′ (u(0)) = 1, we have
φp (u′ (1)) − φp (u′ (0)) < 0. This shows that u′ (1) ̸= u′ (0). The proof is complete.
3. Proof of Theorem 1.3 In what follows, we assume that (1.10)–(1.13) hold. 1 Since lims→0+ s [g (t )]1/p dt = +∞, we have, by Hölder’s inequality, that 1
g (t )dt = +∞.
lim
s→0+
(3.1)
s
Define ρ(s), ϱ(s) : [0, +∞) → [0, +∞) by
ρ(s) =
s
e 1 g (t )dt , 0,
(s > 0) (s = 0),
ϱ(s) =
s
1/p dt
e 1 [g (t )] 0,
,
(s > 0) (s = 0).
Clearly, ρ ′ , ϱ′ > 0 in (0, +∞). By (1.13), ϱ ∈ C 1 [0, +∞) and ϱ′ (0) = ϱ0 . The following lemma is the key of the proof of Theorem 1.3. Lemma 3.1. Let g satisfy (1.12) and (1.13). Then the function
s 0 ρ(t )[ϱ′ (t )]p dt , Θ (s) = ρ(s) 0,
(s > 0) (s = 0)
W. Zhou et al. / Nonlinear Analysis 75 (2012) 3994–4005
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belongs to C 1 [0, +∞) and satisfies
Θ ′ (s) + g (s)Θ (s) = [ϱ′ (s)]p , Θ (s) 6 [ϱ(s)] ,
∀s > 0,
(3.2)
∀s > 0.
p
(3.3)
Proof. Clearly, Θ ∈ C 1 (0, +∞). By direct calculations, one can verify (3.2). Since ρ ′ > 0 in (0, +∞), we have, by the definition of Θ , s
0 6 Θ (s) 6
[ϱ′ (t )]p dt ,
∀s ∈ (0, 1),
0
hence lims→0+ Θ (s) = 0. This shows that Θ (s) is continuous at s = 0. Using [ϱ′ (s)]p = g (s)[ϱ(s)]p and ρ ′ (s) = g (s)ρ(s), we have
s Θ ( s) =
0
ρ(t )g (t )[ϱ(t )]p dt = ρ(s)
s 0
ρ ′ (t )[ϱ(t )]p dt , ρ(s)
∀s ∈ (0, 1),
integrating by parts yields, for s ∈ (0, 1),
Θ ( s) =
1
[ρ(t )(ϱ(t )) ] | dt − p p
ρ(s)
s 0
s
ρ(t )ϱ (t )[ϱ(t )] ′
p−1
dt
0
= [ϱ(t )]p −
s
p
ρ(s)
ρ(t )ϱ′ (t )[ϱ(t )]p−1 dt 0
6 [ϱ(s)]p .
This proves (3.3). It remains to show that Θ ′ (s) is continuous at s = 0. By (3.3) proved above, we arrive at 06
Θ (s) s
6
[ϱ(s)]p s
,
∀s > 0,
so that, by ϱ′ (0) = ϱ0 , we have that Θ ′ (0) = 0. Using [ϱ′ (s)]p = g (s)[ϱ(s)]p and ρ ′ (s) = g (s)ρ(s), we derive from (3.2) that
lim Θ ′ (s) = lim [ϱ′ (s)]p − g (s)Θ (s)
s→0+
s→0+
p
g (s)
s
ρ(t )[ϱ′ (t )]p dt = lim [ϱ (s)] 1 − ρ(s)[ϱ′ (s)]p s→0+ s ρ(t )[ϱ′ (t )]p dt 0 ′ p = lim [ϱ (s)] 1 − , ρ(s)[ϱ(s)]p s→0+ ′
0
using l’Hospital’s rule and noticing lims→0+ g (s) = +∞, we have
ρ(s)[ϱ′ (s)]p lim Θ (s) = ϱ 1 − lim ′ p−1 + [ϱ(s)]p ρ(s)g (s) s→0+ pρ(s)ϱ (s)[ϱ(s)] s→0+ [ϱ′ (s)]p p = ϱ0 1 − lim ′ p−1 + [ϱ ′ (s)]p s→0+ pϱ (s)[ϱ(s)]
p 0
′
= ϱ0p 1 − lim
s→0+
1
p[g (s)]−(p−1)/p
+1
= 0. This shows that Θ ′ (s) is continuous at s = 0. The proof is complete.
We also need the following embedding theorem (cf. [27, Lemma 3]). Lemma 3.2. We have
λ1
1
|w|p dt 6 0
1
|w ′ |p dt ,
∀w ∈ W01,p [0, 1] \ {0}.
(3.4)
0
Moreover, the equality in (3.4) holds if and only if w is an eigenfunction corresponding to the first eigenvalue λ1 of problem (1.14).
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W. Zhou et al. / Nonlinear Analysis 75 (2012) 3994–4005
Proof of Theorem 1.3. Assume on the contrary that there exists a solution u of problem (1.1) and (1.2). By Lemma 3.1, we have that Θ (u(t )) ∈ C 1 [0, 1] with Θ (u(1)) = Θ (u(0)) = 0. By (3.1) and by a proof similar to that of sufficiency part of Theorem 1.2, we also have that u′ (1) = u′ (0) = 0.
(3.5)
Multiplying (1.1) by Θ (u) and integrating over (0, 1), we obtain
1 |u′ |p Θ ′ (u) + h(t , u)|u′ |p Θ (u) − f (t , u, u′ )Θ (u) dt = 0, 0
by h(t , u) > g (u), we have 1
|u′ |p [Θ ′ (u) + g (u)Θ (u)]dt 6
1
f (t , u, u′ )Θ (u)dt , 0
0
from this and Lemma 3.1 it follows that 1
|u′ |p [ϱ′ (u)]p dt 6
1
f (t , u, u′ )[ϱ(u)]p dt , 0
0
that is 1
|[ϱ(u)]′ |p dt 6 0
1
f (t , u, u′ )[ϱ(u)]p dt . 0
By Lemma 2.4 and (1.10), we derive that f (t , u, u′ ) 6 β eγ , hence 1
|[ϱ(u)]′ |p dt 6 β eγ
0
1
[ϱ(u)]p dt .
(3.6)
0
If β eγ < λ1 , it follows from (3.4) and (3.6) that 1
[ϱ(u)]p dt = 0. 0
This implies that u ≡ 0, which is a contradiction. If β eγ = λ1 , using the second conclusion of Lemma 3.2 and noticing (3.6) yield that ϱ(u) must be an eigenvalue function corresponding to the eigenvalue λ1 of problem (1.14), so that ϱ(u) satisfies
′ φp ([ϱ(u)]′ ) + λ1 [ϱ(u)]p−1 = 0,
t ∈ (0, 1).
Integrating the equation over (0, 1) yields
φp ([ϱ(u)] )(1) − φp ([ϱ(u)] )(0) = −λ1 ′
′
1
ϱ(u)
p−1
dt .
0
Using ϱ′ (0) = ϱ0 and (3.5), we find that the left side in the above equality equals zero, hence, 1
p−1 ϱ(u) dt = 0.
0
This implies that u ≡ 0, which is a contradiction. The proof is complete. Acknowledgments The authors would like to express their deep thanks to the referees for their careful reading of the manuscript and important comments which improve the paper. References [1] G.I. Barenblatt, M. Bertsch, A.E. Chertock, V.M. Prostokishin, Self-similar intermediate asymptotic for a degenerate parabolic filtration–absorption equation, Proc. Natl. Acad. Sci. 18 (2000) 9844–9848. [2] M. Bertsch, M. Ughi, Positivity properties of viscosity solutions of a degenerate parabolic equation, Nonlinear Anal. TMA 14 (1990) 571–592. [3] E. Chasseigne, J.L. Vázquez, The pressure equation in the fast diffusion range, Rev. Mat. Iberoam. 19 (3) (2003) 873–917. [4] Z. Yao, W. Zhou, Nonuniqueness of solutions for a singular diffusion problem, J. Math. Anal. Appl. 325 (2007) 183–204. [5] S. Staněk, Positive solutions of singular Dirichlet boundary value problems with time and space singularities, Nonlinear Anal. TMA 71 (2009) 4893–4905.
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