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Positive solutions for a fourth order p -Laplacian boundary value problem
Nonlinear Analysis 74 (2011) 2612–2623
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Positive solutions for a fourth order p-Laplacian boundary value problem✩ Jiafa Xu ∗ , Zhilin Yang Department of Mathematics, Qingdao Technological University, No 11 Fushun Road, Qingdao, Shandong Province, PR China
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Article history: Received 12 July 2010 Accepted 16 December 2010 MSC: 34B18 47H07 47H11 45M20 26D15
abstract In this paper, we study the existence, multiplicity and uniqueness of positive solutions for the fourth order p-Laplacian boundary value problem
′′ p−1 ′′ ′′ (|u | u ) = f (t , u), u(2i) (0) = u(2i) (1) = 0,
i = 0, 1.
Here p > 0 and f ∈ C ([0, 1] × R+ , R+ ) (R+ := [0, ∞)). Based on a priori estimates achieved by utilizing properties of concave functions, we use fixed point index theory to establish our main results. © 2010 Elsevier Ltd. All rights reserved.
Keywords: p-Laplacian equation Positive solution Fixed point index Upper and lower solution Iterative sequence
1. Introduction In this paper, we study the existence, multiplicity and uniqueness of positive solutions for the fourth order p-Laplacian boundary value problem
(|u′′ |p−1 u′′ )′′ = f (t , u), t ∈ (0, 1), u(2i) (0) = u(2i) (1) = 0, i = 0, 1,
(1.1)
where p > 0 and f ∈ C ([0, 1] × R+ , R+ ) (R+ := [0, ∞)). Note that by a positive solution of (1.1) we mean a function u ∈ C 2 [0, 1] ∩ C 4 (0, 1) that solves (1.1) and satisfies |u′′ |p−1 u′′ ∈ C 2 (0, 1) and u(t ) > 0, t ∈ (0, 1). Second order differential equations with the p-Laplacian operator arise in modeling different physical and natural phenomena, which can be encountered in, for instance, non-Newtonian mechanics, nonlinear elasticity, glaciology, population biology, combustion theory, and nonlinear flow laws, see [1,2]. This explains why many papers have been published on existence of solutions for differential equations with the p-Laplacian operator; see, for instance, [3–14] and references therein. Fourth order boundary value problems, including those with the p-Laplacian operator, have their origin in beam theory [15,16], ice formation [17,18], fluids on lungs [19], brain warping [20,21], designing special curves on surfaces [22,20], etc. In beam theory, more specifically, a beam with a small deformation, a beam of a material which ✩ Supported by the NNSF of China (Grant 10871116 and 10971179) and the NSF of Shandong Province of China (Grant ZR2009AL014).
∗
Corresponding author. E-mail addresses: [email protected], [email protected] (J. Xu), [email protected], [email protected] (Z. Yang).
0362-546X/$ – see front matter © 2010 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2010.12.016
J. Xu, Z. Yang / Nonlinear Analysis 74 (2011) 2612–2623
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satisfies a nonlinear power-like stress and strain law and a beam with two-sided links which satisfies a nonlinear powerlike elasticity law, can be described by fourth order differential equations along with their boundary value conditions. For more background and applications, we refer the reader to the work by Timoshenko [23] on elasticity, the monograph by Soedel [24] on deformation of structure, and the work by Dulácska [25] on the effects of soil settlement. Due to their wide applications, the existence and multiplicity of positive solutions for fourth order p-Laplacian boundary value problems has also attracted increasing attention over the last decades; see [26–40] and references therein. The singular fourth-order p-Laplacian four-point boundary value problem
(|u′′ |p−1 u′′ )′′ = f (t , u(t )), 0 < t < 1, u(0) = u(1) − au(ξ ) = u′′ (0) = u′′ (1) − bu′′ (η) = 0
(1.2)
arises in beam theory and was studied by Zhang and Liu [36], in which the method of upper and lower solutions was used in establishing their main results on positive solutions of the above problem. Here p > 1, 0 < ξ , η < 1, f ∈ C ((0, 1) × (0, ∞), (0, ∞)) may be singular at t = 0 and/or t = 1 and u = 0. In [37], Zhang and Liu also considered the existence of positive solutions for (1.2), but with the original boundary value conditions in (1.2) replaced by the multi-point boundary conditions m−2
u(0) −
−
m−2
ai u(ξi ) = u(1) = u′′ (0) −
i=1
−
bi u′′ (ηi ) = u′′ (1) = 0,
(1.3)
i=1
where m ≥ 3, ai , bi , ξi , ηi ∈ (0, 1) (i = 1, 2, . . . , m − 2) are constants with i=1 ai < 1 and i=1 bi < 1, and f ∈ C ((0, 1) × R+ , R+ ) may be singular at t = 0 and/or t = 1. By virtue of monotone iterative techniques, they established a necessary and sufficient condition of the pseudo-C 3 [0, 1] positive solutions for their problem. In [31], Guo et al. discussed the existence and multiplicity of positive solutions for the fourth-order quasilinear singular differential equation
∑m−2
(|u′′ |p−2 u′′ )′′ = λg (t )f (u),
0 < t < 1,
∑m−2
(1.4)
sharing the boundary value conditions in (1.1), where λ is a positive parameter; f ∈ C ([0, ∞), (0, ∞)) is nondecreasing on [0, ∞), with f (u) > δ um for all u ∈ [0, ∞), where δ > 0 and m > p − 1 are two constants; g ∈ C ((0, 1), (0, ∞)) and g (t ) ̸≡ 0 on any subinterval of (0, 1). Applying fixed point index theory and the method of upper and lower solutions, they proved that there is a threshold λ∗ < ∞ for which (1.4) admits at least two positive solutions if 0 < λ < λ∗ , one positive solution if λ = λ∗ , and no positive solution if λ > λ∗ . Our problem (1.1) merely involves the simply supported boundary value conditions and a nonsingular nonlinearity f . Nevertheless, our methodology and results in this paper are entirely different from those in the papers cited above. We observe that if p = 1, then (1.1) reduces to the Lidstone problem
u(4) = f (t , u), t ∈ (0, 1), u(2i) (0) = u(2i) (1) = 0, i = 0, 1,
(1.5)
which can describe deformations of an elastic beam, with both ends simply supported, in an equilibrium state, and thus has been extensively studied (see [41–48] and references therein). It is natural to ask what connections do exist between (1.1) and (1.5). Some connections between them will be shown by repeatedly invoking Hölder’s inequalities in our proofs. We will use fixed point index theory to establish our main results based on a priori estimates achieved by utilizing some properties of concave functions, properties including Hölder’s inequalities and our new inequality (2.3) below. This paper is organized as follows. Section 2 contains some preliminary results. Section 3 is devoted to the existence of positive solutions for (1.1) and Section 4 to the multiplicity of positive solutions for (1.1). In Section 5, we discuss the uniqueness of positive solutions for (1.1), and prove that the unique positive solution can be uniformly approximated by an iterative sequence beginning with any function u which is continuous, nonnegative and not identically vanishing on [0, 1] (see Theorem 5.2). Section 6, finally, contains some remarks that serve to show some connections between (1.1) and (1.5). 2. Preliminaries Let E := C [0, 1],
P := {u ∈ E : u(t ) ≥ 0, ∀t ∈ [0, 1]},
Then (E , ‖ · ‖) is a real Banach space and P a cone on E. Let G(t , s) :=
t (1 − s), s(1 − t ),
0 ≤ t ≤ s ≤ 1, 0 ≤ s ≤ t ≤ 1.
‖u‖ := max |u(t )|. t ∈[0,1]
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J. Xu, Z. Yang / Nonlinear Analysis 74 (2011) 2612–2623
Then it is easy to see that (1.1) is equivalent to the fixed point equation u = Au, where A : P → P is defined by
(Au)(t ) :=
1
∫
1
∫
G(t , s)
G(s, τ )f (τ , u(τ ))dτ
1/p
ds.
(2.1)
0
0
Note that if f ∈ C ([0, 1] × R+ , R+ ), then A : P → P is a completely continuous operator. Lemma 2.1 ([49]). Let Ω ⊂ E be a bounded open set and A : Ω ∩ P → P is a completely continuous operator. If there exists v0 ∈ P \ {0} such that v − Av ̸= λv0 for all v ∈ ∂ Ω ∩ P and λ ≥ 0, then i(A, Ω ∩ P , P ) = 0, where i denotes the fixed point index on P. Lemma 2.2 ([49]). Let Ω ⊂ E be a bounded open set with 0 ∈ Ω . Suppose A : Ω ∩ P → P is a completely continuous operator. If v ̸= λAv for all v ∈ ∂ Ω ∩ P and 0 ≤ λ ≤ 1, then i(A, Ω ∩ P , P ) = 1. Lemma 2.3. Let ψ(t ) := sin(π t ). Then ψ ∈ P \ {0} and 1
∫
G(t , s)ψ(t )dt = 0
ψ(s), 2
π
1
∫
1
G(t , s)ψ(s)ds = 0
1
π2
ψ(t ).
(2.2)
Lemma 2.4. Let ϕ ∈ P be concave on [0, 1]. Then 1
∫
2
ϕ(t ) sin(π t )dt ≥
π2
0
‖ϕ‖.
(2.3)
Proof. We first suppose that there is a t0 ∈ (0, 1) such that ϕ(t0 ) = ‖ϕ‖. Then the concavity of ϕ implies 1
∫
ϕ(t ) sin(π t ) dt =
t0
∫
0
ϕ(t ) sin(π t ) dt +
∫
0
=
ϕ
0
≥ ϕ(t0 )
t t0
· t0 +
t0
∫
t0 − t t0
π2
· 0 sin(π t ) dt +
∫
1
ϕ
t sin(π t ) dt +
∫
1
(1 − t ) sin(π t ) dt
1−t 1 − t0
t0
0
2
≥
ϕ(t ) sin(π t ) dt t0
t0
∫
1
· t0 +
t − t0 1 − t0
· 1 sin(π t ) dt
t0
‖ϕ‖,
where we have used the simple fact: t
∫
s sin(π s)ds +
1
∫
0
(1 − s) sin(π s) ds ≥ t
2
π2
,
t ∈ [0, 1].
The remaining situations (with ϕ(0) = ‖ϕ‖ or with ϕ(1) = ‖ϕ‖) can be analogously handled. The proof is completed.
Lemma 2.5. Let ψ(t ) := sin(π t ). Then for every w ∈ P \ {0}, there are two positive numbers bw , aw such that the inequality aw ψ(t ) ≤
1
∫
G(t , s)w(s)ds ≤ bw ψ(t ) 0
holds for all t ∈ [0, 1]. Proof. Let w1 (t ) :=
1 0
G(t , s)w(s)ds. Then w1 ∈ C 2 [0, 1] ∩ P, and
w1 (0) = w1 (1) = ψ(0) = ψ(1) = 0. Now w ∈ P \ {0} implies 1
∫
sw(s)ds > 0,
1
∫
0
(1 − s)w(s)ds > 0, 0
w1 ( t ) > 0, ψ(t )
t ∈ (0, 1).
By l’Hôpital’s rule we have
− w1 (t ) lim = lim + + ψ(t ) t →0 t →0
t 0
sw(s)ds +
1
1
t
0
(1 − s)w(s)ds = π cos(π t )
(1 − s)w(s)ds >0 π
J. Xu, Z. Yang / Nonlinear Analysis 74 (2011) 2612–2623
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and
− w1 ( t ) = lim lim − − ψ(t ) t →1 t →1
t 0
sw(s)ds +
1
1
t
0
(1 − s)w(s)ds = π cos(π t )
sw(s)ds
π
> 0.
w (t )
1 The last inequalities say that ψ( may be viewed as a positive continuous function on [0, 1]. Hence there are two positive t) w1 (t ) numbers bw , aw , for which aw ≤ ψ(t ) ≤ bw holds for all t ∈ [0, 1]. This completes the proof.
Lemma 2.6 (Hölder’s Inequalities). Let θ > 0 and ϕ ∈ P. Then 1
∫
ϕ(t )dt
θ
1
∫ ≤
(ϕ(t ))θ dt
0
0
if θ ≥ 1, and 1
∫
ϕ(t )dt
θ
1
∫ ≥
0
(ϕ(t ))θ dt
0
if 0 < θ ≤ 1. 3. Existence of positive solutions ∗
Let p∗ := max{1, p}, p∗ := min{1, p}, βp := π 4p , αp := π 4p∗ . We denote Bρ := {u ∈ E : ‖u‖ < ρ} for ρ > 0 in the sequel. (H1) (H2) (H3) (H4) (H5)
f ∈ C ([0, 1] × R+ , R+ ). There exist a1 > βp and c > 0 such that f (t , x) ≥ a1 xp − c for all x ∈ R+ and t ∈ [0, 1]. There exist b1 ∈ (0, αp ) and r > 0 such that f (t , x) ≤ b1 xp for all x ∈ [0, r ] and t ∈ [0, 1]. There exist a2 > βp and r > 0 such that f (t , x) ≥ a2 xp for all x ∈ [0, r ] and t ∈ [0, 1]. There exist b2 ∈ (0, αp ) and c > 0 such that f (t , x) ≤ b2 xp + c for all x ∈ R+ and t ∈ [0, 1].
Theorem 3.1. Suppose that (H1), (H2) and (H3) are satisfied. Then (1.1) has at least one positive solution. Proof. Let M1 := {u ∈ P : u = Au + λψ for some λ ≥ 0},
where ψ(t ) := sin(π t ). We claim that M1 is bounded. Indeed, u ∈ M1 implies that u is concave and that u(t ) ≥ (Au)(t ), t ∈ [0, 1]. By definition, we have u( t ) ≥
∫
1
G(t , s)
0
1
∫
G(s, τ )f (τ , u(τ ))dτ
1/p ds
0
for all u ∈ M1 , t ∈ [0, 1]. Now the inequalities 0 < p∗ ≤ 1,
0 < p∗ /p ≤ 1
(3.1)
and 0 ≤ G(t , s) < 1,
0 ≤ G(s, τ ) < 1,
t , s, τ ∈ [0, 1],
(3.2)
along with Lemma 2.6 and (H2), enable us to obtain u (t ) ≥ p∗
∫
1
G(t , s) 0
1
∫
1
∫
1
∫
≥ 0
p∗ ds
G(t , s)G(s, τ )f p∗ /p (τ , u(τ ))dτ ds
0 1
∫
G(s, τ )f (τ , u(τ ))dτ
1/p
0
≥ 0
1
∫
p /p
G(t , s)G(s, τ )(a1∗ up∗ (τ ) − c p∗ /p )dτ ds
0
for all u ∈ M1 and t ∈ [0, 1], the simple inequality
(x + y)p∗ /p ≤ xp∗ /p + yp∗ /p ,
x, y ∈ R+
(3.3)
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J. Xu, Z. Yang / Nonlinear Analysis 74 (2011) 2612–2623
being used. Multiply (3.3) by sin(π t ) and integrate over [0, 1] and use (2.2) to obtain 1
∫
p /p
up∗ (t ) sin(π t ) dt ≥ a1∗ π −4
1
∫
0
up∗ (t ) sin(π t ) dt − 2c p∗ /p π −5 ,
0
so that 1
∫
p /p
up∗ (t ) sin(π t ) dt ≤ 2π −1 c p∗ /p /(a1∗
− π 4 ) := N1
0
for all u ∈ M1 . Recall that every u ∈ M1 is concave. So is up∗ with p∗ ∈ (0, 1]. Now Lemma 2.4 yields
‖up∗ ‖ ≤ N1 π 2 /2 for all u ∈ M1 , which implies the boundedness of M1 , as claimed. Taking R > sup{‖u‖ : u ∈ M1 }, we have u − Au ̸= λψ,
∀u ∈ ∂ BR ∩ P , λ ≥ 0.
Applying Lemma 2.1 we obtain i(A, BR ∩ P , P ) = 0.
(3.4)
Let M2 := {u ∈ Br ∩ P : u = λAu for some λ ∈ [0, 1]}.
We shall prove M2 = {0}. Indeed, if u ∈ M2 , we have u(t ) ≤ (Au)(t ) =
1
∫
G(t , s)
1
∫
0
G(s, τ )f (τ , u(τ ))dτ
1/p
ds,
∀u ∈ Br ∩ P .
0
Now the inequalities p∗ ≥ 1,
p∗ /p ≥ 1
(3.5)
and (3.2), along with Lemma 2.6 and (H3), enable us to obtain p∗
u (t ) ≤
∫
1
G(t , s) 0
∫
1
∫
1
∫
0
1
ds
∗ G(t , s)G(s, τ )f p /p (τ , u(τ ))dτ ds
0 1
≤ 0
G(s, τ )f (τ , u(τ ))dτ
p∗
1/p
0
≤ ∫
1
∫
p∗ /p p∗
G(t , s)G(s, τ )b1
u (τ )dτ ds
(3.6)
0
for all u ∈ M2 , t ∈ [0, 1]. Multiply both sides of the above by sin(π t ) and integrate over [0, 1] and use (2.2) to obtain 1
∫
p∗
u (t ) sin(π t ) dt ≤
p∗ /p b1
0
π
−4
1
∫
∗
up (t ) sin(π t ) dt , 0
so that 1
∫
∗
up (t ) sin(π t ) dt = 0 0
for all u ∈ M2 . This implies M2 = {0} and thus u ̸= λAu,
∀u ∈ ∂ Br ∩ P , λ ∈ [0, 1].
(3.7)
Now Lemma 2.2 yields i(A, Br ∩ P , P ) = 1.
(3.8)
Combining this with (3.4) gives i(A, (BR \ Br ) ∩ P , P ) = 0 − 1 = −1. Hence the operator A has at least one fixed point on (BR \ Br ) ∩ P, and thus (1.1) has at least one positive solution. This completes the proof.
J. Xu, Z. Yang / Nonlinear Analysis 74 (2011) 2612–2623
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Theorem 3.2. Suppose that (H1), (H4) and (H5) are satisfied. Then (1.1) has at least one positive solution. Proof. Let M3 := {u ∈ Br ∩ P : u = Au + λψ for some λ ≥ 0}.
We claim M3 ⊂ {0}. Indeed, if u ∈ M3 , then we have u ≥ Au by definition. (H4) implies u( t ) ≥
∫
1
G(t , s)
1
∫
G(s, τ )a2 u (τ )dτ p
1/p
ds.
0
0
Recall (3.1) and (3.2). Now Lemma 2.6 implies p /p
up∗ (t ) ≥ a2∗
1
∫
1
∫
G(t , s)G(s, τ )up∗ (τ )dτ ds 0
0
for all u ∈ M3 , t ∈ [0, 1]. Multiply both sides of the above by sin(π t ) and integrate over [0, 1] and use (2.2) to obtain 1
∫
u (t ) sin(π t ) dt ≥ p∗
p /p a2∗
π
−4
0
so that
1
∫
up∗ (t ) sin(π t ) dt , 0
1 0
up∗ (t ) sin(π t ) dt = 0 for all u ∈ M3 , whence M3 ⊂ {0}, as claimed. As a result of this, we have
u − Au ̸= λψ,
∀u ∈ ∂ Br ∩ P , λ ≥ 0.
Now Lemma 2.1 yields i(A, Br ∩ P , P ) = 0.
(3.9)
Let M4 := {u ∈ P : u = λAu for some λ ∈ [0, 1]}.
We assert M4 is bounded. Indeed, if u ∈ M4 , then u is concave and u ≤ Au, which can be written in the form u( t ) ≤
∫
1
G(t , s)
1
∫
0
G(s, τ )f (τ , u(τ ))dτ
1/p
ds.
0
Recall (3.2) and (3.5). Now by Lemma 2.6 and (H5), we obtain p∗
u (t ) ≤
∫
1
G(t , s) 0
∫
1
∫
1
∫
0
1
ds
∗ G(t , s)G(s, τ )f p /p (τ , u(τ ))dτ ds
0 1
≤ 0
∗ G(t , s)G(s, τ )(b2 up (τ ) + c )p /p dτ ds
0 1
∫
G(s, τ )f (τ , u(τ ))dτ
p∗
1/p
0
≤ ∫
1
∫
1
∫
0
p∗ /p p∗
G(t , s)G(s, τ )(b3
≤
p∗ /p
u (τ ) + c1
)dτ ds
(3.10)
0
for all u ∈ M4 , b3 ∈ (b2 , αp ) and c1 > 0 being chosen so that ∗ /p
(b2 z + c )p
p∗ /p p∗ /p
≤ b3
z
p∗ /p
+ c1
for all z ∈ R+ . Multiply both sides of (3.10) by sin(π t ) and integrate over [0, 1] and use (2.2) to obtain 1
∫
∗
p∗ /p
up (t ) sin(π t ) dt ≤ b3
π −4
1
∫
0
p∗ /p
∗
up (t ) sin(π t ) dt + 2c1
π −5 ,
0
so that 1
∫
p∗ /p
∗
up (t ) sin(π t ) dt ≤ 2π −1 c1
p∗ /p
/(π 4 − b3
) := N2
0
for all u ∈ M4 . This, together with Lemma 2.6, leads to 1
∫
u(t ) sin(π t ) dt ≤ 0
1
∫
p∗
p∗
u (t )(sin(π t )) dt 0
1/p∗
1/p∗
≤ N2
(3.11)
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J. Xu, Z. Yang / Nonlinear Analysis 74 (2011) 2612–2623
for all u ∈ M4 . Recall u is concave. Now the boundedness of M4 , as asserted, follows by Lemma 2.4 and (3.11). Taking R > sup{‖u‖ : u ∈ M4 }, we have u ̸= λAu,
∀u ∈ ∂ BR ∩ P , λ ∈ [0, 1].
(3.12)
Now Lemma 2.2 yields i(A, BR ∩ P , P ) = 1.
(3.13)
Combining this with (3.9) gives i(A, (BR \ Br ) ∩ P , P ) = 1 − 0 = 1. Hence the operator A has at least one fixed point on (BR \ Br ) ∩ P, and thus (1.1) has at least one positive solution. This completes the proof. 4. Multiplicity of positive solutions In this section, we will study the multiplicity of positive solutions for (1.1). To this end, we need the following conditions. (H6) There are ω > 0 and ϱ ∈ (0, 384/5) such that the inequality f (t , u) < ϱp/p ωp holds for all u ∈ [0, ω] and t ∈ [0, 1]. (H7) There are ω > 0, δ ∈ (0, 1/2) and ζ > 180δ −p∗ such that the inequality f (t , u) > ζ p/p∗ ωp holds for all u ∈ [δω, ω] and t ∈ [δ, 1 − δ]. ∗
Theorem 4.1. Suppose that (H1), (H2), (H4) and (H6) are satisfied. Then (1.1) has at least two positive solutions. Proof. Inequalities (3.2) and (3.5), along with Lemma 2.6 and (H6), enable us to obtain p∗
(Au) (t ) =
∫
1
G(t , s)
0
∫
1
1
∫
G(s, τ )f (τ , u(τ ))dτ
p∗
1/p ds
0 1
∫
≤ 0
∗ G(t , s)G(s, τ )f p /p (τ , u(τ ))dτ ds
0
≤ ϱω
p∗
1
∫
1
∫
0
G(t , s)G(s, τ )dτ ds 0
∗
= ≤
ϱωp (t 4 − 2t 3 + t ) 24
5ϱ 384
ω
p∗
∗
< ωp
for all u ∈ Bω ∩ P and t ∈ [0, 1]. This implies ‖Au‖ < ‖u‖ for all u ∈ Bω ∩ P, and, in turn, u ̸= λAu for all u ∈ ∂ Bω ∩ P, λ ∈ [0, 1]. Now Lemma 2.2 yields i(A, Bω ∩ P , P ) = 1.
(4.1)
On the other hand, in view of (H2) and (H4), we may choose R > ω and r ∈ (0, ω) so that (3.4) and (3.9) hold (see the proofs of Theorems 3.1 and 3.2). Combining (3.4), (3.9) and (4.1), we obtain i(A, (BR \ Bω ) ∩ P , P ) = 0 − 1 = −1,
i(A, (Bω \ Br ) ∩ P , P ) = 1 − 0 = 1.
Hence A has at least two fixed points, one on (BR \ Bω ) ∩ P and the other on (Bω \ Br ) ∩ P, and thus (1.1) has at least two positive solutions. The proof is completed. To prove Theorem 4.2 below, we need an extra cone P1 , defined by
P1 :=
u∈P :
min u(t ) ≥ δ‖u‖ ,
δ≤t ≤1−δ
where δ ∈ (0, 1/2) is given by (H7). Lemma 4.1. A(P1 ) ⊂ P1 .
J. Xu, Z. Yang / Nonlinear Analysis 74 (2011) 2612–2623
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Proof. After some computations, we have G(s, s) ≥ G(t , s) ≥ δ G(s, s),
G(s, τ ) ≥ s(1 − s)G(τ , τ )
(4.2)
for all t ∈ [δ, 1 − δ], s, τ ∈ [0, 1]. Now the preceding inequalities, together with the definition of A, imply
p∗ ∫ min (Au)(t ) = min
δ≤t ≤1−δ
δ≤t ≤1−δ
∫ ≥ δ
1
G(t , s)
=δ
1
G(s, s)
≥δ
1
∫
p∗ ds
G(s, τ )f (τ , u(τ ))dτ
1/p
p∗ ds
0
∫
1
G(s, s)
∫
1
∫
G(s, τ )f (τ , u(τ ))dτ
1/p
p∗ ds
0
0
p∗
G(s, τ )f (τ , u(τ ))dτ
1/p
0
0
0
p∗
1
∫
1
G(z , s)
1
∫
G(s, τ )f (τ , u(τ ))dτ
1/p
p∗ ds
0
0
= δ p∗ (Au)p∗ (z ) for all u ∈ P and z ∈ [0, 1], and thus minδ≤t ≤1−δ (Au)(t ) ≥ δ‖Au‖ for all u ∈ P. Therefore, we have A(P ) ⊂ P1 and, in particular, A(P1 ) ⊂ P1 . This completes the proof. Theorem 4.2. Suppose that (H1), (H3), (H5) and (H7) are satisfied. Then (1.1) has at least two positive solutions. Proof. By Lemma 4.1, we have A(P1 ) ⊂ P1 . Observe that ω ≥ u(t ) ≥ δω for all u ∈ ∂ Bω ∩ P1 , ω > 0 being determined by (H7). Combining this with (4.2) and (H7), one infers
(‖Au‖)
p∗
[
max (Au)(t )
≥
]p∗
t ∈[δ,1−δ]
= ≥ δ p∗
1
∫
≥ δ p∗
30
1
∫
0
δ p∗
G(s, τ )f (τ , u(τ ))dτ
1/p
p∗ ds
0
G(s, s)G(s, τ )f p∗ /p (τ , u(τ ))dτ ds
0 1
∫
1
1
∫
0
∫
0
=
G(t , s)
max t ∈[δ,1−δ]
≥
1
∫
G(s, s)s(1 − s)G(τ , τ )f p∗ /p (τ , u(τ ))dτ ds
0 1
∫
G(τ , τ )ζ ωp∗ dτ 0
ζ ωp∗ δ p∗ 180
> ωp∗
for all u ∈ ∂ Bω ∩ P1 , and thus ‖Au‖ > ‖u‖ for all u ∈ ∂ Bω ∩ P1 , so that u − Au ̸= λψ,
∀u ∈ ∂ Bω ∩ P1 , λ ≥ 0.
Now Lemma 2.1 yields i(A, Bω ∩ P1 , P1 ) = 0.
(4.3)
On the other hand, in view of (H3) and (H5), we may choose R > ω and r ∈ (0, ω) so that (3.7) and (3.12) hold (see the proofs of Theorems 3.1 and 3.2). Naturally, (3.7) and (3.12) still hold, if P is replaced by P1 . Lemma 2.2 then yields i(A, Br ∩ P1 , P1 ) = 1,
i(A, BR ∩ P1 , P1 ) = 1.
Combining the above with (4.3), we obtain i(A, (BR \ Bω ) ∩ P1 , P1 ) = 1 − 0 = 1,
i(A, (Bω \ Br ) ∩ P1 , P1 ) = 0 − 1 = −1.
Hence A has at least two fixed points, one on (BR \ Bω ) ∩ P1 and the other on (Bω \ Br ) ∩ P1 , and thus (1.1) has at least two positive solutions. The proof is completed.
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J. Xu, Z. Yang / Nonlinear Analysis 74 (2011) 2612–2623
5. Uniqueness of positive solutions and iterative convergence In this section, we will establish the uniqueness of positive solutions for (1.1) when f grows p-sublinearly at 0 and at ∞, i.e. (H4) and (H5), together with (H9) below, are satisfied. Moreover, we shall prove that the unique positive solution can be uniformly approximated by an iterative sequence {An u} beginning with any function u ∈ P \ {0} (see Theorem 5.2). We now list our hypotheses in this section. (H8) f (t , u) is increasing in u, that is, the inequality f (t , u1 ) ≤ f (t , u2 ) holds for all u1 , u2 ∈ R+ satisfying u1 ≤ u2 . (H9) f (t , λu) > λp f (t , u) for each λ ∈ (0, 1), u > 0 and t ∈ [0, 1]. Theorem 5.1. If (H1), (H4), (H5), (H8) and (H9) hold, then (1.1) has exactly one positive solution. Proof. By Theorem 3.2, (1.1) has at least one positive solution. It then remains to prove the uniqueness of positive solutions for (1.1). Indeed, if u1 and u2 are two positive solutions of (1.1), then ui ( t ) =
1
∫
G(t , s)
1
∫
G(s, τ )f (τ , ui (τ ))dτ
1/p ds
0
0
and wi (t ) := ( 0 G(t , s)f (s, ui (s))ds)1/p ∈ P \ {0} (i = 1, 2). By Lemma 2.5, there are four positive numbers bi , ai for which a ai ψ ≤ ui ≤ bi ψ (i = 1, 2) and thus u2 ≥ b2 u1 . Let
1
1
µ0 := sup{µ > 0 : u2 ≥ µu1 }. Clearly µ0 > 0 and u2 ≥ µ0 u1 . We claim µ0 ≥ 1. If the claim is false, then µ0 < 1 and u2 ( t ) ≥
1
∫
G(t , s)
1
∫
0
G(s, τ )f (τ , µ0 u1 (τ ))dτ
ds
0 1
∫
1/p
G(t , s)g (s)ds + µ0
= 0
1
∫
G(t , s) 0
1
∫
G(s, τ )f (τ , u1 (τ ))dτ
1/p ds
0
1
∫
G(t , s)g (s)ds + µ0 u1 (t ),
= 0
where g (s) :=
1
∫
G(s, τ )f (τ , µ0 u1 (τ ))dτ
1/p
0
− µ0
1
∫
G(s, τ )f (τ , u1 (τ ))dτ
1/p
Now (H9) implies g ∈ P \ {0}. By Lemma 2.5, there is a positive number ε such that for all t ∈ [0, 1]. This leads to u2 ( t ) ≥
ε
b1
.
0
1 0
G(t , s)g (s)ds ≥ εψ(t ) = ε sin(π t )
u1 (t ) + µ0 u1 (t ),
a contradiction with the definition of µ0 . As a result, µ0 ≥ 1 and thus u2 ≥ u1 . Similarly u1 ≥ u2 . Thus we have u1 = u2 . This completes the proof. To prove Theorem 5.2 below, we have to consider upper solutions and lower solutions of the fixed point equation u = Au.
(5.1)
To this end, we need the following two lemmas. Lemma 5.1. Let (H1) and (H4) hold, with r > 0 being determined by (H4). Then for any ε ∈ (0, r ], vε (t ) := ε(sin(π t ))1/p∗ is a lower solution of (5.1), that is, Avε ≥ vε . Proof. For any ε ∈ (0, r ], we have by (H4) and Lemma 2.6,
(Avε ) (t ) = p∗
∫
1
G(t , s) 0
G(t , s)
p /p
) dτ
1/p
p∗ ds
G(s, τ )a2 ε p (sin(π τ ))p/p∗ dτ
p∗ /p ds
0 1
∫ 0
p /p
G(s, τ )f (τ , ε(sin(π τ ))
1
∫
0
≥ a2∗ ε p∗
1/p∗
0
1
∫ ≥
1
∫
∫
1
G(t , s)G(s, τ ) sin(π τ )dτ ds
0
= a2∗ ε p∗ π −4 sin(π t )
(5.2)
J. Xu, Z. Yang / Nonlinear Analysis 74 (2011) 2612–2623
2621
and thus 1/p
(Avε )(t ) ≥ a2 π −4/p∗ vε (t ) ≥ vε (t ). This completes the proof.
Lemma 5.2. Let (H1) and (H5) hold, with b2 and c determined by (H5). Choose b3 ∈ (b2 , αp ) and c1 > 0 so that ∗ /p
(b2 z + c )p
p∗ /p p∗ /p
≤ b3
p∗ /p
+ c1
z
for all z ≥ 0 and let ∗ 1/p
51/p c1 π −1/p 96−1/p
m0 :=
∗
∗
1/p
.
1 − b3 π −4/p
∗
Then for any m ≥ m0 , wm (t ) := m(sin(π t ))1/p is an upper solution of (5.1), that is, Awm ≤ wm . ∗
Proof. Since 1
∫ 0
1
∫
G(t , s)G(s, τ )dsdτ =
t (1 − t )(1 + t − t 2 ) 24
0
≤
5 sin(π t ) 96π
for all t ∈ [0, 1], it follows from (H5) and Lemma 2.6 that ∗
(Awm )p (t ) ≤
1
∫ 0
∗ ∗ G(t , s)G(s, τ )(b2 mp sinp/p (π τ ) + c )p /p dτ ds
0 1
∫
1
∫
1
∫
p∗ /p
G(t , s)G(s, τ )(b3
≤ 0 0 p∗ /p
∗
p∗ /p
∗
p∗ /p
π −4 mp sin(π t ) + c1
= b3 ≤ b3
p∗ /p
∗
mp sin(π τ ) + c1
24−1 (t 4 − 2t 3 + t )
p∗ /p
π −4 mp sin(π t ) + 5c1
)dτ ds
π −1 96−1 sin(π t )
(5.3)
and thus p∗ /p
(Awm )(t ) ≤ (b3
p∗ /p
∗
π −4 mp + 5c1
1/p
π −1 96−1 )1/p (sin(π t ))1/p
∗ 1/p
∗
∗
≤ (b3 π −4/p m + 51/p c1 π −1/p 96−1/p )(sin(π t ))1/p ≤ wm (t ) ∗
∗
for any m ≥ m0 and t ∈ [0, 1]. This completes the proof.
∗
∗
Theorem 5.2. Let all the conditions in Theorem 5.1 hold and u∗ the unique positive solution of (1.1). Then for any u ∈ P \ {0}, the iterative sequence {An u} converges to u∗ as n → ∞, that is, limn→∞ ‖An u − u∗ ‖ = 0. Proof. Let (Fu)(t ) := f (t , u(t )). Then F : P → P, and (H4) and (H8) imply that F (P \ {0}) ⊂ P \ {0}. Now for any u ∈ P \ {0}, by Lemma 2.5, there is an au > 0 such that 1
∫
G(s, τ )f (τ , u(τ ))dτ ≥ au sin(π s) 0
for all s ∈ [0, 1] and
(Au)p∗ (t ) ≥
1
∫
G(t , s)(au sin(π s))p∗ /p ds
0
≥ apu∗ /p
∫
1
G(t , s) sin(π s) ds
0
= apu∗ /p π −2 sin(π t ) for all t ∈ [0, 1], so that
(Au)(t ) ≥ a1u/p π −2/p∗ (sin(π t ))1/p∗ . 1/p
Note that we may assume εu := au π −2/p∗ ≤ r, where r is determined in (H4). This, along with Lemma 5.1, implies that vεu (t ) := εu (sin(π t ))1/p∗ is a lower solution of (5.1) and Au ≥ vεu . On the other hand, for any u ∈ P \ {0}, letting du := max{f (t , x) : 0 ≤ t ≤ 1, 0 ≤ x ≤ ‖u‖},
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J. Xu, Z. Yang / Nonlinear Analysis 74 (2011) 2612–2623
we have du s(1 − s)
1
∫
G(s, τ )f (τ , u(τ ))dτ ≤
2
0
≤
du sin(π s) 2π
for all s ∈ [0, 1] and thus p∗
(Au) (t ) =
∫
1
G(t , s)
G(s, τ )f (τ , u(τ ))dτ
p∗
1/p ds
0
0
∫
1
∫
1
G(t , s)
≤
du sin(π s)
∗ /p
≤ dup
ds
2π
0 ∗ /p
(2π )−p
p∗
1/p
1
∫
G(t , s) sin(π s) ds 0
∗ /p
= dup
∗ /p
(2π )−p
π −2 sin(π t ),
so that
(Au)(t ) ≤ d1u/p (2π )−1/p π −2/p (sin(π t ))1/p ≤ mu (sin(π t ))1/p ∗
∗
∗
1/p
for all t ∈ [0, 1], where mu := max{m0 , du (2π )−1/p π −2/p } ≥ m0 , m0 being given in Lemma 5.2. Now Lemma 5.2 implies ∗ that wmu (t ) := mu (sin(π t ))1/p is an upper solution of (5.1) with Au ≤ wmu . So far we have shown that vεu ≤ Au ≤ wmu for any u ∈ P \ {0}, with vεu a lower solution of (5.1) and wmu a upper solution of (5.1). (H8) implies that A is an increasing operator and hence An vεu ≤ An+1 u ≤ An wmu for all positive integer n. As is well known [49], as n → ∞, {An vεu } converges to the smallest fixed point of A on [vεu , wmu ] and {An wmu } to the greatest fixed point of A on [vεu , wmu ]. Theorem 5.1 implies that u∗ is the unique fixed point of A on [vεu , wmu ]. Therefore limn→∞ An vεu = limn→∞ An wmu = u∗ , whence limn→∞ An u = u∗ for any u ∈ P \ {0}. The proof is completed. ∗
Remark 5.1. Let f (t , u) := uq , where q ∈ (0, p). Then (H1), (H4), (H5), (H8) and (H9) hold. By Theorems 5.1 and 5.2, (1.1) has exactly one positive solution, and the iterative sequence un+1 (t ) :=
1
∫
G(t , s) 0
1
∫
G(s, τ )uqn (τ )dτ
1/p
ds,
t ∈ [0, 1], n = 0, 1, 2, . . . ,
0
uniformly converges to the unique positive solution u∗ for every function u0 ∈ P \ {0}. 6. Concluding remarks As was mentioned in the Introduction, some connections should exist between (1.1) and (1.5). This will be clearer from Theorems 6.1 and 6.2 below. As the prelude to them, the following conditions are formulated on letting p = 1 in defining αp and βp .
(H2)′ (H3)′ (H4)′ (H5)′
There exist a1 There exist b1 There exist a2 There exist b2
> π 4 and c > 0 such that f (t , x) ≥ a1 xp − c for all x ∈ R+ and t ∈ [0, 1]. ∈ (0, π 4 ) and r > 0 such that f (t , x) ≤ b1 xp for all x ∈ [0, r ] and t ∈ [0, 1]. > π 4 and r > 0 such that f (t , x) ≥ a2 xp for all x ∈ [0, r ] and t ∈ [0, 1]. ∈ (0, π 4 ) and c > 0 such that f (t , x) ≤ b2 xp + c for all x ∈ R+ and t ∈ [0, 1].
Noticing that limp→1 αp = limp→1 βp = π 4 and applying Theorems 3.1 and 3.2, we obtain the following results. Theorem 6.1. Suppose that (H1), (H2)′ and (H3)′ are satisfied. Then there is ε ∈ (0, 1) such that (1.1) has at least one positive solution for all p satisfying |p − 1| < ε . Theorem 6.2. Suppose that (H1), (H4)′ and (H5)′ are satisfied. Then there is ε ∈ (0, 1) such that (1.1) has at least one positive solution for all p satisfying |p − 1| < ε . References [1] R. Glowinski, J. Rappaz, Approximation of a nonlinear elliptic problem arising in a non-Newtonian fluid flow model in glaciology, Math. Model. Number. Anal. 37 (2003) 175–186. [2] J.I. Diaz, F. de Thélin, On a nonlinear parabolic problem arising in some models related to turbulent flows, SIAM J. Math. Anal. 25 (1994) 1085–1111. [3] D. Ji, W. Ge, Multiple positive solutions for some p-Laplacian boundary value problems, Appl. Math. Comput. 187 (2007) 1315–1325. [4] R. Avery, J. Henderson, Existence of three pseudo-symmetric solutions for a one dimensional p-Laplacian, J. Math. Anal. Appl. 277 (2003) 395–404.
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Positive solutions for a fourth order p-Laplacian boundary value problem✩ Jiafa Xu ∗ , Zhilin Yang Department of Mathematics, Qingdao Technological University, No 11 Fushun Road, Qingdao, Shandong Province, PR China
article
info
Article history: Received 12 July 2010 Accepted 16 December 2010 MSC: 34B18 47H07 47H11 45M20 26D15
abstract In this paper, we study the existence, multiplicity and uniqueness of positive solutions for the fourth order p-Laplacian boundary value problem
′′ p−1 ′′ ′′ (|u | u ) = f (t , u), u(2i) (0) = u(2i) (1) = 0,
i = 0, 1.
Here p > 0 and f ∈ C ([0, 1] × R+ , R+ ) (R+ := [0, ∞)). Based on a priori estimates achieved by utilizing properties of concave functions, we use fixed point index theory to establish our main results. © 2010 Elsevier Ltd. All rights reserved.
Keywords: p-Laplacian equation Positive solution Fixed point index Upper and lower solution Iterative sequence
1. Introduction In this paper, we study the existence, multiplicity and uniqueness of positive solutions for the fourth order p-Laplacian boundary value problem
(|u′′ |p−1 u′′ )′′ = f (t , u), t ∈ (0, 1), u(2i) (0) = u(2i) (1) = 0, i = 0, 1,
(1.1)
where p > 0 and f ∈ C ([0, 1] × R+ , R+ ) (R+ := [0, ∞)). Note that by a positive solution of (1.1) we mean a function u ∈ C 2 [0, 1] ∩ C 4 (0, 1) that solves (1.1) and satisfies |u′′ |p−1 u′′ ∈ C 2 (0, 1) and u(t ) > 0, t ∈ (0, 1). Second order differential equations with the p-Laplacian operator arise in modeling different physical and natural phenomena, which can be encountered in, for instance, non-Newtonian mechanics, nonlinear elasticity, glaciology, population biology, combustion theory, and nonlinear flow laws, see [1,2]. This explains why many papers have been published on existence of solutions for differential equations with the p-Laplacian operator; see, for instance, [3–14] and references therein. Fourth order boundary value problems, including those with the p-Laplacian operator, have their origin in beam theory [15,16], ice formation [17,18], fluids on lungs [19], brain warping [20,21], designing special curves on surfaces [22,20], etc. In beam theory, more specifically, a beam with a small deformation, a beam of a material which ✩ Supported by the NNSF of China (Grant 10871116 and 10971179) and the NSF of Shandong Province of China (Grant ZR2009AL014).
∗
Corresponding author. E-mail addresses: [email protected], [email protected] (J. Xu), [email protected], [email protected] (Z. Yang).
0362-546X/$ – see front matter © 2010 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2010.12.016
J. Xu, Z. Yang / Nonlinear Analysis 74 (2011) 2612–2623
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satisfies a nonlinear power-like stress and strain law and a beam with two-sided links which satisfies a nonlinear powerlike elasticity law, can be described by fourth order differential equations along with their boundary value conditions. For more background and applications, we refer the reader to the work by Timoshenko [23] on elasticity, the monograph by Soedel [24] on deformation of structure, and the work by Dulácska [25] on the effects of soil settlement. Due to their wide applications, the existence and multiplicity of positive solutions for fourth order p-Laplacian boundary value problems has also attracted increasing attention over the last decades; see [26–40] and references therein. The singular fourth-order p-Laplacian four-point boundary value problem
(|u′′ |p−1 u′′ )′′ = f (t , u(t )), 0 < t < 1, u(0) = u(1) − au(ξ ) = u′′ (0) = u′′ (1) − bu′′ (η) = 0
(1.2)
arises in beam theory and was studied by Zhang and Liu [36], in which the method of upper and lower solutions was used in establishing their main results on positive solutions of the above problem. Here p > 1, 0 < ξ , η < 1, f ∈ C ((0, 1) × (0, ∞), (0, ∞)) may be singular at t = 0 and/or t = 1 and u = 0. In [37], Zhang and Liu also considered the existence of positive solutions for (1.2), but with the original boundary value conditions in (1.2) replaced by the multi-point boundary conditions m−2
u(0) −
−
m−2
ai u(ξi ) = u(1) = u′′ (0) −
i=1
−
bi u′′ (ηi ) = u′′ (1) = 0,
(1.3)
i=1
where m ≥ 3, ai , bi , ξi , ηi ∈ (0, 1) (i = 1, 2, . . . , m − 2) are constants with i=1 ai < 1 and i=1 bi < 1, and f ∈ C ((0, 1) × R+ , R+ ) may be singular at t = 0 and/or t = 1. By virtue of monotone iterative techniques, they established a necessary and sufficient condition of the pseudo-C 3 [0, 1] positive solutions for their problem. In [31], Guo et al. discussed the existence and multiplicity of positive solutions for the fourth-order quasilinear singular differential equation
∑m−2
(|u′′ |p−2 u′′ )′′ = λg (t )f (u),
0 < t < 1,
∑m−2
(1.4)
sharing the boundary value conditions in (1.1), where λ is a positive parameter; f ∈ C ([0, ∞), (0, ∞)) is nondecreasing on [0, ∞), with f (u) > δ um for all u ∈ [0, ∞), where δ > 0 and m > p − 1 are two constants; g ∈ C ((0, 1), (0, ∞)) and g (t ) ̸≡ 0 on any subinterval of (0, 1). Applying fixed point index theory and the method of upper and lower solutions, they proved that there is a threshold λ∗ < ∞ for which (1.4) admits at least two positive solutions if 0 < λ < λ∗ , one positive solution if λ = λ∗ , and no positive solution if λ > λ∗ . Our problem (1.1) merely involves the simply supported boundary value conditions and a nonsingular nonlinearity f . Nevertheless, our methodology and results in this paper are entirely different from those in the papers cited above. We observe that if p = 1, then (1.1) reduces to the Lidstone problem
u(4) = f (t , u), t ∈ (0, 1), u(2i) (0) = u(2i) (1) = 0, i = 0, 1,
(1.5)
which can describe deformations of an elastic beam, with both ends simply supported, in an equilibrium state, and thus has been extensively studied (see [41–48] and references therein). It is natural to ask what connections do exist between (1.1) and (1.5). Some connections between them will be shown by repeatedly invoking Hölder’s inequalities in our proofs. We will use fixed point index theory to establish our main results based on a priori estimates achieved by utilizing some properties of concave functions, properties including Hölder’s inequalities and our new inequality (2.3) below. This paper is organized as follows. Section 2 contains some preliminary results. Section 3 is devoted to the existence of positive solutions for (1.1) and Section 4 to the multiplicity of positive solutions for (1.1). In Section 5, we discuss the uniqueness of positive solutions for (1.1), and prove that the unique positive solution can be uniformly approximated by an iterative sequence beginning with any function u which is continuous, nonnegative and not identically vanishing on [0, 1] (see Theorem 5.2). Section 6, finally, contains some remarks that serve to show some connections between (1.1) and (1.5). 2. Preliminaries Let E := C [0, 1],
P := {u ∈ E : u(t ) ≥ 0, ∀t ∈ [0, 1]},
Then (E , ‖ · ‖) is a real Banach space and P a cone on E. Let G(t , s) :=
t (1 − s), s(1 − t ),
0 ≤ t ≤ s ≤ 1, 0 ≤ s ≤ t ≤ 1.
‖u‖ := max |u(t )|. t ∈[0,1]
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J. Xu, Z. Yang / Nonlinear Analysis 74 (2011) 2612–2623
Then it is easy to see that (1.1) is equivalent to the fixed point equation u = Au, where A : P → P is defined by
(Au)(t ) :=
1
∫
1
∫
G(t , s)
G(s, τ )f (τ , u(τ ))dτ
1/p
ds.
(2.1)
0
0
Note that if f ∈ C ([0, 1] × R+ , R+ ), then A : P → P is a completely continuous operator. Lemma 2.1 ([49]). Let Ω ⊂ E be a bounded open set and A : Ω ∩ P → P is a completely continuous operator. If there exists v0 ∈ P \ {0} such that v − Av ̸= λv0 for all v ∈ ∂ Ω ∩ P and λ ≥ 0, then i(A, Ω ∩ P , P ) = 0, where i denotes the fixed point index on P. Lemma 2.2 ([49]). Let Ω ⊂ E be a bounded open set with 0 ∈ Ω . Suppose A : Ω ∩ P → P is a completely continuous operator. If v ̸= λAv for all v ∈ ∂ Ω ∩ P and 0 ≤ λ ≤ 1, then i(A, Ω ∩ P , P ) = 1. Lemma 2.3. Let ψ(t ) := sin(π t ). Then ψ ∈ P \ {0} and 1
∫
G(t , s)ψ(t )dt = 0
ψ(s), 2
π
1
∫
1
G(t , s)ψ(s)ds = 0
1
π2
ψ(t ).
(2.2)
Lemma 2.4. Let ϕ ∈ P be concave on [0, 1]. Then 1
∫
2
ϕ(t ) sin(π t )dt ≥
π2
0
‖ϕ‖.
(2.3)
Proof. We first suppose that there is a t0 ∈ (0, 1) such that ϕ(t0 ) = ‖ϕ‖. Then the concavity of ϕ implies 1
∫
ϕ(t ) sin(π t ) dt =
t0
∫
0
ϕ(t ) sin(π t ) dt +
∫
0
=
ϕ
0
≥ ϕ(t0 )
t t0
· t0 +
t0
∫
t0 − t t0
π2
· 0 sin(π t ) dt +
∫
1
ϕ
t sin(π t ) dt +
∫
1
(1 − t ) sin(π t ) dt
1−t 1 − t0
t0
0
2
≥
ϕ(t ) sin(π t ) dt t0
t0
∫
1
· t0 +
t − t0 1 − t0
· 1 sin(π t ) dt
t0
‖ϕ‖,
where we have used the simple fact: t
∫
s sin(π s)ds +
1
∫
0
(1 − s) sin(π s) ds ≥ t
2
π2
,
t ∈ [0, 1].
The remaining situations (with ϕ(0) = ‖ϕ‖ or with ϕ(1) = ‖ϕ‖) can be analogously handled. The proof is completed.
Lemma 2.5. Let ψ(t ) := sin(π t ). Then for every w ∈ P \ {0}, there are two positive numbers bw , aw such that the inequality aw ψ(t ) ≤
1
∫
G(t , s)w(s)ds ≤ bw ψ(t ) 0
holds for all t ∈ [0, 1]. Proof. Let w1 (t ) :=
1 0
G(t , s)w(s)ds. Then w1 ∈ C 2 [0, 1] ∩ P, and
w1 (0) = w1 (1) = ψ(0) = ψ(1) = 0. Now w ∈ P \ {0} implies 1
∫
sw(s)ds > 0,
1
∫
0
(1 − s)w(s)ds > 0, 0
w1 ( t ) > 0, ψ(t )
t ∈ (0, 1).
By l’Hôpital’s rule we have
− w1 (t ) lim = lim + + ψ(t ) t →0 t →0
t 0
sw(s)ds +
1
1
t
0
(1 − s)w(s)ds = π cos(π t )
(1 − s)w(s)ds >0 π
J. Xu, Z. Yang / Nonlinear Analysis 74 (2011) 2612–2623
2615
and
− w1 ( t ) = lim lim − − ψ(t ) t →1 t →1
t 0
sw(s)ds +
1
1
t
0
(1 − s)w(s)ds = π cos(π t )
sw(s)ds
π
> 0.
w (t )
1 The last inequalities say that ψ( may be viewed as a positive continuous function on [0, 1]. Hence there are two positive t) w1 (t ) numbers bw , aw , for which aw ≤ ψ(t ) ≤ bw holds for all t ∈ [0, 1]. This completes the proof.
Lemma 2.6 (Hölder’s Inequalities). Let θ > 0 and ϕ ∈ P. Then 1
∫
ϕ(t )dt
θ
1
∫ ≤
(ϕ(t ))θ dt
0
0
if θ ≥ 1, and 1
∫
ϕ(t )dt
θ
1
∫ ≥
0
(ϕ(t ))θ dt
0
if 0 < θ ≤ 1. 3. Existence of positive solutions ∗
Let p∗ := max{1, p}, p∗ := min{1, p}, βp := π 4p , αp := π 4p∗ . We denote Bρ := {u ∈ E : ‖u‖ < ρ} for ρ > 0 in the sequel. (H1) (H2) (H3) (H4) (H5)
f ∈ C ([0, 1] × R+ , R+ ). There exist a1 > βp and c > 0 such that f (t , x) ≥ a1 xp − c for all x ∈ R+ and t ∈ [0, 1]. There exist b1 ∈ (0, αp ) and r > 0 such that f (t , x) ≤ b1 xp for all x ∈ [0, r ] and t ∈ [0, 1]. There exist a2 > βp and r > 0 such that f (t , x) ≥ a2 xp for all x ∈ [0, r ] and t ∈ [0, 1]. There exist b2 ∈ (0, αp ) and c > 0 such that f (t , x) ≤ b2 xp + c for all x ∈ R+ and t ∈ [0, 1].
Theorem 3.1. Suppose that (H1), (H2) and (H3) are satisfied. Then (1.1) has at least one positive solution. Proof. Let M1 := {u ∈ P : u = Au + λψ for some λ ≥ 0},
where ψ(t ) := sin(π t ). We claim that M1 is bounded. Indeed, u ∈ M1 implies that u is concave and that u(t ) ≥ (Au)(t ), t ∈ [0, 1]. By definition, we have u( t ) ≥
∫
1
G(t , s)
0
1
∫
G(s, τ )f (τ , u(τ ))dτ
1/p ds
0
for all u ∈ M1 , t ∈ [0, 1]. Now the inequalities 0 < p∗ ≤ 1,
0 < p∗ /p ≤ 1
(3.1)
and 0 ≤ G(t , s) < 1,
0 ≤ G(s, τ ) < 1,
t , s, τ ∈ [0, 1],
(3.2)
along with Lemma 2.6 and (H2), enable us to obtain u (t ) ≥ p∗
∫
1
G(t , s) 0
1
∫
1
∫
1
∫
≥ 0
p∗ ds
G(t , s)G(s, τ )f p∗ /p (τ , u(τ ))dτ ds
0 1
∫
G(s, τ )f (τ , u(τ ))dτ
1/p
0
≥ 0
1
∫
p /p
G(t , s)G(s, τ )(a1∗ up∗ (τ ) − c p∗ /p )dτ ds
0
for all u ∈ M1 and t ∈ [0, 1], the simple inequality
(x + y)p∗ /p ≤ xp∗ /p + yp∗ /p ,
x, y ∈ R+
(3.3)
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J. Xu, Z. Yang / Nonlinear Analysis 74 (2011) 2612–2623
being used. Multiply (3.3) by sin(π t ) and integrate over [0, 1] and use (2.2) to obtain 1
∫
p /p
up∗ (t ) sin(π t ) dt ≥ a1∗ π −4
1
∫
0
up∗ (t ) sin(π t ) dt − 2c p∗ /p π −5 ,
0
so that 1
∫
p /p
up∗ (t ) sin(π t ) dt ≤ 2π −1 c p∗ /p /(a1∗
− π 4 ) := N1
0
for all u ∈ M1 . Recall that every u ∈ M1 is concave. So is up∗ with p∗ ∈ (0, 1]. Now Lemma 2.4 yields
‖up∗ ‖ ≤ N1 π 2 /2 for all u ∈ M1 , which implies the boundedness of M1 , as claimed. Taking R > sup{‖u‖ : u ∈ M1 }, we have u − Au ̸= λψ,
∀u ∈ ∂ BR ∩ P , λ ≥ 0.
Applying Lemma 2.1 we obtain i(A, BR ∩ P , P ) = 0.
(3.4)
Let M2 := {u ∈ Br ∩ P : u = λAu for some λ ∈ [0, 1]}.
We shall prove M2 = {0}. Indeed, if u ∈ M2 , we have u(t ) ≤ (Au)(t ) =
1
∫
G(t , s)
1
∫
0
G(s, τ )f (τ , u(τ ))dτ
1/p
ds,
∀u ∈ Br ∩ P .
0
Now the inequalities p∗ ≥ 1,
p∗ /p ≥ 1
(3.5)
and (3.2), along with Lemma 2.6 and (H3), enable us to obtain p∗
u (t ) ≤
∫
1
G(t , s) 0
∫
1
∫
1
∫
0
1
ds
∗ G(t , s)G(s, τ )f p /p (τ , u(τ ))dτ ds
0 1
≤ 0
G(s, τ )f (τ , u(τ ))dτ
p∗
1/p
0
≤ ∫
1
∫
p∗ /p p∗
G(t , s)G(s, τ )b1
u (τ )dτ ds
(3.6)
0
for all u ∈ M2 , t ∈ [0, 1]. Multiply both sides of the above by sin(π t ) and integrate over [0, 1] and use (2.2) to obtain 1
∫
p∗
u (t ) sin(π t ) dt ≤
p∗ /p b1
0
π
−4
1
∫
∗
up (t ) sin(π t ) dt , 0
so that 1
∫
∗
up (t ) sin(π t ) dt = 0 0
for all u ∈ M2 . This implies M2 = {0} and thus u ̸= λAu,
∀u ∈ ∂ Br ∩ P , λ ∈ [0, 1].
(3.7)
Now Lemma 2.2 yields i(A, Br ∩ P , P ) = 1.
(3.8)
Combining this with (3.4) gives i(A, (BR \ Br ) ∩ P , P ) = 0 − 1 = −1. Hence the operator A has at least one fixed point on (BR \ Br ) ∩ P, and thus (1.1) has at least one positive solution. This completes the proof.
J. Xu, Z. Yang / Nonlinear Analysis 74 (2011) 2612–2623
2617
Theorem 3.2. Suppose that (H1), (H4) and (H5) are satisfied. Then (1.1) has at least one positive solution. Proof. Let M3 := {u ∈ Br ∩ P : u = Au + λψ for some λ ≥ 0}.
We claim M3 ⊂ {0}. Indeed, if u ∈ M3 , then we have u ≥ Au by definition. (H4) implies u( t ) ≥
∫
1
G(t , s)
1
∫
G(s, τ )a2 u (τ )dτ p
1/p
ds.
0
0
Recall (3.1) and (3.2). Now Lemma 2.6 implies p /p
up∗ (t ) ≥ a2∗
1
∫
1
∫
G(t , s)G(s, τ )up∗ (τ )dτ ds 0
0
for all u ∈ M3 , t ∈ [0, 1]. Multiply both sides of the above by sin(π t ) and integrate over [0, 1] and use (2.2) to obtain 1
∫
u (t ) sin(π t ) dt ≥ p∗
p /p a2∗
π
−4
0
so that
1
∫
up∗ (t ) sin(π t ) dt , 0
1 0
up∗ (t ) sin(π t ) dt = 0 for all u ∈ M3 , whence M3 ⊂ {0}, as claimed. As a result of this, we have
u − Au ̸= λψ,
∀u ∈ ∂ Br ∩ P , λ ≥ 0.
Now Lemma 2.1 yields i(A, Br ∩ P , P ) = 0.
(3.9)
Let M4 := {u ∈ P : u = λAu for some λ ∈ [0, 1]}.
We assert M4 is bounded. Indeed, if u ∈ M4 , then u is concave and u ≤ Au, which can be written in the form u( t ) ≤
∫
1
G(t , s)
1
∫
0
G(s, τ )f (τ , u(τ ))dτ
1/p
ds.
0
Recall (3.2) and (3.5). Now by Lemma 2.6 and (H5), we obtain p∗
u (t ) ≤
∫
1
G(t , s) 0
∫
1
∫
1
∫
0
1
ds
∗ G(t , s)G(s, τ )f p /p (τ , u(τ ))dτ ds
0 1
≤ 0
∗ G(t , s)G(s, τ )(b2 up (τ ) + c )p /p dτ ds
0 1
∫
G(s, τ )f (τ , u(τ ))dτ
p∗
1/p
0
≤ ∫
1
∫
1
∫
0
p∗ /p p∗
G(t , s)G(s, τ )(b3
≤
p∗ /p
u (τ ) + c1
)dτ ds
(3.10)
0
for all u ∈ M4 , b3 ∈ (b2 , αp ) and c1 > 0 being chosen so that ∗ /p
(b2 z + c )p
p∗ /p p∗ /p
≤ b3
z
p∗ /p
+ c1
for all z ∈ R+ . Multiply both sides of (3.10) by sin(π t ) and integrate over [0, 1] and use (2.2) to obtain 1
∫
∗
p∗ /p
up (t ) sin(π t ) dt ≤ b3
π −4
1
∫
0
p∗ /p
∗
up (t ) sin(π t ) dt + 2c1
π −5 ,
0
so that 1
∫
p∗ /p
∗
up (t ) sin(π t ) dt ≤ 2π −1 c1
p∗ /p
/(π 4 − b3
) := N2
0
for all u ∈ M4 . This, together with Lemma 2.6, leads to 1
∫
u(t ) sin(π t ) dt ≤ 0
1
∫
p∗
p∗
u (t )(sin(π t )) dt 0
1/p∗
1/p∗
≤ N2
(3.11)
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J. Xu, Z. Yang / Nonlinear Analysis 74 (2011) 2612–2623
for all u ∈ M4 . Recall u is concave. Now the boundedness of M4 , as asserted, follows by Lemma 2.4 and (3.11). Taking R > sup{‖u‖ : u ∈ M4 }, we have u ̸= λAu,
∀u ∈ ∂ BR ∩ P , λ ∈ [0, 1].
(3.12)
Now Lemma 2.2 yields i(A, BR ∩ P , P ) = 1.
(3.13)
Combining this with (3.9) gives i(A, (BR \ Br ) ∩ P , P ) = 1 − 0 = 1. Hence the operator A has at least one fixed point on (BR \ Br ) ∩ P, and thus (1.1) has at least one positive solution. This completes the proof. 4. Multiplicity of positive solutions In this section, we will study the multiplicity of positive solutions for (1.1). To this end, we need the following conditions. (H6) There are ω > 0 and ϱ ∈ (0, 384/5) such that the inequality f (t , u) < ϱp/p ωp holds for all u ∈ [0, ω] and t ∈ [0, 1]. (H7) There are ω > 0, δ ∈ (0, 1/2) and ζ > 180δ −p∗ such that the inequality f (t , u) > ζ p/p∗ ωp holds for all u ∈ [δω, ω] and t ∈ [δ, 1 − δ]. ∗
Theorem 4.1. Suppose that (H1), (H2), (H4) and (H6) are satisfied. Then (1.1) has at least two positive solutions. Proof. Inequalities (3.2) and (3.5), along with Lemma 2.6 and (H6), enable us to obtain p∗
(Au) (t ) =
∫
1
G(t , s)
0
∫
1
1
∫
G(s, τ )f (τ , u(τ ))dτ
p∗
1/p ds
0 1
∫
≤ 0
∗ G(t , s)G(s, τ )f p /p (τ , u(τ ))dτ ds
0
≤ ϱω
p∗
1
∫
1
∫
0
G(t , s)G(s, τ )dτ ds 0
∗
= ≤
ϱωp (t 4 − 2t 3 + t ) 24
5ϱ 384
ω
p∗
∗
< ωp
for all u ∈ Bω ∩ P and t ∈ [0, 1]. This implies ‖Au‖ < ‖u‖ for all u ∈ Bω ∩ P, and, in turn, u ̸= λAu for all u ∈ ∂ Bω ∩ P, λ ∈ [0, 1]. Now Lemma 2.2 yields i(A, Bω ∩ P , P ) = 1.
(4.1)
On the other hand, in view of (H2) and (H4), we may choose R > ω and r ∈ (0, ω) so that (3.4) and (3.9) hold (see the proofs of Theorems 3.1 and 3.2). Combining (3.4), (3.9) and (4.1), we obtain i(A, (BR \ Bω ) ∩ P , P ) = 0 − 1 = −1,
i(A, (Bω \ Br ) ∩ P , P ) = 1 − 0 = 1.
Hence A has at least two fixed points, one on (BR \ Bω ) ∩ P and the other on (Bω \ Br ) ∩ P, and thus (1.1) has at least two positive solutions. The proof is completed. To prove Theorem 4.2 below, we need an extra cone P1 , defined by
P1 :=
u∈P :
min u(t ) ≥ δ‖u‖ ,
δ≤t ≤1−δ
where δ ∈ (0, 1/2) is given by (H7). Lemma 4.1. A(P1 ) ⊂ P1 .
J. Xu, Z. Yang / Nonlinear Analysis 74 (2011) 2612–2623
2619
Proof. After some computations, we have G(s, s) ≥ G(t , s) ≥ δ G(s, s),
G(s, τ ) ≥ s(1 − s)G(τ , τ )
(4.2)
for all t ∈ [δ, 1 − δ], s, τ ∈ [0, 1]. Now the preceding inequalities, together with the definition of A, imply
p∗ ∫ min (Au)(t ) = min
δ≤t ≤1−δ
δ≤t ≤1−δ
∫ ≥ δ
1
G(t , s)
=δ
1
G(s, s)
≥δ
1
∫
p∗ ds
G(s, τ )f (τ , u(τ ))dτ
1/p
p∗ ds
0
∫
1
G(s, s)
∫
1
∫
G(s, τ )f (τ , u(τ ))dτ
1/p
p∗ ds
0
0
p∗
G(s, τ )f (τ , u(τ ))dτ
1/p
0
0
0
p∗
1
∫
1
G(z , s)
1
∫
G(s, τ )f (τ , u(τ ))dτ
1/p
p∗ ds
0
0
= δ p∗ (Au)p∗ (z ) for all u ∈ P and z ∈ [0, 1], and thus minδ≤t ≤1−δ (Au)(t ) ≥ δ‖Au‖ for all u ∈ P. Therefore, we have A(P ) ⊂ P1 and, in particular, A(P1 ) ⊂ P1 . This completes the proof. Theorem 4.2. Suppose that (H1), (H3), (H5) and (H7) are satisfied. Then (1.1) has at least two positive solutions. Proof. By Lemma 4.1, we have A(P1 ) ⊂ P1 . Observe that ω ≥ u(t ) ≥ δω for all u ∈ ∂ Bω ∩ P1 , ω > 0 being determined by (H7). Combining this with (4.2) and (H7), one infers
(‖Au‖)
p∗
[
max (Au)(t )
≥
]p∗
t ∈[δ,1−δ]
= ≥ δ p∗
1
∫
≥ δ p∗
30
1
∫
0
δ p∗
G(s, τ )f (τ , u(τ ))dτ
1/p
p∗ ds
0
G(s, s)G(s, τ )f p∗ /p (τ , u(τ ))dτ ds
0 1
∫
1
1
∫
0
∫
0
=
G(t , s)
max t ∈[δ,1−δ]
≥
1
∫
G(s, s)s(1 − s)G(τ , τ )f p∗ /p (τ , u(τ ))dτ ds
0 1
∫
G(τ , τ )ζ ωp∗ dτ 0
ζ ωp∗ δ p∗ 180
> ωp∗
for all u ∈ ∂ Bω ∩ P1 , and thus ‖Au‖ > ‖u‖ for all u ∈ ∂ Bω ∩ P1 , so that u − Au ̸= λψ,
∀u ∈ ∂ Bω ∩ P1 , λ ≥ 0.
Now Lemma 2.1 yields i(A, Bω ∩ P1 , P1 ) = 0.
(4.3)
On the other hand, in view of (H3) and (H5), we may choose R > ω and r ∈ (0, ω) so that (3.7) and (3.12) hold (see the proofs of Theorems 3.1 and 3.2). Naturally, (3.7) and (3.12) still hold, if P is replaced by P1 . Lemma 2.2 then yields i(A, Br ∩ P1 , P1 ) = 1,
i(A, BR ∩ P1 , P1 ) = 1.
Combining the above with (4.3), we obtain i(A, (BR \ Bω ) ∩ P1 , P1 ) = 1 − 0 = 1,
i(A, (Bω \ Br ) ∩ P1 , P1 ) = 0 − 1 = −1.
Hence A has at least two fixed points, one on (BR \ Bω ) ∩ P1 and the other on (Bω \ Br ) ∩ P1 , and thus (1.1) has at least two positive solutions. The proof is completed.
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J. Xu, Z. Yang / Nonlinear Analysis 74 (2011) 2612–2623
5. Uniqueness of positive solutions and iterative convergence In this section, we will establish the uniqueness of positive solutions for (1.1) when f grows p-sublinearly at 0 and at ∞, i.e. (H4) and (H5), together with (H9) below, are satisfied. Moreover, we shall prove that the unique positive solution can be uniformly approximated by an iterative sequence {An u} beginning with any function u ∈ P \ {0} (see Theorem 5.2). We now list our hypotheses in this section. (H8) f (t , u) is increasing in u, that is, the inequality f (t , u1 ) ≤ f (t , u2 ) holds for all u1 , u2 ∈ R+ satisfying u1 ≤ u2 . (H9) f (t , λu) > λp f (t , u) for each λ ∈ (0, 1), u > 0 and t ∈ [0, 1]. Theorem 5.1. If (H1), (H4), (H5), (H8) and (H9) hold, then (1.1) has exactly one positive solution. Proof. By Theorem 3.2, (1.1) has at least one positive solution. It then remains to prove the uniqueness of positive solutions for (1.1). Indeed, if u1 and u2 are two positive solutions of (1.1), then ui ( t ) =
1
∫
G(t , s)
1
∫
G(s, τ )f (τ , ui (τ ))dτ
1/p ds
0
0
and wi (t ) := ( 0 G(t , s)f (s, ui (s))ds)1/p ∈ P \ {0} (i = 1, 2). By Lemma 2.5, there are four positive numbers bi , ai for which a ai ψ ≤ ui ≤ bi ψ (i = 1, 2) and thus u2 ≥ b2 u1 . Let
1
1
µ0 := sup{µ > 0 : u2 ≥ µu1 }. Clearly µ0 > 0 and u2 ≥ µ0 u1 . We claim µ0 ≥ 1. If the claim is false, then µ0 < 1 and u2 ( t ) ≥
1
∫
G(t , s)
1
∫
0
G(s, τ )f (τ , µ0 u1 (τ ))dτ
ds
0 1
∫
1/p
G(t , s)g (s)ds + µ0
= 0
1
∫
G(t , s) 0
1
∫
G(s, τ )f (τ , u1 (τ ))dτ
1/p ds
0
1
∫
G(t , s)g (s)ds + µ0 u1 (t ),
= 0
where g (s) :=
1
∫
G(s, τ )f (τ , µ0 u1 (τ ))dτ
1/p
0
− µ0
1
∫
G(s, τ )f (τ , u1 (τ ))dτ
1/p
Now (H9) implies g ∈ P \ {0}. By Lemma 2.5, there is a positive number ε such that for all t ∈ [0, 1]. This leads to u2 ( t ) ≥
ε
b1
.
0
1 0
G(t , s)g (s)ds ≥ εψ(t ) = ε sin(π t )
u1 (t ) + µ0 u1 (t ),
a contradiction with the definition of µ0 . As a result, µ0 ≥ 1 and thus u2 ≥ u1 . Similarly u1 ≥ u2 . Thus we have u1 = u2 . This completes the proof. To prove Theorem 5.2 below, we have to consider upper solutions and lower solutions of the fixed point equation u = Au.
(5.1)
To this end, we need the following two lemmas. Lemma 5.1. Let (H1) and (H4) hold, with r > 0 being determined by (H4). Then for any ε ∈ (0, r ], vε (t ) := ε(sin(π t ))1/p∗ is a lower solution of (5.1), that is, Avε ≥ vε . Proof. For any ε ∈ (0, r ], we have by (H4) and Lemma 2.6,
(Avε ) (t ) = p∗
∫
1
G(t , s) 0
G(t , s)
p /p
) dτ
1/p
p∗ ds
G(s, τ )a2 ε p (sin(π τ ))p/p∗ dτ
p∗ /p ds
0 1
∫ 0
p /p
G(s, τ )f (τ , ε(sin(π τ ))
1
∫
0
≥ a2∗ ε p∗
1/p∗
0
1
∫ ≥
1
∫
∫
1
G(t , s)G(s, τ ) sin(π τ )dτ ds
0
= a2∗ ε p∗ π −4 sin(π t )
(5.2)
J. Xu, Z. Yang / Nonlinear Analysis 74 (2011) 2612–2623
2621
and thus 1/p
(Avε )(t ) ≥ a2 π −4/p∗ vε (t ) ≥ vε (t ). This completes the proof.
Lemma 5.2. Let (H1) and (H5) hold, with b2 and c determined by (H5). Choose b3 ∈ (b2 , αp ) and c1 > 0 so that ∗ /p
(b2 z + c )p
p∗ /p p∗ /p
≤ b3
p∗ /p
+ c1
z
for all z ≥ 0 and let ∗ 1/p
51/p c1 π −1/p 96−1/p
m0 :=
∗
∗
1/p
.
1 − b3 π −4/p
∗
Then for any m ≥ m0 , wm (t ) := m(sin(π t ))1/p is an upper solution of (5.1), that is, Awm ≤ wm . ∗
Proof. Since 1
∫ 0
1
∫
G(t , s)G(s, τ )dsdτ =
t (1 − t )(1 + t − t 2 ) 24
0
≤
5 sin(π t ) 96π
for all t ∈ [0, 1], it follows from (H5) and Lemma 2.6 that ∗
(Awm )p (t ) ≤
1
∫ 0
∗ ∗ G(t , s)G(s, τ )(b2 mp sinp/p (π τ ) + c )p /p dτ ds
0 1
∫
1
∫
1
∫
p∗ /p
G(t , s)G(s, τ )(b3
≤ 0 0 p∗ /p
∗
p∗ /p
∗
p∗ /p
π −4 mp sin(π t ) + c1
= b3 ≤ b3
p∗ /p
∗
mp sin(π τ ) + c1
24−1 (t 4 − 2t 3 + t )
p∗ /p
π −4 mp sin(π t ) + 5c1
)dτ ds
π −1 96−1 sin(π t )
(5.3)
and thus p∗ /p
(Awm )(t ) ≤ (b3
p∗ /p
∗
π −4 mp + 5c1
1/p
π −1 96−1 )1/p (sin(π t ))1/p
∗ 1/p
∗
∗
≤ (b3 π −4/p m + 51/p c1 π −1/p 96−1/p )(sin(π t ))1/p ≤ wm (t ) ∗
∗
for any m ≥ m0 and t ∈ [0, 1]. This completes the proof.
∗
∗
Theorem 5.2. Let all the conditions in Theorem 5.1 hold and u∗ the unique positive solution of (1.1). Then for any u ∈ P \ {0}, the iterative sequence {An u} converges to u∗ as n → ∞, that is, limn→∞ ‖An u − u∗ ‖ = 0. Proof. Let (Fu)(t ) := f (t , u(t )). Then F : P → P, and (H4) and (H8) imply that F (P \ {0}) ⊂ P \ {0}. Now for any u ∈ P \ {0}, by Lemma 2.5, there is an au > 0 such that 1
∫
G(s, τ )f (τ , u(τ ))dτ ≥ au sin(π s) 0
for all s ∈ [0, 1] and
(Au)p∗ (t ) ≥
1
∫
G(t , s)(au sin(π s))p∗ /p ds
0
≥ apu∗ /p
∫
1
G(t , s) sin(π s) ds
0
= apu∗ /p π −2 sin(π t ) for all t ∈ [0, 1], so that
(Au)(t ) ≥ a1u/p π −2/p∗ (sin(π t ))1/p∗ . 1/p
Note that we may assume εu := au π −2/p∗ ≤ r, where r is determined in (H4). This, along with Lemma 5.1, implies that vεu (t ) := εu (sin(π t ))1/p∗ is a lower solution of (5.1) and Au ≥ vεu . On the other hand, for any u ∈ P \ {0}, letting du := max{f (t , x) : 0 ≤ t ≤ 1, 0 ≤ x ≤ ‖u‖},
2622
J. Xu, Z. Yang / Nonlinear Analysis 74 (2011) 2612–2623
we have du s(1 − s)
1
∫
G(s, τ )f (τ , u(τ ))dτ ≤
2
0
≤
du sin(π s) 2π
for all s ∈ [0, 1] and thus p∗
(Au) (t ) =
∫
1
G(t , s)
G(s, τ )f (τ , u(τ ))dτ
p∗
1/p ds
0
0
∫
1
∫
1
G(t , s)
≤
du sin(π s)
∗ /p
≤ dup
ds
2π
0 ∗ /p
(2π )−p
p∗
1/p
1
∫
G(t , s) sin(π s) ds 0
∗ /p
= dup
∗ /p
(2π )−p
π −2 sin(π t ),
so that
(Au)(t ) ≤ d1u/p (2π )−1/p π −2/p (sin(π t ))1/p ≤ mu (sin(π t ))1/p ∗
∗
∗
1/p
for all t ∈ [0, 1], where mu := max{m0 , du (2π )−1/p π −2/p } ≥ m0 , m0 being given in Lemma 5.2. Now Lemma 5.2 implies ∗ that wmu (t ) := mu (sin(π t ))1/p is an upper solution of (5.1) with Au ≤ wmu . So far we have shown that vεu ≤ Au ≤ wmu for any u ∈ P \ {0}, with vεu a lower solution of (5.1) and wmu a upper solution of (5.1). (H8) implies that A is an increasing operator and hence An vεu ≤ An+1 u ≤ An wmu for all positive integer n. As is well known [49], as n → ∞, {An vεu } converges to the smallest fixed point of A on [vεu , wmu ] and {An wmu } to the greatest fixed point of A on [vεu , wmu ]. Theorem 5.1 implies that u∗ is the unique fixed point of A on [vεu , wmu ]. Therefore limn→∞ An vεu = limn→∞ An wmu = u∗ , whence limn→∞ An u = u∗ for any u ∈ P \ {0}. The proof is completed. ∗
Remark 5.1. Let f (t , u) := uq , where q ∈ (0, p). Then (H1), (H4), (H5), (H8) and (H9) hold. By Theorems 5.1 and 5.2, (1.1) has exactly one positive solution, and the iterative sequence un+1 (t ) :=
1
∫
G(t , s) 0
1
∫
G(s, τ )uqn (τ )dτ
1/p
ds,
t ∈ [0, 1], n = 0, 1, 2, . . . ,
0
uniformly converges to the unique positive solution u∗ for every function u0 ∈ P \ {0}. 6. Concluding remarks As was mentioned in the Introduction, some connections should exist between (1.1) and (1.5). This will be clearer from Theorems 6.1 and 6.2 below. As the prelude to them, the following conditions are formulated on letting p = 1 in defining αp and βp .
(H2)′ (H3)′ (H4)′ (H5)′
There exist a1 There exist b1 There exist a2 There exist b2
> π 4 and c > 0 such that f (t , x) ≥ a1 xp − c for all x ∈ R+ and t ∈ [0, 1]. ∈ (0, π 4 ) and r > 0 such that f (t , x) ≤ b1 xp for all x ∈ [0, r ] and t ∈ [0, 1]. > π 4 and r > 0 such that f (t , x) ≥ a2 xp for all x ∈ [0, r ] and t ∈ [0, 1]. ∈ (0, π 4 ) and c > 0 such that f (t , x) ≤ b2 xp + c for all x ∈ R+ and t ∈ [0, 1].
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