On singular p-Laplacian boundary value problems involving integral boundary conditions

On singular p-Laplacian boundary value problems involving integral boundary conditions

Dang Dinh Hai

and Xiao Wang

Department of Mathematics and Statistics, Mississippi State University Mississippi State, MS 39762, USA

Received 13 January 2019, appeared 9 December 2019 Communicated by Gennaro Infante

Abstract. We prove the existence of positive solutions for thep-Laplacian equations

−(φ(u⁰))⁰ =λf(t,u), t∈(0, 1)

with integral boundary conditions. Here λ is a positive parameter, φ(s) = |s|^p−2s, p > _1, f : (_{0, 1})×(_0,∞) → R is p-superlinear or p-sublinear at ∞ in the second variable and is allowed be singular att=0, 1 andu=0.

Keywords: p-Laplacian, integral boundary conditions, positive solutions.

2010 Mathematics Subject Classification: 34B15, 34B18.

1 Introduction

Consider the one-dimensional p-Laplacian equation

−(φ(u⁰))⁰ = λf(t,u), t∈ (0, 1) (1.1) with boundary conditions

au(0)−bu⁰(0) =

Z ₁

0 g(t)u(t)dt, u⁰(1) =0, (1.2) or

au(0)−bu⁰(0) =

Z ₁

0

g(t)u(t)dt, u(1) =0, (1.3) where φ(s) = |s|^p−2s,p> 1,a> 0,b≥0,g:(0, 1)→[0,∞),f :(0, 1)×(0,∞)→_{R, and}λis a positive parameter.

Equation (1.1) arises in some physical models such as non-Newtonian fluids, chemical reactions, and population biology, see e.g. [2,3,7,8]. The integral boundary conditions occur in thermal conduction, semiconductor and hydrodynamic problems [5,6,11].

BCorresponding author. Email: dang@math.msstate.edu

In [16], Zhang and Feng studied the existence of positive solutions to (1.1)–(1.2)) with b> 0 for a certain range for λwhen f(t,u) =ω(t)F(t,u)is nonnegative and nonsingular in uunder a variety of assumptions involving lim_u→0^{+ F}_u⁽_p^t,u−1⁾ and/or lim_u→_∞ ^F(t,u)

u^p−1 . In particular, they showed in [16, Theorem 1.3] that (1.1)–(1.2) has a positive solution uλ for all λ > 0 if either lim_u→0^{+ F}_u⁽_p^t,u−1⁾ = 0 and lim_u→_∞ ^F(t,u)

u^p−1 = _∞ (p-superlinear) uniformly for t ∈ [0, 1], or lim_u→0^{+ F}_u⁽_p^t,u−1⁾ = _∞ and lim_u→_∞ ^F(t,u)

u^p−1 = 0 (p-sublinear) uniformly for t ∈ [0, 1]. In addition, lim_λ→0ku_λk_∞ = _∞in the former case and lim_λ→0ku_λk_∞ =0 in the latter case. The approach used in [16] was fixed point theory in a cone, and the nonnegativity assumption of f there was essential to ensure that the equivalent fixed point mapping maps the cone of nonnega- tive continuous functions into itself. In this note, we shall establish the existence of positive solutions to (1.1) with boundary condition (1.2) or (1.3) when f(t,u)can be±_∞atu=0 and is either p-superlinear or p-sublinear at ∞, which has not been considered in the literature to the best of our knowledge.

Our results when applied to the model equation

−(φ(u⁰))⁰ = ^λ t^α

c

u^δ +u^ρ

, t∈ (0, 1), (1.4)

where c ∈ _R, α,δ ∈ [0, 1),ρ > 0, give the existence of a large positive solution to (1.4) with boundary conditions (1.2) or (1.3) for λsmall whenρ > p−1, or for λlarge whenρ < p−1.

We refer to [4,9,10,12–15] for results related to (1.1) with integral boundary conditions.

Define q(t) = t if (1.2) holds and q(t) = min(t, 1−t) if (1.3) holds. We shall make the following assumptions:

(A1) g:(0, 1)→[0,∞)is integrable andR1

0 g(t)dt<a.

(A2) f :(0, 1)×(0,∞)→_Ris a Carathéodory function, that is f(.,z)is measurable forz >0 and f(t,·)is continuous for a.e.t∈ (0, 1).

(A3) There exists a constant δ ∈ [0, 1) such that for each k > 0, there exists a function γ_k: (0, 1)→[0,∞)withγ_k/q^δ ∈ L¹(0, 1)such that

|f(t,z)| ≤γ_k(t)z^−δ for a.e.t ∈(0, 1)andz ∈(0,k].

(A4) There existγ∈ L¹(0, 1)withγ>0 a.e. on(0, 1)andν∈ {0,∞} such that

zlim→_∞

f(t,z) γ(t)z^p−1 =ν, uniformly for a.e. t∈ (0, 1).

By a solution of (1.1) with boundary condition (1.2) (resp. (1.3)), we mean a function u ∈ C¹[0, 1] with φ(u⁰)absolutely continuous on [0, 1], and satisfying (1.2) (resp. (1.3)). Our main result is the following.

Theorem 1.1. Let (A1)–(A3) hold.

(i) If (A4) holds with ν = _∞then there exists a constantλ₀ > 0such that for λ< λ₀, (1.1) with boundary condition(1.2)) or(1.3)has a positive solution u_λwithku_λk_∞ →_∞asλ→0⁺. (ii) If (A4) holds withν=0andlimz→_∞ f(t,z) =_∞uniformly for a.e. t∈(0, 1)then there exists a

constantλ˜₀>0such that forλ>λ^˜₀, (1.1)with boundary condition(1.2)or(1.3)has a positive solution u_λ withku_λk_∞ →_∞asλ→_∞.

2 Preliminary result

We shall denote the norm in L^p(0, 1)byk · k_p.

Lemma 2.1. Let h∈ L¹(0, 1)and let (A1) hold. Then the equation

−(φ(u⁰))⁰ =h on(0, 1) (2.1) with boundary condition(1.2)or(1.3)has a unique solution u≡Th∈ C¹[_{0, 1}]. Furthermore,

kuk_∞ ≤ Mφ⁻¹(khk₁), (2.2)

where M=max _a−k^a+_g^b_k

1, 2^p−11

,and the map T: L¹(0, 1)→C[0, 1]is completely continuous.

Proof. Suppose (1.2) holds. By integrating (2.1) and using (1.2), it follows that (2.1) with boundary condition (1.2) has a unique solutionu, given by

u(t) =C+

Z _t

0 φ⁻¹ _{Z 1}

s h

ds, where

C=

bφ⁻¹ R1

0 h +R1

0 g(t)Rt 0φ⁻¹

R1 s h

ds dt a−R1

0 g . (2.3)

Since|C| ≤ ^{(b+kgk1)φ−1(khk1)}

a−kgk₁ , it follows that kuk_∞ ≤ ^a+b

a− kgk₁^φ

−1(khk₁)_,

i.e. (2.2) holds. Since |u|_C1 = kuk_∞+ku⁰k_∞ ≤ (M+1)φ⁻¹(khk₁), it follows that T maps bounded sets in L¹(0, 1)into bounded sets inC¹[0, 1]and hence relatively compact subsets in C[_{0, 1}]. To show continuity of T, let (_hn) ⊂ L¹(_{0, 1}) _andh ∈ L¹(_{0, 1})be such that hn → h in L¹(0, 1). Letun= Thnandu=Th. Then

un(t) =Cn+

Z _t

0 φ⁻¹ Z ₁

s hn

ds,

where Cn is given by (2.3) with h replaced by hn. It is easy to see that Cn → C and hence u_n→uin C[0, 1].

Suppose next that (1.3) holds. By integrating (2.1) and using (1.3), it follows that (2.1) with boundary condition (1.3) has a unique solutionu, given by

u(t) =

Z ₁

t φ⁻¹

−C+

Z _s

0 h

ds, whereC= φ(u⁰(0))∈_Ris the unique solution of H(ξ) =0, where

H(ξ) =

a−

Z ₁

0 g _{Z 1}

0 φ⁻¹

−ξ+

Z _s

0 h

ds−bφ⁻¹(ξ) +

Z ₁

0

_{Z 1}

t g

φ⁻¹

−ξ+

Z _t

0 h

dt.

Note that the existence and uniqueness of C follows from the fact that H is continuous, de- creasing inξ with lim_ξ→_∞H(ξ) = −_∞ and lim_ξ→−_∞H(ξ) = _{∞. Since} H(C)< 0 ifC > khk₁ and H(C)>0 ifC<−khk₁, it follows that|C| ≤ khk₁. Hence

kuk_∞ ≤φ⁻¹(₂khk₁)_.

i.e. (2.2) holds. Since|u|_C1 =kuk_∞+ku⁰k_∞ ≤2

p

p−1φ⁻¹(khk₁), it follows thatT maps bounded sets in L¹(0, 1)into bounded sets in C¹[0, 1]. To show continuity of T, let (h_n)⊂ L¹(0, 1)and h∈ L¹(0, 1)be such thath_n→hin L¹(0, 1). Letu_n=Th_nandu=Th. Then

u_n(t) =

Z ₁

t

φ⁻¹

−C_n+

Z _s

0

h_n

ds

fort ∈[_{0, 1}]_{, where}C_n ∈_Ris the unique solution ofH_n(ξ) =_{0, where} H_n(ξ) =

a−

Z ₁

0 g _{Z 1}

0 φ⁻¹

−ξ+

Z _s

0 h_n

ds−bφ⁻¹(ξ) +

Z ₁

0

_{Z 1}

t g

φ⁻¹

−ξ+

Z _t

0 h_n

dt.

SupposeC_n> C+k|h_n−hk₁. Then

−C_n+

Z _s

0 h_n <−C+

Z _s

0 h

for s ∈ [0, 1], which together with (A1) and the fact that φ⁻¹ is increasing imply H_n(C_n) <

H(C). On the other hand, ifC_n<C− kh_n−hk₁ then

−C_n+

Z _s

0 h_n >−C+

Z _s

0 h

fors ∈ [_{0, 1}], which implies H_n(C_n)> H(C). Hence we reach a contradiction in either case.

Consequently,

|C_n−C| ≤ kh_n−hk₁,

which impliesCn →C asn →∞. Using the formulas for un andu, it is easily seen that (un) converges touinC[0, 1], which completes the proof.

Next, we establish a comparison principle.

Lemma 2.2. Let0 ≤ r₀ < r₁ ≤ 1and let h₁,h₂ ∈ L¹(r₀,r₁)be such that h₁ ≥ h₂on (r₀,r₁). Let u,v∈ C¹[r₀,r₁]satisfy







−(φ(u⁰))⁰ = h₁, −(φ(v⁰))⁰ =h₂ on(r₀,r₁)_, au(r₀)−bu⁰(r₀)−Rr1

r0 g(t)u(t)dt≥av(r₀)−bv⁰(r₀)−Rr1

r0 g(t)v(t)dt, u⁰(r₁)≥v⁰(r₁) or u(r₁)≥v(r₁).

Then u≥v on[r0,r₁].

Proof. Suppose on the contrary that there exists r^∗ ∈ (r₀,r₁) _{such that} u(r^∗) < v(r^∗)_{. Let} (α,β) ⊂ (r₀,r₁) be the largest open interval containing r^∗ such that u < v on (α,β). Then u(α)≤v(α)andu(β)≤v(β). Multiplying the equation

−(φ(u⁰)−φ(v⁰))⁰ =h₁−h₂≥0 on(r₀,r₁) (2.4) byu−vand integrating on(α,β), we obtain

Cα−C_β+

Z _β

α

(φ(u⁰)−φ(v⁰))(u⁰−v⁰) =

Z _β

α

(h₁−h₂)(u−v)≤0, (2.5) whereC_κ = (φ(u⁰(κ))−φ(v⁰(κ))(u(κ)−v(κ))_,κ∈ {_α,β}. We claim thatC_α =_{0 and}C_β ≤_0.

To show C_α = 0, we verify thatu(α) = v(α). If α> r₀ then clearlyu(α) = v(α). Suppose α=r0 andu(r0)<v(r0). We show that this will lead to a contradiction.

Case 1: u⁰(r₁)≥v⁰(r₁).

Then, since φ(u⁰)−φ(v⁰) is nonincreasing, it follows that u⁰ ≥ v⁰ on [r₀,r₁]. Hence min[r₀,r₁](u−v) = (u−v)(r₀), which together with the boundary inequality at r₀ and (A1) imply

0≤b(u⁰−v⁰)(r₀)≤a(u−v)(r₀)−

Z _r1

r0

g(u−v)dt≤

a−

Z _r1

r0

g

(u−v)(r₀)<0, (2.6) a contradiction.

Case 2. u(r₁)≥v(r₁).

Thenu(β) = v(β)and henceu⁰(β)≥ v⁰(β), which implies u⁰ ≥v⁰ on [r₀,β]. In particular, min_[r0,β](u−v) = (u−v)(r₀). If β= r₁ then we reach a contradiction as in case 1. Suppose β < r₁. We shall verify that u ≥ v on [β,r₁]. If not, then there exists an interval (α₀,β₀) ⊂ (_β,r₁)_{such that}u<von(α₀,β₀)_and(u−v)(α₀) = (u−v)(β₀) =_0.

Multiplying (2.4) byu−vand integrating on(α₀,β₀)gives Z _β0

α₀

(φ(u⁰)−φ(v⁰))(u⁰−v⁰) =_0,

which implies u⁰ = v⁰ on [α₀,β₀]. Hence u = v on [α₀,β₀], a contradiction. Thus u−v ≥ 0 on [β,r₁], which, together with min_[r0,β](u−v) = (u−v)(r₀) and (u−v)(r₀) < 0, gives min_[r0,r1](u−v) = (u−v)(r0)and again (2.6) holds, a contradiction.

Thusu(α) = v(α)in both cases. Next, we claim thatC_β ≤ 0. Ifu(r₁)≥ v(r₁)then u(β) = v(β)while ifu⁰(r₁)≥v⁰(r₁)thenu(β) =v(β)ifβ<r₁ andC_β = (φ(u⁰(β))−φ(v⁰(β))(u(β)− v(β))≤0 if β=r₁. This proves the claim. Hence (2.5) gives

Z _β

α

(φ(u⁰)−φ(v⁰))(u⁰−v⁰) =0,

which impliesu⁰ =v⁰on[α,β]and sou=v+con[α,β], wherecis a negative constant. Hence α= r₀ and β= r₁. Using the assumption on the boundary atr₀, we obtain a−Rr₁

r₀ g c≥ 0.

Thus c≥0, a contradiction. Hence u≥von[r₀,r₁], which completes the proof.

Lemma 2.3. Let h∈ L¹(0, 1)with h ≥ 0and let u ∈ C¹[0, 1] withφ(u⁰)absolutely continuous on [0, 1]satisfying

(φ(u⁰))⁰ ≤ h on(0, 1), (2.7) with either

au(0)−bu⁰(0)≥

Z ₁

0 g(t)u(t)dt, u⁰(1)≥0, (2.8) or

au(0)−bu⁰(0)≥

Z ₁

0 g(t)u(t)dt, u(1)≥0. (2.9) Supposekuk_∞ >Lφ⁻¹(khk₁),where L = _a^2ma_−kg⁺_k^b

1 and m=2

2−p p−1

+.Then

u(t)≥ckuk_∞q(t) (2.10)

for t ∈[_{0, 1}]_,where c=1/L.

Proof. By Lemma2.2,u≥von [0, 1], wherevsatisfies (φ(v⁰))⁰ = h on (0, 1) with

av(0)−bv⁰(0) =

Z ₁

0 g(t)v(t)dt, v⁰(1) =0 if (2.8) holds, and

av(0)−bv⁰(0) =

Z ₁

0 g(t)v(t)dt, v(1) =0

if (2.9) holds. Supposekuk_∞ = |u(τ)|for someτ ∈[0, 1]. Thenu(τ)>0. Indeed, ifu(τ)≤ 0 then in view of (2.2), we get

kuk_∞ =−u(τ)≤ −v(τ)≤Mφ⁻¹(khk₁)≤ Lφ⁻¹(khk₁), where Mis defined in Lemma2.1. This contradicts the assumption onkuk_∞.

Supposeτ∈(0, 1). Letw∈C¹[0,τ]be the solution of ((φ(w⁰))⁰ =h on(0,τ),

aw(0)−bw⁰(0) =R1

0 g(t)w(t)dt, w(τ) =kuk_∞. A calculation shows that

w(t) =K_C+

Z _t

0 φ⁻¹

C+

Z _s

0 h

ds (2.11)

fort ∈[0, 1], where

K_C =

bφ⁻¹(C) +R1

0 g(t)Rt

0φ⁻¹ C+Rs 0 h

ds dt

a− kgk₁ ^{, (2.12)}

andC=φ(w⁰(0))is the unique solution of H_τ(ρ) =kuk_∞, where H_τ(ρ) =K_ρ+

Z _τ

0 φ⁻¹

ρ+

Z _s

0 h

ds.

Note that the existence and uniqueness ofC follows from the fact that H_τ is increasing in ρ and limρ→_∞Hτ(ρ) =_{∞, limρ→−∞}Hτ(ρ) =−_∞.

Using Lemma 2.2 with r₀ = _{0 and} r₁ = τ, we deduce that u ≥ w on [_0,τ]_{. If} w⁰(₀) ≤ ₀ thenC≤0 and hence

kuk_∞ = Hτ(C)≤ R1

0 g(t)Rt

0 φ⁻¹ Rs 0 h

ds dt a− kgk₁ +

Z _τ

0 φ⁻¹ Z _s

0 h

ds

≤ ^a

a− kgk₁^φ

−1(khk₁)< Lφ⁻¹(khk₁)_, a contradiction. Hencew⁰(0)>0 i.e.C>0.

Using the inequality(x+y)^r≤2^(r−1)+(x^r+y^r)forx,y≥0 withr = (p−1)⁻¹, we obtain φ⁻¹

φ(w⁰(0)) +

Z _s

0 h

≤m

w⁰(0) +φ⁻¹(khk₁),

wherem=2

_2−p

p−1

+. Hence it follows from (2.11)–(2.12) that kuk_∞ ≤ ^bw

0(0) +mkgk₁ w⁰(0) +φ⁻¹(khk₁)

a− kgk₁ +m

w⁰(₀) +φ⁻¹(khk₁)

=m₁w⁰(0) +M₁φ⁻¹(khk₁), wherem₁ = ^b+ma

a−kgk₁ andM₁= ^ma

a−kgk₁. Consequently, w⁰(0)≥ kuk_∞−M₁φ⁻¹(khk₁)

m₁ ≥ kuk_∞ m₁

1− ^M1 L

= kuk_∞

L , (2.13)

where we have used the assumption kuk_∞ > Lφ⁻¹(khk₁). SinceK_C,h≥0, it follows from (2.11) and (2.13), that

u(t)≥w(t)≥φ⁻¹(C)t =w⁰(0)t≥ kuk_∞t

L (2.14)

fort ∈[0,τ].

Next, we establish a lower bound estimate foru(t)in terms ofkuk_∞on[τ, 1]. By Lemma2.2, u≥zon[τ, 1], wherez∈C¹[τ, 1]satisfies

((φ(z⁰))⁰ = h on (τ, 1), z(τ) =kuk_∞, z⁰(1) =0 if (2.8) holds, and

((φ(z⁰))⁰ =h on(_{τ, 1})_, z(τ) =kuk_∞, z(1) =0 if (2.9) holds.

Suppose first that (2.8) holds. Then z(t) =D+

Z ₁

t φ⁻¹ _{Z 1}

s h

ds,

where D = kuk_∞−R1

τ φ⁻¹ R1 s h

ds. Since L ≥ 2m ≥ 2, it follows from Lemma 2.2 with r0= τ,r₁ =1,b=0,g≡0 that

u(t)≥z(t)≥ kuk_∞−φ⁻¹(khk₁)≥ kuk_∞/2 (2.15) fort ∈[τ, 1]. Next, suppose (2.9) holds. Then

z(t) =

Z ₁

t φ⁻¹

−D−

Z _s

0 h

ds, (2.16)

where D=φ(z⁰(0))is the unique solution of Z ₁

τ

φ⁻¹

D+

Z _s

0 h

ds= −kuk_∞. (2.17)

Sinceh≥0, it follows from (2.17) that

(₁−τ)φ⁻¹(D)≤ −kuk_∞,

which impliesD≤ −φ(kuk_∞). Hence, sincekuk_∞ ≥2mφ⁻¹(khk₁), it follows that

−D−

Z _s

0

h≥ φ(kuk_∞)− khk₁≥

1− ¹ φ(2m)

φ(kuk_∞)≥ ^φ(kuk_∞)

2 . (2.18)

Using (2.16)–(2.18), we obtain

u(t)≥z(t)≥(1/2)^p−11 kuk_∞(₁−t) _(2.19) fort ∈[τ, 1]. Since

L≥2m=

(2 if p≥2,

2^p−11 if 1< p<2 =max

2^p−11, 2 ,

it follows that min L⁻¹, 2⁻¹, 2^1−1p

= L⁻¹, it follows from (2.14), (2.15), and (2.19) that (2.10) holds for the caseτ∈(0, 1).

Ifτ=1 thenw∈ C¹[0, 1]and (2.14) holds fort∈ [0, 1], which implies (2.10) sincet ≥q(t) for t ∈ [0, 1]. Finally, if τ = 0 then z ∈ C¹[0, 1] and (2.15), (2.19) hold for t ∈ [0, 1], which implies (2.10). This completes the proof of Lemma2.3.

3 Proof of the main result

Proof. Let E = C[0, 1] be equipped with k · k_∞. For the rest of the proof, we set ˜γ_k = γ_k/q^δ, where γ_k is defined in (A3), and recall that ˜γ_k ∈ L¹(0, 1). For v ∈ C[0, 1] with kvk_∞ ≤ k for somek≥1, it follows from (A3) that there existsγ_k ∈ L¹(_{0, 1})_withγ_k ≥0 such that

|f(t, ˜v)| ≤γ_k(t)v˜^−δ ≤γ˜_k(t) (3.1) for a.e. t∈(0, 1), where ˜v=max(v,q). Letλ>0. Then, by Lemma2.1, the equation

−(φ(u⁰))⁰ =λf(t, ˜v), t∈(0, 1)

with boundary condition (1.2) or (1.3) has a unique solution u ≡ A_λv ∈ C¹[0, 1]. Let S_λ : E→ L¹(0, 1)be defined byS_λv=λf(t, ˜v). ThenS_λis continuous by the Lebesgue dominated convergence Theorem. By (3.1),S_λ maps bounded sets inEinto bounded sets inL¹(0, 1). Since A_λ = T◦S_λ, where T is defined in Lemma 2.1, it follows that A_λ : E → E is completely continuous.

(i) Suppose (A4) holds withν=_∞.

Let λ ∈ (0, 1)satisfy M(λkγ˜_Lk₁)^p−11 < L, where M and L are defined in Lemma 2.1 and Lemma2.3 respectively. We claim that

(a)If u∈E satisfies u= θA_λu for someθ ∈(_{0, 1}]thenkuk_∞ 6= L.

Indeed, let u ∈ E satisfy u = θA_λu for some θ ∈ (0, 1], and suppose kuk_∞ = L. Then u/θ= T(S_λu)and (2.2) gives

kuk_∞ ≤ Mθφ⁻¹(kS_λuk₁)≤ M(λkγ˜_Lk₁)^p−11 < L, a contradiction, which proves the claim.

Next, we show that

(b)There exists a constant R_λ > L such that if u∈ E satisfies u = A_λu+ξ for someξ ≥0then kuk_∞ 6= R_λ.

Letu∈Esatisfyu= A_λu+ξ for someξ ≥0. Thenu−ξ = A_λuand thereforeusatisfies

−(φ(u⁰))⁰ =λf(t, ˜u) on(0, 1) with

au(0)−bu⁰(0)−

Z ₁

0 g(t)u(t)dt=

a−

Z ₁

0 g

ξ ≥0, u⁰(1) =0 (3.2) if (1.2) holds, and

au(0)−bu⁰(0)−

Z ₁

0 g(t)u(t)dt=

a−

Z ₁

0 g

ξ ≥0, u(1) =ξ ≥0 (3.3) if (1.3) holds.

LetK>0 be such that f(t,z)>0 for a.e. t∈(0, 1)and allz≥K. Forz ∈(0,K), |f(t,z)| ≤ γ_K(t)z^−δ for a.e.t∈ (0, 1)in view of (A3). Consequently,

f(t, ˜u)≥ −γ_K(t)u˜^−δ ≥ −γ˜_K(t) (3.4) for a.e.t∈ (0, 1). Hence Lemma2.3 holds withh=λγ˜_K ifkuk_∞> Lφ⁻¹(λkγ˜_Kk₁).

Letω∈ C¹₁

4,¹₂

satisfy

(−(φ(ω⁰))⁰ =γ(t) on(1/4, 1/2), ω(_1/4) =ω(_1/2) =_0,

and let R₀ >0 be such that(λR₀)^p−11L⁻¹kωk_∞>_4.

Since lim_z→_∞ ^f(t,z)

γ(t)z^p−1 =_∞uniformly fort∈ (0, 1), there existsk₀ >1 such that f(t,z)≥R₀γ(t)z^p−1

for a.e. t ∈(0, 1)and all z≥ k₀. Suppose |uk_∞ = R_λ > Lmax(φ⁻¹(λkγ˜_Kk₁), 4k₀). Then (2.10) holds.

In particular,u≥ckuk_∞q≥qon[0, 1]andu≥(c/4)kuk_∞ ≥k0on[1/4, 1/2]. Hence ˜u≡u and

−(φ(u⁰))⁰ =λf(t,u)≥λR₀γ(t)u^p−1 ≥λR₀

ckuk_∞ 4

p−1

γ(t)

for a.e.t∈ (1/4, 1/2). By Lemma2.2 withr₀ =1/4,r₁=1/2,b=0,g≡0, we obtain u≥(λR0)^p−11(c/4)kuk_∞ω

on [1/4, 1/2], which implies (λR₀)^p−11ckωk_∞ ≤ 4, a contradiction with the choice of R₀ (c= L⁻¹). Hencekuk_∞ 6= R_λ i.e. (b) holds.

Let λ also be small enough so that φ⁻¹(λkγ˜_Kk₁) < 1. Then it follows from Lemma Ain the Appendix that A_λ has a fixed point u_λ ∈ E with ku_λk_∞ > L > Lφ⁻¹(λkγ˜_Kk₁). Hence Lemma 2.3 gives, u_λ ≥ cku_λk_∞q ≥ q on [0, 1] and sou_λ = u˜_λ is a positive solution of (1.1) under boundary condition (1.2) or (1.3). We verify next thatku_λk_∞ →_∞asλ→0⁺. Suppose on the contrary thatku_λk_∞ 6→_∞asλ→0⁺. Then there exist a constantC >0 and a sequence (λn)converging to 0 such thatku_λnk_∞ ≤Cfor alln. By (A3),

|f(t,u_λn)| ≤γ_C(t)u⁻_λ^δ

n ≤γ˜_C(t)

for a.e. t∈(0, 1), from which (2.2) gives

L< ku_λnk_∞ ≤ Mφ⁻¹(λ_nkf(t,u_λn)k₁)≤Mφ⁻¹(λ_nkγ˜_Ck₁),

a contradiction fornlarge. Thus ku_λk_∞ →_∞asλ→0⁺, which completes the proof of (i).

(ii) Suppose (A4) holds withν=0.

Let L₀ > Lφ⁻¹(kγ˜_Kk₁)_{, where} K is defined in (3.4). Let C₀ > (L₀/kω¯k_∞)^p−1, where ¯ω is the solution of

(−(φ(ω¯))⁰ =_{1 on}(_{1/4, 1/2})_,

¯

ω(1/4) =ω¯(1/2) =0.

Since limz→_∞ f(t,z) =_∞uniformly fort ∈(0, 1), there exists a constantc₀ >1 such that

f(t,z)≥C₀ (3.5)

for a.e. t∈(0, 1)andz≥c₀.

Supposeλ>λ^˜₀, where ˜λ₀= ^4c_L^0L

0

p−1

. We claim that

(c)If u∈ E satisfies u= A_λu+ξ for some ξ ≥0thenkuk_∞ 6=λ

1 p−1L₀.

Let u ∈ E satisfy u = A_λu+ξ for some ξ ≥ 0. Then (3.4) and either (3.2) or (3.3) hold. Supposekuk_∞ = λ

p−11L₀. Thenkuk_∞ > Lφ⁻¹(λkγ˜_Kk₁) and therefore Lemma 2.3 with h=λγ˜_Kgives

u(t)≥ckuk_∞q(t) =cλ^p−11L₀q(t)>4c₀q(t)

fort ∈(0, 1). In particular,u≥c₀ on[1/4, 1/2], which, together with (3.5), implies

−(φ(u⁰))⁰ =λf(t, ˜u) =λf(t,u)≥λC0 on (1/4, 1/2). Lemma2.2 then givesu≥(λC₀)^p−11ω¯ on[1/4, 1/2].

Consequently,

λ

p−11L₀=kuk_∞≥ (λC₀)^p−11kω¯k_∞

i.e. C₀ ≤ (L₀/kω¯k_∞)^p−1, a contradiction with the choice of C₀. Hence (c) holds. Next, we verify

(d) There exists R_{λ 1} such that if u ∈ E satisfies u = θA_λu for some θ ∈ (_{0, 1}] then kuk_∞ 6= R_λ.

Letu∈ Esatisfyu= θA_λufor someθ ∈(0, 1]. Then

−(φ(u⁰))⁰ =λθ^p−1f(t, ˜u) on(0, 1).

Using (3.4), we see that (2.7) holds with h = λθ^p−1γ˜_K. Let σ > 1 be such that 0 < f(t,z) ≤ γ(t)z^p−1for a.e.t∈(0, 1)and allz ≥σ.

Let f₁(z) = sup_t∈(0,1) ^f_γ^(t,z_(t)⁾ for z ≥ σ and f₁(z) = f₁(σ) for z ∈ (0,σ). Then f₁ > 0 and therefore (A3) gives

f(t,z)≤γ_σ(t)z^−δ+γ(t)f₁(z) _(3.6) for a.e.t ∈(_{0, 1})_{and all}z>_0.

Supposekuk_∞ = R_λ > max((λkγ˜_Kk₁)^p−11,L,λ

1

p−1L₀). Thenu ≥ q on (0, 1)by Lemma2.3 and therefore (3.6) gives

f(t, ˜u)≤ γ˜_σ(t) +γ(t)f^ˆ₁(kuk_∞)_,

where ˆf₁(z) =sup_0<t≤z f₁(t). This, together with (2.2), implies kuk_∞ ≤ Mφ⁻¹(λkγ˜_σk₁+λkγk₁f^ˆ₁(kuk_∞)), i.e.

kγ˜_σk₁+kγk₁f^ˆ₁(kuk_∞)

kuk_∞^p−1 ≥ ¹

λM^p−1. (3.7)

Since lim_z→_∞ fˆ1(z)

z^p−1 =0, the left side of (3.7) goes to 0 askuk_∞ →_∞, we reach a contradiction if R_λ is large enough. Hence (d) holds.

By Lemma A in the Appendix, A_λ has a fixed point u_λ with ku_λk_∞ ≥ λ

1

p−1L₀. Using Lemma 2.3, we see that u_λ is a positive solution of (1.1) under boundary condition (1.2) or (1.3). Clearly, ku_λk_∞ →_∞asλ→∞, which completes the proof of Theorem1.1.

Acknowledgements

The authors thank the referee for carefully reading the manuscript and providing constructive remarks.

Appendix

We shall state a version of Krasnoselskii’s fixed point theorem in a Banach space. The proof presented here is essentially done in [1, Theorem 12.3], but with no cones involved.

Lemma A. Let E be a Banach space and T : E → E be a completely continuous operator. Suppose there exist h∈ E,h6=0and positive constants r,R with r6=R such that

(a) If y∈ E satisfies y=θTy,θ ∈(0, 1]thenkyk 6=r, (b) If y∈ E satisfies y=Ty+ξh,ξ ≥0thenkyk 6= R.

Then T has a fixed point y ∈E withmin(r,R)<kyk<max(r,R).

Proof. Define H : [_{0, 1}]×E → E by H(_θ,y) = θTy. Then H is completely continuous and H(θ,y)6=y forkyk=r in view of (a).

By the homotopy invariance property,

deg(I−H(1,·),B_r, 0) =deg(I−H(0,·),B_r, 0) =deg(I,B_r, 0) =1, where B_rdenotes the open ball centered at 0 with radiusr inE. Hence

deg(I−T,B_r, 0) =1.

By (b) and the homotopy invariance property,

deg(I−(T+ξh),B_R, 0) =C

for allξ ≥0. We claim that C= 0. Suppose on the contrary thatC6=0. Let M =sup{kTyk: kyk ≤R}and chooseξ = ^M_k⁺_hk^R. Then there existsy∈B_Rsuch thaty=Ty+ξh. Hence

kyk ≥ξkhk − kTyk ≥ξkhk −M= R,

a contradiction which proves the claim. In particular, deg(I−T,B_R, 0) = 0, and Lemma A follows from the excision property of degree theory.

References

[1] H. Amann, Fixed point equations and nonlinear eigenvalue problems in ordered Banach spaces,SIAM Rev.18(1976), 620–709.https://doi.org/10.1137/1018114;MR0415432 [2] C. Atkinson, K. E. Ali, Some boundary value problems for the Bingham model,

J. Non-Newton. Fluid Mech.41(1992), 339–363.https://doi.org/10.1016/0377-0257(92) 87006-w

[3] L. E. Bobisud, Steady state turbulent flow with reaction,Rocky Mountain J. Math.21(1991), 993–1007.https://doi.org/10.1216/rmjm/1181072925;MR1138147

[4] A. Boucherif, Second order boundary value problems with integral boundary condi- tions, Nonlinear Anal. 70(2009), 364–371. https://doi.org/10.1016/j.na.2007.12.007;

MR2468243

[5] J. R. Cannon, The solution of the heat equation subject to the specification of energy, Quart. Appl. Math.21(1963), 155–160.https://doi.org/10.1016/0022-247X(64)90061-7;

MR0160047

[6] R. Y. Chegis, Numerical solution of a heat conduction problem with an integral condition (in Russian),Litovsk. Mat. Sb. 24(1984), No. 4, 209–215.MR0785889

[7] J. I. Diaz,Nonlinear partial differential equations and free boundaries, Pitman, London, 1985.

MR0853732

[8] R. Glowinski, J. Rappaz, Approximation of a nonlinear elliptic problem arising in a non- Newtonian fluid flow model in glaciology, Math. Model. Numer. Anal. 37(2003), 175–186.

https://doi.org/10.1051/m2an:2003012;MR1972657

[9] J. R. Graef, L. Kong, Positive solutions for third order semipositone boundary value problems,Appl. Math. Lett.22(2009), 1154–1160.https://doi.org/10.1016/j.aml.2008.

11.008;MR2532528

[10] G. Infante, Eigenvalues and positive solutions of ODEs involving integral equations, Discrete Contin. Dyn. Syst., Suppl. (2005), 436–442. https://doi.org/10.3934/proc.

2005.2005.436;MR2192701

[11] I. Ionkin, Solution of a boundary value problem in heat conduction theory with nonlocal boundary conditions,Differencial⁰nye Uravnenija13(1977), 294–304.MR0603291

[12] R. A. Khan, The generalized method of qualilinearization and nonlinear boundary value problems with integral boundary condition, Electron. J. Qual. Theory Differ. Equ. 2003, No. 19, 1–15.https://doi.org/10.14232/ejqtde.2003.1.19;MR2039793

[13] L. Kong, Second order singular boundary value problems with integral boundary con- ditions, Nonlinear Anal. 72(2010), 2628–2638. https://doi.org/10.1016/j.na.2009.11.

010;MR2577824

[14] J. R. L. Webb, G. Infante, Positive solutions of nonlocal boundary value problems in- volving integral equations,NoDEA Nonlinear Differential Equations Appl.15(2008), No. 1–2, 45–67.https://doi.org/10.1007/s00030-007-4067-7;MR2408344

[15] Z. Yang, Positive solutions of a second order integral boundary value problem, J. Math.

Anal. Appl. 321(2006), No. 2, 751–765. https://doi.org/10.1016/j.jmaa.2005.09.002;

MR2241153

[16] X. Zhang, M. Feng, Existence of a positive solution for one-dimensional singular p-Laplacian problems and its parameter dependence,J. Math. Anal. Appl.413(2014), 566–

582.https://doi.org/10.1016/j.jmaa.2013.11.038;MR3159788