We study them-point nonlinear boundary-value problem −[p(t)u0(t)]0=λf(t, u(t

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Electronic Journal of Differential Equations, Vol. 2004(2004), No. 119, pp. 1–7.

ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu ftp ejde.math.txstate.edu (login: ftp)

SEMIPOSITONE m-POINT BOUNDARY-VALUE PROBLEMS

NICKOLAI KOSMATOV

Abstract. We study them-point nonlinear boundary-value problem

−[p(t)u⁰(t)]⁰=λf(t, u(t)), 0< t <1, u⁰(0) = 0,

m−2

X

i=1

αiu(ηi) =u(1),

where 0 < η1 < η2 < · · · < ηm−2 < 1, αi > 0 for 1 ≤ i ≤ m−2 and Pm−2

i=1 αi <1, m≥3. We assume that p(t) is non-increasing continuously differentiable on (0,1) andp(t)>0 on [0,1]. Using a cone-theoretic approach we provide sufficient conditions on continuousf(t, u) under which the problem admits a positive solution.

1. Introduction

In this note we consider the nonlinearm-point eigenvalue problem

−[p(t)u⁰(t)]⁰=λf(t, u(t)), 0< t <1, (1.1) u⁰(0) = 0,

m−2

X

i=1

αiu(ηi) =u(1), (1.2)

where 0< η₁< η₂<· · ·< η_m−2<1,α_i>0 for 1≤i≤m−2,Pm−2

i=1 α_i<1. We also assume that the functionp(t) is non-increasing continuously differentiable on (0,1) andp(t)>0 on [0,1]. The inhomogeneous term in (1.1) is allowed to change its sign. Other assumptions onf(t, u(t)) will be made later.

The study of multi-point boundary-value problems was initiated by Il’in and Moi- seev in [7, 8]. Many authors since then considered nonlinear multi-point boundary- value problems (see, e.g., [2, 4, 5, 6, 9, 14, 15, 16, 17] and the references therein).

In particular, Ma studied in [15] positive solutions to the three-point nonlinear boundary-value problem

−u⁰⁰(t) =a(t)f(u(t)), 0< t <1, u(0) = 0, αu(η) =u(1),

2000Mathematics Subject Classification. 34B10, 34B18.

Key words and phrases. Green’s function; fixed point theorem; positive solutions;

multi-point boundary-value problem.

c

2004 Texas State University - San Marcos.

Submitted April 23, 2004. Published October 10, 2004.

1

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where 0< α, 0< η <1 andαη <1. The results of [15] were complemented in the works of Webb [17], Kaufmann [9], Kaufmann and Kosmatov [10], and Kaufmann and Raffoul [11].

Among the studies dealing with semipositone multi-point boundary-value problems, we mention the papers by Cao and Ma [3] and Liu [13]. Cao and Ma considered the boundary-value problem

−u⁰⁰(t) =λa(t)f(u(t), u⁰(t)), 0< t <1, u(0) = 0,

m−2

X

i=1

αiu(ηi) =u(1).

The authors applied the Leray-Schauder fixed point theorem to obtain an interval of eigenvalues for which at least one positive solution exists. Liu applied a fixed point index method to obtain such an interval for

−u⁰⁰(t) =λa(t)f(u(t)), 0< t <1, u⁰(0) = 0, αu(η) =u(1).

Our approach is based on Krasnosel’ski˘ı’s cone-theoretic theorem [12] and enables us to show the existence of a positive solution for the semipositone problem (1.1), (1.2). Other applications of Krasnosel’ski˘ı’s fixed point theorem to semipositone problems can, for example, be found in [1].

2. Preliminaries

We now proceed with the auxiliaries. Consider the equation

−[p(t)u⁰(t)]⁰=g(t), 0< t <1, (2.1) with the boundary conditions (1.2).

For convenience we setα=Pm−2

i=1 αi. Recall thatα <1.

Lemma 2.1. If g∈C[0,1]andg(t)≥0 on [0,1], then u(t) =−

Z t

0

Z t

s

dτ p(τ)

g(s)ds+ 1 1−α

Z 1

0

Z 1

s

dτ p(τ)

g(s)ds

− 1 1−α

m−2

X

i=1

αi

Z ηi

0

Z ηi

s

dτ p(τ)

g(s)ds

(2.2)

is the unique nonnegative solution on[0,1]of the problem (2.1), (1.2).

Proof. Integration of (2.1) from 0 totwith the use of the boundary condition (1.2) at 0 yields

u⁰(t) =− 1 p(t)

Z t

0

g(s)ds≤0.

Integrating again we get u(t) =−

Z t

0

1 p(s)

Z s

0

g(τ)dτ

ds+A=− Z t

0

Z t

s

dτ p(τ)

g(s)ds+A.

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Using the multi-point condition in (1.2) we determine A and obtain (2.2). Since u⁰(t)≤0,

u(t)≥u(1)

= α

1−α Z 1

0

Z 1

s

dτ p(τ)

g(s)ds− 1 1−α

m−2

X

i=1

αi

Z η_i

0

Z η_i

s

dτ p(τ)

g(s)ds

= 1

1−α

m−2

X

i=1

αi

hZ 1

0

Z 1

s

dτ p(τ)

g(s)ds− Z η_i

0

Z η_i

s

dτ p(τ)

g(s)dsi

≥0

on [0,1] and the proof is complete.

For g(t) = 1 on [0,1], we denote by u0(t) the unique solution (2.2). Then we have

C= max

t∈[0,1]u0(t) =u0(0)

= 1

1−α Z 1

0

Z 1

s

dτ p(τ)

g(s)ds− 1 1−α

m−2

X

i=1

αi

Z ηi

0

Z ηi

s

dτ p(τ)

g(s)ds.

The Green’s function for−[p(t)u⁰(t)]⁰= 0 with (1.2) is given by G(t, s) = 1

1−α Z 1

s

dτ p(τ)

− (Rt

s dτ

p(τ), s≤t 0, s > t−

( ₁

1−α

Pm−2

i=1 αiχi(s)Rηi

s dτ

p(τ), s≤η_m−2

0, s > η_m−2,

where

χi(s) =

(1, s≤ηi

0, s > η_i. Note that

max

t∈[0,1]

Z 1

0

G(t, s)ds=C. (2.3)

The integral operatorT:B → B associated with (1.1), (1.2) is defined by T u(t) =

Z 1

0

G(t, s)f(s, u(s))ds A routine argument shows thatT is completely continuous.

Definition 2.2. LetBbe a Banach space and letC ⊂ B be closed and nonempty.

ThenC is said to be a cone if

(1) αu+βv ∈ Cfor allu, v∈ C and for allα, β≥0, and (2) u,−u∈ Cimpliesu≡0.

Our Banach space,B, is the spaceC[0,1] with the normkuk= max_t∈[0,1]|u(t)|.

We will show now that the unique solution (2.2) satisfies min

t∈[0,1]u(t)≥γkuk, (2.4)

where

γ= max

1≤i≤m−2

αi(1−ηi) 1−αiηi

.

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To this end, note that the solution (2.2) is concave, sinceg(t)≥0 andu⁰(t), p⁰(t)≤0 on [0,1]. By concavity and sinceu(1)> αiu(ηi) for each 1≤i≤m−2,

kuk=u(0)

≤u(1) +u(1)−u(ηi) 1−ηi

(0−1)

< u(1) 1−α_iη_i αi(1−ηi)

= 1−αiηi

α_i(1−η_i) min

t∈[0,1]u(t) and hence (2.4) holds.

The estimate (2.4) is used for defining our coneC ⊂ B by C={u(t)∈ B:u(t)≥0 on [0,1], min

t∈[0,1]u(t)≥γkuk}. (2.5) It turns out that our operatorT is cone-preserving. Fixed points ofT are solutions of (1.1), (1.2). The existence of a fixed point ofT follows from a fixed point theorem due to Krasnosel’ski˘ı [12], which we now state.

Theorem 2.3. Let B be a Banach space and let C ⊂ B be a cone in B. Assume that Ω₁,Ω₂ are open with0∈Ω₁,Ω₁⊂Ω₂, and let

T:C ∩(Ω2\Ω1)→ C be a completely continuous operator such that either

(i) kT uk ≤ kuk,u∈ C ∩∂Ω₁, andkT uk ≥ kuk,u∈ C ∩∂Ω₂, or (ii) kT uk ≥ kuk,u∈ C ∩∂Ω₁, andkT uk ≤ kuk,u∈ C ∩∂Ω₂. ThenT has a fixed point inC ∩(Ω2\Ω1).

The following assumptions will stand throughout the remainder of this note:

(A1) f(t, z) is a continuous function on [0,1]×[0,∞)

(A2) There existsM >0 such thatf(t, z) +M ≥0 on [0,1]×[0,∞)

(A3) There exist continuous nonnegative nondecreasing on [0,∞) functionsψa(z) andψb(z) withψb(z)≤f(t, z) +M ≤ψa(z) on [0,1]×[0,∞).

3. Positive solutions We now state our main results.

Theorem 3.1. Let the assumptions (A1)-(A3) be satisfied. Assume, in addition, that

lim

z→0⁺

ψ_a(z)

z = 0 and lim

z→∞

ψ_b(z) z =∞.

Then, for a sufficiently smallλ >0, the problem (1.1), (1.2) has a positive solution.

Proof. Consider the equation

−[p(t)u⁰(t)]⁰ =λfp(t, u(t)−uλ(t)), 0< t <1, (3.1) with the boundary conditions (1.2), where

fp(t, z) =

(f(t, z) +M, z≥0 f(t,0) +M, z≤0

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anduλ(t) =λM u0(t) (u0(t) is given by (2.2) forg≡1). Our objective is to show that the problem (3.1), (1.2) has a positive solution.

Our completely continuous and cone-preserving operator associated with (3.1), (1.2) is defined by

Tλu(t) =λ Z 1

0

G(t, s)fp(s, u(s)−uλ(s))ds Since lim_z→0+ψ_a(z)

z = 0, there existsR1>0 such that ψ_a(z)≤ 1

λCz for allz≤R₁.

Define Ω₁={u∈ B:kuk< R₁}, then foru∈ C ∩∂Ω₁ we have ψ_a(u(s))≤ψ_a(kuk)≤ 1

λCR₁ (3.2)

for all s∈[0,1], since ψa(z) is nondecreasing. Now, if u(s)≥uλ(s) for s∈[0,1], then

fp(s, u(s)−uλ(s)) =f(s, u(s)−uλ(s)) +M ≤ψa(u(s)−uλ(s))≤ψa(u(s)).

Ifu(s)≤uλ(s), then

fp(s, u(s)−uλ(s)) =f(s,0) +M ≤ψa(0)≤ψa(u(s))

(we know thatu(s)≥0 as an element ofC). Combining both cases and using (3.2) and (2.3), we get

kTλuk= max

t∈[0,1]λ Z 1

0

G(t, s)fp(s, u(s)−uλ(s))ds

≤ max

t∈[0,1]λ Z 1

0

G(t, s)ψa(u(s))ds

≤λ max

t∈[0,1]

Z 1

0

G(t, s)ds 1

λCR₁=R₁, that is,kTλuk ≤ kukonC ∩∂Ω1.

Since lim_z→∞^ψ^b_z^(z)=∞, then also lim_z→∞^ψ^b^{(γz−λM C)}_z =∞. Thus, there exists R2>0 large enough (so thatR2> ^{λM C}_γ and R2> R1) such that

ψb(γz−λM C)≥ 1 λCz for allz≥R2. In fact,

ψb(γR2−λM C)≥ 1

λCR2. (3.3)

Define Ω2={u∈ B:kuk< R2}, then foru∈ C ∩∂Ω2 we have u(s)−uλ(s)≥γkuk −λM u0(s)≥γR2−λM C >0.

Now, for alls∈[0,1],

f_p(s, u(s)−u_λ(s)) =f(s, u(s)−u_λ(s)) +M ≥ψ_b(u(s)−u_λ(s))≥ψ_b(γR₂−λM C),

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sinceψb(z) is nondecreasing. Therefore, by (3.3) and (2.3), kTλuk= max

t∈[0,1]λ Z 1

0

G(t, s)fp(s, u(s)−uλ(s))ds

≥ max

t∈[0,1]λ Z 1

0

G(t, s)ψb(γR2−λM C)ds

≥λ max

t∈[0,1]

Z 1

0

G(t, s)ds 1

λCR₂=R₂, that is,kTλuk ≤ kukonC ∩∂Ω2.

Since the assumptions of Theorem 2.3 are satisfied, we conclude that the problem (3.1), (1.2) has a positive solution inC ∩(Ω2\Ω1), which we denote byup.

Letλbe small enough so thatR1> ^{λM C}_γ . Now we haveup(t)≥γkupk ≥γR1>

λM C≥uλ(t) for all t∈[0,1]. Set u(t) =up(t)−uλ(t), then

−[p(t)u⁰(t)]⁰=−[p(t)u⁰_p(t)]⁰−λM

=λf_p(t, u_p(t)−u_λ(t))−λM

=λ(f(t, up(t)−uλ(t)) +M)−λM

=λf(t, u(t)),

which shows thatu(t) is a positive solution of (1.1), (1.2). The proof is complete.

Example. To illustrate our main result, we consider the inhomogeneous term in the form of the function

f(t, z) =−1 +z²(2 + sin (4πz(1 +t³))).

The function f(t, z) is continuous and, setting M = 1, we get f(t, z) +M ≥ 0 on [0,1]×[0,∞). In addition, for ψb(z) =z² andψa(z) = 3z², we have ψb(z)≤ f(t, z) +M ≤ψa(z) and

lim

z→0⁺

ψa(z)

z = 0 and lim

z→∞

ψb(z) z =∞.

Thus, Theorem 3.1 applies.

With only minor adjustments to the argument above one can prove our next theorem.

Theorem 3.2. Let the assumptions (A1)-(A3) be satisfied. Assume, in addition, that

lim

z→0⁺

ψa(z)

z =∞ and lim

z→∞

ψb(z) z = 0.

Then, for a sufficiently smallλ >0, the problem (1.1), (1.2) has a positive solution.

Remark. If problem (1.1), (1.2) has a positive solution for some λ1 >0, there is also a positive solution for eachλ∈(0, λ1].

We say that a functionψ(z) is sublinear if lim

z→0⁺

ψ(z)

z =∞ and lim

z→∞

ψ(z) z = 0.

On the other hand, if lim

z→0⁺

ψ(z)

z = 0 and lim

z→∞

ψ(z) z =∞,

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then the functionψ is called superlinear.

If in the assumption (A3) we takeψa(z) =ψb(z), then the following corollary to Theorems 3.1 and 3.2 becomes immediate.

Corollary 3.3. Let the assumptions (A1)-(A3) be satisfied. Assume, in addition, thatψ_a(z) =ψ_b(z)is either sublinear or superlinear. Then, for a sufficiently small λ >0, the problem (1.1), (1.2) has a positive solution.

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Department of Mathematics and Statistics, University of Arkansas at Little Rock, Little Rock, AR 72204-1099, USA

E-mail address:[email protected]