# Multiple Bounded Positive Solutions to Integral Type BVPs for Singular Second Order ODEs on the Whole Line

Authors: Yuji Liu
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

This paper is concerned with the integral type boundary value problems of the second order differential equations with one-dimensionalp-Laplacian on the whole line. By constructing a suitable Banach space and a operator equation, sufficient conditions to guarantee the existence of at least three positive solutions of the BVPs are established. An example is presented to illustrate the main results. The emphasis is put on the one-dimensional p-Laplacian term [ρ(t)Φ(x’(t))]’ involved with the function ρ, which makes the solutions un-concave.


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## Body

## 1. Introduction

The multipoint boundary-value problems for linear second order ordinary differential equations (ODEs for short) was initiated by Il'in and Moiseev [1]. Since then, more general nonlinear multi-point boundary-value problems (BVPs for short) were studied by several authors, see the text books [2–4] and the references cited therein.Differential equations governed by nonlinear differential operators have been widely studied. In this setting the most investigated operator is the classical one-dimensionalp-Laplacian, that is, Φp(x)=|x|p-2x with p>1. This operator is involved in some models, for example, in non-Newtonian fluid theory, diffusion of flows in porous media, nonlinear elasticity, and theory of capillary surfaces. The related nonlinear differential equation has the form
(1.1)[Φ(x')]′=f(t,x,x′),t∈(-∞,+∞),
where Φ(x)=|x|p-2x with p>1 is a one dimensional p-Laplacian. For a comprehensive bibliography on this subject, see, for example [5–9].In this paper, we consider the more generalized BVP for second order differential equation on the whole line withp-Laplacian coupled with the integral type BCs, that is the BVP
(1.2)[ρ(t)Φ(x′(t))]′+f(t,x(t),x′(t))=0,t∈R,limt→-∞x(t)=:x(-∞)=∫-∞+∞g(s)x(s)ds,limt→+∞x(t)=:x(+∞)=∫-∞+∞h(s)x(s)ds,
where f:R3→R is a nonnegative Caratheodory function, g,h:R→[0,∞) satisfy g,h∈L1(R),ρ∈C(R,(0,∞)), the integrals in mentioned equations are meant in the sense of Riemann-Stieljes, Φ(x)=|x|p-2x with p>1 is called a one dimensional p-Laplacian, whose inverse function is denoted by Φ-1.The purpose is to establish sufficient conditions for the existence of at least three positive solutions of BVP(1.2). The result in this paper generalizes and improves some known ones since the one-dimensional p-Laplacian term [ρ(t)Φ(x'(t))]' involved with the function ρ, which makes the solutions unconcave and there exists no paper concerned with the existence of at least three positive solutions of this kind of integral BVPs on the whole lines. This paper fills the gap.The remainder of this paper is organized as follows: the main result (Theorem2.8) is presented in Section 2, and the example to show the main result is given in Section 3.


## 2. Main Results

In this section, we first present some background definitions in Banach spaces and state an important three fixed point theorem. Then the main results are given and proved.Definition 2.1.
LetX be a real Banach space. The nonempty convex closed subset P of X is called a cone in X if ax∈P for all x∈P and a≥0 and x∈X and -x∈X imply x=0.Definition 2.2.
A mapψ:P→[0,+∞) is a nonnegative continuous concave or convex functional map provided ψ is nonnegative, continuous, and satisfies ψ(tx+(1-t)y)≥tψ(x)+(1-t)ψ(y), or ψ(tx+(1-t)y)≤tψ(x)+(1-t)ψ(y), for all x,y∈P and t∈[0,1].Definition 2.3.
An operatorT:X→X is completely continuous if it is continuous and maps bounded sets into precompact sets.Definition 2.4.
Leta,b,c,d,h>0 be positive constants, α,ψ be two nonnegative continuous concave functionals on the cone P, γ,β,θ be three nonnegative continuous convex functionals on the cone P. Define the convex sets as follows:
(2.1)Pc={x∈P:∥x∥<c},P(γ,α;a,c)={x∈P:α(x)≥a,γ(x)≤c},P(γ,θ,α;a,b,c)={x∈P:α(x)≥a,θ(x)≤b,γ(x)≤c},Q(γ,β;,d,c)={x∈P:β(x)≤d,γ(x)≤c},Q(γ,β,ψ;h,d,c)={x∈P:ψ(x)≥h,β(x)≤d,γ(x)≤c}.Lemma 2.5 (see [10]).
LetX be a real Banach space, P be a cone in X, α,ψ be two nonnegative continuous concave functionals on the cone P,γ,β,θ be three nonnegative continuous convex functionals on the cone P. Assume that there exists a constant M>0 such that
(2.2)α(x)≤β(x),∥x∥≤Mγ(x),∀x∈P.
Furthermore, suppose that h,d,a,b,c>0 are constants with d<a. Let T:Pc¯→Pc¯ be a completely continuous operator. If(C1)
{x∈P(γ,θ,α;a,b,c)∣α(x)>a}≠∅ and α(Tx)>aforeveryx∈P(γ,θ,α;a,b,c);(C2)
{x∈Q(γ,θ,ψ;h,d,c)∣β(x)<d}≠∅ and β(Tx)<dforeveryx∈Q(γ,θ,ψ;h,d,c);(C3)
α(Tx)>a for x∈P(γ,α;a,c) with θ(Tx)>b;(C4)
β(Tx)<d for each x∈Q(γ,β;d,c) with ψ(Tx)<h,then T has at least three fixed points x1, x2 and x3 such that β(x1)<d,α(x2)>a,β(x3)>d,α(x3)<a.Let us list the assumptions(H1)
g,h:R→[0,∞) satisfy ∫-∞+∞‍g(s)ds<1, ∫-∞+∞‍h(s)ds<1 and
(2.3)g(t)(1-∫-∞+∞h(s)ds)-h(t)(1-∫-∞+∞g(s)ds)≥0,t∈R.(H2)
ρ∈C(R),ρ(t)>0 for t∈R with ∫-∞0‍Φ-1(1/ρ(t))dt=∫0+∞‍Φ-1(1/ρ(t))dt<+∞.(H3)
f(t,c,0)≢0 on any finite subinterval of R for each c∈R, f:R3→R is a Carathédory function, that is,(i)
t→f(t,x,(1/Φ-1(ρ(t)))y) is measurable for any (x,y)∈R2,(ii)
(x,y)→f(t,x,(1/Φ-1(ρ(t)))y) is continuous for a.e. t∈R,(iii)
for eachr>0, there exists nonnegative function ϕr∈L1(R) such that max{|u|,|v|}≤r implies
(2.4)|f(t,u,1Φ-1(ρ(t))v)|≤ϕr(t),a.e.t∈R.Choose(2.5)X={andthereexistthelimitslimt→-∞x(t),x∈C1(R):limt→+∞x(t),limt→-∞Φ-1(ρ(t))x′(t)andlimt→+∞Φ-1(ρ(t))x′(t)}.
For x∈X, define the norm of x by
(2.6)∥x∥=max{supt∈R|x(t)|,supt∈RΦ-1(ρ(t))x′(t)}.
One can prove that X is a Banach space with the norm ||x|| for x∈X.Letx∈X. Consider the following auxiliary BVP
(2.7)[ρ(t)Φ(y′(t))]′+f(t,x(t),x′(t))=0,t∈R,y(-∞)=∫-∞+∞g(s)y(s)ds,y(+∞)=∫-∞+∞h(s)y(s)ds.Lemma 2.6.
Suppose that (H1)–(H3) hold. Ify∈C1(R) such that [ρΦ(y')]'∈L1(R) is a solution of BVP(2.7), then(i)
y is bounded and nonnegative on R;(ii)
y(t) is concave with respect to τ=∫-∞t‍Φ-1(1/ρ(s))ds/∫-∞+∞‍Φ-1(1/ρ(s))ds;(iii)
fork>0, it holds that mint∈[-k,k]y(t)≥μsupt∈Ry(t) with μ=∫-∞-k‍Φ-1(1/ρ(s))ds/2∫-∞+∞‍Φ-1(1/ρ(s))ds;(iv)
there exists a unique constantAx∈[-∫-∞+∞‍f(s,x(s),x'(s))ds,0] such that
(2.8)∫-∞+∞g(t)(1-∫-∞+∞h(s)ds)-h(t)(1-∫-∞+∞g(s)ds)1-∫-∞+∞g(s)ds×∫-∞tΦ-1(1ρ(s))Φ-1(Ax+∫s+∞f(u,x(u),x′(u))du)dsdt+∫-∞+∞Φ-1(1ρ(s))Φ-1(Ax+∫s+∞f(u,x(u),x'(u))du)ds=0.Proof.
Sincex∈X, we get
(2.9)r=max{supt∈R|x(t)|,supt∈RΦ-1(ρ(t))x′(t)}<+∞.
Then there exists a nonnegative function ϕr∈L1(-∞,+∞) such that(2.10)0≤f(t,x(t),x′(t))=f(t,x(t),1Φ-1(ρ(t))Φ-1(ρ(t))x′(t))≤ϕr(t),t∈R.
Then
(2.11)∫-∞+∞f(s,x(s),x′(s))dsisconvergent.
(i) We know there exist the limitslimt→-∞ρ(t)Φ(y'(t)) and limt→+∞ρ(t)Φ(y'(t)). We claim that there exists τ0∈R such that y'(τ0)=0. In fact, if y'(t)>0 for all t∈R, we get y(-∞)<y(t)<y(+∞) for all t∈R. It follows that
(2.12)y(-∞)≥∫-∞+∞g(s)dsy(-∞),y(+∞)≤∫-∞+∞h(s)dsy(+∞).
Then (H1) implies y(-∞)≥0≥y(+∞), which contradicts to y'(t)>0 for all t∈R. Similarly we can prove that y'(t)<0 does not hold. Then there exists τ0∈R such that y'(τ0)=0.
Since[ρ(t)Φ(y'(t))]'=-f(t,x(t),x'(t))≤0, we know that ρ(t)Φ(y'(t)) is decreasing on R. Then ρ(t)Φ(y'(t))≥0 for all t≤τ0 and ρ(t)Φ(y'(t))≤0 for all t≥τ0. Hence
(2.13)yisincreasingfort∈(-∞,τ0]anddecreasingfort∈[τ0,+∞).
One sees that
(2.14)y′(t)={-Φ-1(1ρ(t)∫τ0tf(u,x(u),x′(u))du)=:-G(t),t≥τ0,Φ-1(1ρ(t)∫tτ0f(u,x(u),x′(u))du)=:H(t),t≤τ0.
Since ∫-∞+∞‍Φ-1(1/ρ(t))dt<+∞ and ∫-∞+∞‍f(t,x(t),x'(t))dt<+∞, we see that
(2.15)both∫t+∞G(s)ds,∫-∞tH(s)dsareconvergent.
Then we get that
(2.16)y(t)={y(+∞)+∫t+∞G(s)ds,t≥τ0,y(-∞)+∫-∞tH(s)ds,t≤τ0.
This tells us that y is bounded on R.
It follows from (2.7) and (2.16) that
(2.17)y(-∞)=∫-∞τ0g(s)dsy(-∞)+∫τ0+∞g(s)dsy(+∞)+∫-∞τ0g(t)∫-∞tH(s)dsdt+∫τ0+∞g(t)∫t+∞G(s)dsdt,y(+∞)=∫-∞τ0h(s)dsy(-∞)+∫τ0+∞h(s)dsy(+∞)+∫-∞τ0h(t)∫-∞tH(s)dsdt+∫τ0+∞h(t)∫t+∞G(s)dsdt.
Then
(2.18)(1-∫-∞τ0‍g(s)ds)y(-∞)-∫τ0+∞g(s)dsy(+∞)=∫-∞τ0g(t)∫-∞ttH(s)dsdt+∫τ0+∞g(t)∫t+∞G(s)dsdt,-∫-∞τ0h(s)dsy(-∞)+(1-∫τ0+∞h(s)ds)y(+∞)=∫-∞τ0h(t)∫-∞tH(s)dsdt+∫τ0+∞h(t)∫t+∞G(s)dsdt.
Hence
(2.19)y(-∞)=|∫-∞τ0g(t)∫-∞tH(s)dsdt+∫τ0+∞g(t)∫t+∞G(s)dsdt-∫τ0+∞‍g(s)ds∫-∞τ0h(t)∫-∞tH(s)dsdt+∫τ0+∞h(t)∫t+∞G(s)dsdt1-∫τ0+∞‍h(s)ds||1-∫-∞τ0g(s)ds-∫τ0+∞g(s)ds-∫-∞τ0h(s)ds1-∫τ0+∞h(s)ds|,y(+∞)=|1-∫-∞τ0g(s)ds∫-∞τ0g(t)∫-∞tH(s)dsdt+∫τ0+∞g(t)∫t+∞G(s)dsdt-∫-∞τ0h(s)ds∫-∞τ0h(t)∫-∞tH(s)dsdt+∫τ0+∞h(t)∫t+∞G(s)dsdt||1-∫-∞τ0g(s)ds-∫τ0+∞g(s)ds-∫-∞τ0h(s)ds1-∫τ0+∞h(s)ds|.
Since(2.20)|1-∫-∞τ0g(s)ds-∫τ0+∞g(s)ds-∫-∞τ0h(s)ds1-∫τ0+∞h(s)ds|=|1-∫-∞+∞g(s)ds-∫τ0+∞g(s)ds1-∫-∞+∞h(s)ds1-∫τ0+∞h(s)ds|>0,
we get from (2.19) that y(-∞)≥0 and y(+∞)≥0. Hence (2.16) implies that
(2.21)y(t)≥0,∀t∈R.
(ii) We prove thaty(t) is concave with respect to τ on R. It is easy to see that τ∈C(R,(0,1)) and
(2.22)dτdt=Φ-1(1ρ(t))1∫-∞+∞Φ-1(1/ρ(s))ds>0.
Thus
(2.23)dydt=dydτdτdt=dydτΦ-1(1ρ(t))1∫-∞+∞Φ-1(1/ρ(s))ds.
It follows that
(2.24)ρ(t)Φ(dydt)=Φ(dydτ)Φ(1∫-∞+∞Φ-1(1/ρ(s))ds).
Hence
(2.25)[ρ(t)Φ(dydt)]′=Φ′(dydτ)d2ydτ2dτdtΦ(1∫-∞+∞Φ-1(1/ρ(s))ds).
So
(2.26)d2ydτ2=Φ(∫-∞+∞‍Φ-1(1ρ(s))ds)[ρ(t)Φ(dy/dt)]′Φ'(dy/dτ)(dτ/dt).
Since [ρ(t)Φ(y'(t))]′≤0,Φ′(y)>0(y≠0) and (dτ/dt)>0, we get that (d2y/dτ2)≤0. Hence y(t) is concave with respect to τ on R.
(iii) Now, we prove that(2.27)mint∈[-k,k]y(t)≥μsupt∈Ry(t).
Since dτ/dt>0 for all t∈R, there exists the inverse function of τ=τ(t). Denote the inverse function of τ=τ(t) by t=t(τ).
It follows from (2.13) that supt∈Ry(t)=y(τ0). One sees
(2.28)mint∈[-k,k]y(t)=min{y(-k),y(k)}.
If min{y(-k),y(k)}=y(k)=y(t(τ(k))), note τ(k)<1, then for t∈[-k,k], one has
(2.29)y(t)≥y(t(τ(k)))=y(t(1-τ(k)+τ(τ0)1+τ(τ0)τ(k)1-τ(k)+τ(τ0)+τ(k)1+τ(τ0)τ(τ0))).
Noting that 1>τ(k) and y(t) is concave with respect to τ, then, for t∈[-k,k],
(2.30)y(t)≥1-τ(k)+τ(τ0)1+τ(τ0)y(t(τ(k)1-τ(k)+τ(τ0)))+τ(k)1+τ(τ0)y(t(τ(τ0)))≥∫-∞kΦ-1(1ρ(s))ds12∫-∞+∞Φ-1(1/ρ(s))dsy(τ0)=μsupt∈Ry(t).
Similarly, if min{y(-k),y(k)}=y(-k)=(y(t(τ(-k))), note τ(-k)<1, for t∈[-k,k], one has
(2.31)y(t)≥y(t(τ(-k)))=y(t(1+τ(τ0)-τ(-k)1+τ(τ0)τ(-k)1+τ(τ0)-τ(-k)+τ(-k)1+τ(τ0)τ(τ0)))≥1+τ(τ0)-τ(-k)1+τ(τ0)y(t(τ(-k)1+τ(τ0)-τ(-k)))+τ(-k)1+τ(τ0)y(t(τ(τ0)))≥∫-∞-kΦ-1(1ρ(s))ds12∫-∞+∞Φ-1(1/ρ(s))dsy(τ0)>μsupt∈Ry(t).
Hence (2.27) holds.
(iv) Finally, we prove the uniqueness ofAx. Define
(2.32)Hx(c)=∫-∞+∞g(t)(1-∫-∞+∞h(s)ds)-h(t)(1-∫-∞+∞g(s)ds)1-∫-∞+∞g(s)ds×∫-∞tΦ-1(1ρ(s))Φ-1(c+∫s+∞f(u,x(u),x′(u))du)dsdt+∫-∞+∞Φ-1(1ρ(s))Φ-1(c+∫s+∞f(u,x(u),x′(u))du)ds.
Then (H1) implies that Hx∈C(R,R) is increasing on R and Hx(0)≥0. Let
(2.33)c¯=-∫-∞+∞f(u,x(u),x′(u))du,
then Hx(c¯)≤0. By mean value theorem, there exists an unique Ax∈[c¯,0] satisfying Hx(Ax)=0. Then (iv) holds. This completes the proof of the lemma.Choose1≥k>0 and
(2.34)μ=∫-∞-kΦ-1(1/ρ(s))ds2∫-∞∞Φ-1(1/ρ(s))ds.
Define the cone P⊆X by
(2.35)P={x∈X:x(t)≥0,t∈R,mint∈[-k,k]x(t)≥μmaxt∈Rx(t)}.
It is easy to see that P is a cone in X.Define the operatorT:P→X by
(2.36)(Tx)(t)=∫-∞+∞g(u)∫-∞uΦ-1(1/ρ(s))Φ-1(Ax+∫s+∞f(r,x(r),x′(r))dr)dsdu1-∫-∞+∞g(s)ds+∫-∞tΦ-11ρ(s)Φ-1(Ax+∫s+∞f(r,x(r),x′(r))dr)ds,
where Ax satisfies
(2.37)∫-∞+∞g(u)(1-∫-∞+∞h(s)ds)-h(u)(1-∫-∞+∞g(s)ds)1-∫-∞+∞g(s)ds×∫-∞uΦ-1(1ρ(s))Φ-1(Ax+∫s+∞f(r,x(r),x′(r))dr)dsdu+∫-∞+∞Φ-1(1ρ(s))Φ-1(Ax+∫s+∞‍f(r,x(r),x′(r))dr)ds=0.
It follows from Lemma 2.6(iv) that T:P→X is well defined and Ax∈[-∫-∞+∞‍f(s,x(s),x'(s))ds,0]. It is easy to show that
(2.38)[ρ(t)Φ((Tx)′(t))]′+f(t,x(t)x′(t))=0,t∈R,(Tx)(-∞)=∫-∞+∞g(s)(Tx)(s)ds,(Tx)(+∞)=∫-∞+∞h(s)(Tx)(s)ds.
It follows from Lemma 2.6(i) and (iii) that Tx∈P for all x∈P. Then T:P→P is well defined.Lemma 2.7.
Suppose (H1)–(H3) hold. ThenT is completely continuous.Proof.
First, we prove thatT is continuous.
We claim that the functionAx:P→[0,+∞) is continuous in x.
Let{xn}∈P with xn→x0∈P as n→∞ in P. Let {Axn}(n=1,2,…) be the constants determined by
(2.39)∫-∞+∞g(t)(1-∫-∞+∞h(s)ds)-h(t)(1-∫-∞+∞g(s)ds)1-∫-∞+∞g(s)ds×∫-∞tΦ-1(1ρ(s))Φ-1(Axn+∫s+∞f(r,xn(r),xn′(r))dr)dsdt+∫-∞+∞Φ-1(1ρ(s))Φ-1(Axn+∫s+∞f(r,xn(r),xn′(r))dr)ds=0
Since xn→x0 in P as n→∞, there exists an r>0 such that ∥xn∥≤r. The fact f is a Carathédory function means that there exists a nonnegative function ϕr∈L1(-∞,+∞) such that
(2.40)0≤f(t,xn(t),xn′(t))=f(t,xn(t),1Φ-1(ρ(t))Φ-1(ρ(t))xn′(t))≤ϕr(t),t∈R.
Then
(2.41)∫-∞+∞f(s,xn(s),xn′(s))dsisconvergent.
So
(2.42)Axn∈[-∫-∞+∞f(s,xn(s),xn′(s))ds,0]⊆[-∫-∞+∞ϕr(s)ds,0],
which means that {Axn} is uniformly bounded.
Suppose that{Axn} does not converge to Ax0. By the bounded property, we know that there exist two subsequences {Axnk(1)} and {Axnk(2)} of {Axnk} with Axnk(1)→c1 and Axnk(2)→c2 and c1≠c2.
By the construction ofAxn, (n=1,2,…), we have
(2.43)∫-∞+∞g(t)(1-∫-∞+∞h(s)ds)-h(t)(1-∫-∞+∞g(s)ds)1-∫-∞+∞g(s)ds×∫-∞tΦ-1(1ρ(s))Φ-1(Axnk(1)+∫s+∞‍f(u,xnk(u),xnk′(u))du)dsdt+∫-∞+∞Φ-1(1ρ(s))Φ-1(Axnk(1)+∫s+∞f(u,xnk(u),xnk′(u))du)ds=0.
Let k→∞, using Lebesgue's dominated convergence theorem, the above equality implies
(2.44)∫-∞+∞g(t)(1-∫-∞+∞h(s)ds)-h(t)(1-∫-∞+∞g(s)ds)1-∫-∞+∞g(s)ds×∫-∞t‍Φ-1(1ρ(s))Φ-1(c1+∫s+∞f(u,x0(u),xo′(u))du)dsdt+∫-∞+∞Φ-1(1ρ(s))Φ-1(c1+∫s+∞f(u,x0(u),xo′(u))du)ds=0.
By Lemma 2.6 (iv), we get c1=Ax0. Similarly, c2=Ax0. Thus c1=c2, a contradiction. So, for any xn→x0, one has Axn→Ax0, which means Ax:P→R is continuous.
SinceAx is continuous, together with the continuity of (x,y)→f(t,x,(1/Φ-1(ρ(t)))y), we get that T is continuous.
Second, we show thatT is maps bounded subsets into bounded sets.
LetD⊆P be bounded. Then, there exists r>0 such that D⊆{x∈P:∥x∥≤r}. Hence
(2.45)r=max{supt∈R|x(t)|,supt∈RΦ-1(ρ(t))x′(t)}<+∞.
Then there exists a nonnegative function ϕr∈L1(-∞,+∞) such that
(2.46)0≤f(t,x(t),x′(t))=f(t,x(t),1Φ-1(ρ(t))Φ-1(ρ(t))x′(t))≤ϕr(t),t∈R.
Then
(2.47)∫-∞+∞f(t,x(t),x′(t))dt≤∫-∞+∞ϕr(t)dt:=L.
Thus |Ax|≤L for all x∈D. Therefore,
(2.48)(Tx)(t)=∫-∞+∞g(t)∫-∞tΦ-1(1/ρ(s))Φ-1(Ax+∫s+∞f(u,x(u),x′(u))du)dsdt1-∫-∞+∞g(s)ds+∫-∞tΦ-1(1ρ(s))Φ-1(Ax+∫s+∞f(u,x(u),x′(u))du)ds≤(∫-∞+∞g(t)∫-∞t‍Φ-1(1/ρ(s))dsdt1-∫-∞+∞g(s)ds+∫-∞tΦ-1(1ρ(s))ds)Φ-1(2L)≤(∫-∞+∞g(t)∫-∞t‍Φ-1(1/ρ(s))dsdt1-∫-∞+∞g(s)ds+∫-∞+∞Φ-1(1ρ(s))ds)Φ-1(2L)=:M1.
On the other hand, we have
(2.49)Φ-1(ρ(t))|(Tx)′(t)|=|Φ-1(Ax+∫t∞f(u,x(u),x'(u))du)|≤Φ-1(2L)=:M2.
Then
(2.50)∥(Tx)∥=max{supt∈R(Tx)(t),supt∈RΦ-1(ρ(t))(Tx)′(t)}<∞.
So, {Tx:x∈D} is bounded.
Third, given a bounded setD⊆P, we prove that both {Tx:x∈D} and {Φ-1(ρ(t))(Tx)':x∈D} are equicontinuous on each finite subinterval on R.
Then, there existsr>0 such that D⊆{x∈P:∥x∥≤r}. Hence
(2.51)r=max{supt∈R|x(t)|,supt∈RΦ-1(ρ(t))x′(t)}<+∞.
Then there exists a nonnegative function ϕr∈L1(-∞,+∞) such that
(2.52)0≤f(t,x(t),x′(t))=f(t,x(t),1Φ-1(ρ(t))Φ-1(ρ(t))x′(t))≤ϕr(t),t∈R.
Then
(2.53)|Ax+∫t+∞‍f(t,x(t),x′(t))dt|≤∫-∞+∞‍ϕr(t)dt:=L.
For anyϵ>0, since Φ-1 is uniformly continuous on [-L,L], there exists δ1>0 such that
(2.54)|Φ-1(u)-Φ-1(v)|<ϵ,u,v∈[-L,L],|u-v|<δ1.
For any a,b∈R,t1,t2∈[a,b] and x∈D with t1≤t2, we have
(2.55)|Ax+∫t1+∞f(t,x(t),x′(t))dt-Ax-∫t2+∞f(t,x(t),x′(t))dt|≤∫t1t2ϕr(s)ds.
Then there exists δ>0 such that
(2.56)|Ax+∫t1+∞f(t,x(t),x′(t))dt-Ax-∫t2+∞f(t,x(t),x′(t))dt|<δ1,t1,t2∈[a,b],|t1-t2|<δ.
Hence t1,t2∈[a,b],|t1-t2|<δ imply that
(2.57)|Φ-1(ρ(t1))(Tx)(t1)-Φ-1(ρ(t2))(Tx)(t2)|=|Φ-1(Ax+∫t1+∞‍f(t,x(t),x′(t))dt)-Φ-1(Ax+∫t2+∞f(t,x(t),x′(t))dt)|<ϵ.
It follows that {Tx:x∈D} is equicontinuous on each finite subinterval on R.
On the other hand, we have(2.58)|(Tx)(t1)-(Tx)(t2)|=∫t1t2‍Φ-1(1ρ(s))Φ-1(Ax+∫s+∞‍f(u,x(u),x′(u))du)ds≤∫t1t2‍Φ-1(1ρ(s))dsΦ-1(2L)→0uniformlyast1→t2.
Then {Φ-1(ρ(t))(Tx)':x∈D} is equicontinuous on each finite subinterval on R.
At last given a bounded setD⊆P, we show that both {Tx:x∈D} and {Φ-1(ρ(t))(Tx)':x∈D} are equiconvergent at ±∞, respectively.
Then, there existsr>0 such that D⊆{x∈P:∥x∥≤r}. Hence
(2.59)r=max{supt∈R|x(t)|,supt∈RΦ-1(ρ(t))x'(t)}<+∞.
Then there exists a nonnegative function ϕr∈L1(-∞,+∞) such that
(2.60)0≤f(t,x(t),x′(t))=f(t,x(t),1Φ-1(ρ(t))Φ-1(ρ(t))x′(t))≤ϕr(t),t∈R.
Then
(2.61)|Ax+∫t+∞f(t,x(t),x′(t))dt|≤∫-∞+∞ϕr(t)dt:=L.
For anyϵ>0, since Φ-1 is uniformly continuous on [-L,L], there exists δ1>0 such that
(2.62)|Φ-1(u)-Φ-1(v)|<ϵ,u,v∈[-L,L],|u-v|<δ1.
Since
(2.63)|ρ(t)Φ((Tx)'(t))-Ax|=∫t∞f(u,x(u),x′(u))du≤∫t∞ϕr(u)du→0
uniformly as t→∞, we get that there exists T>0 such that
(2.64)|ρ(t)Φ((Tx)′(t))-Ax|<δ1,t≥T.
Hence t>T implies that
(2.65)|Φ-1(ρ(t))(Tx)′(t)-Φ-1(Ax)|=|Φ-1(ρ(t)Φ((Tx)′(t)))-Φ-1(Ax)|<ϵ.
Furthermore, we get
(2.66)|(Tx)(t)-∫-∞+∞g(t)∫-∞tΦ-1(1/ρ(s))Φ-1(Ax+∫s+∞‍f(u,x(u),x′(u))du)dsdt1-∫-∞+∞g(s)ds+∫-∞+∞Φ-1(1ρ(s))Φ-1(Ax+∫s+∞f(u,x(u),x′(u))du)ds∫-∞+∞g(t)1-∫-∞+∞g(s)|=∫t+∞Φ-1(1ρ(s))Φ-1(Ax+∫s∞f(u,x(u),x′(u))du)ds≤∫t+∞Φ-1(1ρ(s))dsΦ-1(2L)→0uniformlyast→+∞.
Hence {Tx:x∈D} and {Φ-1(ρ(t))(Tx)':x∈D} are equiconvergent at +∞.
Similarly we can how that{Tx:x∈D} and {Φ-1(ρ(t))(Tx)':x∈D} are equiconvergent at -∞. We omit the details.
Therefore,T:P→P is equiconvergent at ±∞. So the operator T:P→P is completely continuous.
Define the functionals onP by
(2.67)γ(y)=supt∈RΦ-1(ρ(t))|y′(t)|,y∈P,β(y)=supt∈R|y(t)|,y∈P,θ(y)=supt∈R|y(t)|,y∈P,α(y)=mint∈[-k,k]|y(t)|,y∈P,ψ(y)=mint∈[-k,k]|y(t)|,y∈P.
It is easy to see that α,ψ are two nonnegative continuous concave functionals on the cone P, γ,β,θ are three nonnegative continuous convex functionals on the cone P and α(y)≤β(y) for all y∈P.
Fore1,e2,c>0 and 1≥k>0, define
(2.68)M=max{∫-∞+∞(1/Φ-1(ρ(s)))ds1-∫-∞+∞g(s)ds,1},L1=∫-k0Φ-1(1ρ(s))Φ-1(2|s|)ds,L2=∫0kΦ-1(1ρ(s))Φ-1(2s)ds,Q=min{Φ(c)4+π,Φ(c)4+πΦ(1-∫-∞+∞‍g(s)ds∫-∞+∞‍(1/Φ-1(ρ(s)))ds)},W=Φ(e2μ)max{Φ(1L1),Φ(1L2)},E=Φ(e1)4+π,δ(t)={1t2,|t|≥1,1|t|,|t|≤1,μ=∫-∞-kΦ-1(1/ρ(s))ds2∫-∞+∞Φ-1(1/ρ(s))ds.Theorem 2.8.
Suppose that (H1)–(H3) hold. Given positive constantse1,e2,c and k∈(0,1), let Q,W, and E be as above. If
(2.69)c≥e2μ>e2>e1>0,Q≥W
and(A1)
f(t,u,(1/Φ-1(ρ(t)))v)≤δ(t)Q  for all   t∈R,u∈[0,c],v∈[-c,c];(A2)
f(t,u,(1/Φ-1(ρ(t)))v)≥δ(t)W  for all  t∈[-k,k],u∈[e2,e2/μ],v∈[-c,c];(A3)
f(t,u,(1/Φ-1(ρ(t)))v)≤δ(t)E  for all  t∈R,u∈[0,e1],v∈[-c,c];
then BVP(1.2) has at least three positive solutions x1,x2,x3 such that
(2.70)supt∈Rx1(t)<e1,mint∈[-k,k]x2(t)>e2,supt∈Rx3(t)>e1,mint∈[-k,k]x3(t)<e2.Proof.
We prove that all conditions in Lemma2.5 are satisfied.
(i) By the definitions, it is easy to show thatα,ψ are two nonnegative continuous concave functionals on the cone P, γ,β,θ are three nonnegative continuous convex functionals on the cone P and α(y)≤β(y) for all y∈P. One sees x∈P is a positive solution of BVP (1.2) if and only if x is a solution of the operator equation x=Tx.
(ii) Fory∈P, we have
(2.71)|y(t)|=|∫-∞ty′(s)ds+y(-∞)|≤∫-∞t|y′(s)|ds+|y(-∞)|≤∫-∞t1Φ-1(ρ(s))|Φ-1(ρ(s))y′(s)|ds+∫-∞+∞g(s)|y(s)|ds≤∫-∞+∞1Φ-1(ρ(s))dssupt∈RΦ-1(ρ(t))|y′(t)|∫-∞+∞g(s)dssupt∈R|y(t)|.
It follows that
(2.72)supt∈R|y(t)|≤∫-∞+∞1Φ-1(ρ(s))dssupt∈RΦ-1(ρ(t))|y′(t)|∫-∞+∞g(s)dssupt∈R|y(t)|.
Then
(2.73)supt∈R|y(t)|≤∫-∞+∞(1/Φ-1(ρ(s)))ds1-∫-∞+∞g(s)dssupt∈RΦ-1(ρ(t))|y′(t)|.
Hence
(2.74)∥y∥=max{supt∈R|y(t)|,supt∈RΦ-1(ρ(t))|y′(t)|}≤max{∫-∞+∞(1/Φ-1(ρ(s)))ds1-∫-∞+∞g(s)ds,1}supt∈RΦ-1(ρ(t))|y′(t)|.
It follows that ||y||≤Mγ(y) for all y∈P.
(iii) Corresponding to Lemma2.5,
(2.75)c=c,h=μe1,d=e1,a=e2,b=e2μ.
Now, we prove that all other conditions of Lemma 2.5 hold. One sees that 0<d<a. The remainder is divided into five steps.
Step1.Prove that T:Pc¯→Pc¯.
Fory∈Pc¯, we have ||y||≤c. Then 0≤y(t)≤c and -c≤Φ-1(ρ(t))y'(t)≤c for all t∈R. So (A1) implies that
(2.76)f(t,y(t),y′(t))=f(t,y(t),1Φ-1(ρ(t))Φ-1(ρ(t))y′(t))≤δ(t)Q,t∈R.
We have
(2.77)Φ-1(ρ(t))|(Ty)′(t)|=|Φ-1(Ay+∫t+∞f(u,y(u),y′(u))du)|≤Φ-1(∫-∞+∞|f(u,y(u),y′(u))|du)≤Φ-1(∫-∞+∞δ(s)Qdr)=Φ-1(Q)Φ-1(4+π)≤c.
Similarly to (ii), we can show that
(2.78)0≤(Ty)(t)≤Msupt∈RΦ-1(ρ(t))|(Ty)′(t)|≤MΦ-1(∫-∞+∞‍δ(s)Qdr)=MΦ-1(Q)Φ-1(4+π)≤c.
It follows that
(2.79)∥Ty∥=max{supt∈R|(Ty)(t)|,supt∈RΦ-1(ρ(t))|(Ty)′(t)|}≤c.
Then T:Pc¯→Pc¯.
Step2. Prove that
(2.80){y∈P(γ,θ,α;a,b,c)∣α(y)>a}={y∈P(γ,θ,α;e2,e2μ,c)∣α(y)>e2}≠∅
and α(Ty)>e2foreveryy∈P(γ,θ,α;e2,e2/μ,c).
Choosey(t)=e2/2μ for all t∈R. Then y∈P and
(2.81)α(y)=e22μ>e2,θ(y)=e22μ≤e2μ,γ(y)=0<c.
It follows that {y∈P(γ,θ,α;a,b,c)∣α(y)>a}≠∅.
Fory∈P(γ,θ,α;a,b,c), one has that
(2.82)α(y)=mint∈[-k,k]y(t)≥e2,θ(y)=supt∈Ry(t)≤e2μ,γ(y)=supt∈R|y′(t)|≤c.
Then
(2.83)e2≤y(t)≤e2μ,t∈[-k,k],Φ-1(ρ(t))|y′(t)|≤c.
Thus (A2) implies that
(2.84)f(t,y(t),y′(t))≥δ(t)W,t∈[-k,k].
Similarly to Lemma 2.6(i), we know that there exists τ0∈(0,1) such that (Ty)'(τ0)=0. Then
(2.85)(Ty)(t)={(Ty)(+∞)+∫t+∞Φ-1(1ρ(s))Φ-1(∫τ0sf(u,y(u),y′(u))du)ds,t≥τ0,(Ty)(-∞)+∫-∞tΦ-1(1ρ(s))Φ-1(∫sτ0f(u,y(u),y′(u))du)ds,t≤τ0.
Since
(2.86)α(Ty)=mint∈[-k,k](Ty)(t)≥μsupt∈R(Ty)(t)=μ(Ty)(τ0),
if τ0≥0, we get
(2.87)α(Ty)≥μ[(Ty)(-∞)+∫-∞τ0Φ-1(1ρ(s))Φ-1(∫sτ0f(u,y(u),y′(u))du)ds]>μ[∫-k0Φ-1(1ρ(s))Φ-1(∫s0f(u,y(u),y′(u))du)ds]≥μ[∫-k0Φ-1(1ρ(s))Φ-1(∫s0δ(u)Wdu)ds]≥μ[∫-k0‍Φ-1(1ρ(s))Φ-1(2|s|)ds]Φ-1(W)≥e2.
If τ0<0, we get
(2.88)α(Ty)≥μ[(Ty)(+∞)+∫τ0+∞Φ-1(1ρ(s))Φ-1(∫τ0sf(u,y(u),y′(u))du)ds]>μ[∫τ0kΦ-1(1ρ(s))Φ-1(∫τ0sf(u,y(u),y′(u))du)ds]≥μ[∫0kΦ-1(1ρ(s))Φ-1(∫0sδ(u)Wdu)ds]≥μ[∫0kΦ-1(1ρ(s))Φ-1(2t)ds]Φ-1(W)≥e2.
This completes Step 2.
Step3.Prove that {y∈Q(γ,θ,ψ;h,d,c)∣β(y)<d}={y∈Q(γ,θ,ψ;μe1,e1,c)|β(y)<e1}≠∅ and
(2.89)β(Ty)<e1foreveryy∈Q(γ,θ,ψ;h,d,c)=Q(γ,θ,ψ;μe1,e1,c).
Choosey(t)=μe1. Then y∈P, and
(2.90)ψ(y)=μe1≥h,β(y)=θ(y)=μe1<e1=d,γ(y)=0≤c.
It follows that {y∈Q(γ,θ,ψ;h,d,c)∣β(y)<d}≠∅.
Fory∈Q(γ,θ,ψ;h,d,c), one has that
(2.91)ψ(y)=mint∈[-k,k]y(t)≥h=μe1,θ(y)=supt∈Ry(t)≤d=e1,γ(y)=supt∈R|y′(t)|≤c.
Hence we get that
(2.92)0≤y(t)≤e1,t∈R;-c≤Φ-1(ρ(t))y′(t)≤c,t∈R.
Then (A3) implies that
(2.93)f(t,y(t),y′(t))≤δ(t)E,t∈R.
So
(2.94)β(Ty)=Msupt∈RΦ-1(ρ(t))|(Ty)′(t)|≤M|Φ-1(Ay+∫t+∞f(u,y(u),y′(u))du)|≤MΦ-1(∫-∞+∞|f(u,y(u),y′(u))|du)≤MΦ-1(∫-∞+∞δ(u)Edu)≤MΦ-1(4+π)Φ-1(E)=e1=d.
This completes Step 3.
Step4.Prove that α(Ty)>a for y∈P(γ,α;a,c) with θ(Ty)>b;
Fory∈P(γ,α;a,c)=P(γ,α;e2,c) with θ(Ty)>b=e2/μ, we have that α(y)=mint∈[-k,k]y(t)≥e2 and γ(y)=supt∈RΦ-1(ρ(t))|y'(t)|≤c and supt∈R(Ty)(t)>e2/μ. Then
(2.95)α(Ty)=mint∈[-k,k](Ty)(t)≥μβ(Ty)>μe2μ=e2=a.
This completes Step 4.
Step5. Prove that β(Ty)<d for each y∈Q(γ,β;d,c) with ψ(Ty)<h.
Fory∈Q(γ,β;d,c) with ψ(Ty)<d, we have γ(y)=supt∈RΦ-1(ρ(t))|y'(t)|≤c and β(y)=supt∈Ry(t)≤d=e1 and ψ(Ty)=mint∈[-k,k](Ty)(t)<h=e1μ. Then
(2.96)β(Ty)=supt∈R(Ty)(t)≤1μmint∈[-k,k](Ty)(t)<1μe1μ=e1=d.
This completes the Step 5.
Then Lemma2.5 implies that T has at least three fixed points x1, x2, and x3 such that
(2.97)β(x1)<e1,α(x2)>e2,β(x3)>e1,α(x3)<e2.
Hence BVP(1.2) has three decreasing positive solutions x1,x2 and x3 such that (2.70) holds. The proof is complete.


## 3. Examples

Now, we present an example, whose three positive solutions cannot be obtained by theorems in known papers, to illustrate the main results.Example 3.1.
Consider the following BVP(3.1)[e3t2(x′(t))3]′+f(t,x(t),x′(t))=0,t∈R,x(-∞)=0,x(+∞)=0.
Corresponding to BVP(1.2), one sees that  ϕ(x)=x3,ϕ-1(x)=x1/3, g(t)=h(t)≡0, ρ(t)=e3t2, f:R×R×R→[0,∞) is nonnegative and continuous and is defined by
(3.2)f(t,x,y)=δ(t)(f0(x)+g0(et2|y|)),δ(t)={1t2,|t|≥1,1|t|,|t|≤1.Choosek=1, e1=50,e2=250,c=40000. By direct computation, we see that Q,W, and E are given by
(3.3)M=max{∫-∞+∞(1/Φ-1(ρ(s)))ds1-∫-∞+∞g(s)ds,1}=π2,L1=∫-k0Φ-1(1ρ(s))Φ-1(2|s|)ds=21/3∫-10‍e-s2|s|1/6ds>0.06,L2=∫0kΦ-1(1ρ(s))Φ-1(2s)ds=21/3∫01e-s2s1/6ds>0.06,μ=∫-∞-kΦ-1(1/ρ(s))ds2∫-∞+∞‍Φ-1(1/ρ(s))ds=∫-∞-1e-s2dsπ>0.37,Q=min{Φ(c)4+π,Φ(c)4+πΦ(1-∫-∞+∞g(s)ds∫-∞+∞(1/Φ-1(ρ(s)))ds)}=64×10124+π>8.53×1012,W=Φ(e2μ)max{Φ(1L1),Φ(1L2)}<16.673×675.73<3.43×1011,E=Φ(e1)4+π=25004+π>333.33.
One can show that
(3.4)c≥e2μ>e2>e1>0,Q≥W.
Suppose that
(3.5)f0(x)={166.67,x∈[0,50],166.67+44.36×1011-166.67250-50(x-50),x∈[50,250],44.36×1011,x∈[250,40000],44.36×1011ex-40000,x≥40000,|g0(y)|≤10,∀y∈R.From(3.6)f(t,x,e-t2y)=δ(t)[f0(x)+g0(y)],
it is easy to show that(A1)
f(t,u,e-t2v)≤8.53×1012δ(t) for all t∈R,u∈[0,400000],v∈[-40000,40000];(A2)
f(t,u,e-t2v)≥3.43×1011δ(t) for all t∈[-1,1],u∈[250,1000],v∈[-40000,40000];(A3)
f(t,u,e-t2v)≤333.33δ(t) for all t∈R,u∈[0,50],v∈[-40000,40000];then Theorem2.8 implies that BVP(3.1) has at least three positive solutions x1,x2,x3 such that
(3.7)supt∈Rx1(t)<50,mint∈[-1,1]x2(t)>250,supt∈Rx3(t)>50,mint∈[-1,1]x3(t)<250.Remark 3.2.
Example3.1 implies that there is a large number of functions that satisfy the conditions of Theorem 2.8. In addition, the conditions of Theorem 2.8 are also easy to check.



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*Source: 352159-2012-09-25.xml*