# Solvability of a Third-Order Singular Generalized Left Focal Problem in Banach Spaces

Authors: Youwei Zhang
Journal: International Journal of Analysis
 (2014)

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## Abstract

We consider the existence of positive solution for a third-order singular generalized left focal boundary value problem with full derivatives in Banach spaces. Green’s function and its
properties, explicit a priori, estimates will be presented. By means of the theories of the fixed point in cones,
we establish some new and general results on the existence of single and multiple positive solutions to the
third-order singular generalized left focal boundary value problem. Our results are generalizations and
extensions of the results of the focal boundary value problem. An example is included to illustrate the
results obtained.


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## Body

## 1. Introduction

Third-order differential equation describes many phenomena in applied mathematics, physical science, aeronautics, and applied mechanics such as the study of steady flows produced by free jets, wall jets, liquid jets, the flow past a stretching plate, and Blasius flow [1–6]. For two-dimensional flow of a fluid with small viscosity adhering to the flat plate, the simplified version, which has been derived by Blasius, only uses two unknowns u and v. Laminar flat plate flow across a flat plate can be expressed using a boundary-layer equation defined by Blasius. Simplified momentum equation is given by
(1)u∂u∂x+v∂v∂y=v∂2u∂y2,u(x,0)=v(x,0)=0,u(x,∞)=U∞,
where u=u(x,y) is the velocity measured, v=v(x,y) is the transverse velocity component, x is length of the plate, y is the distance away from the plate, and U∞ is free-stream velocity. Assume that the leading edge of the plate is x=0 and the plate is infinity long; this equation can be simplified as
(2)-U∞22xςf′f′′+U∞22x(ςf′-f)=vU∞2xvf′′′,
where f=f(ς) is the dimensionless stream function. For further simplification, we have the Blasius differential equation
(3)f′f′′+2f′′′=0,
where the boundary conditions are ς=0,f=0, and f′=0 or ς=∞,f′=0. For other related application results of the problem we refer to [7–10]. Recently, increasing attention is paid to the question of the solution for the third-order focal boundary value problem, especially for right focal problem. The main means are the Leray-Schauder continuation theorem, iteration of monotone mapping, upper and lower solutions method, fixed point theory, and so on. Many applications of the above tools of various nonlocal boundary conditions include recent works [11–13] and the reference therein. It is well known that a powerful tool for proving the existence of the solution to the focal boundary value problem is the fixed point theorem. In many cases, it is possible to find single or multiple solutions for the given problem. We would like to mention some results of [14, 15]. Anderson [14] has investigated the existence of a solution to a third-order generalized right focal problem
(4)x′′′(t)=f(t,x(t)),t1<t<t3,x(t1)=x′(t2)=0,γx(t3)+δx′′(t3)=0,hhhhhhhhhhhhhhhhhhhhhhhht1<t2<t3,
for certain constraints placed on nonnegative f; the main approach is the Leggett and Williams fixed point theorem. Minhós [15] has proved an Ambrosetti-Prodi type result for the third-order fully nonlinear right focal problem
(5)u′′′(t)+f(t,u(t),u′(t),u′′(t))=sp(t),hhhhhhhhhhhhhhhhhhhhllhhh0<t<1,u(0)=A,au′(0)-bu′′(0)=B,cu′(1)+du′′(1)=C,
where f:[0,1]×R→R and p:[0,1]→R+ are continuous functions and s∈R. The author has studied the existence, nonexistence, and multiplicity of solutions as variable s. Further, we may refer to recent contributions of Agarwal [16] and Agarwal et al. [17], Cabada and Heikkilä [18], and Wong and Agarwal [19] for the right focal problem. Accordingly, a third-order left focal problem for differential equation plays a role in many related fields too, even less attention has been focused, however, on third-order left focal problem. This paper will consider the existence of a positive solution for the following third-order singular generalized left focal boundary value problem with full derivatives in Banach spaces
(Ps)u′′′(t)+h(t)f(t,u(t),u′(t),u′′(t))=0,hhhhhhhhhhhhhhhhhhhhh0<t1<t<t2,au(t1)-bu′′(t1)=0,u′(η)=0,t1<η<t2,cu(t2)+du′(t2)=0,
where a,b,c,d,t1,η,t2,h, and f satisfy (H1)
a,b,c,d∈]0,∞[ and k=2bc+2ad(t2-η)+ac(t2-t1)(t1+t2-2η)>0,(H2)
t1<η<t2 with t2-η>η-t1,(H3)
h∈C(]t1,t2[,[0,∞[) may be singular at t=t1 and/or t=t2 and ∫t1t2‍h(s)ds<∞,(H4)
f∈C([t1,t2]×[0,∞[×]-∞,∞[2,[0,∞[).The novelty of our problem is to show the existence of solution of the left focal boundary value problem; we will concentrate on the solvability of the third-order singular generalized left focal problem for differential equation. To the best of our knowledge, it is the first time that the third-order left focal problem has been discussed successfully. As one might expect, the technique of finding Green’s function for the corresponding homogeneous left focal problem, as the kernel of an integral equation of Hammerstein type, has proven to be an effective method in most of third-order left focal problems. For this, we will employ known sign properties of Green’s function and its partial derivative cleverly. Note that the solutionu of the problem (Ps) lacks concavity, which is slightly different from right focal problem. In order to overcome this difficulty, we will establish the appropriate a priori estimates of the solution. On the other hand, the condition on nonlinearity required in this paper is different from those of existing papers, which motivated us to consider the above left focal problem, since f may be superlinear (sublinear). As usual, the function f in the problem (Ps) is called superlinear if f_0=0 and f¯∞=∞; it is called sublinear when f_∞=0 and f¯0=∞, where
(6)f_0=lim⁡inf⁡μ→0+min⁡t∈[t1,t2]f(t,μ,ν,ϖ)μ,f¯∞=lim⁡sup⁡μ→∞max⁡t∈[t1,t2]f(t,μ,ν,ϖ)μ,f_∞=lim⁡inf⁡μ→∞min⁡t∈[t1,t2]f(t,μ,ν,ϖ)μ,f¯0=lim⁡sup⁡μ→0+max⁡t∈[t1,t2]f(t,μ,ν,ϖ)μ,
for ν,ϖ∈]-∞,∞[. Based on the known results of the singular multipoint boundary value problem and high-order right focal problem for differential equation, this study is devoted to proving the existence of solution for the third-order singular generalized left focal problem (Ps), and the full derivatives are involved in the nonlinear term explicitly. Under the suitable limit and growth conditions on the nonlinearity f in cones, we will establish some new and general results on the existence of single and multiple positive solutions to the problem (Ps); our results develop some results of the third-order focal problem and improve the results of third-order singular problem. The author believes that those results are useful in many applications for temporary new fields. Now we state the fixed point theorems.Lemma 1 (see [20, 21]).
LetK be a cone in a Banach space E. Assume that K(V1),K(V2) are open subsets of E with 0∈K(V1),K(V1)¯⊂K(V2). If T:K∩(K(V2)¯∖K(V1))→K is a completely continuous operator such that either (A1)
∥Tx∥≤∥x∥ for all x∈K∩∂K(V1) and ∥Tx∥≥∥x∥ for all x∈K∩∂K(V2) or(A2)
∥Tx∥≥∥x∥ for all x∈K∩∂K(V1) and ∥Tx∥≤∥x∥ for all x∈K∩∂K(V2).
Then,T has a fixed point in K∩(K(V2)¯∖K(V1)).Letγ and θ be nonnegative continuous convex functionals on a cone K,  let α be a nonnegative continuous concave functional on K,  let ψ be a nonnegative continuous functional on K, and let m1,m2,m3, and m4 be positive numbers. Define the convex sets as follows:
(7)P(γ,m4)={u∈K:γ(u)<m4},P(γ,α,m2,m4)={u∈K:m2≤α(u),γ(u)≤m4},P(γ,θ,α,m2,m3,m4)={u∈K:m2≤α(u),θ(u)≤m3,γ(u)≤m4},
and a closed set
(8)Q(γ,ψ,m1,m4)={u∈K:m1≤ψ(u),γ(u)≤m4}.Lemma 2 (see [22]).
LetK be a cone in a real Banach space E. Let γ and θ be nonnegative continuous convex functionals on K,  let α be a nonnegative continuous concave functional on K, and let ψ be a nonnegative continuous functional on K satisfying ψ(λu)≤λψ(u) for 0≤λ≤1 such that for some positive numbers ϵ and m4(9)α(u)≤ψ(u),∥u∥≤ϵγ(u),
for all u∈P(γ,m4)¯. Suppose that T:P(γ,m4)¯→P(γ,m4)¯ is completely continuous and there exist positive numbers m1, m2, and m3 with m1<m2 such that (B1)
{u∈P(γ,θ,α,m2,m3,m4):α(u)>m2}≠∅ and α(Tu)>m2 for u∈P(γ,θ,α,m2,m3,m4),(B2)
α(Tu)>m2 for u∈P(γ,α,m2,m4) with θ(Tu)>m3,(B3)
0∉Q(γ,ψ,m1,m4) and ψ(Tu)<m1 for u∈Q(γ,ψ,m1,m4) with ψ(u)=m1.
Then,T has at least three fixed points u1,u2,u3∈P(γ,m4)¯ such that
(10)γ(ui)≤m4fori=1,2,3,m2<α(u1),m1<ψ(u2)withα(u2)<m2,ψ(u3)<m1.


## 2. Preliminaries

We will employ several lemmas; these lemmas are based on the corresponding linear third-order generalized left focal boundary value problem.Lemma 3.
Let(H1), (H2) hold. Green’s function G(t,s) for the boundary value problem
(11)u′′′(t)=0,t1<t<t2,au(t1)-bu′′(t1)=0,u′(η)=0,t1<η<t2,cu(t2)+du′(t2)=0
is given as
(12)G(t,s)={s∈[t1,η]{x1(t,s),s≤t,y1(t,s),t≤s,s∈[η,t2]{x2(t,s),s≤t,y2(t,s),t≤s,
where
(13)x1(t,s)=1k(b+a2(t-t1)2-a(η-t1)(t-t1))×(c(t2-s)2+2d(t2-s)-2(c(t2-t1)+d)(η-s))+(t-t1)(η-s)-12(t-s)2,y1(t,s)=1k(b+a2(t-t1)2-a(η-t1)(t-t1))×(c(t2-s)2+2d(t2-s)-2(c(t2-t1)+d)(η-s))+(t-t1)(η-s),x2(t,s)=1k(b+a2(t-t1)2-a(η-t1)(t-t1))×(c(t2-s)2+2d(t2-s))-12(t-s)2,y2(t,s)=1k(b+a2(t-t1)2-a(η-t1)(t-t1))×(c(t2-s)2+2d(t2-s)).Proof.
First, we check thatG(t,s) is defined for each (t,s)∈[t1,t2]×[t1,t2]. Obviously, x1(s,s)=y1(s,s), x2(s,s)=y2(s,s). At s=η,
(14)x1(t,η)=1k(b+a2(t-t1)2-a(η-t1)(t-t1))×(c(t2-η)2+2d(t2-η))-12(t-η)2=x2(t,η),y1(t,η)=1k(b+a2(t-t1)2-a(η-t1)(t-t1))×(c(t2-η)2+2d(t2-η))=y2(t,η).
Next, we check that G(t,s) satisfies the boundary conditions of problem (11). For convenience, we note that
(15)∂∂tG(t,s)={s∈[t1,η]{ak(t-η)(c(t2-s)2+2d(t2-s)-2(c(t2-t1)+d)(η-s)(t2-s)2)+(η-t),s≤t,ak(t-η)(c(t2-s)2+2d(t2-s)-2(c(t2-t1)+d)(η-s)(t2-s)2)+(η-s),t≤s,s∈[η,t2]{ak(t-η)×(c(t2-s)2+2d(t2-s))-(t-s),s≤t,ak(t-η)(c(t2-s)2+2d(t2-s)),t≤s,(16)∂2∂t2G(t,s)={s∈[t1,η]{ak(c(t2-s)2+2d(t2-s)-2(c(t2-t1)+d)(η-s)(t2-s)2)-1,s≤t,ak(c(t2-s)2+2d(t2-s)-2(c(t2-t1)+d)(η-s)(t2-s)2),t≤s,s∈[η,t2]{ak(c(t2-s)2+2d(t2-s))-1,s≤t,ak(c(t2-s)2+2d(t2-s)),t≤s.
Fors∈[t1,η], by (12) and (16), we have
(17)aG(t1,s)-b∂2∂t2G(t1,s)=ay1(t1,s)-b∂2∂t2y1(t1,s)=abk(c(t2-s)2+2d(t2-s)-2(c(t2-t1)+d)(η-s))-abk(c(t2-s)2+2d(t2-s)-2(c(t2-t1)+d)(η-s))=0;
for s∈[η,t2], we get
(18)aG(t1,s)-b∂2∂t2G(t1,s)=ay2(t1,s)-b∂2∂t2y2(t1,s)=abk(c(t2-s)2+2d(t2-s))-abk(c(t2-s)2+2d(t2-s))=0.
Fors∈[t1,η], by (12) and (15), we have (∂/∂t)G(η,s)=(∂/∂t)x1(η,s)=0; for s∈[η,t2], we get (∂/∂t)G(η,s)=(∂/∂t)y2(η,s)=0.
Fors∈[t1,η], by (12) and (15), we have
(19)cG(t2,s)+d∂∂tG(t2,s)=cx1(t2,s)+d∂∂tx1(t2,s)=ck(b+a2(t2-t1)2-a(η-t1)(t2-t1))×(c(t2-s)2+2d(t2-s)-2(c(t2-t1)+d)(η-s))+c(t2-t1)(η-s)-c2(t2-s)2+adk(t2-η)×(c(t2-s)2+2d(t2-s)-2(c(t2-t1)+d)(η-s))+d(η-t2)=c2(t2-s)2+d(t2-s)-(c(t2-t1)+d)(η-s)+c(t2-t1)(η-s)-c2(t2-s)2+d(η-t2)=0;
for s∈[η,t2], we get
(20)cG(t2,s)+d∂∂tG(t2,s)=cx2(t2,s)+d∂∂tx2(t2,s)=ck(b+a2(t2-t1)2-a(η-t1)(t2-t1))×(c(t2-s)2+2d(t2-s))-c2(t2-s)2+adk(t2-η)×(c(t2-s)2+2d(t2-s))+d(s-t2)=c2(t2-s)2+d(t2-s)-c2(t2-s)2+d(s-t2)=0.
Hence, G(t,s) satisfies the boundary conditions of problem (11).Lemma 4.
Let(H1), (H2) hold. If b>(a/2)(t2-t1)2, then Green’s function as given in (12) satisfies
(21)G(t,s)≥0,G(t,s)≥τ1G(η,s),|∂∂tG(t,s)|≥τ2∂∂tG(η,s),hhhhhhhh(t,s)∈[t1,t2]2,
where
(22)τ1=min⁡⁡{τ11,τ12}∈]0,1[,τ11=min⁡{min⁡s∈[t1,η]x1(t1,t1)x1(η,s),min⁡s∈[η,t2]x2(t2,s)x2(η,s)},τ12=min⁡{min⁡s∈[t1,η]y1(η,η)y1(t1,s),min⁡s∈[η,t2]y2(η,s)y2(t2,s)},τ2=min⁡{min⁡s∈[t1,η]|(∂/∂t)G(s,s)(∂/∂t)G(t1,s)|,min⁡s∈[η,t2]|(∂/∂t)G(s,s)(∂/∂t)G(t2,s)|}∈]0,1[.Proof.
Conditionb>(a/2)(t2-t1)2 implies that b>(a/2)(η-t1)2 and b+(a/2)(t2-t1)2-a(η-t1)(t2-t1)>0 for t1<η<t2. To prove the nonnegative property of Green’s function, combining with (15) and (16), we divide it into two cases.
Case  1. For s≤t,
(23)∂∂tG(t,s)=∂∂tx1(t,s)=ak(t-η)×(c(t2-s)2+2d(t2-s)-2(c(t2-t1)+d)(η-s)(t2-s)2)+(η-t)=c(t-η)k(a(s-t1)2-2b)≥0,∂2∂t2G(t,s)=∂2∂t2x1(t,s)=ak(c(t2-s)2+2d(t2-s)-2(c(t2-t1)+d)(η-s)(t2-s)2)-1≤0
imply that x1(t,s) is nondecreasing and concave in t on [t1,η]; it follows that G(η,s)≥G(t,s) for each (t,s)∈[t1,η]×[t1,η]:
(24)G(t1,s)=x1(t1,s)=bk(c(t2-s)2+2d(t2-s)-2(c(t2-t1)+d)(η-s)(t2-s)2)-12(t1-s)2=bk(c(t2-s)2+2d(t2-s)-2(c(t2-t1)+d)(η-s)(t2-s)2)≥0,∂∂tG(t,s)>0,∂∂tG(s,s)>0,∂∂tG(t1,s)>0.
Likewise,
(25)∂∂tG(t,s)=∂∂tx2(t,s)=ak(t-η)(c(t2-s)2+2d(t2-s))-(t-s)=(t-η)(ak(c(t2-s)2+2d(t2-s))-1)+(s-η)≤0,∂2∂t2G(t,s)=∂2∂t2x2(t,s)=ak(c(t2-s)2+2d(t2-s))-1=1k(ac(t2-s)2+2ad(t2-s)-k)≤0
guarantee that x2(t,s) is nonincreasing and concave in t on [η,t2]; we have G(η,s)≥G(t,s) for each (t,s)∈[η,t2]×[η,t2]:
(26)G(t2,s)=x2(t2,s)=1k(b+a2(t2-t1)2-a(η-t1)(t2-t1))×(c(t2-s)2+2d(t2-s))-12(t2-s)2=12k(2bc+ac(t2-t1)(t1+t2-2η)-k)(t2-s)2+dk(2b+a(t2-t1)(t1+t2-2η))(t2-s)≥ad(t2-s)k(s-η)(t2-η)≥0,∂∂tG(t,s)<0,∂∂tG(s,s)>0,∂∂tG(t2,s)<0.
From the above argument, we obtain that
(27)G(t,s)≥0,(28)G(t,s)≥min⁡{min⁡s∈[t1,η]x1(t1,t1)x1(η,s),min⁡s∈[η,t2]x2(t2,s)x2(η,s)}G(η,s);(29)|∂∂tG(t,s)|≥min⁡s∈[η,t2]|(∂/∂t)G(s,s)(∂/∂t)G(t2,s)|∂∂tG(η,s),
we have (29), since (∂/∂t)G(t,s)≥min⁡s∈[t1,η]((∂/∂t)G(s,s)/(∂/∂t)G(t1,s))(∂/∂t)G(η,s)=0.
Case 2. For t≤s,
(30)∂∂tG(t,s)=∂∂ty1(t,s)=ak(t-η)(c(t2-s)2+2d(t2-s)-2(c(t2-t1)+d)(η-s)(t2-s)2)+(η-s)=t-ηk×(ac(t2-s)2+2ad(t2-s)-2a(c(t2-t1)+d)(η-s)(t2-s)2)+(η-s)≤t-s≤0,∂2∂t2G(t,s)=∂2∂t2y1(t,s)=ak(c(t2-s)2+2d(t2-s)-2(c(t2-t1)+d)(η-s)(t2-s)2)≥0
imply that y1(t,s) is nonincreasing and convex in t on [t1,η]; it follows that G(t,s)≥G(η,s) for each (t,s)∈[t1,η]×[t1,η]:
(31)G(η,s)=y1(η,s)=1k(b-a2(η-t1)2)×(c(t2-s)2+2d(t2-s)-2(c(t2-t1)+d)(η-s)(t2-s)2)+(η-t1)(η-s)≥0,∂∂tG(t,s)<0,∂∂tG(s,s)>0,∂∂tG(t1,s)<0.
Likewise,
(32)∂∂tG(t,s)=∂∂ty2(t,s)=ak(t-η)(c(t2-s)2+2d(t2-s))≥0,∂2∂t2G(t,s)=∂2∂t2y2(t,s)=ak(c(t2-s)2+2d(t2-s))≥0
guarantee that y2(t,s) is nondecreasing and convex in t on [η,t2]; it follows that G(t,s)≥G(η,s) for each (t,s)∈[η,t2]×[η,t2]:
(33)G(η,s)=y2(η,s)=1k(b-a2(η-t1)2)×(c(t2-s)2+2d(t2-s))≥0,∂∂tG(t,s)>0,∂∂tG(s,s)>0,∂∂tG(t2,s)>0.
So we get
(34)G(t,s)≥0;(35)G(t,s)≥min⁡{min⁡s∈[t1,η]y1(η,η)y1(t1,s),min⁡s∈[η,t2]y2(η,s)y2(t2,s)}G(η,s),(36)|∂∂tG(t,s)|≥min⁡s∈[t1,η]|(∂/∂t)G(s,s)(∂/∂t)G(t1,s)|∂∂tG(η,s),
we have (35), since (∂/∂t)G(t,s)≥min⁡s∈[η,t2](((∂/∂t)G(s,s))/((∂/∂t)G(t2,s)))(∂/∂t)G(η,s)=0.
It is easy to show thatτ1,τ2∈]0,1[. Therefore, we obtain that G(t,s)≥0 for each (t,s)∈[t1,t2]×[t1,t2]; (27)–(36) imply the desired conclusion.


## 3. A Priori Estimates

Lemma 5.
Let(H1), (H2) hold. If g∈C[t1,t2] with g≥0, then a solution of the problem
(37)u′′′(t)=g(t),0<t1<t<t2,au(t1)-bu′′(t1)=0,u′(η)=0,t1<η<t2,cu(t2)+du′(t2)=0
satisfies
(38)u(t)≥0,t∈[t1,t2],min⁡t∈[ξ,ω]u(t)≥τ1max⁡t∈[t1,t2]|u(t)|,
where [ξ,ω]⊆[t1,t2], (ξ<ω).Proof.
A solution of problem (37) is u(t)=∫t1t2‍G(t,s)g(s)ds, where G(t,s) is defined by (12); it is not difficult to show that the results hold by Lemma 4 and we omit the proof process.Lemma 6.
Let(H1), (H2) hold. If g∈C[t1,t2] with g≥0, then a solution of problem (37) satisfies
(39)max⁡t∈[t1,t2]u(t)≤τ3max⁡t∈[t1,t2]|u′(t)|,max⁡t∈[t1,t2]|u′(t)|≤τ4max⁡t∈[t1,t2]|u′′(t)|,
where
(40)τ3=max⁡{b+(a/2)(t2-t1)2a(t2-η)((b-a2(η-t1)2)(c(t2-t1)2+2d(t2-t1))+k(η-t1)2(b-a2(η-t1)2))×((η-t1)(k-a(c(t2-t1)2+2d(t2-t1))))-1,(b-(a/2)(η-t1)2)(c(t2-η)2+2d(t2-η))(t2-η)(k-a(c(t2-η)2+2d(t2-η))),(b(c(t2-t1)2+2d(t2-t1)))×(inf⁡s∈[t1,η]⁡((t2-s)2a(η-t1)×(c(t2-s)2+2d(t2-s)-2(c(t2-t1)+d)(η-s)(t2-s)2)+k(s-η)(t2-s)2)inf⁡s∈[t1,η]⁡(a(η-t1))-1,b+(a/2)(t2-t1)2a(t2-η)},τ4=t2-η.Proof.
According to the nonnegative solutionu of problem (37), we establish a priori estimates as follows. For t1≤s≤t≤η, we can check that max⁡t∈[t1,η]u(t)=x1(η,s),max⁡t∈[t1,η]|u′(t)|=(∂/∂t)x1(t1,s),max⁡t∈[t1,η]|u′′(t)|=-(∂2/∂t2)x1(η,s); note that (a/k)(c(t2-s)2+2d(t2-s)-2(c(t2-t1)+d)(η-s))-1≤0, so
(41)max⁡t∈[t1,η]⁡u(t)max⁡t∈[t1,η]⁡|u′(t)|≤((b-a2(η-t1)2)×(c(t2-s)2+2d(t2-s)-2(c(t2-t1)+d)(η-s))+k(η-t1)2(b-a2(η-t1)2))×((η-t1)(k-a(c(t2-s)2+2d(t2-s)-2(c(t2-t1)+d)(η-s)(t2-s)2)))-1≤((b-a2(η-t1)2)×(c(t2-t1)2+2d(t2-t1))+k(η-t1)2(b-a2(η-t1)2))×((η-t1)(k-a(c(t2-t1)2+2d(t2-t1))))-1,max⁡t∈[t1,η]⁡u′(t)max⁡t∈[t1,η]⁡|u′′(t)|≤((η-t)(k-a(c(t2-s)2+2d(t2-s)-2(c(t2-t1)+d)(η-s)(t2-s)2)))×(k-a(c(t2-s)2+2d(t2-s)-2(c(t2-t1)+d)(η-s)(t2-s)2))-1≤η-t1.
Forη≤s≤t≤t2, we can check that max⁡t∈[η,t2]u(t)=x2(η,s),max⁡t∈[η,t2]|u′(t)|=-(∂/∂t)x2(t2,s),max⁡t∈[η,t2]|u′′(t)|=-(∂2/∂t2)x2(t2,s); notice that (a/k)(c(t2-s)2+2d(t2-s))-1≤0 and thus
(42)max⁡t∈[η,t2]⁡u(t)max⁡t∈[η,t2]⁡|u′(t)|≤(b-(a/2)(η-t1)2)(c(t2-η)2+2d(t2-η))(t2-η)(k-a(c(t2-s)2+2d(t2-s)))≤(b-(a/2)(η-t1)2)(c(t2-η)2+2d(t2-η))(t2-η)(k-a(c(t2-η)2+2d(t2-η))),max⁡t∈[η,t2]⁡u′(t)max⁡t∈[η,t2]⁡|u′′(t)|≤k(t-η)-a(t-η)(c(t2-s)2+2d(t2-s))k-a(c(t2-s)2+2d(t2-s))≤t2-η.
Fort1≤t≤s≤η, we can check that max⁡t∈[t1,η]u(t)=y1(t1,s),max⁡t∈[t1,η]|u′(t)|=-(∂/∂t)y1(t1,s),max⁡t∈[t1,η]|u′′(t)|=(∂2/∂t2)y1(t1,s); note that (a/k)(c(t2-s)2+2d(t2-s)-2(c(t2-t1)+d)(η-s))-1≤0 and so
(43)max⁡t∈[t1,η]⁡u(t)max⁡t∈[t1,η]⁡|u′(t)|≤(b(c(t2-t1)2+2d(t2-t1)))×(a(η-t1)(c(t2-s)2+2d(t2-s)-2(c(t2-t1)+d)(η-s)(t2-s)2)+k(s-η)(t2-s)2)-1≤(b(c(t2-t1)2+2d(t2-t1)))×(inf⁡s∈]t1,η]((t2-s)2a(η-t1)×(c(t2-s)2+2d(t2-s)-2(c(t2-t1)+d)(η-s)(t2-s)2(t2-s)2(t2-s)2)+k(s-η)(t2-s)2)inf⁡s∈]t1,η])-1,max⁡t∈[t1,η]⁡u′(t)max⁡t∈[t1,η]⁡|u′′(t)|≤(a(η-t1)(c(t2-s)2+2d(t2-s)-2(c(t2-t1)+d)(η-s)(t2-s)2)+k(s-η)(η-t1)(t2-s)2)×(a(c(t2-s)2+2d(t2-s)-2(c(t2-t1)+d)(η-s)(t2-s)2))-1≤η-t1.
Forη≤t≤s≤t2, we can check that max⁡t∈[η,t2]u(t)=y2(t2,s),max⁡t∈[η,t2]|u′(t)|=(∂/∂t)y2(t2,s),max⁡t∈[η,t2]|u′′(t)|=(∂2/∂t2)y2(η,s); then
(44)max⁡t∈[η,t2]⁡u(t)max⁡t∈[η,t2]⁡|u′(t)|=b+(a/2)(t2-t1)2-a(η-t1)(t2-t1)a(t2-η)≤b+(a/2)(t2-t1)2a(t2-η),max⁡t∈[η,t2]⁡u′(t)max⁡t∈[η,t2]⁡|u′′(t)|≤a(t2-η)(c(t2-s)2+2d(t2-s))a(c(t2-s)2+2d(t2-s))=t2-η.
From the above argument, we can obtain the desired result (39).LetE be the Banach space C2[t1,t2] with the norm ∥u∥=max⁡⁡{∥u∥0,∥u′∥0,∥u′′∥0}, where ∥·∥0=max⁡t∈[t1,t2]|·(t)|. Set a cone in E(45)K={u∈E:uisnonnegativeon⁡[t1,t2],andsatisfiesau(t1)-bu′′(t1)=0,u′(η)=0,cu(t2)+du′(t2)=0,min⁡t∈[ξ,ω]u(t)≥τ1max⁡t∈[t1,t2]|u(t)|}.In view of the assumptions (H1) and (H2), take
(46)u0(t)=-m-t3+λ2λ1m-t2+λ3λ1m-t+(t13-6bat1+(2ba-t12)λ2λ1-t1λ3λ1)m-,
where λ1=2η(act1-act2-ad)-act12+2bc+act22+2adt2, λ2=3η2(act1-act2-ad)-act13+6bct1+act23+3adt22,  λ3=3η2(-act12+act22+2adt2)+2η(act13-6bct1-act23-3adt22),  and m- is an arbitrary positive constant. By some simplifying, we obtain that λ1,λ2≥0. Set a constant
(47)ϑ:=max⁡{-t13+λ2λ1t12+λ3λ1t1+λ4,-η3+λ2λ1η2+λ3λ1η+λ4,-t-3+λ2λ1t-2+λ3λ1t-+λ4,-t23+λ2λ1t22+λ3λ1t2+λ4},
where t-=(λ2+3λ1λ3+λ22)/3λ1,λ4=t13-(6b/a)t1+((2b/a)-t12)(λ2/λ1)-t1(λ3/λ1). It is not difficult to check that u0(t) is nonnegative on [t1,t2] and satisfies au0(t1)-bu0′′(t1)=0, u0′(η)=0, cu0(t2)+du0′(t2)=0. Under the assumption min⁡⁡{-ξ3+(λ2/λ1)ξ2+(λ3/λ1)ξ+λ4,-ω3+(λ2/λ1)ω2+(λ3/λ1)ω+λ4,-η3+(λ2/λ1)η2+(λ3/λ1)η+λ4}≥τ1ϑ, we have min⁡t∈[ξ,ω]u0(t)≥τ1max⁡t∈[t1,t2]|u0(t)|; that is, u0∈K, so K∖{0}≠∅. By assumption (H3), both 0<∫t1t2‍G(t1,s)h(s)ds<∞,0<∫t1t2‍G(η,s)h(s)ds<∞ and 0<∫t1t2‍G(t2,s)h(s)ds<∞ hold. Thus, we can define an operator T by
(48)(Tu)(t):=∫t1t2‍G(t,s)h(s)f(s,u(s),u′(s),u′′(s))ds,hhhhhhhhhhhhhhhhhhhhhhhhhhlllhhhhhlt∈[t1,t2].Lemma 7.
Let(H1)–(H4) hold. Then, T:K→K is completely continuous.Proof.
Since conditions(H1)–(H4) hold, by Lemma 5, we observe that (Tu)(t)≥0 for t∈[t1,t2]; a(Tu)(t1)-b(Tu)′′(t1)=0, (Tu)′(η)=0, c(Tu)(t2)+d(Tu)′(t2)=0,min⁡t∈[ξ,ω](Tu)(t)≥τ1max⁡t∈[t1,t2]|(Tu)(t)| are satisfied, so Tu∈K. Since h(t) may be singular at t=t1 and/or t=t2, we will take the arguments to show that the operator T is completely continuous. Assume that un,u0∈K(n=1,2,…) satisfy ∥un-u0∥→0(n→∞); then there exists R>0 such that ∥un∥≤R for arbitrary positive integer n. One has that
(49)∥Tun-Tu0∥0=max⁡t∈[t1,t2]|Tun(t)-Tu0(t)|≤max⁡t∈[t1,η]|∫t1t2‍G(t,s)h(s)×(f(s,un,un′,un′′)-f(s,u0,u0′,u0′′))ds∫t1t2‍|+max⁡t∈[η,t2]|∫t1t2‍G(t,s)h(s)×(f(s,un,un′,un′′)-f(s,u0,u0′,u0′′))ds∫t1t2‍|≤max⁡t∈[t1,η]|∫t1t‍G(η,s)h(s)×(f(s,un,un′,un′′)-f(s,u0,u0′,u0′′))ds+∫tη‍G(t1,s)h(s)×(f(s,un,un′,un′′)-f(s,u0,u0′,u0′′))ds∫t1t2‍|+max⁡t∈[η,t2]|∫ηt‍G(η,s)h(s)×(f(s,un,un′,un′′)-f(s,u0,u0′,u0′′))ds+∫tt2‍G(t2,s)h(s)×(f(s,un,un′,un′′)-f(s,u0,u0′,u0′′))ds∫t1t2‍|≤∫t1η‍G(η,s)h(s)×|f(s,un,un′,un′′)-f(s,u0,u0′,u0′′)|ds+∫t1η‍G(t1,s)h(s)×|f(s,un,un′,un′′)-f(s,u0,u0′,u0′′)|ds+∫ηt2‍G(η,s)h(s)×|f(s,un,un′,un′′)-f(s,u0,u0′,u0′′)|ds+∫ηt2‍G(t2,s)h(s)×|f(s,un,un′,un′′)-f(s,u0,u0′,u0′′)|ds,∥(Tun)′-(Tu0)′∥0=max⁡t∈[t1,t2]|(Tun)′(t)-(Tu0)′(t)|≤max⁡t∈[t1,η]|∫t1t2‍∂∂tG(t,s)h(s)×|f(s,un,un′,un′′)-f(s,u0,u0′,u0′′)|ds∫t1t2‍∂∂t|+max⁡t∈]η,t2]|∫t1t2‍∂∂tG(t,s)h(s)×|f(s,un,un′,un′′)-f(s,u0,u0′,u0′′)|ds∫t1t2‍∂∂t|≤max⁡t∈[t1,η]∫t1t2‍|∂∂tG(t,s)|h(s)×|f(s,un,un′,un′′)-f(s,u0,u0′,u0′′)|ds+max⁡t∈]η,t2]∫t1t2‍|∂∂tG(t,s)|h(s)×|f(s,un,un′,un′′)-f(s,u0,u0′,u0′′)|ds,∥(Tun)′′-(Tu0)′′∥0=max⁡t∈[t1,t2]|(Tun)′′(t)-(Tu0)′′(t)|≤max⁡t∈[t1,η]|∫t1t2‍∂2∂t2G(t,s)h(s)×|f(s,un,un′,un′′)-f(s,u0,u0′,u0′′)|ds∫t1t2‍∂2∂t2|+max⁡t∈]η,t2]|∫t1t2‍∂2∂t2G(t,s)h(s)×|f(s,un,un′,un′′)-f(s,u0,u0′,u0′′)|ds∫t1t2‍∂2∂t2|≤max⁡t∈[t1,η]∫t1t2‍|∂2∂t2G(t,s)|h(s)×|f(s,un,un′,un′′)-f(s,u0,u0′,u0′′)|ds+max⁡t∈]η,t2]∫t1t2‍|∂2∂t2G(t,s)|h(s)×|f(s,un,un′,un′′)-f(s,u0,u0′,u0′′)|ds.
In view of the continuity of f(t,μ,ν,ϖ) in μ,ν,ϖ, we can obtain that, for any t∈[t1,t2],∥Tun-Tu0∥=max⁡⁡{∥Tnu-Tu∥0,∥(Tnu)′-(Tu)′∥0,∥(Tn)′′u-(Tu)′′∥0}→0(n→∞). This means that the operator T:K→K is continuous. It follows from the properties of the function h that we can choose two sequences {αn}n=1∞⊂[t1,η[, {βn}n=1∞⊂]η,t2]. Obviously αn<βn for any n≥1 and αn→t1,βn→t2 as n→∞, respectively. Define
(50)hn(t)={inf⁡{h(t),h(αn)},t∈[t1,αn],h(t),t∈[αn,βn],inf⁡{h(βn),h(t)},t∈[βn,t2],
and an operator sequence Tn by
(51)(Tnu)(t)=∫t1t2‍G(t,s)hn(s)f(s,u(s),u′(s),u′′(s))ds,hhhhhhhhhhhhhhhlhhhhhhhhhhhhhhhhhhht∈[t1,t2].
Clearly, hn(t):[t1,t2]→[0,∞[ is a piecewise continuous function; the operator Tn:K→K is well defined. Further, we have that Tn:K→K is completely continuous.
Letr>0,Br:={u∈K:∥u′′∥0≤r} and Mr=max⁡⁡{f(t,μ,ν,ϖ):μ∈K,t∈[t1,t2],μ∈[0,τ3τ4r],ν∈[-τ4r,τ4r],ϖ∈[-r,r]}. We will show that Tn approaches T uniformly on Br as n→∞. From the absolute continuity of integral, we obtain
(52)lim⁡n→∞max⁡t∈[t1,t2]∫l(n)‍G(t,s)h(s)ds=0,
where l(n)=[t1,αn]∪[βn,t2]. Green’s functions G(t,s) and (52) imply that, for any u∈Br,t∈[t1,αn]; we have
(53)∥Tnu-Tu∥0=max⁡t∈[t1,αn]|∫t1t2‍G(t,s)(hn(s)-h(s))f(s,u,u′,u′′)ds|≤Mrmax⁡t∈[t1,αn]∫t1t2‍G(t,s)|hn(s)-h(s)|ds≤Mrmax⁡t∈[t1,αn]∫l(n)‍G(t,s)h(s)ds⟶0(n⟶∞).
For any u∈Br,t∈[βn,t2], we get that ∥Tnu-Tu∥0→0(n→∞) and, for any u∈Br,t∈[αn,βn], there is ∥Tnu-Tu∥0=0(n≥1).
Similarly, for anyu∈Br,t∈[t1,αn],t∈[αn,βn], or t∈[βn,t2], we can obtain that ∥(Tnu)′-(Tu)′∥0→0(n→∞) and ∥(Tnu)′′-(Tu)′′∥0→0(n→∞), respectively.
The above argument implies that∥Tnu-Tu∥→0(n→∞); that is, the sequence Tn uniformly approximates T on any bounded subset of K. Hence, T:K→K is completely continuous.


## 4. Single Solution

Now, we are ready to present the sufficient conditions for the existence of at least one positive solution to the problem(Ps).Theorem 8.
Suppose(H1)–(H4) hold; f(t,0,0,0)≢0 on any subinterval of [t1,t2] and(C1)
there exist a constantL and a function F(t) which is integrable on [t1,t2], satisfying
(54)hhhhhhhhhhhhhhhf2(t,μ,ν,ϖ)≤F(t),(t,μ,ν,ϖ)∈[t1,t2]×[L,∞[×]-∞,∞[2,(C2)
there exist constantsm, M(m<M) such that [m,M]⊆[ξ,ω] and satisfy
(55)lim⁡inf⁡μ→0+min⁡t∈[m,M]hf(t,μ,ν,ϖ)μ=∞.
Then, the boundary value problem(Ps) has at least one positive solution.Proof.
By the definition of operatorT and Lemma 7, it suffices to show that the conditions of Lemma 1 hold with respect to T. Furthermore, from the fact that G(t,s) is Green’s function, it is not difficult to prove that a fixed point of T is coincident with the solution of the boundary value problem (Ps), so we concentrate on the existence of the fixed point of the operator T.
First of all, we show that(56)lim⁡v→∞Φ(v)v=0,
where
(57)Φ(v)∶=sup⁡u∈∂K(v)max⁡⁡{max⁡t∈[t1,t2]|∫t1t2‍G(t,s)h(s)×f(s,u,u′,u′′)ds∫t1t2‍|,max⁡t∈[t1,t2]|∫t1t2‍∂∂tG(t,s)h(s)×f(s,u,u′,u′′)ds∫t1t2‍|,max⁡t∈[t1,t2]|∫t1t2‍∂2∂t2G(t,s)h(s)×f(s,u,u′,u′′)ds∫t1t2‍|max⁡t∈[t1,t2]|∫t1t2‍G(t,s)h(s)},
and K(v)={u∈K:∥u∥0≤v,v∈]0,∞[}. For convenience, we introduce notation as follows:
(58)L(t)={t∈[t1,t2]:u(t)≤L},M1=max⁡{h2(s)ds∫t1t2(max⁡⁡{|∂∂tG(t1,s)|,‍)1/2(∫t1t2‍(max⁡⁡{G(t1,s),G(η,s),G(t2,s)})2×h2(s)ds∫t1t2‍)1/2,(∫t1t2‍(max⁡⁡{|∂∂tG(t1,s)|,∂∂tG(η,s),|∂∂tG(t2,s)|})2×h2(s)ds∫t1t2(max⁡⁡{|∂∂tG(t1,s)|,‍)1/2,(∫t1t2‍(max⁡{|∂2∂t2G(t1,s)|,|∂2∂t2G(η,s)|,|∂2∂t2G(t2,s)|})2×h2(s)ds∫t1t2(max⁡⁡{|∂∂tG(t1,s)|,‍)1/2(∫t1t2‍(max⁡⁡{G(t1,s),G(η,s),G(t2,s)})2},M2=max⁡(t,u,u′)∈[t1,t2]×[0,L]×]-∞,∞[2f2(t,u(t),u′(t),u′′(t)),M3=∫t1t2‍F(s)ds,M4=min⁡{min⁡s∈[m,M]G(η,s),min⁡s∈[m,M]|∂2∂t2G(η,s)|}.
By the well-known Hölder’s inequality, for any t∈[t1,t2] and u∈∂K(v), by using hypothesis (C1), we get
(59)max⁡t∈[t1,t2]|∫t1t2‍G(t,s)h(s)f(s,u,u′,u′′)ds|≤∫t1t2‍max⁡{G(t1,s),G(η,s),G(t2,s)}×h(s)f(s,u,u′,u′′)ds≤(∫t1t2‍(max⁡{G(t1,s),G(η,s),G(t2,s)})2h2(s)ds)1/2×(∫t1t2‍f2(s,u,u′,u′′)ds)1/2≤M1(∫[t1,t2]-L(t)‍f2(s,u,u′,u′′)ds+∫L(t)‍f2(s,u,u′)ds)1/2≤M1(∫t1t2‍F(s)ds+∫L(t)‍f2(s,u,u′,u′′)ds)1/2≤M1(M2(t2-t1)+M3)1/2,max⁡t∈[t1,t2]|∫t1t2‍∂∂tG(t,s)h(s)f(s,u,u′,u′′)ds|≤∫t1t2‍max⁡{|∂∂tG(t1,s)|,∂∂tG(η,s),|∂∂tG(t2,s)|}×h(s)f(s,u,u′,u′′)ds≤(∫t1t2‍(max⁡{|∂∂tG(t1,s)|,∂∂tG(η,s),|∂∂tG(t2,s)|})2×h2(s)ds∫t1t2‍(max⁡{|∂∂tG(t1,s)|,∂∂tG(η,s),|∂∂tG(t2,s)|})2)1/2(∫t1t2‍f2(s,u,u′,u′′)ds)1/2≤M1(M2(t2-t1)+M3)1/2,max⁡t∈[t1,t2]|∫t1t2‍∂2∂t2G(t,s)h(s)f(s,u,u′,u′′)ds|≤∫t1t2‍max⁡{|∂2∂t2G(t1,s)|,|∂2∂t2G(η,s)|,|∂2∂t2G(t2,s)|}h(s)f(s,u,u′,u′′)ds≤(∫t1t2‍(max⁡{|∂2∂t2G(t1,s)|,|∂2∂t2G(η,s)|,|∂2∂t2G(t2,s)|})2×h2(s)ds∫t1t2‍(max⁡{|∂2∂t2G(t1,s)|,|∂2∂t2G(η,s)|,|∂2∂t2G(t2,s)|})2)1/2(∫t1t2‍f2(s,u,u′,u′′)ds)1/2≤M1(M2(t2-t1)+M3)1/2.
Consequently, lim⁡v→∞(Φ(v)/v)≤lim⁡v→∞(M1(M2(t2-t1)+M3)1/2/v)=0.
Next, we verify that(60)lim⁡v→0+Ψ(v)v=∞,
where
(61)Ψ(v)≔inf⁡u∈∂K(v)max⁡{max⁡t∈[t1,t2]|∫t1t2‍G(t,s)h(s)f(s,u,u′,u′′)ds|,max⁡t∈[t1,t2]|∫t1t2‍∂∂tG(t,s)h(s)×f(s,u,u′,u′′)ds∫t1t2‍|,max⁡t∈[t1,t2]|∫t1t2‍∂2∂t2G(t,s)h(s)×f(s,u,u′,u′′)dsmax⁡t∈[t1,t2]∫t1t2‍|}.
In fact, hypothesis (C2) implies that, for any u∈K, n∈N, there exists ϵn>0 such that ϵn→0 and
(62)hhhhhhhhhhhhhhhhhlhf(t,μ,ν,ϖ)μ≥n,(t,μ,ν,ϖ)∈[m,M]×]0,ϵn[×]-∞,∞[2.
Then by (62) and Lemma 4, we have
(63)max⁡t∈[t1,t2]|∫t1t2‍G(t,s)h(s)f(s,u,u′,u′′)ds|=max⁡t∈[ξ,ω]∫t1t2‍G(t,s)h(s)f(s,u,u′,u′′)ds≥τ1∫t1t2‍G(η,s)h(s)f(s,u,u′,u′′)ds≥τ1M4min⁡t∈[m,M]u(t)∫mM‍h(s)f(s,u,u′,u′′)uds≥τ12M4∥u∥0∫mM‍h(s)f(s,u,u′,u′′)uds,max⁡t∈[t1,t2]|∫t1t2‍∂∂tG(t,s)h(s)f(s,u,u′,u′′)ds|=max⁡t∈[t1,t2]∫t1t2‍|∂∂tG(t,s)|h(s)f(s,u,u′,u′′)ds≥τ2∫t1t2‍∂∂tG(η,s)h(s)f(s,u,u′,u′′)ds≥τ2M4min⁡t∈[m,M]u(t)∫mM‍h(s)f(s,u,u′,u′′)uds≥τ1τ2M4∥u∥0∫mM‍h(s)f(s,u,u′,u′′)uds,max⁡t∈[t1,t2]|∫t1t2‍∂2∂t2G(t,s)h(s)f(s,u,u′,u′′)ds|=max⁡t∈]t1,t2]∫t1t2‍|∂2∂t2G(t,s)|h(s)f(s,u,u′,u′′)ds=∫t1t2‍|∂2∂t2G(η,s)|h(s)f(s,u,u′,u′′)ds≥M4min⁡t∈[m,M]u(t)∫mM‍h(s)f(s,u,u′,u′′)uds≥M4∥u∥0∫mM‍h(s)f(s,u,u′,u′′)u(s)ds.
The argument guarantees that, for v∈]0,ϵn[,
(64)Ψ(v)v=1vinf⁡u∈∂K(v)max⁡{max⁡t∈[t1,t2]|∫t1t2‍G(t,s)h(s)f(s,u,u′,u′′)ds|,max⁡t∈[t1,t2]|∫t1t2‍∂∂tG(t,s)×h(s)f(s,u,u′,u′′)ds∫t1t2‍∂∂t|,max⁡t∈[t1,t2]|∫t1t2‍∂2∂t2G(t,s)×h(s)f(s,u,u′,u′′)ds∫t1t2‍∂2∂t2|}≥1vinf⁡u∈∂K(v)max⁡⁡{∫mM‍h(s)f(s,u,u′,u′′)uτ12M4∥u∥0×∫mM‍h(s)f(s,u,u′,u′′)uds,τ1τ2M4∥u∥0∫mM‍h(s)f(s,u,u′,u′′)uds,M4∥u∥0∫mM‍h(s)f(s,u,u′,u′′)uds}=max⁡{1vinf⁡u∈∂K(v)τ12M4∥u∥0∫mM‍h(s)f(s,u,u′,u′′)uds,1vinf⁡u∈∂K(v)τ1τ2M4∥u∥0∫mM‍h(s)f(s,u,u′,u′′)uds,1vinf⁡u∈∂K(v)M4∥u∥0∫mM‍h(s)f(s,u,u′,u′′)uds}≥max⁡{τ12M4∫mM‍h(s)f(s,u,u′,u′′)uds,τ1τ2M4∫mM‍h(s)f(s,u,u′,u′′)uds,M4∫mM‍h(s)f(s,u,u′,u′′)uds∫mM‍h(s)f(s,u,u′,u′′)u}≥max⁡{τ12M4(M-m)n,τ1τ2M4(M-m)n,M4(M-m)nτ12M4(M-m)}≥max⁡{τ1,τ2}τ1M4(M-m)n.
Hence, (60) holds.
Finally, from the definitions ofΦ and Ψ, it is easy to check that
(65)Φ(v)=sup⁡u∈∂K(v)∥Tu∥,Ψ(v)=inf⁡u∈∂K(v)∥Tu∥.
Note that (56) and (60) imply that there exist V1, V2 such that V2≫V1 and Φ(V2)/V2≤1, Ψ(V1)/V1≥1, respectively. Thus, it follows that
(66)∥Tu∥≥∥u∥,u∈∂K(V1),∥Tu∥≤∥u∥,u∈∂K(V2).
By virtue of (66), the hypotheses of Lemma 1 are satisfied; we infer that there exists u*∈K such that Tu*=u*, and V1≤∥u*∥≤V2. In addition, condition (H2) guarantees that the fixed point is positive, so the problem (Ps) has at least one positive solution. This completes the proof.Remark 9.
Since the definition of the operatorT, a slight modification of M4=min⁡⁡{min⁡s∈[m,M]G(η,s)h(t),min⁡s∈[m,M]|(∂2/∂t2)G(η,s)h(t)|} and(C2)
there exist constantsm, M(m<M) such that [m,M]⊆[ξ,ω] and satisfy
(67)liminf⁡μ→0+min⁡t∈[m,M]f(t,μ,ν,ϖ)μ=∞
which guarantee that the problem(Ps) has at least one positive solution.Furthermore,f may be superlinear or sublinear; we can give the following existence results for the problem (Ps) and the proof is omitted.Theorem 10.
Suppose(H1)–(H4) hold, f(t,0,0,0)≢0 on any subinterval of [t1,t2] and (D1)
f_0=0 and f¯∞=∞, or(D2)
f_∞=0 and f¯0=∞.
Then, the problem(Ps) has at least one positive solution.


## 5. Multiple Solutions

Let the nonnegative continuous convex functionalsγ and θ, the nonnegative continuous concave functional α, and the nonnegative continuous functional ψ be defined on the cone K by
(68)γ(u)=max⁡t∈[t1,t2]|u′′(t)|,θ(u)=max⁡t∈[t1,t2]|u(t)|,α(u)=min⁡t∈[ξ,ω]u(t),ψ(u)=max⁡t∈[t1,t2]|u′(t)|.To prove our main results, we recommend notation(69)N1=max⁡{∫t1η‍-∂2∂t2x1(η,s)h(s)ds,∫t1η‍∂2∂t2y1(t1,s)h(s)ds}+max⁡⁡{∫ηt2‍-∂2∂t2x2(t2,s)h(s)ds,∫ηt2‍∂2∂t2y2(η,s)h(s)ds},N2=τ1(max⁡{∫t1η‍x1(η,s)h(s)ds,∫t1η‍y1(t1,s)h(s)ds}+max⁡{∫ηt2‍x2(η,s)h(s)ds,∫ηt2‍y2(t2,s)h(s)ds}),N3=max⁡{∫t1η‍∂∂tx1(t1,s)h(s)ds,∫t1η‍-∂∂ty1(t1,s)h(s)ds}+max⁡{∫ηt2‍-∂∂tx2(t2,s)h(s)ds,∫ηt2‍∂∂ty2(t2,s)h(s)ds}.Now, we are ready to apply Avery and Peterson fixed point theorem to the operatorT to give the sufficient conditions for the existence of at least three positive solutions to the problem (Ps).Theorem 11.
Let(H1)–(H4) hold; f(t,0,0,0)≢0 on any subinterval of [t1,t2]. Suppose that there exist numbers m1, m2, m4 such that 0<m1<m2≤τ12ϑ/max⁡⁡{|6t1-(2λ2/λ1)|,|6t2-(2λ2/λ1)|}m4,min⁡{-ξ3+(λ2/λ1)ξ2+(λ3/λ1)ξ+λ4,-ω3+(λ2/λ1)ω2+(λ3/λ1)ω+λ4,-η3+(λ2/λ1)η2+(λ3/λ1)η+λ4}≥τ1ϑ, and(E1)
f(t,μ,ν,ϖ)≤m4/N1 for (t,μ,ν,ϖ)∈[t1,t2]×[0,τ3τ4m4]×[-τ4m4,τ4m4]×[-m4,m4],(E2)
f(t,μ,ν,ϖ)>m2/N2 for (t,μ,ν,ϖ)∈[ξ,ω]×[m2,m2τ1-2]×[-τ4m4,τ4m4]×[-m4,m4],(E3)
f(t,μ,ν,ϖ)≤m1/N3 for (t,μ,ν,ϖ)∈[t1,t2]×[0,τ3τ4m4]×[-τ4m4,m1]×[-m4,m4].Then, the (Ps) has at least three positive solutions u1,u2,u3∈K such that
(70)max⁡t∈[t1,t2]|ui′′(t)|≤m4fori=1,2,3,m2<min⁡t∈[ξ,ω]u1(t),m1<max⁡t∈[t1,t2]u2′(t)withmin⁡t∈[ξ,ω]u2(t)<m2,max⁡t∈[t1,t2]u3′(t)<m1.Proof.
By the definition of operatorT and its properties, it suffices to show that the conditions of Lemma 2 hold with respect to T. Furthermore, from the fact that G(t,s) is the Green function, it is not difficult to prove that a fixed point of T is coincident with the solution of the boundary value problem (Ps), so we concentrate on the existence of the fixed point of the operator T.
The functionals are defined as (68); we also note that α, γ, θ, and ψ are continuous nonnegative functionals on K satisfying ψ(λu)≤λψ(u) for 0≤λ≤1 such that, for some positive numbers ϵ and m4, α(u)≤θ(u)=ψ(u) and ∥u∥0≤ϵγ(u), for u∈P(γ,m4)¯, where ϵ≤1. If u∈P(γ,m4)¯, then γ(u)=max⁡t∈[t1,t2]|u′′(t)|≤m4. We get max⁡t∈[t1,t2]|u(t)|≤τ3max⁡t∈[t1,t2]|u′(t)|≤τ3τ4m4 by (39). It follows from assumption (E1) that f(t,u(t),u′(t),u′′(t))≤m4/N1. On the other hand, for u∈K, we have proved that Tu∈K in Lemma 7 and thus
(71)γ(Tu)=max⁡t∈[t1,t2]|(Tu)′′(t)|≤max⁡t∈[t1,t2]|∫t1η‍∂2∂t2G(t,s)h(s)f(s,u,u′,u′′)ds|+max⁡t∈[t1,t2]|∫ηt2‍∂2∂t2G(t,s)h(s)f(s,u,u′,u′′)ds|=max⁡{∫t1η‍-∂2∂t2x1(η,s)h(s)f(s,u,u′,u′′)ds,∫t1η‍∂2∂t2y1(t1,s)h(s)f(s,u,u′,u′′)ds}+max⁡{∫ηt2‍-∂2∂t2x2(t2,s)h(s)f(s,u,u′,u′′)ds,∫ηt2‍∂2∂t2y2(η,s)h(s)f(s,u,u′,u′′)ds}≤m4N1(max⁡{∫t1η‍-∂2∂t2x1(η,s)h(s)ds,∫t1η‍∂2∂t2y1(t1,s)h(s)ds}+max⁡{∫ηt2‍-∂2∂t2x2(t2,s)h(s)ds,∫ηt2‍∂2∂t2y2(η,s)h(s)ds})=m4.
Therefore, T:P(γ,m4)¯→P(γ,m4)¯.
Firstly, to verify the condition(B1) of Lemma 2 holds, we choose u0 which is defined by (46); (47) gives the expression of ϑ. Set m-=m2/(τ12ϑ); we have
(72)α(u0)=min⁡t∈[ξ,ω]|u0(t)|=m-min⁡{-ξ3+λ2λ1ξ2+λ3λ1ξ+λ4,-ω3+λ2λ1ω2+λ3λ1ω+λ4,-η3+λ2λ1η2+λ3λ1η+λ4}≥τ1m-ϑ=m2τ1>m2,θ(u0)=max⁡t∈[t1,t2]|u0(t)|=m-ϑ=m2τ12>m2,(73)γ(u0)=max⁡t∈[t1,t2]|u0′′(t)|=m-max⁡{|6t1-2λ2λ1|,|6t2-2λ2λ1|}=max⁡{|6t1-(2λ2/λ1)|,|6t2-(2λ2/λ1)|}ϑm2τ12≤m4.
Equation (72) implies that α(u0)≥τ1θ(u0); that is, min⁡t∈[ξ,ω]u0(t)≥τ1max⁡t∈[t1,t2]|u0(t)|. So u0∈K; it holds u0∈P(γ,θ,α,m2,m2τ1-2,m4) and {P(γ,θ,α,m2,m2τ1-2,m4)∣α(u0)>m2}≠∅. Hence, if u∈P(γ,θ,α,m2,m2τ1-2,m4), then m2≤u(t)≤m2τ1-2, |u′′(t)|≤m4 for t∈[ξ,ω]. In view of assumption (E2), we get
(74)α(Tu)=min⁡t∈[ξ,ω](Tu)(t)=min⁡{∫t1η‍x1(ξ,s)h(s)f(s,u,u′,u′′)ds,∫t1η‍y1(ω,s)h(s)f(s,u,u′,u′′)ds,∫ηt2‍x2(ω,s)h(s)f(s,u,u′,u′′)ds,∫ηt2‍y2(ξ,s)h(s)f(s,u,u′,u′′)ds,∫t1η‍y1(η,s)h(s)f(s,u,u′,u′′)ds,∫ηt2‍y2(η,s)h(s)f(s,u,u′,u′′)ds}≥τ1(max⁡⁡{∫t1η‍x1(η,s)h(s)f(s,u,u′,u′′)ds,∫t1η‍y1(t1,s)h(s)f(s,u,u′,u′′)ds}+max⁡⁡{∫ηt2‍x2(η,s)h(s)f(s,u,u′,u′′)ds,∫ηt2‍y2(t2,s)h(s)f(s,u,u′,u′′)ds})≥m2N2τ1(max⁡{∫t1η‍x1(η,s)h(s)ds,∫t1η‍y1(t1,s)h(s)ds}+max⁡⁡{∫ηt2‍x2(η,s)h(s)ds,∫ηt2‍y2(t2,s)h(s)ds})=m2.
Thus, α(Tu)>m2 for each u∈P(γ,θ,α,m2,m2τ1-2,m4).
Next, we verify that the condition(B2) of Lemma 2 is satisfied. In fact, if u∈P(γ,α,m2,m4) with θ(Tu)>m2τ1-2, then
(75)α(Tu)=min⁡t∈[ξ,ω](Tu)(t)≥τ1max⁡t∈[t1,t2](Tu)(t)=τ1θ(Tu)>m2.
So α(Tu)>m2 for any u∈P(γ,α,m2,m4) with ϑ(Tu)>m2τ1-2.
Finally, we assert that the condition(B3) of Lemma 2 is fulfilled. It is clear that 0∉Q(γ,ψ,m1,m4). Suppose that u∈Q(γ,ψ,m1,m4) with ψ(u)=m1; by assumption (E3), we have
(76)ψ(Tu)=max⁡t∈[t1,t2]|(Tu)′(t)|≤max⁡t∈[t1,t2]∫t1η‍|∂∂tG(t,s)|h(s)f(s,u,u′,u′′)ds+max⁡t∈[t1,t2]∫ηt2‍|∂∂tG(t,s)|h(s)f(s,u,u′,u′′)ds=max⁡{∫t1η‍∂∂tx1(t1,s)h(s)f(s,u,u′,u′′)ds,∫t1η‍-∂∂ty1(t1,s)h(s)f(s,u,u′,u′′)ds}+max⁡{∫ηt2‍-∂∂tx2(t2,s)h(s)f(s,u,u′,u′′)ds,∫ηt2‍∂∂ty2(t2,s)h(s)f(s,u,u′,u′′)ds}≤m1N3(max⁡{∫t1η‍∂∂tx1(t1,s)h(s)ds,∫t1η‍-∂∂ty1(t1,s)h(s)ds}+max⁡{∫ηt2‍-∂∂tx2(t2,s)h(s)ds,∫ηt2‍∂∂ty2(t2,s)h(s)ds})=m1.
Hence, ψ(Tu)≤m1 for u∈Q(γ,ψ,m1,m4) with ψ(u)=m1.
Hence, the hypotheses of Lemma2 are satisfied. Therefore, we can conclude that the operator T has three positive fixed points u1,u2,u3∈P(γ,m4)¯. In addition, condition (A2) guarantees that those fixed points are positive, so the problem (Ps) has at least three positive solutions u1(t), u2(t), u3(t) satisfying max⁡t∈[t1,t2]|ui′′(t)|≤m4, for i=1,2,3, m2<min⁡t∈[ξ,ω]u1(t),  and m1<max⁡t∈[t1,t2]u2′(t) with min⁡t∈[ξ,ω]u2(t)<m2,max⁡t∈[t1,t2]u3′(t)<m1.


## 6. Applications

The known Falkner-Skan equation(77)u′′′(t)+12(ϱ+1)u′′(t)u(t)+ρ(1-u′(t)2)=0,
whose solution is the similarity solution of the two-dimensional incompressible laminar boundary-layer equations [4, 5]. When ϱ=1, which arises in the study of two-dimensional incompressible viscous flow past a thin semi-infinite flat plate ρπ, 0≤ρ≤1. The special case ρ=0 is Blasius’s equation, in which the wedge reduces to a flat plate. The special case ρ=1/2 is called Homann's equation, in which the wedge reduces to a flat plate. The special case ρ=1 is called Hiemenz’s equation. ρ>0 corresponds to flow toward the wedge, otherwise, to flow away from the wedge. When ϱ≠1, taking ϱ=ρ, Blasius flow over a flat plate with a sharp edge as ρ=0, flow over a wedge with half angle θ1/2=ρπ/(ρ+1), 0<θ1/2<π/2 as 0<ρ<1, Hiemenz flow toward a plane stagnation point as ρ=1, flow into a corner with θ1/2>π/2 as 1<ρ<2, and no corresponding simple ideal flow as ρ>2 exist.Now, we give a general equation of (77), subject to the boundary conditions. But it is even more difficult to solve, because it is singular; the solution u(t) denotes the physically relevant solutions for ϱ=1 and the dimensionless stream function for ϱ≠1 of (77), respectively.Consider(78)u′′′(t)+h(t)f(t,u(t),u′(t),u′′(t))=0,hhhhhhhhhhhhhhhhhhhhllhhh0<t<1,u(0)-u′′(0)=u′(14)=u(1)+u′(1)=0,where(79)h(t)=1t(1-t),f(t,μ,ν,ϖ)={1+t2(μ2/3+2πarctan(1+ν2ϖ4)),0≤μ≤1,1+t2(1+2πarctan(1+ν2ϖ4)),μ>1.It is easy to check that hypotheses(C1), (C2) hold. Taking L=1, [m,M]=[1/3,2/5]⊆[ξ,ω]=[1/4,1/2],F(t)=1, by some calculations, we have M1=(33/16)2π, M2=1, M3=1, M4=17/36, τ1=8/11, and
(80)G(t,s)={s∈[0,14]{116(2t2-t+4)(s2+2)+t(14-s)-12(t-s)2,s≤t,116(2t2-t+4)(s2+2)+t(14-s),t≤s,s∈[14,1]{116(2t2-t+4)(s2-4s+3)-12(t-s)2,s≤t,116(2t2-t+4)(s2-4s+3),t≤s;
then f2(t,μ,ν,ϖ)≤1 for (t,μ,ν,ϖ)∈[0,1]×[1,∞[×]-∞,∞[2 and lim⁡inf⁡μ→0+min⁡t∈[1/4,1/2](hf(t,μ,ν,ϖ)/μ)=∞. By Theorem 8, problem (78) has at least one positive solution. These results accord with ϱ=1; the solutions of (77) are known to exist [23].On the other hand, let(81)f(t,μ,ν,ϖ)={10-2t+34μ2+(νϖ1010)3,0<μ<3.78,10-2t+34×3.782+(νϖ1010)3,μ≥3.78.It is easy to check that hypotheses(E1)–(E3) hold and f(t,0,0,0)≢0 on [0,1]. By some calculations, we have k=4, τ1=8/11, τ3=8.992, τ4=3/4, λ1=4, λ2=29/8, λ3=-15/16, λ4=29/16, ϑ=29/16, min⁡⁡{1.795,1.797,1.795}≥τ1ϑ≈1.318, N1≈2.238, N2≈0.730, N3≈0.505. If we choose m1=1, m2=2, m4=105, then f(t,μ,ν,ϖ) satisfies
(82)f(t,μ,ν,ϖ)≤m4N1≈4.468×104,(t,μ,ν,ϖ)∈[0,1]×[0,674400]×[-7.5×104,7.5×104]×[-105,105],f(t,μ,ν,ϖ)>m2N2≈2.740,(t,μ,ν,ϖ)∈[14,12]×[2,3.78]×[-7.5×104,7.5×104]×[-105,105],f(t,μ,ν,ϖ)≤m1N3≈1.980,(t,μ,ν,ϖ)∈[0,1]×[0,1]×[-7.5×104,1]×[-105,105].
Then all hypotheses of Theorem 11 hold. Hence, problem (78) has at least three positive solutions u1(t), u2(t), and u3(t) such that max⁡0≤t≤1|ui′′(t)|≤104(i=1,2,3), 1<min⁡1/4≤t≤3/4u1(t), 1/2<max⁡0≤t≤1u2′(t) with min⁡1/4≤t≤3/4u2(t)<1, max⁡0≤t≤1u3′(t)<1/2 hold. These results coincide with ϱ≠1; the solutions of (77) are the dimensionless stream functions.



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*Source: 638976-2014-05-26.xml*