# On a Constructive Approach for Derivative-Dependent Singular Boundary Value Problems

Authors: R. K. Pandey; Amit K. Verma
Journal: International Journal of Differential Equations
 (2011)

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## Abstract

We present a constructive approach to establish existence and uniqueness of solution of singular boundary value problem-(p(x)y′(x))′=q(x)f(x,y,py′) for 0<x≤b,y(0)=a,α1y(b)+β1p(b)y′(b)=γ1. Here p(x)>0 on (0,b) allowing p(0)=0. Further q(x) may be allowed to have integrable discontinuity at x=0, so the problem may be doubly singular.


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## Body

## 1. Introduction

Consider the following singular boundary value problem:(1.1)My≡-(p(x)y′(x))′=q(x)f(x,y(x),p(x)y′(x)),0<x≤b,(1.2)y(0)=a,α1y(b)+β1p(b)y′(b)=γ1,
where α1>0, β1≥0, γ1 is any finite constant. We assume that p(x) and q(x) satisfy the following conditions:(A1)
p(x)>0 in (0,b], p∈C[0,b]∩C1(0,b) and ∫0b(dt/p(t))<∞,(A2)
q(x)>0 in (0,b] and ∫0bq(t)dt<∞.In this work we establish existence and uniqueness of solution of the singular problem (1.1)-(1.2). We use monotone iterative method. For this we require an appropriate iterative scheme. In this regard Cherpion et al. [1] suggest the following approximation scheme:(1.3)-αn+1′′+k̃(x)αn+1′+l̃(x)αn+1=f(x,αn,αn′)+k̃(x)αn′+l̃(x)αn,αn+1(0)=αn+1(1)=0,
for the regular boundary value problem -y′′=f(x,y,y′),y(0)=y(1)=0. They also suggest that (1.3) with l̃(x)=0 or with constant k̃ and l̃ does not work for the Dirichlet boundary condition.Thus, for our problem we consider the following iterative scheme:(1.4)Lyn+1=F(x,yn,pyn′),0<x≤b,yn+1(0)=0,α1yn+1(b)+β1p(b)yn+1′(b)=γ1,
where
(1.5)Ly=-(p(x)y′(x))′-μ(x)q(x)p(x)y′(x)-λk(x)q(x)y(x),F(x,y,py′)=q(x)f(x,y,py′)-μ(x)q(x)p(x)y′(x)-λk(x)q(x)y(x).
We assume that k(x) and μ(x) satisfy the following conditions.(A3)
k(x)∈C[0,b], ∃mk, Mk∈R such that 0<mk≤k(x)≤Mk.(A4)
μ(x)∈Lq1(0,b), that is,∫0bq(x)μ(x)dx<∞.(A5)
Further, we assume that the homogeneous boundary value problemLy=0, y(0)=0, and α1y(b)+β1p(b)y′(b)=0 has only trivial solution.For∫0b(dt/p(t))<∞, several researchers ([2–5]) suggest to reduce the singular problem to regular problem by a change of variable. But in [6] it is suggested that a direct consideration of singular problems provide better results.Further, the following sign restrictions are imposed by several researchers ([4, 5, 7–9]):  (i)
yf(x,y,0)>0, |y|>M0, where M0 is a constant ([7–9]) or(ii)
f(x,M1,0)≥0≥f(x,-M2,0),M1,M2≥0 ([4, 5]).But such sign restrictions are quite restrictive as the simple differential equationy′′=2 fails to satisfy the sign restrictions (i) and (ii) ([7]).In the present work we consider the singular boundary value problem (SBVP) directly and do not impose any sign restriction. Further, we do not assume that the pointx=0 is a regular singular point as assumed in [6, 9]. We use iterative scheme (1.4) to establish existence and uniqueness of the solution of the problem. With the help of nonnegativity of Green's function, existence uniqueness of linear singular boundary value problem (LSBVP) is established.This paper is divided in four sections. In Section2, we show that singular point x=0 is of limit circle type; hence, spectrum is pure point spectrum with complete set of orthonormal eigenfunctions. In Section 3, we prove the existence uniqueness of the corresponding LSBVP. Finally, in Section 4, using the results of Section 3, we establish the existence uniqueness of solutions of the nonlinear problem (1.1)-(1.2).


## 2. Eigenfunction Expansion

LetLs2(0,b) be a Hilbert space with weight s(x)=k(x)q(x) and the inner product defined as(2.1)〈u,v〉=∫0bs(t)u(t)v(t)¯dt.
Conditions on p, q, μ, and k guarantee that the singular point x=0 is of limit circle type (Weyl's Theorem, [10, page 438]. Thus, we have pure point spectrum ([11, page 125]. Next, from the Lagrange's identity, it is easy to see that all the eigenvalues are real, simple, positive, and eigenfunctions are orthogonal. Let the eigenvalues be 0<λ0<λ1<λ2<⋯, and let the corresponding eigenfunctions be ψ0,ψ1,ψ2,…, respectively. Next, we transform Ly=0 by changing variable z=g(x)y, where g(x)=e∫0xμ(t)q(t)dt, to(2.2)-(p(x)z′(x))′+r(x)z(x)=λs(x)z(x),0<x≤b,
where r(x)=(1/2){μ(x)q(x)p(x)}′+(1/4){μ(x)q(x)}2p(x). Now following the analysis of Theorem 2.7, (i), (ii) and Theorem 2.17 of [11] for the operator (1/s)(r+M), where M is defined by (1.1), the following results can be established.Theorem 2.1.
Letf(x) be the primitive of an absolutely continuous function, and let
(2.3)1s(r+M)f∈Ls2(0,b),f(b)sin⁡α-p(b)f′(b)cos⁡α=0,whereαis real,lim⁡x→0p(x)Wx[f,ψ]=0,
for every nonreal λ, where ψ(x,λ)∈Ls2(0,b) is a solution of (2.2) and W[f,ψ] is the wronskian of f and ψ. Then
(2.4)f(x)=∑n=0∞cnψn(x),(0≤x≤b),
being the series absolutely and uniformly convergent on [0,b].Theorem 2.2.
Letf∈Ls2(0,b). Then
(2.5)∫0bs(x){f(x)}2dx=∑n=0∞cn2.Theorem 2.3.
Letf∈Ls2(0,b), and let Φ(x,λ) be the solution of
(2.6)-(p(x)z′(x))′+r(x)z(x)-λs(x)z(x)=s(x)f(x),0<x≤b,
satisfying α11z(b)+β1p(b)z′(b)=0, where α11=α1-(1/2)β1μ(b)p(b)q(b). Then for λ not equal to any of the values of λn, one has
(2.7)Φ(x,λ)=∑n=0∞cnψnλ-λn,
where the series is absolutely convergent.Remark 2.4.
Since∥·∥s on Ls2(0,b) is equivalent to ∥·∥q on Lq2(0,b), we can apply Theorems 2.1–2.3 in Lq2(0,b) also.


## 3. Linear Singular Sturm-Liouville’s Problem

In this section we apply Theorem 1.1 of   [12] to the differential operator L and generate two linearly independent solutions of the linear problem. Further, with the help of these solutions, Green's function is constructed, and nonnegativity of the Green's function is established.Theorem 3.1.
Letp(x), q(x), k(x), and μ(x) satisfy (A1), (A2), (A3), and (A4), respectively. Then the initial value problems (IVPs)
(3.1)Ly=0,0<x≤b,y(0)=a0,lim⁡x→0+p(x)y′(x)=b0,(3.2)Ly=0,0<x≤b,y(b)=c0,p(b)y′(b)=d0
have a solution in Ls2(0,b) or equivalently in Lq2(0,b) (Remark 2.4).


### 3.1. Green's Function

Green's functionG(x,t,λ) for the differential operator L can be defined as(3.3)G(x,t,λ)=1p(x)W(ψ,ϕ){-ψ(t)ϕ(x),if0<t≤x,-ψ(x)ϕ(t),ifx≤t≤b,
where ϕ=S{α1y1(x)+β1y2(x)}, S=1/α12+β12, y1(b)=0, p(b)y1′(b)=-1, y2(b)=1, p(b)y2′(b)=0, and ψ is a nontrivial solution of IVP (3.1) with a0=0, b0=1. From (A5), it is easy to conclude that p(x)W(ψ,ϕ)|x=b≠0; thus, ϕ and ψ are linearly independent.Next we establish nonnegativity of Green's function. For this we need to establish following results.Lemma 3.2.
Ify(x) satisfies Ly(x)=q(x)f(x)≥0, for 0<x≤b, y(0)=0, and α1y(b)+β1p(b)y′(b)=γ1≥0, where p(x), q(x), μ(x), and k(x) satisfy (A1), (A2), (A3), and (A4), respectively, then y(x)≥0 provided that λ≤0.Proof.
We divide the proof in two cases as follows.
Case i.
Whenλ<0. On contrary assume that there exists a point c∈(0,b) such that y(c)<0. Then from the continuity of the solutions there exists a point d∈(0,b) such that y(d)<0, y′(0)=0, and y′′(d)≥0. Now at the point d, we have
(3.4)-p(d)y′′(d)-p′(d)y′(d)-μ(d)q(d)p(d)y′(d)-λk(d)q(d)y(d)<0,
which is a contradiction. Hence, y(x)≥0 when λ<0.
Case ii.
Whenλ=0. Using the same notations as in the previous case, we have y′(x)>0 for x>d and y′(x)<0 for x<d.
Now, we consider the interval[d,x0]⊂(0,b) where y′(d)=0, y<0 in [d,x0], and y′>0 in (d,x0]. Then
(3.5)P=∫dx0{p(t)g(t)(y′(t))2-s(t)g(t)f(t)y(t)}dt>0.
Integrating the first term by parts, we get
(3.6)P=p(x0)g(x0)y′(x0)y(x0)<0,
which is again a contradiction. Thus, y(x)≥0 when λ≤0.Lemma 3.3.
Consider the following differential equation:(3.7)Ly(x)=0,0<x≤b,
where p(x), q(x), μ(x), and k(x) satisfy (A1), (A2), (A3), and (A4), respectively, with the boundary conditions:
(3.8)y(0)=0,α1p(b)+β1p(b)y′(b)=γ1.
Then LSBVP (3.7)-(3.8) has a unique solution given by
(3.9)y(x)=γ1ψ(x)α1ψ(b)+β1p(b)ψ′(b),
provided that λ is none of the eigenvalues of the corresponding eigenvalue problem and ψ satisfies (3.1). Moreover, y(x)≥0 if γ1≥0 and 0<λ<λ0, where λ0 is the first positive zero of α1ψ(b,λ)+β1p(b)ψ′(b,λ).Proof.
From Theorem3.1, it is easy to see that the unique solution of (3.7)-(3.8) can be written as
(3.10)y(x)=γ1ψ(x)α1ψ(b)+β1p(b)ψ′(b),
provided that α1ψ(b,λ)+β1p(b)ψ′(b,λ)≠0; that is, λ is none of the eigenvalue of the corresponding eigenvalue problem (Section 2). Since ψ(0)=0, limx→0+p(x)ψ′(x)=1, and ψ(x,λ) does not change sign for 0<λ<λ0, we get that y(x)≥0 for 0<λ<λ0, provided that γ1≥0.Lemma 3.4.
For the linear differential operator associated with(3.11)Ly(x)=q(x)f(x),0<x≤b,(3.12)y(0)=0,α1y(b)+β1p(b)y′(b)=γ1,
with f∈Lq2(0,b), the generalized Green's function for the corresponding homogeneous boundary value problem is given by
(3.13)G(x,t,λ)=∑n=0∞ψ(x,λn)ψ(t,λn)λn-λ,
where ψ(x,λi) are the normalized eigenfunctions corresponding to the eigenvalue λi. G(x,t,λ) satisfies the homogeneous boundary condition provided that λ≠λ0,λ1,…. Solution of the non-homogeneous LSBVP (3.11)-(3.12) is
(3.14)y(x)=γ1ψ(x,λ)α1ψ(b)+β1p(b)ψ′(b)+∫0bq(t)f(t)G(x,t,λ)dt.
The series on the right is absolutely convergent.Proof.
The solutiony(x) of (3.11)-(3.12) can be written as sum of the solution of (3.11) with boundary condition y(0)=0,α1y(b)+β1p(b)y′(b)=0 and solution of (3.7) with boundary condition y(0)=0, α1y(b)+β1p(b)y′(b)=γ1,
(3.15)y(x)=γ1ψ(x,λ)α1ψ(b,λ)+β1p(b)ψ′(b,λ)+∫0bq(t)f(t)G(x,t,λ)dt,
where G(x,t,λ) is Green's function defined by (3.3). Now using the analysis of ([11, page 38], it is easy to show that the generalized Green's function is given by
(3.16)G(x,t,λ)=∑n=0∞ψ(x,λn)ψ(t,λn)λn-λ,
and absolute convergence of the series on the right-hand side follows from the analysis of ([11, page 38]. This completes the proof.Lemma 3.5.
Iff∈Lq2(0,b), γ1≥0 and f≥0, then solution of (3.11)-(3.12) is nonnegative provided that 0<λ<λ0.Proof.
We first show thatG(x,t)≥0 for all 0≤x,t≤b if 0<λ<λ0. Fixing t, G(x,t) satisfies LG(x,t)=0,0<x≤t-, where ′≡∂/∂x. Since G(0,t)=0, α1G(t-,t)+β1p(t-)G′(t-,t)≥0 for 0<λ<λ0, from Lemma 3.3  G(x,t)≥0 for 0≤x≤t-, provided that 0<λ<λ0. By the symmetry, continuity, and G(t,t)≥0 for 0<λ<λ0, it follows that G(x,t)≥0 for 0≤x,t≤b, provided that 0<λ<λ0. The result follows.Corollary 3.6.
Ify(x) satisfies Ly(x)=q(x)f(x)≥0 for 0<x≤b and y(0)=0, α1y(b)+β1p(b)y′(b)=γ1≥0, then y(x)≥0, provided that λ<λ0.Proof.
The proof follows from Lemmas3.2 and 3.5.Corollary 3.7.
The solution of the boundary value problem in Lemma3.4 is unique.Proof.
The proof follows from Corollary3.6.


## 3.1. Green's Function

Green's functionG(x,t,λ) for the differential operator L can be defined as(3.3)G(x,t,λ)=1p(x)W(ψ,ϕ){-ψ(t)ϕ(x),if0<t≤x,-ψ(x)ϕ(t),ifx≤t≤b,
where ϕ=S{α1y1(x)+β1y2(x)}, S=1/α12+β12, y1(b)=0, p(b)y1′(b)=-1, y2(b)=1, p(b)y2′(b)=0, and ψ is a nontrivial solution of IVP (3.1) with a0=0, b0=1. From (A5), it is easy to conclude that p(x)W(ψ,ϕ)|x=b≠0; thus, ϕ and ψ are linearly independent.Next we establish nonnegativity of Green's function. For this we need to establish following results.Lemma 3.2.
Ify(x) satisfies Ly(x)=q(x)f(x)≥0, for 0<x≤b, y(0)=0, and α1y(b)+β1p(b)y′(b)=γ1≥0, where p(x), q(x), μ(x), and k(x) satisfy (A1), (A2), (A3), and (A4), respectively, then y(x)≥0 provided that λ≤0.Proof.
We divide the proof in two cases as follows.
Case i.
Whenλ<0. On contrary assume that there exists a point c∈(0,b) such that y(c)<0. Then from the continuity of the solutions there exists a point d∈(0,b) such that y(d)<0, y′(0)=0, and y′′(d)≥0. Now at the point d, we have
(3.4)-p(d)y′′(d)-p′(d)y′(d)-μ(d)q(d)p(d)y′(d)-λk(d)q(d)y(d)<0,
which is a contradiction. Hence, y(x)≥0 when λ<0.
Case ii.
Whenλ=0. Using the same notations as in the previous case, we have y′(x)>0 for x>d and y′(x)<0 for x<d.
Now, we consider the interval[d,x0]⊂(0,b) where y′(d)=0, y<0 in [d,x0], and y′>0 in (d,x0]. Then
(3.5)P=∫dx0{p(t)g(t)(y′(t))2-s(t)g(t)f(t)y(t)}dt>0.
Integrating the first term by parts, we get
(3.6)P=p(x0)g(x0)y′(x0)y(x0)<0,
which is again a contradiction. Thus, y(x)≥0 when λ≤0.Lemma 3.3.
Consider the following differential equation:(3.7)Ly(x)=0,0<x≤b,
where p(x), q(x), μ(x), and k(x) satisfy (A1), (A2), (A3), and (A4), respectively, with the boundary conditions:
(3.8)y(0)=0,α1p(b)+β1p(b)y′(b)=γ1.
Then LSBVP (3.7)-(3.8) has a unique solution given by
(3.9)y(x)=γ1ψ(x)α1ψ(b)+β1p(b)ψ′(b),
provided that λ is none of the eigenvalues of the corresponding eigenvalue problem and ψ satisfies (3.1). Moreover, y(x)≥0 if γ1≥0 and 0<λ<λ0, where λ0 is the first positive zero of α1ψ(b,λ)+β1p(b)ψ′(b,λ).Proof.
From Theorem3.1, it is easy to see that the unique solution of (3.7)-(3.8) can be written as
(3.10)y(x)=γ1ψ(x)α1ψ(b)+β1p(b)ψ′(b),
provided that α1ψ(b,λ)+β1p(b)ψ′(b,λ)≠0; that is, λ is none of the eigenvalue of the corresponding eigenvalue problem (Section 2). Since ψ(0)=0, limx→0+p(x)ψ′(x)=1, and ψ(x,λ) does not change sign for 0<λ<λ0, we get that y(x)≥0 for 0<λ<λ0, provided that γ1≥0.Lemma 3.4.
For the linear differential operator associated with(3.11)Ly(x)=q(x)f(x),0<x≤b,(3.12)y(0)=0,α1y(b)+β1p(b)y′(b)=γ1,
with f∈Lq2(0,b), the generalized Green's function for the corresponding homogeneous boundary value problem is given by
(3.13)G(x,t,λ)=∑n=0∞ψ(x,λn)ψ(t,λn)λn-λ,
where ψ(x,λi) are the normalized eigenfunctions corresponding to the eigenvalue λi. G(x,t,λ) satisfies the homogeneous boundary condition provided that λ≠λ0,λ1,…. Solution of the non-homogeneous LSBVP (3.11)-(3.12) is
(3.14)y(x)=γ1ψ(x,λ)α1ψ(b)+β1p(b)ψ′(b)+∫0bq(t)f(t)G(x,t,λ)dt.
The series on the right is absolutely convergent.Proof.
The solutiony(x) of (3.11)-(3.12) can be written as sum of the solution of (3.11) with boundary condition y(0)=0,α1y(b)+β1p(b)y′(b)=0 and solution of (3.7) with boundary condition y(0)=0, α1y(b)+β1p(b)y′(b)=γ1,
(3.15)y(x)=γ1ψ(x,λ)α1ψ(b,λ)+β1p(b)ψ′(b,λ)+∫0bq(t)f(t)G(x,t,λ)dt,
where G(x,t,λ) is Green's function defined by (3.3). Now using the analysis of ([11, page 38], it is easy to show that the generalized Green's function is given by
(3.16)G(x,t,λ)=∑n=0∞ψ(x,λn)ψ(t,λn)λn-λ,
and absolute convergence of the series on the right-hand side follows from the analysis of ([11, page 38]. This completes the proof.Lemma 3.5.
Iff∈Lq2(0,b), γ1≥0 and f≥0, then solution of (3.11)-(3.12) is nonnegative provided that 0<λ<λ0.Proof.
We first show thatG(x,t)≥0 for all 0≤x,t≤b if 0<λ<λ0. Fixing t, G(x,t) satisfies LG(x,t)=0,0<x≤t-, where ′≡∂/∂x. Since G(0,t)=0, α1G(t-,t)+β1p(t-)G′(t-,t)≥0 for 0<λ<λ0, from Lemma 3.3  G(x,t)≥0 for 0≤x≤t-, provided that 0<λ<λ0. By the symmetry, continuity, and G(t,t)≥0 for 0<λ<λ0, it follows that G(x,t)≥0 for 0≤x,t≤b, provided that 0<λ<λ0. The result follows.Corollary 3.6.
Ify(x) satisfies Ly(x)=q(x)f(x)≥0 for 0<x≤b and y(0)=0, α1y(b)+β1p(b)y′(b)=γ1≥0, then y(x)≥0, provided that λ<λ0.Proof.
The proof follows from Lemmas3.2 and 3.5.Corollary 3.7.
The solution of the boundary value problem in Lemma3.4 is unique.Proof.
The proof follows from Corollary3.6.


## 4. Nonlinear Sturm-Liouville’s Problem

In this section, we establish the existence uniqueness of solution of the nonlinear problem (1.1)-(1.2). For this, first we prove that the sequences generated by (1.4) are monotonic sequences (Lemmas 4.2 and 4.3). Then using the bound for py′ (Lemmas 4.9 and 4.10), the uniform convergence of these sequences to a solution of the nonlinear problem is established (Theorem 4.11). Finally the uniqueness of the solution is established in Theorem 4.14.The nonlinear boundary value problem(4.1)-(p(x)y′(x))′=q(x)f(x,y,py′),0<x≤b,y(0)=a,α1y(b)+β1p(b)y′(b)=γ2
can be transformed to (4.2)-(p(x)u′(x))′=q(x)f(x,u+a,pu′),0<x≤b,u(0)=0,α1u(b)+β1p(b)u′(b)=γ2-aα1,
with u=y-a. Further, the functions f(x,u+a,pu′) and f(x,y,py′) satisfy the same Lipschitz condition, so we may work with the boundary value problem(4.3)-(p(x)y′(x))′=q(x)f(x,y,py′),0<x≤b,y(0)=0,α1y(b)+β1p(b)y′(b)=γ1.
Next, we define upper solution u0(x) and lower solution v0(x) such that u0≥v0, which work as initial iterates for our constructive approach.Definition 4.1.
A functionu0(x)∈C[0,b]∩C2(0,b] is an upper solution if
(4.4)-(p(x)u0′(x))′≥q(x)f(x,u0,pu0′),0<x≤b,u0(0)=0,α1u0(b)+β1p(b)u0′(b)≥γ1,
and a function v0(x)∈C[0,b]∩C2(0,b] is a lower solution if
(4.5)-(p(x)v0′(x))′≤q(x)f(x,v0,pv0′),0<x≤b,v0(0)=0,α1v0(b)+β1p(b)v0′(b)≤γ1.Lemma 4.2.
Ifλ<0, λk(x)≤K1, |μ(x)|≤L1, Ly≥0 for 0<x≤b, y(0)=0, and α1y(b)+β1p(b)y′(b)≥0, then
(4.6)(K1-λk(x))y-(μ(x)+L1(sign⁡y′))py′≥0,0<x≤b,
provided that
(4.7)1+λ∫0bdtp(t)g(t)∫0bs(t)g(t)dt-sup⁡(pgb)∫0bdtp(t)g(t)>0,0<x≤b,(K1-λk(x))-(L1-|μ(x)|)Φ(p,q,s,g)≥0,0<x≤b,
hold. Here,
(4.8)Φ(p,q,s,g)=sup⁡(pgb)I(p,q,s,g)-λ∫0bs(t)g(t)dt,(4.9)I(p,q,s,g)=(1+λ∫0bdtp(t)g(t)∫0bs(t)g(t)dt-sup⁡(pgb)∫0bdtp(t)g(t))-1.Proof.
The solution of the equationLy=qf≥0, y(0)=0, and α1y(b)+β1p(b)y′(b)=γ1≥0 is given by (3.15) where G(x,t,λ) is defined by (3.3). Substituting y(x) from (3.15) into (4.6), it is easy to see that we require the following inequalities in order to complete the proof
(4.10)(K1-λk(x))ψ-(μ(x)+L1(sign⁡y′))pψ′≥0,0<x≤b,(4.11)(K1-λk(x))ϕ-(μ(x)+L1(sign⁡y′))pϕ′≥0,0<x≤b.
Hereψ satisfies the IVP at x=0; that is, Lψ=0, ψ(0)=0 and limx→0+p(x)ψ(x)=1, and ϕ satisfies the IVP at x=b; that is, Lϕ=0, ϕ(b)=β1 and p(b)ϕ′(b)=-α1. The solutions ψ and ϕ cannot have either point of maxima (at the point of maxima the Lψ=0 or Lϕ=0 will be contradicted) or point of minima (since to have minima, maxima is bound to occur). So, finally we have ψ′(x)≥0 and ϕ′(x)≤0 on [0,b]. As |μ(x)|≤L1, it is enough to prove the following inequalities:
(4.12)(K1-λk(x))ψ-(L1-|μ(x)|)pψ′≥0,0<x≤b,(4.13)(K1-λk(x))ϕ+(L1-|μ(x)|)pϕ′≥0,0<x≤b.
Next, we prove the inequality (4.12), and the other one can be proved in a similar manner.
By the mean value theorem, there existτ∈(0,b) such that ψ(b)=bψ′(τ). Writing Lψ=0 in the following form:
(4.14)-(p(x)g(x)ψ(x)′)′-λs(x)g(x)ψ(x)=0,
and integrating it first from τ to x and then x to b, we get that
(4.15)p(x)ψ′(x)≤ψ(x)Φ(p,q,s,g)on⁡0<x≤b.
Here Φ(p,q,s,g) is given by (4.8). Now, the result follows from (4.7), (4.12), and (4.15).Lemma 4.3.
If0<λ<λ0, λk(x)≤K1, |μ(x)|≤L1, Ly≥0 for 0<x≤b, y(0)=0, and α1y(b)+β1p(b)y′(b)≥0, then
(4.16)(K1-λk(x))y-(μ(x)+L1(sign⁡y′))py′≥0,0<x≤b,
provided that
(4.17)1-sup⁡(pgb)∫0bdtpg>0,0<x≤b,(K1-λk(x))(1-sup⁡(pgb)∫0bdtpg)-(L1-|μ(x)|)sup⁡(pgb)≥0,0<x≤b
or
(4.18)1-λ∫0bs(t)g(t)dt∫0bdtp(t)g(t)>0,(K1-λk(x))(1-λ∫0bsgdt∫0bdtpg)-(L1-|μ(x)|)λ∫0bsgdt≥0,0<x≤b,
hold.Proof.
Similar to the proof of Lemma4.2, we need to establish two inequalities (4.10)-(4.11) for 0<λ<λ0. Here ψ and ϕ cannot have the point of minima in (0,b), because at the point of minima, the differential equation Lψ=0 or Lϕ=0 will be contradicted. So either ψ′(x)≥0 and ϕ′(x)≤0 or ψ(x) and ϕ(x) both are concave downwards on [0,b]. Thus we can divide the proof in two cases:
Case i.
ψ and ϕ both are concave downwards.
We prove forψ as similar analysis provides result for ϕ. Let the point of maxima be x0∈(0,b). Then ψ′(x)>0 for x<x0 and ψ′(x)<0 for x>x0. On both sides of x0, the inequality (4.10) will be reduced into the following two inequalities:
(4.19)(K1-λk(x))ψ-(L1+μ(x))pψ′≥0,ψ′(x)≥0,(K1-λk(x))ψ+(L1-μ(x))pψ′≥0,ψ′(x)≤0.
For a point x on the left side of x0, we integrate (4.14) from x to x0 twice and get
(4.20)p(x)ψ′(x)≤λψ(x)∫0bs(t)g(t)dt1-λ∫0bs(t)g(t)dt∫0b(1/p(t)g(t))dt.
Similarly for any point x on the right side of x0, we get
(4.21)-p(x)ψ′(x)≤λψ(x)∫0bs(t)g(t)dt1-λ∫0bs(t)g(t)dt∫0b(1/p(t)g(t))dt.
Now, the result follows from the fact that |μ(x)|≤L1 and from (4.18) to (4.21).
Case ii.
Whenψ′(x)≥0 and ϕ′(x)≤0.
To establish the inequality (4.16), we require to establish the inequalities (4.12)-(4.13). We prove the inequality (4.12), and the proof for (4.13) is quite similar. By the mean value theorem, there exists τ∈(0,b) such that ψ(b)=bψ′(τ). Integrating (4.14) first from τ to x and then from x to b, we get
(4.22)p(x)ψ′(x)≤ψ(x)sup⁡(pg/b)1-sup⁡(pg/b)∫0b(dt/pg),
and the result follows from (4.12), (4.17), and (4.22). This completes the proof.Lemma 4.4.
Ifun is an upper solution of (4.3) and un+1 is defined by (1.4)–(1.5), then un≥un+1 for λ<λ0.Proof.
Letw=un-un+1. w satisfies Lw=-(pun′)′-qf(x,un,pyn′)≥0,0<x≤b,w(0)=0,α1w(b)+β1p(b)w′(b)≥0, and the result follows from Corollary 3.6.Proposition 4.5.
Letu0 be an upper solution of (4.3), and let f(x,y,py′) satisfy the following(F1)
f(x,y,py′) is continuous on(4.23)D0={(x,y,py′):[0,b]×[v0,u0]×R},(F2)
∃K1≡K1(D0) such that for all (x,y,v),(x,w,v)∈D0,(4.24)K1(y-w)≤f(x,y,v)-f(x,w,v)fory≥w,and(F3)
∃0≤L1≡L1(D0) such that for all (x,y,v1),(x,y,v2)∈D0,(4.25)|f(x,y,v1)-f(x,y,v2)|≤L1|v1-v2|,
and (4.7), (4.17), or (4.18) hold. Then the functions un defined by (1.4)–(1.5) are such that, for all n∈ℕ, (i) un is upper solution of (4.3) and (ii) un≥un+1.Proof.
Sinceu0 is an upper solution from Lemma 4.4, we have u0≥u1. Assume that the claim is true for n-1; that is, un-1 is an upper solution and un-1≥un.
Letw=un-1-un. We have
(4.26)-(pun′)′-qf(x,un,pun′)≥q{(K1-λk(x))w-(μ(x)+L1(sign⁡w′))pw′},
and from Lemmas 4.2 and 4.3 we get -(pun′)′-qf(x,un,pun′)≥0,0<x≤b.
Thus,un is an upper solution for all n∈ℕ. From Lemma 4.4 we have un≥un+1. Hence, the result follows.
Similar results (Lemma4.6, Proposition 4.7) follow for lower solutions.Lemma 4.6.
Ifvn is a lower solution of (4.3) and vn+1 is defined by (1.4)–(1.5) then vn≤vn+1 for λ<λ0.Proposition 4.7.
Letv0 be a lower solution of (4.3), let f(x,y,py′) satisfies (F1)–(F3) and (4.7), (4.17), or (4.18) hold. Then the functions vn defined by (1.4)–(1.5) are such that, for all n∈ℕ, (i) vn is lower solution of (4.3) and (ii) vn≤vn+1.Proposition 4.8.
Iff(x,y,py′) satisfies(F4)
f(x,u0,pu0′)-f(x,v0,pv0′)-μ(x)(pu0′-pv0′)-λk(x)(u0-v0)≥0 for 0<x≤b such that λk(x)≤K1 and |μ(x)|≤L1,and in addition let (F1)–(F3) and (4.7), (4.17), or (4.18) hold, then for all n∈ℕ the functions un and vn defined by (1.4)–(1.5) satisfy vn≤un.Proof.
Letwi=ui-vi, then wi satisfies Lwi=q(x)hi-1 for all i∈N such that
(4.27)hi(x)=f(x,ui,pui′)-f(x,vi,pvi′)-μ(x)(pui′-pvi′)-λk(x)(ui-vi),0<x≤b.
Since v0≤u0, we prove that v1≤u1. Since w1 is solution of Lw1=qh0≥0, w1(0)=0 and α1w1(0)+β1p(b)w1′(b)=0, from Corollary 3.6 we have w1≥0. Let n≥2, let hn-2≥0, and un-1≥vn-1, then we prove that hn-1≥0 and un≥vn. Consider
(4.28)hn-1=f(x,un-1,pun-1′)-f(x,vn-1,pvn-1′)-μ(x)pwn-1′-λk(x)wn-1≥(K1-λk(x))wn-1-(μ+L(sign⁡wn-1′))pwn-1′.
Since wn-1 is a solution of Lwn-1=qhn-2≥0, wn-1(0)=0, and α1wn-1(0)+β1p(b)wn-1′(b)=0; hence, from Lemmas 4.2 and 4.3, we have hn-1≥0. Thus, from Corollary 3.6 on Lwn=qhn-1≥0, wn(0)=0 and α1wn(0)+β1p(b)wn′(b)=0, we have wn≥0, that is, un≥vn. This completes the proof.Lemma 4.9.
Iff(x,y,py′) satisfies(F5)
for all(x,y,v)∈D0, |f(x,y,v)|≤φ(|v|) where φ:[0,∞)→(0,∞) is continuous and satisfies(4.29)∫0bq(s)ds<∫l0∞dsφ(s),
where l0=sup[0,b]|p(x)u0(x)/b|, then there exists R0>0 such that any solution of
(4.30)-(py′)′≥qf(x,y,py′),0<x≤b,(4.31)y(0)=0,α1y(b)+β1p(b)y′(b)≥γ1,
with y∈[v0,u0] for all x∈[0,b], satisfies ∥py′∥∞<R0.Proof.
We divide the proof in three parts.
Case i.
If solution is not monotone throughout the interval, then we consider the interval(x0,x]⊂(0,b) such that y′(x0)=0 and y′(x)>0 for x>x0. Integrating (4.30) from x0 to x we get
(4.32)∫0py′dsφ(s)≤∫0bq(s)ds.
From (F5) we can choose R0>0 such that
(4.33)∫0py′dsφ(s)≤∫0bq(s)ds≤∫l0R0dsφ(s)≤∫0R0dsφ(s),
which gives
(4.34)p(x)y′(x)≤R0.
Now we consider the case in which y′(x)<0 for x<x0, y′(x0)=0, and proceeding in the similar way we get
(4.35)-p(x)y′(x)≤R0,
and the result follows.Case ii.
Ify is monotonically increasing in (0,b), that is, y′>0 in (0,b), then by the mean value theorem there exists a point τ∈(0,b) such that
(4.36)y′(τ)=y(b)-y(0)b≤|u0b|.
Now, integrating (4.30) from τ to x, we get
(4.37)∫0py′dsφ(s)≤∫0bq(t)dt+∫0l0dsφ(s).
Further, from (F5) we can choose R0 such that
(4.38)∫0py′dsφ(s)≤∫0bq(s)ds+∫0l0dsφ(s)≤∫0R0dsφ(s).
which gives p(x)y′(x)≤R0.Case iii.
Ify is monotonically decreasing in (0,b); that is, y′<0 in (0,b), then argument similar to Case  ii yields
(4.39)∫0-py′dsφ(s)≤∫0bq(s)ds+∫0l0dsφ(s)≤∫0R0dsφ(s),
and we get
(4.40)-p(x)y′(x)≤R0,
and the result follows.Lemma 4.10.
Iff(x,y,py′) satisfies (F5), then there exists R0>0 such that any solution of
(4.41)-(py′)′≤qf(x,y,py′),0<x≤b,y(0)=0,α1y(b)+β1p(b)y′(b)≤γ1,
with y∈[v0,u0] for all x∈[0,b], satisfies ∥py′∥∞<R0.Proof.
Proof follows from the analysis of Lemma4.9.Theorem 4.11.
Letu0 and v0 be upper and lower solutions. Let f(x,y,py′) satisfy (F1) to (F5) and (4.7), (4.17), or (4.18) hold. Then, boundary value problem (4.3) has at least one solution in the region D0. If λ<λ0 is chosen such that λk(x)≤K1 and |μ(x)|≤L1, where λ0 is the first positive eigenvalue of the corresponding eigenvalue problem, then the sequences {un} and {vn} generated by (1.4)–(1.5) with initial iterate u0 and v0 converge monotonically and uniformly towards solutions ũ(x) and ṽ(x) of (4.3). Any solution z(x) in D0 must satisfy ṽ(x)≤z(x)≤ũ(x).Proof.
From Lemmas4.2–4.10, Propositions 4.5–4.8, and we get two monotonic sequences {un} and {vn} which are bounded by u0 and v0; respectively, and by Dini's Theorem their uniform convergence is assured. Let {un} and {vn} converge uniformly to ũ and ṽ.
By Lemmas4.9 and 4.10, it is easy to see that the sequences {pun′} and {pvn′} are uniformly bounded. Now, from(4.42)|pyn′(x1)-pyn′(x2)|=|∫x1x2(pyn′)′dt|,
uniform convergence of {yn}, properties (A1)–(A4), and (F1), it is easy to prove that {pyn′} is equicontinuous. Hence, by Arzela-Ascoli's Theorem there exist a uniform convergent subsequence {pynk′} of {pyn′}. Since limit is unique so original sequence will also converge uniformly to the same limit say py′. It is easy to see that, if yn→ỹ, then pyn′→pỹ′. Therefore sequences {pun′} and {pvn′} converge uniformly to pũ′ and pṽ′, respectively.
LetG(x,t) be Green's function for the linear boundary value problem Lyn=0, yn(0)=0, and α1yn(b)+β1p(b)yn′(b)=0. Then solution of (1.4)–(1.5) can be written as(4.43)yn=Cx2+∫0bG(x,t){F(t,yn-1,pyn-1′)+H(t)}dt,
where H(t)=2C(tp′(t)+p(t))+2Ctμ(t)q(t)p(t)+λCt2s(t) and C=γ1/(α1b2+2β1bp(b)).
Now, uniform convergence of{yn}, {pyn′} and continuity of f(x,y,py′) imply that {(1/q)F(x,yn,py′n)} converges uniformly in [0,b]. Hence, {(1/q)F(x,yn,pyn′)} converges in the sense of mean in Lq2(0,b). Taking limit as n→∞ and using Lemma 2.4 ([11, page 27]), we get(4.44)y=Cx2+∫0bG(x,t){F(t,y,py′)+H(t)}dt,
which is the solution of the boundary value problem (4.3).
Any solutionz(x) in D0 plays the role of u0(x). Hence z(x)≥ṽ(x). Similarly, z(x)≤ũ(x). This completes the proof.Remark 4.12.
The case whenλ=0 corresponds to the case when f(x,y,py′)≡f(x,py′). In such cases the boundary value problem (4.3) can be reduced to two initial value problems -z′=qf(x,z), z(0)=-α1 and py′=z, y(0)=β1. From the assumptions on p(x), q(x), and f(x,y,py′), one can easily conclude existence uniqueness of solutions of the nonlinear boundary value problem.Remark 4.13.
Suppose, in addition to the hypothesis of Theorem4.11, |f(x,y,py′)|≤N0 in D0. Then lower solution v0 and upper solution u0 may be obtained as solution of the following linear boundary value problems:
(4.45)-(pv0′)′+N0q(x)=0,0<x≤b,v0(0)=0,α1v0(b)+β1p(b)v0′(b)=γ1,-(pu0′)′-N0q(x)=0,0<x≤b,u0(0)=0,α1u0(b)+β1p(b)u0′(b)=γ1.Theorem 4.14.
Suppose thatf(x,y,py′) satisfies (F1), (F3), and ∃ constants K1(D0)<λ0 such that
(4.46)K1(u-v)≤f(x,u,py′)-f(x,v,py′).
Then the boundary value problem (4.3) has unique solution.Proof.
Letu and v be two solutions of (4.3), then we get
(4.47)-(p(u-v)′)′=q(x){f(x,u,pu′)-f(x,v,pv′)},0<x≤b,or-(p(u-v)′)′+L1q(x)(pu′-pv′)-K1q(x)(u-v)≥0,0<x≤b,(u-v)(0)=0,α1(u-v)(b)+β1p(b)(u-v)′(b)=0.
Since K1<λ0, from Corollary 3.6 we get u-v≥0 or u≥v. Similarly v≥u. Therefore, the solution of (4.3) is unique.



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*Source: 261963-2011-11-13.xml*