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Multiplicity of positive solutions of superlinear semipositone singular Neumann problems
Boundary Value Problems volume 2014, Article number: 217 (2014)
Abstract
Introduction
Neumann boundary value problems have been studied by many authors. We are mainly interested in the semipositone case. This paper deals with the existence and multiplicity of positive solutions of a superlinear semipositone singular Neumann boundary value problem.
Preliminaries
The proof of our main results relies on a nonlinear alternative of LeraySchauder type, the method of upper and lower solutions and on a wellknown fixed point theorem in cones.
Main results
We obtained the existence of at least two different positive solutions.
1 Introduction
We will be concerned with the existence and multiplicity of positive solutions of the superlinear singular Neumann boundary value problem in the semipositone case
Here the type of perturbations g(x,u) may be singular near u=0 and g(x,u) is superlinear near u=+\mathrm{\infty}. From the physical point of view, g(x,u) has an attractive singularity near u=0 if
and the superlinearity of g(x,u) means that
By the semipositone case of (1.1), we mean that g(x,u) may change sign and satisfies F(x,u)=g(x,u)+M\ge 0 where M>0 is a constant.
It is well known that the existence of positive solutions of boundary value problems has been studied by many authors in [1]–[6] and references therein. They mainly considered the case of p(x)\equiv 1 and q(x)\equiv 0. In [7], the authors studied positive solutions of Neumann boundary problems of second order impulsive differential equations in the positone case, based on a nonlinear alternative principle of LeraySchauder type and a wellknown fixed point theorem in cones. This paper attempts to study the existence and multiplicity of positive solutions of second order superlinear singular Neumann boundary value problems in the semipositone case. The techniques we employ here involve a nonlinear result of LeraySchauder, the wellknown fixed point theorem in cones and the method of upper and lower solutions. We prove that problem (1.1) has at least two different positive solutions. Moreover, we do not take the restrictions p(x)\equiv 1 or q(x)\equiv 0.
Throughout this paper, we assume that the perturbed part g(x,u) satisfies the following hypotheses:
(H_{1}): g(x,u)\in C(I\times {R}^{+},{R}^{+}), p(x)\in {C}^{1}(I), q(x)\in C(I), p(x)>0, q(x)>0.
(H_{2}): There exists a constant M>0 such that F(x,u)=g(x,u)+M\ge 0 for all x\in I and u\in (0,\mathrm{\infty}).
In Section 2, we perform a study of the sign of the Green’s function of the corresponding linear problems
In detail, we construct the Green’s function G(x,y) and give a sufficient condition to ensure G(x,y) is positive. This fact is crucial for our arguments. We denote
We also use \omega (x) to denote the unique solution of (1.2) with h(x)=1, \omega (x)={\int}_{0}^{1}G(x,y)\phantom{\rule{0.2em}{0ex}}dy. In Section 3, we state and prove the main results of this paper.
2 Preliminaries
For the reader’s convenience we introduce some results of Green’s functions. Let Q=I\times I, {Q}_{1}=\{(x,y)\in Q0\le x\le y\le 1\}, {Q}_{2}=\{(x,y)\in Q0\le y\le x\le 1\}.
Considering the homogeneous boundary value problem
and let G(x,y) be the Green’s function of problem (2.1). Then G(x,y) can be written as
where m and n are linearly independent, and m, n and ω satisfy the following lemma.
Lemma 2.1
[7]
Suppose that (H_{1}) holds and problem (2.1) has only zero solution, then there exist two functionsm(x)andn(x)satisfying:

(i)
m(x)\in {C}^{2}(I,R) is increasing and m(x)>0, x\in I;

(ii)
n(x)\in {C}^{2}(I,R) is decreasing and n(x)>0, x\in I;

(iii)
Lm\equiv {(p(x){m}^{\prime})}^{\prime}+q(x)m=0, m(0)=1, {m}^{\prime}(0)=0;

(iv)
Ln\equiv {(p(x){n}^{\prime})}^{\prime}+q(x)n=0, n(1)=1, {n}^{\prime}(1)=0;

(v)
\omega \equiv p(x)({m}^{\prime}(x)n(x)m(x){n}^{\prime}(x)) is a positive constant.
Lemma 2.2
[7]
The Green’s functionG(x,y)defined by (2.2) has the following properties:

(i)
G(x,y) is continuous in Q;

(ii)
G(x,y) is symmetrical on Q;

(iii)
G(x,y) has continuous partial derivatives on {Q}_{1}, {Q}_{2};

(iv)
For each fixed y\in I, G(x,y) satisfies LG(x,y)=0 for x\ne y, x\in I. Moreover, {G}_{x}^{\prime}(0,y)={G}_{x}^{\prime}(1,y)=0 for y\in (0,1).

(v)
For x=y, {G}_{x}^{\prime} has discontinuity point of the first kind, and
{G}_{x}^{\prime}(y+0,y){G}_{x}^{\prime}(y0,y)=\frac{1}{p(y)},\phantom{\rule{1em}{0ex}}y\in (0,1).
Lemma 2.3
[8]
Suppose that conditions in Lemma 2.1hold andh:I\to Ris continuous. Then the problem
has a unique solution, which can be written as
Next we state the theorem of fixed points in cones, which will be used in Section 3.
Theorem 2.1
[9]
Let X be a Banach space and K (⊂X) be a cone. Assume that{\mathrm{\Omega}}_{1}, {\mathrm{\Omega}}_{2}are open subsets of X with0\in {\mathrm{\Omega}}_{1}, {\overline{\mathrm{\Omega}}}_{1}\subset {\mathrm{\Omega}}_{2}, and let
be a continuous and compact operator such that either

(i)
\parallel Tu\parallel \ge \parallel u\parallel, u\in K\cap \partial {\mathrm{\Omega}}_{1} and \parallel Tu\parallel \le \parallel u\parallel, u\in K\cap \partial {\mathrm{\Omega}}_{2}; or

(ii)
\parallel Tu\parallel \le \parallel u\parallel, u\in K\cap \partial {\mathrm{\Omega}}_{1} and \parallel Tu\parallel \ge \parallel u\parallel, u\in K\cap \partial {\mathrm{\Omega}}_{2}.
Then T has a fixed point inK\cap ({\overline{\mathrm{\Omega}}}_{2}\mathrm{\setminus}{\mathrm{\Omega}}_{1}).
In applications below, we take X=C(I) with the supremum norm \parallel \cdot \parallel and define
One may readily verify that K is a cone in X. Now suppose that F:I\times R\to [0,\mathrm{\infty}) is continuous and define an operator T:X\to X by
for u\in X and x\in [0,1].
Lemma 2.4
T is well defined and maps X into K. Moreover, T is continuous and completely continuous.
3 Main results
In this section we establish the existence and multiplicity of positive solutions to (1.1). Since we are mainly interested in the attractivesuperlinear nonlinearities g(x,u) in the semipositone case, we assume that the hypotheses of the following theorem are satisfied.
Theorem 3.1
Suppose that (H_{1}) and (H_{2}) hold. Furthermore, assume the following:
(H_{3}): There exist continuous, nonnegative functionsf(u)andg(u)such that
andf(u)>0is nonincreasing andh(u)/f(u)is nondecreasing inu\in (0,\mathrm{\infty}).
(H_{4}): There existsr>\frac{M\parallel \omega \parallel}{\sigma}such that\frac{r}{f(\sigma rM\parallel \omega \parallel )\{1+\frac{h(r)}{f(r)}\}}>\parallel \omega \parallel.
(H_{5}): There exists a constantA>M, \epsilon >0such that
Then problem (1.1) has at least one positive solutionv\in C(I)with0<\parallel v+M\omega \parallel <r.
Before we present the proof of Theorem 3.1, we state and prove some facts.
First, it is easy to see that we can take c>0 and {n}_{0}>1 such that
Lemma 3.1
Suppose that (H_{1})(H_{5}) hold, then\alpha (x)=(M+c)\omega (x)is a strict lower solution to the problem
where{F}_{n}(x,u)=F(x,max\{u,\frac{1}{n}\}), (x,u)\in I\times R.
Proof
It is easy to see that {\alpha}^{\prime}(0)=(M+c){\omega}^{\prime}(0)=0 and {\alpha}^{\prime}(1)=(M+c){\omega}^{\prime}(1)=0.
Since \alpha (x)M\omega (x)=c\omega (x)\ge c\sigma \parallel \omega \parallel >\frac{1}{{n}_{0}}\ge \frac{1}{n}, and using (3.1), we have \epsilon >\alpha (x)M\omega (x)=c\omega (x)\ge \frac{1}{n}>0.
By assumption (H_{5}), we have
This implies that \alpha (x) is a strict lower solution to (3.3). □
Lemma 3.2
Suppose that (H_{1})(H_{5}) hold. Then the problem
has at least one positive solution{\beta}_{n}(x)with\parallel {\beta}_{n}\parallel <r.
Proof
The existence is proved using the LeraySchauder alternative principle together with a truncation technique.
Since (H_{4}) holds, we have
Consider the family of problems
where \lambda \in I and {f}_{n}(u)=f(max\{u,1/n\}), (x,u)\in I\times R. {f}_{n}(u) is nonincreasing.
Problem (3.5) is equivalent to the following fixed point problem in C[0,1]
where {T}_{n} is defined by
We claim that any fixed point β of (3.6) for any \lambda \in [0,1] must satisfy \parallel \beta \parallel \ne r. Otherwise, assume that β is a solution of (3.6) for some \lambda \in [0,1] such that \parallel \beta \parallel =r. Note that {f}_{n}(x,u)\ge 0. By Lemma 3.1, for all x, \beta (x)M\omega (x)\ge \sigma rM\parallel \omega \parallel \ge 1/n. Hence, for all x,
Then we have, for all x,
Therefore,
This is a contradiction and the claim is proved. □
From this claim, the nonlinear alternative of LeraySchauder guarantees that problem (3.6) (with \lambda =1) has a fixed point, denoted by {\beta}_{n}, in {B}_{r}, i.e., problem (3.4) has a positive solution {\beta}_{n} with \parallel {\beta}_{n}\parallel <r. (In fact, it is easy to see that {\beta}_{n}(x)\ge 1/n with \parallel {\beta}_{n}\parallel \ne r.)
Lemma 3.3
Suppose that (H_{1})(H_{5}) hold, then{\beta}_{n}(x)is an upper solution of problem (3.3).
Proof
By Lemma 3.2 we know that {\beta}_{n}(x) is a solution to equation (3.4).
If {\beta}_{n}(x)M\omega (x)\ge \frac{1}{n}, then
If {\beta}_{n}(x)M\omega (x)\le \frac{1}{n}, then
Since {\beta}_{n}^{\prime}(0)={\beta}_{n}^{\prime}(1)=0, we have
This implies that {\beta}_{n}(x) is an upper solution of problem (3.3). □
Lemma 3.4
Suppose that (H_{1})(H_{5}) hold, then{\beta}_{n}(x)\ge \alpha (x) (n>{n}_{0}).
Proof
Let z(x)=\alpha (x){\beta}_{n}(x), we will prove z(x)\le 0. If this is not true for n>{n}_{0}, there exists {x}_{0}\in [0,1] such that z({x}_{0})=maxz(x)>0, {z}^{\prime}({x}_{0})=0, {z}^{\u2033}({x}_{0})\le 0. Then {(p({x}_{0}){z}^{\prime}({x}_{0}))}^{\prime}\le 0.
Since \alpha ({x}_{0})M\omega ({x}_{0})=c\omega ({x}_{0})\ge c\sigma \parallel \omega \parallel >\frac{1}{{n}_{0}}\ge \frac{1}{n}, \alpha ({x}_{0})M\omega ({x}_{0})\le c\parallel \omega \parallel <\epsilon, and {f}_{n}(u) is nonincreasing, we have
and
This is a contradiction and completes the proof of Lemma 3.4. □
Proof of Theorem 3.1
To show (1.1) has a positive solution, we will show
has a solution u\in C(I), u(x)>M\omega (x), x\in I.
If this is true, then v(x)=u(x)M\omega (x) is a positive solution of (1.1) since
As a result, we will only concentrate our study on (3.14).
By Lemmas 3.13.4 and the upper and lower solutions method, we know that (3.3) has a solution {u}_{n} with (M+c)\omega (x)=\alpha (x)\le {u}_{n}(x)\le {\beta}_{n}(x)<r. Thus we have {u}_{n}(x)M\omega (x)\ge c\sigma \parallel \omega \parallel, {u}_{n}(x)\le {\beta}_{n}(x)<r.
By the fact that {u}_{n} is a bounded and equicontinuous family on [0,1], the ArzelaAscoli theorem guarantees that {\{{u}_{n}\}}_{n\in {N}_{0}} has a subsequence {\{{u}_{{n}_{k}}\}}_{k\in N}, which converges uniformly on [0,1] to a function u\in C[0,1]. Then u satisfies u(x)M\omega (x)\ge c\sigma \parallel \omega \parallel, u(x)<r for all x. Moreover, {u}_{{n}_{k}} satisfies the integral equation
Letting k\to \mathrm{\infty}, we arrive at
where the uniform continuity of F(x,u(x)M\omega (x)) on [0,1]\times [c\sigma \parallel \omega \parallel ,r] is used. Therefore, u is a positive solution of (3.14).
Finally, it is not difficult to show that \parallel u\parallel <r. Assume otherwise: note that F(x,u)\ge 0. By Lemma 2.4, for all x, u(x)\ge 1/n and r\ge u(x)M\omega (x)\ge \sigma rM\parallel \omega \parallel \ge 1/n. Hence, for all x,
Then we have for all x,
Therefore,
This is a contradiction and completes the proof of Theorem 3.1. □
Corollary 3.1
Let us consider the following boundary value problem
where\alpha >0, \beta >0andk:[0,1]\to Ris continuous, \mu >0is chosen such that
hereH=\parallel k\parallel. Then problem (3.17) has a positive solutionu\in C[0,1].
Proof
We will apply Theorem 3.1 with M=\mu H and
Clearly, (H_{1})(H_{3}) and (H_{5}) are satisfied.
Set
Since T(\frac{M\parallel \omega \parallel}{\sigma})=0, T(\mathrm{\infty})=0, then there exists r\in (\frac{M\parallel \omega \parallel}{\sigma},\mathrm{\infty}) such that
This implies that there exists r\in (\frac{M\parallel \omega \parallel}{\sigma},\mathrm{\infty}) such that \mu <\frac{r{(\sigma rM\parallel \omega \parallel )}^{\alpha}}{\parallel \omega \parallel \{1+{r}^{\alpha +\beta}+2H{r}^{\alpha}\}}, so (H_{4}) is satisfied.
Since \beta >1. Thus all the conditions of Theorem 3.1 are satisfied, so the existence is guaranteed. □
Next we will find another positive solution to problem (1.1) by using Theorem 2.1.
Theorem 3.2
Suppose that conditions (H_{1})(H_{5}) hold. In addition, it is assumed that the following two conditions are satisfied:
(H_{6}): F(x,u)=g(x,u)+M\ge {f}_{1}(u)+{h}_{1}(u)for some continuous nonnegative functions{f}_{1}(u)and{h}_{1}(u)with the properties that{f}_{1}(u)>0is nonincreasing and{h}_{1}(u)/{f}_{1}(u)is nondecreasing.
(H_{7}): There existsR>rsuch that\frac{R}{\sigma {f}_{1}(R)\{1+\frac{{h}_{1}(\sigma RM\parallel \omega \parallel )}{{f}_{1}(\sigma RM\parallel \omega \parallel )}\}}<\parallel \omega \parallel.
Then, besides the solution u constructed in Theorem 3.1, problem (1.1) has another positive solution\tilde{v}\in C[0,1]withr<\parallel \tilde{v}+M\omega \parallel \le R.
Proof
To show (1.1) has a positive solution, we will show (3.14) has a solution \tilde{u}\in C[0,1] with \tilde{u}(x)>M\omega (x) for x\in [0,1] and r\le \parallel \tilde{u}\parallel \le R.
Let X=C[0,1] and K be a cone in X defined by (2.5). Let
and define the operator T:K\cap ({\overline{\mathrm{\Omega}}}_{R}\setminus {\mathrm{\Omega}}_{r})\to K by
where G(x,y) is as in (2.2).
For each \tilde{u}\in K\cap ({\overline{\mathrm{\Omega}}}_{R}\setminus {\mathrm{\Omega}}_{r})r\le \parallel \tilde{u}\parallel \le R, we have 0<\sigma rM\parallel \omega \parallel \le \tilde{u}(x)M\omega (x)\le R. Since F:[0,1]\times [\sigma rM\parallel \omega \parallel ,R]\to [0,\mathrm{\infty}) is continuous, it follows from Lemma 2.4 that the operator T:K\cap ({\overline{\mathrm{\Omega}}}_{R}\setminus {\mathrm{\Omega}}_{r})\to K is well defined, is continuous and completely continuous.
First we show
In fact, if \tilde{u}\in K\cap \partial {\mathrm{\Omega}}_{r}, then \parallel \tilde{u}\parallel =r and \tilde{u}(x)\ge \sigma r>M\parallel \omega \parallel for x\in I. So we have
This implies \parallel T\tilde{u}\parallel <\parallel \tilde{u}\parallel, i.e., (3.20) holds.
Next we show
To see this, let \tilde{u}\in K\cap \partial {\mathrm{\Omega}}_{R}, then \parallel \tilde{u}\parallel =R and \tilde{u}(x)\ge \sigma R>M\parallel \omega \parallel for x\in I. As a result, it follows from (H_{6}) and (H_{7}) that, for x\in I,
Now (3.20), (3.21) and Theorem 2.1 guarantee that T has a fixed point \tilde{u}\in K\cap ({\overline{\mathrm{\Omega}}}_{R}\setminus {\mathrm{\Omega}}_{r}) with r\le \parallel \tilde{u}\parallel \le R. Clearly, this \tilde{u} is a positive solution of (3.14). This completes the proof of Theorem 3.2. □
Let us consider again example (3.17) in Corollary 3.1 for the superlinear case, i.e., \alpha >0, \beta >1 and k:[0,1]\to R is continuous, \mu >0 is chosen such that (3.18) holds, here H=\parallel k\parallel. Then problem (3.17) has a positive solution \tilde{u}\in C[0,1]. Clearly, (H_{1})(H_{6}) are satisfied.
Since \beta >1, then (H_{7}) is satisfied for R large enough because when R\to \mathrm{\infty},
Thus all the conditions of Theorem 3.2 are satisfied, so the existence is guaranteed.
Corollary 3.2
Assume that\alpha >0, \beta >1andk:I\to Ris continuous, \mu >0is chosen such that (3.18) holds. TakeH=\parallel k\parallel. Then problem (3.17) has at least two different positive solutions.
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Acknowledgements
The authors express their thanks to the referee for his valuable suggestions. The work was supported by the National Natural Science Foundation of China (No: 11171350).
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Li, Q., Cong, F., Li, Z. et al. Multiplicity of positive solutions of superlinear semipositone singular Neumann problems. Bound Value Probl 2014, 217 (2014). https://doi.org/10.1186/s1366101402170
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DOI: https://doi.org/10.1186/s1366101402170
Keywords
 positive solutions
 superlinear
 semipositone
 singular
 Neumann problem