Abstract
We establish a fixed point theorem for decreasing and convex operators in a probabilistic Banach space partially ordered by a normal cone. We give an application to the study of the existence and uniqueness of positive solutions to a two-point boundary value problem.
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1 Introduction and preliminaries
The concept of probabilistic Banach spaces was introduced by Serstnev [1] by adapting the idea of Menger [2] to linear spaces. Fixed point theory in such spaces was studied and developed by many authors (see [3–8] and the references mentioned therein).
This paper deals with the existence and uniqueness of fixed points for a certain class of convex and decreasing operators defined in a probabilistic Banach space partially ordered by a cone.
For the sake of convenience, we first give some definitions and known results from the existing literature. For more details, we refer to [4, 9].
Definition 1.1
A function \(f: \mathbb{R}\to\mathbb{R}\) is said to be a distribution function if it satisfies the following conditions:
-
(i)
f is non-decreasing;
-
(ii)
f is left-continuous;
-
(iii)
\(\inf_{t\in\mathbb{R}}f(t)=0\) and \(\sup_{t\in\mathbb{R}}f(t)=1\).
We denote by D the set of all distribution functions.
Definition 1.2
A triangular norm, briefly a T-norm, is a mapping \(T: [0,1]\times[0,1]\to[0,1]\) that is continuous and such that for every \(a,b,c,d\in[0,1]\),
-
(i)
\(T(a,1)=a\);
-
(ii)
\(T(a,b)=T(b,a)\);
-
(iii)
\(c\geq a, d\geq b\Longrightarrow T(c,d)\geq T(a,b)\);
-
(iv)
\(T(T(a,b),c)=T(a,T(b,c))\).
As standard examples, \(T_{m}(a,b)=\min\{a,b\}\) and \(T_{p}(a,b)=ab\) on \([0,1]\) are T-norms.
Definition 1.3
Let X be a real vector space, T be a T-norm and \(N:X\to D\) be a given mapping. We say that N is a probabilistic norm on X if the following conditions hold:
-
(i)
\(N_{x}(0)=0\) for every \(x\in X\);
-
(ii)
\(N_{x}(t)=1\) for all \(t>0\) iff \(x=0\);
-
(iii)
\(N_{\alpha x}(t)=N_{x} (\frac{t}{|\alpha|} )\) for all \(x\in X\) and \(\alpha\in\mathbb{R}\backslash\{0\}\);
-
(iv)
\(N_{x+y}(s+t) \geq T(N_{x}(s),N_{y}(t))\) for all \(x,y\in X\) and \(s,t\geq0\).
In this case, the triplet \((X,N,T)\) is said to be a probabilistic normed space.
In the above definition, for \(x \in X\), the distribution function \(N(x)\) is denoted by \(N_{x}\) and \(N_{x}(t)\) is the value \(N_{x}\) at \(t\in \mathbb{R}\).
Example 1.4
Let \((X,\|\cdot\|)\) be a normed linear space. For all \(x\in X\), define the mapping
Then \((X,N,T_{p})\) and \((X,N,T_{m})\) are probabilistic normed spaces.
Example 1.5
Let \((X,\|\cdot\|)\) be a normed linear space. For all \(x\in X\), define the mapping
Then \((X,N,T_{p})\) is a probabilistic normed space.
Now, let us recall some topological properties of probabilistic normed spaces.
Definition 1.6
Let \((X,N,T)\) be a probabilistic normed space. A sequence \(\{x_{n}\}\) in X is said to be convergent to a point \(x\in X\) if for any \(\varepsilon>0\) and \(\lambda> 0\), there exists a positive integer N such that
for every \(n\geq N\).
Definition 1.7
Let \((X,N,T)\) be a probabilistic normed space. A sequence \(\{x_{n}\}\) in X is said to be Cauchy if for any \(\varepsilon >0\) and \(\lambda> 0\), there exists a positive integer N such that
for every \(n,m\geq N\).
Definition 1.8
Let \((X,N,T)\) be a probabilistic normed space. It is said to be a Banach probabilistic normed space (or complete) if every Cauchy sequence in X is convergent to a point in X.
Definition 1.9
Let \((X,N,T)\) be a probabilistic normed space. A subset A of X is said to be closed if every convergent sequence in A converges to an element of A.
Definition 1.10
Let \((X,N,T)\) be a probabilistic Banach space. A nonempty subset \(P\subseteq X\) is a cone if it satisfies the following conditions:
-
(i)
P is closed and convex;
-
(ii)
if \(p\in P\), \(tp\in P\) for every \(t\geq0\);
-
(iii)
if both p and −p are in P, then \(p=0\).
Let ⪯ be the partial order on X induced by the cone P in X. That is,
Thus X becomes a partially ordered probabilistic Banach space. If \(x,y\in X\), the notation \(x\prec y\) means that \(x\preceq y\) and \(x\neq y\).
Definition 1.11
Let \((X,N,T)\) be a probabilistic Banach space. A cone P in X is said to be normal if there is some constant \(K>0\) (normal constant) such that
Definition 1.12
Let \((X,N,T)\) be a probabilistic Banach space and \(P\subseteq X\) be a cone in X. Let C be a convex subset in X. An operator \(A: C \to X\) is called a convex operator if
for all \(x,y \in C\), \(x\preceq y\) and \(t\in[0,1]\).
Definition 1.13
Let \((X,N,T)\) be a probabilistic Banach space and \(P\subseteq X\) be a cone in X. An operator \(A: C\to X\) is said to be a decreasing operator if
The paper is organized as follows. In Section 2, we study the existence and uniqueness of positive fixed points for a certain class of decreasing and convex operators \(A: P\to P\). In Section 3, we study the existence and uniqueness of positive solutions to the nonlinear functional equation \(x=x_{0}+Bx\), where \(x_{0}\in P\) and \(B: P\to P\) is a given operator satisfying certain conditions. Section 4 contains a Banach version of our man result established in Section 2. Finally, in Section 5, we present an application of our main result to the study of the existence and uniqueness of positive solutions to a nonlinear differential equation of second order with two-point boundary value problem.
2 Main result and proof
Before stating our main result, we need some lemmas.
Lemma 2.1
Let \((X,N,T)\) be a probabilistic Banach space and \(\{x_{n}\}\) be a sequence in X that converges to some \(x\in X\). Then any subsequence of \(\{x_{n}\}\) converges to x.
Proof
Let \(\{x_{\varphi(n)}\}\) be a subsequence of \(\{x_{n}\}\), where \(\varphi: \mathbb{N}\to\mathbb{N}\) is a mapping satisfying
for every \(n\in\mathbb{N}\). Let \(\varepsilon>0\) and \(\lambda>0\). Since \(\{x_{n}\}\) converges to \(x\in X\), there is some positive integer N such that
for every \(n\geq N\). On the other hand,
Then
for every \(n\geq N\). This proves that \(\{x_{\varphi(n)}\}\) converges to x. □
Lemma 2.2
(see [9])
Let \((X,N,T)\) be a probabilistic Banach space. Let \(\{x_{n}\}\) and \(\{ y_{n}\}\) be two sequences in X and \(\{\alpha_{n}\}\) be a real sequence. The following properties hold:
-
(i)
if \(\{x_{n}\}\) converges to \(x\in X\) and \(\{y_{n}\}\) converges to \(y\in X\), then \(\{x_{n}+y_{n}\}\) converges to \(x+y\);
-
(ii)
if \(\{\alpha_{n}\}\) converges to some \(\alpha\in\mathbb {R}\) and \(\{x_{n}\}\) converges to some \(x\in X\), then \(\{\alpha_{n}x_{n}\}\) converges to αx.
Lemma 2.3
Let \((X,N,T)\) be a probabilistic Banach space and \(P\subseteq X\) be a normal cone in X with normal constant \(K>0\). Let \(\{u_{n}\}\) be a sequence in X such that
for every \(m,n\geq N\), where \(a\in X\) and \(\{\xi_{n}\}\) is a real sequence such that \(\xi_{n}\to0\) as \(n\to\infty\). Then \(\{u_{n}\}\) is a Cauchy sequence in the probabilistic Banach space \((X,N,T)\).
Proof
Let \(\varepsilon>0\) and \(\lambda>0\). Without restriction of the generality, we may assume that \(\xi_{n}\neq0\) for every \(n\geq N\). Since P is a normal cone, we have
for every \(m,n\geq N\). On the other hand, since \(N_{a}\in D\), we have
Then there is some \(t^{*}\in\mathbb{R}\) such that
Since \(\xi_{n}\to0\) as \(n\to\infty\), there is some positive integer \(N'\) such that
for every \(n\geq N'\). Since \(N_{a}\) is non-decreasing, we get
for every \(n\geq N'\). Finally, using (2.1) and (2.2), we obtain
for every \(n,m\geq\max\{N,N'\}\). This proves that \(\{u_{n}\}\) is a Cauchy sequence in the probabilistic Banach space \((X,N,T)\). □
Lemma 2.4
Let \((X,N,T)\) be a probabilistic Banach space and \(P\subseteq X\) be a normal cone in X with normal constant \(K>0\). Let us consider two sequences \(\{u_{n}\}\) and \(\{v_{n}\}\) in X such that
for every \(n\geq N\). Then
Proof
Let \(\varepsilon>0\) and \(\lambda>0\). Since \(\{v_{n}\}\) converges to 0, there exists some \(N'\in\mathbb{N}\) such that
for any \(n\geq N'\). On the other hand, since P is normal, we have
for any \(n\geq N\). Thus we proved that
for any \(n\geq\max\{N,N'\}\), which implies that \(\{u_{n}\}\) converges to 0. □
The following result is an immediate consequence of Lemma 2.2 and Lemma 2.4.
Lemma 2.5
Let \((X,N,T)\) be a probabilistic Banach space and \(P\subseteq X\) be a normal cone in X with normal constant \(K>0\). Let us consider three sequences \(\{u_{n}\}\), \(\{v_{n}\}\) and \(\{w_{n}\}\) in X such that
for every \(n\geq N\). Then
Now, we are ready to state and prove our main result.
Let \((X,N,T)\) be a probabilistic Banach space and \(P\subseteq X\) be a normal cone in X with normal constant \(K>0\). We denote by \(\mathcal{A}\) the set of operators \(A: P\to P\) satisfying the following conditions:
- (\(\mathcal{A}1\)):
-
\(0\prec A0\);
- (\(\mathcal{A}2\)):
-
A is a convex and decreasing operator;
- (\(\mathcal{A}3\)):
-
there exist \(\gamma\in(0,1)\) and \(m_{0},n_{0}\in\mathbb {N}\) with \(n_{0}>m_{0}\) such that
$$ A^{2m_{0}+2}0-A^{2m_{0}}0\succeq\gamma \bigl(A^{2m_{0}+3}0-A^{2m_{0}}0\bigr) $$(2.3)and
$$ A^{2n_{0}}0\succeq\frac{1}{2} \bigl(A^{2m_{0}+1}0+A^{2m_{0}}0 \bigr). $$(2.4)
Theorem 2.6
Let \(A\in\mathcal{A}\). Then
-
(i)
A has a unique fixed point \(x^{*} \in P\);
-
(ii)
for any initial value \(x_{0}\in P\), the Picard sequence \(\{ x_{n}\}\) in X defined by
$$x_{n} =Ax_{n-1}, \quad n\geq1 $$converges to \(x^{*}\);
-
(iii)
we have the estimates
$$\begin{aligned}& N_{x_{2(m_{0}+n)}-x^{*}}(t) \\& \quad \geq T \biggl(N_{A0} \biggl( \frac {(n-2-n_{0}+m_{0})t}{K^{2}} \biggr),N_{A0} \biggl(\frac {(n-2-n_{0}+m_{0})t}{K^{2}} \biggr) \biggr) \end{aligned}$$(2.5)for every \(n>n_{0}+2-m_{0}\), \(t\in\mathbb{R}\), and
$$\begin{aligned}& N_{x_{2(m_{0}+n)+1}-x^{*}}(t) \\& \quad \geq T \biggl(N_{A0} \biggl( \frac {(n-1-n_{0}+m_{0})t}{K^{2}} \biggr),N_{A0} \biggl(\frac {(n-1-n_{0}+m_{0})t}{K^{2}} \biggr) \biggr) \end{aligned}$$(2.6)for every \(n>n_{0}+1-m_{0}\), \(t\in\mathbb{R}\).
Proof
Let us consider the sequence \(\{u_{n}\}\) in P defined by
and
Since A is a decreasing operator, we have
Using (2.7) and (2.9), we obtain
For every \(n\in\mathbb{N}\), we define the set
Clearly \(\mathcal{S}_{n}\neq\emptyset\) since \(\gamma\in\mathcal{S}_{n}\) for every \(n\in\mathbb{N}\). For every \(n\in\mathbb{N}\), let
Let \(n\in\mathbb{N}\) be fixed. By the definition of \(s_{n}\), there exists a sequence \(\{a_{p}\}\subset\mathcal{S}_{n}\) such that \(a_{p}\to s_{n}\) as \(p\to\infty\). Thus we have
for every \(p\in\mathbb{N}\). This means that
for every \(p\in\mathbb{N}\). Since P is closed, letting \(p\to\infty \), using Lemma 2.2 and (2.9), we obtain
Now, we shall prove that \(e_{n}\to0\) as \(n\to\infty\). Using inequalities (2.8), (2.9), (2.10) and the fact that A is a decreasing convex operator, for every \(n\geq n_{0}-m_{0}-1\), we have
Thus we proved that
for every \(n\geq n_{0}-m_{0}-1\). Using that A is convex and decreasing and inequality (2.8), we obtain
which implies that
for every \(n\geq n_{0}-m_{0}-1\). This implies that
for every \(n\geq n_{0}-m_{0}-1\). So, for every \(n\geq n_{0}-m_{0}-1\), we have
that is,
Now, let us consider the function \(f:[0,1]\to[0,1]\) defined by
Since
for every \(n\geq n_{0}-m_{0}-1\) and f is a non-decreasing function, we obtain
Thus we have
for every \(n\geq n_{0}-m_{0}+1\). Letting \(n\to\infty\) in (2.11), we get
that is,
Using (2.10), we deduce that for all \(n, p \in\mathbb{N}\), we have
Since P is a normal cone and \(e_{n}\to0\) as \(n\to\infty\), using Lemma 2.3, we obtain that \(\{u_{2(m_{0}+n)}\}\) is a Cauchy sequence. Since P is closed and \((X,N,T)\) is a complete probabilistic normed space, there is \(x^{*}\in P\) such that \(\{u_{2n}\}\) converges to \(x^{*}\). Using (2.12), Lemma 2.2 and Lemma 2.4, we deduce that
On the other hand, using (2.9), we have
for every \(n,p\in\mathbb{N}\). This implies that
for every \(n,p\in\mathbb{N}\). Fix \(n\in\mathbb{N}\) and letting \(p\to \infty\), from (2.13) and since P is closed, we get
that is,
Similarly, we can observe that
Thus we proved that
for all \(n\in\mathbb{N}\). Using the fact that A is a decreasing operator, (2.14) yields
for all \(n\in\mathbb{N}\). Letting \(n\to\infty\) in the above inequality and using (2.13), we obtain
that is, \(x^{*}=Ax^{*}\). Thus we proved that \(x^{*}\in P\) is a fixed point of the operator A.
Now, let \(x_{0}\in P\) be an arbitrary point. Let us consider the Picard sequence \(\{x_{n}\}\) in P defined by
Since A is a decreasing operator, using that \(u_{0}\preceq x_{0}\) and \(u_{0}\preceq x_{1}\), by induction we obtain
for every \(n\in\mathbb{N}\). Letting \(n\to\infty\) in (2.16) and (2.17), using (2.13) and Lemma 2.5, we obtain
which implies that \(\{x_{n}\}\) converges to \(x^{*}\).
Let us prove that \(x^{*}\) is the unique fixed point of the operator A. Suppose that \(y^{*}\in P\) is another fixed point of A. Let \(x_{0}=y^{*}\) and consider the Picard sequence \(\{x_{n}\}\) in P defined by (2.15). Then we have
Letting \(n\to\infty\) in (2.16) and (2.17), using (2.13), we obtain \(x^{*}=y^{*}\). Thus we proved that \(x^{*}\) is the unique fixed point of A.
Let us prove estimate (2.5). Let \(t\in\mathbb{R}\), we have
By (2.16), we have
Since P is a normal cone, we get
Using (2.14) and (2.9), we have
Since P is a normal cone, we get
It follows from (2.18), (2.19) and (2.20) that
On the other hand, by (2.9) and (2.10), we have
Using (2.9), (2.12) and the above inequality, we obtain
Using (2.11) and the above inequality, we obtain
for every \(n>n_{0}+2-m_{0}\). Since P is a normal cone, we have
for every \(n>n_{0}+2-m_{0}\). Now, (2.5) follows immediately from (2.21) and (2.22). The proof of estimate (2.6) follows using similar arguments as above. This ends the proof. □
3 Positive solutions for the nonlinear functional equation: \(x=x_{0}+Bx\)
In this section, from our main theorem (Theorem 2.6), we deduce an existence and uniqueness result for the nonlinear operator equation on ordered probabilistic Banach spaces
where \(x_{0}\in P\) and \(B: P\to P\) is a given operator satisfying certain conditions. There have appeared a series of research results concerning this kind of nonlinear operator Eq. (3.1) because of the crucial role played by nonlinear equations in applied science as well as in mathematics (see [10–13]).
Let \((X,N,T)\) be a probabilistic Banach space and \(P\subseteq X\) be a normal cone in X with normal constant \(K>0\). We denote by \(\mathcal{B}\) the set of operators \(B: P\to P\) satisfying the following conditions:
- (\(\mathcal{B}1\)):
-
\(B0=0\);
- (\(\mathcal{B}2\)):
-
B is a convex and decreasing operator.
We have the following result.
Theorem 3.1
Let \(B\in\mathcal{B}\) and \(x_{0}\in P\) such that \(0\prec x_{0}\). Then the operator Eq. (3.1) has a unique solution \(x^{*}\in P\).
Proof
Let us define the operator \(A: P\to P\) by
Obviously, \(x\in P\) is a solution to Eq. (3.1) if and only if x is a fixed point of A. We have just to prove that the operator A satisfies the required conditions of Theorem 2.6, that is, A belongs to the class of operators \(\mathcal{A}\).
• Condition (\(\mathcal{A}1\)). Since \(x_{0}\succ0\), using (\(\mathcal{B}1\)), we have
Thus condition (\(\mathcal{A}1\)) is satisfied.
• Condition (\(\mathcal{A}2\)). Let \(x,y\in P\) such that \(x\preceq y\). Using the fact that B is a decreasing operator, we obtain
which implies that
that is,
Thus A is a decreasing operator.
Now, let \(t\in[0,1]\), \(x,y\in P\) such that \(x\preceq y\). Since B is a convex operator, we have
which implies that
Thus we have
Then A is a convex operator. Condition (\(\mathcal{A}2\)) is then satisfied.
• Condition (\(\mathcal{A}3\)). Let \(n_{0}=1\) and \(m_{0}=0\). Since \(B0=0\), we have
On the other hand,
Clearly, we have
Now, we have
Moreover, for any \(\gamma\in(0,1)\), we have
Thus we have
Since \(\gamma\in(0,1)\) and B is a decreasing operator, we have
and
Then we get
for every \(\gamma\in(0,1)\).
Hence \(A\in\mathcal{A}\) and the result follows from Theorem 2.6. □
4 The case of Banach spaces
Let \((X,\|\cdot\|)\) be a Banach space and \(P\subset X\) be a cone in X. Let us suppose that P is a normal cone with normal constant \(K>0\), that is,
Let \(A: P\to P\) be an operator satisfying the following conditions:
-
(i)
\(0\prec A0\);
-
(ii)
A is a convex and decreasing operator;
-
(iii)
there exist \(\gamma\in(0,1)\) and \(m_{0},n_{0}\in\mathbb{N}\) with \(n_{0}>m_{0}\) such that
$$A^{2m_{0}+2}0-A^{2m_{0}}0\succeq\gamma\bigl(A^{2m_{0}+3}0-A^{2m_{0}}0 \bigr) $$and
$$A^{2n_{0}}0\succeq\frac{1}{2} \bigl(A^{2m_{0}+1}0+A^{2m_{0}}0 \bigr). $$
Let us consider the probabilistic Banach space \((X,N,T_{m})\), where
for every \(x\in X\).
Lemma 4.1
P is a normal cone in the probabilistic Banach space \((X,N,T_{m})\) with normal constant K.
Proof
At first, let us prove that P is also a cone in the probabilistic Banach space \((X,N,T_{m})\). Indeed, we have just to prove that P is also closed in \((X,N,T_{m})\). Let \(\{x_{n}\}\) be a sequence in P such that \(\{x_{n}\}\) converges to some \(x\in X\) in \((X,N,T_{m})\). Let \(\varepsilon>0\), by the definition of the convergence in a probabilistic normed space, there exists some \(N\in\mathbb{N}\) such that
for every \(n\geq N\). Then we have
for every \(n\geq N\), which is equivalent to
for every \(n\geq N\). Thus \(\{x_{n}\}\) converges to x with respect to \(\| \cdot\|\). Since P is closed with respect to the topology of the norm \(\|\cdot\|\), we have \(x\in P\). Then we have proved that P is also closed in the probabilistic Banach space \((X,N,T_{m})\).
Now, let us prove that P is normal in \((X,N,T_{m})\). Let \(x,y\in X\) such that
Since P is a normal cone in the Banach space \((X,\|\cdot\|)\) with normal constant K, we have
Then, for every \(t>0\), we have
If \(t\leq0\), obviously we have
Thus, for every \(t\in\mathbb{R}\), we have
This proves that P is a normal cone in \((X,N,T_{m})\) with normal constant K. □
Now, using Theorem 2.6 and Lemma 4.1, we obtain the following fixed point result in Banach spaces ([14], Theorem 2.1).
Corollary 4.2
Suppose that conditions (i)-(iii) are satisfied. Then
-
(I)
A has a unique fixed point \(x^{*} \in P\);
-
(II)
for any initial value \(x_{0}\in P\), the Picard sequence \(\{ x_{n}\}\) in X defined by
$$x_{n} =Ax_{n-1},\quad n\geq1 $$converges to \(x^{*}\);
-
(III)
we have the estimates
$$\bigl\Vert x_{2(m_{0}+n)}-x^{*}\bigr\Vert \leq\frac{K^{2}\|A0\|}{n-2-n_{0}+m_{0}} $$for every \(n>n_{0}+2-m_{0}\), and
$$\bigl\Vert x_{2(m_{0}+n)+1}-x^{*}\bigr\Vert \leq\frac{K^{2}\|A0\|}{n-1-n_{0}+m_{0}} $$for every \(n>n_{0}+1-m_{0}\).
5 An application to a two-point boundary value problem
In this section, we present an application of Theorem 2.6 to a two-point boundary value problem.
Let \(a: [0,1]\to\mathbb{R}\) be a given function satisfying the following conditions:
- (a1):
-
a is a continuous function;
- (a2):
-
\(a(x(1-x))=a(x)\) for every \(x\in[0,1]\);
- (a3):
-
\(0< m\leq a(x)\leq M\) for every \(x\in[0,1]\).
Clearly, the set of functions \(a: [0,1]\to\mathbb{R}\) satisfying the above conditions is not empty.
Example 5.1
Let \(a: [0,1]\to\mathbb{R}\) be a positive constant function. Then a satisfies conditions (a1)-(a4).
Example 5.2
Let \(a: [0,1]\to\mathbb{R}\) be the function defined by
where \(\alpha>0\) and \(\beta\geq0\) are constants. Then a satisfies conditions (a1)-(a4) with
Let \(f:[0,\infty)\to\mathbb{R}\) be a function satisfying the following conditions:
- (f1):
-
f is a continuous function, \(f\geq0\);
- (f2):
-
f is a decreasing and convex function;
- (f3):
-
\(f(0)=1\), \(0< f (\frac{M}{8} )<1\);
- (f4):
-
\(f(\gamma x)\geq\frac{1}{2}\) for every \(x\in [0,\frac {M}{8} ]\), where
$$\gamma=1-\frac{1}{6}f \biggl(\frac{M}{8} \biggr) \biggl(1-f \biggl(\frac {m}{2} \biggr) \biggr). $$
The set of functions \(f:[0,\infty)\to\mathbb{R}\) satisfying the above conditions is not empty.
Example 5.3
Let \(m=M=1\), that is,
Let \(f: [0,\infty)\to\mathbb{R}\) be the function defined by
Then f satisfies conditions (f1)-(f4) with \(\gamma=\frac{1\text{,}943}{2\text{,}079}\).
Now, let us consider the following two-point boundary value problem:
Let \((X,N,T_{m})\) be the probabilistic Banach space, where \(X=C([0,1])\) is the set of real continuous functions in \([0,1]\) and \(N:X\to D\) is given by
Let
Then P is a normal cone in the probabilistic Banach space \((X,N,T_{m})\). The partial order ⪯ induced by the cone P in the set X is defined by
We have the following result.
Theorem 5.4
The boundary value problem (5.1) has a unique positive solution \(u^{*}\in P\).
Proof
The Green function associated to (5.1) is given by
Then problem (5.1) is equivalent to the integral equation
Let us consider the nonlinear operator \(A: P\to P\) defined by
We have to prove that A has a unique fixed point in P. Theorem 2.6 will be used for the proof.
Clearly, the operator A is convex and decreasing with respect to the partial order ⪯.
For every \(x\in[0,1]\), we have
Then we have
Moreover, we have
which implies that
The above inequality yields
On the other hand, for every \(x\in[0,1]\), we have
Using the fact that f is a decreasing and convex function, we obtain
Similarly, using condition (a2), we have
Thus we have
for every \(x\in[0,1]\). Moreover, using (5.2), we get
for every \(x\in[0,1]\). Now, (5.3) and (5.4) yield
for every \(x\in[0,1]\). Thus we have
Using condition (f4), we obtain
for every \(x\in[0,1]\). Finally, using (5.2), for all \(x\in [0,1]\), we have
where \(\epsilon=f (\frac{M}{8} )\in(0,1)\).
Now, the desired result follows from Theorem 2.6 with \(m_{0}=0\) and \(n_{0}=2\). □
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This project was funded by the National Plan for Science, Technology and Innovation (MAARIFAH), King Abdulaziz City for Science and Technology, Kingdom of Saudi Arabia, Award Number (12-MAT 2913-02).
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Jleli, M., Samet, B. Positive fixed points for convex and decreasing operators in probabilistic Banach spaces with an application to a two-point boundary value problem. Fixed Point Theory Appl 2015, 84 (2015). https://doi.org/10.1186/s13663-015-0338-x
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DOI: https://doi.org/10.1186/s13663-015-0338-x