Existence and uniqueness of a positive solutions for the product of operators

1.   Introduction

In this paper we prove some fixed point theorems for the problem

However, these kind of theorems are related to some "quadratic" problems. Let us mention the quadratic integral equation

Special cases of Eq (1.2) were investigated in connection with the applications of some kind of problems in the theories of radiative transfer, neutron transport, and the kinetic theory of gases ^[4]. A more general problem (motivated by some practical interests in plasma physics) was investigated in ^[21]. See ^[11,24] for other applications.

So far, two methods have been proposed to solve Eq (1.1). In the former, the measure of non-compactness technique (see ^{[2,8,9,10,14,17]}) is used to prove the existence of a solution for Eq (1.1), and in the latter, Dhage ^[13] used the combining Schauder's fixed point theorem and Banach's contraction principle to prove the existence of a solution for Eq (1.1), also see ^[5]. Some authors have combined the two methods and have proved the existence of a solution for Eq (1.1).

In this paper, we prove the existence of a positive solution for the Eq (1.1) in which the operators $A$, $B$, and $C$ are concave (or convex) or monotone. Also, we give a successively sequence to approximate it. But what is our motivation to prove the existence of a positive solution for the Eq (1.1) when the operators $A$, $B$, and $C$ are concave (or convex) or monotone? The mentioned methods have not provided a way to approximate the solution for the Eq (1.1). Also, in the case that the operators $A$, $B$, and $C$ are concave (or convex) or monotone, we do not have to suppose the continuity, compactness, and upper-lower assumptions for the operators $A$, $B$, and $C$. These assumptions play an important roles in order to prove the existence of positive solutions for nonlinear differential and integral equations and they are difficult to verify for real problems. Furthermore, there exist more extensively applied of the positive solution of nonlinear differential and integral equations in practical issues (see ^{[3,12,28,29,30,31]}).

The start of proving the existence of a positive solution for differential and integral equations can be found in the Picard investigation (see ^[25], p.129–138). Authors in ^[18,19,20] generalized theorems for abstract operator equations with special positive operators called $u_0$-concave. After that, ordered concavity (convexity) and $\alpha$-concavity (convexity) were introduced by Amann ^[1] in 1976 and Potter ^[26] in 1977. In ^{[7,22,23,32,33,34]}, some others type of concave operators were investigated.

The paper is organized as follows: In Section 2, we introduce some of the preliminaries needed for the next sections. In Section 3, we prove some existential results for the Eq (1.1). Furthermore, we provide some examples that satisfy the main results. In Section 4, we prove the existence of positive solutions for nonlinear quadratic integral equations by theorems provided in the main results section. Section 5 is devoted to concluding and proposing new ideas.

2.   Some basic definitions and notations

Throughout this paper, we assume that $E$ is a real Banach algebra. Which means, $E$ is a real Banach space in which an operation of multiplication is defined, subject to the following properties (for all $x, y, z \in E, \lambda \in \mathbb{R}$):

$(1)$ $(xy)z = x(yz)$;

$(2)$ $x(y + z) = xy + xz$ and $(x + y)z = xz + yz$;

$(3)$ $\lambda(xy) = (\lambda x)y = x(\lambda y)$;

$(4)$ $||xy|| \leq ||x|| ||y||$.

Now let us recall the concepts of cone and partial order for a Banach algebra. A subset $P$ of $E$ is called a cone of $E$ if

$(1)$ $P$ is a non-empty closed and $\theta \in P$;

$(2)$ $\lambda P + \gamma P \subseteq P$ for all non-negative real numbers $\lambda, \gamma$;

$(3)$ $P^{2} = P \cdot P \subseteq P$;

$(4)$ $P \cap (-P) = \{\theta\}$,

where $\theta$ denotes the null of $E$. For a given cone $P \subseteq E$, we can define a partial ordering $\leq$ with respect to $P$ by $x \leq y$ if and only if $y - x \in P$. $x < y$ will stand for $x \leq y$ and $x \neq y$. The cone $P$ is called normal if there is a number $M > m_0$ such that for all $x, y \in E$,

The last positive number satisfying the above inequality is called the normal constant of $P$. In the following, we always assume that $P$ is a cone in $E$ and, $\leq$ is the partial ordering with respect to $P$. We call such space ordered Banach algebra and denote it by (OBA).

If $x_{1}, x_{2} \in E$, the set $[x_{1}, x_{2}] = \{x \in E \; |\; x_{1} \leqslant x \leqslant x_{2} \}$ is called the order interval between $x_{1}$ and $x_{2}$. An operator $A : E \rightarrow E$ is called increasing (decreasing) if $x \leqslant y$ implies $Ax \leqslant Ay$ ($Ax \geqslant Ay$) where $x, y \in E$. For $h > \theta$ (i.e. $h \geqslant \theta$ and $h \neq \theta$), set

It is easy to notice that $P_{h} \subseteq P$.

Lemma 2.1. (^[15]) The two following assumptions are equivalent:

$(1)$ $P$ is a normal cone,

$(2)$ $x_{n} \leqslant z_{n} \leqslant y_{n}$ ($n = 1, 2, 3, ...$) and $||x_{n} - x|| \to 0$, $||y_{n} - x|| \to 0$, imply that $||z_{n}- x|| \to 0$.

Definition 2.1. (^[16]) Let $\alpha$ be a real number such that $0 \leq \alpha < 1$. An operator $A : P \rightarrow P$ is called an $\alpha$-concave ((-$\alpha$)-convex) if it satisfies,

Theorem 2.2. (^[6]) Assume that $P$ is a normal cone and the operator $T$ satisfies:

(D1) $T : P_{h} \to P_{h}$ is an increasing self-map in $P_{h}$;

(D2) For any $x \in P_{h}$ and $t \in (0, 1)$, there exists $\beta(t) \in (0, 1)$ such that $T(tx) \geqslant t^{\beta(t)}Tx$;

(D3) For every $x_{0} \in P$, there is a constant $l \geq 0$ suchthat $x_{0} \in [\theta, lh]$.

Then, operator equation $x = Tx+ x_{0}$ has a unique solution in $P_{h}$.

3.   Main results

Now the main results could be stated and proved.

Theorem 3.1. Let $P$ be a normal cone, $A : P \rightarrow P$ is an increasing $\alpha$-concave operator, $B: P \rightarrow P$ is an increasing $\gamma_{1}$-concave operator, and $C : P \rightarrow P$ is anincreasing $\gamma_{2}$-concave operator such that $\gamma_{1}+\gamma_{2} = \gamma \leqslant 1$. Also, suppose that

(i) there exists $h > \theta$ such that $h \cdot h \in P_{h}$, and $Ah, Bh, Ch \in P_{h}$;

(ii) there exists a constant $\delta_{0} > 0$ such that for all $x\in P$, we have $Ax \geqslant \delta_{0} Bx \cdot Cx$.

Then, the operator Eq (1.1) has a unique solution $x^{*}$ in $P_{h}$. Moreover, for the constructing successivelysequence $y_{n} = Ay_{n-1} + By_{n-1} \cdot Cy_{n-1} (n = 1, 2, \cdots)$ and for any initial value $y_{0} \in P_{h}$, we have $y_{n} \to x^{*}$ as $n \to \infty$.

Proof. Since $Ah, Bh, Ch \in P_{h}$, there exist constants $\lambda_{1}, \lambda_{2}, \mu_{1}, \mu_{2}, \upsilon_{1}, \upsilon_{2} > 0$ such that $\lambda_{1}h \leqslant Ah \leqslant \lambda_{2}h, \mu_{1}h \leqslant Bh \leqslant \mu_{2}h, \upsilon_{1}h \leqslant Ch \leqslant \upsilon_{2}h$. We have

By $(i)$, there exist $r, s > 0$ such that $rh \leqslant h \cdot h \leqslant sh$. We get

Hence, we can write $K_{1} h \leqslant Ah + Bh \cdot Ch \leqslant J_{1} h$, where $K_{1} = \lambda_{1} + \mu_{1}\upsilon_{1}r$ and $J_{1} = \lambda_{2} + \mu_{2}\upsilon_{2}s$. Thus, $Ah+ Bh \cdot Ch \in Ph$. Define the operator $T = A + B \cdot C$ by $Tx = Ax+ Bx \cdot Cx$. Then $T : P \rightarrow P$ and $Th \in P_{h}$. Next, we show that $T : P_{h} \rightarrow P_{h}$. By (2.1), for any $t \in (0, 1)$ and $x \in P$, we have

For any $x \in P_{h}$, we can choose a sufficiently small number $t_{0} \in (0, 1)$ such that

Note that $T : P \rightarrow P$ is an increasing self-map and by (3.1),

where $J_{2} = \frac{\lambda_{2}}{t^{\alpha}_{0}} + \frac{\mu_{2}\upsilon_{2}}{t^{\gamma}_{0}}s$. Also,

where $K_{2} = \lambda_{1}t^{\alpha}_{0} + \mu_{1}\upsilon_{1}t^{\gamma}_{0}r$. Thus $Tx \in P_{h}$. Henece, $T : P_{h} \to P_{h}$. Moreover, $A : P_{h} \to P_{h}, B : P_{h} \to P_{h}$ and $C : P_{h} \to P_{h}$. In the following, we show that for any $t \in (0, 1)$, there exists $\beta_{0}(t) \in (\alpha, 1)$ with respect to $t$, such that for all $x \in P_{h}$,

By $(ii)$, there exists $\delta_{0} > 0$ such that $Ax \geqslant \delta_{0}Bx \cdot Cx$. Consider the following function:

It is easy to prove that $f$ is non-negative in $(0, 1)$. Especially, for any $t \in (0, 1)$ we have $t^{\beta} > t$ and $t^{\alpha} > t^{\beta}$. Furthermore, for fixing $t \in (0, 1)$, we have $lim_{\beta \to 1^{-}}f (t) = 0$. So, there exists $\beta_{0}(t) \in (\alpha, 1)$ with respect to $t$ such that

Hence, we have

Then, we can get

Consequently, for any $t \in (0, 1)$ and $x \in P_{h}$ we have

Therefore,

Let $x_{0} = \theta$. Application of Theorem 2.2 implies that the equation $Tx = x$ has a unique solution $x^{*}$ in $P_{h}$. It can be concluded that the operator Eq (1.1) has a unique solution $x^{*}$ in $P_{h}$. Now we can construct the successively sequence $y_{n} = Ay_{n-1} + By_{n-1} \cdot Cy_{n-1} (n = 1, 2, \cdots)$ for any initial point $y_{0} \in P_{h}$. Since $y_{0} \in P_{h}$ and $Ty_{0} \in P_{h}$, we can choose a sufficiently small number $t_{0} \in (0, 1)$ such that

Note that $0 < \beta_{0}(t_{0}) < 1$, and we can also take a positive integer $k$ such that

Put $u_{0} = t^{k}_{0}y_{0}, v_{0} = \frac{1}{t^{k}_{0}}y_{0}$. Evidently, $u_{0}, v_{0} \in P_{h}$ and $u_{0} \leqslant y_{0} \leqslant v_{0}$. By the monotonicity of $T$, we have $Tu_{0} \leqslant Tv_{0}$. Furthermore, by combining (3.2) and (3.3) we have

By (3.4), one can obtains that $k\beta_{0}(t_{0}) + 1 < k$. Thus

Therefore, $Tu_{0} \geqslant t_{0}^{ k\beta_{0}(t_{0}) + 1 }y_{0} > t_{0}^{k}y_{0} = u_{0}$. By (3.2),

Thus,

The application of (3.6) implies $Tv_{0} \leqslant \frac{1}{ t_{0}^{ k\beta_{0}(t_{0}) + 1 } } y_{0} \leqslant \frac{1}{t^{k}_{0}}y_{0} = v_{0}$. Thus, $u_{0} \leqslant Tu_{0} \leqslant Tv_{0} \leqslant v_{0}$. For $n = 1, 2, \cdots$, let $u_{n} = Tu_{n-1}, v_{n} = Tv_{n-1}$. Then, $u_{n} \leqslant y_{n} \leqslant v_{n}$ ($n = 1, 2, \cdots$). Similar to the proof of Theorem 1.3 of ^[35], there exists $y^{*} \in P_{h}$ such that $Ty^{*} = y^{*}$ and $\lim_{n \to \infty}u_{n} = \lim v_{n} = y^{*}$. Then, by Lemma 2.1, $y_{n} \to y^{*} (n \to \infty)$. Since the fixed point of the operator $T$ in $P_{h}$ is unique, we have $x^{*} = y^{*}$.

Example 3.1. Consider the Banach space $E = C[0, 1]$ with the supremum norm. $E$ is a Banach algebra by the multiplication $(x \cdot y)(t) = x(t) \cdot y(t)$. Assume that $P = \{x(t) \in C[0, 1]\; |\; x(t) \geqslant 0, \; \forall t \in [0, 1]\}$. Then $E$ is an (OBA) by the cone $P$. Let us define operators $A, B, C : P \longrightarrow P$ as the following

Assume that $h(t) = 1$. We can prove that all of the assumptions of Theorem 3.1 are satisfied and the operator $T = A + B \cdot C$ has a unique positive solution.

Example 3.2. Consider the Banach space $E = L_{\infty}[0, 1]$ with the $L_{\infty}$ norm. $E$ is a Banach algebra by the multiplication $(x \cdot y)(t) = x(t) \cdot y(t)$. Assume that $P = \{x(t) \in L_{\infty}[0, 1]\; |\; x(t) \geqslant 0, \; \forall t \in [0, 1]\}$. Then $E$ is an (OBA) by the cone $P$. For any $x \in P$ and $t \in [0, 1]$, let us define

It is easy to prove that for any $x \in P$, $A(x), B(x), C(x) \in P$. Also we have $h > 0$. Therefore, $A, B, C, h: P \longrightarrow P$. It is easily noticed that, $A$ is an $\alpha$-concave operator (for some $0 < \alpha < 1$), $B$ is a $\gamma_{1}$ concave operator (for some $0 \leqslant \gamma_{1} < 1$), and $C$ is a $\gamma_{2}$ concave operator (for some $0 \leqslant \gamma_{2} < 1$) such that $\gamma_{1} + \gamma_{1} \leqslant 1$. Now let us consider that

Then, $x(t) \in P$ and for $t = \frac{1}{n}$ we have

Therefore, there is no $\delta_{0} > 0$ such that for all $t \in [0, 1]$ we have $Ax(t) \geqslant \delta_{0} Bx(t) \cdot Cx(t)$. Hence, the assumption $(ii)$ of Theorem 3.1 is not satisfied. It is easy to prove that the rest of the assumptions of Theorem 3.1 are satisfied and $T = A + B \cdot C$ has a unique positive solution.

Example 3.3. Consider $E$, that is defined in Example 3.1. Let us define operators $A, B, C : P \longrightarrow P$ as the following

Suppose that $h(t) = \frac{\pi}{4}$. We have $A, B, C : P \longrightarrow P$ and $h \cdot h > 0$. Also, for all $x \in P$, $Ax \geqslant Bx \cdot Cx$. None of the functions $A, B$, and $C$ are concave. But the operator $T = A + B \cdot C$ has a unique positive solution.

Example 3.4. Consider the Banach space $E = C[0, 1]$ with the supremum norm. Let us consider the multiplication

for any $x, y \in E$. $E$ is a Banach algebra(^[27]). Assume that, $P = \{x(t) \in C[0, 1]\; |\; x(t) \geqslant 0, \; \forall t \in [0, 1]\}$. Then, $E$ is an (OBA) by the cone $P$. Now suppose that, $h(t) = 1 > 0$. We have $h \cdot h(x) = x$. Then, $h \cdot h \notin P_{h}$.

Example 3.5. Consider $E$ that is defined in Example 3.4. Let $h(t) = 1 > 0$. Let us define the operators $A, B, C : P \longrightarrow P$ as the following

We can prove that all of the assumptions of Theorem 3.1, except $h \cdot h > 0$, are satisfied. Operator $T = A + B \cdot C$ has no positive solution.

Let the operators $A, B, C : P \longrightarrow P$ be defined as the following

It is easy to prove that all of the assumptions of the Theorem 3.1, except the assumption $h \cdot h > 0$, are satisfied. The equation $T = A + B \cdot C$ has a unique positive solution.

Lemma 3.2. Assume that, $P$ is a normal cone and the operator $A$ satisfies the following conditions:

(D1) $A : P_{h} \to P_{h}$ is decreasing in $P_{h}$;

(D2) For any $x \in P_{h}$ and $t \in (0, 1)$, there exists $\alpha(t) \in (0, 1)$ such that $A(tx) \leqslant t^{-\alpha(t)}Ax$.

Then, there exist $u_{0}, v_{0} \in P_{h}$ such that $u_{0} < v_{0}$, $u_{0} \leqslant Av_{0} \leqslant Au_{0} \leq v_{0}$.

Proof. Since $Ah \in P_{h}$, we can select a sufficiently small number $t_{0}$ such that

Note that $0 < \alpha(t_{0}) < 1$, and we can choose a positive integer $k$ such that

Set $u_{0} = t_{0}^{k}h$ and $v_{0} = \frac{1}{t_{0}^{k}}h$. Evidently, $u_{0}, v_{0} \in P_{h}$ and $u_{0} < v_{0}$. By the monotonicity of $A$, $Au_{0} \geqslant Av_{0}$. Furthermore, by (D2) and (3.8) we have,

By (3.10), we get that $k \alpha(t_{0}) + 1 < k$. Thus,

Hence,

By (D2),

Thus,

Application of (3.9) implies that $Av_{0} \geqslant t_{0}^{k\alpha(t_{0})+1}h \geqslant t_{0}^{k}h = u_{0}$. So we have

Theorem 3.3. Assume that $P$ is normal cone, the operator $T$ satisfies (D1) and(D2) of Lemma 3.2, and there is a constant $l \geqslant 0$ such that $x_{0} \in[\theta, lh]$.

Then, the operator equation $Tx + x_{0} = x$ has a unique solutionin $P_{h}$.

Proof. For all $x \in P_{h}$, we have $Tx \in P_{h}$. Then, there exist real numbers $\lambda, \mu > 0$ such that $\lambda h \leqslant Tx \leqslant \mu h$. Thus,

Hence,

Define the operator $F$ by,

By (3.15), and considering the monotonicity of the operator $T$, the operator $F: P_{h} \to P_{h}$ is decreasing. Furthermore, for all $x \in P_{h}$ and $t \in (0, 1)$, we have

Lemma 3.2 implies that, there exist $u_{0}, v_{0} \in P_{h}$ such that

Construct the successively sequences

By the monotonicity of $F$, we have $v_{1} = Fv_{0} \leqslant Fu_{0} = u_{1}$. Similarly, we have

By continuing this process, for $n = 1, 2, \cdots$, we get

Therefore, $\{u_{2n}\}$, $\{v_{2n+1}\}$ are the increasing and $\{u_{2n+1}\}$, $\{v_{2n}\}$ are decreasing sequences. By (3.20), for $n = 1, 2, \cdots$, we have

Assume that,

Thus, for $n = 1, 2, \cdots$, we have $u_{2n} \geqslant t_{2n}v_{2n}$ and $v_{2n+1} \geqslant t_{2n+1}v_{2n+1}$. Then,

Therefore, $t_{n+1} \geqslant t_{n}$, i.e. $t_{n}$ is an increasing sequence such that $t_{n} \subseteq (0, 1]$. If $t_{n} \to t^{*}$ as $n \to \infty$, then $t^{*} = 1$. Otherwise, $0 < t^{*} < 1$. We distinguish two cases:

Case (i): There exists an integer $N$ such that $t_{N} = t^{*}$. In this case, we know $t_{n} = t^{*}$, for all $n \geqslant N$. So for $n \geqslant N$, we have

i.e. $(t^{*})^{\alpha(t^{*})}u_{2n+1} \leqslant v_{2n+1}$. By the definition of $t_{2n+1}$, we have $t_{2n+1} = t^{*} \geqslant (t^{*})^{\alpha(t^{*})} > t^{*}$, which is a contradiction.

Case (ii): For all integer $n$, $t_{n} < t^{*}$. Then,

By the definition of $t_{n}$, we have $t_{2n+1} \geqslant t_{2n}(t^{*})^{\alpha(t^{*})-1}$. If $n \to \infty$, we get $t^{*} \geqslant (t^{*})^{\alpha(t^{*})} > t^{*}$, which is a contradiction. Thus, $lim_{n \to \infty} t_{n} = 1$. For any natural number $p$, we have

Since $P$ is normal, we have

where $N$ is the normal constant. Hence, we can claim that $u_{2n}$ and $v_{2n}$ are Cauchy sequences. Since $E$ is a complete space, there exist $u^{*}$ and $v^{*}$ such that $u_{2n} \to u^{*}$, $v_{2n} \to v^{*}$ as $n \to \infty$. By (3.20), we know that $u_{2n} \leqslant u^{*} \leqslant v^{*} \leqslant v_{2n}$ where $u^{*}, v^{*} \in P_{h}$. Then

Furthermore,

Thus, $u^{*} = v^{*}$. Let $x^{*} = u^{*} = v^{*}$. Also, by (3.20), we have

Then $v_{2n+1} \to x^{*}$ and $u_{2n+1} \to x^{*}$ as $n \to \infty$. By the inequality $u_{2n} \leqslant x^{*} \leqslant v_{2n}$ for $n = 1, 2, \cdots$, we have

If $n \to \infty$, we get $x^{*} = Fx^{*}$. That is, $x^{*}$ is a fixed point of $F$ in $P_{h}$. In the following, we prove that $x^{*}$ is the unique fixed point of $F$ in $P_{h}$. Let $\bar{x}$ be any fixed point of $F$ in $P_{h}$. Set $r_{1} = \sup\{r > 0 \; |\; rx^{*} \leqslant \bar{x} \leqslant \frac{1}{r}x^{*}\}$. Evidently, $0 < r_{1} < \infty$ and $r_{1}x^{*} \leqslant \bar{x} \leqslant \frac{1}{r_{1}}x^{*}$. Next, we prove that $r_{1} \geqslant 1$. If $0 < r_{1} < 1$,

However, by $r_{1}^{\alpha(r_{1})}x^{*} \leqslant \bar{x}$, $r_{1}^{\alpha(r_{1})} \leqslant r_{1}$. Since $r_{1}^{\alpha(r_{1})} > r_{1}$, we get a contradiction. Hence, $r_{1} \geqslant 1$ and we get $\bar{x} \geqslant r_{1}x^{*} \geqslant x^{*}$. Similarly, we can prove that $x^{*} \geqslant \bar{x}$, and $x^{*} = \bar{x}$. Therefore, $F$ has a unique fixed point $x^{*}$ in $P_{h}$. That is to say, $Tx + x_{0} = x$ has a unique solution in $P_{h}$.

Comment 3.4. In ^[26], Theorem 3.3 is proved by Hilbert'sprojective metric method where, $\alpha$ is constant function. But, the successively sequence that converges to the fixed pointhas not been obtained.

Theorem 3.5. Consider that $P$ is a normal cone, $A, B : P \rightarrow P$ aredecreasing operators, and $C : P \rightarrow P$ is a decreasing(-$\alpha$)-convex operator. Assume that

(i) there exists $h > \theta$ such that $Ah \in P_{h}, Bh \in P_{h}$ and $Ch \in P_{h}$;

(ii) $h \cdot h \in P_{h}$.

Then, the operator Eq (1.1) has a unique solution $x^{*}$ in $P_{h}$. Moreover, for the constructing successivelysequence $y_{n} = Ay_{n-1} + By_{n-1} \cdot Cy_{n-1}, n = 1, 2, ...$ and for any initial value $y_{0} \in P_{h}$, we have $y_{n}\to x^{*}$ as $n \to \infty$.

Proof. Since $Ah, Bh, Ch \in P_{h}$, there exist constants $\lambda_{1}, \lambda_{2}, \mu_{1}, \mu_{2}, \upsilon_{1}, \upsilon_{2} > 0$ such that $\lambda_{1}h \leqslant Ah \leqslant \lambda_{2}h, \mu_{1}h \leqslant Bh \leqslant \mu_{2}h, \upsilon_{1}h \leqslant Ch \leqslant \upsilon_{2}h$. Similar to the proof of the Theorem 3.1, we can prove that, if the operator $T = A + B \cdot C$ is defined by $Tx = Ax+ Bx \cdot Cx$. Then, $T : P \rightarrow P$ and $Th \in P_{h}$. Next, we show that $T : P_{h} \rightarrow P_{h}$. By (2.1) and the monotonicity of $A, B$, we have

For any $x \in P_{h}$, we can select a sufficiently small number $t_{0} \in (0, 1)$ such that

$T : P \rightarrow P$ is decreasing and by (3.23) we have,

By (ii), there exist $s > 0$ such that $\frac{1}{s}h \leqslant h \cdot h \leqslant sh$. So, we have $Tx \geqslant J_{2}h$ where $J_{2} = \lambda_{2} + \mu_{2}\upsilon_{2}\frac{1}{s}t_{0}^{\alpha}$. Also,

Hence, we have $Tx \leqslant K_{2}h$, where $K_{2} = \lambda_{1} + \mu_{1}\upsilon_{1}t_{0}^{-\alpha}s$. Thus, $Tx \in P_{h}$. So, $T : P_{h} \to P_{h}$. Moreover, $A : P_{h} \to P_{h}, B : P_{h} \to P_{h}$, and $C : P_{h} \to P_{h}$. On the other hand, for any $t \in (0, 1)$ and $x \in P_{h}$,

Then,

Therefore, $T$ is the (-$\alpha$)-convex operator. Let, $x_{0} = \theta$. Application of Theorem 3.3 implies that the equation $Tx = x$ has a unique solution $x^{*}$ in $P_{h}$. That is, the operator Eq (1.1) has a unique solution $x^{*}$ in $P_{h}$. Now, we construct successively the sequence $y_{n} = Ay_{n-1} + By_{n-1} \cdot Cy_{n-1} (n = 1, 2, \cdots)$ for any initial point $y_{0} \in P_{h}$. Since $y_{0} \in P_{h}$ and $Ty_{0} \in P_{h}$, we can choose a sufficiently small number $t_{0} \in (0, 1)$ such that

Since $0 < \alpha(t_{0}) < 1$, we can also take a positive integer $k$ such that

Set $u_{0} = t^{k}_{0}y_{0}, v_{0} = \frac{1}{t^{k}_{0}}y_{0}$. Let $u_{n+1} = Tu_{n}$, $v_{n+1} = Tv_{n} (n = 1, 2, \cdots)$. By Theorem 3.3, $u_{n} \to x^{*}$ and $v_{n} \to x^{*}$(as $n \to \infty$). By (3.25), we have $u_{0} \leqslant y_{1} \leqslant v_{0}$. Let us define $y_{n+1} = Ty_{n}$. Since $T$ is monotone decreasing and by (3.20), we get

Then by Lemma 2.1, $y_{2n} \to x^{*}$, $y_{2n+1} \to x^{*}$ as $n \to \infty$. Thus, for $\varepsilon > 0$, there exists an integer $N$ such that for $n \geqslant N$,

Therefore, (3.27) show that $y_{n} \to x^{*}$ as $n \to \infty$.

4.   Applications to nonlinear integral equation

Theorem 4.1. Assume that $E = C[0, 1]$, $P = \{x(t) \in C[0, 1]\; |\; x(t) \geqslant 0, \; \forall t \in [0, 1]\}$,

(H1) $g(t, x) : [0, 1] \times [0, \infty) \rightarrow [0, \infty), u(t, x) : [0, 1] \times [0, \infty) \rightarrow [0, \infty)$ and $f(t, x) : [0, 1] \times [0, \infty) \rightarrow [0, \infty)$ are increasing operators with respect to $x$;

(H2) there exists $h > \theta$ in $P$ such that $g(t, h), u(t, h)\in P_{h}$ and $h \cdot h \in P_{h}$;

(H3) $g(t, x)$ is an $\alpha$-concave operator, $f(t, x)$ is a $\gamma_{1}$-concave operator, and $u(t, x)$ is a $\gamma_{2}$-concave operator with respect to $x$ such that $\gamma_{1}+\gamma_{2} = \gamma \leqslant 1$;

(H4) $G(t, s)$ is non-negative for any $t, s \in [0, 1]$. Also, forany fixed $t \in [0, 1]$, the function $G(t, s)$ is boundedin $[0, 1]$ and for any fixed $s \in [0, 1]$, we have $G(t, s) \in P_{h}$;

(H5) there exists $\delta_{0} > 0$ such that for any $t \in[0, 1]$ and any $y \in [0, \infty)$ we have

Then the problem

has a unique positive solution $x^{*}$ in $P_{h}$. Moreover, forany $x_{0} \in P_{h}$ and for the constructing successivelysequence

we have $||x_{n} - x^{*}|| \to 0$ as $n \to \infty$.

Proof. Let us define the operators $A : P \rightarrow E$, $B : P \rightarrow E$, and $C : P \rightarrow E$ as the following:

It can easily be noticed that $x^{*}$ is a solution of the problem (4.1) if $x^{*} = Ax^{*} + Bx^{*} \cdot Cx^{*}$. By $(H1)$ and $(H4)$, we notice that $A : P \rightarrow P$, $B : P \rightarrow P$ and $C : P \rightarrow P$. By $(H3)$, for any $\lambda \in (0, 1)$ and $x \in P$, we have

Then, $C$ is a $\gamma_{1}$-concave operator with respect to $x$. By $(H2)$, $Ah \in P_{h}$ and $Bh \in P_{h}$. By $(H4)$, for any $s \in [0, 1]$ there exist $\lambda(s), \mu(s) > 0$ such that

Since $G(t, s)$ is bounded, $\lambda(s), \mu(s)$ are bounded positive real numbers(for any $s \in [0, 1]$). Therefore,

and

Thus, $Ch \in P_{h}$. Hence, the condition $(i)$ of Theorem 3.1 is satisfied. By $(H2)$, the condition $(iii)$ of Theorem 3.1 is satisfied and by $(H5)$ the condition $(ii)$ of Theorem 3.1 is satisfied. Then, Theorem 4.1 follows from Theorem 3.1.

Example 4.1. Assume that $E = C[0, 1]$, $P = \{x(t) \in C[0, 1]\; |\; x(t) \geqslant 0, \; \forall t \in [0, 1]\}$. Let us define

for $t \in [0, 1]$ and $x \in P$. It is easy to prove that $g$ is $\alpha-$concave (for $\alpha = \frac{1}{2}$). Also, $f$ is a $\gamma_{1}$-concave operator (for $\gamma_{1} = 0$) and $u$ is a $\gamma_{2}$-concave operator (for $\gamma_{2} = 1$) with respect to $x$. Suppose that $G(t, s) = \frac{e^{-ts}}{1+ts}$ (for $t, s \in [0, 1]$). It is easy to see that $g$, $u$, $f$, and $G$ are satisfied in all assumptions of Theorem 4.1 for $h(t) = 1$. Hence, the problem

has a unique positive solution.

Theorem 4.2. Assume that $E = C[0, 1]$, $P = \{x(t) \in C[0, 1]\; |\; x(t) \geqslant 0, \; \forall t \in [0, 1]\}$,

(H1) $g(t, x) : [0, 1] \times [0, \infty) \rightarrow [0, \infty), u(t, x) : [0, 1] \times [0, \infty) \rightarrow [0, \infty)$ and $f(t, x) : [0, 1] \times [0, \infty) \rightarrow [0, \infty)$ are decreasing with respect to $x$;

(H2) there exists $h > \theta$ in $P$ such that $g(t, h), u(t, h)\in P_{h}$, and $h \cdot h \in P_{h}$;

(H3) $f(t, x)$ is (-$\alpha$)-convex with respect to $x$;

(H4) $G(t, s)$ is non-negative for any $t, s \in [0, 1]$, for anyfixed $t \in [0, 1]$, the function $G(t, s)$ is boundedin $[0, 1]$, and for any fixed $s \in [0, 1]$, we have $G(t, s) \in P_{h}$.

Then, the problem

has a unique positive solution $x^{*}$ in $P_{h}$. Moreover, forany $x_{0} \in P_{h}$ and for the constructing successivelysequence

we have $||x_{n} - x^{*}|| \to 0$ as $n \to \infty$.

Proof. Let us define the operators $A : P \rightarrow E$, $B : P \rightarrow E$ and $C : P \rightarrow E$ as the following:

It is easily noticed that if $x^{*} = Ax^{*} + Bx^{*} \cdot Cx^{*}$, $x^{*}$ is a solution of the problem (4.1). By $(H1)$ and $(H4)$, we have $A : P \rightarrow P$, $B : P \rightarrow P$ and $C : P \rightarrow P$. By $(H3)$, for any $\lambda \in (0, 1)$ and $x \in P$ we have,

Hence, $C$ is an (-$\alpha$)-convex operator. By $(H2)$, we have $Ah \in P_{h}$ and $Bh \in P_{h}$. By $(H4)$, for any $s \in [0, 1]$, there exist $\lambda(s)$ and $\mu(s) > 0$ such that

Since $G(t, s)$ is bounded, $\lambda(s)$ and $\mu(s)$ are bounded positive real numbers (for any $s \in [0, 1]$). Therefore, we have

and

Thus, $Ch \in P_{h}$. Hence, the assumption $(i)$ of Theorem 3.5 is satisfied. By $(H2)$, the assumption $(iii)$ of Theorem 3.5 is satisfied. Then, Theorem 4.2 follows from Theorem 3.5.

Example 4.2. Assume that $E = C[0, 1]$, $P = \{x(t) \in C[0, 1]\; |\; x(t) \geqslant 0, \; \forall t \in [0, 1]\}$. Let us define,

for $t \in [0, 1]$ and $x \in P$. It can be proved that $f$ is (-$\alpha$)-convex (for $\alpha = \frac{1}{2}$). Also $u, g$ are convex sub-homogeneous in $x$. Suppose that $G(t, s) = \frac{e^{-ts}}{1+ts}$(for $t, s \in [0, 1]$). We can see that $g$, $u$, $f$, and $G$ are satisfied in all assumptions of Theorem 4.2 for $h(t) = 1$. Then the problem

has a unique positive solution.

5.   Conclusions

In this paper, we firstly proved the existence of a positive solution for the Eq (1.1) and approximated it by the constructing successively sequence, where $A$ is an $\alpha$-concave operator, $B: P \rightarrow P$ is an increasing $\gamma_{1}$-concave operator and $C : P \rightarrow P$ is an increasing $\gamma_{2}$-concave operator such that $\gamma_{1} + \gamma_{2} = \gamma \leqslant 1$.

Secondly, we proved the existence of a positive solution for the Eq (1.1) and approximated it by the constructing successively sequence, where $A, C$ are decreasing operators and $C$ is a (-$\alpha$)-convex operator.

Thirdly, we proved the existence a positive solution for some nonlinear integral equations and approximated it by the constructing successively sequence(especially in the case of quadratic integral equation).

Remark 5.1. It is suggested that the Theorems 3.1 and 3.5 be proved without the assumption $h \cdot h > \theta,$ and also the Theorem 3.1 be proved without assumption $(ii)$. Another interesting topic can be the comparison of the results of Theorems 3.1 and 3.5 with the results of theorems that are proven by the measure of non-compactness ^[2] and Dhage's techniques ^[13].

Acknowledgments

The authors would like to thank the anonymous referee for his/her comments that helped us improve this article.

Conflict of interest

Authors state no conflicts of interest.