Received 18 April 2016; accepted 26 June 2016; published 29 June 2016
In several papers among them  -  , integral equations with nonsigular kernels have been studied. In  -  Darwish et al. introduced and studied the quadratic Volterra equations with supremum. Also, Banaś et al. and Darwish   -  studied quadratic integral equations of arbitrary orders with singular kernels. In  , Darwish generalized and extended Banaś et al.  results to the perturbed quadratic integral equations of arbitrary orders with singular kernels.
In this paper, we will study the q-perturbed quadratic integral equation with supremum
where, , and.
By using Darbo fixed point theorem and the monotonicity measure of noncompactness due to Banaś and Olszowy  , we prove the existence of monotonic solution to Equation (1) in.
2. q-Calculus and Measure of Noncompactness
First, for a real parameter, we define a q-real number by
and a q-analog of the Pochhammer symbol (q-shifted factorial) is defined by
Also, the q-analog of the power is given by
Notice that, exists and we will denote it by.
More generally, for we define
Notice that. Therefore, if, then.
The q-gamma function is defined by
where Or, equivalently, and satisfies
Next, the q-derivative of a function f is given by
and the q-derivative of higher order of a function f is defined by
The q-integral of a function f defined on the interval is defined by
If f is given on the interval and then
The operator is defined by
The fundamental theorem of calculus satisfies for and, i.e., , and if f is continuous at, then.
The following four formulas will be used later in this paper
where denotes the q-derivative with respect to variable t.
Notice that, if and, then.
Definition 1.  Let f be a function defined on. The fractional q-integral of the Riemann-Liouville type of order is given by
Notice that, for, the above q-integral reduces to (11).
Definition 2.  The fractional q-derivative of the Riemann-Liouville type of order is given by
where denotes the smallest integer greater than or equal to.
In q-calculus, the derivative rule for the product of two functions and integration by parts formulas are
Lemma 1. Let and f be a function defined on. Then the following formulas are verified:
Lemma 2.  For, using q-integration by parts, we have
Second, we recall the basic concepts which we need throughout the paper about measure of noncompactness.
We assume that is a real Banach space with zero element and we denote by the closed ball with radius r and centered x, where.
Now, let and denote by and Conv X the closure and convex closure of X, respectively. Also, the symbols and stands for the usual algebraic operators on sets.
Moreover, the families and are defined by and respectively.
Definition 3.  Let If the following conditions
5) if is a sequence of closed subsets of with and
then hold. Then, the mapping is said to be a measure of noncompactness in E.
Here, is the kernel of the measure of noncompactness.
Our result will establish in C(I) the Banach space of all defined, continuous and real functions on with.
Next, we defined the measure of noncompactness related to monotonicity in, see   .
We fix a bounded subset of. For and denotes the modulus of continuity of the function y given by
Moreover, we let
Notice that, all functions in Y are nondecreasing on I if and only if.
Now, we define the map on as
Clearly, μ verifies all conditions in Definition 3 and, therefore it is a measure of noncompactness in  .
Definition 4.Let Let be a continuous operator. Suppose that maps bounded sets onto bounded ones. If there exists a bounded with, then is said to be satisfies the Darbo condition with respect to a measure of noncompactness.
If, then is called a contraction operator with respect to.
Theorem 1.  Let be a bounded, convex and closed subset of E. If is a Contraction operator with respect to. Then has at least one fixed point belongs to Q.
3. Existence Theorem
Let us consider the following suggestions:
a1) is continuous and
a2) The superposition operator F generated by the function f satisfies for any nonnegative function y the condition, where c is the same constant as in a1).
a3) is a continuous operator which satisfies the Darbo condition for the measure of noncompactness with a constant. Also, if.
a5) The function is continuous on and nondecreasing and separately. Moreo-
a6) is a continuous operator and there is a nondecreasing function such that for any. Moreover, for every function which is nonnegative on I, the function is nonnegative and nondecreasing on I.
a7) such that
Before, we state and prove our main theorem, we define the two operators and on as follows
respectively. Finding a fixed point of the operator defined on the space is equivalent to solving Equation (1).
Theorem 2. Assume the suggestions (a1)-(a7) be verified, then Equation (1) has at least one solution which is nondecreasing on I.
Proof. We divide the proof into seven steps for better readability.
Step 1: We will show that the operator maps into itself.
For this, it is sufficient to show that if. Fix and let and with. We have
Notice that, we have used
Notice that, since the function k is uniformly continuous on, then when we have that.
Thus, and therefore,
Step 2: applies into itself.
Now, , we have
Therefore, if we get from assumption a7) the following
Therefore, maps into itself.
We define the subset of by
It is clear that is closed, convex and bounded.
Step 3: applies the set into itself.
By this facts and suggestions a1), a4) and a6), we obtain transforms into itself.
Step 4: The operator is continuous on.
To prove this, we fix to be a sequence in with. We will show that.
Thus, we have,
As and are continuous operators, such that
Also, such that
Furthermore, such that
Now, take, then (38) gives us that
This shows that is continuous in.
Step 5: In recognition of with respect to the quantity.
Now, we take Let us fix an arbitrarily number and choose and with. We will be supposed that because no generality will be loss. Then, by using our suggestions and inequality (31), we get
The last estimate implies
Since the function k is uniformly continuous on and the function f is continuous on, then the last inequality gives us that
Step 6: In recognition of with respect to the quantity d.
Here, we fix an arbitrary and with. Then, by our assumption, we obtain our suggestions, we have
Now, we will prove that
We find that
But, because is increasing with respect to t, then
and, since is negative for then
Inequalities (50) and (51) imply that
This inequality and (47) gives us
The above estimate implies that
Step 7: is contraction with respect to the measure of noncompactness.
Inequalities (46) and (54) give us that
Inequality (57) enables us to use Theorem 1, then there are solutions to Equation (1) in.
This finishes our proof.
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