Explicit bounds on certain integral inequalities via conformable fractional calculus

In this paper, we present some explicit upper bounds for integral inequalities with the help of Katugampola-type conformable fractional calculus. The results have been obtained to cover the previous published studies for Gronwall–Bellman and Bihari like integral inequalities.


Introduction and preliminaries
In the history of development calculus, integral inequalities have been thought of as a key factor in the theory of differential and integral equations. For instance, Gronwall, Bellman and Bihari have great contribution in the literature (Bellman, 1943;Bihari, 1965;Dragomir, 1987Dragomir, , 2002Gronwall, 1919;Pachpatte, 1995). However, in non-integer order of situations, the bounds provided by the above authors are not feasible.

PUBLIC INTEREST STATEMENT
Differential and integral inequalities play a vital role in the study of existence, uniqueness, boundedness, stability and other qualitative properties of solutions of differential and integral equations. One can hardly imagine these theories without the well-known Gronwall inequality and its non-linear version Bihari inequality. In addition to this, fractional calculus has a number of fields of application such as control theory, computational analysis and engineering. Thus, a number of new definitions have been introduced in academia to provide the best method for fractional calculus. In this paper, we presented a retarded Gronwall-Bellman-and Bihari-like conformable fractional integrals inequalities using the Katugampola conformable fractional calculus.
In this study, we presented a retarded Gronwall-Bellman-and Bihari-like conformable fractional integrals inequalities using the Katugampola conformable fractional calculus. In detail, Katugampola conformable derivatives for ∈ 0, 1 and t ∈ [0, ∞) given by provided the limits exist (for detail see, Katugampola, 2014). If f is fully differentiable at t, then A function f is −differentiable at a point t ≥ 0 if the limit in (1.1) exists and is finite. This definition yields the following results.
where the integral is the usual Riemann improper integral, and ∈ (0, 1].
We will also use the following important results, which can be derived from the results above.
Lemma 1 Let the conformable differential operator D be given as in (1.1), where ∈ (0, 1] and t ≥ 0, and assume the functions f and g are -differentiable as needed. Then, In this paper, using the Katugampola-type conformable fractional calculus, we introduced retarded Gronwall-Bellman-and Bihari-like conformable fractional integrals inequalities.

Main findings and cumulative results
Throughout this paper, all the functions which appear in the inequalities are assumed to be realvalued and all the integrals involved exist on the respective domains of their definitions, and C M, S and C 1 M, S denote the class of all continuous functions and the first-order conformable derivative, respectively, defined on set M with range in the set S. Additionally, R denotes the set of real numbers such that ℝ + = [0, ∞), ℝ 1 = [1, ∞) and ℚ = [0, T) are the given subset of ℝ.
where Proof Let us first assume that m > 0. Define the non-decreasing positive function z(t) by the righthand side of (2.1). Then, u(t) ≤ z(t) and z(0) = m, and as r(t) ≤ t. Then, the solution of the above fractional order differential equation by taking integration from 0 to t, we get Since u(t) ≤ z(t), we get the desired inequality, that is where ✷ Theorem 3 Let x, y ∈ C(ℚ, ℝ + ), r ∈ C 1 (ℚ, ℚ), assume that r is non-decreasing with r(t) ≤ t for t ≥ 0.
Proof Let us first assume that n > 0. Define the non-decreasing positive function z(t) by the righthand side of (2.4). Then, u(t) ≤ z(t) and z(0) = n, and as in the same steps with above proof, we get Then, the solution of the above fractional order differential equation by taking integration from 0 to t, we get Now using the result of Theorem 2, we obtain In other words, we get Since u(t) ≤ z(t), we get the desired inequality, that is where X(t) and Y(t) are defined in (2.3). ✷ Theorem 4 Let x, y ∈ C(ℚ, ℝ + ), r ∈ C 1 (ℚ, ℚ), assume that r is non-decreasing with r(t) ≤ t for t ≥ 0.
If u ∈ C(ℚ, ℝ + ) satisfies where m ≥ 0 and q > 1 are constant, then where X(t) and Y(t) are defined in 2.3.
Proof Let us first assume that m > 0. Define the non-decreasing positive function z(t) by the righthand side of (2.7). Then, u q (t) ≤ z(t) and z(0) = m, and as in the same steps with the above proof, we get Then, the solution of the above fractional order differential equation by taking integration from 0 to t, we get Since u q (t) ≤ z(t), we get the desired inequality, that is where X(t) and Y(t) are defined in 2.3. ✷ Theorem 5 Let x, y ∈ C(ℚ, ℝ + ), r ∈ C 1 (ℚ, ℚ), i ∈ C(ℝ + , ℝ + ), assume that r and are non-decreasing with r(t) ≤ t for t ≥ 0 and i ( ) > 0 for > 0, respectively. If u ∈ C(ℚ, ℝ + ) satisfies where m ≥ 0 is constant, then where and  −1 is the inverse function defined by so that for all t > 0.
Proof Let us first suppose that m > 0. Define the non-decreasing positive function z(t) by the righthand side of (2.8). Then, u(t) ≤ z(t) and z(0) = m, and as in the same steps with the above proofs, we get Then, from the definition of , we have Then, taking -th order of conformable derivative of (z(t)), we obtain Then, by taking integration from 0 to t, we get (2.11) (z(t)) = � z(t) 0 1 max( 1 (s), 2 (s)) d s.
Proof The proof of Theorem 6 can be done following the similar steps of proof of Theorems 5 and 3.