Solution to Fractional Schrödinger and Airy Differential Equations via Integral Transforms

Abstract

Due to the need and the necessity to express a physical phenomenon in terms of an effective and comprehensive analytical form, this paper is devoted to study of Airy functions, which arise from the Airy differential equations, by means of integral transforms. Illustrative examples are also provided. The result reveals that the integral transforms are very useful tools to solve differential equations.

Some Approaches for Solving Multiplicative Second-Order Linear Differential Equations with Variable Exponentials and Multiplicative Airy’s Equation

“…Given that the Airy's equation is linear, the analytical solution at the origin may be discovered using the power series solution approach. Their use in the approximate solution of differential equations with a simple turning point, the approximate solution of integrals with converging saddle points, and the mathematical modeling of physical processes is becoming more and more common [25][26][27][28][29][30][31][32]. However, this equation has not been examined in the multiplicative analysis.…”

Discrete fractional solutions to the k‐hypergeometric differential equation

“…The Airy differential equation, which is a special case of Schrödinger's equation for a particle, confined within a triangular potential well with constant force. The Airy function also plays an important role in microscopy and astronomy, which describes the pattern due to diffraction and interference by a point source of light 38–41 . The fractional Airy differential equation has been studied in Aghili and Zeinali 41 …”

New trends in Laplace type integral transforms with applications

“…The authors have already studied several methods to evaluate series, integrals and solve fractional differential equations, specially the popular Laplace transform method, [1], [2], [3], [4], [5], [6], [7], [8] and this work is a completion for their previous researches. Proof: See [14].…”

“…At this point, we evaluate the complex integral by virtue of Titchmarsh theorem [3] f (t) = e −λt 1 π…”