Properties of the resolvent of a linear Abel integral equation: implications for a complementary fractional equation

• Main Author: Leigh Becker
• Format: Article
• Language: English
• Published: University of Szeged 2016-09-01
• Series: Electronic Journal of Qualitative Theory of Differential Equations
• Subjects: abel integral equations; fixed points; fractional differential equations; mittag-leffler functions; resolvents; riemann–liouville operators; singular kernels; volterra integral equations
• Online Access: http://www.math.u-szeged.hu/ejqtde/periodica.html?periodica=1&paramtipus_ertek=publication&param_ertek=5093

New and known properties of the resolvent of the kernel of linear Abel integral equations of the form \begin{equation} \tag{A$_\lambda$} x(t) = f(t) - \lambda \int_0^t (t - s)^{q-1}x(s)\,ds, \end{equation} where $\lambda > 0$ and $q \in (0,1)$, are assembled and derived here. First, a priori bounds on potential solutions of the resolvent equation \begin{equation} \tag{R$_\lambda $} R(t) = {\lambda }t^{q-1} - {\lambda }\int_0^t (t - s)^{q-1}R(s)\,ds \end{equation} are obtained. Second, it is proven - using these bounds, Banach's contraction mapping principle, new continuation and translation results, and Schaefer's fixed point theorem - that (R$_\lambda $) has a unique continuous solution on $(0,\infty)$, which is called the resolvent in the literature and denoted here by $R(t)$. Third, both known and new properties of $R(t)$ are derived. Fourth, $R(t)$ is shown to be completely monotone and the unique continuous solution of the initial value problem of fractional order $q$: \begin{equation} \tag{I$_\lambda $} D^{q}x(t) = -\lambda \Gamma(q) x(t), \qquad \lim_{t \to 0^+} t^{1-q}x(t) = \lambda , \end{equation} where $D^{q}$ denotes the Riemann-Liouville fractional differential operator. Finally, the resolvent integral function $\int_0^t R(s)\,ds$ is shown to be the unique continuous solution of an integral equation closely related to (R$_\lambda$). Closed-form expressions for it and $R(t)$ are derived.