Dear All,
Can I transform a linear fractional Volterra integro-differential equation into a fractional differential equation? If yes, then how?
The equations are written in the attached file.
Thank you very much in advance for your help.
Sarah
Dear Sarah,
I think that you have to impose some extra conditions on f, k and even to the notion of the fractional derivative to obtain the corresponding fractional differential equation (FDE). Even in the simplest cases it is not clear what is the corresponding FDE. Let us consider a couple of examples.
First, take k(x,t)=1 and f is differentiable. Then differentiating
(1) D^\alpha y(x)=f(x)+\int_0^x k(x,t)y(t)dt
yields
(2) DD^\alpha y(x)=f'(x)+y(x).
But since the fractional integration and differentiation does not commute in general, DD^\alpha is not necessarily a fractional derivative of order \alpha+1. If D^\alpha denotes the Riemann-Liouville fractional derivative, then (2) corresponds to FDE:
(3) D^(\alpha+1) y(x)-y(x)=f'(x).
However, if D^\alpha denotes e.g. the Caputo fractional derivative, then DD^\alpha is not necessarily D^(1+\alpha).
Second, take k(x,t)=c(x-t)^(\beta-1) for some constant c. If c=1/\Gamma(\beta), then the integral term in (1) corresponds to the Riemann-Liouville fractional integral of order \beta, which is denoted as I^\beta y. If the fractional derivative D^\beta f exists and D^\beta is the left inverse of I^\beta, then (1) converts into
(4) D^\beta D^\alpha y(x)=D^\beta f(x)+y(x).
But for the same reason as in the first case, D^\beta D^\alpha is not D^(\alpha+\beta) in general.
Hopefully you will find this useful in your further considerations.
Best regards, Jukka
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