I'm trying to compute the Riemann-Liouville fractional derivative of order $\alpha$ of the Bessel function $J_{0}(\sqrt{t})$ with $0 < \alpha < 1.$
$\mathbf{Some\,\,terminology}$
The Riemann-Liouville fractional derivative of a function $f(t)$ is given by
$$I_t^{\alpha}(f(t)) = \frac{1}{\Gamma(\alpha)}\int_{0}^{t}(t-z)^{\alpha-1}f(z)dz, \text{ where}\, \alpha > 0.$$
The Riemann-Liouville fractional derivative is defined as the derivative of RL fractional integral, i.e. $$\mathcal{D}_t^{\alpha}(f(t)) = \frac{1}{\Gamma(1 – \alpha)}\frac{d}{dt}\int_{0}^{t}(t-z)^{-\alpha}f(z)dz, \text{where} \,\,0 < \alpha \leq 1. $$
In order to compute $\mathcal{D}_t^{\alpha}\big(J_0(\sqrt{t})\big),$ I can simply plug $J_0(\sqrt{t}) = \sum_{m=0}^\infty \frac{(-1)^m}{(m!)^2}\left(\frac{\sqrt{t}}{2}\right)^{2m}$ into the above formula to obtain
$$\frac{1}{\Gamma(1 – \alpha)}\sum_{m=0}^{\infty}\bigg(\frac{1}{2}\bigg)^{2m}\frac{(-1)^m}{(m!)^2}\frac{d}{dt}\int_{0}^{t}(t-z)^{-\alpha}z^{m}dz.$$ I don't know how to simplify the above expression. Any help highly appreciated. Also is there any way to compute RL fractional derivative of bessel function using laplace transforms.
Best Answer
Using the proposed series expansion: \begin{align} \mathcal{D}_t^{\alpha}\left( J_0(\sqrt t) \right) &=\frac{1}{\Gamma(1 - \alpha)}\sum_{m=0}^{\infty}\left(\frac{1}{2}\right)^{2m}\frac{(-1)^m}{(m!)^2}\frac{d}{dt}\int_{0}^{t}(t-z)^{-\alpha}z^{m}\,dz\\ &=\frac{1}{\Gamma(1 - \alpha)}\sum_{m=0}^{\infty}\left(\frac{1}{2}\right)^{2m}\frac{(-1)^m}{(m!)^2}\frac{d}{dt}t^{m+1-\alpha}\int_{0}^{1}(1-x)^{-\alpha}x^{m}\,dx\\ &=\frac{1}{\Gamma(1 - \alpha)}\sum_{m=0}^{\infty}\left(\frac{1}{2}\right)^{2m}\frac{(-1)^m}{(m!)^2}\left( m+1-\alpha \right)t^{m-\alpha}\int_{0}^{1}(1-x)^{-\alpha}x^{m}\,dx \end{align} where $z=tx$ was changed in the integral, as proposed by @metamorphy. This integral corresponds to the Beta function integral representation: \begin{equation} \int_{0}^{1}(1-x)^{-\alpha}x^{m}\,dx=B(1-\alpha,1+m) =\frac{\Gamma(1-\alpha)\Gamma(1+m)}{\Gamma(2-\alpha+m)} \end{equation} and thus, \begin{align} \mathcal{D}_t^{\alpha}\left( J_0(\sqrt t) \right) &=t^{-\alpha}\sum_{m=0}^{\infty}\frac{(-1)^m}{(m!)^2}\left( m+1-\alpha \right)\left(\frac{\sqrt t}{2}\right)^{2m}\frac{\Gamma(1+m)}{\Gamma(2-\alpha+m)}\\ &=t^{-\alpha}\sum_{m=0}^{\infty}\frac{(-1)^m}{m!\Gamma(m-\alpha+1)}\left(\frac{\sqrt t}{2}\right)^{2m} \end{align} This series can be rearranged to nearly match a Bessel function expansion: $$J_\nu(x)\sum_{m=0}^\infty \frac{(-1)^m}{m!\Gamma(m+\nu+1)}\left(\frac x2\right)^{2m+\nu}$$ with $\nu=-\alpha$, \begin{align} \mathcal{D}_t^{\alpha}\left( J_0(\sqrt t) \right) &=2^{-\alpha} t^{-\frac{\alpha}{2}} \sum_{m=0}^{\infty}\frac{(-1)^m}{m!\Gamma(m-\alpha+1)}\left(\frac{\sqrt t}{2}\right)^{2m-\alpha}\\ &=2^{-\alpha} t^{-\frac{\alpha}{2}} J_{-\alpha}\left(\sqrt{t}\right) \end{align}
Alternatively, the integral \begin{equation} I(t)=\int_{0}^{t}(t-z)^{-\alpha}J_0(\sqrt z)\,dz \end{equation} can be considered as a convolution operation for the Laplace transform for the functions $z^{-\alpha}$ and $J_0(\sqrt z)$. Their Laplace transforms are respectively $\Gamma(1-\alpha) p^{\alpha-1}$ (classical integral representation of the Gamma function) and $p^{-1}e^{-1/(4p)}$ (see Ederlyi TI 4.14(25) p. 185). Then \begin{equation} I(t)=\mathcal L^{-1}\left[\Gamma(1-\alpha) p^{\alpha-2}e^{-1/(4p)}\right] \end{equation} This inverse transform can be found in Ederlyi TI 5.6(40) p.245: \begin{equation} I(t)=\Gamma(1-\alpha)2^{1-\alpha}t^{\frac{1-\alpha}{2}}J_{1-\alpha}\left( \sqrt{t} \right) \end{equation} Finally, after differentiation, \begin{equation} \mathcal{D}_t^{\alpha}\left( J_0(\sqrt t) \right) =2^{-\alpha} t^{-\frac{\alpha}{2}} J_{-\alpha}\left(\sqrt{t}\right) \end{equation} as expected from the above calculation.