content_image/387816.jpg content_image/387796.jpg content_image/387799.jpg content_image/387800.jpg content_image/387801.jpg content_image/387803.jpg content_image/387802.jpg content_image/387797.jpg content_image/387798.jpg content_image/387817.jpg content_image/387818.jpg content_image/387804.jpg content_image/387805.jpg
<img src='content_image/387816.jpg'>

Here △$_{2}$ is the two dimensional Laplacian operator. Us- ing the fractional cylindrical Laplacian operator defined above, this equation becomes 

<img src='content_image/387796.jpg'>

By separation of variables u ( t, r, θ ) = R ( r )Θ( θ ) T ( t ), it can be decomposed to 

<img src='content_image/387799.jpg'>

The first two equations are simple. The third equation is a fractional generalization of the Bessel equation [11, 12]. It can be solved by fractional series expansion. Since Bessel equation is singular at r = 0. We must use such ansatz: R = r αρ ∑ ∞ m =0 c$_{m}$r $^{αm}$. Substitute it into the above equation, we get 

<img src='content_image/387800.jpg'>

<img src='content_image/387801.jpg'>

<img src='content_image/387803.jpg'>

To have a starting term, c$_{0}$ ̸ = 0, so 

<img src='content_image/387802.jpg'>

<img src='content_image/387797.jpg'>

<img src='content_image/387798.jpg'>

<img src='content_image/387817.jpg'>

<img src='content_image/387818.jpg'>

and c$_{1}$ = 0. 

By the recursive relation, a solution is implied 

where d$_{0}$ = 1, 

<img src='content_image/387804.jpg'>

and ρ satisfies Eq.(33). 

Eq.(33) in one-order calculus ( α = 1) is simply ρ = ± ν . We meshed in Fig. 4 the surface defined by the equation (33). After solving the equation with ν = 3 for ρ when α varies, we plotted in Fig. 5 R$_{ρ}$ ( r ) belonging to different values of α . 

<img src='content_image/387805.jpg'>

