# Numerical method for space fractional derivative in 1 dimension

I am very new to the subject of fractional derivatives which arise while characterizing the anomalous transport of passive scalar in turbulence.

I have found an equation of the following form, to describe such anomalous transport, which I have to solve numerically, $$\partial_{t}f(x,t)=kD^{\alpha}_{|x|}f(x,t)+g(x,t)$$ where $$k$$ is some constant, $$D^{\alpha}_{|x|}$$ is symmetric Riesz fractional derivative of order $$\alpha$$ and $$g(x,t)$$ is a source/sink term. The numerical algorithm that I can find are too mathematically abstract for me, therefore it would be of great help if anyone could sketch out the basic ideas behind constructing numerical algorithms for such equations.

• If you put a passively advected test particle in your turbulent field, just plot its RMS deviation vs. time in the log-log scale, and see what power law matches $\langle (\delta x)^2 \rangle \sim t^{\alpha}$. Commented Apr 6, 2023 at 23:54
• The problem with the equation is that it is complicated. There are no simple methods to solve it. The other problem is that if you don't understand how a method works, how would you verify that you correctly implemented it? Commented Apr 7, 2023 at 2:40
• Thanks for the comment @WolfgangBangerth. I am struggling to grasp the algorithms because they involve complex mathematical terms that are making it more difficult for me to understand. I was simply wondering if there is a way to discretize the equation using an analogous approach, similar to how we discretize a 1D Partial Differential Equation (PDE) with Finite Volume technique. Commented Apr 7, 2023 at 19:32
• @Sayan The short answer is no. That's because the definition of a derivative $d/dx$ is pretty straightforward and understandable to the average high school student. The definition of the fractional derivative $D^\alpha$ is not straightforward at all, with multiple competing definitions out in the literature that do not in fact agree. Do you feel like you actually understand what the fractional derivative is, and how it is defined? I suspect that if you struggle with the discretization, you already struggle with the definition. Commented Apr 7, 2023 at 21:38
• @WolfgangBangerth, I saw the definitions and they were defined with infinite series. Commented Apr 8, 2023 at 3:58

Fractional derivatives are, in principle, very difficult to analyze due to their cumbersome conclusions. However, it is worth noting that the numerical solution algorithms for this problem are exactly the same as in the field of integer differentiation.

First of all, it should be noted that three different definitions are used for numerical methods:

1. Riemann-Liouville: $$D_{a+}^\alpha f(x) = \frac{1}{\Gamma(m-\alpha)}\frac{d^m}{dx^m}\int_a^x{\frac{f(s)}{(x-s)^{1+\alpha-m}}ds}; m=\lceil\alpha \rceil$$
2. Grunwald-Letnikov: $$\left. D_{a+}^\alpha f(x) \right|_{x_0} = \lim_{h \rightarrow 0} \frac{1}{h^\alpha} \sum_{i=0}^{N}{(-1)^{i}{\alpha \choose i}f(x-ih)}; Nh=x_0-a$$
3. Caputo: $$D_{a+}^\alpha f(x) = \frac{1}{\Gamma(m-\alpha)}\int_a^x{\frac{\frac{d^m}{dx^m}f(s)}{(x-s)^{1+\alpha-m}}ds}; m=\lceil\alpha \rceil$$

The easiest to learn is the Grunwald-Letnikov derivative, because it is already represented numerically, if you do not take into account the limit. For the rest of the definitions, integrals and differentials are integers and are calculated through approximation by polynomials (for example, the Simpson method). The derivation of formulas in this case is difficult, but real.

As a result, everything somehow boils down to solving either a matrix equation ($$Ax=b$$) or stochastic algorithms. As far as I know, the Meerschaert have made the most progress in this. However, there are really a lot of articles on this topic on the Internet, so it shouldn't be difficult to find them.

At the moment, I am doing research on this topic myself, so I can invite you to get acquainted with my project on GitHub with the implementation of some methods for solving fractional differential equations.

The following literature may be useful: