On a solution of a fractional integral equation

Question

I am looking for the solution of $$\frac{d^\alpha}{d x^\alpha}f(x)=g(x)f(x),$$ where $\alpha \in (0,1)$ and $\frac{d^\alpha}{d x^\alpha}$ is the Caputo derivative.

A series of Jumarie's papers, "2005On the solution of the stochastic differential equation of exponential growth driven by fractional Brownian motion", "2005On the representation of fractional Brownian motion as an integral with respect to and $(dt)^\alpha$" and "2006Modified Riemann-Liouville derivative and fractional Taylor series of nondifferentiable functions further results", and many other later papers, the author claim the solution is $$ f(x)=f(0)E_\alpha\left(\alpha \int_0^x (x-y)^{\alpha-1}g(y)dy\right), $$ where $E_\alpha$ is the Mittag-Leffler function. Apart from the idea provided in these papers, do we have other ways to prove it?

By the way, the author also apply one result of the Mittag-Leffler function, $$ E_\alpha\left((x+y)^\alpha\right)=E_\alpha(x^\alpha)E_\alpha(y^\alpha). $$ May I ask what is the condition for this to be true? At least, Wolfram Mathematica does not support this equation.

Many thanks in advance.

Answer 1

My Solving Idea

I would solve it by assuming that we can split $f$ into a homogeneous solution $f_{h}$ and a particular solution $f_{p}$ such that $f\left( x \right) = c \cdot f_{h}\left( x \right) + f_{p}\left( x \right)$, where $c$ is a constant.

With the assumption and the fact that the Fractional Differential Equation (FDE) is linear and homogeneous, we can solve $$ \begin{align*} \operatorname{D}\limits^{\alpha}\left( f\left( x \right) \right) &= g\left( x \right) \cdot f\left( x \right)\\ \end{align*} $$ for $f_{h}$ and $f_{p}$.

Solving for the Homogeneous Solution $f_{h}$

Derivation of the solution

$\small{\text{assuming that g is a constant}}$ Composing the characteristic equation gives: $$ \begin{align*} \operatorname{D}\limits^{\alpha}\left( \exp\left( \lambda \cdot x \right) \right) - g\left( x \right) \cdot \exp\left( \lambda \cdot x \right) &= 0\\ \lambda^{\alpha} \cdot \exp\left( \lambda \cdot x \right) - g\left( x \right) \cdot \exp\left( \lambda \cdot x \right) &= 0\\ \left( \lambda^{\alpha} - g\left( x \right) \right) \cdot \exp\left( \lambda \cdot x \right) &= 0\\ \lambda^{\alpha} - g\left( x \right) &= 0\\ \lambda^{\alpha} &= g\left( x \right)\\ \lambda &= \sqrt[\alpha]{\left| g\left( x \right) \right|} \cdot \left( \operatorname{cis}\left( \arg\left( g\left( x \right) \right) \right) \right)^{\frac{1}{\alpha}}\\ \lambda &= \sqrt[\alpha]{\left| g\left( x \right) \right|} \cdot \operatorname{cis}\left( \frac{1}{\alpha} \cdot \arg\left( g\left( x \right) \right) \right)\\ \lambda_{k} &= \sqrt[\alpha]{\left| g\left( x \right) \right|} \cdot \operatorname{cis}\left( \frac{1}{\alpha} \cdot \left( \operatorname{Arg}\left( g\left( x \right) \right) + 2 \cdot k \cdot \pi \right) \right)\\ \lambda_{k} &= \sqrt[\alpha]{\left| g\left( x \right) \right|} \cdot \operatorname{cis}\left( \frac{\operatorname{Arg}\left( g\left( x \right) \right)}{\alpha} + \frac{2}{\alpha} \cdot k \cdot \pi \right)\\ \end{align*} $$

where k is an intiger ($k \in \mathbb{Z}$), $\arg$ is the Complex Argument Function, $\operatorname{Arg}\left( z \right)$ is the Argument of $z$ wich is the closest to $0$ and $\operatorname{cis}$ is the CIS Function (another name for the complex exponential) given by $\operatorname{cis}\left( z \right) = \cos\left( z \right) + i \cdot \sin\left( z \right) = \exp\left( z \cdot i \right)$ (where $i^{2} = -1$) according to Euler's Formula.

This means that the roots of the equation are $\lambda_{k} = \sqrt[\alpha]{\left| g\left( x \right) \right|} \cdot \cos\left( \frac{\operatorname{Arg}\left( g\left( x \right) \right)}{\alpha} + \frac{2}{\alpha} \cdot k \cdot \pi \right) + \sqrt[\alpha]{\left| g\left( x \right) \right|} \cdot \sin\left( \frac{\operatorname{Arg}\left( g\left( x \right) \right)}{\alpha} + \frac{2}{\alpha} \cdot k \cdot \pi \right) \cdot i$ aka if $g\left( x \right) \ne 0$ $\lambda^{\alpha} - g\left( x \right) = \prod\limits_{k = 1}^{\left| \mathbb{S} \right|}\left[ \lambda + \sqrt[\alpha]{\left| g\left( x \right) \right|} \cdot \operatorname{cis}\left( \frac{\operatorname{Arg}\left( g\left( x \right) \right)}{\alpha} + \frac{2}{\alpha} \cdot k \cdot \pi \right) \right]$ where $\mathbb{S}$ is the set of all roots of the equation and $\left| \mathbb{S} \right|$ is the cardinality of $\mathbb{S}$. This gives the multiplicity of the all roots $= 1$ for $\alpha \ne 0$. That means that if $\alpha \in \mathbb{N} \implies \mathbb{S} = \alpha$.

With $f_{h}\left( x \right) = Q_{0}\left( x \right) \cdot \exp\left( \Re\left( \lambda_{k} \right) \cdot x \right) \cdot \cos\left( \Im\left( \lambda_{k} \right) \cdot x \right) + P_{0}\left( x \right) \cdot \exp\left( \Re\left( \lambda_{k} \right) \cdot x \right) \cdot \sin\left( \Im\left( \lambda_{k} \right) \cdot x \right)$ where $\Re$ is the Real Part and $\Im$ is the Imaginary Part we'll get $$ \begin{align*} f_{h,\, k}\left( x \right) &= c_{2 \cdot k} \cdot \exp\left( \Re\left( \lambda_{k} \right) \cdot x \right) \cdot \cos\left( \Im\left( \lambda_{k} \right) \cdot x \right) + c_{2 \cdot k + 1} \cdot \exp\left( \Re\left( \lambda_{k} \right) \cdot x \right) \cdot \sin\left( \Im\left( \lambda_{k} \right) \cdot x \right)\\ f_{h}\left( x \right) &= \sum\limits_{k = 1}^{\left| \mathbb{S} \right|}\left[ c_{2 \cdot k} \cdot \exp\left( \Re\left( \lambda_{k} \right) \cdot x \right) \cdot \cos\left( \Im\left( \lambda_{k} \right) \cdot x \right) + c_{2 \cdot k + 1} \cdot \exp\left( \Re\left( \lambda_{k} \right) \cdot x \right) \cdot \sin\left( \Im\left( \lambda_{k} \right) \cdot x \right) \right]\\ f_{h}\left( x \right) &= \sum\limits_{k = 1}^{\left| \mathbb{S} \right|}\left[ f_{h,\, k}\left( x \right) \right]\\ \end{align*} $$ where all $c$s are constansts.

Criticism of the Solution

However, this solution has a problem. For each non-real $\alpha$ ($\alpha \notin \mathbb{Q}$) it gives infinitely many roots as a solution, making $\left| \mathbb{S} \right|$ tends to infinity and thus the homogeneous solution for this $\alpha$ does not takes a closed form (if $g\left( x \right) \ne 0$). However, for all real $\alpha$ ($\alpha \in \mathbb{Q}$) it takes a closed form.

Solving for the Particular Solution $f_{p}$

Since it is a homogeneous FDE, it has no particular solution aka $f_{p} = 0$.

Proving $f\left( x \right) = f\left( 0 \right) \cdot \operatorname{E}_{\alpha}\left( \alpha \cdot \int\limits_{0}^{x} \left( x - y \right)^{\alpha - 1} \cdot g\left( y \right)\, \operatorname{d}y \right)$

Criticism of the Solution

The solution excludes quite a few trivial functions $f\left( x \right)$, like all $f\left( x \right)$ that are not defined for $x = 0$, e.g $f\left( x \right) = \ln\left( x \right)$, $f\left( x \right) = \frac{1}{x}$, $f\left( x \right) = x^{x}$, ...

But that also means that we cannot prove it if we have just disproved it for a few examples as long as I'm not wrong here.

Conditions for $\operatorname{E}_{\alpha}\left( \left( x + y \right)^{\alpha} \right) = \operatorname{E}_{\alpha}\left( x^{\alpha} \right) \cdot \operatorname{E}_{\alpha}\left( y^{\alpha} \right)$ to beeing True

There are a few simple conditions that must be met in order for the expressions to remain defined, e.g. if $x = -y$ or $x = 0 ~\vee~ 0 = y$ then $\alpha > 0$ (since division by $0$ is undefined and $0^{0}$ is also undefined) and $\alpha \in \mathbb{R}^{+}$ or $\Im\left( \alpha \right) = 0$ (since $0$ to the power of any or most imaginary unit / units is not defined). I don't see any other conditions. However, I am unable to derive it or provide a general proof of it. This is probably material for another question on stackexchange.