Are there a solutions for $\frac{\mathrm{d}^\alpha f}{\mathrm{d}x^\alpha} = xe^x$?

Question

Are there solutions for the fractional differential equation $$\frac{\mathrm{d}^\alpha f}{\mathrm{d}x^\alpha} = xe^x$$ for $0 < \alpha < 1$? I have been using the Caputo definition $$ \frac{\mathrm{d}^\alpha f}{\mathrm{d}x^\alpha} = \frac{1}{\Gamma(1-\alpha)} \int_0^x \frac{f'(x')}{(x-x')^\alpha} \mathrm{d} x' $$ but I'm interested in solutions also with other definitions.

With integer order derivatives $$ \frac{\mathrm{d}^n f}{\mathrm{d}x^n} = xe^x$$ the solution is $f = (x-n)e^x$, so I guessed that for fractional order the solution might be $f \propto (x-\alpha)e^x$. However, plugging this into the Caputo definition doesn't really work, as one ends up with a lower incomplete Gamma function $\gamma$. I now realize that this is due to the exponential function having the property $$\frac{\mathrm{d}^\alpha e^x}{\mathrm{d}x^\alpha} \propto e^x \gamma(\alpha, x) \neq e^x .$$

Is there a way to obtain a solution or does this property of the exponentials mean that there is no solution?

Answer 1

Let's ask a more general question: Let $g$ be analytic at $0$ and $\forall n \in \{ m \in \mathbb{N}_{0} \mid m < \lceil \alpha \rceil \}{:}\, g^{( n )}( 0 ) = 0$. What is the inverse of this Caputo fractional derivative of $g$ such that ${_{0}^{\text{C}}\operatorname{D}}_{u}^{\alpha}\left[ f( u ) \right]( x ) = g( x )$ (in the ROC of it's series)?

An answer is $\fbox{$f( x ) = \sum_{k = \lceil \alpha \rceil}^{\infty}\left( g_{k} \cdot \frac{\Gamma( k + 1 )}{\Gamma( k + \alpha + 1 )} \cdot x^{k + \alpha} \right) + P( x )$ with $\deg( P ) < \lceil \alpha \rceil$}$ (I included a proof at the end). You can generalize this to all analytic functions but the term get's way more complicated and we would have much more work.

What does this mean to you? We have $$g( x ) := x \cdot \exp( x ) = x \cdot \sum_{k = 0}^{\infty}\left( \tfrac{1}{k!} \cdot x^{k} \right) = \sum_{k = 0}^{\infty}\left( \tfrac{1}{k!} \cdot x^{k + 1} \right)$$ aka $g( 0 ) = 0$. So for $0 < \alpha < 1$ aka $\lceil \alpha \rceil = 1$ we have \begin{align*} f( x ) &= \sum\limits_{k = 1}^{\infty}\left( \frac{\Gamma( k + 1 )}{k! \cdot \Gamma( k + \alpha + 1 )} \cdot x^{k + \alpha} \right) + \text{constant}\\ &= \sum\limits_{k = 1}^{\infty}\left( \frac{1}{\Gamma( k + \alpha + 1 )} \cdot x^{k + \alpha} \right) + \text{constant}\\ &= \sum\limits_{k = 0}^{\infty}\left( \frac{1}{\Gamma( k + \alpha + 2 )} \cdot x^{k + \alpha + 1} \right) + \text{constant}\\ &= x^{\alpha + 1} \cdot \sum\limits_{k = 0}^{\infty}\left( \frac{1}{\Gamma( k + \alpha + 2 )} \cdot x^{k} \right) + \text{constant}. \end{align*}

We can simplify this via incomplete Gamma functions or the general Mittag-Leffler function, to:

$${\huge \fbox{$\fbox{$f( x ) = x^{\alpha + 1} \cdot \operatorname{E}_{1,\, \alpha + 2}( x ) + \text{constant}$}$}}$$

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Proof.

We know ${_{0}^{\text{C}}\operatorname{D}}_{u}^{\alpha}\left[ u^{\beta} \right]( x ) = \begin{cases} 0, &\text{for } \lceil \alpha \rceil > \beta \in \mathbb{N}_{0}\\ \frac{\Gamma( \beta + 1 )}{\Gamma( \beta - \alpha + 1 )} \cdot u^{\beta - \alpha}, &\text{for } \lceil \alpha \rceil < \beta \end{cases}$, so:

\begin{align*} f( x ) &= \sum\limits_{k = \lceil \alpha \rceil}^{\infty}\left( g_{k} \cdot \frac{\Gamma( k + 1 )}{\Gamma( k + \alpha + 1 )} \cdot x^{k + \alpha} \right) + P( x )\\ \implies {_{0}^{\text{C}}\operatorname{D}}_{u}^{\alpha}\left[ f( u ) \right]( x ) &= {_{0}^{\text{C}}\operatorname{D}}_{u}^{\alpha}\left[ \sum\limits_{k = \lceil \alpha \rceil}^{\infty}\left( g_{k} \cdot \frac{\Gamma( k + 1 )}{\Gamma( k + \alpha + 1 )} \cdot u^{k + \alpha} \right) + P( u ) \right]( x )\\ &= {_{0}^{\text{C}}\operatorname{D}}_{u}^{\alpha}\left[ \sum\limits_{k = \lceil \alpha \rceil}^{\infty}\left( g_{k} \cdot \frac{\Gamma( k + 1 )}{\Gamma( k + \alpha + 1 )} \cdot u^{k + \alpha} \right) \right]( x ) + {_{0}^{\text{C}}\operatorname{D}}_{u}^{\alpha}\left[ P( u ) \right]( x )\\ &= {_{0}^{\text{C}}\operatorname{D}}_{u}^{\alpha}\left[ \sum\limits_{k = \lceil \alpha \rceil}^{\infty}\left( g_{k} \cdot \frac{\Gamma( k + 1 )}{\Gamma( k + \alpha + 1 )} \cdot u^{k + \alpha} \right) \right]( x )\\ &= \sum\limits_{k = \lceil \alpha \rceil}^{\infty}\left( g_{k} \cdot \frac{\Gamma( k + 1 )}{\Gamma( k + \alpha + 1 )} \cdot {_{0}^{\text{C}}\operatorname{D}}_{u}^{\alpha}\left[ u^{k + \alpha} \right]( x ) \right)\\ &= \sum\limits_{k = \lceil \alpha \rceil}^{\infty}\left( g_{k} \cdot \frac{\Gamma( k + 1 )}{\Gamma( k + \alpha + 1 )} \cdot \frac{\Gamma( k + \alpha + 1 )}{\Gamma( k + 1 )} \cdot x^{k} \right)\\ &= \sum\limits_{k = \lceil \alpha \rceil}^{\infty}\left( g_{k} \cdot x^{k} \right) \end{align*}

And $g \text{ is analytic at } 0 \implies \exists ( g_{n} )_{n \in \mathbb{N}_{0}}{:}\, g( x ) = \sum_{k = 0}^{\infty}\left( g_{k} \cdot x^{k} \right)$ and $$\left( \forall n \in \{ m \in \mathbb{N}_{0} \mid m < \lceil \alpha \rceil \}{:}\, g^{( n )}( 0 ) = 0 \right)\\ \implies \left( \forall n \in \{ m \in \mathbb{N}_{0} \mid m < \lceil \alpha \rceil \}{:}\, \sum_{k = n}^{\infty}\left( g_{k} \cdot \frac{k!}{( k - n )!} \cdot 0^{k - n} \right) = 0 \right)\\ \implies \left( \forall n \in \{ m \in \mathbb{N}_{0} \mid m < \lceil \alpha \rceil \}{:}\, g_{n} = 0 \right),$$ so $g( x ) = \sum_{k = \lceil \alpha \rceil}^{\infty}\left( g_{k} \cdot x^{k} \right)$. Combining gives us $${_{0}^{\text{C}}\operatorname{D}}_{u}^{\alpha}\left[ f( u ) \right] = \sum_{k = \lceil \alpha \rceil}^{\infty}\left( g_{k} \cdot x^{k} \right) = g( x ). \qquad \text{q.e.d.}$$

Answer 2

For the Caputo derivative using Taylor series, because $\alpha \in (0,1)$ we can swap the integral and sum and we can say

$$D^\alpha (xe^x) = \sum_{n=0}^\infty \frac{n+1}{n!\cdot\Gamma(1-\alpha)}\int_0^x\frac{(x-x')^n}{{x'}^\alpha}dx' = \sum_{n=0}^\infty \frac{n+1}{\Gamma(n+2-\alpha)}x^{n+1-\alpha}$$

by dominated convergence.

For other fractional derivatives like the Riemann-Liouville variant and its equivalents, one should be able to use Feynman's trick instead $xe^x = \partial_a(e^{ax})|_{a=1}$ to evaluate those integrals.