# Solutions to this fractional differential equation

So we all know that $\frac d{dx}e^x=e^x$ and that the $n$th derivative of $e^x$ is still $e^x$, but upon entering fractional calculus, this is ruined. Let $D^\alpha$ be the $\alpha$th derivative with respect to $x$. Then, as we can see, when $\alpha\in[0,1)$,

$$D^\alpha e^x=-\frac1{\Gamma(1-\alpha)}e^x\gamma(\alpha,x)\ne e^x$$

where we use the lower incomplete gamma function.

Which raises the interesting question:

What are the solutions to the following fractional differential equation? $$D^\alpha f(x)=f(x)$$

where we have

$$D^\alpha f(x)=\frac1{\Gamma(n-\alpha)}\int_0^x\frac{f^{(n)}(t)}{(x-t)^{\alpha+1-n}}\ dt$$

with $n=\lfloor\alpha+1\rfloor$.

$f$ may be a function of $\alpha$.

• This does of course depend on the differintegral used: with the fixed limit $a=-\infty$, you do get $D_{-\infty}^{\alpha}e^{x} = e^{x}$ Commented May 14, 2017 at 13:18
• @Chappers Yes, but I have given the differintegral I want to use for this problem. Commented May 14, 2017 at 13:19
• @LowGPA It's the well-known Gamma function Commented May 31, 2017 at 22:27
• I guess $\gamma$ denotes Euler–Mascheroni constant? Commented May 31, 2017 at 22:30
• @LowGPA It's the lower incomplete gamma function. See the link in the first paragraph. Commented May 31, 2017 at 22:31

The first part is not a strict answer to the question, but not far. The full answer is added in second part.

Consider the series expansion : $$e^x=\sum_{k=0}^\infty \frac{x^k}{k!} = \sum_{k=0}^\infty \frac{x^k}{\Gamma(k+1)} \qquad |x|<1$$

Compare to the Mittag-Leffler function : $$E_\alpha(x)=\sum_{k=0}^\infty \frac{x^k}{\Gamma(\alpha k+1)}$$ http://mathworld.wolfram.com/Mittag-LefflerFunction.html $$\text{Or }\qquad E_\alpha(x^\alpha)=\sum_{k=0}^\infty \frac{x^{\alpha k}}{\Gamma(\alpha k+1)}$$

This function matches the exponential function in particular case $$\alpha=1$$.

It is of interest to see what is the fractional derivative of $$\left(E_\alpha(x^\alpha)-1\right)$$. We will see latter why the first term of the series is considered apart. $$\frac{d^\alpha}{dx^\alpha}\left(E_\alpha(x^\alpha)-1\right)=\sum_{k=1}^\infty \frac{1}{\Gamma(\alpha k+1)}\frac{d^\alpha }{dx^\alpha}(x^{\alpha k})$$ $$\frac{d^\alpha }{dx^\alpha}(x^{\alpha k})=\frac{\Gamma(\alpha k+1)}{\Gamma\left(\alpha (k-1)+1\right)}x^{\alpha(k-1)}$$

$$\frac{d^\alpha}{dx^\alpha}\left(E_\alpha(x^\alpha)-1\right)=\sum_{k=1}^\infty \frac{1}{\Gamma(\alpha k+1)}\frac{\Gamma(\alpha k+1)}{\Gamma\left(\alpha (k-1)+1\right)}x^{\alpha(k-1)}$$

$$\frac{d^\alpha}{dx^\alpha}\left(E_\alpha(x^\alpha)-1\right)=\sum_{k=1}^\infty \frac{x^{\alpha(k-1)}}{\Gamma\left(\alpha (k-1)+1\right)}=\sum_{h=0}^\infty \frac{x^{\alpha h}}{\Gamma\left(\alpha h+1\right)}$$

$$\frac{d^\alpha}{dx^\alpha}\left(E_\alpha(x^\alpha)-1\right)=E_\alpha(x^\alpha)$$

$$\frac{d^\alpha}{dx^\alpha}E_\alpha(x^\alpha)=E_\alpha(x^\alpha)+\frac{d^\alpha}{dx^\alpha}(1)$$

This is close to the expected equation $$\quad \frac{d^\alpha}{dx^\alpha}f(x)=f(x)\qquad \text{with} \quad f(x)=E_\alpha(x^\alpha)$$ But there is an extra term $$\frac{d^\alpha}{dx^\alpha}(1)=\frac{x^{-\alpha}}{\Gamma(1-\alpha)}$$

This is the difference compared to the case $$\alpha=1$$ of the exponential : $$\frac{d^1}{dx^1}e^x=e^x+\frac{d^1}{dx^1}(1)=e^x$$ The first term in the series expansion of $$e^x$$ is constant$$=1$$. So its derivative is $$0$$, which is not the case for the fractional derivative of order different from $$1$$.

In fact, this difference comes from the definition of the lower bound $$=0$$ in the Riemann-Liouville operator for fractional differ-integration.

In order to have a full solution, the Mittag-Leffler function has to be extended. Instead to limit the series to the terms with $$k\geq 0$$ consider all terms from $$k=-\infty$$ to $$+\infty$$. $$f(x)=\sum_{k=-\infty}^\infty \frac{x^{\alpha k}}{\Gamma(\alpha k+1)}$$ The same calculus as above shows that $$f(x)$$ is a formal solution of the fractional differential equation $$\frac{d^\alpha}{dx^\alpha}f(x)=f(x)$$

Note :

Also, this is valid for the exponential function and $$\alpha=1$$ since $$\quad \frac{1}{k!}=\frac{1}{\Gamma(k+1)}=0 \quad\text{in}\quad k<0 \quad\to\quad e^x=\sum_{k=-\infty}^\infty \frac{x^k}{k!}$$ .

• Your double sum does not converge for $\alpha\not\in\mathbb{Z}$, as the reflection formula $$\Gamma(x)\Gamma(1-x)=\pi\csc(\pi x)$$ guarantees the terms will blow up in magnitude to $\infty$ as either $k\rightarrow \infty$ or $k\rightarrow -\infty$ Commented May 20, 2022 at 23:18

This is a (heuristic) answer for $\alpha=1/k$ for $k\in \mathbb{N}$. And assuming $Df$ exists. Note that I believe (see edit in the end) that we could extend this answer to any $\alpha$.

Let $\alpha = 1/2$. Then, by definition of the fractional derivative, we have $$D^\alpha D^\alpha f = Df$$ Substituting on the LHS what the differential equation implies we obtain $$D^\alpha f = D f$$

More general, $$D^{(k-1)\alpha}f=D f \quad (*)$$

I can only think of the function $f\equiv 0$ which satisfies this and the given differential equation.

Edit: We could continue with equation $(*)$ by plugging in repetedly the diff eq on the LHS to obtain $$D^{(k-2)\alpha}f=D f$$ $$D^{(k-3)\alpha}f=D f$$ $$\ldots$$ $$f=D f$$ Since we know two solutions for the very last equation and since one of them does not satisfy the fract diff eq given in the question, we conclude that indeed $f\equiv 0$ is the only soluton.

I believe that this is also true for any $\alpha$ if we assume that $f$ has a derivative up to a certain order: Start with rational $\alpha=p/q$ then start the argument with $(D^\alpha)^q f$.

• Sorry, forgot to mention that $f$ could be a function of $\alpha$, or else I think $f=0$ is the only solution. Commented May 14, 2017 at 20:44
• @SimplyBeautifulArt do you mean $x=g(\alpha)$ for some $g$? But doesn't that cause troubles for the definition of the fract derivative? For example would $D^{1/3}D^{2/3}f=Df$ still hold?
– Elsa
Commented May 15, 2017 at 2:18
• No, I mean the solution should be of the form $f_\alpha(x)$, a function depending on both $\alpha$ and $x$. Commented May 15, 2017 at 10:53
• @SimplyBeautifulArt so the fract diff eq would be $D^\alpha f_\alpha = f_0$ where the RHS is explained by $f=D^0 f$? If so, I do not see the meaning of this. If it is not this, I do not see why my answer wouldn't cover that case....I didn't forbid $f$ to have a parameter equal to $\alpha$ in my derivations....
– Elsa
Commented May 15, 2017 at 13:00
• No, it would be $D^\alpha f_\alpha(x)=f_\alpha(x)$. Commented May 15, 2017 at 15:19

IDK if $\alpha<1$

For special cases of $\alpha >1$,

$f^\alpha=f$ gives:

$f=\sum_{k\leq\alpha} C_ke^{e^{\frac{{2ki\pi}}{\alpha}}x}+\sum_{n\leq\alpha} C_ne^{e^{\frac{{2i\pi}}{n}}x}$ where $k,n\in\mathbb{N}^*$

• Only for $\alpha\in\mathbb N$. $\alpha\in\mathbb R$ behaves much differently. Commented May 31, 2017 at 22:27
• This is for $\alpha\in \mathbb{R}$ Commented May 31, 2017 at 22:29
• No... I can assure you its not. It already fails $\alpha=1/2$. See the link in the first paragraph of my post. Commented May 31, 2017 at 22:30
• Did I not mention "IDK if $\alpha<1$" "for $\alpha>1$"? Commented May 31, 2017 at 22:31
• Oh, my bad. But it follows directly that this would fail for $\alpha=3/2$ then. Commented May 31, 2017 at 22:33