Solutions to this fractional differential equation

Solution 1:

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

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

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

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

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

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 ADDITION :

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 $$\quad \frac{1}{k!}=\frac{1}{\Gamma(k+1)}=0 \quad\text{in}\quad k<0 \quad\to\quad e^x=\sum_{k=-\infty}^\infty \frac{x^k}{k!} $$ .