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The Mittag-Leffler function $E_{\alpha}(x)$ has an important property: $$ \frac{\partial^{\alpha}}{\partial x^{\alpha}} E_{\alpha}(x^{\alpha}) = E_{\alpha}(x^{\alpha}). $$ I tried to find an analogue of such function that satisfies $$ \frac{\partial^{\alpha}}{\partial t^{\alpha}} G_{\alpha}((a+bt)^{\alpha}) = CG((a+bt)^{\alpha}). $$ I looked through some papers on fractional differentiating but didn't find such analogue. Then I tried to construct it myself. I found fractional derivative of order $\alpha$ of $(a+bt)^{n \alpha}$. If it was something like $$ (a+bt)^{\alpha(n-1)} F(n,\alpha,b)H(t) $$ I'll be able to construct a power series with convinient coefficents like in the case of the Mittag-Leffler function. But in general this fractional derivative is a series, that can't be represented in such manner. Then I think there may exist some pseudodifferential operator $F_{\alpha}$ (PsDO) for which there exists a series by powers of $(a+bt)^{\alpha}$ invariant under action of $F_{\alpha}$. I would be very happy if you can help me to find such series and such operator that generalize both Mittag-Leffler function and fractional differentiation operator.

Edited This problem is related to specification of PsDO that sends $(a+bt)^{\alpha n}$ to $(a+bt)^{\alpha(n-1)} F(n,\alpha,b) H(t)$.

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1 Answer 1

The first equation leads to some discussion. In fact, it is valid if one assumes that x>0 and the Caputo derivative. The real eigenfunction of that equation is an exponential. Any way, to solve your problem, you must assume that t>-a/b and apply the shift invariant property of the derivative that, if a,b>0, is only valid with the Grunwald-Letnikov derivative. If you want to discuss this send me a mail. mdo@fct.unl.pt

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