2
$\begingroup$

It is mentioned in some papers (Appendix in this paper, for example) that the (formal) solution of the fractional drift (or transport) equation $$ \partial_{t}^{\alpha}u(t,x)+\partial_{x}u(t,x)=0\quad\text{and}\quad u(0,x)=f(x)\quad\text{($x\in\mathbb{R}$, $t\ge0$)} $$ is given as $$ u(t,x)=\int_{0}^{\infty}\frac{1}{t^{\alpha}}\phi\left(-\alpha,1-\alpha;-\frac{y}{t^{\alpha}}\right)f(x-y)dy. $$ Here $0<\alpha<1$, $\partial_{t}^{\alpha}$ denotes the fractional differential operator in Caputo sense defined by $$ \partial_{t}^{\alpha}g(t):=\frac{1}{\Gamma(1-\alpha)}\int_{0}^{t}\frac{g'(s)}{(t-s)^{\alpha}}ds=\frac{1}{\Gamma(1-\alpha)}\frac{d}{dt}\int_{0}^{t}\frac{g(s)-g(0)}{(t-x)^{\alpha}}ds $$ and $\phi$ is the Wright function defined by $$ \phi(\beta,\gamma;z):=\sum_{k=0}^{\infty}\frac{z^{k}}{k!\Gamma(\beta k+\gamma)}. $$

This solution is given by taking the Laplace transform of the equation.

I would like to ask whether this function is indeed a solution of the fractional drift equation via direct calculations or not.

The problem I'm stacking is just one; I cannot calculate $\partial_{t}^{\alpha}u$. As you immediately think, I have tried to calculate as follows: $$ \partial_{t}^{\alpha}u=\int_{0}^{\infty}\sum_{k=0}^{\infty}\frac{(-y)^{k}}{k!\Gamma(-\alpha k+1-\alpha)}\partial_{t}^{\alpha}t^{-\alpha(k+1)} f(x-y)dy. $$ However, $\partial_{t}^{\alpha}t^{-\alpha(k+1)}$ does not converge. For the reason, I'm finding the other way to calculate but I don't find.

Thank you for advance.

Improve this question
$\endgroup$
9
  • $\begingroup$ $\partial_{t}^{\alpha}\frac{t^{-\alpha(k+1)}}{(-\alpha(k+1))!}=\frac{t^{-\alpha(k+2)}}{(-\alpha(k+2))!}$in the usual interpretation of such a fractional derivative. $\endgroup$
    – Tom Copeland
    May 17, 2016 at 16:21
  • $\begingroup$ Taking the derivative w.r.t. x of the convolution integral, integrating by parts, and comparing to the derivative w.r.t. t should give the desired results. $\endgroup$
    – Tom Copeland
    May 17, 2016 at 16:30
  • $\begingroup$ @TomCopeland How is the equality derived? I recognize that the (left side) Caputo derivative of negative power functions is undefined... $\endgroup$
    – user
    May 17, 2016 at 17:15
  • $\begingroup$ The fractional integro-derivatives of the divided powers for $t \ge 0$ are designed to rep the operation $\partial_{t}^{\alpha}H(t) \frac{t^{-\alpha(k+1)}}{(-\alpha(k+1))!}=H(t)\frac{t^{-\alpha(k+2)}}{(-\alpha(k+2))!}$ where $H(t)\frac{t^{-n-1}}{(-n-1)!}$ for $n \ge 0$ are essentially the Dirac delta function or its $n$'th derivative and $H(t)$ is the Heaviside step function. See any survey paper on fractional derivatives (e.g., links in math.stackexchange.com/questions/1537/…). $\endgroup$
    – Tom Copeland
    May 17, 2016 at 19:37
  • $\begingroup$ Fot $t \gt 0$, dividing by $(-n-1)!$ gives zero for $n$ a positive integer. $\endgroup$
    – Tom Copeland
    May 17, 2016 at 19:42

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.