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Currently I am working with fractional derivative and encountered Caputo's definition:

$\frac{\partial^\alpha}{\partial x^\alpha}y(x = x_0) = \frac{1}{\Gamma(1-\alpha)}\int_{-\infty}^{x_0} (x_0-x)^{-\alpha} \frac{dy}{dx} dx$

where $\alpha$ is some fractional value -- for the purpose of what I'm doing, $\alpha \in [0,1]$.

However, I am having trouble confirming that the Caputo derivative corresponds to a first-order derivative at $\alpha = 1$ and returns the function $y(x_0)$ itself at $\alpha = 0$.

For $\alpha = 0$, $\Gamma (1-0) = 1!= 1$, while the integral equals $y|_{x = -\infty}^{x_0}$. Is there an inherent assumption that the function $y(x = -\infty) = 0$?

For $\alpha = 1$, $\Gamma(1-1) = \infty$, I am guessing we could produce $dy/dx$ using some form of L'Hôpital approach, but do not have the prowess to make it work.

It would be wonderful if someone could show me how to prove these two extreme cases! Thanks in advance!

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The provided formula only works for $\alpha\in(-\infty,1)$, with the typo that $t$ should be $x_0$. In general, the Caputo fractional derivative is defined as

$$_a^C\mathbb D_t^\alpha f(t)=\frac1{\Gamma(n-\alpha)}\int_a^t\frac{f^{(n)}(\tau)}{(t-\tau)^{\alpha+1-n}}~\mathrm d\tau$$

for an integer $n>\alpha$. In the case of $\alpha=1$, we can take $n=2$ and see that we have

$$_a^C\mathbb D_t^1f(t)=\frac1{\Gamma(2-1)}\int_a^xf''(\tau)~\mathrm d\tau=f'(t)-f'(a)$$

which, as you've noted, gives an extra term dependent on the integer derivatives of $f$ at the lower bound. This is intended, and is in constrast to the Riemann-Liouville integral which takes the integer derivatives out of the integral

$$_a^{RL}\mathbb D_t^\alpha f(t)=\frac1{\Gamma(n-\alpha)}\frac{\mathrm d^n}{\mathrm dt^n}\int_a^t\frac{f(\tau)}{(t-\tau)^{\alpha+1-n}}~\mathrm d\tau$$

so that the extra terms cancel out.

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