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When comparing the differences between the Riemann-Liouville definition of the fractional derivative and the Caputo definition, one line from this paper confused me:

In his 1967 paper, Caputo reformulated the definition of the Riemann–Liouville fractional derivative, by switching the order of the ordinary derivative with the fractional integral operator. By doing so, the Laplace transform of this new derivative depends on integer order initial conditions, differently from the initial conditions when we use the Riemann–Liouville fractional derivative, which involve fractional order conditions.

I am not sure how the laplace transform of the RL fractional derivative differs from the Caputo definition, and why it is significant to compare the two of them. I can see how we have "integer order initial conditions" in the Caputo definition, as we are integer differentiating first, but how does this affect the laplace transform of the function? How does it change it's uses and computation? Can we assign a value to the laplace transform of each of these expressions?

If possible please explain in simple terms so a 16 year old can understand! Many thanks in advance.

Definitions of both the RL type fractional derivative and Caputo type for reference:

$$_a^{RL}D_t^n = \frac{d^p}{dx^p}\left(\frac{1}{\Gamma(m)}\int^t_a(t-x)^{m-1}f(x)dx\right)$$ Where $n$ is the fractional order of differentiation ($n\in\mathbb{R}_+$) and $f(x)$ is our integrand;
$p\in\mathbb{N}:={p=⌈n⌉}$, $m\in\mathbb{R}:=m=1-n$
Function outputted in terms of t, $a$ and $t$ are our integration bounds. $$_a^CD_t^n=\frac{1}{\Gamma(m)}\int_a^t(t-x)^{m-1}f^p(x)dx$$ Where the variables are the same, and $f^p(x)$ indicates the $p^{th}$ derivative of $f(x)$ in respect to x.

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After alot of playing around I think I can explain the answer well: The definition for the laplace transform of the $n^{th}$ derivative of $f(t)$ is as follows (in sigma notation):

$$\mathscr{L}\{f^n(t)\}(s)=s^nF(s) - \sum_{k=1}^{n}(s^{n-k})(f^{k-1}\Bigr|_{\substack{t=0}})$$

We can apply this same definition soundly to the Riemann-Liouville and Caputo definitions for the fractional derivative:

$$\mathscr{L}\{^{RL}_0D^\alpha_t\}(s)=s^nF(s) - \sum_{k=1}^{n}(s^{k-1})(^{RL}_0D^{{\alpha}-k}_t\Bigr|_{\substack{t=0}})$$

$$\mathscr{L}\{^C_0D^\alpha_t\}(s)=s^nF(s) - \sum_{k=1}^{n}(s^{{\alpha}-k})(^C_0D^{k-1}_t\Bigr|_{\substack{t=0}})$$ Where $\alpha$ is the fractional order of differentiation and $n=\lceil\alpha\rceil$

The difference between these two definitions is the swap of exponent and differentiation order in the Caputo definition. This is due to the integer differentiation being undertaken first, causing more similarity to the integer-order definition at the top. The issue with the RL definition is how it's Laplace transform initial conditions are skewed due to the fractional integration.

So why is this significant? The best example I know of is highlighted in Michele Caputo's original 1967 paper "Linear Models of Dissipation whose Q is almost Frequency Independant-II"; Caputo uses his definition of the fractional derivative to aid computation of fractional differential equations. Referencing the Laplace transform definitions above, we can see the inital conditions used in the RL definition are still fractional derivatives $(\alpha\in\mathbb{R})$, whereas Caputo's definition uses integer derivatives $(k\in\mathbb{N})$.

In other words, initial conditions such as $f'(0)=1$ used in computation of differential equations are integer derivative defined, not fractional derivative defined; applications of this are now much easier.

Caputo goes on to use this definition in a fractional-order differential equation on dissapation in a plane wave, which I will not pretend to understand. The point is, the pure maths is much easier to apply to fractional differential equations in this form.

Hopefully future people find these definitions of laplace transforms for fractional-order derivatives useful, and can understand why Caputo's definition was significant: mainly in applications through differential equations and mathematically through Laplace transforms.

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