Singularity of Riemann-Liouville Fractional Derivative
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I have recently started studying fractional order calculus and it seems that the definition of the fractional order differentiation, in the Riemann-Liouville (RL) sense, has singularities. To explain my point, let $alpha in (0,1)$ and $f(t)in mathbb{R}$ be a differentiable function. We have
begin{align}
D^{alpha} f(t) = frac{1}{Gamma(1-alpha)}frac{d}{dt}int_{0}^t f(tau)(t-tau)^{-alpha} dtau.
end{align}
Performing integration by parts and carrying out the differentiation, we further have
begin{align}
D^{alpha} f(t) = frac{f(0)t^{-alpha}}{Gamma(1-alpha)}+frac{1}{Gamma(1-alpha)}int_{0}^t dot{f}(tau)(t-tau)^{-alpha} dtau.
end{align}
The term $t^{-alpha}$ brings about a singularity at $t=0$. So, can we say that (i) for a differentiable function the RL derivative of order $alphain (0,1)$ is singular at $t=0$ unless $f(0) = 0$; (ii) if $f(0) = 0$, then the RL derivative corresponds to that of the Caputo?
integration derivatives fractional-calculus
asked Jan 6 at 16:22
MehranMehran
300110
$endgroup$
add a comment |
$begingroup$
I have recently started studying fractional order calculus and it seems that the definition of the fractional order differentiation, in the Riemann-Liouville (RL) sense, has singularities. To explain my point, let $alpha in (0,1)$ and $f(t)in mathbb{R}$ be a differentiable function. We have
begin{align}
D^{alpha} f(t) = frac{1}{Gamma(1-alpha)}frac{d}{dt}int_{0}^t f(tau)(t-tau)^{-alpha} dtau.
end{align}
Performing integration by parts and carrying out the differentiation, we further have
begin{align}
D^{alpha} f(t) = frac{f(0)t^{-alpha}}{Gamma(1-alpha)}+frac{1}{Gamma(1-alpha)}int_{0}^t dot{f}(tau)(t-tau)^{-alpha} dtau.
end{align}
The term $t^{-alpha}$ brings about a singularity at $t=0$. So, can we say that (i) for a differentiable function the RL derivative of order $alphain (0,1)$ is singular at $t=0$ unless $f(0) = 0$; (ii) if $f(0) = 0$, then the RL derivative corresponds to that of the Caputo?
integration derivatives fractional-calculus
asked Jan 6 at 16:22
MehranMehran
300110
$endgroup$
$begingroup$
I believe the power on the $(t-tau)^{-alpha}$ should change and there should be another coefficient?
$endgroup$
– Simply Beautiful Art
Feb 10 at 23:57
$begingroup$
@Simply Beautiful Art, I've followed the definition of RL derivative, $D^{alpha} = (d/dt)D^{-(1-alpha)}$. First, integration by parts and then, differentiation by Leibniz rule. Could you please be more specific?
$endgroup$
– Mehran
Feb 12 at 9:12
$begingroup$
Never mind, mistake on my end. I think this should be fine then.
$endgroup$
– Simply Beautiful Art
Feb 12 at 19:52
add a comment |
$begingroup$
I have recently started studying fractional order calculus and it seems that the definition of the fractional order differentiation, in the Riemann-Liouville (RL) sense, has singularities. To explain my point, let $alpha in (0,1)$ and $f(t)in mathbb{R}$ be a differentiable function. We have
begin{align}
D^{alpha} f(t) = frac{1}{Gamma(1-alpha)}frac{d}{dt}int_{0}^t f(tau)(t-tau)^{-alpha} dtau.
end{align}
Performing integration by parts and carrying out the differentiation, we further have
begin{align}
D^{alpha} f(t) = frac{f(0)t^{-alpha}}{Gamma(1-alpha)}+frac{1}{Gamma(1-alpha)}int_{0}^t dot{f}(tau)(t-tau)^{-alpha} dtau.
end{align}
The term $t^{-alpha}$ brings about a singularity at $t=0$. So, can we say that (i) for a differentiable function the RL derivative of order $alphain (0,1)$ is singular at $t=0$ unless $f(0) = 0$; (ii) if $f(0) = 0$, then the RL derivative corresponds to that of the Caputo?
integration derivatives fractional-calculus
asked Jan 6 at 16:22
MehranMehran
300110
$endgroup$
I have recently started studying fractional order calculus and it seems that the definition of the fractional order differentiation, in the Riemann-Liouville (RL) sense, has singularities. To explain my point, let $alpha in (0,1)$ and $f(t)in mathbb{R}$ be a differentiable function. We have
begin{align}
D^{alpha} f(t) = frac{1}{Gamma(1-alpha)}frac{d}{dt}int_{0}^t f(tau)(t-tau)^{-alpha} dtau.
end{align}
Performing integration by parts and carrying out the differentiation, we further have
begin{align}
D^{alpha} f(t) = frac{f(0)t^{-alpha}}{Gamma(1-alpha)}+frac{1}{Gamma(1-alpha)}int_{0}^t dot{f}(tau)(t-tau)^{-alpha} dtau.
end{align}
The term $t^{-alpha}$ brings about a singularity at $t=0$. So, can we say that (i) for a differentiable function the RL derivative of order $alphain (0,1)$ is singular at $t=0$ unless $f(0) = 0$; (ii) if $f(0) = 0$, then the RL derivative corresponds to that of the Caputo?
integration derivatives fractional-calculus
integration derivatives fractional-calculus
asked Jan 6 at 16:22
MehranMehran
300110
asked Jan 6 at 16:22
MehranMehran
300110
asked Jan 6 at 16:22
MehranMehran
300110
asked Jan 6 at 16:22
MehranMehran
300110
asked Jan 6 at 16:22
MehranMehran
300110
300110
$begingroup$
I believe the power on the $(t-tau)^{-alpha}$ should change and there should be another coefficient?
$endgroup$
– Simply Beautiful Art
Feb 10 at 23:57
$begingroup$
@Simply Beautiful Art, I've followed the definition of RL derivative, $D^{alpha} = (d/dt)D^{-(1-alpha)}$. First, integration by parts and then, differentiation by Leibniz rule. Could you please be more specific?
$endgroup$
– Mehran
Feb 12 at 9:12
$begingroup$
Never mind, mistake on my end. I think this should be fine then.
$endgroup$
– Simply Beautiful Art
Feb 12 at 19:52
add a comment |
$begingroup$
I believe the power on the $(t-tau)^{-alpha}$ should change and there should be another coefficient?
$endgroup$
– Simply Beautiful Art
Feb 10 at 23:57
$begingroup$
@Simply Beautiful Art, I've followed the definition of RL derivative, $D^{alpha} = (d/dt)D^{-(1-alpha)}$. First, integration by parts and then, differentiation by Leibniz rule. Could you please be more specific?
$endgroup$
– Mehran
Feb 12 at 9:12
$begingroup$
Never mind, mistake on my end. I think this should be fine then.
$endgroup$
– Simply Beautiful Art
Feb 12 at 19:52
$begingroup$
I believe the power on the $(t-tau)^{-alpha}$ should change and there should be another coefficient?
$endgroup$
– Simply Beautiful Art
Feb 10 at 23:57
$begingroup$
I believe the power on the $(t-tau)^{-alpha}$ should change and there should be another coefficient?
$endgroup$
– Simply Beautiful Art
Feb 10 at 23:57
$begingroup$
@Simply Beautiful Art, I've followed the definition of RL derivative, $D^{alpha} = (d/dt)D^{-(1-alpha)}$. First, integration by parts and then, differentiation by Leibniz rule. Could you please be more specific?
$endgroup$
– Mehran
Feb 12 at 9:12
$begingroup$
@Simply Beautiful Art, I've followed the definition of RL derivative, $D^{alpha} = (d/dt)D^{-(1-alpha)}$. First, integration by parts and then, differentiation by Leibniz rule. Could you please be more specific?
$endgroup$
– Mehran
Feb 12 at 9:12
$begingroup$
Never mind, mistake on my end. I think this should be fine then.
$endgroup$
– Simply Beautiful Art
Feb 12 at 19:52
$begingroup$
Never mind, mistake on my end. I think this should be fine then.
$endgroup$
– Simply Beautiful Art
Feb 12 at 19:52
add a comment |
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$begingroup$
I believe the power on the $(t-tau)^{-alpha}$ should change and there should be another coefficient?
$endgroup$
– Simply Beautiful Art
Feb 10 at 23:57
$begingroup$
@Simply Beautiful Art, I've followed the definition of RL derivative, $D^{alpha} = (d/dt)D^{-(1-alpha)}$. First, integration by parts and then, differentiation by Leibniz rule. Could you please be more specific?
$endgroup$
– Mehran
Feb 12 at 9:12
$begingroup$
Never mind, mistake on my end. I think this should be fine then.
$endgroup$
– Simply Beautiful Art
Feb 12 at 19:52