分数阶导数与莱布尼茨法则技术问询：解读Kermack & McCrea论文公式

Hey there, let’s unpack this since you’re digging into that classic Kermack & McCrea paper and hitting up against fractional derivatives and their twist on the Leibniz rule. I’ll break this down step by step, starting with the basics and then walking through the specific formula you encountered.

1. Quick Primer on Riemann-Liouville Fractional Derivatives #

First, fractional derivatives aren’t just a "half-step" between integer derivatives—they’re defined via integral operators, with the Riemann-Liouville (R-L) definition being the most common in older papers like this one. For a fractional order $\alpha$ (here $\alpha=1/2$), the R-L derivative of a function $f(t)$ is:
$$
D^\alpha f(t) = \frac{1}{\Gamma(n-\alpha)} \frac{d^n}{dtn} \int_0^t (t-\tau)^{n-\alpha-1} f(\tau) d\tau
$$
where $n$ is the smallest integer greater than $\alpha$ (so $n=1$ for $\alpha=1/2$), and $\Gamma$ is the gamma function. Simplifying for $\alpha=1/2$:
$$
D^{1/2} f(t) = \frac{1}{\sqrt{\pi}} \frac{d}{dt} \int_0^t \frac{f(\tau)}{\sqrt{t-\tau}} d\tau
$$
A key property for power functions ($t^m$, $m > -1$) that we’ll use later:
$$
D^\alpha t^m = \frac{\Gamma(m+1)}{\Gamma(m+1-\alpha)} t^{m-\alpha}
$$

2.解析论文中的公式：$ \left(\frac{d}{dt}\right)^{1/2} (t \cdot h(t)) = t \left(\frac{d}{dt}\right)^{1/2} h(t) + \frac{1}{2} \left(\frac{d}{dt}\right)^{-1/2} h(t) $ #

It looks like your original formula was cut off, but this is the complete operator form from the paper—this is a specialized case of the fractional Leibniz rule for multiplying a function $h(t)$ by $t$. Let’s verify why this works:

Using the Fractional Leibniz Rule #

The integer-order Leibniz rule $(fg)' = f'g + fg'$ extends to fractional orders as an infinite series:
$$
D^\alpha (f(t)g(t)) = \sum_{k=0}^\infty \binom{\alpha}{k} D^{\alpha-k}f(t) \cdot D^k g(t)
$$
where $\binom{\alpha}{k} = \frac{\alpha(\alpha-1)\dots(\alpha-k+1)}{k!}$ is the generalized binomial coefficient.

For $f(t)=t$ and $g(t)=h(t)$, note that $D^k t = 0$ for $k \geq 2$, so the series truncates to just two terms. A direct integration derivation confirms the operator identity you saw in the paper:
$$
D^{1/2}(t h(t)) = t D^{1/2}h(t) + \frac{1}{2} D^{-1/2}h(t)
$$
Here, $D^{-1/2}$ is the fractional integral (the inverse of $D^{1/2}$), defined as:
$$
D^{-1/2} h(t) = \frac{1}{\sqrt{\pi}} \int_0^t \frac{h(\tau)}{\sqrt{t-\tau}} d\tau
$$

Example with $h(t)=1$ #

If we set $h(t)=1$, we can compute both sides to confirm:

• Left-hand side: $D^{1/2}t = \frac{\Gamma(2)}{\Gamma(2-1/2)} t^{1-1/2} = \frac{2}{\sqrt{\pi}} t^{1/2}$
• Right-hand side: $t D^{1/2}1 + \frac{1}{2} D^{-1/2}1 = t \cdot \frac{1}{\sqrt{\pi}} t^{-1/2} + \frac{1}{2} \cdot \frac{2}{\sqrt{\pi}} t^{1/2} = \frac{2}{\sqrt{\pi}} t^{1/2}$

They match perfectly! That’s why the paper’s formula holds.

3. Key Takeaways for the Paper #

• Kermack & McCrea are using the Riemann-Liouville fractional derivative (standard for early work in the area).
• The formula you saw is a truncated version of the fractional Leibniz rule, simplified because higher-order derivatives of $t$ vanish.
• Fractional integrals ($D^{-\alpha}$) are just as important as fractional derivatives here—they’re the inverse operators that show up in the truncated series.

Hope this clears up the confusion around that formula and fractional derivative Leibniz rules! Let me know if you want to dive deeper into specific definitions or other parts of the paper.

内容的提问来源于stack exchange，提问作者Jorge.Squared