积分变换

性质1 (拉普拉斯变换与梅林变换的关系)

$$ \begin{equation} \mathcal{M}\{\mathscr{L}[f(t)](s)\}(1-p) = \Gamma(1-p)\mathcal{M}[f(t)](p). \end{equation} $$

Proof.

$$ \begin{equation} \begin{aligned} \mathcal{M}[\mathscr{L}[f(t)](s)](1-p) =&\int_{0}^{\infty}s^{-p}\int_0^{\infty}\mathrm{e}^{-st}f(t)\mathrm{d}t\mathrm{d}s \\ =&\int_0^{\infty}f(t)\int_0^{\infty}s^{-p}\mathrm{e}^{-st}\mathrm{d}s\mathrm{d}t \\ =&\int_{0}^{\infty}f(t)\int_{0}^{\infty}\left(\frac{z}{t}\right)^{-p}\mathrm{e}^{-z}\frac{\mathrm{d}z}{t}\mathrm{d}t \qquad z=st \\ =&\Gamma(1-p)\int_0^{\infty}t^{p-1}f(t)\mathrm{d}t \\ =&\Gamma(1-p)\mathcal{M}[f(t)](p). \end{aligned} \end{equation} $$

性质2 (卷积的拉普拉斯变换) 函数 $f(t)$ 和 $g(t)$ 的卷积 $(f \ast g)(t)=\displaystyle\int_0^tf(t-\tau)g(\tau)\mathrm{d}\tau$ 的拉普拉斯变换

$$ \begin{equation} \mathscr{L}[f\ast g](s)=\mathscr{L}[f](s)\mathscr{L}[g](s). \end{equation} $$

Proof.

$$ \begin{equation} \begin{aligned} \mathscr{L}[f\ast g](s) =&\int_{0}^{\infty}\mathrm{e}^{-st}\int_0^tf(t-\tau)g(\tau)\mathrm{d}\tau\mathrm{d}t \\ =&\int_0^{\infty}g(\tau)\int_{\tau}^{\infty}\mathrm{e}^{-st}f(t-\tau)\mathrm{d}t\mathrm{d}\tau \\ =&\int_0^{\infty}g(\tau)\int_0^{\infty}\mathrm{e}^{-s(u+\tau)}f(u)\mathrm{d}u\mathrm{d}\tau \qquad u=t-\tau \\ =&\left(\int_0^{\infty}\mathrm{e}^{-s\tau}g(\tau)\mathrm{d}\tau\right)\left(\int_0^{\infty}\mathrm{e}^{-su}f(u)\mathrm{d}u\right) \\ =&\mathscr{L}[f](s)\mathscr{L}[g](s). \end{aligned} \end{equation} $$

性质3 (导函数的拉普拉斯变换) 函数 $f(t)$ 的 $n$ 阶导函数 $f^{(n)}(t)$ 的拉普拉斯变换为

$$ \begin{equation} \mathscr{L}[f^{(n)}](s) = s^n\mathscr{L}[f](s) -\sum_{k=0}^{n-1}s^{n-1-k}f^{(k)}(0). \end{equation} $$

Proof.

$$ \begin{equation} \begin{aligned} \mathscr{L}[f^{(n)}](s) =&\int_0^{\infty}\mathrm{e}^{-st}f^{(n)}(t)\mathrm{d}t \\ =&s\int_0^{\infty}\mathrm{e}^{-st}f^{(n-1)}(t)\mathrm{d}t - f^{(n-1)}(0) \\ =&s^n\int_0^{\infty}\mathrm{e}^{-st}f(t)\mathrm{d}t-\sum_{k=0}^{n-1}s^{n-1-k}f^{(k)}(0) \\ =&s^n\mathscr{L}[f](s) - \sum_{k=0}^{n-1}s^{n-1-k}f^{(k)}(0). \end{aligned} \end{equation} $$

性质4 (导函数的傅里叶变换) 速降函数 $f(x)$ 的 $n$ 阶导函数 $f^{(n)}(x)$ 的傅里叶变换为

$$ \begin{equation} \mathscr{F}[f^{(n)}](k) = (-\mathrm{i}k)^n\mathscr{F}[f](k). \end{equation} $$

Proof.

$$ \begin{equation} \begin{aligned} \mathscr{F}[f^{(n)}](k) =&\int_{\mathbb{R}}\mathrm{e}^{\mathrm{i}kx}f^{(n)}(x)\mathrm{d}x\\ =&-\mathrm{i}k\int_{\mathbb{R}}\mathrm{e}^{\mathrm{i}kx}f^{(n-1)}(x)\mathrm{d}x \\ =&(-\mathrm{i}k)^n\int_{\mathbb{R}}\mathrm{e}^{\mathrm{i}kx}f(x)\mathrm{d}x \\ =&(-\mathrm{i}k)^n\mathscr{F}[f](k). \end{aligned} \end{equation} $$

性质5 (傅里叶变换及其逆变换的平移性质)

$$ \begin{equation} \mathscr{F}[f(x-x_0)](k) = \mathrm{e}^{\mathrm{i}kx_0}\mathscr{F}[f](k), \end{equation} $$

$$ \begin{equation} \mathscr{F}^{-1}[\mathrm{e}^{\mathrm{i}kx_0}\mathscr{F}[f](k)](x) = f(x-x_0). \end{equation} $$

Proof.

$$ \begin{equation} \begin{aligned} \mathscr{F}[f(x-x_0)](k) =&\int_{\mathbb{R}}\mathrm{e}^{\mathrm{i}kx}f(x-x_0)\mathrm{d}x \\ =&\int_{\mathbb{R}}\mathrm{e}^{\mathrm{i}k(x'+x_0)}f(x')\mathrm{d}x' \\ =&\mathrm{e}^{\mathrm{i}kx_0}\mathscr{F}[f](k). \end{aligned} \end{equation} $$

例子 (幂函数的拉普拉斯变换) 设幂函数 $f(t)=t^{\alpha}$, 其中 $\alpha>-1$, 则

$$ \begin{equation} \tilde{f}(s)= \int_0^{\infty}\mathrm{e}^{-st}t^{\alpha}\mathrm{d}t=\int_0^{\infty}\mathrm{e}^{-z}\left(\frac{z}{s}\right)^{\alpha}\frac{\mathrm{d}z}{s} = \frac{\Gamma(1+\alpha)}{s^{1+\alpha}}. \end{equation} $$

例子 ($\mathrm{e}^{-a\vert x\vert}$ 的傅里叶变换) 设 $f(x)=\mathrm{e}^{-a\vert x\vert}$, 其中 $a>0$, 则

$$ \begin{equation} \begin{aligned} \mathscr{F}[f](k) &= \int_{\mathbb{R}}\mathrm{e}^{\mathrm{i}kx}\mathrm{e}^{-a\vert x\vert }\mathrm{d}x \\ =& \int_{\mathbb{R}}(\cos{(kx)}+\mathrm{i}\sin{(kx)})\mathrm{e}^{-a\vert x\vert}\mathrm{d}x \\ =& 2\int_0^{\infty}\cos{(kx)}\mathrm{e}^{-ax}\mathrm{d}x, \end{aligned} \end{equation} $$

而

$$ \begin{equation} \begin{aligned} &\int_0^{\infty}\cos{(kx)}\mathrm{e}^{-ax}\mathrm{d}x \\ =& \frac{1}{k}\int_0^{\infty}\mathrm{e}^{-ax}\mathrm{d}\sin{(kx)} \\ =&\frac{1}{k}\left[\mathrm{e}^{-ax}\sin{(kx)}\left.\right\vert_{x=0}^{x=\infty}+a\int_0^{\infty}\mathrm{e}^{-ax}\sin{(kx)}\mathrm{d}x\right] \\ =&\frac{a}{k}\int_0^{\infty}\mathrm{e}^{-ax}\sin{(kx)}\mathrm{d}x \\ =&\frac{a}{k^2} \left[-\mathrm{e}^{-ax}\cos{(kx)}\left.\right\vert_{x=0}^{x=\infty}-a\int_0^{\infty}\mathrm{e}^{-ax}\cos{(kx)}\mathrm{d}x\right] \\ =&\frac{a}{k^2}-\frac{a^2}{k^2}\int_0^{\infty}\cos{(kx)}\mathrm{e}^{-ax}\mathrm{d}x, \end{aligned} \end{equation} $$

于是

$$ \begin{equation} \int_0^{\infty}\cos{(kx)}\mathrm{e}^{-ax}\mathrm{d}x = \frac{a}{k^2+a^2}. \end{equation} $$

所以

$$ \begin{equation} \mathscr{F}[f](k) =\frac{2a}{k^2+a^2}. \end{equation} $$

H 函数

例子 设函数 $f(t)$ 的拉普拉斯变换 $\tilde{f}(s)=s^{\mu}\mathrm{e}^{-as^{\nu}}$, 其中 $a>0$, 求 $f(t)$.
首先利用梅林变换和拉普拉斯变换的关系, 可以得到

$$ \begin{equation} \Gamma(1-p)\mathcal{M}[f](p) = \int_0^{\infty}s^{-p}s^{\mu}\mathrm{e}^{-as^{\nu}}\mathrm{d}s. \end{equation} $$

对上式积分进行变量替换 $z = as^{\nu}$, $s = (z/a)^{1/\nu}$, $\mathrm{d}s = (z/a)^{1/\nu-1}/(a\nu)\mathrm{d}z$ 可得

$$ \begin{equation} \begin{aligned} \mathcal{M}[f](p) &= \frac{1}{\Gamma(1-p)}\int_0^{\infty}\frac{1}{a\nu}\left(\frac{z}{a}\right)^{\frac{\mu-p}{\nu}}\left(\frac{z}{a}\right)^{\frac{1}{\nu}-1}\mathrm{e}^{-z}\mathrm{d}z \\ &=\frac{1}{\Gamma(1-p)}\frac{1}{\nu a^{\frac{\mu+1-p}{\nu}}}\int_0^{\infty}z^{\frac{\mu+1-p}{\nu}-1}\mathrm{e}^{-z}\mathrm{d}z \\ &=\frac{1}{\nu a^{\frac{\mu+1-p}{\nu}}}\frac{\Gamma(\frac{\mu+1-p}{\nu})}{\Gamma(1-p)}. \end{aligned} \end{equation} $$

所以

$$ \begin{equation} \begin{aligned} f(t) &= \frac{1}{\nu a^{\frac{\mu+1}{\nu}}}\frac{1}{2\pi \mathrm{i}}\int_{c-\mathrm{i}\infty}^{c+\mathrm{i}\infty}\frac{\Gamma(\frac{\mu+1-s}{\nu})}{\Gamma(1-s)}\left(\frac{t}{a^{\frac{1}{\nu}}}\right)^{-s}\mathrm{d}s \\ &=\frac{1}{\nu a^{\frac{\mu+1}{\nu}}}\frac{1}{2\pi \mathrm{i}}\int_{c-\mathrm{i}\infty}^{c+\mathrm{i}\infty}\frac{\Gamma(\frac{\mu+1+s}{\nu})}{\Gamma(1+s)}\left(\frac{a^{\frac{1}{\nu}}}{t}\right)^{-s}\mathrm{d}s \\ &= \frac{1}{\nu a^{\frac{\mu+1}{\nu}}}H_{11}^{10}\left[\left.\frac{a^{\frac{1}{\nu}}}{t}\right\vert_{(\frac{\mu+1}{\nu}, \frac{1}{\nu})}^{(1,1)} \right]. \end{aligned} \end{equation} $$

进一步, 可利用留数定理对上述积分进行展开. 对于伽马函数 $\Gamma(z)$ 而言, $\Gamma(n)=\infty$, $n=0$, $-1$, $-2$, $\cdots$, 并且奇点只有一阶极点 $\{0, -1, -2, \cdots\}$, 所以上述积分的被积函数的奇点, 即一阶极点为 $(\mu+1+s_n)/\nu = -n$, $s_n = -1 - \mu - \nu n$, $n=0$, $1$, $2$, $\cdots$, 于是留数

$$ \begin{equation} \begin{aligned} \textnormal{Res}(s_n) &= \lim_{s\to s_n}(s-s_n)\frac{\Gamma\left(\frac{\mu+1+s}{\nu}\right)}{\Gamma(1+s)}\left(\frac{a^{\frac{1}{\nu}}}{t}\right)^{-s} \\ &= \frac{1}{\Gamma(-\mu-\nu n)}\left(\frac{a^{\frac{1}{\nu}}}{t}\right)^{1+\mu+\nu n} \lim_{s\to s_n}(s-s_n)\Gamma\left(\frac{\mu+1+s}{\nu}\right) \\ &= \frac{\nu}{n!\Gamma(-\mu-\nu n)}\left(\frac{a^{\frac{1}{\nu}}}{t}\right)^{1+\mu}\left(-\frac{a}{t^{\nu}}\right)^{n}, \end{aligned} \end{equation} $$

这是因为

$$ \begin{equation} \begin{aligned} &\lim_{s\to s_n}(s-s_n)\Gamma\left(\frac{\mu+1+s}{\nu}\right)\\ =&\lim_{s\to -1-\mu - \nu n}(s+1+\mu+\nu n)\Gamma\left(\frac{\mu+1+s}{\nu}\right) \\ =& \lim_{s\to -1-\mu-\nu n}\frac{(s+1+\mu+\nu n)\Gamma\left(\frac{\mu+1+s}{\nu}+1\right)}{\frac{\mu+1+s}{\nu}}\\ =& \lim_{s\to -1-\mu-\nu n}\frac{(s+1+\mu+\nu n)\Gamma\left(\frac{\mu+1+s}{\nu}+n+1\right)}{\frac{\mu+1+s}{\nu}\cdots\left(\frac{\mu+1+s}{\nu}+n\right)} \\ =& \lim_{s\to -1-\mu-\nu n}\frac{\Gamma\left(\frac{\mu+1+s}{\nu}+n+1\right)}{\frac{\mu+1+s}{\nu}\cdots\left(\frac{\mu+1+s}{\nu}+n-1\right)}\nu \\ =&\frac{\nu}{(-n)\cdots (-1)} \\ =&\frac{(-1)^n\nu}{n!}. \end{aligned} \end{equation} $$

所以 $f(t)$ 还可以表示为下述形式

$$ \begin{equation} f(t) = \frac{1}{t^{1+\mu}}\sum_{n=0}^{\infty}\frac{1}{n!\Gamma(-\mu-\nu n)}\left(-\frac{a}{t^{\nu}}\right)^{n}. \end{equation} $$

于是, 若定义 Mainardi 函数如下,

$$ \begin{equation} f_{\mu,\nu}(t;a)=\frac{1}{t^{1+\mu}}\sum_{n=0}^{\infty}\frac{1}{n!\Gamma(-\mu-\nu n)}\left(-\frac{a}{t^{\nu}}\right)^{n}, \end{equation} $$

那么

$$ \begin{equation} \mathscr{L}^{-1}\{s^{\mu}\mathrm{e}^{-as^{\nu}}\}(t)=f_{\mu,\nu}(t; a). \end{equation} $$

分数阶积分和导数

性质6 (Caputo 导数的拉普拉斯变换) 由上述卷积和导函数的拉普拉斯变换, $({}_0^CD_t^{\alpha}f)(t)$ 的拉普拉斯变换为

$$ \begin{equation} \mathscr{L}[({}_0^CD_t^{\alpha}f)(t)](s) = s^{\alpha}\mathscr{L}[f](s) - \sum_{k=0}^{n-1}s^{\alpha-1-k}f^{(k)}(0). \end{equation} $$

从属过程(subodinator)

性质7 (拉普拉斯指数) 设 $T(t)$ 是一个$ \alpha$-稳定的从属过程, 则其拉普拉斯指数为 $-s^{\alpha}$, 即

$$ \begin{equation} \tilde{g}(s,t) = \mathbb{E}\mathrm{e}^{-sT(t)}=\int_0^{\infty}\mathrm{e}^{-sT}g(x,t)\mathrm{d}T = \mathrm{e}^{-ts^{\alpha}}, \end{equation} $$

其中 $g(x,t)$ 是 $T(t)$ 的概率密度函数.
性质8 (scaling property) 设 $T(t)$ 是一个 $\alpha$-稳定的从属过程, 则

$$ \begin{equation} T(t) \overset{\mathrm{d}}{=} t^{1/\alpha}T(1). \end{equation} $$

Proof.

$$ \begin{equation} \mathbb{E}\mathrm{e}^{-sT(t)}=\mathrm{e}^{-ts^{\alpha}}=\mathrm{e}^{-1\dot (t^{1/\alpha}s)^{\alpha}} = \mathbb{E}\mathrm{e}^{-(t^{1/\alpha}s)T(1)}. \end{equation} $$

由上述性质, 可以得到

$$ \begin{equation} \begin{aligned} g(x,t) &= \partial_x\int_{0}^xg(u,t)\mathrm{d}u \\ &= \partial_x \mathbb{P}(T(t)\leqslant x) \\ &= \partial_x \mathbb{P}(t^{1/\alpha}T(1)\leqslant x)\\ &= \partial_x \mathbb{P}(T(1)\leqslant t^{-1/\alpha}x) \\ &= \partial_x \int_0^{t^{-1/\alpha}x}g_{\alpha}(u)\mathrm{d}u \\ &= t^{-1/\alpha}g_{\alpha}(t^{-1/\alpha}x), \end{aligned} \end{equation} $$

其中 $g_{\alpha}(x)$ 为 $T(1)$ 的概率密度函数, 其拉普拉斯变换为 $\tilde{g}_{\alpha}(s)=\mathrm{e}^{-s^{\alpha}}$. 进一步,

$$ \begin{equation} g_{\alpha}(x) =\frac{1}{\alpha} H\left[\frac{1}{x}\left.\right\vert_{(\frac{1}{\alpha}, \frac{1}{\alpha})}^{(1,1)} \right]. \end{equation} $$

逆从属过程

设 $h(u,t)$ 是 $E(t)$ 的概率密度函数, 则根据定义, 有

$$ \begin{equation} \begin{aligned} h(u,t) &=\partial_u \mathbb{P}(E(t)\leqslant u) \\ &=\partial_u \mathbb{P}(T(u)\geqslant t) \\ &=-\partial_u \int_0^t g(x,u)\mathrm{d}x \\ &=-\partial_u \int_0^t u^{-1/\alpha}g_{\alpha}(u^{-1/\alpha}x)\mathrm{d}x \\ &=-\partial_u\int_0^{u^{-1/\alpha}t}g_{\alpha}(y)\mathrm{d}y \qquad y=u^{-1/\alpha}x \\ &= \frac{t}{\alpha}u^{-1-1/\alpha}g_{\alpha}(u^{-1/\alpha}t) \\ &=\frac{t}{\alpha^2}u^{-1-1/\alpha}H\left[\frac{u^{1/\alpha}}{t}\left.\right\vert_{(\frac{1}{\alpha}, \frac{1}{\alpha})}^{(1,1)} \right], \end{aligned} \end{equation} $$

$$ \begin{equation} \tilde{h}(u,s) = -\partial_u \frac{\tilde{g}(s,u)}{s}=-\partial_u s^{-1}\mathrm{e}^{-us^{\alpha}}=s^{\alpha-1}\mathrm{e}^{-us^{\alpha}}. \end{equation} $$

欠扩散方程

$$ \begin{equation}\tag{EQ}\label{eq} \partial_t^{\alpha}P(x,t) = D\partial_x^2P(x,t),\qquad P(x,0)=\delta(x-x_0). \end{equation} $$

其中 $\partial_t^{\alpha}={}_0D_t^{\alpha}$ 是 Caputo 导数, $0<\alpha<1$, $D$ 为扩散系数.

对方程两边做拉普拉斯-傅里叶变换, 得

$$ \begin{equation} s^{\alpha}\hat{\tilde{P}}(k,s)-\mathrm{e}^{\mathrm{i}kx_0}s^{\alpha-1} = -Dk^2\hat{\tilde{P}}(k,s), \end{equation} $$

即

$$ \begin{equation} \hat{\tilde{P}}(k,s)= \frac{\mathrm{e}^{\mathrm{i}kx_0}s^{\alpha-1}}{s^{\alpha}+Dk^2}. \end{equation} $$

根据上面的例子, 有

$$ \begin{equation} \tilde{P}(x,s) = \frac{1}{2\sqrt{D}s^{1-\alpha/2}}\mathrm{e}^{-\frac{s^{\alpha/2}}{\sqrt{D}}\vert x-x_0\vert}, \end{equation} $$

$$ \begin{equation} \begin{aligned} P(x,t) &= \frac{1}{\alpha\vert x-x_0\vert}H_{11}^{10}\left[\left.{\left(\frac{\vert x-x_0\vert}{\sqrt{D}}\right)^{2/\alpha}} {t^{-1}}\right\vert_{(1,\frac{2}{\alpha})}^{(1,1)}\right] \\ &=\frac{1}{2\sqrt{D}}f_{-1+\alpha/2, \alpha/2}\left(t;\frac{|x-x_0|}{\sqrt{D}} \right). \end{aligned} \end{equation} $$

$\alpha=1$ 时, 我们有

$$ \begin{equation} \begin{aligned} P(x,t) &= \frac{1}{2\sqrt{D}}f_{-1/2, 1/2}\left(t;\frac{|x-x_0|}{\sqrt{D}} \right)\\ &=\frac{1}{2\sqrt{D}}\frac{1}{t^{1/2}}\sum_{n=0}^{\infty}\frac{1}{n!\Gamma(1/2-(1/2)n)}\left(-\frac{|x-x_0|}{\sqrt{D}}\frac{1}{t^{1/2}}\right)^n \\ &=\frac{1}{\sqrt{4Dt}}\sum_{n=0}^{\infty}\frac{1}{n!\Gamma((1-n)/2)}\left(-\frac{|x-x_0|}{\sqrt{Dt}}\right)^n. \end{aligned} \end{equation} $$

因为当 $n$ 为奇数时, $1/\Gamma((1-n)/2)=0$, 所以

$$ \begin{equation} \begin{aligned} P(x,t)&=\frac{1}{\sqrt{4Dt}}\sum_{n=0}^{\infty} \frac{1}{(2n)!\Gamma((1-2n)/2)}\left(-\frac{|x-x_0|}{\sqrt{Dt}}\right)^{2n}\\ &=\frac{1}{\sqrt{4Dt}}\sum_{n=0}^{\infty} \frac{1}{(2n)!\Gamma(1/2-n)}\left(\frac{|x-x_0|^2}{{Dt}}\right)^{n}. \end{aligned} \end{equation} $$

又因为

$$ \begin{equation} \begin{aligned} &\Gamma(1/2-n)\\ =&\frac{\Gamma{(1/2-n+1)}}{1/2-n} \\ =& \frac{\Gamma{(1/2-n+n)}}{(1/2-n)(1/2-n+1)\cdots(1/2-n+n-1)} \\ =& \frac{\Gamma(1/2)(-2)^n}{1\times 3\times\cdots\times(2n-1)} \\ =& \frac{\sqrt{\pi}(-2)^n}{(2n-1)!!}, \end{aligned} \end{equation} $$

所以

$$ \begin{equation} \begin{aligned} \frac{1}{(2n)!\Gamma{(1-n/2)}} =&\frac{(2n-1)!!}{\sqrt{\pi}(2n)!(-2)^n}\\ =&\frac{1}{\sqrt{\pi}(2n)!!(-2)^n}\\ =&\frac{1}{\sqrt{\pi}(n!2^n)(-2)^n}\\ =&\frac{1}{\sqrt{\pi}n!(-4)^n}, \end{aligned} \end{equation} $$

于是

$$ \begin{equation} \begin{aligned} P(x,t)&=\frac{1}{\sqrt{4Dt}}\sum_{n=0}^{\infty}\frac{1}{\sqrt{\pi}n!(-4)^n}\left(\frac{|x-x_0|^2}{Dt}\right)^n\\ &=\frac{1}{\sqrt{4\pi Dt}}\sum_{n=0}^{\infty}\frac{1}{n!}\left(-\frac{|x-x_0|^2}{4Dt}\right)^n \\ &=\frac{1}{\sqrt{4\pi Dt}}\mathrm{e}^{-\frac{(x-x_0)^2}{4Dt}}. \end{aligned} \end{equation} $$

复合布朗运动

假设方程 ($\ref{eq}$) 的解 $P(x,t)$ 具有如下形式

$$ \begin{equation}\tag{C}\label{c} P(x,t) = \int_0^{\infty}n(s,t)P_0(x,s)\mathrm{d}x, \end{equation} $$

其中 $P_0(x,t)$ 是扩散方程

$$ \begin{equation} \partial_t P_0(x,t) = D\partial_x^2P_0(x,t), \qquad P_0(x,0)=\delta(x-x_0), \end{equation} $$

的解.
下面说明 $n(s,t)$ 就是前面所说的逆从属过程的概率密度函数, 即 $h$. 首先对 ($\ref{c}$) 式两边对 $x$ 积分, 然后进行拉普拉斯变换 $t\to u$, 得

$$ \begin{equation} \int_{\mathbb{R}}P(x,t)\mathrm{d}x = \int_0^{\infty}n(s,t)\int_{\mathbb{R}}P_0(x,s)\mathrm{d}x\mathrm{d}s, \end{equation} $$

$$ \begin{equation} 1 = \int_0^{\infty}n(s,t)\mathrm{d}s, \end{equation} $$

$$ \begin{equation} \frac{1}{u} = \int_0^{\infty}\tilde{n}(s,u)\mathrm{d}s. \end{equation} $$

对方程 ($\ref{eq}$) 进行拉普拉斯变换, 得

$$ \begin{equation} u\tilde{P}(x,u)-\delta(x-x_0)=Du^{1-\alpha}\partial_x^2\tilde{P}(x,u), \end{equation} $$

$$ \begin{equation} \begin{aligned} &u\int_{0}^{\infty}\tilde{n}(s,u)P_0(x,s)\mathrm{d}s - \delta(x-x_0)\\ =&Du^{1-\alpha}\int_0^{\infty}\tilde{n}(s,u)\partial_x^2P_0(x,s)\mathrm{d}s\\ =&u^{1-\alpha}\int_0^{\infty}\tilde{n}(s,u)\partial_s P_0(x,s)\mathrm{d}s \\ =&u^{1-\alpha}\left(\tilde{n}(s, u)P_0(x,s)|_{s=0}^{s=\infty}-\int_{0}^{\infty}\partial_s\tilde{n}(s,u)P_0(x,s)\mathrm{d}s\right), \end{aligned} \end{equation} $$

其中, $\tilde{n}(\infty,u)=0$, $P_0(x,\infty)$ 是经典热扩散方程的稳态解, 即 Boltzmann 分布. 所以上式就可以写作

$$ \begin{equation} \begin{aligned} &\int_0^{\infty}\left(u\tilde{n}(s,u)+u^{1-\alpha}\partial_s\tilde{n}(s,u)\right)P_0(x,s)\mathrm{d}s\\ =& \left(1-u^{1-\alpha}\tilde{n}(0,u)\right)\delta(x-x_0), \end{aligned} \end{equation} $$

要想这个式子成立, 只能是两边都为零, 即

$$ \begin{equation} \partial_s\tilde{n}(s,u) = -u^{\alpha}\tilde{n}(s,u), \end{equation} $$

$$ \begin{equation} \tilde{n}(0,u)=u^{\alpha-1}, \end{equation} $$

解之得

$$ \begin{equation} \tilde{n}(s,u) = u^{\alpha-1}\mathrm{e}^{-u^{\alpha}s}. \end{equation} $$

这个和 $\tilde{h}$ 是一样的, 所以说 $n$ 是逆从属过程的概率密度, 即欠扩散方程 ($\ref{eq}$) 是复合布朗运动 $B(E(t))$ 所对应的扩散方程.