Laplace transform of $f(t)= \frac{d_1}{d_2}\frac{1}{\sqrt{4 \pi Qt}}\frac{d_2-d_1}{t}\exp\left(-\frac{(d_2-d_1)^2}{4Qt}\right)$?

Question

I came across the following Laplace transform of $f(t)$ in a journal article:

$$f(t)= \frac{d_1}{d_2}\frac{1}{\sqrt{4 \pi Qt}}\frac{d_2-d_1}{t}\exp\left(-\frac{(d_2-d_1)^2}{4Qt}\right).$$

The solution provided in the article is

$$\mathcal{L}\left\{ f(t) \right\}=\frac{d_1}{d_2}\exp\left(-\frac{d_2 -d_1}{\sqrt{Q}} \sqrt{s}\right).$$

I am not able to figure out how they arrived at the above solution. My Laplace transform skills aren't that great!

I'd appreciate it if someone could make me understand how this solution has arrived.

Thanks in advance!

Answer 1

Writing $$ p_t(x,y)=\frac{1}{\sqrt{4\pi t}}\exp\Big(-\frac{(x-y)^2}{4t}\Big) $$ for the heat kernel in one dimension we can use a known result. Namely that the resolvent $$ R(x,y,\lambda)=\int_0^\infty e^{-\lambda t}p_t(x,y)\,dt $$ equals $$ R(x,y,\lambda)=\frac{e^{-\sqrt{\lambda}|x-y|}}{2\sqrt{\lambda}} $$ (I think this is found for example in [1]). A bit more general: \begin{align} \tilde{R}(x,y,\lambda)&:=Q\int_0^\infty e^{-\lambda t}p_{Qt}(x,y)\,dt=\int_0^\infty e^{-(\lambda/Q)\,u}p_{u}(x,y)\,du\\[3mm] &=R(x,y,\lambda/Q)=\frac{e^{-\sqrt{\lambda/Q}\,|x-y|}}{2\sqrt{\lambda/Q}}\,.\tag{1} \end{align}

The function $f(t)$ in OP is seen to be $$\tag{2} f(t)=-2\frac{d_1}{d_2}Q\frac{\partial}{\partial x}p_{Qt}(d_1,d_2). $$ Therefore, its Laplace transform equals $$\tag{3} -2\frac{d_1}{d_2}\frac{\partial}{\partial x}\tilde{R}(d_1,d_2,\lambda)=\frac{d_1}{d_2}e^{-\sqrt{\lambda/Q}\,(d_2-d_1)}\,. $$

[1] R.L. Schilling, L. Partzsch, Brownian Motion. An Introduction to Stochastic Processes. de Gruyter Graduate, Berlin 2012.