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

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!

• Indeed, the command of Mathematica 13 LaplaceTransform[ d1/d2/Sqrt[4*PiQt]/t*(d2 - d1)*Exp[-(d2 - d1)^2/t/4/Q], t, s] results in $$\frac{\text{d1} (\text{d2}-\text{d1}) \sqrt{\frac{Q}{(\text{d1}-\text{d2})^2}} e^{-\frac{1}{\sqrt{\frac{Q}{s (\text{d1}-\text{d2})^2}}}}}{\text{d2} \sqrt{Q}} .$$ I leave its simplification on your own. Mar 16 at 8:11

## 1 Answer

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.

• This could be a matter of my lacking in Laplace transform, but I am unable to understand how you derived the Laplace transform in the last equation. Mar 17 at 4:58
• If you accept equations (1) and (2) I think that (3) is trivial. Isn't it ? Mar 17 at 9:48
• I am afraid, but I can't follow. Mar 21 at 2:47
• If you accept (1) it follows from (2) that you have to take the partial derivative of (1) and multiply it by $-2\frac{d_1}{d_2}$. This is the Laplace transform you wanted to calculate. Mar 21 at 9:27