# 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.

• 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

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 $$\tag{3} -2\frac{d_1}{d_2}\frac{\partial}{\partial x}\tilde{R}(d_1,d_2,\lambda)=\frac{d_1}{d_2}e^{-\sqrt{\lambda/Q}\,(d_2-d_1)}\,.$$
• 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