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To solve a fluid diffusion problem, I need to calculate an inverse Laplace transform. The integral, according to the Inverse Laplace theorem, has the form: \begin{equation} \label{i_laplace_1} p(r,t) = \frac{p_{c}}{2\pi i}\lim_{n \rightarrow \infty}\int_{1-i\beta_{n}}^{1+i\beta_{n}}\frac{e^{st}}{s}\frac{K_{0}(\sqrt{\frac{s}{\kappa}} r)}{K_{0}(\sqrt{\frac{s}{\kappa}} a)} d s \end{equation} where $p_{c},r,a$ can be seen as constants. I choose a contour as in attached Figure 1 to avoid the branch point $z=0$ during the computation. The integral is then instead calculated on the contour including $C_{R},C_{\rho},\Gamma_{1},\Gamma_{2}$. On $C_{\rho}$, in several textbooks and papers I saw an approximation as followed: \begin{equation} \label{i_laplace_form} \lim_{\rho \rightarrow 0}\int_{C_{\rho}} \frac{e^{st}}{s}\frac{K_{0}(\sqrt{\frac{s}{\kappa}} r)}{K_{0}(\sqrt{\frac{s}{\kappa}} a)} ds \approx \lim_{\rho \rightarrow 0}\int_{C_{\rho}} \frac{e^{st}}{s} d s = 2\pi i \end{equation} However, I still do not fully understand how good is such an approximation and whether it is mathematically rigorous. Can you give me some insightful discussion?

Hoang Nguyen
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  • Have you tried writing out the bessel functions in their Laurant series around $s=0$? – Alex R. Aug 08 '18 at 17:14
  • Hi Alex, I did. As a real-valued function, $K_0(x)$ converges to $-\ln(x)$, which blows up as $x \rightarrow 0$. However, in the complex plane, such a function is multi-valued, and even though defined in the principal branch cut, z = 0 is still a branch point, so I doubt that an asymptotic behavior of $K_0$ around that point exists. – Hoang Nguyen Aug 08 '18 at 17:36
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    Have a look at this: https://math.stackexchange.com/questions/480235/inverse-laplace-transform-of-bar-p-d-frack-0-sqrts-r-dsk-0-sqrts/481946#481946 – Ron Gordon Aug 08 '18 at 17:52
  • Thanks for the answer, Ron. I have a question related to the integral on $C_4$. Can you tell me more details how did you compute this, and is this an exact or approximate solution?

    My concern in this step is, as a real-valued function, $K_0(x)$ converges to $−ln(x)$, which blows up as $x \rightarrow 0$. However, in the complex plane, such a function is multi-valued, and even though defined in a branch cut, $z = 0$ is still a branch point, so I doubt that an asymptotic behavior of $K_0$ around that point exists.

     – Hoang Nguyen Aug 08 '18 at 19:03
  • @HoangNguyen: on $C_4$, there is a ratio of two logs of very small numbers which approaches one. (Recall that we are integrating a ratio of Bessels.) The branch cut is defined by the given contour. – Ron Gordon Aug 08 '18 at 19:09

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