Reaction–diffusion dynamics in an attractive stepwise-linear potential energy curve under the Gaussian-sink Action

Abstract

The reaction–diffusion dynamics through a Gaussian sink in the presence of an attractive stepwise-linear potential energy curve has been investigated. Exact dynamics has been evaluated for perfect absorption via a centrally placed point sink. An analytical form of the mean first passage time distribution has been provided. Moreover, the stability of a Brownian particle versus the noise intensity in the considered model system has been investigated. We have introduced the Fredholm integral method to solve the Smoluchowski equation in the Laplace domain, obtaining an exact semi-analytical solution for the linear potential energy curve with a general delocalized sink. The knowledge of the Greens function of the equation in the absence of a sink is required. We discuss the result by using Gaussian sink in two different physical contexts with finding different observables like average rate constant in electronic relaxation in solution and quantum yields in photosynthetic systems or doped molecular crystal.

Appendix

Determination of $\mathcal {L}^{-1}\{\frac{e^{-\alpha \sqrt{1+ s}}}{s}\}$:

The classical inversion formula is

$$\begin{aligned} {} \mathcal {L}^{-1}[F(z)](t)=\frac{1}{2 \pi i}\int _{c-i \infty }^{c+i\infty }\mathrm{d}z F(z)e^{zt}=\frac{1}{2 \pi i}\int _{c-i \infty }^{c+i\infty }\mathrm{d}z\frac{e^{-\alpha \sqrt{1+ z}}}{z} e^{zt}\quad \alpha >0. \end{aligned}$$

(A1)

Fig. 13

Schematic diagram of the Bromwich contour

To perform the above integration, we consider the Bromwich contour in Fig. 13 which contains branch points on the negative real axis due to $\sqrt{1+z}$ term. From the residue theorem

$$\begin{aligned} \oint _{C} \mathrm{d}z\frac{e^{-\alpha \sqrt{1+ z}}}{z} e^{zt}= 2\pi i \sum \text {All residues}, \end{aligned}$$

(A2)

where C is a keyhole-type contour (Fig. 13) about the negative real axis. We have also defined $\arg z \in (-\pi ,\pi ]$ so that we could avoid the branch cut, which exists in $-1 \ge {Re}z >-\infty$. There are six parts of this contour, namely $C_j$, $j\in \{ 1,2,3,4,5,6\}$, as shown in Fig. 13. On $C_4$, $z=-1+\epsilon e^{i\theta }$, and thus, the integral vanishes for $\epsilon \rightarrow 0$. Now, consider the integral on $C_2$ and $C_6$. Here, $z=R e^{i \theta }$ ($R\rightarrow \infty$) and we get

$$\begin{aligned} \left[ \int _{C_2}+\int _{C_6}\right] \mathrm{d}z\frac{e^{-\alpha \sqrt{1+ z}}}{z} e^{zt}=\int _{\pi /2}^{\pi }f(\theta ) d\theta +\int _{-\pi }^{-\pi /2}f(\theta ) d\theta =0, \end{aligned}$$

(A3)

where $f(\theta )$ is a function of the variable $\theta$ as R is constant on these arcs. Then, we are left with the following by the Cauchy’s integral theorem with simple pole at $z=0$:

$$\begin{aligned} \left[ \int _{C_3}+\int _{C_5}+\int _{C_1}\right] \mathrm{d}z\frac{e^{-\alpha \sqrt{1+ z}}}{z} e^{zt}=2\pi ie^{-\alpha }. \end{aligned}$$

(A4)

On $C_3$, $z=x e^{i\pi }$ ($x \in (-\infty ,-1]$), and then,

$$\begin{aligned} \sqrt{1+z}=\sqrt{-e^{i\pi }+x e^{i\pi }}=i\sqrt{x-1}. \end{aligned}$$

(A5)

Therefore, we could write the corresponding integral as

$$\begin{aligned} \int _{-\infty }^{-1}\mathrm{d}x \frac{e^{-i \alpha \sqrt{x-1}-xt}}{x}. \end{aligned}$$

(A6)

Similarly, $z=x e^{-i\pi }$ ($x \in [-1,-\infty )$) for the integral on $C_5$, and the integral could be written as

$$\begin{aligned} \int _{-1}^{-\infty }\mathrm{d}x \frac{e^{i \alpha \sqrt{x-1}-xt}}{x}. \end{aligned}$$

(A7)

By considering both the integral on $C_3$ and $C_5$, we may get

$$\begin{aligned} {} \begin{aligned}&\int _{-\infty }^{-1}\mathrm{d}x \frac{e^{-i \alpha \sqrt{x-1}-xt}}{x}+\int _{-1}^{-\infty }\mathrm{d}x \frac{e^{i \alpha \sqrt{x-1}-xt}}{x}\\&\quad = 2i\int _{1}^{\infty } \mathrm{d}x \frac{e^{-x t}}{x} \sin {(\alpha \sqrt{x-1})}\\&\quad =2i\int _{0}^{\infty } \mathrm{d}p \frac{e^{-(p+1) t}}{p+1} \sin {(\alpha \sqrt{p})}. \end{aligned} \end{aligned}$$

(A8)

In the last step, we parameterize by $p=x-1$. We apply Feynman’s integral technique to find the last integral. Let’s consider

$$\begin{aligned} {} I(t)=\int _{0}^{\infty } \mathrm{d}p \frac{e^{-(p+1) t}}{p+1} \sin {(\alpha \sqrt{p})}. \end{aligned}$$

(A9)

Then,

$$\begin{aligned} \frac{\mathrm{d}I}{\mathrm{d}t}=-\int _{0}^{\infty } \mathrm{d}p e^{-(p+1) t}\sin {(\alpha \sqrt{p})} =-\frac{\sqrt{\pi }\alpha e^{-\frac{\alpha ^2}{4t}-t}}{2 t^{3/2}}. \end{aligned}$$

(A10)

Again we could solve for I(t) by integrating the above equation w.r.t the variable t, and after solving, we obtain

$$\begin{aligned} I(t)=\frac{\pi }{2}\left[ e^{-\alpha }\left( 1+ \text {Erf}\left[ \frac{\alpha }{2\sqrt{t}}-\sqrt{t}\right] \right) +e^\alpha \left( -1+ \text {Erf}\left[ \frac{\alpha }{2\sqrt{t}}+\sqrt{t}\right] \right) \right] + c(\alpha ). \end{aligned}$$

(A11)

In the above, $c(\alpha )$ is an integral constant. By calculating I(0) from Eq. (A9) and putting $t=0$ in Eq. (A11), we find that $c(\alpha )=0$. Now, the integral in Eq. (A1) could be evaluated from the integral on $C_1$ in the limit $R\rightarrow \infty$. Thus, from Eqs. (A4) and (A8), we could find the inverse Laplace transform which is

$$\begin{aligned} \begin{aligned} \mathcal {L}^{-1}[F(z)](t)&=\lim _{R\rightarrow \infty } \frac{1}{2 \pi i} \int _{c-i R}^{c+i R}\mathrm{d}z\frac{e^{-\alpha \sqrt{1+ z}}}{z} e^{zt}\\&=\frac{1}{2}e^{-\alpha }\left[ \text {Erfc}\left[ \frac{\alpha }{2\sqrt{t}}-\sqrt{t}\right] + e^{\alpha }\text {Erfc}\left[ \frac{\alpha }{2\sqrt{t}}+\sqrt{t}\right] \right] . \end{aligned} \end{aligned}$$

(A12)