I am studying a new formula that extracts solution to the diffusion-Smoluchowski equation and is rooted on the theory of complex calculus. Namely, the formula looks like
\begin{equation}
P(t) = \mathcal{N} \oint_{\text{Br}} J \cfrac{{\rm e}^{\tfrac{Ht}{4D}} {\rm e}^{-\tfrac{2 \Delta V + 2 \int_{x_0}^{x_1} \sqrt{H + V'(x)^2} \mathop{dx}}{4D}}}{\sqrt{\sqrt{H+ V'(x_0)^2} \sqrt{H + V'(x_1)^2}}} \cfrac{\mathop{dH}}{4D}.
\end{equation}
where our $H$ is the Laplace parameter, $\mathcal{N}$ is a normalisation constant, $J$ is a Jacobian, $V(x)$ is the potential, $\Delta V = V(x_1) - V(x_0)$ and we are doing a Bromwich integral. I have done it for the zero potential and linear potential case. However, I am struggling with the quadratic case $V(x) = \tfrac{1}{2} x^2$ (Ornstein-Uhlenbech solution). It is analytically tractable for quadratic potentials but I would need two cuts and this is what is confusing me. I will show what I have done for the other cases, and then any guidance to extend this to quadratic potentials would help me out. For instance,
consider when the potential is zero (typical diffusion with no mechanical forcing):
At first, let’s take the potential $V(x) = 0$, i.e.,\ assume there is no external force. In this case, one expects to recover the fundamental solution to the heat equation, and indeed we do. So,
\begin{equation}
P_{V = 0}(t) = \oint_{\text{Br}} \cfrac{{\rm e}^{\tfrac{Ht}{4D}} {\rm e}^{-\tfrac{2 \sqrt{H}(x_1 - x_0) }{4D}}}{4D\sqrt{H}} \mathop{dH}.
\end{equation}
The normalisation constant is unity for this case. I approach solving this integral by using complex analysis and the theory of branch points.
Let \begin{equation} \oint_{\Gamma} \cfrac{{\rm e}^{\tfrac{zt}{4D}} {\rm e}^{-\tfrac{2 \sqrt{z}(x_1 - x_0) }{4D}}}{4D\sqrt{z}} \mathop{dz} = \oint_{\Gamma} f(z) \mathop{dz} \hspace{2mm} \text{where} \hspace{2mm} f(z) = \cfrac{{\rm e}^{\tfrac{zt}{4D}} {\rm e}^{-\tfrac{2 \sqrt{z}(x_1 - x_0) }{4D}}}{4D\sqrt{z}}. \end{equation}
This has a branch point at $z = 0$ qnd I choose the branch cut to span to negative infinity (principle branch). We use the keyhole contour. It has 6 components. $\Gamma_1$ spans the imaginary axis and is shifted to the right of the point at zero. $\Gamma_2, \Gamma_4$ are semi-circles with radius $R$, $\Gamma_4$ is a circle of radius $\epsilon$ about the branch point and $\Gamma_3, \Gamma_5$ are the lines just above/below the cut from $-\infty$ to 0 and $0$ to $-\infty$ respectively. It is straightforward to show that the arced segments contribute nothing to the overall integral in the limiting process $R \rightarrow \infty, \epsilon \rightarrow 0$. The semicircular arcs vanish by Jordan’s lemma and then the contribution from $\Gamma_4$ vanishes by using the parameterisation $z = \epsilon \exp(i \theta)$. By Cauchy's residue theorem, we know that \begin{equation} \left[\oint_{\Gamma_1} + \oint_{\Gamma_3} + \oint_{\Gamma_5} \right] f(z) \mathop{dz} = 0 \end{equation} since the branch point is excluded from the interior of the contour. The integral we want to compute is $\cfrac{1}{2 \pi i} \oint_{\Gamma_1} f(z) \mathop{dz}$. On $\Gamma_3$, let’s take $z = p {\rm e}^{i \pi} = - p$. Then, $\mathop{dz} = - \mathop{dp}$ and $\sqrt{z} = i \sqrt{p}$. Since the integral is from $- \infty$ to 0 on this segment, we have \begin{equation} \oint_{\Gamma_3} f = - \int_{\infty}^{0} \cfrac{{\rm e}^{-\tfrac{pt}{4D}} {\rm e}^{-\tfrac{i\sqrt{p}(x_1 - x_0) }{2D}}}{4Di\sqrt{p}} \mathop{dp} = \int_{0}^{\infty} \cfrac{{\rm e}^{-\tfrac{pt}{4D}} {\rm e}^{-\tfrac{i\sqrt{p}(x_1 - x_0) }{2D}}}{4Di\sqrt{p}} \mathop{dp}. \end{equation} Similarly, on $\Gamma_5$ lets follow a similar parameterisation and take $z = p {\rm e}^{- i \pi}$. Note, the minus sign in the exponent represents moving below the branch cut along the line. Now, $\sqrt{z} = - i \sqrt{p}$. Since the integral is from 0 to $- \infty$ on this segment, we have \begin{equation} \oint_{\Gamma_5} f = - \int_{0}^{\infty} \cfrac{{\rm e}^{-\tfrac{pt}{4D}} {\rm e}^{\tfrac{i\sqrt{p}(x_1 - x_0) }{2D}}}{-4Di\sqrt{p}} \mathop{dp} = \int_{0}^{\infty} \cfrac{{\rm e}^{-\tfrac{pt}{4D}} \cdot {\rm e}^{\tfrac{i\sqrt{p}(x_1 - x_0) }{2D}}}{4Di\sqrt{p}} \mathop{dp}. \end{equation} Hence, from Cauchy’s residue formula, we have that our inverse Laplace transform is given by \begin{equation} \cfrac{1}{2 \pi i} \oint_{\Gamma_1} f(z) \mathop{dz} = - \cfrac{1}{2 \pi i } \left[\oint_{\Gamma_3} + \oint_{\Gamma_5} \right] f(z) \mathop{dz}. \end{equation} From here, we can solve the integral explicitly. We shall use the substitution $p = u^2$ to transform our exponent into a quadratic, making it a Gaussian integral, enabling the use of the result \begin{equation} \int_{-\infty}^{\infty} \exp( - \alpha x^2) \mathop{dx} = \sqrt{\cfrac{\pi}{\alpha}}. \end{equation} So, \begin{align} \cfrac{1}{2 \pi i} \oint_{\gamma_1} f(z) \mathop{dz} & = \cfrac{1}{ \pi } \int_{0}^{\infty} \cfrac{{\rm e}^{- \tfrac{pt}{4D}}}{4D \sqrt{p}} \cos \left(\tfrac{\sqrt{p}}{2D} (x_1 - x_0) \right) \mathop{dp} \\ & = \cfrac{1}{\pi} \Re \left\{ \int_{0}^{\infty} \cfrac{{\rm e}^{- \tfrac{pt}{4D}}}{4D \sqrt{p}} \cdot {\rm e}^{\tfrac{i \sqrt{p}(x_1 - x_0)}{2D}} \mathop{dp}\right\} \\ & = \cfrac{1}{4 \pi D} \Re \left\{ \int_{-\infty}^{\infty} {\rm e}^{- \tfrac{u^2 t}{4D} + \tfrac{iu}{2D}(x_1-x_0)} du \right\} \\ & = \cfrac{1}{4 \pi D} \Re \left\{\int_{-\infty}^{\infty} {\rm e}^{-\tfrac{t}{4D}[(u - \tfrac{i}{t}(x_1 - x_0))^2 + \tfrac{1}{t^2}(x_1-x_0)^2]} \mathop{du} \right\} \\ & = \cfrac{1}{\sqrt{4 \pi D t}} {\rm e}^{- \tfrac{(x_1-x_0)^2}{4Dt}}, \end{align}
which is exactly what we hoped. The case $V(x) = \alpha x$ follows very similarly. The case $V(x) = \tfrac{1}{2}x^2$ causes me some trouble. For simplicity, I think it is best if we take $x_0 = 0$ for now. In such a case, one has an $\operatorname{arcsinh}(x_1/\sqrt(z))$ term appearing from the integral, and the Jacobain (which gives a $\tfrac{1}{2} \operatorname{arsinh}$ factor) as well, and I do not know how to handle it and it becomes quite a formidable looking integral. In this case, the integral becomes \begin{equation} P(t) = \mathcal{N} \oint_{\text{Br}} \cfrac{{\rm e}^{\tfrac{Ht}{4D}} {\rm e}^{-\tfrac{x_1^2 + (H-2D) \operatorname{arcsinh}(x_1/\sqrt{H})+ x_1 \sqrt{H+ x_1^2}}{4D}}}{\sqrt{\sqrt{H} \sqrt{H + x_1^2}}} \cfrac{\mathop{dH}}{4D}. \end{equation}
As a little comment, I am quite sure this is the right formula and what is fascinating is how the branch points of the nasty hyperbolic function align perfectly with the other exponent term and the denominators; this must be why it will offer some analytic tractability for this quadratic case.
