I am trying to solve the Fokker Plank equation: \begin{equation} \frac{\partial{P}}{\partial S}=\frac{1}{2}\frac{\partial^2P}{\partial x^2}~. \end{equation} This describes trajectories of particles in a random walk. If the starting point of the random walk is at $S_0=0$ and $x_0=x(S_0)=1$, then the boundary conditions for $P(x,S)$ are $P(x_0,S_0)=\delta(x_0)$, where $\delta$ is the Dirac-delta function. Another boundary condition is $P(x_c, S)=0$ where $x_c\equiv 1-\alpha S$ is a critical value of $x$, and $\alpha>0$ is a fixed parameter. To solve this equation I let $\gamma\equiv x_c-x$, and carry out Fourier transform of $P$, so that $$\tilde{P}(\omega, S)=\int \mathrm{d}\gamma\, \mathrm{e}^{\mathrm{i}\omega\gamma}P(\gamma, S)~,$$ which upon using in the Fokker plank equation (first equation) gives $$\frac{\partial\tilde{P}}{\partial S}=-\frac{\omega^2}{2}\tilde{P}~,$$ and has a solution $$\tilde{P}(\omega, S)=c(\omega)\exp\left(-\frac{\omega^2}{2}S\right).$$ The condition that at $\gamma=0 (\delta=\delta_c)$, $P$ vanishes guarentees that $c(\omega)$ is an odd function. Thus, transforming back, I have $$P(x, S)=\int_0^{\infty}\mathrm{d}\omega\,c(\omega)\sin(\omega\gamma)\exp\left(-\frac{\omega^2}{2}S\right)~,$$ which upon using th boundary condition $P(x_0,S_0)=\delta(x_0)$ immediately gives $$c(\omega)=\frac{2}{\pi}\sin(\omega\gamma_0)~,$$ where $\gamma_0=x_0-x_c$. Finally I have $$P(x, S)=\frac{2}{\pi}\int_0^{\infty}\mathrm{d}\omega\,\sin(\omega\gamma_0)\sin(\omega\gamma)\exp\left(-\frac{\omega^2}{2}S\right)~, $$ which gives $$P(x, S)=\frac{1}{\sqrt{2\pi S}}\left[\exp\left(-\frac{(x-1)^2}{2S}\right)-\exp\left(-\frac{[2(x_c-1)-(x-1)]^2}{2S}\right)\right].$$
I have a problem here. If I substitute this expression for $P(x, S)$ in the original PDE, the left hand side is not equal to right hand side unless $x_c$ is independent of $S$. In my case $x_c=1-\alpha S$, so is dependent on $S$. Any idea where I went wrong? Thanks.