Solving this diffusion equation I have the following diffusion equation  \begin{equation}
\frac{\partial{P}}{\partial N}=\frac{1}{2}\frac{\partial^2P}{\partial x^2}-a\frac{\partial P}{\partial x}\,,
\quad\mbox{where}\ a > 0\ \mbox{is some real parameter.}
\end{equation}
I am given with two boundary conditions:


*

*$P\left(x = 1, N\right) = 0$

*$P\left(x, N = 0\right) = \delta\left(x\right)$, where $\delta\left(x\right)$ is the Dirac delta function.


Does any one know how to proceed to solve for $P\left(x,N\right)$ ?. The boundary condition is making the whole problem difficult. 
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\partiald{\mrm{P}\pars{x,N}}{N} =
{1 \over 2}\,\partiald[2]{\mrm{P}\pars{x,N}}{x} -
a\,\partiald{\mrm{P}\pars{x,N}}{x}\,,\qquad a > 0}$

A 'plane wave' $\ds{\exp\pars{\ic kx - \ic\omega_{k}N}}$ particular solution
satisfies
\begin{align}
0 & = \pars{-\ic\omega_{k} + {1 \over 2}\,k^{2} + \ic ak}
\exp\pars{\ic kx - \ic\omega_{k}N}\implies
\omega_{k} = -\,{1 \over 2}\,k^{2}\,\ic + ak
\end{align}
such that
$\ds{\exp\pars{\ic kx - \ic\omega_{k}N} =
\exp\pars{\ic k\bracks{x - aN} - {1 \over 2}\,k^{2}N}}$. The general solution is given by
\begin{align}
\mrm{P}\pars{x,N} & =
\int_{-\infty}^{\infty}\mrm{A}\pars{k}
\exp\pars{\ic k\bracks{x - aN} - {1 \over 2}\,k^{2}N}\,{\dd k \over 2\pi}
\end{align}

$\ds{\large\left.1\right)\ \mrm{P}\pars{x = 1,N} = 0}$. The solution is a trivial one. Namely, $\bbx{\ds{\mrm{P}\pars{x,N} = 0}}$.

$\ds{\large\left.2\right)\ \mrm{P}\pars{x,N = 0} = \delta\pars{x}}$.
\begin{align}
\delta\pars{x} & =
\int_{-\infty}^{\infty}\mrm{A}\pars{k}
\exp\pars{\ic kx}\,{\dd k \over 2\pi} \implies \mrm{A}\pars{k} = 1
\end{align}
Then,
\begin{align}
\mrm{P}\pars{x,N} & =
\int_{-\infty}^{\infty}
\exp\pars{\ic k\bracks{x - aN} - {1 \over 2}\,k^{2}N}\,{\dd k \over 2\pi}
\\[5mm] & =
\int_{-\infty}^{\infty}
\exp\pars{-\,{1 \over 2}\,N\bracks{k^{2} - 2\,{x - aN \over N}\,k\,\ic}}
\,{\dd k \over 2\pi}
\\[5mm] & =
\int_{-\infty}^{\infty}
\exp\pars{-\,{1 \over 2}\,N\braces{%
\bracks{k - {x - aN \over N}\,k\,\ic}^{2} + {\bracks{x - aN}^{\,2} \over N^{2}}}}
\,{\dd k \over 2\pi}
\\[5mm] & =
\exp\pars{-\,{\bracks{x - aN}^{\,2} \over 2N}}
\int_{-\infty}^{\infty}\exp\pars{-\,{N k^{2} \over 2}}\,{\dd k \over 2\pi}
\\[5mm] & =
\bbx{\ds{{1 \over \root{2\pi N}}\,
\exp\pars{-\,{\bracks{x - aN}^{\,2} \over 2N}}}}
\end{align}
