Diffusion equation with advection and decay I'm trying to solve the following initial value problem.
$\begin{cases} u_t + u_x - u_{xx} &= -u, \quad \text{on} \quad \mathbb R \times \mathbb R_+\\
u(x,0) &= \frac{1}{4\pi} e^\frac{-x^2}{4}, \quad x \in \mathbb R
\end{cases}$
I know how to solve the heat equation $u_t = u_{xx}$, but I have no idea how to get started with this.
So far -inspired by the initial value- I have tried substituting and checking various forms of $\frac{e^\frac{-x^2}{4t}}{\sqrt{4{\pi}t}}$ to no avail.
Is there a general technique that can be used in this case?
 A: You should be able to transform the above equation into the heat equation by the following substitution:
$$ u(x,t) = w(x,t) e^{\frac{1}{2}x-(1+\frac{1}{4})t}$$
In fact (if $\alpha\neq 0$), the general
\begin{align} 
 u_t - \alpha u_{xx} + cu_x = -\lambda u \tag 1 \label{eq1}
\end{align}
equation can be seen to be equivalent to the (one dimensional) heat equation
$$w_t = \alpha w_{xx}$$ using the substitution
$$u(x,t) = w(x,t) \operatorname{exp}\left (\frac{c}{2\alpha}x - \left (\lambda + \frac{c^2}{4\alpha}\right )t\right)$$
I'll include a quick proof here for the sake of completeness:
$$A(x,t) := \operatorname{exp}\left (\frac{c}{2\alpha}x - \left (\lambda + \frac{c^2}{4\alpha}\right )t\right)\quad\text{so}\quad u = wA$$
Then 
\begin{align}
\color {red}{u_t} &= w_tA - \left(\lambda + \frac{c^2}{4\alpha} \right)wA \\
\color {green}{cu_x} &= c\left(w_x A + \frac{c}{2\alpha}wA \right)\\
\color {blue}{\alpha u_{xx}} &= \alpha \left( w_{xx}A + \frac{c}{\alpha}w_xA + \frac{c^2}{4\alpha^2}wA \right)
\end{align}
Substituting back into $(\ref{eq1})$ and dividing the equation by the nonzero $A$, we get
\begin{align}
-\lambda w &= \color {red}{w_t - \lambda w - \frac{c^2}{4\alpha}w} \color  {blue} {- \alpha w_{xx} - cw_x - \frac{c^2}{4\alpha}w} \color{green}{+ cw_x + \frac{c^2}{2\alpha}w}\\
0 &= \color {red}{w_t} - \color {blue}{\alpha w_{xx}}
\end{align}
