I'm looking to solve $\frac{\partial N}{\partial t} = rN - \beta \frac{\partial N}{\partial x} + D \frac{\partial^2 N}{\partial x^2}$ on $-\infty < x < \infty$ and $t>0$ where the initial distribution is a point source release of heat at $x =0, t =0$.
The question has the hint that the solution to the heat equation $\frac{\partial n}{\partial t} = D\frac{\partial^2 n}{\partial x^2}$ with the same point source release is $n(x,t) = \frac{n_0}{2\sqrt{\pi D t}} e^{-\frac{x^2}{4Dt}}$.
I am wondering on how to use this hint. I tried to guess that $N(x,t) = e^{rt} n(x,t)$ with the goal that the PDE becomes the heat equation, but substituting that into the PDE gives $\frac{\partial n}{\partial t} = -\beta \frac{\partial n}{\partial x} + D\frac{\partial^2 n}{\partial x^2}$.
Is there a better guess that allows me to transform it into the heat equation?
Use Fourier Transform.
$$N(x,t)=\frac{1}{\sqrt{2\pi}}\int dk \psi(k, t) e^{ikx}$$
At $t=0$ we have
$$\psi(k, 0)=\frac{1}{\sqrt{2\pi}}$$
The PDE also yields
$$\psi(k , t)=e^{t(r-i\beta k -Dk^2)}\psi(k, 0)$$
Taking the inverse transform we get
$$N(x,t)=\frac{1}{\sqrt{4\pi Dt}}\exp\Big(tr-\frac{(x-\beta t)^2}{4Dt}\Big)$$
Based on this answer, now I can use the hint :)))) The $r$ term disappears by the substitution $$N'\equiv N e^{-rt}$$ Then to make the $\beta$ term disappear, use the moving frame $$x'\equiv x-\beta t$$