# Solving PDE by using known solution to the heat equation

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?

• Try substitution $N(x,t)=e^{at+bx}n(x,t)$ and determine $a$ and $b$ that would eliminate all the unwanted terms. Jun 18 '19 at 17:06
• @Artem thank you! Jun 18 '19 at 17:27

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 $$x'\equiv x-\beta t$$