calculation of diffusion front distance

Question

I have diffusion 1D equation:$$\frac{1}{\gamma}\frac{\partial P}{\partial t}= \frac{\partial^2 P}{\partial x^2}$$ Estimation of front distance is $$x=\sqrt{\gamma t}$$ aand I don't understand how it's derived. Can anyone help me to derive the last formula?

Answer 1

A heuristic proof.

1D diffusion equation,

$$\frac {\partial P }{\partial t}=\gamma \ \frac {\partial ^{2}P }{\partial x^2}\tag{1}$$

(initialized by a Dirac peak) has solution :

$$P(x,t)={\frac {1}{\sqrt{4\pi \gamma t}}}e^{-{\frac {x^{2}}{4\gamma t}}}.\tag{2}$$

(just plug expression (2) into (1)).

When you compare(2) to the standard expression of the pdf of Laplace-Gauss distribution with mean $0$ and standard deviation $\sigma$ :

$$f(x)={\frac {1}{\sigma\sqrt {2\pi}}}e^{-{\frac {x^{2}}{2 \sigma^2}}},\tag{3}$$

you see that $2 \gamma t$ is nothing else that $\sigma^2$ ; thus

$$\sigma=\sqrt{2\gamma t}\approx 1.4 \sqrt{\gamma t} \tag{3}$$

Now, recall the meaning of standard deviation $\sigma$ : around $68\%$ of events are closer to the mean $0$ that one $\sigma$" ( https://en.wikipedia.org/wiki/68%E2%80%9395%E2%80%9399.7_rule). Thus, a fifty-fifty separation (instead of $ \approx 68 \% /32 \%$) will be obtained by lowering the threshold, passing from $1.4$ to approx. $1$ in the formula above. Just handwaving for this last part, I admit, but in fact, one can obtain this result in a rigorous way...

Remark : have a look at this nice document about Brownian motion : https://en.wikipedia.org/wiki/Brownian_motion.