Consider $1,..,N$ indistinguishable particles in $\mathbb{R}^2$ and let them evolve according to a brownian motion and proliferation. Let $u: \mathbb{R}_+ \times \mathbb{R}^2 \rightarrow \mathbb{R}_0^+$ be the particle density at position $x \in \mathbb{R}^2$ and time $t$. This leads to the FKPP equation. In most papers the authors normalize $u$ in such a way that at $t=0$ $ \int u(x,0) dx = 1 $. Then they can use $u$ as a probability density and do some calculations. But in my opinion for later times $t$ this is not fulfilled anymore due to proliferation events, in particular $ \int u(x,t) dx > 1$ for $ t>0$. What does one do in this situations? In mean field systems the normalization is made by the so called mean field scaling $ \frac{1}{N}$. I can exclude the case that there is just a replacement of $ \frac{1}{N}$ by $ \frac{1}{N(t)}$ as this is not a known scaling.
4
-
$\begingroup$ $u(x,t)$ is not a probability density, it is a particle density: the number of particles at time $t$ in the interval $(x,x+dx)$ equals $u(x,t)dx$; so the integral $\int u(x,t)dx$ can become larger than unity, it just means the total number of particles is greater than one. $\endgroup$– Carlo BeenakkerCommented Dec 5, 2017 at 13:24
-
$\begingroup$ Thanks for your comment. I totally agree with you. But how can I then use $u(x,t)$ for calculating the expectation value? Or to put it differently if we assume in addition a pair force how can I calculate the mean field force? If one considers for example the vlasov equation it is no problem as $u(t,x)$ satisfies conservation of mass. $\endgroup$– Jack_Stiller10Commented Dec 5, 2017 at 14:32
-
$\begingroup$ if you want the "average per particle" you just divide by $\int u(x,t)dt$; if you want, say the total energy you don't divide; it all depends on what you want, the problem is perfectly well defined. $\endgroup$– Carlo BeenakkerCommented Dec 5, 2017 at 14:44
-
$\begingroup$ Ok thanks. I want to use $u(x,t)$ as a probability density. Following your comment you mean I should normalize the following way? $\tilde{u} = \frac{1}{N \int u(x,t) dt} $ to achieve $ \int \tilde{u}(t,x)dx =1$ for all $t$ and thus having a probability density. $\endgroup$– Jack_Stiller10Commented Dec 5, 2017 at 21:02
Add a comment
|