Diffusion in an infinite domain

Suppose that a large enough number $N$ of animals is released at a point in space (which we conventionally indicate with $x = 0$) and that the organisms can disperse without any barrier to their diffusion (see Fig. 3B). First we note that we can introduce the function

$$p(x,t) = c(x,t)/N.$$

It represents the fraction of organisms per unit length that at time $t$ is at distance $x$ from the release site. More precisely

$p(x,t) dx =$ fraction of the $N$ organisms that at time $t$ are located between $x$ and $x + dx$ .

Figure 3: (A) One-dimensional problem with absorbing barrier (Dirichlet problem); (B) A certain number of organisms is released in a point (conventionally located at $x = 0$) of a one-dimensional infinite space and randomly disperses without finding any barrier to its movement.

It should be noted that $p(x,t)$ has the same properties as a probability density, in particular the property

$$\int\limits_{ - \infty }^{ + \infty } {p(x,t)dx = 1}$$

at any time t.

One can easily verify by direct substitution into Eq. 14 that the solution of the problem is nothing but a Gaussian distribution with respect to space, characterized by a time-increasing variance; more precisely

$$c(x,t) = N p(x,t)$$

with

$$p(x,t) = \frac{1}{{\sqrt {4\pi Dt} }}\exp \left( { - \frac{1}{2}\frac{{{x^2}}}{{2Dt}}} \right).$$ (5.3)

As time varies, the mean value - which is also the mode and the median - of $p(x,t)$ is always null because, without the transport term, organisms evenly spread to the right and to the left. Instead, the variance grows linearly with time, and hence the concentration profile of individuals becomes flatter and flatter (see Fig. 4). In particular, the standard deviation is given by

$${\sigma _x} = \sqrt {2Dt}$$

and thus increases with the root of time.

It is to be noted that for $t = 0$ the variance vanishes, so the $N$ organisms are all concentrated in the origin. Therefore the solution meets the initial conditions. In fact, since organisms are all simultaneously released at the same point, $c_0(x)$ is a pulse located in $x = 0$. One can also compute the average absolute distance where organisms are located at time $t$ after the release. With simple calculations one gets

$$E\left[ {\left\vert x \right\vert} \right] = \int\limits_{ - \i... ... + \infty } {xp(x,t)dx = \sqrt {\frac{{4Dt}}{\pi }} } \cong 1.128\sqrt {Dt}.$$

The diffusion equation is easily generalized to the case when we consider more than one spatial dimension. In particular when organisms disperse in a plane, described by the spatial coordinates $x$ and $y$, and the diffusion coefficient is the same in all directions (isotropic diffusion), the diffusion equation is

$$\frac{{\partial c}}{{\partial t}} = D\left( {\frac{{{\partial ^2}c}}{{\partial {x^2}}} + \frac{{{\partial ^2}c}}{{\partial {y^2}}}} \right.)$$ (5.4)

The solution to the problem of the release of $N$ organisms in a point of a plane is just a bivariate normal distribution. More precisely, if we introduce the fraction of individuals per unit area $p(x,y,t)$ defined as

$p(x,y,t) dx dy$ = fraction of the $N$ organisms that at time $t$ are located in the square of area $dx dy$ whose lower left corner has coordinates $(x,y)$

the solution of Eq. 16 is given by

$$c(x,y,t) = N p(x,y,t)$$

with

$$p(x,y,t) = \frac{1}{{4\pi Dt}}\exp \left( { - \frac{1}{2}\frac{{{x^2} + {y^2}}}{{2Dt}}} \right).$$

At each time instant $t$ the fraction $p(x,y,t)$ has precisely the shape of a symmetrical bell (see Fig. 5A). Contour lines are therefore circles (Fig. 5B).

Figure 4: Time behaviour of the solution to the diffusion problem when an initial number of organisms is released at $x = 0$. The graph shows various snapshots of the solution at different instants.

It is possible to prove (see the book by Pielou, 1977) that the fraction $f_R$ of the population that lies outside the contour line of radius $R$ at time $t$ is given by

$$f_R = \exp \left( { - \frac{{{R^2}}}{{4Dt}}} \right).$$

Therefore, the radius of the contour line that contains for example 99% of the population at a given instant $t$ satisfies the equation

$$0.01 = \exp \left( { - \frac{{{R^2}}}{{4Dt}}} \right)$$

and thus is given by

$$R = 2\sqrt {\ln 100} \sqrt {Dt} = 4.292\sqrt {Dt}.$$

It increases with the root of time, while the area that contains a certain fraction of the population ($\pi R^2$) grows linearly with time.

Figure 5: Solution to the diffusion problem in the plane when an initial number of organisms is released at the point of coordinates $x = 0$, $y = 0$. (A) Shape of $p(x,y,t)$ with D = 2 and $t$ = 0.5; (B) corresponding contour lines.