Team:Valencia Biocampus/Demonstration/Diffusion1

Proof of a Group Behavior Diffusion Model from a Random Walk Model

In addition to last proof, we will make the assumption that the worm can now move in two dimensions ($x,y$).

As before, time step is $\;q\;$, and space step (either in $x$ or in $y$ coordinate) is $\;h\;$. Each time step $\;q\;$, the worm can move a distance $\;h\;$ either up, down, left or right with probabilities $\;u\;$, $\;d\;$, $\;l\;$ and $\;r\;$, respectively. Nevertheless, now we will consider that this probabilities will be dependent on position, so we must rename them as $\;u(x,y)\;$, $\;d(x,y)\;$, $\;l(x,y)\;$ and $\;r(x,y)\;$. The probability of remaining at the same position is, obviously, $\;1 - u(x,y) - d(x,y) - l(x,y) - r(x,y)\;$. The probability of being in position ($x,y$) at time $\;t\;$, $\;P(x,y,t)\;$ is:

$$ P(x,y,t)\;=\;P(x,y,t-q)\;\left(1 - l(x,y) - r(x,y) - u(x,y) - d(x,y)\right) + P(x-h,y,t-q)\;r(x-h,y) + \\ P(x+h,y,t-q)\;l(x+h,y) + P(x,y-h,t-q)\;u(x,y-h) + P(x,y+h,t-q)\;d(x,y+h)$$
Then, expanding each term as a Taylor series, centered in $\;t\;$, $\;x\;$ and $\;y\;$, neglecting the higher order terms $\;O(h^3)\;$ and $\;O(q^2)\;$, respectively:

$$ P\;=\;\left(P - q\;\frac{\partial P}{\partial t}\right)\;\left(1 - l - r - u - d\right) + \left(P - q\;\frac{\partial P}{\partial t} - h\;\frac{\partial P}{\partial x} + \frac{h^2}{2}\;\frac{\partial^2 P}{\partial x^2}\right)\;\left(r - h\;\frac{\partial r}{\partial x} + \frac{h^2}{2}\;\frac{\partial^2 r}{\partial x^2}\right) + \\ \left(P - q\;\frac{\partial P}{\partial t} + h\;\frac{\partial P}{\partial x} + \frac{h^2}{2}\;\frac{\partial^2 P}{\partial x^2}\right)\;\left(l + h\;\frac{\partial l}{\partial x} + \frac{h^2}{2}\;\frac{\partial^2 l}{\partial x^2}\right) + \left(P - q\;\frac{\partial P}{\partial t} - h\;\frac{\partial P}{\partial y} + \frac{h^2}{2}\;\frac{\partial^2 P}{\partial y^2}\right)\;\left(u - h\;\frac{\partial u}{\partial y} + \frac{h^2}{2}\;\frac{\partial^2 u}{\partial y^2}\right) + \\ \left(P - q\;\frac{\partial P}{\partial t} + h\;\frac{\partial P}{\partial y} + \frac{h^2}{2}\;\frac{\partial^2 P}{\partial y^2}\right)\;\left(d - h\;\frac{\partial d}{\partial y} + \frac{h^2}{2}\;\frac{\partial^2 d}{\partial y^2}\right) + O(h^3) + O(q^2) + O(h\;q)$$
Which, after operating, results in:

$$ q\;\frac{\partial P}{\partial t}\; = \;h\;\left(\frac{\partial l}{\partial x} - \frac{\partial r}{\partial x}\right)\;f + h\;\left(\frac{\partial d}{\partial y} - \frac{\partial u}{\partial y}\right)\;f + \frac{h^2}{2}\left(\frac{\partial^2 r}{\partial x^2} + \frac{\partial^2 l}{\partial x^2}\right)\;f + \frac{h^2}{2}\left(\frac{\partial^2 d}{\partial y^2} + \frac{\partial^2 u}{\partial y^2}\right)\;f + h\;\left(l - r\right)\;\frac{\partial f}{\partial x} + h\;\left(d - u\right)\;\frac{\partial f}{\partial y} + \\ h^2\;\left(\frac{\partial l}{\partial x} + \frac{\partial r}{\partial x}\right)\;\frac{\partial f}{\partial x} + h^2\;\left(\frac{\partial d}{\partial y} + \frac{\partial u}{\partial y}\right)\;\frac{\partial f}{\partial y} + \frac{h^2}{2}\;\left(l + r\right)\;\frac{\partial^2 f}{\partial x^2} + \frac{h^2}{2}\;\left(d + u\right)\;\frac{\partial^2 f}{\partial y^2} + O(h^3) + O(q^2) + O(h\;q)$$
Now we define:

$$ \nu_x\;=\;\lim\limits_{h,\;q,\;\beta_x\rightarrow 0}\frac{\beta_x\;h}{q} $$ $$ \nu_y\;=\;\lim\limits_{h,\;q,\;\beta_y\rightarrow 0}\frac{\beta_y\;h}{q} $$ $$ a_{x,x}\;=\;\lim\limits_{h,\;q\rightarrow 0}\frac{\alpha_x\;h^2}{2q} $$ $$ a_{y,y}\;=\;\lim\limits_{h,\;q\rightarrow 0}\frac{\alpha_y\;h^2}{2q} $$
Where $\;\alpha_x \;=\; r + l\;$, $\;\alpha_y \;=\; u + d\;$, $\;\beta_x \;=\; r - l\;$ and $\;\beta_y \;=\; u - d\;$.

We now take the equation, divide it by $\;q\;$ and take the limit as $\;h,q,\beta_x,\beta_y\rightarrow 0\;$ such that $\;\beta_x h/q\;$, $\;\beta_y h/q\;$, $\;\alpha_x h^2/2q\;$ and $\;\alpha_y h^2/2q\;$ all tend to a constant, obtaining:

$$ \frac{\partial f}{\partial t}\;=\;\frac{\partial^2 f a_{x,x}}{\partial x^2} + \frac{\partial^2 f a_{y,y}}{\partial y^2} - \frac{\partial f \nu_x}{\partial x} - \frac{\partial f \nu_y}{\partial y} $$
Or, what is the same:

$$ \frac{\partial f}{\partial t}\;=\;\nabla·\left(\nabla\left(\underline{\underline{D}} f\right)\right) - \nabla·\left(\underline{\nu} f\right) $$
With:

$$ \underline{\underline{D}} \; = \begin{bmatrix} a_{x,x}(x,y) & 0\\ 0 & a_{y,y}(x,y)\\ \end{bmatrix} \;\;\;\;\;\;\;\;\;\;\;\; \underline{\nu} \; = \begin{bmatrix} \nu_x(x,y)\\ \nu_y(x,y)\\ \end{bmatrix} $$
Nevertheless, in our case of study, we dealed with $\;D\;$ as a constant, so $\;a_{x,x}(x,y)\;=\;a_{y,y}(x,y)\;=\;D\;$, so we can tanke it out from the Laplacian, obtaining what we were waiting anxiously:

$$ \frac{\partial f}{\partial t}\;=\;D\;\nabla^2 f - \nabla·\left(\underline{\nu} f\right) $$
Although, there may exist an analytical solution for this differential equation, it has not been found yet as far as we are concerned; that's why we decided to solve it with numerical tools.