Team:Valencia Biocampus/Demonstration/Diffusion3

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== Proof of a Group Behavior Diffusion Model from a Random Walk Model ==
== Proof of a Group Behavior Diffusion Model from a Random Walk Model ==
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Considerations for the Random Walk:
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To prove that, we will make the assumption that the worm only moves in one dimension ($x$) without loss of generality.
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Step lenghts ($l_t$) in the order of a pixel in size. That implies, $ \Delta t $ as small as possible.
 
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Perfect Random Walk, with uniform probabilistic distributions either for $ v_t $, $\dot{\theta_t}$ and $\delta$.<br/>
 
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Discretizing the whole space into pixels, and assuming, the worm can, either occupy one or not, we can assure that, at each time step, it can only move in four different directions: up, down, right or left from its position. As we considered that each random variable follows an <i>uniform probabilistic distribution</i>, it is equipossible to move in any of these directions, with a probability of $ \frac{1}{4} $ each.  
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At each time step $\;q\;$ it either moves a distance $\;h\;$ to the left with probability $\;l\;$, a distance $\;h\;$ to the right with probability $\;r\;$, or stays in the same position with probability $\;1−r−l\;$ (the isotropic random walk has $\;r\;=\;l\;=\;1/2$, so it cannot rest motionless). We also define the probability that a worm is at a position $\;x\;$ at time $\;t\;$ by $\;P(x,t)\;$. One time step earlier, at time $\;t − q\;$, the walker must have been at position $\;x − δ\;$ and then moved to the right, or at position $\;x + δ\;$ and then moved to the
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left, or at position $\;x\;$ and then not moved at all. Thus:
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Now, we can compute, the probability that the worm is at position $(x_m,y_m)$ at the iteration $n+1$ as follows:
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$$ P(x,t)\;=\;P(x,t-q)\;\left(1 - l - q\right) + P(x-h,t-q)\;r + P(x+h,t-q)\;l $$
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Assuming that $\;q\;$ and $\;h\;$ are so small, that are negligible compared to $\;t\;$ and $\;x\;$ respectively, we can expand de function as a Taylor series, around $\;t\;$ and $\;x\;$. Notice that higher terms than $\;q^2\;$ and than $\;h^3\;$ have been included in $\;O(q^2)\;$ and $\;O(h^3)\;$, respectively:
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$$ P_{n+1}(x_m,y_m)\;=\;\frac{1}{4}\;\left(P_{n}(x_m,y_{m+1}) + P_{n}(x_{m+1},y_m) + P_{n}(x_m,y_{m-1}) + P_{n}(x_{m-1},y_m)\right) $$
 
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$$ P\;=\;\left(P - q\;\frac{\partial P}{\partial t}\right)\;\left(1 - l - r\right) + \left(P - q\;\frac{\partial P}{\partial t} - h\;\frac{\partial P}{\partial x} + \frac{h^2}{2}\;\frac{\partial^2 P}{\partial x^2}\right)\;r + \left(P - q\;\frac{\partial P}{\partial t} + h\;\frac{\partial P}{\partial x} + \frac{h^2}{2}\;\frac{\partial^2 P}{\partial x^2}\right)\;l + O(h^3) + O(q^2)$$
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Rearranging this gives:
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<a href=""><img src="https://static.igem.org/mediawiki/2013/3/34/Cel.png" width="240" height="230" alt="Allowed directions"/></a>
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<span style="font-size:10px">Possible movements of a worm beeing at any of the blue pixels</span>
 
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$$ \frac{\partial P}{\partial t}\;=\;\frac{\alpha\;h^2}{2\;q}\;\frac{\partial^2 P}{\partial x^2} - \frac{\beta\;h}{q}\;\frac{\partial P}{\partial x} + O(h^3) + O(q^2)$$
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If we now subtract $ P_{n}(x_m,y_m) $ from both sides:
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Where $\;\alpha\;=\;r + l\;$ and $\;\beta\;=\;r - l\;$. We now let $\;h,\;q,\;\beta\;\rightarrow\;0\;$ in such a way that the following
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limits are finite:
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$$ P_{n+1}(x_m,y_m) - P_{n}(x_m,y_m)\;=\;\frac{1}{4}\;\left(P_{n}(x_m,y_{m+1}) + P_{n}(x_{m+1},y_m) + P_{n}(x_m,y_{m-1}) + P_{n}(x_{m-1},y_m) - 4\;P_{n}(x_m,y_m)\right) $$
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$$ D\;=\;\alpha\;\lim\limits_{h,\;q,\;\beta\rightarrow 0}\frac{h^2}{2\;q} $$
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$$ P_{n+1}(x_m,y_m) - P_{n}(x_m,y_m)\;=\;\frac{1}{4}\;\left(P_{n}(x_{m+1},y_m) - 2\;P_{n}(x_m,y_m) + P_{n}(x_{m-1},y_m) + P_{n}(x_m,y_{m+1}) - 2\;P_{n}(x_m,y_m) + P_{n}(x_m,y_{m-1})\right) $$
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$$ v\;=\;\lim\limits_{h,\;q,\;\beta\rightarrow 0}\frac{h\;\beta}{q} $$
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And multiply and divide by $q$ (time step) and $h^2$ (space step squared):
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So we can neglect $\;O(h^3)\;$ and $\;O(q^2)\;$, resting:
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$$ \frac{P_{n+1}(x_m,y_m) - P_{n}(x_m,y_m)}{q}\;=\;\frac{h^2}{4\;q}\;\left(\frac{P_{n}(x_{m+1},y_m) - 2\;P_{n}(x_m,y_m) + P_{n}(x_{m-1},y_m)}{h^2} + \frac{P_{n}(x_m,y_{m+1}) - 2\;P_{n}(x_m,y_m) + P_{n}(x_m,y_{m-1})}{h^2}\right) $$
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$$ \frac{\partial P}{\partial t}\;=\;D\;\frac{\partial^2 P}{\partial x^2} - v\;\frac{\partial P}{\partial x} $$
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And, defining the constant $D\;=\;\frac{h^2}{4\;q}$:
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Considerations:
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If we set $\;r\;=\;l\;=\;1/2\;$ as in the isotropic random walk, then $\;\beta\;=\;0\;$, so $\;u\;=\;0\;$, giving as a result the non-biased Diffusion Equation:
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$$ \frac{P_{n+1}(x_m,y_m) - P_{n}(x_m,y_m)}{q}\;=\;D\;\left(\frac{P_{n}(x_{m+1},y_m) - 2\;P_{n}(x_m,y_m) + P_{n}(x_{m-1},y_m)}{h^2} + \frac{P_{n}(x_m,y_{m+1}) - 2\;P_{n}(x_m,y_m) + P_{n}(x_m,y_{m-1})}{h^2}\right) $$
 
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That can be approximated as a Diffusion Equation, by recalling the definition for the first and second derivates:
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$$ \frac{\partial P}{\partial t}\;=\;D\;\frac{\partial^2 P}{\partial x^2}$$
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In this case, $\;v\;$ is constant for all the space, not as in the case that concerns us, where $\;v\;$ depends on the gradient of the attractant, normally distributed (with Gaussian Distributions) in the space of interest. So, with $\;v\;$ constant, it is possible to obtain an analytical solution, given by <i> Montroll & Shlesinger </i> (1984), with initial condition $\;P(x,0)\;=\;\delta(x)\;$, that is:
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$$ P(x,t)\;=\;\frac{1}{\sqrt{4 \pi D t}}\;e^{-\left(x - v t\right)^2/\left(4 D t\right)}$$
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<a href=""><img src="https://static.igem.org/mediawiki/2013/a/a7/Diffus.png" width="700" height="300" alt="Allowed directions"/></a>
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<span style="font-size:10px">Plots of $\;P(x,t)\;$ for different $\;v\;$, and different time instants: left, $\;D$ = $1\;$ and $\;v$ = $1\;$; right, $\;D$ = $1\;$ and $\;v$ = $2\;$</span>
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$$\frac{\partial P}{\partial t} = D \; \left(\frac{\partial^2 P}{\partial x^2} + \frac{\partial^2 P}{\partial y^2}\right) $$
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Latest revision as of 08:53, 4 October 2013

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

To prove that, we will make the assumption that the worm only moves in one dimension ($x$) without loss of generality.

At each time step $\;q\;$ it either moves a distance $\;h\;$ to the left with probability $\;l\;$, a distance $\;h\;$ to the right with probability $\;r\;$, or stays in the same position with probability $\;1−r−l\;$ (the isotropic random walk has $\;r\;=\;l\;=\;1/2$, so it cannot rest motionless). We also define the probability that a worm is at a position $\;x\;$ at time $\;t\;$ by $\;P(x,t)\;$. One time step earlier, at time $\;t − q\;$, the walker must have been at position $\;x − δ\;$ and then moved to the right, or at position $\;x + δ\;$ and then moved to the left, or at position $\;x\;$ and then not moved at all. Thus:

$$ P(x,t)\;=\;P(x,t-q)\;\left(1 - l - q\right) + P(x-h,t-q)\;r + P(x+h,t-q)\;l $$
Assuming that $\;q\;$ and $\;h\;$ are so small, that are negligible compared to $\;t\;$ and $\;x\;$ respectively, we can expand de function as a Taylor series, around $\;t\;$ and $\;x\;$. Notice that higher terms than $\;q^2\;$ and than $\;h^3\;$ have been included in $\;O(q^2)\;$ and $\;O(h^3)\;$, respectively:

$$ P\;=\;\left(P - q\;\frac{\partial P}{\partial t}\right)\;\left(1 - l - r\right) + \left(P - q\;\frac{\partial P}{\partial t} - h\;\frac{\partial P}{\partial x} + \frac{h^2}{2}\;\frac{\partial^2 P}{\partial x^2}\right)\;r + \left(P - q\;\frac{\partial P}{\partial t} + h\;\frac{\partial P}{\partial x} + \frac{h^2}{2}\;\frac{\partial^2 P}{\partial x^2}\right)\;l + O(h^3) + O(q^2)$$
Rearranging this gives:

$$ \frac{\partial P}{\partial t}\;=\;\frac{\alpha\;h^2}{2\;q}\;\frac{\partial^2 P}{\partial x^2} - \frac{\beta\;h}{q}\;\frac{\partial P}{\partial x} + O(h^3) + O(q^2)$$
Where $\;\alpha\;=\;r + l\;$ and $\;\beta\;=\;r - l\;$. We now let $\;h,\;q,\;\beta\;\rightarrow\;0\;$ in such a way that the following limits are finite:

$$ D\;=\;\alpha\;\lim\limits_{h,\;q,\;\beta\rightarrow 0}\frac{h^2}{2\;q} $$ $$ v\;=\;\lim\limits_{h,\;q,\;\beta\rightarrow 0}\frac{h\;\beta}{q} $$
So we can neglect $\;O(h^3)\;$ and $\;O(q^2)\;$, resting:

$$ \frac{\partial P}{\partial t}\;=\;D\;\frac{\partial^2 P}{\partial x^2} - v\;\frac{\partial P}{\partial x} $$
Considerations:

  • If we set $\;r\;=\;l\;=\;1/2\;$ as in the isotropic random walk, then $\;\beta\;=\;0\;$, so $\;u\;=\;0\;$, giving as a result the non-biased Diffusion Equation:

    $$ \frac{\partial P}{\partial t}\;=\;D\;\frac{\partial^2 P}{\partial x^2}$$
  • In this case, $\;v\;$ is constant for all the space, not as in the case that concerns us, where $\;v\;$ depends on the gradient of the attractant, normally distributed (with Gaussian Distributions) in the space of interest. So, with $\;v\;$ constant, it is possible to obtain an analytical solution, given by Montroll & Shlesinger (1984), with initial condition $\;P(x,0)\;=\;\delta(x)\;$, that is:

    $$ P(x,t)\;=\;\frac{1}{\sqrt{4 \pi D t}}\;e^{-\left(x - v t\right)^2/\left(4 D t\right)}$$

Allowed directions
Plots of $\;P(x,t)\;$ for different $\;v\;$, and different time instants: left, $\;D$ = $1\;$ and $\;v$ = $1\;$; right, $\;D$ = $1\;$ and $\;v$ = $2\;$






Okopinska A. (2002) Fokker-Planck equation for bistable potential in the optimized expansion. Physical review E, Volume 65, 062101