Difference between revisions of "Team:Paris Saclay/Modeling"

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<p>Mathematical models and computer simulations provide a great way to describe the function and operation of BioBrick Parts and Devices. Synthetic Biology is an engineering discipline, and part of engineering is simulation and modeling to determine the behavior of your design before you build it. Designing and simulating can be iterated many times in a computer before moving to the lab. This award is for teams who build a model of their system and use it to inform system design or simulate expected behavior in conjunction with experiments in the wetlab.</p>
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The novel encapsulation method based on silica used in our project is a challenging
  
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It has already been proven that bacteria contained in alginate beads which are themselves been wrapped in a silica monolith are able to survive and reproduce.
Here are a few examples from previous teams:
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It is however interesting to study to what extent the porous medium used to allow diffusion of molecules is permissive, given the thickness of silica around the alginate beads that contain our working bacteria.
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<ul>
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%%START OF MODELING
<li><a href="https://2014.igem.org/Team:ETH_Zurich/modeling/overview">ETH Zurich 2014</a></li>
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<li><a href="https://2014.igem.org/Team:Waterloo/Math_Book">Waterloo 2014</a></li>
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The diffusion of particules is based on Fick's first law which is given by:
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\begin{equation}
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\textbf{j} = -D \, \bf{\nabla} n
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\label{eq:fick}
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\end{equation}
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In this equation, $\textbf{j}$ is the diffusion flux, $D$ the diffusion coefficient and $n$ concentration of particles.
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It can be coupled with the continuity equation stating the number  second law
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physical containment method based on silica presented
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%%%%Start of LATEX
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Diffusion de particules - Loi de Fick
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Equation de continuité -> conservation nombre de particules
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\begin{equation}
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\partial_t n = \mathbf{\nabla} \cdot \mathrm{j} \, (+\sigma)
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\label{eq:continuity_eq}
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\end{equation}
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Eq. \ref{eq:fick} + Eq. \ref{eq:continuity_eq} : $\partial_t n = D \, \triangle n \, (+\sigma)$
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Expression du laplacien en coordonnées sphériques
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\begin{equation}
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\triangle a = \frac{1}{r^2} \frac{\partial}{\partial r} \Big(r^2 \frac{\partial a}{\partial r} \Big) + \frac{1}{r^2 \sin \theta} \frac{\partial}{\partial \theta} \Big(\sin \theta \frac{\partial a}{\partial \theta} \Big) + \frac{1}{r^2 \sin \theta} \frac{\partial^2 \varphi}{\partial \varphi^2}
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\end{equation}
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Let's suppose that in the beads $n(r,\theta,\varphi,t) = n(r,t)$ $\rightarrow$ spherical symmetry, no dependence on angles
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Steady diffusion
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\begin{equation}
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\frac{1}{r^2} \frac{\partial}{\partial r} \Big[ r^2 \frac{\partial n}{\partial r} \Big] =0 \rightarrow 2r\frac{\mathrm{d}n}{\mathrm{d}r} + r^2 \frac{\mathrm{d}^2n}{\mathrm{d}r^2} = 0
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\end{equation}
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Let's consider $m=\frac{\mathrm{d}n}{\mathrm{d}r}$ $\rightarrow  2rm + r^2 \frac{\mathrm{d}m}{\mathrm{d}t}=
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2 r +r^2 \frac{\mathring{d}(\log m)}{\mathrm{d}r} =0$
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$\rightarrow \frac{\mathrm{d}(\log m)}{\mathrm{d}r} = -\frac{2}{r} \rightarrow \log m = -2 \log r+ cte \rightarrow m\frac{alpha}{r^2}=\frac{\mathrm{d}n}{\mathrm{d}r}$
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\begin{equation}
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n(r) = -\frac{\alpha}{r} + \beta
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\label{steady_eq}
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\end{equation}
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When $r\rightarrow + \infty \, n=n_\infty $ densité de particule loin du centre de la bille / densité du milieu ext.
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\ref{steady_eq} $\rightarrow n(+\infty) = \beta = n_\infty \rightarrow n(r) = -\frac{\alpha}{r} + n_\infty$
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Problem : determination of $\alpha$. Quelles conditions aux limites choisir pour une solution physique ?
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Unsteady diffusion : $\partial_t n= \frac{D}{r^2} \, \partial_r(r^2 \partial_r n)$
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\begin{equation}
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\frac{1}{D} \frac{\partial n}{\partial t} = \frac{1}{r^2} \frac{\partial}{\partial t} \Big(r^2 \frac{\partial n}{\partial r}\Big) = -\lambda^2
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\label{eq:unsteady_lambda}
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\end{equation}
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it is necessarily negative,necessary condition for the decrease of $n$ in time
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On pose $n(r,t)=T(t)*R(r)$
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\ref{eq:continuity_eq} $\rightarrow$ $\frac{1}{D} R(r) \frac{\mathrm{d}T(t)}{\mathrm{d}t} = \frac{1}{r^2} \frac{\mathrm{d}}{\mathrm{d}r} \Big[ r^2 T(t) \frac{\mathrm{d}R(r)}{\mathrm{d}r}\Big]$
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$\frac{1}{DT} \frac{\mathrm{d}T}{\mathrm{d}t} = \frac{1}{r^2 R} \frac{\mathrm{d}}{\mathrm{d}r} \Big[ r^2 \frac{\mathrm{d}R(r)}{\mathrm{d}r}\Big]$
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Temporal part
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\ref{eq:unsteady_lambda} $\rightarrow$ $\frac{1}{DT} \frac{\mathrm{d}T}{\mathrm{d}t} = - \lambda^2$
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$ \rightarrow$ $\frac{DT}{T} = - \lambda^2 D \mathrm{d}t \rightarrow T(t) = \mathrm{e}^{-\lambda^2 D t}$
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Radial part
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\ref{eq:unsteady_lambda} $\rightarrow$  $\frac{1}{r^2 R} \frac{\mathrm{d}}{\mathrm{d}r} \Big[ r^2 \frac{\mathrm{d}R(r)}{\mathrm{d}r}\Big] = -\lambda^2$
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$\rightarrow r^2 \frac{\mathrm{d}^2R}{\mathrm{d}r^2} + 2 r \frac{\mathrm{d}R}{\mathrm{d}r} + \lambda^2 r^2 R =0 \,$ Eq diff de Bessel -> résolution avec méthode de Newmann
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General formula of the differential equation:
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\begin{equation}
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x^2 y''+ (2p+1)xy' + (\alpha^2x^{2r} + \beta^2)y = 0 \, \, \, \mathrm{with} \, y=y(x)
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\end{equation}
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Solution of the differential equation is :
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\begin{equation}
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y=x^{-p}[C_1 J_{^q/_r}(\frac{\alpha}{r}x^r)+ C_2 Y_{^q/_r}(\frac{\alpha}{r}x^r)]
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\end{equation}
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with $q=\sqrt{p^2-\beta^2}$
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Here we have $2p+1 =2 \rightarrow p=1/2$
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$\alpha^2x^{2r} + \beta^2 = \lambda^2 x^2 \rightarrow \beta=0, r=1, \alpha = \lambda$
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The relations we had before then becomes :
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\begin{equation}
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R(r) = \frac{1}{\sqrt{r}} [C_1 J_{^1/_2}(\lambda r)+ C_2 Y_{^1/_2}(\lambda r)]
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\end{equation}
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$J_{^1/_2}$ and $Y_{^1/_2}$ are respectively Bessel functions of the first and second kind
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$(5) +(6) \rightarrow \qquad n(r,t)=\frac{\mathrm{e}^{-\lambda D^2 t}}{\sqrt{r}} \Big[C_1 J_{^1/_2}(\lambda r)+ C_2 Y_{^1/_2}(\lambda r) \Big]$
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$J_{^1/_2}(x) = \sqrt{\frac{2}{\pi x}} \sin x \qquad$ and $Y_{^1/_2}(x)=-J_{^{-1}/_2}(x) = \sqrt{\frac{2}{\pi x}} \cos x]$
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$n(r,t) = \frac{\mathrm{e}^{-\lambda D^2 t}}{\sqrt{r}} \, \sqrt{\frac{2}{\lambda \pi}} \Big[C_1 \sin(\lambda r) - C_2 \cos(\lambda r) \Big]$
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Revision as of 01:13, 19 September 2015

Modeling

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