Team:Grenoble-EMSE-LSU/Project/Modelling

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<p> which gives : $Y=\frac{a}{bI_0+r}+Be{-(bI_0+r)t} $</p>
<p> which gives : $Y=\frac{a}{bI_0+r}+Be{-(bI_0+r)t} $</p>
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<p> $Y$ is then stabilizing, let's look $C$ :</p>
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<p> $Y$ tends toward a steady state value, $\frac{a}{bI_0+r}$. Let's see how C </p>
<p>$\frac{dC}{dt}=C(r-kI_0Y)$</p>
<p>$\frac{dC}{dt}=C(r-kI_0Y)$</p>
<p>$\frac{dC}{dt}=C\left(r-\frac{kI_0a}{bI_0+r}+Be^{-(bI_0+r)t}\right)$</p>
<p>$\frac{dC}{dt}=C\left(r-\frac{kI_0a}{bI_0+r}+Be^{-(bI_0+r)t}\right)$</p>
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<p> and so, for $I_0=\frac{r^2}{ak-rb}$, we have : $\lim_{t\to\infty}\frac{dC}{dt}(t)=0$</p>
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<p> Thus, in the specific case where $I_0=\frac{r^2}{ak-rb}$, we have : $\lim_{t\to\infty}\frac{dC}{dt}(t)=0$</p>
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<p> The resolution of this equation have shown the possibility to stabilize the system.</p>   
<p> The resolution of this equation have shown the possibility to stabilize the system.</p>   

Revision as of 17:42, 30 September 2013

Grenoble-EMSE-LSU, iGEM


Grenoble-EMSE-LSU, iGEM

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