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Two-proteins-x-and-y-control-each-other-through-mutual-repression-The-dynamic-model-for-the-system-consists-of-the-following-system-of-ODEs-calculate-the-steady-state-values-of-these-two-protein




Question Number 207151 by Wuji last updated on 07/May/24
Two proteins (x and y) control each  other through mutual repression. The  dynamic model for the system consists  of the following system of ODEs.   calculate the steady state values of   these two proteins. comment on the   stability of the steady state and the type of   phase portrait expected for this system.  consider x,y≥0  (dx/dt)=(y/(1+y))−2    (dy/dt)=(x/(1+x))−y
$$\mathrm{Two}\:\mathrm{proteins}\:\left(\mathrm{x}\:\mathrm{and}\:\mathrm{y}\right)\:\mathrm{control}\:\mathrm{each} \\ $$$$\mathrm{other}\:\mathrm{through}\:\mathrm{mutual}\:\mathrm{repression}.\:\mathrm{The} \\ $$$$\mathrm{dynamic}\:\mathrm{model}\:\mathrm{for}\:\mathrm{the}\:\mathrm{system}\:\mathrm{consists} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{following}\:\mathrm{system}\:\mathrm{of}\:\mathrm{ODEs}.\: \\ $$$$\mathrm{calculate}\:\mathrm{the}\:\mathrm{steady}\:\mathrm{state}\:\mathrm{values}\:\mathrm{of}\: \\ $$$$\mathrm{these}\:\mathrm{two}\:\mathrm{proteins}.\:\mathrm{comment}\:\mathrm{on}\:\mathrm{the}\: \\ $$$$\mathrm{stability}\:\mathrm{of}\:\mathrm{the}\:\mathrm{steady}\:\mathrm{state}\:\mathrm{and}\:\mathrm{the}\:\mathrm{type}\:\mathrm{of}\: \\ $$$$\mathrm{phase}\:\mathrm{portrait}\:\mathrm{expected}\:\mathrm{for}\:\mathrm{this}\:\mathrm{system}. \\ $$$$\mathrm{consider}\:\mathrm{x},\mathrm{y}\geqslant\mathrm{0} \\ $$$$\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{\mathrm{y}}{\mathrm{1}+\mathrm{y}}−\mathrm{2}\:\:\:\:\frac{\mathrm{dy}}{\mathrm{dt}}=\frac{\mathrm{x}}{\mathrm{1}+\mathrm{x}}−\mathrm{y} \\ $$
Answered by Berbere last updated on 07/May/24
(dx/dt)=(y/(1+y))−2x=f(x,y);(dy/dt)=(x/(1+x))−y=g(x,y)  f(x,y)=g(x,y)=0⇔ { (((y/(1+y))−2x=0)),(((x/(1+x))−y=0)) :}⇔ { (((y/(1+y))−2x=0)),((x(1−y)=y)) :}  ⇒(y/(1+y))−2(y/(1−y))=0⇒y(1−y)−2y(1+y)=0  −3y^2 −y⇒y(3y+1)=0⇒y=0;y≥0  y=0⇒x=0  Jacobien MatriX J= ((((∂f/∂x)       (∂g/∂x))),(((∂f/∂y)       (∂g/∂y))) )(x,y)=(0,0)  J= (((−2       1)),((    1        −1)) )  det(X−zI_2 )=0⇒X^2 +X−3=0X∈{((−1−(√(13)))/2),((−1+(√(13)))/2)}  Re(((−1+(√(13)))/2))>0 No stability of Equilibre
$$\frac{{dx}}{{dt}}=\frac{{y}}{\mathrm{1}+{y}}−\mathrm{2}{x}={f}\left({x},{y}\right);\frac{{dy}}{{dt}}=\frac{{x}}{\mathrm{1}+{x}}−{y}={g}\left({x},{y}\right) \\ $$$${f}\left({x},{y}\right)={g}\left({x},{y}\right)=\mathrm{0}\Leftrightarrow\begin{cases}{\frac{{y}}{\mathrm{1}+{y}}−\mathrm{2}{x}=\mathrm{0}}\\{\frac{{x}}{\mathrm{1}+{x}}−{y}=\mathrm{0}}\end{cases}\Leftrightarrow\begin{cases}{\frac{{y}}{\mathrm{1}+{y}}−\mathrm{2}{x}=\mathrm{0}}\\{{x}\left(\mathrm{1}−{y}\right)={y}}\end{cases} \\ $$$$\Rightarrow\frac{{y}}{\mathrm{1}+{y}}−\mathrm{2}\frac{{y}}{\mathrm{1}−{y}}=\mathrm{0}\Rightarrow{y}\left(\mathrm{1}−{y}\right)−\mathrm{2}{y}\left(\mathrm{1}+{y}\right)=\mathrm{0} \\ $$$$−\mathrm{3}{y}^{\mathrm{2}} −{y}\Rightarrow{y}\left(\mathrm{3}{y}+\mathrm{1}\right)=\mathrm{0}\Rightarrow{y}=\mathrm{0};{y}\geqslant\mathrm{0} \\ $$$${y}=\mathrm{0}\Rightarrow{x}=\mathrm{0} \\ $$$${Jacobien}\:{MatriX}\:{J}=\begin{pmatrix}{\frac{\partial{f}}{\partial{x}}\:\:\:\:\:\:\:\frac{\partial{g}}{\partial{x}}}\\{\frac{\partial{f}}{\partial{y}}\:\:\:\:\:\:\:\frac{\partial{g}}{\partial{y}}}\end{pmatrix}\left({x},{y}\right)=\left(\mathrm{0},\mathrm{0}\right) \\ $$$${J}=\begin{pmatrix}{−\mathrm{2}\:\:\:\:\:\:\:\mathrm{1}}\\{\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:−\mathrm{1}}\end{pmatrix} \\ $$$${det}\left({X}−{zI}_{\mathrm{2}} \right)=\mathrm{0}\Rightarrow{X}^{\mathrm{2}} +{X}−\mathrm{3}=\mathrm{0}{X}\in\left\{\frac{−\mathrm{1}−\sqrt{\mathrm{13}}}{\mathrm{2}},\frac{−\mathrm{1}+\sqrt{\mathrm{13}}}{\mathrm{2}}\right\} \\ $$$${Re}\left(\frac{−\mathrm{1}+\sqrt{\mathrm{13}}}{\mathrm{2}}\right)>\mathrm{0}\:{No}\:{stability}\:{of}\:{Equilibre} \\ $$$$ \\ $$$$ \\ $$

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