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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

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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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