Biological Differential Equation

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SUMMARY

The discussion centers on a system of coupled differential equations defined by the equations \(\frac{dR_1}{dt} = R_1^0 \cdot g\left( \frac{R_2}{K_R} \right) - R_1\) and \(\frac{dR_2}{dt} = R_2^0 \cdot g\left( \frac{R_1}{K_R} \right) - R_2\). The function \(g\left( \frac{R_i}{K_r} \right)\) is specified as \(\frac{1 + f\cdot \left[ \frac{R_i}{K_R} \right]^2 }{1 + \left[ \frac{R_i}{K_r} \right]^2 }\), where \(f << 1\). The user successfully determined that manipulating \(R_1^0\) and \(R_2^0\) can yield either one or three solutions that satisfy both equations, with equilibrium points identified as the intersections of the steady state solutions.

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Kreizhn
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Homework Statement


I have a system of coupled differential equations of the form
[tex]\frac{dR_1}{dt} = R_1^0 \cdot g\left( \frac{R_2}{K_R} \right) - R_1[/tex]
[tex]\frac{dR_2}{dt} = R_2^0 \cdot g\left( \frac{R_1}{K_R} \right) - R_1[/tex]
where
[tex]g\left( \frac{ R_i}{K_r} \right) = \frac{ 1 + f\cdot \left[ \frac{ R_i}{K_R} \right]^2 }{1 + \left[ \frac{R_i}{K_r} \right]^2 }[/tex]
where f << 1 is a constant, [itex]R_1^0[/itex] is the steady state level of [itex]R_1[/itex] in the absence of [itex]R_2[/itex] and vice versa.

I need to show (graphically) that if we are free to manipulate [itex]R_1^0, R_2^0[/itex] then this can lead to one or three solutions that simultaneously satisfy both equations.

The Attempt at a Solution


It seems to me that an obvious choice for a single solution would be to set [itex]R_i^0 =0[/itex] which will decouple the systems and make them decreasing exponentials. However, other than this I am unsure how to determine that there are three solutions, let alone what it means to do this graphically.
 
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Nevermind, solved it.

The equilibrium points are the intersections of the steady state solutions.
 

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