Reading the paper here, I'm confused about the production of figure 4d. The method of production is detailed in the supplementary information under the heading "Theoretical phase diagram".

Specifically, I'm confused about the use of theta and what it means. I have tried reproducing the figure using MATLAB and the parameters in the paper to see whether I can understand it better, but nothing quite works out.

To be clear, the paper obtains a steady state result for a series of ODEs: $$ y = \frac{\alpha (1+(βy)^2)}{ρ+(βy)^2} $$ Then rearranges as a cubic: $$ y^3 −αy^2 + \frac{ρ}{ β^2} y − \frac{α }{ β^2 } = 0 $$ Apparently, a general cubic with two identical roots has the form: $$ (y-a)(y-a)(y-aθ) = y^3 - (2 + θ)ay^2 + (1+2θ)a^2 y - θa^3 $$ The paper then equates coefficients: \begin{align} ρ &= (1 + 2θ)(1 + 2/θ)\\ aβ &= \frac{(2 + θ)^{3/2}}{θ^{1/2}}\tag{did you mean?} \end{align} And claims that these are the parametric equations that describe the graph in question. When plotting these equations, I get nothing like the graph. I can't really see where it comes from.

The graph in question and some equations are reproduced here on the last page: http:// web.mit.edu/biophysics/sbio/PDFs/L7_notes.pdf

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  • $\begingroup$ I don't have access to nature so I have no idea what the figure looks like. You need to post the figure. $\endgroup$ – dustin May 4 '15 at 15:21

The equation in the OP has been updated to $(y-a)^2(y-a\theta)$ which yields the correct expansion. Therefore, by equating the coefficients, we obtain \begin{align} \alpha &= (2+\theta)a\tag{1}\\ \rho/\beta^2 &= a^2(1+2\theta)\tag{2}\\ \alpha/\beta^2 &= a^3\theta\tag{3} \end{align} Using equations (1) to (3), we get the desired $\rho$ and $\alpha\beta$. Now the authors plotted $\frac{\alpha\beta}{\rho}$ and $\frac{1}{\rho}$.

By plotting $\frac{\alpha\beta}{\rho}$ and $\frac{1}{\rho}$, we can re-create the plot in Mathematica with

ParametricPlot[{1/((1 + 2 x) (1 + 2/x)), (2 + x)^(3/2)/(
  Sqrt[x] (1 + 2 x) (1 + 2/x))}, {x, 0.01, 50 \[Pi]}, 
 PlotRange -> {{0, .15}, {0, 0.6}}, AspectRatio -> 3/4]

enter image description here

If you dont have Mathematica, here is some Python code that will plot it as well.

import numpy as np
import matplotlib.pyplot as plt

fig = plt.figure()
t = np.linspace(0.01, 25*np.pi, 5000)
x = 1/((1+2*t)*(1+2/t))
y = (2+t)**1.5/(np.sqrt(t)*(1+2*t)*(1+2/t))


Here is the Python plot which has smoother lines compared to Mathematica's plot.

enter image description here

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  • $\begingroup$ Thanks! I thought something was up... Assuming their maths had been correct and the equations for rho and alpha-beta were correct, how would you go about using them? I'm a little confused about the theta? $\endgroup$ – user2290362 May 4 '15 at 16:29
  • $\begingroup$ The plot on the last page of web.mit.edu/biophysics/sbio/PDFs/L7_notes.pdf $\endgroup$ – user2290362 May 4 '15 at 16:51
  • $\begingroup$ Sorry could you please add a link to the wolfram alpha graph? $\endgroup$ – user2290362 May 4 '15 at 17:13
  • $\begingroup$ @user2290362 I plotted it in Python since Wolfram will only plot it if you have the full version of Mathematica. $\endgroup$ – dustin May 4 '15 at 17:30

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