Coloring decay

Recursion

[math]\displaystyle{ q+1 }[/math] color, [math]\displaystyle{ d }[/math]-degree

• recursion:

[math]\displaystyle{ f(x,y)=\frac{qy^d}{x^d+qy^d} }[/math]

• fixed point:

[math]\displaystyle{ \hat{x}=\hat{y}=\frac{q}{q+1} }[/math]

• partials:

[math]\displaystyle{ \begin{align} \frac{\partial f(x,y)}{\partial x}&=-\frac{dqx^{d-1}y^d}{(x^d+qy^d)^2}=-f(x,y)(1-f(x,y))\frac{d}{x}\\ \frac{\partial f(x,y)}{\partial y}&=\frac{dqx^{d}y^{d-1}}{(x^d+qy^d)^2}=f(x,y)(1-f(x,y))\frac{d}{y} \end{align} }[/math]

at the fixed point [math]\displaystyle{ \frac{\partial f(\hat{x},\hat{y})}{\partial x}=-\frac{d}{q+1} }[/math] and [math]\displaystyle{ \frac{\partial f(\hat{x},\hat{y})}{\partial y}=\frac{d}{q+1} }[/math].

• ratio:

[math]\displaystyle{ \begin{align} \alpha(x,y) &=\left[-\frac{\partial f(x,y)}{\partial x}\frac{1}{\Phi(x)}+\frac{\partial f(x,y)}{\partial y}\frac{1}{\Phi(y)}\right]\Phi\left(f(x,y)\right)\\ &=-\alpha_1(x,y)+\alpha_2(x,y) \end{align} }[/math]

where

[math]\displaystyle{ \begin{align} \alpha_1(x,y) &= \frac{\partial f(x,y)}{\partial x}\frac{\Phi\left(f(x,y)\right)}{\Phi(x)}\\ \alpha_2(x,y) &= \frac{\partial f(x,y)}{\partial y}\frac{\Phi\left(f(x,y)\right)}{\Phi(y)} \end{align} }[/math]

• uniqueness:

at the fixed point [math]\displaystyle{ \hat{x}=\hat{y}=\frac{q}{q+1} }[/math], [math]\displaystyle{ \alpha(\hat{x},\hat{y})=\frac{2d}{q+1} }[/math]. The critical boundary of uniqueness is [math]\displaystyle{ 2d=q+1 }[/math].

Cancelation

Let the system be right at the critical boundary, i.e. [math]\displaystyle{ 2d=q+1 }[/math]. Then [math]\displaystyle{ \alpha(\hat{x},\hat{y})=1 }[/math].

[math]\displaystyle{ \begin{align} \left.\frac{\partial\alpha(x,y)}{\partial x}\right|_{x=y=\frac{q}{q+1}} &= \left.\left[-\frac{\partial^2 f(x,y)}{\partial x^2}\frac{1}{\Phi(x)}+\frac{\partial f(x,y)}{\partial x}\frac{\Phi'(x)}{(\Phi(x))^2}+\frac{\partial^2 f(x,y)}{\partial y\partial x}\frac{1}{\Phi(y)}\right]\Phi(f(x,y))\right|_{x=y=\frac{q}{q+1}}\\ &\quad\,\,+\left.\left[-\frac{\partial f(x,y)}{\partial x}\frac{1}{\Phi(x)}+\frac{\partial f(x,y)}{\partial y}\frac{1}{\Phi(y)}\right]\Phi'\left(f(x,y)\right)\frac{\partial f(x,y)}{\partial x}\right|_{x=y=\frac{q}{q+1}}\\ &= -\frac{\partial^2 f(\hat{x},\hat{y})}{\partial x^2}+\frac{\partial^2 f(\hat{x},\hat{y})}{\partial y\partial x}-\frac{\Phi'(\hat{x})}{\Phi(\hat{x})}. \end{align} }[/math]

However,

[math]\displaystyle{ \begin{align} \left.\frac{\partial\alpha(x,y)}{\partial y}\right|_{x=y=\frac{q}{q+1}} &= \left.\left[-\frac{\partial^2 f(x,y)}{\partial x\partial y}\frac{1}{\Phi(x)}+\frac{\partial^2 f(x,y)}{\partial y^2}\frac{1}{\Phi(y)}-\frac{\partial f(x,y)}{\partial y}\frac{\Phi'(y)}{(\Phi(y))^2}\right]\Phi(f(x,y))\right|_{x=y=\frac{q}{q+1}}\\ &\quad\,\,+\left.\left[-\frac{\partial f(x,y)}{\partial x}\frac{1}{\Phi(x)}+\frac{\partial f(x,y)}{\partial y}\frac{1}{\Phi(y)}\right]\Phi'\left(f(x,y)\right)\frac{\partial f(x,y)}{\partial y}\right|_{x=y=\frac{q}{q+1}}\\ &= \left[-\frac{\partial^2 f(\hat{x},\hat{y})}{\partial x\partial y}\frac{1}{\Phi(\hat{x})}+\frac{\partial^2 f(\hat{x},\hat{y})}{\partial y^2}\frac{1}{\Phi(\hat{x})}-\frac{1}{2}\frac{\Phi'(\hat{x})}{(\Phi(\hat{x}))^2}\right]\Phi(\hat{x}) +\left[\frac{1}{2}\frac{1}{\Phi(\hat{x})}+\frac{1}{2}\frac{1}{\Phi(\hat{x})}\right]\frac{\Phi'(\hat{x})}{2}\\ &= -\frac{\partial^2 f(\hat{x},\hat{y})}{\partial x\partial y}+\frac{\partial^2 f(\hat{x},\hat{y})}{\partial y^2}. \end{align} }[/math]

Potential

The recursion and derivatives are:

[math]\displaystyle{ \begin{align} f &= f(x,y) =\frac{qy^d}{x^d+qy^d}\\ f_x &= \frac{\partial f}{\partial x} =-f(1-f)\frac{d}{x}\\ f_y &= \frac{\partial f}{\partial y} =f(1-f)\frac{d}{y}. \end{align} }[/math]

Then

[math]\displaystyle{ \begin{align} \frac{f_{x^2}}{f_x} &= \frac{f_x}{f}-\frac{f_x}{1-f}-\frac{1}{x} =\frac{f_x}{f(1-f)}(1-2f)-\frac{1}{x} =-\frac{d}{x}(1-2f)-\frac{1}{x}\\ \frac{f_{xy}}{f_x} &= \frac{f_y}{f}-\frac{f_y}{1-f} =\frac{f_y}{f(1-f)}(1-2f) =-\frac{d}{y}(1-2f)\\ \frac{f_{xy}}{f_y} &= \frac{f_x}{f}-\frac{f_x}{1-f} =\frac{f_x}{f(1-f)}(1-2f) =-\frac{d}{x}(1-2f)\\ \frac{f_{y^2}}{f_y} &= \frac{f_y}{f}-\frac{f_y}{1-f}-\frac{1}{y} =\frac{f_y}{f(1-f)}(1-2f)-\frac{1}{y} =-\frac{d}{y}(1-2f)-\frac{1}{y} \end{align} }[/math]

Recall that

[math]\displaystyle{ \begin{align} \alpha_1(x,y) &= f_x\cdot\frac{\Phi\left(f\right)}{\Phi(x)},\\ \alpha_2(x,y) &= f_y\cdot\frac{\Phi\left(f\right)}{\Phi(y)}. \end{align} }[/math]

Then

[math]\displaystyle{ \begin{align} \frac{\partial\alpha_1(x,y)}{\partial x} &= \alpha_1(x,y)\left[\frac{f_{x^2}}{f_x}+\frac{\Phi'(f)}{\Phi(f)}f_x-\frac{\Phi'(x)}{\Phi(x)}\right]\\ &= f_x\cdot\frac{\Phi\left(f\right)}{\Phi(x)} \left[\frac{f_x}{f(1-f)}\left(1-2f+\frac{\Phi'(f)}{\Phi(f)}f(1-f)\right)-\left(\frac{1}{x}+\frac{\Phi'(x)}{\Phi(x)}\right)\right]\\ &= -\frac{f_x}{x}\cdot\frac{\Phi\left(f\right)}{\Phi(x)}\left[d\left(1-2f+\frac{\Phi'(f)}{\Phi(f)}f(1-f)\right)+\left(1+\frac{\Phi'(x)}{\Phi(x)}x\right)\right]\\ &= -\frac{f_x}{x}\cdot\frac{\Phi\left(f\right)}{\Phi(x)}\left[\frac{(df(1-f)\Phi(f))'}{\Phi(f)}+\frac{(x\Phi(x))'}{\Phi(x)}\right]\\ \frac{\partial\alpha_2(x,y)}{\partial x} &= \alpha_2(x,y)\left[\frac{f_{xy}}{f_y}+\frac{\Phi'(f)}{\Phi(f)}f_x\right]\\ &= f_y\cdot\frac{\Phi\left(f\right)}{\Phi(y)}\frac{f_x}{f(1-f)}\left[1-2f+\frac{\Phi'(f)}{\Phi(f)}f(1-f)\right]\\ &= -\frac{f_y}{x}\cdot\frac{\Phi\left(f\right)}{\Phi(y)}\cdot d\left[1-2f+\frac{\Phi'(f)}{\Phi(f)}f(1-f)\right]\\ &= -\frac{f_y}{x}\cdot\frac{\Phi\left(f\right)}{\Phi(y)}\cdot\frac{(df(1-f)\Phi(f))'}{\Phi(f)}\\ \frac{\partial\alpha_1(x,y)}{\partial y} &= \alpha_1(x,y)\left[\frac{f_{xy}}{f_x}+\frac{\Phi'(f)}{\Phi(f)}f_y\right]\\ &= f_x\cdot\frac{\Phi\left(f\right)}{\Phi(x)}\frac{f_y}{f(1-f)}\left[1-2f+\frac{\Phi'(f)}{\Phi(f)}f(1-f)\right]\\ &= \frac{f_x}{y}\cdot\frac{\Phi\left(f\right)}{\Phi(x)}\cdot d\left[1-2f+\frac{\Phi'(f)}{\Phi(f)}f(1-f)\right]\\ &= \frac{f_x}{y}\cdot\frac{\Phi\left(f\right)}{\Phi(x)}\cdot\frac{(df(1-f)\Phi(f))'}{\Phi(f)}\\ \frac{\partial\alpha_2(x,y)}{\partial y} &= \alpha_2(x,y)\left[\frac{f_{y^2}}{f_y}+\frac{\Phi'(f)}{\Phi(f)}f_y-\frac{\Phi'(y)}{\Phi(y)}\right]\\ &= f_y\cdot\frac{\Phi\left(f\right)}{\Phi(y)} \left[\frac{f_y}{f(1-f)}\left(1-2f+\frac{\Phi'(f)}{\Phi(f)}f(1-f)\right)-\left(\frac{1}{y}+\frac{\Phi'(y)}{\Phi(y)}\right)\right]\\ &= \frac{f_y}{y}\cdot\frac{\Phi\left(f\right)}{\Phi(y)}\left[d\left(1-2f+\frac{\Phi'(f)}{\Phi(f)}f(1-f)\right)-\left(1+\frac{\Phi'(y)}{\Phi(y)}y\right)\right]\\ &= \frac{f_y}{y}\cdot\frac{\Phi\left(f\right)}{\Phi(y)}\left[\frac{(df(1-f)\Phi(f))'}{\Phi(f)}-\frac{(y\Phi(y))'}{\Phi(y)}\right] \end{align} }[/math]

Therefore,

[math]\displaystyle{ \begin{align} \frac{\partial \alpha(x,y)}{\partial x} &= -\frac{\partial\alpha_1(x,y)}{\partial x}+\frac{\partial\alpha_2(x,y)}{\partial x}\\ &= \frac{\Phi(f)}{x} \left[ \frac{f_x}{\Phi(x)} \left( \frac{(df(1-f)\Phi(f))'}{\Phi(f)}+\frac{(x\Phi(x))'}{\Phi(x)} \right) -\frac{f_y}{\Phi(y)}\frac{(df(1-f)\Phi(x))'}{\Phi(f)} \right]\\ &= \frac{\Phi(f)}{x} \left[ \left(\frac{f_x}{\Phi(x)}-\frac{f_y}{\Phi(y)}\right)\frac{(df(1-f)\Phi(f))'}{\Phi(f)} +\frac{f_x}{\Phi(x)} \frac{(x\Phi(x))'}{\Phi(x)} \right]\\ &= -\frac{df(1-f)\Phi(f)}{x} \left[ \left(\frac{1}{x\Phi(x)}+\frac{1}{y\Phi(y)}\right)\frac{(df(1-f)\Phi(f))'}{\Phi(f)} +\frac{1}{x\Phi(x)} \frac{(x\Phi(x))'}{\Phi(x)} \right]\\ \frac{\partial \alpha(x,y)}{\partial y} &= -\frac{\partial\alpha_1(x,y)}{\partial y}+\frac{\partial\alpha_2(x,y)}{\partial y}\\ &= \frac{\Phi(f)}{y} \left[ \frac{f_y}{\Phi(y)} \left( \frac{(df(1-f)\Phi(x))'}{\Phi(f)}-\frac{(y\Phi(y))'}{\Phi(y)} \right) -\frac{f_x}{\Phi(x)}\frac{(df(1-f)\Phi(f))'}{\Phi(f)} \right]\\ &= \frac{\Phi(f)}{y} \left[ \left(\frac{f_y}{\Phi(y)}-\frac{f_x}{\Phi(x)}\right)\frac{(df(1-f)\Phi(f))'}{\Phi(f)} -\frac{f_y}{\Phi(y)}\frac{(y\Phi(y))'}{\Phi(y)} \right]\\ &= \frac{df(1-f)\Phi(f)}{y} \left[ \left(\frac{1}{y\Phi(y)}+\frac{1}{x\Phi(x)}\right)\frac{(df(1-f)\Phi(f))'}{\Phi(f)} -\frac{1}{y\Phi(y)}\frac{(y\Phi(y))'}{\Phi(y)} \right]. \end{align} }[/math]

Consider the following condition:

[math]\displaystyle{ 2(dz(1-z)\Phi(z))'=(z\Phi(z))',\, }[/math]

which implies

[math]\displaystyle{ [2dz(1-z)\Phi(z)-z\Phi(z)]'=0 }[/math] thus [math]\displaystyle{ 2dz(1-z)\Phi(z)-z\Phi(z)=C\, }[/math], which gives us that

[math]\displaystyle{ \Phi(z)=\frac{1}{2dz(1-z)-z}. }[/math]