Reduction 3: to \(N_x,N_y,\varrho_x,\varrho_y\) In this case our aim is to retain only information on the number and typical size of crystal distribution, so we eliminate the dimer concentrations x, y, using $$ \lambda_x = \frac\varrho_x2 N_x , \quad \lambda_y = \frac\varrho_y2 N_y , \quad x = \frac2 N_x^2\varrho_x ,
\quad y = \frac2 N_y^2\varrho_y . $$ (5.46)These transformations selleck kinase inhibitor reformulate the governing Eqs. 5.1–5.6 to $$ \frac\rm d N_x\rm d t = \frac12 \mu (\varrho -R) + \beta N_x – 2 (\mu\nu+\beta) \fracN_x^2\varrho_x – \frac2\xi N_x^3\varrho_x ,\\ $$ (5.47) $$ \frac\rm d N_y\rm d t = \frac12 \mu (\varrho – R) + \beta N_y – 2 (\mu\nu+\beta) \fracN_y^2\varrho_y – \frac2\xi N_y^3\varrho_y , \\ $$ (5.48) $$ \frac\rm d \varrho_x\rm d t = (\varrho-R)(\mu+\alpha N_x) – \frac4\mu\nu
N_x^2\varrho_x EPZ004777 price , \\ $$ (5.49) $$ \frac\rm d \varrho_y\rm d t = (\varrho-R)(\mu+\alpha N_y) – \frac4\mu\nu N_y^2\varrho_y , $$ (5.50)where \(R := \varrho_x + GSK1838705A cell line \varrho_y\). We now transform to total concentrations (N, R) and relative chiralities (ϕ and ζ) via $$ N_x = \frac12 N (1+\phi) , \quad N_y = \frac12 N (1-\phi) , \quad \varrho_x = \frac12 R (1+\zeta) , \quad \varrho_y = \frac12 R (1-\zeta) , $$ (5.51)together with \(c = \frac12 (\varrho – R)\), to obtain $$ \frac\rm d R\rm d t = (\varrho-R)(2\mu+ \alpha N) – \frac4\mu\nu N^2(1+\phi^2-2\phi\zeta)R (1-\zeta^2) , \\ $$
(5.52) $$ \beginarrayrll \frac\rm d N\rm d t & = & \mu (\varrho – R) + \beta N \\ && – \fracN^2R(1-\zeta^2) \left[ 2(\mu\nu+\beta) (1+\phi^2-2\phi\zeta) + \xi N (1+3\phi^2-3\phi\zeta-\phi^3\zeta) \right] , \\ \endarray $$ (5.53) $$ \beginarrayrll \frac\rm d\phi\rm d t &=& \beta\phi – \frac1N\frac\rm d N\rm d t\phi \\&& – \fracNR(1-\zeta^2) \left[ 2(\beta+\mu\nu)(2\phi-\zeta-\phi^2\zeta) + \xi N (3\phi-\zeta+\phi^3-3\phi^2\zeta) \right] , \\ \endarray $$ (5.54) $$ \frac, \zeta\rm d t = \frac\alpha (\varrho-R) N \phiR – \frac1R\frac\rm d R\rm d t \zeta – \frac4\mu\nu N^2 (2\phi-\zeta-\phi^2\zeta)R^2 (1-\zeta^2) . $$ (5.55)We now analyse this system in more detail, since this set of equations conserves mass, and is easier to analyse than Eqs. 5.33–5.35 due to the absence of square roots. We consider the two asymptotic limits (β ≪ 1 and α ∼ ξ ≫ 1) in which, at steady-state, the majority of mass is in the form of clusters. The Symmetric Steady-State Putting ζ = 0 = ϕ, we find the symmetric steady-state is given by $$ 0 = (\varrho-R)(2\mu+\alpha N) – \frac4\mu\nu N^2R , \\ $$ (5.56) $$ 0 =f \mu (\varrho-R) + \beta N – 2(\mu\nu+\beta)\fracN^2R – \frac\xi N^3R . $$ (5.