Fig. 12 Graph of concentrations \(N_x,N_y,\varrho_x,\varrho_y,c\) against time on a logarithmic

time for the asymptotic limit 1, with initial conditions N x  = 0.2 = N y , \(\varrho_x=0.45\), \(\varrho_y=0.44\), other parameters given by α = 1 = ξ = μ, β = 0.01 , \(\varrho=8\). Since model equations are in nondimensional form, the time units are arbitrary Asymptotic Limit 2: α ∼ ξ ≫ 1 In this case we retain the assumptions Seliciclib in vitro that \(\mu,\nu=\cal O(1)\), however, we now RG-7388 in vitro impose \(\beta=\cal O(1)\) and α ∼ ξ ≫ 1. For a steady-state, we require the scalings \(N =\cal O(1/\sqrt\xi)\) and \(\varrho-R=\cal O(1/\xi^3/2)\). Specifically, solving Eqs. 5.56 and 5.57 we find $$ N \sim \sqrt\frac\beta\varrho\xi , \qquad R \sim \varrho – \frac4\mu\nu\alpha\varrho \sqrt\frac\beta\varrho\xi , $$ (5.64)hence the dimer concentrations \(c = \frac12 (\varrho-R) \sim N^3 = \cal O(1/\xi^3/2)\) and \(z = 2 N^2/\varrho \sim N^2 = \cal O(1/\xi)\). More precisely, \(c\sim MK5108 mouse (2\mu\nu/\alpha)\sqrt\beta/\varrho\xi\) and z ∼ 2β/ξ, in contrast with the previous asymptotic scaling which gave z ∼ N 2). To determine the timescales for crystal growth and dissolution, we

use Eq. 5.64 to define $$ N \sim n(t) \sqrt\beta \varrho/\xi , \quad R \sim \varrho – \frac4\mu\nu r(t)\alpha \varrho \sqrt\frac\beta\varrho\xi , $$ (5.65)and so rewrite the governing Eqs. 5.52 and 5.53 as $$ \frac\rm d n\rm d t = \beta n \left( 1 – n^2 – \frac2 n (\beta+\mu\nu)\sqrt\varrho\xi\beta \right) , \\ $$ (5.66) $$ \frac\rm d r\rm d t = \alpha \sqrt\frac\beta\varrho\xi \left( n^2 -r – \frac2\mu r\alpha \sqrt\frac\xi\beta\varrho \right) . $$ (5.67)Here, the former equation for n(t) corresponds to the slower timescale, with a rate β, the rate of equilibration of r(t) being \(\alpha \sqrt\beta\varrho/\xi\). The stability of the symmetric state is determined by $$ \fracRN \frac\rm d \rm d t \left( \beginarrayc \phi(t) \\ \zeta(t) \endarray \right) = \left( \beginarraycc -2 \sqrt\beta\varrho\xi

& \sqrt\beta\varrho\xi Endonuclease \\ -4\mu\nu \sqrt\beta / \xi \varrho & 4\mu\nu \endarray \right) \left( \beginarrayc \phi \\ \zeta \endarray \right) . $$ (5.68)This matrix has one large negative eigenvalue (\(\sim -2\sqrt\beta\varrho\xi\)) and one (smaller) positive eigenvalue (∼4μν); the former corresponds to (1, 0) T hence the decay of ϕ, whilst the latter corresponds to the eigenvector (1, 2) T . Hence the system (Eq. 5.68) has the solution $$ \left( \beginarrayc \phi \\ \zeta \endarray \right) \sim C \left( \beginarrayc 1 \\ 2 \endarray \right) \exp \left( 4 \mu \nu t \sqrt \frac\beta\varrho\xi \right) . $$ (5.69)The chiralities evolve on two timescales, the faster being 2β corresponding to the stable eigenvalue of Eq. 5.