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Next: On at different values Up: Steady-state analysis: variables on Previous: Equilibrium existence

Local stability analysis

Variables zero at the steady state can be discarded from the analysis.

Using

$\displaystyle \dot{\frac{x^2}{ax^2+bx+c}}=\frac{bx^2 +2cx}{\left( {ax^2+bx+c} \right)^2},
$

one derives the diagonal term of the Jacobian (with $ b=2\alpha\left(D-x_i\right)$ and $ c=\alpha \left(D-x_i\right)^2$):

$\displaystyle J_{i,i}=-1 + 2 \sigma \alpha x_i \frac{D \left(D-x_i \right)}{\left( \alpha D^2 + x_i^2 \right)^2}
$

Using the steady state equation 9,

$\displaystyle J_{i,i}=-1 + \frac{2 \alpha}{\sigma x_i} \left( D \left(D-x_i \right)\right)=-1 +$    
$\displaystyle \frac{2}{\sigma}\left( \sigma -x_i - \alpha D \right)$    

$\displaystyle J_{i,i}=1-\frac{2 }{\sigma} \left( x_i+\alpha D \right)=1-\frac{2}{\sigma}\left( \alpha + x_i\left(1+k\alpha \right) \right)
$

The diagonal terms are negative for

$\displaystyle x_i>\frac{\sigma / 2 -\alpha}{1+k\alpha}
$

The off-diagonal terms are given by

$\displaystyle J_{i,j}=- \sigma x_{i}^2 \frac{ 2\alpha x_j + 2 \alpha \left( D - x_j \right)}{\left( \alpha D^2 + x_i^2 \right)^2}
$

$\displaystyle J_{i,j}=- 2 \sigma \alpha x_{i}^2 \frac{ D}{\left( \alpha D^2 + x_i^2 \right)^2}
$

$\displaystyle J_{i,j}= -\frac{2\alpha}{\sigma}D
$

$\displaystyle J_{i,j}-J_{i,i}=-1+2\frac{x_i}{\sigma}
$

Thus, a necessary condition for the equilibrium to be stable is

$\displaystyle \forall \bar{x}_i \mathrm{st} \bar{x}_i\ne 0, \bar{x}_i>\sigma/2$ (12)

This is possible if and only if $ \alpha<1/k^2$ and $ \sigma>2\frac{k\alpha+\sqrt{\alpha}}{1-k^{2} \alpha}$.

Condition 12 is stronger than the requirement for the diagonal element to be negative (and is thus also a sufficient condition), and can never be met by variables equal to the lower solution of equations 10 or 11 .

Thus, for any value of the transcription strength $ \sigma $ and for any number of coexistant variables $ k$, sufficiently low values of $ \alpha $ make the equilibrium stable. If there is a stable equilibrium with $ k$ variables on, there is also a stable equilibrium with $ p$ variables on, for $ 1<p<k$. For sufficiently large $ \sigma $, the necessary condition $ \alpha<1/k^2$ becomes sufficient for stability (see Figure 5 for an illustration of the validity of this condition).


next up previous
Next: On at different values Up: Steady-state analysis: variables on Previous: Equilibrium existence