# Thread: a messy looking ODE system

1. ## a messy looking ODE system

I have 6 ODES:

$\displaystyle W\frac{dl_E}{dt} = -l_E\rho_F + \psi_Ll_B$
$\displaystyle \frac{dl_B}{dt} = \rho_Fl_E - (\psi_L + \chi_L)l_B$
$\displaystyle \frac{dl_I}{dt} = \chi_Ll_B - \omega_Ll_I$
$\displaystyle \sigma\frac{d\rho_F}{dt} = \gamma_{rr}\sigma p_I - \chi_0p_F - ml_E\rho_F + m\psi_Ll_B - \frac{m\chi_Ll_B\rho_F}{1 - \rho_F}$
$\displaystyle \sigma\frac{d\rho_I}{dt} = \frac{\gamma_S\sigma}{K + c} + \chi_0f\rho_F + f(1 + \frac{\rho_F}{1 - \rho_F})m\chi_Ll_B - \sigma\gamma_{rr}\rho_I$
$\displaystyle \frac{dc}{dt} = \gamma(\omega_LR_Ll_I) - \lambda(c - 1)$

for which I'm trying to find the steady states, so setting each equation equal to zero:

$\displaystyle -l_E\rho_F + \psi_Ll_B = 0$....(1)
$\displaystyle \rho_Fl_E - (\psi_L + \chi_L)l_B = 0$....(2)
$\displaystyle \chi_Ll_B - \omega_Ll_I = 0$....(3)
$\displaystyle \gamma_{rr}\sigma p_I - \chi_0p_F - ml_E\rho_F + m\psi_Ll_B - \frac{m\chi_Ll_B\rho_F}{1 - \rho_F} = 0$....(4)
$\displaystyle \frac{\gamma_S\sigma}{K + c} + \chi_0f\rho_F + f(1 + \frac{\rho_F}{1 - \rho_F})m\chi_Ll_B - \sigma\gamma_{rr}\rho_I = 0$....(5)
$\displaystyle \gamma(\omega_LR_Ll_I) - \lambda(c - 1) = 0$....(6)

Then:

From (3) $\displaystyle l_I = \frac{\chi_Ll_B}{\omega_L}$....(7)

From (6) $\displaystyle c = 1 + K_1l_I$ where $\displaystyle K_1 = \frac{\gamma\omega_LR_L}{\lambda}$....(8)

Assume $\displaystyle \chi_0 = 0$

From (1) $\displaystyle l_E = \frac{\psi_Ll_B}{\rho_F}$....(9)

From (5) $\displaystyle \rho_I = \frac{1}{\sigma\gamma_rr}(\frac{\gamma_S\sigma}{K + c} + f(1 + \frac{\rho_F}{1 - \rho_F}))$....(10)

Subbing (10) into (4):

$\displaystyle \gamma_{rr}\sigma(\frac{1}{\sigma\gamma_rr}(\frac{ \gamma_S\sigma}{K + c} + f(1 + \frac{\rho_F}{1 - \rho_F}))) - m\psi_Ll_B + m\psi_Ll_B - \frac{m\chi_Ll_B\rho_F}{1 - \rho_F} = 0$

Subbing (7) into (8):

$\displaystyle c = 1 K_1 \frac{\chi_Ll_B}{\omega_L}$

$\displaystyle \gamma_{rr}\sigma(\frac{1}{\sigma \gamma_{rr}}(\frac{\gamma_S\sigma}{K + 1 + K_1\frac{\chi_Ll_B}{\omega_L}} + f(1 + \frac{\rho_F}{1 - \rho_F}))) - m\psi_Ll_B + m\psi_Ll_B - \frac{m\chi_Ll_B\rho_F}{1 - \rho_F} = 0$

From (2) $\displaystyle l_B = \frac{\rho_Fl_E}{\psi_L + \chi_L}$....(11)

$\displaystyle \gamma_{rr}\sigma(\frac{1}{\sigma\gamma_{rr}}(\fra c{\gamma_S\sigma}{K + 1 + \frac{K_1\rho_Fl_E\chi_L}{\omega_L(\psi_L + \chi_L)}} + f(1 + \frac{\rho_F}{1 - \rho_F}))) - \frac{m\chi_L\rho_F^2l_E}{(1 - \rho_F)(\psi_L + \chi_L)} = 0$....(*)

and due to the nature of the question I can assume $\displaystyle l_E = 0$ and hence have an equation for $\displaystyle \rho_F$.

Was just hoping that someone could please check this over for me, and also tell me if it possible to solve (*) in Maple, (never used it before so not sure).