Asymptotic expansion on 3 nonlinear ordinary differential equations

1. Jul 27, 2014

wel

The 3 nonlinear differential equations are as follows

\epsilon \frac{dc}{dt}=\alpha I + \ c (-K_F - K_D-K_N s-K_P(1-q)), \nonumber

\frac{ds}{dt}= \lambda_b P_C \ \epsilon \ c (1-s)- \lambda_r (1-q) \ s, \nonumber

\frac{dq}{dt}= K_P (1-q) \frac{P_C}{P_Q} \ \ c - \gamma \ q, \nonumber

I want to use asymptotic expansion on $c, s$ and $q$.
And values of parameters are:

$K_F = 6.7 \times 10^{-2},$

$K_N = 6.03 \times 10^{-1}$

$K_P = 2.92 \times 10^{-2}$,

$K_D = 4.94 \times 10^{-2}$,

$\lambda_b= 0.0087$,

$I=1200$

$P_C = 3 \times 10^{11}$

$P_Q = 2.304 \times 10^{9}$

$\gamma=2.74$

$\lambda_{b}=0.0087$

$\lambda_{r}= 835$

$\alpha=1.14437 \times 10^{-3}$

For initial conditions:

c_0(0)= c(0) = 0.25 \nonumber

s_0(0)= cs(0) = 0.02 \nonumber \nonumber

q_0(0)=q(0) = 0.98 \nonumber \nonumber

and

c_i(0)= 0, \ i>0\nonumber

s_i(0)= 0, \ i>0 \nonumber \nonumber

q_i(0)=0, i>0. \nonumber \nonumber

=> i started with the expansions :

c= c_0+ \epsilon c_1 + \epsilon^2 c_2+......... \nonumber

s= s_0+ \epsilon s_1 + \epsilon^2 s_2+......... \nonumber

q= q_0+ \epsilon q_1 + \epsilon^2 q_2+......... \nonumber

we are only interseted in up to fisrt power of $\epsilon$.
so, we should get total 6 approximate differential equations to get answer for
$\frac{dc_0}{dt}, \frac{ds_0}{dt}, \frac{dq_0}{dt}, \frac{dc_1}{dt}, \frac{ds_1}{dt}$and $\frac{dq_1}{dt}$

but i think $\frac{dc_1}{dt}$ will disappear while expanding and equating the up to first power of $\epsilon$, do i need to go further up to $\epsilon{^2}$ because $\frac{dc_1}{dt}$is very important to find and we need 6 approximate differetial equations in total. what can i do? please some one help me.

2. Jul 31, 2014

pasmith

Because you have $\epsilon$ multiplying the highest order derivative which appears, your problem is singular. Thus you will need to do a matched expansion.

Near $t = 0$ you must look at $$C(t) = c(\epsilon t) = C_0(t) + \epsilon C_1(t), \\ Q(t) = q(\epsilon t) = Q_0(t) + \epsilon Q_1(t), \\ S(t) = s(\epsilon t) = S_0(t) + \epsilon S_1(t),$$ so that $$\dot C = \alpha I+ C(−K_F−K_D−K_NS−K_P(1−Q)) \\ \dot S = \epsilon(\lambda_b P_C \epsilon C(1−S)−\lambda_r (1−Q) S), \\ \dot Q = \epsilon\left(K_P(1−Q)\frac{P_C}{P_Q} C−\gamma Q\right).$$ Away from $t = 0$ you look at $c$, $q$ and $s$. At first order in $\epsilon$ you get $$\frac{dc_0}{dt} = c_1(-K_F - K_D - K_Ns_0 - K_P(1 - q_0)) + c_0(-K_Ns_1 -K_Pq_1)$$ which is an algebraic equation for $c_1$, since the value of $\frac{dc_0}{dt}$ is fixed by the constraint (obtained from the leading order terms) that $$\alpha I+ c_0(−K_F−K_D−K_Ns_0−K_P(1−q_0)) = 0.$$

You match the two expansions by requiring that $$\lim_{t \to \infty} C_i(t) = \lim_{t \to 0} c_i(t)$$ etc.