Asymptotic expansion on 3 nonlinear ordinary differential equations

[wel]

The 3 nonlinear differential equations are as follows
\begin{equation}
\epsilon \frac{dc}{dt}=\alpha I + \ c (-K_F - K_D-K_N s-K_P(1-q)), \nonumber
\end{equation}
\begin{equation}
\frac{ds}{dt}= \lambda_b P_C \ \epsilon \ c (1-s)- \lambda_r (1-q) \ s, \nonumber
\end{equation}
\begin{equation}
\frac{dq}{dt}= K_P (1-q) \frac{P_C}{P_Q} \ \ c - \gamma \ q, \nonumber
\end{equation}
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:

\begin{equation}
c_0(0)= c(0) = 0.25 \nonumber
\end{equation}
\begin{equation}
s_0(0)= cs(0) = 0.02 \nonumber \nonumber
\end{equation}
\begin{equation}
q_0(0)=q(0) = 0.98 \nonumber \nonumber
\end{equation}
and
\begin{equation}
c_i(0)= 0, \ i>0\nonumber
\end{equation}
\begin{equation}
s_i(0)= 0, \ i>0 \nonumber \nonumber
\end{equation}
\begin{equation}
q_i(0)=0, i>0. \nonumber \nonumber
\end{equation}

=> i started with the expansions :
\begin{equation}
c= c_0+ \epsilon c_1 + \epsilon^2 c_2+... \nonumber
\end{equation}
\begin{equation}
s= s_0+ \epsilon s_1 + \epsilon^2 s_2+... \nonumber
\end{equation}
\begin{equation}
q= q_0+ \epsilon q_1 + \epsilon^2 q_2+... \nonumber
\end{equation}
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.

 

[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), \\<br /> Q(t) = q(\epsilon t) = Q_0(t) + \epsilon Q_1(t), \\<br /> S(t) = s(\epsilon t) = S_0(t) + \epsilon S_1(t), so that <br /> \dot C = \alpha I+ C(−K_F−K_D−K_NS−K_P(1−Q)) \\<br /> \dot S = \epsilon(\lambda_b P_C \epsilon C(1−S)−\lambda_r (1−Q) S), \\<br /> \dot Q = \epsilon\left(K_P(1−Q)\frac{P_C}{P_Q} C−\gamma Q\right).<br /> Away from t = 0 you look at c, q and s. At first order in \epsilon you get <br /> \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)<br /> 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 <br /> \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 <br /> \lim_{t \to \infty} C_i(t) = \lim_{t \to 0} c_i(t)<br /> etc.