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Asymptotic expansion on 3 nonlinear ordinary differential equations

  1. Jul 27, 2014 #1

    wel

    User Avatar
    Gold Member

    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 [itex]c, s[/itex] and [itex]q[/itex].
    And values of parameters are:

    [itex]K_F = 6.7 \times 10^{-2},[/itex]

    [itex]K_N = 6.03 \times 10^{-1}[/itex]

    [itex]K_P = 2.92 \times 10^{-2}[/itex],

    [itex]K_D = 4.94 \times 10^{-2}[/itex],

    [itex]\lambda_b= 0.0087[/itex],

    [itex]I=1200[/itex]

    [itex]P_C = 3 \times 10^{11}[/itex]

    [itex]P_Q = 2.304 \times 10^{9}[/itex]

    [itex]\gamma=2.74 [/itex]

    [itex]\lambda_{b}=0.0087[/itex]

    [itex]\lambda_{r}= 835[/itex]

    [itex]\alpha=1.14437 \times 10^{-3}[/itex]

    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 [itex]\epsilon[/itex].
    so, we should get total 6 approximate differential equations to get answer for
    [itex]\frac{dc_0}{dt}, \frac{ds_0}{dt}, \frac{dq_0}{dt}, \frac{dc_1}{dt}, \frac{ds_1}{dt}[/itex]and [itex]\frac{dq_1}{dt}[/itex]

    but i think [itex]\frac{dc_1}{dt}[/itex] will disappear while expanding and equating the up to first power of [itex]\epsilon[/itex], do i need to go further up to [itex]\epsilon{^2}[/itex] because [itex]\frac{dc_1}{dt}[/itex]is very important to find and we need 6 approximate differetial equations in total. what can i do? please some one help me.
     
  2. jcsd
  3. Jul 31, 2014 #2

    pasmith

    User Avatar
    Homework Helper

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

    Near [itex]t = 0[/itex] you must look at [tex]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), [/tex] so that [tex]
    \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).
    [/tex] Away from [itex]t = 0[/itex] you look at [itex]c[/itex], [itex]q[/itex] and [itex]s[/itex]. At first order in [itex]\epsilon[/itex] you get [tex]
    \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)
    [/tex] which is an algebraic equation for [itex]c_1[/itex], since the value of [itex]\frac{dc_0}{dt}[/itex] is fixed by the constraint (obtained from the leading order terms) that [tex]
    \alpha I+ c_0(−K_F−K_D−K_Ns_0−K_P(1−q_0)) = 0.[/tex]


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