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Math Help - How to rescale model of diff eqs?

  1. #1
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    How to rescale model of diff eqs?

    Course: Self-Study
    Textbook: Mathematical Models in Population Biology and Epidemiology

    In the model considered here the population is divided into susceptibles (S), infectives (I), isolated or quarantined individuals (Q), and recovered individuals (R), for whom permanent immunity is assumed. Let N denote the total population i.e. $N=S+I+Q+R$, and let $A = S + I + R$ denote the active (nonisolated) individuals. The model takes the form:

    $$\begin{align}
    \frac{dS}{dt}&=\mu N-\mu S− \sigma S \frac{I}{A}\\
    \frac{dI}{dt}&=-(\mu + \gamma)I+\sigma S \frac{I}{A}\\
    \frac{dQ}{dt}&=-(\mu+\xi)Q+\gamma I\\
    \frac{dR}{dt}&=−\mu R+\xi Q \\
    A&=S+I+R
    \end{align}$$

    All newborns are assumed to be susceptible. $\mu$ is the per capita mortality rate, $\sigma$ is the per capita infection rate of an average susceptible individual provided everybody else is infected, $\gamma$ is the rate at which individuals leave the infective class, and $\xi$ is the rate at which individuals leave the isolated class; all are positive constants.

    Question: Rescale the model by: $\tau=\sigma t$, $u=\frac{S}{A}$, $y=\frac{I}{A}$, $q=\frac{Q}{A}$, $z=\frac{R}{A}$. Rearrange your new model as follows:

    $$\begin{align}
    \dot{y}&=y(1-\nu -\theta -y-z+\theta y-(\nu +\zeta)q)\\
    \dot{q}&=(1+q)(\theta y-(\nu+\zeta)q) \\
    \dot{z}&=\zeta q-\nu z +z(\theta y-(\nu+\zeta)q).
    \end{align}$$

    Express the new parameters in terms of the old parameters. Check that all the new parameters and variables are dimensionless.

    I would like some help on rescaling the original model because I have no idea on what it means in this case. I originally thought that I could just multiply by $\frac{1}{A}$ because that would 'scale' the system I have, but when doing so I cannot rearrange it to look like the second model due to the added parameters $\nu$, $\theta$, and $\zeta$. Any guidance would be appreciated.
    Last edited by MadSoulz; May 21st 2014 at 12:21 PM.
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  2. #2
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    Re: How to rescale model of diff eqs?

    toying with this I find it's not a 5 minute solution. There's nothing magic though, you just have to run your diff eq system through the substitutions, carefully deriving the new state variable time derivatives, and churn that all through the algebra mill.

    I suppose that those new parameters that show up with be "obvious" when the numbers are crunched.

    I'll take a further look at it this evening.
    Thanks from MadSoulz
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    Re: How to rescale model of diff eqs?

    Quote Originally Posted by romsek View Post
    toying with this I find it's not a 5 minute solution. There's nothing magic though, you just have to run your diff eq system through the substitutions, carefully deriving the new state variable time derivatives, and churn that all through the algebra mill.

    I suppose that those new parameters that show up with be "obvious" when the numbers are crunched.

    I'll take a further look at it this evening.
    Many thanks for your time. I really appreciate it. For the moment I've skipped over this section and moved on to the next, but I will return to it shortly.
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  4. #4
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    Re: How to rescale model of diff eqs?

    The key to solving this problem is to recognize that \frac{dI}{dt}=\frac{dI}{d\tau}\frac{d\tau}{dt}= \dot{I} \sigma.

    So since y=\frac{I}{A}, solving for I and deriving it gives us \frac{dI}{dt}=(\dot{A}y+A\dot{y})\sigma=\frac{dA}{  dt}y+A\dot{y}\sigma.

    \frac{dI}{dt}=(\mu Q-\gamma I+\xi Q)y+A\dot{y}\sigma, since A=S+I+R.

    By equating both of our \frac{dI}{dt} equations, we have -(\mu+\gamma)I+\sigma S\frac{I}{A}=(\mu Q-\gamma I+\xi Q)y+A\dot{y}\sigma and we can solve for \dot{y} and our new parameters become obvious.

    \begin{align*}\dot{y}&=y\left(1-\frac{\mu}{\sigma}-\frac{\gamma}{\sigma}-y-z+\frac{\gamma}{\sigma}y-\left(\frac{\mu}{\sigma}+\frac{\xi}{\sigma}\right)  q\right) \\ &=y(1-\nu -\theta -y-z+\theta y-(\nu +\zeta)q)\end{align*}

    Thus, \nu=\frac{\mu}{\sigma}, \theta=\frac{\gamma}{\sigma}, and \zeta=\frac{\xi}{\sigma}
    Thanks from romsek
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