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Differential equations - population model

  1. Jun 12, 2005 #1
    "Suppose that a certain population satisfies the initial value problem

    [tex]\frac{dy}{dt}=r(t)y - k, \qquad y(0)=y_0[/tex]

    where the growth rate [tex]r(t)[/tex] is given by

    [tex]r(t)=\frac{1+\sin t}{5}[/tex]

    and [tex]k[/tex] represents the rate of predation.

    (a) Suppose that [tex]k=\frac{1}{5}[/tex]. Plot [tex]y[/tex] versus [tex]t[/tex] for several values of [tex]y_0[/tex] between 1/2 and 1.

    (b) Estimate the critical initial population [tex]y_c[/tex] below which the population will become extinct.

    (c) Choose other values of [tex]k[/tex] and find the corresponding [tex]y_c[/tex] for each one.

    (d) Use that data you have found in parts (a) and (b) to plot [tex]y_c[/tex] versus [tex]k[/tex]."

    I'm trying to apply the Method of Integrating Factors, but I'm stuck. Here's what I have:

    [tex]\frac{dy}{dt}- r(t)y = - k[/tex]

    [tex]\mu = \exp \left[ -\frac{1}{5} \int \left( 1 + \sin t \right) \: dt \right] = \exp \left( \frac{\cos t}{5} - \frac{t}{5} \right)[/tex]

    [tex]y(t) = \exp \left( \frac{t}{5} - \frac{\cos t}{5} \right) \int -k \exp \left( \frac{\cos t}{5} - \frac{t}{5} \right) \: dt[/tex]

    [tex]y(t) = -k \exp \left( \frac{t}{5} - \frac{\cos t}{5} \right) \underbrace{\int \exp \left( \frac{\cos t}{5} - \frac{t}{5} \right) \: dt} _{\mbox{?}}[/tex]

    Any help is highly appreciated.
     
  2. jcsd
  3. Jun 12, 2005 #2

    Pyrrhus

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    the integral you underlined cannot be expressed in elemental functions, but you can still work the rest of the problems.
     
    Last edited: Jun 12, 2005
  4. Jun 13, 2005 #3
    I can get the vector field of this differential equation without any problem. Still, I can't plot [tex]y(t)[/tex] for several values of [tex]y_0[/tex]---part (a)---because I can't find [tex]y(t)[/tex] in the first place. I've attached a Mathematica 5 notebook that shows it.

    Thanks
     

    Attached Files:

  5. Jun 13, 2005 #4

    saltydog

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    Dude, you got Mathematica and you're sayin' you can't? What about NDSolve? Here, I know it by heart by now cus' I use it all the time:

    sol1=NDSolve[{y'[t]-(1+Sin[t])/5 y[t]== -k,y(0)=ystart},y,{t,0,whatever}];

    f[x_]:=Evaluate[y[x] /. sol1];

    Edit: Tell you what Thiago, what about writing a complete Mathematica program (only because the integral is non-elementary) that does all the calculating, searching for the critical values, then outputs a nice plot of y0 vs. k? Whatever. Anyway, why does this happen? That would make a good e don't you think?
     
    Last edited: Jun 13, 2005
  6. Jun 13, 2005 #5

    saltydog

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    Is it just me or is this a tough problem to code for other people? I don't know. I'm kinda slow. Thiago, I've attached a plot of what I came up with. Are you getting the same thing? If you got it, good. If not and want help, just post a message. Still can't say why it behaves that way. You? Don't suppose you want to work on e do you?
     

    Attached Files:

  7. Jun 14, 2005 #6

    saltydog

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    e

    Alright, I got e. Thiago, I don't want to be doing your homework alright. In fact, e is not even your homework so I think I'm ok.

    First instead of integrating indefinitely, do so definitely from [itex]t_0[/itex] to t:

    [tex]\int_{t=0,y=y_0}^{t=t,y=y} d[e^{1/5(Cos(t)-t)}y]=-k\int_{t_0}^t e^{1/5(Cos(t)-t)}[/tex]

    I know, the t's, the y's, y0, is confussing on the left hand side but you know what I mean.

    This yields:

    [tex]y(t)=e^{-1/5(Cos(t)-t)}[y_0e^{1/5}-k\int_0^t e^{1/5(Cos(u)-u)}du][/tex]

    Now,

    [tex]e^{-1/5(Cos(t)-t)}[/tex]

    is always greater than 0, so y(t) will reach zero only when the expression in brackets reaches zero. That is, when:

    [tex]y_0=ke^{-1/5}\int_0^t e^{1/5(Cos(u)-u)}du[/tex]

    Now hear's the important point: The critical point is when [itex]y_0[/itex] just equals that expression and since the integral is an increasing function which reaches a limit, We can say the following about the critical point [itex]y_c[/itex]

    [tex]y_c=ke^{-1/5}\int_0^{\infty} e^{1/5(Cos(u)-u)}du[/tex]

    Numerically, the integral is approximately 5.089. Thus we have the following expression for the critical point as a function of k:

    [tex]y_c=ak[/tex]

    with:
    [tex] a\approx 5.089 e^{-1/5}[/tex]

    The three plots exhibit this. The first plot is the one I generated above using a brute force approach to search for the critical points. The second one is the expression for [itex]y_c[/itex]. The third one is a superposition of both. That is, they are identical.
     

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