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System of DE's, pray/predator model

  1. Jan 15, 2012 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    Consider the pray-predator model where the predator has an alternative way of surviving:
    [itex]\dot x_1=\alpha _1x_1(\beta _1-x_1 )+\gamma _1 x_1 x_2[/itex]
    [itex]\dot x_2=\alpha _2x_2(\beta _2-x_2 )-\gamma _2 x_1 x_2[/itex].
    1)Show that the change of coordinates [itex]\beta _i y_i (t)= x_i \left ( \frac{t}{\alpha _i \beta _i } \right )[/itex] leads to the following system of DE's:
    [itex]\dot y_1 =y_1 (1-y_1)+a_1 y_1 y_2[/itex]
    [itex]\dot y_2 =y_2 (1-y_2)-a_2 y_1 y_2[/itex]
    Where [itex]a_1=\frac{\gamma _1 \beta _2 }{\alpha_1 \beta _1}[/itex] and [itex]a_2=\frac{\gamma _2 \beta _1 }{\alpha _2 \beta _2}[/itex]
    2)What are the stable equilibrium populations when i)[itex]0<a_2<1[/itex], ii)[itex]a_2>1[/itex]?
    3)Suppose that [itex]a_ 1= 3 a_2[/itex] where [itex]a_2[/itex] is the a measure of the agressivity of the predator. What is the value of [itex]a_2[/itex] if the instinct of the predator consists of maximizing its population of stable equilibrium?
    2. Relevant equations

    Chain rule.

    3. The attempt at a solution
    For 1)
    From the first DE of the first system of DE's I must reach [itex]\dot y_1=y_1 (1-y_1)+\frac{\gamma _1 \beta _2 }{\alpha _1 \beta _1} y_1y_2[/itex].
    But using the change of coordinates, I get that [itex]\dot x_1=\alpha _1 \beta _1 ^2 \dot y_1 \Rightarrow \dot y_1 = \frac{\dot x_1 }{\alpha _1 \beta _1 ^2}=\frac{x_1 (\beta _1 -x_1)}{\beta _1 ^2}+\frac{\gamma _1 x_1 x_2}{\alpha _1 \beta _1 ^2}[/itex]. Now I don't know how to get rid of the x's and replace them by the y's. Probably with the change of coordinates but I don't see how. Any tip is appreciated.
     
  2. jcsd
  3. Jan 16, 2012 #2

    fzero

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    Just use [itex]x_i = \beta_i y_i[/itex]. The difference in the argument of the functions only affects the terms containing derivatives.
     
  4. Jan 17, 2012 #3

    fluidistic

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    Oh nice, thank you. I reach the desired result, so part 1) is solved. Just curious though (my math seems too weak), would what you say also be true if the argument was for example quadratic in t? Like [itex]\beta _i y_i (t)= x_i \left ( \frac{t^2}{\alpha _i \beta _i } \right )[/itex] for instance.
     
  5. Jan 17, 2012 #4

    fluidistic

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    Now trying 2).
    x is the pray population and y the predator's one.
    So what they ask me are the equilibrium points of the systems of DE's
    [itex]\dot x_1=\alpha _1x_1(\beta _1-x_1 )+\gamma _1 x_1 x_2[/itex]
    [itex]\dot x_2=\alpha _2x_2(\beta _2-x_2 )-\gamma _2 x_1 x_2[/itex]
    and
    [itex]\dot y_1 =y_1 (1-y_1)+a_1 y_1 y_2[/itex]
    [itex]\dot y_2 =y_2 (1-y_2)-a_2 y_1 y_2[/itex].
    I first tried to get the predator's equilibrium populations. This gave me [itex]y_1=0[/itex] or [itex]y_1=1+a_1 y_2[/itex] and [itex]y_2=0[/itex] or [itex]y_2=1-a_2y_1[/itex]. So there are 4 possible critical points. For instance [itex](y_1,y_2)=(0,0)[/itex], [itex](y_1,y_2)=(0,1)[/itex] and [itex](y_1,y_2)=(1,0)[/itex] are all critical points. The first one being the least interesting I think, it's when there's no population at all. But the 4th critical point, probably the most interesting is causing me troubles. It's when [itex](y_1,y_2)=(1+a_1y_2,1-a_2y_1)[/itex]. I reached that this is equivalent to [itex](y_1,y_2)=\left [ \frac{1+a_1}{1+a_1a_2} , 1-a_2 \left ( \frac{1+a_1}{1+a_1a_2} \right ) \right ][/itex]. To check out if this is indeed a critical point, I plug this back into the system of DE and equate to 0. I reach a condition on the constants... namely that [itex]1+a_1[/itex] must equal [itex]\left ( \frac{1+a_1}{1+a_1a_2} \right ) (1+a_1a_2)[/itex]. I'm surprised I find such a constraint, I'm probably missing something? And I still didn't use the condition(s) on [itex]a_2[/itex].
     
  6. Jan 17, 2012 #5

    fzero

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    Yes, the change of argument doesn't affect the nonderivative terms. But anything other than a linear change of argument will lead to explicit time dependence in the new equation via the chain rule.

    Yes, just simplify [itex]\left ( \frac{1+a_1}{1+a_1a_2} \right ) (1+a_1a_2)[/itex].
     
  7. Jan 18, 2012 #6

    fluidistic

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    Ok thanks fzero.
    The critical points are therefore for the pray:
    [itex](0,0)[/itex], [itex](0, \beta _2)[/itex], [itex](\beta _1 , 0)[/itex] and [itex]\left [ \frac{\left ( \beta _1 + \frac{\gamma _1 \beta _2 }{\alpha _1} \right ) }{\left ( 1+ \frac{\gamma _1 \gamma _2 }{\alpha _1 ^2 } \right ) }, \beta _2 - \frac{\gamma _2 \left ( \beta _1 + \frac{\gamma _1 \beta _2 }{\alpha _1} \right ) }{\alpha _1 \left ( 1+ \frac{\gamma _1 \gamma _2 }{\alpha _1 ^2 } \right ) } \right ][/itex].
    For the predator:
    [itex](0,0)[/itex], [itex](0,1)[/itex], [itex](1,0)[/itex] and [itex]\left [ \frac{1+a_1}{1+a_1a_2} , 1-a_2 \left ( \frac{1+a_1}{1+a_1a_2} \right ) \right ][/itex].
    Now I think I should associate the critical points of the pray to the critical points of the predator, using the given change of coordinates. And maybe make the use of [itex]a_1[/itex] and [itex]a_2[/itex] for the pray's 4th critical point. So far I do not see why if [itex]a_2[/itex] is smaller than 1 or greater than 1 can affect the critical points.
     
  8. Jan 18, 2012 #7

    fzero

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    That should be easy. It's trivial to see how they're related for the first 3 points.

    I don't recall the details of the behavior at these points, but note that the 4th critical point is equal to the 3rd when [itex]a_2=1[/itex]. Because this is a degenerate case, we need to study the behavior in the two regions separately.
     
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