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A system of differential equations

  1. Sep 9, 2006 #1
    Q. Suppose two populations are governed by the equations

    [tex]
    \mathop x\limits^ \bullet = x - \frac{1}{4}x^2 - \frac{1}{4}y^2
    [/tex]

    [tex]
    \mathop y\limits^ \bullet = xy - y - \frac{1}{2}y^2
    [/tex]

    (i) Show that the relevant non-zero critical points are (4,0) and (2,2).
    (ii) Find the equation of the particular trajectory that starts at (x,y) = (3.9,0.1)
    (iii) Find and sketch the trajectories in the neighbourhood of (2,2).

    The first one is just solving the simultaneous equations [tex]\mathop x\limits^ \bullet = \mathop y\limits^ \bullet = 0[/tex].

    I can't find a way to do (ii) properly though because the answer I obtain doesn't appear to allow me to find a relationship between x and y which is independent of t (ie. the required trajectory).

    My attempt at (ii) is as follows. I linearize the system near the critical point (x,y) = (4,0) by setting x = 4 + X and y = Y. Plugging this back into the original system (top of this message) I obtain

    [tex]
    \begin{array}{l}
    \mathop X\limits^ \bullet = - X - Y \\
    \mathop Y\limits^ \bullet = 3Y \\
    \end{array}
    [/tex]

    So the linear system associated with the original system in a neighbourhood of the critical point (x,y) = (4,0) is

    [tex]
    \left[ {\begin{array}{*{20}c}
    {\mathop X\limits^ \bullet } \\
    {\mathop Y\limits^ \bullet } \\
    \end{array}} \right] = \left[ {\begin{array}{*{20}c}
    { - 1} & { - 1} \\
    0 & 3 \\
    \end{array}} \right]\left[ {\begin{array}{*{20}c}
    X \\
    Y \\
    \end{array}} \right]
    [/tex]

    Let C denote the coefficient matrix of the linear system. I found the eigenvalues of C to be [tex]\lambda _1 = - 1[/tex] and [tex]\lambda _2 = 3[/tex] with corresponding eigenvectors

    [tex]
    \mathop {v_1 }\limits^ \to = \left[ {\begin{array}{*{20}c}
    1 \\
    0 \\
    \end{array}} \right]
    [/tex]

    and

    [tex]
    \mathop {v_2 }\limits^ \to = \left[ {\begin{array}{*{20}c}
    1 \\
    { - 4} \\
    \end{array}} \right]
    [/tex]

    respectively.

    So the general solution is:

    [tex]
    \left[ {\begin{array}{*{20}c}
    X \\
    Y \\
    \end{array}} \right] = A\left[ {\begin{array}{*{20}c}
    1 \\
    0 \\
    \end{array}} \right]e^{ - t} + B\left[ {\begin{array}{*{20}c}
    1 \\
    { - 4} \\
    \end{array}} \right]e^{3t}
    [/tex]

    I need the particular trajectory that starts at (x,y) = (3.9,0.1)...in other words at (X,Y) = (-0.1,0.1) since x = 4 + X and y = Y.

    Using (X,Y) = (-0.1,0.1) at t = 0 I get

    [tex]
    \left[ {\begin{array}{*{20}c}
    X \\
    Y \\
    \end{array}} \right] = 0.075\left[ {\begin{array}{*{20}c}
    { - 1} \\
    0 \\
    \end{array}} \right]e^{ - t} - 0.025\left[ {\begin{array}{*{20}c}
    1 \\
    { - 4} \\
    \end{array}} \right]e^{3t}
    [/tex]

    There certainly doesn't look like a relationship between X and Y which is independent of t which means that I can't find the trajectory. So I must have gone wrong somewhere. I've gone over my working many times but I don't know what I'm missing. Can someone please help me out with part (ii)?
     
    Last edited: Sep 9, 2006
  2. jcsd
  3. Sep 9, 2006 #2
    What do you mean by trajectory? Is t the trajectory? I don't see t mentioned in the equations you gave for the populations so I don't understand what you're talking about.
     
  4. Sep 9, 2006 #3
    A trajectory is a path through x-y space as t increases. You are plotting it on an x-y graph, but it's not x(y) or y(x). You are interested in how the point (x,y) changes as t increases, but you don't plot t on the same plot. The trajectory will probably be some sort of curve in the plane, and if you want you can draw arrows on it to show the direction of evolution as t increases.
     
  5. Sep 9, 2006 #4
    I still can't solve the problem and I'm not sure if people understand exactly what I am after so I will give an example. If the eigenvalues of the coefficient matrix were purely imaginary then the general solution of the system would consist of cosines and sines only. Plugging in the initial conditions would give me expressions for x and y in terms of cosines and sines only.

    I would then consider the expressions for x^2 and y^2, multiply one of them by an appropriate factor and then add them. This will generally lead to an equation relating x and y which is in fact an ellipse and that would be a trajectory. Because all of the 't dependence' is eliminated.

    I don't know how I'm supposed to find an explicit relationship (which is independent of any other variables) between x and y with the solutions that I've found. Can someone please help me out?

    Edit: t is a dummy variable...you could call it u,f,d or whatever you want. It arises because you're solving a system of equations.
     
  6. Sep 10, 2006 #5

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    From
    [tex]\frac{dx}{dt}= x-\frac{1}{4}x^2- \frac{1}{4}y^2[/tex]
    and
    [tex]\frac{dy}{dt}= xy- y- \frac{1}{2}y^2[/tex]
    we have
    [tex]\frac{dy}{dx}= \frac{xy- y- \frac{1}{2}y^2}{x- \frac{1}{4}x^2- \frac{1}{4}y^2}[/tex]
    [tex] = \frac{y(4x- 4- 2y)}{4x- x^2- y^2}[/tex]
     
  7. Sep 10, 2006 #6

    saltydog

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    Science Advisor
    Homework Helper

    Benny, t is "time". The populations are functions of time, that is, x(t) and y(t). You don't see time in the ODEs since they'er "homogeneous", that is, the differential equation is not a function of time. You could plot x(t) and y(t) independently but usually the "phase portrait" (y as a fuction of x) gives important information about the dynamics of the system.
     
  8. Sep 10, 2006 #7

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    I rather suspect that Benny understands all that! The problem, rather, is how to solve the equation
    [tex](4y+ 2y^2- 4xy)dx+ (4x- x^2- y^2)dy= 0[/tex]
    that you get after eliminating t.
     
  9. Sep 10, 2006 #8
    Hmm at least one positive thing has come out of this so far. After reading through the replies I've realised that once I linearise a non-linear system I can just divide y' by x' and solve the resulting ODE to find the trajectories rather than solving the linearised system as I had done previously. :rolleyes:

    I suspect that if I have the correct general solution to the linearised system then I should be able to use it to find the trajectory fairly easily. I've come across other examples like the following where I'm given the general solution of the linearised system.

    [tex]
    \begin{array}{l}
    \mathop x\limits^ \bullet = 2xy - 4y - 8 \\
    \mathop y\limits^ \bullet = 4y - x^2 \\
    \end{array}
    [/tex]

    The general solution of the linearised system near (4,2) after setting X = x-4 and Y = y-2 is:

    [tex]
    \left[ {\begin{array}{*{20}c}
    X \\
    Y \\
    \end{array}} \right] = c_1 \left[ {\begin{array}{*{20}c}
    1 \\
    1 \\
    \end{array}} \right]e^{8t} + c_2 \left[ {\begin{array}{*{20}c}
    1 \\
    2 \\
    \end{array}} \right]e^{12t}
    [/tex]

    From the above information the question asks for "the equation of the trajectory that starts at (x,y) = (3.9,2.1)"...ie. at (X,Y) = (-0.1,0.1).

    I find that c_1 = -0.2 and c_2 = 0.1. From what I can see there isn't a relationship between X and Y which I can extract from the solution. Oh well...
     
  10. Sep 10, 2006 #9

    saltydog

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    Science Advisor
    Homework Helper

    If I mis-understood Benny, then let him say so Hall. His comment above most certainly looked like he did not to any objective person reading it. I wished only to be helpful, not subversive like you.
     
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