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Homework Help: Dynamics - Non-Linear systems

  1. Apr 30, 2008 #1
    http://img132.imageshack.us/img132/5153/njga7.jpg [Broken]

    Question 8 is the question I need, but 7 has relevance for it.

    I have my equilibrium points, and am stuck on two of them with the same problem.

    1 of my points (H,P) = (0,L) - No hunters, but prey.

    I put this into my jacobian matrix, and I get two eigenvalues (E)

    E1 = Lb-s , E2 = -r

    E2 is a negative eigenvalue but what is E1? L, b and s are all stated as positive constants, so how do I know if it is negative or positive?

    Without knowing it I can't class the stability of the point, so I can't plot it in the phase plane.

    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Apr 30, 2008 #2


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    I've made a diagram of the phase plane and it seems to me that you need to look at two cases:
    L < (s/b) and L > (s/b) [well, there's also the special case L = (s/b)...] . So you don't know what the relation between the positive constants is; rather, you need to look at both possibilities.

    EDIT: The dynamics will definitely be different in each case. For L > (s/b), there are five equilbrium points, instead of four.
  4. Apr 30, 2008 #3
    Four equilibrium points you say? :P

    I saw 3

    (0,0), (0,L) and ((r/a)(1-s/bL) , s/b)

    So I actually have 4 where L > s/b and L < s/b

    What is the other if you don't mind me asking?
    Last edited: Apr 30, 2008
  5. Apr 30, 2008 #4


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    Sorry, I knew I should have drawn the graph in color. The fourth intersection I was seeing is between two dP/dt = 0 lines. OK, so three or four points...

    I suspect they want you to investigate the L = (s/b) case as well, since it represents a bifurcation between the two dynamical situations.
    Last edited: Apr 30, 2008
  6. Apr 30, 2008 #5
    Thanks, any chance you could scan your phase plane for me? :P

    I haven't had any practise in drawing them up until now, and I get confused in the arrow directions of the trajectories and how they are chosen. AS well as when some points are curved, or straight.
  7. Apr 30, 2008 #6


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    Sorry for the delay -- I had to get home, draw up the diagrams, and use my scanner here. (I won't get into my whole sad tale, but I don't have one where I work; on the other hand, my ancient pre-millennial home computer doesn't interact well with either the attachment window here or with many of the free image-hosts. So I did finally get this onto ImageCave...)


    Once I sorted out my diagram and marked out the isoclines correctly, I now see that there are three or two equilibrium points. (The case you were describing, I gather at this point, is the L > (s/b) case; the other case has one less equilibrium point.)

    You were also asking about how to map out the direction or flow field, I think, so I'll describe that, too. Here's how I deal with it, without using computing aids.

    For dH/dt , we established that it is zero for H = 0 or P = (s/b). So we will re-write its differential equation as

    dH/dt = ( -s + bP ) · H = b · H · ( P - [s/b] ) .

    We know that H is non-negative and all the constants are positive, so (s/b) > 0 . Thus,
    for P > (s/b) , dH/dt >= 0 and for P < (s/b) , dH/dt <= 0 . On the phase plane, then, the H-component of the flow vectors points to the right above the line P = (s/b) and to the left below that line. Moreover, the H-component increases with increasing vertical distance from the line.

    For dP/dt, there is a zero at P = 0 and an isocline

    P = L - (La/r) · H ,

    which is a straight line with intercepts (0, L) and (r/a, 0) . The differential equation can be expressed as

    dP/dt = { r · P · ( 1 - [P/L] ) } - ( a · H · P )

    = (rP/L) · { ( L - [aL/r] · H) - P } .

    The expression in parentheses, "()", is the equation for the P-isocline , from which P is subtracted. This means that if P is "above" the isocline, the expression in curly brackets, "{}", is negative; otherwise, the expression is positive. Since the constants are all positive, (rP/L) > 0, so dP/dt <= 0 above the isocline and dP/dt >= 0 below it. (I'm writing these sentences a little redundantly to try to avoid confusion.) Thus, the P-component of the flow vectors points down above the isocline and up for vectors below it, again with their magnitudes increasing with increasing distance from the slanted isocline.

    The diagrams have these flow vector components sketched for each case. Rather than dwell on what happens in each region, we can for now just look at what happens in the vicinity of the equilibria.

    We'll look at the L < (s/b) case first. There are two equilibrium points, (0, 0) and (0, L). The H-component of the flow field points toward both of them. For (0, 0), though, the P-component is "upward", so the flow near that point slips around it, making it a saddle point. For (0, L), the P-component below the slanted isocline is upward and is downard above it, so all P-components point to (0, P), making it a stable node (or attractor).

    In the L > (s/b) case, those two points are also present, but there is the additional equilibrium point you described. The point (0, 0) remains a saddle point. Since (0, L) is now above the H-isocline P = (s/b), the H-component of the flow field points away from it, so here it is now also a saddlepoint. For our new equilibrium point, however, when we check the four regions surrounding it, we find that the flow vectors "orbit" it in a clockwise direction. That makes it either a center or a stable spiral point. To resolve that further will require evaluating the eigenvalues there. (I'd look into that further, but I have a frightening amount of material to prepare for my students in the next week -- finals are less than two weeks from now...)

    BTW, what book are these problems from?
    Last edited: Apr 30, 2008
  8. May 2, 2008 #7


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    Have you realised the equation of 7 can be solved analytically?

    I'm not sure of 8 yet.
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