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Phase Plane Portrait Plotting

  1. Sep 12, 2011 #1
    1. The problem statement, all variables and given/known data
    Given x' = y - 2
    y' = [itex]\frac{1}{4}[/itex]x-[itex]\frac{1}{2}[/itex]

    Draw the phase plane.

    3. The attempt at a solution
    First I found what I believe are the nullclines by taking x'=0 and y'=0.
    x' = 0 = y -2
    so y = 2.
    y' = 0 = [itex]\frac{1}{4}[/itex]x-[itex]\frac{1}{2}[/itex]
    so x = 2.
    This gives us a stationary point at (2,2).

    I understand that I need to plug in points in each quadrant to see what the slopes do, but what I'm confused about is how to find the apparent lines that the points don't cross.
    I plotted this in pplane and found two linear diagonals which seemed to be lines of attraction.

    How would I find those lines algebraically? I thought maybe x'=y' but that would only give me y = [itex]\frac{1}{4}[/itex]x+[itex]\frac{3}{2}[/itex], which is possible, but there is another line with a negative slope as well.

    Any thoughts? Thanks!
     
  2. jcsd
  3. Sep 12, 2011 #2

    dynamicsolo

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    The two lines x = 2 and y = 2 divide the phase plane into quadrants. The line y = 2 marks all the points where dx/dt = 0 and x = 2 marks the points where dy/dt = 0. We have a parameter t ("time") , so we will want to work with the parametric form of the slope in the phase plane [itex]\frac{dy}{dx} = \frac{y'}{x'} [/itex] .

    If you look at your differential equations, you see that y' > 0 for x > 2 and x' > 0 for y > 2 . So the "first quadrant" with x > 2 , y > 2 has y' > 0 and x' > 0 , so the slope of the direction field lines in that quadrant [itex]\frac{dy}{dx} [/itex] will be positive. You can now make similar reasoning for each of the other quadrants about the point (2, 2). Also, what happens to the slope of the direction field on the lines x = 2 or y = 2 ? What sort of "equilibrium point" is (2, 2)? (You have a warning about this point, considering what [itex]\frac{dy}{dx} [/itex] becomes there.)

    Concerning details of what the direction field does, how does the rational function [itex]\frac{dy}{dx} = \frac{\frac{x}{4}-\frac{1}{2}}{y-2} = \frac{x-2}{4y-8} [/itex] behave as x becomes large (either positively or negatively)? as y becomes large? as both variables do?
     
    Last edited: Sep 12, 2011
  4. Sep 12, 2011 #3
    Well, the steady point, (2,2) will be a saddle solution. Not only because it can be seen due to the slopes in the quadrants, but also because the eigenvalues of the matrix formed by x',y' are +1/2 and -1/2 - thus giving us a saddle point.

    Given your dy/dx example, as x -> positive infinity, x gets infinitely big; when x -> negative infinity x gets infinitely small. Either direction for y will bring the dy/dx term to 0.

    I'm pretty sure I'm understanding that part, but what I'm confused about is the lines that the saddle solutions tend towards. We have y=2,x=2 which divides everything into quadrants, but the slopes push solutions towards to linearly sloping lines that pass through the origin. How would we solve for those?
     
  5. Sep 12, 2011 #4

    dynamicsolo

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    I would be more comfortable if you said that as "x --> negative infinity, the numerator becomes infinitely negative". "Infinitely small" suggests an approach to zero...


    For the slope of the direction vectors dy/dx , what does the ratio tend to for very large x and y in each of the "quadrants"? (Oh, and the lines won't cross at the origin; they'll cross at the saddle point.) Another way of getting the directions of those lines is that they are given by the components eigenvectors corresponding to the positive and negative eigenvalues.
     
  6. Sep 12, 2011 #5
    Sorry about the confusing terms, but I did mean to say negative infinity as opposed to 0.

    Given the eigenvalues, the eigenvectors come out to y = 1/2*x and y = -1/2*x. I'm assuming I did the math right. That does make sense, I believe. I had thought that could be the case originally, but since there are infinitely many eigenvectors I wasn't sure which set to use.
    ie, when [itex]\lambda[/itex]=1/2, the eigenvector comes to s([itex]\frac{1/2}{-1}[/itex]) = ([itex]\frac{x}{y}[/itex]) and when [itex]\lambda[/itex]=-1/2 I got the eigenvectors as t([itex]\frac{1/2}{1}[/itex]) = ([itex]\frac{x}{y}[/itex]) where s and t are just arbitrary constants.

    Ps: I'm not quite sure how to do the matrix formatting properly so please ignore the division symbol in them.
     
  7. Sep 12, 2011 #6

    dynamicsolo

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    Just on a point of terminology -- there are not infinitely many eigenvectors; they are only two (all scalar multiples are considered to be equivalent for this purpose).

    Because the equations are linear, we get away with a bit here. Since the "equilibrium point" is at ( 2, 2 ), we should make the substitution u = x - 2 , v = y - 2 , which shifts the graph to "center" the flow on ( u, v ) = ( 0, 0 ) . This lets us write the matrix for the system of equations as u' = v and v' = (1/4)u . So the eigenvalues do come out as +1/2 and -1/2 , with the corresponding eigenvectors [ 2 1 ] and [ 2 -1 ].

    The negative eigenvalue is telling us that the flow is "attracted" ([itex]\lambda = -1/2[/itex]) along the line v = (-1/2) u and "repelled" ([itex]\lambda = +1/2[/itex]) along v = (+1/2)u . These lines cross at the origin in the uv-plane or at ( 2, 2 ) in the xy-plane. (I don't know if you need the equations for those lines, but you can get them using "point-slope" form; they are akin to the asymptotes of a hyperbola.)

    [I am not clear now why my earlier expression for dy/dx suggests that the slopes of the direction field ought to tend to +/- (1/4) ; I have to re-examine that...]
     
  8. Sep 12, 2011 #7
    That makes a lot of sense now. Thank you for walking me through that!
     
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