1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Singular Points

  1. Jan 17, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the singular points for:

    x' = ax - bxy
    y' = bxy - cy


    2. Relevant equations



    3. The attempt at a solution

    ax - bxy = 0
    bxy - cy = 0

    implies

    x = 0 or y = a/b

    y = 0 or x = c/b

    the the critical points are (o,o) or (c/b, a/b)

    why is (0, a/b) and (c/b, 0) NOT a critical point??
     
  2. jcsd
  3. Jan 18, 2008 #2

    siddharth

    User Avatar
    Homework Helper
    Gold Member

    Because, [itex]x'[/itex] and [itex]y'[/itex] are not 0 for those values of x and y. Substitute them to check.
     
  4. Jan 18, 2008 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Why would you think they would be? They clearly don't satisfy the equations you give.

    When you have a system of non-linear equations with several different solutions, the x and y are not independent- you can't just combine any value of x with any value of y: it is the specific x, y pair that satisfies the equations.
     
  5. Jan 18, 2008 #4
    Oh I see my mistake.

    I thought that they were dependent.
     
  6. Jan 18, 2008 #5
    I have been working with the equations:

    x' = rx - sxy/(1+tx)

    y' = sxy/(1+tx) - wy


    I find that the only singular point is (0,0)

    is this correct?


    I got another point, ( w/(s-w) , r/s ), however, that cannot be a critical point because when substituting, x' and y' fail to be 0.
     
  7. Jan 19, 2008 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Since (w/(s-w), r/s) does not satisfy the equations for a critical point, how did you get that?

    The equations for a critical point are x'= rx- sxy/(1+ tx)= 0 and y'= sxy/(1+ tx)- wy= 0. The first gives sxy/(1+ tx)= rx and the second sxy/(1+ tx)= wy. Since the left sides of both equations are the same, rx= wy. Every point on the line y= (r/w)x (or, if w= 0, the line x= 0) is a singular point.
     
  8. Jan 19, 2008 #7
    x'= rx- sxy/(1+ tx)= 0

    x(r-sy/(1 + tx)) = 0

    x= 0 and (r-sy/(1 + tx)) = 0

    since x = 0 r-sy = 0

    therefore, y = r/s


    similarly we do that for y'= sxy/(1+ tx)- wy= 0 and we get y = 0 and x = w/(s-w)



    when I substitute y= (r/w)x into x'= rx- sxy/(1+ tx), x' is NOT zero. So how could this line be a line of singular points?
     
  9. Jan 19, 2008 #8
    also, what are the differences between equilibrium solutions and singular points?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Singular Points
  1. Singular Points of DE (Replies: 4)

  2. Singular point (Replies: 8)

Loading...