Solving ODE for rotational, irrotational vector field

Discussion in 'Mathematica' started by Murray Eisenberg, Nov 12, 2009.

  1. The vector field

    F[{x_,y_}] := {y/(x^2 + y^2), -(x/(x^2 + y^2))}

    is smooth (except at the origin, where it's undefined). And, as is
    well-known, the trajectories of the vector differential equation

    {x'[t],y'[t]} == F[{x[t],y[t]}]

    are circular about the origin. And yet this vector field is
    "irrotational", i.e., its curl is {0,0,0} everywhere the field is defined.

    I'd like to show that the trajectories really are circular by explicitly
    finding them. So I tried finding the solutions of the differential
    equation like this:


    One gets two solutions. In fact, if you include initial conditions, e.g.,

    D[{x[t], y[t]}, t] == F[{x[t], y[t]}]], {x[0], y[0]} == {1,
    1}}, {x[t], y[t]}, t]

    .... you still get two solutions. The components of each solution involve
    Tan and ArcTan, so I assume that's why there are two pieces. But when I
    piece them together by doing ParametricPlot of both on the same axes, I
    don't get circles: I get semi-circles along with the x-axis, which
    clearly seems to be wrong.

    Can anybody shed light on this mathematically or Mathematicaly? In
    particular, are the domains of solutions not {-Infinity,Infinity}?

    Murray Eisenberg
    Mathematics & Statistics Dept.
    Lederle Graduate Research Tower phone 413 549-1020 (H)
    University of Massachusetts 413 545-2859 (W)
    710 North Pleasant Street fax 413 545-1801
    Amherst, MA 01003-9305
    Murray Eisenberg, Nov 12, 2009
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  2. If you use the intial condition {x[0], y[0]} == {1, -1}} you get the
    other half of the circles.

    If t passes Pi/2, x jumps from + to - Sqrt[2] while y remains at 0.
    Perhaps Mathematica should have drawn it as an exclusion.

    Cheers -- Sjoerd
    Sjoerd C. de Vries, Nov 13, 2009
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  3. Murray Eisenberg

    dh Guest

    Hi Murray,

    the problem you get comes from the temporal behaviour that blows up near

    the origin.

    However, it is easy to show that the trajectories are circular, you

    simply have to eliminate t in your equations:

    dx/dt= Fx and dy/dt= Fy.


    dx= Fx dt and dy= Fy dt


    dy/dx=Fx/Fy= - x/y or y dy = - x dx


    y^2 = -x^2 + c


    y= +/- Sqrt[c-x^2]

    what describes a circle where r^2 = c.

    dh, Nov 13, 2009
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