# Solving ODE for rotational, irrotational vector field

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

1. ### Murray EisenbergGuest

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, y} == {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

2. ### Sjoerd C. de VriesGuest

If you use the intial condition {x, y} == {1, -1}} you get the
other half of the circles.

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

Cheers -- Sjoerd

Sjoerd C. de Vries, Nov 13, 2009

3. ### dhGuest

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.

Therefore

dx= Fx dt and dy= Fy dt

and

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

Integrating:

y^2 = -x^2 + c

or

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

what describes a circle where r^2 = c.

Daniel

dh, Nov 13, 2009