Re: Solving ODE for rotational, irrotational vector field

• To: mathgroup at smc.vnet.net
• Subject: [mg104931] Re: Solving ODE for rotational, irrotational vector field
• From: dh <dh at metrohm.com>
• Date: Fri, 13 Nov 2009 05:57:20 -0500 (EST)
• References: <hdgqbm\$ilg\$1@smc.vnet.net>

```Murray Eisenberg wrote:

> 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}?

>

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

```

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