Fluid flowing in the plane has the velocity (y + 2x - 2x^{3} - 2xy^{2}, - x) at (x, y). Sketch the flow lines near the origin. What happens to an individual particle as t → ∞ ?

**Solution**

The key insight is that the unit circle centre the origin is one possible trajectory. [The velocity is (y + 2x - 2xr^{2}, - x) = (y, -x) on r = 1.] Hence fluid starting inside this circle remains inside it and fluid starting outside it remains outside. Differentiating r^{2} = x^{2} + y^{2} gives r dr/dt = 2x^{2}(1 - r^{2}). Hence if r < 1, then dr/dt is positive, so points inside the unit circle move outwards towards it. Similarly, if r > 1, then dr/dt is negative, so points outside the unit circle move inwards towards it.

To find the behaviour close to the origin, we drop the cubic terms. The equations are then linear and can be solved to get x = (at + b)e^{t}, y = (-at + a - b)e^{t}. This gives S shaped curves centred on the origin at a 45^{o} angle touching the line y = -x.

© John Scholes

jscholes@kalva.demon.co.uk

15 Feb 2002