Limit of the function $V(x,y)=x^4-x^2+2xy+y^2$ Let $$V(x,y)=x^4-x^2+2xy+y^2$$ Consider the coupled d.e.'s:$$\frac {\mathrm d x} {\mathrm d t} = - \frac {\partial V} {\partial x}, \qquad \frac {\mathrm d y} {\mathrm d t} = - \frac {\partial V} {\partial y}$$ If $x=1$ and $y=- \frac 1 2$ at $t=0$, where does the solution tend to as $t \to \infty$? Hints would be appreciated. I found the critical points of V, sketched a countour plot, showed V is a non-increasing function of $t$. I can't see a nice way of doing this, everything I try becomes horribly messy.

The limit must be one of the three critical points. At these points you have $\frac{dx}{dt} = \frac{dy}{dt} = 0$. You've calculated these to be (-1, 1), (0,0), and (1,-1).

The point (0,0) is a repellor, the other two points are attractors. Looking at the contour plot you made (better yet, use a direction field or Mathematica's StreamPlot), you can see that (1, -1/2) is in the basin of attraction of (1,-1).

So the limit is (1,-1).