Behavior of a gradient system Let $\frac{d}{dt}(x,y) = -\nabla g(x,y)$ be a planar system, where $g(x,y)=x^{2}y^{4}$. Show that for $t \to \infty$, $\Phi(t,(x,y))\to 0$ for all $(x,y)\in \mathbb{R}^{2}$ \ $(\{x=0\} \cup \{y=0\})$. 
Al right, what I've tried is: 
lets write this as a system, 
$\dot x = -2xy^{4}$
$\dot y = -4x^{2}y^{3}$
Is it enough to consider the Lyapunov function $V(x,y)=x^{2}+y^{2}$ for $(x,y)=(0,0)$ to say that all trajectories converge to $0$ for all $(x,y)\in \mathbb{R}^{2}$ \ $(\{x=0\} \cup \{y=0\})$ when $t\to \infty$? (I guess that $\Phi(t,(x,y))$ is the function for trajectories given certain $t$ and certain $(x,y)$, but not so sure really).
 A: EDIT:
I have noticed it later, sorry for that. But the time derivative of the $V$ in this answer is not negative definite, it is negative semi-definite (i.e. zero also on coordinate axes) and in fact when I plot the phase plot of the system, not all the initial points converge to origin. So the claim in the question is wrong, not all the initial conditions except the ones on coordinate axes converge to origin. 
We define a function on $(x,y)$ plane $V(x,y) = x^2+y^2$. This function has positive value everywhere in $(x,y)$ plane except at the origin $(x,y)=(0,0)$, where it is zero. Then we find the time derivative of $V$:
$\frac{dV}{dt} = \frac{\partial V}{\partial x} \frac{dx}{dt} + \frac{\partial V}{\partial y} \frac{dy}{dt} = 2x(-2xy^4) + 2y(-4x^2y^3) = -4x^2y^4 -8x^2y^4$
which is negative everywhere on the (x,y) plane except at the origin and on the coordinate axes(x=0 or y=0) where it is zero. 
Now, $V$ is positive everywhere except at origin and it's time derivative is negative everywhere except at the coordinate axes and the origin meaning it will decrease all the time if it is not at the origin. Therefore whichever $(x_0,y_0)$ you choose which is not on the coordinate axes (x or y or both are zero), $V$ is going to decrease towards origin as time goes to infinity and the corresponding $(x,y)$ pair will converge to origin.
To analyze the case when $x=0$ or $y=0$, note that in that case both derivative of $x$ and $y$ are both zero, and so it will remain there forever.
A: We can find a first integral $F(x, y)$ for the system by solving $\nabla F \cdot \nabla g = 0$. When $g = X(x) Y(y)$, a solution is
$$F = \int \frac Y {Y'} dy - \int \frac X {X'} dx.$$
Now since $y^2 - 2 x^2$ is a first integral, a trajectory that doesn't start on the lines $y^2 - 2x^2 = 0$ cannot converge to the origin.
