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Consider the function $V(x,y)=x^4-x^2+2xy+y^2$ and the coupled differential equations $\frac{dx}{dt}=-\frac{\partial V}{\partial x}, \frac{dy}{dt}=-\frac{\partial V}{\partial y}$.

Show that $V(x(t),y(t))$ is a non-increasing function of $t$. If $x=1$ and $y=-\frac12$ at $t=0$ where does the solution tend to as $t \rightarrow \infty$?

I'm fine with showing that $V$ is non-increasing (just the chain rule) but I can't get anything after this, I've tried setting up some differential equations to solve but nothing works.

(We also had to sketch the contours of $V$ earlier in the question)

Thank you

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    Can you explain to me what does $\frac{dx}{dt}=-\frac{\partial V}{\partial x}$ mean? This is nonsensical to me. If on LHS one has $x'(1)$, what would the RHS be? – 2017-01-08
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    @GitGud: I believe that if one adds the arguments then the equation should be $\frac{dx}{dt} (t) = -\frac{\partial V}{\partial x} \big( x(t), y(t) \big)$. – 2017-01-12

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