Sinks and Sources in Gradient Fields Can you please critique my reasoning here? I know that there is a leap I am making that is incorrect, or I don't fully understand a critical piece.
I am looking at the Laplacian of $f(x,y)=2yx - xy^4$. I calculated $\Delta f(x,y) = -12xy^2$.
I am trying to interpret what this tells me about the function. I know that the Laplacian is just $\Delta f(x,y)=\nabla \cdot \nabla f(x,y)$ which is the divergence of $\nabla f(x,y)$. 
Here is where I am struggling...points where the divergence is positive represent sources in the vector field, and points where the divergence is negative represent sinks in the vector field. Sinks in $\nabla f(x,y)$ correspond to local maximums of $f(x,y)$ while sources corresponds to local minimums. 
My question is, there are infinitely many points where $-12xy^2$ is positive and infinitely many where $-12xy^2$ is negative, though I know these points are not all local extrema of $f(x,y)=2yx-xy^4$, so there are some inaccuracies in my logic above.
Am I even remotely interpreting the divergence relating to maximums and minimums correctly here? Any help would be greatly appreciated. 
 A: The words "source" and "sink" are somewhat unfortunate in describing a phenomenon spread out over a region. When we have a flow field ${\bf v}$ then ${\rm div}\, {\bf v}({\bf p})>0$ means that for a small disk (in 2D) or ball (in 3D) centered at ${\bf p}$ there is more fluid flowing out per second on one side of this disk than is flowing in  on the other side of this disk. If this is true for ${\bf p}$ this is true for all points near ${\bf p}$. The reason could be that it rains in the neighborhood of ${\bf p}$ and that this additional amount of fluid has to flow away somehow in addition to the basic underground flow.
Note that local extrema ${\bf p}$ of some scalar function $f$ are signaled by $\nabla f({\bf p})={\bf 0}$, and not by the divergence of ${\bf v}:=\nabla f$ at ${\bf p}$ being positive, resp., negative.
A: The Laplacian is just the trace of the Hessian matrix and not sufficient, in general, to get the local maxima and minima for a function. To accomplish this, you have to consider the entire Hessian matrix and  - what is more important - its determinant to apply the second partial derivative test. 
First of all, you need to determine the critical points for the function: the points where the gradient is zero. In this case the critical points are $(2y-y^4,2x-4xy^3)=(y(2-y^3),x(2-4y^3))$. Now, you have to compute the Hessian matrix:
$$
H(x,y)=\begin{pmatrix}f_{xx}(x,y)&f_{xy}(x,y)\\f_{yx}(x,y)&f_{yy}(x,y)\end{pmatrix}=\begin{pmatrix}0&2-4y^3\\2-4y^3&-12xy^2\end{pmatrix}. 
$$
Thus $\det(H)=-(2-4y^3)^2$, what means the determinant can be only negative or zero. We have to evaluate the determinant only for critical points, so the determinant cannot be zero because this would imply that the $x$ component of the gradient is not zero, thus not a critical point. Then all critical points for this function are saddle points.
