Example 1

Let's return to the potential u=xy.

In[41]:=

  Clear[u,v,x,y]
  u[x_,y_]:=x y

The flow has velocity v=grad u.

In[42]:=

  v[x_,y_]:=grad2[u][x,y]
  v[x,y]

Out[42]=

  {y, x}

The flow is irrotational because it comes from a potential, that is, curl v=curl(grad u)=0.

In[43]:=

  curl2[v][x,y]

Out[43]=

We had the streamlines from above.

In[44]:=

  stream=ContourPlot[x^2-y^2,{x,-5,5},{y,-5,5},ContourShading->False,
  AspectRatio->Automatic];

The velocity is also divergence free (solenoidal). There are no sources or sinks (conservation of "mass").

In[45]:=

  div2[v][x,y]

Out[45]=

This implies u=xy solves Laplace's equation:
div(grad u)=div v=0

Does every potential satisfy Laplace's equation?