Example 3
Example 3
In[51]:=
Clear[u,v]
u[x_,y_]:=x^2+y^2
The flow has velocity v=grad u.
In[52]:=
v[x_,y_]:=grad2[u][x,y]
v[x,y]
Out[52]=
{2 x, 2 y}
The flow is irrotational because it comes from a potential, that is, curl v=curl(grad u)=0. This is a freebie.
In[53]:=
curl2[v][x,y]
Out[53]=
0
Find the streamlines.
In[54]:=
Clear[x,y]
DSolve[x'[t]/(2 x[t])==y'[t]/(2 y[t]),y[t],t]
Out[54]=
-2 C[1] + Log[x[t]]
{{y[t] -> E }}
In[55]:=
Clear[x,y];
DSolve[{x'[t]/(2 x[t])==y'[t]/(2 y[t])},y,t];
y=Evaluate[y[t]/.Flatten[%][[1]]]/.x[t]->x;
y
Out[55]=
-2 C[1] + Log[x] E
So . . .
This simplifies to y=c x, or, for the purposes of contour plotting, y/x=c.
In[56]:=
Clear[x,y]
stream=ContourPlot[y/x,{x,-5,5},{y,-5,5},ContourShading->False,
AspectRatio->Automatic];
The plot is an example of not paying attention and hoping Mathematica knows what it is doing. This is a poor plot. Why? Because y/x has problems at x=0 and we asked for the range {x,-5,5}. The PlotPoints option can help you here.
In[57]:=
Clear[x,y]
stream=ContourPlot[y/x,{x,-5,5},{y,-5,5},ContourShading->False,
AspectRatio->Automatic,PlotPoints->100];
Are there sources or sinks?
Are there sources or sinks? ...
Recall that the net flow across a circle of radius r in this vector fields
is
(by the 2D Divergence theorem):
Double integral of div v over the circle. Since div v=4 this is simply 4 times the area of the circle (4pi r^2).
This potential, u=x^2+y^2, satisfies Poisson's equation with f=4.
In[59]:=
div2[grad2[u]][x,y]
Out[59]=
4