Equipotential Curves and Streamlines

  Clear[pot,v,x,y,z]
  pot[x_,y_]:=x y

  v=grad2[pot][x,y]

Plot the equipotential curves xy=c.

  equipot=ContourPlot[pot[x,y],{x,-5,5},{y,-5,5}, ContourShading->False,
  ContourStyle->Dashing[{.02}],AspectRatio->Automatic];

Find the streamlines.

  Clear[x,y]
  DSolve[x'[t]/y[t]==y'[t]/x[t],y[t],t]

  Clear[x,y];
  DSolve[{x'[t]/y[t]==y'[t]/x[t]},y,t];
  y=Evaluate[y[t]/.Flatten[%][[1]]]/.x[t]->x;
  y

So . . .

y=-Sqrt[x^2+c]. This implies y^2=x^2+c or
x^2-y^2=c

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

  Show[equipot,stream];

Notice anything?

Let's look at another one.

Consider the gradient field {2xy,x^2-y^2}. Does it come from a potential? If so, what is the potential?

Consider the gradient field {2xy,x^2-y^2}. Does it ...

  Clear[equipot,stream]
  equipot=ContourPlot[pot[x,y],{x,-5,5},{y,-5,5}, ContourShading->False,
  ContourStyle->Dashing[{.02}],AspectRatio->Automatic];

Let's find streamlines. Let's help Mathematica out by separating the variables.

  Clear[x,y]
  DSolve[x'[t]/x[t]==2 y'[t] y[t]/(x[t]^2-y[t]^2),y[t],t]

  Clear[x,y]
  DSolve[{x'[t]/x[t]==2 y'[t] y[t]/(x[t]^2-y[t]^2)},y,t];
  y=Evaluate[y[t]/.Flatten[%][[1]]]/.x[t]->x;
  y

So . . .

y=-Sqrt[(1/3 x^3+c)/x]. This implies y^2=1/3 x^2+c/x or
1/3 x^2-y^2+c/x=0. Multiply through by x:
1/3 x^3-xy^2=c

  Clear[x,y]
  stream=ContourPlot[1/3 x^3-x y^2,{x,-5,5},{y,-5,5},ContourShading->False,
  AspectRatio->Automatic];

  Show[equipot,stream];

Notice anything yet? Look closer at the "straighter" parts.

Notice anything yet? Look closer at the ...

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