# Why are these flow lines cut short?

I want to plot some flow lines for a certain vector field, but Mathematica doesn't seem to plot the entire flow line, only a small piece of it. How can I fix this?

VectorPlot[{Re[Sqrt[x + I y]], Im[Sqrt[x + I y]]},
{x, -1, 1},
{y, -1, 1},
VectorPoints -> Fine,
StreamPoints -> Coarse,
StreamStyle -> Red,
StreamScale -> None] ### Update:

I feel dumb for not realizing the discontinuity, but I'll mention that what I really want are flow lines of slope fields, not vector fields. I didn't see how to do this in Mathematica "directly," and hence used VectorPlot. How can I plot the lines of a slope field?

• It plots from -1 to 1 in each direction. What do you want it to do? – bill s Mar 27 '13 at 14:39
• If you mean the discontinuity along the negative x-axis, it's because the vector field is discontinuous and changes direction. – Michael E2 Mar 27 '13 at 15:13
• Chris, please register an account now so you can edit and manage your posts. – Szabolcs Mar 27 '13 at 15:29
• Could you explain what a "slope field" is? Is it the same as a covector field (a one-form)? Or a section of the circle bundle associated with the tangent bundle? Or a section of that bundle with opposite directions identified at each point? – whuber Mar 27 '13 at 18:54
• @whuber: The last option you mentioned, so there's no notion of forwards/backwards direction, just of tangency. – user62206 Mar 28 '13 at 17:38

You could specify seed-points for streamlines directly using StreamPoints: only those stremlines will be plotted which pass through these points. Here I set up two point sets: one at $y = 1$ and another at $y = -1$. These share identical $x$ coordinates so they meet at the discontinuity at $y = 0$. One can control the streamline lenghts by the third argument to StremPoints (second argument controls the minimal distance between streamlines).

points = Join[{#, -1.} & /@ Range[-1, 1, .2], {#, 1.} & /@ Range[-1, 1, .2]];
VectorPlot[{Re[Sqrt[x + I y]], Im[Sqrt[x + I y]]}, {x, -1, 1}, {y, -1,
1}, VectorPoints -> Fine, StreamPoints -> {points, Automatic, 2},
StreamStyle -> Red, StreamScale -> None] Here is a small Manipulate interface to get a grip on how the streampoints are positioned and how length and distance can be controlled. Here a circular layout is used for the streamline-points to be able to capture some missing lines (thanks whuber!).

Manipulate[
points = Table[{Cos@t, Sin@t}*r, {t, 0., 2. \[Pi], d}];
VectorPlot[{Re[Sqrt[x + I y]], Im[Sqrt[x + I y]]}, {x, -1,
1}, {y, -1, 1}, VectorPoints -> Fine,
StreamPoints -> {points, Automatic, l}, StreamStyle -> Red,
StreamScale -> None,
Epilog -> {Green, PointSize@Large, Point@points}],
{{d, \[Pi]/20., "point spacing"}, 0, 1, \[Pi]/40,
Appearance -> "Labeled"},
{{l, 2, "stream length"}, 0, 2, Appearance -> "Labeled"},
{{r, 1, "radius"}, 0, 1, Appearance -> "Labeled"}
] • +1 How would you cope with the streamlines for positive $x$--the ones that ought to converge at the origin but instead are terminated within a distance of about $0.1$ of it? – whuber Mar 27 '13 at 18:49
• Thanks! This is exactly what I needed. – user62206 Mar 28 '13 at 18:58

For a slope field $dy/dx = f(y,x)$, I suppose you have to make do with something like a vector field of the form {1, f[x,y]} or Normalize@{1, f[x,y]}. We might as well normalize the vector, since magnitude doesn't matter in a slope field. Ordinarily, the value of the slope field at a point indicates the tangent line to an integral curve without regard to a direction of motion. I don't think Mathematica can represent that (or at least not conveniently). You have to use a vector.

In your case, it seems better to treat $x$ as a function of $y$ and for $dx/dy = g(x,y)$, use Normalize@{g[x,y], 1}, where g[x,y] is the quotient of the x and y components of your vector field:

VectorPlot[
Evaluate@Normalize@{Re[Sqrt[x + I y]]/Im[Sqrt[x + I y]], 1},
{x, -1, 1}, {y, -1, 1},
VectorPoints -> Fine, StreamPoints -> Coarse,
StreamStyle -> Red, StreamScale -> None, 