The StreamPlot command has an option for the number of StreamPoints.
When I try the first it does not appear to allow you to choose anything other than $10$ (for example, I want to plot $1000$ streamlines)
StreamPlot[{10 x y - 1/2 x - 1/10 x^2, y - y^2 - x y}, {x, 0, 3}, {y,
0, 1/2}, StreamPoints -> 10]
When I try a second variant, it is a slight improvement as
StreamPlot[{10 x y - 1/2 x - 1/10 x^2, y - y^2 - x y}, {x, 0, 3}, {y,
0, 1/2}, ImageSize -> Large, StreamPoints -> Fine]
However, how can I get MMA to add as many, say $1000$ streamlines, in the first quadrant using StreamPlot?
Perhaps there are other approaches using NDSolve and the like, but why can this not be done with StreamPlot?
VectorPoints? Or something along the lines of theContoursoption forContourPlot? $\endgroup$Show[StreamPlot[{10. x y - .5 x - .1 x^2, y - y^2 - x y},{x, 0., 3.}, {y, 0., #}, StreamPoints -> 1, StreamStyle -> Arrowheads[.015]] & /@ Subdivide[0.01, .5, 100], PlotRange -> All]. Granted this is only for (still at most) 100 lines. I tried 1000, but it just didn't look very good. Also note, that the resulting plot only has0 < y < .3. $\endgroup$LineIntegralConvolutionPlot[{10 x y - 1/2 x - 1/10 x^2, y - y^2 - x y}, {x, 0, 3}, {y, 0, 1/2}, RasterSize -> 300]? $\endgroup$