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Mathematics Graphics and Visualization Wolfram Language

Plot of a Vector Field and Nullclines

Max Benczek

Posted 9 years ago

Hello, I'm writing my thesis and I need to put some pictures in it. I need to plot a vector field (which i found how to do) and in it the nullclines for the field. The plot should look like the one in the attachment on the upper right. In that plot one can see the vector field for the system x'=y-x^2 ; y'=x-2 and the respective nullclines which are y=x^2 and x=2. I would be very grateful if anyone could help me and tell me how I can plot that! Best Regards

POSTED BY: Max Benczek

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Shenghui Yang

Shenghui Yang, WOLFRAM

Posted 9 years ago

You can use the `VectorPlot` function to plot the vector field. Use the `ParametricPlot` function to plot the line `x == 2` and the parabola `y == x^2` Show[ ParametricPlot[{{u,u^2},{2,3u^2}},{u,-3,3}], VectorPlot[{y-x^2,x-2},{x,-3,3},{y,-1,5}, VectorScale->{Tiny,Small,None}],PlotRange->{-1,4}]

POSTED BY: Shenghui Yang

Marco Thiel

Marco Thiel, University of Aberdeen - Dept. of Physics/Mathematics

Posted 9 years ago

Hi, what about this Show[VectorPlot[{y - x^2, x - 2}, {x, -3, 3}, {y, -1, 9}, Epilog -> {PointSize[0.03], Point[{x, y} /. First[Solve[{y - x^2 == 0, x - 2 == 0}]]]}, Frame -> False, Axes -> True], ContourPlot[{y - x^2 == 0, x - 2 == 0}, {x, -3, 3}, {y, -1, 9}, ContourStyle -> Red]] where the length of the arrows represents the magnitude/speed. If you prefer the arrows of equal length like in your book this might work: Show[VectorPlot[{y - x^2, x - 2}, {x, -3, 3}, {y, -1, 9}, Epilog -> {PointSize[0.03], Point[{x, y} /. First[Solve[{y - x^2 == 0, x - 2 == 0}]]]}, Frame -> False, Axes -> True, VectorScale -> {Small, Small, None}, VectorPoints -> Coarse, ImageSize -> Medium], ContourPlot[{y - x^2 == 0, x - 2 == 0}, {x, -3, 3}, {y, -1, 9}, ContourStyle -> Red]] Cheers, Marco

Hi,

what about this

Show[VectorPlot[{y - x^2, x - 2}, {x, -3, 3}, {y, -1, 9}, Epilog -> {PointSize[0.03], Point[{x, y} /. First[Solve[{y - x^2 == 0, x - 2 == 0}]]]}, 
  Frame -> False, Axes -> True], ContourPlot[{y - x^2 == 0, x - 2 == 0}, {x, -3, 3}, {y, -1, 9}, ContourStyle -> Red]]

enter image description here

where the length of the arrows represents the magnitude/speed. If you prefer the arrows of equal length like in your book this might work:

Show[VectorPlot[{y - x^2, x - 2}, {x, -3, 3}, {y, -1, 9}, Epilog -> {PointSize[0.03], Point[{x, y} /. First[Solve[{y - x^2 == 0, x - 2 == 0}]]]}, 
  Frame -> False, Axes -> True, VectorScale -> {Small, Small, None}, VectorPoints -> Coarse, ImageSize -> Medium], 
 ContourPlot[{y - x^2 == 0, x - 2 == 0}, {x, -3, 3}, {y, -1, 9}, ContourStyle -> Red]]

enter image description here

Cheers,

Marco

POSTED BY: Marco Thiel

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