Skip to main content

plotting - StreamPlot segmented lines


I am plotting this vector field:



Q[k_, N_] := Show[ContourPlot[3 - 3 BesselJ[0, r[x, y] a^-1], {x, -N, N}, {y, -N, N}, 
ContourStyle -> Opacity[0.3]], 

StreamPlot[{Sin[θ[x, y]], -Cos[θ[x, y]]}, {x, -N, N}, {y, -N, N}, 
PlotRange -> {Full, Full}, RegionFunction -> 
Function[{x, y}, a^2 - a^2/100 < x^2 + y^2 <=  a^2 + a^2/100], 
StreamStyle -> Red, StreamPoints -> {pts, Automatic, Scaled[2]}, 
StreamScale -> None],

StreamPlot[{Abs[(1 - a/r[x, y])^(1/2)] + 

a^(1/2)/r[x, y]^(1/2) Cos[θ[x, y] k], 
a^(1/2)/r[x, y]^(1/2) Sin[θ[x, y] k]}, {x, -N, N}, {y, -N, 
N}, PlotRange -> {Full, Full}, 
RegionFunction -> Function[{x, y}, x^2 + y^2 > a^2], 
StreamStyle -> {Blue, Thick, "Line"}, PerformanceGoal -> "Quality", 
StreamScale -> Full], 

ImageSize -> Large]

with



r[x_, y_] := Sqrt[x^2 + y^2];
θ[x_, y_] := ArcTan[x, y];

a = 3;
pts = Flatten[
Table[{x, y}, {x, -a - a/10, a + a/10, 0.1}, {y, a - a/10, a + a/10, 0.1}], 1];

But I can't find a set of StreamPoints to avoid the segmentation of the lines near the x-axes.


Q[1/2,10]


enter image description here


Any ideas? Thanks!



Answer



r[x_, y_] := Sqrt[x^2 + y^2];
θ[x_, y_] := ArcTan[x, y];

a = 3;
pts = Flatten[Table[{x, y}, {x, -a - a/10, a + a/10, 0.1}, {y, a - a/10, a + a/10, 0.1}], 1];

Q[k_, n_] := Module[{pts1}, pts1 = Table[{0, x}, {x, -2 n, 2 n, 1/2}];

  Show[ContourPlot[ 3 - 3 BesselJ[0, r[x, y] a^-1], {x, -n, n}, {y, -n, n}, 
                   ContourStyle -> Opacity[0.3]], 
       StreamPlot[{Sin[θ[x, y]], -Cos[θ[x, y]]}, {x, -n, n}, {y, -n, n}, PlotRange -> {Full, Full}, 
                 RegionFunction -> Function[{x, y}, a^2 - a^2/100 < x^2 + y^2 <= a^2 + a^2/100], 
                 StreamStyle -> Red, StreamPoints -> {pts, Automatic, Scaled[2]},  
                 StreamScale -> None], 
       StreamPlot[{Abs[(1 - a/r[x, y])^(1/2)] +  a^(1/2)/r[x, y]^(1/2) Cos[θ[x, y] k], 
                      a^(1/2)/r[x, y]^(1/2) Sin[θ[x, y] k]}, {x, -2 n, 2 n}, {y, -2 n, 2 n}, 
                 RegionFunction -> Function[{x, y}, x^2 + y^2 > a^2 && Abs@x < n && Abs@y < n], 
                 StreamPoints -> {pts1, Automatic, Scaled[2]}, StreamStyle -> {Blue, Thick, "Line"}, 

                 PerformanceGoal -> "Quality", StreamScale -> Full], 
   ImageSize -> Large]]

Q[1/2, 10]

Mathematica graphics


Comments

Post a Comment

Popular posts from this blog

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...
Post a Comment
Read more

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...
Post a Comment
Read more

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]
Post a Comment
Read more