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How to plot on a curved plane? [duplicate]

4

$begingroup$

This question already has an answer here:

• 
  Phase portrait on a cylinder
  
  2 answers
  
  

I'm ploting the phase space of a pendulum problem using a symplectic Euler scheme.

$qquad H = frac12p^2 - cos q$, where $dotp=-sin q$ and $dotq=p$

h=0.2; (*time step*)
p[0]=0.0; (*initial conditions*)
q[0]=0.5;

p[i_] := p[i] = p[i - 1] - h*Sin[q[i - 1]];
q[i_] := q[i] = q[i - 1] + h*p[i - 1] - h^2*Sin[q[i - 1]];

ListPlot[Table[p[i], q[i], i, 0, 100], Frame -> True]

gives

Since the vector field is $2π$-periodic in q, it is natural to consider q as a variable on the circle $S^1$, I'd expect it to look something like

Any suggest how to do it?

plotting

share|improve this question

edited Apr 6 at 5:56

m_goldberg

88.6k873200

asked Apr 6 at 2:13

GvxfjørtGvxfjørt

986

$endgroup$

marked as duplicate by Michael E2 plotting
Users with the  plotting badge can single-handedly close plotting questions as duplicates and reopen them as needed.

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Apr 8 at 2:38

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There is no such thing as a "curved plane". If want to plot on a 2-manifold, please give a description of the manifold in Wolfram Language code.
  $endgroup$
  – m_goldberg
  Apr 6 at 5:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MichaelE2 it is indeed a duplicate, I'm happy with the solution provided, feel free to close it.
  $endgroup$
  – Gvxfjørt
  Apr 8 at 1:46
  

  
  
  
  

add a comment | 

4

$begingroup$

This question already has an answer here:

• 
  Phase portrait on a cylinder
  
  2 answers
  
  

I'm ploting the phase space of a pendulum problem using a symplectic Euler scheme.

$qquad H = frac12p^2 - cos q$, where $dotp=-sin q$ and $dotq=p$

h=0.2; (*time step*)
p[0]=0.0; (*initial conditions*)
q[0]=0.5;

p[i_] := p[i] = p[i - 1] - h*Sin[q[i - 1]];
q[i_] := q[i] = q[i - 1] + h*p[i - 1] - h^2*Sin[q[i - 1]];

ListPlot[Table[p[i], q[i], i, 0, 100], Frame -> True]

gives

Since the vector field is $2π$-periodic in q, it is natural to consider q as a variable on the circle $S^1$, I'd expect it to look something like

Any suggest how to do it?

plotting

share|improve this question

edited Apr 6 at 5:56

m_goldberg

88.6k873200

asked Apr 6 at 2:13

GvxfjørtGvxfjørt

986

$endgroup$

marked as duplicate by Michael E2 plotting
Users with the  plotting badge can single-handedly close plotting questions as duplicates and reopen them as needed.

StackExchange.ready(function()
if (StackExchange.options.isMobile) return;

$('.dupe-hammer-message-hover:not(.hover-bound)').each(function()
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');

$hover.hover(
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$hover.showInfoMessage('',
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dismissable: false,
relativeToBody: true
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function()
StackExchange.helpers.removeMessages();

);
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Apr 8 at 2:38

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There is no such thing as a "curved plane". If want to plot on a 2-manifold, please give a description of the manifold in Wolfram Language code.
  $endgroup$
  – m_goldberg
  Apr 6 at 5:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MichaelE2 it is indeed a duplicate, I'm happy with the solution provided, feel free to close it.
  $endgroup$
  – Gvxfjørt
  Apr 8 at 1:46
  

  
  
  
  

add a comment | 

$begingroup$

This question already has an answer here:

• 
  Phase portrait on a cylinder
  
  2 answers
  
  

I'm ploting the phase space of a pendulum problem using a symplectic Euler scheme.

$qquad H = frac12p^2 - cos q$, where $dotp=-sin q$ and $dotq=p$

h=0.2; (*time step*)
p[0]=0.0; (*initial conditions*)
q[0]=0.5;

p[i_] := p[i] = p[i - 1] - h*Sin[q[i - 1]];
q[i_] := q[i] = q[i - 1] + h*p[i - 1] - h^2*Sin[q[i - 1]];

ListPlot[Table[p[i], q[i], i, 0, 100], Frame -> True]

gives

Since the vector field is $2π$-periodic in q, it is natural to consider q as a variable on the circle $S^1$, I'd expect it to look something like

Any suggest how to do it?

plotting

share|improve this question

edited Apr 6 at 5:56

m_goldberg

88.6k873200

asked Apr 6 at 2:13

GvxfjørtGvxfjørt

986

$endgroup$

h=0.2; (*time step*)
p[0]=0.0; (*initial conditions*)
q[0]=0.5;

p[i_] := p[i] = p[i - 1] - h*Sin[q[i - 1]];
q[i_] := q[i] = q[i - 1] + h*p[i - 1] - h^2*Sin[q[i - 1]];

ListPlot[Table[p[i], q[i], i, 0, 100], Frame -> True]

share|improve this question

edited Apr 6 at 5:56

m_goldberg

88.6k873200

asked Apr 6 at 2:13

GvxfjørtGvxfjørt

986

share|improve this question

edited Apr 6 at 5:56

m_goldberg

88.6k873200

asked Apr 6 at 2:13

GvxfjørtGvxfjørt

986

edited Apr 6 at 5:56

m_goldberg

88.6k873200

edited Apr 6 at 5:56

m_goldberg

88.6k873200

asked Apr 6 at 2:13

GvxfjørtGvxfjørt

986

asked Apr 6 at 2:13

GvxfjørtGvxfjørt

986

1 Answer
1

active

oldest

votes

7

$begingroup$


h = 0.2;
p[0, a_] := 0;
q[0, a_] := a
p[i_, a_] := p[i, a] = p[i - 1, a] - h*Sin[q[i - 1, a]];
q[i_, a_] :=
q[i, a] = q[i - 1, a] + h*p[i - 1, a] - h^2*Sin[q[i - 1, a]];


plots = Table[
ListPointPlot3D[
Table[Sin[q[i, a]], Cos[q[i, a]], p[i, a], i, 0, 100],
PlotStyle -> PointSize[0.008],
PlotRange -> -1, 1, -1, 1, -3, 3], a, 0.5, 3, 0.5];

Show[plots,
Graphics3D[Opacity[0.1], Cylinder[0, 0, -3, 0, 0, 3]]]

share|improve this answer

answered Apr 6 at 6:10

ulviulvi

1,176612

$endgroup$

add a comment | 

7

$begingroup$


h = 0.2;
p[0, a_] := 0;
q[0, a_] := a
p[i_, a_] := p[i, a] = p[i - 1, a] - h*Sin[q[i - 1, a]];
q[i_, a_] :=
q[i, a] = q[i - 1, a] + h*p[i - 1, a] - h^2*Sin[q[i - 1, a]];


plots = Table[
ListPointPlot3D[
Table[Sin[q[i, a]], Cos[q[i, a]], p[i, a], i, 0, 100],
PlotStyle -> PointSize[0.008],
PlotRange -> -1, 1, -1, 1, -3, 3], a, 0.5, 3, 0.5];

Show[plots,
Graphics3D[Opacity[0.1], Cylinder[0, 0, -3, 0, 0, 3]]]

share|improve this answer

answered Apr 6 at 6:10

ulviulvi

1,176612

$endgroup$

add a comment | 

7

$begingroup$


h = 0.2;
p[0, a_] := 0;
q[0, a_] := a
p[i_, a_] := p[i, a] = p[i - 1, a] - h*Sin[q[i - 1, a]];
q[i_, a_] :=
q[i, a] = q[i - 1, a] + h*p[i - 1, a] - h^2*Sin[q[i - 1, a]];


plots = Table[
ListPointPlot3D[
Table[Sin[q[i, a]], Cos[q[i, a]], p[i, a], i, 0, 100],
PlotStyle -> PointSize[0.008],
PlotRange -> -1, 1, -1, 1, -3, 3], a, 0.5, 3, 0.5];

Show[plots,
Graphics3D[Opacity[0.1], Cylinder[0, 0, -3, 0, 0, 3]]]

share|improve this answer

answered Apr 6 at 6:10

ulviulvi

1,176612

$endgroup$

add a comment | 

$begingroup$


h = 0.2;
p[0, a_] := 0;
q[0, a_] := a
p[i_, a_] := p[i, a] = p[i - 1, a] - h*Sin[q[i - 1, a]];
q[i_, a_] :=
q[i, a] = q[i - 1, a] + h*p[i - 1, a] - h^2*Sin[q[i - 1, a]];


plots = Table[
ListPointPlot3D[
Table[Sin[q[i, a]], Cos[q[i, a]], p[i, a], i, 0, 100],
PlotStyle -> PointSize[0.008],
PlotRange -> -1, 1, -1, 1, -3, 3], a, 0.5, 3, 0.5];

Show[plots,
Graphics3D[Opacity[0.1], Cylinder[0, 0, -3, 0, 0, 3]]]

share|improve this answer

answered Apr 6 at 6:10

ulviulvi

1,176612

$endgroup$

share|improve this answer

answered Apr 6 at 6:10

ulviulvi

1,176612

answered Apr 6 at 6:10

ulviulvi

1,176612

answered Apr 6 at 6:10

ulviulvi

1,176612