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My first question: When using desolve_system_rk4(), is there a way to see the plots of the dependent variables vs. the independent variable? Thanks.

asked 2015-03-24 00:13:06 +0100

GeoffH gravatar image

updated 2015-03-24 01:30:14 +0100

calc314 gravatar image
 x,y,t=var('x y t')
#x=function('x',t)
P=desolve_system_rk4([x*(1-y),-y*(1-x)],[x,y],ics=[0,0.5,2],ivar=t,end_points=20)
P=P+desolve_system_rk4([x*(1-y),-y*(1-x)],[x,y],ics=[0,0.3,2],ivar=t,end_points=20)
Q=[ [i,j] for i,j,k in P]
LP=list_plot(Q)
# Q=[ [j,k] for i,j,k in P]
Q=[ [j,k] for i,j,k in P]
LP=list_plot(Q)
show(LP)
plot(x);plot(y) #Doesn't work
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answered 2015-03-24 01:34:34 +0100

calc314 gravatar image

The command desolve_system_rk4 returns a list of lists of the form [time pt, x value, y value]. You can use a list comprehension as below to pull out coordinate pairs. Then, you can plot using the line command.

x,y,t=var('x y t')
P=desolve_system_rk4([x*(1-y),-y*(1-x)],[x,y],ics=[0,0.5,2],ivar=t,end_points=20)
tx = [[q[0],q[1]] for q in P]
ty =  [[q[0],q[2]] for q in P]
line(tx)+line(ty,color='red')
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wow...I'm amazed at how helpful that is. I didn't understand how desolve_system_rk4 worked, so I couldn't work out how to get the tx and ty curves. Now I have a much better idea of both...thanks!

GeoffH gravatar imageGeoffH ( 2015-03-24 05:05:31 +0100 )edit
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Then you should accept the answer by clicking on the check mark.

rws gravatar imagerws ( 2015-03-24 07:58:32 +0100 )edit

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Asked: 2015-03-24 00:13:06 +0100

Seen: 937 times

Last updated: Mar 24 '15