2
$\begingroup$ 
H[t_] = {{0, 1}, {1, t}};
tmax=5;
fun[X_?ArrayQ, t_] := -I (H[t].X - X.H[t]);

sol = NDSolve[{σ'[t] == fun[σ[t], t], σ[0] == 
Table[If[i == 1 \[And] j == 1, 1, 0], {i, 1, 2}, {j, 1, 
  2}]}, σ[t], {t, 0, tmax}]

Flatten[Evaluate[σ[t] /. sol /. {t -> tmax/100}], 1] // MatrixForm

 Plot[Table[Flatten[Evaluate[σ[t] /. sol], 1][[i, i]], {i, 1, 2}], {t, 0,tmax}, PlotRange -> All]

Table plot

How to change the color of two lines in plot... and why 

Flatten[Evaluate[σ[t] /. sol], 1][[1, 1]]

gives 

t

while 

 Flatten[Evaluate[σ[t] /. sol], 1][[2, 2]]

gives

 {InterpolatingFuction[{{0.,5.}},<>}[t]}[[2,2]]
Improve this question
$\endgroup$
2
  • $\begingroup$ Why would you write H[t_] = {{0, 1}, {1, 1}}; ? what is t doing here? $\endgroup$
    – Nasser
    Jan 4, 2014 at 0:18
  • $\begingroup$ I can change H[t_] and consider that it is time dependent. You can take it like H[t_] = {{0, 1}, {1, t}}; $\endgroup$
    – santosh
    Jan 4, 2014 at 0:21

2 Answers 2

Reset to default
2
$\begingroup$

Using the solutions you have derived:

y[x_] := Diagonal[\[Sigma][t] /. sol /. t -> x];
u[x_] := Chop[y[x][[1]]]
v[x_] := Chop[y[x][[2]]]

Then plotting:

Plot[{u[t], v[t]}, {t, 0, 5}, 
 PlotStyle -> {{Thick, Blue}, {Thick, Red}}]

gives:

enter image description here

Improve this answer
$\endgroup$
4
  • $\begingroup$ Thanks for your answer. For my First question...if number of evaluated functions are very large means {i,1,n}, where n considers as 20 then its very inconvenient to write in your way... $\endgroup$
    – santosh
    Jan 5, 2014 at 20:14
  • $\begingroup$ Also consider H[t] is 20 x 20 matrix... $\endgroup$
    – santosh
    Jan 5, 2014 at 23:17
  • $\begingroup$ @santosh Sorry for too narrow focus. I agree for larger systems this would be very invonvenient. In that case I would tabulete the results and use ListPlot. You could still use Diagonal to extract diagonal elements of matrix but that is up to you. $\endgroup$
    – ubpdqn
    Jan 6, 2014 at 2:33
  • $\begingroup$ Is it possible to only evaluate Diagonal element inside the NDsolve to make it faster? something like...sol = NDSolve[{σ'[t] == fun[σ[t], t], σ[0] == Table[If[i == 1 [And] j == 1, 1, 0], {i, 1, 2}, {j, 1, 2}]}, Diagonal[σ[t]], {t, 0, tmax},DependentVariables-> σ[t]] $\endgroup$
    – santosh
    Jan 6, 2014 at 23:29
3
$\begingroup$

You need to evaluate the data series separately. E.g.

Plot[{
  Flatten[σ[t] /. sol, 1][[1, 1]],
  Flatten[σ[t] /. sol, 1][[2, 2]]},
 {t, 0, tmax}, PlotRange -> All]

enter image description here

See here for explanation. (Your data is in an unevaluated form within Plot.)

Addendum

For the second part of your question, you have this :-

Flatten[Evaluate[σ[t] /. sol], 1][[1, 1]]

t

Presumably you want the curve data: Evaluate is unnecessary, but you need values for t, i.e.

ListPlot[Table[
  Flatten[σ[t] /. sol, 1][[1, 1]],
  {t, 0, tmax, 0.1}], PlotRange -> All]

enter image description here

Further to OP's comment

While using Plot would be ideal for automatic x-axis (t) sample intervals - i.e. more dense where y changes more rapidly - a suitable solution using the original function is to use a sample interval of 0.01 in Table and use ListPlot or ListLinePlot:-

ListLinePlot[Table[
  Flatten[σ[t] /. sol, 1][[i, i]],
  {i, 1, 2}, {t, 0, tmax, 0.01}],
 PlotRange -> All, PlotStyle -> {Red, Green}]

enter image description here

Improve this answer
$\endgroup$
3
  • $\begingroup$ Thanks for your answer. For my First question...if number of evaluated functions are very large means {i,1,n}, where n considers as 20 then its very inconvenient to write in your way... $\endgroup$
    – santosh
    Jan 5, 2014 at 20:14
  • $\begingroup$ Also consider H[t] is 20 x 20 matrix... $\endgroup$
    – santosh
    Jan 5, 2014 at 23:18
  • $\begingroup$ That should work ok with Table and ListLinePlot. $\endgroup$
    – Chris Degnen
    Jan 5, 2014 at 23:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.