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The following function generates a stream plot for t1 and t2.

StreamPlot[{k, (-L1 k Sin[t1[t]])/(L2 Sin[t2[t]])},
           {t1[t], -Pi/2, Pi/2}, {t2[t], -Pi/2, Pi/2}

I know that each streamline gives me a set of values of t1 and t2 for which

L1 Cos[t1[t]] + L2 Cos[t2[t]] == rvar

is constant, for any values of L1 and L2.

I want to extract the values of a streamline for a particular value of rvar. How can I do this?

A solution for the differential equation is also something that can help. Thanks

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  • $\begingroup$ Have a look at StreamPoints. $\endgroup$ – rmw Sep 23 '18 at 14:19
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Another approach is to use ContourPlot. For instance,

cp = With[{L1 = 1, L2 = 3, rvar = 3}, 
    ContourPlot[L1 Cos[t1[t]] + L2 Cos[t2[t]] == rvar, 
    {t1[t], -Pi/2, Pi/2}, {t2[t], -Pi/2, Pi/2}, ContourStyle -> Red, 
    FrameLabel -> Automatic, ImageSize -> Large, LabelStyle -> {Black, Bold, Medium}]]

enter image description here

and the values of the points making up the curve are given by

pts = cp[[1, 1]];

which can be visualized by

ListPlot[pts, ImageSize -> Large, LabelStyle -> {Black, Bold, Medium}]

enter image description here

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Just use NDSolve. It is meant precisely for this, i.e. numerical solution of differential equations. StreamPlot is for plotting and won't be accurate anyway.

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  • $\begingroup$ I tried using that, but I'm unable to put it in the correct form i think. How can i put two variables as such? $\endgroup$ – Ashwin Kumar Sep 23 '18 at 14:38
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Here is an NDSolveValue approach. First, some example parameters:

L1 = 1;
L2 = 3;
rvar = 3;

Your equation:

eqn = L1 Cos[t1[t]]+L2 Cos[t2[t]]==rvar

To use NDSolveValue, we need an initial condition, for which I will use FindInstance:

initial = {t1[t], t2[t]} /. First @ FindInstance[
    {eqn, -π < t1[t] < π, -π < t2[t] < π},
    {t1[t], t2[t]},
    Reals
]

{-π/2, 0}

We have two dependent variables, so we need another ODE. The obvious choice is to parametrize by the arc-length. So, the ODE is:

sol = NDSolveValue[
    {
    eqn, t1'[t]^2 + t2'[t]^2 == 1, 
    t1[0] == initial[[1]], t2[0] == initial[[2]],
    WhenEvent[EuclideanDistance[{t1[t],t2[t]}, initial]<.002, end = t; "StopIntegration"]
    },
    {t1, t2},
    {t, 0, 18}
];

Probably a better detector of when the solution overlaps itself could be used, but the above did the job.

Visualization:

plot = ParametricPlot[Through @ sol @ t, {t, 0, end}]

enter image description here

And a table of values:

pts = Table[Through @ sol @ t, {t, Subdivide[0, end, 30]}];

Visualization:

Show[plot, ListPlot[pts, PlotStyle->Red]]

enter image description here

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