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[✓] Size step is effectively zero?

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Hello, I am trying to plot a double pendulum. But when I try to plot the results, this error occurs:

"NDSolve::ndsz: At t == 0.5517583468854775`, step size is effectively zero; singularity or stiff system suspected."

Any idea what I've got wrong?

Lagrangian = (m1 + m2)/2*l1^2*\[Theta]1'[t]^2 + 
   m2/2*l2^2*\[Theta]2[t]^2 + 
   m2*l1*l2*\[Theta]1'[t]*\[Theta]2'[t]*
    Cos[\[Theta]1[t] + \[Theta]2[t]] + (m1 + m2)*g*l1*
    Cos[\[Theta]1[t]] + m2*l2*g*Cos[\[Theta]2[t]];

g = 9.81;
l1 = 1;
l2 = 1;
m1 = 1;
m2 = 1;

eq1 = D[D[Lagrangian, \[Theta]1'[t]], t] - D[Lagrangian, \[Theta]1[t]];
eq2 = D[D[Lagrangian, \[Theta]2'[t]], t] - D[Lagrangian, \[Theta]2[t]];

sol = NDSolve[{eq1 == 0, 
    eq2 == 0, \[Theta]1[0] == Pi/10, \[Theta]1'[0] == 
     0, \[Theta]2[0] == 0, \[Theta]2'[0] == 
     0}, {\[Theta]1, \[Theta]2}, {t, 0, 10}];

Plot[{Evaluate[\[Theta]1[t] /. sol], 
  Evaluate[\[Theta]2[t] /. sol]}, {t, 0, 10}, PlotRange -> A]
POSTED BY: Jonas Hamp
Answer
1 month ago

Is it possible that second Lagrangian term was supposed to be m2/2*l2^2*\[Theta]2'[t]^2 (note derivative of theta_2)? With that change the integration does not go singular and the plot seems reasonable.

POSTED BY: Daniel Lichtblau
Answer
1 month ago

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