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I am trying to solve the following equation

σ = 0.05;
f = 0.5;
μ = 0.1;
a[0] == 1
γ[0] == 0


γ'[t] == σ - 3/8 a[t]^2 + f/(2 a[t]) Cos[γ[t]]

a'[t] == -μ a[t] + f/2  Sin[γ[t]]

Unfortunately Mathematica is unable to do so. Is there a way to obtain the much needed solution ?

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  • $\begingroup$ Does this give something close to what you need: funcs = NDSolveValue[{\[Gamma]'[t] == \[Sigma] - 3/8 a[t]^2 + f/(2 a[t]) Cos[\[Gamma][t]], a'[t] == -\[Mu] a[t] + f/2 Sin[\[Gamma][t]], a[0] == 1, \[Gamma][0] == 0}, {\[Gamma], a}, {t, 0, 30}]; Plot[Evaluate@Through@funcs@t, {t, 0, 30}]? $\endgroup$ – kglr Mar 1 '15 at 16:34
  • $\begingroup$ I am looking for exactly that.Thank you. $\endgroup$ – Kerem Sönmez Mar 1 '15 at 16:48
  • $\begingroup$ posted the comment as answer. Welcome to mma.se. $\endgroup$ – kglr Mar 1 '15 at 17:02
  • $\begingroup$ Welcome to Mathematica.SE! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Read the faq! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! $\endgroup$ – user9660 Mar 1 '15 at 17:06
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funcs = NDSolveValue[{γ'[t] == σ - 3/8 a[t]^2 + f/(2 a[t]) Cos[γ[t]], 
   a'[t] == -μ a[t] + f/2 Sin[γ[t]],
   a[0] == 1, γ[0] == 0}, {γ, a}, {t, 0, 30}]; 

Plot[Evaluate@Through@funcs@t, {t, 0, 30}]

enter image description here

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I believe that it is this solution that you want to:

DSolve[Derivative[1][a][t] == -0.1 a[t] + 0.25 Sin[γ[t]], {a[t]}, {t}]

enter image description here

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