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I have been for long time now try to solve this equation

q1[t_] = D[x[t], t, t] - 2 (1/y[t]) D[y[t], t] D[x[t], t]

With its complex conjugate

q2[t_] = Conjugate[D[x[t], t, t]] - 2 (1/Conjugate[y[t]]) 
         Conjugate[D[y[t], t]] Conjugate[D[x[t], t]]

In y[t] and x[t] , where:

y[t] D[x[t], t] Conjugate[D[x[t], t]] = Exp[-100 t] 

I tried NDSolve with arbitrary initial conditions, but nothing worked with me. According to the above constraint, I should get both y and x varying with time.

Any help how to make this work, even with solving analytically by DSolve or Reduce ?

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    $\begingroup$ Pleased show what you have tried so far.IT IS unclear which ode you intend to solve. $\endgroup$ – Ulrich Neumann Oct 8 '19 at 19:41
  • $\begingroup$ Well, with all due respect, as a member for 5 years, you should have format this question better. This time I've done it for you, but please take some time to learn to format the code properly. Also, have a look at this. $\endgroup$ – xzczd Oct 9 '19 at 7:24
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We write the first equation in the form

eq1 = {D[x[t], t, t] - 2 D[Log[y[t]], t] D[x[t], t] == 0

In the second equation, we take the logarithm and differentiate, then we have

eq2=D[Log[y[t]], t] == -100 - D[Log[Abs[D[x[t], t]]^2], t]

Further suppose that x[t] is real. We substitute expression D[Log[y[t]], t] from the second equation into the first and add the initial data, then we get

eq1 = {x''[t] + 2 (100 + D[Log[x'[t]^2], t]) x'[t] == 0, x[0] == 1, 
   x'[0] == 1};
sol = NDSolveValue[eq1, x, {t, 0, 1}, WorkingPrecision -> 50]
LogLogPlot[sol[t], {t, 0, 1}, AxesLabel -> {"t", "x"}]

LogLogPlot[Exp[-100 t]/sol'[t]^2, {t, 0, 1}, PlotRange -> All, 
 PlotPoints -> 200, AxesLabel -> {"t", "y"}]

Figure 1

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  • $\begingroup$ Thanks so much. @AlexTrounev $\endgroup$ – Dr. phy Oct 9 '19 at 3:46
  • $\begingroup$ @Dr.phy You're welcome! $\endgroup$ – Alex Trounev Oct 9 '19 at 11:41

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