0
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I've been trying to solve a set of differential equations that involve the square of a derivative, similar to another post I made here, and trigonometric functions, as seen here \begin{equation} \frac{d\tau}{ds} = \sin\sigma \left( E \cos\sigma - p_\tau \sin \sigma \right) \end{equation} \begin{equation} \left(\frac{d\sigma}{ds}\right)^2 = \sin^2\sigma \left[ \left( E \cos\sigma - p_\tau \sin \sigma \right)^2 - M^2 \right] \end{equation}

I followed an entirely analogous process to a suggestion that was given in that post, but I was faced with two problems:

  1. Using the boundary conditions I chose, I only get a plot that ranges in $\tau$ from (approximately) $[-1,5]$, when the expected result covers all $\tau$.

  2. I have only been able to solve the equations and get a plot for $E=0$. If $E \neq 0$, then DSolve runs for several hours on end (and is currently still doing so).

The code I implemented was the following(in the code $y$ corresponds to $\sigma$ and $t$ to $\tau$):

(*Boundary Conditions*)
E0 = 0;
M = 5;

sol = DSolve[{t'[y]^2 == (E0*Cos[y] - p*Sin[y])^2/((E0*Cos[y] - p*Sin[y])^2 - M^2), t[1] == 2}, t, y]

Table[Plot[Evaluate[t[y] /. sol], {y, -1, Pi}, PlotLabel -> p], {p, 
  5.9, 6, 0.001}]

Plot[Evaluate[t[y] /. sol /. p -> 5.942], {y, -1, Pi}, 
 FrameLabel -> {"y", "t"}, PlotStyle -> Red, Frame -> True]

The expected results are:

enter image description here

Where left corresponds to $E=0$ and right to $E^2 > M^2$.

Thank you very much for the help.

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  • 1
    $\begingroup$ What green and blue lines mean on these pictures? $\endgroup$ Sep 15, 2023 at 12:02
  • $\begingroup$ The blue lines can be ignored, they are there only as visual aid, but aren't part of the solution. The green lines correspond to the boundaries $\sigma = 0, \pi$. $\endgroup$
    – MarcelRomp
    Sep 15, 2023 at 12:38

2 Answers 2

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In this post we answer the question how to extend periodic solution t[y] up to $4 \pi$ and how to compute solution in a case of E^2>M^2. Solution in a case of E=0 is given by

E0 = 0;
M = 5;

sol = DSolve[{t'[
      y]^2 == (E0*Cos[y] - p*Sin[y])^2/((E0*Cos[y] - p*Sin[y])^2 - 
       M^2), t[1] == 2}, t, y]

Visualization

cp = Plot[Evaluate[t[y] /. sol /. p -> 5.942], {y, -Pi, Pi}, 
  FrameLabel -> {"y", "t"}, PlotStyle -> Red, Frame -> True, 
  PlotLabel -> Row[{"E = ", E0}]];

fig1 = Show[
  Show[MapAt[Translate[#, {0, 2 Pi}] &, cp, 1], cp, 
   MapAt[Translate[#, {0, -2 Pi}] &, cp, 1], PlotRange -> All], 
  Graphics[{Green, Thick, Line[{{0, -8}, {0, 13}}], 
    Line[{{Pi, -8}, {Pi, 13}}]}]]

Figure 1

Solution in a case of E0=6 is given by

E0 = 6;
M = 5;

sol1 = DSolve[{t'[
      y]^2 == (E0*Cos[y] - p*Sin[y])^2/((E0*Cos[y] - p*Sin[y])^2 - 
       M^2), t[1] == 2}, t, y]
(*{{t -> Function[{y}, (144 Cos[1] Cos[y] Sqrt[
        14 - p^2 - 36 Cos[2] + p^2 Cos[2] + 12 p Sin[2]] Sqrt[
        14 - p^2 - 36 Cos[2 y] + p^2 Cos[2 y] + 12 p Sin[2 y]] - 
       24 p Cos[y] Sin[1] Sqrt[
        14 - p^2 - 36 Cos[2] + p^2 Cos[2] + 12 p Sin[2]] Sqrt[
        14 - p^2 - 36 Cos[2 y] + p^2 Cos[2 y] + 12 p Sin[2 y]] - 
       24 p Cos[1] Sqrt[
        14 - p^2 - 36 Cos[2] + p^2 Cos[2] + 12 p Sin[2]] Sin[y] Sqrt[
        14 - p^2 - 36 Cos[2 y] + p^2 Cos[2 y] + 12 p Sin[2 y]] + 
       4 p^2 Sin[1] Sqrt[
        14 - p^2 - 36 Cos[2] + p^2 Cos[2] + 12 p Sin[2]] Sin[y] Sqrt[
        14 - p^2 - 36 Cos[2 y] + p^2 Cos[2 y] + 12 p Sin[2 y]] + 
       6 Sqrt[2]
         ArcTanh[(-11 - 
          6 I p + (-25 I + 6 p + I p^2) Tan[1])/((-6 I + 
            p) Sqrt[-11 + 12 p Tan[1] - (-25 + p^2) Tan[1]^2])] Cos[
         1] Cos[y] Sqrt[-(6 Cos[1] - p Sin[1])^2] Sqrt[
        14 - p^2 - 36 Cos[2 y] + p^2 Cos[2 y] + 12 p Sin[2 y]]
         Sqrt[-11 + 12 p Tan[1] + 25 Tan[1]^2 - p^2 Tan[1]^2] - 
       6 Sqrt[2]
         ArcTanh[(
         11 - 6 I p + 
          I (-25 + 6 I p + p^2) Tan[1])/((6 I + p) Sqrt[-11 + 
           12 p Tan[1] - (-25 + p^2) Tan[1]^2])] Cos[1] Cos[
         y] Sqrt[-(6 Cos[1] - p Sin[1])^2] Sqrt[
        14 - p^2 - 36 Cos[2 y] + p^2 Cos[2 y] + 12 p Sin[2 y]]
         Sqrt[-11 + 12 p Tan[1] + 25 Tan[1]^2 - p^2 Tan[1]^2] - 
       Sqrt[2] p ArcTanh[(-11 - 
          6 I p + (-25 I + 6 p + I p^2) Tan[1])/((-6 I + 
            p) Sqrt[-11 + 12 p Tan[1] - (-25 + p^2) Tan[1]^2])] Cos[
         1] Sqrt[-(6 Cos[1] - p Sin[1])^2] Sin[y] Sqrt[
        14 - p^2 - 36 Cos[2 y] + p^2 Cos[2 y] + 12 p Sin[2 y]]
         Sqrt[-11 + 12 p Tan[1] + 25 Tan[1]^2 - p^2 Tan[1]^2] + 
       Sqrt[2] p ArcTanh[(
         11 - 6 I p + 
          I (-25 + 6 I p + p^2) Tan[1])/((6 I + p) Sqrt[-11 + 
           12 p Tan[1] - (-25 + p^2) Tan[1]^2])] Cos[
         1] Sqrt[-(6 Cos[1] - p Sin[1])^2] Sin[y] Sqrt[
        14 - p^2 - 36 Cos[2 y] + p^2 Cos[2 y] + 12 p Sin[2 y]]
         Sqrt[-11 + 12 p Tan[1] + 25 Tan[1]^2 - p^2 Tan[1]^2] - 
       6 Sqrt[2]
         ArcTanh[(-11 - 
          6 I p + (-25 I + 6 p + I p^2) Tan[y])/((-6 I + 
            p) Sqrt[-11 + 12 p Tan[y] - (-25 + p^2) Tan[y]^2])] Cos[
         1] Cos[y] Sqrt[
        14 - p^2 - 36 Cos[2] + p^2 Cos[2] + 12 p Sin[2]]
         Sqrt[-(6 Cos[y] - p Sin[y])^2]
         Sqrt[-11 + 12 p Tan[y] + 25 Tan[y]^2 - p^2 Tan[y]^2] + 
       6 Sqrt[2]
         ArcTanh[(
         11 - 6 I p + 
          I (-25 + 6 I p + p^2) Tan[y])/((6 I + p) Sqrt[-11 + 
           12 p Tan[y] - (-25 + p^2) Tan[y]^2])] Cos[1] Cos[y] Sqrt[
        14 - p^2 - 36 Cos[2] + p^2 Cos[2] + 12 p Sin[2]]
         Sqrt[-(6 Cos[y] - p Sin[y])^2]
         Sqrt[-11 + 12 p Tan[y] + 25 Tan[y]^2 - p^2 Tan[y]^2] + 
       Sqrt[2] p ArcTanh[(-11 - 
          6 I p + (-25 I + 6 p + I p^2) Tan[y])/((-6 I + 
            p) Sqrt[-11 + 12 p Tan[y] - (-25 + p^2) Tan[y]^2])] Cos[
         y] Sin[1] Sqrt[
        14 - p^2 - 36 Cos[2] + p^2 Cos[2] + 12 p Sin[2]]
         Sqrt[-(6 Cos[y] - p Sin[y])^2]
         Sqrt[-11 + 12 p Tan[y] + 25 Tan[y]^2 - p^2 Tan[y]^2] - 
       Sqrt[2] p ArcTanh[(
         11 - 6 I p + 
          I (-25 + 6 I p + p^2) Tan[y])/((6 I + p) Sqrt[-11 + 
           12 p Tan[y] - (-25 + p^2) Tan[y]^2])] Cos[y] Sin[1] Sqrt[
        14 - p^2 - 36 Cos[2] + p^2 Cos[2] + 12 p Sin[2]]
         Sqrt[-(6 Cos[y] - p Sin[y])^2]
         Sqrt[-11 + 12 p Tan[y] + 25 Tan[y]^2 - 
         p^2 Tan[y]^2])/(2 (-6 Cos[1] + p Sin[1]) Sqrt[
       14 - p^2 - 36 Cos[2] + p^2 Cos[2] + 
        12 p Sin[2]] (-6 Cos[y] + p Sin[y]) Sqrt[
       14 - p^2 - 36 Cos[2 y] + p^2 Cos[2 y] + 
        12 p Sin[2 y]])]}, {t -> 
   Function[{y}, (-144 Cos[1] Cos[y] + 24 p Cos[y] Sin[1] + 
       24 p Cos[1] Sin[y] - 4 p^2 Sin[1] Sin[y] + 
       6 Sqrt[2]
         ArcTanh[(-11 - 
          6 I p + (-25 I + 6 p + I p^2) Tan[1])/((-6 I + 
            p) Sqrt[-11 + 12 p Tan[1] - (-25 + p^2) Tan[1]^2])] Cos[
         1] Cos[y] Sqrt[-((6 Cos[1] - p Sin[1])^2/(
         14 - p^2 - 36 Cos[2] + p^2 Cos[2] + 12 p Sin[2]))]
         Sqrt[-11 + 12 p Tan[1] + 25 Tan[1]^2 - p^2 Tan[1]^2] - 
       6 Sqrt[2]
         ArcTanh[(
         11 - 6 I p + 
          I (-25 + 6 I p + p^2) Tan[1])/((6 I + p) Sqrt[-11 + 
           12 p Tan[1] - (-25 + p^2) Tan[1]^2])] Cos[1] Cos[
         y] Sqrt[-((6 Cos[1] - p Sin[1])^2/(
         14 - p^2 - 36 Cos[2] + p^2 Cos[2] + 12 p Sin[2]))]
         Sqrt[-11 + 12 p Tan[1] + 25 Tan[1]^2 - p^2 Tan[1]^2] - 
       Sqrt[2] p ArcTanh[(-11 - 
          6 I p + (-25 I + 6 p + I p^2) Tan[1])/((-6 I + 
            p) Sqrt[-11 + 12 p Tan[1] - (-25 + p^2) Tan[1]^2])] Cos[
         1] Sqrt[-((6 Cos[1] - p Sin[1])^2/(
         14 - p^2 - 36 Cos[2] + p^2 Cos[2] + 12 p Sin[2]))]
         Sin[y] Sqrt[-11 + 12 p Tan[1] + 25 Tan[1]^2 - p^2 Tan[1]^2] +
        Sqrt[2] p ArcTanh[(
         11 - 6 I p + 
          I (-25 + 6 I p + p^2) Tan[1])/((6 I + p) Sqrt[-11 + 
           12 p Tan[1] - (-25 + p^2) Tan[1]^2])] Cos[
         1] Sqrt[-((6 Cos[1] - p Sin[1])^2/(
         14 - p^2 - 36 Cos[2] + p^2 Cos[2] + 12 p Sin[2]))]
         Sin[y] Sqrt[-11 + 12 p Tan[1] + 25 Tan[1]^2 - 
         p^2 Tan[1]^2] - 
       6 Sqrt[2]
         ArcTanh[(-11 - 
          6 I p + (-25 I + 6 p + I p^2) Tan[y])/((-6 I + 
            p) Sqrt[-11 + 12 p Tan[y] - (-25 + p^2) Tan[y]^2])] Cos[
         1] Cos[y] Sqrt[-((6 Cos[y] - p Sin[y])^2/(
         14 - p^2 - 36 Cos[2 y] + p^2 Cos[2 y] + 12 p Sin[2 y]))]
         Sqrt[-11 + 12 p Tan[y] + 25 Tan[y]^2 - p^2 Tan[y]^2] + 
       6 Sqrt[2]
         ArcTanh[(
         11 - 6 I p + 
          I (-25 + 6 I p + p^2) Tan[y])/((6 I + p) Sqrt[-11 + 
           12 p Tan[y] - (-25 + p^2) Tan[y]^2])] Cos[1] Cos[
         y] Sqrt[-((6 Cos[y] - p Sin[y])^2/(
         14 - p^2 - 36 Cos[2 y] + p^2 Cos[2 y] + 12 p Sin[2 y]))]
         Sqrt[-11 + 12 p Tan[y] + 25 Tan[y]^2 - p^2 Tan[y]^2] + 
       Sqrt[2] p ArcTanh[(-11 - 
          6 I p + (-25 I + 6 p + I p^2) Tan[y])/((-6 I + 
            p) Sqrt[-11 + 12 p Tan[y] - (-25 + p^2) Tan[y]^2])] Cos[
         y] Sin[1] Sqrt[-((6 Cos[y] - p Sin[y])^2/(
         14 - p^2 - 36 Cos[2 y] + p^2 Cos[2 y] + 12 p Sin[2 y]))]
         Sqrt[-11 + 12 p Tan[y] + 25 Tan[y]^2 - p^2 Tan[y]^2] - 
       Sqrt[2] p ArcTanh[(
         11 - 6 I p + 
          I (-25 + 6 I p + p^2) Tan[y])/((6 I + p) Sqrt[-11 + 
           12 p Tan[y] - (-25 + p^2) Tan[y]^2])] Cos[y] Sin[
         1] Sqrt[-((6 Cos[y] - p Sin[y])^2/(
         14 - p^2 - 36 Cos[2 y] + p^2 Cos[2 y] + 12 p Sin[2 y]))]
         Sqrt[-11 + 12 p Tan[y] + 25 Tan[y]^2 - 
         p^2 Tan[y]^2])/(2 (6 Cos[1] - p Sin[1]) (-6 Cos[y] + 
         p Sin[y]))]}}*)

Visualization

cp1 = Plot[Evaluate[t[y] /. sol1 /. p -> 9.8], {y, -Pi, Pi}, 
  FrameLabel -> {"y", "t"}, PlotStyle -> Red, Frame -> True, 
  PlotLabel -> Row[{"E = ", E0}]]

fig2 = Show[
  Show[MapAt[Translate[#, {0, 2 Pi}] &, cp1, 1], cp1, 
   MapAt[Translate[#, {0, -2 Pi}] &, cp1, 1], PlotRange -> All], 
  Graphics[{Green, Thick, Line[{{0, -8}, {0, 13}}], 
    Line[{{Pi, -8}, {Pi, 13}}]}]]

Figure 2

Update 1 We can compare numerical solution proposed by Ulrich Neumann with exact solution as follows

erg[p_?NumericQ, E0_?NumericQ, M_?NumericQ] := 
 NDSolve[{t'[\[Sigma]]^2 == (E0*Cos[\[Sigma]] - 
        p*Sin[\[Sigma]])^2/((E0*Cos[\[Sigma]] - p*Sin[\[Sigma]])^2 - 
       M^2), t[1] == 2}, t, {\[Sigma], -Pi, Pi}, 
  Method -> {"StiffnessSwitching", 
    Method -> {"ExplicitRungeKutta", Automatic}}, 
  WorkingPrecision -> 30]

Visualization in a case of E0=6 with cp1 computed above

Show[cp1,Plot[Evaluate[Chop[t[y] /. erg[4911/500, 6, 5]]], {y, -Pi, Pi}, 
  PlotStyle -> {{Blue, Dashed}, {Blue, Dashed}}] // Quiet]

Figure 3

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  • $\begingroup$ Nice solution too (+1)! Unfortunately Mathematica v12.2 can't reproduce sol1, that's why I tried numerical solution $\endgroup$ Sep 16, 2023 at 7:09
  • $\begingroup$ @UlrichNeumann Oh, sorry, I forgot to add exact solution. $\endgroup$ Sep 16, 2023 at 9:05
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Try numerical solution NDSolve[...,Method->"StiffnessSwitching"].

Solutions seem to be complex, is this intended?

erg[p_?NumericQ, E0_?NumericQ, M_?NumericQ]  := 
NDSolve [{t'[\[Sigma]]^2 == (E0*Cos[\[Sigma]] -p*Sin[\[Sigma]])^2/((E0*Cos[\[Sigma]] - p*Sin[\[Sigma]])^2 - M^2), t[1] == 2}, t, {\[Sigma], -Pi, Pi}, 
Method -> "StiffnessSwitching"]

Plot[Evaluate[ReIm[t[y] /. erg[5.942, 0, 5]]], {y, -Pi, Pi}]

enter image description here

Plot[Evaluate[ReIm[t[y] /. erg[5.942, 6, 5]]], {y, -Pi, Pi}]

enter image description here

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  • $\begingroup$ The solutions being complex could be the problem, because they should belong to the real domain. Is there a way to only search for real solutions? $\endgroup$
    – MarcelRomp
    Sep 15, 2023 at 12:56
  • $\begingroup$ I don't know how to force real solutions. But have a look at your ode, t'[\[Sigma]]^2 sometimes is negativ. Check your ode! $\endgroup$ Sep 15, 2023 at 13:06
  • $\begingroup$ @UlrichNeumann This is nice solution (+1). There are real solutions computed with special choice of E, M, p. See my answer. $\endgroup$ Sep 16, 2023 at 4:18

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