Set of differential equations with trigonometric functions and a derivative squared

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 $$$$\frac{d\tau}{ds} = \sin\sigma \left( E \cos\sigma - p_\tau \sin \sigma \right)$$$$ $$$$\left(\frac{d\sigma}{ds}\right)^2 = \sin^2\sigma \left[ \left( E \cos\sigma - p_\tau \sin \sigma \right)^2 - M^2 \right]$$$$

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:

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

Thank you very much for the help.

• What green and blue lines mean on these pictures? Commented Sep 15, 2023 at 12:02
• 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 $\sigma = 0, \pi$. Commented Sep 15, 2023 at 12:38

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}}]}]]


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}}]}]]


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]


• Nice solution too (+1)! Unfortunately Mathematica v12.2 can't reproduce sol1, that's why I tried numerical solution Commented Sep 16, 2023 at 7:09
• @UlrichNeumann Oh, sorry, I forgot to add exact solution. Commented Sep 16, 2023 at 9:05

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}]


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


• 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? Commented Sep 15, 2023 at 12:56
• 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! Commented Sep 15, 2023 at 13:06
• @UlrichNeumann This is nice solution (+1). There are real solutions computed with special choice of E, M, p. See my answer. Commented Sep 16, 2023 at 4:18