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I've been trying to solve the following differential equation: \begin{equation}\left( \frac{d \tau}{d \sigma} \right)^2 = \frac{(E_0 \cosh\sigma-p_\tau\sinh\sigma)^2}{(E_0 \cosh\sigma-p_\tau\sinh\sigma)^2-M^2}\end{equation} For different values of the constants $E_0, M \text{ and } p_\tau$.

For $E_0$, I managed to a figure that resembles two hyperbolas, I probably have to tweak $p_\tau$ to get the full figure.

However, when dealing with $E_0 \neq 0$, things get trickier, particularly for negative values of $\sigma$, I get this: enter image description here

The code I implemented was

E0 = 5;
M = 1;
p = 1;

(*Plus or minus roots after removing the square*)
pode = \[Tau]'[\[Sigma]] == (E0*Cosh[\[Sigma]] - p*Sinh[\[Sigma]])/
    Sqrt[((E0*Cosh[\[Sigma]] - p*Sinh[\[Sigma]])^2 - M^2)];
node = \[Tau]'[\[Sigma]] == -(E0*Cosh[\[Sigma]] - p*Sinh[\[Sigma]])/
    Sqrt[((E0*Cosh[\[Sigma]] - p*Sinh[\[Sigma]])^2 - M^2)];

ic1 = \[Tau][1] == 2;

psol = DSolve[{pode, ic1}, \[Tau], \[Sigma]]
nsol = DSolve[{node, ic1}, \[Tau], \[Sigma]]

psolution = Re[\[Tau][\[Sigma]] /. psol]
nsolution = Re[\[Tau][\[Sigma]] /. nsol]

I chose to leave the square roots and solve for two ODE's, instead of just 1 with squared, because the solution yields many branches, and only some of them are physically significant, so I wanted to analyze the branches this way.

And finally for the plot

Plot[{psolution, nsolution}, {\[Sigma], -10, 10}, 
FrameLabel -> {"\[Sigma]", "\[Tau]"}, Frame -> True, 
PlotStyle -> {Blue, Green}, 
PlotLegends -> {"Positive Root", "Negative Root"}]

Thank you for the help.

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    $\begingroup$ what exactly is the question you are asking? Are you saying the answer by DSolve is not correct? $\endgroup$
    – Nasser
    Sep 21, 2023 at 3:44
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    $\begingroup$ or are you asking about the plot having swiggly lines at the edges? this could be a plotting/numerical issue. $\endgroup$
    – Nasser
    Sep 21, 2023 at 3:55

1 Answer 1

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These is a simple integral problem

   f = 
   \[Integral](
         g = ((E0*Cosh[\[Sigma]] - p*Sinh[\[Sigma]])/
          Sqrt[((E0*Cosh[\[Sigma]] - p*Sinh[\[Sigma]])^2 - M^2)] // 
                  TrigToExp // FullSimplify)) 
    \[DifferentialD]\[Sigma] /. 
       {ArcTanh[a_] - ArcTanh[b_] :>  ArcTanh[FullSimplify[(a - b)/(1 - a b)]]}

     ff = Sqrt[f^2 // FullSimplify] // PowerExpand

$$\frac{1}{2} \tanh ^{-1}\left(\frac{2 e^{\sigma } \sqrt{2 e^{2 \sigma } \left(\text{E0}^2-2 M^2-p^2\right)+e^{4 \sigma } (\text{E0}-p)^2+(\text{E0}+p)^2} (\text{E0} \sinh (\sigma )-p \cosh (\sigma ))}{e^{4 \sigma } (\text{E0}-p)^2+(\text{E0}+p)^2-2 M^2 e^{2 \sigma }}\right)$$

Up to signs, thats the solution

    D[ff, \[Sigma]]^2 == g^2 // FullSimplify
    True

Fine tuning for branch cuts depending in the relative values of the constants can be done by comparing plots with plots of NDSolveValue results for fixed constants.

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