Why does the ODE $$ \theta^{\prime} = (1-\theta^2)^{3/2}, \quad \theta(0) = 0, $$

have the two (different) solutions

$$\theta_1(t) = \frac{t}{\sqrt{1+t^{2}}}, \quad \theta_2(t) = \frac{-t}{\sqrt{1+t^{2}}}, $$

for $t \in (-\delta, \delta)$, whereby $0<\delta <1$?

How is this possible, since the right-hand side of the ODE is locally Lipschitz in a small neighbourhood around $0$ ?

Please make the universe work again.


$\theta_1$ is indeed a solution of the ODE, but $\theta_2$ is not.

$$\theta_2^\prime(t) = -\frac{1}{(1+t^2)^{3/2}} \neq \frac{1}{(1+t^2)^{3/2}} =\left(1-\theta_2^2(t)\right)^{3/2}$$

  • $\begingroup$ Thanks, Mathematica and Wolfram Alpha are doing it wrong. $\endgroup$ – Joker123 Sep 15 '19 at 16:39
  • 2
    $\begingroup$ @Joker123 MMA/WA assumes by default the complex field and thus finds solutions for both branches of the square root, that is, it recognizes the fraction in the exponent and "cleverly" transforms the equation to solve $θ′^2=(1-θ)^3$. Then both solutions are indeed correct. In MMA there should be options to restrict to real interpretations of the equation, in WA these options are well hidden, if they exist. $\endgroup$ – Lutz Lehmann Sep 15 '19 at 18:31

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