ODE - how to solve this ODE obtained from a geodesic problem?

Question

I am stuck at solving this ODE that I obtained by trying to compute geodesics:

$$ \begin{equation} g(\theta) \frac{d^2 \theta}{dt^2} + \frac{1}{2} \frac{\partial g(\theta)}{\partial \theta} \left( \frac{d\theta}{dt} \right)^2 = 0. \end{equation} $$

where $$g(\theta) = \frac{3(\theta^2+1)}{(\theta-1)^2(\theta+1)^2},$$ $$\frac{\partial g(\theta)}{\partial \theta} = \frac{-6\theta(\theta^2 + 3)}{(\theta-1)^3 (\theta+1)^3}$$

Any help? (It has been several years that I have not been confronted to differential equations... Thank you)

We can assume $\theta(0) = \theta_0$ and $\theta(1) = \theta_1$ as initial conditions.

In SAGEMath, I obtain:

-1/2*(sqrt(2)*arcsinh(2*(theta(t) - 1)/abs(2*theta(t) + 2)) + sqrt(2)*arcsinh(2*(theta(t) + 1)/abs(2*theta(t) - 2)) - 2*arcsinh(theta(t)))/_K1 == _K2 + t

Answer 1

Sage gave you the solution (in implicit form). Maple's is essentially the same:

$$ {\rm arcsinh} \left(\theta\right)-\frac{\sqrt {2}}{2}{\rm arctanh} \left( \,{\frac {1+\theta}{\sqrt {2\,{\theta}^{2}+2}}}\right)+\frac{\sqrt {2}}{2}{\rm arctanh} \left({\frac {1-\theta}{\sqrt {2\,{\theta}^{2} +2}}}\right)-{c_1}\,t-{c_2}=0 $$

Write this as $f(\theta) = c_1 t + c_2$. The boundary conditions give you the values of $c_1$ and $c_2$: $c_2 = f(\theta_0)$, $c_1 = f(\theta_1) - f(\theta_0)$.

The graph of $f(\theta)$ looks like this:

Note the singularities at $\theta = \pm 1$, where $f(\theta) \to \pm \infty$. On each of the intervals $(-\infty, -1)$, $(-1,1)$ and $(1,+\infty)$ the function is monotone, so there will be a unique solution to your initial value problem if $\theta_0$ and $\theta_1$ are both in one of those intervals. If $\theta_0$ and $\theta_1$ are in different intervals, there is no solution. You might have guessed that from the fact that $g(\theta)$ is singular at $\theta = \pm 1$.

Answer 2

Your equation is equivalent to

$$ \frac{d}{dt} \left[\sqrt{g(\theta)} \dot{\theta}\right] = 0 $$

which gives

$$ g(\theta) \dot{\theta} = C $$

this is a separable ODE and can be solved implicitly for $t$ as a function of $\theta$ by partial fractions. If you are lucky the function becomes easily invertible.

Answer 3

The calculus below shows how to solve the ODE.

This leads to an integral which can be expressed on a closed form : it is the formula already given by SAGE. So, I didn't rewrite it.

The result is $t$ as a function of $\theta$. I don't think that the inverse function $\theta(t)$ can be expressed on a closed form.