Let $\psi : [0,\infty] \to \mathbb R$ be a strictly positive, continuously differentiable function, and consider the non-linear ODE $$\ddot x = - \frac{1}{4} \frac{\psi'(x)}{\psi(x)} \left( \dot x^2 - \frac{1}{\psi(x)} \right),$$ with initial conditions $x(0) = 0$ and $\dot x(0) = 1$. I would like to find an explicit choice of $\psi(x)$ which satisfies $$\lim_{x\to\infty} \psi(x) = \infty,$$ and for which the solution $x(t)$ is strictly increasing and asymptotically linear (i.e. $x(t) = O(t)~$). If the ODE could be solved explicitly, that would be even better.

I have used Mathematica to experiment with $\psi(x) = x+1$. Based on the numerical evidence, this example seem to work, but I don't know how to rigorously analyze the solution.

Can you come up with a function $\psi(x)$ for which the above holds? If not, could you point me toward a reference which discusses how to rigorously show the properties above in the absence of an explicit solution?

Edit: This is the equation governing the $x$-position of geodesics in the plane $\mathbb R^2$ under the conformal Riemannian metric $g_{ij}(x,y) = \sqrt{\psi(x)} \delta_{ij}(x).$


2 Answers 2


Unfortunately, no such $\psi(t)$ and $x(t)$ may exist exactly as you want them. From the equation, assuming $\psi\to\infty$, and $\dot x\to c>0$, we have that the logarithmic derivative of $\psi$ is $O(\ddot x)$, and integrating we get $\log(\psi)=$ $O(\dot x)=O(1)$ so $\psi=O(1)$ too, a contradiction.

rmk. Should you consider weaker requirements, I would first make an Ansatz on $x(t)$ and then solve the first order ODE for $\psi.$


WARNING: Unchecked computations appear in the discussion below.

First, I don't get the same equation that you do. I would prefer to write the metric as $e^u (dx^2 + dy^2)$, where $u$ is a function of $x$ only. To derive the equation for a constant speed geodesic, I would find the Euler-Lagrange equation for the functional $\gamma \mapsto \int |\gamma'(t)|^2 e^{u(x(t))}dt$. Since $\gamma$ has constant speed, we can assume that $|\gamma'|^2 = \ell^2$ is a constant. A straightforward calculation gives me the following ODE for the $x$ co-ordinate of $\gamma$:

$$x'' = u'(x)(-(x')^2 + \frac{1}{2}\ell^2)$$

In principle, this can be solved using the usual approach of reducing it to a first order ODE by writing in the form

$$ F'(x')x'' = u'(x)x' $$

and integrating. Then you can try solving the first order ODE. I didn't see how to do this.

But there is a trivial linear solution: $t \mapsto (ct, ct)$, where $c$ is a constant equal to $\ell/\sqrt{2}$ above, which appears to hold no matter what the function $u$ is. This seems surprising, so you shouldn't take my calculations seriously unless you check them carefully first.


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