A non-linear coupled second order ODE for a geodesic

Question

Let $g = (dx)^2 + e^x (dy)^2$ be a Riemannian metric on $\mathbb{R}^2$. I want to find the equation of the geodesic. To do this, I know that I can solve a system of coupled non-linear second order ODEs. After calculating the Christoffel symbols, I have arrived at the following system for $\gamma(t)=(\gamma^1(t),\gamma^2(t)) =(u(t),v(t))$: $$ u''-\frac{1}{2}e^u(v')^2 =0 $$ $$ v''+u'v'=0 $$ I'm not very familiar with ODEs beyond the basic techniques that one might learn in a first course. I started by guessing $u=-k\log t$ for some $k>0$, but this turned out to not work. Can someone help me solve this system? This is not for homework. I wanted to do an example to help me better understand parallel transport, but it ended up being too complicated. Thanks.

EDIT: Intuitively, I feel like a geodesic from $(0,0)$ to $(1,1)$ will travel in a path that sticks closely to the $y$-axis for most of its time, because $e^x$ makes $g$ "lengthier" the bigger the value of $x$. This led me to guess a solution that is like $\log t$, but I don't know if this intuition is correct.

Per AlexanderJ93's solution, I have graphed the parameterization on Desmos for $a=b=c=1$ and got a curve that is roughly what I expected in the previous edit.

Answer 1

I'm going to focus on the ODE system (for more on the geometry, see Ted Shifrin's comment).

We want to solve the nonlinear system

$$ \begin{align}u'' - \frac{1}{2}v'^2\exp u & = 0 \\ v'' + u'v' & = 0\end{align}$$

Notice that the lowest order of $v$ is actually $v'$, so let's have $w = v'$. This gives us a second equation of

$$ w' + u'w = 0 $$

Multiply through by $\exp(u)$ (standard integrating factor method) to get

$$ \exp(u)w' + u'\exp(u) w = [w\exp(u)]' = 0 $$

and so we can integrate and solve

$$ w = a \exp(-u) $$

Put this into the first equation to get

$$ u'' - \frac{a^2}{2}\exp(-u) = 0 $$

Multiply through by $u'$ to get

$$ u'u'' - \frac{a^2}{2}u'\exp(-u) = \frac{1}{2}\left[u'^2 + a^2\exp(-u)\right]' = 0 $$

Integrate this to get the separable ODE

$$ u' = \sqrt{b^2-a^2\exp(-u)} $$

where $b^2 \geq a^2\exp(-u) \geq 0$. Integrate again

$$ \frac{2}{b}\tanh^{-1}\left(\frac{1}{b}\sqrt{b^2-a^2\exp(-u)}\right) = c + t $$

which we can solve explicitly as

$$ u = 2\log\left(\frac{a}{2b}\text{cosh}\left(c+bt\right)\right) $$

(replacing $b$ with $2b$ and $c$ with $bc$ for neatness). Then,

$$v = \int a\exp(-u) \text dt = \frac{4b^2}{a}\int \text{sech}(c+bt) \text dt $$

which we can integrate to

$$ v = \frac{4b^2}{ac}\tan^{-1}\sinh(b+ct) = \frac{4b^2}{ac}\text{gd}(b+ct) $$

where $\text{gd}$ is the Gudermannian function. Thus,

$$ \gamma(t) = \left( 2\log\left(\frac{a}{2b}\text{cosh}\left(c+bt\right)\right) , \frac{4b^2}{ac}\text{gd}(b+ct) \right) $$

Answer 2

Haven't checked the system is correct, but by writing $u(0) = u_{0}$, $u'(0) = u_{0}'$, and similarly for $v$, we can solve if $v_{0}' \neq 0$ by writing the second as $\frac{v''}{v'} = -u'$ and integrating to $v' = (e^{u_{0}}v_{0}') e^{-u}$. Substituting into the first gives $u'' - \frac{1}{2}(e^{u_{0}}v_{0}')^{2} e^{-u} = 0$. Multiplying by $u'$ and integrating gives $$ (u')^{2} + (e^{u_{0}}v_{0}')^{2} e^{-u} = (u_{0}')^{2} + e^{u_{0}} (v_{0}')^{2}, $$ which has the form $(u')^{2} = c_{2} - c_{1}^{2} e^{-u}$. This can be integrated and solved explicitly for $u$ (for example, by substituting $e^{u/2} = (c_{1} \cosh x)/\sqrt{c_{2}}$) and then for $v$, though the details are probably easier to write out oneself than to follow.