Is a geodesic always regular?

Assume that $(M, \nabla)$ is a smooth manifold equipped with an affine connection. In our geometry lectures, we defined a geodesic as a curve $\gamma: I \rightarrow M$ satisfying the following system of differential equations:

$$\ddot\gamma_k(t)+\sum_{i,j=1}^{n}\Gamma_{ij}^{k}(\gamma_1(t), \dots, \gamma_n(t))\dot\gamma_i(t)\dot\gamma_j(t)=0, \;\; k=1,2, \dots, n,$$

where $\gamma_i$ denotes the $i$-th coordinate of $\gamma$ and $\Gamma_{ij}^k$ are the Christoffel symbols, all in suitable local coordinates.

What I fail to see is how the above system of equation ensures that the curves $\gamma$ are regular (which should hold even if they are defined on maximal interval). More specifically, our teacher claims that either the geodesic is "trivial", i.e. a constant map (mapping the whole interval to one point), or it is a regular curve. Can it be seen somewhat directly from the equations above, or how can it be proven?

Thanks in advance for any help.

PS: This question is very similar to this question. However, I do not know how can the datum of the affine connection $\nabla$ be "translated" into the notion of Riemannian metric (or if the notions are somewhat fundamentally different). Therefore the answer there does not really help me.

Edit: My definition of Chritoffel symbols is that they are the bunch of unique functions such that $$\nabla_{\frac{\partial}{\partial u_i}}\frac{\partial}{\partial u_i}=\sum_{k=0}^{n}\Gamma_{ij}^k\frac{\partial}{\partial u_k}, \;\; i,j, \in \{1,2, \dots, n\}.$$

• How do you define the Christoffel symbols without a metric? – Mateus Sampaio Nov 25 '13 at 22:16
• @Mateus Sampaio: By the equations $\nabla_{\frac{\partial}{\partial u_i}}\frac{\partial}{\partial u_i}=\sum_{k=0}^{n}\Gamma_{ij}^k\frac{\partial}{\partial u_k}$. – Pavel Čoupek Nov 25 '13 at 22:28
• What about the definition of a regular curve? – Mateus Sampaio Nov 25 '13 at 22:40
• @Mateus Sampaio: The tangent vector of the curve is nonzero (at any point). I believe one can equivalently say that $\dot\gamma(t)\neq 0$, i.e. one can compute it in the above local coordinates. – Pavel Čoupek Nov 25 '13 at 22:50
• The notion of the metric is not needed so, but all you have to do to understand the other solution is to switch $\langle \frac{d \gamma}{dt}, \frac{d \gamma}{dt} \rangle$ to $\sum_k \dot{\gamma}_k (t)^2$, because that's is just the lenght of the vector $\dot{\gamma}$. – Mateus Sampaio Nov 25 '13 at 22:57

So looking at why a geodesic is a regular curve?, it seems that regular simply means $\dot\gamma \ne 0$ on the whole curve. There is a theorem that says that if $F:\mathbb R^n \to \mathbb R^n$ is smooth (actually locally Lipschitz should be enough), then if we try to solve the ode: $$\dot x = F(x), \quad x(t_0) = x_0$$ and extend it both forwards and backwards in time to a function $x:(t_0-t_1,t_0+t_2) \to \mathbb R^n$ (where $t_1>0$ and $t_2>0$ might be $\infty$), then that solution, if it exists, is unique. This means that if at any time $F(x(t_0)) = 0$, then $x(t) = x(t_0)$ is a solution, and hence the only solution, along the whole interval.
By using an atlas, it should be possible to extend this to manifolds: $$F:M\times TM \to T(M\times TM), \quad F(x,y) = \left(y,\sum_{ijk} \Gamma^i_{jk}(x) y_j y_k\frac\partial{\partial x_i}\right) .$$
Also, if the function $F$ is smooth (i.e. if the Christoffel symbols are smooth), then you can show that the solution $x(t)$ is also smooth.
• Also, the theorem says that the only way that $x(t)$ cannot be extended in the positive direction beyond $t=T$ is because $\|x(t)\| \to \infty$ as $t \to T$. But I don't think this is needed for what I am saying here. – Stephen Montgomery-Smith Nov 26 '13 at 3:26