Consider the following equation on a 2D sphere with Riemannian metric.
$\partial_t u+\nabla_uu+\nabla p=0$
where the unknown are $u:[0,T]\times S^2\to TS^2$ and $p:[0,T]\times S^2\to[0,\infty)$.
My question is pretty basic: what does it mean to differentiate $u$ along $t$? Should one see it as a covariant derivative of some kind, or simply the classical derivative in $t$ in some local coordinates (taking for instance the spheric ones, letting $u=u_\phi\frac{\partial}{\partial\phi}+u_\theta\frac{\partial}{\partial\theta}$ and simply putting $\partial_tu=\partial_tu_\phi\frac{\partial}{\partial\phi}+\partial_tu_\theta\frac{\partial}{\partial\theta}$)?
Thank you a lot!
After a while I founded an answer to that, so I would like to share it with the community.
Basically the question was: supposing a time dependant vector filed $v(t,\cdot)\in\mathscr{X}(M)$, in which sense do we conceive $\partial_t v$?
Observe that $v:\mathbb{R}\times M\to TM$ has the simple, but important property that for a fixed $x\in M$ $v(t,x)\in T_xM$ for every $t$. Therefore, since the fibre does not change through time we can differentiate $v$ with respect to $t$ fixing $x$: $Dv(x,t):T_t\mathbb{R}\to T_{v(x,t)}T_xM$. Since $x$ is fixed one can identify $T_{v(x,t)}T_xM$ with $T_xM$, and therefore we have $Dv(x,t)[\frac{\partial}{\partial t}]=(v(x,t),\partial_tv(x,t))=\partial_tv(x,t)$ where $\partial_t$ is the differentiation component wise in local coordinates described in the question.
To summarise: a priori $\partial_tv$ belongs to $TTM$, but since the fibre does not change trough time for a fixed $x$, one can simply conceive $\partial_tv\in\mathscr{X}(M)$ taking advantage of the identification $T_{v(x,t)}T_xM\approx T_xM$.