Radially Accelerated Observers on Schwarzschild Spacetime I'll first provide a little bit of context, and then proceed to the question. Feel free to jump ahead to the question if you feel so, I'll try to keep all the absolutely essential information over there. You do not need to address anything on the context, but I believe it can be helpful to know what I was doing when the issue arose.
Context
A while ago I saw the question How can we explain the position of Mann's planet when travelling on Miller's planet in Interstellar movie? and it made me want to compute the time dilation relative to the stationary spaceship for three astronauts: a stationary astronaut (which is just a computation of the gravitational redshift), a free falling astronaut (gravitational redshift and a kinematic correction), and an astronaut who's returning to the ship. Interstellar uses a Kerr black hole, but I'm doing computations on Schwarzschild for simplicity.
To consider the astronaut returning, I decided to consider an observer moving with an acceleration proportional to the one needed to keep someone static. The intuition was "If I need to maintain my ship's rockets at this power to keep the ship stationary, if I double their power I should start moving away from the black hole". There would also be the bonus that setting the proportionality constant to $0$ or $1$ should recover my previous results, and hence I'd have a simple way to check the computations. However, something weird happened once I started actually doing the math.
Question
The equation of motion for an observer with acceleration $a^\mu$ is
$$u^\mu \nabla_\mu u^\nu = a^\nu,$$
where $u^\mu$ is the four-velocity. If we desire the four-velocity to stay normalized during the trajectory (and, as far as I know, we always do), then we must have
$$
\begin{align}
0 &= u^\mu \nabla_\mu (u^\nu u_\nu), \\
&= 2 u^\nu u^\mu \nabla_\mu u_\nu, \\
&= 2 u^\nu a_\nu. \tag{1}
\end{align}
$$
Now suppose we want to consider an observer on Schwarzschild spacetime moving radially with acceleration $$a^\mu = \frac{\alpha M}{r^2} \left(\frac{\partial}{\partial r}\right)^\mu,$$ where $M$ is the black hole's mass and $\alpha$ is some arbitrary proportionality constant. Notice that $\alpha = 0$ leads to a freely-falling observer and $\alpha = 1$ leads to a static observer.
Now, given this acceleration, Eq. (1) implies that either $\alpha = 0$ or $\dot{r} = 0$. However, the second case consists of a static observer, i.e., $\alpha = 1$. Hence, it seems that $\alpha = 2$, for example, is not possible.
Assuming I did not do anything wrong (please correct me if that is the case), what is the physical meaning behind this?
 A: After reading the comment by A.V.S. and thinking for a while, I decided to tackle a simpler problem: observers moving in a single direction in Minkowski spacetime. More specifically, instead of considering observers while the acceleration vector I proposed earlier in Schwarzschild, let us think of observers in Minkowski spacetime with acceleration given by
$$a^\mu = \alpha \left(\frac{\partial}{\partial x}\right)^\mu.$$
The argument in the original post still applies in this situation and leads to the conclusion that either $\alpha = 0$ or $\dot{x} = 0$. The special symmetries of Minkowski spacetime ends up forcing $\alpha = 0$ in both situations.
Let us then ignore this fact for a while and just try to solve for the four-velocity anyway. For this acceleration, the equations of motion will end up being
$$\left\lbrace\begin{aligned}
\ddot{t} &= 0, \\
\ddot{x} &= \alpha.
\end{aligned}\right.$$
Solving them leads to
$$\left\lbrace\begin{aligned}
t(\tau) &= t_0 + \dot{t}_0 \tau, \\
x(\tau) &= x_0 + \dot{x}_0 \tau + \frac{\alpha}{2} \tau^2.
\end{aligned}\right.$$
Let us pick, for simplicity, $x_0 = t_0 = 0$ and also set $\dot{x}_0 = 0$. The condition $u^\mu u_\mu = -1$ then implies $\dot{t}_0 = 1$. Hence, we get to
$$\left\lbrace\begin{aligned}
t(\tau) &= \tau, \\
x(\tau) &= \frac{\alpha}{2} \tau^2.
\end{aligned}\right.$$
So far, it might seem that we are just doing calculations blindly, but notice that with so simple equations we can write the spatial coordinate in terms of the time coordinate. We get
$$x(t) = \frac{\alpha}{2} t^2,$$
which means that the velocity as measured in this particular choice of frame (defined by $x_0 = t_0 = 0$ and $\dot{x}_0 = 0$) is
$$\frac{\text{d} x}{\text{d} t} = \alpha t,$$
which will surpass the speed of light within coordinate time $t = \frac{c}{\alpha}.$
Hence, as suggested by A.V.S.'s comment, only a few specific observers can have accelerations which do not depend on time. As an observer accelerates through space, they also need to accelerate through time so that inertial observers still measure their speeds to be less than the speed of light. In other words, the physical interpretation for the problem I was facing is that causality forbids arbitrary accelerations in a single spatial direction through spacetime.
