What is the velocity of point with constant acceleration in its proper frame? This is posed as an example in the first chapter of  Landau and Lifschitz's Classical Theory of Fields.   The answer given in the text is based on the following analysis:

In its proper frame its 4-acceleration $w^i = (0,w/c^2,0,0)$, from the definition of $w^i$ as along the $x^1$-axis only.   Thus, the scalar product $w^i\cdot w_i = -c^2/4$.

Expressing this scalar in the lab reference frame, where the particle's velocity along its $x$-axis is $v$, I find the equation $$ (du^0/dS)^2 - (du^1/dS)^2  = - c^2/4 , $$   where $u^i$ is the 4-velocity.
Expressed in the lab system, the above equation, according to Landau, reduces to an easily integrated equation for the $x$-component of the velocity: $d\gamma\cdot v/dt=w$, since $w$ is the constant acceleration in the proper frame.   I find no such ease, because the differential element of proper time, $dS=c\,dt/\gamma$, is a function of $v$ in the lab system, and $u^i$ are also functions of $v$.   Has anyone else tried to reproduce Landau's analysis?
 A: Ok let me try to make some points clear.
First, in the instantaneous rest frame of the particle its speed is zero, so its 4-velocity reads $u^\mu = (1,0,0,0)$. Note that $u^2 = 1$ which, differentiating, implies that 4-velocity and 4-acceleration are orthogonal $u^i  w_i = 0$, in every frame. So in the proper frame we must have $w^0 = 0$ and $w^a = {d v^a \over d t}$, $a = 1,2,3$, where $v^a$ is the 3-velocity and $t$ the time (which is also the proper time, because we are in the proper frame of the particle).
If the particle "feels" constant acceleration this means that in its proper frame we must have ${d v^a \over d t} = $ constant. We can turn this into a covariant statement for the 4-acceleration $w^i$. Namely, if ${d v^a \over d t} = $ constant, we have, in the proper frame, $w^i w_i = -\left({d v^a \over d t}\right)^2 = $ constant (remember that $w^0 = 0$). Now, $w^i w_i$ = constant must hold in every frame, because its a Lorentz invariant quantity. This is how you can define uniformly accelerated motion in a covariant way.
Letting
\begin{equation}
w^i w_i = - a^2 = \text{constant} 
\end{equation}
we can solve this in a generic frame assuming motion in the $x$ axis. So we take $w^i = (w^0,w^1,0,0)$. Plugging into the above we find
\begin{equation}
(w^0)^2 - (w^1)^2 = - a^2 \implies \qquad w^0 = a \sinh{ \eta}, \qquad w^1 = a \cosh{ \eta}
\end{equation}
Now we have to relate this to the 4-velocity $u^i$. Assuming $u^i = (u^0, u^1, 0, 0)$, and imposing the mass-shell constraint $u^2 = 1$ we similarly get $u^0 = \cosh{\mu}$ and $u^1 = \sinh{\mu}$.
Finally,
\begin{equation}
w^0 = {d u^0 \over d S} = {d \mu \over d S} \sinh{\mu}, \qquad w^1 = {d u^1 \over d S} = {d \mu \over d S} \cosh{\mu}.
\end{equation}
So we conclude
\begin{equation}
{d \mu \over d S} = a = \text{constant} \implies \mu = a S \qquad \text{and} \qquad \eta = \mu = a S
\end{equation}
So we have
\begin{equation}
u^i(S) = (\cosh{(a S)}, \sinh{(a S)}, 0, 0).
\end{equation}
This can be integrated further to find the space-time position $x^i(S)$,
\begin{equation}
x^i(S) = {1 \over a} (\sinh{(a S)}, \cosh{(a S)}, 0, 0)
\end{equation}
this solution is historically called hyperbolic motion.
