Coordinate transformations of the metric tensor 
Let's have metric (it describes the space-time of uniformly accelerating observer in Minkowski space-time)
  $$
ds^2 = v^2du^2 - dv^2.  \qquad (.0)
$$
  I need to find expressions for $u = f(x, t), v = g(x, t)$, which leads to
  $$
ds^2 = dt^2 - dx^2, \quad u = f(x, t), \quad v = g(x, t).
$$
  How to get $f, g$? 

My attempt.
I substituted $u = f(x, t), v = g(x, t)$ in $ds^2 = u^2dv^2 - du^2$ and then, by equating it to $dt^2 - dx^2$, got system of PDE:
$$
g^2 (\partial_{x}f)(\partial_{t}f) - (\partial_{x}g) (\partial_{t}g) = 0, \quad g^{2}(\partial_{x}f)^{2} - (\partial_{x}g)^{2} = -1 , \quad g^{2}(\partial_{t}f)^{2} - (\partial_{t}g)^{2} = 1 .
$$
It may be "simplified" to
$$
(\partial_{x}g)^{2} - (\partial_{t}g)^{2} = 1 , \quad g^{2}\left((\partial_{t}f)^{2} - (\partial_{x}f)^{2}\right) = 1. \qquad (.1)
$$
How to solve it? Or maybe there is more simple method to get an expressions for $f, g$?
Edit.
I got an equation 
$$
\partial_{tt}f - \partial_{xx}f = 0 . \qquad (.2)
$$ 
But I can't get some conditions, which can help me to choose some partial solution. By the other words, I can't use $(.1)$ in the right way. So, what conditions may I use to solve $(.2)$?
Edit.
I got the solution by getting a solution for transition from Minkowski space-time to original space-time which metric is given by $(.0)$. I used method of separation of variables.
 A: The solution is $t = v \sinh( u), x = v \cosh(u)$, and you easily invert the formulae: $$u = \operatorname{arctanh}\left(\frac{t}{x}\right),\ \ v = \sqrt{x^2-t^2}.$$ 
A: Starting from $(\partial_{x}g)^{2} - (\partial_{t}g)^{2} = 1 $, we may use a change of variable : $x^+= x+t, x^-= x - t$, with $\partial_x = (\partial_x^+ + \partial_x^-) , \partial_t = (\partial_x^+ -
 \partial_x^- )$
This gives : $ 4\partial_{x^+} g \partial_{x^-} g = 1$. From the symmetry of this equation, we may try a solution of kind $g (x^+x^-)$, this will gives : $4x^+x^-(g' (x^+x^-))^2=1$, that is $g' (x^+x^-) = \pm \frac{1}{2 \sqrt{x^+x^-}}$, that is : $g(x^+x^-) = \pm \sqrt{x^+x^-}$.So finally : 
$$g = \pm \sqrt{x^2-t^2} \tag{1}$$
By replacing $g$ by its value, we obtain : $\partial_x f = \pm \frac{t}{x^2-t^2}, \partial_t f = \mp \frac{x}{x^2-t^2} $,
where the 2 signs must be different (because of $g^2 (\partial_{x}f)(\partial_{t}f) = (\partial_{x}g) (\partial_{t}g) < 0$), so finally :
$$\partial_x f = \pm (- \frac{t}{x^2-t^2}), \partial_t f = \pm ( \frac{x}{x^2-t^2}) \tag{2}$$
Now, by looking at : $ds^2=dx^2-dt^2 = g^2 df^2-dg^2$, we see that multiplying $x$ and $t$ by a real value $\lambda$, multiply $ds^2$ by $\lambda^2$, but $g^2$ and d$g^2$ are also multiplied by $\lambda^2$, so $f$ is necessarily invariant, so it must be a function of $\frac{t}{x}$, $f = f(\frac{t}{x})$, so we get : 
$$\partial_x f = - \frac{t}{x^2}f'(\frac{t}{x}), \partial_t f = + \frac{1}{x} f'(\frac{t}{x})\tag{3}$$
By comparing equations $(2)$ and $(3)$, we get : 
$$f'(\frac{t}{x}) = \pm \frac{1}{1 - \frac{t^2}{x^2}}\tag{4}$$
So, finally : 
$f = \pm Arctanh(\frac{t}{x})\tag{5}$
A: An other possibility is analytic continuation.
By choosing $U = iu, T = it$, we see that we have to go from a metrics $-(dT^2+dx^2)$ to a metric $-(v^2dU^2+dv^2)$.
Clearly, the second metrics is simply the first metrics (which is the euclidean metrics, up to a sign) in polar coordinates.
So, we may choose : $x = v \cos U, t = v \sin U$
So, we have : $v^2 = T^2+x^2, \tan U = \frac{t}{X}$
By analytic continuation, this gives : $v^2=x^2-t^2, \tanh u = \frac{t}{x}$
A: My understanding is that Minkowski spacetime is only applicable to inertial frames, i.e. frames that are not accelerating. Your observer is accelerating relative to Minkowski spacetime, which implies your observer is accelerating relative to another observer in an inertial (Minkowski) frame. Are you referring to proper acceleration, which is relative to the inertial observer, or coordinate acceleration which depends on the choice of coordinate system and hence choice of observer. An accelerating observer in Minkowski spacetime is a contradiction and equations will reflect this.We could start with three spatial axes $x_1,x_2,x_3$ scaled in light seconds and a temporal axis, $x_0$, scaled in q x seconds with $q^2 = -1$, then $(ds)^2 = -(dx_0)^2 + (dx_1)^2 + (dx_2)^2 + (dx_3)^2$ which may be written
$$
(ds)^2=
\begin{pmatrix}
 dx_0 & dx_1 & dx_2 & dx_3
\end{pmatrix}
\begin{pmatrix}
 -1 & 0 & 0 & 0\\
 0 & 1 & 0 & 0\\
 0 & 0 & 1 & 0\\
 0 & 0 & 0 & 1
\end{pmatrix}
\begin{pmatrix}
 dx_0 \\
 dx_1\\
 dx_2\\
 dx_3
\end{pmatrix}
$$
Relationship between coordinate time $x_0$ and proper time $\tau$. Define $\tau^2 \equiv - s^2$ then $(d\tau)^2 = -(ds)^2 = (dx_0)^2 - (dx_1)^2 - (dx_2)^2 - (dx_3)^2$ =$(dx_0)^2(1-[(\frac{dx_1}{dx_0})^2+(\frac{dx_2}{dx_0})^2 + (\frac{dx_3}{dx_0})^2])$, write $v^2 =[( \frac{dx_1}{dx_0})^2+(\frac{dx_2}{dx_0})^2 + (\frac{dx_3}{dx_0})^2]$ and $t=x_0$ then $d\tau^2 = (1-v^2)dt^2$. The proper time $\tau_{t_1}^{t_2}$ that elapses between coordinate times $t_1$ and $t_2$ is $$\tau_{t_1}^{t_2}= \int_{t_1}^{t_2}(1-v^2)^{\frac{1}{2}}dt$$ and because an inertial frame requires $v$ to be constant $$\Delta\tau = (1-v^2)^{\frac{1}{2}}\Delta t$$This describes an observer in a Minkowski space, my interpretation is that you want to describe an observer accelerating relative to the Minkowski observer. This requires replacing the above diagonal matrix with a metric tensor so that ds becomes an interval of spacetime in an accelerating frame. To do this requires defining what is meant by acceleration.
