Classical wave equation and relativity Let it be $f=f(r,t)$ any function representing the "shape" or "outline" of an object.  In classical physics $f$ obeys the "classic" wave equation.
\begin{equation}
\frac{∂^2f}{∂t^2}-v^2\nabla^2 f=0
\end{equation}
If one imagines the object to be a pool ball then $f$ would precisely describe the translational motion of the ball to right or left with velocity $v$. However, the relativistic form of this exact equation (with $0\leqslant v \leqslant c$) do not exists, since:


*

*If a phenomenon obeys classic wave equation then $v$ must be equal in all reference frames, which obvious is not the case of a pool ball.

*Considering previous paragraph (paragraph 1.) and the postulate that the speed of light must be equal on all reference frames, follows that classic wave equation holds only in those cases where $v=c$ (Maxwell equations for example). 


My question is: Is there a problem with my reasoning and wave equation holds as it is. If not, what is the equivalent relativistic form of this equation.
 A: The premise that $\frac{\partial^2f}{\partial t^2}-v^2 \frac{\partial^2f}{\partial x^2}=0$ implies anything about what transformation to use is false.
If these were surface waves on water in a Newtonian world, then the correct transformation to use would in fact be a Galilean transformation, and you'd get a wave equation + a transport term. Plug in $\bar{f}(x+u t,t)=f(x,t)$ to see that here, where a subscript x denotes differentiation with respect to the first argument and subscript t with respect to the second argument:
$$\bar{f}_{tt}-(v^2-u^2) \bar{f}_{xx}+2u\bar{f}_{xt}=0$$
If these were surface waves on water in a relativistic world, then the correct transformation to use would be a Lorentz transformation. If $v<c$, then this would transform weirdly. Plug in $\bar{f}(\gamma x+\gamma u t,\gamma t+\gamma u x/c^2)=f(x,t)$ to find:
$$\frac{c^4 - u^2 v^2}{c^4-u^2 c^2} \bar{f}_{tt}+2  u \frac{c^4 - v^2c^2}{c^4- u^2c^2}\bar{f}_{xt}-c^2 \frac{v^2 c^2-u^2c^2}{c^4-u^2 c^2}\bar{f}_{xx}=0$$
It just so happens that if $v=c$, this simplifies to $\bar{f}_{tt}-c^2 \bar{f}_{xx}=0$.
Mathematically, the Lorentz transformation with "speed of light" $v$ is a symmetry of the partial differential equation. Physically, that doesn't prove that the symmetry is the correct one to use. All three partial differential equations here are the correct ones to use in one context or another.
I'm not a good historian of physics, but I believe this is what Einstein is credited with over Lorentz. It was clear for a while that mathematically, Maxwell's equations are invariant under a Lorentz transformation. What Einstein did is give the correct physical interpretation.
A: 
However, the relativistic form of this exact equation (with $0\leqslant v \leqslant c$) do not exists, since:
  
  
*
  
*If a phenomenon obeys classic wave equation then $v$ must be equal in all reference frames, which obvious is not the case of a pool ball.
  
*Considering previous paragraph (paragraph 1.) and the postulate that the speed of light must be equal on all reference frames, follows that classic wave equation holds only in those cases where $v=c$ (Maxwell equations for example). 
  

It's not clear to me what you mean by relativistic form of the wave equation:
\begin{equation}
\frac{∂^2f}{∂t^2}-v^2\nabla^2 f=0.
\end{equation}
If you mean the form of the equation that is consistent with the special theory of relativity, then the thing to bear in mind is that the form is not important; this equation can be consistent with the special theory of relativity even if $v$ depends on the frame, provided $f$,$v$ are correctly transformed.
In any given frame the pattern has shape and velocity that are constant in time, so the pattern can be expressed as function of $\mathbf x - \mathbf v t$:
$$
f(\mathbf x,t) = \Phi(\mathbf x - \mathbf vt)
$$
Function $f(\mathbf x,t)$ solves the wave equation, for any function $\Phi$. 
Uniformly moving rigid pattern is uniformly moving and rigid in all inertial frames. In non-relativistic theory the pattern $\Phi$ is the same in all inertial frames. In a relativistic theory it is not, since it undergoes the Lorentz contraction.
This means that in order to use the wave equation to describe the pattern in a different reference frame, one has to use appropriate $\Phi$ and $v$ . If the pattern is moving with constant velocity and the shape does not change, this is always possible.
A: Your ball does not obey a classic wave equation even though its outline moves with constant speed. You would at least have to the consider Schroedinger's equation for the wave function. This, however, is not a wave equation but has a first order derivative with respect to time and its solutions do not preserve shape in time.
Your ball is simply an extended object moving at speed v relative to its inertial reference frame. If you measure it in a different inertial reference frame you just have to use the Lorentz transformation for coordinates and time. So you get, e.g. a longitudinal diameter contraction and the speed adds to the reference frame speed according to the velocity addition of special relativity.
When you look at the general solution of classical (sub c) wave propagation in a reference frame according to you wave equation, you simply apply the Lorentz transformation to obtain the shape and time behavior in a different reference frame.   
