All smooth solutions to this PDE? I need someone who's better at PDEs than me to help me find all solutions to the following partial differential equation 
$\left(\frac{\partial T}{\partial t}\right)^2 - \frac{1}{t^2}\left(\frac{\partial T}{\partial r}\right)^2 = 1$
It is easy to see that this PDE has two solutions: $T = t$ and $T=t\cosh(r)$. I want to determine if there can be any others. 
I have checked all separable functions (i.e. $T(t,r)$ of the form $T= A(t)B(r)$), but this leads only to the two solutions above.
I also tried finding the curves $r(t)$ in the $(t,r)$ plane on which $T$ is constant. On these curves we have $r'(t) = -\frac{\partial T/ \partial t}{\partial T/ \partial r}$. I tried to combine this with the PDE equation above to see if I could solve for $r(t)$, but I couldn't make any progress there. Any help is appreciated! Thanks!
 A: A very partial answer :
$$\left(\frac{\partial T}{\partial t}\right)^2 - \frac{1}{t^2}\left(\frac{\partial T}{\partial r}\right)^2 = 1 \qquad\qquad [1]$$
Change of function :  $\qquad \cosh\left(f(r,t)\right) =\frac{\partial T}{\partial t} \qquad\implies\qquad t\:\sinh\left(f(r,t) \right)=\frac{\partial T}{\partial r}$
$$\frac{\partial^2 T}{\partial r \partial t }=\sinh\left(f\right)\frac{\partial f}{\partial r}=t\:\cosh\left(f\right)\frac{\partial f}{\partial t}+\sinh\left(f\right)$$
$$\sinh\left(f\right)\frac{\partial f}{\partial r}-t\:\cosh\left(f\right)\frac{\partial f}{\partial t}=\sinh\left(f\right)\qquad\qquad [2]$$
Solving with the method of characteristics. The set of differential equations is :
$$\frac{dr}{\sinh\left(f\right)}=-\frac{dt}{t\cosh\left(f\right)}=\frac{df}{\sinh\left(f\right)}$$
From $\frac{dr}{\sinh\left(f\right)}=\frac{df}{\sinh\left(f\right)}$ a first equation of characteristic curve is : $\quad \frac{f(r,t)}{r}=c_1$
From $ -\frac{dt}{t\cosh\left(f\right)}=\frac{df}{\sinh\left(f\right)}$ a second equation of characteristic curve is : $t\:\sinh\left(f(r,t)\right)=c_2$
The general solution of the PDE $[2]$ , expressed on implicit form, is :
$$\Phi\left(\frac{f}{r} \: ,\: t\:\sinh\left(f\right)\right)=0$$
where $\phi$ is any differentiable function of two variables.
Unfortunately, in the general case, this implicit equation cannot be solved for $f$ in order to express $f(r,t)$ on explicit form.
Supposing that we chose some particular function $\Phi$ which could be solved for $f$, this particular functions $f(r,t)$ would be convenient.
Then $T(r,t)$ could be obtained by integration :
$$T(r,t)=\int \cosh\left(f(r,t)\right)dt \:\:=\:\: t\:\int \sinh\left(f(r,t)\right)dr $$
A: Using the change of variables:
$$t=e^{x+y}\ ;\ r=y-x$$
your PDE translates into
$$u_x\cdot u_y=e^{2x+2y}$$
So finding solutions for the above PDE will give us solutions to your equation, for example:
$$u(x,y)=e^{x+y}+A\Rightarrow T(r,t)=t+A$$
$$u(x,y)=\frac{1}{2}(Be^{2x}+B^{-1}e^{2y})+A\Rightarrow T(r,t)=\frac{t}{2}[Be^r+(Be^r)^{-1}]+A$$
It is possible to simplify the equation even further by doing another change of variables:
$$z=\frac{e^{2x}}{2}\ ;\ w=\frac{e^{2y}}{2}$$
and get the equation $v_zv_w=1$.
