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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!

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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 : $$T(r,t)=\int \cosh\left(f(r,t)\right)dt \:\:=\:\: t\:\int \sinh\left(f(r,t)\right)dr $$

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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 $v_zv_w=1$.

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