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


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 $v_zv_w=1$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.