I have a problem in understanding a passage from the nots of a professor of us.
The starting problem is this PDE:
$$ \dfrac{\partial^2 t}{\partial u^2} = \frac{1}{h}\frac{\partial}{\partial h}\left(h^2 \frac{\partial t}{\partial h}\right)$$
Where $ t = t(u, h)$.
Then he says that by the following change of variable $$ \begin{cases} \xi = 2\sqrt{h} - u \\\\ \eta = 2\sqrt{h} + u \end{cases} $$
We should get
$$\frac{\partial^2 t}{\partial\xi\partial\eta} = 0$$
Simply.
Yet I have done the calculations five times, and what I end up with is:
$$\frac{\partial^2 t}{\partial\xi\partial\eta} = \frac{-2}{\eta + \xi}\left(\frac{\partial t}{\partial \xi} + \frac{\partial t}{\partial\eta}\right)$$
I am not asking anyone of you to do the complete calculation, for I know it's so tedious... But I really cannot get out of this. Any hint? It could also be that he wrote wrong? Or the change of variable is not correct?
Thank you in advance!
Note that the canonical form may include first-order terms as well. Using the proposed change of variable, we compute the partial derivatives by using the chain rule: \begin{aligned} t_u &= t_\xi\xi_u + t_\eta\eta_u = -t_\xi + t_\eta \\ t_h &= t_\xi\xi_h + t_\eta\eta_h = (t_\xi + t_\eta) h^{-1/2} . \end{aligned} The product rule gives $h^{-1}(h^2t_h)_h = 2t_h + ht_{hh} $. Similarly to above, the second derivatives satisfy \begin{aligned} t_{uu} &= -t_{u\xi} + t_{u\eta} = t_{\xi\xi} - 2t_{\xi\eta} + t_{\eta\eta} \\ t_{hh} &= (t_{h\xi} + t_{h\eta}) h^{-1/2} - \tfrac12 (t_\xi + t_\eta) h^{-3/2} \\ &= (t_{\xi\xi} + 2t_{\xi\eta} + t_{\eta\eta})/h - \tfrac12 (t_\xi + t_\eta) h^{-3/2} . \end{aligned} Using $\frac1{4}(\xi + \eta) = \sqrt{h}$, we are now left with $$ t_{\xi\eta} - \tfrac32 (\xi+\eta)^{-1} (t_\xi + t_\eta) = 0\, , $$ which doesn't yield $t_{\xi\eta} = 0$. With this 2nd-order hyperbolic PDE ($h>0$), integration of $\frac{\text dh}{\text du} = \pm\sqrt{h}$ leads indeed to the characteristic variables $\xi= \pm 2\sqrt{h} - u$ and $\eta = \pm2\sqrt{h} + u$. Please let me know if you spot any mistake.