Salutations, I have been trying to solve this nonlinear wave equation that is a partial differential equation of the form (just for academical curiosity): $$u_{tt}=a(e^{\lambda u}u_x)_x$$ As I read in this document (link) (page 3, equation No. 7), it can be solved using this form of solution: $$ \begin{align} u(x,t) & =\phi(x)+\psi(t)\\ & \Updownarrow \\ u_{tt} & =\psi_{tt} \\ u_x & =\phi_x \end{align}$$ Then, replacing in non-linear PDE, I got this procedure: $$ \begin{split} \psi_{tt} & = a[e^{\lambda\left(\phi+\psi\right)}\phi_x]_x \\ & =a\left(e^\left(\lambda\phi\right)e^\left(\lambda\psi\right)\phi_x\right)_x \\ &=ae^\left(\lambda\psi\right)\left(e^\left(\lambda\phi\right)\phi_x\right)_x \\ & \Updownarrow \\ \frac{1}{e^\left(\lambda\psi\right)}\psi_{tt} &=a\left(e^\left(\lambda\phi\right)\phi_x\right)_x =m \end{split}$$ The previous procedure led to the following equations: $$ \begin{align} a\left(e^\left(\lambda\phi\right)\phi_x\right)_x &=m \label{1}\tag{1}\\ \psi_{tt} &=me^\left(\lambda\psi\right)\label{2}\tag{2} \end{align} $$ Approaching equation \eqref{1}, I got this procedure: $$ \begin{split} a\left(e^\left(\lambda\phi\right)\phi_x\right)_x=m \implies \int a\left(e^\left(\lambda\phi\right)\phi_x\right)_x \,dx & = \int \frac{m}{a} \,dx\\ e^\left(\lambda\phi\right)\phi_x & =\frac{mx}{a}+c_1\\ \int e^\left(\lambda\phi\right) \,d\phi &= \int \left(\frac{mx}{a}+c_1\right) \,dx \\ \frac{1}{\lambda}e^\left(\lambda\phi\right)&=\frac{m}{2a}x^2+c_1x+c_2 \end{split} $$ Finally, $$\phi(x)=\frac{1}{\lambda}\ln\left(\frac{\lambda m}{2a}x^2+\lambda c_1x+\lambda c_2\right)$$
Next, when I approached equation \eqref{2}, I got this procedure: $$ \begin{split} \frac{\partial^2 \psi}{\partial t^2} &= me^\left(\lambda\psi\right) \\ \int \frac{\partial^2 \psi}{\partial t^2} \frac{d\psi}{dt} \,dt &=m\int e^\left(\lambda\psi\right) \frac{d\psi}{dt}\,dt\\ \frac{1}{2}\left(\frac{d\psi}{dt}\right)^2 &= me^\left(\lambda\psi\right)+c \end{split} $$ Then, $$ \int \frac{1}{\sqrt{2me^\left(\lambda\psi\right)+c_1}}\,d\psi = t+k $$ I´ve taken trigonometric substitutions and even I´ve used the hyperbolic inverse function and I´ve got these results for integral in left side: $$ \begin{align} \frac{-2}{\lambda\sqrt{c_1}}\ln\left(\frac{\sqrt{c_1} + \sqrt{2me^\left(\lambda\psi\right)+c_1}}{\sqrt{2c_1-2me^\left(\lambda*\psi\right)}}\right)\label{I}\tag{I}&\\ \frac{-2}{\lambda\sqrt{2me^\left(\lambda\psi\right)+c_1}} & \label{II}\tag{II} \\ \frac{-1}{\lambda\sqrt c_1}\ln\left(\frac{\sqrt{c_1} +w}{\sqrt{c_1} -w}\right)&\text{ where }w=\sqrt{2me^\left(\lambda\psi\right)+c_1}\label{3}\tag{II} \end{align} $$
I require help for finding the solution to equation $(2)$ because none of possible solutions guide to exact solution shown in this paper in spanish (link) (page 3, section 2).
I would be very thankful with any guidance or starting steps and/or explanations to find the correct procedure and exact solution.
Thanks for your time and your attention.
