I’d like to solve the O.D.E.:
$$(1+\Phi_{x})\,\Phi_{xx}=\left(-1+\sqrt{1+\Phi^2}\right)\Phi,$$
where $\Phi=\Phi(x)$, $x\in(-1,1)$. It can be written as
$$ \frac{\Bbb d}{\Bbb dx}\left(\Phi_{x}+\frac{1}{2}\Phi_{x}^2\right)=\left(-1+\sqrt{1+\Phi^2}\right)\Phi, $$ which leads to $$ \Phi_{x}^2+2\,\Phi_{x}-\int_{-1}^{x}f\big(\Phi(t)\big)\,\Bbb dt=0 $$ where $f\big(\Phi(t)\big)=2\left(-1+\sqrt{1+\Phi^2(t)}\right)\Phi(t)$. This gives $$ \Phi_{x}=-1+\sqrt{1+\int_{-1}^{x}f\big(\Phi(t)\big)\,\Bbb dt}. $$ In other words, we obtain an integro-differential equation. Any idea to start with proving the existence of solutions (numerically or analytically)? Or it is easier to directly deal with the original nonlinear 2nd-order O.D.E. (an exact solution is available)? Thanks!
$$\left(1+\frac{\Bbb d\Phi}{\Bbb dx}\right)\frac{\Bbb d^2\Phi}{\Bbb dx^2}=\left(-1+\sqrt{1+\Phi^2}\right)\Phi$$ Let $\frac{\Bbb d\Phi}{\Bbb dx}=F(\Phi)$ then $\frac{\Bbb d^2\Phi}{\Bbb dx^2}=\frac{\Bbb dF}{\Bbb d\Phi}\frac{\Bbb d\Phi}{\Bbb dx}=\frac{\Bbb dF}{\Bbb d\Phi}F(\Phi)$ $$(1+F)F\frac{\Bbb dF}{\Bbb d\Phi}=\left(-1+\sqrt{1+\Phi^2}\right)\Phi$$ This can be integrated: $$\frac{1}{2}F^2+\frac{1}{3}F^3=\int\left(-1+\sqrt{1+\Phi^2}\right)\Phi \Bbb d\Phi$$ A particular solution of $(1+F)F\frac{\Bbb dF}{\Bbb d\Phi}=\left(-1+\sqrt{1+\Phi^2}\right)\Phi$ is: $$F=-1+\sqrt{1+\Phi^2}$$ $\frac{\Bbb d\Phi}{\Bbb dx}=-1+\sqrt{1+\Phi^2}$ $$x=\int \frac{\Bbb d\Phi}{-1+\sqrt{1+\Phi^2}}=\frac{-1+\sqrt{1+\Phi^2}}{\Phi}-\sinh^{-1}(\Phi)+C$$ This doesn’t provide all the solutions.