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I'd like to solve the ODE

$$(1+\Phi_{x})\,\Phi_{xx}=(-1+\sqrt{1+\Phi^2})\,\Phi,$$

where $\Phi=\Phi(x)$, $x\in(-1,1)$. It can be written as

$$ \frac{d}{dx}\Big(\Phi_{x}+\frac{1}{2}\Phi_{x}^2\Big)=(-1+\sqrt{1+\Phi^2})\,\Phi, $$ which leads to $$ \Phi_{x}^2+2\,\Phi_{x}-\int_{-1}^{x}f(\Phi(t))\,dt=0 $$ where $f(\Phi(t))=2\big(-1+\sqrt{1+\Phi^2(t)}\big)\,\Phi(t)$. This gives $$ \Phi_{x}=-1+\sqrt{1+\int_{-1}^{x}f(\Phi(t))\,dt}. $$ 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 ODE (an exact solution is available)? Thanks!

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$$(1+\frac{d\Phi}{dx})\frac{d^2\Phi}{dx^2}=(-1+\sqrt{1+\Phi^2})\Phi$$ Let $\frac{d\Phi}{dx}=F(\Phi)$ then $\frac{d^2\Phi}{dx^2}=\frac{dF}{d\Phi}\frac{d\Phi}{dx}=\frac{dF}{d\Phi}F(\Phi)$ $$(1+F)F\frac{dF}{d\Phi}=(-1+\sqrt{1+\Phi^2})\Phi$$ This can be integrated : $$\frac{1}{2}F^2+\frac{1}{3}F^3=\int(-1+\sqrt{1+\Phi^2})\Phi d\Phi$$ A particular solution of $(1+F)F\frac{dF}{d\Phi}=(-1+\sqrt{1+\Phi^2})\Phi$ is : $$F=-1+\sqrt{1+\Phi^2}$$ $\frac{d\Phi}{dx}=-1+\sqrt{1+\Phi^2}$ $$x=\int \frac{d\Phi}{-1+\sqrt{1+\Phi^2}}=\frac{-1+\sqrt{1+\Phi^2}}{\Phi}-\sinh^{-1}(\Phi)+C$$ This doesn't provides all the solutions.

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  • $\begingroup$ Thanks so much for providing this idea. Indeed, It works. About the integration $$\int(-1+\sqrt{1+\Phi^2})\Phi d\Phi,$$ it seems to be $\frac{1}{3} \left(\Phi^2+1\right)^{3/2}-\frac{\Phi^2}{2}$? $\endgroup$ – LCH Dec 27 '15 at 10:10
  • $\begingroup$ You are right. This is an obvious mistake : I forgot the $\Phi$ in the integral. So, the end of my answer is deleted. $\endgroup$ – JJacquelin Dec 27 '15 at 10:37

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