I am looking for a symbolic closed-form solution to the following BVP.
$$\frac{d^2 \phi}{dx^2} = c_1 \left ( e^{c_2(\phi(x)+c_3)} - e^{c_4(\phi(x)+c_3)} \right )$$
wherein $c_1, c_2, c_3$ and $c_4$ are real-valued scalar constants.
The boundary conditions are $\phi'(0) = 0$ & $\phi'(1) = 1$ in the unit domain $(0,1)$
I tried solving this using Mathematica's DSolve function, but was unsuccessful in doing so. Then, I tried solving for a linearised version of the RHS using a Taylor expansion about an operating point $\phi^*$. This worked and produced a symbolic answer.
I do realise that non-linear problems are hard and do not always have a general solution. But in this case, I do know that we have a solution (from theory, and also was able to compute it numerically with a relative error/tolerance of $10^{-7}$).
Furthermore, even though we have a double Neumann BC (flux BC) for this elliptic/BVP problem, we have a unique solution due to the fact that the source term is implicit, i.e. it contains the solution variable $\phi(x)$.
Any help regarding solving this problem in any suitable computer algebra package is highly appreciated.
Try the substitution, $$\phi = ln z$$ $$\phi_x = z^{-1}z_x$$ $$\phi_{xx}=-z^{-2}(z_x)^2 + z^{-1}z_{xx}$$
Then, $$-z^{-2}(z_x)^2 + z^{-1}z_{xx}=A_1z^{b_1}+A_2z^{b_2}$$ $$z_{xx}-z^{-1}(z_x)^2=A_1z^{b_3}+A_2z^{b_4}$$
Let $w(z)=(z_x)^2$, $$w_zz_x = 2z_xz_{xx}.$$ Then, $$\frac{1}{2}w_z-z^{-1}w=A_1z^{b_3}+A_2z^{b_4}$$
This is a linear first order equation that you can lookup a general solution. For instance: http://eqworld.ipmnet.ru/en/solutions/ode/ode0103.pdf