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Originally, I intended to solve the following pde: $$\frac{1}{r}\frac{\partial}{\partial r}\bigg(r \theta^{\beta}\frac{\partial \theta}{\partial r}\bigg) +\frac{\partial }{\partial z}\bigg(\theta^{\beta} \frac{\partial \theta}{\partial z} \bigg)=0 ;\ 0\leq r \leq r_0; \ 0 \leq z \leq l$$

with the following BCs: $$\theta(r,0) = 1 \text{ ; } \theta(r,l) = \theta_0 \text{ (a constant)}$$ $$\frac{\partial \theta}{\partial r}\bigg\rvert_{(0,z)}=\frac{\partial \theta}{\partial r}\bigg\rvert_{(r_0,z)}=0$$ where $\beta$ is some constant.

I employed variable separation method, assuming the solution to be of the form $\theta(r,z) = R(r)Z(z)$

This lead to the following ODEs: $$R'' + \frac{\beta R'^2}{R} + \frac{R'}{r} - \lambda^2 R =0 \qquad ; \qquad Z'' + \frac{\beta Z'^2}{Z} + \lambda^2 Z =0$$

$\lambda^2$ being separation constant

Now, How do I handle these non-linear ODEs to find closed-form solution(does it exists)?

Any tricks/suggestion would be greatly appreciated.

Edit: As suggest by @Professor Vector, we can use variable transform in the equation and BCs and solve it like this

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Wouldn't it be simpler to solve the linear PDE $$\frac{1}{r}\frac{\partial}{\partial r}\bigg(r \frac{\partial \eta}{\partial r}\bigg) +\frac{\partial }{\partial z}\bigg( \frac{\partial \eta}{\partial z} \bigg)=0 ;\ 0\leq r \leq r_0; \ 0 \leq z \leq l$$ for $$\eta(r,z)=\frac{\theta(r,z)^{\beta+1}}{\beta+1}?$$ Edit: as @MrYouMath pointed out, we want to use $\eta=\ln\theta$ in the case $\beta=-1$.

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  • $\begingroup$ @MrYouMath Thanks for pointing that out! For the sake of completeness, I've included it in my answer. $\endgroup$ – Professor Vector Oct 8 '17 at 13:13
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This is an old question with a good answer already. However, there is no answer to the original question. Here is a method for solving the pair of nonlinear second order ODEs.


The first equation is $$RR''+\beta R'^2+\frac{1}{r}R'+(-\lambda^2)R=0.$$ Using the transformation $\phi(r) = R^{\beta+1}(r)$, we get the linear equation $$\phi''+\frac{1}{r}\phi'+\gamma\phi=0,$$ where $\gamma=-\lambda^2(\beta+1)$. Using series or some other method, you can determine that $\phi(r) = c_1J_0(\gamma^{1/2}r)+c_2Y_0(\gamma^{1/2}r)$, where $J_0$ and $Y_0$ are Bessel functions. Thus, $$R(r;c_1,c_2) = \left[c_1J_0(\gamma^{1/2}r)+c_2Y_0(\gamma^{1/2}r)\right]^{\frac{1}{\beta+1}}$$


The second equation is $$Z'' + \frac{\beta}{Z}Z'^2+\lambda^2 Z.$$ Using the transformation $\psi(Z) = Z'^2$, we get the linear equation $$ \psi'+\frac{2\beta}{Z}\psi+2\lambda^2 Z = 0. $$ This is a first order linear equation, whose solution can be found via integrating factor, giving $\psi(Z) = c_3 Z^{-2\beta}-\lambda^2Z^{2-2\beta}$. Substituting back in for $\psi$ gives us a separable first order equation, $$ Z' = \pm\sqrt{c_3 Z^{-2\beta}-\lambda^2Z^{2-2\beta}}, $$ which has an implicit solution given by $$ z = c_4 \pm \frac{Z\sqrt{1-\frac{\lambda^2Z^2}{c_3}}F_{(2,1)}(\frac{1}{2},\frac{\beta+1}{2};\frac{\beta+3}{2};\frac{\lambda^2Z^2}{c_3})}{(\beta+1)\sqrt{c_3Z^{-2\beta}(1-\frac{\lambda^2Z^2}{c_3})}} = G(Z;c_3,c_4),$$

where $F_{(2,1)}$ is the ordinary hypergeometric function.

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