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


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 $\beta=-1$.

| cite | improve this answer | |
  • $\begingroup$ @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

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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.