# Solution to a second order semilinear elliptic PDE

I am trying to find a solution (if one exists) to this PDE where (r,z) are cylindrical coordinates. $$u_{rr}+u_{zz}+u_{r}/r-u/r^2 = u^2u_{r}/r+u(u_{r})^2+u(u_z)^2$$ u is a dimensionless function (valid for all of space) whose value has the range between 0 and 1. u is the azimuthal component of a velocity field, so maybe it vanishes as r goes to zero, unless the solution is a vortex spike (tornado) as r goes to zero. I really don't know what the solution should look like, except that it should be symmetric about the z = 0 plane, i.e. u(r,+z) = u(r,-z), and that the solution must satisfy a certain integral whose value is experimentally known. After a great deal of work using classical symmetry analysis, I only found the trivial solution u = 0. Nonclassical symmetry analysis seems to be way too difficult to use, even with Mathematica symmetry programs. According to the book "Solving Nonlinear PDEs with Maple and Mathematica", "it is not possible to numerically solve elliptic PDEs via the predefined [Mathematica] functions PDSolve and NDSolve." I have not been able to find any Lagrangian or Hamiltonian for this problem. Perhaps there are none. According to P. J. Olver "Applications of Lie Groups to DEs" Second Edition, page 333, the Frechet derivative of the system of PDEs must be self-adjoint in order for Euler-Lagrange equations to exist. I don't think the Frechet derivative is self-adjoint for this case. Is anyone familiar with Dr. Shijun Liao's new Homotopy Analysis Method for solving nonlinear PDEs using Mathematica? I have tried so many things, and none of them seem to work. Seems like I need a miracle.

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Matlab PDE toolbox is a better tool for numerical solution of PDE than either Maple or Mathematica. It can solve linear and some nonlinear PDE, but unfortunately yours does not appear to fit their model. Maybe you should step back and consider questions such as: (1) do you have other boundary conditions? (2) Does the solution minimize some functional? (3) What do you expect to get from numerical solution of you had one? – user31373 Jul 28 '12 at 20:50
Are you sure about the last term in your equation? Shouldn't it be $u(u_r)^2$ rather than $u^2(u_r)$? – Jon Jul 30 '12 at 13:18

Let us consider the more general case $$\Delta y=\frac{1}{r}y^2y_r+y(y_r^2+y_z^2)$$ with $$\Delta =\frac{1}{r}\frac{\partial}{\partial r}r\frac{\partial}{\partial r}+\frac{1}{r^2}\frac{\partial^2}{\partial\theta^2}+\frac{\partial^2}{\partial z^2}$$ the Laplace operator in cylindrical coordinates. We now assume that the solution has the form $$y(r,z,\theta)=u(r,z)e^{im\theta}$$ with $m\in\mathbb{Z}$. We will get $$\frac{1}{r}\frac{\partial}{\partial r}r\frac{\partial u}{\partial r}+\frac{\partial^2u}{\partial z^2}-\frac{m^2}{r^2}u=\left[u\left(\frac{\partial u}{\partial r}\right)^2+u\left(\frac{\partial u}{\partial z}\right)^2+\frac{1}{r}\frac{\partial u}{\partial r}u^2\right]e^{2im\theta}$$ that holds provided $\theta=\pi$. This recovers OP equation for $m=1$. I am able to find a solution for $m=0$ and can be written down as $$u(r,z)=\frac{1}{\sqrt{r^2+z^2}}.$$

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Thanks for your elegant work. I guess that by adding the one simple term (for m = 1 ) the problem becomes much more difficult, i.e. the original problem is tough to solve. Additionally I have not been able to find any Lagrangian or Hamiltonian for this. However, if a solution is found, it can be substituted into an integral to compare with experimental results known to a high degree. There is a new very general technique by Dr. Shijun Liao called the Homotopy Analysis Method which works for nonlinear PDEs using Mathematica. When all else fails, I will try this. – Skybobcat Aug 7 '12 at 20:28