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This is the simplest equation arising from a fascinating (to me) and obscure vector field theory of mathematical physics first developed in 1962, and for which no solutions have ever been found. The following PDE is the corrected version (June 13, 2017) from the original post (February 25, 2015), which had a sign error. This PDE is the result of 47 years of work for me, so any help would be appreciated since I am not a professional mathematician or physicist. $$u_{rr}+u_{zz}+u_{r}/r-u/r^2+u(u_r ^2+u_z ^2+uu_r/r)=0 $$where (r,z) are cylindrical coordinates and u(r,z) is a dimensionless azimuthal velocity field defined for all (Euclidean) space, with conditions \begin{cases}0\leq r\lt \infty\\-\infty\lt z\lt\infty\\0\leq u(r,z)\lt\infty\\u(r,z)=u(r,-z)\\ \lim_{r\to0}u(r,z)=0\\ \lim_{r^2+z^2\to\infty}u(r,z)=0\\\end{cases}Lie symmetry analysis yields no valuable results. I have unsuccessfully tried to find an ansatz that would simplify the PDE, and possibly lead to a closed form solution. Numerical analysis cannot be used because the domain is unbounded. I do not think there is a Lagrangian for this PDE. So my question is this: Do solutions exist for this PDE and conditions? And if so, how does one go about finding them? I have an undergraduate degree in physics, so my skills with PDEs are modest. Most texts on elliptic PDEs are beyond my abilities, so any help would be greatly appreciated!

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  • $\begingroup$ Do you know if the PDE should be the Euler-Lagrange equation for some minimization problem? That might help. $\endgroup$ Sep 4, 2014 at 6:10
  • $\begingroup$ @Joonas Ilmavirta I do not think this PDE is an Euler-Lagrange equation. $\endgroup$ Sep 4, 2014 at 15:44
  • $\begingroup$ The above PDE is the 2017 corrected equation which has a + sign for the last term. The excellent analysis given below is not valid because it was based upon the previously posted incorrect 2014 version of the PDE with a - sign in font of the last term (my mistake). So according to @Willie Wong, the above corrected PDE may have a nontrivial solution (subject to more analysis), and I believe it does have only one solution although I have not yet found it. I will post the answer if and when I find it. $\endgroup$ Jul 30, 2022 at 18:39

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Multiply your equation by $u$ and integrate over $\mathbb{R}^3$ in azimuthal coordinates (so the volume form is $r~\mathrm{d} r~\mathrm{d}z ~\mathrm{d}\theta$ you get

$$ 0 = 2\pi \iint_{r,z} (u_{rr} + u_{zz} + u_r/r) ur - u^3 u_r - r u^2(u_r^2 + u_z^2 + 1/r^2) \mathrm{d}r \mathrm{d}z $$

Integrate by parts the first term (assuming that it is integrable) using the boundary conditions you get

$$ 0 = \iint_{r,z} - (u_r^2 + u_z^2) r - \frac14 \partial_r(u^4) - r u^2(u_r^2 + u_z^2 + 1/r^2) \mathrm{d}r \mathrm{d}z $$

The middle term integrates to zero since $u^4 = 0$ when $r = 0$ and $r \to \infty$. The first and last terms are manifestly negative. Hence the only solution is $u \equiv 0$.


Equivalently, extend your function to a rotationally symmetric function on $\mathbb{R}^3$: that is $u = u(r,z,\theta) = u(r,z)$.

Your equation can be re-written as

$$ \triangle u - \frac{u^2}{r} \partial_r u = u(u_r^2 + u_z^2 + 1/r^2) > 0$$

as $u$ is non-negative. And by assumption that $u = 0$ when $r = 0$ indicates that under a regularity assumption $u^2 / r$ is bounded above and below. So we can apply the maximum principle, which with the boundary condition that $u\to 0$ as $r^2 + z^2 \to \infty$ shows that $u$ must vanish identically.


If the sign of the last term in your equation is $+$ and not $-$, then neither of the above argument would work, and then the equation can potentially have a solution (requires more analysis). (Basically, as it stands you have a repulsive self-interaction term which prevents a bound state from existing. Changing the sign makes the self-interaction attractive which then may allow for a bound state.)

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    $\begingroup$ That was very elegant and simple. I worked for a year on that PDE, and you found the unique solution to be trivial in a flash. I wish the sign in the last term was + and not -, but alas, I am sure it is -. $\endgroup$ Sep 4, 2014 at 15:40
  • $\begingroup$ When you say "Integrate by parts the first term...you get...". I don't seem to be able to get that. Can you show me how to do that? $\endgroup$ Sep 5, 2014 at 4:56
  • $\begingroup$ I suppress the boundary terms as they are all zero $$\int u_{zz} ur \mathrm{d} z = - \int (u_z)^2 r \mathrm{d}z $$ and $$ \int u_{rr} u r \mathrm{d}r = \int - u_r \partial_r(ur) \mathrm{d}r = - \int (u_r)^2 r \mathrm{d}r - \int u_r u \mathrm{d} r $$ the last term cancels the $u_r/r * u r$ term that appears. $\endgroup$ Sep 5, 2014 at 7:36
  • $\begingroup$ Amazing how that all fits together so smoothly. Thank you. $\endgroup$ Sep 5, 2014 at 21:25

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