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I had asked a similar question about existence of solutions to a system of linear PDEs, and the answer made me realize that I needed to change things to try to make the construction work. So for the same construction, I now have the following system of 6 PDEs to solve, and I have certain "initial" or "boundary" conditions as before. Call the hypersurfaces $H_1 = \{r_1=0\}$ and $H_2 = \{-r_1^2 + r^2_2 = \epsilon\}$

The setting is $\mathbb{R}^4$ with polar coordinates $(r_1, \theta_1, r_2, \theta_2)$. I want to solve for (or show the existence of) functions $\beta, \gamma, \alpha, \mu$, on a subset of $\nu (H_1 \cap H_2)$, such that they satisfy the following:

\begin{align*} 2r_1\beta+(r_1^2 - 1)\frac{\partial \beta}{\partial r_1} + r_1\frac{\partial \alpha}{\partial \theta_1} &= r_1\\ (r_1^2 - 1)\frac{\partial \beta}{\partial r_2}+r_2\frac{\partial \mu}{\partial \theta_1}&=0\\ (r_1^2-1)\frac{\partial \beta}{\partial \theta_2} + r_2^2\frac{\partial \gamma}{\partial \theta_1} &= 0\\ r_2^2\frac{\partial \gamma}{\partial r_1}+r_1\frac{\partial \alpha}{\partial \theta_2}&=0\\ 2\gamma r_2 + r_2^2\frac{\partial \gamma}{\partial r_2} + r_2\frac{\partial \mu}{\partial \theta_2} &= r_2\\ r_1\frac{\partial \alpha}{\partial r_2} - r_2\frac{\partial \mu}{\partial r_1} &=0 \end{align*}

As for "initial" or "boundary" conditions, I need $\beta = \gamma = 1$ and $\alpha = \mu = 0$ when $r_1 = \delta$, and $\beta = o(r_1)$ near $r_1 = 0$. A solution is not expected to exist all over $H_1 \cap H_2$, but I want to see if solutions can be found that are defined for some values of $\theta_2$ along $r_1 = 0$.

I am not able to work out a general form for the solutions as was suggested in the answer to the previous question, possibly because all the functions $\beta, \gamma, \alpha, \mu$ and their partials are now "coupled".

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    $\begingroup$ Here's a hint: Set $\bar\alpha = r_1\,\alpha$, $\bar\beta = ({r_1}^2-1)(\beta-\tfrac12)$, $\bar\mu = r_2\,\mu$, and $\bar\gamma = {r_2}^2(\gamma-\tfrac12)$ and consider the equations that you get for $\bar\alpha$, $\bar\beta$, $\bar\mu$, and $\bar\gamma$. $\endgroup$
    – Robert Bryant
    Commented Sep 24, 2021 at 14:36
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    $\begingroup$ Also, are you sure that you have the signs right? For example, if one flipped the signs of $\beta$ and $\gamma$, all the equations for the barred quantities would be differences of partials except for the third equation, which would still be a sum. If the equations were all differences, then the general solution would be easy to write down, and the solution would have geometric significance, but with the flipped sign, the system looks very contrived. $\endgroup$
    – Robert Bryant
    Commented Sep 24, 2021 at 16:22
  • $\begingroup$ The equations come from here: let $V = \beta (r_1-\frac{1}{r_1}) \partial_{r_1} + \gamma r_2 \partial_{r_2} + \alpha \partial_{\theta_1} + \mu \partial_{\theta_2}$, and $\omega = r_1 d r_1 d\theta_1 + r_2 d r_2 d\theta_2$. When is $\mathcal{L}_{V}\omega = \omega$? I checked a few times, I think the signs are correct. The idea is: try and extend the (Liouville) vector field $V_1 = (r_1-\frac{1}{r_1}) \partial_{r_1} + r_2 \partial_{r_2}$ over some points in $H_1 \cap H_2$. I don't know if the solution should have any geometric significance at all, and $V$ is indeed just a guess. It may not work. $\endgroup$
    – Agniva Roy
    Commented Sep 28, 2021 at 19:19
  • $\begingroup$ As it turns out, it doesn't seem to work. The general solution for $\beta$ looks like $\beta$ = $\frac{1}{r_1^2-1}(\frac{r_1^2}{2} - \frac{1}{2} - \frac{\partial H + \partial F + \partial G}{\partial \theta_1})$, where $H$ is a function of $r_1$ and $\theta_1$, $F$ of $\theta_2$ and $\theta_1$, and $G$ of $r_2$ and $\theta_1$. Near $r_1 = 0$, for $\beta$ to be $o(r_1)$, $ \frac{\partial H + \partial F + \partial G}{\partial \theta_1}$ must be independent of $\theta_1$, but since the functions must be periodic restricted to $\theta_1$, no solution exists anywhere for all values of $\theta_1$. $\endgroup$
    – Agniva Roy
    Commented Oct 4, 2021 at 15:16

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