I am looking to find a solution, or even just prove the existence of one, to the following system of linear PDEs. They come up in a construction I am trying to work out in symplectic geometry. Here $(r\_1, \theta\_1, r\_2, \theta\_2)$ are the polar coordinates on $\mathbb{R}^4$, and $\beta$ and $\gamma$ are smooth functions in terms of those coordinates, that I would like to solve for:


$$2\gamma + r\_2\frac{\partial \gamma}{\partial r\_2} = 1$$
$$\frac{\partial \beta}{\partial \theta\_2}(r\_1^2-1) - r\_2^2\frac{\partial \gamma}{\partial \theta\_1} = 0$$
$$2\beta r\_1 + \frac{\partial \beta}{\partial r\_1}(r\_1^2-1) = r\_1$$,


and I need the conditions that $\beta = \gamma = 1$ when $r\_1^2 \geq \delta > 0$, and $\beta = r\_1$ when $r\_1 = 0$ and $\theta\_2 = 0, 0.5$. For certain reasons, I know that a solution (i.e. functions $\beta$ and $\gamma$) will not exist if I want $\beta = r\_1$ whenever $r\_1 = 0$, but I want to know if they might exist around two values of $\theta\_2$.



Existence of solution to a system of linear PDEs with boundary conditions
The general solution of your equations in a simply connected domain on which $r\_2\not=0$ and $r\_1\not=\pm1$ is
$$
\beta = \frac12 + \frac1{{(r\_1}^2{-}1)}\,
\left(\frac{\partial a}{\partial\theta\_1}+b(\theta\_1,r\_2)\right)
\quad\text{and}\quad
\gamma= \frac12 + \frac1{{r\_2}^2}\,
\left(\frac{\partial a}{\partial\theta\_2}+c(\theta\_2,r\_1)\right),
$$
where $a = a(\theta\_1,\theta\_2)$ is a function of $\theta\_1$ and $\theta\_2$ only, $b$ is a function of $\theta\_1$ and $r\_2$ only, and $c$ is a function of $\theta\_2$ and $r\_1$ only.


To see this, set $\bar\beta = ({r\_1}^2{-}1)\bigl(\beta-\tfrac12\bigr)$ and $\bar\gamma = {r\_2}^2\bigl(\gamma-\tfrac12\bigr)$ and note that the given equations imply the constant coefficient linear equations
$$
\frac{\partial\bar\beta}{\partial r\_1} 
= \frac{\partial\bar\gamma}{\partial r\_2} 
= \frac{\partial\bar\beta}{\partial\theta\_2}-\frac{\partial\bar\gamma}{\partial\theta\_1} = 0,
$$
which are easily solved.


I do not see how you can choose $a$, $b$, and $c$ so that your 'boundary conditions' are satisfied. Because the general solution depends on three functions of two variables, one would expect to be able to specify (initial) conditions along a surface in the domain, but not along a hypersurface (which is what boundaries generally are).


