# Local solution for linear PDE with "saddle" degeneration at origin?

Does a PDE of the form $$y \frac{\partial u}{\partial x}+x \frac{\partial u}{\partial y} = c(x,y) \, u +d(x,y)$$ necessarily have a solution near $(x,y)=(0,0)$?

My attempt: The method of characteristics reduces this to the simple ODE $u'=c u +d$ along the characteristic curves, i.e. the integral curves of the vector field $$v(x,y,u)=(y,x,c(x,y)u+d(x,y))$$ on $\mathbb{R}^3_{(x,y,u)}$. It's easy to see that these characteristic curves all lie over hyperbolae in the $xy$-plane (as in the figure below). The problem is that the origin is a sort of "degenerate" characteristic curve. Here's my hope: Use the union of the $x$- and $y$-axes as a singular initial hypersurface by assigning $u=0$ for $x=0$ and $y=0$. Then use integrating factors to solve $u'=cu+d$ along all characteristic curves that meet either the $x$- or $y$-axis. This should give a solution everywhere except the four open rays in the set $X=\{y=\pm x\} \setminus \{(0,0)\}$, which do not meet the $x$- or $y$-axis. Then one can extend this solution to $X$ using continuity, but now you'd need to prove smoothness somehow.

To state the obvious: if $c(0,0) = 0$ and $d(0,0) \ne 0$, and $u$ and its partial derivatives exist at $(0,0)$, then the PDE can't be true at $(0,0)$: the left side is $0$ and the right is nonzero.
EDIT: In your specific case it seems there can still be obstructions. Consider $g(x,y) = 1 + ax + by$, so the equation is $$y u_x + x u_y = (-ay - bx) u + ax + by$$ Writing $u(x,y) = \sum_{i\ge 0} \sum_{j\ge 0} v_{ij} x^i y^j$, I get from the Taylor coefficients of total degree $\le 2$ $$\left[ \begin {array}{cccccc} b&0&1&0&0&0\\ a&1&0&0 &0&0\\ 0&b&0&0&1&0\\ 0&a&b&2&0&2 \\ 0&0&a&0&1&0\end {array} \right] \pmatrix{v_{00} \cr v_{10}\cr v_{01}\cr v_{20}\cr v_{11}\cr v_{02}\cr} = \pmatrix{a \cr b\cr 0\cr 0\cr 0\cr}$$ which is an inconsistent system if $a^2 \ne b^2$.
• Ah, thanks. I forgot to specify that $c(0,0)=d(0,0)=0$. The actual form of the equation I'm dealing with has $c(x,y)=-x g_y(x,y) - y g_x(x,y)$ and $d(x,y)=g(x,y)-1$ for some $g$ with $g(0,0)=1$. And everything in sight is smooth. Do you know if there are other obstructions in this case? Commented Oct 13, 2016 at 22:03
• Not a $C^2$ solution. Commented Oct 16, 2016 at 7:31