Consider the following PDE: \begin{equation} p \frac{\partial f(p, q)}{\partial p}-q \frac{\partial f(p, q)}{\partial q}=g(p, q),\tag{$\star$} \end{equation} where $g$ is a flat function at the point (0,0).

Let $X$ denote the the vector field $p \frac{\partial}{\partial p}-q \frac{\partial}{\partial q}$. The equation $\star$ can be also written as $\mathcal{L}_Xf=g,$ where $\mathcal{L}_Xf$ stands for the Lie derivative of $f$ along the vector field $X.$

I need to prove that there exists at least one $C^\infty$ smooth solution to the equation $\star$ in some neighbourhood of the point $(0,0).$ I also want this solution to be flat at the point $(0,0).$

The method of characteristics doesn't really work in this situation since (0,0) is a singular point of the vector field $X.$ The characteristics of $\star$ are given by the level sets of the function $pq.$ So it is easy to find a smooth solution in the domain $\mathbb R^2 \setminus \{(p, q)|pq=0\}.$ For example, the initial data can be given as: $f(p, q)=0$ if $p=q$ or $p=-q$.

But it is unclear whether or not this solution can be extended to some neighbourhood of the point $(0,0).$

I was thinking about this problem for quite a while but with no success so far.

  • $\begingroup$ For example, with $g(p,q) = p^2 q^2$, $f(p,q) = p^2 q^2 \ln |p|$ is a solution in a neighbourhood of $(0,0)$, but it is not $C^\infty$ and I doubt that any solution for this $g$ can be made $C^\infty$ in a neighbourhood of $(0,0)$. $\endgroup$ – Robert Israel Nov 28 '18 at 15:13
  • $\begingroup$ @RobertIsrael that's right but the function $g(p, q)=p^2q^2$ is not flat at the point (0,0). en.wikipedia.org/wiki/Flat_function $\endgroup$ – Ilia Nov 28 '18 at 15:19
  • $\begingroup$ What does flat mean? $\endgroup$ – Deane Yang Nov 28 '18 at 15:54
  • 2
    $\begingroup$ @DeaneYang, "flat" probably means that all the partial derivatives of $g(p,q)$ (of any order) vanish at $p=q=0$. $\endgroup$ – Igor Khavkine Nov 28 '18 at 19:55
  • $\begingroup$ @IgorKhavkine, yeah, that's right. $\endgroup$ – Ilia Nov 28 '18 at 21:13

The answer is 'yes, a smooth, flat solution $f$ exists when $g$ is smooth and flat'.

Here is one way to show this: I'll first do the case in which $g$ is even, i.e., $g(-p,-q)=g(p,q)$ and, for convenience, I'll assume that $g$ is defined on the entire $pq$-plane. (See the remark at the end about the local case.)

Let $(u,v) = (p^2{-}q^2,\,2pq)$, and note that there is a (unique) function $\bar g$ on the $uv$-plane such that $g(p,q) = \bar g(u,v)$ and that $\bar g$ is smooth and flat at $(u,v)=(0,0)$. I will look for a solution of the form $f(p,q) = \bar f(u,v)$. By the Chain Rule, the equation $(\star)$ then becomes $$ 2\sqrt{u^2{+}v^2} \, \frac{\partial \bar f}{\partial u} = \bar g, $$ so $$ \frac{\partial \bar f}{\partial u}(u,v) = \frac{\bar g(u,v)}{2\sqrt{u^2{+}v^2}}. $$ The right hand side is smooth and flat at $(u,v) = (0,0)$, so one has a solution in the form $$ \bar f(u,v) = \int_0^u \frac{\bar g(t,v)}{2\sqrt{t^2{+}v^2}}\,\mathrm{d}t\,. $$ This $\bar f$ is flat and gives a solution to the problem, in fact, the unique solution that satisfies $\bar f(0,v) = 0$.

In the general case, one can write $g = g_0 + g_1$, where $g_0$ is even and $g_1$ is odd, i.e., $g_1(-p,-q) = -g_1(p,q)$. So, to finish, by the linearity of the equation, it only remains to solve the equation when $g$ is odd. This can be done by using the above solution on the half-planes $p>0$ and $p<0$ and being a little careful about the matching. However, the right way to think about it in the odd case is that $\bar g$ is actually a section of a nontrivial flat line bundle over the punctured $uv$-plane, and the above integral is then taken using parallel translation in the flat line bundle along segements of the form $\sigma(t) = (tu,v)$ for $0\le t\le 1$, thereby yielding a section $\bar f$ of this nontrivial line bundle over the punctured $uv$-plane. This defines a solution $f(p,q)$ of the equation on the punctured $pq$-plane that is odd and that vanishes to infinite order at $(p,q)=(0,0)$, i.e., it is a flat smooth solution, as desired.

For the local problem, one just assumes that $g$ is defined on an open neighborhood defined by $|p^2+q^2|<\epsilon^2$ for some $\epsilon>0$ and the line integrals in the formula will then still work to give the desired solution $f$ on the same neighborhood.

Added remark: It's probably worth pointing out that the above result can be applied to prove that the equation $(\star)$ can be solved for a smooth $f$ for a given $g$ if and only if $g$ is smooth and all of its 'balanced partials' (e.g., $g$, $g_{pq}$, $g_{ppqq}$, $g_{pppqqq}$, etc.) vanish at $(p,q) = (0,0)$. This can be done using the above result plus the theorem that any formal Taylor series is the actual Taylor series of some smooth function.

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