# Existence of solution for the PDE $p \frac{\partial f(p, q)}{\partial p}-q \frac{\partial f(p, q)}{\partial q}=g(p, q)$

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 $$(0,0).$$

• 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 $(0,0)$. – Robert Israel Nov 28 '18 at 15:13
• @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 – Ilia Nov 28 '18 at 15:19
• What does flat mean? – Deane Yang Nov 28 '18 at 15:54
• @DeaneYang, "flat" probably means that all the partial derivatives of $g(p,q)$ (of any order) vanish at $p=q=0$. – Igor Khavkine Nov 28 '18 at 19:55
• @IgorKhavkine, yeah, that's right. – 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.