The existence of differential operator of the form $AB=0$ We define $\mathcal A$ is a differential operator of order $n$ with variable coefficients if
$$ \mathcal A:=\sum_{|\alpha|\leq n}A_\alpha (x) D^\alpha $$
where $\alpha$ is an muti-index and $A_\alpha(x)$ is a matrix in suitable dimension and $D^\alpha$ is the $\alpha$-th differential operator. 
Next, take $\mathbb U$ and $\mathbb V$ to be finite dimensional inner product spaces and we denote by $\mathcal D'(\mathbb R^n,\mathbb U)$ the space of distributions valued in $\mathbb U$. Then we can think $\mathcal A$ is mapping from $\mathcal D'(\mathbb R^n,\mathbb U)$ to $\mathcal D'(\mathbb R^n,\mathbb V)$.
We also define, for two differential operators $\mathcal A$ and $\mathcal P$, that $\mathcal Q\mathcal P$ is exact if and only if the kernel of $\mathcal Q$ is the image of $\mathcal P$. (which exclude the case $\mathcal Q\equiv 0$ or $\mathcal P\equiv 0$ and also the case that image of $\mathcal P$ is only a subset of Kernel of $\mathcal Q$)
My questions are:
(1): given any differential operator $\mathcal P$ of order $n$, $n\in\mathbb N$, is there exists a number $m\in \mathbb N$ such that there exists a differential operator $\mathcal Q$ with order $m$ and $\mathcal Q\mathcal P$ is exact?
(2): similarly, given any differential operator $\mathcal Q$ of order $m$, $m\in\mathbb N$, is there exists a number $n\in \mathbb N$ such that there exists a differential operator $\mathcal P$ with order $n$ and $\mathcal Q\mathcal P$ is exact?
PS: this question come from the observation from this paper, section $4$, theorem 4.1. Note that in theorem 4.1 part (i), it says $\mathcal Q\mathcal P=0$ is different then what I have here. In theorem 4.1 part (i) it only means that the kernel of $\mathcal Q$ contains the image of $\mathcal P$, but I want them to be equal.
PPS: In that paper, example 4.2 gives an counterexample that there exists a 1st order operator $\mathcal P$ such that there does not exist a 1st order operator $\mathcal Q$ such that $\mathcal Q\mathcal P=0$. But in my question I don't want to limit the order of operator. If there are no 1st order $\mathcal Q$, could there be a 2nd order $\mathcal Q$ such that $\mathcal Q\mathcal P$ is exact?
Thank you!
 A: I think that the answer to (2), generally speaking is 'no'.  For example, consider the case of $\mathcal{A}$ being the Cauchy-Riemann operator:
$$
\mathcal{A}(u,v) = (u_x - v_y,\ u_y+v_x).
$$
The kernel of $\mathcal{A}$ on $\mathbb{R}^2$ is the set of $(u,v)$ such that $u+iv$ is a holomorphic function of $z = x + i y$.  Now, if $\mathcal{B}$ were a linear operator on some space of smooth functions to pairs of smooth functions, in order for it to be exact, the image of $\mathcal{B}$ would have to, in particular, consist only of real-analytic functions, but this is clearly not possible since, if $\mathcal{B}$ is nonzero, its image will necessarily contain compactly supported smooth functions.
The answer to (1) is also 'no', without further assumptions.  For example, let $\mathcal{B}(u) = xu$ where $u$ is a function of a single variable $x$.  The image of $\mathcal{B}$ (an operator of order zero) is all smooth functions vanishing at the origin.  It is clear that this image is not the kernel of any differential operator $\mathcal{A}$ of the specified type.
