Let $\operatorname{Op}_h(x,D)(a)$ denote the Weyl-quantisation of a symbol $a$. Is there an explicit way to invert this pseudo-differential operator in an asymptotic series? By this I mean, can we find $(x,\xi) \mapsto \sum_{n=0}^{N} c_n(x,\xi)h^n$ such that $$a \sharp \left(\sum_{n=0}^{N} c_n h^n \right)=1 + O(h^{(n+1)})$$ for any $N \in \mathbb{N}$ explicitly?

The only thing that I can see is how to do this step by step. For example, the zero-th order term is given by taking $c_0:=a^{-1}$. Then, I would have to look how to subpress the next higher-order term and so on. But is there also a closed way to write down the inverse expansion?

If anything is unclear, please let me know.


1 Answer 1


You need some ellipticity condition to start off. Let us assume as you do that $$ c=c(x,\xi, h), \quad \vert\partial_x^\alpha\partial_{\xi}^\beta c\vert\le C_{\alpha\beta} h^{\vert\beta\vert}, \quad\text{i.e.}\quad\forall (\alpha, \beta), \sup_{(x,\xi, h)\in \mathbb R^n\times\mathbb R^n\times (0,1]}\vert\partial_x^\alpha\partial_{\xi}^\beta c\vert h^{-\vert\beta\vert}<+\infty. \tag{$\ast$}$$ Assuming that $\inf_{(x,\xi, h)\in \mathbb R^{2n}\times(0,1]}{\vert c(x,\xi,h)\vert}>0$, you see that the function $b_0=1/c$ is a symbol of order 0 (say is in $S^0$), i.e. satisfies $(\ast)$ (this is the ellipticity assumption). Then you write $$ c\sharp b_0=cb_0 +hc_1=1+hc_1, \quad c_1\in S^0\quad\text{depending on $c, b_0$.} $$ Then you write $$ c\sharp (b_0+hb_1)=1+h(c_1+c\sharp b_1)=1+h(c_1+cb_1)+h^2c_2,\quad c_2\in S^0\quad\text{depending on $c, b_0, b_1$.} $$ Using again the ellipticity, you choose $b_1=-c_1/c=-c_1b_0$ which belongs to $S^0$ and you get $$ c\sharp (b_0+hb_1)=1+h^2c_2, c_2\in S^0\quad\text{depending on $c, b_0, b_1$.} $$ It is easy to go on: you write $$ c\sharp (b_0+h b_1+h^2 b_2)=1+h^2 (c_2+c\sharp b_2)=1+h^2(c_2+cb_2)+h^3c_3, $$ you choose $b_2=-c_2/c$ and you get $c\sharp (b_0+h b_1+h^2 b_2)=1+h^3c_3,$ and so on. The induction procedure is easy to write.


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