# Inverse of pseudo differential operator

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.

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.