I'm interested in the analyticity of Frobenius-like series solutions to a PDE in $z=(z_1,\ldots ,z_N)\in\mathbb{C}^N$ with regular singular behavior at $z_\alpha=0$ for all $\alpha=1,\ldots, N$.
For $i=(i_1,\ldots,i_N)\in\mathbb{Z}_{+}$ define the order $|i|=\sum_{\alpha=1}^{N}i_\alpha$ differential operator $\mathcal{D}_{z}^{i}=\prod_{\alpha=1}^N z_\alpha^{i_\alpha}\frac{\partial^{i_\alpha}}{\partial z_{\alpha}^{i_{\alpha}}}$. Consider an order $n$ PDE of the form:
$(*)$ $\sum_{i}P^{i}(z)\mathcal{D}_{z}^{i}y=0$
where $|i|\le n$ and where $P^{i}(z)$ is analytic in a polydisc $\Delta:\{|z_\alpha|<R_\alpha\}$. Furthermore, assume that $\frac{1}{P^{i}(z)}$ is also analytic in $\Delta$ for all $i$ of the form $(0,\ldots,n,\ldots,0)$ i.e. for each $\alpha=1,\ldots,N$ the multiplicative inverse of the coefficient of $z_\alpha^{n}\frac{\partial^{n}y}{\partial z_{\alpha}^{n}}$ in $(*)$ is also analytic in $\Delta$.
Consider a Frobenius series solution of $(*)$ of the form
$Y(z,\rho)=\sum_{k_1,\ldots,k_N\ge 0}y_k z^{\rho+k}$, $y_0=1$
for $z^k=\prod_{\alpha=1}^N z_\alpha^{k_\alpha}$ etc where $\rho\in\mathbb{C}^N$ satisfies a degree $n$ indicial polynomial equation. For a given root $\rho=r$ and assuming that $\rho+k$ is not an indicial root for all $k>0$, it is straightforward to show that a unique formal series solution $Y(z,r)$ exists.
Question: is $\sum_{k_1,\ldots,k_N\ge 0}y_k z^{k}=Y(z,r)z^{-r}$ analytic on $\Delta$?
