Let $X$ be a compact complex manifold, and $D=\sum_{i=1}^{r} D_i$ be a simple normal crossing divisor on $X$. Let $\alpha$ be a logarithmic $(p,q)$-form, namely, on an open subset $U$, we can write $$\alpha=\frac{dz^1}{z^1}\wedge\alpha_1+\alpha_2,$$ where $D_1$ is defined by $\left\{z^1=0\right\}$ on $U$, $\alpha_1\in A^{0,q}(U,\Omega_X^{p}(\log (D-D_1))$, and $\alpha_2$ doesn't contain $dz^1.$
Question 1: I am wondering that if there exists some way to eliminate the log pole of a log $(p,q)$-form (especially for general $q\geq 1$!) along $D_1$ (and then similarly for other components)? More precisely, can we construct a log form $\beta,$ such that $\alpha-\beta$ doesn't possess any log pole along $D_1$?
Here are some thoughts: We first consider the simplest case $r=1$, i.e., $D=D_1$ is a smooth divisor. Recall that the Poincare residue map is defined by: $$\textrm{Res}_{D} (\alpha)=\alpha_1|_\left\{z_1=0\right\}=:\gamma.$$ This map is globally defined. And in our case, $\gamma$ is smooth on $D.$ Using the tubular neighborhood theorem plus the partition of unity to extend $\gamma$ smoothly to a form $\widetilde \gamma$ on $X$ and letting $$\widetilde \zeta :=\widetilde \gamma\wedge\frac{dz^1}{z^1}\in A^{0,q}(X,\Omega_X^p(\log D).$$ Then, one can check that $\alpha$ and $\widetilde \zeta$ have the same residue along $D_1$. And our goal is then to prove that $$\alpha-\widetilde \zeta\in A^{p,q}(X).$$
However, since now we are considering the case $q\geq 1$, the antiholomorphic part also may can play role? For example, suppose that locally one has $$\alpha-\widetilde \zeta={\bar z^1} \frac{dz^1}{z^1}\wedge d\bar z^1, (**)$$ then it also has residue zero but still has log pole. So, here appears my another questions.
Question 2: For a general log $(p,q)$-form $\alpha$, using the rough method I mentioned, can we extend the corresponding $\gamma$ well enough to avoid similar situations (**) that occurred, so that we can guarantee the difference doesn't have a log pole along $D_1$? If it is not always possible, can we add some extra assumptions on $\alpha$ to make it true, for example, being $\bar\partial$-closed or/and $\partial$-exact? Or, any other approach to make it possible?
Any suggestions and references would be greatly appreciated. Thanks a lot!
