I've asked this question on stack exchange before but no one could help me so I wish I can get some help here. Let's first start with the definition of the canonical solution:

Consider $\frac{\partial w}{\partial \bar{z}}=u$. We say $w_0$ is a canonical solution if $\frac{\partial w_0}{\partial \bar{z}}=u$, and $w_0 \perp \mathcal{H}(\mathbb{D})=\{ f \in \mathcal{O}(\mathbb{D}): f \in L^2(\mathbb{D}) \}$, where $\mathcal{O}(\mathbb{D})$ is the space of holomorphic functions on the unit disk. And $w_0 \perp \mathcal{H}(\mathbb{D}) \iff \int_{\mathbb{D}} w_0(\zeta)\cdot \overline{h(\zeta)} dV_{\zeta}=0$, for any $h \in \mathcal{H}(\mathbb{D})$.

Here, we are finding $w_0$ as follows:

Let $$ w_1(z)= \frac{1}{2\pi i} \int_{\mathbb{D}} \frac{u(\zeta)}{\zeta -z} d\overline{\zeta} \wedge d{\zeta}. $$ Define $$ Pw_1(z):= \int_{\mathbb{D}} K(z,\zeta) w_1(\zeta) dV_{\zeta}, $$ where $K(z,\zeta)$ is the Bergman kernel on the unit disk which's been calculated here. Now, we define $w_0:= w_1 - Pw_1$, and I want to prove this is a canonical solution.

What I've done so far is that if $w_1$ is holomorphic, then by the definition of the Bergman kernel, $Pw_1(z)=w_1(z)$ and hence, $w_0=0$ which is the trivial case.

So, suppose $w_1$ is not holomorphic. In this case, doing some calculations and using the Cauchy-Green's formula, I figured that we can actually write $$ w_0=w_1-Pw_1= -\frac{1}{2\pi i} \int_{\mathbb{D}} \frac{(1-\lvert \zeta \rvert ^2) u(\zeta)}{(1-\overline{\zeta}z)(\zeta-z)} d\zeta \wedge d\overline{\zeta}. $$ So, I actually need to prove that this is a canonical solution but I don't know how to proceed from here. Any hint/help is appreciated.