# 微积分网课代修|函数代写Function theory代考|MAEN5060 Construction of the Bergman Kernel by means of an Orthonormal System

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## 微积分网课代修|偏微分方程代写Partial Differential Equation代考|Construction of the Bergman Kernel by means of an Orthonormal System

Now we will endeavor to write down the Bergman kernel for the disk by means of an orthonormal basis for $A^{2}(D)$, that is, by applying Proposition 1.2.8. For a general $\Omega$, it can be rather difficult to actually write down a complete orthonormal system for $A^{2}(\Omega)$. Fortunately, the unit disk $D \subseteq \mathbb{C}$ has enough symmetry that we can actually pull this off.

It is not difficult to see that $\left{z^{j}\right}_{j=0}^{\infty}$ is an orthogonal system for $A^{2}(D)$. That is, the elements are pairwise orthogonal; but they are not normalized to have unit length. We may confirm the first of these assertions by noting that, if $j \neq k$, then
\begin{aligned} \left\langle z^{j}, z^{k}\right\rangle &=\iint_{D} z^{j} \overline{z^{k}} d A(z)=\int_{0}^{1} \int_{0}^{2 \pi} r^{j} e^{i j \theta} r^{k} e^{-i k \theta} d \theta r d r \ &=\int_{0}^{1} r^{j+k+1} d r \int_{0}^{2 \pi} e^{i(j-k) \theta} d \theta=0 \end{aligned}
The system $\left{z^{j}\right}$ is complete: If $\left\langle f, z^{j}\right\rangle=0$ for every $j$, then $f$ will have a null power series expansion and hence be identically zero. It remains to normalize these monomials so that we have a complete orthonormal system.
We calculate that
\begin{aligned} \iint_{D}\left|z^{j}\right|^{2} d A(z) &=\int_{0}^{1} \int_{0}^{2 \pi} r^{2 j} d \theta r d r \ &=2 \pi \int_{0}^{1} r^{2 j+1} d r \ &=\pi \cdot \frac{1}{j+1} \end{aligned}

## 微积分网课代修|偏微分方程代写Partial Differential Equation代考|Construction of the Bergman Kernel by way of Differential Equations

It is actually possible to obtain the Bergman kernel of a domain in the plane from the Green’s function for that domain (see $[\mathrm{EPS}]$ ). Let us now summarize the key ideas. Chapter 8 , especially Section $8.2$, contains more detailed information about the Green’s function. Unlike the first two Bergman kernel constructions, the present one will work for any domain with $C^{2}$ boundary.
First, the fundamental solution for the Laplacian in the plane is the function
$$\Gamma(\zeta, z)=\frac{1}{2 \pi} \log |\zeta-z|$$
(in Proposition 8.1.4 we shall prove this assertion). This means that $\triangle_{\zeta} \Gamma(\zeta, z)=\delta_{z}$. [Observe that $\delta_{z}$ denotes the Dirac “delta mass” at $z$ and $\Delta_{\zeta}$ is the Laplacian in the $\zeta$ variable.] Here the derivatives are interpreted in the sense of distributions. In more prosaic terms, the condition is that
$$\int \Gamma(\zeta, z) \cdot \Delta \varphi(\zeta) d \xi d \eta=\varphi(z)$$
for any $C^{2}$ function $\varphi$ with compact support. We write, as usual, $\zeta=\xi+i \eta$.

## 微积分网课代修|偏微分方程代写Partial Differential Equation代考|Construction of the Bergman Kernel by means of anOrthonormal System

$$\left\langle z^{j}, z^{k}\right\rangle=\iint_{D} z^{j} z^{k} d A(z)=\int_{0}^{1} \int_{0}^{2 \pi} r^{j} e^{i j \theta} r^{k} e^{-i k \theta} d \theta r d r \quad=\int_{0}^{1} r^{j+k+1} d r \int_{0}^{2 \pi} e^{i(j-k) \theta} d \theta=0$$

$$\iint_{D}\left|z^{j}\right|^{2} d A(z)=\int_{0}^{1} \int_{0}^{2 \pi} r^{2 j} d \theta r d r \quad=2 \pi \int_{0}^{1} r^{2 j+1} d r=\pi \cdot \frac{1}{j+1}$$

## 微积分网课代修|偏微分方程代写Partial Differential Equation代考|Construction of the Bergman Kernel by way of DifferentialEquations

$$\Gamma(\zeta, z)=\frac{1}{2 \pi} \log |\zeta-z|$$
(在命题 8.1.4 中，我们将证明这个断言) 。这意味着 $\triangle_{\zeta} \Gamma(\zeta, z)=\delta_{z}$. [观察到 $\delta_{z}$ 表

$$\int \Gamma(\zeta, z) \cdot \Delta \varphi(\zeta) d \xi d \eta=\varphi(z)$$