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Find the Green function of the first quadrant $x_1>0, x_2>0$.
HINT: Use the Green function of the half space $\Omega:=\left\{x\in\mathbb{R}^n : x_n > 0\right\}$ which is given by\begin{align*} G_n(x,y)&:=E_n(x_1-y_1,\cdots,x_{n-1}-y_{n-1},x_n-y_n)\\ &\quad\ -E_n(x_1-y_1,\cdots,x_{n-1}-y_{n-1},x_n+y_n). \end{align*}

How I can find the Green function of the first quadrant? The difference to the half space is that I have another boundary…

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For $1 \leqslant k \leqslant n$, define$$ τ_k(x_1, \cdots, x_n) = (x_1, \cdots, x_{k - 1}, -x_k, x_{k + 1}, \cdots, x_n). \quad \forall x_1, \cdots, x_n \in \mathbb{R} $$ And for $1 \leqslant k_1 < \cdots < k_m \leqslant n$, define$$ τ_{k_1, \cdots, k_m} = τ_{k_1} \circ \cdots \circ τ_{k_m}. $$ Thus the Green function of $\{(x_1, \cdots, x_n) \in \mathbb{R}^n \mid x_n > 0\}$ is $G_n(x, y) = E_n(x - y) - E_n(x - τ_n(y))$. Now define\begin{align*} H_n(x, y) &= (E_n(x - y) - E_n(x - τ_1(y))) - (E_n(x - τ_2(y)) - E_n(x - τ_1(τ_2(y))))\\ &= E_n(x - y) - E_n(x - τ_1(y)) - E_n(x - τ_2(y)) + E_n(x - τ_{1, 2}(y)) \end{align*} for $x, y \in D = \{(x_1, \cdots, x_n) \in \mathbb{R}^n \mid x_1, x_2 > 0\}$. It is easy to see that$$ Δ_x H_n(x, y) = -δ(x - y). \quad \forall x, y \in D $$ Note that for any $1 \leqslant k \leqslant n$ and $x, y \in \mathbb{R}^n$,$$ τ_k(τ_k(x)) = x, \quad τ_k(x + y) = τ_k(x) + τ_k(y), \quad E_n(x) = E_n(-x) = E_n(τ_k(x)), $$ thus\begin{align*} H_n(x, y) &= E_n(x - y) - E_n(x - τ_1(y)) - E_n(x - τ_2(y)) + E_n(x - τ_{1, 2}(y))\\ &= E_n(x - y) - E_n(τ_1(x) - y) - E_n(τ_2(x) - y) + E_n(τ_{1, 2}(x) - y). \end{align*}

For $x^* \in \partial D \cap \{x_1 = 0\}$, because $τ_1(x) → τ_1(x^*) = x^*,\ τ_2(x) → τ_2(x^*)\ (x → x^*,\ x \in D)$, then $τ_{1, 2}(x) = τ_2(τ_1(x)) → τ_2(x^*)\ (x → x^*,\ x \in D)$, which implies $\lim\limits_{\substack{x → x^*\\x \in D}} H_n(x, y) = 0$. Analogously, $\lim\limits_{\substack{x → x^*\\x \in D}} H_n(x, y) = 0$ for $x^* \in \partial D \cap \{x_2 = 0\}$. Thus, $H_n(x, y)|_{x \in \partial D} = 0$. Therefore, $H_n(x, y)$ is the Green function of $D$.

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