Dirichlet Green function of a Euclidean ball

Work under the standing hypothesis [F9]. Let $n\ge 3$, $a\in {\mathbb{R}}^n$, $R>0$ and $\mathrm{\Omega }=B_R(a)$; let $\mathrm{\Phi }$ be the kernel of [F1]. For $y\in \mathrm{\Omega }$ with $y\mathrm{\ne }a$ put $y^{\ast }:=I_R(y)=a+R^2(y-a)\mathrm{/}\mathrm{\mid }y-a{\mathrm{\mid }}^2$ and ${\kappa }_y:=(R\mathrm{/}\mathrm{\mid }y-a\mathrm{\mid }{)}^{n-2}$; by [F5], $\mathrm{\mid }{y}^{\ast }-a\mathrm{\mid }=R^2\mathrm{/}\mathrm{\mid }y-a\mathrm{\mid }>R$, so $y^{\ast }\mathrm{\notin }\bar{\mathrm{\Omega }}$. Define $G(x,y):=\mathrm{\Phi }(x-y)-{\kappa }_y\mathrm{\Phi }(x-y^{\ast })$ for $x\in \mathrm{\Omega }\setminus \{y\}$ when $y\mathrm{\ne }a$, and $G(x,a):=\mathrm{\Phi }(x-a)-\mathrm{\Phi }(R)$ for $x\in \mathrm{\Omega }\setminus \{a\}$. Now put $u:=x-a$ and $v:=y-a\mathrm{\ne }0$, so that $x-y^{\ast }=(x-a)-R^{2}v\mathrm{/}\mathrm{\mid }v{\mathrm{\mid }}^2=u-R^{2}v\mathrm{/}\mathrm{\mid }v{\mathrm{\mid }}^2$ and $x-y=u-v$: expanding the square $\mathrm{\mid }u-R^{2}v\mathrm{/}\mathrm{\mid }v{\mathrm{\mid }}^2{\mathrm{\mid }}^2=\mathrm{\mid }u{\mathrm{\mid }}^2-2R^2⟨u,v⟩\mathrm{/}\mathrm{\mid }v{\mathrm{\mid }}^2+R^4\mathrm{/}\mathrm{\mid }v{\mathrm{\mid }}^2$ and multiplying by $\mathrm{\mid }v{\mathrm{\mid }}^2$ gives $\mathrm{\mid }y-a{\mathrm{\mid }}^2\mathrm{\mid }x-y^{\ast }{\mathrm{\mid }}^2=\mathrm{\mid }x-a{\mathrm{\mid }}^2\mathrm{\mid }y-a{\mathrm{\mid }}^2-2R^2⟨u,v⟩+R^4$, while $R^2\mathrm{\mid }x-y{\mathrm{\mid }}^2=R^2\mathrm{\mid }x-a{\mathrm{\mid }}^2-2R^2⟨u,v⟩+R^2\mathrm{\mid }y-a{\mathrm{\mid }}^2$; subtracting yields the first algebraic identity below, and the same expansion with the roles of $x$ and $y$ exchanged yields the second, since $\mathrm{\mid }v{\mathrm{\mid }}^2\mathrm{\mid }u{\mathrm{\mid }}^2=\mathrm{\mid }u{\mathrm{\mid }}^2\mathrm{\mid }v{\mathrm{\mid }}^2$, $\mathrm{\mid }{x}^{\ast }-a\mathrm{\mid }=R^2\mathrm{/}\mathrm{\mid }x-a\mathrm{\mid }$ and $x^{\ast },y^{\ast }$ are defined symmetrically. $\mathrm{\mid }y-a{\mathrm{\mid }}^2\mathrm{\mid }x-y^{\ast }{\mathrm{\mid }}^2-R^2\mathrm{\mid }x-y{\mathrm{\mid }}^2=(R^2-\mathrm{\mid }x-a{\mathrm{\mid }}^2)(R^2-\mathrm{\mid }y-a{\mathrm{\mid }}^2),\mathrm{\mid }y-a{\mathrm{\mid }}^2\mathrm{\mid }x-y^{\ast }{\mathrm{\mid }}^2=\mathrm{\mid }x-a{\mathrm{\mid }}^2\mathrm{\mid }y-x^{\ast }{\mathrm{\mid }}^2\mathrm{.}$

Correctors. Fix $y\in \mathrm{\Omega }$ with $y\mathrm{\ne }a$. Since $y^{\ast }\mathrm{\notin }\bar{\mathrm{\Omega }}$, the translate $x\mapsto \mathrm{\Phi }(x-y^{\ast })$ is smooth with vanishing Laplacian on a neighbourhood of the closed ball $\bar{\mathrm{\Omega }}$ by [F2], so $H_y:={\kappa }_y\mathrm{\Phi }(\cdot -y^{\ast })$ lies in $C^2(\bar{\mathrm{\Omega }})$ and is harmonic on $\mathrm{\Omega }$; by construction $G(x,y)=\mathrm{\Phi }(x-y)-H_y(x)$ for $x\in \mathrm{\Omega }\setminus \{y\}$. For the centre put $H_a:=\mathrm{\Phi }(R)$, the constant corrector: it is $C^2$ on $\bar{\mathrm{\Omega }}$, harmonic, and $G(x,a)=\mathrm{\Phi }(x-a)-H_a(x)$ by definition.

Boundary values of the correctors. If $y\mathrm{\ne }a$ and $\mathrm{\mid }x-a\mathrm{\mid }=R$, the first identity of step 1.1 gives $\mathrm{\mid }y-a{\mathrm{\mid }}^2\mathrm{\mid }x-y^{\ast }{\mathrm{\mid }}^2=R^2\mathrm{\mid }x-y{\mathrm{\mid }}^2$, hence $\mathrm{\mid }x-y^{\ast }\mathrm{\mid }=(R\mathrm{/}\mathrm{\mid }y-a\mathrm{\mid })\mathrm{\mid }x-y\mathrm{\mid }$; with the formula of [F1] this yields ${\kappa }_y\mathrm{\Phi }(x-y^{\ast })=R^{n-2}(\mathrm{\mid }y-a\mathrm{\mid }\mathrm{\mid }x-y^{\ast }\mathrm{\mid }{)}^{2-n}\mathrm{/}((n-2){\omega }_{n-1})=R^{n-2}(R\mathrm{\mid }x-y\mathrm{\mid }{)}^{2-n}\mathrm{/}((n-2){\omega }_{n-1})=\mathrm{\mid }x-y{\mathrm{\mid }}^{2-n}\mathrm{/}((n-2){\omega }_{n-1})=\mathrm{\Phi }(x-y)$. For $y=a$ and $\mathrm{\mid }x-a\mathrm{\mid }=R$ we have $H_a(x)=\mathrm{\Phi }(R)=\mathrm{\Phi }(x-a)$ by [F1]. So $H_y=\mathrm{\Phi }(\cdot -y)$ on $\mathrm{\partial }\mathrm{\Omega }$ in both cases.

Positivity. Let $x,y\in \mathrm{\Omega }$ be distinct. If $y\mathrm{\ne }a$, then $\mathrm{\mid }x-y\mathrm{\mid }>0$, and the first identity of step 1.1 together with $R^2-\mathrm{\mid }x-a{\mathrm{\mid }}^2>0$, $R^2-\mathrm{\mid }y-a{\mathrm{\mid }}^2>0$ gives $\mathrm{\mid }y-a{\mathrm{\mid }}^2\mathrm{\mid }x-y^{\ast }{\mathrm{\mid }}^2=R^2\mathrm{\mid }x-y{\mathrm{\mid }}^2+(R^2-\mathrm{\mid }x-a{\mathrm{\mid }}^2)(R^2-\mathrm{\mid }y-a{\mathrm{\mid }}^2)>R^2\mathrm{\mid }x-y{\mathrm{\mid }}^2$, so $\mathrm{\mid }y-a\mathrm{\mid }\mathrm{\mid }x-y^{\ast }\mathrm{\mid }>R\mathrm{\mid }x-y\mathrm{\mid }>0$; because $n\ge 3$ makes the exponent $2-n$ negative and $r\mapsto r^{2-n}$ strictly decreasing, ${\kappa }_y\mathrm{\Phi }(x-y^{\ast })=R^{n-2}(\mathrm{\mid }y-a\mathrm{\mid }\mathrm{\mid }x-y^{\ast }\mathrm{\mid }{)}^{2-n}\mathrm{/}((n-2){\omega }_{n-1})<R^{n-2}(R\mathrm{\mid }x-y\mathrm{\mid }{)}^{2-n}\mathrm{/}((n-2){\omega }_{n-1})=\mathrm{\Phi }(x-y)$, that is $G(x,y)>0$. If $y=a$, then $0<\mathrm{\mid }x-a\mathrm{\mid }<R$ and strict decrease of $r\mapsto r^{2-n}$ gives $G(x,a)=\mathrm{\Phi }(x-a)-\mathrm{\Phi }(R)>0$.

Symmetry. Let $x,y\in \mathrm{\Omega }\setminus \{a\}$. The second identity of step 1.1 gives $\mathrm{\mid }y-a\mathrm{\mid }\mathrm{\mid }x-y^{\ast }\mathrm{\mid }=\mathrm{\mid }x-a\mathrm{\mid }\mathrm{\mid }y-x^{\ast }\mathrm{\mid }$, so ${\kappa }_y\mathrm{\Phi }(x-y^{\ast })=R^{n-2}(\mathrm{\mid }y-a\mathrm{\mid }\mathrm{\mid }x-y^{\ast }\mathrm{\mid }{)}^{2-n}\mathrm{/}((n-2){\omega }_{n-1})=R^{n-2}(\mathrm{\mid }x-a\mathrm{\mid }\mathrm{\mid }y-x^{\ast }\mathrm{\mid }{)}^{2-n}\mathrm{/}((n-2){\omega }_{n-1})={\kappa }_x\mathrm{\Phi }(y-x^{\ast })$; since $\mathrm{\Phi }$ depends only on the norm, $\mathrm{\Phi }(x-y)=\mathrm{\Phi }(y-x)$, hence $G(x,y)=G(y,x)$. For the case of the centre, $x\in \mathrm{\Omega }\setminus \{a\}$: $G(a,x)=\mathrm{\Phi }(a-x)-{\kappa }_x\mathrm{\Phi }(a-x^{\ast })$ and $\mathrm{\mid }{x}^{\ast }-a\mathrm{\mid }=R^2\mathrm{/}\mathrm{\mid }x-a\mathrm{\mid }>R$ gives ${\kappa }_x\mathrm{\Phi }(a-x^{\ast })=(R\mathrm{/}\mathrm{\mid }x-a\mathrm{\mid }{)}^{n-2}(R^2\mathrm{/}\mathrm{\mid }x-a\mathrm{\mid }{)}^{2-n}\mathrm{/}((n-2){\omega }_{n-1})=R^{2-n}\mathrm{/}((n-2){\omega }_{n-1})=\mathrm{\Phi }(R)$, whence $G(a,x)=\mathrm{\Phi }(a-x)-\mathrm{\Phi }(R)=\mathrm{\Phi }(x-a)-\mathrm{\Phi }(R)=G(x,a)$.

Harmonicity, continuity and zero trace. If $y\mathrm{\ne }a$, [F2] makes $x\mapsto \mathrm{\Phi }(x-y)$ smooth and harmonic on ${\mathbb{R}}^n\setminus \{y\}\supseteq \mathrm{\Omega }\setminus \{y\}$, and $H_y$ is smooth harmonic on $\mathrm{\Omega }$ by step 2.1; hence $G(\cdot ,y)=\mathrm{\Phi }(\cdot -y)-H_y$ is smooth and harmonic on $\mathrm{\Omega }\setminus \{y\}$, and the same holds for $y=a$ with the constant $H_a$. For the continuous extension: fix $y$ and let $x\to z\in \mathrm{\partial }\mathrm{\Omega }$ with $x\in \mathrm{\Omega }$. By [F1] and continuity of $\mathrm{\Phi }$ off the origin, $\mathrm{\Phi }(x-y)\to \mathrm{\Phi }(z-y)$ and ${\kappa }_y\mathrm{\Phi }(x-y^{\ast })\to {\kappa }_y\mathrm{\Phi }(z-y^{\ast })=\mathrm{\Phi }(z-y)$ by step 2.2 applied at the boundary point $z$; for $y=a$, $\mathrm{\Phi }(x-a)\to \mathrm{\Phi }(z-a)=\mathrm{\Phi }(R)$ because $\mathrm{\mid }z-a\mathrm{\mid }=R$. Hence $G(x,y)\to 0$ for every $z\in \mathrm{\partial }\mathrm{\Omega }$, and $G(\cdot ,y)$ extends continuously to $\bar{\mathrm{\Omega }}\setminus \{y\}$ with zero boundary trace.

Distributional identity. Let $\varphi \in C_c^{\mathrm{\infty }}(\mathrm{\Omega })$ and let $\psi \in C_c^{\mathrm{\infty }}({\mathbb{R}}^n)$ be its extension by zero. By the definitions of [F9], $(\mathrm{\Delta }{T}_{G(\cdot ,y)})(\varphi )=T_{G(\cdot ,y)}(\mathrm{\Delta }\varphi )={\int }_{\mathrm{\Omega }}G(x,y)\mathrm{\Delta }\varphi (x)dx$, so $⟨-\mathrm{\Delta }{T}_{G(\cdot ,y)},\varphi ⟩=-{\int }_{{\mathbb{R}}^n}\mathrm{\Phi }(x-y)\mathrm{\Delta }\psi (x)dx+{\int }_{\mathrm{\Omega }}{H}_y(x)\mathrm{\Delta }\varphi (x)dx$. The first term equals $\psi (y)=\varphi (y)$ by [F4], since $-{\mathrm{\Delta }}_x{T}_{\mathrm{\Phi }(\cdot -y)}={\delta }_y$ on ${\mathbb{R}}^n$. For the second term: $H_y\in C^2(\bar{\mathrm{\Omega }})$ is harmonic and $\varphi$ vanishes on a neighbourhood of $\mathrm{\partial }\mathrm{\Omega }$, so the second Green identity [F8] with $u=\varphi$, $v=H_y$ gives ${\int }_{\mathrm{\Omega }}{H}_y\mathrm{\Delta }\varphi dx={\int }_{\mathrm{\Omega }}\varphi \mathrm{\Delta }{H}_{y}dx+{\int }_{\mathrm{\partial }\mathrm{\Omega }}(H_y{\mathrm{\partial }}_{\nu }\varphi -\varphi {\mathrm{\partial }}_{\nu }{H}_y)dS=0$, both boundary terms vanishing because $\varphi$ and its first derivatives are zero near $\mathrm{\partial }\mathrm{\Omega }$. Hence $⟨-\mathrm{\Delta }{T}_{G(\cdot ,y)},\varphi ⟩=\varphi (y)={\delta }_y(\varphi )$ for every test function $\varphi$, that is $-{\mathrm{\Delta }}_x{T}_{G(\cdot ,y)}={\delta }_y$ in ${\mathcal{D}}^{\mathrm{\prime }}(\mathrm{\Omega })$.

Steps 2.1, 2.2 and 3.1 verify the corrector clause and the zero-trace clause of the Dirichlet Green definition [F3] for the ball $\mathrm{\Omega }=B_R(a)$ and the kernel $G$ of step 1.1, step 3.2 verifies its distributional clause, and step 2.3 gives strict positivity while step 2.4 gives symmetry; so $G$ is the positive symmetric Dirichlet Green function of $B_R(a)$. Symmetry also follows independently from the published theorem [F7], whose hypotheses hold because $\mathrm{\Omega }$ is a bounded $C^1$ domain by [F6] and the correctors $H_y$ of step 2.1 lie in $C^2(\bar{\mathrm{\Omega }})$.