Poisson kernel of a Euclidean ball

Work under [F6]. Put $u:=x-a$ and $w:=y-a$, so that $\mathrm{\mid }u\mathrm{\mid }<R$ and $\mathrm{\mid }w\mathrm{\mid }=R$; for $x\mathrm{\ne }a$ put ${\kappa }_x:=(R\mathrm{/}\mathrm{\mid }u\mathrm{\mid }{)}^{n-2}$ and $x^{\ast }:=a+R^{2}u\mathrm{/}\mathrm{\mid }u{\mathrm{\mid }}^2$, the inversion of $x$, while for $x=a$ the corrector is the constant $H_a=\mathrm{\Phi }(R)$ by [F3]. By [F3] the corrector for the pole $x$ is $H_x(z)={\kappa }_x\mathrm{\Phi }(z-x^{\ast })$ when $x\mathrm{\ne }a$; its value at $y$ is well defined because $x^{\ast }\mathrm{\notin }{B}_R(a)$ and $y\in \mathrm{\partial }{B}_R(a)$ are different points, and $\mathrm{\Phi }$ is smooth there by [F1]. Also $D_z\mathrm{\Phi }(z-x){\mathrm{\mid }}_{z=y}$ is defined because $y\mathrm{\ne }x$.

Magnitude identity. Let $x\mathrm{\ne }a$. From $x^{\ast }-a=R^{2}u\mathrm{/}\mathrm{\mid }u{\mathrm{\mid }}^2$ and $y-x^{\ast }=(y-a)-R^{2}u\mathrm{/}\mathrm{\mid }u{\mathrm{\mid }}^2$ we get $\mathrm{\mid }y-x^{\ast }{\mathrm{\mid }}^2=\mathrm{\mid }w{\mathrm{\mid }}^2-2R^2⟨w,u⟩\mathrm{/}\mathrm{\mid }u{\mathrm{\mid }}^2+R^4\mathrm{/}\mathrm{\mid }u{\mathrm{\mid }}^2$; multiplying by $\mathrm{\mid }u{\mathrm{\mid }}^2$ and using $\mathrm{\mid }w{\mathrm{\mid }}^2=R^2$ gives $\mathrm{\mid }u{\mathrm{\mid }}^2\mathrm{\mid }y-x^{\ast }{\mathrm{\mid }}^2=R^2\mathrm{\mid }u{\mathrm{\mid }}^2-2R^2⟨u,w⟩+R^4=R^2\mathrm{\mid }x-y{\mathrm{\mid }}^2$, because $\mathrm{\mid }x-y{\mathrm{\mid }}^2=\mathrm{\mid }u-w{\mathrm{\mid }}^2=\mathrm{\mid }u{\mathrm{\mid }}^2-2⟨u,w⟩+R^2$. Hence $\mathrm{\mid }u\mathrm{\mid }\mathrm{\mid }y-x^{\ast }\mathrm{\mid }=R\mathrm{\mid }x-y\mathrm{\mid }>0$.

Vector identity. Let $x\mathrm{\ne }a$. Adding and subtracting $u$ and using $x^{\ast }-a=R^{2}u\mathrm{/}\mathrm{\mid }u{\mathrm{\mid }}^2$ gives $(x-y)+\frac{\mathrm{\mid }u{\mathrm{\mid }}^2}{R^2}(y-x^{\ast })=(u-w)+\frac{\mathrm{\mid }u{\mathrm{\mid }}^2}{R^2}(w-\frac{R^{2}u}{\mathrm{\mid }u{\mathrm{\mid }}^2})=u-w+\frac{\mathrm{\mid }u{\mathrm{\mid }}^{2}w}{R^2}-u=-(1-\frac{\mathrm{\mid }u{\mathrm{\mid }}^2}{R^2})(y-a)=-\frac{R^2-\mathrm{\mid }x-a{\mathrm{\mid }}^2}{R^2}(y-a)\mathrm{.}$

The gradient of each term of [F2] at $z=y$. By [F5] and [F1], the gradient of $\mathrm{\Phi }$ is $\mathrm{\nabla }\mathrm{\Phi }(\zeta )=-\mathrm{\mid }\zeta {\mathrm{\mid }}^{-n}\zeta \mathrm{/}{\omega }_{n-1}$ for $\zeta \mathrm{\ne }0$, since $\mathrm{\nabla }\mathrm{\mid }\zeta {\mathrm{\mid }}^{2-n}=(2-n)\mathrm{\mid }\zeta {\mathrm{\mid }}^{-n}\zeta$ and the prefactor is $1\mathrm{/}((n-2){\omega }_{n-1})$; hence $D_z\mathrm{\Phi }(z-x){\mathrm{\mid }}_{z=y}=-(\mathrm{\mid }y-x{\mathrm{\mid }}^{-n}(y-x))\mathrm{/}{\omega }_{n-1}=\mathrm{\mid }x-y{\mathrm{\mid }}^{-n}(x-y)\mathrm{/}{\omega }_{n-1}$, and $D_z[{\kappa }_x\mathrm{\Phi }(z-x^{\ast })]{\mathrm{\mid }}_{z=y}=-{\kappa }_x\mathrm{\mid }y-x^{\ast }{\mathrm{\mid }}^{-n}(y-x^{\ast })\mathrm{/}{\omega }_{n-1}$. By step 2.1, ${\kappa }_x\mathrm{\mid }y-x^{\ast }{\mathrm{\mid }}^{-n}=(R\mathrm{/}\mathrm{\mid }u\mathrm{\mid }{)}^{n-2}(\mathrm{\mid }u\mathrm{\mid }\mathrm{/}R{)}^n\mathrm{\mid }x-y{\mathrm{\mid }}^{-n}=(\mathrm{\mid }u{\mathrm{\mid }}^2\mathrm{/}{R}^2)\mathrm{\mid }x-y{\mathrm{\mid }}^{-n}$.

The boundary-slot derivative. Subtracting the two expressions of step 3.1 and using step 2.2, $D_z(\mathrm{\Phi }(z-x)-H_x(z)){\mathrm{\mid }}_{z=y}=\frac{1}{{\omega }_{n-1}}\mathrm{\mid }x-y{\mathrm{\mid }}^{-n}[(x-y)+\frac{\mathrm{\mid }u{\mathrm{\mid }}^2}{R^2}(y-x^{\ast })]=-\frac{R^2-\mathrm{\mid }x-a{\mathrm{\mid }}^2}{{\omega }_{n-1}{R}^2\mathrm{\mid }x-y{\mathrm{\mid }}^n}(y-a)$ for $x\mathrm{\ne }a$.

The case of the centre. For $x=a$ the corrector is the constant $H_a=\mathrm{\Phi }(R)$ of [F3], so $D_z(\mathrm{\Phi }(z-a)-H_a){\mathrm{\mid }}_{z=y}=\mathrm{\nabla }\mathrm{\Phi }(y-a)=-R^{-n}(y-a)\mathrm{/}{\omega }_{n-1}$ by the gradient computation of step 3.1; dotting with $\nu (y)=(y-a)\mathrm{/}R$ gives ${\mathrm{\partial }}_{{\nu }_y}G(a,y)=-R^{-n}\cdot R\mathrm{/}{\omega }_{n-1}=-R^{1-n}\mathrm{/}{\omega }_{n-1}$ and $P_{R,a}(a,y)=R^{1-n}\mathrm{/}{\omega }_{n-1}$, which is exactly the formula $(R^2-\mathrm{\mid }a-a{\mathrm{\mid }}^2)\mathrm{/}(R{\omega }_{n-1}\mathrm{\mid }a-y{\mathrm{\mid }}^n)=R^2\mathrm{/}(R{\omega }_{n-1}{R}^n)$.

The Poisson kernel. Dotting step 4.1 with $\nu (y)=(y-a)\mathrm{/}R$ from [F4] gives ${\mathrm{\partial }}_{{\nu }_y}G(x,y)=-\frac{(R^2-\mathrm{\mid }x-a{\mathrm{\mid }}^2)(y-a)\cdot (y-a)}{{\omega }_{n-1}{R}^3\mathrm{\mid }x-y{\mathrm{\mid }}^n}=-\frac{R^2-\mathrm{\mid }x-a{\mathrm{\mid }}^2}{R{\omega }_{n-1}\mathrm{\mid }x-y{\mathrm{\mid }}^n}$, because $(y-a)\cdot (y-a)=R^2$; hence by [F2], $P_{R,a}(x,y)=-{\mathrm{\partial }}_{{\nu }_y}G(x,y)=(R^2-\mathrm{\mid }x-a{\mathrm{\mid }}^2)\mathrm{/}(R{\omega }_{n-1}\mathrm{\mid }x-y{\mathrm{\mid }}^n)$ for $x\mathrm{\ne }a$.

Steps 5.1 and 4.2 give $P_{R,a}(x,y)=(R^2-\mathrm{\mid }x-a{\mathrm{\mid }}^2)\mathrm{/}(R{\omega }_{n-1}\mathrm{\mid }x-y{\mathrm{\mid }}^n)$ for every $x\in B_R(a)$ and $y\in \mathrm{\partial }{B}_R(a)$. This explicit expression is continuous on $B_R(a)\times \mathrm{\partial }{B}_R(a)$: numerator and denominator are continuous there and the denominator $R{\omega }_{n-1}\mathrm{\mid }x-y{\mathrm{\mid }}^n$ is nonzero at every point of the product because an interior point $x$ and a boundary point $y$ are never equal, so $\mathrm{\mid }x-y\mathrm{\mid }>0$. Hence the negative boundary-slot normal derivative of the ball Green function is the continuous function displayed in the statement.