$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\dag}{\dagger}$ $\newcommand{\const}{\mathrm{const}}$ $\newcommand{\arcsinh}{\operatorname{arcsinh}}$ ##[7.3. Green function](id:sect-7.3) --------------------- > 1. [Newton shell theorem](#sect-7.3.1) > 2. [Green function](#sect-7.3.2) > 2. [Green function. II](#sect-7.3.3) ###[Newton shell theorem](id:sect-7.3.1) Let $n\ge 3$. Consider spherically symmetric density $f$ (thus depending on $r=(x\_1^2+x\_2^2+\ldots+x\_n^2)^{\frac{1}{2}}$ only). Then it creates a density which is also spherically symmetric. Assume first that $f=0$ in $B(0,R)$ and consider $u$ in $B(0,R)$. Here $u$ must be harmonic and then due to mean-value theorem $u=u(0)$ is constant (in particular $\nabla u=0$). More precisely \begin{equation} u= -\frac{1}{n-2} \int\_R^\infty \rho f(\rho)\,d\rho \label{eq-7.3.1} \end{equation} where we replaced lower limit $0$ by $R$ since $f(\rho)=0$ as $r\< R$. Assume now that $f=0$ in $\\{r\ge R\\}$ and consider $u$ there. Then $u=r^{2-n} A+B$. **[Exercise 1.](id:exercise-7.3.1)** Prove that if $f$ is spherically symmetric and $f=0$ in $\\{r\ge R\\}$ then $u=r^{2-n} A+B$ there. To have potential at infinity equal to $0$ we must take $B=0$. Further plug it into ([7.2.2](./S7.2.html#mjx-eqn-eq-7.2.2)) with $\Omega=B(0,R)$. The left-hand expression becomes the total charge $\int\_{B(0,R)} f(x)\,dV$ while the right-hand expression becomes $-(n-2)\sigma\_n A$ and therefore \begin{equation} u= -\frac{1}{(n-2)\sigma\_n}r^{2-n}\int \_{B(0,R)} f\,dx = -\frac{1}{n-2}r^{2-n}\int \_0^R \rho^{n-1} f(\rho)\,d\rho. \label{eq-7.3.2} \end{equation} Then in the general case we to calculate $u$ we break $f=f\_1+f\_2$ where $f\_1=0$ for $r\ge R$ and $f\_2=0$ for $r\le R$ and then calculate $u$ as $r=R$ and finally set $R=r$: \begin{equation} u= -\frac{1}{n-2} \int\_r^\infty \rho f(\rho)\,d\rho-\frac{1}{n-2}r^{2-n}\int \_0^r f(\rho)\,d\rho. \label{eq-7.3.3} \end{equation} In particular, if **[Theorem 1.](id:thm-7.3.1)** Let $f$ be spherically symmetric. Then a. If $f=0$ as $r\ge R$ then $u$ coincides with a potential created by the same mass was concentrated in the origin. b. If $f=0$as $r\le R$ then $u=\const$ there. **[Remark 1.](id:remark-7.3.1)** Statement (b) is often expressed by a catch phrase "There is no gravity in the cavity". In particular if $f(r)=\const$ as $R\_1\le r\le R\_2$ and $f(r)=0$ as $r\le R\_1$ or $r\ge R\_2$ , then all calculations are easy. Amazingly, the same is true for a shell between two proportional ellipsoids (Ivory theorem). ###[Green function. I](id:sect-7.3.2) Recall ([7.2.7](./S7.2.html#mjx-eqn-eq-7.2.7))--([7.2.8](./S7.2.html#mjx-eqn-eq-7.2.8)) \begin{multline} u(y)=\int\_{\Omega} G^0(x,y)\Delta u(x)\,dV \\\\ + \int\_\Sigma \bigl(-u(x)\frac{\partial G^0}{\partial \nu\_x}(x,y)+ G^0(x,y)\frac{\partial u}{\partial \nu}(x)\bigr)\, dS \qquad \tag{7.2.7}\label{eq-7.3.26-7} \end{multline} with \begin{equation} G^0(x,y)=\left\\{\begin{aligned} -&\frac{1}{(n-2)\sigma\_n}|x-y|^{2-n}&& n\ne 2,\\\\ -&\frac{1}{4\pi}|x-y|^{-1} &&n=3,\\\\ &\frac{1}{2\pi}\log |x-y| &&n=2,\\\\ &\frac{1}{2} |x-y| &&n=1 \end{aligned}\right. \tag{7.2.8}\label{eq-7.3.26-8} \end{equation} where we changed notation $G^0$ instead of $G$ as we will redefine $G$ later. Let $g(x,y)$ be solution to problem \begin{align} &\Delta\_x g(x,y)=0\qquad&&\text{in }\\ \Omega,\label{eq-7.3.4}\\\\[3pt] &g(x,y)=-G^0(x,y) &&\text{as }\\ x\in\Sigma\label{eq-7.3.5} \end{align} and condition $u\to 0$ as $|x|\to \infty$ if $\Omega$ is unbounded. In virtue of (\ref{eq-7.3.4}) and ([7.2.5](./S7.2.html#mjx-eqn-eq-7.2.5)) \begin{multline} 0=\int\_{\Omega} g(x,y)\Delta u(x)\,dV + \\\\ \int\_\Sigma \bigl(-u(x)\frac{\partial g}{\partial \nu\_x}(x,y)+ g(x,y)\frac{\partial u}{\partial \nu}(x)\bigr)\, dS. \qquad \label{eq-7.3.6} \end{multline} Adding to (\ref{eq-7.3.26-7}) we get \begin{multline} u(y)=\int\_{\Omega} G(x,y)\Delta u(x)\,dV + \\\\ \int\_\Sigma \bigl(-u(x)\frac{\partial G}{\partial \nu\_x}(x,y)+ G(x,y)\frac{\partial u}{\partial \nu}(x)\bigr)\, dS \qquad \label{eq-7.3.7} \end{multline} with \begin{equation} G(x,y):=G^0(x,y)+g(x,y). \label{eq-7.3.8} \end{equation} So far we have not used (\ref{eq-7.3.5}) which is equivalent to $G(x,y)=0$ as $x\in \Sigma$. But then \begin{equation\*} u(y)=\int\_{\Omega} G(x,y)\Delta u(x)\,dV - \int\_\Sigma \frac{\partial G}{\partial \nu\_x}(x,y)u(x)\, dS \end{equation\*} and therefore \begin{equation} u(y)=\int\_{\Omega} G(x,y)f(x)\,dV - \int\_\Sigma \frac{\partial G}{\partial \nu\_x}(x,y)\phi(x)\, dS \label{eq-7.3.9} \end{equation} is a solution to \begin{align} &\Delta u=f \qquad&&\text{in }\\ \Omega,\label{eq-7.3.10}\\\\[3pt] &u=\phi &&\text{on }\ \Sigma.\label{eq-7.3.11} \end{align} Similarly if we replace (\ref{eq-7.3.5}) by Robin boundary condition \begin{equation} \bigl(\frac{\partial g}{\partial \nu\_x}-\alpha g\bigr)(x,y)= -\bigl(\frac{\partial G^0}{\partial \nu\_x}-\alpha G^0\bigr) (x,y) \qquad \text{as }\ x\in\Sigma \label{eq-7.3.12} \end{equation} we get $\bigl(\frac{\partial G}{\partial \nu\_x}-\alpha G\bigr)(x,y)=0$ as $x\in \Sigma$ and rewriting (\ref{eq-7.3.7}) as \begin{multline} u(y)=\int\_{\Omega} G(x,y)\Delta u(x)\,dV + \\\\ \int\_\Sigma \bigl(-u(x) \bigl[ \frac{\partial G}{\partial \nu\_x}- \alpha G \bigr] (x,y)+ G(x,y)\bigl[ \frac{\partial u}{\partial \nu}- \alpha u\bigr]\bigr)\, dS \qquad \label{eq-7.3.13} \end{multline} we get \begin{equation} u(y)=\int\_{\Omega} G(x,y)f(x)\,dV + \int\_\Sigma G(x,y)\psi(x)\, dS \label{eq-7.3.14} \end{equation} for solution of the (\ref{eq-7.3.10}) with the boundary condition \begin{equation} \frac{\partial u}{\partial \nu}-\alpha u=\psi \qquad \text{on }\ \Sigma. \label{eq-7.3.15} \end{equation} Finally, if we take $g$ satisfying mixed boundary condition \begin{align} & g(x,y)=-G^0(x,y) && \text{as }\\ x\in \Sigma',\label{eq-7.3.16}\\\\ & \bigl(\frac{\partial g}{\partial \nu\_x}-\alpha g\bigr)(x,y)= -\bigl(\frac{\partial G^0}{\partial \nu\_x}-\alpha G^0\bigr) (x,y) &&\text{as }\ x\in\Sigma'' \qquad\label{eq-7.3.17} \end{align} we get \begin{multline} u(y)=\int\_{\Omega} G(x,y)f(x)\,dV + \int\_{\Sigma'} G(x,y)\psi(x)\, dS - \\\\ \int\_{\Sigma''} \frac{\partial G(x,y)}{\partial \nu\_x}\phi(x)\, dS\qquad \label{eq-7.3.18} \end{multline} for solution of the (\ref{eq-7.3.10}) with the boundary condition \begin{align} &u=\phi && \text{on }\ \Sigma',\label{eq-7.3.19}\\\\ &\frac{\partial u}{\partial \nu}-\alpha u=\psi && \text{on }\ \Sigma'', \label{eq-7.3.20} \end{align} where $\Sigma=\Sigma'\cup\Sigma''$ and $\Sigma'\cap\Sigma''=\emptyset$. **[Definition 1.](id:definition-7.3.1)** $G(x,y)$ (defined for corresponding boundary problem) is called *Green's function*. **[Remark 2.](id:remark-7.3.2)** It is similar (by usage) to heat kernel $\frac{1}{2\sqrt{kt}}e^{-\frac{|x-y|^2}{4kt}}$. One can prove **[Theorem 2.](id:thm-7.3.2)** \begin{equation}G(x,y)=G(y,x). \label{eq-7.3.21} \end{equation} ###[Green function. II](id:sect-7.3.3) Consider now purely Neumann problem in the connected domain. We cannot solve $\Delta g=0$ with the boundary condition $\frac{\partial g}{\partial \nu\_x}=-\frac{\partial G^0}{\partial \nu\_x}$ as such problem requires one solvability condition to the right-hand expression and boundary value \begin{equation} \int\_\Omega f\, dV+ \int \_\Sigma \psi\,dS =0 \label{eq-7.3.22} \end{equation} and this condition fails as $\int\_\Sigma \frac{\partial G^0}{\partial \nu\_x}\, dS\ne 0$ (one can prove it). To fix it we consider $g$ satisfying \begin{align} &\Delta\_x g(x,y)=c\qquad&&\text{in }\ \Omega,\label{eq-7.3.23}\\\\[3pt] &\frac{\partial g}{\partial \nu\_x}(x,y)= -\frac{\partial G^0}{\partial \nu\_x}(x,y) &&\text{as }\ x\in\Sigma \label{eq-7.3.24} \end{align} with unknown constant $c$; chosing it correctly one can satisfy solvability condition. Then \begin{equation\*} u(y)=\int\_{\Omega} G(x,y)f(x)\,dV + \int\_{\Sigma} G(x,y)\psi(x)\, dS + c \int \_V u\,dx \end{equation\*} and therefore \begin{equation} u(y)=\int\_{\Omega} G(x,y)f(x)\,dV + \int\_{\Sigma} G(x,y)\psi(x)\, dS +C \label{eq-7.3.25} \end{equation} gives us solution if it exists (and it exists under solvability condition (\ref{eq-7.3.24})). This solution is defined up to a constant. --------------- [$\Leftarrow$](./S7.2.html) [$\Uparrow$](../contents.html) [$\downarrow$](./S7.P.html) [$\Rightarrow$](../Chapter8/S8.1.html)