Suppose $\\Omega\\subset\\R^n$ is a domain with $C^1$ boundary, given a $C^1$ vector field $\\mathbf w\\in C^1(\\bar\\Omega)$, we have the \\iemph{divergence theorem}
\n\\[
\n\\int_\\Omega\\div \\mathbf w\\rd x=\\int_{\\pt\\Omega}\\mathbf w\\cdot\\n\\rd S,
\n\\]
\nwhere $x\\in\\Omega$ and $\\n$ is the outward unit normal vector of $\\pt\\Omega$.
\n\\begin{prob}
\n Show that the divergence theorem also holds for $\\mathbf w\\in C(\\bar \\Omega)\\cap C^1(\\Omega)$.
\n\\end{prob}
\n
\n\\begin{rem}
\n For any function $u\\in C(\\bar \\Omega)\\cap C^1(\\Omega)$, define $\\mathbf w$ by $\\mathbf w\\eqdef(0,\\ldots,u,\\ldots,0)$, i.e., the $i$’s components is $u$ and others are $0$, then the divergence theorem shows that
\n \\[
\n \\int_\\Omega (D_iu)(x)\\rd x=\\int_{\\Omega}\\div \\mathbf w\\rd x=\\int_{\\pt\\Omega}\\mathbf w\\cdot\\n\\rd S=\\int_{\\pt\\Omega}u(x)\\nu_i\\rd S,
\n \\]
\n where $\\n=(\\nu_1,\\nu_2,\\ldots, \\nu_n)$ is the exterior unit normal vector of $\\pt\\Omega$. This formula can also be obtained by integration by parts.
\n\\end{rem}
\nNow, take $\\mathbf w$ as the \\iemph{gradient} of a function $u\\in C^1(\\bar\\Omega)\\cap C^2(\\Omega)$, i.e., $\\mathbf w=\\D u=\\left(\\frac{\\pt u}{\\pt x_1}, \\frac{\\pt u}{\\pt x_2},\\ldots,\\frac{\\pt u}{\\pt x_n}\\right)$, and since $\\div(\\D u)=\\sum_{i=1}^n\\frac{\\pt}{\\pt x_i}\\left(\\frac{\\pt u}{\\pt x_i}\\right)=\\Delta u$, we obtain
\n\\[
\n\\int_\\Omega\\Delta u\\rd x=\\int_{\\pt\\Omega}\\frac{\\pt u}{\\pt\\n}\\rd S\\eqdef\\int_{\\pt\\Omega}\\D u\\cdot\\n\\rd S.
\n\\]
\nIf $u,v\\in C^1(\\bar\\Omega)\\cap C^2(\\Omega)$, take $\\mathbf w$ as $u\\D v$ and $v\\D u$ in the above formula, since $\\div(u\\D v)=u\\Delta v+\\D u\\D v$, then we have, respectively
\n\\begin{equation}\\label{eq:1}
\n \\int_\\Omega u\\Delta v\\rd x=\\int_{\\pt \\Omega} u\\frac{\\pt v}{\\pt\\n}\\rd S-\\int_\\Omega \\D u\\cdot\\D v\\rd x,
\n\\end{equation}
\n\\begin{equation}\\label{eq:2}
\n \\int_\\Omega v\\Delta u\\rd x=\\int_{\\pt \\Omega} v\\frac{\\pt u}{\\pt\\n}\\rd S-\\int_\\Omega \\D u\\cdot\\D v\\rd x.
\n\\end{equation}
\nBoth \\eqref{eq:1} and \\eqref{eq:2} are called the \\iemph{Green’s first formula}. Subtraction \\eqref{eq:2} from \\eqref{eq:1}, we obtain
\n\\begin{equation}\\label{eq:3}
\n \\int_\\Omega (u\\Delta v-v\\Delta u)\\rd x=\\int_{\\pt\\Omega}\\left(u\\frac{\\pt v}{\\pt\\n}-v\\frac{\\pt u}{\\pt\\n}\\right)\\rd S,
\n\\end{equation}
\nwhich is called the \\iemph{Green’s second formula}.<\/p>\n
Now, we will use Green’s formula to prove Green’s identity, which play an important role in discussion of harmonic functions.
\n\\begin{thm}
\n Suppose $\\Omega$ is a bounded domain in $\\R^n$ with $C^1$ boundary and $u\\in C^1(\\bar\\Omega)\\cap C^2(\\Omega)$, then for any $x\\in\\Omega$ we have the following \\iemph{Green’s identity}(also called \\iemph{Green’s representation})
\n \\begin{equation}\\begin{split}
\n u(x)&=\\int_\\Omega\\Gamma(x-y)\\Delta_y u(y)\\rd y\\\\
\n &\\qquad-\\int_{\\partial\\Omega}\\left(
\n \\Gamma(x-y)\\frac{\\pt u}{\\pt\\n_y}(y)
\n -u(y)\\frac{\\pt\\Gamma}{\\pt \\n_y}(x-y)\\right)\\rd S_y,
\n \\end{split}\\end{equation}
\n where $\\n_y$ is the exterior unit normal vector of $\\pt\\Omega$.
\n\\end{thm}
\n\\begin{rem}
\n In particular, If $u$ is a harmonic function in $\\Omega$, then
\n \\[
\n u(x)=\\int_{\\pt\\Omega}\\left(u\\frac{\\pt\\Gamma(x-y)}{\\pt\\n_y}
\n -\\Gamma(x-y)\\frac{\\pt u}{\\pt\\n_y}\\right)\\rd S_y,
\n \\]
\n called the \\iemph{fundamental integral formula} of harmonic functions. It states that for harmonic function $u\\in C^1(\\pt\\Omega)\\cap C^2(\\Omega)$, its value at any point $x\\in\\Omega$ can be represented by $u$ and its normal derivative on $\\pt\\Omega$.
\n\\end{rem}
\n\\begin{proof}
\n Firstly, we should note that: for any fixed $x\\in\\Omega$, $\\Gamma(x-y)$ has a singular point at $y=x$. Thus, directly apply the Green’s second formula is not allowed. The idea is replace $\\Omega$ by $\\Omega\\setminus B_r(x)$, $B_r(x)\\subsubset\\Omega$ is a ball centered at $x$ with radius $r$, and let $r$ limits to $0$ to obtain the desired result.<\/p>\n
Just as we have analyzed, fix $x\\in\\Omega$, take $u=\\Gamma=\\Gamma(x-\\cdot)$ and $v=u$ in \\eqref{eq:3}, we get Suppose $\\Omega\\subset\\R^n$ is a domain …<\/p>\n
\n \\[\\begin{split}
\n \\int_{\\Omega\\setminus B_r(x)}\\Gamma\\Delta u-u\\Delta\\Gamma\\rd y
\n &=\\int_{\\pt(\\Omega\\setminus B_r(x))}\\left(
\n \\Gamma\\frac{\\pt u}{\\pt\\n_y}-
\n u\\frac{\\pt\\Gamma}{\\pt\\n_y}\\right)\\rd S_y\\\\
\n &=\\int_{\\pt\\Omega}\\left(
\n \\Gamma\\frac{\\pt u}{\\pt\\n_y}-
\n u\\frac{\\pt\\Gamma}{\\pt\\n_y}\\right)\\rd S_y
\n +\\int_{\\pt B_r(x)}\\left(
\n \\Gamma\\frac{\\pt u}{\\pt\\n_y}-
\n u\\frac{\\pt\\Gamma}{\\pt\\n_y}\\right)\\rd S_y,
\n \\end{split}\\]
\n where $\\n_y$ is the unit exterior normal vector of $\\pt(\\Omega\\setminus B_r(x))$. Take $r\\to 0$, and note that $\\Delta\\Gamma=0$ in $\\Omega\\setminus B_r(x)$, we have
\n \\begin{equation}\\label{eq:4}
\n \\int_\\Omega\\Gamma\\Delta u\\rd y=\\int_{\\pt\\Omega}\\left(
\n \\Gamma\\frac{\\pt u}{\\pt\\n_y}-
\n u\\frac{\\pt\\Gamma}{\\pt\\n_y}\\right)\\rd S_y
\n +\\lim_{r\\to 0}\\int_{\\pt B_r(x)}\\left(
\n \\Gamma\\frac{\\pt u}{\\pt\\n_y}-
\n u\\frac{\\pt\\Gamma}{\\pt\\n_y}\\right)\\rd S_y.
\n \\end{equation}
\n Since $u$ is $C^2$ in $\\bar B_r(x)$, there exists a constant $M>0$, such that $|\\Delta u(y)|\\leq M$ for any $y\\in B_r(x)$. Now,
\n \\[\\begin{split}
\n \\int_{\\pt B_r(x)} \\Gamma\\frac{\\pt u}{\\pt \\n_y}\\rd S_y
\n &=\\Gamma(r)\\int_{\\pt B_r(x)}\\frac{\\pt u}{\\pt \\n_y}\\rd S_y\\\\
\n &=-\\Gamma(r)\\int_{\\pt B_r(x)}\\frac{\\pt u}{\\pt\\mathbf r_y}\\rd S_y\\\\
\n &=-\\Gamma(r)\\int_{B_r(x)}\\Delta u\\rd y,
\n \\end{split}\\]
\n and
\n \\[\\begin{split}
\n \\left|\\int_{\\pt B_r(x)}\\Gamma\\frac{\\pt u}{\\pt\\n_y}\\rd S_y\\right|
\n &=\\left|-\\Gamma(r)\\int_{B_r(x)}\\Delta u\\rd y\\right|\\\\
\n &\\leq\\left|\\Gamma(r)\\right|\\int_{B_r(x)}\\left|\\Delta u\\right|\\rd y\\\\
\n &\\leq M\\left|\\Gamma(r)r^{n-1}\\right|\\omega_n\\\\
\n &=\\begin{cases}
\n \\frac{M}{2}r|\\ln r| ,&n=2,\\\\
\n \\frac{M}{n(n-2)}r, &n\\geq3.
\n \\end{cases}
\n \\end{split}\\]
\n Thus,
\n \\begin{equation}\\label{eq:5}
\n \\lim_{r\\to0}\\int_{\\pt B_{r}(x)}\\Gamma\\frac{\\pt u}{\\pt\\n_y}\\rd S_y=0.
\n \\end{equation}
\n Lastly,
\n \\[\\begin{split}
\n \\int_{\\pt B_r(x)} u\\frac{\\pt\\Gamma}{\\pt\\n_y}\\rd S_y
\n &=\\int_{\\pt B_r(x)}u \\frac{1}{n\\omega_n}r^{1-n}
\n \\langle -\\frac{x-y}{r}, \\n_y\\rangle\\rd S_y\\\\
\n &=\\int_{\\pt B_r(x)}u \\frac{1}{n\\omega_n}r^{1-n}
\n \\langle -\\frac{x-y}{r}, \\frac{x-y}{r}\\rangle\\rd S_y\\\\
\n &=-\\frac{1}{n\\omega_n r^{n-1}}\\int_{\\pt B_r(x)} u\\rd S_y,
\n \\end{split}\\]
\n thus,
\n \\begin{equation}\\label{eq:6}
\n \\lim_{r\\to0}\\int_{\\pt B_r(x)}u\\frac{\\pt\\Gamma}{\\pt\\n_y}\\rd S_y=-u(x).
\n \\end{equation}
\n Plugs \\eqref{eq:6} and \\eqref{eq:7} into \\eqref{eq:5}, we shall reach the desired result.
\n\\end{proof}
\n\\begin{prob}
\n Show in detail that \\eqref{eq:7} holds.
\n\\end{prob}
\n\\begin{rem}
\n Comparing with Remark 1<\/a>, we have
\n \\[
\n \\int_{\\pt\\Omega}\\frac{\\pt\\Gamma}{\\pt\\n_y}(x-y)\\rd S_y=1,
\n \\]
\n for any $x\\in\\Omega$. This can be obtained by letting $u\\equiv1$ in \\ref{thm:1}.
\n\\end{rem}
\n\\begin{quote}
\n \\iemph{PDE} is analysis, and analysis mainly are inequality, thus the ability of analysis is totally depend on the ability to estimate inequalities.
\n\\end{quote}<\/p>\n","protected":false},"excerpt":{"rendered":"