Green's Theorem
Theorem
Let’s assume $u, v \in C^2( \bar{U})$. Then, the following expressions hold:
(i) $\displaystyle \int_{U} \Delta u dx=\int_{\partial U} \dfrac{\partial u}{\partial \nu}dS$
(ii) $\displaystyle \int_{U} Dv \cdot Du dx = -\int_{U} u \Delta v dx+\int_{\partial U}\dfrac{\partial v}{\partial \nu}udS$
(iii) $\displaystyle \int_{U} (u\Delta v - v\Delta u )dx = \int_{\partial U} \left( \dfrac{\partial v}{\partial \nu}u - \dfrac{\partial u}{\partial \nu} v\right)dS$
These are collectively referred to as Green’s formula.
- $\Delta$ is Laplacian
- $D$ is Gradient
- $\nu$ is Outward unit normal vector
Proof
Let’s assume $u, v \in C^1(\bar{U})$. Then, the following formula holds:
$$ \int_{U} u_{x_{i}}vdx = -\int_{U} uv_{x_{i}}dx + \int_{\partial U} uv\nu^{i} dS\quad (i=1,\dots , n) $$
(i)
By substituting $u$ with $u_{x_{i}}$, and $v$ with $1$ in the partial integration formula, we obtain the following formula.
$$ \int_{U} u_{x_{i} x_{i}}dx = \int_{\partial U} u_{x_{i}}\nu^{i} dS \quad (i=1,\cdots , n) $$
Summing up for all $i=1,\cdots, n$ gives:
$$ \int_{U} (u_{x_{1} x_{1}}+\cdots +u_{x_{n} x_{n}} )dx = \int_{\partial U}( u_{x_{1}}\nu^{1} +\cdots u_{x_{n}}\nu^n)dS $$
By the definition of Laplacian and $\dfrac{\partial u}{\partial \nu}:=\boldsymbol{\nu}\cdot Du$, the following is true.
$$ \int_{U} \Delta u dx=\int_{\partial U} \dfrac{\partial u}{\partial \nu}dS $$
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(ii)
By substituting $v$ with $v_{x_{i}}$ in the partial integration formula, we obtain the following formula.
$$ \int_{U} u_{x_{i}}v_{x_{i}}dx = -\int_{U} uv_{x_{i}x_{i}}dx + \int_{\partial U} uv_{x_{i}}\nu^{i} dS \quad (i=1,\cdots , n) $$
Summing up for all $i=1,\cdots ,n$ gives:
$$ \int_{U} (u_{x_{1}}v_{x_{1}}+\cdots +u_{x_{n}}v_{x_{n}} )dx = -\int_{U} u(v_{x_{1}x_{1}}+\cdots v_{x_{n} x_{n}})dx + \int_{\partial U} ( v_{x_{1}}\nu^1 +\cdots v_{x_{n}}\nu^n )udS $$
Which simplifies to:
$$ \int_{U} Du\cdot Dvdx = -\int_{U} u\Delta vdx + \int_{\partial U} \dfrac{\partial v}{\partial \nu}u dS $$
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(iii)
Switching the places of $u$ and $v$ in (ii) gives the following formula.
$$ \int_{U} Du \cdot Dv dx = -\int_{U} v \Delta u dx+\int_{\partial U}\dfrac{\partial u}{\partial \nu}vdS $$
Subtracting (ii) from the above formula gives:
$$ 0= -\int_{U} ( v \Delta u -u\Delta v) dx+\int_{\partial U} \left( \dfrac{\partial u}{\partial \nu}v -\dfrac{\partial v}{\partial \nu}u \right)dS $$
Which simplifies to:
$$ -\int_{U} ( v \Delta u -u\Delta v) dx=\int_{\partial U} \left( \dfrac{\partial v}{\partial \nu}u -\dfrac{\partial u}{\partial \nu}v \right)dS $$
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