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Greenの定理の証明

Last updated at Posted at 2019-05-09

これが一番楽だと思います(多分)

##証明

任意の閉領域$\Omega$について、その境界を$\Gamma$とし、$\boldsymbol{n}$を$\Gamma$上の外向き法線ベクトルとする。この時、まず次のGaussの発散定理が成り立つ。

  \int_\Omega \nabla \cdot \boldsymbol{A} d\Omega 
    = \int_\Gamma \boldsymbol{A} \cdot \boldsymbol{n} d\Gamma

また、$f$,$g$をスカラー関数とすると、ナブラの分配法則$\nabla(fg) = g\nabla f +f\nabla g$より、次式が成り立つ。

  \int_\Omega \nabla \cdot [\nabla f(\boldsymbol{r}) g(\boldsymbol{r})] d\Omega
    =  \int_\Omega [\nabla^2 f(\boldsymbol{r})] g(\boldsymbol{r}) d\Omega
       +\int_\Omega \nabla f(\boldsymbol{r}) \cdot \nabla g(\boldsymbol{r}) d\Omega

上式の左辺にGaussの発散定理を適用すると、次のようになる。

  \int_\Gamma [\nabla f(\boldsymbol{r}) g(\boldsymbol{r})] \cdot \boldsymbol{n} d\Gamma
    =  \int_\Omega [\nabla^2 f(\boldsymbol{r})] g(\boldsymbol{r}) d\Omega
       +\int_\Omega \nabla f(\boldsymbol{r}) \cdot \nabla g(\boldsymbol{r}) d\Omega

この式を、Greenの定理という。

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