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4.3 エネルギー保存則


  $\displaystyle v_{i,j+1}^2 + w_{i,j+1}^2$ $\displaystyle =$ $\displaystyle \frac{1}{4}
\left\{
\left[
(v_{i+1,j}+v_{i-1,j})^2
+2\lambda(v_{i...
..._{i-1,j})(w_{i+1,j}-w_{i-1,j})
+\lambda^2(w_{i+1,j}-w_{i-1,j})^2
\right]\right.$
      $\displaystyle \left.
+
\left[
\lambda^2(v_{i+1,j}-v_{i-1,j})^2
+2\lambda(v_{i+1,j}-v_{i-1,j})(w_{i+1,j}+w_{i-1,j})
+(w_{i+1,j}+w_{i-1,j})^2
\right]\right\}$
    $\displaystyle =$ $\displaystyle \frac{1}{4}\left\{
(1+\lambda^2)(v_{i+1,j}^2+v_{i-1,j}^2)
+2(1-\lambda^2)v_{i+1,j}v_{i-1,j}\right.$
      $\displaystyle +
(1+\lambda^2)(w_{i+1,j}^2+w_{i-1,j}^2)
+2(1-\lambda^2)w_{i+1,j}w_{i-1,j}$
      $\displaystyle \left.+4\lambda(v_{i+1,j}w_{i+1,j}-v_{i-1,j}w_{i-1,j})
\right\}$
    $\displaystyle =$ $\displaystyle \frac{1}{4}\left\{
(1+\lambda^2)(v_{i+1,j}^2+v_{i-1,j}^2+w_{i+1,j}^2+w_{i-1,j}^2)
+2(1-\lambda^2)(v_{i+1,j}v_{i-1,j}+w_{i+1,j}w_{i-1,j})\right.$
      $\displaystyle \left.+4\lambda(v_{i+1,j}w_{i+1,j}-v_{i-1,j}w_{i-1,j})
\right\}.$


  $\displaystyle 4\sum_{i=1}^{N-1}\left(v_{i,j+1}^2+w_{i,j+1}^2\right)$ $\displaystyle =$ $\displaystyle (1+\lambda^2)
\left(v_{0,j}^2+v_{1,j}^2+v_{N-1,j}^2+v_{N,j}^2
+w_{0,j}^2+w_{1,j}^2+w_{N-1,j}^2+w_{N,j}^2\right)$
      $\displaystyle +2(1+\lambda^2)\sum_{i=2}^{N-2}(v_{i,j}^2+w_{i,j}^2)$
      $\displaystyle +2(1-\lambda^2)\sum_{i=1}^{N-1}
(v_{i+1,j}v_{i-1,j}+w_{i+1,j}w_{i-1,j})$
      $\displaystyle +4\lambda
(v_{N,j}w_{N,j}+v_{N-1,j}w_{N-1,j}
-v_{1,j}w_{1,j}-v_{0,j}w_{0,j})$

特に $ \lambda =1$ の場合

  $\displaystyle \sum_{i=1}^{N-1}\left(v_{i,j+1}^2+w_{i,j+1}^2\right)$ $\displaystyle =$ $\displaystyle \frac{1}{2}
\left(v_{0,j}^2+v_{1,j}^2+v_{N-1,j}^2+v_{N,j}^2
+w_{0,j}^2+w_{1,j}^2+w_{N-1,j}^2+w_{N,j}^2\right)$
      $\displaystyle +\sum_{i=2}^{N-2}(v_{i,j}^2+w_{i,j}^2)$
      $\displaystyle +(v_{N,j}w_{N,j}+v_{N-1,j}w_{N-1,j}
-v_{1,j}w_{1,j}-v_{0,j}w_{0,j})$
    $\displaystyle =$ $\displaystyle \sum_{i=0}^{N}(v_{i,j}^2+w_{i,j}^2)$
      $\displaystyle -\frac{1}{2}
\left(v_{0,j}^2+v_{1,j}^2+v_{N-1,j}^2+v_{N,j}^2
+w_{0,j}^2+w_{1,j}^2+w_{N-1,j}^2+w_{N,j}^2\right)$
      $\displaystyle -(v_{1,j}w_{1,j}+v_{0,j}w_{0,j}-v_{N,j}w_{N,j}-v_{N-1,j}w_{N-1,j})$
    $\displaystyle =$ $\displaystyle \sum_{i=0}^{N}(v_{i,j}^2+w_{i,j}^2)$
      $\displaystyle -\frac{1}{2}
\left[(v_{0,j}+w_{0,j})^2+(v_{1,j}+w_{1,j})^2
+(v_{N-1,j}-w_{N-1,j})^2+(v_{N,j}-w_{N,j})^2
\right]$

それゆえ

$\displaystyle v_{0,j+1}^2+w_{0,j+1}^2+v_{N,j+1}^2+w_{N,j+1}^2=
\frac{1}{2}
\lef...
...j})^2+(v_{1,j}+w_{1,j})^2
+(v_{N-1,j}-w_{N-1,j})^2+(v_{N,j}-w_{N,j})^2
\right]
$

であればエネルギー保存則が成り立つ。

簡単のため同次 Dirichlet 境界条件の場合

$\displaystyle w_{0,j+1}^2+w_{N,j+1}^2=
\frac{1}{2}
\left[w_{0,j}^2+(v_{1,j}+w_{1,j})^2
+(v_{N-1,j}-w_{N-1,j})^2+w_{N,j}^2
\right]
$

$\displaystyle w_{0,j+1}^2
=\frac{1}{2}
\left[w_{0,j}^2+(v_{1,j}+w_{1,j})^2\right],\quad
w_{N,j+1}^2=
\frac{1}{2}
\left[(v_{N-1,j}-w_{N-1,j})^2+w_{N,j}^2
\right]
$


next up previous contents
Next: 参考文献 Up: 4. 桂田メモ Previous: 4.2 Friedrichs の差分法
Masashi Katsurada
平成14年11月29日