Caiyou Yuan / 离散分部求和举例

Created Tue, 13 Oct 2020 15:55:59 +0800 Modified Tue, 14 Apr 2026 15:04:31 +0800
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定义离散算子, $$ \mathcal{L}_h^x u_{i,j} = \frac{2}{x_{i+1} - x_{i-1}} \left( \frac{u_{i+1,j} - u_{i,j}}{x_{i+1} - x_{i}} -\frac{u_{i,j} - u_{i-1,j}}{x_{i} - x_{i-1}} \right) $$

$$ \mathcal{L}_h^y u_{i,j} = \frac{2}{y_{j+1} - y_{j-1}} \left(\frac{u_{i,j+1} - u_{i,j}}{y_{j+1} - y_{j}} - \frac{u_{i,j} - u_{i,j-1}}{y_{j} - y_{j-1}} \right) $$

定义离散内积\((\cdot, \cdot)_h\),

$$(u, v)_h = \sum_{i=1}^{N-1}\sum_{j=1}^{N-1}u_{i,j}v_{i,j} \left(\frac{x_{i+1} - x_{i-1}}{2}\right) \left(\frac{y_{j+1} - y_{j-1}}{2}\right)$$

故, $$ \begin{aligned} (-\mathcal{L}_h^x u, u)_h & = -\sum_{i=1}^{N-1}\sum_{j=1}^{N-1} \left(\frac{u_{i+1,j} - u_{i,j}}{x_{i+1} - x_{i}}- \frac{u_{i,j} - u_{i-1,j}}{x_{i} - x_{i-1}}\right)u_{i,j} \left(\frac{y_{j+1} - y_{j-1}}{2}\right)\\ &= -\sum_{j=1}^{N-1}\left(\frac{y_{j+1} - y_{j-1}}{2}\right) \sum_{i=1}^{N-1} \left(\frac{u_{i+1,j} - u_{i,j}}{x_{i+1} - x_{i}}- \frac{u_{i,j} - u_{i-1,j}}{x_{i} - x_{i-1}}\right)u_{i,j} \end{aligned} $$

注意到, $$ \begin{aligned} \sum_{i=1}^{N-1} \left(\frac{u_{i+1,j} - u_{i,j}}{x_{i+1} - x_{i}}- \frac{u_{i,j} - u_{i-1,j}}{x_{i} - x_{i-1}}\right)u_{i,j} &= \sum_{i=1}^{N-1} \left(\frac{u_{i+1,j} - u_{i,j}}{x_{i+1} - x_{i}}\right)u_{i,j}- \sum_{i=1}^{N-1} \left(\frac{u_{i,j} - u_{i-1,j}}{x_{i} - x_{i-1}}\right)u_{i,j}\\ &= \sum_{i=1}^{N-1} \left(\frac{u_{i+1,j} - u_{i,j}}{x_{i+1} - x_{i}}\right)u_{i,j}- \sum_{i=0}^{N-2} \left(\frac{u_{i+1,j} - u_{i,j}}{x_{i+1} - x_{i}}\right)u_{i+1,j}\\ &= -\sum_{i=1}^{N-1} \frac{(u_{i+1,j} - u_{i,j})^2}{x_{i+1} - x_{i}} -\frac{(u_{1,j} - u_{0,j})}{x_{1} - x_{0}}u_{1,j} +\frac{(u_{N,j} - u_{N-1,j})}{x_{N} - x_{N-1}}u_{N,j} \end{aligned} $$

结合边界条件 \(u_{0,j} = u_{N, j} = 0\), 结果即简化为 $$ \sum_{i=1}^{N-1} \left(\frac{u_{i+1,j} - u_{i,j}}{x_{i+1} - x_{i}}- \frac{u_{i,j} - u_{i-1,j}}{x_{i} - x_{i-1}}\right)u_{i,j}= -\sum_{i=0}^{N-1} \frac{(u_{i+1,j} - u_{i,j})^2}{x_{i+1} - x_{i}} $$

故, $$ \begin{aligned} (-\mathcal{L}_h^x u, u)_h &= \sum_{j=1}^{N-1}\left(\frac{y_{j+1} - y_{j-1}}{2}\right) \sum_{i=0}^{N-1} \frac{(u_{i+1,j} - u_{i,j})^2}{x_{i+1} - x_{i}}\\ &= \sum_{i=0}^{N-1}\sum_{j=1}^{N-1} \frac{y_{j+1} - y_{j-1}}{2(x_{i+1} - x_{i})}(u_{i+1,j} - u_{i,j})^2 \end{aligned} $$

同理, $$ \begin{aligned} (-\mathcal{L}_h^y u, u)_h &= \sum_{i=1}^{N-1}\sum_{j=0}^{N-1} \frac{x_{i+1} - x_{i-1}}{2(y_{j+1} - y_{i})}(u_{i,j+1} - u_{i,j})^2 \end{aligned} $$

故, $$ \begin{aligned} (-\mathcal{L}_h u, u)_h &= (-\mathcal{L}_h^x u, u)_h + (-\mathcal{L}_h^y u, u)_h \\ &= \sum_{i=0}^{N-1}\sum_{j=1}^{N-1} \frac{y_{j+1} - y_{j-1}}{2(x_{i+1} - x_{i})}(u_{i+1,j} - u_{i,j})^2+ \sum_{i=1}^{N-1}\sum_{j=0}^{N-1} \frac{x_{i+1} - x_{i-1}}{2(y_{j+1} - y_{i})}(u_{i,j+1} - u_{i,j})^2 \end{aligned} $$