FDM-2:差分推导

1. 泰勒展开公式：

前向泰勒展开:

\[u(x+\Delta x) = u(x) + \Delta x \frac{\partial u}{\partial x} + \frac{\Delta x^2}{2} \frac{\partial^2 u}{\partial x^2} + \frac{\Delta x^3}{6} \frac{\partial^3 u}{\partial x^3} + \frac{\Delta x^4}{24} \frac{\partial^4 u}{\partial x^4} + \mathcal{O}(\Delta x^5)\]

后向泰勒展开:

\[u(x-\Delta x) = u(x) - \Delta x \frac{\partial u}{\partial x} + \frac{\Delta x^2}{2} \frac{\partial^2 u}{\partial x^2} - \frac{\Delta x^3}{6} \frac{\partial^3 u}{\partial x^3} + \frac{\Delta x^4}{24} \frac{\partial^4 u}{\partial x^4} + \mathcal{O}(\Delta x^5).\]

2. 一阶导数中心差分 \(\frac{\partial u}{\partial x}\)：

将两式相减消去偶数阶导数项：

\[u(x+\Delta x) - u(x-\Delta x) = 2 \Delta x \frac{\partial u}{\partial x} + \frac{\Delta x^3}{3} \frac{\partial^3 u}{\partial x^3} + \mathcal{O}(\Delta x^5)\]

解出一阶导数：

\[\frac{\partial u}{\partial x} = \frac{u(x+\Delta x) - u(x-\Delta x)}{2\Delta x} + \frac{\Delta x^2}{6} \frac{\partial^3 u}{\partial x^3} + \mathcal{O}(\Delta x^4)\]

即：

\[\frac{\partial u}{\partial x} = \frac{u(x+\Delta x) - u(x-\Delta x)}{2\Delta x} + \mathcal{O}(\Delta x^2)\]

3. 二阶导数中心差分 \(\frac{\partial^2 u}{\partial x^2}\)：

将两式相加消去奇数阶导数项：

\[u(x+\Delta x) + u(x-\Delta x) = 2u(x) + \Delta x^2 \frac{\partial^2 u}{\partial x^2} + \frac{\Delta x^4}{12} \frac{\partial^4 u}{\partial x^4} + \mathcal{O}(\Delta x^6).\]

解出二阶导数：

\[\frac{\partial^2 u}{\partial x^2} = \frac{u(x+\Delta x) - 2u(x) + u(x-\Delta x)}{\Delta x^2} + \frac{\Delta x^2}{12} \frac{\partial^4 u}{\partial x^4} + \mathcal{O}(\Delta x^4).\]

即：

\[\frac{\partial^2 u}{\partial x^2} = \frac{u(x+\Delta x) - 2u(x) + u(x-\Delta x)}{\Delta x^2} + \mathcal{O}(\Delta x^2)\]

4. 离散格式（网格表示）：

一阶导数离散:

\[\left. \frac{\partial u}{\partial x} \right|_{i,j} = \frac{u_{i+1,j} - u_{i-1,j}}{2\Delta x}\]

二阶导数离散:

\[\left. \frac{\partial^2 u}{\partial x^2} \right|_{i,j} = \frac{u_{i+1,j} - 2u_{i,j} + u_{i-1,j}}{\Delta x^2}\]

• 实际计算中，\(\mathcal{O}(\Delta x^2)\) 提醒我们：网格尺寸 \(\Delta x\) 减半，误差将减少约 4 倍。