计算一阶导数的四阶中心差分格式

本文地址：https://goodgoodstudy.blog.youkuaiyun.com/article/details/112008150

原理：利用指定点 $y_t$ 周围的四个点$(y_{t-2},y_{t-1},y_{t+2},y_{t+2})$构造 拉格朗日插值曲线：
$$\begin{array}{rl}y(x)=& y_{t-2}\frac{(x-x_{t-1})(x-x_{t+1})(x-x_{t+2})}{(x_{t-2}-x_{t-1})(x_{t-2}-x_{t+1})(x_{t-2}-x_{t+2})}\\ & \\ & +y_{t-1}\frac{(x-x_{t-2})(x-x_{t+1})(x-x_{t+2})}{(x_{t-1}-x_{t-2})(x_{t-1}-x_{t+1})(x_{t-1}-x_{t+2})}\\ & \\ & +y_{t+1}\frac{(x-x_{t-2})(x-x_{t-1})(x-x_{t+2})}{(x_{t+1}-x_{t-2})(x_{t+1}-x_{t-1})(x_{t+1}-x_{t+2})}\\ & \\ & +y_{t+2}\frac{(x-x_{t-2})(x-x_{t-1})(x-x_{t+1})}{(x_{t+2}-x_{t-2})(x_{t+2}-x_{t-1})(x_{t+2}-x_{t+1})}\end{array}$$

分母是可以直接写出来的，记 $h\equiv x_t-x_{t-1}$ 为自变量间隔：
$$\begin{array}{rl}y(x)=& y_{t-2}\frac{(x-x_{t-1})(x-x_{t+1})(x-x_{t+2})}{(-h)(-3h)(-4h)}\\ & \\ & +y_{t-1}\frac{(x-x_{t-2})(x-x_{t+1})(x-x_{t+2})}{(h)(-2h)(-3h)}\\ & \\ & +y_{t+1}\frac{(x-x_{t-2})(x-x_{t-1})(x-x_{t+2})}{(3h)(2h)(-h)}\\ & \\ & +y_{t+2}\frac{(x-x_{t-2})(x-x_{t-1})(x-x_{t+1})}{(4h)(3h)(h)}\\ & \\ =& [-y_{t-2}(x-x_{t-1})(x-x_{t+1})(x-x_{t+2})\\ & +2y_{t-1}(x-x_{t-2})(x-x_{t+1})(x-x_{t+2})\\ & -2y_{t+1}(x-x_{t-2})(x-x_{t-1})(x-x_{t+2})\\ & +y_{t+2}(x-x_{t-2})(x-x_{t-1})(x-x_{t+1})]\frac{1}{12h^3}\end{array}$$

接下来就是求该三次函数在 $x_t$ 处的导数 $y^{\prime }(x_t)$
y′(x)={−yt−2[(x−xt+1)(x−xt+2)+(x−xt−1)(x−xt+2)+(x−xt−1)(x−xt+1)]+2yt−1[(x−xt+1)(x−xt+2)+(x−xt−2)(x−xt+2)+(x−xt−2)(x−xt+1)]−2yt+1[(x−xt−1)(x−xt+2)+(x−xt−2)(x−xt+2)+(x−xt−2)(x−xt−1)]+yt+2[(x−xt−1)(x−xt+1)+(x−xt−2)(x−xt+1)+(x−xt−2)(x−xt−1)]}112h3 \begin{aligned} y'(x) =& \{ -y_{t-2}[(x-x_{t+1})(x-x_{t+2})+(x-x_{t-1})(x-x_{t+2})+(x-x_{t-1})(x-x_{t+1})]\\ \\ & + 2y_{t-1}[(x-x_{t+1})(x-x_{t+2})+(x-x_{t-2})(x-x_{t+2})+(x-x_{t-2})(x-x_{t+1})]\\ \\ & -2y_{t+1}[(x-x_{t-1})(x-x_{t+2})+(x-x_{t-2})(x-x_{t+2})+(x-x_{t-2})(x-x_{t-1})]\\ \\ & + y_{t+2}[(x-x_{t-1})(x-x_{t+1})+(x-x_{t-2})(x-x_{t+1})+(x-x_{t-2})(x-x_{t-1})] \}\frac{1}{12h^3} \end{aligned} y′(x)=​{−yt−2​[(x−xt+1​)(x−xt+2​)+(x−xt−1​)(x−xt+2​)+(x−xt−1​)(x−xt+1​)]+2yt−1​[(x−xt+1​)(x−xt+2​)+(x−xt−2​)(x−xt+2​)+(x−xt−2​)(x−xt+1​)]−2yt+1​[(x−xt−1​)(x−xt+2​)+(x−xt−2​)(x−xt+2​)+(x−xt−2​)(x−xt−1​)]+yt+2​[(x−xt−1​)(x−xt+1​)+(x−xt−2​)(x−xt+1​)+(x−xt−2​)(x−xt−1​)]}12h31​​
所以
y′(xt)={−yt−2[(xt−xt+1)(xt−xt+2)+(xt−xt−1)(xt−xt+2)+(xt−xt−1)(xt−xt+1)]+2yt−1[(xt−xt+1)(xt−xt+2)+(xt−xt−2)(xt−xt+2)+(xt−xt−2)(xt−xt+1)]−2yt+1[(xt−xt−1)(xt−xt+2)+(xt−xt−2)(xt−xt+2)+(xt−xt−2)(xt−xt−1)]+yt+2[(xt−xt−1)(xt−xt+1)+(xt−xt−2)(xt−xt+1)+(xt−xt−2)(xt−xt−1)]}112h3={−yt−2[(−h)(−2h)+(h)(−2h)+(h)(−h)]+2yt−1[(−h)(−2h)+(2h)(−2h)+(2h)(−h)]−2yt+1[(h)(−2h)+(2h)(−2h)+(2h)(h)]+yt+2[(h)(−h)+(2h)(−h)+(2h)(h)]}112h3=(yt−2−8yt−1+8yt+1−yt+2)112h \begin{aligned} y'(x_t) =& \{ -y_{t-2}[(x_t-x_{t+1})(x_t-x_{t+2})+(x_t-x_{t-1})(x_t-x_{t+2})+(x_t-x_{t-1})(x_t-x_{t+1})]\\ \\ & + 2y_{t-1}[(x_t-x_{t+1})(x_t-x_{t+2})+(x_t-x_{t-2})(x_t-x_{t+2})+(x_t-x_{t-2})(x_t-x_{t+1})]\\ \\ & -2y_{t+1}[(x_t-x_{t-1})(x_t-x_{t+2})+(x_t-x_{t-2})(x_t-x_{t+2})+(x_t-x_{t-2})(x_t-x_{t-1})]\\ \\ & + y_{t+2}[(x_t-x_{t-1})(x_t-x_{t+1})+(x_t-x_{t-2})(x_t-x_{t+1})+(x_t-x_{t-2})(x_t-x_{t-1})] \}\frac{1}{12h^3} \\ \\ =& \{ -y_{t-2}[(-h)(-2h)+(h)(-2h)+(h)(-h)]\\ \\ & + 2y_{t-1}[(-h)(-2h)+(2h)(-2h)+(2h)(-h)]\\ \\ & -2y_{t+1}[(h)(-2h)+(2h)(-2h)+(2h)(h)]\\ \\ & + y_{t+2}[(h)(-h)+(2h)(-h)+(2h)(h)] \}\frac{1}{12h^3} \\ \\ =& ( y_{t-2} -8y_{t-1} +8y_{t+1}- y_{t+2} )\frac{1}{12h} \\ \end{aligned} y′(xt​)===​{−yt−2​[(xt​−xt+1​)(xt​−xt+2​)+(xt​−xt−1​)(xt​−xt+2​)+(xt​−xt−1​)(xt​−xt+1​)]+2yt−1​[(xt​−xt+1​)(xt​−xt+2​)+(xt​−xt−2​)(xt​−xt+2​)+(xt​−xt−2​)(xt​−xt+1​)]−2yt+1​[(xt​−xt−1​)(xt​−xt+2​)+(xt​−xt−2​)(xt​−xt+2​)+(xt​−xt−2​)(xt​−xt−1​)]+yt+2​[(xt​−xt−1​)(xt​−xt+1​)+(xt​−xt−2​)(xt​−xt+1​)+(xt​−xt−2​)(xt​−xt−1​)]}12h31​{−yt−2​[(−h)(−2h)+(h)(−2h)+(h)(−h)]+2yt−1​[(−h)(−2h)+(2h)(−2h)+(2h)(−h)]−2yt+1​[(h)(−2h)+(2h)(−2h)+(2h)(h)]+yt+2​[(h)(−h)+(2h)(−h)+(2h)(h)]}12h31​(yt−2​−8yt−1​+8yt+1​−yt+2​)12h1​​

这就是计算一阶导数的四阶中心差分格式！