两点三次Hermite插值的求解过程

两点三次Hermite插值是指：已知被插函数$f(x)$在区间$[x_0,x_1]$的两个端点的函数值和导数值为

$$y_0=f(x_0),y_1=f(x_1),m_0=f^{\prime}(0),m_1=f^{\prime}(x_1)$$

需要寻找一个三次多项式$H_3(x)$满足

$$H_3(x_0)=y_0,H_3(x_1)=y_1,H_3^{\prime}(x_0)=m_0,H_3^{\prime}(x_1)=m_1$$

记区间$[x_0,x_1]$上三次多项式空间为$\mathbb{P}_3$，首先求其一组基函数$\{\alpha_0,\alpha_1,\beta_0,\beta_1\}$在$[x_0,x_1]$上的表达式，使之满足

$$\begin{aligned}&\alpha_{0}(x_0)=1&&\alpha_1(x_0)=0&&\beta_0(x_0)=0&&\beta_1(x_0)=0&\\&\alpha_0(x_1)=0&&\alpha_1(x_1)=1&&\beta_0(x_1)=0&&\beta_1(x_1)=0&\\&\alpha_0^{\prime}(x_0)=0&&\alpha_1^{\prime}(x_0)=0&&\beta_0^{\prime}(x_0)=1&&\beta_1(x_0)=0&\\&\alpha_0^{\prime}(x_1)=0&&\alpha_1^{\prime}(x_1)=0&&\beta_0^{\prime}(x_1)=0&&\beta_1^{\prime}(x_1)=1&\end{aligned}$$

显然基函数一定都是三次多项式。

由$\alpha_0(x_1)=0$和$\alpha_0^{\prime}(x_1)=0$，有$(x-x_1)^2 \mid \alpha_0(x)$，则$\alpha_0(x)=(ax+b)(x-x_1)^2$。

由$\alpha_0(x_0)=1$，有$(ax_0+b)(x_0-x_1)^2=1$；由$\alpha_0^{\prime}(x_0)=0$，有$a(x_0-x_1)+2(ax+b)=0$。

于是，$a=-\frac{2}{(x_0-x_1)^3},b=\frac{3x_0-x_1}{(x_0-x_1)^3}$，经调整后得

$$\alpha_0(x)=(1+2\frac{x-x_0}{x_1-x_0})(\frac{x-x_1}{x_0-x_1})^2=(1+2l_1(x))\cdot l_0^2(x)$$

同理有

$$\alpha_1(x) = (1+2\frac{x-x_1}{x_0-x_1})(\frac{x-x_0}{x_1-x_0})^2=(1+2l_0(x))\cdot l_1^2(x)$$

由$\beta_0(x_1)=0$和$\beta_0^{\prime}(x_1)=0$，有$(x-x_1)^2 \mid \beta_0(x)$；由$\beta_0(x_0) = 0$，有$(x-x_0) \mid \beta_0(x)$，因此有$\beta_0(x)=c(x-x_0)(x-x_1)^2$。

由$\beta_0^{\prime}(x_0)=1$，有$c=\frac{1}{(x_0-x_1)^2}$，于是有

$$\beta_0(x)=(x-x_0)(\frac{x-x_1}{x_0-x_1})^2=(x-x_0)\cdot l_0^2(x)$$

同理有

$$\beta_1(x)=(x-x_1)(\frac{x-x_0}{x_1-x_0})^2=(x-x_1)\cdot l_1^2(x)$$

因此

$$H_3(x)=y_0\alpha_0(x)+y_1\alpha_1(x)+m_0\beta_0(x)+m_1\beta_1(x)$$