【数值计算】五次样条插值推导

函数形式

其中。

已知条件

经过的个点：，满足，，其中。

四阶及以下导数连续：，其中。

边界条件：，，，。

推导过程

改写一下的形式：

满足

$\begin{cases} p_{k}(x_{k}) = y_{k} \\ p_{k}(x_{k+1}) = y_{k+1} \\ p^{'}_{k}(x_{k}) = y^{'}_{k} \\ p^{'}_{k}(x_{k+1}) = y^{'}_{k+1} \\ p^{''}_{k}(x_{k}) = y^{''}_{k} \\ p^{''}_{k}(x_{k+1}) = y^{''}_{k+1} \\ \end{cases}$

则有

$\begin{cases} f_{k}(x_{k})=1,&f_{k+1}(x_{k})=0,&g_{k}(x_{k})=0,&g_{k+1}(x_{k})=0,&h_{k}(x_{k})=0,&h_{k+1}(x_{k})=0\\ f_{k}(x_{k+1})=0,&f_{k+1}(x_{k+1})=1,&g_{k}(x_{k+1})=0,&g_{k+1}(x_{k+1})=0,&h_{k}(x_{k+1})=0,&h_{k+1}(x_{k+1})=0\\ f^{'}_{k}(x_{k})=0,&f^{'}_{k+1}(x_{k})=0,&g^{'}_{k}(x_{k})=1,&g^{'}_{k+1}(x_{k})=0,&h^{'}_{k}(x_{k})=0,&h^{'}_{k+1}(x_{k})=0\\ f^{'}_{k}(x_{k+1})=0,&f^{'}_{k+1}(x_{k+1})=0,&g^{'}_{k}(x_{k+1})=0,&g^{'}_{k+1}(x_{k+1})=1,&h^{'}_{k}(x_{k+1})=0,&h^{'}_{k+1}(x_{k+1})=0\\ f^{''}_{k}(x_{k})=0,&f^{''}_{k+1}(x_{k})=0,&g^{''}_{k}(x_{k})=0,&g^{''}_{k+1}(x_{k})=0,&h^{''}_{k}(x_{k})=1,&h^{''}_{k+1}(x_{k})=0\\ f^{''}_{k}(x_{k+1})=0,&f^{''}_{k+1}(x_{k+1})=0,&g^{''}_{k}(x_{k+1})=0,&g^{''}_{k+1}(x_{k+1})=0,&h^{''}_{k}(x_{k+1})=0,&h^{''}_{k+1}(x_{k+1})=1 \end{cases}$

解得

$\begin{cases} f_{k}(x) = -6x^5+15x^4-10x^3+1\\ f_{k+1}(x) = 6x^5-15x^4+10x^3\\ g_{k}(x) = -3x^5+8x^4-6x^3+x\\ g_{k+1}(x) = -3x^5+7x^4-4x^3\\ h_{k}(x) = -0.5x^5+1.5x^4-1.5x^3+0.5x^2\\ h_{k+1}(x) = 0.5x^5-x^4+0.5x^3 \end{cases}$

则

$p_{k}(x) = (-6y_{k}+6y_{k+1}-3s_{k}y^{'}_{k}-3s_{k}y^{'}_{k+1}-0.5s^{2}_{k}y^{''}_{k}+0.5s^{2}_{k}y^{''}_{k+1})t^{5}_{k}+(15y_{k}-15y_{k+1}+8s_{k}y^{'}_{k}+7s_{k}y^{'}_{k+1}+1.5s^{2}_{k}y^{''}_{k}-s^{2}_{k}y^{''}_{k+1})t^{4}_{k}\\ +(-10y_{k}+10y_{k+1}-6s_{k}y^{'}_{k}-4s_{k}y^{'}_{k+1}-1.5s^{2}_{k}y^{''}_{k}+1.5s^{2}_{k}y^{''}_{k+1})t^{3}_{k}+0.5s^{2}_{k}y^{''}_{k}t^{2}_{k}+s_{k}y^{''}_{k}t_{k}+y_{k}$

其中：，。

由和，分别可推出以下两条等式

$-\frac{168}{s^{3}_{k-1}}y^{'}_{k-1} -(\frac{192}{s^{3}_{k-1}}+\frac{192}{s^{3}_{k}})y^{'}_{k} -\frac{168}{s^{3}_{k}}y^{'}_{k+1} -\frac{24}{s^{2}_{k-1}}y^{''}_{k-1} +(\frac{36}{s^{2}_{k-1}}-\frac{36}{s^{2}_{k}})y^{''}_{k}+\frac{24}{s^{2}_{k}}y^{''}_{k+1} =\frac{360}{s^{4}_{k-1}}y_{k-1} +(\frac{360}{s^{4}_{k}}-\frac{360}{s^{4}_{k-1}})y_{k} -\frac{360}{s^{4}_{s_k}}y_{k+1}$

$-\frac{24}{s^{2}_{k-1}}y^{'}_{k-1} +(\frac{36}{s^{2}_{k}}-\frac{36}{s^{2}_{k}})y^{'}_{k} +\frac{24}{s^{2}_{k}}y^{'}_{k+1} -\frac{3}{s^{2}_{k}}y^{''}_{k-1} +(\frac{9}{s_{k-1}}+\frac{9}{s_{k}})y^{''}_{k} -\frac{3}{s_{k}}y^{''}_{k+1} =\frac{60}{s^{3}_{k-1}}y_{k-1} -(\frac{60}{s^{3}_{k-1}}+\frac{60}{s^{3}_{k}})y_{k} +\frac{60}{s^{3}_{k}}y_{k+1}$

求解出、、、、、代入即为所求的五次样条解析式。

【数值计算】五次样条插值推导

函数形式

已知条件

推导过程

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