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Interpolation, C1 matching, and cubic reproduction for Hermite pieces

Statement

For the data in Piecewise cubic Hermite interpolant, each piece has the prescribed endpoint values and slopes: $P_{i} (x_{i}) = y_{i}$ , $P_{i} (x_{i + 1}) = y_{i + 1}$ , $P_{i}^{'} (x_{i}) = m_{i}$ , and $P_{i}^{'} (x_{i + 1}) = m_{i + 1}$ . The resulting interpolant is $C^{1}$ on $[x_{0}, x_{N}]$ (with the corresponding one-sided derivatives at the endpoints). On a single interval, if the four prescribed endpoint data are the values and derivatives of a polynomial $q$ of degree at most three, then $P_{i} = q$ on that interval.

Facts & Assumptions

Given: Strictly increasing real nodes, values, and prescribed slopes as in Piecewise cubic Hermite interpolant.

[F1]

Piecewise cubic Hermite interpolant: on a cell, $P_{i} = y_{i} H_{00} (t) + y_{i + 1} H_{01} (t) + h_{i} m_{i} H_{10} (t) + h_{i} m_{i + 1} H_{11} (t)$ with $t = (x - x_{i}) / h_{i}$ and the four displayed cubic basis polynomials.

[F2]

Sum, product, quotient, and chain rules from difference quotients: for every integer $m \geq 1$ , $(x^{m})^{'} = m x^{m - 1}$ .

Proof

1.1F1algebra

The basis values are $H_{00} (0) = 1$ , $H_{01} (0) = H_{10} (0) = H_{11} (0) = 0$ , and $H_{01} (1) = 1$ , $H_{00} (1) = H_{10} (1) = H_{11} (1) = 0$ . Substitution into [F1] gives $P_{i} (x_{i}) = y_{i}$ and $P_{i} (x_{i + 1}) = y_{i + 1}$ .

1.2F1F2algebra

Differentiating the displayed basis using [F2] gives $(H_{00}^{'}, H_{01}^{'}, H_{10}^{'}, H_{11}^{'}) (0) = (0, 0, 1, 0)$ and $(H_{00}^{'}, H_{01}^{'}, H_{10}^{'}, H_{11}^{'}) (1) = (0, 0, 0, 1)$ . Since $d t / d x = 1 / h_{i}$ , differentiation of [F1] therefore gives $P_{i}^{'} (x_{i}) = m_{i}$ and $P_{i}^{'} (x_{i + 1}) = m_{i + 1}$ .

2.1step 1.1step 1.2

At a shared node $x_{i}$ , the left and right pieces both have value $y_{i}$ by step 1.1 and both have derivative $m_{i}$ by step 1.2. Thus the piecewise function and its first derivative match there; the one-sided endpoint statements follow from the same step 1.2 calculation.

3.1F1F2algebra∎

Write a degree-at-most-three polynomial in the normalized coordinate as $q (t) = a_{0} + a_{1} t + a_{2} t^{2} + a_{3} t^{3}$ . For $q = 1$ , its endpoint values are $(1, 1)$ and scaled slopes are $(0, 0)$ ; for $q = t$ , they are $(0, 1)$ and $(1, 1)$ ; for $q = t^{2}$ , they are $(0, 1)$ and $(0, 2)$ ; and for $q = t^{3}$ , they are $(0, 1)$ and $(0, 3)$ , by [F2] and the chain rule under $t = (x - x_{i}) / h_{i}$ . Substitution in [F1] gives $H_{00} + H_{01} = 1$ , $H_{01} + H_{10} + H_{11} = t$ , $H_{01} + 2 H_{11} = t^{2}$ , and $H_{01} + 3 H_{11} = t^{3}$ . The formula is linear in endpoint values and slopes, so it reproduces their linear combination $q$ .

Depends on

Used by

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Sources

Carl R. de Boor, A Practical Guide to Splines, revised edition, Chapter IV, pp. 39–40 (standard reference, not scraped)