Intro ...

Cubic Hermite Interpolation

Intro ...

We look for a curve

$$ c(t) = a_0 + a_1 t + a_2 t^2 + a_3 t^3 $$

that satisfies

$$ \begin{cases} c(0) = p_0, \quad c^{\prime}(0) = v_0 \\ c(1) = p_1, \quad c^{\prime}(1) = v_1 \end{cases}, $$

or we should solve the following system

$$ \begin{cases} p_0 = a_0 \\ v_0 = a_1 \\ p_1 = a_0 + a_1 + a_2 + a_3 \\ v_1 = a_1 + 2a_2 + 3a_3 \end{cases}. $$

which gives us the curve parameters we want in terms of the given data

$$\begin{cases} a_0 = p_0 \\ a_1 = v_0 \\ a_2 = 3p_1 - 3p_0 - 2v_0 - v_1 \\ a_3 = -2p_1 + 2p_0 + v_0 + v_1 \end{cases}. $$

If we rearrange the terms for $c(t)$ we get

$$ c(t) = (1 - 3t^2 + 2t^3)p_0 + (t - 2t^2 + t^3)v_0 + (-t^2 + t^3)v_1 + (3t^2 - 2t^3)p_1 $$

or in basis function form

$$ c(t) = H_0^3(t)p_0 + H_1^3(t)v_0 + H_2^3(t)v_1 + H_3^3(t)p_1, $$

where

$$ \begin{cases} H_0^3(t) = 1 - 3t^2 + 2t^3 \\ H_1^3(t) = t - 2t^2 + t^3 \\ H_2^3(t) = -t^2 + t^3 \\ H_3^3(t) = 3t^2 - 2t^3 \end{cases}. $$

Plotting the Cubic Hermite basis functions

Examples

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