Interpolation

Hermite's Interpolating Polynomials

Here, we implement a two point Hermite interpolation function which passes through the function and its first derivative for the interval $[0,1]$. The polynomial is determined by four constraint equations, matching the function and its derivative at the points $0$ and $1$.

Start by defining the 3rd degree polynomial and its derivative by

\[\begin{aligned} g(x) &= a_0 + a_1 x + a_2 x^2 + a_3 x^3 , \\ g'(x) &= a_1 + 2 a_2 x + 3 a_3 x^2 , \end{aligned}\]

and apply the constraints

\[\begin{aligned} g(0) &= f_0 & & \Rightarrow & a_0 &= f_0 , \\ g(1) &= f_1 & & \Rightarrow & a_0 + a_1 + a_2 + a_3 &= f_1 , \\ g'(0) &= f'_0 & & \Rightarrow & a_1 &= f'_0 , \\ g'(1) &= f'_1 & & \Rightarrow & a_1 + 2 a_2 + 3 a_3 &= f'_1 . \\ \end{aligned}\]

Solving for $a_0, a_1, a_2, a_3$ leads to

\[\begin{aligned} a_0 &= f_0 , & a_1 &= f'_0 , & a_2 &= - 3 f_0 + 3 f_1 - 2 f'_0 - f'_1 , & a_3 &= 2 f_0 - 2 f_1 + f'_0 + f'_1 , \end{aligned}\]

so that the polynomial $g(x)$ reads

\[g(x) = f_0 + f'_0 x + (- 3 f_0 + 3 f_1 - 2 f'_0 - f'_1) x^2 + (2 f_0 - 2 f_1 + f'_0 + f'_1) x^3 .\]

The function and derivative values can be factored out, so that $g(x)$ can be rewritten as

\[g(x) = f_0 (1 - 3 x^2 + 2 x^3) + f_1 (3 x^2 - 2 x^3) + f'_0 (x - 2 x^2 + x^3) + f'_1 (- x^2 + x^3) ,\]

or in generic form as

\[g(x) = f_0 a_0(x) + f_1 a_1(x) + f'_0 b_0(x) + f'_1 b_1(x) ,\]

with basis functions

\[\begin{aligned} a_0 (x) &= 1 - 3 x^2 + 2 x^3 , & b_0 (x) &= x - 2 x^2 + x^3 , \\ a_1 (x) &= 3 x^2 - 2 x^3 , & b_1 (x) &= - x^2 + x^3 . \end{aligned}\]

The derivative $g'(x)$ accordingly reads

\[g'(x) = f_0 a'_0(x) + f_1 a'_1(x) + f'_0 b'_0(x) + f'_1 b'_1(x) ,\]

with

\[\begin{aligned} a'_0 (x) &= - 6 x + 6 x^2 , & b'_0 (x) &= 1 - 4 x + 3 x^2 , \\ a'_1 (x) &= 6 x - 6 x^2 , & b'_1 (x) &= - 2 x + 3 x^2 . \end{aligned}\]

The basis functions $a_0$and $a_1$ are associated with the function values at $x_0$ and $x_1$, respectively, while the basis functions $b_0$ and $b_1$ are associated with the derivative values at $x_0$ and $x_1$. The basis functions satisfy the following relations,

\[\begin{aligned} a_i (x_j) &= \delta_{ij} , & b_i (x_j) &= 0 , & a'_i (x_j) &= 0 , & b'_i (x_j) &= \delta_{ij} , & i,j &= 0, 1 , \end{aligned}\]

where $\delta_{ij}$ denotes the Kronecker-delta, so that

\[\begin{aligned} g(0) &= f_0 , & g(1) &= f_1 , & g'(0) &= f'_0 , & g'(1) &= f'_1 . \end{aligned}\]