A “Parabolic-ends” cubic spline is a spline whose end segments are second-degree curves or parabolas.
It is a special case of the “Fixed-third” spline, where the third derivatives at the endpoints are set to zero.
Conditions: $S_1'''(x_1) = 0, \ S_{n-1}'''(x_n) = 0$.
From these conditions, the following expressions for the coefficients are obtained:
- $6d_1 = 0$
- $6d_{n-1} = 0$
Thus:
$$ \boxed { d_1 = 0 } \tag{IV.1} $$$$ \boxed { d_{n-1} = 0 } \tag{IV.2} $$From (5), (VI.1):
$$ \boxed{ c_1 - c_2 = 0 } \tag{IV.3} $$From (5), (VI.2):
$$ \boxed{ -c_{n-1} + c_n = 0 } \tag{IV.4} $$Thus, we obtain two missing equations for the matrix:
- $c_1 - c_2 = 0$
- $-c_{n-1} + c_n = 0$
These are entered into the matrix:
$$ \begin{bmatrix} 1 & -1 & 0 & 0 & \dots & 0 & 0 \\ h_1 & 2h_1 + 2h_2 & h_2 & 0 & \dots & 0 & 0 \\ 0 & h_2 & 2h_2 + 2h_3 & h_3 & \dots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \dots & h_{n-2} & 2h_{n-2} + 2h_{n-1} & h_{n-1} \\ 0 & 0 & 0 & \dots & 0 & -1 & 1 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ c_3 \\ \vdots \\ c_{n-1} \\ c_n \end{bmatrix} = \begin{bmatrix} 0 \\ R_1 \\ R_2 \\ \vdots \\ R_{n-2} \\ 0 \end{bmatrix} $$where $R_k = 3\left(\frac{\delta_{k+1}}{h_{k+1}} - \frac{\delta_k}{h_k}\right)$.
The matrix can now be solved using the tridiagonal matrix algorithm.
After finding the coefficients $c_k$, the other coefficients $b_k$ and $d_k$ can be computed using formulas (6) and (5) respectively.
Special case: $n = 2$. #
In the case where there are two points (and therefore only one segment), the equations become identical:
$$ \begin{cases} c_1 - c_2 = 0 \\ -c_1 + c_2 = 0 \end{cases} \Rightarrow \begin{cases} c_1 - c_2 = 0 \\ c_1 - c_2 = 0 \end{cases} $$Thus, we get only one equation instead of two.
To resolve this situation, we equate the absolute values of $c_1$ and $c_2$ (which is fictitious).
Then $c_1$ and $c_2$ are equal to zero.