A [[Clamp|Clamped]] [[Spline]] $S(x)$ is one such that the [[Derivative]] of the two end points of the spline is constrained to given values, $y'_{\text{left}},y'_{\text{right}}$.
### Constraining given points & specific slopes
Given points $(x_{1},y_{1}), \dots,(x_{n}y_{n})$ we wish our spline to [[Interpolation|interpolate]] through, if we know $y_{i}'$ (the slope for each $x_{i}$), then each piecewise of the spline would be defined as:
$ \begin{align}
\huge
S_{i}(x) &= \frac{y'_{i}}{\Delta x_{i}^{2}}\pa{x-x_{i}}^{2}(x_{x_{i+1}})^{2} \\
&+
\frac{y_{i+1}'}{\Delta x_{i}^{2}}(x-x_{i})^{2}(x-x_{i+1}) \\
&+
\small
\frac{1}{2}(y_{i}+y_{i+1}) + m_{i}\pa{
x- \frac{1}{2}(x_{i}+x_{i+1})
}\pa{
1- \frac{2}{\Delta x_{i}^{2}}
(x-x_{i})(x-x_{i+1})}
\end{align}
$
Where $m_{i}= \frac{\Delta y_{i}}{\Delta x_{i}}$.
$\large \begin{align}
S_{i}''(x) &= \frac{y_{i}'}{\Delta x_{i}^{2}}\pa{6x-2x_{i}-4x_{i+1}} \\
&+ \frac{y'_{i+1}}{\Delta x_{i}^{2}}\pa{
6x-2x_{i+1}-4x_{i}
} \\
&- \frac{6m_{i}}{\Delta x_{i}^{2}}\pa{2x-(x_{i}+x_{i+1})}
\end{align}$
$\huge \begin{align}
S_{i}''(x_{i}) &= -\frac{4}{\Delta x_{i}} y_{i}' - \frac{2}{\Delta x_{i}}y_{i+1}' + \frac{6}{\Delta x_{i}} m_{i} \\
S_{i}''(x_{i+1}) &= \frac{2}{\Delta x_{i}} y_{i}' + \frac{4}{\Delta x_{i}}y_{i+1}' - \frac{6}{\Delta x_{i}} m_{i}
\end{align}$
Because we want our spline to be 2nd order [[Continuous]] $S\in\mathcal C^{2}$, we impose this restriction:
$\large \begin{align}
S''_{i-1}(x_{i} )&= S_{i}''(x_{i})\\
\small
\frac{1}{\Delta x_{i-1}}y'_{i-1} + \frac{2}{\Delta x_{i-1}}y_{i}' -
\frac{3}{\Delta x_{i}}m'_{i-1} &=
\small
- \frac{2}{\Delta x_{i}}y_{i}' - \frac{1}{\Delta x_{i}} y'_{i+1} + \frac{3}{\Delta x}m_{i}
\end{align}
$
Assuming we are working with [[Clamped Spline|Clamped Splines]],
$\begin{align}
\Delta x_{0}^{-1} y_{0}' + 2\Delta x_{0}^{-1} y_{1}' - 3\Delta x_{0}^{-1}m_{0}
&=
-2 \Delta x_{1}^{-1} y_{1}' - \Delta x_{1}^{-1}y_{2}' + 3 \Delta x_{1}^{-1}m_{1}
\\
\end{align}$
$\small
\begin{cases}
4 \Delta x_{0}^{-1}y_{1}' &+ \Delta x_{1}^{-1}y_{2}' &
&=
3\left(
\tiny
\Delta x_{0}^{-1} m_{0} + \Delta x_{1}^{-1} m_{1}
- \frac{1}{3}\Delta x_{0}^{-1} y'_{\text{left}}
\right)
\\
\Delta x_{1}^{-1} y_{1}' &+ 2(\Delta x_{1}^{-1} &+ \Delta x_{2}^{-1})y_{2}' + \Delta x_{2}^{-1}y_{3}'
&= 3\pa{\Delta x_{1}^{-1}m_{1}+ \Delta x_{2}^{-1}m_{2}} \\
\Delta x_{2}^{-1}y_{2}' &+ 2(\Delta x_{2}^{-1} + \Delta x_{3}^{-1}) y_{3}' &+ \Delta x_{3}^{-1}y_{4}'
&= 3\pa{ \Delta x_{2}^{-1}m_{2} + \Delta x_{3}^{-1}m_{3} }
\end{cases}
$
You can then solve this [[System of Linear Equations]].
>[!info]
>![[Pasted image 20260204154046.png|invert]]