Cubic Spline Interpolation practice Going over practice problems for our final exam. I'm stuck on a problem involving cubic splines. In fact, I don't even know where to begin.
I need to find the natural cubic spline $S(t)$ at $t_0=0, t_1=1, t_2=2, t_3=3$ where $$S(t_0)=0\\ S(t_1)=2\\ S(t_2)=1\\ S(t_3)=0$$
As @ja72 has pointed out, there are 3 intervals, thus 3 splines. There are conditions we impose on these splines so that we get the desired curve. 
$$
\begin{align}
S_1(t_0) &= 0\\
S_1(t_1) &= 2\\
S_2(t_2) &= 1\\
S_3(t_3) &= 0\\
S_1(t_1) &= S_2(t_1)\\
S_2(t_2) &= S_3(t_2)\\
S_1'(t_1) &= S_2'(t_1)\\
S_2'(t_2) &= S_3'(t_2)\\
S_1''(t_1) &= S_2''(t_1)\\
S_2''(t_2) &= S_3''(t_2)\\
S_1''(t_0) &= 0\\
S_3''(t_3) &= 0
\end{align}
$$
Now, I can plug in the exact $t_j$ values and $S_i(t_j)$ values directly into these 12 equations and should wind up with a linear system, 12 coeffs (4 from each spline). 
So I went ahead and plugged all of the $t$ values into the condition equations. I was left with a 12 x 12 matrix. I used MatLab to solve the system and got the coefficients that @ja72 got in their final solution. My professor said we'll have to solve the matrices by hand. What is the best way to solve a 12 x 12 matrix this way?
 A: Typically a cubic spline is given for a set of points such that for each interval a cubic function is fitted to the points, with matching slopes and curvature, as well as the endpoints, have zero curvature (2nd derivative). This is the definition of a natural spline.
In your case, you have 4 points and 3 intervals. The number of unknowns is 12, as for each interval there are 4 unknowns (the cubic part).
Your constraints are:

*

*Points on given nodes (4 equations)

*Interval end points match (2 equations)

*Interval end slopes match (2 equations)

*Interval end curvature match (2 equations)

*End point cuvature zero (2 equations)

In total, you have 12 equations with 12 unknowns.
See http://www.physics.utah.edu/~detar/phys6720/handouts/cubic_spline/cubic_spline/node1.html for more details. Also wikipedia has a nice write-up.
For your case the solution is
$$ y(x)=\begin{cases}
\frac{2x(7-2x^{2})}{5} & x=0\ldots1\\
\frac{5x^{3}-27x^{2}+41x-9}{5} & x=1\ldots2\\
\frac{-x^{3}+9x^{2}-31x+39}{5} & x=2\ldots3
\end{cases} $$
Details
The general equation for a cubic spline is an interpolation using the values $Y_i$ and the 2nd derivatives $Y''_i$ at each node. Once the 2nd derivatives are known, then the spline is solved for and the following equation can be used for interpolation between nodes $i$ and $i+1$.
This is done with a parameter $t=0\ldots 1$ that spans the interval
$$\begin{aligned}
  x_i & = X_i + t\,h \\
  y_i & = (1-t) Y_i + y\,Y_{i+1} + \tfrac{h^2}{6} \left( (1-t) (t^2-2 t) Y''_i + t (t^2-1) Y''_{i+1} \right) \end{aligned} $$
The system of equations to find a spline is formed using the following procedure:

*

*Start Point Either the spline in a natural spline with $Y''_1 = 0$ or it has a specified slope $Y'_1$ at which point the following equation must be used
$$ Y''_2 + 2 Y''_1 = \tfrac{6}{h^2} \left( Y_2 - Y_1 - h Y'_1 \right) \tag{i = 1}$$


*Internal Points Here the slope needs to match at nodes and this is ensured with the following equation for $i=2 \ldots n-1$
$$ Y''_{i-1} + Y''_{i+1} + 4 Y''_i = \tfrac{6}{h^2} \left( Y_{i-1} + Y_{i+1} - 2 Y_i \right) \tag{i = 2 to n-1}$$


*End Point Again the spline is either natural with $Y''_n = 0$ or with the specified slope $Y'_n$, and the following equation must be used.
$$ Y''_{n-1} +2 Y''_n = \tfrac{6}{h^2} \left( Y_{n-1} - Y_n + h Y'_n \right) \tag{i = n}$$
For your example, I assume you want a natural spline so only the interior points are considered, resulting in these two equations
$$\begin{aligned}
  4 Y''_2 + Y''_3 & = \tfrac{6}{h^2} \left( Y_3 - 2 Y_2 + Y_1 \right) \\
  Y''_2 + 4 Y''_3 & = \tfrac{6}{h^2} \left( Y_4 - 2 Y_3 + Y_1 \right)
\end{aligned}$$
Using $h=1$ and $Y=[0,2,1,0]$ and solved for $Y''_2$ and $Y''_3$.
NOTE: I have assumed constant interval with $h = X_{i}-X_{i-1}$, but that above can be adapted to a varying grid also.
