# Does not-a-knot cubic spline interpolation always have a tridiagonal system of equations when solving for the second derivatives?

I've been looking at the question Not-a-knot cubic spline interpolation using tridiagonal solver, for much the same reasons as the original asker - to figure out how to implement not-a-knot spline interpolation using a tridiagonal solver.

The definitions from that answer/question are as follows: $$i=0,1,...,n-2,n-1$$ $$h_i = x_{i+1} - x_i$$ $$\delta_i=\dfrac{y_{i+1}^{\prime\prime}-y_i^{\prime\prime}}{6h_i}$$ $$s_i=\dfrac{y_{i+1}-y_i}{x_{i+1}-x_i}$$

While the answer is very helpful and makes a lot of sense, I'm running into an issue when the answerer solved for the second derivatives instead of the first, which uses the equation: $$S_i(x)=y_i+\beta_i\left(x-x_i\right)+\frac{y_i^{\prime\prime}}{2}\left(x-x_i\right)^2+\delta_i\left(x-x_i\right)^3$$ As they proved in their answer, by combining $$h_0 y_0^{\prime\prime}+2(h_0+h_1)y_1^{\prime\prime}+h_1 y_2^{\prime\prime}=6(s_1-s_0)$$ with the not-a-knot boundary conditions, you get the following equation: $$(h_0^2-h_1^2)y_0^{\prime\prime}+(2h_0^2+3h_0 h_1+h_1^2)y_1^{\prime\prime}=6h_0(s_1-s_0)$$ However, in the case that the first three points are equally spaced, or that $$h_0 = h_1$$, then the equation resolves to $$hy_1'' = (s_1 - s_0)$$ where $$h=h_0=h_1$$. It's possible to get a second equation by substituting the value of $$y_1''$$ into one of the earlier equations: $$hy_0''+hy_2'' = 4(s_1-s_0)$$

While I think these equations would fully define the system, they aren't tridiagonal, correct? Is there a tridiagonal solution when solving for the second derivatives, under the circumstances $$h_0=h_1$$ (or $$h_{n-2}=h_{n-1}$$)?

$$\begin{pmatrix}0&1&0\\1&2&1\\0&1&0\end{pmatrix}$$