Coefficient Matrix Of Cubic Spline Interpolation.

[SherlockOhms]

Homework Statement

I'm trying to derive the coefficient matrix (a) of a parabolically terminated cubic spline. This is the matrix of coefficients $a_i \rightarrow a_n$ where n is the number of data points provided.

With this matrix you can find all the other coefficients (b and c) that define the spline using: $$b_i = \frac{y_i}{h_i} - a_i h_i$$ where $y_i$ is the corresponding value of the given set of y values and $h_i$ is defined as $x_{i + 1} - x_i$.

The $a$ coefficients (n in total) satisfy the following n - 2 linear equations: $$h_1 a_i + 2(h_{i + 1} + h_i)a_{i + 1} + h_{i + 1}a_{i + 1} = \frac{y_{i + 2} - y_{i + 1}}{h_{i + 1}} - \frac{y_{i + 1} - y_i}{h_i}$$

So, you have n coefficients and n - 2 equations. Therefore, you need 2 extra equations to be able to solve the system.

For a natural cubic spline these end conditions are $a_1$ = 0 and $a_n$ = 0. So, you can then write the system in the form: $Ma = d$. Where, $M$ is a tridiagonal matrix of $h_i$ along it's sub diagonal (n - 1 in length), $h_{i + 1}$ along it's super diagonal (n - 1 in length) and $2(h_{i + 1} + h_i)$ along it's main diagonal (n in length). Thus an n x n matrix is produced. Then $a$ is the coefficient matrix of $a_i \rightarrow a_n$ (n x 1) and $d$ is the n x 1 vector filled with $\frac{y_{i + 2} - y_{i + 1}}{h_{i + 1}} - \frac{y_{i + 1} - y_i}{h_i}$. Incorporating the extra conditions in, the first and last rows of matrix M will be 1 0 0 ... and ...0 0 0 1 respectively. As well as the first and last rows of the vector $d$ being 0. This ensures that $a_1$ and $a_n$ = 0.

So, my question is, when writing this linear system for a parabolically terminated cubic spline in matrix form, who's end conditions are $-a_1 + a_2 = 0$ and $-a_{n-1} + a_n = 0$, how do the first and last rows of M and d change? I've outlined above how the matrices are set up for the given conditions $a_1$ = 0 and $a_n$ = 0, and I just wnat to know how they change with these new conditions.

Homework Equations

All above.

The Attempt at a Solution

All above.

 

[Ray Vickson]

Homework Statement

I'm trying to derive the coefficient matrix (a) of a parabolically terminated cubic spline. This is the matrix of coefficients $a_i \rightarrow a_n$ where n is the number of data points provided.

With this matrix you can find all the other coefficients (b and c) that define the spline using: $$b_i = \frac{y_i}{h_i} - a_i h_i$$ where $y_i$ is the corresponding value of the given set of y values and $h_i$ is defined as $x_{i + 1} - x_i$.

The $a$ coefficients (n in total) satisfy the following n - 2 linear equations: $$h_1 a_i + 2(h_{i + 1} + h_i)a_{i + 1} + h_{i + 1}a_{i + 1} = \frac{y_{i + 2} - y_{i + 1}}{h_{i + 1}} - \frac{y_{i + 1} - y_i}{h_i}$$

So, you have n coefficients and n - 2 equations. Therefore, you need 2 extra equations to be able to solve the system.

For a natural cubic spline these end conditions are $a_1$ = 0 and $a_n$ = 0. So, you can then write the system in the form: $Ma = d$. Where, $M$ is a tridiagonal matrix of $h_i$ along it's sub diagonal (n - 1 in length), $h_{i + 1}$ along it's super diagonal (n - 1 in length) and $2(h_{i + 1} + h_i)$ along it's main diagonal (n in length). Thus an n x n matrix is produced. Then $a$ is the coefficient matrix of $a_i \rightarrow a_n$ (n x 1) and $d$ is the n x 1 vector filled with $\frac{y_{i + 2} - y_{i + 1}}{h_{i + 1}} - \frac{y_{i + 1} - y_i}{h_i}$. Incorporating the extra conditions in, the first and last rows of matrix M will be 1 0 0 ... and ...0 0 0 1 respectively. As well as the first and last rows of the vector $d$ being 0. This ensures that $a_1$ and $a_n$ = 0.

So, my question is, when writing this linear system for a parabolically terminated cubic spline in matrix form, who's end conditions are $-a_1 + a_2 = 0$ and $-a_{n-1} + a_n = 0$, how do the first and last rows of M and d change? I've outlined above how the matrices are set up for the given conditions $a_1$ = 0 and $a_n$ = 0, and I just wnat to know how they change with these new conditions.

Homework Equations

All above.

The Attempt at a Solution

All above.

You have already said how the first and last row change: instead of a1 = 0 you have a1-a2=0, and instead of a_n = 0 you have a_(n-1)-a_n = 0.

 

[SherlockOhms]

Think I've already got it sorted. Thanks for the reply.