1
$\begingroup$

enter image description here

Thoughts: Since we have information up to the second derivative, we could use linear, quadratic, and cubic splines to solve the problem. I am stumped on how to find the coefficients, however.

(More thoughts: A linear spline is just connected line segments, so the coeffs would come from a line equation of sorts... right?)

$\endgroup$
1
$\begingroup$

You need a $5$th-order polynomial in every segment $[x_i,x_{i+1}]$ to fit the function values $f_i$, $f_{i+1}$, first-order derivatives $f_i'$, $f_{i+1}'$, and second-order derivatives $f_i''$, $f_{i+1}''$. To find the coefficients quickly, we need a good basis. For example, it's easy to find a linear function

$$y_1(x)=f_i+\frac{f_{i+1}-f_i}{x_{i+1}-x_i}(x-x_i)$$

that produces the correct function values $f_i,f_{i+1}$ at $x_i,x_{i+1}$. To get the derivatives right, we add two more cubic functions

$$y_3(x)=y_1(x)+a_3(x-x_i)(x-x_{i+1})^2+b_3(x-x_i)^2(x-x_{i+1}).$$

The good thing about the basis is that neither of them change the function values $f_i,f_{i+1}$, and that $a_3(x-x_i)(x-x_{i+1})^2$ has zero derivative at $x_{i+1}$ and $b_3(x-x_i)^2(x-x_{i+1})$ has zero derivative at $x_i$. So the coefficients can be determined independently from the derivatives $f_i',f_{i+1}'$. Once $y_3(x)$ is found, we find the $5$th-degree polynomial

$$y_5(x)=y_3(x)+a_5(x-x_i)^2(x-x_{i+1})^3+b_5(x-x_i)^3(x-x_{i+1})^2$$

in the same manner. The already satisfied conditions $f_i,f_{i+1}$ and $f_i',f_{i+1}'$ are not affected. The $a_5$ term has zero second-order derivative at $x_{i+1}$ and the $b_5$ term has zero second-order derivative at $x_i$. So the coefficients $a_5,b_5$ can again be determined from $f_i''$ and $f_{i+1}''$ independently.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.