Cardinal cubic B-spline

A cardinal cubic B-spline is a degree-three Cardinal B-spline. At the three interior integer knots, its nonzero normalized values are $1/6$, $4/6$, and $1/6$.

For the centered Cardinal cubic B-spline $\beta$ with knot values $1/6,4/6,1/6$, write $s(x)={\sum }_{j=-1}^{m+1}{c}_j\beta (x/h-j)$ on $[0,mh]$. interpolation gives $c_{i-1}+4c_i+c_{i+1}=6u(ih)$ for $i=0,\dots ,m$. Since ${\beta }^{\prime \prime }(0)=-2$ and ${\beta }^{\prime \prime }(\pm 1)=1$, the natural endpoint conditions $s^{\prime \prime }(0)=s^{\prime \prime }(mh)=0$ give the displayed extra two rows. These supply the two equations missing from interpolation alone.

For knot spacing $h=1/2$ on $[0,1]$, the five translates indexed $-3,-2,-1,0,1$ span the $C^2$ piecewise cubic space. With normalized knot values $1/6,2/3,1/6$, nodal interpolation gives $(c_{-3}+4c_{-2}+c_{-1})/6=u(0)$ and the two shifted equations at $h,2h$. Left-end Hermite data give $(c_{-1}-c_{-3})/(2h)=u^{\prime }(0)$ and $(c_{-3}-2c_{-2}+c_{-1})/h^2=u^{\prime \prime }(0)$. These five equations uniquely determine the coefficients. Derivative data need existence or separate prescription; continuity of $u$ alone does not supply them.

For the unit-spaced cubic cardinal B-spline on $[0,4]$, the four scalar cubic Bernstein basis coefficient rows are $(0,0,0,1/6)$, $(1/6,1/3,2/3,2/3)$, $(2/3,2/3,1/3,1/6)$ and $(1/6,0,0,0)$. Each row applies on its interval with local parameter in $[0,1]$. They result from the compactly supported $C^2$ truncated-cube combination $\frac{1}{6}{\sum }_{j=0}^4(-1{)}^j(\genfrac{}{}{0ex}{}{4}{j})(x-j{)}_{+}^3$. Each graph piece is a cubic Bézier curve with abscissa controls $j,j+1/3,j+2/3,j+1$.

Sampling consecutive Cardinal cubic B-splines at the centers of their supports gives diagonal values $2/3$ and adjacent values $1/6$, with all other values zero. Finite truncation gives the displayed B-spline collocation matrix, including its endpoint rows without wrapping or adding boundary conditions. Its inverse has alternating signs. The exact inverse norm of midpoint cubic spline collocation is less than three for each finite dimension and approaches three as the dimension grows.

Conjugate the collocation matrix by the diagonal matrix with entries $(-1{)}^i$. The resulting negative-off-diagonal matrix has a nonnegative inverse, as a convergent Neumann series shows. The absolute inverse row sums are therefore the entries of its inverse applied to the all-ones vector. They solve $4u_i-u_{i-1}-u_{i+1}=6$ with $u_0=u_{n+1}=0$, whose solution is displayed. The maximum occurs at the central row or central two rows and equals the inverse operator norm on ${\ell }^{\infty }$. Consequently the associated spline interpolation operator has norm at most three independently of dimension.