A sufficient monotonicity square for cubic Hermite interpolation

For $d\mathrm{\ne }0$, differentiating [F1] by [F3], using $y_1-y_0=hd$ and $dt\mathrm{/}dx=1\mathrm{/}h$, gives $p^{\mathrm{\prime }}(x)\mathrm{/}d=\alpha +(6-4\alpha -2\beta )t+(3\alpha +3\beta -6)t^2$. Indeed, $H_{01}^{\mathrm{\prime }}=6t-6t^2$, $H_{10}^{\mathrm{\prime }}=1-4t+3t^2$, and $H_{11}^{\mathrm{\prime }}=-2t+3t^2$; substitution produces those coefficients. Direct substitution in [F1] also gives $p(x_0)=y_0$ and $p(x_1)=y_1$.

For $0\le t\le 1$ put ${\lambda }_{00}=(1-\alpha \mathrm{/}3)(1-\beta \mathrm{/}3)$, ${\lambda }_{30}=(\alpha \mathrm{/}3)(1-\beta \mathrm{/}3)$, ${\lambda }_{03}=(1-\alpha \mathrm{/}3)(\beta \mathrm{/}3)$, and ${\lambda }_{33}=(\alpha \mathrm{/}3)(\beta \mathrm{/}3)$. These are nonnegative and sum to one. The polynomial in step 1.1 is affine in $(\alpha ,\beta )$, so it equals the convex combination with these weights of its values at $(0,0),(3,0),(0,3),(3,3)$, namely $6t(1-t)$, $3(1-t{)}^2$, $3t^2$, and $3(1-2t{)}^2$. Each of these four polynomials is nonnegative on $[0,1]$.

step 2.1 gives $p^{\mathrm{\prime }}(x)\mathrm{/}d\ge 0$ throughout the cell. If $d>0$, $p^{\mathrm{\prime }}\ge 0$; if $d<0$, $p^{\mathrm{\prime }}\le 0$. For any two points $u<v$ in the cell, [F2] transfers that derivative sign to $p(v)-p(u)$, so $p$ is monotone; step 1.1 gives its endpoint values $y_0,y_1$, so every value lies between them. If $d=0$, then $y_1=y_0$ and [F1] reduces to $p=y_0(H_{00}+H_{01})=y_0$, since $m_0=m_1=0$ and $H_{00}+H_{01}=1$.