Bicubic interpolation

Remarks

The sixteen conditions determine the sixteen coefficients. Indeed, $p$ is the tensor product of one-dimensional cubic Hermite interpolation in $x$ and in $y$, so it exists and is unique. Example: for $f(x, y) = x^3 y^3$, the data are $f = f_x = f_y = 0$ at every corner except $(1, 1)$, where $f = 1$, $f_x = f_y = 3$ and $f_{xy} = 9$; the interpolant is $x^3 y^3$, since it has the required form. Non-example: the function $\sin(\pi x) \sin(\pi y)$ has zero value at all corners, but it is not a polynomial of the required form, so it is not a bicubic interpolant of any data. On a grid, along a shared edge the value and the normal derivative are cubics determined by data at the two common corners only. So adjacent patches join with continuous value and first derivatives. The variant used for images replaces the derivatives by differences of neighboring pixel values, or uses a cubic convolution kernel on a $4 \times 4$ neighborhood.