I am trying to simulate the image warp effect, that is used in Adobe Photoshop.
The rectangular image is warped according to a cubic Bézier surface (in 2D, all Z components are 0). Having any Bézier surface, vertical distortion $d \in[0,1]$ can be applied to it.
Left: input bézier surface, $d=0$, Right: output surface, $d=0.8$
Do you have any idea, what is done to the Bézier surface (16 points), when converting from the version on the left to the output on the right?
Clearly the $y$-coordinates of the Bezier patch control points are being left unchanged, and the $x$-values are being "tapered". When I say "tapered", I mean that the upper edge of the patch is being shrunk inwards, the lower edge is being expanded outwards, and the mid-height curve is being left unchanged.
I don't really know what the "$d=0.8$" value means. You say $d=0$ corresponds to no tapering, and I'd guess that $d=1$ is very strong tapering, where the top edge of the patch gets shrunk to a point (which means that the lower edge will double in width).
Let's assume that the left-hand patch has corners at $(-1,-1)$, $(-1,1)$, $(1,-1)$, $(1,1)$. If my guess about $d$ is correct, then the required transformation is: $$ T(x,y) = \big( (1-dy)x, y\big) $$ It's easy to confirm that: $$ T(x,0) = (x,0) \quad \text{for all } x $$ $$ T(1,-1) = (1+d, -1) $$ $$ T(1,1) = (1-d, 1) $$ and so on.