Can we consider W(x;0) to be identity warp because it's $\Delta \boldsymbol{p} = 0$ (not $ \boldsymbol{p} = 0 $)
mpotoole
We are assuming that our warping function $W$ has the property that $W(x;0) = x$ (which is our identity warp).
For an affine warp $W_x(x;p) = p_1 x + p_2 y + p_3$ and $W_y(x;p) = p_4 x + p_5 y + p_6$, it's clear that $W(x;0) = 0 \neq x$. But we can re-parameterize our affine warp such that the property $W(x;0) = x$ holds true, by using the following warp instead: $W_x(x;p) = (p_1 + 1) x + p_2 y + p_3$ and $W_y(x;p) = p_4 x + (p_5 + 1) y + p_6$.
Can we consider W(x;0) to be identity warp because it's $\Delta \boldsymbol{p} = 0$ (not $ \boldsymbol{p} = 0 $)
We are assuming that our warping function $W$ has the property that $W(x;0) = x$ (which is our identity warp).
For an affine warp $W_x(x;p) = p_1 x + p_2 y + p_3$ and $W_y(x;p) = p_4 x + p_5 y + p_6$, it's clear that $W(x;0) = 0 \neq x$. But we can re-parameterize our affine warp such that the property $W(x;0) = x$ holds true, by using the following warp instead: $W_x(x;p) = (p_1 + 1) x + p_2 y + p_3$ and $W_y(x;p) = p_4 x + (p_5 + 1) y + p_6$.