Calculate the Derivative of Gradient Field of an Image

Question

I meet a confusing thing in image processing recently....

Assume the image $x \in \mathbb{R}^n$, with its derivative (difference) matrix: $D^+ = \begin{bmatrix} D_h \\ Dv \end{bmatrix} \in \mathbb{R}^{2n\times n}$ ($+$ means forward difference), also equal to $\nabla$. Therefore, it is natural to define the divergence: $\triangle = \nabla \cdot \nabla$.

I have seen some papers use $div = \triangle = D_h^-D_h^+ + D_v^-D_v^+ \in \mathbb{R}^{n\times n}$, where $-$ denotes the backward difference.

Here is my question: assume I want to calculate the $\frac{\partial\|\nabla x -p\|_2^2}{\partial x}$ where $p\in \mathbb{R}^{2n\times 1}$ is a vector not related to $x$, what is the result? I have seen some authors use $\nabla\cdot (\nabla x -p)$.

However, if writing the $\nabla$ as matrix form $D$ as I have introduced before, $D^T$ is exactly adjoint of gradient, not backward difference. Hence $-\triangle x$ would appear! So what is the right formula? Could anyone tell me?

Answer 1

Consider the expansion of the term below $$\|\nabla x -p\|_2^2 = (\nabla x -p)^T(\nabla x -p)$$ $$\|\nabla x -p\|_2^2 = (x^T\nabla^T -p^T)(\nabla x -p)$$ $$\|\nabla x -p\|_2^2 = (x^T\nabla^T\nabla x - x^T\nabla^T p - p^T\nabla x +p^Tp)$$

Now consider the following basic definition:

$$\frac{\partial(A x)}{\partial x} = A^T$$ $$\frac{\partial(x^TA)}{\partial x} = A$$

Now applying the above two definitions together with the expansion of the objective above to differentiate the objective we have

$$\frac{\partial\|\nabla x -p\|_2^2}{\partial x} = 2\nabla^T\nabla x - 2\nabla^Tp$$

$$\frac{\partial\|\nabla x -p\|_2^2}{\partial x} = 2\nabla^T(\nabla x - p)$$

since $\nabla^T = \nabla$, therefore we have

$$\frac{\partial\|\nabla x -p\|_2^2}{\partial x} = 2\nabla(\nabla x - p)$$ the constant 2 is just a constant, so the result we have is consistent with the ones that authors are using, its simply a consequence of vector differentiation