Calculate TV-norm by FEM

Mathematical definition

Let $\Omega\subset\mathbb{R}^{n}$ be a domain with Lipschitz boundary, $u(x)$ is a scalar function, total-variaional norm is defined as follow
$\|u\|_{TV} := \int_{\Omega} \sqrt{\nabla u(x)\cdot\nabla u(x)} dx.$
Instead of the above definition, we can introduce the norm with fractional derivative operator
$\|u\|_{TV^{s}} := \int_{\Omega} \sqrt{A^{s}u(x) A^{s}u(x)} dx,$
where $A := \sqrt{Id - \Delta}$ . Let $f(u) := \|u\|_{TV^{s}}$ , we calculate $f'(u)$ in the following. Taking $\epsilon > 0$ be a small positive real number, $v(x)$ is a perturbation of the function $u(x)$ , then we have
$\begin{align} f'(u)v = & \lim_{\epsilon \rightarrow 0} \frac{f(u+\epsilon v) - f(u)}{\epsilon} \\ = & \lim_{\epsilon\rightarrow 0}\frac{1}{\epsilon}\int_{\Omega} \sqrt{A^{s}(u+\epsilon v) A^{s}(u+\epsilon v)} - \sqrt{A^{s}u A^{s}u} dx \\ = & \int_{\Omega}\frac{1}{\sqrt{A^{s}u A^s u}} A^{s}u A^{s}v dx \end{align}$
Hence, we obtain
$f'(u)\cdot = \int_{\Omega}\frac{1}{\sqrt{A^{s}u A^s u}} A^{s}u A^{s}\cdot dx$
Since $A^s u(x)$ is in the denominator, we, instead, usually use the following functional
$f(u) := \int_{\Omega}\sqrt{A^s u(x) A^s u(x) + \epsilon}dx.$
We easily know that
$f'(u)v = \int_{\Omega}\frac{1}{\sqrt{A^{s}u A^s u + \epsilon}} A^{s}u A^{s}v dx.$

Computer code

In the following, we use bold font to represent the finite-dimensional vectors and matrixes. Let $\mathbf{M}\in \mathbb{R}^{N\times N}$ be the matrix such that
$\int_{\Omega}u(x)v(x)dx = \lim_{N\rightarrow \infty} (\mathbf{u}, \mathbf{v})_{\mathbf{M}} = \lim_{N\rightarrow \infty} \mathbf{u}^{T}\mathbf{M}\mathbf{v}.$
We can assemble to formulate the corresponding matrix $\mathbf{A}^{s}$ and
$\int_{\Omega}\frac{1}{\sqrt{A^{s}u A^s u + \epsilon}} A^{s}u v dx \approx \mathbf{u_f}^{T}\mathbf{M},$
where $v$ is the test function, $\mathbf{u_f}$ is finite-dimensional approximation of the function
$\frac{1}{\sqrt{A^{s}u A^s u + \epsilon}} A^{s}u.$
Here, $\epsilon$ is a small number which prevent the division zero error.
Then, we can calculate the following matrix
$\mathbf{u_f}^{T}\mathbf{M}\mathbf{A}^{s}$
which is an approximation of the operator $f'(u)$ , and this approximation maps $\mathbb{R}_{\mathbf{M}}^{N}$ to $\mathbb{R}$ . $\mathbb{R}_{\mathbf{M}}^{N}$ means $N$ -dimensional Euclidean space with inner product $(\mathbf{u}, \mathbf{v})_{\mathbf{M}} = \mathbf{u}^{T}\mathbf{M}\mathbf{v}$ .

2020