Fractional Norm on Boundary

math

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We consider a polygon $K\subset \mathbb R^2$ and the boundary $\partial K = \cup_i f_i$. We give equivalent formulation of $H^{1/2}$ norm on $\partial K$.

Form 1: Harmonic Extension

For a function $u$ harmonic in $K$, i.e., $\Delta u = 0$ in $K$, we have the norma equivalence

$$\|u\|_{1/2,\partial K,E}:=\|\nabla u\|_{K}.$$

Form 2: Sobolev-Slobodeckij Intergral Form

An equivalent Sobolev-Slobodeckij $1/2$-norm:

$$\|u\|_{1/2,\partial K,\int} := \int_{\partial K} \int_{\partial K} \frac{|u(x) - u(y)|^2}{|x-y|^{d}} ds(x) ds(y).$$

Form 2: Hilbert Scales (Interpolation)

Let $A_{\partial K} = - \Delta_{\partial K}$ be the Laplace-Beltrami operator defined on the boundary. Then $A_{\partial K}$ defines a norm on the subspace of $H^{1/2}(\partial K)$ with average $\int_{\partial K} v = 0$. On such subspace, we the define

$$\|u\|_{1/2,\partial K,I} := (A_{\partial K}^{1/2} u, u)_{\partial K}^{1/2}.$$

Theorem

All these definitions are equivalent.

Question: how the dependence of the domain? Check references.

The compactness argument does give a proof but the dependence of the domain is unclear.

A stability result

Consider the elliptic equation $\Delta u = 0$ in $K$ and $u = g$ on $\partial K$. Then the stability result is

$$\|\nabla u\|_{K}\leq C \|g\|_{1/2,\partial K}.$$
Now the question is: which $1/2$ norm? Using $\|g\|_{1/2,\partial K,E}$ is trivial as this is the definition. How to prove the stability using the integral form $\|g\|_{1/2,\partial K,\int}$? Check the elliptic equation or boundary element method?

Remark

For the stabilization, the ideal choice is the harmonic extension one. But it is not computable. Or the key of VEM is NOT to solve the harmonic extension inside.

When using the integral form, we further split to the broken $1/2$-norm which results in a graph Laplacian with constant coefficients. Then we need to prove the inequality for $\|u\|_{1/2,\partial K}$ and the broken half norm $\sum_i \|u\|_{1/2,f_i}$.

It turns out the last one is a good balance.

• Computation. Restrict to $\partial K$, the function is piecewise linear and $A_{\partial K}$ can be assembled as a grapha Laplacian matrix with weight of the neighboring face size. Then just take sqrt in MATLAB.
• Proof. The coercivity and continunity comes naturally since we are computing the half norm. But we do need the norm equivalence which could depend on the shape of domain.

Connection with broken 1/2 and weighted L2 norms

To prove the lower or upper bound, it is reduced to the generalized eigenvalue estimate. For example, the broken norm is $A_0= diag (-1,2,-1)$ and $A_{\partial K} = diag(-1/h_i,1/h_i+1/h_{i+1}, -1/h_{i+1})$
Then we can study the generlized eigenvalue problem

$$A_{\partial K}^{1/2} x = \lambda A_0 x.$$
If we assume the neighboring face size is comparable, the so-called locally quasi-uniform, then we can scale $A_{\partial K}$ by a diagonal matrix and the estimate should depends on the size of these matrices and thus the number of faces on boundary.

If the boundary mesh is not local quasi-uniform, consider the artifical partition and a $|\log h|$ term appears.

The key is the relation of $\|u\|_{1/2,\partial K}$ and the broken half norm $\sum_i \|u\|_{1/2,f_i}$. The ideal one is the identity

$$\|u\|_{1/2,\partial K}^2 = \sum_i \|u\|_{1/2,f_i}^2$$ which is true for non-negative integer Sobolev norms, e.g. $L^2(\partial K) = \sum_i L^2(f_i)$. For fractional norms, the differential operator is defined globally and thus $(-\Delta_i)^{1/2}$ is not simply the restriction of $(-\Delta_{\partial K})^{1/2}|_{f_i}$. We should go back to the definition of fractional norm on a domain $D$: $$|u|_{1/2,D}^2 = \int_D\int_D \frac{|u(x) - u(y)|^2}{|x-y|^2}dx\ dy.$$

With the decomposition $\partial K = \cup_i f_i$:

$$|u|_{1/2,\partial K}^2 = \sum_i \sum_j \int_{f_i}\int_{f_j} \frac{|u(x) - u(y)|^2}{|x-y|^2}dx\ dy.$$
If we denote $a_{ij} = \left (\int_{f_i}\int_{f_j} \frac{|u(x) - u(y)|^2}{|x-y|^2}dx\ dy\right )^{1/2},$ and $A = (a_{ij})$, then $|u|_{1/2,\partial K}^2 = \|A\|_{F}^2$. The diagonal of the matrix $A$ is the broken half norms.

So we get the inequality for $u\in H^{1/2}(f_i) \cap H^{1/2}(\partial K)$

$$\sum_i \|u\|_{1/2,f_i}^2\leq \|u\|_{1/2,\partial K}^2.$$
The essential difficulty is to estimate the cross term. Before getting into detail, notice that we are dealing with piecewise linear and continuous function $u$ on the boundary. So $\|u\|_{1/2,f_i}^2$ is computable:
$$|u|_{1/2,f_i}^2 = k_i^2 |f_i|^2 = |u(x_i) - u(x_{i+1})|^2,$$ where $k_i$ is the slope of $u$ along $f_i = [x_i, x_{i+1}]$. The $L^2$ norm is
$$\|u\|_{0,f_i}^2 = |f_i|\left (2u(x_i)^2 + 2u(x_{i+1})^2 + 2u(x_{i})u(x_{i+1})\right )/6.$$

We should modify the stabilization to

$$S(u,v) = \sum_i\|u\|_{1/2,f_i}\|v\|_{1/2,f_i}.$$ Then the upper bound $$S(u,u) = \sum_i\|u\|_{1/2,f_i}^2 \leq \|u\|_{1/2,\partial K}^2 = \|\nabla u\|_K^2,$$ holds with constant $1$.

The lower bound which leads to the coercivity is more delicate. We can make assumption on $K$: on the polygon $K$, the inequality

$$|u|_{1/2,\partial K}^2\leq C\sum_i |u|_{1/2,f_i}^2$$ for $u\in \mathbb B_1(\partial K)$ (continuous and piecewise linear functions).

The rest is devoted to prove the above inequality and see what kind of conditions should be imposed to the polygon.

A thought on the bound of enumerator

$$\begin{align} |u(x) - u(y)|&\leq |u(x) - u(x_i)|+|u(x_i) - u(x_{i+1})|+\ldots + |u(x_j) - u(y)|\\ &= k_i (x-x_i) + k_{i+1}|f_{i+1}| + k_j(x_j - y). \end{align}$$
Then estimate the double integral $$\int_{f_i}\int_{f_j} \frac{|x-x_i|^2}{|x-y|^2}dx\ dy.$$

1. Convex and shape regular polygons. Easy. Cite W. Cao's result.
2. Allow the neighboring edge length change dramatically by adding an artifical grid to decompose larger edge to the case 1. A logrithmic factor $\log (h_{\max}/h_{\min})$ might needed.
3. Allow the concave case as long as $d(f_i,f_j)>\delta$ is uniformly bounded below. The factor $|f_i||f_j|/\delta^2$ might needed. Rule out the cusp and samll angles.
4. For interface problems, the element contains a small angle is always a triangle and stabilization is not needed. Rectangle with a large aspect ratio is allowed. So good enough for the mesh generated for interface problems.