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Goal

I wish to prove that it is possible to relate the Sobolev norms on an arbitrary triangle $K$ to Sobolev norms on a reference triangle $\hat{K}$.

Preliminaries

To this end: Let $F \colon \hat{K} \to K$ be an invertible affine map given by $F(\hat{x}) = B\hat{x} + c$. For a function $\hat{v} \in C^2(\hat{K})$, we define the corresponding function $v \in C^2(K)$ by $$ v(x) = (\hat{v}\circ F^{-1})(x). $$

I am interested in giving a bound for the Sobolev semi-norm $|v|_{2, K}$ in terms of $|\hat{v}|_{2, \hat{K}}$ and the matrix $B$.

I use the definitions

$$ |v|_{2, K} = \left(\int_{K} \sum_{|\alpha| = 2} |\partial^\alpha v(x)|^2 dx \right)^{1/2} $$ where $\partial^\alpha v$ is the the mixed partial derivatives of order $|\alpha| = 2$.

What I have tried

I started by trying to bound $|\partial^\alpha v(x)|$ in terms of the directional derivatives

$$ |\partial^\alpha v(x)| \leq \sup_{\|\xi_1\|, \|\xi_2\| \leq 1} |\nabla ((\nabla v(x))\cdot \xi_1)\cdot \xi_2)| $$ and then use that since $v(x) = \hat{v}(\hat{x})$ this equals

$$ |\partial^\alpha v(x)| \leq \sup_{\|\xi_1\|, \|\xi_2\| \leq 1} |\nabla ((\nabla v(x))\cdot \xi_1)\cdot \xi_2)| \\ = \sup_{\|\xi_1\|, \|\xi_2\| \leq 1} |\nabla ((\nabla v(x))\cdot B^{-1}\xi_1)\cdot B^{-1}\xi_2)| \\ \leq \sup_{\|\xi_1\|, \|\xi_2\| \leq 1} |\nabla ((\nabla v(x))\cdot \xi_1)\cdot \xi_2)|\| B^{-1} \|^{2}, $$

however - I get lost in the notation. Am I on the right path, and is there any better notation for working with derivatives in this fashion?

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Yes, it is possible to estimate $|v|_{2,K}$ with $|\hat v|_{2,\hat K}$. It is even true for the general case $|v|_{m,K}$ and in literature it is known as the "Transformation rule", for example see "Finite Elements" by D. Braess.

$$|v|_{m,K} \leq C \|B^{-1}\|^m |\det B|^{1/2} |\hat v|_{m,\hat K}, \qquad \forall v \in H^m(K)$$

We have by the chain rule $$|\partial^\alpha v|_{0,\hat K} \leq C \|B^{-1}\|^m \sum_{|\beta|=m} |\partial^\beta \hat v ~\circ ~F^{-1}|_{0,\hat K} \leq C \|B^{-1}\|^m |\det B|^{1/2} |\partial^\alpha \hat v|_{0,K}$$ and summing over $\alpha$ for $|\alpha|=m$ gives the desired result. Your notation is fine (I am not aware of a better one) and indeed you were on the right track.

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