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I'm learning mixed FEM using the book "Mixed Finite Element Methods and Applications" by Daniele Boffi, Franco Brezzi and Michel Fortin. I'm learning the simplicial approximation of H(curl) space (chapter 2.5.3,2.5.4) which is defined by $$H(curl,\Omega)=\{u:u\in L^2(\Omega)^3, \text{curl } u\in L^2(\Omega)^3\},$$ and its trace is defined by $n\times u$, where $n$ is unit outer normal vector.

The author mentioned that the trace is only in $H^{-1/2}(\partial\Omega)$ which is not enough for interpolation on edges or faces. So we hope to work in a slightly smaller space, which is $$ X(\Omega):=\{u:u\in L^p(\Omega)^3, \text{curl } u\in L^p(\Omega)^3,n\times u\in L^p(\partial\Omega)^2 \}, p>2.$$

The author in chapter 2.5.4 states that if they choose $u\in H(curl)\cap H^{1/2+\epsilon}$, then $u\in X(\Omega)$, for some $p=2+\delta(\epsilon)$.

I don't really understand how to lift $\text{curl } u$ to $p>2$, since curl has only part of derivatives involved, and I don't know how to apply Sobolev embeddings, or if I should use them. Anyone has some ideas help me with the problem? Thanks!

Arctic Char
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MikeMichael_maths
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  • It does not look right. $u$ has its curl (part of the gradient) in $L^2$, and the rest of the gradient is only half derivative (+$\varepsilon$) in $L^2$. There is no way that this controls the $L^p$ norm of the curl. That would almost imply that $H^1$ (which has a stronger norm than $H(curl)\cap H^{1/2+\varepsilon}$) embeds into $W^{1,p}$, p>2. This is not true, simply by scaling. – Lorenzo Pompili Dec 15 '24 at 08:05
  • Sometimes it happens that you have better estimates for the curl compared to the full gradient, but the curl is definitely not in L^p, p>2 in general, if the function is just in $H^1(\Omega)$. – Lorenzo Pompili Dec 15 '24 at 08:16
  • Thanks! I think your comments make sense... but 'curl' is quite special, like all gradients are in its kernel, so is there any possibility that the 'curl' coincidentally has extra regularity, like when $u\in H^1$, then $\text{curl } u\in H^\varepsilon$ maybe? Then by sobolev embedding it can be $L^p$ for some $p>2$. – MikeMichael_maths Dec 15 '24 at 13:49

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$ \let\eps\varepsilon $ Let me write more details to motivate what I said. $u\in H^1(\Omega)$ can't imply $\nabla\times u\in L^p(\Omega)$ for any $p>2$. To see this, assume $\Omega$ is a ball centered at zero without loss of generality. By contradiction, assume this is true. This would imply the inequality $$ \|\nabla\times u\|_{L^p(B_1)}\leq C (\|u\|_{L^2(B_1)}+\|\nabla u\|_{L^2(B_1)}) $$ for some constant $C$ (all explicitly definable operators on Banach spaces are always continuous, see this answer and the comments therein). In particular, the inequality is true for test functions $u\in C^\infty_c(B)$. For fixed $u$, consider the rescaled function $u_{\eps}(x):=u(\eps^{-1}x)$ for $\eps\leq 1$ (extending the function as zero where not defined). You can verify that $$\|u_\eps\|_{L^2(B_1)}=\eps^{3/2}\|u\|_{L^2(B_1)},$$ $$\|\nabla u_\eps\|_{L^2(B_1)}=\eps^{1/2}\|\nabla u\|_{L^2(B_1)},$$ $$ \|\nabla\times u_\eps\|_{L^p(B_1)}=\eps^{3/p-1} \|\nabla\times u\|_{L^p(B_1)}. $$ But $u_\eps$ has to satisfy the above inequality, so by substitution you get for any $\eps\leq 1$ $$ \|\nabla\times u\|_{L^p(B_1)}\leq C (\eps^{3/2-3/p+1}\|u\|_{L^2(B_1)}+\eps^{3/2-3/p}\|\nabla u\|_{L^2(B_1)}). $$ The two powers of $\eps$ on the right-hand side are positive, so this implies $$ \|\nabla\times u\|_{L^p(B_1)}=0. $$ This is not true in general, so the inequality above has to be false.

The above is called scaling argument. It is often useful to tell whether you can expect an estimate to be true or not.

In principle it could be that the curl enjoys some additional properties (I vaguely remember something on that line but I can't find a reference at the moment). But the estimate you can expect needs to have the right scaling

Lorenzo Pompili
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