since $-\Delta\mathbf{m}=\mathrm{curl}^2\mathbf{m}-\nabla\mathrm{div}\mathbf{m}$ integration by parts yields: $$\begin{aligned}\int_\Omega-\Delta\mathbf{m}\cdot\mathbf{z}dx&=\int_\Omega\operatorname{curl}\mathbf{m}\cdot\operatorname{curl}\mathbf{z}+\operatorname{div}\mathbf{m}\operatorname{div}\mathbf{z}dx\\&-\int_\Gamma(\operatorname{curl}\mathbf{m}\times\boldsymbol{n})\cdot\mathbf{z}dS-\int_\Gamma\operatorname{div}\mathbf{m}(\mathbf{z}\cdot\boldsymbol{n})dS,\end{aligned}$$ I know the Green's theorem has been used, but I'm not sure how this $\int_{\Omega}\operatorname{curl}\mathbf{m}\cdot\operatorname{curl}\mathbf{z}$ and $\int_\Gamma(\operatorname{curl}\mathbf{m}\times n)\cdot\mathbf{z}dS$ came about.
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These terms come from a variant of Green's formula applied to the $\mathrm{curl}^2\mathbf{m}$ term, see this or this question. – Korf Aug 16 '23 at 11:09
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$curl^2m=\nabla \times \nabla \times m$? – p yz Aug 16 '23 at 13:38
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Yes, it is a you write. Or in the original notation $\mathrm{curl}^2\mathbf{m} = \mathrm{curl}(\mathrm{curl} (\mathbf{m}))$. The power of $2$ is meant as double application of curl. – Korf Aug 16 '23 at 14:27
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Thank you, I understand. – p yz Aug 17 '23 at 06:54