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when we consider a Immersed Membrane Without Shear, we can define a regularized elastic energy as $$\mathcal{E}(\varphi)=\int_{\Omega} E(|\nabla \varphi|) \frac{1}{\varepsilon} \zeta\left(\frac{\varphi}{\varepsilon}\right) d x$$ then after some operations we get

**Theorem 3.**4 The temporal variation of $E$ given by the principle of virtual work satisfies $$ \partial_t \mathcal{E}=-\int_{\Omega} F \cdot u d x $$ and corresponds to the following force: $$ F=\nabla\left(E(|\nabla \varphi|) \frac{1}{\varepsilon} \zeta\left(\frac{\varphi}{\varepsilon}\right)\right)-\operatorname{div}\left(E^{\prime}(|\nabla \varphi|)|\nabla \varphi| n \otimes n \frac{1}{\varepsilon} \zeta\left(\frac{\varphi}{\varepsilon}\right)\right) . $$ where we recall that the normal is defined by $n=\frac{\nabla \varphi}{|\nabla \varphi|}$. in the expression of that force, we absorb the gradient part into the pressure term of N-S equations, but then it says that in the case without shear force, we can rewrite the force as $$ F=\operatorname{div}\left(E^{\prime}(|\nabla \varphi|)|\nabla \varphi|(\mathbb{I}-n \otimes n) \frac{1}{\varepsilon} \zeta\left(\frac{\varphi}{\varepsilon}\right)\right) $$, so I don't know how we get $$ F=\operatorname{div}\left(E^{\prime}(|\nabla \varphi|)|\nabla \varphi|(\mathbb{I}-n \otimes n) \frac{1}{\varepsilon} \zeta\left(\frac{\varphi}{\varepsilon}\right)\right) $$, may be it is the divergence of a term like $pI$ in N-S equation?can anyone helps me, thank you

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Notice two things:

  • $\nabla \left(E \left( \left| \nabla \phi \right| \right) \right) = E'(|\nabla\phi|)\nabla \left| \nabla\phi \right|$
  • $\nabla \left( E' \left( \left| \nabla \phi \right| \right) \frac{1}{\varepsilon} \zeta \left( \frac{\phi}{\varepsilon} \right) \right) = \nabla \cdot \left( E' \left( \left| \nabla \phi \right| \right) \frac{1}{\varepsilon} \zeta \left( \frac{\phi}{\varepsilon} \right) \mathcal{I} \right)$, where $\mathcal{I}$ is the identity tensor.

Using these two properties, you can write

\begin{split} F &= \nabla \left( E \left( \left| \nabla \phi \right| \right) \frac{1}{\varepsilon} \zeta \left( \frac{\phi}{\varepsilon} \right) \right) - \mathrm{div}\left( E'(|\nabla\phi|)\left|\nabla\phi \right| n \otimes n\frac{1}{\varepsilon}\zeta\left(\frac{\phi}{\varepsilon}\right) \right) \\ &= \nabla \left( E' \left( \left| \nabla \phi \right| \right) \left| \nabla \phi \right| \frac{1}{\varepsilon} \zeta \left( \frac{\phi}{\varepsilon} \right) \right) - \mathrm{div}\left( E'(|\nabla\phi|)\left|\nabla\phi \right| n \otimes n\frac{1}{\varepsilon}\zeta\left(\frac{\phi}{\varepsilon}\right) \right) \\ &= \mathrm{div} \left( E' \left( \left| \nabla \phi \right| \right) \left| \nabla \phi \right| \frac{1}{\varepsilon} \zeta \left( \frac{\phi}{\varepsilon} \right) \mathcal{I} \right) - \mathrm{div}\left( E'(|\nabla\phi|)\left|\nabla\phi \right| n \otimes n\frac{1}{\varepsilon}\zeta\left(\frac{\phi}{\varepsilon}\right) \right) \\ &= \mathrm{div}\left( E'(|\nabla\phi|) \left| \nabla\phi \right| \left(\mathcal{I} - n \otimes n\right)\frac{1}{\varepsilon}\zeta\left(\frac{\phi}{\varepsilon}\right) \right) \end{split}

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  • $\begingroup$ could you please teach me how to derive the second row from the first row? thank you $\endgroup$
    – 吴yuer
    Sep 18, 2023 at 4:56
  • $\begingroup$ to derive second row from the first, i have omitted the terms involving $\zeta' \nabla \phi$ - it looks like these terms can be absorbed into the pressure as mentioned in the book. $\endgroup$ Sep 19, 2023 at 7:28

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