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$\def\di\{\mbox{div}} \def\n{\nabla}$ I'd like help to undertand the small proof of the theorem 5.2 here (dowload). How can I see that \begin{equation} 0 = \int_{\{\xi > 0\}}\di \left \{ \Bigl( \dfrac{1}{2} \langle A \n \xi, \n \xi \rangle + \Gamma \Bigr) \psi - \langle \psi , \n \xi \rangle A \n \xi \right\}dx \end{equation}

My thougths are that

\begin{eqnarray*} 0& =& \lim_{\gamma \rightarrow 0} \dfrac{E_0(\xi_\gamma , \Omega) - E_0(\xi,\Omega)}{\gamma} \\ &=& \lim_{\gamma \rightarrow 0} \Bigr( \int_{\{\xi >0\}} \left \{ \dfrac{1}{2} \langle A \n \xi, \xi \rangle + \Gamma \right \} \di(\psi)dx \\ & &+ \int_{\{\xi >0\}} \left \{ \dfrac{1}{2} \langle (DA) \psi \n \xi,\n \xi \rangle - \langle A \n \xi , D \psi \n \xi \rangle + \langle \n \Gamma , \psi \rangle \right \}dx \Bigr)\\ &=&\int_{\{\xi > 0\}}\di \left \{ \Bigl( \dfrac{1}{2} \langle A \n \xi, \n \xi \rangle + \Gamma \Bigr) \psi - \langle \psi , \n \xi \rangle A \n \xi \right\}dx \end{eqnarray*}

Then I imagine that zero is the limit when $ \gamma \rightarrow 0$ in the expression below

\begin{array}{l} \int_{\{\xi > 0\}}\di \left\{ \Bigl( \dfrac{1}{2} \langle A \n \xi, \n \xi \rangle + \Gamma \Bigr) \psi - \langle \psi , \n \xi \rangle A \n \xi \right\}dx = \\ \int_{\{\xi > 0\}}\Bigl( \dfrac{1}{2} \langle A \n \xi, \n \xi \rangle + \Gamma \Bigr) \di (\psi)dx + \int_{\{\xi > 0\}}\langle \n \Bigl( \dfrac{1}{2} \langle A \n \xi, \n \xi \rangle + \Gamma \Bigr), \n \psi \rangle dx \\ -\int_{\{\xi > 0\}}\langle \psi , \n \xi \rangle \di (A \n \xi )dx-\int_{\{\xi > 0\}} \langle \n \Bigr( \langle \psi , \n \xi \rangle \Bigl), A \n \xi \rangle dx =\\ \int_{\{\xi > 0\}} \Bigl( \dfrac{1}{2} \langle A \n \xi, \n \xi \rangle + \Gamma \Bigr) \di (\psi) dx + \\ \int_{\{\xi > 0\}} \dfrac{1}{2} \langle( D(A \n \xi) \n \xi ,\n \psi \rangle dx + \int_{\{\xi > 0\}} \dfrac{1}{2}\langle A \n \xi D^{2} \xi,\n \psi \rangle dx + \int_{\{\xi > 0\}} \langle \Gamma, \n \psi \rangle dx \\ -\int_{\{\xi > 0\}} \langle \psi , \n \xi \rangle \di (A \n \xi ) dx - \int_{\{\xi > 0\}} \langle D \psi \n \xi, A \n \xi \rangle dx -\int_{\{\xi > 0\}} \langle D² \xi \psi , A \n \xi \rangle dx \end{array}

Then I must to show that \begin{eqnarray*} \int_{\{\xi >0\}} \dfrac{1}{2} \langle (DA) \psi \n \xi,\n \xi \rangle dx &=& \int_{\{\xi > 0\}} \dfrac{1}{2} \langle( D(A \n \xi) \n \xi ,\n \psi \rangle dx + \int_{\{\xi > 0\}} \dfrac{1}{2}\langle A \n \xi D^{2} \xi,\n \psi \rangle dx\\ &-& \int_{\{\xi > 0\}} \langle \psi , \n \xi \rangle \di (A \n \xi ) dx -\int_{\{\xi > 0\}} \langle D² \xi \psi , A \n \xi \rangle dx \end{eqnarray*}

Am I riht? Im particular, I don't get this equality, I don't know what means $DA$. The signal $D$ is used to denote the jacobian. But $A$ is a $n\times n$ matrix. How can I do this?

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Could you post a different link to the article, possible the springer 'description' page instead of directly to the fulltext? – icurays1 Nov 12 '12 at 18:32
I get an "error occurred" message in my browser, when clicking on the link, just now. – amWhy Nov 12 '12 at 18:33
Ok.I will try correct this now. – user29999 Nov 12 '12 at 18:35
Please, try to do dowlnload now. If not get. Please. say me. – user29999 Nov 12 '12 at 18:42

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