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I have this in my notes: $\int_\Omega{(\partial_i^2 u)(\partial_k^2 u)} = \int_{\partial\Omega}{(\partial_i u)(\partial_k^2 u)\cdot n \mathbf{d}S} - \int_\Omega{(\partial_i u)(\partial_i \partial_k^2 u)}$ but I can't see how that boundary term becomes that. Can anyone tell me what theorem/result is used to get that?

(This is a line in the topic of elliptic regularity that I am doing atm, if that helps at all..)

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up vote 2 down vote accepted

I think there is a summation in the integral. For example, $\partial_i^2 u$ should be $\sum_{i=1}^n\partial_i^2 u=\Delta u$, and $\partial_k^2 u$ should be $\sum_{k=1}^n\partial_k^2 u=\Delta u$, etc. With this understanding, the formula you are asking should be $$\int_\Omega(\Delta u)^2=\int_{\partial\Omega}(\Delta u)\nabla u\cdot n dS-\int_\Omega\nabla u\cdot\nabla(\Delta u).$$ To derive this, one can apply the divergence theorem to the vector field $X=(\Delta u)\nabla u$. That is: $$\int_\Omega\nabla\cdot X=\int_{\partial\Omega}X\cdot ndS.$$ Then $$LHS=\int_\Omega\nabla\cdot X=\int_\Omega\nabla\cdot ((\Delta u)\nabla u)=\int_\Omega\nabla u\cdot\nabla(\Delta u)+\int_\Omega(\Delta u)^2$$ because $\nabla\cdot\nabla u=\Delta u$. On the other hand, $$RHS=\int_{\partial\Omega}X\cdot ndS=\int_{\partial\Omega}(\Delta u)\nabla u\cdot ndS.$$

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Thanks. Sorry for the late acceptance. – Court Nov 26 '11 at 11:58
No problem. It's good that it helps. – Paul Nov 26 '11 at 12:46

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