# How to integrate twice of this viscous term?

I am reading a paper, and I do not understand why the author said the following term when integrated twice will become,

$\int\limits_\Omega {{\rm{d}}\Omega {{\bf{\psi }}^{\bf{u}}}\cdot\nabla \cdot\left( {2\nu D\left( {{\bf{u}}} \right)} \right)} = - \int\limits_\Gamma {{\rm{d}}\Gamma {{\bf{\psi }}^{\bf{u}}}\cdot\left( {2\nu D\left( { {\bf{u}}} \right)} \right)\cdot{\bf{n}}} + \int\limits_\Gamma {{\rm{d}}\Gamma {\bf{n}}\cdot\left( {2\nu D\left( {{{\bf{\psi }}^{\bf{u}}}} \right)} \right)\cdot {\bf{u}}} - \int\limits_\Omega {{\rm{d}}\Omega {\bf{u}}\cdot\nabla \cdot\left( {2\nu D\left( {{{\bf{\psi }}^{\bf{u}}}} \right)} \right)}$

Where both $\bf{u}$ and ${\bf{\psi }}^{\bf{u}}$ are vector field (velocity and adjoint velocity), and the strain rate $D\left( {{\bf{u}}}\right)=\frac{1}{2}\left( {\nabla {\bf{v}} + {\bf{v}}\nabla } \right)$.

Could anyone help me, see if it is correct and how to integrate it?

Thanks a lot!

Here is the link of this paper,

Please see How equation 12 is integrated into equation 13.

The term appears in deriving adjoint equation that is very similar to the primal navier stokes equation during shape optimization.

And frozen turbulence assumption is used.

$\nabla \cdot\left( {2\nu D\left( {{\bf{u}}} \right)} \right)$ is what is called the diffusion term (for incompressible flow).

Thanks :)

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The terms don't make sense as you've written them--- what is D(u)? Is it a differential operator? What is it applied to? What is the double dot-product in the left hand side? You should write it out with explicit index contractions, or link the paper. – Ron Maimon May 27 '12 at 5:18
I have just added more details to the question. :) – Daniel May 27 '12 at 6:40
I do not have time to work it out, but it looks like the term inside the integral is first rewritten in two terms: one with everything in the divergence operator, subtracted with the difference with the initial term. On this first term you can apply Stokes theorem to obtain a surface integral. – Bernhard May 27 '12 at 7:52
Hmm, so how to apply chain rule to this term $\vec a \cdot \nabla \cdot \nabla \vec b$, so that I can move the $\vec a$ into the $\nabla$ operator? Thanks – Daniel May 29 '12 at 19:14
@Daniel en.wikipedia.org/wiki/… So it should be $\nabla\cdot(a\cdot\nabla b)$. Do you also have some kind of continuity? Btw: the link you provide does not work (anymore) – Bernhard Jun 3 '12 at 14:45