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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,

http://web.cos.gmu.edu/~rlohner/pages/publications/papers/reno04adj.pdf

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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    I have just added more details to the question. :)2012-05-27

1 Answers 1

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I looked up the definition of the symmetric gradient and if (5.14) of the following paper is correct, it should be $D=\frac{1}{2}\left(\nabla+\nabla^T\right)$

Transforming the integrand as follows: $\psi^{n}\nabla\left(2\nu D\left(u\right)\right)=\nabla\left(2\nu\psi^{u}D\left(u\right)\right)-2\nu\nabla\psi^{u}D\left(u\right) $ perform the first integration:

$\int_{\Omega}d\Omega\psi^{u}\nabla\left(2\nu D\left(u\right)\right)=\int_{\Gamma}d\Gamma\psi^{u}\left(2\nu D\left(u\right)\right)n-\int_{\Omega}d\Omega2\nu\nabla\psi^{u}D\left(u\right)$

Now expand $D$ $\int_{\Omega}d\Omega2\nu\nabla\psi^{u}D\left(u\right)=\int_{\Omega}d\Omega2\nu\frac{1}{2}\nabla\psi^{u}\left(\nabla u+\nabla^{T}u\right)$

The first term is transformed: $\nabla\psi^{u}\nabla u=\nabla\left(\nabla\psi^{u}u\right)-u\nabla\left(\nabla\psi^{u}\right)$ Which leads to $\int_{\Omega}d\Omega2\nu\nabla\psi^{u}\nabla u=\int_{\Gamma}2\nu\nabla\psi^{u}un-\int_{\Omega}d\Omega 2\nu u\nabla\left(\nabla\psi^{u}\right)$ Now the task is completed if it is possible to show that $\nabla\psi^{u}\nabla^{T}u=\nabla\left(\nabla^{T}\psi^{u}u\right)-\nabla\left(\nabla^{T}\psi\right)u$

Differentiating $\nabla\left(\nabla^{T}\psi^{u}u\right)$ leads to

$\nabla\left(\nabla^{T}\psi^{u}u\right)=\nabla\left((\psi^{u})^{T}\nabla u\right)=\nabla(\psi^{u})^{T}\nabla u+(\psi^{u})^{T}\nabla\left(\nabla u\right)=\nabla\left(\nabla^{T}\psi\right)u+\nabla^{T}\psi^{u}\nabla u$

If this is not quite the right approach, it should not be too far from one.

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    $D(\boldsymbol{u})$ is really a tensor of the second rank. RHS is a scalar, as i see it. There might be a need here to fall back on brute force and write out the integrand in components.2012-05-29