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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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There are a lot of symbols and terms here without context. It would help if you added a link to the paper you found this equation in, which may have other assumptions and conditions that you did not think to include. –  Rahul May 27 '12 at 5:42
    
I have just added more details to the question. :) –  Daniel May 27 '12 at 6:40
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1 Answer

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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I am always wondering if there is any rule of $D$ operator, such as chain rule and any other rules that are applicable to $\nabla$ operator, etc.? :( –  Daniel May 27 '12 at 18:22
    
I am sure such rules can be derived. In the same paper I referenced there is at least one example. I am not feeing particularly confident about the transposed Laplacian though and in fact since D appears to be a tensor operation, care must be taken with regards to the nature of the expressions under the integrals –  Valentin May 27 '12 at 21:29
    
And in your equation: $$\nabla\psi^{u}\nabla^{T}u=\nabla\left(\nabla^{T}\psi^{u}u\right)-\nabla\left( \nabla^{T}\psi\right)u$$ What would it mean of "$\psi^{u}u$"? Since both $\psi^{\bf{u}}$ and $\bf{u}$ are vectors... –  Daniel May 29 '12 at 15:51
    
I probably should have written $$\nabla \left(\left(\nabla^T \psi^u \right)u \right)$$ –  Valentin May 29 '12 at 17:07
    
And when you wrote, $$\psi^{n}\nabla\left(2\nu D\left(u\right)\right)=\nabla\left(2\nu\psi^{u}D\left(u\right)\right)-2\nu\nabla‌​\psi^{u}D\left(u\right),$$ it should be actually something like $${\psi ^{\bf{u}}} \cdot \nabla \cdot \left( {2\nu D\left( {\bf{u}} \right)} \right) = blahblah$$, Since $\nabla \cdot \left( {2\nu D\left( {\bf{u}} \right)} \right)$ is kind of laplace operator, right? And so the result would be a vector and then ${\psi ^{\bf{u}}} \cdot \nabla \cdot \left( {2\nu D\left( {\bf{u}} \right)} \right)$ would be a scalar. But how about the right side??? –  Daniel May 29 '12 at 18:10
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