Scalar product of biharmonic friction with velocity components

[twilder]

TL;DR Summary

Find $\underline u \cdot [(A\nabla^4 \underline u)]$

I know that taking the scalar product of the harmonic (Laplacian) friction term with $\underline u$ is

$$\underline u \cdot [\nabla \cdot(A\nabla \underline u)] = \nabla \cdot (\underline u A \nabla \underline u) - A (\nabla \underline u )^2 $$

where $\underline u = (u,v)$ and $A$ is a constant.

I am unsure how to expand the biharmonic version,

$$\underline u \cdot [(A\nabla^4 \underline u)] = \cdots $$

I have got to this stage so far,

$$\underline u \cdot [\nabla \cdot (A^{1/2}\nabla(\nabla \cdot (A^{1/2} \nabla \underline u)))] = \nabla \cdot [\underline u A^{1/2}\nabla(\nabla \cdot (A^{1/2}\nabla \underline u))]- A^{1/2}\nabla(\nabla \cdot (A^{1/2}\nabla \underline u))\cdot \nabla \underline u$$

There doesn't appear to be any clear references for this type of expansion in textbooks or online (that I have come across). Does anyone have any ideas how I could go further with this?

 

[jvicens]

I can understand your confusion with expanding the biharmonic version of the friction term. The biharmonic operator is a fourth-order differential operator and can be written as the product of two second-order operators, which can make it more challenging to work with.

One approach to expanding this term is to first rewrite it in terms of the Laplacian operator, using the identity that relates the biharmonic operator to the Laplacian operator:

$$\nabla^4 \underline u = \nabla^2 (\nabla^2 \underline u) = \nabla \cdot (\nabla \cdot (\nabla \cdot (\nabla \underline u)))$$

Substituting this into the original equation, we get:

$$\underline u \cdot [(A\nabla^4 \underline u)] = \underline u \cdot [A \nabla \cdot (\nabla \cdot (\nabla \cdot (\nabla \underline u)))]$$

Using the product rule for the scalar product, we can expand this further:

$$\underline u \cdot [A \nabla \cdot (\nabla \cdot (\nabla \cdot (\nabla \underline u)))] = A (\nabla \cdot (\nabla \cdot (\nabla \cdot (\nabla \underline u)))) \cdot \underline u + \underline u \cdot [A \nabla \cdot (\nabla \cdot (\nabla \cdot (\nabla \underline u)))]$$

Now, we can use the identity you provided in the original post, to expand the second term on the right-hand side:

$$\underline u \cdot [A \nabla \cdot (\nabla \cdot (\nabla \cdot (\nabla \underline u)))] = A (\nabla \cdot (\nabla \cdot (\nabla \cdot (\nabla \underline u)))) \cdot \underline u + \nabla \cdot (\underline u A \nabla (\nabla \cdot (\nabla \underline u))) - A (\nabla \underline u)^2$$

This is the final expansion of the biharmonic version of the friction term. It is similar to the expansion of the harmonic version, but with an