I found the heuristical relation between velocity of the fluid and pressure given by
1/2 ∂iuk∂kui + 1/2 ∂juk∂kui - ν /2∂k∂k(∂jui + ∂iuj) - λ /4(∂jui + ∂iuj) + ∂i∂jp = 0
For an incompressible fluid, the kinetic energy can be computed exactly with this equation along with Navier-Stokes equations, taking no-slip boundary conditions (see Article Exact solutions for restricted incompressible Navier–Stokes ...
). However this equation gives rise to biharmonic equation for pressure∂i∂j∂i∂jp = 0
by taking twice spatial partial derivative and also multiplaying the equation by δij.
I am not sure how you are working to draw your conclusion but if you start by the momentum equation written in terms of the acceleration:
a + Grad p = a*
taking the divergence the classical Poisson equation writes with a source term in the form
Div Grad p = Lap p = Div a* ,
consequently you should demonstrate that Div Grad ( Div a* )= 0 to be able to write Lap (Lap p) =0. Isn't that?
That seems to me not so straightforward ... In literature it is also quite debated the fact that you cannot commute the spatial operators on the frontier of a bounded domain.
Could you show that ∆ [∇ · (u · ∇u)] = 0, at least on a 2D case?
Well, suppose that we have an incompressible fluid that holds Navier-Stokes equations in a domain Ω in R3, with no-slip boundary conditions (u=0 in ∂Ω) and the equation above. Then, it can be proved that ∇ · (u · ∇u) is a square-integrable function in Ω (see equation (50) in Article Exact solutions for restricted incompressible Navier–Stokes ...
and use Cauchy–Schwarz inequality). So it can be expanded as a linear combination of harmonic functions (spherical harmonics expansion). Hence, ∇ · (u · ∇u) satisfies the Laplace equation.
Manuel Garcia Casado
Let me have some time to work on, but it seems you are advoking the general orthogonality requirement as requested in the Hodge decomposition of the momentum. What about more general BCs. for the acceleration?
And, let me say my doubt, you statement implies that ∇ · (u · ∇u) can be assumed to be only pointwise linear in spatial coordinates. That would sound quite a strong result.
Have you tried to use the stream function in 2D to show that the source term ∆ [∇·(u ·∇u)] is zero in any case?
A simple case what I have found in 2D is to take the stream function as
ψ(x,y)= 1/6 x3 + 1/2 y2 x + A x +B y +C
Then
∇·(u ·∇u) = 2 (∂x∂yψ)2 - 2 ∂x∂xψ ∂y∂yψ = 2 y2 - 2 x2
In this case, ∆ [∇·(u ·∇u)] = 0.
Manuel Garcia Casado
you are simply assuming a specific functional expression of the velocity field that satisfies your assumption. You have to demonstrate that ∆ [∇·(u ·∇u)] = 0 is always true for any type of function.
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