Navier-Stokes equation, about pressure... I'm a computer science student writting a dissertation about fluid simulation on real time applications. I'm trying to understand a few things regarding the pressure term:
1) When talking about the Helmholtz-Hodge decomposition, my books say that you can decompose any vector field into a divergence-free field, and a gradient of a scalar fied, and that scalar field "turns out" to be the pressure. How do they get to that conclusion?
(http://http.developer.nvidia.com/GPUGems/gpugems_ch38.html chapter 38.2.4 section "The Helmholtz-Hodge Decomposition")
2) In many books or articles when computing pressure people say to take the divergence of the momentum equation, that ends up in a poisson equation. I don't understand why they take the divergence of everything, probably due to my lack of understanding of how poisson equation works or something related.
(http://http.developer.nvidia.com/GPUGems/gpugems_ch38.html chapter 38.2.4 section "The Helmholtz-Hodge Decomposition - Second realization" , in this article they don't take the divergence of the momentum equation but it's similar and with the same purpose)
Keep in mind that I'm not a physics student so I kinda had to learn everything myself because in my university we don't really see much of this things, so probably my confusion is due to the lack of some basic concept that I haven't seen yet. Any help is appreciated, thanks.
 A: In the incompressible limit, the divergence vanishes, which is reason enough to take it. Moreover, the time derivative of the divergence vanishes, meaning that the divergence was also zero just a moment ago. That puts a constraint on the pressure field in the form of the Poisson equation.
From a physical point of view this feels odd. What is it that obliges the pressure to behave in this way? The key is the realise that the Mach number $q/a$ is vanishingly small, not because $q$ is vanishingly small(very dull solution) but because $a$ is unboundedly large, meaning that acoustic waves have unlimited time in which to reach equilibrium.
This is very much a handwaving explanation. A rigorous mathematical demonstration of passage to the incompressible limit is non-trivial.  A reference is Klainerman, Sergiu, and Andrew Majda. "Compressible and incompressible fluids." Communications on Pure and Applied Mathematics 35.5 (1982): 629-651.
A: here is a way to see (2). assuming the fluid to be incompressible , mass conservation tells you that $div\  u = 0.$  the newtons law for inviscid fluid is $\frac{du}{dt} = -\frac 1\rho\nabla P.$  take the divergence(dot product with $\nabla$) gives you the poisson equation $\Delta P = 0.$
A: Answering (1) requires understanding what the Navier-Stokes equations represent and how they are derived.  The Wikipedia article for the Navier-Stokes equations shows this derivation from the Cauchy momentum equation for continua, in which pressure shows up explicitly.  The Navier-Stokes equations thus have this built into them, and when they are constrained for incompressible fluids, then it reappears explicitly as a term equivalent to internal or thermodynamic work.  It represents energy imparted to the flow from the fluid itself.
People can and do write whole books on these equations and their solutions. The assumption of incompressibility usually holds when the viscosity isn't an issue, and the speeds are low, less than about 1/4 of the mach number of the fluid. Other assumptions you will run into are laminar flow or no-slip boundary conditions. All of these are to make the equations, which in their full generality are nonlinear and difficult, more tractable.
