Editing Navier–Stokes equations (section)

====Strong form====
Consider the incompressible Navier–Stokes equations for a [[Newtonian fluid]] of constant density <math display="inline">\rho</math> in a domain
<math display="block"> \Omega \subset \mathbb R^d \quad (d=2, 3)</math>
with boundary
<math display="block"> \partial \Omega = \Gamma_D \cup \Gamma_N ,</math>
being <math display="inline">\Gamma_D</math> and <math display="inline">\Gamma_N</math> portions of the boundary where respectively a [[Dirichlet boundary condition|Dirichlet]] and a [[Neumann boundary condition]]  is applied (<math display="inline">\Gamma_D \cap \Gamma_N = \emptyset</math>):<ref name="Quarteroni">{{cite book |author-link1= Alfio Quarteroni |last1=Quarteroni |first1=Alfio |title=Numerical models for differential problems |publisher=Springer |isbn=978-88-470-5522-3 |edition=Second|date=2014-04-25 }}</ref>
<math display="block">
\begin{cases}
\rho \dfrac{\partial \mathbf{u}}{\partial t} + \rho (\mathbf{u} \cdot \nabla) \mathbf{u} - \nabla \cdot \boldsymbol{\sigma} (\mathbf{u}, p) = \mathbf{f} & \text{ in } \Omega \times (0, T) \\
\nabla \cdot \mathbf{u} = 0  & \text{ in } \Omega \times (0, T) \\
\mathbf{u} = \mathbf{g} & \text{ on } \Gamma_D \times (0, T) \\
 \boldsymbol{\sigma} (\mathbf{u}, p) \hat{\mathbf{n}} = \mathbf{h} & \text{ on } \Gamma_N \times (0, T) \\
\mathbf{u}(0)= \mathbf{u}_0 & \text{ in } \Omega \times \{ 0\}
\end{cases}
</math>
<math display="inline">\mathbf{u}</math> is the fluid velocity, <math display="inline">p</math> the fluid pressure, <math display="inline">\mathbf{f}</math> a given forcing term, <math>\hat{\mathbf{n}}</math> the outward directed unit normal vector to <math display="inline">\Gamma_N</math>, and <math display="inline">\boldsymbol{\sigma}(\mathbf{u}, p)</math> the [[viscous stress tensor]] defined as:<ref name="Quarteroni" /> 
<math display="block">\boldsymbol{\sigma} (\mathbf{u}, p) = -p \mathbf{I} + 2 \mu \boldsymbol{\varepsilon}(\mathbf{u}).</math>
Let <math display="inline">\mu</math> be the dynamic viscosity of the fluid, <math display="inline">\mathbf{I}</math> the second-order [[Identity matrix|identity tensor]] and <math display="inline">\boldsymbol{\varepsilon}(\mathbf{u})</math> the [[strain-rate tensor]] defined as:<ref name="Quarteroni" /> 
<math display="block">
\boldsymbol{\varepsilon} (\mathbf{u}) = \frac{1}{2} \left(\left( \nabla \mathbf{u} \right) + \left( \nabla \mathbf{u} \right)^\mathrm{T}\right).
</math>
The functions <math display="inline">
\mathbf{g}
</math> and <math display="inline">
\mathbf{h}
</math> are given Dirichlet and Neumann boundary data, while <math display="inline">
\mathbf{u}_0
</math> is the [[initial condition]]. The first equation is the momentum balance equation, while the second represents the [[conservation of mass|mass conservation]], namely the [[continuity equation]]. 
Assuming constant dynamic viscosity, using the vectorial identity
<math display="block"> \nabla \cdot \left( \nabla \mathbf{f} \right)^\mathrm{T} = \nabla ( \nabla \cdot \mathbf{f} )</math>
and exploiting mass conservation, the divergence of the total stress tensor in the momentum equation can also be expressed as:<ref name="Quarteroni" /> 
<math display="block">
\begin{align}
\nabla \cdot \boldsymbol{\sigma} (\mathbf{u}, p)
& = \nabla \cdot \left(-p \mathbf{I} + 2 \mu \boldsymbol{\varepsilon}(\mathbf{u}) \right) \\
& = - \nabla p + 2 \mu \nabla \cdot \boldsymbol{\varepsilon}(\mathbf{u}) \\
& = - \nabla p + 2 \mu \nabla \cdot \left [ \tfrac{1}{2} \left(\left(\nabla \mathbf{u} \right) + \left(\nabla \mathbf{u} \right)^\mathrm{T}\right) \right] \\
& = -\nabla p + \mu \left(\Delta \mathbf{u} + \nabla \cdot \left(\nabla \mathbf{u} \right)^\mathrm{T} \right) \\
& =  -\nabla p + \mu \bigl( \Delta \mathbf{u} + \nabla  \underbrace{(\nabla \cdot \mathbf{u})}_{=0} \bigr)
= -\nabla p + \mu \, \Delta \mathbf{u}.
\end{align}
</math>
Moreover, note that the Neumann boundary conditions can be rearranged as:<ref name="Quarteroni" /> 
<math display="block">
 \boldsymbol{\sigma}(\mathbf{u}, p) \hat{\mathbf{n}} = \left(-p \mathbf{I} + 2 \mu \boldsymbol{\varepsilon}(\mathbf{u})\right)\hat{\mathbf{n}} = -p \hat{\mathbf{n}} + \mu \frac{\partial \boldsymbol u}{\partial \hat{\mathbf{n}}}.
</math>