A proof explaining the properties and bounds of the equations, such as Navier–Stokes existence and smoothness, is one of the important unsolved problems in mathematics.
[1] The Navier–Stokes equations are based on the assumption that the fluid, at the scale of interest, is a continuum – a continuous substance rather than discrete particles.
Another necessary assumption is that all the fields of interest including pressure, flow velocity, density, and temperature are at least weakly differentiable.
The first term on the right-hand side of the equation is the ordinary Eulerian derivative (the derivative on a fixed reference frame, representing changes at a point with respect to time) whereas the second term represents changes of a quantity with respect to position (see advection).
A continuity equation may be derived from conservation principles of: A continuity equation (or conservation law) is an integral relation stating that the rate of change of some integrated property φ defined over a control volume Ω must be equal to the rate at which it is lost or gained through the boundaries Γ of the volume plus the rate at which it is created or consumed by sources and sinks inside the volume.
Validity is retained if φ is a vector, in which case the vector-vector product in the second term will be a dyad.
[3] Using the formula for the divergence of a dyad, we then have Note that the gradient of a vector is a special case of the covariant derivative, the operation results in second rank tensors;[3] except in Cartesian coordinates, it is important to understand that this is not simply an element by element gradient.
Rearranging : The leftmost expression enclosed in parentheses is, by mass continuity (shown before), equal to zero.
A shorter though less rigorous way to arrive at this result would be the application of the chain rule to acceleration: where u = (u, v, w).
The generic density of the momentum source s seen previously is made specific first by breaking it up into two new terms, one to describe internal stresses and one for external forces, such as gravity.
Note that the mechanical pressure p is equal to the negative of the mean normal stress:[4] The motivation for doing this is that pressure is typically a variable of interest, and also this simplifies application to specific fluid families later on since the rightmost tensor τ in the equation above must be zero for a fluid at rest.
For completion, one must make hypotheses on the forms of τ and p, that is, one needs a constitutive law for the stress tensor which can be obtained for specific fluid families and on the pressure.
The formulation for Newtonian fluids stems from an observation made by Newton that, for most fluids, In order to apply this to the Navier–Stokes equations, three assumptions were made by Stokes: The above list states the classic argument[5] that the shear strain rate tensor (the (symmetric) shear part of the velocity gradient) is a pure shear tensor and does not include any inflow/outflow part (any compression/expansion part).
This means that its trace is zero, and this is achieved by subtracting ∇ ⋅ u in a symmetric way from the diagonal elements of the tensor.
The value of λ, which produces a viscous effect associated with volume change, is very difficult to determine, not even its sign is known with absolute certainty.
The equation of state to use depends on context (often the ideal gas law), the conservation of energy will read: Here, h is the specific enthalpy, T is the temperature, and Φ is a function representing the dissipation of energy due to viscous effects: With a good equation of state and good functions for the dependence of parameters (such as viscosity) on the variables, this system of equations seems to properly model the dynamics of all known gases and most liquids.
For the special (but very common) case of incompressible flow, the momentum equations simplify significantly.
Many salt solutions and molten polymers are non-Newtonian fluids, as are many commonly found substances such as ketchup, custard, toothpaste, starch suspensions, paint, blood, and shampoo.
In that case (take component 3 to be zero): The vector function ψ is still defined via: but this must simplify in some way also since the flow is assumed 2D.
If orthogonal coordinates are assumed, the curl takes on a fairly simple form, and the equation above expanded becomes: Examining this equation shows that we can set ψ1 = ψ2 = 0 and retain equality with no loss of generality, so that: the significance here is that only one component of ψ remains, so that 2D flow becomes a problem with only one dependent variable.
The equation for ψ can simplify since a variety of quantities will now equal zero, for example: if the scale factors h1 and h2 also are independent of x3.