Navier-Stokes Problem
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## Analysis of Navier-Stokes equations

The mathematics is not possible to find a solution of the Navier-Stokes equations up to now. Could be found any reasons?

To show a possible reason, in the following the shear stress of the Navier-Stokes equations will be analysed. This occurs applying the Navier-Stokes model in an area of a simple flow field.

### Needed basics and formula.

For this proof the following condition will be used:

The 3. Newtonian law „actio = reactio“ describes mechanical systems.

This means, that the forces and momenta at the section plane are equal in absolut value and directly opposed. The method of sections confirms to this law. The inner forces and momenta come at the section plane become outer ones. If a System is separated into subsystems and the physical model is applied to them, the addition of the forces and momenta must be equal to the forces and momenta applying to the origin system.

First the occurring shear stresses will be described. This based on the derivation of the book “Prandtl's Essentials of Fluid Mechanics“ [1] from the chapter 5 „Fundamental Equations of Fluid Mechanics“ in the subchapter 5.2 „Navier-Stokes Equations“.

The occurring stresses at a volume element in a flow are shown in figure 1.

Figure 1: Shear vectors at the system boundaries [1]

The normal stresses can be converted with the term [1]

into:

, , [1].

Newtonian fluids will be described by following equations [1]:

Normal stresses:

1.0.0

1.0.1

1.0.2

Shear stresses [1]:

1.1.0

1.1.1

1.1.2

with the symmetry condition [1]:

, , 1.2.0

### Specification of the flow field

To simplify subjects and without loss of generality a laminar and steady flow will be considered with

with 2.0.0

The flow field will have only a velocity component in the x-direction. Outer forces won't be taken into account in this example.

The equations 1.0.0 to 1.0.2 will be simplified as:

2.1.0

2.1.1

2.1.2

The normal stresses yield to -p. They have no significance in this treatment and won't be considered in the following part.

The equations 1.1.0 to 1.1.2 can be expressed as:

2.2.0

2.2.1

2.2.2

The shear stresses exist only in the xz-pane in this example.

### Shear stresses at the origin system

In figure 2 the shear stresses at the system which described in equations 2.2.1, 2.3.1 and 1.2.0 are plotted. The shear stresses only depend on the velocity gradient in x-direction.

Figure 2: Stresses in a simple flow field

### Separation into subsystems

For the examination the considered physical model will be orthographically separated to the flow direction into two subsystems “1” and “2”. The generated subsystems have a common section plane.

Figure 3: System separation

### Shear stresses at the subsystems

At both subsystems the shear stresses will be determined. The equations and the physical model above are also valid for the subsystems. Due to the configuration of the flow field in 2.0.0 and with the equations 2.2.1 and 1.2.0 the shear tresses can be described as:

5.0.0

5.0.1

5.0.2

5.0.3

Figure 4: Shear stresses of the subsystems

### Superposing the subsystems

After determination the shear stresses at the subsystems, they will be superposed. This is shown in figure 5. At the section plane of both subsystems the following stresses must be considered:

6.0.0

6.0.1

Figure 5: Superposition of subsystem "1" and "2"

The resulting stress follows to:

6.1.0

This is shown in figure 6.

Figure 6: Resulting inner stress

At the section plane of the subsystems yields a resulting difference stress dτxz. This phenomenon doesn't occur at a separation parallel to the x-direction.

### Result

The resulting shear stress dτxz doesn't exist in the origin system.

That means, that the flow model of “Navier-Stokes” can't meet the requirement of the 3. Newtonian law “Actio = Reactio“. The sums of the subsystems stresses are not equal to the particular stresses of the complete system.

From mathematically sight it can be interpreted as a discontinuous of the shear stresses at the systems section planes. The shear stresses always contain a constant component being independent from the direction of their normal. This will be caused by the symmetry condition.

As shown above the physical model of “Navier-Stokes” violates the 3. Newtonian law. The symmetry condition of the shear stresses can be detected as the reason. Due to this, the components of the shear stresses behaviour orthographically to the flow direction is discontinuous. This caused that no complete solutions can be found for this system of equations.

### Formula symbols

μ : dynamic viscosity [Ns/m2]

ρ : density [kg/m3]

σ : normal stress [N/m2]

τ : shear stress [N/m2]

p : pressure [N/m2]

u : component of velocity x-direction [m/s]

v : component of velocity y-direction [m/s]

w : component of velocity z-direction [m/s]

: velocity vector

Double index:

1. denotes direction of the surface normal

2. denotes direction of vector

### Mathematical appendix

In a system from the stresses yield:

With the normal stresses can be written as

, , .

That yields:

### References

1. Prandtl's Essentials of Fluid Mechanics SE /Springer