A type D breakdown of the Navier Stokes equation in d=3 spatial dimensions

In this paper a type D breakdown of the Navier Stokes (NS) in d=3 is demonstrated. The element of the breakdown also occurs in the Euler equation. We consider the fact that in d=2 Ladyzhenskaya found a generalized type B solution. The discussion revolves around the notion, also found in quantum spin theory, that in the behavior of a system can be quite different from the behavior in d=2 dimensions. Concerning applications, our resolution of the problem implies that e.g. hydrology problems formulated as a NS equation can only be solved in computational approximation.


Introduction
The Navier Stokes (NS) equation is basic to many geophysical fluid mechanical problems. We mention here e.g. an application to groundwater flow (Masciopinto & Palmiotta, 2013) and coastal dynamics sediment transport giving dam breakdown (Morichon, Desombre, & Simian, 2013). There are however also many applications of the NS equation outside the field of geophysics. Basic physical aspects leading to the NS equation can be found in textbooks such as Prandtl's Essentials of Fluid Mechanics text (Oertel, 2004, p. 118 and further).
The problem of a possible exact solution of the NS equation in ℝ d , with d = 3, is still open today.
We cite here the transformation of the NS equation to a simpler form (Kozachok, 2013) and the work of Otelbayev (2013). Moreover, although Cafarelli c.s. produced an important inequality (Cafarelli, 1984) which is useful in the study of the NS equation, it is out of scope of the present paper. We also mention that for a finite time interval [0, T) in ℝ 3 , there is an exact solution that blows up when T increases to untraceable large size (Fefferman, 2000). The exploding temporal behavior is a clue to a possible breakdown.

PUBLIC INTEREST STATEMENT
In the present age, data analysis and mathematical modelling is becoming increasingly more important. Therefore it can be handy to know which mathematical problems have an exact solution and which problems do not. The Navier Stokes equation is used in many fields related to fluid dynamics. We mention for instance, acoustics, aerodynamics and space weather.

Aim of present work
In the paper we aim for a breakdown of type D. The requirements for that type of breakdown in d = 3 can be found in Fefferman (2000). The key characteristic of the type D breakdown lies in Du Dt . Because the Euler equation shares this form with the NS equation, the type D breakdown also occurs in the Euler equation. From a historical perspective, the NS equation is a viscous extension of the Euler equation Papanastasiou, Georgiou, and Alexandrou (2000) and Landau and Lifschitz (1987). This is also the starting point of the study of the linear non-stationary problem (Ladyzhenskaya, 1969, p. 81). Apparently, because d = 2 has an exact generalized solution, the addition of viscosity in fluid dynamics is most likely not causing a breakdown in d = 2. Moreover, although out of scope to the present paper, the cause of type D breakdown also resides in the Navier-Poisson equation as well (Papanastasiou et al., 2000, pp. 182-185).
The demonstration of type D breakdown starts with introducing the NS equation and the conditions to satisfy such a breakdown. Subsequently, particular velocity vectors are introduced and it is demonstrated that a breakdown force can match the initial velocity vectors. We find that, as required in a type D breakdown, no pressure function and velocity vector combination can be found in this case. An interesting extension of a possible breakdown in d = 3, namely the Oldroyd model in a viscoelastic NS equation Hynd (2013), will be discussed in a future study.
The breakdown velocity vectors are subsequently employed in a compound form and it is noted that, although the sum of the compound form is approximative, any selected point in time can be associated to a velocity breakdown equation. The reason is the fact that between two different points on the real axis, one has the continuum. In the limit (which only can be approximated) the compound form is exact.
The criterion for force is a force vector function which is independent of time and is incidentally also divergence free.
In d = 2 we have a generalized solution (Ladyzhenskaya, 1969). How does this relate to a type D breakdown in d = 3. In an analogy, we may note a case outside of continuum mechanics. In anyon spin statistics (Wilckzek, 1982), the reduction of spatial dimensions from d = 3 to d = 2 gives a profound change in the spin behavior of particles. In Fefferman (2000) the differences between a d = 2 and d = 3 NS equation are briefly mentioned.
Finally, as a practical consequence of the demonstrated type D breakdown, only a weak solution or an approximative numerical solution is allowed in the use of the incompressible viscous NS equation in d = 3. We mention fields such as geo-hydrology, hydro-meteorology, aerodynamics and space weather. In the latter case, for instance, ion drift studies relate the NS equation to electromagnetic phenomena (Shukla, 1982).

Preliminaries
In the first place let us select any function u 0 k (x) ∈ C N (ℝ 3 ), 1 < N ∼ large and, k = 1, 2, 3 with the following characteristics. The u 0 (x) vector in ℝ 3 , must be divergence free. Moreover, according to Equation (8) of Fefferman (2000) we must observe spatial periodicity, (1.3) for e j unit vectors, (e j ) m = j,m , with, j, m = 1, 2, 3. So the set of vector functions, can be defined. Of course we may select any function from  . Moreover, requirements for the force vector f in (1.1) are captured in the set with j, k = 1, 2, 3. Note, that N in both (2.1) and (2.2) is in ℕ with 1 < N. Again, = ( 1 , 2 , 3 ) with the k together with m, finite elements of ℕ. So, f ∈  where we must see arbitrary ∈ ℕ 3 , m ∈ ℕ, both finite, and K ∈ ℝ. In addition, n = x n = (1.1)

Initial definitions
Subsequently, let us define a vector function The c is a constant vector (c ≠ 0), independent of x ∈ ℝ 3 , t ∈ ℝ, t ≥ 0 and h.

Requirements (2) (3) and (10) (Fefferman, 2000)
Equation (3) of Fefferman (2000) does not apply as required in a type D breakdown, i.e. we don't have From Equation (2.3) it easily follows that, as required in a type D breakdown looking at (3) in Fefferman (2000), Fefferman (2000). More generally, however, we may write for the div and so in (2.6) there is the vanishing double sum of the divergence free u 0 k,j is the Kronecker delta. Looking again at (2.6) and at (2.5), we may conclude that in our case, Equation (2) of Fefferman (2000) breaks down.
Moreover, e j shifts do not hold for v h , for all x ∈ ℝ 3 . This is obviously true for t ≠ h looking at definition (2.3). It also holds, because of the linear x term in (2.3), for t = h. In addition, noting that for u 0 ∈  it follows e.g. u 0 (x + 2e j ) = u 0 (x + e j ) = u 0 (x). Hence, it can be observed from (2.3), that   (2000). To wrap this up, we don't have Fefferman's for our v h (x, t) and j = 1, 2, 3, as required in a D breakdown.

This leads to
In the third place, in  v h we must have The three results (2.12), (2.14) and (2.15) then give, for t = h, h ∈ [0, ∞), (2.16)

Or, after rewriting and noting that
The gradient of the scalar pressure, p = p(x, t) with for arbitrary p, gives, curl grad p(x, t) = 0.

Selection of f, requirement (9) (Fefferman, 2000)
If we note that f must be in , defined in (2.2), then we can select u 0 from  , defined in (2.1) such that a stationary f = f crl is given by and is in , see (2.2). Note that the f crl must be divergence free. We have f crl (x + e j ) = f crl (x). Note also that condition (9) of Fefferman (2000) for stationary force, suppressing the notation of the t = 0 in the function, is The constant K is no longer of importance to the upper limit in (2.19). For completeness, the original condition is given in (1.4) and used in the definition in Equation (2.2). If t = 0, then substituting this on both sides of the inequality expressing the condition we find is

Subset criteria for u 0 selection and their use in the breakdown
It is furthermore assumed that: • u 0 is not a vector function of absolute constants, • u 0 is not a gradient vector of an arbitrary proper function (x) and • ∇ × ∇ × u 0 ∕ ≡0 . This is possible because we have C N ℝ 3 u 0 functions 1 < N ∼ large.
The previous three points are the constitutive description of a subset of functions of  , given in (2.1). A fourth point is added subsequently. Looking at (2.18) and at (2.19), we may note that the function u 0 must be bounded in order to satisfy f crl ∈  i.e. its equivalent in (2.19) in combination with the f crl (x) = ∇ × u 0 (x) +  c, u 0 (x) of (2.18). We then denote this series of additional conditions on u 0 with  ′ ⊂  and select u 0 ∈  � . Of course, the definition of  is such that we are able to have a non empty subset  ′ . Moreover, because of  c, u 0 (x + e j ) =  c, u 0 (x) and f ∈ , it follows from from (2.17) that Hence, if p is possible in the NS equation with (u 0 , f ), the gradient p cannot be spatial periodic when v h in (2.4) is not spatial periodic. This agrees with the fact that v h obeys the type D requirement of not being a solution of Equation (10) of Fefferman (2000).
(2.17) If we then substitute f crl given in (2.18) into Equation (2.17), noting that we may write f crl (x) = f (x, h), then, because the left hand side of (2.17), i.e. (c + x) + D c, u 0 (x) = f (x, h) − (∇p)(x, h), does not contain h dependence, it follows that for arbitrary h ∈ [0, ∞) and c an absolute constant vector, Hence, it suffices to inspect p(x, h) = p(x). We note that ∇ × c = ∇ × x = 0. Hence, Because of the use of C N ℝ 3 , 1 < N ∼ large, functions, we are allowed to take the curl on left and right hand side of (2.22) and note that because u 0 ∈  � , Equation (2.23) cannot be fulfilled. This is so because curl on grad results for all p into the zero vector on the left hand side of this Equation (2.23) while on the right hand side ∇ × ∇ × u 0 (x) is not the zero vector by necessity. It was assumed that u 0 ∈  � , which entails, ∇ × ∇ × u 0 (x)∕ ≡0.
Hence, with f crl ∈  and u 0 ∈  � it is possible to define a function v h in (2.4), using a u 0 from  ′ with arbitrary h ∈ [0, ∞) such that it is impossible to find a p = p(x, t) function. Hence it is not possible to find (p, v h ) that solves the NS Equation (1.1) with a valid (u 0 , f crl ). This represents requirement (1) of Fefferman (2000).
Also, in anticipation of the discussion, in d = 2, the curl is a scalar. See e.g. Aris (1989, p. 223). Hence, the curl geometry in d = 2 is quite different from the one in d = 3.

Euler equation
From the previous analysis of the NS equation a differential form is obtained with,  c, ≡ 1 + c ⋅ ∇ − ∇ 2 . In the Euler equation there is no viscosity, i.e. = 0, (Papanastasiou et al., 2000). The differential form then is equal to  c, ≡ 1 + c ⋅ ∇. In that particular case we can also select, in a way similar to (2.18), using = 0, a f crl (x) force vector function from . This leads us to the same impossibility as presented in (2.23). The impossibility amounts to, on the one hand the d = 3 gradient of the pressure, which vanishes under d = 3 curl operation and ∇ × ∇ × u 0 (x)∕ ≡0 for a u 0 ∈  � on the other hand. The reason for a similar breakdown is that the NS and the Euler equation share the total differential Du Dt form.

Compound form
The previous analyses of both the NS and the Euler equation can be generalized with the use of the set of intervals Λ defined by The global function u can subsequently be based on Λ and is defined by Here, (t) = 1, when, t ∈ and (t) = 0, when, t ∉ . Hence, we can see that u( The definition of the form in (2.25) resembles the use of discreteness in the Feynman path integral, Feynman and Hibbs (1965, pp. 30-39) and Kumano-go (2008).

Conclusion and discussion
The definition of u in (2.25) is an increasingly better -and unstoppable-approximation of the possibility that u(x, t) = v h (x, t) and t in [h, h + ). The analysis can be maintained because of the continuum between h and h + , with 0 < → 0. Hence, conclusions for v h have their impact on u. If no v h can be found then no u can be found either. Therefore, it is sufficient to only look at the D breakdown yes-or-no possibility of v h defined in (2.3) and (2.4).
Furthermore, the curl in d = 3 is an operation, curl:ℝ 3 → ℝ 3 . The difference with d = 2 is that in that case the curl is a scalar (Aris, 1989). Note btw that if one employs the 3-curl to compute the curl of a d = 2 vector in a space spanned by e.g. e 1 and e 2 , by extension such that e.g. u 3 = 0, then, ∇ × u, projects out of the space spanned by e 1 and e 2 . Hence, such an extension is not a d = 2 curl.
Nevertheless it can be noted that in a possible exact breakdown type D for d = 2, home made curl operators can perhaps provide sufficient tooling. An interesting example is based on the home-made outer product, × 2 . It is defined for v = (v 1 , v 2 ) and u = (u 1 , u 2 ) by v × 2 u = e 1 v 2 − e 2 v 1 v 1 u 2 − v 2 u 1 . With it we define a curl in d = 2 for j = x j and grad 2 = ( 1 , 2 ), as and w = w(x 1 , x 2 , t) ∈ ℝ 2 . The curl is obtained from our × 2 definition when v 1 → 1 and v 2 → 2 is substituted. Note the required div 2 curl 2 w = 0 and curl 2 grad 2 w = 0. Furthermore, the operation Another point that can be raised against curl 2 = grad 2 × 2 is that for parity transformation ( 1 , 2 ) → (− 1 , − 2 ), by definition we must have (− grad 2 )× 2 = grad 2 × 2 . Consequently, the d = 2 breakdown equation, grad 2 p(x) = grad 2 × 2 u 0 (x) − (c + x) shows an inconsistency between p(−x) = sp(x) and u 0 (−x) = su 0 (x), s = ±1, on the one hand and curl 2 curl 2 u 0 (x)∕ ≡0 on the other hand. This awkward behavior is no big surprise because an official curl 2 :ℝ 2 → ℝ 2 is non-existent.
To return to our main conclusion. There is a type D breakdown in d = 3 and it does not disagree with the prove (Ladyzhenskaya, 1969) that there is a generalized solution in d = 2 space. This is a valid statement despite the possibility of a most likely fit-for-purpose home made operator curl 2 .
Interestingly, the qualitative difference between d = 2 and d = 3 spatial dimensions occurs elsewhere too. There is a remarkable difference in behavior, also observed experimentally (Laughlin, 1983), of quantum spin statistics in d = 2 versus d = 3. This difference is between complete freedom of spin statistics of 'particles' in a d = 2 space, i.e. the so called anyon spin, opposed to the spin statistics of particles that are allowed the freedom of a d = 3 space. We see a discrete number (e.g. 2 for electrons) of possibilities for spin states in d = 3. The research into this difference was initiated by Leinaas and Myrheim (1977) and Wilckzek (1982) and further explained and deepened by e.g. Lerda (1992) and Kitaev (2006). Apparently adding or subtracting a spatial degree of freedom provides significant changes of behavior of a system. This apparently goes for systems under the rule of quantum mechanics as well as for systems under the rule of continuum mechanics.
Concerning the C N set of functions, the question is raised whether a possible rejection of the breakdown presented in this paper can be based on its use of 1 < N ∼ large. The reader is invited to capture, with genuine differentiation operations, the difference between N very large and, growing if necessary but finite, as opposed to infinite continuous differentiable. The set of functions,  ′ , probably also has an untraceable large cardinality. The breakdown of (p, u) with C N , with N → ∞ requirement can be replaced with N 'untraceable large' and, growing when considered necessary. In the final analysis, the NS problem with its requirements is rooted in the physics of fluids. Note however that the present analysis also does not escape completely from the illusion: "infinity is known curl 2 w = grad 2 × 2 w