Global Journal of Researches in Engineering, I: Numerical Methods, Volume 23 Issue 1

Here it is clear that there exists a solution of PNS that is not smooth in time for the first and higher derivatives. The Full Equation Proof for the Periodic Navier Stokes Equations Integrating the Navier Stokes equations over an epsilon ball we obtain, ∫ ; 1 + 2 + 4 = −∫ ; 3 (IV) The first part of 3 becomes, 3 = ∇ ∙ ∇( Ξ ) , where ∇( Ξ ) = 1 � 32 ∇ 1 2 + �⃗ 3 3 � where is the pressure and is the density of the fluid. Dividing Eq.(IV) by the measure or volume of the ball of radius epsilon centered at point . ; we know since Ξ is continuous everwhere on the 3-Torus (since integrals are continuous in inverting gradient), and in particular at the the center of the epsilon ball (note higher order derivatives of 3 blowup, not 3 and pressure), then, (V) However using the Fet theory, we can see that the integral on the RHS of Eq.(IV) divided by the volume of the ball is related to the integral over the ball centered at of 1 4 2 (Ξ( ) − Ξ(a) ). Using Eq.(V), we obtain a difference of exactly zero so that we are left with, lim →0 1 | ; | ∫ ; 1 + 2 + 4 = 0 ∫ ; 1 + 2 + 4 = 0 (VI) Eq.(VI) is the PDE we obtained previously and occurs at an arbitrarily small epsilon ball centered at each cell of the lattice of the 3-Torus. In reference [5], we showed that, 1 + 2 + 4 = 3Φ( ) Φ( ) ≥ 0 and 1 + 2 + 4 ≥ 0 Also Ξ 2 = ‖ ‖ � �⃗ � = � 3 �⃗ ∙ � �⃗ ⨂Δ 3 �� � �⃗ � where Ξ 1 = Ξ + Ξ 2 Recall that the three velocities are isotropic and they are continuous on ; and Ξ 2 is continuous on the epsilon ball. Also Ξ 2 is independent of for Ḧ lder continuous functions at = 1/3. Also if we specify the time, the solution is a Ḧ lder continuous function in the data with a Ḧ lder constant equal to one half. Global Journal of Researches in Engineering Volume XxXIII Issue I Version I 58 Year 2023 © 2023 Global Journals ( ) I Exploring Finite-Time Singularities and Onsager’s Conjecture with Endpoint Regularity in the Periodic Navier Stokes Equations lim →0 1 | ; | ∫ ; Ξ ( ) = Ξ (a) Since the negative pressure gradients are greater than or equal to zero being reciprocal Weierstrass P functions and 32 ≥ 0 and 1 and 2 cancel in the space ℑ( 3 , ) when integrating on the six faces of surface of a cell of 3 , we have that,