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

and Λ( 1 , 2 , 3 , ) is given as, Λ( 1 , 2 , 3 , ) = 2 0 ( ) ( 1 , 2 , 3 ) 3 ( 1 , 2 , 3 , ) ∂ 3 ∂ 1 + 2 0 ( ) ( 1 , 2 , 3 ) 3 ( 1 , 2 , 3 , ) ∂ 3 ∂ 2 − 3 3 � ∂ 3 ∂ 3 � ( 1 , 2 , 3 , ) + 2 �� ∂ 3 ∂ 3 � ( 1 , 2 , 3 , ) + 3 ∂ ∂ 3 � where ⃗ = � 1 , , 2 , � is the forcing vector and ⃗ = ( 1 , 2 , 3 ) is the velocity in each cell of the 3-Torus. For the three forcing terms, set them equal to products of reciprocals of degenerate Weierstrass P functions shifted in spatial coordinates from the center ( , , ) , = 1. . . Here the ( , , ) is the center of each cell of the lattice belonging to the flat torus. Upon substituting the Weierstrass P functions and their reciprocals (unity divided by P-function) into Eq.(1) together with the forcing terms given by Λ, it can be observed that in the equation that terms in it are multiplied by reciprocal Weierstrass P functions which touch the centers of the cells of the lattice, thus simplifying Eq.(1). The initial condition in 3 at = 0 is instead of a product of reciprocal degenerate Weierstrass P functions for forcing, is a sum of these functions. The parameter in the degenerate Weierstrass P function, if chosen to be small gives a ball, = { ∈ ℝ 3 : �| |� 2 = (| 1 | 2 + | 2 | 2 + | 3 | 2 ) 1 2 ≤ } Here we are in Cartesian space ℝ 3 with 2-norm 2 . Since the terms are squared in length in the initial condition for 3 we require to multiply by dynamic viscosity to obtain units of velocity. In the above, the forcing is taken to be different than the gradient of pressure. Introducing the space ℑ( 3 , ) = � ∈ ℝ + , 3 ∈ � 3 ; � : 2 1 1 + 2 = 0 & 1 + 2 + = 0, ∀ 1 , 2 ∈ × ( ⊂ ℝ )& 2 = 12 & 3 ( 1 , 2 , 3 , ) ∈ 0 ( 3 )� , where � 3 ; � is the 1-dimensional -ball centered at 3 , i=1,2,…N, ranging through the expanding lattice generated by the flat torus.. The point 3 coincides with the center point ( , , ) , where ⃑ = ( 1 − , 2 − , 3 − ), = 1,2, … . = �( − 1) 1 3 + 2 3 1 � 2 3 3 1 + �( − 1) 2 3 + 2 3 2 � 2 3 3 2 − 3 � 3 3 1 2 1 3 + 3 3 2 2 2 3 − 3 1 3 3 1 − 3 2 3 3 2 � + 3 3 1 1 2 3 3 2 = �( − 1) 1 3 + 2 3 1 � 2 3 3 1 + �( − 1) 2 3 + 2 3 2 � 2 3 3 2 + 3 � 3 1 3 3 1 + 3 2 3 3 2 � + � 2 � 2 3 3 2 � 3 2 � 3 + � 1 � 2 3 3 1 � 3 1 � 3 Next the sum of the two first vorticities is used together with the vorticity sum set to the sum of the first two components of the equivalent expression which is twice the angular velocity, 1 + 2 = 2 2 3 − 2 1 3 − 2 3 ( 2 − 1 ) 1 2 + 2 2 + 3 2 Thus using the definition of vorticity we have the following equation in the space ℑ( 3 , ) , The 3 points are along segments parallel to the 3 -axis, throughout the lattice. For points belonging to the space ℑ( 3 , ) , the following part of Eq.(1) is exactly zero: That is = 0 on the subspace ℑ( 3 , ). 1 , 2 are linearly dependent in this space. In the second equivalent expression for , in the space ℑ( 3 , ) , = 0. © 2023 Global Journals Global Journal of Researches in Engineering Volume XxXIII Issue I Version I 49 Year 2023 ( ) I Exploring Finite-Time Singularities and Onsager’s Conjecture with Endpoint Regularity in the Periodic Navier Stokes Equations + 3 3 2 2 2 3 3