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The Navier-Stokes Equation and Helmholtz Decomposition

General Mathematics 2024-01-22 v5

Abstract

This work explores Navier-Stokes equation with no gravitational forces. In short, it shows that any smooth solution that decays quickly must take the form u(x,t)14πCurl(R3Curl(u(x,t))xxdV)=0t1ρGrad(Γ(x,s))ds. \textbf{u}(x,t)- \dfrac{1}{4\pi}\textbf{Curl}\Biggl( \int_{\mathbb{R}^3}^{}{\dfrac{\textbf{Curl} (\textbf{u} (x^\prime,t))}{|x-x^\prime|}}dV^\prime\Biggr) = -\int_{0}^{t}{\dfrac{1}{\rho} \textbf{Grad}\big(\Gamma(x,s)\big)}ds. Consequently, any curl free solution must be written as u(x,t)=1ρGrad(0tΓ(x,s)ds)\textbf{u}(x,t) = -\dfrac{1}{\rho} \textbf{Grad}\biggl(\int_{0}^{t}{\Gamma(x,s) ds}\biggr) with Γ\Gamma a known function which is related to the heat equation. Even further it shows if there exist a value kNk\in \mathbb{N} such that curlk((u)u)(x,t)=0\textbf{curl}^k\biggl((\textbf{u}\cdot \nabla )\textbf{u}\biggr)(x,t)=\textbf{0} for all ttt^\prime\le t then u(x,t)=Hk+1(ξ1,ξ2,ξ3,t)0t1ρGrad(Γ(x,s))ds,     t[t,)\textbf{u}(x,t) = \textbf{H}^{k+1}(\xi_1,\xi_2,\xi_3,t) -\int_{0}^{t}{\dfrac{1}{\rho} \textbf{Grad}\big(\Gamma(x,s)\big)}ds, ~~~~~ t\in [t^\prime,\infty) with ξi(x,t):=R3α(xy,tν)vik(x,0)dy,     vik(x,0)=(curlk(u(x,0)))i,     1i3\xi_i(x,t):= \int_{\mathbb{R}^3}^{}{\alpha(x-y,\dfrac{t}{\nu})v^k_i(x,0)}dy, ~~~~~ v^k_i(x,0) = \biggl(\textbf{curl}^k(\textbf{u}(x,0))\biggr)_i, ~~~~~ 1\le i\le 3 and Hk\textbf{H}^k the kthk^{th} application of Helmholtz operator. Hence, if there is another solution where the non-linear term is infinitly curlable then the solution is not unique. If the solution is unique, then this is the only possible solution.

Keywords

Cite

@article{arxiv.2302.14852,
  title  = {The Navier-Stokes Equation and Helmholtz Decomposition},
  author = {Roy Burson},
  journal= {arXiv preprint arXiv:2302.14852},
  year   = {2024}
}

Comments

I would like to give special thanks to all my professors especially Dr. Enakousta for helping me anytime I needed it

R2 v1 2026-06-28T08:52:16.957Z