The Navier-Stokes Equation and Helmholtz Decomposition
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arXiv
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| Format: | Preprint |
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2023
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| _version_ | 1866917570684125184 |
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| author | Burson, Roy |
| author_facet | Burson, Roy |
| contents | 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
$$ \textbf{u}(x,t)- \dfrac{1}{4π}\textbf{Curl}\Biggl( \int_{\mathbb{R}^3}^{}{\dfrac{\textbf{Curl} (\textbf{u} (x^\prime,t))}{|x-x^\prime|}}dV^\prime\Biggr) = -\int_{0}^{t}{\dfrac{1}ρ \textbf{Grad}\big(Γ(x,s)\big)}ds.$$ Consequently, any curl free solution must be written as $$\textbf{u}(x,t) = -\dfrac{1}ρ \textbf{Grad}\biggl(\int_{0}^{t}{Γ(x,s) ds}\biggr)$$ with $Γ$ a known function which is related to the heat equation. Even further it shows if there exist a value $k\in \mathbb{N}$ such that $$\textbf{curl}^k\biggl((\textbf{u}\cdot \nabla )\textbf{u}\biggr)(x,t)=\textbf{0}$$ for all $t^\prime\le t$ then $$\textbf{u}(x,t) = \textbf{H}^{k+1}(ξ_1,ξ_2,ξ_3,t) -\int_{0}^{t}{\dfrac{1}ρ \textbf{Grad}\big(Γ(x,s)\big)}ds, ~~~~~ t\in [t^\prime,\infty)$$ with $$ξ_i(x,t):= \int_{\mathbb{R}^3}^{}{α(x-y,\dfrac{t}ν)v^k_i(x,0)}dy, ~~~~~ v^k_i(x,0) = \biggl(\textbf{curl}^k(\textbf{u}(x,0))\biggr)_i, ~~~~~ 1\le i\le 3$$ and $\textbf{H}^k$ the $k^{th}$ 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. |
| format | Preprint |
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arxiv_https___arxiv_org_abs_2302_14852 |
| institution | arXiv |
| publishDate | 2023 |
| record_format | arxiv |
| spellingShingle | The Navier-Stokes Equation and Helmholtz Decomposition Burson, Roy General Mathematics 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 $$ \textbf{u}(x,t)- \dfrac{1}{4π}\textbf{Curl}\Biggl( \int_{\mathbb{R}^3}^{}{\dfrac{\textbf{Curl} (\textbf{u} (x^\prime,t))}{|x-x^\prime|}}dV^\prime\Biggr) = -\int_{0}^{t}{\dfrac{1}ρ \textbf{Grad}\big(Γ(x,s)\big)}ds.$$ Consequently, any curl free solution must be written as $$\textbf{u}(x,t) = -\dfrac{1}ρ \textbf{Grad}\biggl(\int_{0}^{t}{Γ(x,s) ds}\biggr)$$ with $Γ$ a known function which is related to the heat equation. Even further it shows if there exist a value $k\in \mathbb{N}$ such that $$\textbf{curl}^k\biggl((\textbf{u}\cdot \nabla )\textbf{u}\biggr)(x,t)=\textbf{0}$$ for all $t^\prime\le t$ then $$\textbf{u}(x,t) = \textbf{H}^{k+1}(ξ_1,ξ_2,ξ_3,t) -\int_{0}^{t}{\dfrac{1}ρ \textbf{Grad}\big(Γ(x,s)\big)}ds, ~~~~~ t\in [t^\prime,\infty)$$ with $$ξ_i(x,t):= \int_{\mathbb{R}^3}^{}{α(x-y,\dfrac{t}ν)v^k_i(x,0)}dy, ~~~~~ v^k_i(x,0) = \biggl(\textbf{curl}^k(\textbf{u}(x,0))\biggr)_i, ~~~~~ 1\le i\le 3$$ and $\textbf{H}^k$ the $k^{th}$ 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. |
| title | The Navier-Stokes Equation and Helmholtz Decomposition |
| topic | General Mathematics |
| url | https://arxiv.org/abs/2302.14852 |