The Navier-Stokes Equation and Helmholtz Decomposition

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 $$\mathbf{\text{u}}(x,t)-{\displaystyle \frac{1}{4\pi }}\mathbf{\text{Curl}}\left({\int }_{{\mathbb{R}}^3}^{}{\displaystyle \frac{\mathbf{\text{Curl}}(\mathbf{\text{u}}(x^{\mathrm{\prime }},t))}{\mathrm{\mid }x-x^{\mathrm{\prime }}\mathrm{\mid }}}d{V}^{\mathrm{\prime }}\right)=-{\int }_0^t{\displaystyle \frac{1}{\rho }}\mathbf{\text{Grad}}(\mathrm{\Gamma }(x,s))ds\mathrm{.}$$ Consequently, any curl free solution must be written as $$\mathbf{\text{u}}(x,t)=-{\displaystyle \frac{1}{\rho }}\mathbf{\text{Grad}}\left({\int }_0^t\mathrm{\Gamma }(x,s)ds\right)$$ with $\mathrm{\Gamma }$ 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 $${\mathbf{\text{curl}}}^k((\mathbf{\text{u}}\cdot \mathrm{\nabla })\mathbf{\text{u}})(x,t)=\mathbf{\text{0}}$$ for all $t^{\mathrm{\prime }}\le t$ then $$\mathbf{\text{u}}(x,t)={\mathbf{\text{H}}}^{k+1}({\xi }_1,{\xi }_2,{\xi }_3,t)-{\int }_0^t{\displaystyle \frac{1}{\rho }}\mathbf{\text{Grad}}(\mathrm{\Gamma }(x,s))ds,t\in [t^{\mathrm{\prime }},\mathrm{\infty })$$ with $${\xi }_i(x,t):={\int }_{{\mathbb{R}}^3}^{}\alpha (x-y,{\displaystyle \frac{t}{\nu }})v_i^k(x,0)dy,v_i^k(x,0)=({\mathbf{\text{curl}}}^k(\mathbf{\text{u}}(x,0)){)}_i,1\le i\le 3$$ and ${\mathbf{\text{H}}}^k$ 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.