K}(1+\mid x \mid)^{-K} \quad \rm{on }\;R^{N}\;\rm{ for\; any }\;\alpha \;\rm{ and \;}K.$$

and $$\label{eqn6}
\mid\partial_{x}^{\alpha}\partial_{t}^{\beta}\vec{f}(x,t)\mid\;\leq\;C_{\alpha \beta K}(1+\mid x \mid +t)^{-K} \quad \rm{on }\;R^{N}\times[0,\infty)\; \rm{ for \;any }\;\alpha,\beta \;\rm{ and \;}K.$$

To start the process of solution let us add ($ - \sum_{n=1}^{N} u_{n}\frac{\partial u_{k}}{\partial x_{n}}\;$) to both sides of the equations (\[eqn1\]). Then we have:

$$\label{eqn7}
\frac{\partial u_{k}}{\partial t}\;=\;\nu\,\Delta\,u_{k}\;-\;\frac{\partial p}{\partial x_{k}}\;+\;f_{k}(x,t)- \; \sum_{n=1}^{N} u_{n}\frac{\partial u_{k}}{\partial x_{n}}\;\;\;\;\;\;\;\;\;\;\;\; (x\in R^{N},\;\;t\geq 0,\;\;{1\leq k \leq N})$$

$$\label{eqn8}
\emph{div}\,\vec{u}\;=\; \sum_{n=1}^{N} \frac{\partial u_{n}}{\partial x_{n}}\; =\;0\;\;\;\;\;\;\;\;\;\;  (x\in R^{N},t\geq 0)$$

$$\label{eqn9}
\vec{u}(x,0)\; = \; \vec{u}^{0}(x)\;\;\;\;\;\;\;\;\;\; (x\in R^{N})$$

$$\label{eqn10}
u_{k}(x,t) \rightarrow 0\;\;, \;\;\frac{\partial u_{k}}{\partial x_{n}}\;\rightarrow\;0\;\; \rm{as} \;\;\mid x \mid \;\rightarrow\; \infty\;\;\;( {1\leq k \leq N} ,\;\; {1\leq n \leq N}) \;\;\;$$

$$\label{eqn11}
\mid\partial_{x}^{\alpha}\vec{u}^{0}(x)\mid\;\leq\;C_{\alpha K}(1+\mid x \mid)^{-K} \quad\rm{on }\;R^{N}\;\rm{ for \; any }\;\alpha\;\rm{ and }\;K.$$

$$\label{eqn12}
\mid\partial_{x}^{\alpha}\partial_{t}^{\beta}\vec{f}(x,t)\mid\;\leq\;C_{\alpha \beta K}(1+\mid x \mid +t)^{-K} \quad\rm{on }\;R^{N}\times[0,\infty)\;\rm{ for \; any }\;\alpha,\beta\;\rm{ and }\;K.$$

Let us denote

$$\label{eqn13}
\tilde{f}_{k}(x,t)\; = \; f_{k}(x,t) \; - \; \sum_{n=1}^{N} u_{n}\frac{\partial u_{k}}{\partial x_{n}}\;\;\;\;\;\;({1\leq k \leq N})$$

or we can present it in the vector form:

$$\label{eqn14}
\vec{\tilde{f}}(x,t)\; = \; \vec{f}(x,t) \; - \;(\;\vec{u}\;\cdot\;\nabla\;)\;\vec{u}\;$$

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Solution of the system of equations $\textbf {(\ref{eqn7})}\;$ - $\;\textbf {(\ref{eqn14})}$
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Let us assume that all operations below are valid. The validity of 