condition $(\ref{eqn3})\;\; (\vec{u}^{0} (x_{1}, x_{2}, x_{3})\in \overrightarrow{TS})$:

$$\label{eqn202}
\vec{u}(x_{1}, x_{2}, x_{3}, t)|_{t = 0}\;=\;\vec{u}^{0}(x_{1}, x_{2}, x_{3})$$

Let us put {$\vec{u}(x_{1}, x_{2}, x_{3}, t)$, p($x_{1}, x_{2}, x_{3},$ t)} into equations $(\ref{eqn1})$ , $(\ref{eqn2})$ and apply Fourier and Laplace transforms to the result identities considering initial condition $(\ref{eqn3})$. After all required operations (as in sections 2 and 3) we receive that vector-function $\vec{u}(x_{1}, x_{2}, x_{3}, t)$ satisfies integral equation:

$$\label{eqn203}
\vec{u}\;=\;\bar{\bar{S}}\;\cdot\;\vec{f}\;- \bar{\bar{S}}\cdot(\vec{u}\cdot\nabla)\vec{u}\;+\;\bar{\bar{B}}\cdot\vec{u}^{0}\; = \bar{\bar{S}}^{\nabla}\cdot\vec{u}$$

Then vector-function grad p $\in \overrightarrow{TS}$ is defined by equations $(\ref{eqn1})$ where vector-function $\vec{u}$ is received from equation $(\ref{eqn203})$.

Here $\vec{f} \in \overrightarrow{TS}$, $\vec{u}^{0} \in \overrightarrow{TS}$ and $\bar{\bar{S}},\;\bar{\bar{B}},\;\;\bar{\bar{S}}^{\nabla}\;$ are matrix integral operators.

Vector-functions $ \bar{\bar{S}}\cdot\vec{f},\;\;\bar{\bar{B}}\cdot\vec{u}^{0},\;\; \bar{\bar{S}}\cdot(\vec{u}\cdot\nabla)\vec{u}\;$ are also belong $\overrightarrow{TS}$ since Fourier transform maps perfect space $\overrightarrow{TS}$ onto $\overrightarrow{TS}$.

Going from the other side, let us assume that $\vec{u}(x_{1}, x_{2}, x_{3}, t)\in \overrightarrow{TS}$ is continuous in t $\in$ \[0,$\infty$) solution of integral equation $(\ref{eqn203})$. Integral-operators $S_{ij}\cdot(\vec{u}\cdot\nabla)\vec{u}$ are continuous in t $\in$ \[0,$\infty$) \[see $(\ref{eqn160})\;-\;(\ref{eqn162})$\]. From here we receive that according to $(\ref{eqn203})$ $\\$

$\vec{u}(x_{1}, x_{2}, x_{3}, 0)  = \vec{u}^{0}(x_{1}, x_{2}, x_{3})$\

and also that $\vec{u}(x_{1}, x_{2}, x_{3}, t)$ is differentiable by t $\in$ \[0,$\infty$). As described before, the Fourier transform maps perfect space $\overrightarrow{TS}$ on itself. Hence, {$\vec{u}(x_{1}, 