  The linear Navier-Stokes equations in three dimensions are given by:
$u_{it}(x,t)-\rho \triangle u_i(x,t)-p_{x_i}(x,t)=$ $w_i(x,t)$ , $div
\textbf{u}(x,t)=0,i=1,2,3$ with initial conditions:
$\textbf{u}|_{(t=0)\bigcup\partial\Omega}=0$. The Green function to the
Dirichlet problem $\textbf{u}|_{(t=0)\bigcup\partial\Omega}=0$ of the equation
$u_{it}(x,t)-\rho\triangle u_i(x,t)=f_i(x,t)$ present as:
$G(x,t;\xi,\tau)=Z(x,t;\xi,\tau)+V(x,t;\xi,\tau).$ Where
$Z(x,t;\xi,\tau)=\frac{1}{8\pi^{3/2}(t-\tau)^{3/2}}\cdot
e^{-\frac{(x_1-\xi_1)^2+(x_2-\xi_2)^2+(x_3-\xi_3)^2}{4(t-\tau)}}$ is the
fundamental solution to this equation and $V(x,t;\xi,\tau)$ is the smooth
function of variables $(x,t;\xi,\tau)$. The construction of the function
$G(x,t;\xi,\tau)$ is resulted in the book [1 p.106]. By the Green function we
present the Navier-Stokes equation as:
$u_i(x,t)=\int_0^t\int_{\Omega}\Big(Z(x,t;\xi,\tau)+V(x,t;\xi,\tau)\Big)\frac{dp(\xi,\tau)}{d\xi}d\xi
d\tau +\int_0^t\int_{\Omega}G(x,t;\xi,\tau)w_i(\xi,\tau)d\xi d\tau$. But $div
\textbf{u}(x,t)=\sum_1^3 \frac{du_i(x,t)}{dx_i}=0.$ Using these equations and
the following properties of the fundamental function: $Z(x,t;\xi,\tau)$:
$\frac{dZ(x,t;\xi,\tau)}{d x_i}=-\frac{d Z(x,t; \xi,\tau)}{d \xi_i},$ for the
definition of the unknown pressure p(x,t) we shall receive the integral
equation. From this integral equation we define the explicit expression of the
pressure: $p(x,t)=-\frac{d}{dt}\triangle^{-1}\ast\int_0^t\int_{\Omega}\sum_1^3
\frac{dG(x,t;\xi,\tau)}{dx_i}w_i(\xi,\tau)d\xi
d\tau+\rho\cdot\int_0^t\int_{\Omega}\sum_1^3\frac{dG(x,t;\xi,\tau)}{dx_i}w_i(\xi,\tau)d\xi
d\tau.$ By this formula the following estimate:
$\int_0^t\sum_1^3\Big\|\frac{\partial p(x,\tau)}{\partial
x_i}\Big\|_{L_2(\Omega)}^2 d
\tau<c\cdot\int_0^t\sum_1^3\|w_i(x,\tau)\|_{L_2(\Omega)}^2 d\tau$ holds.
