f}{\partial V};\\
\label{DPN1}
0&=&-\frac1W P_V P_W+\frac V{2W^2}P_V^2+\frac{W''W}V-\frac{V'W'W}{V^2}
   +\frac{(W')^2}{2V}-\frac V2;\\
\label{DNr1}
\dot N^r&=&0;\\
\label{DPNr1}
0&=&(P_V)'V-P_W W';\\
\label{DV1}
\dot V&=&\frac N W P_W-\frac{N V}{W^2}P_V;\\
\label{DPV1}
\dot P_V&=&\frac N{2W^2}P_V^2+\frac{N'W'W}{V^2}
   +\frac{N(W')^2}{2V^2}-\frac N2;\\
\label{DW1}
\dot W&=&\frac N W P_V;\\
\label{DPW1}
\dot P_W&=&\frac N{W^2}P_V P_W-\frac{N V}{W^3}P_V^2
   +\frac{N''W}V-\frac{N'V'W}{V^2}+\frac{N'W'}V+\frac{N W''}V-\frac{N V'W'}{V^2}.\end{aligned}$$

### The Schwarzschild solution.

The Schwarzschild solution [@Schw] is probably the most famous spherically symmetric solution to the Einstein equations. First of all, we choose the gauge condition (\[g.c.1.1\]) in the form $$\label{g.c.1.2}
N=\frac1V.$$

Following to Schwarzschild [@Schw], we shall seek for a solution that does not depend on time. Then, Eq. (\[eLNr1\]) is identically satisfied, and Eqs. (\[eLN1\]), (\[eLV1\]) under the condition (\[g.c.1.2\]) look like $$\begin{aligned}
\label{eLN2}
0&=&\frac V2-\frac{(W')^2}{2V}-\frac{W W''}V+\frac{V'W'W}{V^2};\\
\label{eLV2}
0&=&\frac V2-\frac{(W')^2}{2V}+\frac{V'W'W}{V^2}.\end{aligned}$$ From (\[eLN2\]), (\[eLV2\]) we get $$\label{Schw.W}
W''=0;\qquad
W(r)=C_1r+C_2.$$ One can put $C_1=1$, $C_2=0$, so that the length of a circle with a center in coordinate origin would be $2\pi r$. Then $W(r)=r$. Eq. (\[eLV2\]) takes the form $$\label{eLV2_1}
0=\frac V2-\frac1{2V}+\frac{V'r}{V^2}.$$ Its solution can be written as $$\label{Schw.sol}
V(r)=\frac1{\sqrt{1-\displaystyle\frac{2GM}r}},$$ where $M$ is associated with a Schwarzschild mass. Eq. (\[eLW1\]) is identically satisfied. It is easy to see that the same solution follows from Eqs. (\[DN1\]) – (\[DPW1\]). In our consideration we have started from 