(\[g\_0\]) is presented in fig. \[fig1\]. Comparing the numerical results and the simplified solution (\[w1\]) we see a good agreement (see fig. \[fig1\]). The plateau for a wide range of velocities is formed after a short time, $t=0.1$s, and it remains almost unchanged up to the end of the calculation. For the time $t>0.1$s, the drift of the Langmuir wave spectrum toward smaller phase velocities becomes observable. At $t=0.5$s, the maximum phase velocity is half of the initial beam velocity.

Plasma density increasing with distance
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An increasing plasma density leads to a shift toward larger phase velocities. For $v>v_0$ we have a negative derivative at the edge of the electron distribution function, and electrons absorb waves with the corresponding phase velocities. Absorption of waves then leads to acceleration of particles. This process continues until all the waves generated during the beam relaxation are absorbed by the electrons.

In the case of increasing density we are unable to find an exact solution, but we can find the solution for $t\rightarrow \infty$. Using conservation of energy ([@Ryutov69]) $$\begin{aligned}
\label{e_cons}
  \omega _{pe}\int_0^{u(t)}\frac{W(v,t)}{v^2}dv +
  \int_0^{u(t)}\frac{mn'}{2u(t)}v^2dv \cr
  =\frac m2 \int_0^{v_0}g_0(v)v^2dv\end{aligned}$$ and that the fact $W(v,t)=0$ at $t\rightarrow \infty$ we can find the maximum velocity 