Such a function can be chosen as the solution to the differential equation $$\label{2c}
-\f{\dot{\mu}(t)}{\mu^2(t)}+\f{\ep(t)}{\mu(t)}=\f{A|\dot{\ep}(t)|}{\ep(t)},$$ where $A:=2||y-z_0||$. Denote $\rho(t):=\f{1}{\mu(t)}.$ Then $$\dot{\rho}(t)+\ep(t)\rho(t)=\f{A|\dot{\ep}(t)|}{\ep(t)}$$ and $$\label{3c}
\rho(t)=\left[A\il^t_0\f{|\dot{\ep}(s)|}{\ep(s)}e^{\il^s_0\ep(\tau)d\tau}ds
+\f{1}{\mu(0)}\right]
e^{-\il^t_0\ep(\tau)d\tau}.$$ Since by Lemma \[stupid\] $e^{\il^t_0\ep(\tau)d\tau}\to\infty$ as $t\to\infty$ one can apply L’Hôspital’s rule to obtain from (\[3c\]) and condition 4 that $$\lim_{t\to +\infty}\rho(t)=\lim_{t\to +\infty}\f{|\dot{\ep}(t)|}{\ep^2(t)}=0.$$ Therefore $\mu(t)$ tends to $+\infty$. To complete the proof one can take $\mu(0)$ sufficiently small for the second inequality in (\[1c\]) to hold. By Theorem \[t1\] one concludes that $||z(t)-x(t)||\to 0$ as $t\to +\infty$ and by Lemma \[l2\] that $||x(t)-y||\to 0$ as $t\to +\infty$. Therefore it follows from the estimate: $$||z(t)-y||\le ||z(t)-x(t)||+||x(t)-y||,$$ that $||z(t)-y||\to 0$ as $t\to +\infty$. $\Box$

From the proof it is clear that one can get the estimate $||z(t)-x(t)||\le\f{1}{\mu(t)}\to 0$ as $t\to +\infty$, and for the term $||x(t)-y||$ one can get the rate of convergence if some additional assumptions are made on $F$ or on $z_0$ (see Propositions \[prop1\],  \[prop2\], and also Remark \[prem\]).

The result obtained in Theorem \[t5\] is similar to the result in [@al2]. The assumptions in [@al2] are slightly different. The method of investigation in [@al2] is based on a linear differential inequality which is a particular case of (\[ric4\]) with $\s(t)\equiv 0$. This 