there exists $T\geq 0$ such that $\nu(t)\in C^1[T,+\infty)$ and $$\label{1b}
\nu(t)>0,\q  -\f{\dot{\nu}(t)}{\nu^2(t)}\leq C,\q \hbox{for}\q t\in [T,+\infty).$$ Then $$\label{2b}
\lim_{t\to+\infty}\il^t_0\nu(\tau)d\tau = +\infty.$$ \[stupid\]

[**Proof**]{} One can integrate (\[1b\]) $$-\il^t_T \f{\dot{\nu}(\tau)}
{\nu^2(\tau)}d\tau\leq \il^t_T Cd\tau,\q t\in [T,+\infty)$$ and get $$\f{1}{\nu(t)}\leq C(t-T) + \f{1}{\nu(T)}.$$ Without loss of generality we can assume that $C>0$, and then $$\nu(t)\geq\f{1}{ C(t-T) + \f{1}{\nu(T)}}.$$ Integrating this inequality one gets (\[2b\]) and completes the proof. $\Box$

Lemmas  \[l1\] - \[l3\] and Lemmas  \[l9\] -  \[stupid\] imply the following result.

Assume that:

1.  problem (\[noneq\]) has a unique solution $y$;

2.  $F$ is w-continuous and monotone;

3.  $F$ is continuously Fréchet differentiable and $$\label{N1}
    ||F'(x)||\le N_1, \q   \forall x\in H;$$

4.  $\ep(t)>0$ is continuously differentiable and tends to zero monotonically as $t\to
    +\infty$, and $\lim_{t\to +\infty}\f{\dot{\ep}(t)}{\ep^2(t)}=0$.

Then, for $\P$ defined by (\[4.1\]), Cauchy problem (\[caupr\]) has a unique solution $z(t)$ all for $t\in [0,+\infty)$ and $$\lim_{t\to\infty}||z(t)-y||=0.$$ \[t5\]

[**Proof**]{} In order to verify the assumptions of Theorem \[t1\] we use estimate (\[star1\]) to conclude that $\a(t)=\s(t)=0$ and $\g(t)=\ep(t)$ in formula (\[semon\]). By (\[xdot\]) $\b(t)=||\dot{x}(t)||$ because $\a(t)=0$. By (\[est8\]) $$\
\b(t)=||\dot{x}(t)||\le\f{|\dot{\ep}(t)|}{\ep(t)}||y-z_0||.$$ To apply Theorem \[t1\] one has to find a function $\mu(t) \in C^1[0,+\infty)$ satisfying (\[ric1\]) that is $$\label{1c}
\f{|\dot\ep(t)|}{\ep(t)}||y-z_0||\leq \f{1}{2\mu(t)}\left(\ep(t)-\f{
\dot\mu(t)}{\mu(t)}\right),\q \mu(0)||x(0)-z_0||<1.$$ 