as $t\to +\infty$ and provide a decay of a nonnegative solution to integral inequality (\[ric4\]) even if $\s(t)$ tends to infinity. Moreover in the first and the third examples $v(t)$ tends to zero as $t\to
+\infty$ when $\g(t)\to 0$ and $\s(t)\to +\infty$.

Let $\P(h,t)$ be Fréchet differentiable with respect to $h$ and satisfy the following condition:

there exists a differentiable function $x(t)$, $x:[0,+\infty)\to H$, such that for any $h\in H,\, t\in[0,+\infty)$ $$\label{semon}
(\P(h,t),h-x(t))\le \a(t)||h-x(t)||-\g(t)||h-x(t)||^2+\s(t)||h-x(t)||^3,$$ where $\alpha(t)$ is a continuous function, $\alpha(t)\ge 0$, $\g(t)$ and $\s(t)$ satisfy conditions (\[ric1\]) of Lemma \[ric\] with $$\label{xdot}
\b(t):=||\dot{x}(t)||+\a(t),\q v(0):=||z_0-x(0)||,\q v(t):=||z(t)-x(t)||,$$ and $\mu(t)$ tends to $+\infty$ as $t\to +\infty$.

Then problem (\[caupr\]) has a unique solution $z(t)$ defined for all $t\in
[0,\infty)$, and $$\label{solut}
||z(t)-x(t)||<\frac 1 {\mu(t)},\q \lim_{t\to+\infty}||z(t)-x(t)||=0.$$ \[t1\]

One can choose the regularizing operator $\P(h,t)$ in (\[caupr\]) such that condition (\[semon\]) holds in the case when $F^{\pr *}(h)F'(h)$ is not boundedly invertible (see Sect. 3). \[nobound\]

[**Proof**]{} [**of Theorem \[t1\]**]{} Since $\P(h,t)$ is Fréchet differentiable with respect to $h$ there exists the solution to problem (\[caupr\]) on the maximal interval $[0,T_1)$ of the existence of the solution to (\[caupr\]). One has to show that $T_1=+\infty$. Assume $T_1<+\infty$. Since $H$ is a real Hilbert space 