one has: $$\label{eneq}
\f{1}{2}\f{d}{dt}||z(t)-x(t)||^2=(\dot{z}-\dot{x},z(t)-x(t))
=(\P(z(t),t),z(t)-x(t))-(\dot{x},z(t)-x(t)).$$ Therefore from (\[semon\]) and (\[xdot\]) one obtains $$\label{est1}
\f{1}{2}\f{d}{dt}||z(t)-x(t)||^2\le
-\g||z(t)-x(t)||^2+\s(t)||z(t)-x(t)||^3+\b(t)||z(t)-x(t)||.$$ Denote $$v(t):=||z(t)-x(t)||.$$ From (\[est1\]) one has: $$v(t)\dot{v}(t)\le -\g(t)v^2(t)+\s(t)v^3(t)+\b(t)v(t).$$ If $v>0$, one gets: $$\label{ineq11}
\dot{v}(t)\le -\g(t)v(t)+\s(t)v^2(t)+\b(t).$$ If $v=0$ on some interval, then inequality (\[ineq11\]) is satisfied trivially because $\beta(t)\ge 0$. Thus (\[ineq11\]) holds for all $t>0$.

By Lemma \[ric\] one obtains $$\label{appr}
||z(t)-x(t)||\le\f{1}{\mu(t)},\q \t{ for } \q t\in[0,T_1).$$ From (\[appr\]) one concludes that $z(t)$ does not leave the ball $B_1$ centered at $x(t)$ with radius $(\min_{t\in [0,T_1]}\mu(t))^{-1}$ $ > 0$. Since $\max_{0\leq t \leq T_1}||x(t)||<\infty$, one concludes that $\sup_{0\leq t < T_1}||z(t)||<\infty$. Therefore there exists a sequence $\{t_n\}\to T_1$ such that $\{z(t_n)\}$ converges weakly to some $\tilde{z}$. From equation (\[caupr\]) one derives the uniform boundedness of the norm $||\dot{z}(t)||$ on $[0,T_1)$. Thus there exists $\lim_{t\to T_1}||z(t)-\tilde{z}||=0$. Since the conditions for the uniqueness and local solvability of the Cauchy problem for equation (\[caupr\]) with initial condition $z(T_1)=\tilde{z}$ are satisfied, one can continue the solution to (\[caupr\]) through $T_1$. This contradicts the assumption of maximality of $T_1$, thus $T_1=+\infty$. Moreover, from (\[ric5\]) one gets: $$\label{est2}
\lim_{t\to+\infty}||z(t)-x(t)||\le \lim_{t\to+\infty}\f{1}{\mu(t)}=0.$$ $\Box$

Regularized Continuous Methods\
for Monotone Operators
===============================

In this section we apply the regularization procedure described in Sect. 2 to 