respect to multiplication of $\ep(t)$ by a positive constant, one can choose $\ep(0)$ sufficiently large in order to satisfy condition 4. Such a choice can be done for an arbitrary $z_0$. \[yz0\]

[**Proof**]{} [**of Theorem \[t41\]**]{} Since $x(t)\equiv y$ in our case, one gets $\beta(t)=\a(t)$. Let us choose $\mu(t):=\f{\lambda}{\ep(t)}$. Conditions of Theorem \[t1\] can be written as follows: $$\label{gnc1}
\f{N_1N_2}{\ep(t)}+\f{N_2}{4\sqrt{\ep(t)}}\le
\f{\lambda}{2\ep(t)}\left(\g-\f{|\dot{\ep}(t)|}{\ep(t)}\right),$$ $$\label{gnc2} 
\a(t)\le\f{\ep(t)}{2\lambda}\left(\g-\f{|\dot{\ep}(t)|}{\ep(t)}\right),\q
\lambda||y-z_0||<\ep(0).$$ Inequality (\[gnc1\]) is equivalent to the following one: $$\label{gnc1a}
\lambda\ge \f{4N_1N_2+N_2\sqrt{\ep(t)}}{2\left(\g-\f{|\dot{\ep}(t)|}{\ep(t)}\right)}.$$ Take $$\label{lmd}
\lambda:=\f{4N_1N_2+N_2\sqrt{\ep(0)}}{2C_0}.$$ Then (\[gnc1a\]) follows from (\[lmd\]), from the monotonicity of $\ep(t)$ and from (\[c0min\]). Inequality (\[sbg\]) implies that $$\label{4.18c}
\f{4N_1N_2+N_2\sqrt{\ep(0)}}{2C_0}<\f{C_0}{2\max_{t\in
[0,+\infty)}\left\{\f{\a(t)}{\ep(t)}\right\}}$$ If $\lambda$ is defined by (\[lmd\]), then one has: $$\max_{t\in
[0,+\infty)}\left\{\f{\a(t)}{\ep(t)}\right\}<\f{C_0}{2\lambda}.$$ From (\[c0min\]) it follows that the first inequality in (\[gnc2\]) holds. Finally, one obtains from (\[sbg\]) that $$\f{4N_1N_2+N_2\sqrt{\ep(0)}}{2C_0}<\f{\ep(0)}{||y-z_0||},$$ which implies the second inequality in (\[gnc2\]). Since $x(t)\equiv y$ and $\mu(t):=\f{\lambda}{\ep(t)}$, by (\[est2\]) and (\[lmd\]) one concludes that (\[n\]) holds. $\Box$

Consider now a variant of Gauss-Newton continuous method with $\P$ defined as follows: $$\label{P2}
\P(h,t):=-T^{-1}_{\ep(t)}(h)\{F^{\pr *}(h)F(h)+\ep(t)(h-z_0)\}.$$ Here $F$ is not assumed compact. The following lemma is a consequence of Theorem 2.3 in [@ars] (see also Theorem 2.4 in [@bns]).

Assume that:

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

2.  $\P$ 