and inequalities (\[2ep\]), (\[TT\]), one obtains: $$\ep(t)(P_{\ep(t)}(\xi)T^{-1}_{\ep(t)}(h)(h-y),h-y)=
\ep(t)(P_{\ep(t)}(\xi)T^{-1}_{\ep(t)}(\xi)(h-y),h-y)$$ $$+\ep(t)(P_{\ep(t)}(\xi)
\{T^{-1}_{\ep(t)}(h)-T^{-1}_{\ep(t)}(\xi)\}(h-y),h-y)\leq \f{1}{2}||h-y||^2$$ $$+||P_{\ep(t)}(\xi)T^{-1}_{\ep(t)}(\xi)||\,||T(h)-T(y)||\,
||h-y||^2+||P_{\ep(t)}(\xi)T^{-1}_{\ep(t)}(\xi)\{T(y)-T(\xi)\}||\,||h-y||^2$$ $$\label{4.8}
\leq\left(\f{1}{2}+C\right)||h-y||^2+\f{N_1N_2}{\ep(t)}
||h-y||^3.$$ Define: $$\label{g}
\g(t)\equiv\g:=\f{1}{2}-C;$$ $$\label{s}
\sigma(t):=\f{N_1N_2}{\ep(t)}+\f{N_2}{4\sqrt{\ep(t)}};$$ $$\label{a}
\a(t):=||(P_{\ep(t)}(\xi)-I)(y-z_0)||.$$ These functions are positive if (\[sup\]) holds. If $\epsilon (t)\to 0$ then $\sigma(t)\to +\infty$ and $\alpha(t)\to 0$ as $t\to +\infty$. Comparing (\[4.7\]) - (\[a\]) with (\[semon\]) and applying Lemma \[ric\] one completes the proof. $\Box$

Suppose that the assumptions of Lemma \[l41\] are satisfied and:

1.  $P(y-z_0)=0$, where $P$ is an orthonormal projector onto the null-space of $T(\xi)$;

2.  $\ep(t)>0$ is continuously differentiable, monotonically tends to $0$, and $$\label{c0min}
    C_0:=\min_{t\in [0,+\infty)}\left\{\g-\f{|\dot{\ep}(t)|}{\ep(t)}\right\}>0,$$ with $\g$ defined in (\[g\]);

3.  $C_{\a}:=\ep(0)\max_{t\in [0,+\infty)}\left\{\f{\a(t)}{\ep(t)}\right\}
    <\infty$;

4.  $\ep(0)$ is chosen so that $$\label{sbg}
    \f{4N_1N_2+N_2\sqrt{\ep(0)}}{2C_0}<\ep(0)\min\left\{\f{C_0}{2C_\a},
    \f{1}{||y-z_0||}\right\}.$$

Then Cauchy problem (\[caupr\]) has a unique solution $z(t)$ for $t\in
[0,\infty)$ and $$\label{n}
||z(t)-y||\le \f{2C_0}{4N_1N_2+N_2\sqrt{\ep(0)}}\ep(t).$$ \[t41\]

If $T(\xi)$ is injective then Condition 1 is satisfied automatically. \[nlsp\]

From (\[a\]) one can see that $\a$ depends on $y$ and cannot be known a priori. However it follows from condition 1 of Theorem \[t41\] that $\a(t)\to 0$ as $t\to
+\infty$. Therefore in numerical applications of this scheme one should try different functions $\ep(t)$ (and different points $z_0$) to satisfy condition 3. Then, since $C_0$ and $C_\a$ are invariant with 