Let *H* and *V* be two real Hilbert spaces. Assume and *V* is dense subspace in *H* and the injection of *V* into *H* is continuous. If *H* is identified with its dual space, then we may write *V* ⊂ *H* ⊂ *V** densely, and the corresponding injections are continuous. The norm on *V* (resp. *H*) will be denoted by || · || (resp. |· |). The duality pairing between the element *v*
_{1} of *V** and the element *v*
_{2} of *V* is denoted by (*v*
_{1}
*, v*
_{2}), which is the ordinary inner product in *H* if *v*
_{1}
*, v*
_{2} ∈ *H*. For the sake of simplicity, we may consider

\parallel u{\parallel}_{*}\le \phantom{\rule{2.77695pt}{0ex}}\mid u\mid \phantom{\rule{2.77695pt}{0ex}}\le \phantom{\rule{2.77695pt}{0ex}}\parallel u\parallel ,\phantom{\rule{1em}{0ex}}u\in V

where || · ||_{*} is the norm of the element of *V**. If an operator *A* is bounded linear from *V* to *V** and generates an analytic semigroup, then it is easily seen that

H=\left\{x\in {V}^{*}:{\int}_{0}^{T}\parallel A{e}^{tA}x{\parallel}_{*}^{2}dt<\infty \right\},

for the time *T >* 0. Therefore, in terms of the intermediate theory we can see that

{{\left(V,{V}^{*}\right)}_{\frac{1}{2}}}_{,2}=H

where {{\left(V,{V}^{*}\right)}_{\frac{1}{2}}}_{,2} denotes the real interpolation space between *V* and *V**.

We note that a nonlinear operator *A* is said to be hemicontinuous on *V* if

w-\underset{t\to 0}{lim}\phantom{\rule{2.77695pt}{0ex}}A\left(x+ty\right)=Ax

for every *x, y* ∈ *V* where "*w -* lim" indicates the weak convergence on *V**. Let *A* : *V* → *V** be given a single-valued, monotone operator and hemicontinuous from *V* to *V** such that

**(A1)** *A*(0) = 0, (*Au - Av, u - v*) ≥ *ω*
_{1} ||*u - v*||^{2}
*- ω*
_{2} |*u - v|*
^{2},

**(A2)** ||*Au*||_{*} ≤ *ω*
_{3}(||*u*|| + 1)

for every *u, v* ∈ *V* where *ω*
_{2} ∈ ℝ and *ω*
_{1}
*, ω*
_{3} are some positive constants.

Here, we note that if 0 ≠ *A*(0), then we need the following assumption:

\left(Au,u\right)\ge {\omega}_{1}\parallel u{\parallel}^{2}-{\omega}_{2}\mid u{\mid}^{2}

for every *u* ∈ *V* . It is also known that *A* is maximal monotone, and *R*(*A*) = *V** where *R*(*A*) denotes the range of *A*.

Let the controller *B* is a bounded linear operator from a Banach space *L*
^{2}(0*, T*; *U*) to *L*
^{2}(0*, T*; *H*) where *U* is a Banach space.

For each *t* ∈ [0*, T*], we define *x*
_{
t
} : [ *-h*, 0] → *H* as

{x}_{t}\left(s\right)=x\left(t+s\right),\phantom{\rule{1em}{0ex}}-h\le s\le 0.

We will set

\Pi ={L}^{2}\left(-h,0;V\right)\phantom{\rule{1em}{0ex}}\mathsf{\text{and}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}{\mathbb{R}}^{+}=\left[0,\infty \right).

Let ℒ and ℬ be the Lebesgue *σ*-field on [0, ∞) and the Borel *σ*-field on [*- h*, 0] respectively. Let *k* : ℝ^{+} × ℝ^{+} × ∏ → *H* be a nonlinear mapping satisfying the following:

**(K1)** For any *x*. ∈ ∏ the mapping *k*(·, ·*, x*.) is strongly ℒ × ℬ -measurable;

**(K2)** There exist positive constants *K*
_{0}, and *K*
_{1} such that

\begin{array}{c}\mid k\left(t,s,x.\right)-k\left(t,s,y.\right)\mid \phantom{\rule{2.77695pt}{0ex}}\le {K}_{1}\parallel x.-y.\parallel {}_{\Pi},\\ \mid k\left(t,s,0\right)\mid \phantom{\rule{2.77695pt}{0ex}}\le {K}_{0}\end{array}

for all (*t, s*) ∈ ℝ^{+} × [ *-h*, 0] and *x*., *y*. ∈ ∏.

Let *g* : ℝ^{+} × ∏ × *H* → *H* be a nonlinear mapping satisfying the following:

**(G1)** For any *x* ∈ ∏, *y* ∈ *H* the mapping *g*(*·, x., y*) is strongly ℒ -measurable;

**(G2)** There exist positive constants *L*
_{0}
*, L*
_{1}, and *L*
_{2} such that

\begin{array}{c}\mid g\left(t,x.,y\right)-g\left(t,\widehat{x}.,\u0177\right)\mid \le {L}_{1}\parallel x.-\widehat{x}.{\parallel}_{\Pi}+{L}_{2}\mid y-\u0177\mid ,\\ \mid g\left(t,0,0\right)\mid \le {L}_{0}\end{array}

for all *t* ∈ ℝ^{+}, *x*, \widehat{x}.\in \Pi, and *y*, \u0177\in H.

**Remark 2.1**. *The above operator g is the semilinear case of the nonlinear part of quasi-linear equations considered by Yong and Pan*[13].

*For x* ∈ *L*
^{2}(*-h, T*; *V* )*, T >* 0 *we set*

G\left(t,x\right)=g\left(t,{x}_{t},{\int}_{0}^{t}k\left(t,s,{x}_{s}\right)ds\right).

*Here, as in*[13], *we consider the Borel measurable corrections of x*(·).

**Lemma 2.1**. *Let x* ∈ *L*
^{2}(- *h, T*; *V* )*. Then, the mapping t* ↦ *x*
_{
t
} *belongs to C*([0*, T* ]; ∏) *and*

\parallel x.{\parallel}_{{L}^{2}\left(0.T;\Pi \right)}\le \sqrt{T}\parallel x{\parallel}_{{L}^{2}\left(-h,T;V\right)}.

(2.1)

*Proof*. It is easy to verify the first paragraph and (2.1) is a consequence of the estimate:

\begin{array}{cc}\hfill \parallel x.{\parallel}_{{L}^{2}\left(0.T;\Pi \right)}^{2}& \le {\int}_{0}^{T}\parallel {x}_{t}{\parallel}_{\Pi}^{2}dt\le {\int}_{0}^{T}{\int}_{-h}^{0}\parallel x\left(t+s\right){\parallel}^{2}dsdt\hfill \\ \le {\int}_{0}^{T}dt{\int}_{-h}^{T}\parallel x\left(s\right){\parallel}^{2}ds\le T\parallel x{\parallel}_{{L}^{2}\left(-h,T;V\right)}^{2}.\hfill \end{array}

**Lemma 2.2**. *Let x* ∈ *L*
^{2}(- *h, T*; *V* )*, T >* 0*. Then, G*(·*, x*) ∈ *L*
^{2}(0*, T*; *H*) *and*

\begin{array}{c}\parallel G\left(\cdot ,x\right){\parallel}_{{L}^{2}\left(0,T;H\right)}\le {L}_{0}\sqrt{T}+{L}_{2}{K}_{0}{T}^{3\u22152}\u2215\sqrt{3}\\ +\left({L}_{1}\sqrt{T}+{L}_{2}{K}_{1}{T}^{3\u22152}\u2215\sqrt{2}\right)\parallel x{\parallel}_{{L}^{2}\left(-h,T;V\right)}.\end{array}

(2.2)

*Moreover, if x*
_{1}
*, x*
_{2} ∈ *L*
^{2}(- *h, T*; *V* )*, then*

\parallel G\left(\cdot ,{x}_{1}\right)-G\left(\cdot ,{x}_{2}\right){\parallel}_{{L}^{2}\left(0,T;H\right)}\le \left({L}_{1}\sqrt{T}+{L}_{2}{K}_{1}{T}^{3\u22152}\u2215\sqrt{2}\right)\parallel {x}_{1}-{x}_{2}{\parallel}_{{L}^{2}\left(-h,T;V\right)}.

(2.3)

*Proof*. It follows from (K2) and (2.1) that

\begin{array}{c}\parallel {\int}_{0}^{\cdot}k\left(\cdot ,s,{x}_{s}\right)ds{\parallel}_{{L}^{2}\left(0,T;H\right)}\le \phantom{\rule{2.77695pt}{0ex}}\parallel {\int}_{0}^{\cdot}k\left(\cdot ,s,0\right)ds{\parallel}_{{L}^{2}\left(0,T;H\right)}\\ +\parallel {\int}_{0}^{\cdot}\left(k\left(\cdot ,s,{x}_{s}\right)-k\left(\cdot ,s,0\right)\right)ds{\parallel}_{{L}^{2}\left(0,T;H\right)}\\ \le {K}_{0}{T}^{3\u22152}\u2215\sqrt{3}+{\left\{{\int}_{0}^{T}\mid {\int}_{0}^{t}{K}_{1}\parallel {x}_{s}{\parallel}_{\Pi}ds{\mid}^{2}dt\right\}}^{1\u22152}\\ \le {K}_{0}{T}^{3\u22152}\u2215\sqrt{3}+{\left\{{\int}_{0}^{T}{K}_{1}^{2}t{\int}_{0}^{t}\parallel {x}_{s}{\parallel}_{\Pi}^{2}dsdt\right\}}^{1\u22152}\\ \le {K}_{0}{T}^{3\u22152}\u2215\sqrt{3}+{K}_{1}T\u2215\sqrt{2}\parallel {x}_{\cdot}{\parallel}_{{L}^{2}\left(0,T;\Pi \right)}\\ \le {K}_{0}{T}^{3\u22152}\u2215\sqrt{3}+{K}_{1}{T}^{3\u22152}\u2215\sqrt{2}\parallel x{\parallel}_{{L}^{2}\left(-h,T;V\right)}\end{array}

and hence, from (G2), (2.1), and the above inequality, it is easily seen that

\begin{array}{c}\parallel G\left(\cdot ,x\right){\parallel}_{{L}^{2}\left(0,T;H\right)}\phantom{\rule{2.77695pt}{0ex}}\le \phantom{\rule{2.77695pt}{0ex}}\parallel G\left(\cdot ,0\right)\parallel +\parallel G\left(\cdot ,x\right)-G\left(\cdot ,0\right)\parallel \\ \le {L}_{0}\sqrt{T}+{L}_{1}\parallel {x}_{\cdot}{\parallel}_{{L}^{2}\left(0,T;\Pi \right)}+{L}_{2}\parallel {\int}_{0}^{\cdot}k\left(\cdot ,s,{x}_{s}\right)ds{\parallel}_{{L}^{2}\left(0,T;H\right)}\\ \le {L}_{0}\sqrt{T}+{L}_{1}\sqrt{T}\parallel x{\parallel}_{{L}^{2}\left(-h,T;V\right)}\\ +{L}_{2}\left({K}_{0}{T}^{3\u22152}\u2215\sqrt{3}+{K}_{1}{T}^{3\u22152}\u2215\sqrt{2}\parallel x{\parallel}_{{L}^{2}\left(-h,T;V\right)}\right).\end{array}

Similarly, we can prove (2.3).

Let us consider the quasi-autonomous differential equation

\left\{\begin{array}{c}{x}^{\prime}\left(t\right)+Ax\left(t\right)=f\left(t\right),\phantom{\rule{1em}{0ex}}0<t\le T,\\ x\left(0\right)={\varphi}^{0}\end{array}\right.

(E)

where *A* satisfies the hypotheses mentioned above. The following result is from Theorem 2.6 of Chapter III in [1].

**Proposition 2.1**. *Let Φ*
^{0} ∈ *H and f* ∈ *L*
^{2}(0*, T*; *V** )*. Then, there exists a unique solution x of (E) belonging to*

C\left(\left[0,T\right];H\right)\cap {L}^{2}\left(0,T;V\right)\cap {W}^{1,2}\left(0,T;{V}^{*}\right)

*and satisfying*

\mid x\left(t\right){\mid}^{2}+{\int}_{0}^{t}\parallel x\left(s\right){\parallel}^{2}ds\le {C}_{1}\left(\mid {\varphi}^{0}{\mid}^{2}+{\int}_{0}^{t}\parallel f\left(s\right){\parallel}_{*}^{2}ds\right),

(2.4)

{\int}_{0}^{t}\parallel \frac{dx\left(s\right)}{ds}{\parallel}_{*}^{2}dt\le {C}_{1}\left(\mid {\varphi}^{0}{\mid}^{2}+{\int}_{0}^{t}\parallel f\left(s\right){\parallel}_{*}^{2}ds\right)

(2.5)

*where C*
_{1}
*is a constant*.

Acting on both sides of (E) by *x*(*t*), we have

\frac{1}{2}\frac{d}{dt}\mid x\left(t\right){\mid}^{2}+{\omega}_{1}\parallel x\left(t\right){\parallel}^{2}\le {\omega}_{2}\mid x\left(t\right){\mid}^{2}+\left(f\left(t\right),x\left(t\right)\right).

As is seen Theorem 2.6 in [1], integrating from 0 to *t*, we can determine the constant *C*
_{1} in Proposition 2.1.

We establish the following result on the solvability of the equation (SE).

**Theorem 2.1**. *Let A and the nonlinear mapping g be given satisfying the assumptions mentioned above. Then, for any* (*Φ*
^{0}
*, Φ*
^{1}) ∈ *H* × *L*
^{2}(- *h*, 0; *V* ) *and f* ∈ *L*
^{2}(0*, T*; *V**)*, T >* 0*, the following nonlinear equation*

\left\{\begin{array}{c}{x}^{\prime}\left(t\right)+Ax\left(t\right)=G\left(t,x\right)+f\left(t\right),\phantom{\rule{1em}{0ex}}0<t\le T,\\ x\left(0\right)={\varphi}^{0},\phantom{\rule{1em}{0ex}}x\left(s\right)={\varphi}^{1}\left(s\right),\phantom{\rule{1em}{0ex}}-h\le s\le 0\end{array}\right.

(2.6)

*has a unique solution x belonging to*

{L}^{2}\left(-h,T;V\right)\cap {W}^{1,2}\left(0,T;{V}^{*}\right)\subset C\left(\left[0,T\right];H\right)

*and satisfying that there exists a constant C*
_{2}
*such that*

\parallel x{\parallel}_{{L}^{2}\left(-h,T;V\right)\cap {W}^{1,2}\left(0,T;V*\right)}\le {C}_{2}\left(1+\mid {\varphi}^{0}\mid +\parallel {\varphi}^{1}{\parallel}_{{L}^{2}\left(-h,0;V\right)}+\parallel f{\parallel}_{{L}^{2}\left(0,T;V*\right)}\right).

(2.7)

*Proof*. Let *y* ∈ *L*
^{2}(0*, T*; *V* ). Then, we extend it to the interval (*-h*, 0) by setting *y*(*s*) = *Φ*
^{1}(*s*) for *s* ∈ (*-h*, 0), and hence, *G*(·*, y*(·)) ∈ *L*
^{2}(0*, T*; *H*) from Lemma 2.2. Thus, by virtue of Proposition 2.1, we know that the problem

\left\{\begin{array}{c}{x}^{\prime}\left(t\right)+Ax\left(t\right)=G\left(t,y\right)+f\left(t\right),\phantom{\rule{1em}{0ex}}0<t,\\ x\left(0\right)={\varphi}^{0},\phantom{\rule{1em}{0ex}}x\left(s\right)={\varphi}^{1}\left(s\right)\phantom{\rule{1em}{0ex}}-h\le s\le 0\end{array}\right.

(2.8)

has a unique solution *x*
_{
y
} ∈ *L*
^{2}(0*, T*; *V* ) ∩ *W*
^{1,2}(0*, T*; *V** ) corresponding to *y*. Let us fix *T*
_{0}
*>* 0 so that

{\omega}_{1}^{-1}{e}^{{\omega}_{2}{T}_{0}}\left({L}_{1}{\sqrt{T}}_{0}+{L}_{2}{K}_{1}{T}_{0}^{3\u22152}\u2215\sqrt{2}\right)<1.

(2.9)

Let *x*
_{
i
} , *i* = 1, 2, be the solution of (2.8) corresponding to *y*
_{
i
} . Multiplying by *x*
_{1}(*t*) *- x*
_{2}(*t*), we have that

\begin{array}{c}\left({\u1e8b}_{1}\left(t\right)-{\u1e8b}_{2}\left(t\right),{x}_{1}\left(t\right)-{x}_{2}\left(t\right)\right)+\left(A{x}_{1}\left(t\right)-A{x}_{2}\left(t\right),{x}_{1}\left(t\right)-{x}_{2}\left(t\right)\right)\\ =\left(G\left(t,{y}_{1}\right)-G\left(t,{y}_{2}\right),{x}_{1}\left(t\right)-{x}_{2}\left(t\right)\right),\end{array}

and hence it follows that

\begin{array}{l}\frac{1}{2}\frac{d}{dt}\mid {x}_{1}(t)-{x}_{2}(t){\mid}^{2}+{\omega}_{1}\parallel {x}_{1}(t)-{x}_{2}(t){\parallel}^{2}\\ \le {\omega}_{2}\mid {x}_{1}(t)-{x}_{2}(t){\mid}^{2}+\parallel G(t,{y}_{1}))-G(t,{y}_{2}){\parallel}_{*}\parallel {x}_{1}(t)-{x}_{2}(t)\parallel .\end{array}

From Lemma 2.2 and integrating over [0,*t*], it follows

\begin{array}{c}\frac{1}{2}\mid {x}_{1}\left(t\right)-{x}_{2}\left(t\right){\mid}^{2}+{\omega}_{1}{\int}_{0}^{t}\parallel {x}_{1}\left(s\right)-{x}_{2}\left(s\right){\parallel}^{2}ds\\ \le \frac{1}{2c}{\int}_{0}^{t}\parallel G\left(s,{y}_{1}\right)-G\left(s,{y}_{2}\right){\parallel}_{*}^{2}ds\\ +\frac{c}{2}{\int}_{0}^{t}\parallel {x}_{1}\left(s\right)-{x}_{2}\left(s\right){\parallel}^{2}ds+{\omega}_{2}{\int}_{0}^{t}\mid {x}_{1}\left(s\right)ds-{x}_{2}\left(s\right)ds{\mid}^{2}ds,\end{array}

where *c* is a positive constant satisfying 2*ω*
_{1}
*- c >* 0. Here, we used that

ab\le \frac{{a}^{p}}{p}+\frac{{b}^{q}}{q},\phantom{\rule{1em}{0ex}}{p}^{-1}+{q}^{-1}=1\left(1<p<\infty \right)

for any pair of nonnegative numbers *a* and *b*. Thus, from (2.3) it follows that

\begin{array}{c}\mid {x}_{1}\left(t\right)-{x}_{2}\left(t\right){\mid}^{2}+\left(2{\omega}_{1}-c\right){\int}_{0}^{t}\parallel {x}_{1}\left(s\right)ds-{x}_{2}\left(s\right)ds{\parallel}^{2}ds\\ \le {c}^{-1}{\left({L}_{1}{\sqrt{T}}_{0}+{L}_{2}{K}_{1}{T}_{0}^{3\u22152}\u2215\sqrt{2}\right)}^{2}{\int}_{0}^{t}\parallel {y}_{1}\left(s\right)-{y}_{2}\left(s\right){\parallel}^{2}ds\\ +2{\omega}_{2}{\int}_{0}^{t}\mid {x}_{1}\left(s\right)-{x}_{2}\left(s\right){\mid}^{2}ds.\end{array}

By using Gronwall's inequality, we get

\begin{array}{c}\mid {x}_{1}\left({T}_{0}\right)-{x}_{2}\left({T}_{0}\right){\mid}^{2}+\left(2{\omega}_{1}-c\right){\int}_{0}^{{T}_{0}}\parallel {x}_{1}\left(s\right)-{x}_{2}\left(s\right){\parallel}^{2}ds\\ \le {c}^{-1}{\left({L}_{1}{\sqrt{T}}_{0}+{L}_{2}{K}_{1}{T}_{0}^{3\u22152}\u2215\sqrt{2}\right)}^{2}{e}^{2{\omega}_{2}{T}_{0}}{\int}_{0}^{{T}_{0}}\parallel {y}_{1}\left(s\right)-{y}_{2}\left(s\right){\parallel}^{2}ds.\end{array}

Taking *c* = *ω*
_{1}, it holds that

\begin{array}{l}\parallel {x}_{1}-{x}_{2}{\parallel}_{{L}^{2}(0,{T}_{0};V)}\le {\omega}_{1}^{-1}{e}^{{\omega}_{2}{T}_{0}}({L}_{1}{\sqrt{T}}_{0}\\ +{L}_{2}{K}_{1}{T}_{0}^{3/2}/\sqrt{2})\parallel {y}_{1}-{y}_{2}{\parallel}_{{L}^{2}(0,{T}_{0};V)}.\end{array}

Hence, we have proved that *y* ↦ *x* is a strictly contraction from *L*
^{2}(0*, T*
_{0}; *V* ) to itself if the condition (2.9) is satisfied. It shows that the equation (2.6) has a unique solution in [0*, T*
_{0}].

From now on, we derive the norm estimates of solution of the equation (2.6). Let *y* be the solution of

\left\{\begin{array}{c}{y}^{\prime}\left(t\right)+Ay\left(t\right)=f\left(t\right),\phantom{\rule{1em}{0ex}}0<t\le {T}_{0},\\ y\left(0\right)={\varphi}^{0}.\end{array}\right.

(2.10)

Then,

\frac{d}{dt}\left(x\left(t\right)-y\left(t\right)\right)+\left(Ax\left(t\right)-Ay\left(t\right)\right)=G\left(t,x\right),

by multiplying by *x*(*t*) - *y*(*t*) and using the assumption (A1), we obtain

\begin{array}{c}\frac{1}{2}\frac{d}{dt}\mid x\left(t\right)-y\left(t\right){\mid}^{2}+{\omega}_{1}\parallel x\left(t\right)-y\left(t\right){\parallel}^{2}\\ \le {\omega}_{2}\mid x\left(t\right)-y\left(t\right){\mid}^{2}+\parallel G\left(t,x\right){\parallel}_{*}\parallel x\left(t\right)-y\left(t\right)\parallel .\end{array}

By integrating over [0*, t*] and using Gronwall's inequality, we have

\begin{array}{l}\parallel x-y{\parallel}_{{L}^{2}(0,{T}_{0};V)}\le {\omega}_{1}^{-1}{e}^{{\omega}_{2}{T}_{0}}\parallel G(\eta ,x){\parallel}_{{L}^{2}(0,{T}_{0};{V}^{*})}\\ \le {\omega}_{1}^{-1}{e}^{{\omega}_{2}{T}_{0}}\{{L}_{0}{\sqrt{T}}_{0}+{L}_{2}{K}_{0}{T}_{0}^{3/2}/\sqrt{3}\\ +({L}_{1}{\sqrt{T}}_{0}+{L}_{2}{K}_{1}{T}_{0}^{3/2}/\sqrt{2})(\parallel x{\parallel}_{{L}^{2}(0,{T}_{0};V)}+\parallel {\varphi}^{1}{\parallel}_{{L}^{2}(-h,0;V})\},\end{array}

and hence, putting

N={\omega}_{1}^{-1}{e}^{{\omega}_{2}{T}_{0}}\mathsf{\text{and}}L={L}_{1}{\sqrt{T}}_{0}+{L}_{2}{K}_{1}{T}_{0}^{3\u22152}\u2215\sqrt{2},

it holds

\begin{array}{l}\parallel x{\parallel}_{{L}^{2}(0,{T}_{0};V)}\le \frac{N}{1-NL}({L}_{0}{\sqrt{T}}_{0}+{L}_{2}{K}_{0}{T}_{0}^{3/2}/\sqrt{3})\\ +\frac{1}{1-NL}\parallel y{\parallel}_{{L}^{2}(0,{T}_{0};V)}+\frac{NL}{1-NL}\parallel {\varphi}^{1}{\parallel}_{{L}^{2}(-h,0;V})\\ \le \frac{N}{1-NL}({L}_{0}{\sqrt{T}}_{0}+{L}_{2}{K}_{0}{T}_{0}^{3/2}/\sqrt{3})\\ +\frac{{C}_{1}}{1-NL}(\mid {\varphi}^{0}\mid +\parallel f{\parallel}_{{L}^{2}(0,{T}_{0};{V}^{*})})\\ +\frac{NL}{1-NL}\parallel {\varphi}^{1}{\parallel}_{{L}^{2}(-h,0;V)}\\ \le {C}_{2}(1+\mid {\varphi}^{0}\mid +\parallel {\varphi}^{1}{\parallel}_{{L}^{2}(-h,0;V)}+\parallel f{\parallel}_{{L}^{2}(0,{T}_{0};{V}^{*})})\end{array}

(2.11)

for some positive constant *C*
_{2}. Since the condition (2.9) is independent of initial values, the solution of (2.6) can be extended to the internal [0*, nT*
_{0}] for natural number *n*, i.e., for the initial value (*x*(*nT*
_{0}), *x*
_{
n
}
*T*
_{
0
} ) in the interval [*nT*
_{0}, (*n* + 1)*T*
_{0}], as analogous estimate (2.11) holds for the solution in [0, (*n* + 1)*T*
_{0}].

**Theorem 2.2**. *If* (*Φ*
^{0}, *Φ*
^{1}) ∈ *H* × *L*
^{2}(*-h*, 0, *V* )) *and f* ∈ *L*
^{2}(0*, T*; *V** )*, then x* ∈ *L*
^{2}(*-h, T*; *V* ))∩ *W*
^{1,2}(0*, T*; *V** )*, and the mapping*

H\times {L}^{2}(-h,0;V)\times {L}^{2}(0,T;{V}^{*})\ni ({\varphi}^{0},{\varphi}^{1},f)\mapsto x\in {L}^{2}(-h,T;V))\cap {W}^{1,2}(0,T;{V}^{*})

*is continuous*.

*Proof*. It is easy to show that if (*Φ*
^{0}
*, Φ*
^{1}) ∈ *H* × *L*
^{2}(*-h*, 0; *V* )) and *f* ∈ *L*
^{2}(0*, T*; *V** ) for every *T >* 0, then *x* belongs to *L*
^{2}(- *h, T*; *V* )∩*W*
^{1,2}(0*, T*; *V**). Let

\left({\varphi}_{i}^{0},{\varphi}_{i}^{1},{f}_{i}\right)\in H\times {L}^{2}\left(-h,0;V\right)\times {L}^{2}\left(0,{T}_{1};{V}^{*}\right)

and *x*
_{
i
} be the solution of (2.6) with \left({\varphi}_{i}^{0},{\varphi}_{i}^{1},{f}_{i}\right) in place of \left({\varphi}^{0},{\varphi}^{1},f\right) for *i* = 1,2.

Then, in view of Proposition 2.1 and Lemma 2.2, we have

\begin{array}{c}\frac{1}{2}\frac{d}{dt}\mid {x}_{1}\left(t\right)-{x}_{2}\left(t\right){\mid}^{2}+{\omega}_{1}\parallel {x}_{1}\left(t\right)-{x}_{2}\left(t\right){\parallel}^{2}\\ \le {\omega}_{2}\mid {x}_{1}\left(t\right)-{x}_{2}\left(t\right){\mid}^{2}+\parallel G\left(t,{x}_{1}\right)-G\left(t,{x}_{2}\right){\parallel}_{*}\parallel {x}_{1}\left(t\right)-{x}_{2}\left(t\right)\parallel \\ +\parallel {f}_{1}\left(t\right)-{f}_{2}\left(t\right){\parallel}_{*}\parallel {x}_{1}\left(t\right)-{x}_{2}\left(t\right)\parallel \end{array}

(2.12)

If *ω*
_{1}
*- c/* 2 *>* 0, we can choose a constant *c*
_{1}
*>* 0 so that

{\omega}_{1}-c\u22152-{c}_{1}\u22152>0

and

\begin{array}{l}\parallel {f}_{1}(t)-{f}_{2}(t)){\parallel}_{*}\parallel {x}_{1}(t)-{x}_{2}(t)\parallel \le \frac{1}{2{c}_{1}}\parallel {f}_{1}(t)-{f}_{2}(t){\parallel}_{*}^{2}\\ +\frac{{c}_{1}}{2}\parallel {x}_{1}(t)-{x}_{2}(t){\parallel}^{2}.\end{array}

Let *T*
_{1}
*< T* be such that

2{\omega}_{1}-c-{c}_{1}-{c}^{-1}{e}^{2{\omega}_{2}{T}_{1}}{\left({L}_{1}{\sqrt{T}}_{1}+{L}_{2}{K}_{1}{T}_{1}^{3\u22152}\u2215\sqrt{2}\right)}^{2}>0.

Integrating on (2.12) over [0*, T*
_{1}] and as is seen in the first part of proof, it follows

\begin{array}{l}(2{\omega}_{1}-c-{c}_{1})\parallel {x}_{1}-{x}_{2}{\parallel}_{{L}^{2}(0,{T}_{1};V)}^{2}\le {e}^{2{\omega}_{2}{T}_{1}}\{\mid {\varphi}_{1}^{0}-{\varphi}_{2}^{0}{\mid}^{2}\\ +\frac{1}{c}\parallel G(t,{x}_{1})-G(t,{x}_{2}){\parallel}_{{L}^{2}(0,{T}_{1};{V}^{*})}^{2}+1{c}_{1}\parallel {f}_{1}-{f}_{2}{\parallel}_{{L}^{2}(0,{T}_{1};{V}^{*})}^{2}\}\\ \le {e}^{2{\omega}_{2}{T}_{1}}\{\mid {\varphi}_{1}^{0}-{\varphi}_{2}^{0}{\mid}^{2}\\ +\frac{1}{c}({L}_{1}{\sqrt{T}}_{1}+{L}_{2}{K}_{1}{T}_{1}^{3/2}/\sqrt{2}{)}^{2}\parallel {x}_{1}-{x}_{2}{\parallel}_{{L}^{2}(-h,{T}_{1};V)}^{2}\\ +\frac{1}{{c}_{1}}\parallel {f}_{1}-{f}_{2}{\parallel}_{{L}^{2}(0,{T}_{1};{V}^{*})}^{2}\}.\end{array}

Putting that

{N}_{1}=2{\omega}_{1}-c-{c}_{1}-{c}^{-1}{e}^{2{\omega}_{2}{T}_{1}}{\left({L}_{1}{\sqrt{T}}_{1}+{L}_{2}{K}_{1}{T}_{1}^{3\u22152}\u2215\sqrt{2}\right)}^{2}

we have

\begin{array}{c}\parallel {x}_{1}-{x}_{2}{\parallel}_{{L}^{2}\left(0,{T}_{1};V\right)}\le \frac{{e}^{{\omega}_{2}{T}_{1}}}{{N}_{1}^{1\u22152}}\left(\mid {\varphi}_{1}^{0}-{\varphi}_{2}^{0}\mid +\frac{1}{{c}_{1}}\parallel {f}_{1}-{f}_{2}{\parallel}_{{L}^{2}\left(0,{T}_{1};V*\right)}\right)\\ \phantom{\rule{1em}{0ex}}+\frac{{c}^{-1\u22152}{e}^{{\omega}_{2}{T}_{1}}\left({L}_{1}\sqrt{{T}_{1}}+{L}_{2}{K}_{1}{T}_{1}^{3\u22152}\u2215\sqrt{2}\right)}{{N}_{1}^{1\u22152}}\parallel {\varphi}_{1}^{1}-{\varphi}_{2}^{1}{\parallel}_{{L}^{2}\left(-h,0;V\right)}.\end{array}

(2.13)

Suppose that

({\varphi}_{n}^{0},{\varphi}_{n}^{1},{f}_{n})\to ({\varphi}^{0},{\varphi}^{1},f)\text{in}H\times {L}^{2}(-h,0;V))\times {L}^{2}(0,T;{V}^{*}),

and let *x*
_{
n
} and *x* be the solution (2.6) with \left({\varphi}_{n}^{0},{\varphi}_{n}^{1},{f}_{n}\right) and \left({\varphi}^{0},{\varphi}^{1},f\right) respectively.

By virtue of (2.13) with *T* being replaced by *T*
_{1}, we see that

{x}_{n}\to x\text{\u212fin}{L}^{2}(-h,{T}_{1};V))\cap {W}^{1,2}(0,{T}_{1};{V}^{*})\subset C([0,{T}_{1}];H).

This implies that \left({x}_{n}\left({T}_{1}\right),{\left({x}_{n}\right)}_{{T}_{1}}\right)\to \left(x\left({T}_{1}\right),{x}_{{T}_{1}}\right) in *H* × *L*
^{2} (-*h*, 0; *V*). Hence, the same argument shows that

{x}_{n}\to x\phantom{\rule{1em}{0ex}}\mathsf{\text{in}}\phantom{\rule{1em}{0ex}}{L}^{2}\left({T}_{1},min\left\{2{T}_{1},T\right\};V\right)\cap {W}^{1,2}\left({T}_{1},min\left\{2{T}_{1},T\right\};{V}^{*}\right).

Repeating this process, we conclude that

{x}_{n}\to x\phantom{\rule{1em}{0ex}}\mathsf{\text{in}}\phantom{\rule{1em}{0ex}}{L}^{2}\left(-h,T;V\right)\cap {W}^{1,2}\left(0,T;{V}^{*}\right).

**Remark 2.2**. *For x* ∈ *L*
^{2}(0*, T*; *V* )*, we set*

G\left(t,x\right)={\int}_{0}^{t}k\left(t-s\right)g\left(s,x\left(s\right)\right)ds

*where k belongs to L*
^{2}(0*, T*) *and g* : [0*, T*] × *V* → *H be a nonlinear mapping satisfying*

\mid g\left(t,x\right)-g\left(t,y\right)\mid \le L\parallel x-y\parallel

*for a positive constant L. Let x ∈ L*
^{2}(0*, T*; *V* )*, T >* 0*. Then, G*(·*, x*) *∈ L*
^{2}(0*, T*; *H*) *and*

\parallel G\left(\cdot ,x\right){\parallel}_{{L}^{2}\left(0,T;H\right)}\le L\parallel k{\parallel}_{{L}^{2}\left(0,T\right)}\sqrt{T}\parallel x{\parallel}_{{L}^{2}\left(0,T;V\right)}.

*Moreover, if x*
_{1}
*, x*
_{2}
*∈ L*
^{2}(0*, T*; *V* )*, then*

\parallel G\left(\cdot ,{x}_{1}\right)-G\left(\cdot ,{x}_{2}\right){\parallel}_{{L}^{2}\left(0,T;H\right)}\le L\parallel k\parallel \sqrt{T}\parallel {x}_{1}-{x}_{2}{\parallel}_{{L}^{2}\left(0,T;V\right)}.

*Then, with the condition that*

{\omega}_{1}^{-1}{e}^{{\omega}_{2}{T}_{0}}L\parallel k\parallel \sqrt{{T}_{0}}<1

*in place of the condition (2.9), we can obtain the results of Theorem 2.1*.