At the beginning for the investigation of the posed problem we set the following expression in order to obtain of the a priori estimations

for a. e. *t* ∈ (0, *T*), i.e. *y* = *A*^{−1} *x* is contained in the bounded subset of the space *y* ∈ *C*^{1}(0, *T*; *V*) ∩ *C*^{0}(0, *T*; *Y*), consequently, we obtain that if the weak solution *x*(*t*) exists then it belongs to a bounded subset of the space *C*^{0}(0, *T*; *X*) ∩ *C*^{1}(0, *T*; *V*^{∗}).

#### Proof of Theorem 2.2

In order to prove of the solvability theorem we will use the Faedo-Galerkin approach. Let the system
$\begin{array}{}{\displaystyle {\left\{{y}^{k}\right\}}_{k=1}^{\mathrm{\infty}}\subset Y}\end{array}$
be total in *Y* such that it is complete in the spaces *Y*, *V*, and also in the spaces *X*, *H*. We will seek out of the approximative solutions *y*_{m}(*t*), and consequently *x*_{m}(*t*), in the form

$$\begin{array}{}{\displaystyle {x}_{m}\left(t\right)\equiv A{y}_{m}\left(t\right)=\stackrel{m}{\sum _{k=1}}{c}_{i}\left(t\right)A{y}^{k}\text{\hspace{0.17em}or\hspace{0.17em}}{x}_{m}\left(t\right)\in span\left\{{y}^{1},...,{y}^{m}\right\}}\end{array}$$

as the solutions of the considered problem, where *c*_{i}(*t*) are the unknown functions that will be defined as solutions of the following Cauchy problem for system of ODE

$$\begin{array}{}{\displaystyle \frac{{d}^{2}}{d{t}^{2}}\u3008{x}_{m},{y}^{j}\u3009+\u3008F\left({x}_{m}\right),A{y}^{j}\u3009=\u3008g\left({x}_{m},B{y}_{mt}\right),{y}^{j}\u3009,\phantom{\rule{1em}{0ex}}j=1,2,...,m}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{x}_{m}\left(0\right)={x}_{0m},\phantom{\rule{thinmathspace}{0ex}}{x}_{tm}\left(0\right)={x}_{1m},}\end{array}$$

where *x*_{0m} and *x*_{1m} are contained in *span*{*y*^{1},…, *y*^{m}}, *m* = 1, 2, …, moreover,

$$\begin{array}{}{\displaystyle {x}_{0m}\u27f6{x}_{0}\phantom{\rule{1em}{0ex}}\text{in\hspace{0.17em}}{\left[X,Y\right]}_{\frac{1}{2}}\subseteq V;\phantom{\rule{1em}{0ex}}{x}_{1m}\u27f6{x}_{1}\phantom{\rule{1em}{0ex}}\text{in\hspace{0.17em}}X,m\nearrow \mathrm{\infty}.}\end{array}$$

Thus we obtain the following problem

$$\begin{array}{}\frac{{d}^{2}}{d{t}^{2}}\u3008{x}_{m},{y}^{j}\u3009+\u3008F\left({x}_{m}\right),A{y}^{j}\u3009=\u3008g\left({x}_{m},B{y}_{mt}\right),{y}^{j}\u3009,\phantom{\rule{1em}{0ex}}j=1,2,...,m\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\u3008{x}_{m}\left(t\right),{y}^{j}\u3009\left|{\phantom{\rule{thinmathspace}{0ex}}}_{t=0}\right.=\u3008{x}_{0m},{y}^{j}\u3009,\phantom{\rule{thinmathspace}{0ex}}\frac{d}{dt}\u3008{x}_{m}\left(t\right),{y}^{j}\u3009\left|{\phantom{\rule{thinmathspace}{0ex}}}_{t=0}\right.=\u3008{x}_{1m},{y}^{j}\u3009\end{array}$$(10)

that is solvable by virtue of estimates (9) on (0, *T*) for any *m* = 1,2,…, *j* = 1,2,… and *T* > 0. Hence we set

$$\begin{array}{}\frac{{d}^{2}}{d{t}^{2}}\u3008{x}_{m},z\u3009+\u3008F\left({x}_{m}\right),Az\u3009=\u3008g\left({x}_{m},B{y}_{mt}\right),z\u3009\end{array}$$(11)

for any *z* ∈ *Y* and *m* = 1,2,….

Consequently, with use of the known procedure ([7, 8, 9]) we obtain, *y*_{mt} ∈ *C*^{0} (0, *T*; *V*), *y*_{m} ∈ *C*^{0} (0, *T*; *Y*) and *x*_{m} ∈ *C*^{0} (0, *T*; *X*), *x*_{mt} ∈ *C*^{0} (0, *T*; *V*^{∗}), moreover, they are contained in the bounded subset of these spaces for any *m* = 1, 2,…. Hence from (9) we get

$$\begin{array}{}{x}_{mtt}\in {C}^{0}\left(0,T;{Y}^{\ast}\right)\phantom{\rule{thinmathspace}{0ex}}\text{or}\phantom{\rule{thinmathspace}{0ex}}{x}_{m}\in {C}^{2}\left(0,T;{Y}^{\ast}\right),\phantom{\rule{thinmathspace}{0ex}}({V}^{\ast}\subset {Y}^{\ast}).\end{array}$$

Thus we obtain, that the sequence
$\begin{array}{}{\left\{{x}_{m}\right\}}_{m=1}^{\mathrm{\infty}}\end{array}$ of the approximated solutions of the problem is contained in a bounded subset of the space

$$\begin{array}{}{C}^{0}\left(0,T;X\right)\cap {C}^{1}\left(0,T;{V}^{\ast}\right)\cap {C}^{2}\left(0,T;{Y}^{\ast}\right)\end{array}$$

or
$\begin{array}{}{\left\{{x}_{m}\right\}}_{m=1}^{\mathrm{\infty}}\end{array}$ such that for a. e. *t* ∈ (0, *T*) the following inclusions take place
$\begin{array}{}{\left\{{y}_{m}\left(t\right)\right\}}_{m=1}^{\mathrm{\infty}}\end{array}$ ⊂ *Y* ⊂ *X* ⊂ *H*,
$\begin{array}{}{\left\{{y}_{mt}\left(t\right)\right\}}_{m=1}^{\mathrm{\infty}}\subset V,{\left\{{y}_{mtt}\left(t\right)\right\}}_{m=1}^{\mathrm{\infty}}\subset {X}^{\ast}.\end{array}$ So we have

$$\begin{array}{}{\left\{{y}_{m}\left(t\right)\right\}}_{m=1}^{\mathrm{\infty}}\subset {C}^{0}\left(0,T;Y\right)\cap {C}^{1}\left(0,T;V\right)\cap {C}^{2}\left(0,T;{X}^{\ast}\right).\end{array}$$

Therefore,
$\begin{array}{}{\left\{{y}_{m}\left(t\right)\right\}}_{m=1}^{\mathrm{\infty}}\end{array}$ possess a precompact subsequence in
$\begin{array}{}{C}^{1}\left(0,T;{\left[{X}^{\ast},Y\right]}_{\frac{1}{2}}\right)\end{array}$ and in *C*^{1} (0, *T*; *V*), as
$\begin{array}{}{\left[{X}^{\ast},Y\right]}_{\frac{1}{2}}\subseteq V\end{array}$ by virtue of conditions on *X* and *A* (by virtue of well known results, see, e. g. [10, 11] etc.). From here follows *y*_{m} (*t*) ⟶ *y*(*t*) in *C*^{1} (0, *T*; *V*) for *m* ↗ ∞ (Here and hereafter in order to abate the number of index we don’t change the indexes of subsequences). Then the sequence
$\begin{array}{}{\left\{F\left(A{y}_{m}\left(t\right)\right)\right\}}_{m=1}^{\mathrm{\infty}}\subset {X}^{\ast}\end{array}$ and bounded for a. e. *t* ∈ (0, *T*); the sequence

$$\begin{array}{}{\left\{g\left({x}_{m}\left(t\right),{x}_{mt}\left(t\right)\right)\right\}}_{m=1}^{\mathrm{\infty}}\equiv {\left\{g\left(A{y}_{m}\left(t\right),B{y}_{mt}\left(t\right)\right)\right\}}_{m=1}^{\mathrm{\infty}}\subset H\end{array}$$

and bounded for a. e. *t* ∈ (0, *T*) also, by virtue of the condition (*iii*). Indeed, for any *m* the estimation

$$\begin{array}{}{\u2225g\left(A{y}_{m},B{y}_{mt}\right)\u2225}_{H}\left(t\right)\le {\u2225A{y}_{m}\left(t\right)\u2225}_{H}+{\u2225B{y}_{mt}\left(t\right)\u2225}_{H}+{\u2225g\left(0,0\right)\u2225}_{H}\end{array}$$

holds and, therefore,
$\begin{array}{}{\left\{g\left(A{y}_{m}\left(t\right),B{y}_{mt}\left(t\right)\right)\right\}}_{m=1}^{\mathrm{\infty}}\end{array}$ is contained in a bounded subset of *H* for a. e. *t* ∈ (0, *T*). Consequently,
$\begin{array}{}{\left\{F\left(A{y}_{m}\right)\right\}}_{m=1}^{\mathrm{\infty}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\left\{g\left(A{y}_{m}\left(t\right),B{y}_{mt}\left(t\right)\right)\right\}}_{m=1}^{\mathrm{\infty}}\end{array}$ have weakly converging subsequences to *η*(*t*) and *θ* (*t*) in *X*^{∗} and *H*, respectively, for a. e. *t* ∈ (0, *T*). Hence one can pass to the limit in (11) with respect tom↗ ∞. Then we obtain the following equation

$$\begin{array}{}\frac{{d}^{2}}{d{t}^{2}}\u3008x,z\u3009+\u3008A\eta \left(t\right),z\u3009=\u3008\theta \left(t\right),z\u3009.\end{array}$$(12)

It remained to show the following: if the sequence
$\begin{array}{}{\left\{{x}_{m}\left(t\right)\right\}}_{m=1}^{\mathrm{\infty}}\equiv {\left\{A{y}_{m}\left(t\right)\right\}}_{m=1}^{\mathrm{\infty}}\end{array}$ is weakly converging to *x* (*t*) = *Ay* (*t*) then *η* (*t*) = *F* (*x* (*t*)) and *θ* (*t*) = *g* (*x* (*t*), *By*_{t} (*t*)). In order to show these equations are fulfilled we will use the monotonicity of *F* and the condition (*iii*).

We start by showing *θ* (*t*) = *g* ((*t*), *By*_{t} (*t*)) as *x* ∈ *X* ⊂ *H* and *y*_{t} ∈ *V*, *By*_{t} ∈ *H* therefore *g* (*x*, *By*_{t}) is defined for a. e. *t* ∈ (0, *T*). Consequently, one can consider of the expression

$$\begin{array}{}\u3008g\left(A{y}_{m}\left(t\right),B{y}_{mt}\left(t\right)\right)-g\left(Ay\left(t\right),B{y}_{t}\left(t\right)\right),\hat{\phantom{\rule{thinmathspace}{0ex}}y}\u3009\end{array}$$

for any Ŷ ∈ *C*^{0} (0, *T*; *Y*) ∩ *C*^{1} (0, *T*; *V*). So we set this expression and investigate this for any Ŷ ∈ *C*^{0} (0, *T*; *Y*) ∩ *C*^{1} (0, *T*; *V*); then we have

$$\begin{array}{}\left|\u3008g\left(A{y}_{m}\left(t\right),B{y}_{mt}\left(t\right)\right)-g\left(Ay\left(t\right),B{y}_{t}\left(t\right)\right),\hat{\phantom{\rule{thinmathspace}{0ex}}y}\u3009\right|\le \end{array}$$

$$\begin{array}{}{g}_{1}\left|\u3008A{y}_{m}\left(t\right)-Ay\left(t\right),\hat{\phantom{\rule{thinmathspace}{0ex}}y}\left(t\right)\u3009\right|+{g}_{2}\left|\u3008B{y}_{mt}\left(t\right)-B{y}_{t}\left(t\right),\hat{\phantom{\rule{thinmathspace}{0ex}}y}\left(t\right)\u3009\right|\end{array}$$(13)

that takes place by virtue of the condition (*iii*). Using here the weak convergences of *Ay*_{m} (*t*) ⇀ *Ay* (*t*) and *By*_{mt} (*t*) ⟶ *By*_{t} (*t*) and by passing to the limit in the inequation (13) with respect to *m* : *m* ↗ ∞ we get

$$\begin{array}{}\left|\u3008\theta \left(t\right)-g\left(Ay\left(t\right),B{y}_{t}\left(t\right)\right),\hat{\phantom{\rule{thinmathspace}{0ex}}y}\u3009\right|\le 0\end{array}$$

for any Ŷ ∈ *C*^{0} (0, *T*; *H*). Consequently, the equation *θ* (*t*) = *g* ((*t*), *By*_{t} (*t*)) holds, then the following equation is valid

$$\begin{array}{}\frac{{d}^{2}}{d{t}^{2}}\u3008x,z\u3009+\u3008A\eta \left(t\right),z\u3009=\u3008g\left(x\left(t\right),B{y}_{t}\left(t\right)\right),z\u3009\end{array}$$

for any *z* ∈ *Y*, as
$\begin{array}{}{\left\{{y}^{k}\right\}}_{k=1}^{\mathrm{\infty}}\end{array}$ is complete in *Y* that display fulfilling of equation

$$\begin{array}{}A\eta \left(t\right)=g\left(x\left(t\right),B{y}_{t}\left(t\right)\right)-\frac{{d}^{2}x}{d{t}^{2}}\end{array}$$(14)

in the sense of *Y*^{∗}.

In order to show the equation *η* (*t*) = *F* (*x*(*t*)) one can use the monotonicity of *F*. So the following inequation holds

$$\begin{array}{}\u3008A\circ F\left(Az\right)-A\circ F\left(Ay\right),z-y\u3009=\u3008F\left(Az\right)-F\left(Ay\right),Az-Ay\u3009=\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\u3008F\left(\stackrel{~}{\phantom{\rule{thinmathspace}{0ex}}x}\right)-F\left(x\right),\stackrel{~}{\phantom{\rule{thinmathspace}{0ex}}x}-x\u3009\ge 0\end{array}$$

for any *y*, *z* ∈ *Y*, *Ay* = *x* and *Az* = x̃ by condition (*i*). Then one can write

$$\begin{array}{}0\le \u3008F\left({x}_{m}\right)-F\left(\stackrel{~}{\phantom{\rule{thinmathspace}{0ex}}x}\right),{x}_{m}-\stackrel{~}{\phantom{\rule{thinmathspace}{0ex}}x}\u3009=\u3008F\left(A{y}_{m}\right)-F\left(Az\right),A{y}_{m}-Az\u3009=\end{array}$$

take account here the equation (9)

$$\begin{array}{}\u3008F\left(A{y}_{m}\right),A{y}_{m}\u3009-\u3008\frac{{d}^{2}}{d{t}^{2}}{x}_{m}-g\left({x}_{m},B{y}_{mt}\right),z\u3009-\u3008F\left(Az\right),A{y}_{m}-Az\u3009=\\ \phantom{\rule{2em}{0ex}}\u3008F\left({x}_{m}\right),{x}_{m}\u3009-\u3008\frac{{d}^{2}}{d{t}^{2}}{x}_{m}-g\left({x}_{m},B{y}_{mt}\right),z\u3009-\u3008F\left(\stackrel{~}{\phantom{\rule{thinmathspace}{0ex}}x}\right),{x}_{m}-\stackrel{~}{\phantom{\rule{thinmathspace}{0ex}}x}\u3009.\end{array}$$(15)

Here one can use the well-known inequation

$$\begin{array}{}limsup\u3008F\left({x}_{m}\right),{x}_{m}\u3009\le \u3008\eta ,x\u3009=\u3008\eta ,Ay\u3009=\u3008A\eta ,y\u3009.\end{array}$$

Then passing to the limit in (15) with respect to *m* : *m* ↗ ∞ we obtain

$$\begin{array}{}\phantom{\rule{1em}{0ex}}0\le \u3008A\eta ,y\u3009-\u3008\frac{{d}^{2}}{d{t}^{2}}x-g\left(x,B{y}_{t}\right),z\u3009-\u3008F\left(\stackrel{~}{\phantom{\rule{thinmathspace}{0ex}}x}\right),x-\stackrel{~}{\phantom{\rule{thinmathspace}{0ex}}x}\u3009=\\ \u3008A\eta ,y\u3009-\u3008A\eta ,z\u3009-\u3008F\left(\stackrel{~}{\phantom{\rule{thinmathspace}{0ex}}x}\right),Ay-Az\u3009=\u3008A\eta -A\circ F\left(\stackrel{~}{\phantom{\rule{thinmathspace}{0ex}}x}\right),y-z\u3009\end{array}$$

by virtue of (14).

Consequently, we obtain that the equation *Aη* (*t*) = *A* ∘ *F* (*x*) holds since *z* is arbitrary element of *Y*.

Now it remains to show that the obtained function *x*(*t*) = *Ay*(*t*) satisfies the initial conditions. Consider the following equation

$$\begin{array}{}{\displaystyle \u3008{y}_{mt},A{y}_{m}\u3009\left(t\right)=\underset{0}{\stackrel{t}{\int}}\u3008\frac{{d}^{2}}{d{s}^{2}}A{y}_{m},{y}_{m}\u3009ds+\underset{0}{\stackrel{t}{\int}}\u3008\frac{d}{ds}B{y}_{m},\frac{d}{ds}B{y}_{m}\u3009ds+\u3008{y}_{1m},A{y}_{0m}\u3009=}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\displaystyle \underset{0}{\stackrel{t}{\int}}\u3008\frac{{d}^{2}}{d{s}^{2}}{y}_{m},A{y}_{m}\u3009ds+\underset{0}{\stackrel{t}{\int}}{\u2225\frac{d}{ds}B{y}_{m}\u2225}_{H}^{2}ds+\u3008{y}_{1m},A{y}_{0m}\u3009}\end{array}$$

for *m* = 1, 2,…, here *x*_{m}(*t*) = *Ay*_{m}(*t*). Hence we get: the left side is bounded as far as all added items in the right side are bounded by virtue of the obtained estimations. Therefore, one can pass to limit with respect to *m* as here *y*_{mt} is continous with respect to *t* for any *m*; then *y*_{mt} strongly converges to *y*_{t} and *Ay*_{m} weakly converges to *Ay* in *H*. It must be noted the equation

$$\begin{array}{}{\displaystyle \underset{m\u27f6\mathrm{\infty}}{lim}\stackrel{t}{\underset{0}{\int}}{\u2225\frac{d}{ds}B{y}_{m}\u2225}_{H}^{2}dxds=\underset{0}{\stackrel{t}{\int}}{\u2225\frac{d}{ds}By\u2225}_{H}^{2}ds}\end{array}$$

holds by virtue of the above reasonings that
$\begin{array}{}{\left\{{y}_{m}\left(t\right)\right\}}_{m=1}^{\mathrm{\infty}}\end{array}$ is a precompact subset in *C*^{1} (0, *T*; *V*). Consequently, the left side converges to the expression of
such type, i.e. to 〈*y*_{t}, *Ay*〉 (*t*). The obtained results show that the following convergences are just: *x*_{m} (*t*) = *Ay*_{m} (*t*) ⇀ *Ay* (*t*) = *x* (*t*) in *X*, *x*_{mt} (*t*) = *Ay*_{mt} ⇀ *Ay*_{t} = *x*_{t} (*t*) in *V*^{∗}. From here follows, that the initial conditions are fulfilled in the sense of *X* and *V*^{∗}, respectively.

Thus the existence theorem is fully proved. □

It is clear that all conditions of Theorem 2.2 are fulfilled for this problem under above conditions of this example.

We would like to note the equation with main part of such type hasn’t been studied earlier.

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