In order to prove the existence of the generalized solution we apply Galerkin method.

**Theorem 3** *Assume that the assumptions of Theorem * 2 *hold*, *then the problem* (1.1)-(1.4) *has a unique solution* $u\in H(Q)$.

*Proof* Let

$\{{w}_{k}(x)\}$ be a fundamental system in

$H(0,1)$, such that

$({w}_{k},{w}_{i})={\delta}_{k,i}=\{\begin{array}{cc}1,\hfill & k=i,\hfill \\ 0,\hfill & k\ne i.\hfill \end{array}$

We have to find for each

$n\in {\mathbb{N}}^{\ast}$, the approximate solution of the problem (1.1)-(1.4) which has the following form:

${u}^{(n)}=\sum _{k=1}^{n}{\beta}_{k}(t){w}_{k}(x).$

(4.1)

Denote

$\begin{array}{r}{\phi}^{(n)}(x)=\sum _{k=1}^{n}{\phi}_{k}{w}_{k}(x),\phantom{\rule{2em}{0ex}}{\psi}^{(n)}(x)=\sum _{k=1}^{n}{\psi}_{k}{w}_{k}(x),\\ {\beta}_{k}(0)={\phi}_{k},\phantom{\rule{2em}{0ex}}{\beta}_{k}^{\mathrm{\prime}}(0)={\psi}_{k},\end{array}$

(4.2)

the approximate of the functions

$\phi (x)$ and

$\psi (x)$. Substituting the approximate solution in equation (

1.1), multiplying both sides by

${\int}_{x}^{1}(h(\xi )-h(x)){w}_{i}(\xi )\phantom{\rule{0.2em}{0ex}}d\xi $, then integrating according to

*x* on

$(0,1)$, we get

$\begin{array}{r}{\int}_{0}^{1}({u}_{tt}^{(n)}(x,t)-{a}^{2}(x,t){u}_{xx}^{(n)}(x,t)+c(x,t){u}^{(n)}(x,t))({\int}_{x}^{1}(h(\xi )-h(x)){w}_{i}(\xi )\phantom{\rule{0.2em}{0ex}}d\xi )\phantom{\rule{0.2em}{0ex}}dx\\ \phantom{\rule{1em}{0ex}}={\int}_{0}^{1}f(x,t)({\int}_{x}^{1}(h(\xi )-h(x)){w}_{i}(\xi )\phantom{\rule{0.2em}{0ex}}d\xi )\phantom{\rule{0.2em}{0ex}}dx\\ \phantom{\rule{2em}{0ex}}+{\int}_{0}^{1}{\int}_{0}^{t}\alpha (t-s)K(s,{u}^{(n)}(x,s))\phantom{\rule{0.2em}{0ex}}ds({\int}_{x}^{1}(h(\xi )-h(x)){w}_{i}(\xi )\phantom{\rule{0.2em}{0ex}}d\xi )\phantom{\rule{0.2em}{0ex}}dx.\end{array}$

(4.3)

Substituting (4.1) in (4.3), we get

Integrating by parts in

${L}_{2}(0,1)$ the left-hand side of (4.4) yields

Denote

$\begin{array}{r}{\theta}_{k,i}={({h}^{\mathrm{\prime}}(x){\int}_{x}^{1}{w}_{k}(\xi )\phantom{\rule{0.2em}{0ex}}d\xi ,{\int}_{x}^{1}{w}_{i}(\xi )\phantom{\rule{0.2em}{0ex}}d\xi )}_{{L}_{2}(0,1)},\\ \begin{array}{rl}{\sigma}_{k,i}=& {(2{(a{a}_{x})}_{x}{w}_{k}(x),{\int}_{x}^{1}(h(\xi )-h(x)){w}_{i}(\xi )\phantom{\rule{0.2em}{0ex}}d\xi )}_{{L}_{2}(0,1)}\\ -{((4a{a}_{x}{h}^{\mathrm{\prime}}+{a}^{2}{h}^{(2)}){w}_{k}(x),{\int}_{x}^{1}{w}_{i}(\xi )\phantom{\rule{0.2em}{0ex}}d\xi )}_{{L}_{2}(0,1)}\\ -{a}^{2}{h}^{\mathrm{\prime}}{({w}_{k}(x),{w}_{i}(x))}_{{L}_{2}(0,1)}-{(c{h}^{\mathrm{\prime}}{\int}_{x}^{1}{w}_{k}(\xi )\phantom{\rule{0.2em}{0ex}}d\xi ,{\int}_{x}^{1}{w}_{i}(\xi )\phantom{\rule{0.2em}{0ex}}d\xi )}_{{L}_{2}(0,1)}\\ +{({c}_{x}{\int}_{x}^{1}{w}_{k}(\xi )\phantom{\rule{0.2em}{0ex}}d\xi ,{\int}_{x}^{1}(h(\xi )-h(x)){w}_{i}(\xi )\phantom{\rule{0.2em}{0ex}}d\xi )}_{{L}_{2}(0,1)},\end{array}\\ \begin{array}{rl}{F}_{i}(t)=& {\int}_{0}^{1}f(x,t)({\int}_{x}^{1}(h(\xi )-h(x)){w}_{i}(\xi )\phantom{\rule{0.2em}{0ex}}d\xi )\phantom{\rule{0.2em}{0ex}}dx\\ +\sum _{k=1}^{n}{\int}_{0}^{1}({\int}_{0}^{t}\alpha (t-s)K(s,{\beta}_{k}(s){w}_{k}(x))\phantom{\rule{0.2em}{0ex}}ds)({\int}_{x}^{1}(h(\xi )-h(x)){w}_{i}(\xi )\phantom{\rule{0.2em}{0ex}}d\xi )\phantom{\rule{0.2em}{0ex}}dx.\end{array}\end{array}$

then (4.5) becomes

$\sum _{k=1}^{n}{\beta}_{k}^{\mathrm{\prime}\mathrm{\prime}}(t){\theta}_{k,i}+{\beta}_{k}(t){\sigma}_{k,i}={F}_{i}(t),\phantom{\rule{2em}{0ex}}{\beta}_{k}(0)={\phi}_{k},\phantom{\rule{2em}{0ex}}{\beta}_{k}^{\mathrm{\prime}}(0)={\psi}_{k}.$

Consequently, we obtain a Cauchy system of second-order integro-differential equations with smooth coefficients, so it has one and only one solution that for every *n* there exists a unique sequence ${u}^{(n)}$ that satisfies (4.3). □

**Lemma 4** *The sequence* $({u}^{(n)})$ *is bounded*.

*Proof* Multiplying (4.3) by

${\beta}_{i}^{\mathrm{\prime}}(t)$ then summing with respect to

*i* from 1 to

*n* it yields

Integrating (4.6) over

*t* from 0 to

*τ* we obtain

Thanks to Cauchy inequality,

*ϵ*-inequality, the hypotheses on the operator

*K* to the last term in the right-hand side of (4.7), we get

$\begin{array}{c}2|{\int}_{{Q}^{\tau}}{\int}_{0}^{t}a(t-s)K(s,{u}^{(n)}(x,s))\phantom{\rule{0.2em}{0ex}}ds({\int}_{x}^{1}(h(\xi )-h(x)){u}_{t}^{(n)}(\xi ,t)\phantom{\rule{0.2em}{0ex}}d\xi )\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt|\hfill \\ \phantom{\rule{1em}{0ex}}\le {l}^{2}\tau {\parallel K(t,{u}^{(n)})\parallel}_{{L}^{2}({Q}^{\tau})}^{2}+4{k}_{1}^{2}{\int}_{{Q}^{\tau}}{({\int}_{x}^{1}{u}_{t}(\xi ,t)\phantom{\rule{0.2em}{0ex}}d\xi )}^{2}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt\hfill \\ \phantom{\rule{1em}{0ex}}\le {l}^{2}\tau {\parallel {u}^{(n)}\parallel}_{{L}^{2}({Q}^{\tau})}^{2}+4{k}_{1}^{2}{\int}_{{Q}^{\tau}}{({\int}_{x}^{1}{u}_{t}^{(n)}(\xi ,t)\phantom{\rule{0.2em}{0ex}}d\xi )}^{2}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt.\hfill \end{array}$

Using similar inequalities for the second, the third and the fourth terms in the right-hand side of (4.7), then regrouping the same terms yields

Let

$\tilde{L}=\tilde{M}/m$, where

$\tilde{M}={k}_{1}max(3{C}_{1}+4{A}_{0}{A}_{1}+4{A}_{1}^{2}+2+4{k}_{1},10{A}_{0}{A}_{1}+{A}_{0}^{2}+4{A}_{1}^{2}+\frac{{l}^{2}T}{{k}_{1}},{C}_{0})$

then (4.8) becomes

$\begin{array}{r}{\int}_{0}^{1}[{({\int}_{x}^{1}{u}^{(n)}(\xi ,\tau )\phantom{\rule{0.2em}{0ex}}d\xi )}^{2}+{u}^{(n)}{(x,\tau )}^{2}+{({\int}_{x}^{1}{u}_{t}^{(n)}(\xi ,\tau )\phantom{\rule{0.2em}{0ex}}d\xi )}^{2}]\phantom{\rule{0.2em}{0ex}}dx\\ \phantom{\rule{1em}{0ex}}\le \tilde{L}[{\int}_{0}^{1}[{({\int}_{x}^{1}{\phi}^{(n)}(\xi )\phantom{\rule{0.2em}{0ex}}d\xi )}^{2}+{({\phi}^{(n)}(x))}^{2}+{({\int}_{x}^{1}{\psi}^{(n)}(\xi )\phantom{\rule{0.2em}{0ex}}d\xi )}^{2}]\phantom{\rule{0.2em}{0ex}}dx+{\parallel f\parallel}^{2}\\ \phantom{\rule{2em}{0ex}}+{\int}_{{Q}^{\tau}}[{({\int}_{x}^{1}{u}^{(n)}(\xi ,t)\phantom{\rule{0.2em}{0ex}}d\xi )}^{2}+|{u}^{(n)}(x,t){|}^{2}+{({\int}_{x}^{1}{u}_{t}^{(n)}(\xi ,t)\phantom{\rule{0.2em}{0ex}}d\xi )}^{2}]\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt].\end{array}$

(4.9)

Now, we apply Gronwall lemma to get

$\begin{array}{r}{\int}_{0}^{1}[{({\int}_{x}^{1}{u}_{t}(\xi ,\tau )\phantom{\rule{0.2em}{0ex}}d\xi )}^{2}+|u(x,\tau ){|}^{2}+{({\int}_{x}^{1}u(\xi ,\tau )\phantom{\rule{0.2em}{0ex}}d\xi )}^{2}]\phantom{\rule{0.2em}{0ex}}dx\\ \phantom{\rule{1em}{0ex}}\le {e}^{\tau \tilde{L}}({\int}_{0}^{1}[{({\int}_{x}^{1}{\phi}^{(n)}(\xi )\phantom{\rule{0.2em}{0ex}}d\xi )}^{2}+{({\phi}^{(n)}(x))}^{2}+{({\int}_{x}^{1}{\psi}^{(n)}(\xi )\phantom{\rule{0.2em}{0ex}}d\xi )}^{2}]\phantom{\rule{0.2em}{0ex}}dx+{\parallel f\parallel}^{2}).\end{array}$

(4.10)

Integrating (4.10) according to

*τ* on

$[0,T]$ yields

${\parallel {u}^{(n)}\parallel}_{H(Q)}^{2}\le T{e}^{T\tilde{L}}({\parallel {\phi}^{(n)}\parallel}_{H(0,1)}^{2}+{\parallel {\psi}^{(n)}\parallel}^{2}+{\parallel f\parallel}_{{L}_{2}(Q)}^{2}).$

(4.11)

Thus inequality (4.11) implies the boundedness of the sequence ${u}^{(n)}$. □

**Remark 5** We have proved that the sequence $\{{u}^{(n)}\}$ is bounded, so we can extract a subsequence, which we denote by $\{{u}^{({n}_{k})}\}$ that is weakly convergent. Now we prove that its limit is the desired solution of the problem (1.1)-(1.4).

**Lemma 6** *The limit of the subsequence* $\{{u}^{({n}_{k})}\}$ *is the solution of the problem* (1.1)-(1.4).

*Proof* We shall prove that the limit of the subsequence

$\{{u}^{({n}_{k})}\}$ satisfies the identity (2.1). Let

${\theta}_{k}(t)\in {C}^{2}(0,T)$, such that

${\theta}_{k}^{\mathrm{\prime}}(T)=0$, let us prove that identity (2.1) holds for any functions

$v(x,t)={\sum}_{k=1}^{n}{\theta}_{k}(t){w}_{k}(x)\in {H}_{T}(Q)$. Since the set

${S}_{n}=\{v(x,t)={\sum}_{k=1}^{n}{\theta}_{k}(t){w}_{k}(x),{\theta}_{k}(t)\in {C}^{2}(0,T),{\theta}_{k}^{\mathrm{\prime}}(T)=0\}$ is such that

${\bigcup}_{n=1}^{\mathrm{\infty}}{S}_{n}$ is dense in

${H}_{T}(Q)$, it suffices to prove (2.1) for

$v\in {S}_{n}$. Multiplying (4.3) by

${\theta}_{k}^{\mathrm{\prime}}(t)$, summing according to

*k* from 1 to

*n*, then integrating over

*t* from 0 to

*T*, we obtain

Denote by

*u* the weak limit of the subsequence

$\{{u}^{({n}_{k})}\}$ when

*k* tends to infinity. Hence,

$\begin{array}{c}|{\int}_{Q}{\int}_{0}^{t}\alpha (t-s)K(s,{u}^{({n}_{k})}(x,s))\phantom{\rule{0.2em}{0ex}}ds({\int}_{x}^{1}(h(\xi )-h(x)){v}_{t}(\xi ,t)\phantom{\rule{0.2em}{0ex}}d\xi )\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt\hfill \\ \phantom{\rule{2em}{0ex}}-{\int}_{Q}{\int}_{0}^{t}\alpha (t-s)K(s,u(x,s))\phantom{\rule{0.2em}{0ex}}ds({\int}_{x}^{1}(h(\xi )-h(x)){v}_{t}(\xi ,t)\phantom{\rule{0.2em}{0ex}}d\xi )\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt|\hfill \\ \phantom{\rule{1em}{0ex}}=|{\int}_{Q}({\int}_{0}^{t}\alpha (t-s)(K(s,{u}^{({n}_{k})}(x,s))-K(s,u(x,s)))\phantom{\rule{0.2em}{0ex}}ds)\hfill \\ \phantom{\rule{2em}{0ex}}\times ({\int}_{x}^{1}(h(\xi )-h(x)){v}_{t}(\xi ,t)\phantom{\rule{0.2em}{0ex}}d\xi )\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt|\hfill \\ \phantom{\rule{1em}{0ex}}\le 2kl\sqrt{T}{\left({\int}_{Q}{({\int}_{x}^{1}|{v}_{t}|\phantom{\rule{0.2em}{0ex}}d\xi )}^{2}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt\right)}^{1/2}\hfill \\ \phantom{\rule{2em}{0ex}}\times {\left({\int}_{Q}\right|K(s,{u}^{({n}_{k})}(x,s))-K(s,u(x,s)){|}^{2}\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}dx)}^{1/2}\hfill \\ \phantom{\rule{1em}{0ex}}=2kl\sqrt{T}{\left({\int}_{Q}{({\int}_{x}^{1}|{v}_{t}|\phantom{\rule{0.2em}{0ex}}d\xi )}^{2}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt\right)}^{1/2}{\parallel K(t,{u}^{({n}_{k})})-K(t,u)\parallel}_{{L}^{2}(Q)}\hfill \\ \phantom{\rule{1em}{0ex}}\le 2kl\sqrt{T}{\left({\int}_{Q}{({\int}_{x}^{1}|{v}_{t}|\phantom{\rule{0.2em}{0ex}}d\xi )}^{2}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt\right)}^{1/2}{\parallel {u}^{({n}_{k})}-u\parallel}_{{L}^{2}(Q)}\u27f60.\hfill \end{array}$

Finally, by passing to the limit in (4.12), we get that the limit *u* satisfies (2.1). □

**Example 7** Consider the following boundary value problem for hyperbolic integro-differential equation for

$0<x<1$,

$0<t\le T$:

$\begin{array}{r}\frac{{\partial}^{2}u}{\partial {t}^{2}}(x,t)-\frac{1}{8}\frac{{\partial}^{2}u}{\partial {x}^{2}}(x,t)+\frac{1}{2}{\pi}^{2}u(x,t)\\ \phantom{\rule{1em}{0ex}}=2{\pi}^{2}cos(2\pi x){e}^{-\pi t}-(t+\frac{1}{\pi}[{e}^{-\pi t}-1])\frac{sin(2\pi x)}{2{\pi}^{2}}\\ \phantom{\rule{2em}{0ex}}+{\int}_{0}^{t}(t-s)({\int}_{0}^{x}u(r,s)\phantom{\rule{0.2em}{0ex}}dr)\phantom{\rule{0.2em}{0ex}}ds,\end{array}$

(4.13)

subject to the initial conditions

$u(x,0)=cos(2\pi x),\phantom{\rule{2em}{0ex}}{u}_{t}(x,0)=-\pi cos(2\pi x),\phantom{\rule{1em}{0ex}}0<x<1$

(4.14)

and the weighted integral condition

where $h(x)=1-cos(\pi x)$. It is easy to prove that assumptions (H1)-(H3) are satisfied, then from Theorems 2 and 3, and we deduce that the problem (4.13)-(4.16) has a unique generalized solution in the sense of Definition 1. Moreover, the function $u(x,t)={e}^{-\pi t}cos(2\pi x)$ is the solution of this problem.