**Theorem 1**

If the assumptions (A1)-(A3) be satisfied, then the (1)-(4) problem has a unique solution.

**Proof**

Let an iteration for Fourier coefficients of (5) are:

\(u_{0}^{{(N+1)}}(t)={\varphi _0}+{\psi _0}t+\frac{2}{\pi }\int_{0}^{t} {\int_{0}^{\pi } {(t - \tau ){r^{(N)}}(\tau )f(\xi ,\tau ,{u^{(N)}})d\xi d\tau } } ,\)

\(\begin{gathered} u_{{ck}}^{{(N+1)}}(t)={\varphi _{ck}}\cos 2kt+\frac{{{\psi _{ck}}}}{{2k}}\sin 2kt \hfill \\ +\frac{1}{{\pi k}}\int_{0}^{t} {\int_{0}^{\pi } {{r^{(N)}}(\tau )f(\xi ,\tau ,{u^{(N)}})\cos 2k\xi \sin 2k(t - \tau )d\xi d\tau } } , \hfill \\ \end{gathered}\)

\(\begin{gathered} u_{{sk}}^{{(N+1)}}(t)={\varphi _{sk}}\cos 2kt+\frac{{{\psi _{sk}}}}{{2k}}\sin 2kt \hfill \\ +\frac{1}{{\pi k}}\int_{0}^{t} {\int_{0}^{\pi } {{r^{(N)}}(\tau )f(\xi ,\tau ,{u^{(N)}})\sin 2k\xi \sin 2k(t - \tau )d\xi d\tau } } , \hfill \\ \end{gathered}\)

$${r^{(N+1)}}(t)=\frac{{E^{\prime\prime}(t) - \pi \sum\limits_{{k=1}}^{\infty } {\left( {2k} \right)\left( {{\varphi _{sk}}\cos 2kt+\frac{{{\psi _{sk}}}}{{2k}}\sin 2kt+\frac{1}{{2k}}\int_{0}^{t} {{r^{(N)}}(\tau ){f_{sk}}(\tau )\sin 2k(t - \tau )d\tau } } \right)} }}{{\int_{0}^{\pi } {xf(x,t,{u^{(N)}})dx} }}$$

8

From the conditions of the theorem, \({u^{(0)}}(t) \in B,\,t \in [0,T]\) is obvious.

Let us analyse Fourier coefficients one by one for \(N=0\)

Using Cauchy, Bessel inequalities, after that Lipschitz condition, we get

\(\begin{gathered} \left\| {{u^{(1)}}(t)} \right\|=\mathop {\hbox{max} }\limits_{{0 \leqslant t \leqslant T}} \frac{{\left| {u_{0}^{{(1)}}(t)} \right|}}{2}+\sum\limits_{{k=1}}^{\infty } {\left[ {\mathop {\hbox{max} }\limits_{{0 \leqslant t \leqslant T}} \left| {u_{{ck}}^{{(1)}}(t)} \right|+\mathop {\hbox{max} }\limits_{{0 \leqslant t \leqslant T}} \left| {u_{{sk}}^{{(1)}}(t)} \right|} \right]} \leqslant \hfill \\ \frac{{\left\| {{\varphi _0}} \right\|+\left\| {{\psi _0}} \right\|\left| T \right|}}{2}+\sum\limits_{{k=1}}^{\infty } {\left\| {{\varphi _{ck}}} \right\|+\left\| {{\varphi _{sk}}} \right\|} +\frac{\pi }{{2\sqrt 6 }}\sum\limits_{{k=1}}^{\infty } {\left\| {{\psi _{ck}}} \right\|+\left\| {{\psi _{sk}}} \right\|} + \hfill \\ \left( {\frac{{\sqrt 3 T\sqrt T \left( {T+\sqrt 2 \pi } \right)}}{3}} \right)\left\| {{r^{(0)}}(t)} \right\|\left\| {b(x,t)} \right\|\left\| {{u^{(0)}}(t)} \right\|+\left( {\frac{{\sqrt 3 T\sqrt T \left( {T+\sqrt 2 \pi } \right)}}{3}} \right)\left\| {{r^{(0)}}(t)} \right\|M. \hfill \\ \end{gathered}\)

Hence from the conditions of the theorem \({u^{(1)}}(t) \in B,\,t \in [0,T].\)

If considered \(4k\int_{0}^{\pi } {\varphi (x)\sin 2kxdx} =2\int_{0}^{\pi } {\varphi ^{\prime}(x)\cos 2kxdx}\) for the inverse coefficient (7) will be rewritten as

\({r^{(1)}}(t)=\frac{{E^{\prime\prime}(t) - \pi \sum\limits_{{k=1}}^{\infty } {{{\varphi ^{\prime}}_{ck}}\cos 2kt - \pi \sum\limits_{{k=1}}^{\infty } {{\psi _{sk}}\sin 2kt} - 2\sum\limits_{{k=1}}^{\infty } {\int_{0}^{t} {\int_{0}^{\pi } {{r^{(0)}}(\tau )f(\tau )\sin 2k\xi \sin 2k(t - \tau )d\xi d\tau } } } } }}{{\int_{0}^{\pi } {xf(x,t,{u^{(0)}})dx} }}\)

Applying Cauchy, Bessel inequalities, after that Lipschitz condition, we get

\(\left\| {{r^{(1)}}(t)} \right\| \leqslant \frac{1}{{{M_0}\frac{{{\pi ^2}}}{2}}}\left( \begin{gathered} \left( {E^{\prime\prime}(t)} \right]+\pi \left\| {{{\varphi ^{\prime}}_{ck}}} \right\|+\pi \left\| {{\psi _{sk}}} \right\|+2T\sqrt T \pi \left\| {{r^{(0)}}(t)} \right\|\left\| {b(x,t)} \right\|\left\| {{u^{(0)}}(t)} \right\| \hfill \\ +2T\sqrt T \pi \left\| {{r^{(0)}}(t)} \right\|M \hfill \\ \end{gathered} \right)\)

It means that \({r^{(1)}}(t) \in B,\,t \in [0,T].\) The same estimations for the step N,

\(\begin{gathered} \left\| {{u^{(N+1)}}(t)} \right\|=\mathop {\hbox{max} }\limits_{{0 \leqslant t \leqslant T}} \frac{{\left| {u_{0}^{{(N)}}(t)} \right|}}{2}+\sum\limits_{{k=1}}^{\infty } {\left[ {\mathop {\hbox{max} }\limits_{{0 \leqslant t \leqslant T}} \left| {u_{{ck}}^{{(N)}}(t)} \right|+\mathop {\hbox{max} }\limits_{{0 \leqslant t \leqslant T}} \left| {u_{{sk}}^{{(N)}}(t)} \right|} \right]} \leqslant \hfill \\ \frac{{\left\| {{\varphi _0}} \right\|+\left\| {{\psi _0}} \right\|\left| T \right|}}{2}+\sum\limits_{{k=1}}^{\infty } {\left\| {{\varphi _{ck}}} \right\|+\left\| {{\varphi _{sk}}} \right\|} +\frac{\pi }{{2\sqrt 6 }}\sum\limits_{{k=1}}^{\infty } {\left\| {{\psi _{ck}}} \right\|+\left\| {{\psi _{sk}}} \right\|} + \hfill \\ \left( {\frac{{\sqrt 3 T\sqrt T \left( {T+\sqrt 2 \pi } \right)}}{3}} \right)\left\| {{r^{(N)}}(t)} \right\|\left\| {b(x,t)} \right\|\left\| {{u^{(N)}}(t)} \right\|+\left( {\frac{{\sqrt 3 T\sqrt T \left( {T+\sqrt 2 \pi } \right)}}{3}} \right)\left\| {{r^{(N)}}(t)} \right\|M, \hfill \\ \end{gathered}\)

and

\(\left\| {{r^{(N+1)}}(t)} \right\| \leqslant \frac{1}{{{M_0}\frac{{{\pi ^2}}}{2}}}\left( \begin{gathered} \left( {E^{\prime\prime}(t)} \right]+\pi \left\| {{{\varphi ^{\prime}}_{ck}}} \right\|+\pi \left\| {{\psi _{sk}}} \right\| \hfill \\ +2T\sqrt T \pi \left\| {{r^{(N)}}(t)} \right\|\left\| {b(x,t)} \right\|\left\| {{u^{(N)}}(t)} \right\|+2T\sqrt T \pi \left\| {{r^{(N)}}(t)} \right\|M \hfill \\ \end{gathered} \right)\)

We have \({r^{(N+1)}}(t) \in B\) as well according to \({u^{(N)}}(t) \in B,\,t \in [0,T].\)and theorem we obtain \({u^{(N+1)}}(t) \in B,\,t \in [0,T].\)

For convergence we applied same methods

$$\begin{gathered} \left\| {{u^{^{{(N+1)}}}}(t) - {u^{^{{(N)}}}}(t)} \right\| \leqslant \hfill \\ {\left( {\frac{{\sqrt 3 T\sqrt T (T+\sqrt 2 \pi )}}{3}+SM} \right)^N}\left\| {{r^{(N)}}(t)} \right\|...\left\| {{r^{(2)}}(t)} \right\|\left\| {{r^{(1)}}(t)} \right\| \hfill \\ A\frac{1}{{\sqrt {N!} }}{\left( {\int_{0}^{t} {\int_{0}^{\pi } {{b^2}(\xi ,\tau )d\xi d\tau } } } \right)^{\frac{N}{2}}} \hfill \\ \end{gathered}$$

9

As a result \({u^{(N+1)}} \to {u^{(N)}}\) when \(N \to \infty \,\,\), hence \({r^{(N+1)}} \to {r^{(N)}}\,\). Let us show that

$$\mathop {\lim }\limits_{{N \to \infty }} {u^{(N+1)}}(t)=u(t),\,\,\,\mathop {\lim }\limits_{{N \to \infty }} {r^{(N+1)}}(t)=r(t)$$

.

First examine the difference between the exact and approach inverse coefficients,

\(\begin{gathered} r(t) - {r^{(N+1)}}(t)=\frac{{ - \pi \sum\limits_{{k=1}}^{\infty } {\int_{0}^{t} {\int_{0}^{\pi } {r(\tau )f(\xi ,\tau ,u)\sin 2k\xi \sin 2k(t - \tau )d\xi d\tau } } } }}{{\int_{0}^{\pi } {xf(\xi ,\tau ,u)dx} }} - \hfill \\ \frac{{ - \pi \sum\limits_{{k=1}}^{\infty } {\int_{0}^{t} {\int_{0}^{\pi } {{r^{(N+1)}}(\tau )f(\xi ,\tau ,{u^{(N+1)}})\sin 2k\xi \sin 2k(t - \tau )d\xi d\tau } } } }}{{\int_{0}^{\pi } {xf(\xi ,\tau ,{u^{(N+1)}})dx} }} \hfill \\ \end{gathered}\)

then add and subtract \(\int_{0}^{t} {\int_{0}^{\pi } {r(\tau )f(\xi ,\tau ,{u^{(N+1)}})d\xi d\tau } }\), and finally we applied consecutively Cauchy, Bessel inequalities and Lipschitz condition,

$$\left\| {r(t) - {r^{(N+1)}}(t)} \right\| \leqslant S\left\| {r(t)} \right\|\left\| {b(x,t)} \right\|\left\| {u(t) - {u^{(N+1)}}(t)} \right\|$$

10

By taking the difference between the exact solution and the sequential solution, if apply Gronwall’s inequality, we find

\(\begin{gathered} \left\| {u(t) - {u^{(N+1)}}(t)} \right\| \leqslant \left( {\frac{{\sqrt 3 T\sqrt T (T+\sqrt 2 \pi )}}{3}} \right){\left( {\frac{{\sqrt 3 T\sqrt T (T+\sqrt 2 \pi )}}{3}+SM} \right)^N} \hfill \\ \left\| {{r^{(N)}}(t)} \right\|\left\| {{r^{(N)}}(t)} \right\|...\left\| {{r^{(2)}}(t)} \right\|\left\| {{r^{(1)}}(t)} \right\|A\frac{1}{{\sqrt {N+1!} }}{\left( {\int_{0}^{t} {\int_{0}^{\pi } {{b^2}(\xi ,\tau )d\xi d\tau } } } \right)^{\frac{{N+1}}{2}}} \times \hfill \\ \exp \left( {2\frac{{\sqrt 3 T\sqrt T (T+\sqrt 2 \pi )}}{3}+SM} \right)\left\| {r(t)} \right\|\left\| {b(x,t)} \right\| \hfill \\ \end{gathered}\)

or

\(\left\| {u(t) - {u^{(N+1)}}(t)} \right\| \leqslant 0 \cdot \exp \left( {2\frac{{\sqrt 3 T\sqrt T (T+\sqrt 2 \pi )}}{3}+SM} \right)\left\| {r(t)} \right\|\left\| {b(x,t)} \right\|\)

Thus \({u^{(N+1)}}(t) \to u(t),\,\,{r^{(N+1)}}(t) \to r(t)\) when \(N \to \infty\).

Let consider we have two solutions \((u,r)\) and \((v,q)\) of (1)-(4). It follows the same procedures as above,

$$\left\| {r(t) - q(t)} \right\| \leqslant S\left\| {r(t)} \right\|\left\| {b(x,t)} \right\|\left\| {u(t) - v(t)} \right\|$$

11

Here

\(S=\frac{{\frac{{2\sqrt T }}{{{M_*}}}}}{{\left( {1 - \frac{{2\sqrt T }}{{{M_*}}}M} \right)}}\)

By using (10) in (11), we get

$$\left\| {u(t) - v(t)} \right\| \leqslant \left( {\frac{{\sqrt 3 T\sqrt T (T+\sqrt 2 \pi )}}{3}+SM} \right)\left\| {r(t)} \right\|\left\| {b(x,t)} \right\|\left\| {u(t) - v(t)} \right\|$$

12

If Gronwall inequality is applied to the last inequality

$$\left\| {u(t) - v(t)} \right\| \leqslant 0 \times \exp \left( {\frac{{\sqrt 3 T\sqrt T (T+\sqrt 2 \pi )}}{3}+SM} \right){\left( {\int_{0}^{t} {\int_{0}^{\pi } {{r^2}(\tau ){b^2}(\xi ,\tau )d\xi d\tau } } } \right)^{\frac{1}{2}}}$$

13

Then \(u(t)=v(t)\), therefore \(r(t)=q(t)\). The proof is completed.