Since system (2.5) forms a Riesz basis in \(L_{2}(0,1)\) and systems (2.5)–(2.6) form a system of biorthogonal functions in \(L_{2}(0,1)\), we’ll seek the first component \(u(x,t)\) of classical solution \(\{ u(x,t), a(t), b(t)\}\) of problem (1.1)–(1.3), (1.6), and (1.7) in the form
$$ u(x,t) = \sum_{k = 0}^{\infty } u_{k}(t)X_{k}(x), $$
(3.1)
where
$$ u_{k}(t) = \int _{0}^{1} u(x,t)Y_{k}(x)\,dx \quad (k = 0,1, \ldots). $$
(3.2)
Moreover, \(X_{k}(x)\) and \(Y_{k}(x)\) are defined by relations (2.5) and (2.6), respectively.
Then by applying the method of separation of variables, from (1.1) and (1.2) we have
$$\begin{aligned}& c(t)u'_{0}(t) = F_{0}(t; u, a, b)\quad (0 \le t \le T), \end{aligned}$$
(3.3)
$$\begin{aligned}& c(t)u'_{2k - 1}(t) + \lambda _{k}^{2}u_{2k - 1}(t) = F_{2k - 1}(t; u, a, b) \quad (k = 1,2 \ldots ; 0 \le t \le T), \end{aligned}$$
(3.4)
$$\begin{aligned}& c(t)u'_{2k}(t) + \lambda _{k}^{2}u_{2k}(t) = F_{2k}(t; u, a, b) - 2a \lambda _{k}u_{2k - 1}(t)\quad (k = 1,2 \ldots ; 0 \le t \le T), \end{aligned}$$
(3.5)
$$\begin{aligned}& u_{k}(0) + \delta u_{k}(T) + \int _{0}^{T} p(t)u_{k}(t)\,dt = \varphi _{k}\quad (k = 0,1,2, \ldots ), \end{aligned}$$
(3.6)
where
$$\begin{aligned}& \lambda _{k} = 2k\pi\quad (k = 1,2, \ldots ), \\& F_{k}(t; u, a, b) = f_{k}(t) + b(t)g_{k}(t) + a(t)u_{k}(t),\quad (k = 0,1,2, \ldots ), \\& f_{k}(t) = \int _{0}^{1} f(x,t)Y_{k}(x)\,dx, \qquad g_{k}(t) = \int _{0}^{1} g(x,t)Y_{k}(x)\,dx, \\& \varphi _{k} = \int _{0}^{1} \varphi (x)Y_{k}(x)\,dx \quad (k = 0,1, \ldots ). \end{aligned}$$
Solving problem (3.3)–(3.6), we obtain
$$\begin{aligned}& \begin{aligned}[b] u_{0}(t) &= (1 + \delta )^{ - 1} \biggl( \varphi _{0} - \int _{0}^{T} p(t)u _{0}(t)\,dt - \delta \int _{0}^{T} \frac{1}{c(t)}F_{10}(t; u, a,b)\,dt \biggr) \\ &\quad {}+ \int _{0}^{t} \frac{1}{c(\tau )}F_{0}( \tau ; u, a,b)\,d\tau ,\end{aligned} \end{aligned}$$
(3.7)
$$\begin{aligned}& \begin{aligned}[b] u_{2k - 1}(t) &= \frac{e^{ - \int _{0}^{t} \frac{\lambda _{k}^{2}}{c(s)}\,ds}}{1 + \delta e^{ - \int _{0}^{T} \frac{ \lambda _{k}^{2}}{c(s)}\,ds}} \biggl( \varphi _{2k - 1} - \int _{0}^{T} p(t)u _{2k - 1}(t)\,dt \biggr)\\ &\quad {} + \int _{0}^{t} \frac{1}{c(\tau )}F_{2k - 1}( \tau ; u, a, b)e^{ - \int _{\tau }^{t} \frac{\lambda _{k}^{2}}{c(s)}\,ds} \,d\tau \\ &\quad {}- \frac{\delta e^{ - \int _{0}^{T} \frac{\lambda _{k}^{2}}{c(s)}\,ds}}{1 + \delta e^{ - \int _{0}^{T} \frac{\lambda _{k}^{2}}{c(s)}\,ds}} \int _{0} ^{T} \frac{1}{c(\tau )}F_{2k - 1}( \tau ; u, a, b)e^{ - \int _{\tau } ^{t} \frac{\lambda _{k}^{2}}{c(s)}\,ds} \,d\tau ,\quad (k = 1,2, \ldots ),\end{aligned} \end{aligned}$$
(3.8)
$$\begin{aligned}& \begin{aligned}[b] u_{2k}(t) &= \frac{e^{ - \int _{0}^{t} \frac{\lambda _{k}^{2}}{c(s)}\,ds}}{1 + \delta e^{ - \int _{0}^{T} \frac{\lambda _{k}^{2}}{c(s)}\,ds}} \biggl( \varphi _{2k} - \int _{0}^{T} p(t)u_{2k}(t)\,dt \biggr)\\ &\quad {} + \int _{0}^{t} \frac{1}{c( \tau )}F_{2k}( \tau ; u, a, b)e^{ - \int _{\tau }^{t} \frac{\lambda _{k} ^{2}}{c(s)}\,ds} \,d\tau \\ &\quad {}- \frac{\delta e^{ - \int _{0}^{T} \frac{\lambda _{k}^{2}}{c(s)}\,ds}}{1 + \delta e^{ - \int _{0}^{T} \frac{\lambda _{k}^{2}}{c(s)}\,ds}} \int _{0} ^{T} \frac{1}{c(\tau )}F_{2k}( \tau ; u, a, b)e^{ - \int _{\tau }^{t} \frac{ \lambda _{k}^{2}}{c(s)}\,ds} \,d\tau + q_{k}(t,T), \end{aligned} \end{aligned}$$
(3.9)
where
$$\begin{aligned} q_{k}(t,T) &= - 2a\lambda _{k} \biggl( \varphi _{2k - 1} - \int _{0}^{T} p(t)u _{2k - 1}(t)\,dt \biggr) \\ &\quad {}\times \frac{\delta e^{ - \int _{0}^{t} \frac{\lambda _{k}^{2}}{c(s)}\,ds}}{1 + \delta e^{ - \int _{0}^{T} \frac{\lambda _{k} ^{2}}{c(s)}\,ds}} \biggl[ \int _{0}^{t} \frac{d\tau }{c(\tau )} - \frac{ \delta e^{ - \int _{0}^{T} \frac{\lambda _{k}^{2}}{c(s)}\,ds}}{1 + \delta e^{ - \int _{0}^{T} \frac{\lambda _{k}^{2}}{c(s)}\,ds}} \int _{0}^{T} \frac{d \tau }{c(\tau )} \biggr] \\ &\quad {}- 2a\lambda _{k} \biggl[ \int _{0}^{t} \frac{1}{c(\tau )} \biggl( \int _{0}^{\tau } \frac{1}{c(\xi )}F_{2k - 1}( \xi ; u, a, b)e^{ - \int _{ \xi }^{t} \frac{\lambda _{k}^{2}}{c(s)}\,ds} \,d\tau \biggr)\,d\xi \\ &\quad {} - \frac{\delta e^{ - \int _{0}^{T} \frac{\lambda _{k}^{2}}{c(s)}\,ds}}{1 + \delta e^{ - \int _{0}^{T} \frac{ \lambda _{k}^{2}}{c(s)}\,ds}} \int _{0}^{T} \frac{1}{c(\tau )} \biggl( \int _{0}^{\tau } \frac{1}{c(\xi )}F_{2k - 1}( \xi ; u, a, b)e^{ - \int _{ \xi }^{t} \frac{\lambda _{k}^{2}}{c(s)}\,ds} \,d\xi \biggr)\,d\tau \biggr] \\ &\quad {}+ 2a\lambda _{k}\frac{\delta e^{ - \int _{0}^{T} \frac{\lambda _{k}^{2}}{c(s)}\,ds}}{1 + \delta e^{ - \int _{0}^{T} \frac{ \lambda _{k}^{2}}{c(s)}\,ds}} \int _{0}^{T} \frac{1}{c(\xi )}F_{2k - 1}( \xi ; u, a, b)e^{ - \int _{\xi }^{t} \frac{\lambda _{k}^{2}}{c(s)}\,ds} \,d \xi \\ &\quad {}\times \biggl[ \int _{0}^{t} \frac{1}{c(\tau )}\,d\tau - \frac{\delta e ^{ - \int _{0}^{T} \frac{\lambda _{k}^{2}}{c(s)}\,ds}}{1 + \delta e^{ - \int _{0}^{T} \frac{\lambda _{k}^{2}}{c(s)}\,ds}} \int _{0}^{T} \frac{1}{c( \tau )}\,d\tau \biggr] \quad (k = 1,2, \ldots ). \end{aligned}$$
(3.10)
After substituting expressions \(u_{0}(t)\), \(u_{2k - 1}(t)\), and \(u_{2k}(t)\) (\(k = 1,2, \ldots \)), respectively described by (3.7), (3.8), and (3.9), into (3.1), we have
$$\begin{aligned} u(x,t) &= (1 + \delta )^{ - 1} \biggl( \varphi _{0} - \int _{0}^{T} p(t)u _{0}(t)\,dt - \delta \int _{0}^{T} \frac{1}{c(t)}F_{10}(t; u, a,b)\,dt \biggr) \\ &\quad {} + \int _{0}^{t} \frac{1}{c(\tau )}F_{0}( \tau ; u, a,b)\,d\tau \\ &\quad {}+ \sum_{k = 1}^{\infty } \biggl\{ \frac{e^{ - \int _{0}^{t} \frac{ \lambda _{k}^{2}}{c(s)}\,ds}}{1 + \delta e^{ - \int _{0}^{T} \frac{\lambda _{k}^{2}}{c(s)}\,ds}} \biggl( \varphi _{2k - 1} - \int _{0}^{T} p(t)u_{2k - 1}(t)\,dt \biggr) \\ &\quad {}+ \int _{0}^{t} \frac{1}{c(\tau )}F_{2k - 1}( \tau ; u, a, b)e^{ - \int _{\tau }^{t} \frac{\lambda _{k}^{2}}{c(s)}\,ds} \,d\tau \\ &\quad {}- \frac{e^{ - \int _{0}^{T} \frac{\lambda _{k}^{2}}{c(s)}\,ds}}{1 + \delta e^{ - \int _{0}^{T} \frac{\lambda _{k}^{2}}{c(s)}\,ds}} \int _{0} ^{T} \frac{1}{c(\tau )}F_{2k - 1}( \tau ; u, a, b)e^{ - \int _{\tau } ^{t} \frac{\lambda _{k}^{2}}{c(s)}\,ds} \,d\tau \biggr\} X_{2k - 1} \\ &\quad {} + \sum _{k = 1}^{\infty } \biggl\{ \frac{e^{ - \int _{0}^{t} \frac{\lambda _{k}^{2}}{c(s)}\,ds}}{1 + \delta e^{ - \int _{0}^{T} \frac{\lambda _{k} ^{2}}{c(s)}\,ds}} \biggl( \varphi _{2k} - \int _{0}^{T} p(t)u_{2k}(t)\,dt \biggr) \\ &\quad {}+ \int _{0}^{t} \frac{1}{c(\tau )}F_{2k}( \tau ; u, a, b)e^{ - \int _{\tau }^{t} \frac{\lambda _{k}^{2}}{c(s)}\,ds} \,d\tau \\ &\quad {}- \frac{\delta e^{ - \int _{0}^{T} \frac{\lambda _{k}^{2}}{c(s)}\,ds}}{1 + \delta e^{ - \int _{0}^{T} \frac{\lambda _{k}^{2}}{c(s)}\,ds}} \int _{0}^{T} \frac{1}{c( \tau )}F_{2k}( \tau ; u, a, b)e^{ - \int _{\tau }^{t} \frac{\lambda _{k} ^{2}}{c(s)}\,ds} \,d\tau + q_{k}(t,T) \biggr\} X_{2k}(x). \end{aligned}$$
(3.11)
Using (1.7) gives
$$\begin{aligned}& \begin{aligned}[b] a(t) &= \bigl[h(t)\bigr]^{ - 1}\Biggl\{ \bigl(c(t)h'_{1}(t) - f(x_{1}, t)\bigr)g(x_{2}, t) - \bigl(c(t)h'_{2}(t) - f(x_{2}, t)\bigr)g(x_{1}, t) \\ &\quad {}+ \sum_{k = 1}^{\infty } \lambda _{k}^{2}u_{2k - 1}(t) \bigl(g(x_{2},t)X_{2k - 1}(x_{1}) - g(x_{1},t)X_{2k - 1}(x_{2})\bigr) \\ &\quad {} + \sum_{k = 1}^{\infty } \lambda _{k}^{2}u_{2k}(t) \bigl(g(x_{2},t)X _{2k}(x_{1}) - g(x_{1},t)X_{2k}(x_{2}) \bigr) \Biggr\} , \end{aligned} \end{aligned}$$
(3.12)
$$\begin{aligned}& \begin{aligned}[b] b(t) &= \bigl[h(t)\bigr]^{ - 1}\Biggl\{ h_{1}(t) \bigl(c(t)h'_{2}(t) - f(x_{2}, t)\bigr) - h_{2}(t) \bigl(c(t)h'_{1}(t) - f(x_{1}, t)\bigr) \\ &\quad {}+ \sum_{k = 1}^{\infty } \lambda _{k}^{2}u_{2k - 1}(t) \bigl(h_{1}(t)X_{2k - 1}(x_{2}) - h_{2}(t)X_{2k - 1}(x_{1})\bigr) \\ &\quad {}+ \sum_{k = 1}^{\infty } \lambda _{k}^{2}u_{2k}(t) \bigl(h_{1}(t)X _{2k}(x_{2}) - h_{2}(t)X_{2k}(x_{1}) \bigr) \Biggr\} , \end{aligned} \end{aligned}$$
(3.13)
where
$$ h(t) \equiv h_{1}(t)g(x_{2},t) - h_{2}(t)g(x_{1},t) \ne 0\quad (0 \le t \le T). $$
We substitute relations (3.8) and (3.9) into (3.12) and (3.13), respectively, to obtain
$$\begin{aligned}& a(t) = \bigl[h(t)\bigr]^{ - 1}\Biggl\{ \bigl(c(t)h'_{1}(t) - f(x_{1}, t)\bigr)g(x_{2}, t) - \bigl(c(t)h'_{2}(t) - f(x_{2}, t)\bigr)g(x_{1}, t) \\& \hphantom{a(t) =} {}+ \sum_{k = 1}^{\infty } \lambda _{k}^{2} \biggl[ \frac{e^{ - \int _{0} ^{t} \frac{\lambda _{k}^{2}}{c(s)}\,ds}}{1 + \delta e^{ - \int _{0}^{T} \frac{ \lambda _{k}^{2}}{c(s)}\,ds}} \biggl( \varphi _{2k - 1} - \int _{0}^{T} p(t)u _{2k - 1}(t)\,dt \biggr) \\& \hphantom{a(t) =}{}+ \int _{0}^{t} \frac{1}{c(\tau )}F_{2k - 1}( \tau ; u, a, b)e^{ - \int _{\tau }^{t} \frac{\lambda _{k}^{2}}{c(s)}\,ds} \,d\tau \\& \hphantom{a(t) =} {}- \frac{\delta e^{ - \int _{0}^{T} \frac{\lambda _{k}^{2}}{c(s)}\,ds}}{1 + \delta e^{ - \int _{0}^{T} \frac{ \lambda _{k}^{2}}{c(s)}\,ds}} \int _{0}^{T} \frac{1}{c(\tau )}F_{2k - 1}( \tau ; u, a, b)e^{ - \int _{\tau }^{t} \frac{\lambda _{k}^{2}}{c(s)}\,ds} \,d\tau \biggr] \\& \hphantom{a(t) =}{}\times \bigl(g(x_{2},t)X_{2k - 1}(x_{1}) - g(x_{1},t)X_{2k - 1}(x _{2})\bigr) \\& \hphantom{a(t) =} {}+ \sum_{k = 1}^{\infty } \lambda _{k}^{2} \biggl[ \frac{e^{ - \int _{0} ^{t} \frac{\lambda _{k}^{2}}{c(s)}\,ds}}{1 + \delta e^{ - \int _{0}^{T} \frac{ \lambda _{k}^{2}}{c(s)}\,ds}} \biggl( \varphi _{2k} - \int _{0}^{T} p(t)u_{2k}(t)\,dt \biggr) \\& \hphantom{a(t) =}{}+ \int _{0}^{t} \frac{1}{c(\tau )}F_{2k}( \tau ; u, a, b)e^{ - \int _{\tau }^{t} \frac{\lambda _{k}^{2}}{c(s)}\,ds} \,d\tau \\& \hphantom{a(t) =} {}- \frac{\delta e^{ - \int _{0}^{T} \frac{\lambda _{k}^{2}}{c(s)}\,ds}}{1 + \delta e^{ - \int _{0}^{T} \frac{ \lambda _{k}^{2}}{c(s)}\,ds}} \int _{0}^{T} \frac{1}{c(\tau )}F_{2k}( \tau ; u, a, b)e^{ - \int _{\tau }^{t} \frac{\lambda _{k}^{2}}{c(s)}\,ds} \,d\tau + q_{k}(t,T) \biggr] \\& \hphantom{a(t) =} {}\times \bigl(g(x_{2},t)X_{2k - 1}(x_{1}) - g(x_{1},t)X_{2k - 1}(x _{2})\bigr) \Biggr\} , \end{aligned}$$
(3.14)
$$\begin{aligned}& b(t) = \bigl[h(t)\bigr]^{ - 1}\Biggl\{ h_{1}(t) \bigl(c(t)h'_{2}(t) - f(x_{2}, t)\bigr) - h_{2}(t) \bigl(c(t)h'_{1}(t) - f(x_{1}, t)\bigr) \\& \hphantom{b(t) =} {}+ \sum_{k = 1}^{\infty } \lambda _{k}^{2} \biggl[ \frac{e^{ - \int _{0} ^{t} \frac{\lambda _{k}^{2}}{c(s)}\,ds}}{1 + \delta e^{ - \int _{0}^{T} \frac{ \lambda _{k}^{2}}{c(s)}\,ds}} \biggl( \varphi _{2k - 1} - \int _{0}^{T} p(t)u _{2k - 1}(t)\,dt \biggr) \\& \hphantom{b(t) =}{}+ \int _{0}^{t} \frac{1}{c(\tau )}F_{2k - 1}( \tau ; u, a, b)e^{ - \int _{\tau }^{t} \frac{\lambda _{k}^{2}}{c(s)}\,ds} \,d\tau \\& \hphantom{b(t) =} {}- \frac{\delta e^{ - \int _{0}^{T} \frac{\lambda _{k}^{2}}{c(s)}\,ds}}{1 + \delta e^{ - \int _{0}^{T} \frac{ \lambda _{k}^{2}}{c(s)}\,ds}} \int _{0}^{T} \frac{1}{c(\tau )}F_{2k - 1}( \tau ; u, a, b)e^{ - \int _{\tau }^{t} \frac{\lambda _{k}^{2}}{c(s)}\,ds} \,d\tau \biggr] \\& \hphantom{b(t) =}\times{}\bigl(h_{1}(t)X_{2k - 1}(x_{2}) - h_{2}(t)X_{2k - 1}(x_{1})\bigr) \\& \hphantom{b(t) =} {}+ \sum_{k = 1}^{\infty } \lambda _{k}^{2} \biggl[ \frac{e^{ - \int _{0} ^{t} \frac{\lambda _{k}^{2}}{c(s)}\,ds}}{1 + \delta e^{ - \int _{0}^{T} \frac{ \lambda _{k}^{2}}{c(s)}\,ds}} \biggl( \varphi _{2k} - \int _{0}^{T} p(t)u_{2k}(t)\,dt \biggr) \\& \hphantom{b(t) =}{}+ \int _{0}^{t} \frac{1}{c(\tau )}F_{2k}( \tau ; u, a, b)e^{ - \int _{\tau }^{t} \frac{\lambda _{k}^{2}}{c(s)}\,ds} \,d\tau \\& \hphantom{b(t) =} {}- \frac{\delta e^{ - \int _{0}^{T} \frac{\lambda _{k}^{2}}{c(s)}\,ds}}{1 + \delta e^{ - \int _{0}^{T} \frac{ \lambda _{k}^{2}}{c(s)}\,ds}} \int _{0}^{T} \frac{1}{c(\tau )}F_{2k}( \tau ; u, a, b)e^{ - \int _{\tau }^{t} \frac{\lambda _{k}^{2}}{c(s)}\,ds} \,d\tau + q_{k}(t,T) \biggr] \\& \hphantom{b(t) =}{}\times \bigl(h_{1}(t)X_{2k}(x_{2}) - h_{2}(t)X_{2k}(x _{2})\bigr)\Biggr\} . \end{aligned}$$
(3.15)
Thus, the problem of finding the solutions (1.1)–(1.3), (1.6), and (1.7) reduces to finding a solution of system (3.11), (3.14), and (3.15) with unknown functions \(u(x,t)\), \(a(t)\), and \(b(t)\), respectively.
Lemma 3.1
If
\(\{ u(x,t), a(t), b(t)\}\)
is a classical solution of problem (1.1)–(1.3), (1.6), and (1.7), then the functions
$$\begin{aligned}& u_{0}(t) = \int _{0}^{1} u(x,t)Y_{0}(x)\,dx, \\& u_{2k - 1}(t) = \int _{0}^{1} u(x,t)Y_{2k - 1}(x)\,dx\quad (k = 1,2, \ldots), \\& u_{2k}(t) = \int _{0}^{1} u(x,t)Y_{2k}(x)\,dx\quad (k = 1,2, \ldots) \end{aligned}$$
satisfy system (3.7), (3.8), and (3.9) on the interval
\([0,T]\).
Corollary 3.2
Suppose that system (3.11), (3.14) and (3.15) has a unique solution. Then the problem (1.1)–(1.3), (1.6), and (1.7), cannot have more than one solution; in other words, if problem (1.1)–(1.3), (1.6), and (1.7) has a solution, then it is unique.
Now, consider the operator
$$ \varPhi (u,a,b) = \bigl\{ \varPhi _{1}(u,a,b), \varPhi _{2}(u,a,b), \varPhi _{3}(u,a,b)\bigr\} $$
in the space \(E_{T}^{3}\), where
$$\begin{aligned}& \varPhi _{1}(u,a,b) = \tilde{u}(x,t) = \sum_{k = 0}^{\infty } \tilde{u} _{k}(t)X_{k}(x), \\& \varPhi _{2}(u,a,b) = \tilde{a}(t), \qquad \varPhi _{3}(u,a,b) = \tilde{b}(t), \end{aligned}$$
and the functions \(\tilde{u}_{0}(t)\), \(\tilde{u}_{2k - 1}(t)\), \(\tilde{u}_{2k}(t)\) (\(k = 1, 2,\ldots\)), \(\tilde{a}(t)\), and \(\tilde{b}(t)\) are equal to the right-hand sides of (3.7), (3.8), (3.9), (3.14), and (3.15), respectively.
Assume that the data for problem (1.1)–(1.3) and (1.7) satisfy the following conditions:
-
(A)
\(\varphi (x) \in C^{2}[0,1]\), \(\varphi '''(x) \in L_{2}(0,1)\), \(\varphi (0) = \beta \varphi (1)\), \(\varphi '(0) = \varphi '(1)\), \(\varphi ''(0) = \beta \varphi ''(1)\) (\(\beta \ne \pm 1\));
-
(B)
\(f(x,t) \in C_{x,t}^{2,0}(D_{T})\), \(f_{xxx}(x,t) \in L_{2}(D _{T})\), \(f(0,t) = \beta f(1,t)\), \(f_{x}(0,t) = f_{x}(1,t)\),
$$ f_{xx}(0,t) = \beta f_{xx}(1,t)\quad (\beta \ne \pm 1), (0 \le t \le T); $$
-
(C)
\(g(x,t) \in C_{x,t}^{2,0}(D_{T})\), \(g_{xxx}(x,t) \in L_{2}(D _{T})\), \(g(0,t) = \beta g(1,t)\), \(g_{x}(0,t) = g_{x}(1,t)\),
$$ g_{xx}(0,t) = \beta g_{xx}(1,t) \quad (\beta \ne \pm 1), (0 \le t \le T); $$
-
(D)
\(\delta \ge 0\), \(c(t) \in C[0,T]\), \(c(t) > 0\) (\(0 \le t \le T\)), \(0 \le p(t) \in C[0,T]\), \(h_{i}(t) \in C^{1}[0,T]\) (\(i = 1,2\)), \(h(t) \equiv h_{1}(t)g(x_{2},t) - h_{2}(t)g(x_{1},t) \ne 0\).
From (3.10), it is easy to see that
$$\begin{aligned}& \Biggl( \sum_{k = 1}^{\infty } \bigl(\lambda _{k}^{3} \bigl\Vert q_{k}(t,T) \bigr\Vert _{C[0,T]}\bigr)^{2} \Biggr)^{\frac{1}{2}} \\& \quad \le 4\sqrt{3} \vert a \vert \biggl\Vert \frac{1}{c(t)} \biggr\Vert _{C[0,T]}(1 + \delta ) \Biggl[ \Biggl( \sum_{k = 1}^{\infty } \bigl(\lambda _{k}^{3} \Vert \varphi _{2k - 1} \Vert _{C[0,T]}\bigr)^{2} \Biggr)^{\frac{1}{2}} \\& \qquad {}+ T \bigl\Vert p(t) \bigr\Vert _{C[0,T]} \Biggl( \sum _{k = 1}^{ \infty } \bigl(\lambda _{k}^{3} \bigl\Vert u_{2k - 1}(t) \bigr\Vert _{C[0,T]}\bigr)^{2} \Biggr)^{\frac{1}{2}} \Biggr] \\& \qquad {}+ 6\sqrt{2} \vert a \vert \biggl\Vert \frac{1}{c(t)} \biggr\Vert _{C[0,T]} ^{2}\sqrt{ \bigl\Vert c(t) \bigr\Vert _{C[0,T]}} T(1 + \delta )^{2} \Biggl[ \sqrt{T} \Biggl( \int _{0}^{T} \sum_{k = 1}^{\infty } \bigl(\lambda _{k} ^{3} \bigl\vert f_{2k - 1}(\tau ) \bigr\vert \bigr)^{2} \,d\tau \Biggr)^{ \frac{1}{2}} \\& \qquad {}+ T \bigl\Vert a(t) \bigr\Vert _{C[0,T]} \Biggl( \sum _{k = 1}^{ \infty } \bigl(\lambda _{k}^{3} \bigl\Vert u_{2k - 1}(t) \bigr\Vert _{C[0,T]}\bigr)^{2} \Biggr)^{\frac{1}{2}}\\& \qquad {} + \sqrt{T} \bigl\Vert b(t) \bigr\Vert _{C[0,T]} \Biggl( \int _{0}^{T} \sum_{k = 1}^{\infty } \bigl(\lambda _{k}^{3} \bigl\vert g _{2k - 1}(\tau ) \bigr\vert \bigr)^{2} \,d\tau \Biggr)^{\frac{1}{2}} \Biggr]. \end{aligned}$$
Hence, taking into account inequalities (2.9)–(2.12), we have
$$\begin{aligned}& \Biggl( \sum_{k = 1}^{\infty } \bigl(\lambda _{k}^{3} \bigl\Vert q_{k}(t,T) \bigr\Vert _{C[0,T]}\bigr)^{2} \Biggr)^{\frac{1}{2}} \\& \quad \le 4\sqrt{3} \vert a \vert \biggl\Vert \frac{1}{c(t)} \biggr\Vert _{C[0,T]}(1 + \delta )\Biggl[2\sqrt{2} \bigl\Vert \varphi '''(x) \bigr\Vert _{L_{2}(0,1)} \\& \qquad {}+ T \bigl\Vert p(t) \bigr\Vert _{C[0,T]} \Biggl( \sum _{k = 1}^{ \infty } \bigl(\lambda _{k}^{3} \bigl\Vert u_{2k - 1}(t) \bigr\Vert _{C[0,T]}\bigr)^{2} \Biggr)^{\frac{1}{2}} \Biggr] \\& \qquad {}+ 6\sqrt{2} \vert a \vert \biggl\Vert \frac{1}{c(t)} \biggr\Vert _{C[0,T]} ^{2}\sqrt{ \bigl\Vert c(t) \bigr\Vert _{C[0,T]}} T(1 + \delta )^{2}\Biggl[2 \sqrt{2T} \bigl\Vert f_{xxx}(x,t) \bigr\Vert _{L_{2}(D_{T})} \\& \qquad {} + T \bigl\Vert a(t) \bigr\Vert _{C[0,T]} \Biggl( \sum _{k = 1}^{ \infty } \bigl(\lambda _{k}^{3} \bigl\Vert u_{2k - 1}(t) \bigr\Vert _{C[0,T]}\bigr)^{2} \Biggr)^{\frac{1}{2}} \\& \qquad {}+ 2\sqrt{2T} \bigl\Vert b(t) \bigr\Vert _{C[0,T]} \bigl\Vert g_{xxx}(x,t) \bigr\Vert _{L_{2}(D_{T})} \Biggr]. \end{aligned}$$
(3.16)
Now from (3.7)–(3.9), respectively, we have
$$\begin{aligned}& \begin{aligned}[b] &\bigl\Vert \tilde{u}_{0}(t) \bigr\Vert _{C[0,T]} \\ &\quad \le (1 + \delta )^{ - 1}\bigl( \vert \varphi _{0} \vert + T \bigl\Vert p(t) \bigr\Vert _{C[0,T]} \bigl\Vert u_{0}(t) \bigr\Vert _{C[0,T]}\bigr) \\ &\qquad {}+ \bigl(1 + \delta (1 + \delta )^{ - 1}\bigr) \biggl\Vert \frac{1}{c(t)} \biggr\Vert _{C[0,T]} \biggl[ \sqrt{T} \biggl( \int _{0}^{T} \bigl\vert f_{0}(t) \bigr\vert ^{2}\,dt \biggr)^{\frac{1}{2}} \\ &\qquad {}+ T \bigl\Vert a(t) \bigr\Vert _{C[0,T]} \bigl\Vert u_{0}(t) \bigr\Vert _{C[0,T]}+ \sqrt{T} \bigl\Vert b(t) \bigr\Vert _{C[0,T]} \biggl( \int _{0}^{T} \bigl\vert g_{0}(t) \bigr\vert ^{2}\,dt \biggr) ^{\frac{1}{2}} \biggr],\end{aligned} \end{aligned}$$
(3.17)
$$\begin{aligned}& \begin{aligned}[b] &\Biggl( \sum_{k = 1}^{\infty } \bigl(\lambda _{k}^{3} \bigl\Vert \tilde{u}_{2k - 1}(t) \bigr\Vert _{C[0,T]}\bigr)^{2} \Biggr)^{\frac{1}{2}} \\ &\quad \le \sqrt{5} \Biggl[ \Biggl( \sum_{k = 1}^{\infty } \bigl(\lambda _{k}^{3} \Vert \varphi _{2k - 1} \Vert _{C[0,T]}\bigr)^{2} \Biggr)^{\frac{1}{2}} + T \bigl\Vert p(t) \bigr\Vert _{C[0,T]} \Biggl( \sum_{k = 1}^{\infty } \bigl(\lambda _{k} ^{3} \bigl\Vert u_{2k - 1}(t) \bigr\Vert _{C[0,T]}\bigr)^{2} \Biggr)^{ \frac{1}{2}} \Biggr] \\ &\qquad {}+ \sqrt{5} (1 + \delta )) \biggl\Vert \frac{1}{c(t)} \biggr\Vert _{C[0,T]} \Biggl[ \sqrt{T} \Biggl( \int _{0}^{T} \sum_{k = 1}^{\infty } \bigl( \lambda _{k}^{3} \bigl\vert f_{2k - 1}(\tau ) \bigr\vert \bigr)^{2} \,d\tau \Biggr) ^{\frac{1}{2}} \\ &\qquad {}+ T \bigl\Vert a(t) \bigr\Vert _{C[0,T]} \Biggl( \sum _{k = 1} ^{\infty } \bigl(\lambda _{k}^{3} \bigl\Vert u_{2k - 1}(t) \bigr\Vert _{C[0,T]}\bigr)^{2} \Biggr)^{\frac{1}{2}} \\ &\qquad {}+ T \bigl\Vert b(t) \bigr\Vert _{C[0,T]} \Biggl( \int _{0}^{T} \sum_{k = 1}^{\infty } \bigl(\lambda _{k}^{3} \bigl\Vert g_{2k - 1}(\tau ) \bigr\Vert _{C[0,T]}\bigr)^{2} \,d\tau \Biggr)^{\frac{1}{2}} \Biggr], \end{aligned} \end{aligned}$$
(3.18)
$$\begin{aligned}& \begin{aligned}[b] &\Biggl( \sum_{k = 1}^{\infty } \bigl(\lambda _{k}^{3} \bigl\Vert \tilde{u}_{2k}(t) \bigr\Vert _{C[0,T]}\bigr)^{2} \Biggr)^{\frac{1}{2}}\\ &\quad \le \sqrt{6} \Biggl[ \Biggl( \sum_{k = 1}^{\infty } \bigl(\lambda _{k} ^{3} \Vert \varphi _{2k} \Vert _{C[0,T]}\bigr)^{2} \Biggr)^{\frac{1}{2}}+ T \bigl\Vert p(t) \bigr\Vert _{C[0,T]} \Biggl( \sum_{k = 1}^{\infty } \bigl( \lambda _{k}^{3} \bigl\Vert u_{2k}(t) \bigr\Vert _{C[0,T]}\bigr)^{2} \Biggr) ^{\frac{1}{2}} \Biggr] \\ &\qquad {}+ \sqrt{6} (1 + \delta ) \biggl\Vert \frac{1}{c(t)} \biggr\Vert _{C[0,T]} \Biggl[ \sqrt{T} \Biggl( \int _{0}^{T} \sum_{k = 1}^{\infty } \bigl( \lambda _{k}^{3} \bigl\vert f_{2k}(\tau ) \bigr\vert \bigr)^{2} \,d\tau \Biggr)^{ \frac{1}{2}}\\ &\qquad {} + T \bigl\Vert a(t) \bigr\Vert _{C[0,T]} \Biggl( \sum_{k = 1} ^{\infty } \bigl(\lambda _{k}^{3} \bigl\Vert u_{2k}(t) \bigr\Vert _{C[0,T]}\bigr)^{2} \Biggr)^{\frac{1}{2}} \\ &\qquad {}+ \sqrt{T} \bigl\Vert b(t) \bigr\Vert _{C[0,T]} \Biggl( \int _{0}^{T} \sum_{k = 1}^{\infty } \bigl(\lambda _{k}^{3} \bigl\vert g_{2k}( \tau ) \bigr\vert \bigr)^{2} \,d\tau \Biggr)^{\frac{1}{2}} \Biggr]\\ &\qquad {} + \sqrt{6} \Biggl( \sum_{k = 1}^{\infty } \bigl(\lambda _{k}^{3} \bigl\Vert q _{k}(t,T) \bigr\Vert _{C[0,T]}\bigr)^{2} \Biggr)^{\frac{1}{2}}. \end{aligned} \end{aligned}$$
(3.19)
Taking into consideration (3.16), from (3.17)–(3.19), by virtue of (2.9)–(2.12), we have the following estimates:
$$\begin{aligned}& \begin{aligned}[b] \bigl\Vert \tilde{u}_{0}(t) \bigr\Vert _{C[0,T]} &\le A_{1}(T) + B_{1}(T) \bigl\Vert a(t) \bigr\Vert _{C[0,T]} \bigl\Vert u(x,t) \bigr\Vert _{B_{2,T} ^{3}}\\ &\quad {} + C_{1}(T) \bigl\Vert u(x,t) \bigr\Vert _{B_{2,T}^{3}} + D_{1}(T) \bigl\Vert b(t) \bigr\Vert _{C[0,T]},\end{aligned} \end{aligned}$$
(3.20)
$$\begin{aligned}& \begin{aligned}[b] \Biggl( \sum_{k = 1}^{\infty } \bigl(\lambda _{k}^{3} \bigl\Vert \tilde{u}_{2k - 1}(t) \bigr\Vert _{C[0,T]}\bigr)^{2} \Biggr)^{\frac{1}{2}} &\le A_{2}(T) + B_{2}(T) \bigl\Vert a(t) \bigr\Vert _{C[0,T]} \bigl\Vert u(x,t) \bigr\Vert _{B _{2,T}^{3}} \\ &\quad {}+ C_{2}(T) \bigl\Vert u(x,t) \bigr\Vert _{B_{2,T}^{3}} + D_{2}(T) \bigl\Vert b(t) \bigr\Vert _{C[0,T]},\end{aligned} \end{aligned}$$
(3.21)
$$\begin{aligned}& \begin{aligned}[b] \Biggl( \sum_{k = 1}^{\infty } \bigl(\lambda _{k}^{3} \bigl\Vert \tilde{u}_{2k}(t) \bigr\Vert _{C[0,T]}\bigr)^{2} \Biggr)^{\frac{1}{2}} &\le A_{3}(T) + B_{3}(T) \bigl\Vert a(t) \bigr\Vert _{C[0,T]} \bigl\Vert u(x,t) \bigr\Vert _{B_{2,T}^{3}} \\ &\quad {}+ C_{3}(T) \bigl\Vert u(x,t) \bigr\Vert _{B_{2,T}^{3}} + D_{3}(T) \bigl\Vert b(t) \bigr\Vert _{C[0,T]}, \end{aligned} \end{aligned}$$
(3.22)
where
$$\begin{aligned}& A_{1}(T) = 2(1 + \delta )^{ - 1} \bigl\Vert \varphi (x) \bigr\Vert _{L_{2}(0,1)} + 2\sqrt{T} \bigl(1 + \delta (1 + \delta )^{ - 1} \bigr) \biggl\Vert \frac{1}{c(t)} \biggr\Vert _{C[0,T]} \bigl\Vert f(x,t) \bigr\Vert _{L_{2}(D _{T})}, \\& B_{1}(T) = \bigl(1 + \delta (1 + \delta )^{ - 1}\bigr)T \biggl\Vert \frac{1}{c(t)} \biggr\Vert _{C[0,T]}, \\& C_{1}(T) = (1 + \delta )^{ - 1}T \bigl\Vert p(t) \bigr\Vert _{C[0,T]}, \\& D_{1}(T) = \bigl(1 + \delta (1 + \delta )^{ - 1}\bigr) \biggl\Vert \frac{1}{c(t)} \biggr\Vert _{C[0,T]}\sqrt{T} \bigl\Vert b(t) \bigr\Vert _{C[0,T]} \bigl\Vert g(x,t) \bigr\Vert _{L_{2}(D_{T})}, \\& A_{2}(T) = 2\sqrt{10} \bigl\Vert \varphi '''(x) \bigr\Vert _{L_{2}(0,1)} + 2\sqrt{10T} (1 + \delta ) \biggl\Vert \frac{1}{c(t)} \biggr\Vert _{C[0,T]} \bigl\Vert f_{xxx}(x,t) \bigr\Vert _{L_{2}(D_{T})}, \\& B_{2}(T) = \sqrt{6} T(1 + \delta )) \biggl\Vert \frac{1}{c(t)} \biggr\Vert _{C[0,T]}, \\& C_{2}(T) = \sqrt{6} T \bigl\Vert p(t) \bigr\Vert _{C[0,T]}, \\& D_{2}(T) = 4\sqrt{3T} (1 + \delta ) \biggl\Vert \frac{1}{c(t)} \biggr\Vert _{C[0,T]} \bigl\Vert g_{xxx}(x,t) \bigr\Vert _{L_{2}(D_{T})}, \\& \begin{aligned} A_{3}(T) &= 4\sqrt{3} \bigl\Vert \varphi '''(x) (1 - b - ax) - 3a\varphi ''(x) \bigr\Vert _{L_{2}(0,1)} \\ &\quad {}+ 4\sqrt{3T} (1 + \delta ) \biggl\Vert \frac{1}{c(t)} \biggr\Vert _{C[0,T]} \bigl\Vert f_{xxx}(x,t) (1 - b - ax) - 3af_{xx}(x,t) \bigr\Vert _{L_{2}(D_{T})} \\ &\quad {}+ 12\sqrt{2} \vert a \vert (1 + \delta ) \biggl\Vert \frac{1}{c(t)} \biggr\Vert _{C[0,T]} \bigl\Vert \varphi '''(x) \bigr\Vert _{L _{2}(0,1)} \\ &\quad {}+ 12\sqrt{3} \vert a \vert \biggl\Vert \frac{1}{c(t)} \biggr\Vert _{C[0,T]}^{2}\sqrt{ \bigl\Vert c(t) \bigr\Vert _{C[0,T]}} T\sqrt{T} (1 + \delta )^{2} \bigl\Vert f_{xxx}(x,t) \bigr\Vert _{L_{2}(D_{T})}, \end{aligned} \\& B_{3}(T) = \sqrt{6} T(1 + \delta ) \biggl\Vert \frac{1}{c(t)} \biggr\Vert _{C[0,T]} + 6\sqrt{2} \vert a \vert \biggl\Vert \frac{1}{c(t)} \biggr\Vert _{C[0,T]}^{2}\sqrt{ \bigl\Vert c(t) \bigr\Vert _{C[0,T]}} T^{2}(1 + \delta )^{2}, \\& C_{3}(T) = \sqrt{3} T \bigl\Vert p(t) \bigr\Vert _{C[0,T]} \biggl[ \sqrt{2} + 4 \vert a \vert \biggl\Vert \frac{1}{c(t)} \biggr\Vert _{C[0,T]}(1 + \delta ) \biggr], \\& \begin{aligned} D_{3}(T) &= 6\sqrt{2T} (1 + \delta ) \biggl\Vert \frac{1}{c(t)} \biggr\Vert _{C[0,T]}\biggl( \bigl\Vert g_{xxx}(x,t) (1 - b - ax) - 3ag_{xx}(x,t) \bigr\Vert _{L_{2}(D_{T})} \\ &\quad {}+ \vert a \vert \biggl\Vert \frac{1}{c(t)} \biggr\Vert _{C[0,T]}\sqrt{ \bigl\Vert c(t) \bigr\Vert _{C[0,T]}} T(1 + \delta ) \bigl\Vert g_{xxx}(x,t) \bigr\Vert _{L_{2}(D_{T})} \biggr). \end{aligned} \end{aligned}$$
We conclude from (3.20)–(3.22) that
$$ \begin{aligned}[b] \bigl\Vert \tilde{u}(x,t) \bigr\Vert _{B_{2,T}^{3}} &\le A_{4}(T) + B_{4}(T) \bigl\Vert a(t) \bigr\Vert _{C[0,T]} \bigl\Vert u(x,t) \bigr\Vert _{B_{2,T} ^{3}} \\ &\quad {}+ C_{4}(T) \bigl\Vert u(x,t) \bigr\Vert _{B_{2,T}^{3}} + D_{4}(T) \bigl\Vert b(t) \bigr\Vert _{C[0,T]}, \end{aligned} $$
(3.23)
where
$$\begin{aligned}& A_{4}(T) = A_{1}(T) + A_{2}(T) + A_{3}(T), B_{4}(T) = B_{1}(T) + B _{2}(T) + B_{3}(T), \\& C_{4}(T) = C_{1}(T) + C_{2}(T) + C_{3}(T), D_{4}(T) = D_{1}(T) + D _{2}(T) + D_{3}(T). \end{aligned}$$
Now from (3.14) and (3.15) we have
$$\begin{aligned}& \bigl\Vert \tilde{a}(t) \bigr\Vert _{C[0,T]} \\& \quad \le \bigl\Vert \bigl[h(t)\bigr]^{ - 1} \bigr\Vert _{C[0,T]}\Biggl\{ \bigl\Vert \bigl(c(t)h'_{1}(t) - f(x_{1}, t) \bigr)g(x_{2}, t) - \bigl(c(t)h'_{2}(t) - f(x_{2}, t)\bigr)g(x_{1}, t)) \bigr\Vert _{C[0,T]} \\& \qquad {}+ \Biggl( \sum_{k = 1}^{\infty } \lambda _{k}^{ - 2} \Biggr)^{ \frac{1}{2}} \Biggl[ \Biggl( \sum _{k = 1}^{\infty } \bigl( \lambda _{k}^{3} \Vert \varphi _{2k - 1} \Vert _{C[0,T]} \bigr)^{2} \Biggr) ^{\frac{1}{2}} \\& \qquad {} + T \bigl\Vert p(t) \bigr\Vert _{C[0,T]} \Biggl( \sum_{k = 1} ^{\infty } \bigl( \lambda _{k}^{3} \bigl\Vert u_{2k - 1}(t) \bigr\Vert _{C[0,T]} \bigr)^{2} \Biggr)^{\frac{1}{2}} \\& \qquad {}+ (1 + \delta ) \biggl\Vert \frac{1}{c(t)} \biggr\Vert _{C[0,T]} \Biggl[ \sqrt{T} \Biggl( \int _{0}^{T} \sum_{k = 1}^{\infty } \bigl( \lambda _{k}^{3} \bigl\vert f_{2k - 1}(\tau ) \bigr\vert \bigr) ^{2}\,d\tau \Biggr)^{\frac{1}{2}} \\& \qquad {}+ T \bigl\Vert a(t) \bigr\Vert _{C[0,T]} \Biggl( \sum _{k = 1}^{\infty } \bigl( \lambda _{k}^{3} \bigl\Vert u_{2k - 1}(t) \bigr\Vert _{C[0, T]} \bigr)^{2} \Biggr)^{\frac{1}{2}} \\& \qquad {}+ \sqrt{T} \bigl\Vert b(t) \bigr\Vert _{C[0,T]} \Biggl( \int _{0} ^{T} \sum_{k = 1}^{\infty } \bigl( \lambda _{k}^{3} \bigl\vert g_{2k - 1}( \tau ) \bigr\vert \bigr)^{2}\,d\tau \Biggr)^{\frac{1}{2}} \Biggr] \\& \qquad {}+ \Biggl( \sum_{k = 1}^{\infty } \bigl( \lambda _{k}^{3} \Vert \varphi _{2k} \Vert _{C[0,T]} \bigr)^{2} \Biggr)^{\frac{1}{2}}+ T \bigl\Vert p(t) \bigr\Vert _{C[0,T]} \Biggl( \sum _{k = 1}^{\infty } \bigl( \lambda _{k}^{3} \bigl\Vert u_{2k}(t) \bigr\Vert _{C[0,T]} \bigr) ^{2} \Biggr)^{\frac{1}{2}} \\& \qquad {}+ (1 + \delta ) \biggl\Vert \frac{1}{c(t)} \biggr\Vert _{C[0,T]} \Biggl[ \sqrt{T} \Biggl( \int _{0}^{T} \sum_{k = 1}^{\infty } \bigl( \lambda _{k}^{3} \bigl\vert f_{2k}(\tau ) \bigr\vert \bigr)^{2}\,d \tau \Biggr)^{\frac{1}{2}} \\& \qquad {}+ T \bigl\Vert a(t) \bigr\Vert _{C[0,T]} \Biggl( \sum _{k = 1}^{\infty } \bigl( \lambda _{k}^{3} \bigl\Vert u_{2k}(t) \bigr\Vert _{C[0, T]} \bigr)^{2} \Biggr)^{\frac{1}{2}} \\& \qquad {}+ \sqrt{T} \bigl\Vert b(t) \bigr\Vert _{C[0,T]} \Biggl( \int _{0}^{T} \sum_{k = 1}^{\infty } \bigl( \lambda _{k}^{3} \bigl\vert g _{2k}(\tau ) \bigr\vert \bigr)^{2}\,d\tau \Biggr)^{\frac{1}{2}} \Biggr] \\& \qquad {} + \Biggl( \sum_{k = 1}^{\infty } \bigl(\lambda _{k}^{3} \bigl\Vert q_{k}(t,T) \bigr\Vert _{C[0,T]}\bigr)^{2} \Biggr)^{\frac{1}{2}} \Biggr] \\& \qquad {}\times \bigl\Vert \bigl\vert g(x_{2},t) \bigr\vert + \bigl\vert g(x_{1},t) \bigr\vert \bigr\Vert _{C[0,T]} \Vert ax + b \Vert _{C[0,1]}\Biggr\} , \end{aligned}$$
(3.24)
$$\begin{aligned}& \bigl\Vert \tilde{b}(t) \bigr\Vert _{C[0,T]} \\& \quad \le \bigl\Vert \bigl[h(t)\bigr]^{ - 1} \bigr\Vert _{C[0,T]}\Biggl\{ \bigl\Vert h_{1}(t) \bigl(c(t)h'_{2}(t) - f(x_{2}, t)\bigr) - h_{2}(t) \bigl(c(t)h'_{1}(t) - f(x_{1}, t)\bigr) \bigr\Vert _{C[0,T]} \\& \qquad {}+ \Biggl( \sum_{k = 1}^{\infty } \lambda _{k}^{ - 2} \Biggr)^{ \frac{1}{2}} \Biggl( \sum _{k = 1}^{\infty } \bigl( \lambda _{k}^{3} \Vert \varphi _{2k - 1} \Vert _{C[0,T]} \bigr)^{2} \Biggr) ^{\frac{1}{2}} \\& \qquad {}\times T \bigl\Vert p(t) \bigr\Vert _{C[0,T]} \Biggl( \sum_{k = 1}^{\infty } \bigl( \lambda _{k}^{3} \bigl\Vert u_{2k - 1}(t) \bigr\Vert _{C[0,T]} \bigr)^{2} \Biggr)^{\frac{1}{2}} \\& \qquad {}+ (1 + \delta ) \biggl\Vert \frac{1}{c(t)} \biggr\Vert _{C[0,T]} \Biggl[ \sqrt{T} \Biggl( \int _{0}^{T} \sum_{k = 1}^{\infty } \bigl( \lambda _{k}^{3} \bigl\vert f_{2k - 1}(\tau ) \bigr\vert \bigr)^{2}\,d\tau \Biggr) ^{\frac{1}{2}} \\& \qquad {}+ T \bigl\Vert a(t) \bigr\Vert _{C[0,T]} \Biggl( \sum_{k = 1} ^{\infty } \bigl( \lambda _{k}^{3} \bigl\Vert u_{2k - 1}(t) \bigr\Vert _{C[0, T]} \bigr)^{2} \Biggr)^{\frac{1}{2}} \\& \qquad {}+ T \bigl\Vert b(t) \bigr\Vert _{C[0,T]} \Biggl( \int _{0}^{T} \sum_{k = 1}^{\infty } \bigl( \lambda _{k}^{3} \bigl\vert g_{2k - 1}( \tau ) \bigr\vert \bigr)^{2}\,d\tau \Biggr)^{\frac{1}{2}} \Biggr] \\& \qquad {}+ \Biggl( \sum_{k = 1}^{\infty } \bigl( \lambda _{k}^{3} \Vert \varphi _{2k} \Vert _{C[0,T]} \bigr)^{2} \Biggr)^{\frac{1}{2}}+ T \bigl\Vert p(t) \bigr\Vert _{C[0,T]} \Biggl( \sum _{k = 1}^{\infty } \bigl( \lambda _{k}^{3} \bigl\Vert u_{2k}(t) \bigr\Vert _{C[0,T]} \bigr) ^{2} \Biggr)^{\frac{1}{2}} \\& \qquad {}+ (1 + \delta ) \biggl\Vert \frac{1}{c(t)} \biggr\Vert _{C[0,T]} \Biggl[ \sqrt{T} \Biggl( \int _{0}^{T} \sum_{k = 1}^{\infty } \bigl( \lambda _{k}^{3} \bigl\vert f_{2k}(\tau ) \bigr\vert \bigr)^{2}\,d \tau \Biggr)^{\frac{1}{2}} \\& \qquad {}+ T \bigl\Vert a(t) \bigr\Vert _{C[0,T]} \Biggl( \sum _{k = 1}^{\infty } \bigl( \lambda _{k}^{3} \bigl\Vert u_{2k}(t) \bigr\Vert _{C[0, T]} \bigr)^{2} \Biggr)^{\frac{1}{2}} \\& \qquad {}+ T \bigl\Vert b(t) \bigr\Vert _{C[0,T]} \Biggl( \int _{0}^{T} \sum_{k = 1}^{\infty } \bigl( \lambda _{k}^{3} \bigl\vert g_{2k}(\tau ) \bigr\vert \bigr)^{2}\,d\tau \Biggr)^{\frac{1}{2}} \Biggr] \\& \qquad {}+ \Biggl( \sum_{k = 1}^{\infty } \bigl(\lambda _{k}^{3} \bigl\Vert q_{k}(t,T) \bigr\Vert _{C[0,T]}\bigr)^{2} \Biggr)^{\frac{1}{2}} \bigl\Vert \bigl\vert h_{1}(t) \bigr\vert + \bigl\vert h_{2}(t) \bigr\vert \bigr\Vert _{C[0,T]} \Vert ax + b \Vert _{C[0,1]}\Biggr\} . \end{aligned}$$
(3.25)
Using (3.16), by virtue of (2.9)–(2.12), from the relations (3.24) and (3.25), we obtain
$$\begin{aligned}& \begin{aligned}[b] \bigl\Vert \tilde{a}(t) \bigr\Vert _{C[0,T]} &\le A_{5}(T) + B_{5}(T) \bigl\Vert a(t) \bigr\Vert _{C[0,T]} \bigl\Vert u(x,t) \bigr\Vert _{B_{2,T}^{3}} \\ &\quad {}+ C _{5}(T) \bigl\Vert u(x,t) \bigr\Vert _{B_{2,T}^{3}} + D_{5}(T) \bigl\Vert b(t) \bigr\Vert _{C[0,T]},\end{aligned} \end{aligned}$$
(3.26)
$$\begin{aligned}& \begin{aligned}[b] \bigl\Vert \tilde{b}(t) \bigr\Vert _{C[0,T]} &\le A_{6}(T) + B_{6}(T) \bigl\Vert a(t) \bigr\Vert _{C[0,T]} \bigl\Vert u(x,t) \bigr\Vert _{B_{2,T}^{3}} \\ &\quad {}+ C _{6}(T) \bigl\Vert u(x,t) \bigr\Vert _{B_{2,T}^{3}} + D_{6}(T) \bigl\Vert b(t) \bigr\Vert _{C[0,T]}, \end{aligned} \end{aligned}$$
(3.27)
where
$$\begin{aligned}& A_{5}(T) = \bigl\Vert \bigl[h(t)\bigr]^{ - 1} \bigr\Vert _{C[0,T]}\Biggl\{ \bigl\Vert h_{1}(t) \bigl(c(t)h'_{2}(t) - f(x_{2}, t)\bigr) - h_{2}(t) \bigl(c(t)h'_{1}(t) - f(x _{1}, t)\bigr) \bigr\Vert _{C[0,T]} \\& \hphantom{A_{5}(T) =}{}+ \Biggl( \sum_{k = 1}^{\infty } \lambda _{k}^{ - 2} \Biggr)^{ \frac{1}{2}} \biggl[ 2\sqrt{2} \biggl[ \biggl( 1 + 4\sqrt{3} \vert a \vert \biggl\Vert \frac{1}{c(t)} \biggr\Vert _{C[0,T]}(1 + \delta ) \biggr) \bigl\Vert \varphi '''(x) \bigr\Vert _{L_{2}(0,1)} \\& \hphantom{A_{5}(T) =}{}+ \bigl\Vert \varphi '''(x) (1 - b - ax) - 3a\varphi ''(x) \bigr\Vert _{L_{2}(0,1)}\biggr] + 2 \sqrt{2T} (1 + \delta ) \biggl\Vert \frac{1}{c(t)} \biggr\Vert _{C[0,T]} \\& \hphantom{A_{5}(T) =}{}\times \biggl[ \biggl( 1 + 6\sqrt{2} \vert a \vert \biggl\Vert \frac{1}{c(t)} \biggr\Vert _{C[0,T]} \sqrt{ \bigl\Vert c(t) \bigr\Vert _{C[0,T]}} T(1 + \delta ) \biggr) \bigl\Vert f_{xxx}(x,t) \bigr\Vert _{L_{2}(D_{T})} \\& \hphantom{A_{5}(T) =}{}+ \bigl\Vert f_{xxx}(x,t) (1 - b - ax) - 3af_{xx}(x,t) \bigr\Vert _{L_{2}(D _{T})}\biggr]\biggr]\\& \hphantom{A_{5}(T) =}{}\times \bigl\Vert \bigl\vert g(x_{2},t) \bigr\vert + \bigl\vert g(x_{1},t) \bigr\vert \bigr\Vert _{C[0,T]} \Vert ax + b \Vert _{C[0,1]}\Biggr\} , \\& B_{5}(T) = 2T \bigl\Vert \bigl[h(t)\bigr]^{ - 1} \bigr\Vert _{C[0,T]} \Biggl( \sum_{k = 1}^{\infty } \lambda _{k}^{ - 2} \Biggr)^{\frac{1}{2}}(1 + \delta ) \biggl\Vert \frac{1}{c(t)} \biggr\Vert _{C[0,T]}\\& \hphantom{B_{5}(T) =}{}\times \biggl( 1 + 3 \sqrt{2} \vert a \vert \biggl\Vert \frac{1}{c(t)} \biggr\Vert _{C[0,T]} \sqrt{ \bigl\Vert c(t) \bigr\Vert _{C[0,T]}} T(1 + \delta ) \biggr) \\& \hphantom{B_{5}(T) =}{}\times \bigl\Vert \bigl\vert g(x_{2},t) \bigr\vert + \bigl\vert g(x_{1},t) \bigr\vert \bigr\Vert _{C[0,T]} \Vert ax + b \Vert _{C[0,1]}, \\& C_{5}(T) = 2T \bigl\Vert \bigl[h(t)\bigr]^{ - 1} \bigr\Vert _{C[0,T]} \Biggl( \sum_{k = 1}^{\infty } \lambda _{k}^{ - 2} \Biggr)^{\frac{1}{2}} \bigl\Vert p(t) \bigr\Vert _{C[0,T]} \biggl( 1 + 2\sqrt{3} (1 + \delta ) \biggl\Vert \frac{1}{c(t)} \biggr\Vert _{C[0,T]} \biggr) \\& \hphantom{C_{5}(T) =}{}\times \bigl\Vert \bigl\vert g(x_{2},t) \bigr\vert + \bigl\vert g(x_{1},t) \bigr\vert \bigr\Vert _{C[0,T]} \Vert ax + b \Vert _{C[0,1]}, \\& D_{5}(T) = 2\sqrt{2T} (1 + \delta ) \biggl\Vert \frac{1}{c(t)} \biggr\Vert _{C[0,T]} \bigl\Vert \bigl[h(t)\bigr]^{ - 1} \bigr\Vert _{C[0,T]} \Biggl( \sum_{k = 1} ^{\infty } \lambda _{k}^{ - 2} \Biggr)^{\frac{1}{2}} \\& \hphantom{D_{5}(T) =}{}\times \biggl[ \biggl( 1 + 6\sqrt{2} \vert a \vert \biggl\Vert \frac{1}{c(t)} \biggr\Vert _{C[0,T]} \sqrt{ \bigl\Vert c(t) \bigr\Vert _{C[0,T]}} T(1 + \delta ) \biggr) \bigl\Vert g_{xxx}(x,t) \bigr\Vert _{L_{2}(D_{T})} \\& \hphantom{D_{5}(T) =}{}+ \bigl\Vert g_{xxx}(x,t) (1 - b - ax) - 3ag_{xx}(x,t) \bigr\Vert _{L_{2}(D _{T})}\biggr]\\& \hphantom{D_{5}(T) =}{}\times \bigl\Vert \bigl\vert g(x_{2},t) \bigr\vert + \bigl\vert g(x_{1},t) \bigr\vert \bigr\Vert _{C[0,T]} \Vert ax + b \Vert _{C[0,1]}, \\& A_{6}(T) = \bigl\Vert \bigl[h(t)\bigr]^{ - 1} \bigr\Vert _{C[0,T]}\\& \hphantom{A_{6}(T) =}{}\times\Biggl\{ \bigl\Vert \bigl(c(t)h'_{1}(t) - f(x_{1}, t)\bigr)g(x_{2}, t) - \bigl(c(t)h'_{2}(t) - f(x_{2}, t)\bigr)g(x _{1}, t) \bigr\Vert _{C[0,T]} \\& \hphantom{A_{6}(T) =}{}+ \Biggl( \sum_{k = 1}^{\infty } \lambda _{k}^{ - 2} \Biggr)^{ \frac{1}{2}} \biggl[ 2\sqrt{2} \biggl[ \biggl( 1 + 4\sqrt{3} \vert a \vert \biggl\Vert \frac{1}{c(t)} \biggr\Vert _{C[0,T]}(1 + \delta ) \biggr) \bigl\Vert \varphi '''(x) \bigr\Vert _{L_{2}(0,1)} \\& \hphantom{A_{6}(T) =}{}+ \bigl\Vert \varphi '''(x) (1 - b - ax) - 3a\varphi ''(x) \bigr\Vert _{L_{2}(0,1)}\biggr] + 2 \sqrt{2T} (1 + \delta ) \biggl\Vert \frac{1}{c(t)} \biggr\Vert _{C[0,T]} \\& \hphantom{A_{6}(T) =}{}\times \biggl[ \biggl( 1 + 6\sqrt{2} \vert a \vert \biggl\Vert \frac{1}{c(t)} \biggr\Vert _{C[0,T]} \sqrt{ \bigl\Vert c(t) \bigr\Vert _{C[0,T]}} T(1 + \delta ) \biggr) \bigl\Vert f_{xxx}(x,t) \bigr\Vert _{L_{2}(D_{T})} \\& \hphantom{A_{6}(T) =}{}+ \bigl\Vert f_{xxx}(x,t) (1 - b - ax) - 3af_{xx}(x,t) \bigr\Vert _{L_{2}(D _{T})}\biggr]\biggr]\\& \hphantom{A_{6}(T) =}{}\times \bigl\Vert \bigl\vert h_{1}(t) \bigr\vert + \bigl\vert h_{2}(t) \bigr\vert \bigr\Vert _{C[0,T]} \Vert ax + b \Vert _{C[0,1]}\Biggr\} , \\& B_{6}(T) = 2T \bigl\Vert \bigl[h(t)\bigr]^{ - 1} \bigr\Vert _{C[0,T]} \Biggl( \sum_{k = 1}^{\infty } \lambda _{k}^{ - 2} \Biggr)^{\frac{1}{2}}(1 + \delta ) \biggl\Vert \frac{1}{c(t)} \biggr\Vert _{C[0,T]} \\& \hphantom{B_{6}(T) =}{}\times \biggl( 1 + 3\sqrt{2} \vert a \vert \biggl\Vert \frac{1}{c(t)} \biggr\Vert _{C[0,T]} \sqrt{ \bigl\Vert c(t) \bigr\Vert _{C[0,T]}} T(1 + \delta ) \biggr)\\& \hphantom{B_{6}(T) =}{}\times\bigl\Vert \bigl\vert h_{1}(t) \bigr\vert + \bigl\vert h_{2}(t) \bigr\vert \bigr\Vert _{C[0,T]} \Vert ax + b \Vert _{C[0,1]}, \\& C_{6}(T) = 2T \bigl\Vert \bigl[h(t)\bigr]^{ - 1} \bigr\Vert _{C[0,T]} \Biggl( \sum_{k = 1}^{\infty } \lambda _{k}^{ - 2} \Biggr)^{\frac{1}{2}} \bigl\Vert p(t) \bigr\Vert _{C[0,T]} \biggl( 1 + 2\sqrt{3} (1 + \delta ) \biggl\Vert \frac{1}{c(t)} \biggr\Vert _{C[0,T]} \biggr) \\& \hphantom{C_{6}(T) =}{}\times \bigl\Vert \bigl\vert h_{1}(t) \bigr\vert + \bigl\vert h_{2}(t) \bigr\vert \bigr\Vert _{C[0,T]} \Vert ax + b \Vert _{C[0,1]}, \\& D_{6}(T) = 2\sqrt{2T} (1 + \delta ) \biggl\Vert \frac{1}{c(t)} \biggr\Vert _{C[0,T]} \bigl\Vert \bigl[h(t)\bigr]^{ - 1} \bigr\Vert _{C[0,T]} \Biggl( \sum_{k = 1} ^{\infty } \lambda _{k}^{ - 2} \Biggr)^{\frac{1}{2}} \\& \hphantom{D_{6}(T) =}{}\times \biggl[ \biggl( 1 + 6\sqrt{2} \vert a \vert \biggl\Vert \frac{1}{c(t)} \biggr\Vert _{C[0,T]} \sqrt{ \bigl\Vert c(t) \bigr\Vert _{C[0,T]}} T(1 + \delta ) \biggr) \bigl\Vert g_{xxx}(x,t) \bigr\Vert _{L_{2}(D_{T})} \\& \hphantom{D_{6}(T) =}{}+ \bigl\Vert g_{xxx}(x,t) (1 - b - ax) - 3ag_{xx}(x,t) \bigr\Vert _{L_{2}(D _{T})}\biggr]\\& \hphantom{D_{6}(T) =}{}\times \bigl\Vert \bigl\vert h_{1}(t) \bigr\vert + \bigl\vert h_{2}(t) \bigr\vert \bigr\Vert _{C[0,T]} \Vert ax + b \Vert _{C[0,1]}. \end{aligned}$$
From inequalities (3.23), (3.26), and (3.27), we deduce that
$$\begin{aligned}& \bigl\Vert \tilde{u}(x,t) \bigr\Vert _{B_{2,T}^{3}} + \bigl\Vert \tilde{a}(t) \bigr\Vert _{C[0,T]} + \bigl\Vert \tilde{b}(t) \bigr\Vert _{C[0,T]} \\& \quad \le A(T) + B(T) \bigl\Vert a(t) \bigr\Vert _{C[0,T]} \bigl\Vert u(x,t) \bigr\Vert _{B _{2,T}^{3}} \\& \qquad {}+ C(T) \bigl\Vert u(x,t) \bigr\Vert _{B_{2,T}^{3}} + D(T) \bigl\Vert b(t) \bigr\Vert _{C[0,T]}, \end{aligned}$$
(3.28)
where
$$\begin{aligned}& A(T) = A_{4}(T) + A_{5}(T) + A_{6}(T),\qquad B(T) = B_{4}(T) + B_{5}(T) + B _{6}(T), \\& C(T) = C_{4}(T) + C_{5}(T) + C_{6}(T),\qquad D(T) = D_{4}(T) + D_{5}(T) + D _{6}(T). \end{aligned}$$
Theorem 3.3
If conditions (A)–(D) and the condition
$$ \bigl(B(T) \bigl(A(T) + 2\bigr) + C(T) + D(T)\bigr) \bigl(A(T) + 2\bigr) < 1 $$
(3.29)
hold, then problem (1.1)–(1.3), (1.6), and (1.7) has a unique solution in the ball
\(K = K_{R}( \Vert z \Vert _{E_{T}^{3}} \le R \le A(T) + 2)\)
of the space
\(E_{T}^{3}\).
Remark 3.4
Inequality (3.29) is satisfied for sufficiently small values of T.
Proof
In the space \(E_{T}^{3}\), we consider the equation
$$ z = \varPhi z, $$
(3.30)
where \(z = \{ u, a, b\}\), and the components \(\varPhi _{i}(u,a, b)\) (\(i = 1,2,3\)) of operator \(\varPhi (u,a, b)\) are defined by the right-hand side of Eqs. (3.11), (3.14), and (3.15).
Consider the operator \(\varPhi (u,a, b)\) in the ball \(K = K_{R}\) of the space \(E_{T}^{3}\). Similarly, with the aid of (3.28), we obtain that for any \(z_{1},z_{2},z_{3} \in K_{R}\) the following inequalities hold:
$$\begin{aligned}& \Vert \varPhi z \Vert _{E_{T}^{3}} \\& \quad \le A(T) + B(T) \bigl\Vert a(t) \bigr\Vert _{C[0,T]} \bigl\Vert u(x,t) \bigr\Vert _{B_{2,T}^{3}} + C(T) \bigl\Vert u(x,t) \bigr\Vert _{B_{2,T}^{3}} + D(T) \bigl\Vert b(t) \bigr\Vert _{C[0,T]} \\& \quad \le A(T) + B(T) \bigl(A(T) + 2\bigr)^{2} + C(T) \bigl(A(T) + 2\bigr) + D(T) \bigl(A(T) + 2\bigr) < A(T) + 2, \end{aligned}$$
(3.31)
$$\begin{aligned}& \Vert \varPhi z_{1} - \varPhi z_{2} \Vert _{E_{T}^{3}} \le B(T)R\bigl( \bigl\Vert a_{1}(t) - a_{2}(t) \bigr\Vert _{C[0,T]} + \bigl\Vert u_{1}(x,t) - u_{2}(x,t) \bigr\Vert _{B_{2,T}^{3}}\bigr) \\& \hphantom{\Vert \varPhi z_{1} - \varPhi z_{2} \Vert _{E_{T}^{3}} \le}{}+ C(T) \bigl\Vert u_{1}(x,t) - u_{2}(x,t) \bigr\Vert _{B_{2,T}^{3}} + D(T) \bigl\Vert b_{1}(t) - b_{2}(t) \bigr\Vert _{C[0,T]}. \end{aligned}$$
(3.32)
Then by (3.29), from (3.31) and (3.32) it is clear that operator Φ on the set \(K = K_{R}\) satisfies the conditions of the contraction mapping principle. Therefore, operator Φ has a unique fixed point \(z = \{ u, a, b\}\), in the ball \(K = K_{R}\), which is a solution of Eq. (3.30), i.e., in the ball \(K = K_{R}\) it is the unique solution of the system (3.11), (3.14), and (3.15). Then the function \(u(x,t)\), as an element of space \(B_{2,T}^{3}\), is continuous and has continuous derivatives \(u_{x}(x,t)\) and \(u_{xx}(x,t)\) in \(D_{T}\).
Next, from (3.4) and (3.5), it follows that \(u'_{k}(t)\) (\(k = 1,2, \ldots \)) are continuous on \([0,T]\), and consequently we have
$$\begin{aligned}& \Biggl( \sum_{k = 1}^{\infty } \bigl(\lambda _{k} \bigl\Vert u'_{2k - 1}(t) \bigr\Vert _{C[0,T]}\bigr)^{2} \Biggr)^{\frac{1}{2}} \\& \quad \le \biggl\Vert \frac{1}{c(t)} \biggr\Vert _{C[0,T]}\sqrt{2} \Biggl[ \Biggl( \sum _{k = 1}^{\infty } \bigl( \lambda _{k}^{3} \bigl\Vert u_{2k - 1}(t) \bigr\Vert _{C[0, T]} \bigr) ^{2} \Biggr)^{\frac{1}{2}} \\& \qquad {}+ 2\sqrt{2} \bigl\Vert \bigl\Vert f_{x}(x,t) + a(t)u_{x}(x,t) + b(t)g_{x}(x,t) \bigr\Vert _{C[0,T]} \bigr\Vert _{L_{2}(0,1)} \Biggr] \\& \quad < + \infty , \\& \Biggl( \sum_{k = 1}^{\infty } \bigl(\lambda _{k} \bigl\Vert u'_{2k}(t) \bigr\Vert _{C[0,T]}\bigr)^{2} \Biggr)^{\frac{1}{2}} \\& \quad \le \biggl\Vert \frac{1}{c(t)} \biggr\Vert _{C[0,T]}\sqrt{3} \Biggl[ \Biggl( \sum _{k = 1}^{\infty } \bigl( \lambda _{k}^{3} \bigl\Vert u_{2k}(t) \bigr\Vert _{C[0, T]} \bigr)^{2} \Biggr) ^{\frac{1}{2}} \\& \qquad {}+ 2\sqrt{2} \bigl\Vert \bigl\Vert \bigl(f_{x}(x,t) + a(t)u_{x}(x,t) + b(t)g _{x}(x,t)\bigr) (1 - b - ax) \\& \qquad {} + a\bigl(f(x,t) + a(t)u(x,t) + b(t)g(x,t)\bigr) \bigr\Vert _{C[0,T]} \bigr\Vert _{L_{2}(0,1)} \\& \qquad {}+ 2a \Biggl( \sum _{k = 1}^{\infty } \bigl( \lambda _{k}^{3} \bigl\Vert u_{2k - 1}(t) \bigr\Vert _{C[0,T]}\bigr)^{2} \Biggr) ^{\frac{1}{2}} \Biggr] \\& \quad < + \infty . \end{aligned}$$
Hence we conclude that the function \(u_{t}(t,x)\) is continuous in domain \(D_{T}\).
Further, it is possible to verify that Eq. (1.1) and conditions (1.2), (1.3), (1.6), and (1.7) are satisfied in the usual sense. Consequently, \(\{ u(x,t),a(t), b(t)\}\) is a solution of (1.1)–(1.3), (1.6), and (1.7), and by Corollary 3.2 it is unique in the ball \(K = K_{R}\). The proof is complete. □
From Theorems 1.2 and 3.3, the following assertion follows directly.
Theorem 3.5
Suppose that all assumptions of Theorem 3.3
and the compatibility conditions
$$\begin{aligned}& \int _{0}^{1} \varphi (x)\,dx = 0, \\& h_{i}(0) + \delta h_{i}(T) + \int _{0}^{T} p(t)h_{i}(t)\,dt = \varphi (x _{i})\quad (i = 1, 2) \end{aligned}$$
hold. If
$$\int _{0}^{1} f(x,t)\,dx = 0, \qquad \int _{0}^{1} g(x,t)\,dx = 0\quad (0 \le t \le T), $$
the, problem (1.1)–(1.5) has a unique classical solution in the ball
\(K = K_{R}\).