## 3 Expansion of the Direct Problem (1)-(5) on \([T,\infty)\)
From Lemmas 2 and 3, the series \(\sum\limits_{n=0}^{\infty}{\left|{{\lambda_{n}}}\right|^{2}}{\left|{{A_{n}}} \right|}<\infty\); hence, from (23), we obtain  
\[\frac{{\partial u_{2}(1,t)}}{{\partial t}}=-\sum\limits_{n=0}^{\infty}{\lambda_ {n}^{2}{A_{n}}{e^{-\lambda_{n}^{2}(t-T)}}\sin\lambda_{n}x} \tag{34}\]  
\[\frac{{\partial u_{2}(1,t)}}{{\partial t}}\in C(T,\infty),\text{and there exists,}\lim \limits_{t\to T}\frac{{\partial u_{2}(1,t)}}{{\partial t}}<\infty. \tag{35}\]  
Denote \(u_{2}(1,t)=q_{2}(t)\), from (34) and (35), it follows that \(q_{2}(t)\in H^{1}[T,\infty)\). **Lemma 4. **_Let the function \(\frac{{\partial u_{2}(1,t)}}{{\partial t}}\) be defined by Equation (34). Then, \(d_{1}>0\) such that for any \(t\geq T+1\)_  
\[\left|{\frac{{\partial u_{2}(1,t)}}{{\partial t}}}\right|\leq d_{1}{e^{-(t-T-1 )}}+{A_{0}}{e^{-\lambda_{0}^{2}(t-T)}}.\]  
**Proof. ** From (34) and (35), it follows that  
\[\left|{\frac{{\partial u_{2}(1,t)}}{{\partial t}}}\right|\leq d_{1}\sum \limits_{n=0}^{\infty}{\lambda_{n}^{2}{e^{\lambda_{n}^{2}(t-T)}}},\]  
where \(d_{1}\) some number. Let us assume that \(t\geq T+1\) and \(n>0\)  
\[\lambda_{n}^{2}{e^{-\lambda_{n}^{2}}{e^{-\lambda_{n}^{2}(t-T-1)}}}\leq\lambda _{n}^{2}{e^{-\lambda_{n}^{2}}{e^{-(t-T-1)}}}.\]  
From \(d_{1}n\leq\lambda_{n}\leq d_{2}n\) it follows that, for \(n>0\) and numbers \(d_{1}\), \(d_{2}>0\)  
\[\lambda_{n}^{2}{e^{-\lambda_{n}^{2}}}\leq d_{2}n^{2}{\left[{e^{-d_{1}^{2}}} \right]}^{n}, \tag{36}\]  
from (36), it follows that \(\sum\limits_{n=0}^{\infty}{n^{2}{\left[{e^{-d_{1}^{2}}}\right]}^{n}}<\infty\). Hence there is a number \(d_{3}\) for any \(t\geq T+2\)  
\[\left|{\frac{{\partial u_{2}(1,t)}}{{\partial t}}}\right|\leq d_{3}{e^{-(t-T -1)}}+{A_{0}}{e^{-\lambda_{0}^{2}(t-T)}},\]  
from (35) and Lemma 4, it follows that  
\[\frac{{\partial u_{2}(1,t)}}{{\partial t}}\in C[T,\infty)\cap L_{1}[T,\infty )\cap L_{2}[T,\infty).\]  
Now, let us introduce the notation  
\[h(t)=\left\{ \begin{array}{l}q_{1}(t),\ t\in(0,T]\\ q_{2}(t),\ t\in[T,\infty)\end{array} ,\ u(x,t)=\left\{ \begin{array}{l}u_{1}( x,t),\ t\in(0,T],\\ u_{2}(x,t),\ t\in[T,\infty),\end{array} \right.x\in(0,1), \tag{37}\]  
From (33) and (37), it follows that, for any \(\varepsilon>0\) there is \(\chi_{\varepsilon}(t)\) which is defined as a function such that, for any \(t\geq 0\)  
\[\sup\limits_{0\leq x\leq 1}\left\{{\left|{u(x,t)}\right|},\left|{\frac{{ \partial u(x,t)}}{{\partial x}}}\right|,\left|{\frac{{\partial^{2}u(x,t)}}{{ \partial x^{2}}}}\right|\right\}\leq\chi_{\varepsilon}(t), \tag{38}\]  
where  
\[\chi_{\varepsilon}(t)=\left\{\begin{array}{l}d_{6}(\varepsilon),\Since \(\chi_{\varepsilon}(t)\in L_{1}[0,\infty)\), then the Fourier transform for \(t\) can be used for the combined direct problem (1)-(5) and (18)-(21). The lemma is proofed. \(\Box\)  
From Lemma 1 and Equation (38), we obtain the following theorem. **Theorem 1. **_Let \(\Phi(t)\in C[0,\infty)\) and \(\Phi(t)\) is limited over this line. Then, the following relations are true_  
\[ \begin{array}{l}\int\limits_{0}^{\infty}\frac{\partial u(x,t)}{ \partial x}\Phi(t)dt=\frac{\partial}{\partial x}\left[\int\limits_{0}^{\infty }u(x,t)\Phi(t)dt\right],\\ \int\limits_{0}^{\infty}\frac{\partial^{2}u(x,t)}{\partial x^{2}}\Phi(t)dt= \frac{\partial^{2}}{\partial x^{2}}\left[\int\limits_{0}^{\infty}u(x,t)\Phi( t)dt\right]\end{array} \]  
**Lemma 5. **_Let \(u(x,t)\) be a solution of the combined problem (1)-(5) and (18)-(21). Then, the following relations are true_  
\[\lim\limits_{x\to 0}\int\limits_{0}^{\infty}|u(x,t)-h(t)|dt=\lim\limits_{x \to 1}\int\limits_{0}^{\infty}|u(x,t)-u(1,t)|dt=\lim\limits_{x\to 1}\int\limits_{0}^{ \infty}|u^{\prime}x(x,t)-u^{\prime}x(1,t)|dt=0.\]  
**Proof.