## 3 Expansion of the Direct Problem (1)-(5) on \([T,\infty)\)  
Let us study the following PDE system in the interval \([T,\infty)\). \[\frac{\partial u_{2}(x,t)}{\partial t}=\frac{\partial^{2}u_{2}(x,t)}{\partial x ^{2}},\;0\leq x\leq 1,\;t\in[T,\infty), \tag{18}\]  
\[u_{2}(x,T)=u_{1}(x,T)=f(x),\;0\leq x\leq 1, \tag{19}\]  
\[u_{2}(0,t)=0,\;T\leq t<\infty, \tag{20}\]  
\[\frac{\partial u_{2}}{\partial x}(1,t)+\kappa u_{2}(1,t)=0,\;T\leq t<\infty, \kappa>0. \tag{21}\]  
Assume that  
\[f(x)\in W_{2}^{2}[0,1];f(0)=0,\;f^{\prime}(1)+\kappa f^{\prime}(1)=0, \tag{22}\]  
We obtain the following solution by applying the separation of variables as a way for solving problem (18)-(21)  
\[u_{2}(x,t)=\sum\limits_{n=1}^{\infty}A_{n}e^{-\lambda_{n}^{2}(t-T)}\sin \lambda_{n}x,0\leq x\leq 1,t\geq T, \tag{23}\]  
where \(\cot\lambda_{n}=-\frac{\lambda_{n}}{\kappa}\),  
\[A_{n}=\frac{4}{2\lambda_{n}-\sin 2\lambda_{n}}\int\limits_{0}^{1}f(x)\sin \lambda_{n}xdx. \tag{24}\]  
By integrating the right side of (24) twice, we obtain  
\[A_{n}=\frac{4}{\lambda_{n}(2\lambda_{n}-\sin 2\lambda_{n})}\int\limits_{0}^{1}f^{ \prime\prime}(x)\sin\lambda_{n}xdx, \tag{25}\]  
From (22) and (25), we define a number \(c_{1}>0\) such that for any \(n\)  
\[|A_{n}|\leq\frac{c_{1}}{\lambda_{n}^{2}}. \tag{26}\]  
From (23) and (26), any \(t\geq T+1\)  
\[|u_{2}(x,t)|\leq c_{1}\sum\limits_{n=1}^{\infty}\lambda_{n}^{-2}e^{-\lambda_ {n}^{2}(t-T)}, \tag{27}\]  
\[u_{2}(x,T)=u_{1}(x,T)=f(x),\;0\leq x\leq 1, \tag{19}\]0  
then,  
\[u_{2}(x,T)=u_{1}(x,T)=f(x),\;0\leq x\leq 1, \tag{19}\]1  
Since  
\[u_{2}(x,T)=u_{1}(x,T)=f(x),\;0\leq x\leq 1, \tag{19}\]2Let us consider there exists the numbers \(c_{2}\) and \(c_{3}\) such that for any \(n\)  
\[u_{2}(x,T)=u_{1}(x,T)=f(x),\;0\leq x\leq 1, \tag{19}\]3  
and, it follows from (30) and (31), that  
\[u_{2}(x,T)=u_{1}(x,T)=f(x),\;0\leq x\leq 1, \tag{19}\]4  
then, it follows from (18), (27)-(32) that there is \(c_{4}>0\) known as a number such that for any \(t\geq T+2\)  
\[u_{2}(x,T)=u_{1}(x,T)=f(x),\;0\leq x\leq 1, \tag{19}\]5  
Now, let us examine the behavior \(\frac{\partial u_{2}(1,t)}{\partial t}\),  
**Lemma 3**. : _Let \(A_{n}\) be defined by the formula (24). Then_  
\[u_{2}(x,T)=u_{1}(x,T)=f(x),\;0\leq x\leq 1, \tag{19}\]6  
_where \(f^{(4)}(x)\) is the fourth derivative with respect to \(x\) for function \(f(x)\)._  
Proof. : \(A_{n}\) defined by the Equation (24), and integrating \(\int\limits_{0}^{1}f(x)\sin\lambda_{n}xdx\) in parts twice, we obtain  
\[u_{2}(x,T)=u_{1}(x,T)=f(x),\;0\leq x\leq 1, \tag{19}\]7  
from (3) and (19)  
\[u_{2}(x,T)=u_{1}(x,T)=f(x),\;0\leq x\leq 1, \tag{19}\]8  
Since  
\[ \begin{array}{l}\frac{\sin\lambda_{n}}{\lambda_{n}^{2}}f^{\prime}(1)+\frac{ \cos\lambda_{n}}{\lambda_{n}}f(1)=\frac{\sin\lambda_{n}}{\lambda_{n}^{2}} \frac{\partial u_{2}}{\partial x}(1,t)+\frac{\cos\lambda_{n}}{\lambda_{n}}u_{ 2}(1,t)\\ =\frac{\cos\lambda_{n}}{\lambda_{n}^{2}}\Big{[}\frac{\partial u_{2}}{\partial x }(1,t)+\lambda_{n}u_{2}(1,t)\Big{]}=\frac{\lambda_{n}}{\kappa}\frac{\cos \lambda_{n}}{\kappa\lambda_{n}}\Big{[}\frac{\partial u_{2}}{\partial x}(1,t)+ \kappa u_{2}(1,t)\Big{]}=0,\end{array} \]  
as a result, we obtain  
\[\int\limits_{0}^{1}f(x)\sin\lambda_{n}xdx=-\frac{1}{\lambda_{n}^{2}}\int \limits_{0}^{1}f^{\prime\prime}(x)\sin\lambda_{n}xdx,\]  
Integrating the right part of the previous equation twice in parts, it leads to  
\[-\frac{1}{\lambda_{n}^{2}}\int\limits_{0}^{1}f^{\prime\prime}(x)\sin\lambda_{ n}xdx=\frac{1}{\lambda_{n}^{3}}\big{[}\cos\lambda_{n}f^{\prime\prime}(1)-f^{ \prime\prime}(0)\big{]}-\frac{1}{\lambda_{n}^{4}}\Bigg{[}\sin\lambda_{n}f^{ \prime\prime}(1)-\int\limits_{0}^{1}f^{(4)}(x)\sin\lambda_{n}xdx\Bigg{]},\]  
The lemma is proof.