## 2 Materials and Methods Direct Formulation of the Problem on Interval (\(0,t\))
Differentiating the function \(C_{n}(t)\sin(n\pi x)\) by \(x\) twice and using (11), we obtain  
\(\left(C_{n}(t)\sin(n\pi x)\right)^{\prime\prime}{}_{xx}=\frac{2\sin(n\pi x)}{ (n\pi)^{3}}\int\limits_{0}^{t}e^{-(n\pi)^{2}(t-\tau)}q^{\prime\prime\prime}{}_ {1}(\tau)d\tau+\frac{2\sin(n\pi x)}{(n\pi)}q^{\prime}{}_{1}(t)-\frac{2\sin(n \pi x)}{(n\pi)^{3}}q^{\prime\prime}{}_{1}(t)\), since the number series \(\sum\limits_{n-1}^{\infty}\frac{1}{n^{2}}\), converge according to the Weierstrass criterion, the functional series \(2\sum\limits_{n=0}^{\infty}\frac{\sin(n\pi x)}{(n\pi)}q^{\prime}{}_{1}(t)\), converge absolutely and uniformly on \((0,1)\times(0,T]\). Then, we need to check the convergence for \(\sum\limits_{n=0}^{\infty}\frac{\sin(n\pi x)}{(n\pi)}\) to any \(\varepsilon>0\) in this series, related to the Dirichlet criterion, the convergence is consistently on \([\varepsilon,1-\varepsilon]\). Meanwhile, any \(\varepsilon>0\) series \(\sum\limits_{n-0}^{\infty}\left(C_{n}(t)\sin(n\pi x)\right)^{\prime\prime}{}_{xx}\) converges on \([\varepsilon,1-\varepsilon]\times(0,T]\) and the parts of this series are nonstop, we obtain \[u_{1}(x,0)=0,\;0\leq x\leq 1,\]5 \[u_{1}(x,0)=0,\;0\leq x\leq 1,\]6  
The lemma is proofed. \(\square\)  
Now, let us examine the function \(u_{1}(x,T)\). **Lemma 2**. : _Function \(u_{1}(x,T)\), defined by formulas (9) and (11), belongs to space \(H^{4}[0,1]\)._  
**Proof. ** From (5), (9) and (11) it follows that  
\[u_{1}(x,0)=0,\;0\leq x\leq 1,\]7  
where  
\[u_{1}(x,0)=0,\;0\leq x\leq 1,\]8  
Since the conditions  
\[ \begin{array}{l}\frac{\partial u_{1}(x,T)}{\partial x}=\sum\limits_{n=0}^{ \infty}n\pi C_{n}(T)\cos(n\pi x),\\ \frac{\partial^{2}u_{1}(x,T)}{\partial x^{2}}=-\sum\limits_{n=0}^{\infty} \left(n\pi\right)^{2}C_{n}(T)\sin(n\pi x),\\ \frac{\partial^{2}u_{1}(x,T)}{\partial x^{3}}=-\sum\limits_{n=0}^{\infty} \left(n\pi\right)^{3}C_{n}(T)\cos(n\pi x),\\ \frac{\partial^{4}u_{1}(x,t)}{\partial x^{4}}=\sum\limits_{n=0}^{\infty} \left(n\pi\right)^{4}C_{n}(T)\sin(n\pi x),\end{array}  \tag{16}\]  
are right, then, form (15) and (16) by means of the Weierstrass criterion which leads to the convergence of the series, therefore  
\[u_{1}(x,T),\frac{\partial u_{1}(x,T)}{\partial x},\frac{\partial^{2}u_{1}(x,T )}{\partial x^{2}},\frac{\partial^{3}u_{1}(x,T)}{\partial x^{3}}\in C[0,1]. \tag{17}\]  
We will show that \(\frac{\partial^{4}u_{1}(x,T)}{\partial x^{4}}\in L_{2}[0,1]\). From (16) and (17), we obtain  
\[(n\pi)^{4}C_{n}(T)=\frac{2}{(n\pi)}\int\limits_{0}^{T}e^{-(n\pi)^{2}(T-\tau) }q^{\prime\prime}{}_{1}(\tau)d\tau+2(n\pi)q^{\prime}{}_{1}(T)-\frac{2}{(n\pi )}q^{\prime\prime}{}_{1}(T).\]  
From (12), it follows that  
\[\left|\frac{2}{(n\pi)}\int\limits_{0}^{T}e^{-(n\pi)^{2}(T-\tau)}q^{\prime \prime\prime}{}_{1}(\tau)d\tau\right|\leq\left\|q^{\prime\prime\prime}{}_{1}( t)\right\|_{L_{2}[0,T]}\frac{\sqrt{2}}{(n\pi)^{2}},\]  
First series  
\[\sum\limits_{n=0}^{\infty}\left|\frac{2}{(n\pi)}\int\limits_{0}^{T}e^{-(n\pi) ^{2}(T-\tau)}q^{\prime\prime\prime}{}_{1}(\tau)d\tau\right|\leq\infty.\]  
Second series  
\[2q^{\prime}{}_{1}(T)\sum\limits_{n=0}^{\infty}(n\pi)<\infty.\]Third series  
\[2{q^{\prime\prime}}_{1}(T)\sum\limits_{n=0}^{\infty}\frac{1}{(n\pi)},\]  
absolutely converges on \([0,1]\times(0,T]\) then \(\frac{\partial u_{1}(x,T)}{\partial x^{2}}\in L_{2}[0,1]\). \(\square\)