Let us consider an auxiliary Cauchy problem to construct a solution to problem (\ref{eqn1-1}). The auxiliary problem is comparable with the Dirichlet problem (\ref{eqn1-1}):
\begin{eqnarray}\label{eqn4-1}
\left\{
\begin{array}{lll}
\displaystyle \frac{\partial^2 U(x,t)}{\partial t^2}=\frac{\partial^2 U(x, t)}{\partial x^2}, \quad x\in (0; \pi), \quad t\in (0; 1),\\
\displaystyle U(0, t)=0, \quad t\in [0; 1], \quad U(1, t) =0, \quad t\in [0; 1],\\ \displaystyle U(x, 0)= f(x), \quad x \in [0; \pi], \quad \frac{\partial U(x, 0)}{\partial t}=g(x), \quad x\in [0; \pi],\\ \displaystyle f(x)\in H^4_0[0; \pi], \quad g(x)\in H^3_0[0; \pi]. \\ \end{array}\right.
\end{eqnarray}
Note that such a problem is considered in \cite{Sveshnikov}, p. 259. The following statement is true:
\begin{lemma}\label{hilfUniq}
A solution to problem (\ref{eqn4-1}) exists $\forall x \in [0; \pi], \forall t\in [0;+\infty)$, is unique, is continuous in the initial data and can be represented as a series:
\begin{eqnarray}\label{eqn4-2}
\begin{array}{lll}
\displaystyle U(x,t)=\sum\limits_{n=1}^{\infty}\left\{f_n\cos(nt)+\frac{g_n}{n}\sin(nt)\right\}\sin(nx), \\
\displaystyle f_n=\frac{2}{\pi}\int\limits_0^{\pi}f(x)\sin(nx)dx, \quad g_n=\frac{2}{\pi}\int\limits_0^{\pi}g(x)\sin(nx)dx,
\end{array}
\end{eqnarray}
if $f(x)\in H^4_0[0;\pi],$ $g(x)\in H_0^3[0; \pi].$
\end{lemma}
The proof is given in \cite{Sveshnikov}, pp. 259-260.

The following lemma is true:
\begin{lemma}\label{eqn4-3}
Let a solution to problem (\ref{eqn4-1}) exist, i.e. $f(x)\in H_0^4[0; \pi],$ $g(x)\in H_0^3[0; \pi].$ Then $\displaystyle\frac{\partial^4 U(x, t)}{\partial x^3\partial t}\in L_2[0; \pi]$ $\forall t\in [0; +\infty),$ $\displaystyle\frac{\partial U(x, t)}{\partial t}\in H_0^3[0; \pi]$ $\forall t\in [0;+\infty).$
\end{lemma}
\begin{proof}[Proof:]
We differentiate the series (\ref{eqn4-2}), since the solution (\ref{eqn4-1}) is defined by formula (\ref{eqn4-2}):
\begin{eqnarray}\label{eqn4-4}
\displaystyle\frac{\partial^4 U(x,t)}{\partial x^3\partial t}=\sum\limits_{n=1}^{\infty}n^4\left\{-f_n\sin(nt)+\frac{g_n}{n}\cos(nt)\right\}\cos(nx).
\end{eqnarray}
Since $f(x)\in H^4_0[0; \pi],$ $g(x)\in H_0^3[0; \pi],$ then 
\begin{eqnarray}\label{eqn4-5}
 f_n=\frac{F_n}{n^4}, g_n=\frac{G_n}{n^3}, \exists C_1>0, \sum\limits_{n=1}^{\infty}F_n^2\leq C_1, \sum\limits_{n=1}^{\infty} G_n^2\leq C_1, 
\end{eqnarray} where
\begin{eqnarray*} 
F_n=\frac{2}{\pi}\int\limits_0^{\pi}f^{(4)}(x)\cos(nx)dx, G_n=-\frac{2}{\pi}\int\limits_0^{\pi}g'''(x)\sin(nx)dx.
\end{eqnarray*}
The equality is true:
\begin{eqnarray*}
\displaystyle\frac{\partial^4 U(x,t)}{\partial x^3\partial t}=\sum\limits_{n=1}^{\infty}\left\{F_n\sin(nt)-G_n\cos(nt)\right\}\sin(nx).
\end{eqnarray*}
We obtain, composing the scalar product in the space $L_2[0;\pi],$ and taking into account the norm in this space $||s(x)||_{L_2[0; \pi]}^2\equiv (s(x); s(x)):$ 
\begin{eqnarray*}
 \displaystyle\Bigg|\left(\frac{\partial^4 U(x,t)}{\partial x^3\partial t},\frac{\partial^4 U(x,t)}{\partial x^3\partial t}\right)\Bigg|=\\=\sum\limits_{n=1}^{ \infty}\left\{F_n^2\sin^2(nt)+F_nG_n\sin(2nt)+G_n^2\cos^2(nt)\right\}\leq 2\sum\limits_{n=1}^{\infty}\left\{F_n^2+G_n^2\right\}.
\end{eqnarray*}
This series converges absolutely by (\ref{eqn4-5}), therefore, $\displaystyle\frac{\partial^4 U(x,t)}{\partial x^3\partial t}\in L_2[0; \pi] \forall t\in [0; +\infty).$

The assertion of the lemma follows from this.
\end{proof}