Let us now consider problem (\ref{eqn1-1}) and the series 
\begin{eqnarray}\label{eqn5-1}
u(x, t)=\sum\limits_{n=1}^{\infty}\left\{f_n\cos(nt)+\frac{h_n-f_n\cos n}{\sin n}\sin( nt)\right\}\sin(nx),
\end{eqnarray}
where $f_n$ are given by formula (\ref{eqn4-2}), $\displaystyle h_n=\frac{2}{\pi}\int\limits_0^{\pi}h(x)\sin(nx)dx. $

Note that series (\ref{eqn5-1}) upon formal substitution into all relations (\ref{eqn1-1}) turns them into a true equality.

We have not yet defined the classes of functions $f(x), h(x)$ to solve problem (\ref{eqn1-1}). We turn to the solution to problem (\ref{eqn4-1}): $f(x)\in H_0^4[0;\pi],$ $g(x)\in H_0^3[0;\pi]$ to determine them.  However, series (\ref{eqn5-1}) does not contain the components of the expansion of function $g(x)$ into a Fourier series. We compare series (\ref{eqn5-1}) and series (6) in order to determine components $g_n$
\begin{eqnarray}\label{eqnGn}
\sum\limits_{n=1}^{\infty}\left\{f_n\cos(nt)+\frac{h_n-f_n\cos n}{\sin n}\sin(nt)\right\}\sin(nx)=\\=\sum\limits_{n=1}^{\infty}\left\{f_n\cos(nt)+\frac{g_n}{n}\sin(nt)\right\}\sin(nx).
\end{eqnarray}
If we choose
\begin{eqnarray}\label{eqnGnF}
g_n=\frac{h_n-f_n\cos(n)}{\sin n}\cdot n,
\end{eqnarray}
then the solutions to the Dirichlet (\ref{eqn1-1}) and Cauchy (\ref{eqn4-1}) problems coincide. Let us determine the conditions under which this is possible.

We can use a result from number theory for what follows.
\begin{theorem}\label{Dirichlet}
There are only finitely many relatively prime numbers $n, m\in\mathbb{N}$ such that 
\begin{eqnarray}\label{JapanNum}
\displaystyle\Big|\pi-\frac{m}{n}\Big|\leq\frac{C_{\pi}}{n^{\mu}}, \quad \mu \leq 7.7.
\end{eqnarray}

\end{theorem}
\begin{proof}[Proof:] According to the result of \cite{Japan}, there is a number $C_{\pi}>0$ such that there are only finitely many relatively prime numbers $n, m\in\mathbb{N}$ such that (\ref{JapanNum}) is true.

Let us prove that exists $N$ such that $\forall n, m>N$ $\displaystyle\Big|\pi-\frac{m}{n}\Big|\geq\frac{C_{\pi}}{n^{\mu}}.$ Suppose that this is not so, that is, there are infinitely many numbers $n, m\in\mathbb{N}$ such that (\ref{JapanNum}) is true. Since there are only finitely many such relatively prime numbers, there are infinitely many natural numbers $n, m,$ which have a common factor and  inequality (\ref{JapanNum}) is true for these numbers.

Suppose that pair of numbers $n_0, m_0 \in\mathbb{N}$ is such that (\ref{JapanNum}) is true. Therefore, there are infinitely many numbers $k,$ $n_k=n_0k, m_k=m_0k, k \in\mathbb{N},$ and $\displaystyle\Big|\pi-\frac{m_k}{n_k}\Big|\leq\frac{C_{\pi}}{n_k^{7.7}}$ is true for $n_k, m_k.$ Since $m_0/n_0$ is rational and $\pi$ is transcendental, then $\exists \alpha>0$ such that $\displaystyle\alpha<\Big|\pi-\frac{m_0}{n_0}\Big|.$

Then 
\begin{eqnarray}\label{DirichletEqn}
\displaystyle0<\alpha<\Big|\pi-\frac{m_0}{n_0}\Big|=\Big|\pi-\frac{m_k}{n_k}\Big|\leq\frac{C_{\pi}}{n_0^{7.7}k^{7.7}}.
\end{eqnarray}
Since $k\in\mathbb{N},$ it is always possible to choose $k$ such that $\displaystyle\alpha>\frac{C_{\pi}}{n_0^{7.7}k^{7.7}}.$ This contradicts (\ref{DirichletEqn}), and the set of admissable factors $k$ for numbers $m_0, n_0\in\mathbb{N}$ is finite. The theorem is true since there are finitely many relatively prime numbers $n, m\in\mathbb{N}$ such that (\ref{JapanNum}) is true and there are finitely many of their admissible factors.
\end{proof} 
\begin{lemma}\label{h10} Let $\displaystyle \eta(x)\equiv\sum\limits_{n=1}^{\infty}\frac {h_n-f_n\cos n}{\sin n}n\sin(nx),$ $h(x), f(x) \in H_0^k[0; \pi],$ where $k=\mu+4,$ if $\mu$ is an integer, and $k=[\mu]+5,$ if it is not an integer. Then $\eta(x)\in H^3_0[0; \pi].$
\end{lemma} 
\begin{proof}[Proof:]
Consider series
\begin{eqnarray}\label{eqn5-6} 
\eta(x)=\sum\limits_{n=1}^{\infty}\frac{h_n-f_n\cos n}{\sin n}n\sin(nx).
\end{eqnarray}
Since theorem \ref{Dirichlet} is true, $\exists N,$ that $\forall n, m>N$ $\displaystyle\Big|\pi-\frac{m}{n}\Big|\geq\frac{C_{\pi}}{n^{\mu}},$ then $\displaystyle |\pi n- m|\geq\frac{C_{\pi}}{n^{\mu-1}}.$ Let us use natural number $m$ for summation in series (\ref{eqn5-6}). We have to find how numbers $n$ and $m$ are related to make this summation possible.

Let us choose numbers $n$ so that $\displaystyle 0<|\pi - \frac{m}{n} |<\frac{\pi}{2n}$ or $\displaystyle\frac{\pi}{2}(2n-1)<m<\frac{\pi}{2}(2n+1).$ This inequality is true for any natural $m>1.$  We obtain that the whole set of real numbers is divided by intervals $\displaystyle\left(\frac{\pi}{2}(2n-1); \frac{\pi}{2}(2n+1)\right)$ by enumerating all the  possible natural values of number $n.$  In this case, only irrational numbers $\displaystyle\frac{\pi}{2}(2k-1), k\in\mathbb{N}$ are excluded from the set of real numbers, while all the natural numbers greater than one remain, and each number $m(n)$ falls into one of the partition intervals. Thus, each natural $m>1$ corresponds to number $n,$ and for sufficiently large $n$ a following estimate is true: $n<m(n)<4n.$ This choice of numbers does not contradict the \ref{Dirichlet} theorem discussed above, since condition $\displaystyle 0<|\pi - \frac{m(n)}{n} |<\frac{\pi}{2n}$ does not contradict inequality (\ref{JapanNum}).

The function $\sin(a)$ is increasing provided that  $\displaystyle 0< a< \frac{\pi}{2},$ so the following inequality
\begin{eqnarray*}
|\sin(|\pi n-m(n)|)|\geq \sin\left(\frac{C_{\pi}}{n^{\mu-1}}\right), 
\end{eqnarray*}
is true, whence for sufficiently large $m, n$
\begin{eqnarray*}
|\sin(m(n))|\geq \sin\left(\frac{C_{\pi}}{n^{\mu-1}}\right).
\end{eqnarray*}
Therefore,
\begin{eqnarray*}
 \frac{1}{|\sin(m(n))|}\leq 2\frac{n^{\mu-1}}{C_{\pi}},
\end{eqnarray*}
and
\begin{eqnarray}\label{eqn5-4}
 \frac{1}{|\sin(m)|}\leq 2\frac{m^{\mu-1}}{C_{\pi}}.
\end{eqnarray}
Thus, from (\ref{eqn5-4}) for numbers $n$ starting from some value $N$ the following relation is true
\begin{eqnarray}\label{eqn5-4a}
 \frac{1}{|\sin(n)|}\leq 2\frac{n^{\mu-1}}{C_{\pi}}.
\end{eqnarray}
The estimate $\forall n > N$ follows from (\ref{eqn5-4a}) and (\ref{eqnGnF}) is
\begin{eqnarray}\label{eqn5-5}
\Big|\frac{h_n-f_n\cos n}{\sin n}\Big|\leq\frac{2n^{\mu-1}}{C_{\pi}}|h_n-f_n\cos n|.
\end{eqnarray}
Since only the estimate for $\mu$ is known, we choose $f(x), h(x)\in H^{12}_0[0; \pi].$ This estimate may be greatly overestimated, but we have no other data.

Then the following relation is true.
\begin{eqnarray*} 
\Phi_n=\frac{2}{\pi}\int\limits_0^{\pi}f^{(12)}(x)\sin(nx)dx, \Psi_n =\frac{2}{\pi}\int\limits_0^{\pi}h^{(12)}(x)\sin(nx)dx, \\ \sum\limits_{n=1}^{\infty}\Phi_n^2<C_1, \sum\limits_{n=1}^{\infty}\Psi_n^2<C_1, f_n=\frac{\Phi_n}{n^{12}}, h_n=\frac {\Psi_n}{n^{12}}.
\end{eqnarray*}
The following chain of estimates is correct.
\begin{eqnarray*} |\eta'''(x)|=\Big|\sum\limits_{n=1}^{\infty}\frac{h_n-f_n\cos(n )}{\sin n}(-n^3)\cos(nx)\cdot n\Big|\leq\\ \leq 2\sum\limits_{n=1}^{\infty}\frac{|\Psi_n-\Phi_n\cos(n)|} {n^{12}C_{\pi}\sin(n)}n^4|\cos(nx)|,\\ \sum\limits_{n=1}^{\infty}\frac{2|\Psi_n-\Phi_n\ cos(n)|}{C_{\pi}n^{12}\sin(n) }n^4|\cos(nx)|\leq\\
\leq \sum\limits_{n=1}^{\infty}\frac{2|\Psi_n-\Phi_n\cos(n)|n^{ 7}}{n^{12}}\frac{(n^4)}{C_{\pi}}\leq\sum\limits_{n=1}^{\infty}\frac{2|\Psi_n-\Phi_n\cos(n)|}{n\cdot C_{\pi}}\leq\\
\leq\frac{4}{C_{\pi}}\sum\limits_{n=1}^{\infty}\left\{(\Psi_n-\Phi_n\cos(n))^2+\frac{1}{n^2}\right\}\leq\frac{8}{C_{\pi}}\sum\limits_{n=1}^{\infty}\left\{\Psi_n^2+\Phi_n^2+\frac{1}{n^2}\right\}.
\end{eqnarray*}
This series is convergent because series $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2}$ is convergent,  $\displaystyle\sum_{n=1}^{\infty}\Phi_n^2$ is convergent for $\Phi(x)\in H[0; 1],$ $\displaystyle\sum_{n=1}^{\infty}\Psi_n^2$ is convergent for $\Psi(x)\in H[0; 1].$ Therefore, $\eta(x)\in H_0^3[0; \pi].$
\end{proof}
\begin{lemma}\label{Existance}
Let $f(x), h(x)\in H^{12}_0[0; \pi].$ Then the solution to (\ref{eqn1-1}) exists.
\end{lemma}
\begin{proof}[Proof:]
If $f(x), h(x)\in H_0^{12} [0; \pi],$ then $\displaystyle g(x)\equiv\sum\limits_{n=1}^{\infty}n\frac{h_n-f_n\cos(n)}{\sin n}\sin(nx)\in H^3_0[0; \pi]$ according to lemma \ref{h10}. Solution to (\ref{eqn4-1}) exists and is unique in this case by lemma 2. Moreover, the constructed solution satisfies all the conditions of problem (\ref{eqn1-1}). Therefore, it is also a solution to problem (\ref{eqn1-1}). Thus, a solution to (\ref{eqn1-1}) exists.
\end{proof}