## 5 Solution of the Inverse BVP (1)-(5) and (18)-(21) by the Projection Regularization Method  
Let \(\overline{H}=L_{2}(-\infty;+\infty)+iL_{2}(-\infty;+\infty)\) be the interval on the area of complex numbers, and the set of correction class \(M_{r}\subset\overline{H}\) demarcated by the following  
\[M_{r}=\left\{h(t):h(t)\in\overline{H},\ \int\limits_{0}^{\infty}|h(t)|^{2}dt+ \int\big{|}h^{\prime}(t)\big{|}^{2}dt\leq r^{2}\right\}, \tag{42}\]  
\(r\) known positive number. In order to resolve the problem (1)-(5) and (18)-(21), we present \(F\), as the operator which is mapping from \(\overline{H}\) to \(\overline{H}\) and we named as the operator via the Fourier transform  
\[\hat{h}(\tau)=F[h(t)]=\frac{1}{\sqrt{2\pi}}\int\limits_{0}^{+\infty}h(t)e^{- irt}dt,\ \tau\in\mathbb{R},\ h(t)\in\overline{H}\cap\overline{L}_{1}(-\infty,\infty). \tag{43}\]  
There \(\overline{L}_{1}(-\infty,\infty)\)--interval on the of complex numbers set. Denote by \(\overline{F}\) operator continuation \(F\) in \(\overline{H}\). Following from [PERSON]'s theorem, the operator \(\overline{F}\) has isometric mapping \(\overline{H}\) into \(\overline{H}\). Let \(\hat{h}(\tau)\in\overline{H}\). Then, we have  
\[\overline{F}^{-1}\Big{[}\hat{h}(\tau)\Big{]}=\frac{1}{\sqrt{2\pi}}\lim\limits _{N\to\infty}\int\limits_{-N}^{+N}\hat{h}(\tau)e^{it\tau}d\tau,\ -\infty<t<+\infty, \tag{44}\]  
where the way to the limit has the sense of the convergence of root-mean-square. Using transform \(F\), (1)-(5) and (18)-(21) come down to the following problem  
\[\frac{\partial^{2}\hat{a}(x,\tau)}{\partial x^{2}}=i\tau\hat{a}(x,\tau),\ x\in(0,1),\ - \infty<\tau<+\infty, \tag{45}\]  
\[\hat{a}(0,\tau)=0,\ \hat{a}(x_{0},\tau)=\hat{g}(\tau),-\infty<\tau<+\infty, \tag{46}\]  
where \(\hat{a}(x,\tau)=F[u(x,t)],\ \hat{g}(\tau)=F[g(t)]\). Solutions (45) and (46) are of the form  
\[\hat{a}(x,\tau)=D_{1}(\tau)e^{\mu_{0}\sqrt{\tau}x}+D_{2}(\tau)e^{-\mu_{0} \sqrt{\tau}x},\ \tau>0, \tag{47}\]  
where \(\mu_{0}=\frac{1}{\sqrt{2}}(1+i)\), \(D_{1}(\tau)\) and \(D_{2}(\tau)\) are functions that satisfy (40) and (46). With \(\hat{u}(1,\tau)=\hat{h}(\tau)\) we obtain  
\[\hat{h}(\tau)=\left\{ \begin{array}{ll}\frac{\mathrm{sh}\ \mu_{0}\sqrt{\tau}}{ \mathrm{sh}\ \mu_{0}\sqrt{\tau}x_{0}}\hat{g}(\tau),\ \tau>0,\\ 0,&\tau\leq 0,\end{array} \right. \tag{48}\]  
Therefore, the problem (45) and (46) reduces to the equation  
\[A\hat{h}(\tau)=\hat{g}(\tau),\ 0<\tau<\infty. \tag{49}\]  
Let \(\hat{g}_{0}(\tau)=F[g_{0}(t)],\hat{g}_{\delta}(\tau)=F[g_{\delta}(t)]\) and, from the Formula (41), it follows that  
\[\left\|\hat{g}_{\delta}(\tau)-\hat{g}_{0}(\tau)\right\|_{\overline{H}}\leq \sqrt{2}\delta. \tag{50}\]  
Let \(\hat{M}_{r}\) denote a set of \(\overline{H}\) such that \(\hat{M}_{r}\supset F[M_{r}]\) and \[\hat{M}_{r}=\Bigg{\{}\hat{h}(\tau):\hat{h}(\tau)\in\overline{H},\,\int\limits_{0}^{ \infty}\Big{(}1+\tau^{2}\Big{)}\Big{|}\hat{h}(\tau)\Big{|}d\tau\leq 2\tau^{2} \Bigg{\}}. \tag{51}\]  
Since \(h_{0}(t)\in M_{r}\), then \(\hat{h}_{0}(\tau)\in\hat{M}_{r}\). In order to find the approximation solution for (49)-(51) we use the regularizing family of operators \(\{R_{\alpha}\}\), which are defined by  
\[\hat{h}_{\delta}^{\alpha}(\tau)=R_{\alpha}\delta(\tau)=\left\{ \begin{array}[] {ll}\frac{\mathrm{sh}\,\mu_{0}\sqrt{\tau}}{\mathrm{sh}\,\mu_{0}\sqrt{\tau}x_ {0}}\delta_{\delta}(\tau),\,0\leq\tau\leq\alpha,\\ 0,\hskip 56.905512 pt\tau>\alpha,\end{array} \right. \tag{52}\]  
For selecting a regularization parameter \(\hat{\alpha}=\hat{\alpha}(\hat{g}_{\delta},\delta)\) in Equation (52) from the initial data \((\hat{g}_{\delta},\delta)\), use the equation \(\Big{\|}A\hat{h}_{\delta}^{\alpha}(\tau)-\hat{g}_{\delta}(\tau)\Big{\|}^{2}=1 6\delta^{2}\). Let us describe an estimated solution for (49) by the formulation of \(\hat{h}_{\delta}(\tau)=\hat{h}_{\delta}^{\hat{\alpha}(\hat{g}_{\delta},\delta )}(\tau)\).