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A method for solution of a mixed boundary value problem for a hyperbolic type equation using the operators A. Yu. Trynin Saratov State University
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     IntroductionThe method of separation of variables, the difference scheme methods, and their generalizations are the most popular methods for solving mixed problems for hyperbolic equations. Each of such approaches has its undeniable advantages and disadvantages. The family of methods of separation of variables have great value for the development of analytic approaches to the solution of boundary value problems of fundamental mathematics. The thoroughly developed apparatus of Fourier series, the method of characteristics, the contraction mapping principle, and other analytic methods provide various tools for an intensive study of the most subtle aspects of boundary value problems of this type [1]–[20]. At the same time, the implementation of such methods depends on the solution of some auxiliary problems, for example, the Goursat problem, or the evaluation of the Fourier coefficients of the functions involved in the conditions of the problems. Due to non-triviality of analytic representation of the solution to the Goursat problem, the method of characteristics is used, typically, in the analysis of the Hadamard well-posedness of mixed boundary value problems. The family of Fourier methods, which have great value in analysis of solutions to problems of mathematical physics, requires evaluation of the Fourier coefficients of the functions involved in the conditions of the problem. However, this approach is feasible not for all continuous functions, and cannot be applied even to the function satisfying the boundary value conditions. In addition, when evaluating the Fourier coefficients via numerical integration of rapidly oscillating functions, the computational error of such methods increases considerably, which hinders their use in applied numerical problems. The theory of difference schemes has also been investigated very thoroughly (see, for example, [21]–[27]). This approach to solution of mixed boundary value problems for hyperbolic type equations works quite well in problems of applied mathematics. Most success of numerical mathematics in this field has been achieved by using algorithms based on difference schemes. However, the advantages of the Fourier analysis like signal filtration, noise reduction in initial conditions, etc., remain beyond the scope of such algorithms. In the present paper, we propose a method for solution of mixed boundary value problems for hyperbolic equations that combines the advantages of both above approaches. Mathematically, we solve a mixed boundary value problem with arbitrary initial functions which are assumed to be continuous, but not necessarily satisfying the boundary conditions in the initial conditions and in the inhomogeneity of the equation. Instead of the Fourier coefficients of the functions involved in the conditions of the problem, use is made of their values on the countable set of the zeros of the solutions to auxiliary Cauchy problems. A relative simplicity of numerical implementation of the new method for delivering a generalized solution of problems of mathematical physics is worth pointing out. At each step of the algorithm, a finite set of values of the continuous functions of the problem conditions serves as the domain of definition of the information operator. Numerical implementation of the proposed method depends only on the available tools for solution of ordinary differential equations. In the algorithm, the Fourier coefficients for a countable set of auxiliary functions can be evaluated in advance, which substantially increases the speed of its operation, since in this case the solution is merely obtained by calculation of a linear combination of these functions. In conclusion, we propose a method for finding the Fourier coefficients of auxiliary functions by using the resolvent of a third-order Cauchy differential operator; this method does not involve numerical integration of rapidly oscillating functions. § 1. Formulation of the main resultConsider a mixed boundary value problem where $x\in [0,\pi]$, $t \in [0,T]$, for $T>0$, and the functions $f$, $\varphi$, $\psi$ are continuous, each on its own domain of definition. The functions $f$, $\varphi$, $\psi$ are not assumed to satisfy the boundary conditions (2), (3). The function $q$ is assumed to be of bounded variation.Assuming $\rho_\lambda \geqslant 0$, for each non-negative $\lambda $, we further suppose, exception for some specified cases, that the function $q_\lambda $ depending on the parameter $\lambda$ is an arbitrary element from the ball $V_{\rho_\lambda}[0,\pi]$ of radius $\rho_\lambda =o(\sqrt \lambda/\ln\lambda)$ in the space of functions with bounded variation vanishing at zero, that is, such that
$$
\begin{equation}
V_0^\pi [q_\lambda ]\leqslant \rho_\lambda, \qquad \rho_\lambda =o\biggl(\frac{\sqrt \lambda}{\ln\lambda}\biggr) \quad \text{as }\ \lambda \to \infty,\qquad q_\lambda (0)=0.
\end{equation}
\tag{6}
$$
For any potential $q_\lambda \in V_{\rho_\lambda}[0,\pi]$, as $\lambda \to +\infty$, the zeros of the solution of the Cauchy problem
$$
\begin{equation}
y''+\bigl(\lambda -q_\lambda (x)\bigr)y=0,\qquad y(0,\lambda)=1,\qquad y'(0,\lambda)=h(\lambda),
\end{equation}
\tag{7}
$$
or, under the additional condition
$$
\begin{equation}
V_0^\pi [q_\lambda ]\leqslant \rho_\lambda, \qquad \rho_\lambda =o\biggl(\frac{\sqrt \lambda}{\ln\lambda}\biggr)\quad\text{as }\ \lambda \to \infty,\qquad q_\lambda (0)=0, \qquad h(\lambda)\ne 0,
\end{equation}
\tag{8}
$$
of the Cauchy problem
$$
\begin{equation}
y''+\bigl(\lambda -q_\lambda (x)\bigr)y=0,\qquad y(0,\lambda)=0,\qquad y'(0,\lambda)=h(\lambda),
\end{equation}
\tag{9}
$$
falling into $[0,\pi]$ and arranged in ascending order will be denoted by
$$
\begin{equation}
0\leqslant x_{0,\lambda}< x_{1,\lambda}< \dots <x_{n(\lambda),\lambda}\leqslant\pi\quad (x_{-1,\lambda}<0,\,\, x_{n(\lambda)+1,\lambda}>\pi).
\end{equation}
\tag{10}
$$
Here, $x_{-1,\lambda}<0$, $x_{n(\lambda)+1,\lambda}>\pi$, are the zeros of the continuation of the solution of the Cauchy problem (7) or (9), where the function $q_\lambda $ is defined in some way outside the interval $[0,\pi]$ with preservation of the boundedness of the variation. Under condition (6) or (8), it follows from the oscillation theorem, or the contour integration method, that the number of zeros (10) increases unboundedly, that is, $n(\lambda)\to +\infty$ as $\lambda \to +\infty$. In what follows, we set, for brevity, $n=n(\lambda)$. Consider the functions
$$
\begin{equation}
s_{k,\lambda}(x)=\frac{y(x,\lambda)}{y'(x_{k,\lambda},\lambda)(x-x_{k,\lambda})},
\end{equation}
\tag{11}
$$
where $y(x,\lambda)$ are solutions to the Cauchy problems (7) or (9) with the zeros $x_{k,\lambda}$ arranged as in (10). Definition 1. On the space of functions $f$ continuous on $[0,\pi] $, we define the operators Note that the values of the operators in Definition 1 depend only on those of the function $f(x_{k,\lambda})$ at the zeros $x_{k,\lambda}$ of the functions $y(x,\lambda)$. Along with the theory of S. L. Sobolev, whose foundations were laid in the famous monograph [28] (see also [1]–[4]), and in which the convergence of test functions in the metric of Lebesgue space is used to introduce the concept of a generalized function, a different approach to construction of the theory of generalized functions has been used for a long time, in which the uniform convergence of test functions was employed to produce classes of equivalent sequences associated with a generalized function (see, for example, Ch. 1, § 9 in [3]). This approach is called sequential in the book [29]. We will use a special case of definition of a generalized function in the sense of the sequential approach (see § 1.3 in [29]). Definition 2. The class of equivalent sequences of functions continuously differentiable on a compact subset $K$, that is, the class of functions converging to the same continuous function with respect to the Chebyshev norm $\|f\|_{C(K)}=\max_{x\in K}|f(x)|$, will be called a generalized function defined on the compact set $K$. Remark 1. It is clear that a continuously differentiable function $f$ is equivalent to a generalized function which represents the class of equivalent function sequences converging to this function, since the stationary sequence $f_n\equiv f$ belongs to this class. By a solution of the mixed boundary value problem (1)–(5) we will mean a generalized function in the sense of Definition 2. Replacing $u(x,t)=U(x)V(t)$ in the mixed boundary value problem (1)–(5) and separating the variables in the homogeneous equation corresponding to equation (1), we obtain a system of equations related by the spectral parameter $\widehat\lambda$
$$
\begin{equation}
\begin{gathered} \, U'' + [\widehat\lambda -q(x)]U=0, \\ V'' + \widehat\lambda V=0. \nonumber \end{gathered}
\end{equation}
\tag{16}
$$
Augmenting equation (16) with the boundary conditions for the function $U$ corresponding to the conditions (2), (3), we get the regular Sturm–Liouville problem
$$
\begin{equation}
\begin{gathered} \, U'' + [\widehat\lambda -q(x)]U=0, \nonumber \\ U(0)\cos \alpha +U'(0)\sin \alpha=0, \end{gathered}
\end{equation}
\tag{17}
$$
$$
\begin{equation}
\begin{gathered} \, U(\pi)\cos \beta +U'(\pi)\sin \beta=0. \end{gathered}
\end{equation}
\tag{18}
$$
The properties of such problems are well studied (see, for example, [30]). Let $\widehat{\lambda}_m:=\widehat{\lambda}_m(q,\alpha,\beta)$ and $U_m:=U_m(q,\alpha,\beta,x)$, $m=0,1,\dots$, be, respectively, the eigenvalues and the orthonormal eigenfunctions of the problem (16)–(18). For each $\lambda >0$, the Fourier coefficients for the eigenfunctions of the Sturm–Liouville problem (16)–(18) of functions (11) and the linear function are denoted by
$$
\begin{equation}
\begin{aligned} \, \tau_{k,\lambda,m} &=\langle U_m(q,\alpha,\beta,{\cdot}\,),s_{k,\lambda}\rangle =\int_0^{\pi}U_m(q,\alpha,\beta,\xi)\frac{y(\xi,\lambda)}{y'(x_{k,\lambda},\lambda )(\xi-x_{k,\lambda})}\,d\xi, \\ \tau^{(0)}_m &= \langle U_m(q,\alpha,\beta,{\cdot}\,),1\rangle =\int_0^{\pi}U_m(q,\alpha,\beta,\xi)\,d\xi, \\ \tau^{(1)}_m &=\langle U_m(q,\alpha,\beta,{\cdot}\,),x\rangle =\int_0^{\pi}\xi U_m(q,\alpha,\beta,\xi)\,d\xi. \end{aligned}
\end{equation}
\tag{19}
$$
The set of Fourier coefficients (19), being independent of both the initial conditions (4), (5) and the right-hand side of equation (1), depends only on the parameters of the mixed boundary value problem (viz., on the potential and the boundary conditions), and can be determined in advance for each problem of the form (1)–(5). By
$$
\begin{equation}
\begin{aligned} \, \widehat{AT}_{\lambda,m}[f] &:= \frac{1}{2}\sum_{k=1}^n(\tau_{k-1,\lambda,m} + \tau_{k,\lambda,m})\biggl\{ f(x_{k,\lambda})-\frac{f(\pi)-f(0)}{\pi}x_{k,\lambda}-f(0)\biggr\} \nonumber \\ &\qquad+\frac{f(\pi)-f(0)}{\pi}\tau^{(1)}_m+f(0)\tau^{(0)}_m \end{aligned}
\end{equation}
\tag{20}
$$
we denote the Fourier coefficients of the value of operator (12), for an arbitrary function $f\in C[0,\pi]$. Note that the values of functionals (20) depend only on those of the function $f(x_{k,\lambda})$ at the zeros $x_{k,\lambda}$ of the functions $y(x,\lambda)$. We set
$$
\begin{equation*}
\begin{aligned} \, \nu &=\begin{cases} -e^{-\lambda}\bigl(AT_\lambda (f,0)\cot{\alpha}+AT_\lambda '(f,0)\bigr) &\text{for } \alpha\ne \pi m_1,\, m_1\in \mathbb{Z}, \\ AT_\lambda (f,0) &\text{for } \alpha=\pi m_1,\ m_1\in \mathbb{Z}, \end{cases} \nonumber \\ \widetilde{\nu} &= \begin{cases} -e^{-\lambda}\bigl(AT_\lambda (f,\pi)\cot{\beta}+AT_\lambda '(f,\pi)\bigr) &\text{for } \beta\ne \pi m_2,\, m_2\in \mathbb{Z}, \\ AT_\lambda(f,\pi) &\text{for } \beta=\pi m_2,\, m_2\in \mathbb{Z}, \end{cases} \nonumber \\ \mu &= \frac{\sqrt{3}}{2}\, e^{\lambda}, \end{aligned}
\end{equation*}
\notag
$$
$$
\begin{equation}
\eta(x,\lambda) =\begin{cases} 2\sqrt{\frac{1}{3}}\,\nu \mu x \quad\text{for } x\in\bigl[0,\frac{1}{|\mu|}\bigl(\frac{\sqrt{2}}{3}\bigr)\bigr],\, \alpha\ne \pi m_1,\, m_1\in \mathbb{Z}, \\ \nu\sin^3\Bigl(\mu \Bigl(x+\frac{1}{|\mu|}\Bigl(\arcsin\sqrt{\frac{2}{3}} -\frac{\sqrt{2}}{3}\Bigr)\Bigr)\Bigr) \\ \hphantom{0 \quad} \text{for } x\in\Bigl[\frac{1}{|\mu|}\bigl(\frac{\sqrt{2}}{3}\bigr), \frac{\pi}{|\mu|}-\frac{1}{|\mu|}\Bigl(\arcsin{\sqrt{\frac{2}{3}}} -\frac{\sqrt{2}}{3}\Bigr)\Bigr],\, \alpha\,{\ne}\, \pi m_1,\, m_1{\in}\, \mathbb{Z}, \\ \nu\Bigl(\frac{\pi}{|\mu|}-\frac{1}{|\mu|}\Bigl(\arcsin{\sqrt{\frac{2}{3}}} \,{-}\frac{\sqrt{2}}{3}\Bigr)\Bigr)^{-3}\Bigl(x\,{-}\,\frac{\pi}{|\mu|}+\frac{1}{|\mu|} \Bigl(\arcsin{\sqrt{\frac{2}{3}}}\,{-}\frac{\sqrt{2}}{3}\Bigr)\Bigr)^3 \\ \hphantom{0 \quad} \text{for } x\in\Bigl[0,\frac{\pi}{|\mu|}-\frac{1}{|\mu|} \Bigl(\arcsin{\sqrt{\frac{2}{3}}}-\frac{\sqrt{2}}{3}\Bigr)\Bigr], \, \alpha=\pi m_1,\, m_1\in \mathbb{Z}, \\ 0 \quad\text{for } x\in\Bigl[\frac{\pi}{|\mu|}-\frac{1}{|\mu|} \Bigl(\arcsin{\sqrt{\frac{2}{3}}}-\frac{\sqrt{2}}{3}\Bigr),\pi\Bigr], \end{cases}
\end{equation}
\tag{21}
$$
$$
\begin{equation}
\begin{aligned} \, \widetilde{\eta}(x,\lambda_{n}) &=-\begin{cases} 2\sqrt{\frac{1}{3}}\,\widetilde{\nu_{n}}\mu_{n}(\pi-x)\quad \text{for } x\in\bigl[\pi -\frac{1}{\mu_{n}}\bigl(\frac{\sqrt{2}}{3}\bigr),\pi\bigr],\, \beta\ne \pi m_2,\, m_2{\in}\, \mathbb{Z}, \\ \widetilde{\nu_{n}}\sin^3\Bigl(\mu_{n} \Bigl(\pi-x+\frac{1}{\mu_{n}} \Bigl(\arcsin{\sqrt{\frac{2}{3}}}-\frac{\sqrt{2}}{3}\Bigr)\Bigr)\Bigr) \\ \hphantom{0 \quad} \text{for } x\in\Bigl[\pi-\frac{\pi}{\mu_{n}}+\frac{1}{\mu_{n}}\Bigl(\arcsin{\sqrt{\frac{2}{3}}} -\frac{\sqrt{2}}{3}\Bigr),\pi-\frac{1}{\mu_{n}}\bigl(\frac{\sqrt{2}}{3}\bigr)\Bigr], \\ \qquad \beta\ne \pi m_2,\, m_2\in \mathbb{Z}, \\ \widetilde{\nu_{n}}\Bigl(\frac{\pi}{\mu_{n}}-\frac{1}{\mu_{n}} \Bigl(\arcsin{\sqrt{\frac{2}{3}}}-\frac{\sqrt{2}}{3}\Bigr)\Bigr)^{-3} \\ \qquad\times\Bigl(\pi-\frac{\pi}{\mu_{n}}+\frac{1}{\mu_{n}} \Bigl(\arcsin{\sqrt{\frac{2}{3}}}-\frac{\sqrt{2}}{3}\Bigr)-x\Bigr)^3 \\ \hphantom{0 \quad} \text{for } x\in\Bigl[\pi-\frac{\pi}{\mu_{n}}+\frac{1}{\mu_{n}} \Bigl(\arcsin{\sqrt{\frac{2}{3}}}-\frac{\sqrt{2}}{3}\Bigr),\pi\Bigr],\, \beta=\pi m_2, \, m_2{\in}\, \mathbb{Z}, \\ 0 \quad\text{for } x\in\Bigl[0,\pi-\frac{\pi}{\mu_{n}}+\frac{1}{\mu_{n}} \Bigl(\arcsin{\sqrt{\frac{2}{3}}}-\frac{\sqrt{2}}{3}\Bigr)\Bigr]. \end{cases} \end{aligned}
\end{equation}
\tag{22}
$$
It is worth pointing out that the values of operators (21), (22) depend only on those of the function $f(x_{k,\lambda})$ at the zeros $x_{k,\lambda}$ of the functions $y(x,\lambda)$. We set
$$
\begin{equation}
\begin{aligned} \, \nonumber \sigma_1 &= \begin{cases} 1 &\text{for } (\alpha =\pi m_1,\, m_1\in \mathbb{Z})\wedge f(0)\ne 0, \\ 0 &\text{for } (\alpha \ne \pi m_1,\, m_1\in \mathbb{Z})\vee f(0)=0, \end{cases} \\ \widetilde\sigma_1 &= \begin{cases} 1 &\text{for } (\beta =\pi m_2,\, m_2\in \mathbb{Z})\wedge f(\pi)\ne 0, \\ 0 &\text{for } (\beta \ne \pi m_2,\, m_2\in \mathbb{Z})\vee f(\pi)=0. \end{cases} \end{aligned}
\end{equation}
\tag{23}
$$
For $\lambda >0$, the Fourier coefficients of the eigenfunctions of problem (16)–(18) are denoted by
$$
\begin{equation}
\begin{aligned} \, \widehat{\eta}_{\lambda,m} &=\langle U_m(q,\alpha,\beta,{\cdot}\,),\eta(\,{\cdot}\,,\lambda)\rangle =\int_0^{\pi}U_m(q,\alpha,\beta,\xi)\eta(\xi,\lambda)\,d\xi, \nonumber \\ \widehat{\widetilde\eta}_{\lambda,m} &=\langle U_m(q,\alpha,\beta,{\cdot}\,),\widetilde\eta(\,{\cdot}\,,\lambda)\rangle =\int_0^{\pi}U_m(q,\alpha,\beta,\xi)\widetilde\eta(\xi,\lambda)\,d\xi. \end{aligned}
\end{equation}
\tag{24}
$$
Next, let
$$
\begin{equation}
\mathbb{AT}_{\lambda,j} (f,x)=\sum_{m=0}^{j} \widehat{AT}_{\lambda,m}[f,\eta] U_m(q,\alpha,\beta,x)
\end{equation}
\tag{25}
$$
by the operators associating with a function $f\in C[0,\pi]$ the partial Fourier sums of the function $AT_\lambda (f,x)+\eta (x,\lambda)+\widetilde\eta (x,\lambda)$; here,
$$
\begin{equation}
\begin{aligned} \, \widehat{AT}_{\lambda,m}[f,\eta] &:= \frac{1}{2}\sum_{k=1}^n(\tau_{k-1,\lambda,m} + \tau_{k,\lambda,m})\biggl\{ f(x_{k,\lambda})-\frac{f(\pi)-f(0)}{\pi}x_{k,\lambda}-f(0)\biggr\} \nonumber \\ &\qquad+\frac{f(\pi)-f(0)}{\pi}\tau^{ (1)}_m+f(0)\tau^{(0)}_m +\widehat{\eta}_{\lambda,m} +\widehat{\widetilde\eta}_{\lambda,m} \end{aligned}
\end{equation}
\tag{26}
$$
are the Fourier coefficients. The function $AT_\lambda (f,x)+\eta (x,\lambda)+\widetilde\eta (x,\lambda)$ has absolutely continuous derivative on $[0,\pi]$ and satisfies the boundary conditions (17), (18). Here it is also worth pointing out that the values of operators (25), as well as of functionals (26), depend only on those of the function $f(x_{k,\lambda})$ at the zeros $x_{k,\lambda}$ of the functions $y(x,\lambda)$. The main result of the present paper is as follows. Theorem 1. Let $T>0$, $\varepsilon>0$, let $f$, $\varphi$, $\psi$ be continuous functions (each function is continuous on its own domain), let $q$ have bounded variation, and let $j(\lambda)$ be a function assuming integer or infinite values such that The method proposed in Theorem 1, in comparison with the theory of difference schemes, retains the advantages of the Fourier analysis like filtration and noise removal, etc.. Unlike the classical Fourier and Krylov’s methods, the algorithm of Theorem 1 for the solution of the mixed boundary value problem (1)–(5) is not only capable of expanding the class of admissible sets of functions $f$, $\varphi$, $\psi$ to the spaces of continuous functions on the domains of their definition, but also allows one to use data on these functions only at the nodal points $x_{k,\lambda}$, rather than on a set of full measure. Due to Hadamard well-posedness of the mixed boundary value problem for a hyperbolic equation, the classical solution of problem (1)–(5) with sufficiently smooth conditions, under which such a solution exists, coincides with that obtained by Theorem 1. Remark 2. The coefficients $\widehat{AT}_{\lambda,m}[\,{\cdot}\,,\eta]$ can be evaluated from the set of Fourier coefficients $\tau_{k,\lambda,m}$ of the functions $s_{k,\lambda}$ (see (19)), which is independent of the functions $f$, $\varphi$, $\psi$, and which can be concocted in advance for each boundary value problem (1)–(5). This also applies to the classical solution obtained by the new method. In order not to be bothered with evaluation of integrals of rapidly oscillating functions, we will use the following result for evaluation of $\tau_{k,\lambda,m}$. Proposition 1. Assume that $q_{\lambda}\equiv q$ in the differential equations of the Cauchy problems (7), (9) and in the equation of the Sturm–Liouville problem (16). Then the Fourier coefficients (19) of the functions $s_{k,\lambda}$ (see (11)) with respect to the orthonormal eigenfunctions $U_m$ of the problem (16)–(18) can be evaluated via the Riemann–Stieltjes integral If the potential of the mixed boundary value problem (1)–(5) is continuously differentiable, then the Fourier coefficients of the functions $s_{k,\lambda}$ with respect to the orthonormal eigenfunctions $U_m$ of the Sturm–Liouville problem (16)–(18) can be found using the resolvent of a single differential operator. Proposition 2. If, in the differential equations of the Cauchy problems (7), (9) and in the equation of the Sturm–Liouville problem (16), the potentials are continuously differentiable functions satisfying $q_{\lambda}(x)\equiv -q(x)$, then the Fourier coefficients (19) of the functions $s_{k,\lambda}$ with respect to the orthonormal eigenfunctions $U_m$ of the problem (16)–(18) can be evaluated using the resolvent of the Cauchy differential operator The machinery of numerical solutions of differential equations produces more accurate results than quadrature formulas due to the fact that the errors in the representation of the function by quadrature formulas accumulate at each node of the quadrature formula. The approximation properties of a quadrature formula decrease with increasing number of nodes, At the same time, difference schemes used for representation of a differential operator approximate uniformly the exact solution of the differential equation [22] with increased number of nodes. § 2. Auxiliary resultsBefore proceeding with the proof of Theorem 1 and Propisitions 1 and 2, we will establish some auxiliary results. 2.1. Asymptotic formulas Proposition 3 (see [31], Theorem 1, and [32], Proposition 2). Let $ \rho_\lambda \geqslant0$, $\rho_\lambda =o(\sqrt \lambda\,)$ as $\lambda \to \infty $, and let $V_{\rho_\lambda}[0,\pi]$ be the ball of radius $\rho_\lambda $ in the space of functions of bounded variation vanishing at zero, that is, for any real $\lambda $, Proposition 4 (see [31], Theorem $1'$, and [32], Proposition 3). Let $ \rho_\lambda \geqslant0$, $\rho_\lambda =o(\sqrt \lambda\,)$ as $\lambda \to \infty$, and let $V_{\rho_\lambda}[0,\pi]$ be the ball of radius $\rho_\lambda $ in the space of functions of bounded variation vanishing at zero. Then there exists $\lambda_1 > 4\pi^2 \rho_\lambda^2$ such that, for all $\lambda \geqslant\lambda_1$, for any potential $q_\lambda \in V_{\rho_\lambda}[0,\pi]$, and an arbitrary $x\in [0,\pi]$, the solution of the Cauchy problem (9) satisfies Proposition 5 (see [31], Theorems 2, $2'$, and [32], Proposition 4). Let condition (6) be met. Then, for any potential $q_\lambda \in V_{\rho_\lambda}[0,\pi]$ as $\lambda \to \infty $, the zeros of solutions to the Cauchy problem (7), from $[0,\pi]$ and labelled in ascending order according to (10) satisfy The zeros of the solutions of the Cauchy problem (9) labelled according to (10) with $h(\lambda)\ne 0$, and $q_\lambda$ as in (8) behave asymptotically as
$$
\begin{equation*}
\begin{alignedat}{2} x_{k,\lambda} &=\frac{k}{\sqrt{\lambda}}\, \pi + o\biggl(\frac{\lambda^{-1/2}}{\ln \lambda}\biggr) &\quad &\textit{as }\ \lambda \to \infty, \\ y'(x_{k,\lambda},\lambda) &=h(\lambda)\biggl((-1)^{k} + o\biggl(\frac{1}{\ln \lambda}\biggr)\biggr) &\quad &\textit{as }\ \lambda \to \infty. \end{alignedat}
\end{equation*}
\notag
$$
The convergence to zero in the “little oh” symbol is uniform with respect to $q_\lambda \in V_{\rho_\lambda}[0,\pi]$ and $k$, $ 0\leqslant k \leqslant n$. In order to restore the conditions of problems (7), (9) from the properties of the zeros (10), one can use the results of [33], [34]. We will need another property of the fundamental functions $s_{k,\lambda}$. Lemma 1 (see [32], Lemma 2). Let $\rho_\lambda \geqslant 0$, $\rho_\lambda =o(\sqrt \lambda/\ln \lambda)$ as $\lambda \to \infty$, and $V_{\rho_\lambda}[0,\pi]$ be the ball of radius $\rho_\lambda $ (for the Cauchy problem (9), we also assume that $h(\lambda)\ne 0$). Then there is $\lambda_0$ (depending only on the rate of variation of the radii of the balls $\rho_\lambda$ in (6), or (8)) that, for any potentials $q_\lambda \in V_{\rho_\lambda}[0,\pi]$, for any function $h(\lambda)$, and for all $\lambda >\lambda_0$, the functions $s_{k,\lambda}(x)$ constructed from solutions the of the Cauchy problem (7) and (9) are estimated as 2.2. Differential equation for the functions $s_{k,\lambda}$Proposition 6. Let the function $q_{\lambda}$ have bounded variation on $[0,\pi]$. Then $y(x,\lambda)$ satisfies the differential equation of problems (7), (9), and $y(x_{k,\lambda},\lambda)=0$ if and only if $s_{k,\lambda}$ is a bounded solution of the differential equation Proof. Indeed, let $y(x,\lambda)$ satisfy the differential equation of problems (7), (9). Substituting, we obtain
The other way around. Let $s_{k,\lambda}$ be a bounded solution of the differential equation (34) everywhere on the set $[0,x_{k,\lambda})\cup(x_{k,\lambda},\pi]$. Substring, we verify that the function $y(x,\lambda)={y'(x_{k,\lambda},\lambda)(x-x_{k,\lambda})}s_{k,\lambda}(x)$ satisfies the equation of problems (7), (9). Since $y(x_{k,\lambda})=0$, this relation can be continued to the entire interval $[0,\pi]$. Proposition 6 is proved. 2.3. Some useful operators of the theory of approximation of functionsProposition 7 (see [32], Proposition 1). Let $f\in C[0,\pi]$, and let the functions $q_\lambda $ and $h(\lambda)$ satisfy condition (6) (respectively, (8)) in the case of the Cauchy problem (7) (problem (9)). Then uniformly with respect to $x\in [0,\pi]$ and $q_\lambda \in V_{\rho_\lambda}[0,\pi]$ Proposition 8. Let $ \rho_\lambda \geqslant 0$, $\rho_\lambda =o(\sqrt \lambda/\ln\lambda)$ as $\lambda \to \infty$, and let $V_{\rho_\lambda}[0,\pi]$ be the ball of radius $\rho_\lambda $ in the space of functions of bounded variation which vanish at zero (in the case of the Cauchy problem (9) we also require that $h(\lambda)\ne 0$). Then there exists $\lambda_0$ (depending only on the rate of variation of the radii of the balls $\rho_\lambda $ in (6), or (8)) such that, for any potential $q_\lambda \in V_{\rho_\lambda}[0,\pi]$, any $h(\lambda)$, and any $\lambda >\lambda_0$, the norms of operators (14) and (15) acting from $M[0,\pi]$ into $C[0,\pi]$ and constructed from solutions of the Cauchy problems (7) or (9) are estimated from above as follows: Proof. First, to verify estimate (36) for operator (14), we will proceed as in the case of the Cauchy problem (9). In view of the invariance of operator (13) with respect to multiplication of the function $y(x,\lambda)$ by a non-zero constant, we can assume without loss of generality that $h(\lambda)\equiv 1$.
Consider an arbitrary $x\in [0,\pi]$. Let $k_0$ be the number of the node closest to $x$ (if there are two such nodes, we choose the number of any of them). From the asymptotics of the zeros of the solutions to the Cauchy problem in Proposition 5, we get the estimate
$$
\begin{equation}
|x-x_{k_0,\lambda}|=O\biggl(\frac{\pi}{\sqrt{\lambda}}\biggr).
\end{equation}
\tag{38}
$$
Then the norm of the functional (we will assume that $AT_\lambda$ is represented in the form (13)) which associates with each bounded function $f\in M[0,\pi]$ on $[0,\pi]$ the value of the derivative of the result of the application of the operator (13) at a point $x\in [0,\pi]$ is estimated as follows:
$$
\begin{equation}
\begin{aligned} \, AT_\lambda^{(1)}(x) &\leqslant 2\sum_{k=0}^n|s_{k,\lambda}'(x)|+\frac{2}{\pi} \nonumber \\ &=2\sum_{k=0}^{k_0-1}|s_{k,\lambda}'(x)| +2|s_{k_0,\lambda}'(x)| +2\sum_{k=k_0+1}^n|s_{k,\lambda}'(x)| +\frac{2}{\pi}. \end{aligned}
\end{equation}
\tag{39}
$$
The norm of the functional which associates with each bounded function $f\in M[0,\pi]$ on $[0,\pi]$ the value of the continuous derivative of the value of operator (13), which acts from $M[0,\pi]$ to $C[0,\pi]$ has the form
$$
\begin{equation}
AT_\lambda^{(1)} = \max_{x\in[0,\pi]}AT_\lambda^{(1)}(x).
\end{equation}
\tag{40}
$$
The second term in (39) is estimated using the Lagrange formula, (38), and the asymptotic formulas in Proposition 5
$$
\begin{equation*}
2|s_{k_0,\lambda}'(x)|=2\biggl| \frac{|y'(x,\lambda)(x-x_{k_0,\lambda})-y(x,\lambda)|} {y'(x_{k_0,\lambda},\lambda)(x-x_{k_0,\lambda})^2}\biggr| =o\biggl(\frac{1}{\sqrt{\lambda}}\biggr).
\end{equation*}
\notag
$$
Now from (39) we get the estimate
$$
\begin{equation}
\begin{aligned} \, AT_\lambda^{(1)}(x) &\leqslant 2\sum_{k=0}^{k_0-1}\biggl| \frac{|y'(x,\lambda)|} {y'(x_{k,\lambda},\lambda)(x-x_{k,\lambda})}\biggr|+2\sum_{k=k_0+1}^n\biggl|\frac{|y'(x,\lambda)|} {y'(x_{k,\lambda},\lambda)(x-x_{k,\lambda})}\biggr| \nonumber \\ &\qquad+2\sum_{k=0}^{k_0-1}\biggl| \frac{|y(x,\lambda)|} {y'(x_{k,\lambda},\lambda)(x-x_{k,\lambda})^2}\biggr| +2\sum_{k=k_0+1}^n\biggl| \frac{|y(x,\lambda)|} {y'(x_{k,\lambda},\lambda)(x-x_{k,\lambda})^2}\biggr| \nonumber \\ &\qquad+\frac{2}{\pi}+o\biggl(\frac{1}{\sqrt{\lambda}}\biggr). \end{aligned}
\end{equation}
\tag{41}
$$
From the asymptotic formulas for the zeros of the solution of Cauchy problems in Proposition 5, we have, for sufficiently large $\lambda$,
$$
\begin{equation}
\min_{1\leqslant k \leqslant n}|x_{k,\lambda}-x_{k-1,\lambda}|\geqslant \frac{\pi}{2\sqrt{\lambda}}, \qquad \min(|x-x_{k_0-1,\lambda}|,\,|x-x_{k_0+1,\lambda}|)\geqslant \frac{\pi}{8\sqrt{\lambda}}.
\end{equation}
\tag{42}
$$
By (41), (42) and the asymptotic formulas of Proposition 5, there exists $\lambda_1$ (depending on the rate of variation of the radii of the balls in (6), (8)) such that, for all $\lambda>\lambda_1$,
$$
\begin{equation*}
\begin{aligned} \, T_\lambda^{(1)}(x) &\leqslant 2|y'(x,\lambda)| \mathop{{\sum}'}^n_{k=0}\biggl| \frac{1}{((-1)^k +o(1/\ln\lambda))(x-x_{k,\lambda})} \biggr| \\ &\qquad+2|y(x,\lambda)| \mathop{{\sum}'}^n_{k=0} \biggl| \frac{1}{((-1)^k + o(1/\ln\lambda))(x-x_{k,\lambda})^2} \biggr| +\frac{2}{\pi} +o\biggl(\frac{1}{\sqrt{\lambda}}\biggr). \end{aligned}
\end{equation*}
\notag
$$
Here and in what follows, a prime on the summation symbol means that the sum is not taken for $k\,{=}\,k_0$. If $k_0=0$, then the first term is missing in the sum, and if $k_0=n$, then there is no third term. Hence we have the estimate
$$
\begin{equation}
\begin{aligned} \, &AT_\lambda^{(1)}(x)\leqslant 2 |y'(x,\lambda)|\biggl(1 \,{+}\, \biggl|o\biggl(\frac{1}{\ln{\lambda}}\biggr)\biggr|\biggr) \frac{8\sqrt\lambda}{\pi} \biggl[\int_0^{x-\pi/(8\sqrt\lambda)}\!\!\frac{dt}{x-t} \,{+}\int_{x+\pi/(8\sqrt\lambda)}^\pi \frac{dt}{t\,{-}\,x} \biggr] \nonumber \\ &\quad+2 |y(x,\lambda)|\biggl(1 +\biggl|o\biggl(\frac{1}{\ln{\lambda}}\biggr)\biggr|\Biggr) \frac{8\sqrt\lambda}{\pi} \biggl[\int_0^{x-\pi/(8\sqrt\lambda)} \frac{dt}{(x-t)^2} +\int_{x+\pi/(8\sqrt\lambda)}^\pi \frac{dt}{(t-x)^2} \biggr] \nonumber \\ &\quad+\frac{2}{\pi}+o\biggl(\frac{1}{\sqrt{\lambda}}\biggr). \end{aligned}
\end{equation}
\tag{43}
$$
Since $x(\pi-x)\geqslant 0$ for $x\in[0,\pi]$ and since, for $x\in [0,\pi/(4\sqrt\lambda)]$, by the choice of $k_0$, the first integral in the estimate thus obtained disappears, and further, since there is no second integral for $x\in [\pi-\pi/(4\sqrt\lambda),\pi]$, the sum of the integrals can be estimated by considering two cases:
$$
\begin{equation*}
\begin{aligned} \, &\int_0^{x-\pi/(8\sqrt\lambda)}\frac{dt}{x-t}+\int_{x+\pi/(8\sqrt\lambda)}^\pi \frac{dt}{t-x} \\ &\qquad=-\ln(x-t)\big|_0^{x-\pi/(8\sqrt\lambda)}+\ln(t-x)\big|_{x+\pi/(8\sqrt\lambda)}^\pi \\ &\qquad\leqslant\begin{cases} \ln(\lambda)+\ln{16} &\text{for } x\in \biggl[\dfrac{\pi}{4\sqrt\lambda},\, \pi-\dfrac{\pi}{4\sqrt\lambda}\biggr], \\ \dfrac{1}{2}\ln(\lambda)+\ln{8} &\text{for } x\in \biggl[0,\dfrac{\pi}{4\sqrt\lambda}\biggr] \cup \biggl[\pi-\dfrac{\pi}{4\sqrt\lambda},\pi\biggr]. \end{cases} \end{aligned}
\end{equation*}
\notag
$$
Now let us estimate the sum of the integrals
$$
\begin{equation*}
\int_0^{x-\pi/(8\sqrt\lambda)}\frac{dt}{(x-t)^2} +\int_{x+\pi/(8\sqrt\lambda)}^\pi \frac{dt}{(t-x)^2} =\frac{16\sqrt{\lambda}}{\pi}-\frac{\pi}{x(\pi-x)}.
\end{equation*}
\notag
$$
Since $x(\pi-x)\geqslant 0$ for $x\in[0,\pi]$, since $x(\pi-x)> 0$ for $x\in[\pi/(4\sqrt\lambda),\pi-\pi/(4\sqrt\lambda),\pi]$, since, for $x\in [0,\pi/4\sqrt\lambda)]$, the first integral in the resulting estimate disappears by the choice of $k_0$, and since, for $x\in [\pi-\pi/(4\sqrt\lambda),\pi]$, there is no second integral, we have the following three cases to estimate the sum of the integrals:
$$
\begin{equation*}
\begin{aligned} \, &\int_0^{x-\pi/(8\sqrt\lambda)}\frac{dt}{(x-t)^2}+\int_{x+\pi/(8\sqrt\lambda)}^\pi \frac{dt}{(t-x)^2} \\ &\qquad\leqslant \begin{cases} \dfrac{16\sqrt{\lambda}}{\pi}-\dfrac{\pi}{x(\pi-x)} &\text{for } x\in \biggl[\dfrac{\pi}{4\sqrt\lambda},\,\pi-\dfrac{\pi}{4\sqrt\lambda}\biggr], \\ \dfrac{8\sqrt{\lambda}}{\pi}-\dfrac{1}{\pi-x} &\text{for } x\in \biggl[0,\dfrac{\pi}{4\sqrt\lambda}\biggr], \\ \dfrac{8\sqrt{\lambda}}{\pi}-\dfrac{1}{x} &\text{for } x\in \biggl[\pi-\dfrac{\pi}{4\sqrt\lambda},\pi\biggr]. \end{cases} \end{aligned}
\end{equation*}
\notag
$$
Now by (43) the norm of operator (14) is estimated uniformly with respect to $x\in [0,\pi]$ as follows:
$$
\begin{equation*}
AT_\lambda^{(1)}\leqslant \frac{16\sqrt\lambda}{\pi}\biggl[\ln(\lambda)+\frac{16+\pi\ln{16}}{\pi} \biggr] +o\biggl(\frac{\sqrt{\lambda}}{\ln{\lambda}}\biggr).
\end{equation*}
\notag
$$
Hence there exists a sufficiently large $\lambda_0\geqslant \lambda_1$ such that, for all $\lambda > \lambda_0$, estimate (36) holds in the case of the Cauchy problem (9).
We claim that, for sufficiently large $\lambda$, inequality (36) holds for $s_{k,\lambda}$, $0\leqslant k\leqslant n$ constructed from the solutions of the Cauchy problem (7). To this end, we continue the function
$$
\begin{equation}
q_\lambda (x)= \begin{cases} q_\lambda (x) &\text{for } x\in [0,\pi], \\ 0 &\text{for } x \notin [0,\pi]. \end{cases}
\end{equation}
\tag{44}
$$
Next, we replace the independent variable
$$
\begin{equation}
t=\frac{\pi\bigl(x\sqrt\lambda +\arcsin\sqrt{\lambda/(\lambda+h^2(\lambda))}\,\bigr)} {\pi\sqrt\lambda+\arcsin\sqrt{\lambda/(\lambda+h^2(\lambda))}}.
\end{equation}
\tag{45}
$$
We set
$$
\begin{equation}
\widehat y(t,\widehat \lambda)= y\Biggl(\frac{\pi\sqrt\lambda +\arcsin\sqrt{\lambda/(\lambda+h^2(\lambda))}}{\pi \sqrt\lambda} t-\frac{1}{\sqrt\lambda}\arcsin\sqrt{\frac{\lambda}{\lambda+h^2(\lambda)}},\lambda\Biggr)
\end{equation}
\tag{46}
$$
and define
$$
\begin{equation}
\begin{aligned} \, \widehat q_{\widehat\lambda}(t) &=\Biggl(1+\frac{1}{\pi\sqrt\lambda} \arcsin\sqrt{\frac{\lambda}{\lambda+h^2(\lambda)}}\,\Biggr)^2 \nonumber \\ &\qquad\times q_\lambda \Biggl(\frac{\pi\sqrt\lambda +\arcsin\sqrt{\lambda/(\lambda+h^2(\lambda))}}{\pi \sqrt\lambda} t -\frac{1}{\sqrt\lambda}\arcsin\sqrt{\frac{\lambda}{\lambda+h^2(\lambda)}}\,\Biggr), \end{aligned}
\end{equation}
\tag{47}
$$
here
$$
\begin{equation*}
\widehat\lambda=\Biggl (1+\frac{1}{\pi\sqrt\lambda} \arcsin\sqrt{\frac{\lambda}{\lambda+h^2(\lambda)}}\,\Biggr)^2 \lambda.
\end{equation*}
\notag
$$
Using (45) and employing Picard’s theorem we find that functions (46) solve simultaneously the Cauchy problems
$$
\begin{equation}
\begin{gathered} \, \begin{gathered} \, \widehat y''+\bigl(\widehat\lambda -\widehat q_{\widehat\lambda}(t)\bigr)\widehat y=0, \qquad \widehat y\bigl(t(0),\widehat\lambda\bigr)=y(0,\lambda)=1, \\ \widehat y'\bigl(t(0),\widehat\lambda\bigr) =\Biggl(1+\frac{1}{\pi\sqrt\lambda} \arcsin\sqrt{\frac{\lambda}{\lambda+h^2(\lambda)}}\,\Biggr)h(\lambda) \end{gathered} \end{gathered}
\end{equation}
\tag{48}
$$
and
$$
\begin{equation}
\begin{gathered} \, \begin{gathered} \, \widehat y''+\bigl(\widehat\lambda -\widehat q_{\widehat\lambda}(t)\bigl)\widehat y=0,\qquad \widehat y(0,\widehat\lambda) =y\Biggl(-\frac{1}{\sqrt\lambda}\arcsin\sqrt{\frac{\lambda}{\lambda+h^2(\lambda)}}\, \Biggr)=0, \\ \widehat y'(0,\widehat\lambda)=\sqrt{\lambda+h^2(\lambda)} \Biggl(1+\frac{1}{\pi\sqrt\lambda}\arcsin\sqrt{\frac{\lambda}{\lambda+h^2(\lambda)}}\, \Biggr)=\widehat h (\widehat\lambda). \end{gathered} \end{gathered}
\end{equation}
\tag{49}
$$
Next, by (44) and (47) we have $\sqrt{\widehat\lambda}-1/2\leqslant\sqrt\lambda\leqslant\sqrt{\widehat\lambda}+1/2$, that is $\sqrt{\widehat\lambda}\simeq \sqrt\lambda$. Hence, relation (6) for problem (49) is preserved, since from (44) and (47) we have
$$
\begin{equation}
q_{\widehat\lambda}(0)=0 \quad\text{and}\quad V_0^\pi[\widehat q_{\widehat\lambda}] \leqslant \biggl(1+\frac{1}{2\sqrt\lambda}\biggr)^2 V_0^\pi[q_{\lambda}] = o\biggl(\frac{\sqrt{\lambda}}{\ln{\lambda}}\biggr) = o\Biggl(\frac{\sqrt{\widehat\lambda}}{\ln{\widehat\lambda}}\Biggr).
\end{equation}
\tag{50}
$$
In view of(46), for $t\in[0,\pi]$ and $x\in\bigl[-(1/\sqrt\lambda)\arcsin\sqrt{\lambda/(\lambda+h^2(\lambda))},\pi\bigr]$,
$$
\begin{equation}
s_{k,\lambda}(x)\equiv \frac{y(x,\lambda)}{y'(x_{k,\lambda},\lambda)(x-x_{k,\lambda})} \equiv \frac{\widehat y(t,\widehat\lambda)}{\widehat y'(t_{k,\widehat\lambda},\widehat\lambda) (t-t_{k,\widehat\lambda})} \equiv\widehat s_{k,\widehat\lambda}(t).
\end{equation}
\tag{51}
$$
This completes the proof of (36) since $\widehat s_{k,\widehat \lambda}(t)$ are constructed using the Cauchy problem (49) of the form (9). Let us now prove estimate (37) for the norm of operator (15). Again, we will first consider the case of the Cauchy problem (9). The operator (13) is invariant under multiplication of the function $y(x,\lambda)$ by a non-zero constant, and so we can assume without loss of generality that $h(\lambda)\equiv 1$. Consider an arbitrary $x\in [0,\pi]$. Let $k_0$ be the number of a node closest to $x$ (if there are two such nodes, then we choose the number of any such a node). From the asymptotics of the zeros of solutions to the Cauchy problem in Proposition 5, we get estimate (38). Now the norm of the functional (we represent $\widetilde{AT}_\lambda$ in the form (13)) which associates with each bounded function $f\in M[0,\pi]$ on $[0,\pi]$ the value of the second derivative of the result of the application of operator (13) at $x\in [0,\pi]$ can be estimated as
$$
\begin{equation}
\begin{aligned} \, AT_\lambda^{(2)}(x) &\leqslant 2\sum_{k=0}^n|s_{k,\lambda}''(x)| \nonumber \\ &=2\sum_{k=0}^{k_0-1}|s_{k,\lambda}''(x)| +2|s_{k_0,\lambda}''(x)| +2\sum_{k=k_0+1}^n|s_{k,\lambda}''(x)|. \end{aligned}
\end{equation}
\tag{52}
$$
Next, the norm of operator (15), which associates with each bounded function $f\in M[0,\pi]$ on $[0,\pi]$ the continuous second derivative of the value of operator (13) acting from $M[0,\pi]$ into $C[0,\pi]$ is evaluated as
$$
\begin{equation}
AT_\lambda^{(2)} = \max_{x\in[0,\pi]}AT_\lambda^{(2)}(x).
\end{equation}
\tag{53}
$$
The second term in (52) is estimated via the Lagrange formula, Lemma 1, and the asymptotic formulas in Proposition 5. Hence there exist constants $\lambda_1>0$ and $C_1$ such that, for all $\lambda>\lambda_1$ and $x\in [0,\pi]$,
Now from (39) we get the estimate
$$
\begin{equation}
\begin{aligned} \, AT_\lambda^{(2)}(x) &\leqslant 2\sum_{k=0}^{k_0-1}\biggl| \frac{|y''(x,\lambda)|} {y'(x_{k,\lambda},\lambda)(x-x_{k,\lambda})}\biggr| +2\sum_{k=k_0+1}^n\biggl| \frac{|y''(x,\lambda)|}{y'(x_{k,\lambda},\lambda)(x-x_{k,\lambda})}\biggr| \nonumber \\ &\qquad+4\sum_{k=0}^{k_0-1}\biggl| \frac{|y'(x,\lambda)|} {y'(x_{k,\lambda},\lambda)(x-x_{k,\lambda})^2}\biggr| +4\sum_{k=k_0+1}^n\biggl| \frac{|y'(x,\lambda)|} {y'(x_{k,\lambda},\lambda)(x-x_{k,\lambda})^2}\biggr| \nonumber \\ &\qquad+2\sum_{k=0}^{k_0-1}\biggl| \frac{|y(x,\lambda)|} {y'(x_{k,\lambda},\lambda)(x-x_{k,\lambda})^3}\biggr| \nonumber \\ &\qquad+2\sum_{k=k_0+1}^n\biggl| \frac{|y(x,\lambda)|} {y'(x_{k,\lambda},\lambda)(x-x_{k,\lambda})^3}\biggr| +O({\sqrt{\lambda}}). \end{aligned}
\end{equation}
\tag{54}
$$
From the asymptotic formulas for the zeros of the solution of Cauchy problems in Proposition 5 it follows that (42) holds for all sufficiently large $\lambda$. By (42), (54), and the asymptotic formulas in Proposition 5, there exists $\lambda_2\geqslant \lambda_1$ (which depends only on the rate of variation of the radii of the balls in (6), (8)) such that, for all $\lambda>\lambda_2$,
$$
\begin{equation*}
\begin{aligned} \, AT_\lambda^{(2)}(x) &\leqslant 2|y''(x,\lambda)| \mathop{{\sum}'}^n_{k=0} \biggl| \frac{1}{((-1)^k +o(1/\ln\lambda))(x-x_{k,\lambda})} \biggr| \\ &\qquad +4|y'(x,\lambda)| \mathop{{\sum}'}^n_{k=0} \biggl| \frac{1}{((-1)^k +o(1/\ln \lambda))(x-x_{k,\lambda})^2} \biggr| \\ &\qquad +2|y(x,\lambda)| \mathop{{\sum}'}^n_{k=0} \biggl| \frac{1}{((-1)^k +o(1/\ln\lambda))(x-x_{k,\lambda})^3} \biggr| +O\bigl(\sqrt\lambda\,\bigr). \end{aligned}
\end{equation*}
\notag
$$
The sums are estimated as follows:
$$
\begin{equation*}
\begin{aligned} \, &AT_\lambda^{(2)}(x)\leqslant 2|y''(x,\lambda)|\biggl(1 +\biggl|o\biggl(\frac{1}{\ln\lambda}\biggr)\biggr|\biggr) \frac{8\sqrt\lambda}{\pi} \biggl[ \int_0^{x-\pi/(8\sqrt\lambda)}\!\!\!\frac{dt}{x\,{-}\,t}\,{+}\int_{x+\pi/(8\sqrt\lambda)}^\pi \!\frac{dt}{t\,{-}\,x} \biggr] \\ &\quad+4 |y'(x,\lambda)|\biggl(1 +\biggl|o\biggl(\frac{1}{\ln\lambda}\biggr)\biggr|\biggr) \frac{8\sqrt\lambda}{\pi}\biggl[ \int_0^{x-\pi/(8\sqrt\lambda)}\!\!\frac{dt}{(x-t)^2} +\int_{x+\pi/(8\sqrt\lambda)}^\pi \frac{dt}{(t-x)^2} \biggr] \\ &\quad+2 |y(x,\lambda)|\biggl(1 +\biggl|o\biggl(\frac{1}{\ln{\lambda}}\biggr)\biggr|\biggr) \frac{8\sqrt\lambda}{\pi}\biggl[\int_0^{x-\pi/(8\sqrt\lambda)}\frac{dt}{(x-t)^3} +\int_{x+\pi/(8\sqrt\lambda)}^\pi \frac{dt}{(t-x)^3} \biggr] \\ &\quad +O({\sqrt{\lambda}}). \end{aligned}
\end{equation*}
\notag
$$
Here, we have used the facts that $\max_{x\in[0,\pi]}x(\pi-x)=\pi^2/4$, that $x(\pi-x)> 0$ for $x\in [\pi/(4\sqrt\lambda),\,\pi-\pi/(4\sqrt\lambda)]$, that, for $x\in [0,\pi/(4\sqrt\lambda)]$, the first integral in the resulting estimate disappears by the choice of $k_0$, and that, for $x\in [\pi-\pi/(4\sqrt\lambda),\pi]$, there is no second integral. As a result,
$$
\begin{equation*}
\begin{aligned} \, &\int_0^{x-\pi/(8\sqrt\lambda)}\frac{dt}{x-t}+\int_{x+\pi/(8\sqrt\lambda)}^\pi \frac{dt}{t-x} \\ &\qquad=-\ln(x-t)\big|_0^{x-\pi/(8\sqrt\lambda)}+\ln(t-x)\big|_{x+\pi/(8\sqrt\lambda)}^{\pi} \\ &\qquad\leqslant\begin{cases} \ln(\lambda)+\ln{16} &\text{for } x\in \biggl[\dfrac{\pi}{4\sqrt\lambda},\,\pi-\dfrac{\pi}{4\sqrt\lambda}\biggr], \\ \dfrac{1}{2}\ln(\lambda)+\ln{8} &\text{for } x\in \biggl[0,\dfrac{\pi}{4\sqrt\lambda}\biggr] \cup \biggl[\pi-\dfrac{\pi}{4\sqrt\lambda},\,\pi\biggr]. \end{cases} \end{aligned}
\end{equation*}
\notag
$$
Now let us estimate the sums of the integrals
$$
\begin{equation*}
\begin{gathered} \, \int_0^{x-\pi/(8\sqrt\lambda)}\frac{dt}{(x-t)^2}+\int_{x+\pi/(8\sqrt\lambda)}^\pi \frac{dt}{(t-x)^2} =\frac{16\sqrt{\lambda}}{\pi}-\frac{\pi}{x(\pi-x)}, \\ \int_0^{x-\pi/(8\sqrt\lambda)}\frac{dt}{(x-t)^3}+\int_{x+\pi/(8\sqrt\lambda)}^\pi \frac{dt}{(t-x)^3} \leqslant\frac{8^2{\lambda}}{\pi^2}. \end{gathered}
\end{equation*}
\notag
$$
As a result, using the asymptotic formulas in Proposition 4, we have the estimate
$$
\begin{equation*}
\begin{aligned} \, AT_\lambda^{(2)}(x) &\leqslant |y''(x,\lambda)|\biggl(1 +\biggl|o\biggl(\frac{1}{\ln{\lambda}}\biggr)\biggr| \biggr) \frac{16\sqrt\lambda}{\pi}[\ln(\lambda)+\ln{16}] \\ &\qquad+|y'(x,\lambda)|\biggl(1 +\biggl|o\biggl(\frac{1}{\ln{\lambda}}\biggr)\biggr|\biggr) \biggl[\frac{16\sqrt\lambda}{\pi} \biggr]^2 \\ &\qquad+ |y(x,\lambda)|\biggl(1 +\biggl|o\biggl(\frac{1}{\ln{\lambda}}\biggr)\biggr| \biggr) \frac{16\sqrt\lambda}{\pi}\biggl[\frac{8^2{\lambda}}{\pi^2} \biggr] +O\biggl({\sqrt{\lambda}}\biggr), \end{aligned}
\end{equation*}
\notag
$$
which is uniform with respect to $x\in [0,\pi]$. In all, the norm of the operator is estimated as follows:
$$
\begin{equation*}
AT_\lambda^{(2)}\leqslant \frac{16\lambda}{\pi}\ln(\lambda) +O(\lambda).
\end{equation*}
\notag
$$
Hence there exists a sufficiently large $\lambda_0\geqslant \lambda_2$ such that, for all $\lambda >\lambda_0$, estimate (37) holds in the case of the Cauchy problem (9).
The proof of estimate (37) for sufficiently large $\lambda$ for $s_{k,\lambda}$, $0\leqslant k\leqslant n$ constructed from the solutions to the Cauchy problem (7), is similar to estimate (36) for $s_{k,\lambda}$, $0\leqslant k\leqslant n$, constructed from the solutions of the Cauchy problem (7). This completes the proof of Proposition 8. 2.4. Stability of the mixed boundary value problemLemma 2. The mixed problem (1)–(5) is Hadamard stable. That is, for all positive $T$ and $\varepsilon$, there exists $\delta>0$ such that whenever Proof. Consider the mixed boundary value problem
$$
\begin{equation}
u^{[\varphi]}_{tt} - u^{[\varphi]}_{xx}+ q(x)u^{[\varphi]}=0,
\end{equation}
\tag{55}
$$
$$
\begin{equation}
u^{[\varphi]}(0,t)\cos \alpha +u^{[\varphi]}_{x}(0,t)\sin \alpha=0,
\end{equation}
\tag{56}
$$
$$
\begin{equation}
u^{[\varphi]}(\pi,t)\cos \beta +u^{[\varphi]}_{x}(\pi,t)\sin \beta=0,
\end{equation}
\tag{57}
$$
Here, $\varphi\in C[0,\pi]$. Let us extend function (58) by continuity to the entire $x$-axis. Outside $[0,\pi]$, this function will be defined later via the reflection or symmetry method. The potential $q$ will be extended to the entire $x$-axis first as an even function to $[-\pi,0]$, and then periodically. We thus get the Cauchy problem
$$
\begin{equation}
u^{[\varphi]}_{tt} - u^{[\varphi]}_{xx}+ q(x)u^{[\varphi]}=0,
\end{equation}
\tag{60}
$$
Changing the variables $u^{[\varphi]}(x,t)=\widetilde u (\xi,\eta)$, where $\xi=x-t$, $\eta=x+t$, $x=(\xi+\eta)/2$, $t=(\eta-\xi)/2$, we get the equivalent Cauchy problem for the second canonical form of the same equation
$$
\begin{equation}
L[\widetilde u]=\widetilde{u}_{\xi\eta} - \frac{q((\xi+\eta)/2)}{4}\widetilde{u}=0,
\end{equation}
\tag{63}
$$
$$
\begin{equation}
\widetilde u\biggl(\frac{\xi+\eta}{2},0\biggr)=\varphi \biggl(\frac{\xi+\eta}{2}\biggr),
\end{equation}
\tag{64}
$$
$$
\begin{equation}
-\widetilde u_{\xi}\biggl(\frac{\xi+\eta}{2},0\biggr)+ \widetilde u_{\eta} \biggl(\frac{\xi+\eta}{2},0\biggr)=0.
\end{equation}
\tag{65}
$$
The Goursat problem for the Riemann function $\widetilde{v}(\xi,\eta,\xi_0,\eta_0)$ of our differential expression (63) reads as
$$
\begin{equation*}
L^*[\widetilde{v}]=\widetilde{v}_{\xi\eta} - \frac{q((\xi+\eta)/2)}{4}\widetilde{v}=0,\qquad \widetilde{v}(\xi_0,\eta)=1, \qquad \widetilde{v}(\xi,\eta_0)=1.
\end{equation*}
\notag
$$
Using the Riemann method, we obtain the solution of the Cauchy problem (63)–(65)
$$
\begin{equation*}
\begin{aligned} \, \widetilde{u}(\xi_0,\eta_0) &=\frac{\widetilde{u}(\xi_0,\xi_0)+\widetilde{u}(\eta_0,\eta_0)}{2} -\frac{1}{2}\int_{P(\xi_0,\xi_0)}^{Q(\eta_0,\eta_0)} \bigl(\widetilde{u}(\xi,\xi)\widetilde{v}_{\xi}(\xi,\xi) -\widetilde{v}(\xi,\xi)\widetilde{u}_{\xi}(\xi,\xi)\bigr)\,d\xi \\ &\qquad+\bigl(\widetilde{v}(\eta,\eta)\widetilde{u}_{\eta}(\eta,\eta) -\widetilde{u}(\eta,\eta)\widetilde{v}_{\eta}(\eta,\eta)\bigr)\,d\eta, \end{aligned}
\end{equation*}
\notag
$$
where $P(\xi_0,\xi_0)$ and $Q(\eta_0,\eta_0)$ are the points of intersection of the characteristics passing through the point $(\xi_0,\eta_0)$, with the line $\eta=\xi$.
Returning to the arbitrary variables $x$, $t$ and denoting by $\theta$, $\tau$, $v(\theta,\tau,x,t)$ the variables of the Riemann function, we get the integral representation of the solution of the Cauchy problem (60)–(62)
$$
\begin{equation}
u^{[\varphi]}(x,t)=\frac{\varphi (x-t)+\varphi (x+t)}{2}+\frac{1}{2}\int_{x-t}^{x+t}\varphi (\theta)v_{\tau}(\theta,0,x,t) \,d\theta.
\end{equation}
\tag{66}
$$
To study the stability of the mixed boundary value problem, we obtain an integral representation of its solution by the reflection method. To this end, we first consider the problem on the semi-axis $x\geqslant 0$
$$
\begin{equation}
u^{[\varphi]}_{tt} - u^{[\varphi]}_{xx}+ q(x)u^{[\varphi]}=0,
\end{equation}
\tag{67}
$$
$$
\begin{equation}
u^{[\varphi]}(0,t)\cos \alpha +u^{[\varphi]}_{x}(0,t)\sin \alpha=0,
\end{equation}
\tag{68}
$$
For $\xi\geqslant 0$, consider the function
$$
\begin{equation}
G(\xi)=\frac{\varphi (\xi)}{2}+\frac{1}{2}\int_0^{\xi}\varphi (\theta)v_{\tau}(\theta,0,x,t) \,d\theta.
\end{equation}
\tag{71}
$$
Using the integral representation (66), we have, for $x\geqslant t$,
Let us first consider the case $\alpha \ne \pi k$, $k\in \mathbb Z$. By (72), for any $t\geqslant 0$ and $x\geqslant 0$, the solution of problem (67)–(70) has the representation
$$
\begin{equation}
u^{[\varphi]}(x,t)=\begin{cases} {\displaystyle \dfrac{\varphi(x+t)+\varphi(x-t)}{2} +\dfrac{1}{2}\int_{x-t}^{x+t}\varphi(\theta)v_{\tau}(\theta,0,x,t) \,d\theta} &\text{for } t\,{\leqslant}\, x, \\ {\displaystyle\dfrac{\varphi(x+t)+\varphi(t-x)}{2} +\dfrac{1}{2}\int_0^{x+t}\varphi(\theta)v_{\tau}(\theta,0,x,t) \,d\theta} \\ \ {\displaystyle+\int_0^{t-x} \frac{\varphi(\zeta)}{2}\bigl(v_{\tau}(\zeta,0,x,t)+2\cot \alpha\bigr)e^{-\cot \alpha (x-t+\zeta)}\,d\zeta} \\ \ {\displaystyle+\cot \alpha\int_0^{t-x}\!\biggl(\int_0^{\zeta}\frac{\varphi(\theta)}{2} v_{\tau}(\theta,0,x,t) \,d\theta\biggr) e^{-\cot \alpha (x-t+\zeta)}\,d\zeta} &\text{for } t\,{>}\,x. \end{cases}
\end{equation}
\tag{73}
$$
If instead of the semi-axis we consider the interval $[0,\pi]$, then, under the assumption $\alpha \ne \pi k$, $k\in \mathbb Z$, $\beta \ne \pi m$, $m\in \mathbb Z$, to solve the boundary value problem (55)–(59) for $0\leqslant t \leqslant \pi/2$ by the reflection method we get an integral representation of the solution of problem (55)–(59) on the interval
$$
\begin{equation}
u^{[\varphi]}(x,t)=\begin{cases} {\displaystyle\dfrac{\varphi(x+t)+\varphi(x-t)}{2} +\dfrac{1}{2}\int_{x-t}^{x+t}\varphi(\theta)v_{\tau}(\theta,0,x,t) \,d\theta} \\ \qquad \text{for } t\leqslant \dfrac{\pi}{2}-\biggl|x-\dfrac{\pi}{2}\biggr|,\, x\in[0,\pi], \\ {\displaystyle\dfrac{\varphi(x+t)+\varphi(t-x)}{2} +\dfrac{1}{2}\int_0^{x+t}\varphi(\theta)v_{\tau}(\theta,0,x,t) \,d\theta} \\ \ {\displaystyle+\int_0^{t-x} \dfrac{\varphi(\zeta)}{2}\bigl(v_{\tau}(\zeta,0,x,t)+2\cot \alpha\bigr)e^{-\cot \alpha (x-t+\zeta)}\,d\zeta} \\ \ {\displaystyle+\cot \alpha\int_0^{t-x} \biggl(\int_0^{\zeta}\dfrac{\varphi(\theta)}{2}v_{\tau}(\theta,0,x,t) \,d\theta\biggr) e^{-\cot \alpha (x-t+\zeta)}\,d\zeta} \\ \qquad\text{for } \dfrac{\pi}{2}\geqslant t > x\geqslant 0, \\ {\displaystyle\dfrac{\varphi(x-t)+\varphi(2\pi-t-x)}{2} -\dfrac{1}{2}\int_{\pi}^{2\pi-x-t}\varphi(\theta)v_{\tau}(\theta,0,\pi-x,t) \,d\theta} \\ \ {\displaystyle-\int_{\pi}^{2\pi-x-t} \dfrac{\varphi(\zeta)}{2}\bigl( v_{\tau}(\zeta,0,\pi-x,t)+2\cot \beta\bigr)e^{-\cot \beta (\pi-x-t+\zeta)}\,d\zeta} \\ \ {\displaystyle+\cot \beta\!\int_{\pi}^{2\pi-x-t} \biggl(\int_0^{\zeta}\dfrac{\varphi(\theta)}{2}v_{\tau}(\theta,0,\pi\,{-}\,x,t) \,d\theta\biggr) e^{-\cot \beta (\pi-x-t+\zeta)}\,d\zeta} \\ \qquad\text{for } \dfrac{\pi}{2}\geqslant t >\pi- x\geqslant 0. \end{cases}
\end{equation}
\tag{74}
$$
The derivative of the Riemann function $v_{\tau}$ is continuous with respect to all four variables, and hence, setting
$$
\begin{equation*}
\begin{aligned} \, &M(q,\alpha,\beta) =1+\pi\|v_{\tau}(\,{\cdot}\,,0,\cdot,{\cdot}\,)\|_{C_{[0,\pi]\times[0,\pi] \times [0,\pi/2]}} \\ &\quad+2\max(|{\cot \alpha}|\,|{\cot \beta}|)\exp\bigl(\max(|{\cot \alpha}|\,|{\cot \beta}|)\bigr) \\ &\quad+\frac{\pi^2}{2}\|v_{\tau}(\,{\cdot}\,,0,\cdot,{\cdot}\,)\|_{C_{[0,\pi]\times[0,\pi]\times [0,\pi/2]}}\max(|{\cot \alpha}|\, |{\cot \beta}|)\exp\bigl(\max(|{\cot \alpha}|\, |{\cot \beta}|)\bigr), \end{aligned}
\end{equation*}
\notag
$$
the norm of the solution (74) of boundary value problem (55)–(59) in the Chebyshev space $C_{[0,\pi]\times [0,\pi/2]}$ is estimated as follows:
$$
\begin{equation*}
\|u^{[\varphi]}\|_{C_{[0,\pi]\times [0,\pi/2]}}\leqslant M(q,\alpha,\beta)\|\varphi\|_{C_{[0,\pi]}}.
\end{equation*}
\notag
$$
Now, after a finite number of iterations, we estimate the norm of the solution of the boundary value problem (55)–(59) in Chebyshev space $C_{[0,\pi]\times [0,T]}$ as follows:
$$
\begin{equation}
\|u^{[\varphi]}\|_{C_{[0,\pi]\times [0,T]}}\leqslant M^{[2T/\pi]+1}(q,\alpha,\beta) \|\varphi\|_{C_{[0,\pi]}}.
\end{equation}
\tag{75}
$$
If there exists $k\in \mathbb Z$ such that $\alpha =\pi k$, then we have We extend $q$ to the negative part of the interval $[0,\pi]$ as an even function. Hence the Riemann function is even with respect to $x$. In this case, we extend $\varphi(x)$ to the negative part of the axis as an odd function. The argument in the case $\beta \ne \pi m$, $m\in \mathbb Z$, is similar. This proves Lemma 2 for the mixed boundary value problem of the form (55)–(59).The continuous dependence of the solution to the mixed boundary value problem with of hyperbolic type equation on the initial conditions
$$
\begin{equation*}
\begin{gathered} \, u^{[\varphi]}_{tt} - u^{[\varphi]}_{xx}+ q(x)u^{[\varphi]}=f_n(x,t), \\ u^{[\varphi]}(0,t)\cos \alpha +u^{[\varphi]}_{x}(0,t)\sin \alpha=0, \\ u^{[\varphi]}(x,0)=0, \\ u^{[\varphi]}_t(x,0)=\psi_n(x) \end{gathered}
\end{equation*}
\notag
$$
is secured by the results of [3], XXII, § 3, and [1], Ch. 2, § 19. Lemma 2 is proved. § 3. Solution of the boundary value problemProposition 9. For any positive $\widetilde\varepsilon$, the functions $AT_\lambda (f,x)+\eta (x,\lambda)+\widetilde\eta (x,\lambda)$ satisfy Proof. The function $\eta (x,\lambda)+\widetilde\eta (x,\lambda)$ is twice continuously differentiable on $[0,\pi]$, and its support satisfies
$$
\begin{equation}
\begin{aligned} \, \operatorname{supp}(\eta+\widetilde\eta) &\subset \biggl[0,\frac{\pi}{|\mu|}-\frac{1}{|\mu|}\biggl(\arcsin{\sqrt{\frac{2}{3}}} -\frac{\sqrt{2}}{3}\biggr)\biggr] \nonumber \\ &\qquad\cup \biggl[\pi-\frac{\pi}{|\widetilde{\mu}|} +\frac{1}{|\widetilde{\mu}|}\biggl(\arcsin{\sqrt{\frac{2}{3}}}-\frac{\sqrt{2}}{3}\biggr),\pi\biggr]. \end{aligned}
\end{equation}
\tag{76}
$$
For $\lambda >\ln(4/\sqrt{3})$, by Proposition 8 we have
$$
\begin{equation}
\begin{gathered} \, \biggl[0,\frac{\pi}{|\mu|}-\frac{1}{|\mu|}\biggl(\arcsin\sqrt{\frac{2}{3}} -\frac{\sqrt{2}}{3}\biggr)\biggr]\cap \biggl[\pi-\frac{\pi}{|\widetilde{\mu}|} +\frac{1}{|\widetilde{\mu}|}\biggl(\arcsin\sqrt{\frac{2}{3}}-\frac{\sqrt{2}}{3}\biggr),\pi\biggr]= \varnothing, \nonumber \\ \begin{split} &\|\eta+\widetilde\eta \|_{C\bigl[\sigma_1\bigl(\frac{\pi}{|\mu|} -\frac{1}{|\mu|}\bigl(\arcsin\sqrt{\frac{2}{3}}-\frac{\sqrt{2}}{3}\bigr)\bigr),\, \pi-\widetilde\sigma_1\bigl(\frac{\pi}{|\mu|}+\frac{1}{|\mu|}\bigl(\arcsin\sqrt{\frac{2}{3}} -\frac{\sqrt{2}}{3}\bigr)\bigr)\bigr]} \\ &\qquad \leqslant \max(\nu,\widetilde\nu) =O\bigl(e^{-\lambda}\sqrt{\lambda}\ln\lambda \bigr) \end{split} \end{gathered}
\end{equation}
\tag{77}
$$
for arbitrary boundary conditions of the third kind. In the case of boundary conditions of the first kind, on the entire interval $[0,\pi]$, we have
$$
\begin{equation}
\|\eta+\widetilde\eta \|_{C[0,\pi]} \leqslant \max(\nu,\widetilde\nu) =\operatorname{max}\bigl(|AT_\lambda (f,0)|,|AT_\lambda (f,\pi)|\bigr)O(1).
\end{equation}
\tag{78}
$$
Now the conclusion of Proposition 9 is secured by Proposition 7. 3.1. Estimate of the Fourier coefficients $AT_\lambda(f,x)$ via integration by parts Proposition 10. Let $ \rho_\lambda \geqslant 0$, $\rho_\lambda =o(\sqrt \lambda/\ln\lambda)$ as $\lambda \to \infty$, and let $V_{\rho_\lambda}[0,\pi]$ be the ball of radius $\rho_\lambda $ in the space of functions of bounded variation which vanish at the origin (in the case of the Cauchy problem (9), we also assume that $h(\lambda)\ne 0$). Next, assume that $0<\epsilon<1$, $j(\lambda):=[\lambda^{1+2\epsilon/(1-\epsilon)}]+1$, the function $f$ is continuous on $x\in [0,\pi]$, and the function $q$ has bounded variation. Then the error of uniform approximation by the values of operators (25) satisfies Proof. From the asymptotic formulas (32), (33), we have
We first consider the case where the left-hand boundary value conditions of the Sturm–Liouville problem are the same as in the Cauchy problem (7). Using (32) we have, up to a normalization,
$$
\begin{equation*}
\begin{aligned} \, &\biggl|\int_0^{\pi}AT_\lambda(f,x)U_m(x)\,dx\biggr| \\ &\qquad\leqslant\biggl|\int_0^{\pi}AT_\lambda(f,x)\biggl(\gamma (x,\lambda_m,h)\cos \sqrt{\lambda_m}\, x +\beta (x,\lambda_m,h)\frac{\sin \sqrt{\lambda_m}\, x}{\sqrt{\lambda_m}} \biggr)\,dx\biggr| \\ &\qquad\qquad+\biggl|\int_0^{\pi}AT_\lambda(f,x)\frac{\rho_{\lambda_m}(1+\pi \rho_{\lambda_m})} {2\lambda_m} \biggl(1+\frac{|h(\lambda_m)|}{\sqrt\lambda_m}\biggr)\,dx\biggr| \\ &\qquad=\sqrt{\gamma^2(x,\lambda_m,h)+\frac{\beta^2(x,\lambda_m,h)}{\lambda_m}} \\ &\qquad\qquad\times\biggl|\int_0^{\pi}AT_\lambda(f,x)\bigl(\sin{\phi_{\lambda_m}}\cos \sqrt{\lambda_m}\, x+\cos\phi_{\lambda_m}\sin \sqrt{\lambda_m}\,x \bigr)\,dx\biggr| \\ &\qquad\qquad +\|AT_\lambda(f,{\cdot}\,)\|_{C[0,\pi]}O\biggl(\frac{1}{\lambda_m}\biggr), \end{aligned}
\end{equation*}
\notag
$$
here
$$
\begin{equation*}
\sin\phi_{\lambda_m}=\frac{\gamma (x,\lambda_m,h)}{\sqrt{\gamma^2(x,\lambda_m,h) +\beta^2(x,\lambda_m,h)/\lambda_m}}.
\end{equation*}
\notag
$$
Integrating the Stieltjes integral by parts (see Ch. VIII, § 6, 5 in [35]), we get
$$
\begin{equation*}
\begin{aligned} \, &\biggl|\int_0^{\pi}AT_\lambda(f,x)U_m(x)\,dx\biggr| \\ &\qquad=\biggl|\int_0^{\pi}{AT_\lambda(f,x)}\sin\bigl(\phi_{\lambda_m}+ \sqrt{\lambda_m}\, x \bigr)\,dx\biggr|+\|AT_\lambda(f,{\cdot}\,)\|_{C[0,\pi]}O\biggl(\frac{1}{\lambda_m}\biggr) \\ &\qquad=\biggl|-AT_\lambda(f,x)\frac{\sin'(\phi_{\lambda_m}+ \sqrt{\lambda_m}\, x )}{\lambda_m}\bigg|_0^{\pi}+AT'_\lambda(f,x)\frac{\sin(\phi_{\lambda_m}+ \sqrt{\lambda_m}\, x )}{\lambda_m}\bigg|_0^{\pi} \\ &\qquad\qquad-\frac{1}{\lambda_m}\int_0^{\pi}AT''_\lambda(f,x)\sin\bigl(\phi_{\lambda_m}+ \sqrt{\lambda_m}\, x \bigr)\,dx\biggr| +\|AT_\lambda(f,{\cdot}\,)\|_{C[0,\pi]}O\biggl(\frac{1}{\lambda_m}\biggr). \end{aligned}
\end{equation*}
\notag
$$
We again use the asymptotic formulas in Proposition 3 and return to the eigenfunctions in the first two terms with substitutions
$$
\begin{equation*}
\begin{aligned} \, &\biggl|\int_0^{\pi}AT_\lambda(f,x)U_m(x)\,dx\biggr| \\ &\qquad=\biggl|-AT_\lambda(f,x)\frac{U_m'(x)}{\lambda_m}\bigg|_0^{\pi} +AT'_\lambda(f,x)\frac{U_m(x)}{\lambda_m}\bigg|_0^{\pi}\biggr| \\ &\qquad\quad +\biggl|-\frac{1}{\lambda_m}\int_0^{\pi}AT''_\lambda(f,x)\sin\bigl(\phi_{\lambda_m}+ \sqrt{\lambda_m}\, x \bigr)\,dx\biggr| +\|AT_\lambda(f,{\cdot}\,)\|_{C[0,\pi]}O\biggl(\frac{1}{\lambda_m}\biggr). \end{aligned}
\end{equation*}
\notag
$$
An appeal to (37) shows that
$$
\begin{equation*}
\begin{aligned} \, \biggl|\int_0^{\pi}{AT_\lambda(f,x)}U_m(x)\,dx\biggr| &=\frac{(17\pi\lambda/\pi)\ln\lambda}{\lambda_m}\|f\|_{C[0,\pi]} +\|AT_\lambda(f,{\cdot}\,)\|_{C[0,\pi]}O\biggl(\frac{1}{\lambda_m}\biggr) \\ = \|f\|_{C[0,\pi]}\frac{\lambda \ln\lambda}{\lambda_m}O(1). \end{aligned}
\end{equation*}
\notag
$$
The function $\eta (x,\lambda)+\widetilde\eta (x,\lambda)$ is twice continuously differentiable on $[0,\pi]$. Using (77), (78) employing the estimate $\operatorname{mes}(\operatorname{supp}(\eta+\widetilde\eta))=O(e^{-\lambda})$, and proceeding similarly we have the estimate
$$
\begin{equation}
\bigl| \widehat{AT}_{\lambda,m}[f,\eta]\bigr|= \|f\|_{C[0,\pi]}\frac{\lambda \ln\lambda}{\lambda_m}O(1).
\end{equation}
\tag{81}
$$
Now let us estimate the error of the partial sums of the Fourier series for $AT_\lambda(f,{\cdot}\,)$. The asymptotics of the eigenvalues of the Sturm–Liouville problem (16)–(18) is well known (see, for example, [30], Ch. 1, § 2, (2.12)). Therefore, if, for each $\lambda>0$, we take an eigenvalue number satisfying
$$
\begin{equation*}
\frac{\lambda \ln\lambda}{\lambda_m}\cong \frac{\lambda \ln\lambda}{m^2}\leqslant m^{-1-\epsilon}
\end{equation*}
\notag
$$
with some $0<\epsilon<1$, then the error of approximation of each function $AT_\lambda(f,{\cdot}\,)$ will be uniformly majored by the remainder of the convergent numerical series $\sum_{m=1}^{\infty}m^{-1-\epsilon}$. We next continue the estimate for sufficiently large $\lambda$ and $0<\epsilon<1$.
We set here $\varepsilon:=2\epsilon/(1-\epsilon)>0$. In this case, by of (81), (80) there exists $\lambda_0>0$ such that estimate (79) holds for all $\lambda>\lambda_0$. The case of left boundary conditions is considered similarly as in the Cauchy problem (9). Proposition 10 is proved.3.2. Proof of the main results Proof of Theorem 1. We choose and fix some $\varepsilon>0$. Let $j(\lambda)$ be defined as in (27). Using Propositions 9 and 10, we represent, for any $t\in [0,T]$, the following functions via operator (25):
$$
\begin{equation}
\lim_{\lambda \to \infty}\mathbb{AT}_{\lambda,j(\lambda)} (f(\,{\cdot}\,,t),x) =\lim_{\lambda \to \infty}\sum_{m=0}^{j(\lambda)} \widehat{AT}_{\lambda,m} [f(\,{\cdot}\,,t),\eta]U_m(q,\alpha,\beta,x)=f(x,t),
\end{equation}
\tag{82}
$$
$$
\begin{equation}
\lim_{\lambda \to \infty}\mathbb{AT}_{\lambda,j(\lambda)} (\varphi,x) =\lim_{\lambda \to \infty}\sum_{m=0}^{j(\lambda)} \widehat{AT}_{\lambda,m} [\varphi,\eta]U_m(q,\alpha,\beta,x)=\varphi(x),
\end{equation}
\tag{83}
$$
$$
\begin{equation}
\lim_{\lambda \to \infty}\mathbb{AT}_{\lambda,j(\lambda)} (\psi,x) =\lim_{\lambda \to \infty}\sum_{m=0}^{j(\lambda)} \widehat{AT}_{\lambda,m}[\psi,\eta] U_m(q,\alpha,\beta,x)=\psi(x).
\end{equation}
\tag{84}
$$
Consider the family of mixed problems depending on the parameter $\lambda $:
$$
\begin{equation}
u_{\lambda tt} - u_{\lambda xx}+ q(x)u_{\lambda}=\mathbb{AT}_{\lambda,j(\lambda)} (f(\,{\cdot}\,,t),x),
\end{equation}
\tag{85}
$$
$$
\begin{equation}
u_{\lambda}(0,t)\cos \alpha +u_{\lambda x}(0,t)\sin \alpha=0,
\end{equation}
\tag{86}
$$
$$
\begin{equation}
u_{\lambda}(\pi,t)\cos \beta +u_{\lambda x}(\pi,t)\sin \beta=0,
\end{equation}
\tag{87}
$$
Both functions (88), (89) and the right-hand side of equation (85) have absolutely continuous derivatives with respect to $x$. Each problem (85)–(89) has a unique classical solution, which, after satisfying the initial conditions by the method of separation of variables, decomposes into a Fourier series in the eigenfunctions of the Sturm–Liouville problem (16)–(18) which uniformly converges on $[0,\pi]\times[0,T]$, For $j(\lambda) < \infty$, this series becomes the finite sum
$$
\begin{equation}
\begin{aligned} \, u_{\lambda}(x,t) &=\sum_{m=0}^{j(\lambda)} \biggl( \widehat{AT}_{\lambda,m}[\varphi,\eta] \cos\Bigl(\sqrt{\widehat{\lambda}_m}t\Bigr) + \frac{\widehat{AT}_{\lambda,m}[\psi,\eta]} {\sqrt{\widehat{\lambda}_m}}\sin\Bigl(\sqrt{\widehat{\lambda}_m}t\Bigr) \nonumber \\ &\qquad+\int_0^t\frac{\sin\sqrt{\widehat{\lambda}_m}(t-\tau)}{\sqrt{\widehat{\lambda}_m}} \widehat{AT}_{\lambda,m}[f(\,{\cdot}\,,\tau),\eta]\,d\tau \biggr)U_m(q,\alpha,\beta,x). \end{aligned}
\end{equation}
\tag{90}
$$
By (76)–(78), the measure of the support of the bounded function $\eta+\widetilde\eta$ behaves as $O(e^{-\lambda})$. So, each term
$$
\begin{equation*}
\int_0^t\frac{\sin\sqrt{\widehat{\lambda}_m}(t-\tau)}{\sqrt{\widehat{\lambda}_m}} \widehat{AT}_{\lambda,m}[f(\,{\cdot}\,,\tau),\eta]\,d\tau
\end{equation*}
\notag
$$
in the coefficient of the function $U_m(q,\alpha,\beta,x)$ has the property
$$
\begin{equation*}
\begin{aligned} \, &\int_0^t\frac{\sin\sqrt{\widehat{\lambda}_m}(t-\tau)}{\sqrt{\widehat{\lambda}_m}} \widehat{AT}_{\lambda,m}[f(\,{\cdot}\,,\tau)]\,d\tau \\ &\qquad -\int_0^t\frac{\sin\sqrt{\widehat{\lambda}_m}(t-\tau)}{\sqrt{\widehat{\lambda}_m}} \widehat{AT}_{\lambda,m}[f(\,{\cdot}\,,\tau),\eta]\,d\tau=O(e^{-\lambda}). \end{aligned}
\end{equation*}
\notag
$$
By the Cauchy test for uniform convergence of series, the effect of “correction” of the boundary value conditions $\eta (x,\lambda)+\widetilde\eta (x,\lambda)$ for the right-hand side of equation (1) in the third term of (90) is $O(e^{-\lambda/2})$. Therefore, by (21), (22), (76)–(78), Propositions 9 and 10, and Lemma 2, the solutions of problems (85)–(89) converge uniformly on the rectangle $[\sigma_1\widetilde\varepsilon,\, \pi-\widetilde\sigma_1\widetilde\varepsilon]\times[0,T]$ to the solution of the mixed boundary value problem (1)–(5), In other words, any subsequence $u_{\lambda_n}(x,t)$, $\lambda_n \nearrow \infty$ as $n\to\infty$ belongs to the class of equivalent sequences, which, by Definition 2, is a generalized function, which, in turn, is a generalized solution of the mixed boundary value problem (1)–(5). Theorem 1 is proved. Proof of Proposition 1. In the neighbourhood of any of the zero $x_{k,\lambda}$ (see defined in (10)), using the Lagrange formula, we obtain the estimate
$$
\begin{equation*}
\begin{aligned} \, &s_{k,\lambda}'(x) =\frac{U'_n(x_{k,\lambda})(x-x_{k,\lambda}) +U''_n(x_{k,\lambda})(x-x_{k,\lambda})^2+o((x-x_{k,\lambda})^2)}{(x-x_{k,\lambda})^2} \\ &-\frac{U_n(x_{k,\lambda})\,{+}\,U'_n(x_{k,\lambda})(x\,{-}\,x_{k,\lambda})\,{+}\,(U''_n(x_{k,\lambda})/2) (x\,{-}\,x_{k,\lambda})^2{+}\,o((x\,{-}\,x_{k,\lambda})^2)}{(x\,{-}\,x_{k,\lambda})^2}=o(1). \end{aligned}
\end{equation*}
\notag
$$
First, multiplying the normalized eigenfunction $U_m$ of the Sturm–Liouville problem (16)–(18) by identity (34), and the multiplying (16) by the function $s_{k,\lambda}$, we get
$$
\begin{equation*}
\begin{gathered} \, s_{k,\lambda}''(x)U_m(x) + \frac{2}{(x-x_{k,\lambda})}s_{k,\lambda}'(x)U_m(x) + (\lambda -q(x))s_{k,\lambda}(x)U_m(x)\equiv 0, \\ s_{k,\lambda}(x)U_m''(x) + (\lambda_m - q(x))s_{k,\lambda}(x)U_m(x)\equiv 0. \end{gathered}
\end{equation*}
\notag
$$
Subtracting the second identity from the first one, this gives
$$
\begin{equation*}
\begin{aligned} \, &\bigl(s_{k,\lambda}'(x)U_m(x)- s_{k,\lambda}(x)U_m'(x)\bigr)'+ \frac{2}{(x-x_{k,\lambda})}s_{k,\lambda}'(x)U_m(x) \\ &\qquad + (\lambda -\lambda_m)s_{k,\lambda}(x)U_m(x)\equiv 0. \end{aligned}
\end{equation*}
\notag
$$
Integrating the resulting identity with respect to $x$ from $0$ to $\pi$, we find that
$$
\begin{equation*}
\begin{aligned} \, &\bigl(s_{k,\lambda}'(\pi)U_m(\pi)- s_{k,\lambda}(\pi)U_m'(\pi)\bigr) -\bigl(s_{k,\lambda}'(0)U_m(0)- s_{k,\lambda}(0)U_m'(0)\bigr) \\ &\qquad+\int_0^\pi \frac{2}{(x-x_{k,\lambda})}s_{k,\lambda}'(x)U_m(x)\, dx + (\lambda -\lambda_m) \int_0^\pi s_{k,\lambda}(x)U_m(x)\, dx= 0, \end{aligned}
\end{equation*}
\notag
$$
where each improper integral is taken as a Cauchy principal value. This gives us (29). Proposition 1 is proved. Proof of Proposition 2. Consider the function
$$
\begin{equation}
\Phi_{k,\lambda,m}(x):=\int_{x_{k,\lambda}}^x s_{k,\lambda}(\xi)U_m(\xi)\, d\xi.
\end{equation}
\tag{91}
$$
In the neighbourhood of any zero $x_{k,\lambda}$ (see (10)), in the case of continuously differentiable potential $q_{\lambda}$, from Taylor’s theorem with the Lagrange remainder we get the estimate
$$
\begin{equation*}
\begin{aligned} \, s_{k,\lambda}'(x) &=\frac1{(x-x_{k,\lambda})^2}\biggl(y'(x_{k,\lambda},\lambda)(x-x_{k,\lambda}) +y''(x_{k,\lambda},\lambda)(x-x_{k,\lambda})^2 \\ &\ \, +\frac{y'''(\xi_{k,\lambda},\lambda)}3(x-x_{k,\lambda})^3\biggr) \,{-}\,\frac1{(x-x_{k,\lambda})^2}\biggl(y(x_{k,\lambda},\lambda) \,{+}\,y'(x_{k,\lambda},\lambda)(x\,{-}\,x_{k,\lambda}) \\ &\ \, +\frac{y''(x_{k,\lambda},\lambda)}2(x-x_{k,\lambda})^2 +\frac{y'''(\xi^*_{k,\lambda},\lambda)}3(x-x_{k,\lambda})^3\biggr) \\ &=O(x-x_{k,\lambda}). \end{aligned}
\end{equation*}
\notag
$$
Using this fact and submitting, we verify that function (91) satisfies the conditions of the Cauchy problem (30). From this we get representation (31). Proposition 2 is proved. § 4. Numerical experimentIt is well known (see, for example, [36]–[39]) that Fourier series may diverge on the class of continuous functions. So, the methods for separation of variables cannot be applied if the continuous initial conditions or the inhomogeneity of the equation of a mixed boundary value problem is not representable by their Fourier series. Our numerical experiment was implemented on a basis of the classical example of a Fourier series diverging at a point, which is constructed by the sliding hump method in the form of a lacunary series. Such a series has the property that each of its terms has a harmonic of the Fourier series whose value at the divergence point grows with the harmonic number, the remaining harmonics being bounded. Figure 1 demonstrates that our method is capable of dealing with interferences in initial conditions, which make the Fourier method unfit in principle. In our numerical experiment, the initial condition at $t=0$ (see Figure 1, the dotted line) was taken as the harmonic in the sliding hump method. The figure shows that the corresponding partial sums of the solutions behave differently. The graph of approximation of the solution at $t=2$ by the classical Fourier sums is shown in blue; the red curve is the graph of the partial sum of the same order of solution (28) obtained by the method of values of the operators (25). Therefore, the sliding hump method produces, in the limit, an infinitely divergent series for the classical solution and an exact generalized solution (28) under the initial condition with noise. 
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