2. The existence and main properties of the solution of problem (1.1), (1.2), (1.4), (1.5)

We consider the boundary condition

(2.1)\begin{equation} y (0) \cos \beta + T y(0) \sin \beta = 0,\beta \in [0, {\pi \mathord{\left/ {\vphantom {\pi 2}} \right. } 2}]. \end{equation}

Alongside the problem (1.1)–(1.5) we also consider the eigenvalue problem (1.1), (1.2), (2.1), (1.4), (1.5). The spectral properties of this problem in a more general form of boundary conditions were investigated in [Reference Kapustin and Moiseev29, Reference Kerimov and Aliev30].

It follows from [Reference Kerimov and Aliev30, lemma 2.2 and theorem 2.2] that the following result holds for this problem.

Theorem 2.1 For each $\beta$ and each $\gamma$ the eigenvalues of the boundary value problem (1.1), (1.2), (2.1), (1.4), (1.5) are real and simple, and form an unbounded increasing sequence $\{\lambda _{k} (\beta ,\gamma )\}_{k=1}^{\infty }$ such that

\[ 0 < \lambda_1 (\beta, \gamma) < \lambda_2 (\beta, \gamma) < \ldots < \lambda_k (\beta, \gamma) < \ldots \text{for}\ \beta \in [0, \pi / 2), \]

and

\[ 0 = \lambda_1 (\pi/2,\gamma) < \lambda_2 (\pi / 2,\gamma) < \ldots < \lambda_k ({\pi / 2},\gamma) < \ldots\,. \]

Moreover, the eigenfunction $y_{k,\beta , \,\gamma } (x)$, $k \in \mathbb {N}$, corresponding to the eigenvalue $\lambda _k (\beta , \gamma )$ has $k - 1$ simple zeros in the interval $(0, 1)$.

Let $\tilde H = L_2 (0,1) \oplus \mathbb {C}$ be the Hilbert space with the scalar product

\[ (\tilde y, \tilde v) = (\{y,n\}, \{v,t\}) = \int_0^{1} {u(x)\,\overline {v(x)} } \,\textrm{d}x + |c|^{{-}1} n \bar t. \]

As is known [Reference Aliyev2] that problem (1.1), (1.2), (2.1), (1.4), (1.5) is equivalent the following eigenvalue problem

\[ \tilde L \tilde y = \lambda \tilde y,\tilde y \in D (\tilde L), \]

where $L$ is a self-adjoint bounded below operator in $\tilde H$ defined by

(2.2)\begin{equation} \tilde L \tilde y = L \{y,n\} = \{\ell y,Ty (0)\} \end{equation}

with the domain

\begin{align*} &D (\tilde L ) = \left\{ {\tilde y = \{y,n\} \in \tilde H : y \in W_2^{4} (0,1),\ell y \in L_2 (0,1), y''(0) = 0,}\right.\\ &\left.{y (0) \cos \beta + T y(0) \sin \beta = y (1) \cos \gamma + y''(1) \sin \gamma = 0,n = c y (1)}\right \}. \end{align*}

It is known that the eigenvalues of problem (2.2) are given by the max-min principle [Reference Bolotin20]

\[ \lambda_k (\beta, \gamma) = \mathop {\max }\limits_{\,\tilde V^{(k - 1)} }\quad \mathop {\min }\nolimits_{\scriptstyle \,\,\,\,\tilde y \, \in \mathfrak{L} \atop \scriptstyle (\tilde y,\tilde V^{(k - 1)} ) = 0 } R\,[\tilde y] \]

where $R\,[\tilde y]$ is the Rayleigh quotient

\begin{align*} R\,[\tilde y] &= \frac{(\tilde L \tilde y, \tilde y)}{(\tilde y, \tilde y)} = \frac {\int_0^{1} {\left( {y''^{2} (x) + q(x)y'^{2} (x)} \right)}\,\textrm{d}x + N[y] }{\int_0^{1} {y^{2} (x)\,\textrm{d}x - cy^{2} (1)}}, \\ N [y] &= y^{2} (0) \cot \beta + y'^{2} (1) \cot \gamma, \end{align*}

(we use the convention that if any of the parameters $\beta$ or $\gamma$ is zero, then the boundary value of $y$ at $0$ or $y'$ at $1$ is taken to be zero and the corresponding term in $N [y]$ does not appear), $\mathfrak {L}$ is the set of vectors $\tilde y = \{y,n\} \in \tilde H$ such that the function $y$ satisfies the boundary conditions (1.2), (2.1), (1.4), $\tilde V^{(k-1)}$ is an arbitrary set of linearly independent vectors $\tilde v_j = \{v_j,t_j\}$, $1 \le j \le k-1$, such that the functions $v_j$, $1 \le j \le k-1$, satisfy the boundary conditions (1.2), (2.1), (1.4).

Using this max-min characterization, by following the argument in theorem 9 of [Reference Bolotin20, p. 419] for eigenvalues of (1.1), (1.2), (2.1), (1.4), (1.5) we have the following property.

By virtue of theorem 2.1 and lemma 2.2 for each $\gamma \in [0, {\pi \mathord {\left / {\vphantom {\pi 2}} \right .} 2}]$ we have

(2.3)\begin{equation} \lambda_1 ({\pi \mathord{\left/ {\vphantom {\pi 2}} \right. } 2},\gamma) < \lambda_1 (0,\gamma) < \lambda_2 ({\pi \mathord{\left/ {\vphantom {\pi 2}} \right. } 2},\gamma) < \lambda_2 (0, \gamma) < \,...\, \end{equation}

Proof. Let $\psi _{k} (x, \lambda ), k = 1,2,3,4$, denote the solutions of equation (1.1) normalized for $x = 1$ by the Cauchy conditions

(2.4)\begin{equation} \psi _k^{(s - 1)} (1,\lambda) = \delta _{ks} ,s = 1,2,3, T \psi _k (1,\lambda) = \delta _{k4} , \end{equation}

where $\delta _{ks}$ is the Kronecker delta.

As in [Reference Aliyev, Kerimov and Mekhrabov8, Reference Mekhrabov33], we will seek the solution $y (x, \lambda )$ of (1.1), (1.2), (1.4), (1.5) in the form

(2.5)\begin{equation} y (x,\lambda ) = \sum \limits_{k = 1}^{4} {A_k } \psi _k (x,\lambda), \end{equation}

where $A_{k}, \,k = 1, \,2,3,4$, are some constants.

By (1.4), (1.5) and (2.4) it follows from (2.5) that $A_2 \cos \gamma + A_3 \sin \gamma = 0$, $A_4 - c \lambda A_1 = 0$. Consequently, for the function $y (x, \lambda )$ we have

(2.6)\begin{equation} y(x,\lambda ) = \begin{cases} A_1 \left\{{\psi_1 (x,\lambda) + c \lambda \psi_4 (x,\lambda)}\right\} + A_3 \,\psi_3 (x,\lambda) \quad \textrm{if}\ \gamma = 0, \\ A_1 \left\{{\psi_1 (x,\lambda) + c \lambda \psi_4 (x,\lambda)}\right\}\\ \quad + A_2 \{\psi_2 (x,\lambda) - \psi_3 (x,\lambda)\,\cot \gamma\} \quad \textrm{if}\, \gamma \in (0,{\pi /2}]. \end{cases} \end{equation}

By (1.2) from (2.6) we get

(2.7)\begin{align} &A_1 \left( {\psi''_1 (0,\lambda ) + c \lambda \psi''_4 (0,\lambda )} \right) + A_3 \psi''_3 (0,\lambda ) = 0 \quad \textrm{if}\ \gamma = 0, \notag\\ &A_1 \left( {\psi''_1 (0,\lambda ) + c \lambda \psi''_4 (0,\lambda )} \right) + A_2 \left({\psi''_2 (0,\lambda ) - \psi''_3 (0,\lambda ) \cot \gamma}\right) = 0\quad \textrm{if}\ \gamma \in (0,{\pi \mathord{\left/ {\vphantom {\pi 2}} \right. } 2}]. \end{align}

For brevity, we use the following notations:

(2.8)\begin{align} C_1 (\lambda) &= \psi''_1 (0,\lambda ) + c \lambda \psi''_4 (0,\lambda ),\notag\\ C_2 (\lambda ) &= \begin{cases}\psi''_3 (0,\lambda ) & \textrm{if}\ \gamma = 0, \\ \psi ''_2 (0,\lambda ) - \psi ''_3 \,(0,\lambda )\cot \gamma & \textrm{if} \ \gamma \in (0,{\pi \mathord{\left/ {\vphantom {\pi 2}} \right. } 2}]. \end{cases} \end{align}

It can be seen from (2.7) that to complete the proof of theorem it suffices to show that for each $\lambda \in \mathbb {C}$ the relation

(2.9)\begin{equation} |\,C_1 (\lambda)| + |\,C_2 (\lambda)| > 0 \end{equation}

holds.

If $\lambda > 0$, then by the second part of [Reference Ben Amara and Vladimirov14, lemma 2.1] we get

(2.10)\begin{equation} \begin{aligned}\psi_1 (0,\lambda) > 0, \,\psi'_1 (0,\lambda) < 0, \,\psi''_1 (0,\lambda ) > 0, \,T \psi_1 (0,\lambda) < 0,\\ \psi_2 (0,\lambda) < 0, \,\psi'_2 (0,\lambda) > 0, \,\psi''_2 (0,\lambda ) < 0, \,T \psi_2 (0,\lambda) > 0,\\ \psi_3 (0,\lambda) > 0, \,\psi'_3 (0,\lambda) < 0, \,\psi''_3 (0,\lambda ) > 0, \,T \psi_3 (0,\lambda) < 0,\\ \psi_4 (0,\lambda) < 0, \,\psi'_4 (0,\lambda) > 0, \,\psi''_4 (0,\lambda ) < 0, \,T \psi_4 (0,\lambda) > 0.\end{aligned} \end{equation}

Indeed, by (1.1) and (2.4) for the function $\psi _1 (x,\lambda )$ we have

\[ \mathop {\lim }\limits_{\scriptstyle x \to 1 \atop \scriptstyle x < \,1 } \left( {T\psi _1 (x,\lambda )} \right)^{\prime} = \lambda \mathop {\lim }\limits_{\scriptstyle x \to 1 \atop \scriptstyle x < \,1 } \psi _1 (x,\lambda ) = \lambda > 0, \]

which implies that there exists $(T \psi _1)' (1,\lambda ) = \lambda > 0$. Consequently, there exists $\psi _1^{(4)} (1,\lambda ) = (T \psi _1)' (1,\lambda ) + q(1) \psi ''_1 (1,\lambda ) + q'(1) \psi '_1 (1,\lambda ) = (T \psi _1)' (1,\lambda ) > 0$. Hence $T \psi _1 (x,\lambda ) < 0$ and $\psi '''_1 (x,\lambda ) < 0$ in a sufficiently small left punctured neighbourhood $V_1^{-}$ of the point $x = 1$. Since $x = 1$ is a triple zero of the function $\psi _1' (x,\lambda )$ it follows that $\psi _1'' (x,\lambda ) > 0$, $\psi _1' (x,\lambda ) < 0$ for $x \in V_1^{-}$. Moreover, $\psi _1 (x,\lambda ) > 0$ for $x \in V_1^{-}$. Then it follows from the second part of [Reference Ben Amara and Vladimirov14, lemma 2.1] that $\psi _1 (0,\lambda ) > 0$, $\psi '_1 (0,\lambda ) < 0$, $\psi ''_1 (0,\lambda ) > 0$, $T \psi _1 (0,\lambda ) < 0$. The remaining relations in (2.10) for the functions $\psi _2 (x,\lambda )$, $\psi _3 (x,\lambda )$ and $\psi _4 (x,\lambda )$ are proved similarly.

Let $\lambda > 0$. Then, in view of $c < 0$, by (2.8) we have $C_1 (\lambda ) = \psi ''_1 (0,\lambda ) + c \lambda \psi ''_4 (0,\lambda ) > 0$, and consequently, (2.9) holds.

Now let $\lambda \in \mathbb {C} \,\backslash ({\,0, + \infty })$. If (2.8) fails for some such $\lambda$, then $C_1 (\lambda ) = C_2 (\lambda ) = 0$. Hence the functions $\psi _1 (x,\lambda ) + c \lambda \psi _4 (x,\lambda )$ and $\psi _3 (x,\lambda )$ for $\gamma = 0$, $\psi _2 (x,\lambda ) - \psi _3 (x,\lambda ) \cot \gamma$ for $\gamma \in (0,{\pi \mathord {\left / {\vphantom {\pi 2}} \right .} 2}]$ are solutions of problem (1.1), (1.2), (1.4), (1.5) for such $\lambda$. We consider the function $u (x,\lambda )$ which is defined as follows:

\[ u\,(x,\lambda ) = \begin{cases} \psi_3 (0,\lambda) \left( {\psi_1 (x,\lambda) + c \lambda \psi_4 (x,\lambda)} \right) - \left( {\psi_1 (0,\lambda) + c \lambda \psi_4 (0,\lambda)} \right)\\ \psi_3 (x,\lambda) \ \textrm{for}\ \gamma = 0,\\ \left({\psi_2 \,(0,\lambda ) - \psi_3 \,(0,\lambda ) \cot \gamma}\right) \left( {\psi_1 (x,\lambda) + c \lambda \psi_4 (x,\lambda)} \right)\\ \quad - \left( {\psi_1 (0,\lambda) + c \lambda \psi_4 (0,\lambda)} \right) \\ \left({\psi_2 \,(x,\lambda ) - \psi_3 \,(x,\lambda ) \cot \gamma}\right) \ \textrm{for}\ \gamma \in (0,{\pi \mathord{\left/ {\vphantom {\pi 2}} \right. } 2}]. \end{cases}\]

Note that $u (0, \lambda ) = 0$. Hence the function $u (x, \lambda )$ is an eigenfunction of the eigenvalue problem (1.1), (1.2), (2.1), (1.4), (1.5) for $\beta = 0$ and $\gamma \in [0,{\pi \mathord {\left / {\vphantom {\pi 2}} \right .} 2}]$. Then by theorem 2.1 we have $\lambda > 0$ which contradicts the condition $\lambda \in \mathbb {C} \, \backslash ({\,0, + \infty })$. The proof of this theorem is complete.

Remark 2.4 By (2.6)–(2.8), for each $\lambda \in \mathbb {C}$ the nontrivial solutions $y (x,\lambda )$ of problem (1.1), (1.2), (1.4), (1.5) are nonzero multiples of

(2.11)\begin{align} v (x,\lambda) &= C_2 (\lambda) \left\{{\psi_1 (x,\lambda) + c \lambda \psi_4 (x,\lambda)}\right\}\notag\\ &\quad - C_1 (\lambda) \left\{{ \textrm{sgn} \gamma \, \psi_2 (x,\lambda) - ({-}1)^{1- \textrm{sgn} \gamma} \left({1 + \textrm{sgn} \gamma \, (\cot \gamma - 1) }\right) \,\psi_3 (x,\lambda) }\right\}. \end{align}

As is known (see [Reference Naimark35, Ch. 1, § 2.1]) that for each fixed $x \in [0,1]$ the functions $\psi _{k} (x, \lambda )$, $k = 1,2,3,4$, and their derivatives are entire functions of $\lambda$, and consequently, $v (x,\lambda )$ is also an entire function of $\lambda$ for each fixed $x \in [0,1]$.

Proof. If $y (1,\lambda ) = 0$ for some $\lambda > 0$, then from (1.5) we get $T y (1,\lambda ) = 0$. Since $\gamma \in [0,{\pi \mathord {\left / {\vphantom {\pi 2}} \right .} 2}]$ it follows from (1.4) that $y'(1,\lambda ) y''(1,\lambda ) \le 0$. Then by the second part of [Reference Ben Amara and Vladimirov14, lemma 2.1] we have $y' (0,\lambda ) \, y'' (0,\lambda ) < 0$ which contradicts the condition (1.2).

If $y (0,\lambda ) = 0$ for some $\lambda \le 0$, then $y (x,\lambda )$ is an eigenfunction of problem (1.1), (1.2), (2.1), (1.4), (1.5) for $\beta = 0$ and $\gamma \in [0,{\pi \mathord {\left / {\vphantom {\pi 2}} \right .} 2}]$. Then by theorem 2.1 we have $\lambda > 0$ which contradicts the condition $\lambda \le 0$. The proof of this lemma is complete.

Now, using lemma 2.5, we can normalize the function $y(x,\lambda ),x \in [0,1],\lambda \in \mathbb {R}$, as follows:

(2.12)\begin{equation} y (1,\lambda) = 1 \end{equation}

if $\lambda > 0$, and

(2.13)\begin{equation} y (0, \lambda) = 1 \end{equation}

if $\lambda \le 0$.

For $\lambda \in \mathbb {R}$ we consider the following equation

\[ y(x,\lambda) = 0, \,x \in [\,0,1]. \]

It is obvious that the zeros of this equation are functions of the parameter $\lambda$.

Proof. Let $\lambda _0$ be an arbitrary fixed positive number. If $y (x_0,\lambda _0) = 0$ for $x_0 \in (0,1)$, then it follows from [Reference Ben Amara and Vladimirov14, lemma 2.2] that $y' (x_0,\lambda _0) \ne 0$. If $y (0,\lambda _0) = y' (0,\lambda _0) = 0$, then in view of (1.2), by the first part of [Reference Ben Amara and Vladimirov14, lemma 2.1] we have $y ' (1,\lambda _0) y '' (1,\lambda _0) > 0$ in contradiction with the boundary condition (1.4).

Let $\lambda _0 \le 0$ and $x_0 \in [0, 1)$ such that $y (x_0,\lambda _0) = y' (x_0,\lambda _0) = 0$. Then $y (x,\lambda _0)$ solves the eigenvalue problem defined on $[x_0, 1]$ and determined by equation (1.1) with the boundary conditions $y (x_0) = y' (x_0) = 0$ and (1.4), (1.5). By theorem 2.1 the eigenvalues of this problem are simple and positive which contradicts the condition $\lambda _0 \le 0$.

The continuous differentiability of the zeros contained in $[0,1)$ of the function $y (x, \lambda )$ follows from the well-known implicit function theorem, and the proof of this lemma is complete.

By lemma 2.5, lemma 2.6 implies the following statement.

We consider the function

\[ H(x,\lambda) = \frac{{y (x,\lambda )}}{{T y (x,\lambda )}}\,. \]

By theorem 2.3, remark 2.4 and lemma 2.6 the function $H (x,\lambda )$ is a finite order meromorphic function of $\lambda$ for each fixed $x \in [\,0,1]$.

Let $\mathcal {D}_k = ( {\lambda _{k - 1} (0,\gamma ),\lambda _k (0,\gamma )} )$, $k \in \mathbb {N}$, where $\lambda _{0} (0,\gamma ) = -\, \infty$.

Obviously, the function

\[ F(\lambda ) = \frac{1}{{H(0,\lambda )}} = \frac{{T y (0,\lambda )}}{{y (0,\lambda )}} \]

which is well defined for

\[ \lambda \in \mathcal{D} \equiv \left( {\mathbb{C}\backslash \mathbb{R}} \right) \cup \left( {\bigcup\limits_{k = 1}^{\infty} {\mathcal{D}_k} } \right) \]

and is a meromorphic function of finite order. The eigenvalues $\lambda _k (0,\gamma )$ and $\lambda _k ( {{\pi \mathord {\left / {\vphantom {\pi 2}} \right .} 2}, \gamma }),k = 1,2,.\,.\,.\,,$ of problem (1.1), (1.2), (2.1), (1.4), (1.5) for $\beta = 0$ and $\beta = {\pi \mathord {\left / {\vphantom {\pi 2}} \right .} 2}$ are poles and zeros of function $F (\lambda )$, respectively.

Lemma 2.8 For each $\lambda \in \mathcal {D}$ the relation

(2.14)\begin{equation} \frac{{dF (\lambda )}}{{d\lambda }} ={-}\,\frac{1}{{y^{2} (0,\lambda )}}\left\{ {\int_0^{1} y^{2} (x,\lambda )\,\textrm{d}x - \,c y^{2} (1, \lambda)} \right\} \end{equation}

holds.

Proof. By virtue of equation (1.1) we have

(2.15)\begin{equation} \left( {Ty (x,\mu)} \right)^{\prime} y(x,\lambda ) - \left( {T y (x,\lambda )} \right)^{\prime} y(x,\mu ) = (\mu - \lambda )y(x,\mu )y(x,\lambda). \end{equation}

Integrating equality (2.15) from $0$ to $1$, using the formula for the integration by parts and taking boundary conditions (1.2), (1.4) and (1.5) into account we obtain

(2.16)\begin{align} &-Ty (0, \mu )\,y (0,\lambda ) + T y (0,\lambda )\,y (0,\mu )\notag\\ &\quad =(\mu - \lambda ) \left\{ {\int_0^{1} y(x,\mu )\,y(x,\lambda )\textrm{d}x - c y(1,\mu)\,y(1,\lambda )} \right\}. \end{align}

By (2.16) for $\mu ,\lambda \in \mathcal {D},\mu \ne \lambda$, we have

(2.17)\begin{align} &\frac{Ty(0,\mu )}{y(0,\mu)} - \frac{Ty(0,\lambda)}{y(0,\lambda )}\notag\\ &\quad = - (\mu - \lambda) \frac{1} {y(0,\mu)\,y(0,\lambda)} \left\{ {\int_0^{1} {y(x,\mu )y(x,\lambda )\textrm{d}x} - cy(1,\mu )y(1,\lambda )} \right\}. \end{align}

Dividing both sides of relation (2.17) by $\mu - \lambda$ ($\mu \ne \lambda$) and by passing to the limit as $\mu \to \lambda$ we get (2.14). The proof of this lemma is complete.

Lemma 2.10 The following relation holds:

(2.18)\begin{equation} \mathop {\lim }\limits_{\lambda \to - \,\infty } F (\lambda ) ={+} \,\infty . \end{equation}

Proof. In equation (1.1) we set $\lambda = \rho ^{4}$. By theorem 1 of [Reference Mukhtarov and Aydemir35, Ch. II, § 4.5] in each subdomain $\mathcal {T}$ of the complex $\rho$-plane this equation has four linearly independent solutions $\varphi _{k}(x, \rho ), k = 1,2,3,4$, which are regular with respect to $\rho$ (for sufficiently large $|\,\rho |$) and satisfying the following relations

(2.19)\begin{equation} \varphi_k^{(s)} (x,\rho ) = (\rho \omega _k )^{s} e^{\rho \omega _k x} \,\left\{ {1 + O\left( {{\rho }^{{-}1} } \right)} \right\},\quad k = 1,2,3,4, \enspace s = 0,1,2,3, \end{equation}

where $\omega _{k},k = 1,2,3,4$, are distinct fourth roots of unity.

Let $\lambda < 0$. Then, without loss of generality, we can assume that $\rho$ lies on the bisector of the first quadrant, and the numbers $\omega _{k},k = 1,2,3,4$, are numbered in the following order: $\omega _1 = - 1$, $\omega _2 = i$, $\omega _3 = - i$ and $\omega _4 = 1$.

For brevity, we introduce the notation

\[ [1] = 1 + O (\rho^{{-}1}). \]

Assuming that the initial condition $y(0, \lambda ) = 1$ is imposed, the unique solution of (1.1), (1.2), (1.4), (1.5) together with the initial condition $y(0, \lambda ) = 1$ can be written in the form

\[ y(x, \lambda) = \sum\limits_{k = 1}^{4} {B_k } (\rho )\,\varphi _k (x,\rho ). \]

Writing $B = (B_1,B_2,B_3,B_4)^{T}$, the coefficients $C_k(\rho )$ are solution of the linear algebraic system

\[ M (\rho) \,B(\rho) = (0,0,0,1)^{T}, \]

where the matrix $M (\rho )$ is given by

\[ M (\rho) = \left( \begin{array}{rrrr} {[}1] & -[1] & -[1] & [1]\\ -e^{-\rho}[1] & ie^{ i\rho }[1] & -ie^{-i\rho }[1] & e^{\rho}[1] \\ e^{-\rho}[1] & e^{i\rho} [1] & e^{- i\rho} [1] & e^{\rho}[1] \\ {[}1] & [1] & [1] & [1] \end{array} \right) \]

for $\gamma = 0$, and

\[ M (\rho) = \left( \begin{array}{rrrr} {[}1] & -[1] & -[1] & [1]\\ e^{-\rho}[1] & -e^{ i\rho }[1] & -e^{-i\rho}[1] & e^{\rho}[1]\\ e^{-\rho}[1] & e^{i\rho} [1] & e^{- i\rho} [1] & e^{\rho}[1]\\ {[}1] & [1] & [1] & [1] \end{array} \right) \]

for $\gamma \in (0,{\pi \mathord {\left / {\vphantom {\pi {2].}}} \right .} {2]}}$.

The solution of the system $M (\rho ) \,B (\rho ) = (0,0,0,1)^{T}$ is

\begin{align*} B_1 (\rho) &= \frac{1}{2} [1],\quad B_2 (\rho) ={-}\frac{{\,(1+i)\,e^{-\,i\rho}}}{{2 ((1-i) e^{i\rho} - (1+i)e^{{-}i\rho})}}[1],\\ B_3 (\rho) &= \frac{{(1-i)\,e^{ i\rho}}}{{2 ((1-i) e^{i\rho} - (1+i)e^{{-}i\rho})}}[1],\quad B_4 (\rho) ={-}\frac{2ie^{-\,\rho}}{2 ((1-i) e^{i\rho} - (1+i)e^{{-}i\rho})}[1], \end{align*}

if $\gamma = 0$, and

\begin{align*} B_1 (\rho) &= \frac{1}{2} [1],\quad B_2 (\rho) = \frac{{-\,e^{-\,i\rho}}}{{2 (e^{i\rho} - e^{-\,i\rho})}}[1],\\ B_3 (\rho) &= \frac{{\,e^{ i\rho}}}{{2 (e^{i\rho} - e^{-\,i\rho})}}[1],B_4 (\rho) = \frac{1}{2}\,e^{{-}2\,\rho}[1]. \end{align*}

if $\gamma \in (0,{\pi \mathord {\left / {\vphantom {\pi {2].}}} \right .} {2]}}$. Then for $F (\lambda ) = {{Ty(0,\lambda )} \mathord {\left / {\vphantom {{Ty(0,\lambda )} {y(0,\lambda )}}} \right .} {y(0,\lambda )}}$ we get the following representation

(2.20)\begin{equation} \begin{array}{l} F(\lambda) = \{((1\,{-}\,i) e^{i\rho} - (1\,{+}\,i)e^{{-}i\rho}) [1] - (1\,{+}\,i) \, e^{- i\rho } [1] + (1-i) \,e^{i \rho} [1] + 2i e^{- \rho}[1]\}^{{-}1} \\ \\\,\, \rho^{3} \{-((1-i) e^{i\rho} - (1+i)e^{{-}i\rho}) [1] - (1-i) \, e^{- i\rho } [1] + (1+i) \,e^{i \rho} [1] + 2ie^{-\rho} [1]\} \\ \end{array} \end{equation}

if $\gamma = 0$, and

(2.21)\begin{equation} \begin{array}{l} F (\lambda) = \rho^{3} \{(e^{i\rho} - e^{{-}i\rho}) [1] - \, e^{- i\rho } [1] + \,e^{i \rho} [1] - e^{{-}2\rho}(e^{i\rho} - e^{{-}i\rho}) [1]\}^{{-}1} \\ \\ \,\,\, \{-( e^{i\rho} - e^{{-}i\rho}) [1] + i \, e^{- i\rho } [1] + i \,e^{i \rho} [1] - e^{{-}2\rho}(e^{i\rho} - e^{{-}i\rho}) [1] \} \\ \end{array} \end{equation}

if $\gamma \in (0,{\pi \mathord {\left / {\vphantom {\pi {2].}}} \right .} {2]}}$.

Since $\rho$ lies on the bisector of the first quadrant it follows that $\rho = (1 + i) u$, where $u > 0$, and consequently, $|\rho | = \sqrt 2\,u$. Then, from (2.20) and (2.21) by a straightforward computation, we obtain

(2.22)\begin{align} F(\lambda ) &={-} (1-i)^{{-}1}\rho^{3} [1] ={-} (1-i)^{{-}1}(1+i)^{3} u^{3} [1] = (\sqrt 2)^{{-}1}|\rho|^{3} (1 + O(|\rho|^{{-}1})\notag\\ & = (\sqrt{2})^{{-}1}\,\sqrt[4]{{\,|\lambda |^{3} }} \left({ 1 + O \left( {\left({\sqrt[4]{|\lambda|}\,}\right)^{{-}1} } \right) }\right)\, \textrm{as} \,\lambda \to - \infty. \end{align}

The proof of this lemma is complete.

Lemma 2.11 Let $x \in [0,1)$ and $\lambda > 0$ such that $y (x,\lambda ) = 0$. Then

(2.23)\begin{equation} \frac{{\partial H(x,\lambda )}}{{\partial x}} < 0. \end{equation}

Proof. Let $y (x,\lambda ) = 0$ for some $x \in [0,1)$ and $\lambda > 0$. If $x \in (0,1)$, then it follows from [Reference Ben Amara and Vladimirov14, lemma 2.2] that $y' (x,\lambda ) \,Ty (x,\lambda ) < 0$. If $y (0,\lambda ) = 0$, then in view of (1.2), by the first part of [Reference Ben Amara and Vladimirov14, lemma 2.1] we have $y' (0,\lambda ) Ty (0,\lambda ) < 0$. Therefore, by virtue of (1.1), we get

\[ \frac{{\partial H(x,\lambda )}}{{\partial x}} = \frac{\partial }{\partial x} \left({\frac{{y (x,\lambda )}}{{T y (x,\lambda )}}}\right) = \frac {y' (x,\lambda) \,Ty (x,\lambda) - \lambda y^{2} (x,\lambda)}{\left({T y (x,\lambda)}\right)^{2}} = \frac{y' (x,\lambda )}{T y (x,\lambda )} < 0. \]

The proof of lemma 2.11 is complete.

By $\tau (\lambda )$ we denote the number of zeros of the function $y (x, \lambda )$ contained in $(0, 1)$.

Proof. By corollary 2.7 as $\lambda > 0$ varies the zeros of the function $y(x,\lambda )$ can enter or leave the interval $(0, 1)$ only through the endpoint $x = 0$. Moreover, by lemma 2.6 and the implicit function theorem for every zero $x (\lambda )$ of the function $y (x,\lambda )$ the following relation holds:

\[ x' (\lambda) ={-} \frac{H'_{\lambda} (x,\lambda)}{H'_{x} (x,\lambda)}. \]

If $x (\lambda ) = 0$, then it follows from this and relations (2.14), (2.23) that

\[ x' (\lambda) > 0. \]

Therefore, as $\lambda > 0$ increases the zeros of the function $y (x, \lambda )$ cannot leave the interval $(0, 1)$ through the point $x = 0$. Hence, as $\lambda$, $\mu < \lambda < \nu$, increases the number of zeros of the function $y (x, \lambda )$ cannot decrease, i.e. $\tau (\mu ) \le \tau (\nu )$. The proof of this lemma is complete.

Proof. It is obvious that $\psi _1 (x,0) \equiv 1$. Then by (2.8), (2.13) it follows from (2.11) that $y (x,0) \equiv 1$. Hence, for all $\lambda \in \mathbb {R}$ sufficiently close to zero, the function $y (x,\lambda )$ has no zeros in $(0,1)$. Moreover, by theorem 2.1 we have $\tau (\lambda _k (0,\gamma ) = k - 1$, $k = 1,2,.\,.\,.\,$. Therefore, by lemma 2.12 it follows that if $\lambda > 0$ and $\lambda \in (\lambda _{k-1} (\gamma ,0),\lambda _{k-1} (\gamma ,0)$, then $\tau (\lambda ) = k - 1$. The proof of theorem 2.13 is complete.

It follows from lemma 2.6 that as $\lambda < 0$ varies then the zeros of the function $y(x,\lambda )$ can enter or leave the interval $(0, 1)$ only through the endpoint $x = 1$. To find the number of zeros contained in the interval $(0, 1)$ of the function $y (x,\lambda )$ for $\lambda < 0$ consider the following spectral problem

(2.24)\begin{align} \ell (y) (x) &= \lambda y(x),0 < x < 1,\notag\\ y''(0) &= y' (1) \cos \gamma + y''(1) \sin \gamma = y(1) = Ty (1) = 0.\end{align}

It follows from the second part of [Reference Ben Amara and Vladimirov14, lemma 2.1] that the eigenvalues of problem (2.24) cannot be positive. Let $\eta$ be a real eigenvalue of this problem and $\epsilon > 0$ be the fixed sufficiently small number. The oscillation index of the eigenvalue $\eta$ which denotes by $i (\eta )$ is the difference between the number of zeros of the function $y (x,\lambda )$ for $\lambda \in (\eta - \epsilon ,\eta )$ contained in the interval $(0, 1)$ and the number of the same zeros for $\lambda \in (\eta , \eta + \epsilon )$. This definition directly implies that the number of zeros of the function $y (x,\lambda )$ for $\lambda < 0$ contained in the interval $(0, 1)$ is equal to the sum of the oscillation indices of all eigenvalues of problem (2.24) contained in the interval $(\lambda , 0)$ (see [Reference Ben Amara and Vladimirov13, § 4]).

By following the arguments in theorem 4.1 of [Reference Ben Amara and Vladimirov13] one can justify the following statement.

Theorem 2.14 There exists $\xi < 0$ such that the eigenvalues $\eta _k$, $k =1,2,.\,.\,.\,$, of problem (2.24) are simple, lying on the interval $(-\,\infty , \xi )$, form an unbounded decreasing sequence $\{\eta _k\}_{k=1}^{\infty }$ such that $i (\eta _k) = 1$, $k \in \mathbb {N}$, and

\[ \eta_k ={-} \,4\,\pi^{4} k^{4} + o (k^{4}). \]

Now, based on the above reasoning, we obtain the following formula for the number of zeros contained in $(0, 1)$ of the function $y (x,\lambda )$ for $\lambda < 0$:

(2.25)\begin{equation} \tau (\lambda ) = \sum\limits_{\eta _k \in \,(\lambda ,0)} {i (\eta _k )}. \end{equation}

3. The location of eigenvalues and the oscillatory properties of eigenfunctions of problem (1.1)–(1.5)

Proof. Note that the eigenvalues of problem (1.1)–(1.5) are the roots of the equation

(3.1)\begin{equation} Ty (0,\lambda ) - a \lambda \, y (0,\lambda) = 0. \end{equation}

Let $\lambda$ be the nonreal eigenvalue of this problem. Then $\bar \lambda$ is also eigenvalue of (1.1)–(1.5) because the coefficients $q(x)$, $a$, $c$ and $\gamma$ are real. Moreover, in this case $y(x, \bar \lambda ) = \overline {y(x,\lambda )}$, and consequently, equality (3.1) holds for $\bar \lambda \,$.

Setting $\mu = \bar \lambda$ in relation (2.16) and taking (1.4) into account we get

(3.2)\begin{equation} - a \, ({\overline \lambda} - \lambda) \,|y (0,\lambda)|^{2} = \,({\overline \lambda} - \lambda) \left\{ { \int_0^{1} |y(x,\lambda)|^{2} \textrm{d}x - c |y(1,\lambda)|^{2} } \right\}, \end{equation}

whence, by ${\overline \lambda } \ne \lambda$, implies that

(3.3)\begin{equation} \int_0^{1} |y(x,\lambda)|^{2} \textrm{d}x + a \,|y (0,\lambda)|^{2}- c |y(1,\lambda)|^{2} = 0. \end{equation}

Putting $y (x,\lambda )$ in (1.1)–(1.5), then multiplying both sides of (1.1) by $\overline {y (x,\lambda )}$, integrating this relation from $0$ to $1$, using the formula for the integration by parts, and taking into account conditions (1.2)–(1.5), we obtain

(3.4)\begin{align} &\int_0^{1} {\,|y'' (x,\lambda)|^{2} \textrm{d}x + \int_0^{1} {q(x)\,|y' (x,\lambda )|^{2}}} \textrm{d}x + \mathcal{N} [y (x, \lambda)]\notag\\ &\quad = \lambda \left\{ { \int_0^{1} |y(x,\lambda)|^{2} \textrm{d}x + a \,|y (0,\lambda)|^{2}- c |y(1,\lambda)|^{2} } \right\}, \end{align}

where $\mathcal {N} [y (x,\lambda )] = 0$ for $\gamma = 0$, $\mathcal {N} [y (x,\lambda )] = |y' (1,\lambda )|^{2} \cot \gamma$ for $\gamma \in (0,{\pi \mathord {\left / {\vphantom {\pi {2]}}} \right .} {2]}}$. Hence it follows from (3.3) and (3.4) that

(3.5)\begin{equation} \begin{array}{l} \,\,\int_0^{1} {\,|y'' (x,\lambda)|^{2} \textrm{d}x + \int_0^{1} {q(x)\,|y' (x,\lambda )|^{2}}} \textrm{d}x + \mathcal{N} [y (x, \lambda)] = 0. \end{array} \end{equation}

By the boundary condition (1.3), relation (3.5) implies that $y (x,\lambda ) \equiv 0$, a contradiction.

By the above arguments the entire function on the left-hand side of (3.1) does not vanish for non-real $\lambda$. Consequently, it does not vanish identically. Therefore, its zeros form an at most countable set without finite limit point. The proof of this lemma is complete.

Proof. Let $\lambda$ be an eigenvalue of (1.1)–(1.5). Then by remark 3.2 we have $y (0,\lambda ) \ne 0$. Therefore each root (with regard of multiplicities) of equation (3.1) is a root of the equation

(3.6)\begin{equation} F (\lambda) = a \lambda. \end{equation}

Let $\lambda = \lambda ^{*}$ be a multiple root of (3.6). Then the following relations hold:

(3.7)\begin{equation} F (\lambda^{*}) = a \lambda^{*},F' (\lambda^{*}) = a. \end{equation}

By remark 3.2 and formula (2.14), the second relation of (3.7) implies that

(3.8)\begin{equation} \int_0^{1} {{\,y^{2}(x,\lambda^{*})} \,\textrm{d}x + a \, {y^{2} (0,\lambda^{*})} - c \,y^{2} (1,\lambda^{*}) } = 0. \end{equation}

Since $\lambda ^{*} \in \mathbb {R}$ it follows from (3.4) that

(3.9)\begin{align} &\int_0^{1} {\,y''^{2} (x,\lambda^{*}) \textrm{d}x + \int_0^{1} {q(x)\,y'^{2} (x,\lambda^{*})}} \textrm{d}x + \mathcal{N} [y (x,\lambda^{*})]\notag\\ &\quad =\lambda \left\{ { \int_0^{1} y^{2} (x,\lambda^{*}) \textrm{d}x + a \,y^{2} (0,\lambda^{*})- c \,y^{2} (1,\lambda^{*}) } \right\}, \end{align}

whence, by virtue of (3.8), we get

(3.10)\begin{equation} \int_0^{1} {\,y''^{2} (x,\lambda^{*}) \textrm{d}x + \int_0^{1} {q(x)\,y'^{2} (x,\lambda^{*})}} \textrm{d}x + \mathcal{N} [y (x,\lambda^{*})] = 0. \end{equation}

Consequently, by condition (1.3), from (3.10) we obtain $y (x,\lambda ^{*}) \equiv 0$ which contradicts the condition $y (x,\lambda ^{*}) \not \equiv 0$. The proof of this lemma is complete.

Following the reasoning in [Reference Aliyev and Guliyeva6, lemma 3.3], we can justify the following result.

Lemma 3.4 One has the following representation:

(3.11)\begin{equation} F(\lambda ) = \sum\limits_{k = 1}^{\infty} {\frac{{\lambda c_k }}{{\lambda _k (0,\gamma )\left( {\lambda - \lambda _k (0,\gamma )} \right)}}}\,, \end{equation}

where $c_k = \mathop {\textrm {res}}\limits _{\lambda = \lambda _k (0,\gamma )} F(\lambda )$ and $c_k > 0$, $k =1, \,2,.\,.\,.\,$.

We have the following oscillation theorem for problem (1.1)–(1.5).

Theorem 3.5 For each $\gamma \in [0,{\pi \mathord {\left / {\vphantom {\pi 2}} \right .} 2}]$ the eigenvalues of problem (1.1)–(1.5) form an unbounded nondecreasing sequence $\{\lambda _{k} (\gamma )\}_{k = 1}^{\infty }$ such that

\begin{align*} &\lambda_1 (\gamma) < 0 = \lambda_2 (\gamma) < \lambda_3 (\gamma) < \ldots < \lambda_k (\gamma)< \ldots\ {if}\ a > c - 1,\\ &\lambda_1 (\gamma) = \lambda_2 (\gamma) = 0 < \lambda_3 (\gamma) < \ldots < \lambda_k (\gamma)< \ldots\ {if}\ a = c - 1,\\ &\lambda_1 (\gamma) = 0 < \lambda_2 (\gamma) < \lambda_3 (\gamma) < \ldots < \lambda_k (\gamma)< \ldots\ {if}\ a < c - 1, \end{align*}

$($in the case $c = a + 1$ the eigenvalue $\lambda _1 (\gamma ) = 0$ is double, and it corresponds to the chain consisting of the eigenfunction $y_{1,\gamma } (x)$ and the associated function $y_{2,\gamma } (x))$. The eigenfunction $y_k (x)$, corresponding to the eigenvalue $\lambda _k$, for $k \ge 3$ has exactly $k - 2$ simple zeros in $(0,1)$; moreover, if $a < c - 1$, then the eigenfunctions $y_{1,\gamma } (x)$ and $y_{2,\gamma } (x)$ have no zeros in $(0,1)$, if $a = c - 1$, then the eigenfunction $y_{1,\gamma } (x)$ has no zeros in $(0,1)$, if $a > c - 1$, then $y_{2,\gamma } (x)$ has no zeros in $(0,1)$ and the number of zeros of the eigenfunction $y_{1,\gamma } (x)$ in $(0,1)$ is equal $\sum \limits _{\eta _k \in (\lambda _1 (\gamma ),0)} {i(\eta _k )}$.

Proof. Recall that the eigenvalues of problem (1.1)–(1.5), taking into account their multiplicities, are the roots of equation (3.6). It follows from (3.11) that

\[ F''(\lambda) = 2 \sum\limits_{k = 1}^{\infty} {\frac{{ c_k }}{{\left( {\lambda - \lambda _k (0,\gamma )} \right)^{3}}}},\lambda \in \mathcal{D}. \]

From this we obtain the relation

(3.12)\begin{equation} F''(\lambda) < 0\ \textrm{for}\ \lambda \in \mathcal{D}_1, \end{equation}

i.e. the function $F(\lambda )$ is concave in $\mathcal {D}_1$. Moreover, by (2.14), (2.18) and (3.11) we have

(3.13)\begin{align} F (0) &= 0, F' (0) = c - 1, \end{align}

(3.14)\begin{align} \mathop {\lim }\limits_{\lambda \to - \,\infty } F(\lambda ) &={+} \,\infty,\mathop {\lim }\limits_{\lambda \to \lambda _1 (0,\gamma ) - \,0} F(\lambda ) ={-} \,\infty . \end{align}

Let $f(\lambda ) = F (\lambda ) - a \lambda$. Then it follows from relations (3.12)–(3.14) that

\begin{align*} &f'' (\lambda) < 0 \,\textrm{for }\,\lambda \in \mathcal{D}_1, \\ &f (0) = 0\,\textrm{and}\,f' (0) = c - 1 - a. \end{align*}

Moreover, by (2.22) and (3.14) we have

(3.15)\begin{equation} \mathop {\lim }\limits_{\lambda \to - \,\infty } f (\lambda ) ={-} \,\infty,\mathop {\lim }\limits_{\lambda \to \lambda _1 (0,\gamma ) - \,0} f(\lambda ) ={-} \,\infty . \end{equation}

If $a > c- 1$, then $f' (0) < 0$. Since $f'' (\lambda ) < 0$ in $\mathcal {D}_1$ it follows that $f' (\lambda ) < f' (0) < 0$ for $\lambda \in (0,\lambda _1 (0,\gamma ))$. Then $f (\lambda ) < 0$ for $\lambda \in (0,\lambda _1 (0,\gamma ))$. By the relations $f(0) = 0$ and $f' (0) < 0$ we have $f (\lambda ) > 0$ for all $0 > \lambda$ small. Then it follows from the first relation of (3.15) that there exists $\lambda _{*} (\gamma ) \in (-\,\infty ,0)$ such that $f (\lambda _{*} (\gamma )) = 0$. Hence there exists $\lambda _{**} (\gamma ) \in (\lambda _{*} (\gamma ),0)$ such that $f' (\lambda _{**} (\gamma )) = 0$. Consequently, $f' (\lambda ) > 0$ for $\lambda \in (-\infty ,\lambda _{**} (\gamma ))$ and $f' (\lambda ) < 0$ for $\lambda \in (\lambda _{**} (\gamma ), \lambda _1 (0,\gamma ))$. Therefore, in this case equation (3.6) in the interval $\mathcal {D}_1$ has two simple roots $\lambda _1 (\gamma ) < \lambda _2 (\gamma )$, where $\lambda _1 (\gamma ) = \lambda _{*} (\gamma ) < 0$ and $\lambda _2 (\gamma ) = 0$.

Let $a = c- 1$. Then we have $f' (0) = 0$. Since $f'' (\lambda ) < 0$ in $\mathcal {D}_1$ it follows that $f' (\lambda ) > 0$ for $\lambda \in (-\infty ,0)$ and $f' (\lambda ) < 0$ for $\lambda \in (0, \lambda _1 (0,\gamma ))$. Hence by relations (3.15) $f (\lambda ) < 0$ for $\lambda \in (-\infty ,0) \cup (0, \lambda _1 (0,\gamma ))$. Therefore, in this case $f (0) = f'(0) = 0$, $f'' (0) < 0$ and $f (\lambda ) \ne 0$ for $\lambda \in D_1 \backslash \{0\}$, i.e. equation (3.6) has one double root $\lambda _1 (\gamma ) = \lambda _2 (\gamma ) = 0$ for $a = c - 1$.

If $a < c- 1$, then $f' (0) > 0$. Hence $f' (\lambda ) > f' (0) > 0$ for $\lambda \in (-\,\infty , 0)$. Then $f (\lambda ) < 0$ for $\lambda < 0$. Since $f(0) = 0$ and $f' (0) > 0$ it follows that $f (\lambda ) > 0$ for all $0 < \lambda$ small. Then by virtue of second relation of (3.15) there exists $\lambda ^{*} (\gamma ) \in (0,\lambda _1 (0,\gamma )$ such that $f (\lambda ^{*} (\gamma )) = 0$. Hence there exists $\lambda ^{**} (\gamma ) \in (0,\lambda ^{*} (\gamma ))$ such that $f' (\lambda ^{**} (\gamma )) = 0$. Then $f' (\lambda ) > 0$ for $\lambda \in (-\infty ,\lambda ^{**} (\gamma ))$ and $f' (\lambda ) < 0$ for $\lambda \in (\lambda ^{**} (\gamma ), \lambda _1 (0,\gamma ))$. Thus, in this case equation (3.6) in the interval $\mathcal {D}_1$ has two simple root $\lambda _1 (\gamma ) < \lambda _2 (\gamma )$, where $\lambda _1 (\gamma ) = 0$ and $\lambda _2 (\gamma ) = \lambda ^{*} (\gamma ) > 0$.

By theorem 2.13 and formula (2.25) it follows from the above reasoning that $\tau (\lambda _1 (\gamma )) = \sum _{\eta _k \in \,(\lambda _1 (\gamma ), 0)} {i (\eta _k )}$ and $\tau (\lambda _2 (\gamma )) = 0$ for $a > c - 1$, $\tau (\lambda _1 (\gamma )) = 0$ for $a = c - 1$, and $\tau (\lambda _1 (\gamma )) = \tau (\lambda _2 (\gamma )) = 0$ for $a < c - 1$.

From representation (3.11) we obtain the following relations

(3.16)\begin{equation} \mathop {\lim }\limits_{\lambda \to \, \lambda_{k-1} (0,\gamma) + \,0 } F(\lambda ) ={+} \,\infty,\mathop {\lim }\limits_{\lambda \to \lambda_k (0,\gamma ) - \,0} F(\lambda ) ={-} \,\infty, k = 2,3,.\,.\,.\,. \end{equation}

If equation (3.6) has a root in the interval $\mathcal {D}_k$ for $k \ge 2$, then by lemma 3.3 this root must be simple. Since $F (\lambda )$ is continuous in each of intervals $\mathcal {D}_k$, $k \in \mathbb {N}$, by (3.16) it follows that equation (3.6) has at least one root in each of intervals $\mathcal {D}_k$, $k \ge 2$. Let us show that this equation has only one simple root in $\mathcal {D}_k$ for $k \ge 2$. Indeed, if (3.6) has more than one root, then the two smallest roots $\lambda _{k1}^{*} < \lambda _{k2}^{*}$ satisfy

(3.17)\begin{equation} F' (\lambda_{k1}^{*}) - a < 0\,\textrm{and}\,F' (\lambda_{k2}^{*}) - a > 0. \end{equation}

On the other hand it follows from (3.4) that

\begin{align*} &\int_0^{1} {\,y''^{2} (x,\lambda_{k2}^{*}) \textrm{d}x + \int_0^{1} {q(x)\,y'^{2} (x,\lambda_{k2}^{*})}} \textrm{d}x + \mathcal{N} [y(x,\lambda_{k2}^{*})]\\ &\quad = \lambda_{k2}^{*} \left\{ { \int_0^{1} y^{2}(x,\lambda_{k2}^{*}) \textrm{d}x + a \,y^{2} (0,\lambda_{k2}^{*})- c \,y^{2}(1,\lambda_{k2}^{*}) } \right\}, \end{align*}

whence, by $\lambda _{k2}^{*} > 0$, we get

\[ \int_0^{1} y^{2}(x,\lambda_{k2}^{*}) \textrm{d}x + a \,y^{2} (0,\lambda_{k2}^{*})- c \,y^{2}(1,\lambda_{k2}^{*}) > 0. \]

By (2.14) we obtain from the last relation

\[ F' (\lambda_{k2}^{*}) - a < 0, \]

which contradicts the second relation of (3.17). Therefore, problem (1.1)–(1.5) in each interval $\mathcal {D}_k$, $k \ge 2$, has a unique simple eigenvalue $\lambda _{k + 1} (\gamma )$. Then it follows from theorem 2.13 that $\tau (\lambda _{k + 1} (\gamma )) = k - 1$. The proof of theorem 3.5 is complete.

4. Asymptotic behaviour of eigenvalues and eigenfunctions of problem (1.1)–(1.5)

Let

$$ \nu _\gamma = {{(3 + {\mathop {\textrm sgn}} \gamma )}/ 4},\tilde \nu _\gamma = \nu _\gamma + 1, \,\gamma \in [0,{\pi /{2].}} $$

By [Reference Kerimov and Aliev30, theorem 3.1] we have the following asymptotic formulas

(4.1)\begin{align} \sqrt[4]{{\lambda_k (0, \gamma) }} &= \left({k - \nu_{\gamma}}\right) \pi + O \left( { k^{{-}1} } \right), \end{align}

(4.2)\begin{align} y_{k,0,\gamma} (x) &= \sin \left({k - \nu_{\gamma} }\right) \pi x - (1 - \textrm{sgn} \gamma) ({-}1)^{k} ( {\sqrt 2\,} )^{ - 1} e^{\left({k - \nu_{\gamma} }\right) \,\pi \,(x-1)} + O \left( { k^{{-}1} } \right), \end{align}

where relation ( 4.2) holds uniformly for $x \in [\,0,1]$.

Theorem 4.1 The following asymptotic formulas hold:

(4.3)\begin{align} \sqrt[4]{{\lambda_k (\gamma) }} &= \left({k - \tilde \nu_{\gamma}}\right) \pi + O \left( { k^{{-}1} } \right), \end{align}

(4.4)\begin{align} y_{k,\gamma} (x) &= \sin \left({k - \tilde \nu_{\gamma} }\right) \pi x + (1 - \textrm{sgn} \gamma) ({-}1)^{k} \left( {\sqrt 2\,} \right)^{ - 1} e^{\left({k - \tilde \nu_{\gamma} }\right) \pi (x-1)} + O \left( { k^{{-}1} } \right), \end{align}

where relation (4.4) holds uniformly for $x \in [\,0,1]$.

Proof. Taking (2.19) into account in the boundary conditions (1.2)–(1.4) we get

(4.5)\begin{equation} \sqrt[4]{{\lambda_{k + i} (\gamma) }} = \left({k - \nu_{\gamma} + 1}\right) \pi + O \left( { k^{{-}1} } \right), \end{equation}

where $i$ is some fixed integer. Then using (2.19) and (4.5), and following the arguments in pp. 84–87 of [Reference Mukhtarov and Aydemir35] we obtain the following asymptotic formula

(4.6)\begin{align} y_{k + i,\gamma} (x) &= \sin \left({k - \nu_{\gamma} + 1}\right) \pi x + (1 - \textrm{sgn} \gamma) ({-}1)^{k} \left( {\sqrt 2\,} \right)^{ - 1} e^{\left({k - \nu_{\gamma} + 1 }\right)\pi (x-1)}\notag\\ &\quad + O \left( { k^{{-}1} } \right), \end{align}

which holds uniformly for $x \in [\,0,1]$. Next, using the oscillation properties of the eigenfunctions of problem (1.1)–(1.5) and following the proof of [Reference Kerimov and Aliev30, theorem 3.1] we get $i = 2$. Hence by setting $i = 2$ in (4.5) and (4.6) we obtain (4.3) and (4.4) respectively. The proof of this theorem is complete.

5. Operator interpretation and basis properties of the root functions of problem (1.1)–(1.5)

Let $H = L_{2} (0, 1) \oplus \,\mathbb {C}^{2}$ be the Hilbert space with the scalar product

\[ (\hat y,\hat v)_{H} = (\{ y, m, n\} ,\{ v, s, t\})_{H} = \int_0^{1} {y(x)\,\overline {v(x)\,} } \textrm{d}x + \,|a| ^{ - 1} m \bar s + \,|c| ^{ - 1} n \bar t. \]

In $H$ we define the operator

\[ L \hat y = L \{ y,m,n\} = \{\ell (y), Ty(0),T y(1)\} \]

with the domain

\begin{align*} &D(L) = \left\{ {\{ y} \right.(x),m,n\} \in H \,: \,y \in W_2^{4} (0,1),\ell (y) \in L_2 (0,1),y'' (0) = 0,\\ &y'(1) \cos \gamma + y''(1) \sin \gamma = 0,m = a y(0),n = c y(1)\}, \end{align*}

which dense everywhere in $H$. Then problem (1.1)–(1.5) is equivalent to the eigenvalue problem

(5.1)\begin{equation} L \hat y = \lambda \hat y, y \in D(L). \end{equation}

In this the eigenvalues $\lambda _{k,\gamma },k \in \mathbb {N}$, of problems (1.1)–(1.5) and (5.2) coincide considering their multiplicity, and between the root functions, there is a one-to-one correspondence

\[ y_{k,\gamma} (x) \leftrightarrow \,\hat y_{k,\gamma} = \{ y_{k,\gamma} (x),m_{k,\gamma} ,n_{k,\gamma}\},m_{k,\gamma} = a y_{k,\gamma}(0),n_{k,\gamma} = c y_{k,\gamma} (1),k \in \mathbb{N}. \]

We define the operator $J:H \to H$ by

\[ J \hat y = J \{y, m, n\} = \{y, - m, n\}. \]

Note that operator $J$ generates the Pontryagin space $\Pi _1 = L_2 (0, 1) \oplus \mathbb {C}^{2}$ equipped with an inner product

\[ (\hat y,\hat v)_{\Pi_1} = (J \hat y, \hat v)_{H} = \int_0^{1} {y(x)\,\overline {v(x)\,} } \textrm{d}x + a^{{-}1} m \bar s - \,c^{{-}1} n \bar t, \]

where $\hat y = \{ y, m, n\}$, $\hat v = \{ v, s, t\}$.

Theorem 4.1 implies that

(5.2) \begin{align} y_{k,\gamma} (x) &= y(x, \lambda_k (\gamma))\,\textrm{if}\, a \ne c - 1, \,k \in \mathbb{N}, \,\textrm{and}\, a = c - 1, \,k \ge 3;\notag\\ y_{1,\gamma} (x) &= y(x, \lambda_1 (\gamma)),y_{2,\gamma} (x) = y_{2,\gamma}^{*} (x) + b \,y_{1,\gamma} (x) \ \textrm{if}\ a = c - 1, \end{align}

where $y_{2,\gamma }^{*} (x) = y'_{\lambda } (x, \lambda _1 (\gamma ))$ and $b$ is an arbitrary constant

Let $\{ \hat v^{*}_{k,\gamma } \} _{k = 1}^{\infty }$, $\hat v^{*}_{k,\gamma } = \{ v^{*}_{k,\gamma } ,s^{*}_{k,\gamma } ,t^{*}_{k,\gamma } \}$, is the system of root vectors of the operator $L^{*}$. Then by [Reference Hinton25, formula (7)] we have

(5.3)\begin{align} L \hat y_{k,\gamma} &= \lambda_{k} (\gamma) \,\hat y_{k,\gamma}, \,L^{*} \hat v^{*}_{k,\gamma} = \lambda_k (\gamma) \,\hat v^{*}_{k,\gamma},\textrm{if}\, a \ne c - 1, \,k \in \mathbb{N}, \,\textrm{and}\, a = c - 1, \,k \ge 3; \notag\\ L \hat y_{1,\gamma} &= \lambda_{1} (\gamma) \,\hat y_{1,\gamma},L \hat y_{2,\gamma} = \lambda_{1} (\gamma) \,\hat y_{2,\gamma} + \hat y_{1,\gamma}, L^{*} \hat v^{*}_{1,\gamma} = \lambda_{1} (\gamma) \,\hat v^{*}_{1,\gamma} + \hat v^{*}_{2,\gamma},\notag\\ L^{*} \hat v^{*}_{1,\gamma} &= \lambda_{1} (\gamma)\, \hat v^{*}_{1,\gamma}. \end{align}

In view of (5.2), by (5.3) we obtain

(5.4)\begin{align} \hat v^{*}_{k,\gamma} &= J \hat y_{k,\gamma} \,\textrm{if}\,a \ne c - 1, k \in \mathbb{N}, \,\textrm{and}\, a = c - 1, k \ge 3;\notag\\ \hat v^{*}_{1,\gamma} &= J \hat y^{*}_{2,\gamma} + \tilde b J \hat y_{1,\gamma}, \hat v^{*}_{2,\gamma} = J \hat y_{1,\gamma}, \,\textrm{if}\,c = a + 1, \end{align}

where $\tilde b$ is an arbitrary constant.

By following the arguments in lemma 4.1 of [Reference Aliyev and Mamedova9, pp. 15–16] we can show that the following assertion holds.

Lemma 5.2 Let $\{ \hat v_{k,\gamma } \} _{k = 1}^{\infty }$, $\hat v_{k,\gamma } = \{ v_{k,\gamma }, s_{k,\gamma }, t_{k,\gamma } \}$, be the system that is adjoint to the system $\{ \hat y_{k,\gamma } \} _{k = 1}^{\infty }$. Then

(5.5)\begin{equation} \hat v_{k} = \delta^{{-}1}_{k,\gamma} \,\hat v^{*}_{k,\gamma}, k \in \mathbb{N}, \end{equation}

where $\delta _{k,\gamma } = (y_{k,\gamma }, y_{k,\gamma })_{\Pi _1}$, if $a \ne c -1, k \in \mathbb {N}$, and $a = c - 1, k \ge 3$; $\delta _{1,\gamma } = \delta _{2,\gamma } = (\hat y_{1,\gamma }, \hat y^{*}_{2,\gamma })_{\Pi _1}$ if $a = c - 1$, and $\delta _{k,\gamma } \ne 0$ for $k \in \mathbb {N}$, and $\tilde b = - ( {b + \delta _{1,\gamma }^{-1} \,(\hat y^{*}_{2,\gamma }, \hat y^{*}_{2,\gamma })_{\Pi _1}} )$.

Let

(5.6)\begin{equation} \Delta _{r,l,\gamma} = \left| \begin{array}{l} s_{r,\gamma} \,s_{l,\gamma} \\ t_{r,\gamma} \,\,t_{l,\gamma} \end{array} \right|. \end{equation}

where $r$ and $l$ ($r \ne l$) are arbitrary fixed positive integers.

Proof. The assertions of this theorem for $p = 2$ follow from [Reference Aliyev1, theorems 3.1, 3.2 and corollary 3.1]. The proof of theorem 5.3 for $p \in (1, + \infty ), p \ne 2$, is similar to that of [Reference Kerimov and Aliev30, theorem 5.1] by using asymptotic formulas (4.1)–(4.4). The proof of this theorem is complete.

Using the oscillatory properties of the eigenfunctions of problem (1.1)–(1.5), by theorem 5.3 we can establish sufficient conditions for the system $\{ y_{k,\gamma } (x) \} _{k = 1, k \ne r,l}^{\infty }$ of root functions of this problem to form a basis in $L_{p} (0, 1), \, 1 < p < \infty$.

Proof. By relations (5.4) and (5.5), it follows from (5.6) that

(5.7)\begin{align} \Delta _{r,l,\gamma} &= \left| \begin{array}{l} - \delta _{r,\gamma}^{{-}1}\,m_{r,\gamma} \, - \,\delta_{l,\gamma}^{{-}1}\,m_{l,\gamma} \\ \delta _{r,\gamma}^{{-}1}\,n_{r,\gamma} \delta _{l,\gamma}^{{-}1}\,n_{l,\gamma}\end{array} \right| ={-} \,\delta _{r,\gamma}^{{-}1} \,\delta_{l,\gamma}^{{-}1} \left| \begin{array}{l} m_{r,\gamma} \,m_{l,\gamma} \\ \,n_{r,\gamma} \,\,n_{l,\gamma} \end{array} \right|\notag\\ & = - \,\delta _{r,\gamma}^{{-}1} \,\delta_{l,\gamma}^{{-}1} \left| \begin{array}{l} a \,y_{r,\gamma} (0) \,a \,y_{l,\gamma} (0) \\ c \,y_{r,\gamma} (1)\,\,c\,y_{l,\gamma} (1) \end{array} \right| ={-} a\,c\,\delta _{r,\gamma}^{{-}1} \,\delta_{l,\gamma}^{{-}1} \left| \begin{array}{l} \,y_{r,\gamma} (0) \, \,y_{\,l,\gamma} (0) \\ \,y_{r,\gamma} (1)\,\,y_{\,l,\gamma} (1) \end{array} \right| \end{align}

for $\,r, l \in \mathbb {N}$ in the case $a \ne c - 1$, and for $\,r, l \ge 3$ in the case $a = c - 1$.

By (2.12) and (2.13), relation (5.7) implies that

(5.8)\begin{equation} \Delta _{r,l,\gamma} ={-}\, a\,c\,\delta _{r,\gamma}^{{-}1} \,\delta_{l,\gamma}^{{-}1} \,\left({y_{\,r,\gamma} (0) - y_{\,l,\gamma} (0)}\right) \end{equation}

for $\,r, l \in \mathbb {N}$ in the case $a < c - 1$, for $r,l \ge 2$ in the case $a > c - 1$, and for $r,l \ge 3$ in the case $a = c - 1$, and

(5.9)\begin{equation} \Delta _{1,l,\gamma} ={-}\, a\,c\,\delta _{r,\gamma}^{{-}1} \,\delta_{l,\gamma}^{{-}1} \,\left({1 - y_{\,1,\gamma} (1) \, y_{\,l,\gamma} (0)}\right) \end{equation}

for $l \ge 2$ in the case $a > c - 1$.

Moreover, in the case $a = c - 1$ for $r = 2$, $l \ge 3$, and for $r = 1$ and $l \ge 2$ we have

(5.10)\begin{align} \Delta _{2,l,\gamma} &={-} \,\delta_{2,\gamma}^{{-}1} \delta_{l,\gamma}^{{-}1} \left| \begin{array}{l} m_{1,\gamma} \,m_{\,l,\gamma} \\ \,n_{1,\gamma} \,\,n_{\,l,\gamma} \end{array} \right| ={-} \,\delta_{2,\gamma}^{{-}1} \delta_{l,\gamma}^{{-}1} \left| \begin{array}{l} a\,y_{1,\gamma} (0) \,a \,y_{\,l,\gamma} (0) \\ c\,y_{1,\gamma} (1) \,\,c\,y_{\,l,\gamma} (1) \\ \end{array} \right|\notag\\ & = - a\,c\,\delta_{2,\gamma}^{{-}1} \delta_{l,\gamma}^{{-}1} \left| \begin{array}{cc} y_{1,\gamma} (0) & y_{\,l,\gamma} (0) \\ 1 & 1 \\ \end{array} \right| = \, - a\,c\,\delta _{1,\gamma}^{{-}1} \,\delta_{l,\gamma}^{{-}1} \,\left({y_{\,1,\gamma} (0) - y_{\,l,\gamma} (0)}\right), \end{align}

and

(5.11)\begin{align} \Delta _{1,l,\gamma} &={-}\left| \begin{array}{l} \delta_{1,\gamma}^{{-}1}\,\{m_{2,\gamma}^{*} + \tilde b\, m_{1,\gamma}\} \, \,\delta_{l,\gamma}^{{-}1}\,m_{l,\gamma} \\ \delta _{1,\gamma}^{{-}1}\,\{\,n_{2,\gamma}^{*} + \,\tilde b \,n_{1,\gamma}\} \,\delta _{l,\gamma}^{{-}1}\,n_{l,\gamma} \\ \end{array} \right| ={-}\,\delta _{1,\gamma}^{{-}1} \,\delta_{l,\gamma}^{{-}1} \,\left| \begin{array}{l} m_{2,\gamma}^{*} + \tilde b\, m_{1,\gamma} \,m_{l,\gamma} \\n_{2,\gamma}^{*} + \,\tilde b \,n_{1,\gamma}\,n_{l,\gamma} \\ \end{array} \right|\notag\\ &= - a\,c\,\delta _{1,\gamma}^{{-}1} \,\delta_{l,\gamma}^{{-}1} \,\left| \begin{array}{l} y_{2,\gamma}^{*} (0) + \tilde b\, y_{1,\gamma} (0) \,y_{l,\gamma} (0) \\y_{2,\gamma}^{*} (1) + \tilde b \,y_{1,\gamma} (1)\,y_{\,l,\gamma} (1) \\ \end{array} \right| ={-}\,a\,c\,\delta _{1,\gamma}^{{-}1} \,\delta_{l,\gamma}^{{-}1} \,\left| \begin{array}{l} y_{2,\gamma}^{*} (0) \,y_{\,l,\gamma} (0) \\ y_{2,\gamma}^{*} (1) \,\,\,1 \\ \end{array} \right|\notag\\ & = - a\,c\,\delta _{1,\gamma}^{{-}1} \,\delta_{l,\gamma}^{{-}1} \,\{\,y_{2,\gamma}^{*} (0) - y_{2,\gamma}^{*} (1)\, y_{\,l,\gamma} (0) \}. \end{align}

respectively.

Next, by (2.12) and (2.25), theorem 3.5 implies that

(5.12)\begin{align} &\textrm{sgn} \,y_{k,\gamma} (0) = ({-}1)^{k} \ \textrm{for}\ k \ge 3, \end{align}

(5.13)\begin{align} &y_{1,\gamma} (0) = 1\ \textrm{for}\ a \le c - 1,y_{2,\gamma} (0) > 0 \ \textrm{for}\ a < c - 1,\notag\\ &\textrm{sgn}\,y_{1,\gamma} (1) = ({-}1)^{\tau \,(\lambda_1 (\gamma))}, y_{2,\gamma} (0) = 1\ \textrm{for}\ a > c - 1, \end{align}

where $\tau \,(\lambda _1 (\gamma )) = \sum \limits _{\xi _k \in (\lambda _1 (\gamma ),0)} {i(\xi _k)}$.

Now the statements (i)–(vi) of this theorem follow from (5.8)–(5.13) in view of theorem 5.3. The proof of this theorem is complete.