Let us define the integral operator \(\mathfrak{F}\) by

$$ (\mathfrak{F} u) (x)= \gamma_{1}\gamma_{2} \int _{-\pi}^{0^{-}}G(x,\xi)u(\xi)\,d\xi+\delta _{1}\delta_{2} \int _{0^{+}}^{\pi}G(x,\xi)u(\xi)\,d\xi. $$

(21)

Then the BVTP (1)-(5) converts to the spectral problem for the integral operator \(\mathfrak{F}\) given by

$$(I-\lambda\mathfrak{F})u=0, $$

where *I* is the identity operator. Since the kernel \(G(x,\xi)\) of the integral operator \(\mathfrak{F}\) is symmetric and continuous we can apply the well-known extremal principle (see, for example, [25]). Let \(\{\lambda_{n}\}\) be a sequence of eigenvalues of the integral operator \(\mathfrak{F}\) determined by the extremal principles and \(\{\phi_{n}(x)\}\) be the corresponding sequence of orthonormal eigenfunctions.

### Lemma 4.1

*Let*
\(g\in\oplus C(\Omega)\). *Then*

$$\begin{aligned}& \lim_{m\rightarrow\infty}\Biggl(\gamma_{1} \gamma_{2} \int _{-\pi }^{0^{-}}\Biggl(\mathfrak{F}g-\sum _{i=1}^{m}c_{i}(\mathfrak{F}g) \phi_{i}\Biggr)^{2}\,dx \\& \quad {}+ \delta_{1} \delta_{2} \int _{0^{+}}^{\pi}\Biggl(\mathfrak{F}g-\sum _{i=1}^{m}c_{i}(\mathfrak{F}g) \phi_{i}\Biggr)^{2}\,dx\Biggr)=0, \end{aligned}$$

(22)

*where*
\(c_{i}(\mathfrak{F}g)=\langle\mathfrak{F}g,\phi_{i}\rangle_{H}\)
*denote the Fourier coefficients of*
\(\mathfrak{F}g\)
*with respect to the orthonormal set*
\((\phi_{i})\).

### Proof

Denote \(g_{m}(x)=g(x)-\sum_{i=1}^{m}\langle g,\phi_{i}\rangle_{H}\phi_{i}\). Since \(\{\phi_{n}\}\) is the orthonormal system in *H*, \(\langle g_{m},\phi_{i}\rangle_{H}=0\) for \(i=1,\ldots,m\). From the fact that the eigenvalues \(\lambda_{n}\) are determined by the extremal principle with the corresponding sequence of orthonormal eigenfunctions \(\{\phi_{n}\}\) we have \(\|\mathfrak{F}g_{m}\|_{H}\leq|\lambda_{m+1}|\|g_{m}\|_{H}\). Since \(\lambda_{m+1}\rightarrow0\), \(\|\mathfrak{F}g_{m}\|_{H} \rightarrow0\). Then we have

$$\begin{aligned} \mathfrak{F}g =&\mathfrak{F}g_{m}+ \sum _{i=1}^{m}\langle g,\phi_{i} \rangle_{H}\mathfrak{F}\phi_{i} =\mathfrak{F}g_{m}+ \sum_{i=1}^{m}\lambda_{i}\langle g,\phi_{i}\rangle_{H} \phi_{i} \\ =&\mathfrak{F}g_{m}+\sum_{i=1}^{m} \langle g,\mathfrak{F}\phi_{i}\rangle_{H} \phi_{i}=\mathfrak{F}g_{m}+\sum _{i=1}^{m}\langle \mathfrak{F}g,\phi_{i} \rangle_{H}\phi_{i} \end{aligned}$$

(23)

for arbitrary \(m=1,2,\ldots \) . Letting \(m\rightarrow\infty\) we get

$$ \mathfrak{F}g=\sum_{i=1}^{\infty} \langle \mathfrak{F}g,\phi_{i}\rangle_{H} \phi_{i}, $$

(24)

where the convergence is in the Hilbert space *H*, *i.e.* the equality (22) holds. □

### Corollary 4.2

*If*
\(g\in\oplus C(\Omega)\)
*then the Parseval equality*

$$\|\mathfrak{F}g\|_{H}^{2}=\sum _{i=1}^{\infty}c_{i}^{2}( \mathfrak{F}g) $$

*holds*.

### Corollary 4.3

*The set of orthonormal eigenfunction of the integral operator*
\(\mathfrak{F}\)
*is complete in the range of the integral operator*
\(\mathfrak{F}\)
*given by*

$$R(\mathfrak{F})= \bigl\{ h\in\oplus C(\Omega)| \textit{there exists } g\in \oplus C(\Omega) \textit{ such that } h=\mathfrak{F}g\bigr\} . $$

### Theorem 4.4

*Let the hypotheses and notation of Lemma *
4.1
*hold*. *Then*, *for any*
\(h\in R(\mathfrak{F})\),

$$h=\sum_{i=1}^{\infty}\biggl( \gamma_{1}\gamma_{2} \int _{-\pi }^{0^{-}}h\phi_{i}\, dx+ \delta_{1}\delta_{2} \int _{0^{+}}^{\pi}h\phi_{i}\, dx\biggr) \phi_{i}(x), $$

*where the series converges with respect to the norm*
\(\oplus C(\Omega)\), *i*.*e*. *uniformly on*
\(\Omega=\Omega_{1}\cup\Omega_{2}\).

### Proof

Let \(h=\mathfrak{F}g\). Then for any *n*, *p* we have

$$ \sum_{i=n}^{n+p} \lambda_{i}\langle g,\phi_{i}\rangle_{H} \phi_{i} =\mathfrak{F}\Biggl[\sum_{i=n}^{n+p} \langle g,\phi_{i}\rangle_{H} \phi_{i} \Biggr]. $$

(25)

In view of the fact that the integral operator \(\mathfrak{F}\) is a bounded linear operator in the Banach space \(\oplus C(\Omega)\) we get from (25)

$$ \Biggl\vert \sum_{i=n}^{n+p} \lambda_{i}\langle g,\phi_{i}\rangle_{\mathcal{H}} \phi_{i}\Biggr\vert \leq C\Biggl[\sum_{i=n}^{n+p} \bigl\vert \langle g,\phi_{i}\rangle_{\mathcal{H}}\bigr\vert ^{2}\Biggr]^{1/2} $$

(26)

for some constant *C* independent of *n*. By Bessel’s inequality, the right-hand side of this inequality tends to zero as \(n\rightarrow\infty\) uniformly. Thus the series

$$ \sum_{i=1}^{\infty}\langle \mathfrak{F}g,\phi_{i}\rangle_{H}\phi_{i}(x) $$

(27)

converges in the Banach space \(\oplus C(\Omega)\). Let \(\tilde{h}(x)\) be the sum of the last series. Consequently \(\tilde{h}\in\oplus C(\Omega)\) and

$$ \tilde{h}(x)=\sum_{i=1}^{\infty} \langle \mathfrak{F}g,\phi_{i}\rangle_{H} \phi_{i}(x). $$

(28)

From (23) and (28) it follows that \(\|\mathfrak{F}g-\tilde{h}\|_{H}=0\), *i.e.*
\(h(x)=\tilde{h}(x)\) almost everywhere. Since *h* is also continuous in Ω we have \(h(x)=\tilde{h}(x)\) for all \(x\in\Omega\). Thus

$$ h=\sum_{i=1}^{\infty}\langle h, \phi_{i}\rangle_{H}\phi_{i}, $$

(29)

where the series converges with respect to the norm of \(\oplus C(\Omega)\), *i.e.* uniformly on Ω. □

### Theorem 4.5

*The set of all nonzero eigenvalues of the integral operator*
\(\mathfrak{F}\)
*coincide with the set of the eigenvalues*
\((\lambda_{n})\)
*which are obtained from the extremal principle*.

### Proof

By way of contradiction, suppose there is a nonzero eigenvalue \(\lambda^{*}\) distinct from all eigenvalues \((\lambda_{n})\). Let \(u^{*}\) be the eigenfunction corresponding to the eigenvalue \(\lambda^{*}\). Then from Theorem 4.4 we get

$$ \lambda^{*} u^{*}=\mathfrak{F}u^{*}= \sum_{i=1}^{\infty}\bigl\langle \mathfrak{F}u^{*},\phi_{i}\bigr\rangle _{H} \phi_{i}=\lambda^{*}\sum_{i=1}^{\infty} \bigl\langle u^{*},\phi_{i}\bigr\rangle _{H} \phi_{i}=0 $$

(30)

since \(\langle u^{*},\phi_{i}\rangle_{H}=0\) for all \(i=1,2,\ldots \) by Theorem 2.7. Thus we get a contradiction. □

### Theorem 4.6

*Let*
\(f\in\oplus C^{2}(\Omega)\)
*and satisfy the boundary*-*transmission conditions* (2)-(5). *Then the Fourier series of*
*f*
*with respect to*
\(\{\phi_{i}\}\)
*converges uniformly on*
\(\Omega_{1} \cup\Omega_{2}\), *i*.*e*.

$$ \lim_{n\rightarrow\infty}\Biggl\{ \sup_{x\in\Omega}\Biggl\vert f(x)- \sum_{i=1}^{n}\biggl( \gamma_{1}\gamma_{2} \int _{-\pi}^{0^{-}}f\phi_{i}\,dx+ \delta_{1}\delta_{2} \int _{0^{+}}^{\pi}f\phi_{i}\,dx \biggr)^{2}\phi_{i}(x) \Biggr\vert \Biggr\} =0. $$

### Proof

Let \(f\in\oplus C^{2}(\Omega)\) satisfy the boundary-transmission conditions (2)-(5) and denote \(g=\tau(f)\). Then \(g\in\oplus C(\Omega)\). By virtue of (17) and (21) we have \(f=\mathfrak{F}g\). From Lemma 4.1,

$$ f=\mathfrak{F}g=\sum_{i=1}^{\infty} \langle \mathfrak{F}g,\phi_{i}\rangle_{H} \phi_{i}=\sum_{i=1}^{\infty}\langle f, \phi_{i}\rangle_{H}\phi_{i}, $$

(31)

where the series is convergent in the Banach space \(\oplus C(\Omega)\). □

### Theorem 4.7

*The set of eigenfunctions*
\(\{\phi_{i}(x)\}\)
*is a complete orthonormal set in the Hilbert space*
*H*.

### Proof

Denote by \(\oplus C_{0}^{k}(\Omega)\) the set of all functions \(f\in C^{k}(\Omega)\) which vanishes at some neighborhoods of the points \(x=-\pi\), \(x=0\), and \(x=\pi\). Let \(f\in H\) and \(\epsilon>0\) be given. Then there exists a function \(g\in\oplus C_{0}^{2}(\Omega)\) such that \(\|f-g\|_{\mathcal{H}}<\frac{\epsilon}{3}\) since the set \(\oplus C_{0}^{2}(\Omega)\) is dense in the Hilbert space *H*, *i.e.*
\(\overline{\oplus C_{0}^{2}(\Omega)}=H \) (see, for example, [26]). It is clear that

$$\begin{aligned} \Biggl\Vert f-\sum_{i=1}^{m} \langle f,\phi_{i}\rangle_{H}\phi_{i}\Biggr\Vert _{H} \leq& \Vert f-g\Vert _{H}+ \Biggl\Vert g-\sum_{i=1}^{m} \langle g, \phi_{i}\rangle_{H}\phi_{i}\Biggr\Vert _{H} \\ &{}+\Biggl\Vert \sum_{i=1}^{m} \bigl\langle (g-f),\phi_{i}\bigr\rangle _{H}\phi_{i} \Biggr\Vert _{H} \end{aligned}$$

(32)

for arbitrary *m*. By Bessel’s inequality we have

$$ \Biggl\Vert \sum_{i=1}^{m} \bigl\langle (g-f),\phi_{i}\bigr\rangle _{H} \phi_{i}\Biggr\Vert _{H}^{2} =\sum _{i=1}^{m}\bigl\vert \bigl\langle (g-f), \phi_{i}\bigr\rangle _{H}\bigr\vert ^{2} \leq \Vert f-g\Vert _{H}^{2}< \biggl(\frac{\epsilon}{3} \biggr)^{2} $$

and, by Theorem 4.6, there exists an integer \(n_{0}=n_{0}(\epsilon)\) such that, for \(m>n_{0}\),

$$ \Biggl\Vert g-\sum_{i=1}^{m} \langle g,\phi_{i}\rangle_{H}\phi_{i} \Biggr\Vert _{H}< \frac{\epsilon}{3}. $$

(33)

Finally, from (32) and (33) we get

$$ \Biggl\Vert f-\sum_{i=1}^{m} \langle f,\phi_{i}\rangle_{H}\phi_{i}\Biggr\Vert _{H}< \epsilon $$

for \(m>n_{0}\). The proof is complete. □

Now we are ready to prove the next important result.

### Theorem 4.8

*The set of eigenfunctions*
\((\phi_{i}(x))\)
*of the problem* (1)-(5) *form an orthonormal basis in the Hilbert space*
*H*
*and for any*
\(f\in H\)
*the Parseval equality*

$$ \|f\|_{H}^{2}=\sum_{i=1}^{\infty} \biggl(\gamma_{1}\gamma_{2} \int _{-\pi}^{0^{-}}f\phi_{i}\,dx+ \delta_{1}\delta_{2} \int _{0^{+}}^{\pi}f\phi_{i}\,dx \biggr)^{2} $$

*holds*.

### Proof

Without loss of generality we shall assume that \(\lambda=0\) is not an eigenvalue. Otherwise, we can select a real \(\lambda_{0}\neq0\) such that the problem \(\tau u=\lambda_{0}u\), \(\ell_{i}(u)=t_{i}(u)=0\), \(i=1,2\) has no nontrivial solutions. Then denoting \(\tilde{\lambda}=\lambda-\lambda_{0}\) and \(\tilde{q}(x)=q(x)-\lambda_{0}\) we see that the problem

$$ -u''+\tilde{q}(x)u=\tilde{\lambda}u, \qquad \ell_{i}(u)=t_{i}(u)=0,\quad i=1,2, $$

(34)

has the same properties for the eigenfunctions and eigenvalues as the considered problem (1)-(5). Namely, the pair \((\tilde{\lambda},u(x))\) is the eigen-pair of the problem (34) if and only if the pair \((\lambda,u(x))\) is an eigen-pair of (1)-(5). Clearly, \(\tilde{\lambda}=0\) is not an eigenvalue of the problem (34). Hence, without loss of generality we can assume that \(\lambda=0\) is not an eigenvalue of the considered BVTP (1)-(5). Moreover, if \(\lambda\neq0\), then the pair \((\lambda,u(x))\) is the eigen-pair of the BVTP (1)-(5) if and only if the pair \((\frac{1}{\lambda},u(x))\) is the eigen-pair of the integral operator \(\mathfrak{F}\). Consequently the set \({\phi_{i}}\) form an orthonormal set of eigenfunctions either for \(\mathfrak{F}\) and (1)-(5). Moreover, this set is complete by Theorem 4.7. It is well known that any complete orthonormal set in a Hilbert space forms an orthonormal basis. Consequently, every function \(f\in H\) may be expanded in a Fourier series with respect to the orthonormal set of eigenfunctions \((\phi_{i})\), *i.e.* the equality

$$f=\sum_{i=1}^{\infty}\langle f, \phi_{i}\rangle_{H}\phi_{i} $$

holds, where the series converges with respect to the norm of the Hilbert space *H*. Further, the Parseval equality follows immediately from the last equality. □