We can show that the Green function for problem (1)-(5) is of the form

$$\begin{aligned} G(x,s;\lambda)=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} \frac{\upsilon(s,\lambda)\vartheta(x,\lambda)}{\omega_{(}\lambda)}, & 0< s \leq x< \pi x,s\neq\xi_{i}, i=1,2,\ldots,n+1, \\ \frac{\upsilon(x,\lambda)\vartheta(s,\lambda)}{\omega_{(}\lambda)}, & 0< x \leq s< \pi x,s\neq\xi_{i}, i=1,2,\ldots,n+1, \end{array}\displaystyle \right . \end{aligned}$$

(17)

for \(x, s \in\Omega\) (see, *e.g.*, [26]). It is symmetric with respect to *x* and *s* and is real-valued for real *λ*. Let us show that the function

$$\begin{aligned} y(x,\lambda)=\sum_{k=0}^{n} \frac{1}{a_{k+1}^{2}}\prod_{i=0}^{k} \theta_{i34} \prod_{i=k+1}^{n+1} \theta_{i12} \int _{\xi_{k}+0}^{\xi _{k+1}-0}G(x,s;\lambda)f(s)\,ds, \end{aligned}$$

(18)

called a resolvent, is a solution of the equation

$$\begin{aligned} a(x)y''+\bigl\{ \lambda-q(x)\bigr\} y=f(x) \end{aligned}$$

(19)

(where \(f(x)\neq0\) is a continuous function in each \(\Omega_{i}\) with finite one-hand limits at the endpoints of these intervals) satisfying the boundary-transmission conditions (2)-(5). Without loss of generality, we can assume that \(\lambda=0\) is not an eigenvalue. Otherwise, we take a fixed real number *η* and consider the boundary-value-transmission problem for the differential equation

$$ a(x)y^{\prime\prime}(x,\lambda)+ \bigl\{ (\lambda+\eta)-q(x)\bigr\} y(x,\lambda)=0 $$

(20)

together with the same boundary-transmission conditions (2)-(5) and the same eigenfunctions as for problem (1)-(5). All the eigenvalues are shifted through *η* to the right. It is evident that *η* can be selected so that \(\lambda=0\) is not an eigenvalue of the new problem. Let \(G(x,s;0)=G(x,s)\). Then the function

$$\begin{aligned} y(x,\lambda) =& \sum_{k=0}^{n} \frac{1}{a_{k+1}^{2}}\prod_{i=0}^{k} \theta_{i34} \prod_{i=k+1}^{n+1} \theta_{i12} \int _{\xi_{k}+0}^{\xi _{k+1}-0}G(x,s)f(s)\,ds \end{aligned}$$

(21)

is a solution of the equation \(a(x)y''-q(x)y=f(x)\) satisfying the boundary-transmission conditions (2)-(5). We rewrite (19) in the form

$$\begin{aligned} a(x)y''-q(x)y=f(x)-\lambda y. \end{aligned}$$

(22)

Thus, the homogeneous problem (\(f(x)\equiv0\)) is equivalent to the integral equation

$$\begin{aligned} y(x,\lambda)+\lambda \Biggl\{ \sum_{k=0}^{n} \frac{1}{a_{k+1}^{2}}\prod_{i=0}^{k} \theta_{i34} \prod_{i=k+1}^{n+1} \theta_{i12} \int _{\xi_{k}+0}^{\xi _{k+1}-0}G(x,s)y(s)\,ds \Biggr\} =0. \end{aligned}$$

(23)

Denoting the collection of all the eigenvalues of problem (1)-(4) by \(\lambda_{0}< \lambda_{1}< \lambda_{2}<\cdots<\lambda_{n},\ldots \) and the corresponding normalized eigenfunctions by \(y_{0}, y_{1}, y_{2},\ldots,y_{n},\ldots \) , consider the series

$$Y(x,\xi)=\sum_{n=0}^{\infty}\frac{y_{n}(x)y_{n}(\xi)}{\lambda_{n}}. $$

We can show that \(\lambda_{n}=O(n^{2})\). From this asymptotic formula for the eigenvalues it follows that the series for \(Y(x,\xi)\) converges absolutely and uniformly; therefore, \(Y(x,\xi)\) is continuous in Ω. Consider the kernel

$$K(x,\xi)=G(x,\xi)+ Y(x,\xi)=G(x,\xi)+\sum_{n=0}^{\infty} \frac {y_{n}(x)y_{n}(\xi)}{\lambda_{n}}, $$

which is continuous and symmetric. By a familiar theorem in the theory of integral equations, any symmetric kernel \(K(x,\xi)\) that is not identically zero has at least one eigenfunction [33], that is, there are a number *μ* and a function \(\psi(x)\neq0\) satisfying the equation

$$\begin{aligned} \psi(x)+\mu\Biggl\{ \sum_{k=0}^{n} \frac{1}{a_{k+1}^{2}}\prod_{i=0}^{k} \theta_{i34} \prod_{i=k+1}^{n+1} \theta_{i12} \int _{\xi_{k}+0}^{\xi_{k+1}-0} K(x,\xi)\psi(\xi)\,d\xi\Biggr\} =0 . \end{aligned}$$

(24)

Thus, if we show that the kernel \(K(x,\xi)\) has no eigenfunctions, we obtain \(K(x,\xi)\equiv0\), that is,

$$\begin{aligned} G(x,\xi)=-\sum_{n=0}^{\infty} \frac{y_{n}(x)y_{n}(\xi)}{\lambda_{n}}. \end{aligned}$$

(25)

It follows from equation (23) that

$$\begin{aligned} \sum_{k=0}^{n} \frac{1}{a_{k+1}^{2}}\prod_{i=0}^{k} \theta_{i34} \prod_{i=k+1}^{n+1} \theta_{i12} \int _{\xi_{k}+0}^{\xi_{k+1}-0} G(x,\xi)\psi_{n}(\xi)\,d\xi=-\lambda_{n}^{-1}\psi_{n}(x). \end{aligned}$$

(26)

Therefore,

$$\begin{aligned} \sum_{k=0}^{n} \frac{1}{a_{k+1}^{2}}\prod_{i=0}^{k} \theta_{i34} \prod_{i=k+1}^{n+1} \theta_{i12} \int _{\xi_{k}+0}^{\xi_{k+1}-0} K(x,\xi)\psi_{n}(\xi)\,d\xi=0, \end{aligned}$$

(27)

that is, the kernel \(K(x,\xi)\) is orthogonal to all eigenfunctions of the boundary-value-transmission problem (1)-(5). Let \(y(x)\) be a solution of the integral equation (24). Let us show that \(y(x)\) is orthogonal to all \(\psi_{n}(x)\). In fact, it follows from (24) that

$$\begin{aligned} \sum_{k=0}^{n}\frac {1}{a_{k+1}^{2}} \prod_{i=0}^{k}\theta_{i34} \prod _{i=k+1}^{n+1}\theta_{i12} \int _{\xi_{k}+0}^{\xi _{k+1}-0}y(x)\psi_{n}(x)=0. \end{aligned}$$

Therefore,

$$\begin{aligned} &y(x,\lambda)+\lambda_{0}\Biggl\{ \sum _{k=0}^{n}\frac{1}{a_{k+1}^{2}}\prod _{i=0}^{k}\theta_{i34} \prod _{i=k+1}^{n+1}\theta_{i12} \int _{\xi_{k}+0}^{\xi_{k+1}-0} K(x,\xi)y(\xi)\,d\xi\Biggr\} \\ &\quad=y(x,\lambda)+\lambda_{0}\Biggl\{ \sum _{k=0}^{n}\frac{1}{a_{k+1}^{2}}\prod _{i=0}^{k}\theta_{i34} \prod _{i=k+1}^{n+1}\theta_{i12} \int _{\xi_{k}+0}^{\xi_{k+1}-0} G(x,\xi)y(\xi)\,d\xi\Biggr\} =0, \end{aligned}$$

that is, \(y(x,\lambda)\) is an eigenfunction of the boundary-value-transmission problem (1)-(5). Since it is orthogonal to all \(\psi_{n}(x)\), it is also orthogonal to itself, and, as a consequence, \(y(x,\lambda)=0\) and \(K(x,\xi)=0\). Formula (25) is thus proved.

### Theorem 3.1

(Expansion theorem)

*If*
\(f(x)\)
*has a continuous second derivative in each*
\(\Omega_{i}\) (\(i=1,2,\ldots,n+1\)), *and satisfies the boundary*-*transmission conditions* (2)-(5), *then*
\(f(x)\)
*can be expanded into an absolutely and uniformly convergent series of eigenfunctions of the boundary*-*value*-*transmission problem* (1)-(5) *on* Ω, *that is*,

$$\begin{aligned} f(x)=\sum_{m=0}^{\infty}r_{m} \psi_{m}(x), \end{aligned}$$

(28)

*where*
\(r_{m}=r_{m}(f)\)
*are the Fourier coefficients of f given by*

$$\begin{aligned} r_{m}=\sum_{k=0}^{n} \frac{1}{a_{k+1}^{2}}\prod_{i=0}^{k} \theta_{i34} \prod_{i=k+1}^{n+1} \theta_{i12} \int _{\xi_{k}+0}^{\xi_{k+1}-0} f(x)\psi_{m}(x)\,dx. \end{aligned}$$

(29)

### Proof

Put \(g(x)=a(x)f''-q(x)f\). Then, relying on (18) and (25), we have

$$\begin{aligned} f(x) =& \sum_{k=0}^{n} \frac{1}{a_{k+1}^{2}}\prod_{i=0}^{k} \theta_{i34} \prod_{i=k+1}^{n+1} \theta_{i12} \int _{\xi_{k}+0}^{\xi _{k+1}-0}G(x,\xi)g(\xi)\,d\xi \\ =&-\sum_{m=0}^{\infty}\frac{\psi _{m}(x)}{\lambda_{m}}\sum _{k=0}^{n}\frac{1}{a_{k+1}^{2}}\prod _{i=0}^{k}\theta_{i34} \prod _{i=k+1}^{n+1}\theta_{i12} \int _{\xi_{k}+0}^{\xi _{k+1}-0}\psi_{m}(\xi)g(\xi)\,d\xi \\ \equiv&\sum_{m=0}^{\infty }r_{m} \psi_{m}(x). \end{aligned}$$

(30)

From the orthogonality and normalization of the functions \(\psi_{m}(x)\) we obtain (29). □

### Theorem 3.2

(Modified Parseval equality)

*For any function*
\(f\in \bigoplus_{i=1}^{n+1}L_{2}(\Omega_{i})\), *we have the Parseval equality*

$$\begin{aligned} \sum_{k=0}^{n} \frac{1}{a_{k+1}^{2}}\prod_{i=0}^{k} \theta_{i34} \prod_{i=k+1}^{n+1} \theta_{i12} \int _{\xi_{k}+0}^{\xi _{k+1}-0}f^{2}(x)\,dx=\sum _{m=0}^{\infty}r^{2}_{m}(f). \end{aligned}$$

(31)

### Proof

If \(f(x)\) satisfies the conditions of Theorem 3.1, then (31) follows immediately from the uniform convergence of the series (28). Indeed,

$$\begin{aligned} \sum_{k=0}^{n} \frac{1}{a_{k+1}^{2}}\prod_{i=0}^{k} \theta_{i34} \prod_{i=k+1}^{n+1} \theta_{i12} \int _{\xi_{k}+0}^{\xi _{k+1}-0}f^{2}(x)\,dx=\sum _{m=0}^{\infty}r^{2}_{m}(f). \end{aligned}$$

(32)

Now, suppose that \(f(x)\) is an arbitrary square-integrable function on the intervals \(\Omega_{i}\) (\(i=1,2,\ldots,n+1\)). Slightly modifying the familiar theorem in the theory of real analysis, we can show that there exists a sequence of infinitely differentiable functions \(f_{k}(x)\), converging in mean square to \(f(x)\), such that each function \(f_{k}(x)\) is identically zero in some neighborhoods of the points \(\xi_{i}\) (\(i=0,1,\ldots,n+1\)). From (32) it follows that

$$\begin{aligned} &\sum_{k=0}^{n} \frac{1}{a_{k+1}^{2}}\prod_{i=0}^{k} \theta_{i34} \prod_{i=k+1}^{n+1} \theta_{i12} \int _{\xi_{k}+0}^{\xi_{k+1}-0} \bigl[f_{s}(x)-f_{t}(x) \bigr]^{2}\,dx \\ &\quad=\sum_{m=0}^{\infty }\bigl[r_{m}(f_{s})-r_{m}(f_{t}) \bigr]^{2}, \end{aligned}$$

(33)

where \(r_{m}(f_{s})\) are, as usual, the Fourier coefficients in (29). Since the left-hand side (33) tends to zero as \(s,t \rightarrow\infty\), the right-hand side also tends to zero. By applying the Cauchy-Schwarz inequality we obtain

$$\begin{aligned} \bigl| r_{m}(f)-r_{m}(f_{s})\bigr| \leq \Biggl\{ \sum_{k=0}^{n}\frac{1}{a_{k+1}^{2}}\prod _{i=0}^{k}\theta_{i34} \prod _{i=k+1}^{n+1}\theta_{i12} \int _{\xi_{k}+0}^{\xi_{k+1}-0} \bigl[f(x)-f_{s}(x) \bigr]^{2}\,dx\Biggr\} ^{\frac{1}{2}}. \end{aligned}$$

On the other hand, from the convergence in the mean of \(f_{s}(x)\) to \(f(x)\) it follows that

$$\begin{aligned} \lim_{s\rightarrow\infty}r_{m}(f_{s})= r_{m}(f), \quad m=0,1,2,\ldots. \end{aligned}$$

It follows from (33) that

$$\begin{aligned} \sum_{n=0}^{N} \bigl[r_{m}(f_{s})-r_{m}(f_{t}) \bigr]^{2} \leq\sum_{k=0}^{n} \frac{1}{a_{k+1}^{2}}\prod_{i=0}^{k} \theta_{i34} \prod_{i=k+1}^{n+1} \theta_{i12} \int _{\xi_{k}+0}^{\xi_{k+1}-0} \bigl[f_{s}(x)-f_{t}(x) \bigr]^{2}\,dx \end{aligned}$$

for an arbitrary integer *N*. Passing to the limit as \(s\rightarrow \infty\), we obtain

$$\begin{aligned} \sum_{n=0}^{N} \bigl[r_{m}(f)-r_{m}(f_{t})\bigr]^{2} \leq\sum_{k=0}^{n}\frac{1}{a_{k+1}^{2}}\prod _{i=0}^{k}\theta_{i34} \prod _{i=k+1}^{n+1}\theta_{i12} \int _{\xi_{k}+0}^{\xi_{k+1}-0} \bigl[f(x)-f_{t}(x) \bigr]^{2}\,dx. \end{aligned}$$

Now letting \(N\rightarrow\infty\) gives

$$\begin{aligned} \sum_{n=0}^{\infty} \bigl[r_{m}(f)-r_{m}(f_{t})\bigr]^{2} \leq\sum_{k=0}^{n}\frac{1}{a_{k+1}^{2}}\prod _{i=0}^{k}\theta_{i34} \prod _{i=k+1}^{n+1}\theta_{i12} \int _{\xi_{k}+}^{\xi_{k+1}-} \bigl[f(x)-f_{t}(x) \bigr]^{2}\,dx. \end{aligned}$$

Taking into account the Minkowski inequality, we see that the series \(\sum_{m=0}^{\infty}r^{2}_{m}(f)\) converges. Since

$$\begin{aligned} &\Biggl|\sum_{m=0}^{\infty} \bigl(r_{m}(f)\bigr)^{2}-\sum_{m=0}^{\infty } \bigl(r_{m}(f_{t})\bigr)^{2}\Biggr| \\ &\quad=\Biggl|\sum_{m=0}^{\infty } \bigl[r_{m}(f)-r_{m}(f_{t})\bigr] \bigl[r_{m}(f)+r_{m}(f_{t})\bigr]\Biggr| \\ &\quad\leq \Biggl(\sum_{m=0}^{\infty}\bigl| r_{m}(f)-r_{m}(f_{t})\bigr|^{2} \Biggr)^{\frac{1}{2}} \Biggl(\sum_{m=0}^{\infty} \bigl| r_{m}(f)+r_{m}(f_{t})\bigr|^{2} \Biggr)^{\frac{1}{2}}, \end{aligned}$$

we deduce that \(\sum_{m=0}^{\infty}\{r_{m}(f_{t})\}^{2}\rightarrow\sum_{m=0}^{\infty}r_{m}^{2}(f)\) as \(t\rightarrow\infty\). Moreover, from the convergence in the mean of \(f_{t}(x)\) to \(f(x)\) we derive that

$$\begin{aligned} &\lim_{t\rightarrow\infty}\Biggl(\sum _{k=0}^{n}\frac {1}{a_{k+1}^{2}}\prod _{i=0}^{k} \theta_{i34} \prod _{i=k+1}^{n+1} \theta_{i12} \int _{\xi_{k}+}^{\xi_{k+1}-}f_{t}^{2}(x)\,dx \Biggr) \\ &\quad= \sum_{k=0}^{n}\frac{1}{a_{k+1}^{2}} \prod_{i=0}^{k}\theta_{i34} \prod _{i=k+1}^{n+1}\theta_{i12} \int _{\xi _{k}+}^{\xi_{k+1}-} f^{2}(x)\,dx. \end{aligned}$$

Finally, letting \(t\rightarrow\infty\) in the equality

$$\begin{aligned} \sum_{k=0}^{n} \frac{1}{a_{k+1}^{2}}\prod_{i=0}^{k} \theta_{i34} \prod_{i=k+1}^{n+1} \theta_{i12} \int _{\xi_{k}+}^{\xi_{k+1}-} f_{t}^{2}(x)\,dx = \sum_{m=0}^{\infty}\bigl(r_{m}(f_{t}) \bigr)^{2}, \end{aligned}$$

we obtain (31) for arbitrary \(f\in \bigoplus_{i=1}^{n+1}L_{2}(\Omega_{i})\). The proof is complete. □