# The regularized trace formula for a fourth order differential operator given in a finite interval

## Abstract

In this work, a regularized trace formula for a differential operator of fourth order with bounded operator coefficient is found.

## Introduction

Investigations into the regularized trace formulas of scalar differential operators started with the work [1] firstly. After that work, regularized trace formulas for several differential operators have been studied in some works as [2, 3] and [4]. In [5] a formula for the second regularized trace of the problem generated by a Sturm-Liouville operator equation with a spectral parameter dependent boundary condition is found. The list of the works on this subject is given in [6] and [7]. The trace formulas for differential operators with operator coefficient are investigated in the works [813] and [14]. The boundary conditions in our work are completely different from those in [9].

In this work, we find the following regularized trace formula for a self-adjoint differential operator L of fourth order with bounded operator coefficient:

$$\sum_{m=0}^{\infty} \Biggl[\sum _{n=1}^{\infty} \biggl(\lambda _{mn}-\biggl(m+ \frac{1}{2}\biggr)^{4} \biggr)-\frac{1}{\pi} \int _{0}^{\pi }\operatorname{tr}Q(x)\,dx \Biggr]= \frac{1}{4} \bigl[\operatorname{tr}Q(\pi)-\operatorname{tr}Q(0) \bigr].$$

Here $$\{\lambda_{mn}\}_{n=1}^{\infty}$$ are the eigenvalues of the operator L which belong to the interval $$[(m+\frac{1}{2})^{4}-\|Q\|,(m+\frac{1}{2})^{4}+\|Q\|]$$.

### Notation and preliminaries

Let H be a separable Hilbert space with infinite dimension. Let us consider the operators $$L_{0}$$ and L in the Hilbert space $$H_{1}=L_{2}(0,\pi; H)$$ which are formed by the following differential expressions:

\begin{aligned}[b] &\ell_{0}(y)=y^{{\i}v}(x),\\ &\ell(y)=y^{{\i}v}(x)+Q(x)y(x) \end{aligned}

with the same boundary conditions $$y(0)=y^{\prime\prime}(0)=y^{\prime}(\pi)=y^{\prime\prime\prime}(\pi)=0$$. Suppose that the operator function $$Q(x)$$ in the expression $$\ell(y)$$ satisfies the following conditions:

1. (Q1)

$$Q(x):H \rightarrow H$$ is a self-adjoint kernel operator for every $$x\in[0,\pi]$$. Moreover, $$Q(x)$$ has second order weak derivative in this interval and $$Q^{(i)}(x):H \rightarrow H$$ ($$i=1,2$$) are self-adjoint kernel operators for every $$x\in[0,\pi]$$.

2. (Q2)

$$\|Q\|<\frac{5}{2}$$.

3. (Q3)

There is an orthonormal basis $$\{\varphi_{n}\}_{n=1}^{\infty}$$ of the space H such that

$$\sum_{n=1}^{\infty}\bigl\| Q(x)\varphi_{n} \bigr\| < \infty.$$
4. (Q4)

The functions $$\|Q^{(i)}(x)\|_{\sigma_{1}(H)}$$ are bounded and measurable in the interval $$[0,\pi]$$ ($$i=0,1,2$$).

Here $$\sigma_{1}(H)$$ is the space of kernel operators from H to H as in [15]. Moreover, we denote the norms by $$\|\cdot\|_{H}$$ and $$\|\cdot\|$$ and inner products by $$(\cdot,\cdot)_{H}$$ and $$(\cdot,\cdot)$$ in H and $$H_{1}$$, respectively, and we also denote the sum of eigenvalues of a kernel operator A by $$\operatorname{tr}A=\operatorname{trace}A$$.

Spectrum of the operator $$L_{0}$$ is the set

$$\sigma(L_{0})=\biggl\{ \biggl(\frac{1}{2}\biggr)^{4}, \biggl(\frac{3}{2}\biggr)^{4},\ldots,\biggl(m+\frac {1}{2} \biggr)^{4},\ldots\biggr\} .$$

Every point of this set is an eigenvalue of $$L_{0}$$ which has infinite multiplicity. The orthonormal eigenfunctions corresponding to eigenvalue $$(m+\frac{1}{2})^{4}$$ are in the form

$$\psi_{mn}(x)=\sqrt{\frac{2}{\pi}}\sin\biggl(m+\frac{1}{2} \biggr)x\cdot\varphi_{n}\quad (n=1,2,\ldots).$$
(1.1)

## Some relations between spectrums of operators $$L_{0}$$ and L

Let $$R_{\lambda}^{0}$$, $$R_{\lambda}$$ be resolvents of the operators $$L_{0}$$ and L, respectively. If the operator $$Q:H_{1} \rightarrow H _{1}$$ satisfies conditions (Q2) and (Q3), the following can be proved:

1. (a)

$$QR_{\lambda}^{0} \in\sigma_{1}(H_{1})$$ for every $$\lambda\notin\sigma(L_{0})$$.

2. (b)

Spectrum of the operator L is a subset of the union of pairwise disjoint intervals

$$F_{m}=\biggl[\biggl(m+\frac{1}{2}\biggr)^{4}-\|Q\|, \biggl(m+\frac{1}{2}\biggr)^{4}+\|Q\| \biggr] \quad(m=0,1,2, \ldots), \sigma(L)\subset\bigcup_{m=0}^{\infty}F_{m}.$$
3. (c)

Each point of the spectrum of L, different from $$(m+\frac{1}{2})^{4}$$ in $$F_{m}$$ is an isolated eigenvalue which has finite multiplicity.

4. (d)

The series $$\sum_{n=1}^{\infty} [\lambda_{mn}-(m+\frac {1}{2})^{4} ]$$ ($$m=0,1,2,\ldots$$) are absolutely convergent where $$\{\lambda_{mn}\}_{n=1}^{\infty}$$ are eigenvalues of the operator L in the interval $$F_{m}$$.

Let $$\rho(L)$$ be resolvent set of the operator L; $$\rho(L)={ \mathbf{C}\backslash\sigma(L)}$$. Since $$QR_{\lambda}^{0} \in\sigma_{1}(H_{1})$$ for every $$\lambda\in\rho(L)$$, from the equation $$R_{\lambda}=R_{\lambda}^{0}-R_{\lambda}QR_{\lambda}^{0}$$ we obtain $$R_{\lambda}-R_{\lambda}^{0}\in\sigma_{1}(H_{1})$$.

On the other hand, if we consider the series

$$\sum_{n=1}^{\infty} \biggl[\lambda_{mn}- \biggl(m+\frac{1}{2}\biggr)^{4} \biggr]\quad (m=0,1,2,\ldots)$$

are absolutely convergent then we have

$$\operatorname{tr}\bigl(R_{\lambda}-R_{\lambda}^{0}\bigr)= \sum_{m=0}^{\infty}\sum _{n=1}^{\infty} \biggl[\frac{1}{\lambda_{mn}-\lambda}-\frac{1}{(m+\frac {1}{2})^{4}-\lambda} \biggr]$$

for every $$\lambda\in\rho(L)$$ [10]. If we multiply both sides of this equality with $$\frac{\lambda}{2\pi i}$$ and integrate this equality over the circle $$|\lambda|=b_{p}=(p+1)^{4}$$ ($$p\in\mathbf{N}$$, $$p\geq1$$), then we find

\begin{aligned} &\frac{1}{2\pi i}\int_{|\lambda|=b_{p}}\lambda \operatorname{tr}\bigl(R_{\lambda }-R_{\lambda}^{0}\bigr)\,d\lambda \\ &\quad=\frac{1}{2\pi i}\int_{|\lambda|=b_{p}}\lambda\sum _{m=0}^{p}\sum_{n=1}^{\infty} \biggl[\frac{1}{\lambda_{mn}-\lambda}- \frac{1}{(m+\frac {1}{2})^{4}-\lambda} \biggr]\,d\lambda \\ &\qquad{}+\frac{1}{2\pi i}\int_{|\lambda|=b_{p}}\lambda\sum _{m=p+1}^{\infty}\sum_{n=1}^{\infty} \biggl[\frac{1}{\lambda_{mn}-\lambda}-\frac {1}{(m+\frac{1}{2})^{4}-\lambda} \biggr]\,d\lambda. \end{aligned}
(2.1)

If we consider the relations $$|\lambda_{mn}|< b_{p}$$ ($$m=0,1,2,\ldots,p$$) and $$|\lambda_{mn}|> b_{p}$$ ($$m=p+1, p+2,\ldots$$) for $$n=1,2,3,\ldots$$ , then from (2.1) we get

\begin{aligned} &\frac{1}{2\pi i}\int_{|\lambda|=b_{p}}\lambda \operatorname{tr}\bigl(R_{\lambda }-R_{\lambda}^{0}\bigr)\,d\lambda \\ &\quad=\sum_{m=0}^{p}\sum _{n=1}^{\infty} \biggl[\frac{1}{2\pi i}\int _{|\lambda|=b_{p}}\frac{\lambda \,d\lambda}{\lambda -(m+\frac{1}{2})^{4}}-\frac{1}{2\pi i}\int _{|\lambda|=b_{p}}\frac {\lambda \,d\lambda}{\lambda-\lambda_{mn}} \biggr] \\ &\qquad{}+\sum_{m=p+1}^{\infty}\sum _{n=1}^{\infty} \biggl[\frac {1}{2\pi i}\int _{|\lambda|=b_{p}}\frac{\lambda \,d\lambda}{\lambda-(m+\frac{1}{2})^{4}}-\frac{1}{2\pi i}\int _{|\lambda|=b_{p}}\frac{\lambda \,d\lambda}{\lambda-\lambda_{mn}} \biggr] \\ &\quad=\sum_{m=0}^{p}\sum _{n=1}^{\infty} \biggl[\biggl(m+\frac {1}{2} \biggr)^{4}-\lambda_{mn} \biggr]. \end{aligned}
(2.2)

Moreover, from the formula $$R_{\lambda}=R_{\lambda }^{0}-R_{\lambda}QR_{\lambda}^{0}$$, we obtain the following equality:

$$R_{\lambda}-R_{\lambda}^{0}=-R_{\lambda}^{0}QR_{\lambda}^{0}+R_{\lambda }^{0} \bigl(QR_{\lambda}^{0}\bigr)^{2}-R_{\lambda} \bigl(QR_{\lambda}^{0}\bigr)^{3}.$$
(2.3)

If we put this equality into (2.2), we have

$$\sum_{m=0}^{p}\sum _{n=1}^{\infty} \biggl[\lambda _{mn}-\biggl(m+ \frac{1}{2}\biggr)^{4} \biggr]=M_{p1}+M_{p2}+M_{p}.$$
(2.4)

Here

\begin{aligned}& M_{pj}=\frac{(-1)^{j}}{2\pi i}\int_{|\lambda|=b_{p}}\lambda \operatorname{tr}\bigl[R_{\lambda}^{0}\bigl(QR_{\lambda}^{0} \bigr)^{j}\bigr]\,d\lambda\quad (j=1,2), \end{aligned}
(2.5)
\begin{aligned}& M_{p}=\frac{-1}{2\pi i}\int_{|\lambda|=b_{p}}\lambda \operatorname{tr}\bigl[R_{\lambda}\bigl(QR_{\lambda}^{0} \bigr)^{3}\bigr]\,d\lambda. \end{aligned}
(2.6)

### Theorem 2.1

If the operator function $$Q(x)$$ satisfies condition (Q3), then we have

$$M_{pj}=\frac{(-1)^{j}}{2\pi ij}\int_{|\lambda |=b_{p}} \operatorname{tr}\bigl[\bigl(QR_{\lambda}^{0}\bigr)^{j} \bigr]\,d\lambda.$$

### Proof

It can be shown that the operator function $$QR_{\lambda}^{0}$$ is analytic with respect to the norm in the space $$\sigma_{1}(H_{1})$$ in domain $$\rho(L_{0})=\mathbf{C}\setminus\sigma(L_{0})$$ and

$$\operatorname{tr}\bigl\{ \bigl[\bigl(QR_{\lambda}^{0} \bigr)^{j}\bigr]'\bigr\} =j\operatorname{tr}\bigl[\bigl(QR_{\lambda}^{0} \bigr)'\bigl(QR_{\lambda }^{0}\bigr)^{j-1} \bigr].$$
(2.7)

Considering $$(QR_{\lambda}^{0})'=(QR_{\lambda}^{0})^{2}$$, we can write the formula (2.7) as

$$\operatorname{tr}\bigl\{ \bigl[\bigl(QR_{\lambda}^{0} \bigr)^{j}\bigr]'\bigr\} =j\operatorname{tr}\bigl[R_{\lambda}^{0} \bigl(QR_{\lambda }^{0}\bigr)^{j}\bigr].$$
(2.8)

From (2.5) and (2.8), we obtain

$$M_{pj}=\frac{(-1)^{j+1}}{2\pi ij}\int_{|\lambda|=b_{p}}\lambda \operatorname{tr}\bigl\{ \bigl[\bigl(QR_{\lambda}^{0} \bigr)^{j}\bigr]'\bigr\} \,d\lambda.$$

From here, we find

\begin{aligned} M_{pj}&=\frac{(-1)^{j+1}}{2\pi ij}\int_{|\lambda|=b_{p}} \operatorname{tr}\bigl\{ \bigl[\lambda \bigl(QR_{\lambda}^{0} \bigr)^{j}\bigr]'-\bigl(QR_{\lambda}^{0} \bigr)^{j}\bigr\} \,d\lambda \\ &=\frac{(-1)^{j}}{2\pi ij}\int_{|\lambda|=b_{p}}\operatorname{tr}\bigl[ \bigl(QR_{\lambda}^{0}\bigr)^{j}\bigr]\,d\lambda+ \frac {(-1)^{j+1}}{2\pi ij}\int_{|\lambda|=b_{p}}\operatorname{tr}\bigl\{ \bigl[ \lambda\bigl(QR_{\lambda}^{0}\bigr)^{j} \bigr]'\bigr\} \,d\lambda. \end{aligned}
(2.9)

It can be easily shown that

$$\operatorname{tr}\bigl\{ \bigl[\lambda\bigl(QR_{\lambda}^{0} \bigr)^{j}\bigr]'\bigr\} =\bigl\{ \operatorname{tr}\bigl[ \lambda\bigl(QR_{\lambda}^{0}\bigr)^{j}\bigr]\bigr\} '.$$

Therefore, we have

$$\int_{|\lambda|=b_{p}}\operatorname{tr}\bigl\{ \bigl[\lambda \bigl(QR_{\lambda}^{0}\bigr)^{j}\bigr]' \bigr\} \,d\lambda=\int_{|\lambda|=b_{p}} \bigl\{ \operatorname{tr}\bigl[ \lambda\bigl(QR_{\lambda }^{0}\bigr)^{j}\bigr]\bigr\} '\,d\lambda.$$
(2.10)

The integral on the right-hand side of the last equality can be written as

$$\int_{|\lambda|=b_{p}}\bigl\{ \operatorname{tr}\bigl[\lambda \bigl(QR_{\lambda}^{0}\bigr)^{j}\bigr]\bigr\} '\,d\lambda= \mathop{\int_{|\lambda|=b_{p}}}_{\hphantom{a}\operatorname{Im}\lambda\geq 0}\bigl\{ \operatorname{tr} \bigl[\lambda\bigl(QR_{\lambda}^{0}\bigr)^{j}\bigr]\bigr\} '\,d\lambda +\mathop{\int_{|\lambda|=b_{p}}}_{\hphantom{a}\operatorname{Im}\lambda\leq 0}\bigl\{ \operatorname{tr} \bigl[\lambda\bigl(QR_{\lambda}^{0}\bigr)^{j}\bigr]\bigr\} '\,d\lambda.$$
(2.11)

Let $$\varepsilon_{0}$$ be a constant satisfying the condition $$0<\varepsilon_{0}<b_{p}-(p+\frac{1}{2})^{4}$$. Consider the function $$\operatorname{tr}[\lambda(QR_{\lambda}^{0})^{j}]$$ is analytic in simple connected domains

\begin{aligned}& G_{1}=\{\lambda\in\mathbf{C}:b_{p}-\varepsilon_{0}< \lambda< b_{p}+\varepsilon _{0}, \operatorname{Im}\lambda>- \varepsilon_{0}\}, \\& G_{2}=\{\lambda\in\mathbf{C}:b_{p}-\varepsilon_{0}< \lambda< b_{p}+\varepsilon _{0}, \operatorname{Im}\lambda< \varepsilon_{0}\} \end{aligned}

and

\begin{aligned}& \bigl\{ \lambda\in\mathbf{C}:|\lambda|=b_{p}, \operatorname{Im}\lambda\geq0 \bigr\} \subset G_{1}, \qquad \bigl\{ \lambda\in\mathbf{C}:|\lambda|=b_{p}, \operatorname{Im}\lambda\leq0 \bigr\} \subset G_{2} . \end{aligned}

From (2.11) we obtain

\begin{aligned} \int_{|\lambda|=b_{p}}\bigl\{ \operatorname{tr}\bigl[\lambda \bigl(QR_{\lambda}^{0}\bigr)^{j}\bigr]\bigr\} '\,d\lambda={}&\operatorname{tr}\bigl[-b_{p} \bigl(QR_{-b_{p}}^{0}\bigr)^{j}\bigr] - \operatorname{tr}\bigl[b_{p}(QR_{b_{p}})^{j}\bigr] \\ &{}+\operatorname{tr}\bigl[b_{p}\bigl(QR_{b_{p}}^{0} \bigr)^{j}\bigr] -\operatorname{tr}\bigl[-b_{p} \bigl(QR_{-b_{p}}^{0}\bigr)^{j}\bigr]=0. \end{aligned}
(2.12)

From (2.9), (2.10) and (2.12) we find

$$M_{pj}=\frac{(-1)^{j}}{2\pi ij}\int_{|\lambda|=b_{p}} \operatorname{tr}\bigl[\bigl(QR_{\lambda}^{0}\bigr)^{j} \bigr]\,d\lambda.$$

□

## The formula of the regularized trace of the operator L

In this section, we find a formula for the regularized trace of the operator L. According to Theorem 2.1,

$$M_{p1}=-\frac{1}{2\pi i}\int_{|\lambda|=b_{p}} \operatorname{tr}\bigl[\bigl(QR_{\lambda }^{0}\bigr)\bigr]\,d\lambda.$$
(3.1)

Since $$\{\psi_{mn}\}_{m=0,n=1}^{\infty \infty}$$ is an orthonormal basis of the space $$H_{1}$$, from (3.1) we obtain

\begin{aligned} M_{p1}&=-\frac{1}{2\pi i}\int_{|\lambda|=b_{p}} \sum_{m=0}^{\infty}\sum _{n=1}^{\infty}\bigl(QR_{\lambda }^{0} \psi_{mn},\psi_{mn}\bigr)\,d\lambda \\ &=-\frac{1}{2\pi i}\int_{|\lambda|=b_{p}}\sum _{m=0}^{\infty}\sum_{n=1}^{\infty} \frac{(Q\psi_{mn},\psi_{mn})}{(m+\frac{1}{2})^{4}-\lambda }\,d\lambda \\ & =\sum_{m=0}^{\infty}\sum _{n=1}^{\infty}(Q\psi _{mn},\psi_{mn})\cdot \frac{1}{2\pi i }\int_{|\lambda|=b_{p}}\frac{1}{\lambda- (m+\frac{1}{2} )^{4}}\,d\lambda. \end{aligned}
(3.2)

Considering $$(m+\frac{1}{2})^{4}< b_{p}=(p+1)^{4}$$ for $$m\leq p$$ and $$(m+\frac{1}{2})^{4}>b_{p}=(p+1)^{4}$$ for $$m>p$$, from formula (3.2)

\begin{aligned} M_{p1}&=\sum_{m=0}^{p} \sum_{n=1}^{\infty}(Q\psi_{mn},\psi _{mn})\frac{1}{2\pi i }\int_{|\lambda|=b_{p}} \frac{1}{\lambda- (m+\frac{1}{2} )^{4}}\,d\lambda =\sum_{m=0}^{p}\sum _{n=1}^{\infty}(Q\psi_{mn},\psi _{mn}) \end{aligned}
(3.3)

is obtained.

From (1.1) and (3.3), we have

\begin{aligned} M_{p1}&=\sum_{m=0}^{p} \sum_{n=1}^{\infty}\int_{0}^{\pi} \biggl(Q(x)\sqrt{\frac{2}{\pi}} \sin\biggl(m+\frac{1}{2}\biggr)x\cdot \varphi _{n},\sqrt{\frac{2}{\pi}}\sin\biggl(m+\frac{1}{2} \biggr)x\cdot\varphi_{n} \biggr)_{H}\,dx \\ &=\frac{2}{\pi}\sum_{m=0}^{p}\sum _{n=1}^{\infty}\int_{0}^{\pi} \bigl(Q(x)\varphi_{n}, \varphi_{n}\bigr)_{H} \sin^{2}\biggl(m+\frac {1}{2}\biggr)x\,dx \\ &=\frac{1}{\pi}\sum_{m=0}^{p}\int _{0}^{\pi} \Biggl[\sum_{n=1}^{\infty} \bigl(Q(x)\varphi_{n},\varphi_{n}\bigr)_{H} \Biggr]\bigl(1-\cos (2m+1)x\bigr)\,dx. \end{aligned}
(3.4)

If we consider the formula $$\sum_{n=1}^{\infty }(Q(x)\varphi_{n},\varphi_{n})_{H}=\operatorname{tr}Q(x)$$, then we get

$$M_{p1}=\frac{p+1}{\pi}\int_{0}^{\pi} \operatorname{tr}Q(x)\,dx-\frac{1}{\pi}\sum_{m=0}^{p} \int_{0}^{\pi}\operatorname{tr}Q(x)\cos(2m+1)x\,dx.$$
(3.5)

### Lemma 3.1

If the operator function $$Q(x)$$ satisfies conditions (Q2) and (Q3), then we have

$$\|R_{\lambda}\|< p^{-3}$$

over the circle $$|\lambda|=b_{p}$$.

### Proof

Since the operator function $$Q(x)$$ satisfies conditions (Q2) and (Q3), we have

\begin{aligned}& \{\lambda_{mn}\}_{n=1}^{\infty}\subset \biggl(\biggl(m+ \frac{1}{2}\biggr)^{4}-\|Q\| ,\biggl(m+\frac{1}{2} \biggr)^{4}+\|Q\| \biggr)\quad (m=0,1,2,\ldots), \\& \biggl|\lambda_{mn}-\biggl(m+\frac{1}{2}\biggr)^{4}\biggr|< \|Q\|< \frac{5}{2}\quad (m=0,1,2,\ldots;n=1,2,\ldots). \end{aligned}

If we consider this relation, we get

\begin{aligned} |\lambda_{mn}-\lambda|&= \biggl|\lambda-\biggl(m+ \frac{1}{2}\biggr)^{4}- \biggl(\lambda _{mn}- \biggl(m+\frac{1}{2}\biggr)^{4}\biggr) \biggr| \\ &\geq \biggl|\lambda-\biggl(m+\frac{1}{2}\biggr)^{4} \biggr|- \biggl| \lambda_{mn}-\biggl(m+\frac {1}{2}\biggr)^{4} \biggr| \\ &>|\lambda|-\biggl(m+\frac{1}{2}\biggr)^{4}- \frac{5}{2}=(p+1)^{4}-\biggl(m+\frac {1}{2} \biggr)^{4}-\frac{5}{2} \\ &\geq(p+1)^{4}-\biggl(p+\frac{1}{2}\biggr)^{4}- \frac{5}{2}>2\biggl(p+\frac{1}{2}\biggr)^{3}- \frac {5}{2}>p^{3} \end{aligned}
(3.6)

for $$m\leq p$$ and

\begin{aligned} |\lambda_{mn}-\lambda|&= \biggl|\biggl(m+\frac{1}{2} \biggr)^{4}-\lambda- \biggl(\biggl(m+\frac {1}{2} \biggr)^{4}-\lambda_{mn} \biggr) \biggr| \\ &\geq \biggl|\biggl(m+\frac{1}{2}\biggr)^{4}-\lambda \biggr|- \biggl|\biggl(m+ \frac {1}{2}\biggr)^{4}-\lambda_{mn} \biggr|\geq\biggl(m+ \frac{1}{2}\biggr)^{4}-|\lambda|-\frac{5}{2} \\ &\geq\biggl(p+\frac{3}{2}\biggr)^{4}-(p+1)^{4}- \frac{5}{2}>2(p+1)^{3}-\frac {5}{2}>p^{3} \end{aligned}
(3.7)

for $$m\geq p+1$$.

On the other hand, we can write

$$\| R_{\lambda} \|=\mathop{\max_{m=0,1,\ldots}}_{n=1,2,\ldots}\bigl\{ |\lambda_{mn}- \lambda|^{-1}\bigr\} .$$
(3.8)

From (3.6), (3.7) and (3.8) we get

$$\| R_{\lambda} \|< p^{-3}\quad \bigl(|\lambda|=b_{p}\bigr).$$

□

### Lemma 3.2

If the operator function $$Q(x)$$ satisfies condition (Q3), then we have

$$\bigl\| QR_{\lambda}^{0}\bigr\| _{\sigma_{1}(H_{1})}< 5p^{-2}\sum _{n=1}^{\infty}\bigl\| Q(x)\varphi_{n}\bigr\|$$

over the circle $$|\lambda|=b_{p}$$.

### Proof

Let us show that the series $$\sum_{m=0}^{\infty}\sum_{n=1}^{\infty} \|QR_{\lambda}^{0}\psi_{mn} \|$$ is convergent.

For $$\lambda\notin\sigma(L_{0})$$, we get

\begin{aligned} &\sum_{m=0}^{\infty}\sum _{n=1}^{\infty} \bigl\| QR_{\lambda }^{0} \psi_{mn} \bigr\| \\ &\quad=\sum_{m=0}^{\infty} \sum _{n=1}^{\infty} \biggl|\biggl(m+\frac{1}{2} \biggr)^{4}-\lambda \biggr|^{-1} \|Q\psi _{mn} \| \\ &\quad=\sum_{m=0}^{\infty}\sum _{n=1}^{\infty} \biggl|\biggl(m+\frac {1}{2} \biggr)^{4}-\lambda \biggr|^{-1} \biggl[\int_{0}^{\pi} \biggl\| Q(x)\sqrt{\frac{2}{\pi}}\sin\biggl(m+\frac{1}{2}\biggr)x\cdot \varphi_{n} \biggr\| _{H}^{2}\,dx \biggr]^{\frac{1}{2}} \\ &\quad=\sqrt{\frac{2}{\pi}}\sum_{m=0}^{\infty} \sum_{n=1}^{\infty } \biggl|\biggl(m+\frac{1}{2} \biggr)^{4}-\lambda \biggr|^{-1} \int_{0}^{\pi} \bigl\| Q(x)\varphi_{n} \bigr\| _{H}^{2}\sin^{2} \biggl(m+\frac{1}{2}\biggr)x\,dx \\ &\quad< \sum_{m=0}^{\infty}\sum _{n=1}^{\infty} \biggl|\biggl(m+\frac {1}{2} \biggr)^{4}-\lambda \biggr|^{-1} \biggl[\int_{0}^{\pi} \bigl\| Q(x)\varphi_{n} \bigr\| _{H}^{2}\,dx \biggr]^{\frac{1}{2}} \\ &\quad=\sum_{m=0}^{\infty} \biggl|\biggl(m+ \frac{1}{2}\biggr)^{4}-\lambda \biggr|^{-1}\sum _{n=1}^{\infty} \bigl\| Q(x)\varphi_{n} \bigr\| . \end{aligned}
(3.9)

From this relation we obtain

$$\sum_{m=0}^{\infty}\sum _{n=1}^{\infty} \bigl\| QR_{\lambda }^{0} \psi_{mn} \bigr\| < \infty \quad\bigl(\lambda\notin\sigma(L_{0})\bigr).$$

On the other hand, since the sequence $$\{\psi_{mn}\} _{m=0,n=1}^{\infty \infty}$$ is an orthonormal basis of the space $$H_{1}$$, we get

$$\bigl\| QR_{\lambda}^{0} \bigr\| _{\sigma_{1}(H_{1})}\leq\sum _{m=0}^{\infty}\sum_{n=1}^{\infty} \bigl\| QR_{\lambda}^{0}\psi _{mn} \bigr\|$$
(3.10)

[15]. From (3.9) and (3.10) we obtain

$$\bigl\| QR_{\lambda}^{0} \bigr\| _{\sigma_{1}(H_{1})}\leq\sum _{n=1}^{\infty} \bigl\| Q(x)\varphi_{n} \bigr\| \sum _{m=0}^{\infty} \biggl|\biggl(m+\frac{1}{2} \biggr)^{4}-\lambda \biggr|^{-1}.$$
(3.11)

Furthermore, over the circle $$|\lambda|=b_{p}$$ we get

\begin{aligned} &\sum_{m=0}^{\infty} \biggl|\biggl(m+ \frac{1}{2}\biggr)^{4}-\lambda \biggr|^{-1} \\ &\quad=\sum_{m=0}^{p} \biggl|\biggl(m+ \frac{1}{2}\biggr)^{4}-\lambda \biggr|^{-1}+\sum _{m=p+1}^{\infty} \biggl|\biggl(m+\frac{1}{2} \biggr)^{4}-\lambda \biggr|^{-1} \\ &\quad< \sum_{m=0}^{p} \biggl(|\lambda|- \biggl(m+\frac{1}{2}\biggr)^{4} \biggr)^{-1}+\sum _{m=p+1}^{\infty} \biggl(\biggl(m+\frac{1}{2} \biggr)^{4}-|\lambda | \biggr)^{-1} \\ &\quad=\sum_{m=0}^{p} \biggl((p+1)^{4}-\biggl(m+\frac{1}{2}\biggr)^{4} \biggr)^{-1}+\sum_{m=p+1}^{\infty} \biggl( \biggl(m+\frac {1}{2}\biggr)^{4}-(p+1)^{4} \biggr)^{-1} \\ &\quad< \sum_{m=0}^{p} \biggl((p+1)^{4}-\biggl(p+\frac{1}{2}\biggr)^{4} \biggr)^{-1} \\ &\qquad{}+\sum_{m=p+1}^{\infty} \biggl( \biggl(m+\frac {1}{2}\biggr)^{2}-(p+1)^{2} \biggr)^{-1} \biggl(\biggl(m+\frac {1}{2}\biggr)^{2}+(p+1)^{2} \biggr)^{-1} \\ &\quad< \frac{1}{2}(p+1) \biggl(p+\frac{1}{2}\biggr)^{-3}+ \frac{1}{2}p^{-2}\sum_{m=p+1}^{\infty} \biggl(\biggl(m+\frac{1}{2}\biggr)^{2}-(p+1)^{2} \biggr)^{-1}. \end{aligned}
(3.12)

It can be easily shown that

$$\biggl(m+\frac{1}{2}\biggr)^{2}-(p+1)^{2}> \frac{1}{4}\bigl(m^{2}-p^{2}\bigr)\quad (m\geq p+1)$$
(3.13)

and

$$\sum_{m=p+1}^{\infty}\bigl(m^{2}-p^{2} \bigr)^{-1}< 2p^{-\frac{1}{2}}.$$
(3.14)

From (3.12), (3.13) and (3.14) we get

$$\sum_{m=o}^{\infty} \biggl|\biggl(m+\frac{1}{2} \biggr)^{4}-\lambda \biggr|^{-1}< p^{-2}+4p^{-\frac{5}{2}}< 5p^{-2}.$$
(3.15)

From (3.11) and (3.15) we obtain

$$\bigl\| QR_{\lambda}^{0}\bigr\| _{\sigma_{1}(H_{1})}< 5p^{-2}\sum _{n=1}^{\infty}\bigl\| Q(x)\varphi_{n}\bigr\| \quad \bigl(|\lambda|=b_{p}\bigr).$$

□

### Theorem 3.3

If the operator function $$Q(x)$$ satisfies conditions (Q1), (Q2), (Q3), and (Q4), then we have the formula

\begin{aligned} &\sum_{m=0}^{\infty} \Biggl\{ \sum _{n=1}^{\infty} \biggl[\lambda_{mn}-\biggl(m+ \frac{1}{2}\biggr)^{4} \biggr]-\frac{1}{\pi}\int _{0}^{\pi}\operatorname{tr}Q(x)\,dx \Biggr\} \\ &\quad=\frac{1}{4}\bigl[\operatorname{tr}Q(\pi)-\operatorname{tr}Q(0) \bigr]. \end{aligned}

### Proof

By using Theorem 2.1, Lemma 3.1 and Lemma 3.2, we find

\begin{aligned} |{M_{p2}}| = & {\frac{1}{4\pi}} \biggl|\int _{|\lambda|=b_{p}} \operatorname{tr} \bigl[\bigl(QR_{\lambda}^{0} \bigr)^{2} \bigr]\,d\lambda \biggr| \\ < & {\frac{1}{4\pi}}\int_{|\lambda|=b_{p}} \bigl| \operatorname{tr} \bigl[\bigl(QR_{\lambda}^{0}\bigr)^{2} \bigr] \bigr||\,d\lambda| \\ \leq&{\frac{1}{4\pi}}\int_{|\lambda|=b_{p}}\bigl\| \bigl(QR_{\lambda}^{0}\bigr)^{2}\bigr\| _{\sigma_{1}(H_{1})}|\,d\lambda| \\ \leq& {\frac{1}{4\pi}}\int_{|\lambda|=b_{p}}\bigl\| QR_{\lambda}^{0} \bigr\| \bigl\| QR_{\lambda}^{0} \bigr\| _{\sigma_{1}(H_{1})}|\,d\lambda| \\ \leq& {\frac{\|Q\|}{4\pi}}\int_{|\lambda|=b_{p}}\bigl\| R_{\lambda}^{0}\bigr\| \bigl\| QR_{\lambda}^{0} \bigr\| _{\sigma_{1}(H_{1})}|\,d\lambda| \\ < & c_{1}\int_{|\lambda|=b_{p}}p^{-5}\,d\lambda \\ = & 2\pi\cdot b_{p}\cdot c_{1}\cdot p^{-5}= 2 \pi c_{1}(p+1)^{4}p^{-5}< c_{2}p^{-1}. \end{aligned}
(3.16)

Here c is a positive constant.

By using formula (2.6), Lemma 3.1 and Lemma 3.2, we find

\begin{aligned} |{M_{p}}| = & {\frac{1}{2\pi}} \biggl|\int _{|\lambda|=b_{p}} \lambda \operatorname{tr} \bigl[R_{\lambda} \bigl(QR_{\lambda}^{0}\bigr)^{3} \bigr]\,d\lambda \biggr| \\ \leq& {\frac{b_{p}}{2\pi}}\int_{|\lambda|=b_{p}} \bigl\| R_{\lambda} \bigl(QR_{\lambda}^{0}\bigr)^{3} \bigr\| |\,d\lambda| \\ \leq&{\frac{b_{p}}{2\pi}}\int_{|\lambda|=b_{p}}\| R_{\lambda}\| \bigl\| \bigl(QR_{\lambda}^{0}\bigr)^{3}\bigr\| _{\sigma_{1}(H_{1})}|\,d\lambda| \\ \leq& {\frac{b_{p}}{2\pi}}\int_{|\lambda|=b_{p}}\| R_{\lambda}\| \bigl\| \bigl(QR_{\lambda}^{0}\bigr)^{2}\bigr\| \bigl\| \bigl(QR_{\lambda}^{0}\bigr)\bigr\| _{\sigma_{1}(H_{1})}|\,d\lambda| \\ \leq& {\frac{b_{p}}{2\pi}}\int_{|\lambda|=b_{p}}\| R_{\lambda}\| \| Q\|^{2}\bigl\| R_{\lambda}^{0}\bigr\| ^{2}\bigl\| QR_{\lambda}^{0}\bigr\| _{\sigma_{1}(H_{1})}|\,d\lambda| \\ < & c_{3}\cdot b_{p}^{2}p^{-11} \\ = & c_{3}(p+1)^{8}p^{-11}< c_{4}p^{-3}. \end{aligned}
(3.17)

From (3.16) and (3.17) we get

$$\lim_{p\rightarrow\infty} {M}_{p2}=\lim_{p\rightarrow \infty} {M}_{p}=0.$$
(3.18)

From (2.4) and (3.5) we obtain

\begin{aligned} &\sum_{m=0}^{p} \Biggl\{ \sum _{n=1}^{\infty} \biggl[\lambda _{mn}- \biggl(m+\frac{1}{2}\biggr)^{4} \biggr]-\frac{1}{\pi}\int _{0}^{\pi} \operatorname{tr} Q(x)\,dx \Biggr\} \\ &\quad=-\frac{1}{\pi}\sum_{m=0}^{p} \int_{0}^{\pi}\operatorname{tr} Q(x)\cos (2m+1)x\,dx+M_{p2}+M_{p}. \end{aligned}
(3.19)

From (3.18) and (3.19) we find

\begin{aligned} &\sum_{m=0}^{\infty} \Biggl\{ \sum _{n=1}^{\infty} \biggl[\lambda_{mn}- \biggl(m+\frac{1}{2}\biggr)^{4} \biggr]-\frac{1}{\pi}\int _{0}^{\pi} \operatorname{tr} Q(x)\,dx \Biggr\} \\ &\quad=-\frac{1}{\pi}\sum_{m=0}^{\infty}\int _{0}^{\pi}\operatorname{tr} Q(x) \cos(2m+1)x\,dx. \end{aligned}
(3.20)

Moreover, using conditions (Q1) and (Q4), we get

\begin{aligned} &{-}\frac{1}{\pi}\sum_{m=0}^{\infty} \int_{0}^{\pi}\operatorname{tr} Q(x)\cos(2m+1)x\,dx \\ &\quad=-\frac{1}{2\pi}\sum_{m=1}^{\infty} \biggl[\int_{0}^{\pi}\operatorname{tr} Q(x)\cos mx \,dx-(-1)^{m}\int_{0}^{\pi} \operatorname{tr} Q(x)\cos mx \,dx \biggr] \\ &\quad=-\frac{1}{4} \Biggl\{ \sum_{m=1}^{\infty} \biggl[\frac{2}{\pi}\int_{0}^{\pi} \operatorname{tr} Q(x)\cos mx\,dx \biggr]\cos m0+ \biggl[\frac{1}{\pi}\int _{0}^{\pi }\operatorname{tr}Q(x)\,dx \biggr]\cos0 \Biggr\} \\ &\qquad{}+\frac{1}{4} \Biggl\{ \sum_{m=1}^{\infty} \biggl[\frac{2}{\pi}\int_{0}^{\pi} \operatorname{tr} Q(x)\cos mx\,dx \biggr]\cos m \pi+ \biggl[\frac{1}{\pi}\int _{0}^{\pi}\operatorname{tr} Q(x)\,dx \biggr]\cos0\pi \Biggr\} \\ &\quad=\frac{1}{4}\bigl[\operatorname{tr} Q(\pi)-\operatorname{tr} Q(0) \bigr]. \end{aligned}
(3.21)

From (3.20) and (3.21) we find

$$\sum_{m=0}^{\infty} \Biggl\{ \sum _{n=1}^{\infty} \biggl[\lambda_{mn}-\biggl(m+ \frac{1}{2}\biggr)^{4} \biggr]-\frac{1}{\pi}\int _{0}^{\pi}\operatorname{tr}Q(x)\,dx \Biggr\} = \frac{1}{4}\bigl[\operatorname{tr} Q(\pi)-\operatorname{tr} Q(0)\bigr].$$

The theorem is proved. □

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## Acknowledgements

The authors would like to thank Professor Ehliman Adıgüzelov for his expert assistance and for contributions in detailed editing of the last version of this manuscript.

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Correspondence to Serpil Karayel.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

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Karayel, S., Sezer, Y. The regularized trace formula for a fourth order differential operator given in a finite interval. J Inequal Appl 2015, 316 (2015). https://doi.org/10.1186/s13660-015-0823-0

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• DOI: https://doi.org/10.1186/s13660-015-0823-0

• 47A10
• 34L20

### Keywords

• separable-Hilbert space