Fractional Powers of Anti-Triangular Operator Matrices and Their Applications

Abstract

The concept of fractional powers of operators was initially developed within the framework of functional analysis and has since played a significant role in the study of evolution equations, abstract differential equations, and complex dynamical systems. Meanwhile, the operator matrix approach has emerged as a powerful and widely used tool for analyzing the structural properties of operators. Motivated by these developments, this paper focuses on the explicit computation of fractional powers of anti-triangular operator matrices. Specifically, we first derive an explicit formula for the fractional powers of off-diagonal operator matrices by employing the formal determinant of block operator matrices. Then, based on the Schur factorization, we obtain a representation for the fractional powers of the general anti-triangular case. As applications, the obtained results are further applied to certain differential equations.

1. Introduction

Let $X,Y$ be Banach spaces; $\mathcal{L}(X,Y)$, $\mathcal{B}(X,Y)$, and $\mathcal{C}(X,Y)$ denote the sets of all linear, bounded, and closed linear operators from X to Y, respectively. The main objective of this paper is to compute the fractional powers of anti-triangular operator matrices

$$\mathcal{A}=\left(\begin{array}{cc}A& B\\ D& 0\end{array}\right):\mathcal{D}\left(\mathcal{A}\right)=(\mathcal{D}\left(A\right)\cap \mathcal{D}\left(D\right))\times \mathcal{D}\left(B\right)\subset X\times Y\to X\times Y$$

(1)

with $A\in \mathcal{C}\left(X\right)$, $B\in \mathcal{L}(Y,X)$, and $D\in \mathcal{C}(X,Y)$.

Originally formulated within functional analysis, the theory of fractional powers of operators has become an effective analytical framework for investigating evolution equations, abstract differential equations, and systems with complex dynamical behavior [1,2,3,4]. The study of fractional powers of closed linear operators is an important fundamental topic in the theory of fractional differential equations, with particular emphasis on the characterization of fractional powers of positive linear operators. Foundational contributions in this area were made by Hille and Phillips [5]. Later, Balakrishnan extended the theory to non-negative operators—possibly non-densely defined and not necessarily generating strongly continuous semigroups—by employing an abstract version of the Stieltjes transform [2]. Furthermore, Komatsu developed several integral representations for the fractional powers of this class of operators and systematically investigated the connection between their domains and interpolation spaces [6,7]. Several other studies have also been conducted in this field [8,9,10,11,12,13].

Since the establishment of operator theory, the study of block operator matrices has remained one of its most active and significant topics. These matrices are widely used in various areas, including reaction-diffusion systems, evolution equations, quantum mechanics, control theory, and magnetohydrodynamics [14,15,16,17]. However, research on the fractional powers of block operator matrices remains relatively limited. This is mainly due to the potential non-commutativity among the operator entries, which makes it difficult to obtain explicit expressions for their resolvants.

For lower triangular block operator matrices, $\mathcal{A}=\left(\begin{array}{cc}A& 0\\ B& C\end{array}\right)$, which is associated with the system

$$\left\{\begin{array}{c}{u}_t+Au=f_1\left(u\right),t>0,\hfill \\ v_t+Bu+Cv=f_2(u,v),t>0.\hfill \end{array}\right.$$

In [18], the authors provided computations of ${\mathcal{A}}^{\alpha }$. In [3], the authors first proposed several techniques for explicitly computing fractional powers of $2\times 2$ block operator matrices of the form

$$\left(\begin{array}{cc}{A}_{11}& A_{12}\\ A_{21}& A_{22}\end{array}\right)$$

and further constructed a theoretical framework applicable to various evolutionary models, which was then used to explicit compute the fractional powers of a $4\times 4$ differential operator matrix arising from a weakly coupled system of wave equations. This line of research was subsequently advanced in [19], where a theoretical framework for fractional powers of $3\times 3$ block operator matrices was established together with explicit formulas and applications to several classes of PDEs.

In [20,21], the authors provided representations of the fractional powers of the differential operator matrix ${\mathcal{A}}_{\eta ,\theta }=\left(\begin{array}{cc}0& -I\\ -\Delta & \eta {(-\Delta )}^{\theta }\end{array}\right)$, which is derived from the strongly damped wave equation

$$u_{tt}+\eta {(-\Delta )}^{\theta }{u}_t-\Delta u=f\left(u\right),t>0,$$

in the cases with $(\eta ,\theta )=(0,0)$ and $(\eta ,\theta )=(2,{\displaystyle \frac{1}{2}})$ respectively. In [22], the authors presented a fractional power representation of the differential operator matrix $\mathcal{A}=\left(\begin{array}{ccc}0& -I& 0\\ 0& 0& -I\\ -\Delta & 0& 0\end{array}\right)$, which is derived from a third-order (in time) differential equation

$$u_{ttt}-\Delta u=0,t>0.$$

It is worth emphasizing that the differential operator matrices studied in [20,21,22] are all of anti-triangular form. The presence of zero entries on the diagonal significantly alters the structural and spectral properties of such operator matrices, making them fundamentally different from upper triangular operator matrices. This distinction has attracted considerable attention in the literature [23,24,25,26]. To date, no study has provided a systematic analytical framework for the calculus of fractional powers of anti-triangular operator matrices covering a wide range of possible situations. Existing studies on this topic are relatively scattered and primarily focus on specific cases derived from partial differential equations. Given these circumstances, it is natural to study the representation of fractional powers of anti-triangular operator matrices from a general and systematic perspective. In comparison with existing studies on the representation of fractional powers of block operator matrices, the present work does not directly require the anti-triangular operator matrix to be of positive type; instead, it develops a series of auxiliary results to show such property as a basis of the representation of its fractional powers. By the Frobenius–Schur factorization, an explicit representation of the resolvant of the anti-triangular operator matrix is obtained; then, an analytical framework is further constructed to represent its fractional powers. Finally, we apply the theoretical results to differential operator matrices derived from some evolution systems to illustrative their effectiveness.

The rest of the paper is structured as follows: in Section 2, we begin by introducing the basic theory of operators of positive type and present several related results; subsequently, we derive an explicit expression for the resolvant of $\mathcal{A}$ provided in (1) via the Schur factorization, followed by the establishment of several auxiliary results to verify that $\mathcal{A}$ is of positive type; in Section 3, we obtain a representation for the fractional powers of off-triangular operator matrices (i.e., $A=0$) using the formal determinant of block operator matrices; then, a general representation for the fractional powers of anti-triangular operator matrices is derived in Section 4 by applying the explicit resolvant expression of $\mathcal{A}$ obtained in Section 2.

2. Preliminaries

In this section, we provide the definition and fundamental results of operators of positive type and establish several auxiliary results to support the proofs of the main theorems in the subsequent section. Unless otherwise specified, the operators $\mathcal{A}$, A, B, and D refer to the operators appearing in Equation (1).

 Definition 1 

([1]). Let X be a Banach space and let $A\in \mathcal{C}\left(X\right)$ be densely defined. A is said to be of positive type M if there exists $M\ge 1$ s.t.

(i) 

$(0,\infty )\subset \rho (-A)$.

(ii) 

${(1+s)\left|\right|(s+A)}^{-1}\left|\right|\le M$ for all $s>0$.

The symbol ${\mathcal{P}}_M\left(X\right)$ denotes the class of all linear operators of positive type M in X, and A is of positive type if it belongs to $\mathcal{P}\left(X\right):={\bigcup }_{M\ge 1}{\mathcal{P}}_M\left(X\right)$.

There exists a larger set that is also contained in $\rho (-A)$.

 Remark 1. 

Let $A\in$${\mathcal{P}}_M\left(X\right)$, ${\theta }_M:=arcsin\left({\displaystyle \frac{1}{2M}}\right)$ and $S_{{\theta }_M}:=\{z\in \mathbb{C}:|argz|\le {\theta }_M\}$. Then,

$$E_M:=S_{{\theta }_M}\cup \{z\in \mathbb{C}:|z|\le {\displaystyle \frac{1}{2M}}\}$$

is contained in $\rho (-A)$ and the following estimate holds:

$${(1+|\lambda \left|\right)\left|\right|(\lambda +A)}^{-1}{\left|\right|}_{\mathcal{L}\left(X\right)}\le 2M+1,\lambda \in E_M.$$

Because $E_M\subset \rho (-A)$, there exist $\psi \in (0,{\theta }_M)$ and $r_0\in (0,{\displaystyle \frac{1}{2M}})$ such that

$$S_{\psi }\cup B[0,r_0]\subset \rho (-A),$$

and ${(1+|\lambda \left|\right)\left|\right|(\lambda +A)}^{-1}{\left|\right|}_{\mathcal{L}\left(X\right)}\le M$ holds for all $\lambda \in S_{\psi }\cup B[0,r_0].$ For more details, refer to [1].

 Definition 2 

([18]). Let $A\in$${\mathcal{P}}_M\left(X\right)$ and let Γ be a contour of $S_{\psi }\cup B[0,r_0]$ (see Figure 1 in [18]) oriented from $\infty e^{-i\psi }$ to $\infty e^{i\psi }$. Then,

$$A^{-\alpha }:={\displaystyle \frac{1}{2\pi i}}{\int }_{\Gamma }{(-s)}^{-\alpha }{(s+A)}^{-1}ds$$

can be well-defined for any $\alpha >0$.

 Lemma 1 

([27]). Let $A\in \mathcal{P}\left(X\right).$ Then, $A^{-\alpha }$ is injective and $A^{-\alpha }\in \mathcal{B}\left(X\right)$ for all $\alpha >0$. In this case, its inverse is well-defined and is denoted by $A^{\alpha }:={\left(A^{-\alpha }\right)}^{-1}$.

 Lemma 2 

([1]). Let $A\in$$\mathcal{P}\left(X\right)$.

 (i) 

For $Re\alpha \in (0,1)$,

$$A^{-\alpha }={\displaystyle \frac{sin\left(\pi \alpha \right)}{\pi }}{\int }_0^{\infty }{s}^{-\alpha }{(s+A)}^{-1}ds.$$

 (ii) 

For $Re\alpha \in (0,m+1)$, $m\in N$,

$$A^{-\alpha }={\displaystyle \frac{sin\left(\pi \alpha \right)}{\pi }}{\displaystyle \frac{m!}{(1-\alpha )(2-\alpha )\dots (m-\alpha )}}{\int }_0^{\infty }{s}^{m-\alpha }{(s+A)}^{-m-1}ds.$$

 (iii) 

For $Re\alpha \in (-1,1)$ and $x\in \mathcal{D}\left(A\right)$,

$$A^{\alpha }x={\displaystyle \frac{sin\pi \alpha }{\pi \alpha }}\int s^{\alpha }{(s+A)}^{-2}Axds.$$

 Lemma 3 

([4]). Let $A\in \mathcal{P}\left(X\right).$ Then,

$$\sigma \left(A^{\alpha }\right)={\left[\sigma \left(A\right)\right]}^{\alpha }=\{{\lambda }^{\alpha }:\lambda \in \sigma \left(A\right)\}$$

for each $\alpha >0$.

To compute the fractional powers of the anti-triangular operator matrix $\mathcal{A}$ provided in (1), we make the following assumption on its operator entries and their associated combinations.

 Assumption 1. 

Assume that the following conditions hold.

(i) 

A is of positive type M for some $M\ge 1$.

(ii) 

B is densely defined closable, D is A–bounded.

(iii) 

$(0,\infty )\subset {\Omega }_q$, $B,D$ and ${(\lambda +A)}^{-1}$ commute for all $\lambda \in (0,\infty )$, where

$${\Omega }_q={\{\lambda \in \mathbb{C}:||(\lambda +A)}^{-1}(D{(\lambda +A)}^{-1}B+A)\left|\right|<q\}$$

and $0<q<1$.

(iv) 

$\mathcal{D}\left(\overline{L_{\lambda }}\right)=X$ for all $\lambda \in (0,\infty )$.

(v) 

$\left|\right|\overline{L_{\lambda }}\left|\right|\le M_1$ for all $\lambda \in (0,\infty )$ and some $M_1\ge 0$.

(vi) 

$M_{\lambda }$ is closed for some $\lambda \in (0,\infty )$.

Here,

$$\begin{array}{cc}\hfill & L_{\lambda }:={(\lambda +A)}^{-1}B,\hfill \\ \hfill & M_{\lambda }:=\lambda -D{(\lambda +A)}^{-1}B.\hfill \end{array}$$

(2)

 Remark 2. 

Our approach to studying the fractional powers of $\mathcal{A}$ is to first establish fundamental results in conjunction with the Frobenius–Schur factorization in order to guarantee that $\mathcal{A}$ is of positive type. It is shown that Assumptions (i), (v), and (vi) are natural and necessary in the process of estimating the norm of the resolvant of $\mathcal{A}$ and verifying the closedness of $\mathcal{A}$. In addition, Assumptions (ii) and (iv) are essential, which parallels [17] for analyzing the essential spectrum of block operator matrices via the Frobenius–Schur factorization. Notably, to explicitly represent the resolvant of the corresponding Schur complement under the factorization, we employ the Banach lemma for the first time, motivating the introduction of Assumption (iii). In addition, Remark 3 below provides several illustrative examples of operators satisfying Assumptions (iii) and (v).

 Remark 3. 

Under Assumption 1 (i), condition (v) holds if the operator B is bounded on Y. Indeed, for any $\lambda \in (0,\infty )$, we have

$$\parallel \overline{L_{\lambda }}x\parallel =\parallel \overline{{(\lambda +A)}^{-1}B}{x\parallel \le \parallel (\lambda +A)}^{-1}\parallel \parallel Bx\parallel \le M\parallel B\parallel \parallel x\parallel =:M_1\parallel x\parallel$$

for all $x\in Y$, which implies $\parallel \overline{L_{\lambda }}\parallel \le M_1$; thus, Assumption 1 (v) is satisfied.

When Assumption 1 (i) holds, suppose that A is bounded, $B=I$, and D is A-bounded with $\parallel Dx\parallel \le a\parallel x\parallel +b\parallel Ax\parallel$ for constants $a,b$ satisfying $(a+b)M^2+bM+M\parallel A\parallel <q.$ Moreover, assume that D and ${(\lambda +A)}^{-1}$ commute for all $\lambda \in (0,\infty )$. Then, Assumption 1 (iii) holds. Indeed, for any $x\in X$, we have

$$\begin{array}{cc}\hfill {\parallel D(\lambda +A)}^{-1}x\parallel & {\le a\parallel (\lambda +A)}^{-1}{x\parallel +b\parallel A(\lambda +A)}^{-1}x\parallel \hfill \\ \hfill & \le aM\parallel x\parallel +b\parallel x\parallel +bM\parallel x\parallel =\left(\right(a+b)M+b)\parallel x\parallel ,\hfill \end{array}$$

(3)

meaning that

$${\parallel D(\lambda +A)}^{-1}\parallel \le (a+b)M+b.$$

(4)

Combining (4) with Assumption 1 (i), we obtain

$$\begin{array}{cc}\hfill {\parallel (\lambda +A)}^{-1}(D{(\lambda +A)}^{-1}+A)\parallel & {\le \parallel (\lambda +A)}^{-1}{\parallel \parallel D(\lambda +A)}^{-1}{\parallel +\parallel (\lambda +A)}^{-1}A\parallel \hfill \\ \hfill & \le M\left(\right(a+b)M+b)+M\parallel A\parallel \hfill \\ \hfill & =(a+b)M^2+bM+M\parallel A\parallel <q,\hfill \end{array}$$

which shows that Assumption 1 (iii) is satisfied.

One of the important ingredients of this section is the following proposition.

 Proposition 1. 

Let Assumptions 1 (i), (ii), and (iv) hold. If $M_{\lambda }$ is invertible for all $\lambda \in (0,\infty )$, then $\lambda +\mathcal{A}$ is invertible for all $\lambda \in (0,\infty )$. In this case,

$${(\lambda +\mathcal{A})}^{-1}=\left(\begin{array}{cc}{(\lambda +A)}^{-1}+\overline{L_{\lambda }}{M}_{\lambda }^{-1}D{(\lambda +A)}^{-1}& -\overline{L_{\lambda }}{M}_{\lambda }^{-1}\\ -M_{\lambda }^{-1}D{(\lambda +A)}^{-1}& M_{\lambda }^{-1}\end{array}\right),$$

(5)

where $M_{\lambda }$ and $L_{\lambda }$ are provided in (2).

Proof. 

Because A is of positive type, $\lambda +\mathcal{A}$ has the following decomposition:

$$\begin{array}{cc}\hfill \lambda +\mathcal{A}& =\left(\begin{array}{cc}I& 0\\ D{(\lambda +A)}^{-1}& I\end{array}\right)\left(\begin{array}{cc}\lambda +A& 0\\ 0& M_{\lambda }\end{array}\right)\left(\begin{array}{cc}I& \overline{L_{\lambda }}\\ 0& I\end{array}\right)\hfill \\ \hfill & :={\mathcal{Q}}_1\left(\lambda \right){\mathcal{R}}_1\left(\lambda \right){\mathcal{Q}}_2\left(\lambda \right).\hfill \end{array}$$

(6)

Observe that ${\mathcal{Q}}_1\left(\lambda \right)$ and ${\mathcal{Q}}_2\left(\lambda \right)$ are bounded with bounded inverses on $X\times Y$; thus, it follows that

$${(\lambda +\mathcal{A})}^{-1}=\left(\begin{array}{cc}{(\lambda +A)}^{-1}+\overline{L_{\lambda }}{M}_{\lambda }^{-1}D{(\lambda +A)}^{-1}& -\overline{L_{\lambda }}{M}_{\lambda }^{-1}\\ -M_{\lambda }^{-1}D{(\lambda +A)}^{-1}& M_{\lambda }^{-1}\end{array}\right).$$

(7)

□

The following Lemma can be obtained from Proposition 3.1 in [17] and the resolvant formula.

 Lemma 4. 

Let Assumptions 1 (i) and (ii) hold. Then, ${(\lambda +A)}^{-1}B$ is bounded for some (and hence for all) $\lambda \in (0,\infty )$ iff $D\left(A^{*}\right)\subset D\left(B^{*}\right)$. Moreover, $D{(\lambda +A)}^{-1}$ is bounded for all $\lambda \in (0,\infty )$.

Proof. 

The first assertion is a direct consequence of Proposition 3.1 in [17]. Let $\lambda >0$ and $x\in \mathcal{D}\left(A\right)$. Because D is A-bounded and ${(\lambda +A)}^{-1}$ is bounded, there exist $a,b,c>0$ such that

$$\begin{array}{cc}\hfill {\parallel D(\lambda +A)}^{-1}x\parallel & {\le a\parallel (\lambda +A)}^{-1}{x\parallel +b\parallel A(\lambda +A)}^{-1}x\parallel \hfill \\ \hfill & \le {ac\parallel x\parallel +b\parallel (A+\lambda -\lambda )(\lambda +A)}^{-1}x\parallel \hfill \\ \hfill & \le (ac+b+bc\lambda )\parallel x\parallel .\hfill \end{array}$$

Thus, $D{(\lambda +A)}^{-1}$ is bounded.

Now, let $\mu >0$, $\mu \ne \lambda$. The resolvant identity yields

$$D{(\lambda +A)}^{-1}-D{(\mu +A)}^{-1}=(\lambda -\mu )D{(\lambda +A)}^{-1}{(\mu +A)}^{-1}.$$

Combining the boundedness of the right-hand side with that of $D{(\lambda +A)}^{-1}$, we obtain that $D{(\mu +A)}^{-1}$ is also bounded. This completes the proof. □

 Lemma 5 

([28]). Let $T\in \mathcal{C}(X,Y)$ be densely defined and let $S\in \mathcal{L}(X,Y)$ be T-bounded with a T-bound smaller than 1. Then, $T+S$ is closed if and only if T is closed.

 Proposition 2. 

Let Assumptions 1 (i), (ii), (vi) hold and let $D\left(A^{*}\right)\subset D\left(B^{*}\right)$. Then, $M_{\lambda }$ is closed for all $\lambda \in (0,\infty )$, where $M_{\lambda }$ is provided in (2).

Proof. 

For $\mu \in (0,\infty )$, by applying the resolvant formula, we obtain

$$\begin{array}{cc}\hfill M_{\lambda }-M_{\mu }& =\lambda -D{(\lambda +A)}^{-1}B-\mu +D{(\mu +A)}^{-1}B\hfill \\ & =\lambda -\mu +D{(\mu +A)}^{-1}B-D{(\lambda +A)}^{-1}B\hfill \\ & =\lambda -\mu +(\mu -\lambda )D{(\mu +A)}^{-1}{(\lambda +A)}^{-1}B,\hfill \end{array}$$

then $M_{\mu }$ is closed by Lemmas 4 and 5, and this establishes the desired result. □

Combining Equation (6) with Proposition 2, we immediately obtain the following result.

 Corollary 1. 

Let Assumption 1 (i), (ii), (iv), (vi) hold and $D\left(A^{*}\right)\subset D\left(B^{*}\right)$. Then $\mathcal{A}$ is closed.

 Proposition 3. 

Under Assumption 1 (i) and (iii), $M_{\lambda }$ is invertible for all $\lambda \in (0,\infty )$. In this case,

$$\begin{array}{c}\hfill M_{\lambda }^{-1}=\sum _{n=0}^{\infty }\sum _{k=0}^n{C}_n^k{D}^{n-k}{B}^{n-k}{A}^k{(\lambda +A)}^{k-2n-1},\end{array}$$

(8)

where $M_{\lambda }$ is provided in (2).

Proof. 

Because

$$\begin{array}{cc}\hfill M_{\lambda }=\lambda -D{(\lambda +A)}^{-1}B& =(\lambda +A)-D{(\lambda +A)}^{-1}B-(\lambda +A){(\lambda +A)}^{-1}A\hfill \\ \hfill & =(\lambda +A)(I-D{(\lambda +A)}^{-2}B-{(\lambda +A)}^{-1}A)\hfill \\ \hfill & =(\lambda +A)(I-{(\lambda +A)}^{-1}(D{(\lambda +A)}^{-1}B+A))\hfill \end{array}$$

(9)

and by Assumption 1 (iii) we have

$$∥{(\lambda +A)}^{-1}\left(D{(\lambda +A)}^{-1}B+A\right)∥<q<1,$$

hence,

$$\begin{array}{cc}\hfill M_{\lambda }^{-1}& =\sum _{n=0}^{\infty }{(D{(\lambda +A)}^{-1}B+A)}^n{(\lambda +A)}^{-n-1}\hfill \\ \hfill & =\sum _{n=0}^{\infty }\sum _{k=0}^n{C}_n^k{\left(D{(\lambda +A)}^{-1}B\right)}^{n-k}{A}^k{(\lambda +A)}^{-n-1}\hfill \\ \hfill & =\sum _{n=0}^{\infty }\sum _{k=0}^n{C}_n^k{D}^{n-k}{B}^{n-k}{A}^k{(\lambda +A)}^{k-2n-1}\hfill \end{array}$$

by the Banach lemma. □

 Proposition 4. 

Let Assumptions 1 (i) and (iii) hold. Then, $M_{\lambda }$ is invertible and

$$(1+\lambda )\left|\right|M_{\lambda }^{-1}\left|\right|\le M_2$$

for all $\lambda \in (0,\infty )$ and some $M_2\ge 1$, where $M_{\lambda }$ is provided in (2).

Proof. 

Let $\lambda \in (0,\infty )$ and $M_2={\displaystyle \frac{M}{1-q}}$. Because A is of positive M for some $M\ge 1$, then

$$\begin{array}{cc}\hfill (1+\lambda )\left|\right|M_{\lambda }^{-1}\left|\right|& \le {(1+\lambda )\left|\right|(\lambda +A)}^{-1}\left|\right|\left|\right|(I-{(\lambda +A)}^{-1}(D{(\lambda +A)}^{-1}B+A))\left|\right|\le M_2\hfill \end{array}$$

by (9). □

Combing Proposition 1 with Proposition 4, we obtain the following result.

 Proposition 5. 

Let Assumptions 1 (i)–(v) hold and let D be A-bounded with

$$\begin{array}{c}\hfill \left|\right|Dx\left|\right|\le a\left|\right|x\left|\right|+b\left|\right|Ax\left|\right|\end{array}$$

for certain constants $a,b$ and all $x\in \mathcal{D}\left(A\right)$. Then, $(0,\infty )\subset \rho (-\mathcal{A})$, and

$${(1+\lambda )\left|\right|(\lambda +\mathcal{A})}^{-1}\left|\right|\le M_3$$

holds for some $M_3\ge 1$ and all $\lambda \in (0,\infty )$.

Proof. 

Let $\lambda \in (0,\infty )$. As a direct consequence of Proposition 1 and Proposition 4, we have that $\lambda +\mathcal{A}$ is invertible. From Equation (3), we obtain

$${\left|\right|D(\lambda +A)}^{-1}\left|\right|\le (aM+b+bM);$$

hence,

$$\begin{array}{cc}{(1+\lambda )\left|\right|(\lambda +\mathcal{A})}^{-1}\left|\right|\hfill & \le {(1+\lambda )\left|\right|(\lambda +A)}^{-1}\left|\right|+(1+\lambda )\left|\right|\overline{L_{\lambda }}{M}_{\lambda }^{-1}D{(\lambda +A)}^{-1}\left|\right|\hfill \\ \hfill & +(1+\lambda )\left|\right|\overline{L_{\lambda }}{M}_{\lambda }^{-1}\left|\right|+(1+\lambda )\left|\right|M_{\lambda }^{-1}{\left|\right|\left|\right|D(\lambda +A)}^{-1}\left|\right|\hfill \\ \hfill & +(1+\lambda )\left|\right|M_{\lambda }^{-1}\left|\right|\hfill \\ \hfill & \le M+{\displaystyle \frac{M{M}_1(aM+b+bM)}{1-q}}+{\displaystyle \frac{M{M}_1}{1-q}}+{\displaystyle \frac{M(aM+b+bM)}{1-q}}+{\displaystyle \frac{M}{1-q}}.\hfill \end{array}$$

Consequently, the conclusion follows by taking

$$M_3:=M+{\displaystyle \frac{M{M}_1(aM+b+bM)}{1-q}}+{\displaystyle \frac{M{M}_1}{1-q}}+{\displaystyle \frac{M(aM+b+bM)}{1-q}}+{\displaystyle \frac{M}{1-q}}.$$

□

Combining Corollary 1 with Proposition 5, we obtain the following result, which serves as a theoretical basis for the subsequent calculation of the fractional powers of the anti-triangular operator matrix $\mathcal{A}$ provided in (1).

 Corollary 2. 

Let Assumption 1 hold, $D\left(A^{*}\right)\subset D\left(B^{*}\right)$, and D be A-bounded with

$$\begin{array}{c}\hfill \left|\right|Dx\left|\right|\le a\left|\right|x\left|\right|+b\left|\right|Ax\left|\right|\end{array}$$

for certain constants $a,b$ and all $x\in \mathcal{D}\left(A\right)$. Then, $\mathcal{A}\in \mathcal{P}(X\times X)$.

3. Fractional Powers of Off-Triangular Operator Matrices

This section primarily discusses the fractional powers of the off-triangular operator matrices

$${\mathcal{A}}_0=\left(\begin{array}{cc}0& B\\ D& 0\end{array}\right)$$

with

$$\mathcal{D}\left({\mathcal{A}}_0\right)=\mathcal{D}\left(D\right)\times \mathcal{D}\left(B\right)\subset X\times Y\to X\times Y,$$

where X and Y are Banach spaces.

The objective is to investigate their structural properties, which not only provide insight for studying more general anti-triangular operator matrices in the subsequent section but also play a partial role in the parabolic approximation of certain partial differential equations associated with off-triangular operator matrices.

 Theorem 1. 

Let $B\in \mathcal{C}(Y,X)$ and $D\in \mathcal{C}(X,Y)$ be densely defined, B and D commute, and $-BD$ be of positive type. If

$$\begin{array}{c}\hfill (1+\lambda )\left|\right|D{({\lambda }^2-BD)}^{-1}\left|\right|\le M_4\end{array}$$

and

$$\begin{array}{c}\hfill (1+\lambda )\left|\right|B{({\lambda }^2-BD)}^{-1}\left|\right|\le M_5\end{array}$$

hold for all $\lambda \in (0,\infty )$ and certain $M_4,M_5\ge 1$, then the fractional powers of ${\mathcal{A}}_0$ can be calculated explicitly. For $\alpha \in (0,1)$, we have

$${\mathcal{A}}_0^{-\alpha }=\left(\begin{array}{cc}cos{\displaystyle \frac{\pi \alpha }{2}}{(-BD)}^{-{\displaystyle \frac{\alpha }{2}}}& -sin{\displaystyle \frac{\pi \alpha }{2}}B{(-BD)}^{-{\displaystyle \frac{1+\alpha }{2}}}\\ -sin{\displaystyle \frac{\pi \alpha }{2}}D{(-BD)}^{-{\displaystyle \frac{1+\alpha }{2}}}& cos{\displaystyle \frac{\pi \alpha }{2}}{(-BD)}^{-{\displaystyle \frac{\alpha }{2}}}\end{array}\right).$$

In this case,

$${\mathcal{A}}_0^{\alpha }=\left(\begin{array}{cc}cos{\displaystyle \frac{\pi \alpha }{2}}{(-BD)}^{{\displaystyle \frac{\alpha }{2}}}& sin{\displaystyle \frac{\pi \alpha }{2}}B{(-BD)}^{{\displaystyle \frac{\alpha -1}{2}}}\\ sin{\displaystyle \frac{\pi \alpha }{2}}D{(-BD)}^{{\displaystyle \frac{\alpha -1}{2}}}& cos{\displaystyle \frac{\pi \alpha }{2}}{(-BD)}^{{\displaystyle \frac{\alpha }{2}}}\end{array}\right).$$

Proof. 

For $\lambda \in (0,\infty )$, we have

$$\begin{array}{cc}\hfill \lambda +{\mathcal{A}}_0=& \left(\begin{array}{cc}\lambda & B\\ D& \lambda \end{array}\right).\hfill \end{array}$$

Since ${\lambda }^2\in (0,\infty )\subset \rho \left(BD\right)$, it follows that $\lambda +{\mathcal{A}}_0$ is invertible (see ([29], Problem 71)) and

$$\begin{array}{cc}\hfill {(\lambda +{\mathcal{A}}_0)}^{-1}=& \left(\begin{array}{cc}\lambda {({\lambda }^2-BD)}^{-1}& -B{({\lambda }^2-BD)}^{-1}\\ -D{({\lambda }^2-BD)}^{-1}& \lambda {({\lambda }^2-BD)}^{-1}\end{array}\right).\hfill \end{array}$$

Moreover, since $-BD$ is of positive type, there exist $M_1\ge 1$ such that

$$\begin{array}{cc}\hfill \left|\right|{(\lambda +{\mathcal{A}}_0)}^{-1}\left|\right|& \le \left|\right|\lambda {({\lambda }^2-BD)}^{-1}\left|\right|+\left|\right|B{({\lambda }^2-BD)}^{-1}\left|\right|\hfill \\ & +\left|\right|D{({\lambda }^2-BD)}^{-1}\left|\right|+\left|\right|\lambda {({\lambda }^2-BD)}^{-1}\left|\right|\hfill \\ & \le {\displaystyle \frac{2\lambda }{1+{\lambda }^2}}{M}_1+{\displaystyle \frac{M_4+M_5}{1+\lambda }}.\hfill \end{array}$$

It is evident that

$${\displaystyle \frac{2\lambda }{1+{\lambda }^2}}\le {\displaystyle \frac{c}{1+\lambda }}$$

for all $\lambda \in (0,\infty )$ and some $c\ge 3$; thus, we have

$$\left|\right|{(\lambda +{\mathcal{A}}_0)}^{-1}\left|\right|\le {\displaystyle \frac{c{M}_1+M_4+M_5}{1+\lambda }.}$$

Hence, ${\mathcal{A}}_0$ is the operator of positive type. Then, by Lemma 2 (i),

$${\mathcal{A}}_0^{-\alpha }=\left(\begin{array}{cc}{\displaystyle \frac{sin\pi \alpha }{\pi }}{\int }_0^{\infty }{\lambda }^{-\alpha }\left(\lambda {({\lambda }^2-BD)}^{-1}\right)d\lambda & -{\displaystyle \frac{sin\pi \alpha }{\pi }}{\int }_0^{\infty }{\lambda }^{-\alpha }B{({\lambda }^2-BD)}^{-1}d\lambda \\ -{\displaystyle \frac{sin\pi \alpha }{\pi }}{\int }_0^{\infty }{\lambda }^{-\alpha }D{({\lambda }^2-BD)}^{-1}d\lambda & {\displaystyle \frac{sin\pi \alpha }{\pi }}{\int }_0^{\infty }{\lambda }^{-\alpha }\left(\lambda {({\lambda }^2-BD)}^{-1}\right)d\lambda \end{array}\right)$$

for $0<\alpha <1$.

Note that

$$\begin{array}{cc}\hfill {\displaystyle \frac{sin\pi \alpha }{\pi }}{\int }_0^{\infty }{\lambda }^{-\alpha }\left(\lambda {({\lambda }^2-BD)}^{-1}\right)d\lambda & ={\displaystyle \frac{sin\pi \alpha }{\pi }}{\int }_0^{\infty }{\lambda }^{1-\alpha }{({\lambda }^2-BD)}^{-1}d\lambda \hfill \\ \hfill & ={\displaystyle \frac{sin\pi \alpha }{2\pi }}{\int }_0^{\infty }{\lambda }^{1-\alpha }{\lambda }^{-1}{({\lambda }^2-BD)}^{-1}d\left({\lambda }^2\right)\hfill \\ \hfill & ={\displaystyle \frac{sin\pi \alpha }{2\pi }}{\int }_0^{\infty }{\left({\lambda }^2\right)}^{-{\displaystyle \frac{\alpha }{2}}}{({\lambda }^2-BD)}^{-1}d\left({\lambda }^2\right).\hfill \end{array}$$

Let $t={\lambda }^2$; then, by Lemma 2 (ii),

$$\begin{array}{cc}\hfill {\displaystyle \frac{sin\pi \alpha }{\pi }}{\int }_0^{\infty }{\lambda }^{-\alpha }\left(\lambda {({\lambda }^2-BD)}^{-1}\right)d\lambda & ={\displaystyle \frac{sin\pi \alpha }{2\pi }}{\int }_0^{\infty }{t}^{-{\displaystyle \frac{\alpha }{2}}}{(t-BD)}^{-1}dt\hfill \\ \hfill & =cos{\displaystyle \frac{\pi \alpha }{2}}{(-BD)}^{-{\displaystyle \frac{\alpha }{2}}}\hfill \end{array}$$

for ${\displaystyle \frac{\alpha }{2}}\in (0,1)$, which implies that $\alpha \in (0,2)$.

In addition,

$$\begin{array}{cc}\hfill {\displaystyle \frac{sin\pi \alpha }{\pi }}{\int }_0^{\infty }{\lambda }^{-\alpha }{({\lambda }^2-BD)}^{-1}d\lambda & ={\displaystyle \frac{sin\pi \alpha }{2\pi }}{\int }_0^{\infty }{\lambda }^{-\alpha }{\lambda }^{-1}{({\lambda }^2-BD)}^{-1}d\left({\lambda }^2\right)\hfill \\ \hfill & ={\displaystyle \frac{sin\pi \alpha }{2\pi }}{\int }_0^{\infty }{\left({\lambda }^2\right)}^{-{\displaystyle \frac{1+\alpha }{2}}}{({\lambda }^2-BD)}^{-1}d\left({\lambda }^2\right).\hfill \end{array}$$

Let $t={\lambda }^2$; then, by (ii) of Lemma 2,

$$\begin{array}{cc}\hfill {\displaystyle \frac{sin\pi \alpha }{\pi }}{\int }_0^{\infty }{\lambda }^{-\alpha }{({\lambda }^2-BD)}^{-1}d\lambda & ={\displaystyle \frac{sin\pi \alpha }{2\pi }}{\int }_0^{\infty }{t}^{-{\displaystyle \frac{1+\alpha }{2}}}{(t-BD)}^{-1}dt\hfill \\ \hfill & =sin{\displaystyle \frac{\pi \alpha }{2}}{\displaystyle \frac{sin\pi \left(\frac{1+\alpha }{2}\right)}{\pi }}{\int }_0^{\infty }{t}^{-{\displaystyle \frac{1+\alpha }{2}}}{(t-BD)}^{-1}dt\hfill \\ \hfill & =sin{\displaystyle \frac{\pi \alpha }{2}}{\left(BD\right)}^{-{\displaystyle \frac{1+\alpha }{2}}}\hfill \end{array}$$

for ${\displaystyle \frac{1+\alpha }{2}}\in (0,1)$, which implies that $\alpha \in (-1,1)$.

Since ${\mathcal{A}}_0^{-\alpha }$ is injective by Lemma 1, it follows that

$${\mathcal{A}}_0^{\alpha }=\left(\begin{array}{cc}cos{\displaystyle \frac{\pi \alpha }{2}}{(-BD)}^{{\displaystyle \frac{\alpha }{2}}}& sin{\displaystyle \frac{\pi \alpha }{2}}B{(-BD)}^{{\displaystyle \frac{\alpha -1}{2}}}\\ sin{\displaystyle \frac{\pi \alpha }{2}}D{(-BD)}^{{\displaystyle \frac{\alpha -1}{2}}}& cos{\displaystyle \frac{\pi \alpha }{2}}{(-BD)}^{{\displaystyle \frac{\alpha }{2}}}\end{array}\right).$$

□

Next, we provide an example to illustrate the above result.

 Example 1. 

Consider the wave equation with no damping force and the Dirichlet boundary condition

$$\left\{\begin{array}{c}{u}_{tt}-\Delta u=f\left(u\right),x\in \Omega ,t\ge 0,\hfill \\ u(t,x)=0,x\in \partial \Omega ,t\ge 0,\hfill \end{array}\right.$$

(10)

where $\Omega \subset {\mathbb{R}}^N(N\ge 1)$ is a bounded smooth domain.

Let $X=L^2(\Omega )$ be a given factor space and let

$$\begin{array}{cc}& Au:=(-\Delta )u,\mathcal{D}\left(A\right)=H^2(\Omega )\cap H_0^1(\Omega ).\hfill \end{array}$$

Then, the equation can be reformulated as the system

$$\dot{\mathcal{U}}\left(t\right)+\mathcal{A}\mathcal{U}\left(t\right)=\mathcal{F}\left(u\right),t\ge 0$$

(11)

on the product space $X\times X$, where

$$\begin{array}{cc}& \mathcal{A}:=\left(\begin{array}{cc}0& -I\\ A& 0\end{array}\right),\hfill \\ & \mathcal{U}\left(t\right)=\left(\begin{array}{c}u\\ u_t\end{array}\right)\in \mathcal{D}\left(\mathcal{A}\right)=(H^2(\Omega )\cap H_0^1(\Omega ))\times H_0^1(\Omega ),\hfill \\ & \left(\begin{array}{cc}0& -I\\ A& 0\end{array}\right)\left(\begin{array}{c}u\\ u_t\end{array}\right)=\left(\begin{array}{c}-u_t\\ Au\end{array}\right),\hfill \end{array}$$

and $\mathcal{F}\left(u\right)=\left(\begin{array}{c}0\\ f\left(u\right)\end{array}\right)$. Let $B:=-I$. It is then straightforward to verify that the conditions of Theorem 1 are satisfied. Consequently, the fractional powers of $\mathcal{A}$ can be explicitly calculated and

$${\mathcal{A}}^{-\alpha }=\left(\begin{array}{cc}cos{\displaystyle \frac{\pi \alpha }{2}}{A}^{-{\displaystyle \frac{\alpha }{2}}}& sin{\displaystyle \frac{\pi \alpha }{2}}{A}^{-{\displaystyle \frac{1+\alpha }{2}}}\\ -sin{\displaystyle \frac{\pi \alpha }{2}}{A}^{{\displaystyle \frac{1-\alpha }{2}}}& cos{\displaystyle \frac{\pi \alpha }{2}}{A}^{-{\displaystyle \frac{\alpha }{2}}}\end{array}\right)$$

by Theorem 1, which is consistent with Theorem 5.3 in [3]. Noting that ${\mathcal{A}}^{-\alpha }$ is injective by Lemma 1, it can further be concluded that

$${\mathcal{A}}^{\alpha }=\left(\begin{array}{cc}cos{\displaystyle \frac{\pi \alpha }{2}}{A}^{{\displaystyle \frac{\alpha }{2}}}& -sin{\displaystyle \frac{\pi \alpha }{2}}{A}^{{\displaystyle \frac{\alpha -1}{2}}}\\ sin{\displaystyle \frac{\pi \alpha }{2}}{A}^{{\displaystyle \frac{1+\alpha }{2}}}& cos{\displaystyle \frac{\pi \alpha }{2}}{A}^{{\displaystyle \frac{\alpha }{2}}}\end{array}\right).$$

Since $\sigma \left(\mathcal{A}\right)=\{\pm {\mu }_n^{{\displaystyle \frac{1}{2}}}i,n\in N\}$, it then follows from Lemma 3 that

$$\sigma \left({\mathcal{A}}^{\alpha }\right)=\{\pm {\mu }_n^{\frac{\alpha }{2}}{e}^{i\alpha \pi /2},n\in N\},$$

where ${\left\{{\mu }_n\right\}}_{n\in N}$ is the sequence of eigenvalues of the negative Laplacian operator $-\Delta$.

Because systems (10) and (11) are equivalent, certain problems for system (10) can naturally be studied via the following fractional-order system, which provides a parabolic-type approximation of (11):

$${\displaystyle \frac{d}{dt}}\left(\begin{array}{c}{u}^{\alpha }\\ u_t^{\alpha }\end{array}\right)+{\mathcal{A}}^{\alpha }\left(\begin{array}{c}{u}^{\alpha }\\ u_t^{\alpha }\end{array}\right)=\mathcal{F}\left(u^{\alpha }\right),t\ge 0.$$

For example, in order to solve equation (11) locally in $\mathcal{D}\left(\mathcal{A}\right)$, the nonlinearity $\mathcal{F}$ must not consume any regularity of the solution. This feature restricts the class of nonlinearities for which the problem can be locally solved. To overcome this restriction, one can consider parabolic approximations of problem (11).

In view of Lemma 3 and Theorem 1, spectral properties of ${\mathcal{A}}_0^{\alpha }$ for $\alpha \in (0,1)$ can be established.

 Theorem 2. 

Under the conditions of Theorem 1,

$$\sigma \left({\mathcal{A}}_0^{\alpha }\right)=\{{\lambda }^{\alpha }:{\lambda }^2\in \sigma \left(BD\right)\}$$

for $\alpha \in (0,1)$.

Proof. 

For $\lambda \in \mathbb{C}$, we have

$$\lambda -{\mathcal{A}}_0=\left(\begin{array}{cc}\lambda & -B\\ -D& \lambda \end{array}\right).$$

Since B and D commute, it follows from Problerm 71 in [29] that ${\lambda }^2\in \sigma \left(BD\right).$ if and only if ${\lambda }^2\in \sigma \left(BD\right).$ The conclusion then follows immediately from Lemma 3 and Theorem 1. □

4. Fractional Powers of Anti-Triangular Operator Matrices

Anti-triangular operator matrices naturally arise in the study of certain coupled partial differential equations where the interaction between components is captured specifically by the anti-triangular structure of the operator. The study of their fractional power representations is essential both for analyzing the dynamics of such systems and for developing parabolic-type approximations of the associated equations. In this section, we present the fractional power representation of the general anti-triangular operator matrix $\mathcal{A}$ provided in (1), extending the ideas considered for off-triangular matrices in the previous section.

 Theorem 3. 

Let Assumption 1 hold, $D\left(A^{*}\right)\subset D\left(B^{*}\right)$, and D be A-bounded with

$$\begin{array}{c}\hfill \left|\right|Dx\left|\right|\le a\left|\right|x\left|\right|+b\left|\right|Ax\left|\right|\end{array}$$

for certain constants $a,b$ and all $x\in \mathcal{D}\left(A\right)$. Then, the fractional powers of $\mathcal{A}$ provided in (1) can be calculated explicitly. For $\alpha \in (0,1)$, we have

$${\mathcal{A}}^{-\alpha }=\left(\begin{array}{cc}{\mathcal{A}}_{11}(\alpha ,\lambda )& {\mathcal{A}}_{12}(\alpha ,\lambda )\\ {\mathcal{A}}_{21}(\alpha ,\lambda )& {\mathcal{A}}_{22}(\alpha ,\lambda )\end{array}\right).$$

Here,

$$\begin{array}{cc}\hfill & {\mathcal{A}}_{11}(\alpha ,\lambda )=A^{-\alpha }+{\displaystyle \frac{sin\pi \alpha }{\pi }}\sum _{n=0}^{\infty }\sum _{k=0}^n{z}_2(n,k,\alpha )C_n^k{D}^{n-k+1}{B}^{n-k+1}{A}^k{A}^{-(\alpha +2n-k+2)},\hfill \\ \hfill & {\mathcal{A}}_{12}(\alpha ,\lambda )=-{\displaystyle \frac{sin\pi \alpha }{\pi }}\sum _{n=0}^{\infty }\sum _{k=0}^n{z}_1(n,k,\alpha )C_n^k{D}^{n-k}{B}^{n-k+1}{A}^k{A}^{-(2n-k+1+\alpha )},\hfill \\ \hfill & {\mathcal{A}}_{21}(\alpha ,\lambda )=-{\displaystyle \frac{sin\pi \alpha }{\pi }}\sum _{n=0}^{\infty }\sum _{k=0}^n{z}_1(n,k,\alpha )C_n^k{D}^{n-k+1}{B}^{n-k}{A}^k{A}^{-(2n-k+1+\alpha )},\hfill \\ \hfill & {\mathcal{A}}_{22}(\alpha ,\lambda )={\displaystyle \frac{sin\pi \alpha }{\pi }}\sum _{n=0}^{\infty }\sum _{k=0}^{n}z(n,k,\alpha )C_n^k{D}^{n-k}{B}^{n-k}{A}^k{A}^{-(\alpha +2n-k)},\hfill \\ \hfill & z(n,k,\alpha )={\displaystyle \frac{\pi }{sin\pi (2n-k+\alpha )}}{\displaystyle \frac{(1-2n+k-\alpha )(2-2n+k-\alpha )\dots (-\alpha )}{(2n-k)!}},\hfill \\ \hfill & z_1(n,k,\alpha )={\displaystyle \frac{\pi }{sin\pi (2n-k+1+\alpha )}}{\displaystyle \frac{(-2n+k-\alpha )(1-2n+k-\alpha )\dots (-\alpha )}{(2n-k+1)!}},\hfill \\ \hfill & z_2(n,k,\alpha )={\displaystyle \frac{\pi }{sin\pi (2n-k+2+\alpha )}}{\displaystyle \frac{(-2n+k-1-\alpha )(-2n+k-\alpha )\dots (-\alpha )}{(2n-k+2)!}}.\hfill \end{array}$$

(12)

Proof. 

By Corollary 2, $\mathcal{A}$ is of positive type; then,

$${\mathcal{A}}^{-\alpha }=\left(\begin{array}{cc}{\mathcal{A}}_{11}^{\prime }(\alpha ,\lambda )& {\mathcal{A}}_{12}^{\prime }(\alpha ,\lambda )\\ {\mathcal{A}}_{21}^{\prime }(\alpha ,\lambda )& {\mathcal{A}}_{22}^{\prime }(\alpha ,\lambda )\end{array}\right)$$

holds by Equation (7) and Lemma 2 (i), where

$$\begin{array}{cc}\hfill {\mathcal{A}}_{11}^{\prime }& (\alpha ,\lambda )={\displaystyle \frac{sin\pi \alpha }{\pi }}{\int }_0^{\infty }{\lambda }^{-\alpha }{(\lambda +A)}^{-1}d\lambda \hfill \\ \hfill & +{\displaystyle \frac{sin\pi \alpha }{\pi }}{\int }_0^{\infty }{\lambda }^{-\alpha }{(\lambda +A)}^{-1}B{M}_{\lambda }^{-1}D{(\lambda +A)}^{-1}d\lambda ,\hfill \\ \hfill & {\mathcal{A}}_{12}^{\prime }(\alpha ,\lambda )=-{\displaystyle \frac{sin\pi \alpha }{\pi }}{\int }_0^{\infty }{\lambda }^{-\alpha }{(\lambda +A)}^{-1}B{M}_{\lambda }^{-1}d\lambda ,\hfill \\ \hfill & {\mathcal{A}}_{21}^{\prime }(\alpha ,\lambda )=-{\displaystyle \frac{sin\pi \alpha }{\pi }}{\int }_0^{\infty }{\lambda }^{-\alpha }{M}_{\lambda }^{-1}D{(\lambda +A)}^{-1}d\lambda ,\hfill \\ \hfill & {\mathcal{A}}_{22}^{\prime }(\alpha ,\lambda )={\displaystyle \frac{sin\pi \alpha }{\pi }}{\int }_0^{\infty }{\lambda }^{-\alpha }{M}_{\lambda }^{-1}d\lambda .\hfill \end{array}$$

On the other hand,

$$\begin{array}{c}\hfill \left|\right|{(D{(\lambda +A)}^{-1}B+A)}^n{(\lambda +A)}^{-n-i}\left|\right|\le q^n{\left|\right|(\lambda +A)}^{-i}\left|\right|\le q^n{M}^i\end{array}$$

holds by Assumption 1 (iii) and ${\sum }_{n=0}^{\infty }{q}^n{M}^i$ is convergent; therefore,

$$\sum _{n=0}^{\infty }{(D{(\lambda +A)}^{-1}B+A)}^n{(\lambda +A)}^{-n-i}$$

is uniformly convergent for $i=1,2,3$.

(i) Computation of ${\mathcal{A}}_{22}^{\prime }(\alpha ,\lambda )$. By (8), we have

$$\begin{array}{c}\hfill {\displaystyle \frac{sin\pi \alpha }{\pi }}{\int }_0^{\infty }{\lambda }^{-\alpha }{M}_{\lambda }^{-1}d\lambda ={\displaystyle \frac{sin\pi \alpha }{\pi }}\sum _{n=0}^{\infty }\sum _{k=0}^n{C}_n^k{D}^{n-k}{B}^{n-k}{A}^k{\int }_0^{\infty }{\lambda }^{-\alpha }{(\lambda +A)}^{k-2n-1}d\lambda .\end{array}$$

Let $m=2n-k$. Then,

$$\begin{array}{cc}\hfill {\int }_0^{\infty }& {\lambda }^{-\alpha }{(\lambda +A)}^{k-2n-1}d\lambda ={\int }_0^{\infty }{\lambda }^{m-(m+\alpha )}{(\lambda +A)}^{-m-1}d\lambda \hfill \\ \hfill & ={\displaystyle \frac{\pi }{sin\pi (m+\alpha )}}{\displaystyle \frac{(1-(m+\alpha \left)\right)(2-(m+\alpha \left)\right)\dots (m-(m+\alpha \left)\right)}{m!}}{A}^{-(m+\alpha )}\hfill \\ \hfill & ={\displaystyle \frac{\pi }{sin\pi (2n-k+\alpha )}}{\displaystyle \frac{(1-2n+k-\alpha )(2-2n+k-\alpha )\dots (-\alpha )}{(2n-k)!}}{A}^{-(\alpha +2n-k)}\hfill \end{array}$$

for $m+\alpha \in (0,m+1)$ by Lemma 2 (ii), which implies that $\alpha \in (k-2n,1)$. Denoting

$$z(n,k,\alpha )={\displaystyle \frac{\pi }{sin\pi (2n-k+\alpha )}}{\displaystyle \frac{(1-2n+k-\alpha )(2-2n+k-\alpha )\dots (-\alpha )}{(2n-k)!}},$$

then

$${\mathcal{A}}_{22}^{\prime }(\alpha ,\lambda )={\displaystyle \frac{sin\pi \alpha }{\pi }}\sum _{n=0}^{\infty }\sum _{k=0}^{n}z(n,k,\alpha )C_n^k{D}^{n-k}{B}^{n-k}{A}^k{A}^{-(\alpha +2n-k)}={\mathcal{A}}_{22}(\alpha ,\lambda ).$$

(ii) Computation of ${\mathcal{A}}_{12}^{\prime }(\alpha ,\lambda )$. Observe that

$$\begin{array}{cc}\hfill & -{\displaystyle \frac{sin\pi \alpha }{\pi }}{\int }_0^{\infty }{\lambda }^{-\alpha }{(\lambda +A)}^{-1}B{M}_{\lambda }^{-1}d\lambda \hfill \\ \hfill & =-{\displaystyle \frac{sin\pi \alpha }{\pi }}\sum _{n=0}^{\infty }\sum _{k=0}^n{C}_n^k{D}^{n-k}{B}^{n-k+1}{A}^k{\int }_0^{\infty }{\lambda }^{-\alpha }{(\lambda +A)}^{k-2n-2}d\lambda \hfill \end{array}$$

and

$$\begin{array}{cc}\hfill & {\int }_0^{\infty }{\lambda }^{-\alpha }{(\lambda +A)}^{k-2n-2}d\lambda \hfill \\ \hfill & ={\int }_0^{\infty }{\lambda }^{2n-k+1-(2n-k+1+\alpha )}{(\lambda +A)}^{-(2n-k+1)-1}d\lambda \hfill \\ \hfill & ={\displaystyle \frac{\pi }{sin\pi (2n-k+1+\alpha )}}{\displaystyle \frac{(-2n+k-\alpha )(1-2n+k-\alpha )\dots (-\alpha )}{(2n-k+1)!}}{A}^{-(2n-k+1+\alpha )}\hfill \end{array}$$

(13)

for $2n-k+1+\alpha \in (0,2n-k+2)$ by Lemma 2 (ii), which implies that $\alpha \in (-2n+k-1,1)$. Let

$$z_1(n,k,\alpha ):={\displaystyle \frac{\pi }{sin\pi (2n-k+1+\alpha )}}{\displaystyle \frac{(-2n+k-\alpha )(1-2n+k-\alpha )\dots (-\alpha )}{(2n-k+1)!}};$$

hence,

$${\mathcal{A}}_{12}^{\prime }(\alpha ,\lambda )=-{\displaystyle \frac{sin\pi \alpha }{\pi }}\sum _{n=0}^{\infty }\sum _{k=0}^n{z}_1(n,k,\alpha )C_n^k{D}^{n-k}{B}^{n-k+1}{A}^k{A}^{-(2n-k+1+\alpha )}={\mathcal{A}}_{12}(\alpha ,\lambda ).$$

(iii) Computation of ${\mathcal{A}}_{21}^{\prime }(\alpha ,\lambda )$. Noting that

$$\begin{array}{cc}\hfill -{\displaystyle \frac{sin\pi \alpha }{\pi }}& {\int }_0^{\infty }{\lambda }^{-\alpha }{M}_{\lambda }^{-1}D{(\lambda +A)}^{-1}d\lambda \hfill \\ \hfill & =-{\displaystyle \frac{sin\pi \alpha }{\pi }}\sum _{n=0}^{\infty }\sum _{k=0}^n{C}_n^k{D}^{n-k+1}{B}^{n-k}{A}^k{\int }_0^{\infty }{\lambda }^{-\alpha }{(\lambda +A)}^{k-2n-2}d\lambda ,\hfill \end{array}$$

(14)

it follows from Equation (13) that

$${\mathcal{A}}_{21}^{\prime }(\alpha ,\lambda )=-{\displaystyle \frac{sin\pi \alpha }{\pi }}\sum _{n=0}^{\infty }\sum _{k=0}^n{z}_1(n,k,\alpha )C_n^k{D}^{n-k+1}{B}^{n-k}{A}^k{A}^{-(2n-k+1+\alpha )}={\mathcal{A}}_{21}(\alpha ,\lambda )$$

for $\alpha \in (-2n+k-1,1)$.

(iv) Computation of ${\mathcal{A}}_{11}^{\prime }(\alpha ,\lambda )$. Based on Assumption 1 (i) and Equation (14), it can be concluded that

$${\displaystyle \frac{sin\pi \alpha }{\pi }}{\int }_0^{\infty }{\lambda }^{-\alpha }{(\lambda +A)}^{-1}d\lambda =A^{-\alpha }$$

and

$$\begin{array}{cc}\hfill & {\displaystyle \frac{sin\pi \alpha }{\pi }}{\int }_0^{\infty }{\lambda }^{-\alpha }{(\lambda +A)}^{-1}B{M}_{\lambda }^{-1}D{(\lambda +A)}^{-1}d\lambda \hfill \\ \hfill & ={\displaystyle \frac{sin\pi \alpha }{\pi }}\sum _{n=0}^{\infty }\sum _{k=0}^n{C}_n^k{D}^{n-k+1}{B}^{n-k+1}{A}^k{\int }_0^{\infty }{\lambda }^{-\alpha }{(\lambda +A)}^{k-2n-3}d\lambda .\hfill \end{array}$$

The same strategy can be used to obtain

$$\begin{array}{cc}\hfill & {\int }_0^{\infty }{\lambda }^{-\alpha }{(\lambda +A)}^{k-2n-3}d\lambda \hfill \\ \hfill & ={\displaystyle \frac{\pi }{sin\pi (2n-k+2+\alpha )}}{\displaystyle \frac{(-2n+k-1-\alpha )(-2n+k-\alpha )\dots (-\alpha )}{(2n-k+2)!}}{A}^{-(\alpha +2n-k+2)}\hfill \end{array}$$

for $\alpha \in (-2n+k-2,1)$. Therefore,

$${\mathcal{A}}_{11}^{\prime }(\alpha ,\lambda )=A^{-\alpha }+{\displaystyle \frac{sin\pi \alpha }{\pi }}\sum _{n=0}^{\infty }\sum _{k=0}^n{z}_2(n,k,\alpha )C_n^k{D}^{n-k+1}{B}^{n-k+1}{A}^k{A}^{-(\alpha +2n-k+2)}={\mathcal{A}}_{11}(\alpha ,\lambda ),$$

where

$$z_2(n,k,\alpha )={\displaystyle \frac{\pi }{sin\pi (2n-k+2+\alpha )}}{\displaystyle \frac{(-2n+k-1-\alpha )(-2n+k-\alpha )\dots (-\alpha )}{(2n-k+2)!}}.$$

□

 Remark 4. 

It is worth noting that in order to compute the fractional powers of $\mathcal{A}$ using Lemma 2 (i), (ii) together with its resolvant, the parameter α is initially required to satisfy $0<\alpha <1$. Subsequently, when the problem is reformulated in terms of the fractional powers of the operator entries in the resolvant of $\mathcal{A}$, this requirement on α is relaxed to $-n<\alpha <n+1$ for $n\in \mathbb{N}$, which is consistent with the initial requirement.

We end this section with the following example to illustrate the previous result.

 Example 2. 

Consider the following differential equation:

$$\left\{\begin{array}{c}{u}_{tt}-u+{(\alpha \Delta )}^{\gamma }{u}_t=f\left(u\right),x\in \Omega ,t>0\hfill \\ (1+\beta {(\alpha \Delta )}^{\gamma })u_t=g\left(u\right),x\in \Omega ,t>0\hfill \\ u(0,x)=u_0,u_t(0,x)=v_0,x\in \Omega ,\hfill \\ u(t,x)=0.x\in \partial \Omega ,t>0.\hfill \end{array}\right.$$

(15)

Here, $\Omega \subset {\mathbb{R}}^N(N\ge 3)$ is a bounded domain with a smooth boundary, $f,g\in {\mathcal{C}}^1\left(\mathbb{R}\right)$ satisfy

$$f^{\prime }\left(s\right)\le {c(1+|s|}^{\rho -1}),s\in \mathbb{R}$$

for some

$$0\le \rho \le {\displaystyle \frac{N+2}{N-2}},$$

and

$$\underset{\left|s\right|\to \infty }{lim\; sup}{\displaystyle \frac{f\left(s\right)}{s}}<{\mu }_1,$$

where ${\mu }_1$ denotes the first eigenvalue of $-\Delta$ in $L_2(\Omega )$, $\alpha \in \mathbb{R}$, $\gamma \in (0,1)$, and $\beta \in \{s\in \mathbb{C}:|s|M^2+sM+|\left|A^{\gamma }\right||M<q,M\ge 1,0<q<1\}$.

Let $X_1=H^2(\Omega )\cap H_0^1(\Omega )$ and $X=L_2(\Omega )$ be given factor spaces and let

$$\begin{array}{cc}& A:X_1\to X,\hfill \\ & Au:=\alpha \Delta u.\hfill \end{array}$$

With the above setup, the problem in (15) can be rewritten as the abstract Cauchy problem

$$\left\{\begin{array}{c}\dot{\mathcal{U}}+\mathcal{A}\mathcal{U}=\mathcal{F}\left(u\right),t\ge 0\hfill \\ \mathcal{U}\left(0\right)={\mathcal{U}}_0\hfill \end{array}\right.$$

(16)

on the product space $X\times X$, where

$$\begin{array}{cc}& \mathcal{A}:=\left(\begin{array}{cc}{A}^{\gamma }& -I\\ \beta A^{\gamma }& 0\end{array}\right):\mathcal{D}\left(\mathcal{A}\right)=X_1\times X\to X\times X,\hfill \\ & \mathcal{U}=\left(\begin{array}{c}v\\ u\end{array}\right)\in \mathcal{D}\left(\mathcal{A}\right),{\mathcal{U}}_0=\left(\begin{array}{c}{v}_0\\ u_0\end{array}\right),\hfill \\ & \mathcal{A}\mathcal{U}=\left(\begin{array}{c}{A}^{\gamma }v-u\\ \beta A^{\gamma }v\end{array}\right),\hfill \end{array}$$

and $\mathcal{F}\left(u\right)=\left(\begin{array}{c}f\left(u\right)\\ g\left(u\right)\end{array}\right)$. Because

$$\left|\right|\beta A^{\gamma }x\left|\right|\le \left|\beta \right|\left|\right|A^{\gamma }x\left|\right|$$

for all $x\in X$, it follows from Remark 2 that the conditions in Theorem 3 are satisfied when $A,B,$ and D are replaced by $A^{\gamma },-I$, and $\beta A^{\gamma }$, respectively. Therefore, the fractional powers of $\mathcal{A}$ can be explicitly calculated, and by Theorem 3 we have

$${\mathcal{A}}^{-\alpha }=\left(\begin{array}{cc}{\widehat{\mathcal{A}}}_{11}(\alpha ,\lambda )& {\widehat{\mathcal{A}}}_{12}(\alpha ,\lambda )\\ {\widehat{\mathcal{A}}}_{21}(\alpha ,\lambda )& {\widehat{\mathcal{A}}}_{22}(\alpha ,\lambda )\end{array}\right).$$

Here,

$$\begin{array}{cc}\hfill & {\widehat{\mathcal{A}}}_{11}(\alpha ,\lambda )=A^{-\alpha }+{\displaystyle \frac{sin\pi \alpha }{\pi }}\sum _{n=0}^{\infty }\sum _{k=0}^n{(-\beta )}^{n-k+1}{z}_2(n,k,\alpha )C_n^k{A}^{\gamma (n-k+1)+k}{A}^{-(\alpha +2n-k+2)},\hfill \\ \hfill & {\widehat{\mathcal{A}}}_{12}(\alpha ,\lambda )=-{\displaystyle \frac{sin\pi \alpha }{\pi }}\sum _{n=0}^{\infty }\sum _{k=0}^n{(-1)}^{n-k+1}{\beta }^{n-k}{z}_1(n,k,\alpha )C_n^k{A}^{\gamma (n-k)+k}{A}^{-(2n-k+1+\alpha )},\hfill \\ \hfill & {\widehat{\mathcal{A}}}_{21}(\alpha ,\lambda )=-{\displaystyle \frac{sin\pi \alpha }{\pi }}\sum _{n=0}^{\infty }\sum _{k=0}^n{(-1)}^{n-k}{\beta }^{n-k+1}{z}_1(n,k,\alpha )C_n^k{A}^{\gamma (n-k+1)+k}{A}^{-(2n-k+1+\alpha )},\hfill \\ \hfill & {\widehat{\mathcal{A}}}_{22}(\alpha ,\lambda )={\displaystyle \frac{sin\pi \alpha }{\pi }}\sum _{n=0}^{\infty }\sum _{k=0}^n{(-\beta )}^{n-k}z(n,k,\alpha )C_n^k{A}^{\gamma (n-k)+k}{A}^{-(\alpha +2n-k)},\hfill \\ \hfill & z(n,k,\alpha )={\displaystyle \frac{\pi }{sin\pi (2n-k+\alpha )}}{\displaystyle \frac{(1-2n+k-\alpha )(2-2n+k-\alpha )\dots -\alpha )}{(2n-k)!}},\hfill \\ \hfill & z_1(n,k,\alpha )={\displaystyle \frac{\pi }{sin\pi (2n-k+1+\alpha )}}{\displaystyle \frac{(-2n+k-\alpha )(1-2n+k-\alpha )\dots (-\alpha )}{(2n-k+1)!}},\hfill \\ \hfill & z_2(n,k,\alpha )={\displaystyle \frac{\pi }{sin\pi (2n-k+2+\alpha )}}{\displaystyle \frac{(-2n+k-1-\alpha )(-2n+k-\alpha )\dots -\alpha )}{(2n-k+2)!}},\hfill \end{array}$$

with $z(n,k,\alpha )$, $z_1(n,k,\alpha )$, and $z_2(n,k,\alpha )$ provided in (12). The fractional-order system

$${\displaystyle \frac{d}{dt}}\left(\begin{array}{c}{v}^{\alpha }\\ u^{\alpha }\end{array}\right)+{\mathcal{A}}^{\alpha }\left(\begin{array}{c}{v}^{\alpha }\\ u^{\alpha }\end{array}\right)=\mathcal{F}\left(u^{\alpha }\right),t\ge 0,{\left(\begin{array}{c}{v}^{\alpha }\\ u^{\alpha }\end{array}\right)}_{t=0}=\left(\begin{array}{c}{v}_{0\alpha }\\ u_{0\alpha }\end{array}\right)$$

provides a parabolic-type approximation of (16), which allows for studying properties of the original system (15) such as local well-posedness under weaker regularity assumptions on the nonlinearities f and g.

5. Conclusions

This paper has investigated the fractional powers of anti-triangular operator matrices. First, by employing the method of formal determinants, we derive an explicit expression for the fractional powers of a specific class of anti-triangular operator matrices (the off-diagonal case). Then, an explicit formula for the fractional powers of general anti-triangular operator matrices is further obtained utilizing the Schur factorization technique. Moreover, two illustrative examples are provided to demonstrate the applicability of our results. Our results establish an abstract and operational analytic framework for computing fractional powers of anti-triangular operator matrices.

In future work, we will explore the fractional power representations of certain coupled operator matrices where the elements involved in their domains satisfy some specific relations. Such studies may reveal richer structural phenomena and lead to broader applications in the analysis of complex physical systems. On the other hand, utilizing the fast matrix multiplication techniques [30] could allow for considering the numerical computation of fractional powers of one-sided coupled operator matrices.