The problem of modeling and control of large flexible spacecraft has been a subject of considerable research in recent years. In general, a spacecraft system consists of a rigid bus and several flexible appendages, such as long beams, solar panels, antennae, etc. The flexibility of various components of the spacecraft introduces many unforeseen complexities in the process of system modeling and controller design. To ensure satisfactory performance, it is essential to take into account the distributed nature of the flexible members.

The most natural model for a flexible spacecraft could be given by a hybrid system, i.e. a combination of a finite-dimensional model for the rigid parts, and an infinite-dimensional model for the elastic parts. However, in the commonly used approach of modeling the dynamics of the elastic parts are approximated by considering some finite number of modes. The mechanical system, such as spacecraft with flexible appendages, or robot arm with flexible links, can be modeled as coupled elastic and rigid parts. Many future space applications, such as the space station, rely on lightweight materials and high performance control systems for high precision pointing, tracking, etc., and to achieve high precision demand for such systems, one has to take the dynamic effect of flexible parts into account. Thus, over the last decades there has been a growing interest in obtaining new models for the design, analysis, and control of the research in this area. A wide list of contributions in this area can be found in the literature [1] - [12] .

Let us consider a spacecraft dynamic system descripted by partial differential equations with the initial and free boundary conditions as follows:

$\{\begin{array}{l}\frac{{\partial }^{2}y\left(x,t\right)}{\partial t^2}+\eta \frac{{\partial }^{5}y\left(x,t\right)}{\partial t\partial x^4}+\frac{{\partial }^2}{\partial x^2}\left(p\left(x\right)\frac{{\partial }^{2}y\left(x,t\right)}{\partial x^2}\right)=f\left(t,y\right)\hfill \\ {\frac{{\partial }^{2}y\left(x,t\right)}{\partial x^2}|}_{x=0,l}={\frac{\partial }{\partial x}\left(p\left(x\right)\frac{{\partial }^{2}y\left(x,t\right)}{\partial x^2}\right)|}_{x=0,l}=0\hfill \\ y\left(x,0\right)={\phi }_0\left(x\right),{\frac{\partial y\left(x,t\right)}{\partial t}|}_{t=0}={\psi }_0\left(x\right)\hfill \end{array}$(1.1)

It should be noted that the beam equations discussed in [1] - [5] are different from the system (1.1) above, because system (1.1) stands for the typical beam equation of a flexible spacecraft that has two free ends, where $y\left(x,t\right)$ is the transverse displacement of the point $x$ and at the time $t$, $l$ is the length of the beam, $p\left(x\right)$ is the bending rigidity at the point $x$, $f\left(t,y\right)$ represents the controlled moment of the system.

Since the motion of elastic parts is usually described by a set of partial differential equations with appropriate boundary conditions, and the motion of the rigid parts is described by a set of nonlinear ordinary differential equations. Hence, the motion of the rigid parts coupled with the elastic parts is described generally by a set of coupled nonlinear ordinary differential equations and partial differential equations. We are going to investigate the hybrid spacecraft system later in other articles.

Suppose that $p\left(x\right)\in C^2\left[0,l\right]$, and $0<p_0\le p\left(x\right)\le p_1<+\infty$, where $p_0$ and $p_1$ are constants. Now, we take $L^2\left[0,l\right]$ as a state space, with the inner product and norm as follows:

${\langle f,g\rangle }_0={\displaystyle {\int }_0^{l}f\left(x\right)\overline{g\left(x\right)}\text{d}x},f,g\in L^2\left[0,l\right]$

${\Vert f\Vert }_0^2={\displaystyle {\int }_0^l{\left|f\left(x\right)\right|}^2\text{d}x},f\in L^2\left[0,l\right]$(2.1)

Let $H_1=span\left\{1,x\right\}$, then $L^2\left[0,l\right]=H_1\oplus H_2$, where $H_2$ is the orthogonal complement of $H_1$ in $L^2\left[0,l\right]$. Suppose $P_1$ is the projection operator on $H_1$ and $I-P_1$ is the projection operator on $H_2$, and so the system (1.1) can be rewritten as follows

$\{\begin{array}{l}\frac{{\partial }^{2}y\left(x,t\right)}{\partial t^2}=P_{1}f\left(t,y\right)\hfill \\ y\left(x,0\right)=P_1{\phi }_0,{\frac{\partial y\left(x,t\right)}{\partial t}|}_{t=0}=P_1{\psi }_0\hfill \end{array}$ (2.2)

It is clear that the solution of (2.2) can be described as

$y^{\left(1\right)}\left(x,t\right)=a_1+a_{2}x+a_{3}t+a_{4}tx$(2.3)

where $a_1$, $a_2$, $a_3$ and $a_4$ are determined by $P_1{\phi }_0$, $P_1{\psi }_0$, and $P_{1}f\left(t,y\right)$.

Consider the system (1.1) in $H_2$, we have

$\{\begin{array}{l}\frac{{\partial }^{2}y\left(x,t\right)}{\partial t^2}+\eta \frac{{\partial }^{5}y\left(x,t\right)}{\partial t\partial x^4}+\frac{{\partial }^2}{\partial x^2}\left(p\left(x\right)\frac{{\partial }^{2}y(x,t}{\partial x^2}\right)=\left(I-P_1\right)f\left(t,y\right)\hfill \\ {\frac{{\partial }^{2}y\left(x,t\right)}{\partial x^2}|}_{x=0,l}={\frac{\partial }{\partial x}\left(p\left(x\right)\frac{{\partial }^{2}y\left(x,t\right)}{\partial x^2}\right)|}_{x=0,l}=0\hfill \\ y\left(x,0\right)=\left(I-P_1\right){\phi }_0,{\frac{\partial y\left(x,t\right)}{\partial t}|}_{t=0}=\left(I-P_1\right){\psi }_0\hfill \end{array}$(2.4)

If we denote the solution of (2.4) by $y^{\left(2\right)}\left(x,t\right)$, then the solution of system (1.1) can be described as

$y\left(x,t\right)=y^{\left(1\right)}\left(x,t\right)\oplus y^{\left(2\right)}\left(x,t\right)$(2.5)

It should be noted that the form of $y^{\left(1\right)}\left(x,t\right)$ is ready from (2.3), and $y^{\left(2\right)}\left(x,t\right)$ will play a key role in order to investigate the solution of the system (1.1).

We now define the differential operators $A$ and $T$ as follows:

$\begin{array}{l}A\phi ={\left(p\left(x\right){\phi }^{″}\left(x\right)\right)}^{\prime \text{}\prime },\phi \in D\left(A\right),\hfill \\ D\left(A\right)=\left\{\phi |\phi \in H_2,{\phi }^{″}\left(0\right)={\phi }^{″}\left(l\right)=0,{{\left(p\left(x\right){\phi }^{″}\right)}^{\prime }|}_{x=0,l}=0,{\left(p\left(x\right){\phi }^{″}\left(x\right)\right)}^{\prime \text{}\prime }\in H_2\right\},\hfill \\ T\phi =\eta \phi \text{'}\text{'}\text{'}\text{'}\left(x\right),\phi \in D\left(T\right),D\left(T\right)=D\left(A\right).\hfill \end{array}$

It can be seen from the definitions of $A$ and $T$ that $H_1=span\left\{1,x\right\}$ is the null space of $A$, and both $A$ and $T$ are positively defined self-adjoint operators in $H_2$, and there is the greatest positive number $\lambda$ such that

${\langle A\phi ,\phi \rangle }_0\ge \lambda {\Vert \phi \Vert }_0^2,\phi \in D\left(A\right).$(2.6)

It is easy to show that

$\frac{p_0}{\eta }T\le A\le \frac{p_1}{\eta }T$(2.7)

Integrating by parts with the definitions of $A$ and $T$ as well as the boundary conditions, we have

$\begin{array}{c}{\langle A\phi ,\phi \rangle }_0={\displaystyle {\int }_0^l{\left(p\left(x\right){\phi }^{″}\left(x\right)\right)}^{\prime \text{}\prime }\overline{\phi \left(x\right)}\text{d}x}\\ ={\displaystyle {\int }_0^l{\left(p\left(x\right){\phi }^{″}\left(x\right)\right)}^{\prime }\overline{{\phi }^{\prime }\left(x\right)}\text{d}x}\\ ={\displaystyle {\int }_0^{l}p\left(x\right){\phi }^{″}\left(x\right)\overline{{\phi }^{″}\left(x\right)}\text{d}x}.\end{array}$

It follows from the inequalities $0<p_0\le p\left(x\right)\le p_1<\infty$ that

$p_0{\displaystyle {\int }_0^l{\phi }^{″}\left(x\right)\overline{{\phi }^{″}\left(x\right)}\text{d}x}\le {\langle A\phi ,\phi \rangle }_0\le p_1{\displaystyle {\int }_0^l{\phi }^{″}\left(x\right)\overline{{\phi }^{″}\left(x\right)}\text{d}x}$

That is,

$p_0{\Vert {\phi }^{″}\Vert }_0^2\le {\langle A\phi ,\phi \rangle }_0\le p_1{\Vert {\phi }^{″}\Vert }_0^2.$

Similarly, we have

${\langle T\phi ,\phi \rangle }_0={\langle \eta \phi \text{'}\text{'}\text{'}\text{'},\phi \rangle }_0={\langle \eta {\phi }^{‴},{\phi }^{\prime }\rangle }_0={\langle \eta {\phi }^{″},{\phi }^{″}\rangle }_0=\eta {\Vert {\phi }^{″}\Vert }_0^2$

and

${\Vert {\phi }^{″}\Vert }_0^2=\frac{1}{\eta }{\langle T\phi ,\phi \rangle }_0.$

Hence,

${\langle \frac{p}{\eta }T\phi ,\phi \rangle }_0\le {\langle A\phi ,\phi \rangle }_0\le {\langle \frac{p_1}{\eta }T\phi ,\phi \rangle }_0$

and therefore,

$\frac{p_0}{\eta }T\le A\le \frac{p_1}{\eta }T.$

We can rewrite in terms of the operators $A$ and $T$ the system (2.4) as follows:

$\{\begin{array}{l}\frac{{\text{d}}^{2}y}{\text{d}{t}^2}+\frac{\text{d}}{\text{d}t}\left(Ty\right)+Ay=\left(I-P_1\right)f\left(t,y\right)\hfill \\ y\left(0\right)=\left(I-P_1\right){\phi }_0,{\frac{\text{d}y\left(x,t\right)}{\text{d}t}|}_{t=0}=\left(I-P_1\right){\psi }_0\hfill \end{array}$(2.8)

Let us now introduce a Hilbert space $H=H_2\times H_2$ equipped with general inner product. Set

$y=\left[\begin{array}{l}{y}_1\hfill \\ y_2\hfill \end{array}\right],y_1=A^{\frac{1}{2}}y,y_2=\frac{\text{d}y}{\text{d}t},$

$\mathcal{A}=\left[\begin{array}{cc}0& A^{\frac{1}{2}}\\ -A^{\frac{1}{2}}& -T\end{array}\right],D\left(\mathcal{A}\right)=D\left(A^{\frac{1}{2}}\right)\times D\left(A\right),$

$F\left(t,y\right)=\left[\begin{array}{c}0\\ \left(I-P_1\right)f\left(t,y\right)\end{array}\right],y_0=\left[\begin{array}{l}{y}_1\left(0\right)\hfill \\ y_2\left(0\right)\hfill \end{array}\right]=\left[\begin{array}{c}{A}^{\frac{1}{2}}\left(I-P_1\right){\phi }_0\\ \left(I-P_1\right){\psi }_0\end{array}\right].$

Then the evolution Equation (2.8), or original system (1.1) is equivalent to the following first order evolution equation

$\{\begin{array}{l}\frac{\text{d}y\left(t\right)}{\text{d}t}=\mathcal{A}y\left(t\right)+F\left(t,y\right)\hfill \\ y\left(0\right)=y_0\hfill \end{array}$(2.9)

and the corresponding equation if given by

$\{\begin{array}{l}\frac{\text{d}y\left(t\right)}{\text{d}t}=\mathcal{A}y\left(t\right)\hfill \\ y\left(0\right)=y_0\hfill \end{array}$(2.10)

Theorem 2.1 The linear operator $\mathcal{A}$ in the system (9) is the infinitesimal generator of a $C_0$ semigroup $T\left(t\right)$ satisfying

$\Vert T\left(t\right)\Vert \le M{\text{e}}^{-\delta t}\left(t\ge 0\right)$

where $M$ and $\delta$ are the positive constants.

To prove the Theorem 2.1, we shall first prove the following lemmas.

Lemma 2.1 If $\lambda$ is a complex number with $Re\lambda \ge 0$, then ${\left({\lambda }^2+\lambda T+A\right)}^{-1}$ exists and is bounded.

Proof. It is obviously true for $\lambda =0$. If $\lambda \ne 0$, for any $x\in D\left(A\right)$, let $\lambda =\sigma +i\tau$, $\sigma \ge 0$. We have

$\begin{array}{l}\Vert \left(\lambda +T+\frac{1}{\lambda }A\right)x\Vert \Vert x\Vert \ge \left|\langle \left(\lambda +T+\frac{1}{\lambda }A\right)x,x\rangle \right|\\ =\left|\sigma {\Vert x\Vert }^2+\langle Tx,x\rangle +\frac{\sigma }{{\sigma }^2+{\tau }^2}\langle Ax,x\rangle +i\left[\tau {\Vert x\Vert }^2-\frac{\tau }{{\sigma }^2+{\tau }^2}\langle Ax,x\rangle \right]\right|\\ \ge \langle Tx,x\rangle \ge \omega {\Vert x\Vert }^2,\end{array}$

where $\omega >0$ is the smallest eigenvalue of $T$.

Since $x\in D\left(A\right)$, it can be seen that

$\begin{array}{c}\langle \left(\lambda +T+\frac{1}{\lambda }A\right)x,x\rangle =\sigma {\Vert x\Vert }^2+\langle Tx,x\rangle +\frac{\sigma }{{\sigma }^2+{\tau }^2}\langle Ax,x\rangle \\ +i\left[\tau {\Vert x\Vert }^2-\frac{\tau }{{\sigma }^2+{\tau }^2}\langle Ax,x\rangle \right],\end{array}$

and

$Re\langle -\left(\lambda +T+\frac{1}{\lambda }A\right)x,x\rangle \le -\langle Tx,x\rangle \le -\omega {\Vert x\Vert }^2.$

It follows that the numerical range of $-\left(\lambda +T+\frac{1}{\lambda }A\right)$

$\begin{array}{l}V\left(-\left(\lambda +T+\frac{1}{\lambda }A\right)\right)\\ =\left\{-\langle \left(\lambda +T+\frac{1}{\lambda }A\right)x,x\rangle :\Vert x\Vert =1,x\in D\left(\lambda +T+\frac{1}{\lambda }A\right)\right\}\\ \subseteq \left\{\lambda |Re\lambda \le -\omega \right\}.\end{array}$

This implies that $0\in \rho \left(-\left(\lambda +T+\frac{1}{\lambda }A\right)\right)$ (see [9] ), and so $0\in \rho \left({\lambda }^2+\lambda T+A\right)$. Thus, ${\left({\lambda }^2+\lambda T+A\right)}^{-1}$ exists and is bounded.

Lemma 2.2 If $\lambda$ is complex number with $Re\lambda \ge 0$, $\lambda \ne 0$. ${\left(\frac{1}{\lambda }+\lambda A^{-1}+A^{-\frac{1}{2}}T{A}^{-\frac{1}{2}}\right)}^{-1}$ exists and is bounded.

Proof. First, it should be noted that $A^{-\frac{1}{2}}T{A}^{-\frac{1}{2}}$ can be extended to a bounded linear operator on $H_2$, for every $x\in H_2$, $\lambda =\sigma +i\tau$, $\sigma \ge 0$. Since

$\begin{array}{l}\Vert \left(\frac{1}{\lambda }+\lambda A^{-1}+A^{-\frac{1}{2}}T{A}^{-\frac{1}{2}}\right)x\Vert \Vert x\Vert \ge \left|\langle \left(\frac{1}{\lambda }+\lambda A^{-1}+A^{-\frac{1}{2}}T{A}^{-\frac{1}{2}}\right)x,x\rangle \right|\\ =\left|\frac{\sigma }{{\sigma }^2+{\tau }^2}{\Vert x\Vert }^2+\sigma \langle A^{-1}x,x\rangle +\langle A^{-\frac{1}{2}}T{A}^{-\frac{1}{2}}x,x\rangle +i\left[\frac{-\tau }{{\sigma }^2+{\tau }^2}{\Vert x\Vert }^2+\tau \langle A^{-1}x,x\rangle \right]\right|\\ \ge \frac{\sigma }{{\sigma }^2+{\tau }^2}{\Vert x\Vert }^2+\sigma \langle A^{-1}x,x\rangle +\langle A^{-\frac{1}{2}}T{A}^{-\frac{1}{2}}x,x\rangle \ge \frac{\eta }{p_1}{\Vert x\Vert }^2,\end{array}$

$\frac{1}{\lambda }+\lambda A^{-1}+A^{-\frac{1}{2}}T{A}^{-\frac{1}{2}}$ is invertible. We also see that its image is dense in $H_2$. In fact, if $y_0\in H_2$, and

$\langle \left(\frac{1}{\lambda }+\lambda A^{-1}+A^{-\frac{1}{2}}T{A}^{-\frac{1}{2}}\right)x,y_0\rangle =0,x\in H_2.$

Noticing that $\frac{1}{\lambda }+\lambda A^{-1}+A^{-\frac{1}{2}}T{A}^{-\frac{1}{2}}$ is self-adjoint, we have

$\langle x,\left(\frac{1}{\lambda }+\lambda A^{-1}+A^{-\frac{1}{2}}T{A}^{-\frac{1}{2}}\right)y_0\rangle =0,x\in H_2$

Since $\frac{1}{\lambda }+\lambda A^{-1}+A^{-\frac{1}{2}}T{A}^{-\frac{1}{2}}$ is invertible, $y_0=0$, and therefore the range of $\left(\frac{1}{\lambda }+\lambda A^{-1}+A^{-\frac{1}{2}}T{A}^{-\frac{1}{2}}\right)$ is dense in $H_2$. Thus, ${\left(\frac{1}{\lambda }+\lambda A^{-1}+A^{-\frac{1}{2}}T{A}^{-\frac{1}{2}}\right)}^{-1}$ exists and is bounded.

Lemma 2.3 If $\lambda$ is complex number with $Re\lambda \ge 0$, $\lambda \ne 0$, then resolvent of $A$ can be expressed by

$R\left(\lambda ,\mathcal{A}\right)=\frac{1}{\lambda }\left[\begin{array}{cc}I-\frac{1}{{\lambda }^2}{\left(\frac{1}{{\lambda }^2}+A^{-1}+\frac{1}{\lambda }{A}^{-\frac{1}{2}}T{A}^{-\frac{1}{2}}\right)}^{-1}& \frac{1}{\lambda }{\left(\frac{1}{{\lambda }^2}+A^{-1}+\frac{1}{\lambda }{A}^{-\frac{1}{2}}T{A}^{-\frac{1}{2}}\right)}^{-1}{A}^{-\frac{1}{2}}\\ -\frac{1}{\lambda }{A}^{-\frac{1}{2}}{\left(\frac{1}{{\lambda }^2}+A^{-1}+\frac{1}{\lambda }{A}^{-\frac{1}{2}}T{A}^{-\frac{1}{2}}\right)}^{-1}& A^{-\frac{1}{2}}{\left(\frac{1}{{\lambda }^2}+A^{-1}+\frac{1}{\lambda }{A}^{-\frac{1}{2}}T{A}^{-\frac{1}{2}}\right)}^{-1}{A}^{-\frac{1}{2}}\end{array}\right]$

Proof. We know from Lemma 2 that $R\left(\lambda ,\mathcal{A}\right)$ is a bounded linear operator on $H$ that expression of $R\left(\lambda ,\mathcal{A}\right)$ can be obtained by a direct calculation.

Lemma 2.4 If $\lambda$ is complex number with $Re\lambda \ge 0$ and $\lambda \ne 0$, the family of the operators with the parameter $\lambda$

$\begin{array}{c}F\left(\lambda \right)=\frac{1}{\lambda }{\left(\frac{1}{{\lambda }^2}+A^{-1}+\frac{1}{\lambda }{A}^{-\frac{1}{2}}T{A}^{-\frac{1}{2}}\right)}^{-1}\\ ={\left(\frac{1}{\lambda }+\lambda A^{-1}+A^{-\frac{1}{2}}T{A}^{-\frac{1}{2}}\right)}^{-1}\end{array}$

is uniformly bounded.

Proof. Let $Z_{\lambda }={\left(\frac{1}{\lambda }+\lambda A^{-1}+A^{-\frac{1}{2}}T{A}^{-\frac{1}{2}}\right)}^{-1}x$, $x\in H_2$, then $\left\{\Vert Z_{\lambda }\Vert \right\}$ is bounded for all $\lambda$. Otherwise, there is a ${\lambda }_0$ such that

$\underset{\lambda \to {\lambda }_0}{\text{lim}}\Vert Z_{\lambda }\Vert =+\infty .$

Considering the inner product of the sequence $y_{\lambda }=Z_{\lambda }/\Vert Z_{\lambda }\Vert$ with $\lambda =\sigma +i\tau$, we have

$\begin{array}{l}\langle \left(\frac{1}{\lambda }+\lambda A^{-1}+A^{-\frac{1}{2}}T{A}^{-\frac{1}{2}}\right)y_{\lambda },y_{\lambda }\rangle \\ =\frac{\sigma }{{\sigma }^2+{\tau }^2}+\sigma \langle A^{-1}{y}_{\lambda },y_{\lambda }\rangle +\langle A^{-\frac{1}{2}}T{A}^{-\frac{1}{2}}{y}_{\lambda },y_{\lambda }\rangle \\ +i\left[\frac{-\tau }{{\sigma }^2+{\tau }^2}+\tau \langle A^{-1}{y}_{\lambda },y_{\lambda }\rangle \right].\end{array}$(2.11)

Obviously, the real part of the right hand side (2.11) is greater than $\frac{\eta }{p_0}>0$, on the other hand

$\underset{\lambda \to {\lambda }_0}{\text{lim}}\left(\frac{1}{\lambda }+\lambda A^{-1}+A^{-\frac{1}{2}}T{A}^{-\frac{1}{2}}\right)y_{\lambda }=\underset{\lambda \to {\lambda }_0}{\text{lim}}\frac{x}{\Vert Z_{\lambda }\Vert }=0,$

in which the contradiction occurs. Hence, $\left\{\Vert Z_{\lambda }\Vert \right\}$ is uniformly bounded for every $x\in H_2$, and the result of this lemma turns out by means of the Principle of Uniform Boundedness.

Lemma 2.5 If $\lambda$ is complex number with $Re\lambda \ge 0$, $\lambda \ne 0$, there is a ${\lambda }_0>0$ such that if $\left|\lambda \right|\ge {\lambda }_0$, then ${\left(\frac{1}{\lambda }+T{A}^{-1}+\lambda A^{-1}\right)}^{-1}$ is uniformly bounded.

Proof. For every $x\in H_2$, it is easy to see that

$\begin{array}{l}{\Vert \left(\frac{1}{\lambda }+T{A}^{-1}+\lambda A^{-1}\right)x\Vert }^2\\ =\langle \left(\frac{1}{\lambda }+T{A}^{-1}+\lambda A^{-1}\right)x,\left(\frac{1}{\lambda }+T{A}^{-1}+\lambda A^{-1}\right)x\rangle \\ =\frac{1}{{\left|\lambda \right|}^2}{\Vert x\Vert }^2+\frac{\bar{\lambda }}{\lambda }\langle x,A^{-1}x\rangle +\frac{\lambda }{\bar{\lambda }}\langle A^{-1}x,x\rangle \\ +{\left|\lambda \right|}^2{\Vert A^{-1}x\Vert }^2+\bar{\lambda }\langle T{A}^{-1}x,A^{-1}x\rangle +\lambda \langle A^{-1}x,T{A}^{-1}x\rangle \\ +\frac{1}{\lambda }\langle x,T{A}^{-1}x\rangle +\frac{1}{\bar{\lambda }}\langle T{A}^{-1}x,x\rangle +{\Vert T{A}^{-1}x\Vert }^2\\ \ge \frac{1}{\lambda }\langle x,T{A}^{-1}\rangle +\frac{1}{\bar{\lambda }}\langle T{A}^{-1}x,x\rangle +{\Vert T{A}^{-1}x\Vert }^2.\end{array}$(2.12)

Since $T{A}^{-1}$ is bounded, there is a ${\lambda }_0>0$, such that if $\left|\lambda \right|\ge {\lambda }_0$, the right hand side of the above inequality

$\frac{1}{\lambda }\langle x,T{A}^{-1}x\rangle +\frac{1}{\bar{\lambda }}\langle T{A}^{-1}x,x\rangle +{\Vert T{A}^{-1}x\Vert }^2\ge \frac{1}{4}{\Vert T{A}^{-1}x\Vert }^2\ge \frac{1}{4}{\delta }_0^2{\Vert x\Vert }^2,$(2.13)

where ${\delta }_0>0$, and the last inequality is due to the invertibility of $T{A}^{-1}$. It follows from

$\Vert \left(\frac{1}{\lambda }+T{A}^{-1}+\lambda A^{-1}\right)x\Vert \ge \frac{1}{2}{\delta }_0\Vert x\Vert .$(2.14)

Hence, $\left(\frac{1}{\lambda }+T{A}^{-1}+\lambda A^{-1}\right)$ is invertible.

Next, we shall show by contradiction that the range of $\left(\frac{1}{\lambda }+T{A}^{-1}+\lambda A^{-1}\right)$ is dense in $H_2$. If the range of $\left(\frac{1}{\lambda }+T{A}^{-1}+\lambda A^{-1}\right)$ is not dense in $H_2$, there is a $y_0\in H_2$, $y_0\ne 0$ such that

$\langle \left(\frac{1}{\lambda }+T{A}^{-1}+\lambda A^{-1}\right)x,y_0\rangle =0,x\in H_2$

This implies that,

$\langle \left(\frac{A}{\lambda }+T+\lambda \right)y,y_0\rangle =0,y\in D\left(A\right)$

where $y=A^{-1}x$.

In view of the Lemma 2.1, ${\left(\frac{1}{\lambda }A+T+\lambda \right)}^{-1}$ is a bounded linear operator, and it range is dense in $H_2$. It follows that $y_0=0$, this contradicts that $y_0\ne 0$. Thus the range of $\left(\frac{1}{\lambda }+T{A}^{-1}+\lambda A^{-1}\right)$ is dense in $H_2$. If $\left|\lambda \right|\ge {\lambda }_0$, $Re\lambda \ne 0$, for a fixed $x\in H_2$, let

$Z_{\lambda }={\left(\frac{1}{\lambda }+\lambda A^{-1}+T{A}^{-1}\right)}^{-1}x,\left|\lambda \right|\ge {\lambda }_0$

then it can be shown that $\left\{\Vert Z_{\lambda }\Vert \right\}$ is bounded. Otherwise, there is a sequence $\left\{{\lambda }_n\right\}$ with $\left|{\lambda }_n\right|\ge {\lambda }_0$, and $Re{\lambda }_n\ge 0$ such that

$\underset{n\to \infty }{\text{lim}}\Vert Z_{{\lambda }_n}\Vert =\infty ,$

and

$\left(\frac{1}{{\lambda }_n}+{\lambda }_n{A}^{-1}+T{A}^{-1}\right)\frac{Z_{{\lambda }_n}}{\Vert Z_{{\lambda }_n}\Vert }=\frac{x}{\Vert Z_{{\lambda }_n}\Vert }\to 0,n\to \infty .$(2.15)

Let $y_n=Z_{{\lambda }_n}/\Vert Z_{{\lambda }_n}\Vert$. It follows from (2.14) that

$\Vert \left(\frac{1}{{\lambda }_n}+T{A}^{-1}+{\lambda }_n{A}^{-1}\right)y_n\Vert \ge \frac{{\delta }_0}{2}\Vert y_n\Vert =\frac{{\delta }_0}{2}>0,$

which contradicts (2.15). Hence $\left\{\Vert Z_{\lambda }\Vert \right\}$ is bounded, for every $x\in H_2$. It follows from Principle of Uniform Boundedness that ${\left(\frac{1}{\lambda }+\lambda A^{-1}+T{A}^{-1}\right)}^{-1}$ is uniformly bounded for $\left|\lambda \right|\ge {\lambda }_0$ and $Re\lambda \ge 0$.

Lemma 2.6 Under the condition of the Lemma 5, if $\left|\lambda \right|\ge {\lambda }_0$, $Re\lambda \ge 0$, the family of operators with $\lambda$

$A^{-\frac{1}{2}}{\left(\frac{1}{{\lambda }^2}+A^{-1}+\frac{1}{\lambda }{A}^{-\frac{1}{2}}T{A}^{-\frac{1}{2}}\right)}^{-1}{A}^{-\frac{1}{2}}={\left(\frac{1}{{\lambda }^2}A+\frac{1}{\lambda }T+I\right)}^{-1}$

is uniformly bounded.

Proof. If $\left|\lambda \right|\ge {\lambda }_0$, $Re\lambda \ge 0$, we have from the Lemma 5 that

${\left(\frac{1}{{\lambda }^2}A+\frac{1}{\lambda }T+I\right)}^{-1}=A^{-1}{\left(\frac{1}{{\lambda }^2}+\frac{1}{\lambda }T{A}^{-1}+A^{-1}\right)}^{-1}.$

Thus, the result of the Lemma 6 is concluded by virtue of the Lemma 5.

By virtue of Lemma 2.1 - Lemma 2.6, we can now prove Theorem 2.1.

Proof of Theorem 2.1. Since

$\mathcal{A}=\left[\begin{array}{cc}0& A^{\frac{1}{2}}\\ -A^{\frac{1}{2}}& -T\end{array}\right]$

where $A$ and $T$ are positively defined self-adjoint operators, we can easily verify that ${\left(i\mathcal{A}\right)}^{\text{*}}=i\mathcal{A}$. It follows from the celebrated Stone Theorem in [13] that $\mathcal{A}$ is the infinitesimal generator of a $C_0$ semigroup $T\left(t\right)$ on $H$. On the other hand, we can see that $0\in \rho \left(\mathcal{A}\right)$ by a simple computation gives us

${\mathcal{A}}^{-1}=\left[\begin{array}{cc}-A^{-\frac{1}{2}}T{A}^{-\frac{1}{2}}& -A^{-\frac{1}{2}}\\ A^{\frac{1}{2}}& 0\end{array}\right].$

If $Re\lambda \ge 0$, $\lambda \ne 0$ we can show that the resolvent $R\left(\lambda ,\mathcal{A}\right)$ of $\mathcal{A}$ satisfies

$\Vert R\left(\lambda ,\mathcal{A}\right)\Vert \le \frac{M}{\left|\lambda \right|},1\le M<\infty$(2.16)

In fact, we have seen from Lemma 2.3 that $R\left(\lambda ,\mathcal{A}\right)$ is an analytic function of $\lambda$ on the right half complex plane. According to the analyticity of $R\left(\lambda ,\mathcal{A}\right)$, it

suffices to show that if $\left|\lambda \right|\ge {\lambda }_0>0$, $Re\lambda \ge 0$, then $\Vert R\left(\lambda ,\mathcal{A}\right)\Vert \le \frac{M_1}{\left|\lambda \right|}$. However,

this can be easily obtained by Lemma 2.4 to Lemma 2.6.

Since $\rho \left(\mathcal{A}\right)\supset \left\{\lambda |Re\lambda \ge 0\right\}$ and $\rho \left(\mathcal{A}\right)$ is an open set on the complex plane, there is a constant $ϵ>0$ such that

$\sigma \left(\mathcal{A}\right)\subset \left\{\lambda |Re\lambda \le -ϵ\right\}$

and therefore we can conclude from the stability theorem of analytic semigroup [14] and [15] that there is a constant $\delta >0$ such that

$\Vert T\left(t\right)\Vert \le M{\text{e}}^{-\delta t}\left(t\ge 0\right).$

The proof of Theorem 2.1 is complete.

In this section, let us discuss an optimal control problem of the following system:

$\begin{array}{l}\frac{\text{d}y}{\text{d}t}=\mathcal{A}y\left(t\right)+ℬu\left(\left(y\left(t\right),t\right)\right)\\ y\left(0\right)=y_0\end{array}$ (3.1)

where both state space $\mathscr{H}$ and control space $\mathcal{Y}$ are Hilbert spaces, the state function $y\left(t\right)$ on $\left[0,T\right]$ is valued in $H$, $\mathcal{A}$ is the infinitesimal generator of a $C_0$-semigroup $S\left(t\right)$, $t\ge 0$. $ℬ$ is a bounded linear operator from $L^2\left(\left[0,T\right]:\mathcal{Y}\right)$ to $L^2\left(\left[0,T\right]:\mathscr{H}\right)$, $u\left(y\left(t\right),t\right)$ is a control of the system.

In this section, we shall discuss a specific optimal control, that is, the minimum energy control of the system (3.1). We know that the minimum energy control in an abstract space is, in general, the minimum norm control. So, from mathematics point of view, the existence and uniqueness of the optimal control are essential. If these are true, then how to obtain the optimal control is a significant problem. The main content of this paper is to solve these essential and significant issue.

From the theory of operator semigroup, we see that for every control element $u\left(y(\cdot ),\cdot \right)\in L^2\left(\left[0,T\right]:\mathcal{Y}\right)$, the system (3.1) has an unique mild solution

$y\left(t\right)=S\left(t\right)y_0+{\displaystyle {\int }_0^{t}S\left(t-s\right)ℬ\left(u\left(y\left(s\right),s\right)\right)\text{d}s}$ (3.2)

let $\phi (\cdot )$ be an arbitrary element in $C\left(\left[0,T\right],\mathscr{H}\right)$, and

$\rho ={\inf }_{u\in L^2\left(\left[0,T\right],\mathcal{Y}\right)}\Vert \phi \left(t\right)-S\left(t\right)y_0-{\displaystyle {\int }_0^{t}S\left(t-s\right)ℬu\left(y\left(s,s\right)\right)\text{d}s}\Vert ,$

define the admissible control set of the system (3.1) as follows

$U_{ad}=\left\{u\in L^2\left(\left[0,T\right],\mathcal{Y}\right):\Vert \phi \left(t\right)-S\left(t\right)y_0-{\displaystyle {\int }_0^{t}S}\left(t-s\right)ℬu\left(y\left(s\right),s\right)\Vert \le \rho +ϵ\right\}$ (3.3)

where $ϵ$ is any positive number.

It can be seen from (3.2) that $U_{ad}$ is not empty and contains infinitely many elements related to $\phi$ and $ϵ$. The minimum energy control problem is actually to find the element $u$, satisfying

$\Vert u_0\Vert =\min \left\{\Vert u\Vert :u\in U_{ad}\right\}$ (3.4)

where $u_0$ is said to be a minimum energy control element.

Lemma 3.1 The admissible control set $U_{ad}$ defined by (2.2) is a closed convex set in Hilbert space $L^2\left(\left[0,T\right]:\mathcal{Y}\right)$.

Proof. Convexity. For any $u_1$, $u_2\in U_{ad}$ and a real number $\lambda$, $0<\lambda <1$, it is easy to see from (2.2) that

$\Vert \phi \left(t\right)-S\left(t\right)y_0-{\displaystyle {\int }_0^{t}S}\left(t-s\right)ℬu_i\left(y\left(s\right),s\right)\Vert \le \rho +ϵ,i=1,2$ (3.5)

and hence

$\begin{array}{l}\Vert \phi \left(t\right)-S\left(t\right)y_0-{\displaystyle {\int }_0^{t}S}\left(t-s\right)B\left(\lambda u_1\left(y\left(s\right),s\right)+\left(1-\lambda \right)u_2\left(y\left(s\right),s\right)\right)\text{d}s\Vert \\ \le \lambda \Vert \phi \left(t\right)-S\left(t\right)y_0-{\displaystyle {\int }_0^{t}S}\left(t-s\right)ℬu_1\left(y\left(s\right),s\right)\text{d}s\Vert \\ +\left(1-\lambda \right)\Vert \phi \left(t\right)-S\left(t\right)y_0-{\displaystyle {\int }_0^{t}S}\left(t-s\right)ℬu_2\left(y\left(s\right),s\right)\text{d}s\Vert .\end{array}$ (3.6)

Since $\lambda u_1+\left(1-\lambda \right)u_2\in L^2\left(\left[0,T\right];\mathcal{Y}\right)$, it follows that $\lambda u_1+\left(1-\lambda \right)u_2\in U_{ad}$, this implies that $U_{ad}$ is a convex subset of $L^2\left(\left[0,T\right],\mathcal{Y}\right)$.

Closedness. Suppose $\left\{u_n\right\}\subset U_{ad}$, and ${\text{lim}}_{n\to \infty }\Vert u_n-u^{\text{*}}\Vert =0$. It can be shown that $u^{\text{*}}\in U_{ad}$. In fact, from the definition of $U_{ad}$ we see that

$\Vert \phi \left(t\right)-S\left(t\right)y_0-{\displaystyle {\int }_0^{t}S\left(t-s\right)ℬu_n\left(y\left(s\right),s\right)\text{d}s}\Vert \le \rho +ϵ,n=1,2,\cdots$

Since $S\left(t\right),t\ge 0$ is a $C_0$-semigroup in Hilbert space $\mathscr{H}$, there is a constant $M>0$ such that $\underset{0\le t\le T}{\text{sup}}\Vert S\left(t\right)\Vert \le M$. On the other hand, since $y\left(s\right)$ is differentiable on $\left[0,T\right]$, it is continuous on $\left[0,T\right]$, and hence $\left\{y\left(s\right):s\in \left[0,T\right]\right\}$ is a bounded set in $L^2\left(\left[0,T\right]:\mathcal{Y}\right)$. Thus there is a constant $N>0$ such that $\Vert ℬu\left(y\left(s\right),s\right)\Vert \le N$ $\left(0\le s\le T\right)$ and

$\begin{array}{l}\Vert \phi \left(t\right)-S\left(t\right)y_0-{\displaystyle {\int }_0^{t}S}\left(t-s\right)ℬu^{\text{*}}\left(y\left(s\right),s\right)\text{d}s\Vert \\ \le \Vert \phi \left(t\right)-S\left(t\right)y_0-{\displaystyle {\int }_0^t{u}_n\left(y\left(s\right),s\right)ℬu\left(y\left(s\right),s\right)\text{d}s}\Vert \\ +\Vert {\displaystyle {\int }_0^{t}S}\left(t-s\right)ℬ\left[u_n\left(y\left(s\right),s\right)-u^{\text{*}}\left(y\left(s\right),s\right)\right]\Vert \\ \le \rho +ϵ+M\Vert u_n-u^{\text{*}}\Vert \cdot NT\end{array}$ (3.7)

Letting $n\to \infty$ leads to

$\Vert \phi \left(t\right)-S\left(t\right)y_0-{\displaystyle {\int }_0^{t}S}\left(t-s\right)ℬu^{\text{*}}\left(y\left(s\right),s\right)\text{d}s\Vert \le \rho +ϵ.$

Thus, $u^{\text{*}}\in U_{ad}$, and $U_{ad}$ is a closed set. The proof is complete.

Theorem 3.1 There exists an unique minimum energy control element in the admissible control set $U_{ad}$ of the system (1.1).

Proof. Since $L^2\left(\left[0,T\right]:\mathcal{Y}\right)$ is a Hilbert space, it is naturally a strict convex Banach Space. From the preceding Lemma, we have seen that $U_{ad}$ is a closed convex set in $L^2\left(\left[0,T\right]:\mathcal{Y}\right)$, it follows from [13] that there is an unique element $u_0\in U_{ad}$ such that

$\Vert u_0\Vert =\min \left\{\Vert u\Vert :u\in U_{ad}\right\}$

According to the definition (3.3), $u_0$ is just the desired minimum energy control element of the system (1.1). The proof is complete.

Finally, we shall show that the minimum energy control element can be approached.

Theorem 3.2 Suppose that $u_0$ is the minimum energy control element of the system (1.1), then there exists a sequence $\left\{u_n\right\}\subset U_{ad}$ such that $\left\{u_n\right\}$ converges strongly to $u_0$ in $L^2\left(\left[0,T\right]:\mathcal{Y}\right)$, namely,

$\underset{n\to \infty }{\text{lim}}\Vert u_n-u_0\Vert =0$

Proof. Let $\left\{u_n\right\}$ be a minimizing sequence in the admissible control set $U_{ad}$, then it follows that

$\Vert u_{n+1}\Vert \le \Vert u_n\Vert ,n=1,2,\cdots$ (3.8)

and

${\lim }_{n\to \infty }\Vert u_n\Vert =\inf \left\{\Vert u\Vert :u\in U_{ad}\right\}$ (3.9)

It is obvious that $\left\{u_n\right\}$ is a bounded sequence in $L^2\left(\left[0,T\right];\mathcal{Y}\right)$, and so there is a subsequence $\left\{u_{n_k}\right\}$ of $\left\{u_n\right\}$ such that $\left\{u_{n_k}\right\}$ weakly converges to an element $\tilde{u}$ in $L^2\left(\left[0,T\right];\mathcal{Y}\right)$ (see [16] ).

Since $U_{ad}$ is a closed convex set in $L^2\left(\left[0,T\right];\mathcal{Y}\right)$ (see Lemma 3.1), we see from Mazur’s Theorem in [17] that $U_{ad}$ is a weakly closed set in $L^2\left(\left[0,T\right]:\mathcal{Y}\right)$, thus $\tilde{u}\in U_{ad}$. Combining (3.2) and employing the properties of limits of weakly convergent sequence on norm yield

$\inf \left\{\Vert u\Vert :u\in U_{ad}\right\}\le \Vert \tilde{u}\Vert \le \underset{k\to \infty }{\underset{\_}{\lim }}\Vert u_{n_k}\Vert =\underset{n_k\to \infty }{\lim }\Vert u_{n_k}\Vert =\underset{n\to \infty }{\lim }\Vert u_n\Vert =\inf \left\{\Vert u\Vert ;u\in U_{ad}\right\}.$(3.10)

Thus, we have

${\text{lim}}_{n\to \infty }\Vert u_n\Vert =\Vert \tilde{u}\Vert$ (3.11)

and

$\Vert \tilde{u}\Vert =\inf \left\{\Vert u\Vert :u\in U_{ad}\right\}.$ (3.12)

Since $\left\{u_{n_k}\right\}$ is weakly convergent to $\tilde{u}$, it follows from (3.4) that $\left\{u_{n_k}\right\}$ converges to $\tilde{u}$. Therefore, we see from Theorem 3.2 and (3.4) that $\tilde{u}=u_0$, namely, $\tilde{u}$ is the minimum energy control element. Thus, $\left\{u_{n_k}\right\}$ strongly converges to the minimum energy control element in $L^2\left(\left[0,T\right]:\mathcal{Y}\right)$. Without loss of generality, we can rewrite $\left\{u_{n_k}\right\}$ by $\left\{u_n\right\}$, then the conclusion of theorem is now obtained.