The Riesz basis property of a class of Euler-Bernoulli beam equation

TL;DR

This paper proves that a specific set of eigenvectors for a controlled Euler-Bernoulli beam system forms a Riesz basis, ensuring exponential stability of the system.

Contribution

It establishes the Riesz basis property for eigenvectors of a beam equation under boundary feedback, which was previously unproven.

Findings

Eigenvectors form a Riesz basis

Closed-loop system is exponentially stable

Provides a basis for stability analysis

Abstract

In this paper, we prove that a sequence of generalized eigenvectors of a linear unbounded operator associated with an Euler-Bernoulli beam equation under bending moment boundary feedback forms a Riesz basis for the underlying state Hilbert space. As a consequence, the resulting closed-loop system is exponentially stable.

Equations

$$\left\{\begin{array}[]{l}\displaystyle w_{tt}(x,t)+w_{xxxx}(x,t)=0,\;0<x<1,\;t\geq 0,\cr\vskip 5.69054pt\cr\displaystyle w(0,t)=w_{x}(0,t)=w(1,t)=0,\;t\geq 0,\cr\vskip 5.69054pt\cr\displaystyle w_{xx}(1,t)=u(t),\;t\geq 0,\cr\vskip 5.69054pt\cr\displaystyle w(x,0)=w_{0}(x),\;w_{t}(x,0)=w_{1}(x),\;0\leq x\leq 1,\cr\vskip 5.69054pt\cr\displaystyle y_{m}(t)=w_{xt}(1,t),\;t\geq 0.\end{array}\right.$$

$${\int }_0^1[|w_t(x,t){|}^2+|w_{xx}(x,t){|}^{2}dx\le M{e}^{-\mu t}{\int }_0^1[|w_t(x,0){|}^2+|w_{xx}(x,0){|}^{2}dx?$$

$$\begin{array}[]{l}\mathcal{H}=H_{e}^{2}(0,1)\times L^{2}(0,1),\;H_{e}^{2}(0,1)=\{\phi\in H^{2}(0,1)|\;|\phi(0)=\phi^{\prime}(0)=0\},\end{array}$$

$$\Vert (\varphi ,\psi ){\Vert }_{\mathcal{H}}^2={\int }_0^1[|{\varphi }^{\prime \prime }(x){|}^2+|\psi (x){|}^2]dx,\forall (\varphi ,\psi )\in \mathcal{H}.$$

$$\frac{d}{dt}(w(\cdot ,t),w_t(\cdot ,t))=A(w(\cdot ,t),w_t(\cdot ,t)),$$

$$\left\{\begin{array}[]{l}\displaystyle A(\phi,\psi)=(\psi,-\phi^{\prime\prime\prime\prime}),\;\;\forall(\phi,\psi)\in D(A),\cr\vskip 5.69054pt\cr\displaystyle D(A)=\Big{\{}(\phi,\psi)\in(H^{4}(0,1)\times H^{2}_{e}(0,1))\cap\mathcal{H}:\phi^{\prime\prime}(1)=-k\psi^{\prime}(1),\;\phi(1)=0\Big{\}}.\end{array}\right.$$

$${\tau }_n=(n+1/2)\pi +\mathcal{O}(n^{-1}),{\lambda }_n=i(n+1/2{)}^2{\pi }^2-\frac{2}{k}+\mathcal{O}(n^{-1}).$$

$$\begin{array}[]{l}\displaystyle A^{-1}(\phi,\psi)=\bigg{(}\frac{(3x^{2}-x^{3})}{12}\int_{0}^{1}(1-\xi)^{3}\psi(\xi)d\xi+\frac{(x^{3}-x^{2})}{4}\Big{(}\int_{0}^{1}(1-\xi)\psi(\xi)d\xi-k\phi^{\prime}(1)\Big{)}\cr\vskip 5.69054pt\cr\displaystyle\hskip 85.35826pt-\frac{1}{6}\int_{0}^{x}(x-\xi)^{3}\psi(\xi)d\xi,\phi(x)\bigg{)}.\end{array}$$

$$\left\{\begin{array}[]{l}\phi^{(4)}(x)-\tau^{4}\phi(x)=0,\cr\vskip 5.69054pt\cr\phi(0)=\phi^{\prime}(0)=\phi(1)=0,\;\;\phi^{\prime\prime}(1)=-ik\tau^{2}\phi^{\prime}(1)\end{array}\right.$$

$$\left\{\begin{array}[]{l}\phi^{(4)}(x)-\tau^{4}\phi(x)=0,\cr\vskip 5.69054pt\cr\phi(0)=\phi^{\prime}(0)=0\end{array}\right.$$

$$\varphi (x)=a_1(\cos \tau x-\cosh \tau x)+a_2(\sin \tau x-\sinh \tau x)$$

$$\varphi (x)=(\sin \tau -\sinh \tau )(\cos \tau x-\cosh \tau x)-(\cos \tau -\cosh \tau )(\sin \tau x-\sinh \tau x).$$

$$ik\tau [1-\cos \tau \cosh \tau ]+\cosh \tau \sin \tau -\cos \tau \sinh \tau =0,$$

$$\left\{\begin{array}[]{l}\cos\tau=\mathcal{O}(|\tau|^{-1}),\mbox{ or }\cr\vskip 5.69054pt\cr\displaystyle\cos\tau=-\frac{1}{ik\tau}[\cos\tau\tanh\tau-\sin\tau]+\mathcal{O}(e^{-{\rm Re}\tau}),\mbox{ as ${\rm Im}\tau$ bounded ${\rm Re}\tau\to\infty$}.\end{array}\right.$$

$${\lambda }_n=i(n+1/2{)}^2{\pi }^2-\frac{2}{k}+\mathcal{O}(n^{-1}).$$

$$F_{n}(x)=\left(\begin{array}[]{l}-(-1)^{n}e^{-(n+1/2)\pi(1-x)}+\cos(n+1/2)\pi x-\sin(n+1/2)\pi x+e^{-(n+1/2)\pi x}\cr\vskip 5.69054pt\cr-i(-1)^{n}e^{-(n+1/2)\pi(1-x)}-i\cos(n+1/2)\pi x+i\sin(n+1/2)\pi x+ie^{-(n+1/2)\pi x}\end{array}\right)+\mathcal{O}(n^{-1})$$

$${\tau }^{-2}{\varphi }^{\prime \prime }(x)=(\sin \tau -\sinh \tau )(-\cos \tau x-\cosh \tau x)-(\cos \tau -\cosh \tau )(-\sin \tau x-\sinh \tau x).$$

$$2{\tau }_n^{-2}{e}^{-{\tau }_n}{\varphi }_n^{\prime \prime }(x)=-(-1{)}^n{e}^{-(n+1/2)\pi (1-x)}+\cos (n+1/2)\pi x-\sin (n+1/2)\pi x+e^{-(n+1/2)\pi x}+\mathcal{O}(n^{-1}).$$

$$\left\{\begin{array}[]{l}\displaystyle A_{a}(\phi,\psi)=(\psi,-\phi^{\prime\prime\prime\prime}),\;\;\forall(\phi,\psi)\in D(A_{a}),\cr\vskip 5.69054pt\cr\displaystyle D(A_{a})=\Big{\{}(\phi,\psi)\in(H^{4}(0,1)\times H^{2}_{e}(0,1))\cap\mathcal{H}:\phi^{\prime\prime}(1)=0,\;\phi(1)=0\Big{\}}.\end{array}\right.$$

$$\cosh \omega \sin \omega -\cos \tau \sinh \omega =0,$$

$${\omega }_n=(n+1/2)\pi +\mathcal{O}(n^{-1}).$$

$$G_{n}(x)=\left(\begin{array}[]{l}-(-1)^{n}e^{-(n+1/2)\pi(1-x)}+\cos(n+1/2)\pi x-\sin(n+1/2)\pi x+e^{-(n+1/2)\pi x}\cr\vskip 5.69054pt\cr-i(-1)^{n}e^{-(n+1/2)\pi(1-x)}-i\cos(n+1/2)\pi x+i\sin(n+1/2)\pi x+ie^{-(n+1/2)\pi x}\end{array}\right)+\mathcal{O}(n^{-1}),$$

$$\begin{array}[]{l}\displaystyle\sum_{n>N}^{\infty}\|F_{n}-G_{n}\|_{[L^{2}(0,1)]^{2}}=\sum_{n>N}^{\infty}\|2\tau_{n}^{-2}e^{-\tau_{n}}(\phi_{n}^{\prime\prime}(x),\lambda_{n}\phi_{n}(x))^{\top}\cr\vskip 5.69054pt\cr\displaystyle\hskip 22.76228pt-2\omega_{n}^{-2}e^{-\omega_{n}}(\phi_{an}^{\prime\prime}(x),\lambda_{an}\phi_{an}(x))^{\top}\|_{[L^{2}(0,1)]^{2}}=\sum_{n>N}^{\infty}\mathcal{O}(n^{-2})<+\infty.\end{array}$$

$$\mathrm{Re}⟨A(\varphi ,\psi ),(\varphi ,\psi ){⟩}_{\mathcal{H}}=-k|{\psi }^{\prime }(1){|}^2\le 0,$$

$$\mathrm{Re}⟨A(\varphi ,\psi ),(\varphi ,\psi ){⟩}_{\mathcal{H}}=\mathrm{Re}⟨i{\tau }^2(\varphi ,\psi ),(\varphi ,\psi ){⟩}_{\mathcal{H}}=-k|{\psi }^{\prime }(1){|}^2=0,$$

$$\left\{\begin{array}[]{l}\phi^{(4)}(x)-\tau^{4}\phi(x)=0,\cr\vskip 5.69054pt\cr\displaystyle\phi(0)=\phi^{\prime}(0)=\phi(1)=\phi^{\prime}(1)=\phi^{\prime\prime}(1)=0,\end{array}\right.$$

Full text

Key words. Riesz basis, exponential stability, Euler-Bernoulli beam equation.

1 Introduction

Let us consider the following Euler-Bernoulli beam equation:

[TABLE]

In (1.1), $w(x,t)$ is the transverse displacement of the beam at time $t$ and position $x$, $u$ is the input (control) through bending moment, $y_{m}$ is the output signal (measured angular velocity). We are concerned about the following stabilization problem.

Problem. Given gain $k>0$, does the proportional feedback $u=-ky_{m}$ make the state $(w,w_{t})$ of the system exponentially convergent to zero in the sense that for some $M,\mu>0$,

[TABLE]

We remark that when boundary condition $w(1,t)=0$ of (1.1) is replaced by $w_{xxx}(1,t)=0$, the exponential stability of system under the same proportional feedback $u=-ky_{m}$ was investigated in Guo and Yu (2001). The motivation studying the above Problem is to copy with the Euler-Bernoulli beam with shear force control matched uncertainties (Zhou and Feng (2017)). In this paper, we will prove that the feedback $u=-ky_{m}$ does make system (1.1) exponentially stable in the certain state Hilbert space.

We consider system (1.1) in the energy Hilbert state space defined by

[TABLE]

with the inner product induced norm given by

[TABLE]

Under this setting and with the feedback $u=-ky_{m}$, the closed-loop system of (1.1) can be formulated as

[TABLE]

where the linear operator is defined as

[TABLE]

2 Statement and proof of the main results

Our study in the sequel is focused on the Riesz basis properties of the operator $A$. The first main result in this paper can be stated as follows.

Theorem 2.1**.**

Let $A$ be given by (1.4). Then, there is a sequence of generalized eigenvectors of $A$ which forms a Riesz basis for the state space $\mathcal{H}$. Moreover, $A$ generates an exponential stable $C_{0}$-semigroup on $\mathcal{H}$.

Proof. The proof is broken into several steps as follows.

Step 1. We claim that there is a family of eigenvalues $\{\lambda_{n},\overline{\lambda}_{n}\}$, $\lambda_{n}=i\tau_{n}^{2}$ of $A$ with the following asymptotic expression:

[TABLE]

A direct computation shows that

[TABLE]

By the Sobolev embedding theorem, $A^{-1}$ is compact on $\mathcal{H}$, and thus $\sigma(A)$ only consists of eigenvalues of $A$. It is easily seen that $\lambda=i\tau^{2}\in\sigma(A)$ if and only if there exists $\phi\neq 0$ satisfying

[TABLE]

and the associated egienfunction is $(\phi,\lambda\phi)$. First, the general solution of

[TABLE]

is of the form

[TABLE]

where $a_{1}$, $a_{2}$ are constants. Next, by the condition $\phi(1)=0$, we have $a_{1}=(\sin\tau-\sinh\tau),\;\;a_{2}=-(\cos\tau-\cosh\tau).$Substituting this into (2.3) gives

[TABLE]

The last condition $\phi^{\prime\prime}(1)=-ik\tau^{2}\phi^{\prime}(1)$ yields

[TABLE]

which can be re-written asymptotically as

[TABLE]

By the first equality of (2.6), we get $\tau_{n}=(n+{1}/{2})\pi+\mathcal{O}(n^{-1})$. Substitute $\tau_{n}$ into the second equality of (2.6) to obtain $\mathcal{O}(n^{-1})=-\frac{1}{[ik(n+{1}/{2})\pi]}+\mathcal{O}(n^{-2})$, and so

[TABLE]

Step 2. We claim that there is an eigenfunction $(\phi_{n},\lambda_{n}\phi_{n})^{\top}$ of $A$ corresponding to $\lambda_{n}=i\tau_{n}^{2}$ such that

[TABLE]

and $\lim_{n\to\infty}\|F_{n}(x)\|_{[L^{2}(0,1)]^{2}}=2$, where $F_{n}(x)=2\tau_{n}^{-2}e^{-\tau_{n}}(\phi_{n}^{\prime\prime}(x),\lambda_{n}\phi_{n}(x))^{\top}$. Actually, let $(\phi_{n},\lambda_{n}\phi_{n})$ be the eigenfunction of $A$ corresponding to $\lambda_{n}$, where $\phi_{n}=\phi(x)$ is defined by (2.3) with $\tau=\tau_{n}$. By (2.4), we derive

[TABLE]

Noticing that by (2.1), for any $y>0$ and $0\leq x\leq 1$, $e^{-\tau_{n}y}=e^{-(n+1/2)\pi y}+\mathcal{O}(n^{-1}),\;\sin\tau_{n}x=\sin(n+1/2)\pi x+\mathcal{O}(n^{-1}),\;\cos\tau_{n}x=\cos(n+1/2)\pi x+\mathcal{O}(n^{-1}),$ and letting $\tau=\tau_{n}$ in (2.7), we obtain

[TABLE]

The estimate for $\phi_{n}$ is similar, we omit the detail. By using the Lebesgue’s dominated convergence theorem, it is easy to verify $\lim_{n\to\infty}\|F_{n}(x)\|_{[L^{2}(0,1)]^{2}}=2$.

Step 3. We claim that the eigenfunctions of $A$ form an Riesz basis for $\mathcal{H}$. For this purpose, we introduce the following auxiliary operator $A_{a}$ given by

[TABLE]

By letting $k=0$ in (2.2), we know that $A_{a}$ has compact resolvent. It is easily to verify that the operator $A_{a}$ is skew-adjoint in the state space $\mathcal{H}$, i.e., $A_{a}^{*}=-A_{a}$ and all eigenvalues of $A_{a}$ are located on the imaginary axis and there is a sequence of generalized eigenfunctions of $A_{a}$ forming a Riesz basis for $\mathcal{H}$. Let $\lambda_{a}=i\omega^{2}$ be the eigenvalue of $A_{a}$ and $(\phi_{a},\lambda_{a}\phi_{a})$ be the eigenfunction of $A_{a}$ corresponding to $\lambda_{a}=i\omega^{2}$. By letting $k=0$ in (2.5), we obtain

[TABLE]

which gives

[TABLE]

Similar to the calculation in Step 2, we obtain that the eigenfunction $(\phi_{an},\lambda_{an}\phi_{an})$ of $A_{a}$ have the following asymptotical expression:

[TABLE]

where $G_{n}(x)=2\omega_{n}^{-2}e^{-\omega_{n}}(\phi_{an}^{\prime\prime}(x),\lambda_{n}\phi_{an}(x))^{\top}$. It is easy to see that $\{(\phi_{an},\lambda_{an}\phi_{an})\}_{n=1}^{\infty}\cup$ $\{\mbox{conjugates}\}$ is a Riesz basis for $\mathcal{H}$. It follows that there is an $N>0$ such that

[TABLE]

The same thing is true for conjugates. Therefore, operator $A$ has a sequence of eigenfunctions which quadratically closed to a Riesz basis in the sense of (2.10). By (Guo and Yu, 2001, Theorem 1), we have shown that the eigenfunctions of $A$ form an Riesz basis for $\mathcal{H}$.

Step 4. We claim that $A$ generates an exponential stable $C_{0}$-semigroup on $\mathcal{H}$. Since the eigenfunctions of $A$ form an Riesz basis for $\mathcal{H}$ that is justified by Step 3, the spectrum-determined growth condition holds. In order to show that $e^{At}$ is a exponential stable semigroup, it suffices to prove that ${\rm Re}\lambda<0$ for any $\lambda\in\sigma(A)$. Actually, a simple computation gives

[TABLE]

which implies that for any $\lambda\in\sigma(A)$ must satisfy ${\rm Re}\lambda\leq 0$. Since $A^{-1}$ is compact, we only need to show that there is no eigenvalue on the imaginary axis. Let $\lambda=i\tau^{2}\in\sigma(A)$ with $\tau\in\mathbb{R}^{+}$ and the corresponding eigenfunction $(\phi,\psi)^{\top}\in D(A)$. By (2.11),

[TABLE]

and hence $\psi^{\prime}(1)=0$. Furthermore, $A(\phi,\psi)=i\tau^{2}(\phi,\psi)$ gives that $\psi=i\tau^{2}\phi$ with $\phi$ satisfying

[TABLE]

Now, we show that the above equation admits only zero solution. For this, we prove that there exists at least one zero of $\phi$ in $(0,1)$. Actually, by $\phi(0)=\phi(1)=0$, Rolle’s theorem yields $\phi^{\prime}(\xi_{1})=0$ for some $\xi_{1}\in(0,1)$, which, jointly with $\phi^{\prime}(0)=\phi^{\prime}(1)=0$, implies that $\phi^{\prime\prime}(\xi_{2})=\phi^{\prime\prime}(\xi_{3})=0$ for some $\xi_{2}\in(0,\xi_{1})$, $\xi_{3}\in(\xi_{1},1)$, and so $\phi^{\prime\prime\prime}(\xi_{4})=\phi^{\prime\prime\prime}(\xi_{5})=0$ for some $\xi_{4}\in(\xi_{2},\xi_{3})$, $\xi_{5}\in(\xi_{3},1)$ by the condition $\phi^{\prime\prime}(1)=0$. Thus, there exists a $\xi_{6}\in(\xi_{4},\xi_{5})$ such that $\phi^{(4)}(\xi_{6})=0$, which, together with the first equation of (2.13), gives $\phi(\xi_{6})=0$. Next, we prove that if there are $n$ different zeros of $\phi$ in $(0,1)$, then there at least $n+1$ number of different zeros of $\phi$ in $(0,1)$. Indeed, suppose that $0<\xi_{1}<\xi_{2}<\cdots<\xi_{n}<1$, $\phi(\xi_{j})=0,j=1,2,\ldots,n$. Since $\phi(0)=\phi(1)=0$, if follows from Rolle’s theorem that there exist $\eta_{j},j=1,2,\ldots,n+1$, $0<\eta_{1}<\xi_{1}<\eta_{2}<\xi<2\cdots<\xi_{n}<\eta_{n+1}<1$ such that $\phi^{\prime}(\eta_{j})=0$. By $\phi^{\prime}(0)=\phi^{\prime}(1)=0$, using Rolle’s theorem again, there exist $\alpha_{j},j=1,2,\ldots,n+2$, $0<\alpha_{1}<\eta_{1}<\alpha_{2}<\eta_{2}<\cdots<\xi_{n+1}<\alpha_{n+2}<1$ such that $\phi^{\prime\prime}(\alpha_{j})=0$. It follows from $\phi^{\prime\prime}(1)=0$ that there exist $\beta_{j},j=1,2,\ldots,n+2$, $\alpha_{1}<\beta_{1}<\alpha_{2}<\beta_{2}<\cdots<\alpha_{n+2}<\beta_{n+2}<1$ such that $\phi^{\prime\prime\prime}(\beta_{j})=0$. Using Rolle’s theorem again, we have $\theta_{j},j=1,2,\ldots,n+1$, $\beta_{1}<\theta_{1}<\beta_{2}<\cdots<\beta_{n+1}<\theta_{n+1}<\beta_{n+2}$ such that $\phi^{(4)}(\theta_{j})=0$. Thus, $\phi(\theta_{j})=0$, $j=1,2,\ldots,n+1$. By mathematical induction, there is an infinite number of different zeros $\{x_{j}\}_{j=1}^{\infty}$ of $\phi$ in $(0,1)$. Let $x_{0}\in[0,1]$ be an accumulation point of $\{x_{j}\}_{j=1}^{\infty}$. Obviously, $\phi^{(j)}(x_{0})=0$, $j=0,1,2,3$. Since $\phi$ satisfies the first equation of (2.13), by the uniqueness of the solution of linear ordinary different equation, we have $\phi\equiv 0$. $\Box$

Remark 2.2**.**

The Riesz basis property of $A$ given by (1.4) could be used in dealing with disturbance rejection problem considered in Jin and Guo (2015) for Euler-Bernoulli beam with shear force control (Zhou and Feng (2017)).

Now, we state the second result of this paper as follows:

Theorem 2.3**.**

Let system (1.1) be with the proportional feedback $u=-ky_{m}$. For any initial state $(w(\cdot,0),w_{t}(\cdot,0))\in\mathcal{H}$, the resulting closed loop system of (1.1) admits a unique solution $(w,w_{t})\in C(0,\infty;\mathcal{H})$ satisfying (1.2).

Proof. The results follows directly from Theorem 2.1. $\Box$