1 Introduction

Euler-Bernoulli beam is a widely used model in the engineering. Hence, the stability and control strategy of Euler-Bernoulli beam are important subjects for researching. In the past decades, some results have been investigated. For example, in Refs.[1-4], the authors investigated the stabilization of Euler-Bernoulli beam under collocated boundary feedback control. The exponential stability can be obtained by Lyapunov functions or Riesz basis property. In Refs.[5-8], the authors investigated the non-collocated feedback control issues of Euler beam. However, the time-delay is not studied. As we all know, time-delay is a bad item for the stability of elastic system because it makes the control problems of such systems relatively difficult. Some scholars have paid attention to design stabilizer for time-delay systems.

Recently, there have been some results on dealing with time-delay. For examples, in Refs.[9-12], the authors studied the feedback control strategy to stabilize the system with interior time-delay. In Refs.[13-14], the authors stabilized the system with input and output delay by designing the feedback controller. However, the feedback controller in those papers is designed based on the state predictor which is constructed to covert the system to a new system without time-delay^[14-15]. In this paper, we study the Euler-Bernoulli beam with input delay, the state predictor is not included in the design method, it is different from other researchs. In order to make the system with input delay exponentially stability, we use the traditional feedback control to stabilize the system.

In order to illustrate this control strategy, we study the following Euler-Bernoulli beam:

$ \left\{ {\begin{array}{*{20}{l}} {{w_{xxxx}}(x, t) + {w_{tt}}(x, t) = 0, x \in (0, 1)}\\ {w(0, t) = {w_x}(0, t) = 0}\\ {{w_{xx}}(1, t) = 0}\\ {y(t) = {w_t}(1, t)}\\ {{w_{xxx}}(1, t) = u(t) + \delta u(t - \tau ), |\delta | < 1} \end{array}} \right. $

(1)

where u(t) is input, τ>0 is time-delay.

Therefore, the feedback controller u(t)=w_t(1, t) is given. The system (1) transforms to the next system.

$ \left\{\begin{array}{l}{w_{x x x x}(x, t)+w_{t t}(x, t)=0, x \in(0, 1)} \\ {w(0, t)=w_{x}(0, t)=0} \\ {w_{x x}(1, t)=0} \\ {y(t)=w_{t}(1, t)} \\ {w_{x x x}(1, t)=w_{t}(1, t)+\delta w_{t}(1, t-\tau), |\delta|<1}\end{array}\right. $

(2)

The rest of this paper is arranged as follows. In Section 2, we study the well-posedness of the close-loop system by semi-group theory. In Section 3, the asymptotical expression of eigenvalue of this system is given by spectral analysis. In Section 4, under the feedback control design, we study the exponential stability of the close-loop system. In Section 5, numerical simulation on the dynamical behavior of the system is given to support the results obtained in this paper.

2 Well-Posedness of the Closed-Loop System

In this section, the well-posedness of the system is investigated by semi-group theory. In order to deal with the time delay, the function is given as follows^[16-17]:

$ z(x, t)=w_{t}(1, t-x \tau) $

(3)

Substituting Eq.(3) into system (2). The system is converted into:

$ \left\{ {\begin{array}{*{20}{l}} {{w_{xxxx}}(x, t) + {w_{tt}}(x, t) = 0, x \in (0, 1), t > 0}\\ {\tau {z_t}(x, t) + {z_x}(x, t) = 0}\\ {w(0, t) = {w_x}(0, t) = 0}\\ {{w_{xx}}(1, t) = 0}\\ {y(t) = {w_t}(1, t)}\\ {{w_{xxx}}(1, t) = z(0, t) + \delta z(1, t), |\delta | < 1}\\ {w(x, 0) = {w_0}(x)}\\ {{w_t}(x, 0) = {w_1}(x)} \end{array}} \right. $

(4)

A space is assumed as follows:

$ H_{E}^{2}=\left\{f \in H^{2}(0, 1) | f(0)=f^{\prime}(0)=0\right\} $

where H²(0, 1) is the Sobolev space.

Therefore, the state space is given as:

$ H=H_{E}^{2}(0, 1) \times L^{2}(0, 1) \times L^{2}(0, 1) $

The norm of the space is as follows:

$ \begin{aligned}\|(f, g, h)\|^{2}=& \int_{0}^{1}\left[|g(x)|^{2}+\left|f^{\prime \prime}(x)\right|^{2}+\right.\\ & \tau|\delta||h(x)|^{2} ] \mathrm{d} x \end{aligned} $

We could get that the state space is Hilbert space. Define an operator A in state space,

$ \left\{\begin{array}{c}{A(f(x), g(x), h(x))=\left(g(x), -f^{(4)}(x), -\frac{h^{\prime}(x)}{\tau}\right)} \\ {D(A)=\{(f, g, h) \in H, A(f, g, h) \in H | f^{\prime \prime}(1)=0, \\ f^{\prime \prime \prime}(1)=h(0)+\delta h(1) \} }\end{array}\right. $

(5)

Hence, the system (4) can be written as:

$ \frac{\mathrm{d} Y(t)}{\mathrm{d} t}=A Y(t), t>0 $

(6)

in which

$ Y(t)=\left(w(x, t), w_{t}(x, t), z(x, t)\right) $

Theorem 1   The operator which is defined by Eq.(5) is closed and densely in state space H. 0∈ρ(A) and A^-1 is compact on H. Where ρ(A) is the resolvent set.

Proof   It is easy to obtain that the operator is closed and densely on the state space H^[18]. Hence, the approach is omitted. The solvability of the equation A(f, g, h)=(u, v, w) is studied, where the (u, v, w) is arbitrary on H, and

$ \left\{\begin{array}{l}{-f^{(4)}=v} \\ {g=u} \\ {w=-\frac{h^{\prime}}{\tau}}\end{array}\right. $

(7)

with the boundary conditions

$ \left\{\begin{array}{l}{f(0)=f^{\prime}(0)=0} \\ {h(0)=g(1), f^{\prime \prime}(1)=0} \\ {f^{\prime \prime \prime}(1)=h(0)+\delta h(1)} \\ {u(0)=u^{\prime}(0)=0, w(0)=v(1)}\end{array}\right. $

(8)

Hence, we get

$ \left\{\begin{array}{l} f(x)=-\int_{0}^{x}\left[\int_{0}^{p} \int_{0}^{s} \int_{0}^{q} v(r) \mathrm{d} r \mathrm{d} q \mathrm{d} s\right] \mathrm{d} p+\\ ~~~~~~\frac{c_{2}}{2} x^{2}+\frac{c_{1}}{6} x^{3} \\ u=g \\ h(x)=u(1)-\int_{0}^{x} w(s) \tau \mathrm{d} s \end{array}\right. $

(9)

where

$ \left\{\begin{array}{l}{c_{1}=(1+\delta) u(1)-\int_{0}^{1} w(x) \delta \tau \mathrm{d} x+\int_{0}^{1} v(x) \mathrm{d} x} \\ {c_{2}=\int_{0}^{1}\left[\int_{0}^{s} v(r) \mathrm{d} r\right] \mathrm{d} s-c_{1}}\end{array}\right. $

(10)

Therefore, the Eq.(7) is solvable. The results that 0∈ρ(A) is obtained by closed operator theory. Hence, because

$ D(A) \subset H^{4}(0, 1) \times H_{E}^{2}(0, 1) \times H^{1}(0, 1) $

We get that A^-1 is compact on H by Sobolev's Embedding Theorem. The proof is done.

Theorem 2   A is dissipative operator and composes a C₀ semi-group which is constructed on H.

Proof   For any W=(f, g, h)∈D(A)

$ \begin{aligned}(A W, W)_{H}=& \int_{0}^{1}\left[f^{\prime \prime} g^{\prime \prime}-f^{(4)} g-|\delta| h h^{\prime}\right] \mathrm{d} x=\\ &-g(1) f^{\prime \prime \prime}(1)-\int_{0}^{1}|\delta| h h^{\prime} \mathrm{d} x=\\ &-g(1)(\delta h(1)+h(0))-\\ & \int_{0}^{1}|\delta| h h^{\prime} \mathrm{d} x=\left(\frac{|\delta|}{2}-1\right) h^{2}(0)-\\ & \frac{|\delta|}{2} h^{2}(1)-\delta h(1) h(0)=\\ &-\frac{|\delta|}{2}[h(1)+h(0)]^{2}+\\ &(|\delta|-1) h^{2}(0) \leqslant(|\delta|-1) h^{2}(0) \end{aligned} $

Hence, we could get A is dissipative operator by the condition |δ| < 1. This result and Theorem 1 satisfy the condition of Lumer-Phillips^[19]theorem. Hence, A generates a C₀ semi-group which is contracted on H.

3 Spectral Analysis

In this section, the distribution of spectrum of A is investigated. From the results of Theorem 1, we get that A^-1 is compact, which represents σ(A)=σ_p(A). Hence, asymptotic distribution of eigenvalues of A is the only problem to be studied.

Theorem 3   From Eq.(5), We have:

$ \sigma (A) = {\sigma _P}(A) = \left\{ {\lambda = i{\zeta ^2}|\Delta(\zeta ) = 0, \zeta \ne 0} \right\} $

where

$ \begin{aligned} \Delta(\zeta)=& \zeta(1+\cosh \zeta \cos \zeta)+(1+\\ & \delta e^{{\rm i}\zeta^{2} \tau} ) \mathrm{i}(\cosh \zeta \sin \zeta-\sinh \zeta \cos \zeta) \end{aligned} $

Proof   Let w=(f, g, h), Hence, from Aw=λw, where λ∈C and λ≠0, we have:

$ \left\{\begin{array}{l}{\lambda f=g} \\ {\lambda g=-f^{(4)}} \\ {\lambda h=-\frac{1}{\tau} h^{\prime}}\end{array}\right. $

(11)

With the boundary condition

$ \left\{\begin{array}{l}{f(0)=0, f^{\prime \prime}(1)=0} \\ {f^{\prime}(0)=0, h(0)=g(1)} \\ {f^{\prime \prime \prime}(1)=h(0)+\delta h(1)}\end{array}\right. $

From Eq.(11) and its boundary condition, we have:

$ \left\{\begin{array}{l}{f^{(4)}(x)+\lambda^{2} f(x)=0} \\ {f^{\prime \prime \prime}(1)=\delta h(1)+h(0)} \\ {f(0)=f^{\prime}(0)=f^{\prime \prime}(1)=0}\end{array}\right. $

(12)

To assume λ=iζ² and ζ≠0. Hence, in Eq.(12),

$ f^{4}(x)-\zeta^{4} f(x)=0 $

Considering the boundary condition f(0)=f′(0)=0, we have

$ f(x)=a_{1}(\cosh \zeta x-\cos \zeta x)+a_{2}(\sinh \zeta x-\sin \zeta x) $

Therefore, the boundary condition of Eq.(11) can be written as follows:

$ \left\{ {\begin{array}{*{20}{c}} {{a_2}\zeta (\cosh \zeta + \cos \zeta ) + {a_1}\zeta (\sinh \zeta - \sin \zeta ) = }\\ {\left( {1 + \delta {e^{ - {\rm{i}}\tau {\zeta ^2}}}} \right){\rm{i}} \times \left[ {{a_2}(\sinh \zeta - \sin \zeta ) + } \right.}\\ {{a_1}(\cosh \zeta - \cos \zeta )]}\\ {{a_1}(\cosh \zeta + \cos \zeta ) + {a_2}(\sinh \zeta + \sin \zeta ) = 0} \end{array}} \right. $

(13)

From calculating simple calculation, the determinant Δ(ζ) is obtained as follows:

$ \Delta (\zeta ) = \left( {1 + \delta {e^{ - {\rm{i}}\tau {\zeta ^2}}}} \right)(\cosh \zeta \sin \zeta - \sinh \zeta \cos \zeta ) + \zeta(\cosh \zeta \cos \zeta+1) $

If we want to get the non-trivial solution of Eq.(13), Δ(ζ) must satisfies Δ(ζ)=0 and ζ≠0. Hence, the solution of Eq.(13) is:

$ \begin{array}{c}{a_{1}=-(\sinh \zeta+\sin \zeta)} \\ {a_{2}=\cosh \zeta+\cos \zeta}\end{array} $

Therefore, we have:

$ \begin{aligned} f(x)=&(\cos \zeta+\cosh \zeta)(\sinh \zeta x-\sin \zeta x)-\\ &(\sin \zeta+\sinh \zeta)(\cosh \zeta x-\cos \zeta x) \end{aligned} $

One vector

$ W_{\lambda}=\left(f(x), \lambda f(x), \lambda f(1) e^{-\tau \lambda x}\right) $

which satisfies the Eq.(11) is an eigenvector of A. Hence

$ \sigma(A)=\sigma_{p}(A)=\left\{\lambda={\rm i} \zeta^{2} | \Delta \zeta=0, \zeta \neq 0\right\} $

The proof is completed.

Theorem 4   If operator A is defined by Eq.(5), we have σ(A)={λ_n=iζ_n²}, where

$ \zeta_{n}=\left(n-\frac{1}{2}\right) \pi+o\left(\frac{1}{n}\right) $

When n is large enough, the asymptotic expression of λ_n is:

$ \begin{aligned} \lambda_{n}=& \mathrm{i} \zeta_{n}^{2}=-2\left(1+\delta e^{-\mathrm{i}\left(n-\frac{1}{2}\right)^{2} \pi^{2} \tau^{2}}\right)+\\ &\left(n-\frac{1}{2}\right)^{2} \pi^{2} \mathrm{i}+o\left(\frac{1}{n}\right) \end{aligned} $

Proof   From the result of Theorem 2, we get that the distribution of the spectrum of A is symmetrical in relation to the real axis. Hence, asymptotic zeros of Δζ which should satisfy arg ζ∈(-π/4, π/4). However, we only need to study the ζ when arg ζ∈(-ε, ε) because Δζ has no zero when Δζ∈(-π/4, ε)∪(ε, π/4).

When ζ is large enough, we get:

$ \left\{ {\begin{array}{*{20}{l}} {\cos \zeta = \frac{{(\cos \zeta - \sin \zeta )\left( {\delta {e^{ - {\rm{i}}{\tau ^2}{\zeta ^2}}} + 1} \right){\rm{i}}}}{\zeta } + o\left( {{e^{ - R\zeta }}} \right)}\\ {\cos \zeta = o\left( {\frac{1}{{|\zeta |}}} \right)} \end{array}} \right. $

(14)

From the solution of second equation of Eq.(14), we can see:

$ \zeta_{n}=\left(-\frac{1}{2}+n\right) \pi+o\left(\frac{1}{n}\right) $

(15)

Substituting Eq.(15) into the first equation of Eq.(14), we have:

$ \begin{array}{*{20}{c}} {{{( - 1)}^n}o\left( {\frac{1}{n}} \right) = \frac{{\left( {\delta {e^{ - {\rm{i}}{\tau ^2}{\zeta _n}^2}} + 1} \right){\rm{i}}}}{{{\zeta _n}}} \times }\\ {\left[ {{{( - 1)}^n}\left( {1 + o\left( {\frac{1}{n}} \right) + o\left( {\frac{1}{{{n^2}}}} \right)} \right)} \right]} \end{array} $

which implies

$ o\left( {\frac{1}{n}} \right) = \frac{{ - \left( {\delta {e^{ - {\rm{i}}{\tau ^2}{\zeta _n}^2}} + 1} \right){\rm{i}}}}{{{\zeta _n}}} = \frac{{\left( {\delta {e^{ - {\rm{i}}{\tau ^2}{\zeta _n}^2}} + 1} \right){\rm{i}}}}{{\left( {\frac{1}{2} - n} \right)\pi }} + o\left( {\frac{1}{{{n^2}}}} \right) $

Hence

$ \zeta_{n}=\left(n-\frac{1}{2}\right) \pi+\frac{\left(\delta e^{-\mathrm{i} \tau^{2}\left(n-\frac{1}{2}\right) 2 \pi^{2}}+1\right) \mathrm{i}}{\left(\frac{1}{2}-n\right) \pi}+o\left(\frac{1}{n^{2}}\right) $

Then we have:

$ \begin{aligned} \lambda_{n}=& {\rm i} \zeta_{n}^{2}=-2\left(1+\delta e^{-{\rm i }\tau^{2}\left(n-\frac{1}{2}\right)^2 \pi^{2}}\right)+\\ &\left(n-\frac{1}{2}\right)^{2} \pi^{2} {\rm i}+o\left(\frac{1}{n}\right) \end{aligned} $

The proof is completed.

Therefore, for any τ>0, the real part of eigenvalue of A can be written as follows by the Theorem 4.

$ \begin{aligned} \operatorname{Re} \lambda=&-2\left(\delta \cos \left(\frac{1}{2}-n\right)^{2} \tau^{2} \pi^{2}+1\right)+O\left(\frac{1}{n}\right)<\\ &-2(1-|\delta|)<0 \end{aligned} $

4 Exponential Stability of the Closed Loop System

In this section, the exponential stability of the closed loop system is studied. From the proof of Theorem 2 and Theorem 4, we get that the operator A is dissipative. However, this is not enough to assert the exponential stability of the closed loop system. Hence the exponential stability is studied in this section based on Lyapunov function method^[20-21].

We construct two functions as follows:

$ \begin{aligned} E(t)=& \frac{|\delta|}{2} \tau \int_{0}^{1}\left|w_{t}(1, t-s \tau)\right|^{2} \mathrm{d} s+\\ & \frac{1}{2} \int_{0}^{1}\left[{w_t}^{2}(x, t)+w_{x x}^{2}(x, t)\right] \mathrm{d} x \\ & \rho(t)=\int_{0}^{1} x w_{t}(x, t) w_{x}(x, t) \mathrm{d} x \end{aligned} $

Hence, we obtain:

$ \begin{aligned} \frac{{{\rm{d}}E(t)}}{{{\rm{d}}t}} = &\int_0^1 {\left[ {{w_{xxt}}(x, t){w_{xx}}(x, t) + {w_{tt}}(x, t) \cdot } \right.} \\ &{w_t}(x, t)]{\rm{d}}x + |\delta |\tau \int_0^1 {{w_{tt}}} (1, t - \\ &s\tau ){w_t}(1, t - s\tau ){\rm{d}}s = - {w_t}(1, t) \cdot \\ &{w_{xxx}}(1, t) + \frac{{|\delta |}}{2}w_t^2(1, t) - \\ &\frac{{|\delta |}}{2}w_t^2(1, t - \tau ) \end{aligned} $

$ \begin{aligned} \frac{\mathrm{d} \rho(t)}{\mathrm{d} t}=& \int_{0}^{1} x\left[w_{t t}(x, t) w_{x}(x, t)+\right.\\ & w_{t}(x, t) w_{x t}(x, t) ] \mathrm{d} x=\\ &-\int_{0}^{1} x w_{x}(x, t) w_{x x x x}(x, t) \mathrm{d} x+ \\ &\int_{0}^{1} x w_{t}(x, t) w_{x t}(x, t) \mathrm{d} x= \\ &-\frac{1}{2} \int_{0}^{1}\left[3 w_{x x}^{2}(x, t)+w_{t}^{2}(x, t)\right] \mathrm{d} x+ \\ &-\frac{1}{2} w_{t}^{2}(1, t)-w_{x}(1, t)\left[\delta w_{t}(1, t-\tau)+\right. \\ &w_{t}(1, t) ] \end{aligned} $

Then the following equations are obtained by Cauchy-Schwartz inequality.

$ \begin{array}{l}{-w_{t}(1, t) w_{x}(1, t)=-w_{t}(1, t) \int_{0}^{1} w_{x x}(x, t) \mathrm{d} x \leqslant} \\ ~~~~~{\frac{1}{2}\left[\frac{1}{\gamma} \int_{0}^{1} w_{x x}^{2}(x, t) \mathrm{d} x+\gamma\left|w_{t}(1, t)\right|^{2}\right]} \\ {-\delta w_{t}(1, t-\tau) w_{x}(1, t)=-\delta w_{t}(1, t-} \\ ~~~~~{\tau ) \int_{0}^{1} w_{x x}(x, t) \mathrm{d} x \leqslant \frac{|\delta|}{2}\left[\frac{1}{\gamma} \int_{0}^{1} w_{x x}^{2}(x, t) \mathrm{d} x+\right.} \\ ~~~~~{\gamma\left|w_{t}(1, t-\tau)\right|^{2} ]}\end{array} $

where γ is constant. Thus, we have:

$ \begin{aligned} \frac{\mathrm{d} \rho(t)}{\mathrm{d} t} \leqslant &-\frac{1}{2} \int_{0}^{1}\left[w_{x x}^{2}(x, t)+w_{t}^{2}(x, t)\right] \mathrm{d} x+\\ & \frac{|\delta| \gamma}{2} w_{t}^{2}(1, t-\tau)+\frac{1+|\delta|-2 \gamma}{2 \gamma}\cdot \\ & \int_{0}^{1} w_{x x}^{2}(x, t) \mathrm{d} x+\frac{1+\gamma}{2} w_{t}^{2}(1, t) \end{aligned} $

Then we construct one Lyapunov function

$ V(t)=\varepsilon \rho(t)+E(t)-\varepsilon|\delta| \tau \int_{0}^{1} s w_{t}^{2}(1, t-s \tau) \mathrm{d} s $

In which, ε>0, from the calculation,

$ \begin{aligned} \frac{\mathrm{d} V(t)}{\mathrm{d} t}=& \varepsilon \frac{\mathrm{d} \rho(t)}{\mathrm{d} t}+\frac{\mathrm{d} E(t)}{\mathrm{d} t}-2|\delta| \tau \varepsilon \int_{0}^{1} s w_{tt}(1, t-\\ & s \tau ) w_{t}(1, t-s \tau) \mathrm{d} s=-\delta w_{t}(1, t-\\ & \tau ) w_{t}(1, t)-\frac{|\delta|}{2} w_{t}^{2}(1, t-\tau)+\\ & \frac{|\delta|}{2} w_{t}^{2}(1, t)+\frac{\varepsilon}{2} w_{t}^{2}(1, t)-\\ & \varepsilon w_{x}(1, t)\left[\delta w_{t}(1, t-\tau)+w_{t}(1, t)\right]-\\ &\frac{\varepsilon }{2}\int_0^1 {\left[ {w_t^2(x, t) + 3w_{xx}^2(x, t)} \right]} {\rm{d}}x - \\&w_t^2(1, t) + \varepsilon |\delta |w_t^2(1, t - \tau ) - \\&\varepsilon |\delta |\int_0^1 {w_t^2} (1, t - s\tau ){\rm{d}}s \le - \left[ {1 - \frac{{|\delta |}}{2} - } \right.\\&\varepsilon \frac{{1 + \gamma }}{2}]w_t^2(1, t) - \delta {w_t}(1, t - \tau ){w_t}(1, t) - \\&[\frac{{|\delta |}}{2} - \varepsilon \frac{{\gamma |\delta |}}{2} - \varepsilon |\delta |]w_t^2(1, t - \\&\tau ) - \varepsilon \frac{1}{2}\int_0^1 {\left[ {w_t^2(x, t) + w_{xx}^2(x, t)} \right]} {\rm{d}}x - \\ &\varepsilon |\delta |\int_0^1 {w_t^2} (1, t - s\tau ){\rm{d}}s + \varepsilon \frac{{1 + |\delta | - 2\gamma }}{{2\gamma }} \cdot \\&\int_0^1 {w_{xx}^2} (x, t){\rm{d}}x \le [ - 1 + \frac{{|\delta |}}{2} + \varepsilon \frac{{1 + \gamma }}{2} + \\&\frac{{|\delta |}}{{2\eta }}]w_t^2(1, t) - \varepsilon \frac{1}{2}\int_0^1 {\left[ {w_t^2(x, t) + } \right.} \\&w_{xx}^2(x, t)]{\rm{d}}x - \left[ {\frac{{|\delta |}}{2} - \varepsilon \frac{{\gamma |\delta |}}{2} - \varepsilon |\delta | - } \right.\\&\frac{{|\delta |\eta }}{2}]w_t^2(1, t - \tau ) - |\delta |\varepsilon \int_0^1 {w_t^2} (1, t - \\&s\tau ){\rm{d}}s + \varepsilon \frac{{1 + |\delta | - 2\gamma }}{{2\gamma }}\int_0^1 {w_{xx}^2} (x, t){\rm{d}}x \end{aligned} $

where γ, η are constants. Let γ=(1+|δ|)/2 and η=k/(k+1) when k is large enough, it satisfies

$ \frac{2 k+1}{k}|\delta|-1<0 $

and

$ \begin{array}{*{20}{c}} {0 < \varepsilon < \min \{ [1 - \frac{{(2k + 1)|\delta |}}{{2k}}]\frac{4}{{3 + |\delta |}}, }\\ {\frac{2}{{(k + 1)(5 + |\delta |)}}\} < 1} \end{array} $

Hence, we have:

$ \begin{aligned} \frac{\mathrm{d} V(t)}{\mathrm{d} t}=& \varepsilon \frac{\mathrm{d} \rho(t)}{\mathrm{d} t}+\frac{\mathrm{d} E(t)}{\mathrm{d} t}-2 \varepsilon \tau|\delta| \int_{0}^{1} s w_{tt}(1, t-\\ & s \tau ) w_{t}(1, t-s \tau) \mathrm{d} s \leqslant\left[\frac{(2 k+1)|\delta|}{2 k}+\right.\\ & \varepsilon \frac{3+|\delta|}{4}-1 ] w_{t}^{2}(1, t)+\left[\frac{-|\delta|}{2(k+1)}+\right.\\ & \varepsilon|\delta| \frac{1+|\delta|}{4}-\varepsilon|\delta| ] w_{t}^{2}(1, t-\\ & \tau )-\frac{\varepsilon}{2} \int_{0}^{1}\left[w_{x x}^{2}(x, t)+w_{t}^{2}(x, t)\right] \mathrm{d} x-\\&|\delta| \varepsilon \int_{0}^{1} w_{t}^{2}(1, t-s \tau) \mathrm{d} s \leqslant\\&-|\delta| \varepsilon \int_{0}^{1} w_{t}^{2}(1, t-s \tau) \mathrm{d} s-\\&\varepsilon \frac{1}{2} \int_{0}^{1}\left[w_{t}^{2}(x, t)+w_{x x}^{2}(x, t)\right] \mathrm{d} x=\\&\left[\frac{|\delta| \tau \varepsilon}{2}-|\delta| \varepsilon\right] \int_{0}^{1} w_{t}^{2}(1, t-\\&s \tau ) \mathrm{d} s-E(t) \varepsilon \leqslant\left\{\begin{array}{ll}{-E(t) \varepsilon, } & {0 \leqslant \tau \leqslant 2} \\ {\frac{\varepsilon E(t)}{2}, } & {\tau>2}\end{array}\right.\end{aligned} $

Thus

$ \left| {|\delta |\tau \int_0^1 s w_t^2(1, t - s\tau ){\rm{d}}s - \rho (t)} \right| \le E(t) $

We could get that:

$ 0<E(t)(1-\varepsilon) \leqslant V(t) \leqslant E(t)(1+\varepsilon)\\ \frac{\mathrm{d} V(t)}{\mathrm{d} t} \leqslant\left\{\begin{array}{ll}{\frac{-\varepsilon}{1+\varepsilon} V(t), } & {0 \leqslant \tau \leqslant 2} \\ {\frac{-\varepsilon}{\tau(1+\varepsilon)} V(t), } & {\tau>2}\end{array}\right.\\ V(t) \le \left\{ {\begin{array}{*{20}{l}} {V(0){e^{\frac{{ - \varepsilon t}}{{1 + \varepsilon }}}}, }&{0 < \tau \le 2}\\ {V(0){e^{\frac{{ - \varepsilon t}}{{\tau (1 + \varepsilon )}}}}, }&{\tau > 2} \end{array}} \right. $

Then we know that the system (2) is exponential stability. The control strategy is useful to deal with the input delay.

5 Numerical Simulation

In this section, numerical simulations are given to illustrate the theoretical results.

The Chebyshev spectral method is applied to calculate the displacement numerically for system (2) with feedback control. The parameter δ in system (1) is taken as 0.5.

The step size of time is 0.005. The results of numerical simulation are listed below.

Fig. 1 is the simulation of w(1, t). Fig. 2 is the simulation of w(x, t).

6 Conclusions

In this paper, we investigate the stability of one Euler beam with input delay. The feedback control strategy is used to stabilize the system. The semi-group theory and spectral analysis are used to prove the system is dissipate. Thus, The Lyapunov function is constructed to prove the stability of the system. Then we could get that the feedback control is positive for the stability of the system with time delay. Finally, the numerical simulation is used to show the effect of the feedback controller.