# Boundary Output Feedback Stabilization for a Cascaded-Wave PDE-ODE System with Velocity Recirculation

Authors: Ruicheng Li; Feng-Fei Jin; Baoqiang Yan
Journal: Complexity
 (2021)

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## Abstract

This paper considers the output feedback stabilization for a cascaded-wave PDE-ODE system with velocity recirculation by boundary control. First, we choose a well-known exponentially stable system as its target system and find a backstepping transformation to design a state feedback controller for the original system. Second, we attempt to give an output feedback controller for the original system by introducing the observer. The resulting closed-loop system admits a unique solution which is proved to be exponentially stable. Finally, we give some numerical examples to prove the validity for the theoretical results.


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## Body

## 1. Introduction

In control engineering, ordinary differential equations (ODEs) and partial differential equations (PDEs) are widely used to model such problems. Many researchers have done detailed research on this content in the past decades. Systems modeled by ODEs are common in this respect. On the other hand, there are more and more works that contributed to stabilization for systems described by PDEs in recent years ([1–7]). For instance, aiming at solving the problem of system with variable coefficients and Neumann boundary actuation, Wu et al. used the backstepping approach that converted the wave equation into a same type equation in [8]. Not only that, the coupled PDE-ODE problem was also studied in control systems, where PDEs mainly included heat equations and wave equations. With the development of research on such problems, more results of PDE-ODE emerged. For example, in [9–16], the authors solved the stabilization for heat PDE-ODE by boundary control systems which used the PDE backstepping method. In [10, 13], Tang and Xie considered the systems where ζl,t=Ut in boundary of “PDE part.” The system of ζl,t=Ut−d in “PDE part” was considered in [12]. The stabilization for wave PDE-ODE systems by the PDE backstepping method was solved in [6, 7, 17, 18]. But beyond that, in [19], Meglio solved the problem of stabilizing a linear ODE coupled first-order hyperbolic PDE, which was also based on the backstepping approach. This paper proposed a unified framework of iterative learning control for typical flexible structures under spatiotemporally varying disturbances in [20]. In [21], the control problem was addressed for a hybrid PDE-ODE system that described a nonuniform gantry crane system with constrained tension. In addition, the PDE-ODE models also played a significant role in practical application, such as 3D-printing [22], oil drilling [23], cable elevator [24], battery management [25], traffic [26], and so on.In fact, the backstepping method played an important role in solving many PDEs and PDE-ODE problems, which were introduced by Krstic [27–30]. This approach was dedicated to finding an invertible Volterra integral transformation which mapped the considered plant with the boundary feedback law into a known stable target system. About some PDE systems, Jin and Guo also used the backstepping method to solve the PDEs system of nonlocal terms in [31]. In [32], Hasan and Tang used the backstepping method to model boundary stabilization of the Korteweg–De Vries (KdV) equation with sensors and actuators. For some PDE-ODE systems, the authors also used the backstepping approach to design the controller in [6, 7, 9–18]. In [17], Zhou et al. used directly the known backstepping method (from [30]) to achieve the stabilization of system matched disturbance with boundary control. It can be said that backstepping is a significant method and key move to deal with such PDE and PDE-ODE problems.In this paper, our main focus is on the output feedback stabilization for a cascaded-wave PDE-ODE system subject to boundary control with velocity recirculation:(1)X˙t=DXt+Fζ0,t,t>0,ζttx,t=ζxxx,t+qζt0,t,x∈0,1,t>0,ζx0,t=0,t≥0,ζx1,t=U¯t,t≥0,ζx,0=ζ0x,ζtx,0=ζ1x,X0=X0,x∈0,1,yt=QXt,ζ0,t,ζt0,t,t≥0,where yt is the output (measurement), U¯t∈Lloc20,∞ is the control input of the entire system, X0,ζ0,ζ1 is the initial value, q∈ℝ is a known constant, D∈ℝn×n, and F∈ℝn×1. We suppose that ∑D,F is stabilizable and Q,D is observable. System (1) is considered in the state space ℋ=ℝn×H10,1×L20,1.On some special cases, we consider the following. For system (1) without “X part,” Su et al. gave the backstepping method and the state feedback controller in [33], which was ζttx,t=ζxxx,t+qζt0,t. When q=0, by the backstepping method shown in [30], one can easily design a control law to stabilize system (1). And in [16], Zhou and Xu used backstepping transformation to settle the problem of coupled PDE-ODE where there was no non-local term in heat equation. On such coupling problems, there was few research on PDE-ODE problem with non-local term. Non-local terms, including both boundary terms and strict feedback/Volterra terms, have been the mainstay of the applications of PDE backstepping methods to parabolic PDEs. Equation ζttx,t=ζxxx,t+qutx0,t models the string vibration of an electric guitar with a pickup at the location x0∈0,1. The move of x0 leads to the location change of the pickup. Inspired by it, we let q=0 in ζttx,t=ζxxx,t+qutx0,t. Compared with [30], we add a non-local term qζt0,t to the right end of the wave equation. This makes it more difficult for us to research this problem. In [30], the form of backstepping transformation is “vx,t=ux,t−∫0xkx,yuy,tdy−∫0xlx,yuty,tdy−γxXt”; it does not work to solve the problem in their way. Inspired by (4) in [34], we obtain the new form of backstepping transformation. Using the backstepping and introducing the exponentially stable target system to design state feedback controller are also nice methods in this paper.The paper is organized as follows. We design the state feedback controller for system (1) in Section 2. Section 3 is contributed to designing the output feedback control and proving the exponential stability of the closed-loop system. In Section 4, we give some numerical examples to prove the validity of the proposed controller.


## 2. State Feedback Controller Design

In this section, we are devoted to designing a state feedback controller for system (1). First, we introduce the backstepping approach:(2)Xt=Xt,vx,t=ζx,t−∫0xkx,yζy,tdy−∫0xlx,yζty,tdy−μxζ0,t−γxXt,where the kernel functions kx,y, lx,y, μx, and γx will be defined later. Compared with the backstepping in [30], we add an additional term “−μxζ0,t” in transformation (2). And we hope that turns system (1) into the target system(3)X˙t=D+FGXt+Fv0,t,vttx,t=vxxx,t,vx0,t=0,vx1,t=−c1vt1,t−c2v1,t,vx,0=v0,vtx,0=v1,X0=X0,where c1,c2>0 are tuning parameters and G is chosen such that D+FG is Hurwitz.It is well known that there exists a exponentially stable solution for “v part” in the state space ℋ1=H10,1×L20,1, which is equipped with the norm induced by inner product(4)f1,g1⊤,f2,g2⊤ℋ1=∫01f1′xf2′x+g1xg2xdx+c2f11f21,∀f1,g1⊤,f2,g2⊤∈ℋ1.In order to get these kernel functions, we make the following calculation. Taking derivative of (2) with respect to x, we can obtain(5)vxx,t=ζxx,t−kx,xζx,t−∫0xkxx,yζy,tdy−lx,xζtx,t−∫0xlxx,yζty,tdy−μ′xζ0,t−γ′xXt.Taking derivative of (2) with respect to x twice, we find(6)vxxx,t=ζxxx,t−ddxkx,xζx,t−kx,xζxx,t−kxx,xζx,t−∫0xkxxx,yζy,tdy−ddxlx,xζtx,t−lx,xζxtx,t−lxx,xζtx,t−∫0xlxxx,yζty,tdy−μ″xζ0,t−γ″xXt.In the same way, taking derivative of (2) with respect to t and taking derivative of (2) with respect to t twice and using the partial integral method, we can get the following:(7)X˙t=DXt+Fζ0,t,vtx,t=ζtx,t−∫0xkx,yζty,tdy−∫0xlx,yζtty,tdy−μxζt0,t−γxX˙t=ζtx,t−∫0xkx,yζty,tdy−lx,xζxx,t+lyx,xζx,t−lyx,0ζ0,t−∫0xlyyx,yζy,tdy−qζt0,t∫0xlx,ydy−μxζt0,t−γxDXt−γxFζ0,t,X¨t=D2Xt+DFζ0,t+Fζt0,t,vttx,t=ζttx,t−∫0xkx,yζtty,tdy−∫0xlx,yζttty,tdy−μxζtt0,t−γxX¨t=ζxxx,t+qζt0,t−kx,xζxx,t+kyx,xζx,t−kyx,0ζ0,t−∫0xkyyx,yζy,tdy−qζt0,t∫0xkx,ydy−lx,xζxtx,t+lyx,xζtx,t−lyx,0ζt0,t−∫0xlyyx,yζty,tdy−qζtt0,t∫0xlx,ydy−μxζtt0,t−γxD2Xt−γxDFζ0,t−γxFζt0,t.Then, we have(8)vttx,t−vxxx,t=ddxlx,x+lxx,x+lyx,xζtx,t+ddxkx,x+kxx,x+kyx,xζx,t+∫0xkxxx,y−kyyx,yζy,tdy+∫0xlxxx,y−lyyx,yζty,tdy−kyx,0+γxDF−μ″xζ0,t+γ″x−γxD2Xt−q∫0xlx,ydy+μxζtt0,t+q−q∫0xkx,ydy−lyx,0−γxFζt0,t=0.By the boundary conditionvx0,t=0 in (3) and (5), we obtain(9)vx0,t=−k0,0ζ0,t−l0,0ζt0,t−γ′0Xt−μ′0ζ0,t=−k0,0−μ′0ζ0,t−l0,0ζt0,t−γ′0Xt=0.Hence, we choose the functionskx,y,lx,y,γx,μx to satisfy(10)γ″x=γxD2,γ′0=0,γ0=G,kxxx,y=kyyx,y,kx,x=0,kyx,0=−γxDF−μ″x,lxxx,y=lyyx,y,lx,x=0,lyx,0=q−q∫0xkx,ydy−γxF,q∫0xlx,ydy+μx=0.We can obtainμ0=0 by q∫0xlx,ydy+μx=0. In [30], we find that the solution of γx is as follows:(11)γx=GI0e0D2I0xI0.By (2) and (10), we have ζ0,t=v0,t−μζ0,t−γ0Xt=v0,t+GXt; then, the first equation in (3) is obtained.Next we can letkx,y=k1x−y,lx,y=l1x−y; by “k part” in (10), we can have(12)kyx,0=−k1′x=−γxDF−μ″x,k1x=∫0xγξdξDF+μ″x=∫0xγξdξDF+μ′x−μ′0.Throughq∫0xlx,ydy+μx=0, we get(13)μx=−q∫0xlx,ydy,μ′x=−q∫0xlxx,ydy=−q∫0xl1′x−ydy=ql10−ql1x.Becauseμ′x=−ql1x,lx,x=0, we can acquire l0,0=l10=0,μ′0=0. According to the above formulas, we obtain(14)k1x=∫0xγξdξDF−q∫0xγξdξF+q2x−q2∫0x∫0zk1z−ydydz,(15)l1x=−qx+q∫0x∫0zk1z−ydydz+∫0xγξdξF,(16)μx=−q∫0xlx,ydy=−q∫0xl1x−ydy.According tokx,y=k1x−y,lx,y=l1x−y, simplify the above three formulas in (14)(16). Let δ=z−y,ϑ=x−y; we get ∫0x∫0zk1z−ydydz=∫0x∫0zk1δdδdz, ∫0xl1x−ydy=∫0xl1ϑdϑ, and(17)k1x=∫0xγξdξDF−q∫0xγξdξF+q2x−q2∫0x∫0zk1δdδdz,(18)l1x=−qx+q∫0x∫0zk1δdδdz+∫0xγξdξF=−qx+q∫0xx−δk1δdδ+∫0xγξdξF,(19)μx=−q∫0xl1ϑdϑ.By differentiating (17) with respect to x twice, we transform (17) into a second-order linear ODE and get the solution of k1x:(20)k1x=kqDF−qF+qsinqx−cosqxq∫0xsinqξ⋅γ′ξdξDF−qF+sinqxq∫0xcosqξ⋅γ′ξdξDF−qF.We substitute (20) for (18) and obtain the solution of l1x. In the same way, the solution of μx can also be obtained.Becauseγ⋅∈C10,1, ∫0xγξdξDF∈C20,1. ∫0xγξdξF∈C20,1, q2x∈C20,1, k1δ∈C0,1, so k1⋅∈C20,1, and l1⋅,μ⋅∈C20,1.We suppose that the inverse transformation of (2) exists in the form(21)Xt=Xt,ζx,t=vx,t+∫0xmx,yvy,tdy+∫0xnx,yvty,tdy+hxXt,where mx,y,nx,y,hx are the kernel functions.Derive (21) with respect to x, and we obtain(22)ζxx,t=vxx,t+mx,xvx,t+∫0xmxx,yvy,tdy+nx,xvtx,t+∫0xnxx,yvty,tdy+h′xXt.Differentiating (21) with respect to x twice, we have(23)ζxxx,t=vxxx,t+ddxmx,xvx,t+mx,xvxx,t+mxx,xvx,t+∫0xmxxx,yvy,tdy+ddxnx,xvtx,t+nx,xvxtx,t+nxx,xvtx,t+∫0xnxxx,yvty,tdy−h″xXt.We differentiate (21) with respect to t and get(24)X˙t=D+FGXt+Fv0,t,ζtx,t=vtx,t+∫0xmxx,yvty,tdy+∫0xnx,yvtty,tdy+hxX˙t=vtx,t+∫0xmxx,yvty,tdy+nx,xvxx,t−nyx,xvx,t+nyx,0v0,t+∫0xnyyx,yvy,tdy+hxD+FGXt+hxFv0,t.Differentiating (21) twice on both sides with respect to t yields(25)X¨t=D+FG2Xt+D+FGFv0,t+Fvt0,t,ζttx,t=vttx,t+∫0xmx,yvtty,tdy+∫0xnx,yvttty,tdy+hxX¨t.Then, we have(26)ζttx,t−ζxxx,t=∫0xmyyx,y−mxxx,yvy,tdy+∫0xnyyx,y−nxxx,yvty,tdy−ddxmx,x+mxx,x+myx,xvx,t−ddxnx,x+nxx,x+nyx,xvtx,t+myx,0+hxD+FGFv0,t+nyx,0+hxFvt0,t+hxD+FG2−h″xXt=0.The boundary condition atx=0 in (1) implies(27)ζx0,t=m0,0v0,t+n0,0vt0,t+h′0Xt=0.Hence, we choose the functionsmx,y,nx,y,lx to satisfy(28)h″x=hxD+FG2,h′0=0,h0=G,mxxx,y=myyx,y,mx,x=0,myx,0=−hxD+FGF,nxxx,y=nyyx,y,nx,x=0,nyx,0=−hxF.We obtain the solution ofhx in [30]:(29)hx=GI0e0D+FG2I0xI0.Letmx,y=m1x−y,nx,y=n1x−y; then,(30)myx,0=−m1′x=−hxD+FGF,m1x=∫0xhξdξD+FGF,nyx,0=−n1′x=−hxF,n1x=∫0xhξdξF.Combining the forth equations in (1) and (3), we can obtain the controller U¯ for (1) (31)U¯t=∫01kx1,yζy,tdy+∫01lx1,yζty,tdy+μ′1ζ0,t+γ′1Xt−c1ζt1,t−∫01k1,yζty,tdy+ly1,1ζ1,t−ly1,0ζ0,t−∫01lyy1,yζy,tdy−qζt0,t∫01l1,ydy−μ1ζt0,t−γ1DXt−γ1Fζ0,t−c2ζ1,t−∫01k1,yζy,tdy−∫01l1,yζty,tdy−μ1ζ0,t−γ1Xt=∫01kx1,y+c1lyy1,y+c2k1,yζy,tdy+∫01lx1,y+c1k1,y+c2l1,yζty,tdy+μ′1+c1ly1,0+c1γ1F+c2μ1ζ0,t+c1q∫01l1,ydy+c1μ1ζt0,t−c1ly1,1+c2ζ1,t−c1ζt1,t+c1γ1D+c2γ1+γ′1Xt.For convenience, we describe system (1) under controller (31) as(32)X˙t=DXt+Fζ0,t,ζttx,t=ζxxx,t+qζt0,t,ζx0,t=0,ζx1,t=∫01kx1,y+c1lyy1,y+c2k1,yζy,tdy+∫01lx1,y+c1k1,y+c2l1,yζty,tdy+μ′1+c1ly1,0+c1γ1F+c2μ1ζ0,t+c1q∫01l1,ydy+c1μ1ζt0,t−c1ly1,1+c2ζ1,t−c1ζt1,t+c1γ1D+c2γ1+γ′1Xt.Define the state spaceℋ=ℝn×H10,1×L20,1 for system (32) with the norm(33)X,f,gℋ2=∫01f′x2+gx2dx+k1Xℝn2+k2f02,∀X,f,g∈ℋ,where k1,k2>0.Theorem 1.
Assume thatc1,c2>0andGis chosen such thatD+FGis Hurwitz. For any initial condition such thatX0,ζ⋅,0,ζt⋅,0⊤∈ℋ, closed-loop system (32) admits a unique solutionXt,ζ⋅,t,ζt⋅,t⊤∈C0,∞;ℋ. Moreover, closed-loop system (32) is exponentially stable: there exist two positive constantsM,εdepending onc1,c2andGsuch that(34)Xt,ζ⋅,t,ζt⋅,t⊤ℋ≤Me−εtX0,ζ0,ζ1⊤ℋ,∀t≥0.Proof.
It is a direct result from the equivalence between system (3) and (32). Obviously, we define the operator A1:DA1⟶ℋ1 for the “v part” of target system (3):(35)A1f,g⊤=g,f″⊤,DA1=f,g⊤∈H20,1×H10,1|A1f,g⊤∈ℋ1,f′0=0,f′1=−c1g1−c2f1.
Then the “v part” of system (3) can be written as an abstract evolutionary equation in ℋ1=H10,1×L20,1:(36)ddtv⋅,t,vt⋅,t⊤=A1v⋅,t,vt⋅,t⊤.
As we all know, the operatorA1 can generate an exponentially stable C0-semigroup eA1t on ℋ1 [35]. In other words, there exist M1,α>0 such that(37)eA1t≤M1e−αt,∀t>0.
Next we consider the “X part” of target system (3) on ℝn. Because v0,t∈C0,∞, using the constant variational formula, the solution of “X part” is given by(38)Xt=eD+FGtX0+∫0teD+FGt−sFv0,sds.
BecauseD+FG is Hurwitz, there exist positive constants C5,θ>0 such that(39)eD+FGt≤C5e−θt.
From Theorem1 in [17], the solution of ”X part” is exponentially stable. So, system (3) admits a unique solution and the solution is exponentially stable. According to transformations (2) and (21), we define a bounded invertible operator P:X,f,g⟶X,φ,ψ in ℋ:(40)Xφxψx=Xfx−∫0xkx,yfydy−∫0xlx,ygydy−μxf0−γxXgx−∫0xkx,ygydy−lx,xf′x+lyx,xfx+t1x,Xfxgx=Xφx+∫0xmx,yφydy+∫0xnx,yψydy+hxXψx+∫0xmxx,yψydy+nx,xφ′x−nyx,xφx+t2x,where(41)t1x=−lyx,0f0−∫0xlyyx,yfydy−qg0∫0xlx,ydy−μxg0−γxDX+γxFf0,t2x=nyx,0φ0+∫0xnyyx,yφydy+hxD+FGX+hxFφ0,and kx,y,lx,y,μx,γx,mx,y,nx,y,hx are defined in (14)(16) and (30), respectively.
By the formula above, we have the following expression:(42)X,fx,gx⊤≤P−1X,φx,ψx⊤≤M2X,φx,ψx⊤.
Since the operatorP is bounded and invertible, there exists M2>0 such that P−1<M2. Besides, the “v part” admits a unique solution and the solution is exponentially stable, so the solution of closed-loop system (32) is exponentially stable by (42). There exist M,ε>0 such that (34) holds.


## 3. Observer and Output Feedback Controller Design

In this section, we focus on designing output feedback controller for system (1) on the grounds of the observer. Next, we give the observer for system (1).(43)X^˙t=DX^t+PQX^−QX+Fζ0,t,ζ^ttx,t=ζ^xxx,t+qζt0,t,ζ^x0,t=c3ζ^t0,t−ζt0,t+c4ζ^0,t−ζ0,t,ζ^x1,t=U¯t,ζ^x,0=ζ^0,ζ^tx,0=ζ^1,X^0=X^0,where c3,c4 are positive design parameters.The errorX˜t,ζ˜x,t,ζ˜tx,t=X^t−Xt,ζ^x,t−ζx,t,ζ^tx,t−ζtx,t between system (1) and observer (43) is governed by(44)X˜˙t=D+PQX˜t,ζ˜ttx,t=ζ˜xxx,t,ζ˜x0,t=c3ζ˜t0,t+c4ζ˜0,t,ζ˜x1,t=0,ζ˜x,0=ζ˜0,ζ˜tx,0=ζ˜1,X˜0=X˜0.Define the state spaceℋ=ℝn×H10,1×L20,1 for system (44) with the norm(45)X˜,ζ˜,ζt˜ℋ2=∫01ζ˜t2+ζ˜x2dx+X˜ℝn2+c4f02,∀X˜,ζ˜,ζt˜∈ℋ.Theorem 2.
Suppose thatc3,c4>0. For any initial valueX˜0,ζ˜0,ζ˜1⊤∈ℋ, system44admits a unique exponentially stable solutionX˜t,ζ˜⋅,t,ζ˜t⋅,t⊤∈C0,∞;ℋin the sense that(46)Et≤C1E0e−ω1t,for some positive constantsC1andω1, whereEtis given by(47)Et≔12∫01ζ˜x2x,t+ζ˜t2x,tdx+c42ζ˜21,t+X˜ℝn2.Moreover,(48)jt≤M0e−ω1t,(49)∫t1t2rt2≤M1e−ω1t1+e−ω1t2,∀t2>t1>0,for M0,M1>0 and f⋅∈L20,∞, where(50)ft=∫01kx1,y+c1lyy1,y+c2k1,yζ˜y,tdy+∫01lx1,y+c1k1,y+c2l1,yζ˜ty,tdy+μ′1+c1ly1,0+c1γ1F+c2μ1ζ˜0,t+c1q∫01l1,ydy+c1μ1ζ˜t0,t−c1ly1,1+c2ζ˜1,t−c1ζ˜t1,t+c1γ1D+c2γ1+γ′1X˜t,rt=c1q∫01l1,ydy+c1μ1ζ˜t0,t−c1ζ˜t1,t,jt=ft−rt.Proof.
As we all know, the “ζ˜ part” of system (44) admits a unique exponentially stable solution from [33] and D+PQ is Hurwitz, so system (44) has a unique exponentially stable solution.
Notice that the energyEt−X˜ℝn2 is equivalent to ζ˜⋅,t,ζ˜t⋅,t⊤ℋ12. We only prove (48) and (49). By Cauchy inequality and Hölder inequality, we have(51)jt2≤C∫01ζ˜x2y,tdy+∫01ζ˜t2y,tdy+ζ˜20,t+X˜2t,for some constant C>0.
It is obvious that∫01ζ˜x2y,tdy+∫01ζ˜t2y,tdy+ζ˜20,t+X˜2t is equivalent to Et. Et is exponentially stable, so (48) holds. Next we prove (49).
DifferentiatingEt, we find(52)E˙t=−c3ζ˜t0,t2≤0.
Integrating above equation fromt1 to t2, we obtain(53)∫t1t2ζ˜t0,t2dx=−1c3∫t1t2E˙sds=1c3Et1−Et2≤1c3C1E0e−ω1t1+e−ω1t2.
Define another function(54)ρt=∫01xζ˜tx,tζ˜xx,tdx.
It is clear thatρt≤Et. Taking derivative of (54), we get(55)ρ˙t=12ζ˜t21,t−12∫01ζ˜t2x,t+ζ˜x2x,tdx.
Integrating above equation fromt1 to t2, we have(56)12∫t1t2ζ˜t1,tdt=ρt2−ρt1+12∫t1t2∫01ζ˜t2x,t+ζ˜x2x,tdxdt≤Et2+Et1+∫t1t2Etdt≤C2e−ω1t1+e−ω1t2,for some constant C2>0. Hence, (48) and (49) hold. Especially, when t1=0, ft∈L20,∞.
We give the output feedback controller about system (1):(57)U¯t=∫01kx1,y+c1lyy1,y+c2k1,yζ^y,tdy+∫01lx1,y+c1k1,y+c2l1,yζ^ty,tdy+μ′1+c1ly1,0+c1γ1F+c2μ1ζ^0,t+c1q∫01l1,ydy+c1μ1ζ^t0,t−c1ly1,1+c2ζ^1,t−c1ζ^t1,t+c1γ1D+c2γ1+γ′1X^t.
In terms of control law (57), closed-loop system (1) can be described as(58)X˙t=DXt+Fζ0,t,ζttx,t=ζxxx,t+qζt0,t,ζx0,t=0,ζx1,t=ζ^x1,t,X^˙t=DX^t+PQX^−QXFζ^0,t,ζ^ttx,t=ζ^xxx,t+qζt0,t,ζ^x0,t=c3ζ^t0,t−ζt0,t+c4ζ^0,t−ζ0,t,ζ^x1,t=∫01kx1,y+c1lyy1,y+c2k1,yζ^y,tdy+∫01lx1,y+c1k1,y+c2l1,yζ^ty,tdy+μ′1+c1ly1,0+c1γ1F+c2μ1ζ^0,t+c1q∫01l1,ydy+c1μ1ζ^t0,t−c1ly1,1+c2ζ^1,t−c1ζ^t1,t+c1γ1D+c2γ1+γ′1X^t,ζx,0=ζ0,ζtx,0=ζ1,X0=X0,ζ^x,0=ζ^0,ζ^tx,0=ζ^1,X^0=X^0.
We consider system (58) in the state space ℋ2.Theorem 3.
Assume thatc1,c2>0. For any initial valueX0,ζ⋅,0,ζt⋅,0,X^0,ζ^⋅,0,ζ^t⋅,0⊤∈ℋ2, system58admits a unique solutionXt,ζ⋅,t,ζt⋅,t,X^t,ζ^⋅,t,ζ^t⋅,t⊤∈C0,∞;ℋ2. Moreover, the solution of system58is exponentially stable in the sense that there exist two positive constantsM1,ε1such that(59)Xt,ζ⋅,t,ζt⋅,t,X^t,ζ^⋅,t,ζ^t⋅,t⊤ℋ2≤M1e−ε1tX0,ζ⋅,0,ζt⋅,0,X^0,ζ^⋅,0,ζ^t⋅,0⊤ℋ2,∀t≥0.Proof.
By introducing error variableX˜,ζ˜,ζ˜t, we transform system (58) into (44) and(60)X˙t=DXt+Fζ0,t,ζttx,t=ζxxx,t+qζt0,t,x∈0,1,ζx0,t=0,ζx1,t=∫01kx1,y+c1lyy1,y+c2k1,yζy,tdy+∫01lx1,y+c1k1,y+c2l1,yζty,tdy+μ′1+c1ly1,0+c1γ1F+c2μ1ζ0,t+c1q∫01l1,ydy+c1μ1ζt0,t−c1ly1,1+c2ζ1,t−c1ζt1,t+c1γ1D+c2γ1+γ′1Xt+ft,ζx,0=ζ0,ζtx,0=ζ1,X0=X0.
On account of transformation (3), system (60) can be mapped into the following target system:(61)X˙t=D+FGXt+Fv0,t,vttx,t=vxxx,t,vx0,t=0,vx1,t=−c1vt1,t−c2v1,t+ft,vx,0=v0x,vtx,0=v1x,X0=X0.
For “v part”, we can describe it into an evolutionary equation in ℋ1:(62)ddtv⋅,t,vt⋅,t⊤=A1v⋅,t,vt⋅,t⊤+ℬft,where ℬ=0,δx−1⊤ with δ⋅ being the Dirac distribution. Straightforward computation shows that(63)A1∗ϕ,ψ⊤=−ψ,ϕ″⊤,∀ϕ,ψ⊤∈DA1∗,DA1∗=ϕ,ψ⊤∈H20,1×H10,1|A1∗ϕ,ψ⊤∈ℋ1,ϕ′0=0,ϕ′1=c1ψ1−c2ϕ1.
This is equivalent to saying thatℬ∗A1∗−1 is bounded and for any T>0, there exists MT>0 depending on T, so that the system of the following:(64)vtt∗x,t=vxx∗x,t,vx∗0,t=0,vx∗1,t=−c1vt∗1,t−c2v∗1,t,y0∗=−vt∗1,t,v∗x,0=v0∗x,vt∗x,0=v1∗x,satisfies(65)∫0Tvt∗1,t2≤MTE∗0,where(66)E∗t=12∫01vt∗x,t2+vx∗x,t2dx+c22v∗1,t2.
Actually,(67)A1∗−1ϕ,ψ⊤=∫01∫0yψzdzdy+c2−1∫01ψzdz−c1c2ϕ1−∫0x∫0yψzdzdy,−ϕx⊤,ℬ∗A1∗−1ϕ,ψ⊤=−ϕ1.
Hence,ℬ∗A1∗−1 is bounded on ℋ1. Taking derivative of (66), we have(68)E˙∗t=−c1vt∗1,t2≤0.
Integrating above equation from 0 toT with respect to t, we have(69)∫0Tvt∗1,t2dt≤1c1E∗0.
We have showed thatℬ is admissible for eA1t. Now the solution of (62) can be described as(70)v⋅,t,vt⋅,t⊤=eA1tv0,v1⊤+∫0teA1t−sℬrsds+∫0teA1t−sℬjsds,where ∫0teA1t−sℬjsds is exponentially stable and ℬ is admissible for eA1t. Since eA1t is exponentially stable, the first part on the right hand side in (70) decays exponentially. The exponential decay result of the second part can be obtained by the estimation of (62) in [36], noticing (48), (49), andf⋅∈L20,∞ in Theorem 2.
Finally, the solution to system (60) can be obtained as(71)ζ⋅,t,ζt⋅,t⊤=P−1P0−1v⋅,t,vt⋅,t⊤,which decays exponential from the boundedness of operators P−1,P0−1 and exponential stability of system (61). For “X part,” it admits a unique solution and the solution is exponentially stable by (38). (59) holds.


## 4. Simulation Results

In this section, we give some numerical examples for closed-loop system (58) in order to prove the effectiveness of proposed feedback control law. We let these parameters be q=2,c1=1,c2=1,c3=1.2,c4=0.5 in system (58).And the matrixes are given as follows:(72)D=−11−1−10,P=4−2,Q=−8−2,F=5−2,G=−10−2.We give the initial values:(73)ζ^0x=2cos3x,ζ^1x=−2+3sinx,X^=12,ζ0x=3+2sin3x,ζ1x=−2x,X=34.The solution about system (58) is shown in Figures 1 and 2. From Figure 1, we can see that the solution of “ζ part” in closed-loop system (58) decays rapidly. The image of “X part” eventually tends to zero. In Figure 2, we notice that the result is similar to that in Figure 1. It explains the validity of our output feedback controller (57).Figure 1
Solution of “X,ζ,ζt part” in the closed-loop system (58). (a) Displacement of the ζ-part. (b) Velocity of the ζ-part. (c) X-part.
(a)(b)(c)Figure 2
Solution of “X^,ζ^,ζt^ part” in the closed-loop system (58). (a) Displacement of the ς^-part. (b) Velocity of the ς^-part. (c)X^-part.
(a)(b)(c)


## 5. Concluding Remarks

This paper considers the boundary output feedback stabilization for a cascaded-wave PDE-ODE system with velocity recirculation. Inspired by Krstic and Su et al. [30, 33], we consider the problem of stabilization of a cascaded-wave PDE-ODE system with non-local term. We have used the backstepping method to turn system (1) into target system (3). Next we designed the state feedback controller and the output feedback controller for (1). And Theorems 1–3 have been proved. Finally, as shown in Figures 1 and 2, it is obvious that the output feedback control law (57) can stabilize system (1) exponentially by the measurements. Illustrative simulation results are given to demonstrate the validity of the theoretical analysis and show the performance of the proposed method. We hope to have more innovation and development about stabilization of PDE-ODE systems with more non-local terms in this field afterwards.



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*Source: 9977714-2021-06-04.xml*