# Local Exact Controllability to the Trajectories of Burgers–Fisher Equation

Authors: Xiaofeng Shi
Journal: Mathematical Problems in Engineering
 (2020)

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## Abstract

This paper is addressed to study local exact controllability to the trajectories of the Burgers–Fisher (BF) equation. By using the global Carleman estimate for the second-order parabolic operator, we establish the observable inequality and obtain the exact controllability to the trajectories of the linear system. Then, by local inverse theory, we consider the controllability result for the Burgers–Fisher equation.


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## Body

## 1. Introduction

Burgers–Fisher-type equations describe the interaction between reaction mechanism, convection effect and diffusion process. Due to this, these equations have a wide range of applications in plasma physics, fluid physics, capillary-gravity waves, nonlinear optics, chemical physics and population dynamics [1–3].Burgers–Fisher-type equation is as follows:(1)yt+ayyx=dyxx+by1−y,where a,bandd are the nonnegative numbers.As we know, the approximation numerical methods for the Burgers-type equations have been developed by many researchers, including the moving mesh PDE method [4], the Adomian decomposition method [5, 6], the direct discontinuous Galerkin method [7], and the B-spline quasi-interpolation method [8].Due to the quadratic nonlinearity of the Burgers–Fisher equation, the nonlinear phenomena results in very complex and unfavorable behaviors (e.g., blowing up, shock waves, and chaos). Control by using internal or external actuation has been expected as an effective method to reduce or totally avoid those undesired phenomena.Although the controllability of infinite dimensional systems has been studied extensively, e.g., null controllability [9, 10], local controllability to trajectories [11–14], approximate controllability [15, 16], exact controllability [17, 18], and boundary controllability [19], control of the Burgers–Fisher equation is still in its infancy and remains open. In this paper, we will deal with the local exact controllability to the trajectories of the Burgers–Fisher equation.We consider the controlled system described by the Burgers–Fisher equation:(2)yt−yxx+αyyx−βy1−y=uχω,yt,0=0=yt,1,yxt,0=0=yxt,1,y0,x=y0x,where ut,x is an internal control and ω is a nonempty open interval of [0, 1].Our problem is to guide the solution of (2) to a given trajectory. More accurately, for any given time T>0 and a suitable space X,Y(seen below), if with any initial value y0∈X and the given y¯∈Y satisfying(3)y¯t−y¯xx+αy¯y¯x−βy¯1−y¯=0,y¯t,0=0=y¯t,1,y¯xt,0=0=y¯xt,1,y¯0,x=y¯0x.Then, there exists a controlut,x such that the solution y of (2) satisfies y0,x=y0 and can touch yT,x=y¯T,x.Lettingq=y−y¯, we get a new controlled system:(4)qt−qxx+2βy¯+αy¯x−βq+αy¯qx+αqqx+βq2=uχω,qt,0=0=qt,1,qxt,0=0=qxt,1,q0,x=q0x.It is easy to find that the exact controllability to the trajectories of system (2) is equivalent to the null controllability of system (4).For the convenience of narration, we firstly introduce some notations as a preliminary:(i)
Ly¯y=yt−yxx+2βy¯+αy¯x−βy+αy¯yx is a linear operator.(ii)
Letψx∈C∞0,1, which satisfies ∀x∈0,1,0≤ψ≤1;ψ0=ψ1=0;∀x∈0,1\ω0,ψx>0.For given positive constantsλ and μ, we construct two weight functions:(5)φx,t=eμψ+3tT−t,ax,t=eμψ+3−e5μtT−t,η=λa,θ=eη.(iii)
For simplicity, we abbreviate∫01∫0T⋅ as either ∬⋅ or ∫Q⋅, where Q=0,1×0,T.(iv)
Let(6)χt=1,t∈0,T2,0,t∈3T4,T,ϕt=4T2,t∈0,T2,1tT−t,t∈T2,T,m=T24minx∈0,1ax,T2=e3μ−e5μ,M=T24maxx∈0,1ax,T2=e4μ−e5μ,andF=f:e−λmϕϕ−1/2f∈L10,T;L20,1,U=u:e−λMϕϕ−3/2u∈L20,T;L2ω,Y=y:e−λMϕϕ−3/2y∈C0,T;L2∩L20,T;H1.To end this introductory, let us mention how this work is organized. Sections2 and 3, respectively, establish the well posedness of the linear Burgers–Fisher equation and the nonlinear one. In Section 4, we establish the Carleman estimate for the second parabolic operator (similar estimates can be found in [20–22]). Section 5 is contributed to the null controllability of system (4).


## 2. Well Posedness of Linear BF Equation

Definition 1.
If for anyφ∈H10,1∩L20,1, there exists y∈C0,T;L2∩L20,T;H1, such that(7)yt,φ+∫0tyx,φx+ay+byx,φdt=y0,φ+∫0tf,φdt,then y is called a weak solution of the following system:(8)yt−yxx+ay+byx=f,yt,0=0=yt,1,yxt,0=0=yxt,1,y0,x=y0x.
The main result for the well posedness of linear system (8) is as follows:Proposition 1.
Leta,b∈L20,T;L∞0,1, if y0∈L20,1 and f∈L10,T;L2, then there exists a unique solution of (8): y∈C0,T;L2∩L20,T;H1 and there is a constant C independent of y0,f,a,b,andT, such that(9)yC0,T;L20,1∩L20,T;H10,1≤CeC∫0T2aL∞0,1+bL∞0,1dty0L20,1+fL10,T;L20,1+y0L20,1+fL10,T;L20,11/2.Proposition1 will be proved in the following four sections step-by-step. Firstly, we establish a Galerkin approximate solution for the linear BF equation; secondly, we prove the existence of this solution through a series of mathematical estimates; finally, the uniqueness of the solution is proved.


### 2.1. Galerkin Approximate Solutions

We consider the system(10)yt−yxx+ay+byx=f,yt,0=0=yt,1,yxt,0=0=yxt,1y0,x=y0x.Let(11)Ay=yxx,∀y∈DA=y∈H20,1yt,0=0=yt,1,yxt,0=0=yxt,1.Obviously,A is the second-order operator defined on L20,1. We can find an orthogonal basis φkk=1∞ of L20,1 as the eigenfunctions corresponding to the eigenvalues λk of operator A.Let(12)ym=∑k=1mCmktφk,fm=∑k=1mf,φkφk,where Cmktk=1,2,…,m is the solution of the following ordinary differential equation(13)ymt−ymxx,φk+aym+bymx,φk=fm,φk,Cmk0=y0,φk.According to the classical theory of the ordinary differential equation, equation (13) has a unique solution in the interval 0,Tm.


### 2.2. Energy Estimates

Next, we prove that the solutionym mentioned above is bounded when Tm⟶T.Multiplying the first formula of equation (13) by Cmkt on both sides and taking summation about k from 1 to m, the following is obtained:(14)ymt−ymxx,ym+aym+bymx,ym=fm,ym.That is to say(15)12ddtymL20,12+ymxL20,12=−∫01aym2dx−∫01bymymxdx+fm,ym.Noticing that(16)−∫01aym2dx≤aL∞0,1∫01ym2dx−∫01bymymxdx≤bL∞0,1∫01ymymxdx≤12bL∞0,1∫01ym2dx+12∫01ymx2dxfm,ym≤fmL20,1ymL20,1.Therefore(17)12ddtymL20,12+ymxL20,12≤2aL∞0,1+bL∞0,1ymL20,12+2fmL20,1ymL20,1.According to Gronwall’s inequality, we have(18)ymC0,T;L20,1≤CeC∫0T2aL∞0,1+bL∞0,1dtym0L20,1+fmL20,1.Integrating both sides of (17) about t on 0,T and combining with (18), we can get(19)ymTL20,12+∫0TymxL20,12≤ym0L20,12+∫0T2aL∞0,1+bL∞0,1ymL20,12dt+2∫0TfmL20,1ymL20,1dt≤ym0L20,12+ymC0,T;L20,12∫0T2aL∞0,1+bL∞0,1dt+2fmL10,T;L20,1ymC0,T;L20,1≤ym0L20,12+ymC0,T;L20,1e∫0T2aL∞0,1+bL∞0,1dtfmL10,T;L20,1+ymC0,T;L20,1≤CeC∫0T2‖a‖L∞0,1+‖b‖L∞0,1dtfmL10,T;L20,1+ym0L20,12.Combining (18) and (19), we have(20)ymC0,T;L20,1∩L20,T;H10,1≤CeC∫0T2aL∞0,1+bL∞0,1dtym0L20,1+fmL10,T;L20,1+ym0L20,1+fmL10,T;L20,11/2.Let us denoteH1∗0,1 as the dual space of H10,1 and H1∗ for convenience. Similarly, L∞0,1, L20,1, and so on; also, omit 0,1 for convenience.Noticing that(21)∫0TaymH1∗dt=∫0TSupφH1<aym,φ>H1∗,H1dt=∫0TSupφH1aym,φdt≤∫0TSupφH1aL∞ymL2φL2dt≤C∫0TSupφH1aL∞ymL2φH1dt≤C∫0TaL∞ymL2dt≤CaL10,T;L∞ymC0,T;L2.Similarly(22)∫0TbymxH1∗dt=∫0TSupφH1<bymx,φ>H1∗,H1dt=∫0TSupφH1bymx,φdt≤∫0TSupφH1bL∞ymH1φL2dt≤C∫0TSupφH1aL∞ymL2φH1dt≤C∫0TaL∞ymH1dt≤CbL10,T;L∞ymC0,T;H1≤CbL10,T;L∞ymL20,T;H1.According to the above two estimates and (20), we can obtain(23)∫0TymtH1∗dt=∫0TSupφH1<ymt,φ>H1∗,H1dt=∫0TSupφH1<ymxx−aym−bymx+f,φ>H1∗,H1dt≤CT,ym0,f,a,b,which implies ymtm≥0 is bounded in L10,T;H1∗.


### 2.3. Existence of Weak Solutions

We are in position to use the energy estimates to gain weak solutions. According to (20) and (23), combined with the Lions–Aubin theorem, there exists y∈C0,T;L2∩L20,T;H1 such that(24)ym⟶y,weakly∗inL∞0,T;L2,ym⟶y,weakly  inL20,T;H1,ymt⟶yt,weakly  inL20,T;H1∗,ymt⟶yt,strongly  inL20,T;H1.According to the first formula in (13), for φ∈H10,1, we have(25)ymt,φ+ymx,φx+aym+bymx,φ=fm,φ.Multiplying (25) with χ0,t and integrating with respect to t on 0,T, we obtain(26)∫0Tymt,φχ0,t+ymx,φxχ0,t+aym+bymx,φχ0,tdt=∫0Tfm,φχ0,tdt,where χ0,t is the characteristic function on 0,t.As it is known(27)∫0Taym−ay,φχ0,tdt=∫0taym−ay,φdτ≤∫0Taym−ay,φχ0,tdt≤∫0TaL∞ym−yL2φL2dt=φL2∫0TaL∞ym−yL2dt≤φL2aL20,T;L∞ym−yL20,T;L2,∫0Tbymx−byx,φχ0,tdt=∫0tbymx−byx,φdτ≤∫0Tbymx−byx,φχ0,tdt≤∫0TbL∞ymx−yxL2φL2dt=φL2∫0TbL∞ym−yH1dt≤φL2bL20,T;L∞ym−yL20,T;H1.Therefore(28)∫0Taym,φχ0,tdt⟶∫0Tay,φχ0,tdt,∫0Tbymx,φχ0,tdt⟶∫0Tbyx,φχ0,tdt.Taking the limit on both sides in (26) and combining with (24) and (28), the following can be obtained:(29)∫0Tyt,φχ0,t+yx,φxχ0,t+ay+byx,φχ0,tdt=∫0Tf,φχ0,tdt,i.e., ∀t∈0,T, we get(30)∫0tyt,φ+yx,φx+ay+byx,φdt=∫0tf,φdt,which shows that y satisfies (7).


### 2.4. Uniqueness of Weak Solutions

Ify1 and y2 are two solutions of (8), letting y=y1−y2, then y satisfies(31)yt−yxx+ay+byx=0,yt,0=0=yt,1,yxt,0=0=yxt,1,y0,x=0.When (20) is used for (31) and according to the convergence of solution, we obtain(32)yC0,T;L20,1∩L20,T;H10,1=0.Thus(33)y≡0,that is(34)y1=y2.


## 2.1. Galerkin Approximate Solutions

We consider the system(10)yt−yxx+ay+byx=f,yt,0=0=yt,1,yxt,0=0=yxt,1y0,x=y0x.Let(11)Ay=yxx,∀y∈DA=y∈H20,1yt,0=0=yt,1,yxt,0=0=yxt,1.Obviously,A is the second-order operator defined on L20,1. We can find an orthogonal basis φkk=1∞ of L20,1 as the eigenfunctions corresponding to the eigenvalues λk of operator A.Let(12)ym=∑k=1mCmktφk,fm=∑k=1mf,φkφk,where Cmktk=1,2,…,m is the solution of the following ordinary differential equation(13)ymt−ymxx,φk+aym+bymx,φk=fm,φk,Cmk0=y0,φk.According to the classical theory of the ordinary differential equation, equation (13) has a unique solution in the interval 0,Tm.


## 2.2. Energy Estimates

Next, we prove that the solutionym mentioned above is bounded when Tm⟶T.Multiplying the first formula of equation (13) by Cmkt on both sides and taking summation about k from 1 to m, the following is obtained:(14)ymt−ymxx,ym+aym+bymx,ym=fm,ym.That is to say(15)12ddtymL20,12+ymxL20,12=−∫01aym2dx−∫01bymymxdx+fm,ym.Noticing that(16)−∫01aym2dx≤aL∞0,1∫01ym2dx−∫01bymymxdx≤bL∞0,1∫01ymymxdx≤12bL∞0,1∫01ym2dx+12∫01ymx2dxfm,ym≤fmL20,1ymL20,1.Therefore(17)12ddtymL20,12+ymxL20,12≤2aL∞0,1+bL∞0,1ymL20,12+2fmL20,1ymL20,1.According to Gronwall’s inequality, we have(18)ymC0,T;L20,1≤CeC∫0T2aL∞0,1+bL∞0,1dtym0L20,1+fmL20,1.Integrating both sides of (17) about t on 0,T and combining with (18), we can get(19)ymTL20,12+∫0TymxL20,12≤ym0L20,12+∫0T2aL∞0,1+bL∞0,1ymL20,12dt+2∫0TfmL20,1ymL20,1dt≤ym0L20,12+ymC0,T;L20,12∫0T2aL∞0,1+bL∞0,1dt+2fmL10,T;L20,1ymC0,T;L20,1≤ym0L20,12+ymC0,T;L20,1e∫0T2aL∞0,1+bL∞0,1dtfmL10,T;L20,1+ymC0,T;L20,1≤CeC∫0T2‖a‖L∞0,1+‖b‖L∞0,1dtfmL10,T;L20,1+ym0L20,12.Combining (18) and (19), we have(20)ymC0,T;L20,1∩L20,T;H10,1≤CeC∫0T2aL∞0,1+bL∞0,1dtym0L20,1+fmL10,T;L20,1+ym0L20,1+fmL10,T;L20,11/2.Let us denoteH1∗0,1 as the dual space of H10,1 and H1∗ for convenience. Similarly, L∞0,1, L20,1, and so on; also, omit 0,1 for convenience.Noticing that(21)∫0TaymH1∗dt=∫0TSupφH1<aym,φ>H1∗,H1dt=∫0TSupφH1aym,φdt≤∫0TSupφH1aL∞ymL2φL2dt≤C∫0TSupφH1aL∞ymL2φH1dt≤C∫0TaL∞ymL2dt≤CaL10,T;L∞ymC0,T;L2.Similarly(22)∫0TbymxH1∗dt=∫0TSupφH1<bymx,φ>H1∗,H1dt=∫0TSupφH1bymx,φdt≤∫0TSupφH1bL∞ymH1φL2dt≤C∫0TSupφH1aL∞ymL2φH1dt≤C∫0TaL∞ymH1dt≤CbL10,T;L∞ymC0,T;H1≤CbL10,T;L∞ymL20,T;H1.According to the above two estimates and (20), we can obtain(23)∫0TymtH1∗dt=∫0TSupφH1<ymt,φ>H1∗,H1dt=∫0TSupφH1<ymxx−aym−bymx+f,φ>H1∗,H1dt≤CT,ym0,f,a,b,which implies ymtm≥0 is bounded in L10,T;H1∗.


## 2.3. Existence of Weak Solutions

We are in position to use the energy estimates to gain weak solutions. According to (20) and (23), combined with the Lions–Aubin theorem, there exists y∈C0,T;L2∩L20,T;H1 such that(24)ym⟶y,weakly∗inL∞0,T;L2,ym⟶y,weakly  inL20,T;H1,ymt⟶yt,weakly  inL20,T;H1∗,ymt⟶yt,strongly  inL20,T;H1.According to the first formula in (13), for φ∈H10,1, we have(25)ymt,φ+ymx,φx+aym+bymx,φ=fm,φ.Multiplying (25) with χ0,t and integrating with respect to t on 0,T, we obtain(26)∫0Tymt,φχ0,t+ymx,φxχ0,t+aym+bymx,φχ0,tdt=∫0Tfm,φχ0,tdt,where χ0,t is the characteristic function on 0,t.As it is known(27)∫0Taym−ay,φχ0,tdt=∫0taym−ay,φdτ≤∫0Taym−ay,φχ0,tdt≤∫0TaL∞ym−yL2φL2dt=φL2∫0TaL∞ym−yL2dt≤φL2aL20,T;L∞ym−yL20,T;L2,∫0Tbymx−byx,φχ0,tdt=∫0tbymx−byx,φdτ≤∫0Tbymx−byx,φχ0,tdt≤∫0TbL∞ymx−yxL2φL2dt=φL2∫0TbL∞ym−yH1dt≤φL2bL20,T;L∞ym−yL20,T;H1.Therefore(28)∫0Taym,φχ0,tdt⟶∫0Tay,φχ0,tdt,∫0Tbymx,φχ0,tdt⟶∫0Tbyx,φχ0,tdt.Taking the limit on both sides in (26) and combining with (24) and (28), the following can be obtained:(29)∫0Tyt,φχ0,t+yx,φxχ0,t+ay+byx,φχ0,tdt=∫0Tf,φχ0,tdt,i.e., ∀t∈0,T, we get(30)∫0tyt,φ+yx,φx+ay+byx,φdt=∫0tf,φdt,which shows that y satisfies (7).


## 2.4. Uniqueness of Weak Solutions

Ify1 and y2 are two solutions of (8), letting y=y1−y2, then y satisfies(31)yt−yxx+ay+byx=0,yt,0=0=yt,1,yxt,0=0=yxt,1,y0,x=0.When (20) is used for (31) and according to the convergence of solution, we obtain(32)yC0,T;L20,1∩L20,T;H10,1=0.Thus(33)y≡0,that is(34)y1=y2.


## 3. Well Posedness of Nonlinear BF Equation

Theorem 1.
Fory0∈L20,1, a,b∈L20,T;L∞0,1, and f∈L10,T;W−1,10,1, if there exists a positive number r, such that(35)y0L20,1+fL10,T;W−1,10,1≤r,then the nonlinear system(36)yt−yxx+ay+byx+αyyx+βy2=f,yt,0=0=yt,1,yxt,0=0=yxt,1,y0,x=y0x,has a unique solution: y∈C0,T;H−10,1∩L20,T;L20,1.Proof.
Inspired by [12]. Let y0∈L20,1andf∈L10,T;W−1,10,1, satisfying (35).
We define a mapping:(37)Π:l∈L20,T;L2⟶y∈L20,T;L2,where y is the solution of the following system:(38)yt−yxx+ay+byx=f−αllx+βl2,yt,0=0=yt,1,yxt,0=0=yxt,1,y0,x=y0x.
Comparing (36) and (38), we know that y is the solution of (36) if and only if y is a fixed point of the mapping Π.
Similar to formula (9), we have(39)ΠlL20,T;L20,1≤CeC∫0T2aL∞0,1+bL∞0,1dty0L20,1+f−αllx+βl2L10,T;W−1,10,1+y0L20,1+f−αllx+βl2L10,T;W−1,10,11/2.(40)llxL10,T;W−1,10,1≤12lL20,T;L20,12,l2L10,T;W−1,10,1≤lL20,T;L20,12,
We have(41)ΠlL20,T;L20,1≤Cy0L20,1+fL10,T;W−1,10,1+α2+βlL20,T;L20,12.
For anyR, we note(42)B0,R=l∈L20,T;L20,1lL20,T;L20,1≤R.
It is easy to see that the appropriater,R>0 can be selected to satisfy(43)Cr+α2+βR2<R,then ΠB0,R⊂B0,R.
Now we are in the position to prove thatΠ is a contractive mapping.
Takingl,l¯∈L20,T;L20,1,y^=Πl−Πl¯ is the solution of the following system:(44)yt−yxx+ay+byx=αl¯l¯x−llx+βl¯2−l2,yt,0=0=yt,1,yxt,0=0=yxt,1,y0,x=0.
Similar to (9) and choosing appropriate r,R>0, we can obtain(45)Πl−Πl¯L20,T;L20,1≤Cαl¯l¯x−llx+βl¯2−l2L10,T;W−1,10,1.
Noticing that(46)l¯l¯x−llxL10,T;W−1,10,1≤12l¯2−l2L20,T;L20,1≤12l¯+lL20,T;L20,1l¯−lL20,T;L20,1.
We have(47)Πl−Πl¯L20,T;L20,1≤Cα+2βRl¯−lL20,T;L20,1.
Thus, the mapping is contractive whenCα+2βR≤1. By the Banach fixed-point theorem, Π has a unique fixed point, i.e., Πy=y, which is a solution of (36).


## 4. Carleman Estimate

The main result in this section is given as follows:Theorem 2.
There exist constantsC0andC>0 such that for λ>C0T+T2, ∀y∈DA, we have(48)∫Q1λφθ2yt2+yxx2+λ3φ3θ2y2+λφθ2yx2dxdt≤C∫Qθ2f2dxdt+∫Qω0λ3φ3θ2y2dxdt.In order to prove Theorem2, we need the following conclusion.Proposition 2.
Letyt−yxx=f and u=θy=eηy, then(49)θf=ut+−ηx2−ηt+ηxxu+2ηxux−uxx=I1+I2+Ru,where(50)I1=ut+2ηxux+ηxxu,I2=−ηx2u−uxx,Ru=−ηtu.Proof.
(51)u=θy,y=e−ηu,yt=−e−ηηtu+e−ηut,yx=−e−ηηxu+e−ηux,yxx=e−ηηx2u−e−ηηxxu−2e−ηηxux+e−ηuxx.
Substituting the above formulas intoyt−yxx=f, we achieve (49) and (50).
The proof of Theorem2 will be completed in the following four steps:Step 1: we will get the following estimate:(52)∫QI12+I22+2λ3μ4φ3ψx4u2+λμ2φψx2ux2dxdt≤C∫Qθ2f2dxdt+∫Qλ3μ3φ3u2dxdt.LetI1u,I2u2=∑Ii,ji=1,2,3;j=1,2,whereIi,j denotes the L2 inner product between the ith term of I1u and the jth term of I2u.For allIi,j, integrating by part and using the boundary conditions, we have(53)I1,1=∬−utηx2u=−∫01∫0Tηx2udu=∬u2ηxηxtu+ηx2ut=∬ηxηxtu2,I1,2=−∬utuxx=−∫0T∫01utdux=∫0T∫01uxuxtdx=12∫0T∫01ddtux2dx=0,I2,1=−2∬uxηx3u=−2∫0T∫01ηx3udu=−∫0T∫01ηx3du2=3∫0T∫01ηx2u2ηxx,I2,2=−2∬ηxuxuxx=−2∫0T∫01ηxuxdux=−∫0T∫01ηxdux2=∫0T∫01ηxxux2,I3,1=−∬ηx2ηxxu2,I3,2=−∬ηxxuuxx=∬ηxxux2+∬−12ηxxxxu2.Adding the above equations all together, we have(54)I1u,I2u2=∑i,j=12Ii,j=Iu+Iux,where(55)Iu=∬ηxηxt+2ηx2ηxx−12ηxxxxu2,Iux=∬2ηxxux2.According to the definition ofηx,t, we note that(56)ηxηxt+2ηx2ηxx−12ηxxxxu2=2λ3μ4φ3ψx4+R0u2,2ηxxux2=λμ2φψx2+R1ux2,whereR0≤λ3μ3φ3andR1≤Cλμφ.Noticing thatI1+I2=θf+ηtu,we have(57)I1+I2LQ22=∫QI12+I22+2I1I2dxdt=θf+ηtuLQ22≤C∫Qθ2f2dxdt+∫Qλφ2u2dxdt.Combining (54), (56) and (57), we obtain (52).Step 2: we prove the following estimate:(58)∫QI12+I22+λ3μ3φ3u2+λμφux2dxdt≤C∫Qθ2f2dxdt+∫Qω0λ3μ3φ3u2+λμφux2dxdt.In fact, from (52) we get(59)∫QI12+I22dxdt≤C∫Qθ2f2dxdt+∫Qλ3μ3φ3u2dxdt,∫Qλ3μ4φ3ψx4u2+λμ2φψx2ux2dxdt≤C∫Qθ2f2dxdt+∫Qλ3μ3φ3u2dxdt.Noticing thatψx>0,∀x∈0,1\ω0,we obtain(60)∫Q\Qω0λ3μ4φ3u2+λμ2φux2dxdt≤Cψ∫Qθ2f2dxdt+∫Qλ3μ3φ3u2dxdt=Cψ∫Qθ2f2dxdt+∫Q\Qω0λ3μ3φ3u2dxdt+∫Qω0λ3μ3φ3u2dxdt.In the formula above, if we takeμ0=Cψ+1, then(61)∫Q\Qω0λ3μ3φ3u2+λμφux2dxdt≤C∫Qθ2f2dxdt+∫Qω0λ3μ3φ3u2+λμφux2dxdt,holds forμ≥μ0,that is(62)∫Qλ3μ3φ3u2+λμφux2dxdt≤C∫Qθ2f2dxdt+∫Qω0λ3μ3φ3u2+λμφux2dxdt.From (59) and (62), we can infer that(63)∫QI12+I22dxdt≤C∫Qθ2f2dxdt+∫Qω0λ3μ3φ3u2+λμφux2dxdt.Combining (62) with (63), (58) is easy to be achieved.Step 3: the term∫Qω0λμφux2dxdt can be absorbed in(64)∫QI12+I22+λ3μ3φ3u2+λμφux2dxdt≤C∫Qθ2f2dxdt+∫Qω0λ3μ3φ3u2+λμφux2dxdt,such that the following estimate can be obtained:(65)∫QI12+I22+λ3μ3φ3u2+λμφux2dxdt≤C∫Qθ2f2dxdt+∫Qωλ3μ3φ3u2dxdt.In the following discussion, we consider two nonempty open intervalsω1,ω, satisfying: ω0¯⊂ω1,ω1¯⊂ω and select a nonnegative function χ∈C0∞ω1, which satisfies χ≡1 in ω0.Thus(66)∫Qω0λμφux2dxdt≤C∫Qλμ3φu2dxdt+ε∫Q1λμφuxx2dxdt+Cε∫Qωλ3μ3φ3u2dxdt.According to the definition ofI2=−ηx2u−uxx, then(67)∫Q1λμφuxx2dxdt=∫Q1λμφI2+ηx2u2dxdt≤C∫QI22+λ3μ3φ3u2dxdt.Substituting (67) into (66) and taking enough small ε and sufficient large λ, it is easy to see that (65) holds.Step 4: let us prove the Carleman estimate (48).Similar to formula (67), according to the definition of I1=ut+2ηxux+ηxxu, we have(68)∫Q1λμφut2dxdt=∫Q1λμφI1−2ηxux−ηxxu2dxdt≤C∫QI12+λμ3φu2+λμφux2dxdt.Combining (67) with (68), we have(69)∫Q1λμφut2+uxx2dxdt≤C∫QI12+I22+λ3μ3φ3u2+λμφux2dxdt.According to (65) and (68), we can get(70)∫Q1λμφut2+uxx2+λ3μ3φ3u2+λμφux2dxdt≤C∫Qθ2f2dxdt+∫Qωλ3μ3φ3u2dxdt.By substitutingu with θy, we obtain(71)∫Q1λμφθ2yt2+yxx2+λ3μ3φ3θ2y2+λμφθ2yx2dxdt≤C∫Qθ2f2dxdt+∫Qωλ3μ3φ3θ2y2dxdt.Taking a suitableμ0, when μ≥μ0, the estimate can be reduced to(72)∫Q1λφθ2yt2+yxx2+λ3φ3θ2y2+λφθ2yx2dxdt≤Cμ0∫Qθ2f2dxdt+∫Qωλ3φ3θ2y2dxdt,which implies (48).


## 5. Controllability to the Trajectory

In this section, the controllability of linear system (73) is obtained by the duality of observability-controllability, and the result for the nonlinear system is obtained by means of a local inverse theorem, where the idea can be referred to [13].Firstly, the conclusion for the linear system is as follows.Proposition 3.
∀y0∈L20,1 and f∈F, then there exists a control u∈U, such that(73)Ly¯y=f+uχω,yt,0=0=yt,1,yxt,0=0=yxt,1,y0,x=y0x,has a solution y∈Y, satisfying y⋅,T=0. Furthermore, there exists C>0 such that(74)uU≤CfF+y0xL20,1,yY≤CfF+uU+y0xL20,1.Proposition3 will be proved in three steps.According to the dual theory, in order to get the controllability of system (73), we need an observable inequality of dual operator (76) (for more details, refer [14]).Step 1: we firstly prove the following estimate:(75)∫Qe2λmϕϕ3q2dxdt+∫01q2x,0dx≤C∫Qe2λMϕLy¯∗q2dxdt+∫Qωe2λMϕϕ3q2dxdtConsider the dual operator:(76)Ly¯∗q=−qt−qxx+2βy¯+αy¯x−βq−αy¯qx,qt,0=0=qt,1,qxt,0=0=qxt,1,q0,x=q0x.Similar to (48), the following estimate can be shown:(77)∫Q1λφθ2qt2+qxx2+λ3φ3θ2q2+λφθ2qx2dxdt≤C∫Qθ2Ly¯∗q2dxdt+∫Qωλ3φ3θ2q2dxdt.According to (77), we have(78)∫0T/2∫01e2λmϕϕ3q2dxdt+∫01q2x,0dx≤C∫0T/2∫01q2dxdt+∫01q2x,0dx≤CqL∞0,T2;L20,12≤CχqL∞0,T2;L20,12≤CχqL∞0,T;L20,12≤CχLy¯∗q−χ′qL20,T;L20,12≤CLy¯∗qL20,3T4;L20,12+qL2T2,3T4;L20,12≤C∫03T/4∫01Ly¯∗q2dxdt+∫T/23T/4∫01q2dxdt≤C∫Qe2λMϕLy¯∗q2dxdt+∫Qλ3φ3θ2q2dxdt≤C∫Qe2λMϕLy¯∗q2dxdt+∫Qθ2Ly¯∗q2dxdt∫Qωλ3φ3θ2q2dxdt≤C∫Qe2λMϕLy¯∗q2dxdt+∫Qωλ3φ3θ2q2dxdt.Similarly(79)∫T2T∫01e2λmϕϕ3q2dxdt≤C∫T2T∫01θ2ϕ3q2dxdt≤C∫Qθ2ϕ3q2dxdt≤C∫Qθ2Ly¯∗q2dxdt+∫Qωλ3φ3θ2q2dxdt≤C∫Qe2λMϕLy¯∗q2dxdt+∫Qωλ3φ3θ2q2dxdt.By adding the left and right sides of the two formulas separately, we have(80)∫Qe2λmϕϕ3q2dxdt+∫01q2x,0dx≤C∫Qe2λMϕLy¯∗q2dxdt+∫Qωλ3φ3θ2q2dxdt=C∫Qe2λMϕLy¯∗q2dxdt+∫0T/2∫01λ3φ3θ2q2dxdt+∫1/2T∫01λ3φ3θ2q2dxdt≤C∫Qe2λMϕLy¯∗q2dxdt+∫Qωe2λMϕϕ3q2dxdt.Step 2: we can obtain the following observable inequality:(81)eλmϕϕ1/2qL∞L2+qx,0L20,1≤CeλMϕLy¯∗qL2L2+eλMϕϕ3/2qL2L2ω.Letξ=eλmϕϕ1/2 and ξ′=eλmϕϕ′ϕ−1/2λmϕ+1/2, then ξ′≤Ceλmϕϕ3/2(82)eλmϕϕ1/2qL∞L22=ξqL∞L22≤CLy¯∗ξqL2L22=CξLy¯∗q−ξ′qL2L22≤CξLy¯∗qL2L22+ξ′qL2L22=C∫QξLy¯∗q2dxdt+∫Qξ′q2dxdt=C∫Qe2λmϕϕLy¯∗q2dxdt+∫Qξq2dxdt≤C∫Qe2λm−Mϕϕe2λMϕLy¯∗q2dxdt+∫Qe2λmϕ3ϕ2q2dxdt≤C∫Qe2λMϕLy¯∗q2dxdt+∫Qωe2λMϕϕ3q2dxdt=CeλMϕLy¯∗q2+eλMϕϕL2L2ω3.Step 3: we will prove Proposition3.Note the right side of (81) as(83)q=eλMϕLy¯∗qL2L2+eλMϕϕ3/2qL2L2ω,which defines a norm on the space:(84)Q=q∈C0,T;L2∩L20,T;H1∩L2Qq0,t=q1,t=qx0,t=qx1,t=0,∀t∈0,T.Noticing thatQ¯ is a complete space of Q with respect to the above norm, Q¯ is obviously a Hilbert space, on which the inner product is defined as(85)p,q=∫Qe2λMϕLy¯∗pLy¯∗qdxdt+∫Qωe2λMϕϕ3pqdxdt.Recalling the first formula in (73)(86)Ly¯y=f+uχω.Multiply both sides byq and integrating over Q(87)∫QqLy¯y=∫Qqf+∫Qquχω,that is(88)∫Qqyt−yxx+2βy¯+αy¯x−βy+αy¯yxdxdt=∫Qqfdxdt+∫Qωqudxdt.Integrating by parts, we have(89)∫01yx,Tqx,Tdx−∫01y0xqx,0dx+∫QyLy¯∗qdxdt=∫Qqfdxdt+∫Qωqudxdt,that is(90)∫Qqfdxdt+∫01y0xqx,0dx=∫01yx,Tqx,Tdx+∫QyLy¯∗qdxdt−∫Qωqudxdt.The left of (90) defines a linear function: ℒ:Q⟶R:(91)ℒq=∫Qqfdxdt+∫01y0xqx,0dx,with(92)ℒq≤e−λmϕϕ−1/2fL10,T;L2eλmϕϕ1/2qL∞0,T;L2+y0xL20,1qx,0L20,1≤e−λmϕϕ−1/2fL10,T;L2+y0xL20,1eλmϕϕ1/2qL∞0,T;L2+qx,0L20,1≤Ce−λmϕϕ−1/2fL10,T;L2+y0xL20,1q.Therefore,ℒ is continuous. Then, there exists a unique p on Q, such that(93)ℒq=p,q,∀q∈Q.Thus(94)∫Qqfdxdt+∫01y0xqx,0dx=∫Qe2λMϕLy¯∗pLy¯∗qdxdt+∫Qωe22Mϕϕ3pqdxdt.Lety=e2λMϕLy¯∗pandu=−e2λMϕϕ3p, then (94) can also be written as follows:(95)∫Qqfdxdt+∫01y0xqx,0dx=∫QyLy¯∗qdxdt−∫Qωuqdxdt.Combining (90) and (95), we obtain(96)∫01yx,Tqx,Tdx=0,∀q∈Q,which implies yx,T=0.Furthermore, in the right side of (94), letting q=p and taking(97)Ly¯∗p=e−2λMϕy,p=−e−2λMϕϕ−3u,we have(98)e−λMϕyL20,T;L22+e−λMϕϕ−3/2uL20,T;L2ω2=ℒp≤Ce−λmϕϕ−1/2fL10,T;L2+y0xL20,1p=Ce−λmϕϕ−1/2fL10,T;L2+y0xL20,1eλMϕLy¯∗qL2L2+eλMϕϕ3/2qL2L2ωCe−λmϕϕ−1/2fL10,T;L2+y0xL20,1e−λMϕyL2L2+e−λmϕϕ−3/2fL2L2ω.Therefore(99)e−λMϕyL2L2+e−λMϕϕ−3/2uL2L2ω≤Ce−λmϕϕ−1/2fL10,T;L2+y0xL20,1,that is(100)uU≤CfF+y0xL20,1.In order to obtainyY≤CfF+uU+y0xL20,1, we construct a new system.Letξ=e−λMϕϕ−3/2, then ξy satisfies the following equations:(101)Ly¯ξy=ξf+ξuχωξ′y,ξyt,0=0=ξyt,1,ξyxt,0=0=ξyxt,1,ξyx0,x=ξ0y0x.According to the definition ofϕt and by direct calculation, we obtain(102)ξ′=e−λMϕ−λMϕ′ϕ3/2−32ϕ5/2,ξ′=e−λMϕλMϕ′ϕ3/232ϕ5/2≤Ce−λMϕ.In the following, we estimateξfL10,T;L2,ξuχωL10,T;L2, and ξ′yL10,T;L2:(103)ξfL10,T;L2=e−λMϕϕ−3/2fL10,T;L2=eλm−Mϕϕ−1e−λmϕϕ−1/2fL10,T;L2≤Ce−λmϕϕ−1/2fL10,T;L2=CfF,ξuχωL10,T;L2=e−λMϕϕ−3/2uχωL10,T;L2=e−λMϕϕ−3/2uL10,T;L2ω≤Ce−λMϕϕ−3/2uL20,T;L2ω=CuU,ξ′yL10,T;L2≤Ce−λMϕyL10,T;L2≤Ce−λMϕyL20,T;L2≤CfFy0xL20,1.According to the results of well posedness, we have(104)e−λMϕϕ−3/2yC0,T;L20,1∩L20,T;H10,1=ξyC0,T;L20,1∩L20,T;H10,1≤Cξf+ξuχω+ξ′yL10,T;L2+ξ0y0xL20,1≤CξfL10,T;L2+ξuχωL10,T;L2+ξ′yL10,T;L2+ξ0y0xL20,1≤CfF+uU+y0xL20,1,which implies(105)yY≤CfF+uU+y0xL20,1.The proof of Proposition3 is completed.Next, we consider the controllability of the nonlinear system. We have the following result.Proposition 4.
Let∀ε>0 and y0∈L20,1, satisfying y0L20,1<ε, then there exists a control u∈U, such that(106)Ly¯y+Ny=uχω,Ny=αyyx+βy2yt,0=0=yt,1,yxt,0=0=yxt,1,y0,x=y0x,has a solution y∈Y, satisfying y⋅,T=0.Before the proof, we decompose the nonlinear operator:(107)Ny=αyyx+βy2into(108)Ny=NProy,y+NBily,y,where NProy,y=αyyx and NBily,y=βy2, which have the following properties.Lemma 1.
ForNPro:Y×Y⟶F and NBil:Y×Y⟶F, there exists C>0, such that(109)NProy1,y2F≤Cy1Yy2Y,NBily1,y2F≤Cy1Yy2Y.Proof.
(110)NProy1,y2F=e−λmϕϕ1/2αy1y2xL10,T;L2=α∫0Te−λmϕϕ1/2y1y2xL20,1dt≤αCλ,M∫0Te−2λMϕϕ−3y1y2xL20,1dt≤αCλ,M∫0Te−2λMϕϕ−3y1L∞0,1y2xL20,1dt≤αCλ,M∫0Te−2λMϕϕ−3y1H10,1y2xH10,1dt≤αCλ,M,C′∫0Te−2λMϕϕ−3y1H10,12dt1/2∫0Te−2λMϕϕ−3y2H10,12dt1/2≤Cy1Yy2Y.
Analogously(111)NBily1,y2F=e−λmϕϕ−1/2βy1y2L10,T;L2=β∫0Te−λmϕϕ−1/2y1y2L20,1dt≤βCλ,M∫0Te−2λMϕϕ−3y1y2L20,1dt≤βCλ,M∫0Te−2λMϕϕ−3y1L∞0,1y2L20,1dt≤βCλ,M,C′∫0Te−2λMϕϕ−3y1H10,1y2H10,1dt≤βCλ,M∫0Te−2λMϕϕ−3y1H10,12dt1/2∫0Te−2λMϕϕ−3y2H10,12dt1/2≤Cy1Yy2Y.Lemma 2 (local inverse theorem).
LetEandG be two Banach spaces, Λ:E⟶G satisfying Λ∈C1E,G. If for e^∈E,Λ′e^=g^ and Λ′e^:E⟶G is a surjection. Then, for any g satisfying g−g^<ε, the equation Λe=g must have a solution e.With Lemmas1 and 2, we can prove Proposition 4.Proof.
TakingE=u,y∈U×YLy¯y−uχω∈F, with the norm defined as(112)u,yE=uU2+yY2+Ly¯y−uχωF21/2,and G=F×L20,1.
We define the set-valued mapping:(113)Λ:E⟶Ge=u,y⟼Ly¯y−uχω+Ny,y0x.
Obviously,Λ0,0=0,0 and Λ∈C1E,G.
Lete^=0,0andg^=0,0, then(114)Λ′e^:E⟶G,e=u,y⟼Ly¯y−uχω,y0x.
The conclusion of Proposition3 shows the fact that Λ′e^:E⟶G is a surjection.
According to Lemma2, system (106) has a solution y∈Y and(115)y⋅,T=0.
In this way, null controllability of system (4) is obtained. Equivalently, local exact controllability to the trajectory of system (2) is achieved.



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