Xing LU Tatsien LI

Abstract In this paper the authors first present the definition and some properties of weak solutions to 1-D first order linear hyperbolic systems. Then they show that the constructive method with modular structure originally given in the framework of classical solutions is still very powerful and effective in the framework of weak solutions to prove the exact boundary (null) controllability and the exact boundary observability for first order hyperbolic systems.

Keywords First order linear hyperbolic system, Exact boundary controllability,Exact boundary observability, Constructive method

1 Introduction

1.1 Review on the exact boundary controllability

There are many publications concerning the exact boundary controllability for linear hyperbolic systems. For a kind of 1-D first order linear hyperbolic system, the exact boundary controllability was established by means of the characteristic method by Russell ([17]). Later,the Hilbert Uniqueness Method (HUM for short), a more general and systematic framework was introduced by J.-L. Lions for the study of linear hyperbolic systems, especially, of wave equations ([14–15]), which builds up the relationship between the exact boundary controllability for the system and the exact boundary observability for the adjoint system. Based on the HUM method and Schauder’s fixed point theorem, some results were obtained on the exact boundary controllability for semilinear wave equations by Zuazua ([19–20]). An abstract result on the exact boundary controllability for semilinear equations was established by Lasiecka and Triggiani in [7]. Recently, Coron and Nguyen studied the (null) controllability of a general 1-D linear hyperbolic system using boundary controls on one side in optimal time based on the backstepping method (see [5–6]).

In quasi-linear case there were only few results (see [2–3]) before. Until 2002, Li and Rao proposed a constructive method with modular structure (see [10-11]) and established a complete theory on the exact boundary controllability for the general 1-D first order quasi-linear hyperbolic system without zero eigenvalues

with general non-linear boundary conditions based on the semi-globalC1solution theory (see[8–9]). Under this framework,the local exact boundary controllability was realized by two-sided controls, one-sided controls and two-sided controls with fewer boundary controls, respectively(in the linear case,it gives actually the global controllability). In the case of two-sided controls,the number of boundary controls is equal to the number of unknown variables, namely, the number of all the eigenvalues. In the case of one-sided controls, all the boundary controls act only on the side with more boundary controls,the number of which is reduced to the maximum value of the number of positive eigenvalues and that of negative ones,but the control time must be suitably enlarged. In particular, when the number of positive eigenvalues is equal to that of negative ones, boundary controls can act on either side. In the case of two-sided controls with fewer boundary controls, both the total number of boundary controls and the control time are the same as in the case of one-sided controls, however, the side with fewer boundary controls should be fully controlled, while the rest of boundary controls acts on the other side (see [8]).However, in the last two cases, more strict conditions should be satisfied. For example, for the case of one-sided controls, it requires that the boundary conditions on the side without boundary controls should be reversible. To be explicit, these boundary conditions should be able to be suitably rewritten so that, after adding suitable artificial boundary conditions, the backward problem of this system is well-posed.

The requirement mentioned before doesn’t occur for the system of wave equations,since the system of wave equations with usual boundary conditions is always time reversible, thus the exact boundary controllability is always equivalent to the exact boundary null controllability,namely, the system can drive any given initial dataU0att=0 exactly to the final dataU≡0 att=Tby means of boundary controls. Apparently, this equivalence is not true for the first order hyperbolic system in general,therefore the exact boundary controllability is named as the strong exact boundary controllability, and the exact boundary null controllability is named as the weak exact boundary controllability in[12],respectively. The exact boundary controllability implies the exact boundary null controllability,however,the exact boundary null controllability implies the exact boundary controllability only if the system is time reversible (see [12], also see Remark 4.1). On the other hand, in order to realize the one-sided exact boundary null controllability, controls can act on any given side of the boundary, while on the other side the boundary condition should be homogeneous, thus the number of the boundary controls can be further reduced (see [12]).

As a tool, the duality can be used to study the controllability in the linear case, then the study of observability is also of great importance from this point of view. Russell[17]introduced a kind of observability in the 1-D case that for the backward problem of the first order linear hyperbolic system, the boundary observations can uniquely determine the initial data att=0.It is called as the weak exact boundary observability in [12]. Another kind of observability,which requires that the boundary observations can uniquely determine the final data att=Tfor the backward problem, is called as the strong exact boundary observability in [12]. Both of them can be established by a constructive method for 1-D quasi-linear hyperbolic systems in the framework of classical solutions (see [12]), and the strong exact boundary observability implies naturally the weak exact boundary observability, however, in general, one can not obtain the strong exact boundary observability from the weak exact boundary observability unless the system is time reversible(see[12], also see Remark 5.4). Moreover,in the framework of classical solutions, for some special cases in the linear situation, the relationship between the exact boundary (null) controllability for the original system and the strong (weak) exact boundary observability for the adjoint system in the case of one-sided control was discussed in[12]. We will generalize the corresponding conclusions in the framework of weak solutions.

1.2 Purpose of this paper

In this paper, we will consider the exact boundary controllability for the first order linear hyperbolic system in the framework of weak solutions.

First, to study the controllability from coupled systems of wave equations to first order hyperbolic systems is of great significance, since the latter has much wider connotation than the former. For instance, in 1-D case, a wave equation can be always transformed into a first order hyperbolic system, but the number of positive eigenvalues is always equal to that of negative ones (see [8]). If the wave equation explicitly contains the state variableU, there will be also the zero eigenvalue in the corresponding first order hyperbolic system. But for any given first order hyperbolic system, generally speaking, the number of positive eigenvalues is not necessarily equal to that of negative ones, which often makes trouble in the treatment. In this paper we only consider first order hyperbolic systems without zero eigenvalues. Besides,for wave equations with the usual boundary conditions (Dirichlet, Neumann, Robin, even dissipative boundary conditions), the whole system is always time reversible. However, first order hyperbolic systems are not time reversible in general. In fact, when the number of positive eigenvalues is not equal to that of negative ones, or even when they are equal to each other, but the boundary conditions don’t satisfy certain reversibility, the corresponding backward problem will not be well-posed. Hence, first order hyperbolic systems have not only much more abundant connotations and practical applications than wave equations,but also they have their own difficulties: the methods to treat wave equations may not remain effective for first order hyperbolic systems. Thus it is necessary to carry out a special research on them.

Second, we will study the controllability for first order hyperbolic systems in the framework of weak solutions. In the present paper,we will show that the constructive method with modular structure is very powerful and effective not only in the framework of classical solutions, but also in the framework of weak solutions.

In what follows,we will first discuss the well-posedness for the following 1-D first order linear hyperbolic system in the framework of weak solutions and establish some related estimates,then we study the corresponding exact boundary controllability and exact boundary observability by the constructive method.

The system under consideration is given by

with the boundary conditions

and the initial data

whereU=(u1,···,un)T:(0,+∞)×(0,L)→Rndenotes the state variable,Λ=diag{Λ−,Λ+}is a diagonal matrix of ordern,

withλr<0 (r= 1,···,m) andλs>0 (s=m+1,···,n), the coupling matrixA= (aij) is of ordern. Let=n−m. The boundary coupling matricesG0andG1are of orderm×mandm×, respectively, the boundary control matricesD0andD1are of orderm×M0andm×M1(M0≤1≤m), respectively, both of them are full column-rank matrices. All the matrices mentioned above are with constant elements. Moreover,U= (U−,U+)TwithU−=(u1,···,um)TandU+=(um+1,···,un)T,H=(H−,H+)TwithH−=(h1,···,hM1)TandH+=(hM1+1,···,hM)T(M=M0+M1≤n).

In order to guarantee the well-posedness of the mixed problem of system(1.2),the boundary conditions should satisfy the following requirements:

1. the number of boundary conditions on each boundary is equal to the number of the coming characteristics on this boundary;

2. the boundary conditions on each boundary can be equivalently rewritten in the form that the state variables corresponding to the coming characteristics are explicitly expressed by the state variables corresponding to the departing characteristics.

Boundary conditions (1.3)–(1.4) are just given according to above requirements.

2 L2 Estimates of C1 Solutions

In the framework of classical solutions, by the theory ofC1solutions for the first order hyperbolic system, we have

Lemma 2.1(see[13])For any given T >0, for any given initial data U0∈(C1[0,L])nand any given boundary function H∈(C1[0,T])M, suppose that the conditions of C1compatibility are satisfied at the points(t,x)=(0,0)and(0,L), respectively:

and

where A−and A+are the first m rows and the last m rows of A, respectively, then the mixed problem(1.2)–(1.5)admits a unique C1solution U=U(t,x)on R(T)={(t,x)|0 ≤t≤T,0 ≤x≤L}.

In order to define aL2weak solution and to give its properties, in this section, we first establish someL2estimates on theC1solution to problem (1.2)–(1.5).

We start with anL2estimate ofC1solutions by the characteristic method.

Theorem 2.1For any given T >0, the C1solution U=U(t,x)to the mixed problem(1.2)–(1.5)satisfies

here and hereafter, C(T)>0denotes a different positive constant, depending only on T.

ProofFor simplicity, we denote

which are vectors of ordermandm, respectively, and consider the following system:

Let

Letx=(t) be thesthcharacteristic line passing through the point (t,x) = (0,0) (s=m+1,···,n). For any given∈[0,T1], denotingthe linet=tcan be separated into two parts:

a. Whenβ≤≤L, draw thesth(s=m+1,···,n) characteristic linex=xs(t) passing through the pointit interacts thex-axis at the point (0,αs), namely, we have

Sinceλsis a constant,x=xs(t) as a straight line can be rewritten asx=λs(t−t)+x, thenAlong this characteristic line, thesthcomponentusofUsatisfies

then

hence

Notingfor any givent, whenxruns fromβtoL,αsruns from 0 toL−λst(

On the other hand, letx=be thesthcharacteristic line passing through the point(t,L). For the second integration on the right-hand side of (2.11), notingxs(τ)=λs(τ−t)+x,and changing the order of integration, we have

here and hereafter,Cstands for a different positive constant. By (2.12)–(2.13), it follows from(2.11) that

b. When 0 ≤x≤β, drawing thesth(s=m+1,···,n)characteristic linex=xs(t)passing through the point (t,x), it intersects thet-axis at the point (ts,0), whereThat is to say, we have

Along this characteristic line, noting the boundary condition in (1.3), thesthcomponent ofUsatisfies

where (G0)sis thesthrow ofG0. Integrating (2.16) with respect totalongx=xs(t) fromtstowe have

For any givenr(r=1,···,m), therthcharacteristic linex=xr(t) passing through the point(ts,0) intersects thex-axis at the point (0,αr)=(0,αr(t,x)), namely, we have

Alongx=xr(t), therthcomponenturofUsatisfies

Integrating (2.19) with respect totalongx=xr(t) from 0 tots, we have

Substituting it into (2.17), we get

where (G0)sris the component ofG0on thesthrow andrthcolumn. Similarly, we have

in which, notingwe have

Thus, in a similar way we get

The combination of (2.14) and (2.23) yields

Similarly, we have

Thus, we have

Then, by Gronwall’s inequality, we get

Repeat the procedure above for at mosttimes, we get

Noting thatD0andD1are full column-rank, (2.29) leads to (2.3).

Next, we estimate theL2norm of the trace of the solution on the boundary.

Theorem 2.2For any given T >0, the C1solution U=U(t,x)to the mixed problem(1.2)–(1.5)satisfies

and

ProofWe still consider system (2.5), and prove (2.31) only. Let

For any givent∈[0,T1], for eachs=m+1,···,n,noting (2.32), thesthcharacteristic linex=xs(t) passing through the point (t,L) must intersect thex-axis at a point (0,αs), namely,we have

thus,xs(t) =λs(t−t)+Landαs=αs(t) =L−λst. Along this characteristic line, thesthcomponent ofUsatisfies

Integrating (2.34) with respect totalong this characteristic line from 0 totleads to

then

hence

Noting thatαs(t)=L−λst, whentruns from 0 toT1,αsruns fromLtoL−λsT1(≤L), then the first integration on the right-hand side of (2.37) equals to

As to the second integration on the right-hand side of (2.37), letx=be thesthcharacteristic line passing through the point(T1,L). Notingxs(τ)=λs(τ−t)+L,and changing the order of integration, we have

Therefore, it follows from (2.37) that

then by (2.3) we get

TakingU(T1,x) as the initial data, repeating the procedure above on the interval [T1,2T1],we get

Repeating the above procedure for at mosttimes, fors=m+1,···,n, we get

namely,

Then, by the boundary condition atx=Lin (1.4), we have

Noting thatD0andD1are full column-rank, the combination of (2.44) and (2.45) gives(2.31).

In fact, the result above is true for the trace of the solution at any givenx∈[0,L].

Theorem 2.3Let T >0. Under the hypotheses of Theorem2.1, for any given x∈[0,L],the C1solution U=U(t,x)to the mixed problem(1.2)–(1.5)satisfies

ProofLetU=U(t,x)be theC1solution to the mixed problem(1.2)–(1.5)on[0,T]×[0,L].Changing the role oftandx,U(t,x) is still the uniqueC1solution to the following rightward problem:

and

For any givenx∈[0,L], using Theorem 2.1 for problem (2.47)–(2.48), we have

Noting (2.30) and (2.3) for the mixed problem (1.2)–(1.5), we have

and

then we get (2.46).

Besides, for any givenC1initial dataU0(x) (0 ≤x≤L), the Cauchy problem

with the initial condition

(1.5) admits a uniqueC1solution on the corresponding maximum determinate domainRc, surrounded byt=0 (0 ≤x≤L) and two characteristic lines given by

and

wherex=x(1)(t) is the rightmost characteristic line passing through the point (t,x) = (0,0),andx=x(2)(t) is the leftmost characteristic line passing through the point (t,x) = (0,L).Since the solution depends only on the initial data, as a corollary of Theorem 2.1, we have

Corollary 2.1Let(t∗,x∗)be the intersection point of x=x(1)(t)and x=x(2)(t). For any given t∈[0,t∗], the C1solution U=U(t,x)to Cauchy problem(2.52)and(1.5)satisfies

Similarly, by Theorem 2.3 we have

Corollary 2.2Let(t∗,x∗)be the intersection point of x=x(1)(t)and x=x(2)(t). For any given x∈[0,L], the C1solution U=U(t,x)to Cauchy problem(2.52)and(1.5)satisfies

where T∗=T∗(x)is determined by

Furthermore, letR0be the domain surrounded by t-axis, x-axis and characteristic linex=x(2)(t), and letRLbe the domain surrounded by x-axis,x=Land characteristic linex=x(1)(t), wherex=x(1)(t) andx=x(2)(t) are given by (2.53) and (2.54), respectively. By the theory of one-sided mixed initial-boundary value problem of first order hyperbolic systems([8]), the uniqueC1solutionU=U(t,x) to problem (1.2)–(1.5) onR0is uniquely determined by the initial dataU0and the boundary functionH+(t) onx= 0; while,U=U(t,x) onRLis uniquely determined by the initial dataU0and the boundary functionH−(t) onx=L. Lett=ti(x) (i=1,2) be the inverse function ofx=x(i)(t) (i=1,2) forx∈[0,L]. We have

Theorem 2.4Assume that U=U(t,x)is the unique C1solution to problem(1.2)–(1.5).

(1)For any givenU=U(t,x)satisfies

Moreover, for any given x∈[0,L], U=U(t,x)satisfies

(2)For any given t∈[0,tL1], in which tL1=t1(L), U=U(t,x)satisfies

Moreover, for any given x∈[0,L], U=U(t,x)satisfies

3 Definition and Properties of Weak Solution

3.1 Definition of weak solution

In order to give the definition of weak solution, we first prove

Proposition 3.1For any given T >0, U=U(t,x)∈(C1([0,T]×[0,L]))nis a C1solution to problem(1.2)–(1.5)if and only if U=U(t,x)satisfies

for any givenΦ ∈(C1([0,T]×[0,L]))nsatisfying

ProofAssume that problem(1.2)–(1.5)admits aC1solutionU(t,x)on the domain[0,T]×[0,L] for any givenT >0. Let Φ ∈(C1([0,T]×[0,L]))nsatisfy (3.2). Multiplying ΦTon both sides of (2.52) and integrating by parts, we get

Noting that Φ satisfies (3.2), using the boundary conditions in (1.3)–(1.4), we immediately get(3.1).

On the contrary, let Φ ∈(C1([0,T]×[0,L]))nvanish atx=0,Land att=0,T. By (3.1),we get

SinceU(t,x)∈(C1([0,T]×[0,L]))n, integrating by parts in (3.4), we get

Since the set of functions Φ ∈(C1([0,T]×[0,L]))nmentioned above is dense in (L1((0,T)×(0,L)))n, (3.5) implies that

then

Now, let Φ ∈(C1([0,T]×[0,L]))nsatisfy (3.2). Integrating by parts and noting (3.7), it follows from (3.1) that

Let Φ ∈(C1([0,T]×[0,L]))nvanish atx=0,L, we get

Let B :(C1([0,T]×[0,L]))n→(L1(0,L))nbe defined by

Since B({Φ ∈(C1([0,T]×[0,L]))n,vanishing atx=0,L})is dense in (L1(0,L))n, (3.9)implies

then we have

Similarly, we can prove

and

Then,U=U(t,x) is aC1solution to the mixed problem (1.2)–(1.5).

Thus, (3.1) is an equivalent definition of theC1solution to problem (1.2)–(1.5), which is still valid whenU=U(t,x)∈(L2(0,T;L2(0,L)))n. Therefore, we can give the following

Definition 3.1For any given T >0, U=U(t,x)∈(L2(0,T;L2(0,L)))nis a weak solution to the forward mixed problem(1.2)–(1.5), if(3.1)holds for any givenΦ ∈(C1([0,T]×[0,L]))nsatisfying(3.2).

3.2 Existence and uniqueness of weak solution

In this section, we construct a sequence of classical solutions to approximate the weak solution, then we prove the existence and uniqueness of weak solution and establish some related estimates.

For any given initial dataU0∈(L2(0,L))n, by a mollifier procedure, there existUε0(x) in(C∞0[0,L])nwithε>0 so small that

and

Similarly, for any givenT >0, for any given boundary functionH∈(L2(0,T))M, we can findHε(t)∈(C∞0[0,T])Msuch that

and

Thus, for the mixed problem (1.2)–(1.5) (in whichU0(x) =Uε0(x) andH(t) =Hε(t)), the conditions ofC1compatibility are always satisfied at the points (t,x) = (0,0) and (0,L),respectively. By Lemma 2.1, the corresponding mixed problem (1.2)–(1.5) admits a uniqueC1solutionUε(t,x). By Theorem 2.1, noting (3.14) and (3.16), we have

then, we have

By the linearity, forε1/=ε2>0, we have

Noting (3.15) and (3.17), it follows from (3.20) that {Uε} is a Cauchy sequence in (L2(0,T;L2(0,L)))n, hence there existsU(t,x)∈(L2(0,T;L2(0,L)))n, such that

Noting (3.18),U(t,x) also satisfies (2.3).

Moreover, we have

thenU(t,x) satisfies (2.31). Similarly,U(t,x) satisfies (2.30).

Furthermore,by Theorem 2.3,Corollary 2.1 and Corollary 2.2,U(t,x)satisfies(2.46),(2.55)and (2.56). By Theorem 2.4,U(t,x) satisfies (2.58)–(2.61).

Next, we prove thatU(t,x) is a weak solution to problem (1.2)–(1.5). SinceUε(t,x) ∈C1is aC1solution to the corresponding problem (1.2)–(1.5), by (3.1) we have

for any given Φ ∈(C1([0,T]×[0,L]))nsatisfying (3.2). Noting (3.15), (3.17) and (3.21), and takingε→0, we immediately get the existence of weak solution.

Now we prove the uniqueness of weak solution. The idea of the proof was first proposed in[4,pp.27]for the transport equation,it was also mentioned in[1,pp.231]for first order homogeneous linear hyperbolic systems. Assume thatU1andU2are two weak solutions corresponding to the same initial dataU0∈(L2(0,L))nand the same boundary functionH∈(L2(0,T))M.LetU=U1−U2.By (3.1), for any given Φ ∈(C1([0,T]×[0,L]))nsatisfying (3.2), we have

Let {Fn∈(C10[0,L])n} be a sequence of functions such that

Consider the following problem

Let Ψnbe theC1solution to problem (3.26) and let

We have

Moreover,noting (3.25), we have

Substituting (3.27) into (3.26), we have

Thus, Φn(t,x) satisfies (3.2), then, by (3.24) and noting (3.29), we have

Takingn→+∞, we get

SinceT >0 is arbitrarily given, we haveU(t,·) = 0 for any givent∈[0,T]. The proof is complete.

Therefore, we have

Theorem 3.1For any given T >0, for any given initial data U0∈(L2(0,L))nand any given boundary function H∈(L2(0,T))M, the mixed problem(1.2)–(1.5)admits a unique weak solution U=U(t,x) ∈(L2(0,T;L2(0,L)))n, satisfying estimates(2.3),(2.30)–(2.31),(2.46),(2.55)–(2.56)and(2.58)–(2.61).

Remark 3.1Assume that the conditions (2.1) ofC0compatibility at the points (t,x) =(0,0) and (0,L) are satisfied, respectively. Proceeding a similar procedure as above in the function space (C0([0,T];L2(0,L)))n, the regularity of the weak solution can be improved toU=U(t,x)∈(C0([0,T];L2(0,L)))n.

Remark 3.2In Definition 3.1, the trace of the weak solution is not defined on any givenx∈[0,L], however, in Theorem 3.1, (2.46) indicates that the weak solution on any givenx∈[0,L] is in fact anL2function and itsL2norm can be controlled by theL2norm of initial data and boundary function.

Moreover, (2.55) and (2.56) imply that for the corresponding Cauchy problem, the weak solution on the maximum determinate domainRcdepends only on the initial dataU0.

Estimates (2.58)–(2.61) in Theorem 3.1 imply the following theorem.

Theorem 3.2The weak solution U=U(t,x)∈(L2(0,T;L2(0,L)))nto problem(1.2)–(1.5)on the domain R0(resp. RL), defined in Theorem2.4, is uniquely determined by the initial data U0and the boundary condition H+(t)on x=0 (resp. H−(t)on x=L).

For any givenT >0 and any givenf(t,x) ∈(L2(0,T;L2(0,L)))n, one can consider the inhomogeneous system

with the boundary conditions (1.3)–(1.4) and the initial data (1.5) by the method of operator semigroups (see [16, 18]) to get

Theorem 3.3For any given T >0, for any given initial data U0∈(L2(0,L))n, any given boundary function H∈(L2(0,T))Mand any given f(t,x) ∈(L2(0,T;L2(0,L)))n, the mixed problem(3.33)and(1.3)–(1.5)admits a unique weak solution U=U(t,x)∈(L2(0,T;L2(0,L)))n,satisfying

3.3 Backward problem

In studying the controllability and the observability of weak solutions by the constructive method, in order to construct the weak solution to the forward problem under consideration,a backward problem should be solved beforehand. For this purpose, the weak solution to the backward problem will be discussed in this subsection.

Since the coming characteristics and the departing characteristics change their roles for backward problem, for any givenT >0, we consider the following backward system:

with the final data

where the boundary coupling matriceswith constant elements are any given matrices of orderm×and×m, respectively,H= (H−,H+)TwithH−= (h1,···,hm)TandH+=(hm+1,···,hn)T.

Let Ψ be any given function in (C1([0,T]×[0,L]))n, satisfying

We have

Definition 3.2For any given T >0, U=U(t,x)∈(L2(0,T;L2(0,L)))nis a weak solution to the backward mixed problem(3.35)–(3.36), if we have

for any givenΨ ∈(C1([0,T]×[0,L]))nsatisfying(3.37).

Theorem 3.4For any given T >0, for any given final data UT∈(L2(0,L))nand any given boundary function H∈(L2(0,T))n, the backward problem(3.35)–(3.36)admits a unique weak solution U=U(t,x)∈(L2(0,T;L2(0,L)))n, satisfying

If we taket=T−t, x=L−xin (3.35)–(3.36) andthen, the backward problem(3.35)–(3.36)can be formally transformed into a forward problem previously considered:

with the initial data

We now verify the validity of this consideration. In fact,by Theorem 3.1,problem(3.40)–(3.41)admits a unique weak solutionW=W(t,x)∈(L2(0,T;L2(0,L)))n, satisfying

such that

holds for any given Φ ∈(C1([0,T]×[0,L]))nsatisfying

Lett=T−tand=L−xin (3.43)–(3.44), and let

It easily follows from(3.43)–(3.44)thatU=U(t,x)satisfies(3.38),and Ψ ∈(C1([0,T]×[0,L]))nsatisfies (3.37). Therefore, by Definition 3.2,U=U(t,x) is the weak solution to the backward problem (3.35)–(3.36). Noting (3.42) and (3.45),U=U(t,x) satisfies (3.39).

Remark 3.3By the same manner as estimating the solution to the forward problem,under the assumptions of Theorem 3.4, similar estimates as (2.30)–(2.31), (2.46), (2.55)–(2.56) and(2.58)–(2.61) are valid for the backward problem (3.35)–(3.36), for example, we have

and

Remark 3.4Assume that the backward problem of system(1.2)–(1.4)is solvable, namely,assume that the number of positive eigenvalues is equal to that of negative ones,andGi(i=0,1)are reversible, namely,

Then system (1.2)–(1.4) can be equivalently rewritten as

By Theorem 3.4, the backward problem (3.49) and (3.36) is well-posed. Moreover, by the approximation of classical solutions and the uniqueness of weak solution, it is easy to prove that ifU=(t,x) is the weak solution to problem(1.2)–(1.5), then it is also the weak solution to backward problem (3.49) with the final condition

and vice versa. Hence, system (1.2)–(1.4) is time reversible.

We point out that,making an equivalent algebraic transformation by multiplying a full rank matrix from the left on the boundary condition of the system doesn’t influence the definition and the properties of the weak solution. In the following discussion,this property was used and will be used from time to time.

3.4 Some related results

In order to prove the controllability for system (1.2)–(1.4) and the observability for the adjoint system(4.1)by the constructive method,we prepare some useful lemmas in this section.

First of all, for any givenT >0, assume that Φ =(t,x) ∈(L2(0,T;L2(0,L)))nis a weak solution to the following homogeneous backward system

with

where the boundary coupling matriceswith constant elements are any given matrices of orderrespectively. Assume furthermore that

without loss of generality, we assume thatwhereis an reversible matrix of orderThus the boundary condition onx=0 in (3.50) implies

We have

Theorem 3.5Under the assumptions mentioned above, assume furthermore that(3.52)holds, then the weak solutionΦ =(t,x) ∈(L2(0,T;L2(0,L)))nto the backward problem(3.50)–(3.51)is the weak solution to the forward problem(3.54).

ProofBy Definition 3.1, Φ=Φ(t,x)∈(L2(0,T;L2(0,L)))nis a weak solution to problem(3.54), if

holds for any given Ψ ∈(C1([0,T]×[0,L]))nsatisfying

and

It is easy to check that Φ=Φk(t,x) satisfies

Moreover, noting the boundary condition onx= 0 in (3.50), by (3.52), Φ = Φk(t,x) satisfies(3.53). Thus, Φ=Φk(t,x) is theC1solution to system (3.59) and

with the initial data

By Proposition 3.1, Φk(t,x) satisfies

for any given Ψ ∈(C1([0,T]×[0,L]))nsatisfying (3.56). Takingk→∞in (3.62) and noting(3.57) and (3.58), we get (3.55), then Φ=(t,x) is the weak solution to problem (3.54).

Now we assume thatU=(t,x) ∈(L2(0,T;L2(0,L)))nis a weak solution to problem(1.2)–(1.5). Noticingλi/= 0 (i= 1,···,n), and changing the role oftandxin (1.2), we consider the following rightward problem:

with the boundary condition

and with the initial data

We have

Theorem 3.6Under the assumptions mentioned above, the weak solution U=(t,x) ∈(L2(0,T;L2(0,L)))nto the forward problem(1.2)–(1.5)is also the weak solution to the rightward problem(3.63)–(3.65).

ProofMultiplying ΦTon both sides of (3.63), and using the boundary conditions (3.64),it is easy to check thatU=U(t,x) ∈(L2(0,T;L2(0,L)))nis a weak solution to problem(3.63)–(3.65) if

holds for any given Φ ∈(C1([0,T]×[0,L]))nsatisfying

LetU=(t,x) ∈(L2(0,T;L2(0,L)))nbe the weak solution to problem (1.2)–(1.5). By the uniqueness of weak solution, it suffices to prove thatU=(t,x) satisfies (3.66) for any given Φ ∈(C1([0,T]×[0,L]))nsatisfying (3.67). By the procedure of approximation mentioned in Section 3.2,we can find a sequence ofC1solutionsUk(t,x)of problem(1.2)–(1.5)that converges to(t,x) in (L2(0,T;L2(0,L)))nask→∞, such that

and

ask→∞. It is easy to check thatU=Uk(t,x) solves the rightward problem of system (3.63)with the boundary condition

and the initial data

By Proposition 3.1,Uk(t,x) satisfies

for any given Φ ∈(C1([0,T]×[0,L]))nsatisfying (3.67). Takingk→∞in (3.72) and noting(3.68)–(3.69), we have

for any given Φ ∈(C1([0,T]×[0,L]))nsatisfying (3.67), then(t,x) is the weak solution to problem (3.63)–(3.65).

Similarly to the proof of Theorem 3.6, we have

Theorem 3.7Assume that U=(t,x) ∈(L2(0,T;L2(0,L)))nis a weak solution to the backward problem(3.35)–(3.36), then it is also the weak solution to the rightward problem(3.63)–(3.65).

Similarly, for the leftward problem of system (3.63) with the boundary conditions

and the initial data

we have

Theorem 3.8Under the assumptions mentioned above, the weak solution U=(t,x) ∈(L2(0,T;L2(0,L)))nto problem(1.2)–(1.5)is also the weak solution to the leftward problem(3.63)and(3.74)–(3.75).

Theorem 3.9Assume that U=(t,x) ∈(L2(0,T;L2(0,L)))nis a weak solution to the backward problem(3.35)–(3.36), then it is also the weak solution to the leftward problem(3.63)and(3.74)–(3.75).

4 Exact Boundary (Null) Controllability of Weak Solutions by the Constructive Method

In this paper, we will discuss the exact boundary (resp. null) controllability for system(1.2)–(1.4) and the strong (resp. weak) exact boundary observability for the corresponding adjoint system

with the final data

As mentioned in Introduction, first order hyperbolic systems are not always time reversible,then the exact boundary controllability and the exact boundary null controllability for system(1.2)–(1.4)are not equivalent to each other in general(see[12]), hence they should be discussed separately. Correspondingly,for the adjoint system(4.1),the strong exact boundary observability and the weak exact boundary observability should be taken into consideration respectively,too.

Based on Theorem 3.2 and Theorems 3.5–3.9,the constructive method in the framework of classical solutions(see [8, 12])can also be used in the framework of weak solutions to prove the exact boundary (null) controllability and the strong (weak) exact boundary observability for 1-D first order hyperbolic systems.

In this section and in the next section, we will first list some results on controllability and observability obtained by the constructive method in the framework of weak solutions. The procedure of proofs is similar to that in the framework of classical solutions by the constructive method(see[8,12]). In what follows,we only prove the one-sided exact boundary controllability of weak solutions for system (1.2)–(1.4) and the one-sided strong exact boundary observability of weak solutions for the adjoint system (4.1) to show how the constructive method works.

Lemma 4.1(Two-sided exact boundary controllability)Let T≥T0, where

If M=n, namely, M0= rank(D0) =m and M1= rank(D1) =m, then for any given initial data U0(x) ∈(L2(0,L))nand final data UT(x) ∈(L2(0,L))n, there exists a boundary control H∈(L2(0,T))n, satisfying

here and hereafter, C >0denotes a constant depending only on T, such that the mixed problem(1.2)–(1.5)admits a unique weak solution U=U(t,x), satisfying exactly the final state at the time t=T:

Lemma 4.2(One-sided exact boundary controllability)Assume that the number of positive eigenvalues is not larger than that of negative ones:

Assume furthermore that

Let T≥T0, where

If M=M1= rank(D1) =m, then for any given initial data U0(x) ∈(L2(0,L))nand final data UT(x)∈(L2(0,L))n, for any given boundary function H+(t)∈(L2(0,T))m, there exists a boundary control H−(t)∈(L2(0,T))m, satisfying

such that system(1.2)–(1.4)is exactly controllable at the time t=T.

ProofLet

and

First, for any given initial dataU0(x) ∈(L2(0,L))nand boundary functionH+(t) ∈(L2(0,T))monx= 0, we consider the forward problem (1.2)–(1.3) and (1.5) with the following artificial boundary condition onx=L:

whereF(t)∈(L2(0,T1))mis any given function oft, satisfying

By Theorem 3.1, there exists a unique weak solutionU=Uf(t,x) ∈(L2(0;L2(0,L)))nonRf={(t,x)|0 ≤t≤T1,0 ≤x≤L}, satisfying

and we can determine the value ofU=Uf(t,x) onx=0 as

witha(t)∈(L2(0,T1))nsatisfying

Second, noting (4.7), without loss of generality, we may assume that

whereG01is reversible. Then the boundary condition (1.3) onx= 0 can be equivalently rewritten as

For any given final dataUT(x) ∈(L2(0,L))n, we consider the backward problem (2.52) with final data (4.5), boundary condition (4.17) and the following artificial boundary conditions

and

whereP(t)∈(L2(T2,T))(m−m)andQ(t)∈(L2(T2,T))mare any given function oft, satisfying

and

By Theorem 3.1, there exists a unique weak solutionU=Ub(t,x) ∈(L2(T2,T;L2(0,L)))nonRb={(t,x)|T2≤t≤T,0 ≤x≤L}, satisfying

Then, we can determine the value ofU=Ub(t,x) onx=0 as

withb(t)∈(L2(T2,T))nsatisfying

It is easy to check that botha(t) andb(t) satisfy the boundary condition (1.3) onx= 0.Noting(4.8)and(4.10)–(4.11),we may findc(t)∈(L2(0,T))nsatisfying the boundary condition(1.3) onx=0 and

such that

We now change the status oftandxand consider a rightward problem for system (3.63)with the initial data

and the boundary conditions

and

Again, by Theorem 3.1, there exists a unique weak solutionU=U(t,x)∈(L2(0,T;L2(0,L)))nonR(T)={(t,x)|0 ≤t≤T,0 ≤x≤L}, satisfying

We first prove thatU=U(t,x) satisfies the initial condition (1.5) and the final condition(4.5). In fact,by Theorem 3.6,U=Uf(t,x)is also the solution to the rightward problem(3.63)with (4.28),

and

on the domainRf. However, noting (4.10), by Theorem 3.2, the solution to the rightward problem (3.63) with (4.28), (4.31) and (4.32) is uniquely determined on the domain

by (4.28) and (4.32), which are just the initial data and the boundary condition ont= 0 for the rightward problem (3.63) and (4.27)–(4.29). Hence,

on the domainR0. In particular, we have (1.5).

On the other hand, by Theorem 3.7,U=Ub(t,x) is also the solution to the rightward problem (3.63) with (4.29),

and

on the domainRb. However, noting (4.11), by Theorem 3.2, the solution to the rightward problem (3.63) with (4.35)–(4.34) is uniquely determined on the domain

by (4.29) and (4.35). Similarly, we have

on the domainRb. In particular, we have the final condition (4.5).

We now prove thatU=U(t,x) satisfies

for any given Φ ∈(C1([0,T]×[0,L]))nsatisfying

namely,U=U(t,x) satisfies (1.2)–(1.3) with (1.5) and (4.5) in the sense of weak solutions.(The equivalent integral equation (4.37) is formed by a similar manner of (3.1) in Definition 3.1.)

Regarding the following forward problem

as the ’leftward’ problem of the rightward problem (3.63) and (4.27)–(4.29), noting (1.5) and by Theorem 3.8,U(t,x) is also the weak solution to the forward problem (4.39), we haveW(t,x)=U(t,x), then, by Definition 3.1, we get

for any given Φ ∈(C1([0,T]×[0,L]))n. Let Φ ∈(C1([0,T]×[0,L]))nsatisfy (4.38). NotingU(t,0) =c(t), by its construction,c(t) satisfies the boundary condition (1.3) onx= 0, and noticing (4.5), (4.37) follows from (4.40) immediately.

Finally, we determine the boundary control function by

and, noting (4.25), it follows from Theorem 3.1 that

As a conclusion,for any given initial dataU0(x)∈(L2(0,L))n,any given final dataUT(x)∈(L2(0,L))nand any given boundary functionH+(t)∈(L2(0,T))m,we can construct a functionU=U(t,x) ∈(L2(0,T;L2(0,L)))n, which satisfies exactly (1.2)–(1.3), (1.5) and (4.5) in the sense of weak solutions, and by restrictingU=U(t,x) on the boundaryx=L, we obtain a boundary control functionH−(t) ∈(L2(0,T))mgiven by (4.41). Thus, with this boundary controlH−(t),U=U(t,x) is the weak solution to problem (1.2)–(1.5), which satisfies exactly the final data (4.5) at the timet=T. Hence, system (1.2)–(1.4) is exactly controllable at the timet=T.

Remark 4.1Assumptions (4.6)–(4.7) guarantee that the constructive method with suitable artificial boundary conditions can be applied to system (1.2)–(1.4) for realizing the exact boundary controllability for weak solutions.

Remark 4.2The case of two-sided exact boundary controllability with fewer boundary controls can be similarly discussed by the constructive method.

The constructive method can be also applied to prove the exact boundary null controllability for system (1.2)–(1.4). Taking the one-sided controls case as an example, we have

Lemma 4.3(One-sided exact boundary null controllability)Let T≥(4.8). Assume that H−(t)≡0 (resp. H+(t)≡0), if M=M0=rank(D0)=(resp.M=M1=rank(D1)=m), then for any given initial data U0(x) ∈(L2(0,L))n, there exists a boundary control H+(t)∈(L2(0,T))m(resp. H−(t)∈(L2(0,T))m), satisfying

such that the corresponding mixed problem(1.2)–(1.5)admits a unique weak solution U=U(t,x), satisfying exactly the null final state at the time t=T:

Remark 4.3Under the assumptions of Lemma 4.3, since the system has homogeneous boundary conditions on the side without controls, we can setU≡0 inRb, instead of solving a backward problem as in the proof of Lemma 4.2,hence assumptions(4.6)–(4.7)are not required.

Moreover, for system with homogenous boundary conditions on the side without controls,boundary controls can be acted on either side of the boundary,so that the number of boundary controls can be further reduced tom(≤m).

In general, the one-sided exact boundary controllability and the one-sided exact boundary null controllability are not equivalent. If system (1.2)–(1.4) is exactly controllable, then it must be exactly null controllable. However, in order to deduce the one-sided exact boundary controllability from the one-sided exact boundary null controllability for system(1.2)–(1.4),the number of positive eigenvalues of Λ should be equal to that of negative ones, and the boundary coupling matricesG0andG1should be reversible, namely, we have

Theorem 4.1The one-sided exact boundary controllability and the one-sided exact boundary null controllability are equivalent if the following hypotheses are satisfied:

Remark 4.4Under the assumptions of Theorem 4.1, controls can apply on either side of the boundary, and the number of boundary controls is equal toM=m=m.

5 Strong(weak)Exact Boundary Observability for the Adjoint System by the Constructive Method

For any given final data ΦT∈(L2(0,L))n, the well-posedness of the backward problem(4.1)–(4.2) in the sense of weak solutions is guaranteed by Theorem 3.4. By the constructive method, we have the following result on the strong observability of the adjoint system (4.1).

Lemma 5.1(Two-sided strong exact boundary observability)Assume T≥T0, where T0is given by(4.3). Assume thatΦ = Φ(t,x)is the weak solution to the adjoint problem(4.1)–(4.2). For any given final dataΦT(x) ∈(L2(0,L))n, the boundary observationsΦ−(t,L)andΦ+(t,0)corresponding to the departing characteristics on x=L and on x= 0, respectively,on the interval(0,T)can be used to uniquely determine the final dataΦT(x), and we have the following strong observability inequality:

In particular, if

then(5.1)can be written as

Lemma 5.2(One-sided strong exact boundary observability)Let T≥T0, where T0is given by(4.8). Assume thatΦ=Φ(t,x)is the weak solution to the adjoint problem(4.1)–(4.2).Assume furthermore that(4.6)–(4.7)hold. For any given final dataΦT(x) ∈(L2(0,L))n, the boundary observationΦ−(t,L)corresponding to the departing characteristics on x=L(the side with fewer coming characteristics)on the interval(0,T)can be used to uniquely determine the final dataΦT(x), and we have the following strong observation inequality:

In particular, if

then(5.4)can be written as

ProofAssume that Φ=(t,x)∈(L2(0,T;L2(0,L)))nis a weak solution to the backward problem (4.1)–(4.2) with the final data∈(L2(0,L))n.

Notingλi/=0 (i=1,···,n), the following leftward problem

and

admits a unique weak solution. By Theorem 3.9,the weak solution Φ=(t,x)to the backward problem (4.1)–(4.2) is also the weak solution to the leftward problem (5.7)–(5.8).

Without loss of generality, assume that

Drawing themthand the (m+1)thcharacteristic linesx=xm(t) andx=xm+1(t) passing through the point (0,L) and (T,L), respectively, we have

Noting (4.8), the triangular domainRl, surrounded byx=xm(t),x=xm+1(t) andx=L,intersectsx=0 (see Figure 1).

Figure 1 Triangular domain Rl

By(2.56)in Theorem 3.1 for the Cauchy problem,the solution Φ=(t,x)onRlis uniquely determined by (5.8). Then, noting the boundary conditions in (4.1), there existst=t∗(0

By (4.6)–(4.7), the boundary condition onx=0 in (4.1)implies that there exists an×mmatrixsuch that

then, noting (5.11), we get immediately

In particular, under assumption (5.5),D1is reversible, and Λ−is also reversible, hence it follows from (5.15) that

The proof is complete.

Remark 5.1In Lemma 5.2, assumptions (4.6) and (4.7) guarantee that we can apply the constructive method with suitable boundary observations to get the observability for the adjoint system.

Remark 5.2We can similarly discuss the two-sided strong exact boundary observability with fewer boundary observations.

Remark 5.3The weak exact boundary observability for the adjoint system (4.1) can be similarly studied by means of the constructive method. In this case, the initial data of the solution to the backward adjoint problem (4.1)–(4.2) can be uniquely determined by boundary observations. In one-sided observation case, differently from Lemma 5.2, we don’t need assumptions (4.6)–(4.7) for the weak exact boundary observability, boundary observations can be on either side of the boundary, and the number of boundary observations is equal tomorm, respectively.

Correspondingly, we have

Lemma 5.3(One-sided weak exact boundary observability)Let T≥T0, where T0is given by(4.8). For any given initial dataΦ0(x)∈(L2(0,L))n,

(1)the boundary observationΦ−(t,L)corresponding to all the departing characteristics on x=L on the interval(0,T)can be used to uniquely determine the initial dataΦ0(x)at t=0,and we have the following weak observability inequality:

In particular, if M=M1=rank(D1)=m, then(5.17)can be written as

(2)the boundary observationΦ+(t,0)corresponding to all the departing characteristics on x= 0on the interval(0,T)can be used to uniquely determine the initial dataΦ0(x)at t= 0,and we have the following weak observability inequality:

In particular, if M=M0=rank(D0)=(5.19)can be written as

Remark 5.4Apparently,if system(4.1)possesses the strong exact boundary observability,then it must possess the weak exact boundary observability. Conversely, in order to obtain the one-sided strong exact boundary observability from the one-sided weak exact boundary observability, system (4.1) should satisfy (4.45), namely, it is time reversible. In this case,observations can be on either side of the boundary, and the number of boundary observationsM=m=.

Lemma 5.4If system(1.2)–(1.4)is exactly null controllable under boundary controls H(t)∈(L2(0,T))M(M=M0+M1≤n), then the adjoint system(4.1)satisfies the following weak D0/D1-observability:

ProofBy Definition 3.1,and noting the adjoint system(4.1),if system(1.2)–(1.4)is exactly boundary null controllable, then for any given initial dataU0(x) ∈(L2(0,L))nand null final dataUT(x)≡0, there exist boundary controlsH(t)∈(L2(0,T))M, such that

IfDT0Λ+Φ+(t,0)=0 andDT1Λ−Φ−(t,L)=0, then the left-hand side of (5.22) is equal to zero for any given initial dataU0(x) in (L2(0,L))n, thus Φ(0,t)≡0.The proof is complete.

Combining Lemmas 5.3–5.4, we have

Corollary 5.1If the number of boundary observations can not be reduced for the one-sided weak exact boundary observability for the adjoint system(4.1), then the number of boundary controls can not be reduced for the one-sided exact boundary null controllability for system(1.2)–(1.4).