Download - Positive Linear Observers for Linear Compartmental Systems

POSITIVE LINEAR OBSERVERS FOR LINEAR COMPARTMENTALSYSTEMS∗

J. M. VAN DEN HOF†

SIAM J. CONTROL OPTIM. c© 1998 Society for Industrial and Applied MathematicsVol. 36, No. 2, pp. 590–608, March 1998 007

Abstract. Linear compartmental systems are mathematical systems that are frequently usedin biology and mathematics. The inputs, states, and outputs of such systems are positive, becausethey denote amounts or concentrations of material. For linear dynamic systems the observer problemhas been solved. The purpose of the observer problem is to determine a linear observer such thatthe state can be approximated. The difference between the state and its estimate should convergeto zero. The interpretation in terms of a physical system requires that an estimate of the state bepositive, like the state itself. In this paper conditions on the system matrices are presented thatguarantee that there exists a positive linear observer such that both the error converges to zero andthe estimate is positive.

Key words. compartmental systems, positive linear observers, asymptotic stability

AMS subject classifications. 93D20, 15A48

PII. S036301299630611X

1. Introduction. The purpose of this paper is to derive positive linear observersfor linear compartmental systems.

Compartmental systems are mathematical systems that are frequently used inbiology and mathematics. In addition, a subclass of the class of chemical processescan be modeled as compartmental systems. A compartmental system consists ofseveral compartments with more or less homogeneous amounts of material. The com-partments interact by processes of transportation and diffusion. The dynamics of acompartmental system are derived from mass balance considerations.

In this paper linear compartmental systems consisting of inputs, states, and out-puts will be studied. The outputs of these systems are not the real outputs, i.e.,material leaving the system, but the observations of the amount or concentrations ofmaterial, for example, in one or more compartments. The inputs, states, and outputsare positive, so these systems are called positive linear systems in system theory. Asin linear system theory, the purpose is to determine a linear observer such that thestate x can be approximated by x. The error, x(t) − x(t), should converge to zero.For positive linear systems, the observer provides an approximation of the positivestate. Therefore, the observer should be chosen in such a way that the approximationof the state, x(t), is positive, like the state, x(t), itself.

For linear systems the observer problem has been solved by Luenberger [11]. Seealso [10]. As far as we know, there is no literature on positive observers for positivelinear systems, in which the positivity of x(t) is taken into account. It turns out thatthe existence of a positive linear observer satisfying the above conditions dependslargely on the structure of the system matrices, i.e., the zero/nonzero pattern. Somerelation can be found in the work of Sontag [13, 14].

The outline of the paper is as follows. In section 2 the problem is posed. In sec-tion 3 continuous-time linear compartmental systems are considered, and in section 4the discrete-time case is treated. Concluding remarks are made in section 5.

∗Received by the editors July 8, 1996; accepted for publication (in revised form) January 13, 1997.http://www.siam.org/journals/sicon/36-2/30611.html†CBS (Statistics Netherlands), P.O. Box 4000, NL 2270 JM Voorburg, the Netherlands (jhof@

cbs.nl).

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POSITIVE OBSERVERS FOR LINEAR COMPARTMENTAL SYSTEMS 591

2. Problem formulation. In this section some notation is introduced and theproblem is posed.

The set R+ = [0,+∞) is called the set of the positive real numbers. Let Z+ ={1, 2, . . .} denote the set of positive integers, Zn = {1, . . . , n}, and NI = {0, 1, 2, . . .}.Denote by Rn+ the set of n-tuples of the positive real numbers. The set Rn×m+ willbe called the set of positive matrices of size n by m. Note that Rn+ is not a vectorspace because it does not admit an inverse with respect to addition. For matricesA,B ∈ Rn×m, we will write A ≥ B if aij ≥ bij for all i ∈ Zn, j ∈ Zm, and A > Bif A ≥ B and A 6= B. A matrix A ∈ Rn×n is said to be a Metzler matrix if all itsoff-diagonal elements are in R+; see [9]. Metzler matrices can be characterized asfollows.

PROPOSITION 2.1. A matrix A ∈ Rn×n is a Metzler matrix if and only if thereexists an α ∈ R+ such that (A+ αI) ∈ Rn×n+ .

DEFINITION 2.2. Consider a continuous-time linear dynamic system

x(t) = Ax(t) +Bu(t), x(t0) = x0,y(t) = Cx(t) +Du(t),(2.1)

with x(t) ∈ X ⊂ Rn, u(t) ∈ U ⊂ Rm, y(t) ∈ Y ⊂ Rk, t ∈ T = [t0,∞). Equation (2.1)is said to represent a (continuous-time) positive linear system if for all x0 ∈ Rn+ andfor all u(t) ∈ Rm+ , t ∈ T , we have x(t) ∈ Rn+ and y(t) ∈ Rk+ for t ∈ T ; in other words,X = Rn+, U = Rm+ , and Y = Rk+.

The following proposition provides a characterization of continuous-time positivelinear systems.

PROPOSITION 2.3. A continuous-time linear dynamic system of the form (2.1) isa positive linear system if and only if

B ∈ Rn×m+ , C ∈ Rk×n+ , D ∈ Rk×m+ , and A is a Metzler matrix.

Proof. Suppose first u(t) = 0 for all t ∈ T . For i ∈ Zn, xi(t) ≥ 0 if and only ifxi ≥ 0 whenever xi = 0 and xj ≥ 0 for all j 6= i. This is equivalent to aij ≥ 0 forall j 6= i. Moreover, y(t) = Cx(t) ≥ 0 for x(t) ≥ 0 if and only if C ∈ Rk×n+ . Nowsuppose u(t) 6= 0. For i ∈ Zn, xi(t) ≥ 0 if and only if xi ≥ 0 whenever xj = 0 forall j ∈ Zn. This is equivalent to bir ≥ 0 for r ∈ Zm. Furthermore, if x(t) = 0, theny(t) = Du(t) ≥ 0 if and only if D ∈ Rk×m+ .

For discrete time, the definition of a positive linear system is presented below.DEFINITION 2.4. Consider a discrete-time linear dynamic system

x(t+ 1) = Ax(t) +Bu(t), x(0) = x0,y(t) = Cx(t) +Du(t),(2.2)

with x ∈ X ⊂ Rn, u ∈ U ⊂ Rm, y ∈ Y ⊂ Rk, t ∈ T = NI . Equation (2.2) is saidto represent a (discrete-time) positive linear system if for all x0 ∈ Rn+ and for allu(t) ∈ Rm+ , t ∈ T , we have x(t) ∈ Rn+ and y(t) ∈ Rk+ for t ∈ T ; in other words,X = Rn+, U = Rm+ , and Y = Rk+.

A characterization of discrete-time positive linear systems is as follows.PROPOSITION 2.5. A discrete-time linear system of the form (2.2) is a positive

linear system if and only if

A ∈ Rn×n+ , B ∈ Rn×m+ , C ∈ Rk×n+ , D ∈ Rk×m+ .

The positive linear observer problem is as follows. A positive linear observer fora positive linear system is a positive linear system described by the equations

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592 J. M. VAN DEN HOF

˙x(t) = Hx(t) +Ky(t) + Eu(t), x(t0) = x0,

x(t+ 1) = Hx(t) +Ky(t) + Eu(t), x(t0) = x0,

for the continuous-time case and the discrete-time case, respectively, which yields anestimate x(t) of the state x(t) at time t ∈ T of system (2.1), (2.2), respectively. As inlinear system theory, the observer has to satisfy the following two conditions:

1. x(t0) = x(t0) implies x(t) = x(t) for all t ≥ t0 and for all input functionsu(t), t ≥ t0;

2. x(t) should converge to x(t) for t→∞, for all input functions u(t), t ≥ t0.For linear systems the problem of finding an observer satisfying 1 and 2 has beencompletely solved [11]. The solution is

˙x(t) = (A−KC)x(t) +Ky(t) +Bu(t),x(t+ 1) = (A−KC)x(t) +Ky(t) +Bu(t),

respectively, with K ∈ Rn×k such that A −KC is asymptotically stable; i.e., for thecontinuous-time case, σ(A −KC) ⊆ {λ ∈ C | Re(λ) < 0}, and for the discrete-timecase, σ(A −KC) ⊆ {λ ∈ C | |λ| < 1}. Here σ(A) denotes the spectrum of A. Thenecessary and sufficient conditions for the existence of a matrix K ∈ Rn×k such thatA − KC is asymptotically stable depend on the matrices A and C; i.e., the pair(A,C) should be detectable. Equivalent conditions for detectability can be foundin, for example, [3, pp. 259 and 293], respectively. The interpretation in terms of aphysical system requires that an estimate x(t) be, like x(t), positive. So a positivelinear observer for a positive linear system should also satisfy the following condition:

3. x(t) ∈ Rn+, for all t ≥ t0, if x(t0) ∈ Rn+, y(t) ∈ Rk+, and u(t) ∈ Rm+ for allt ≥ t0.

This third condition is satisfied if and only if K ∈ Rn×k+ and, for the continuous-timecase, A−KC is a Metzler matrix, or for the discrete-time case, A−KC ∈ Rn×n+ . Thisfollows from Propositions 2.3 and 2.5, respectively. Now detectability of (A,C) definedin [3] cannot be used, because then it may be possible that K /∈ Rn×k+ . Of course,detectability is a necessary condition but is not sufficient. Therefore, new necessaryand sufficient conditions on A and C have to be found. The problem considered inthis paper is stated below.

Problem 2.6.Continuous time. Formulate necessary and sufficient conditions on a Metzler

matrix A ∈ Rn×n and a positive matrix C ∈ Rk×n+ such that there exists aK ∈ Rn×k+ , K 6= 0, with

1. A−KC a Metzler matrix;2. σ(A−KC) ⊆ {λ ∈ C | Re(λ) < 0}.

Discrete time. Formulate necessary and sufficient conditions on positive ma-trices A ∈ Rn×n+ and C ∈ Rk×n+ such that there exists a K ∈ Rn×k+ , K 6= 0,with

1. A−KC ∈ Rn×n+ ;2. σ(A−KC) ⊆ {λ ∈ C | |λ| < 1}.

These problems will be solved for linear compartmental systems, which form asubclass of positive linear systems.

3. Continuous time. In this section conditions for the existence of a positivelinear observer for continuous-time linear compartmental systems will be derived.First results from the theory on compartmental systems will be presented.

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F0i

qi

Fji

Fij

Ii

FIG. 3.1. One compartment with possible flows.

3.1. Continuous-time compartmental systems. A compartmental systemis a system consisting of a finite number of subsystems, which are called compart-ments. Each compartment is kinetically homogeneous; i.e., any material entering thecompartment is instantaneously mixed with the material of the compartment. Com-partmental systems are dominated by the law of conservation of mass. They also formnatural models for other areas of applications that are subject to conservation laws.

Consider an n-compartmental system. The behavior of the ith compartment canbe represented as in Figure 3.1. In this figure, qi denotes the amount of materialconsidered in compartment i. The arrows represent the flows into and out of the com-partment. Ii ≥ 0 is the flow into compartment i from outside the system, called theinflow. Fij ≥ 0 and Fji ≥ 0 represent the flow from compartment j into compartmenti and the flow from compartment i into compartment j, respectively. Finally, F0i ≥ 0is the outflow to the environment from compartment i. The mass balance equationsfor every compartment can be written as

qi =∑j 6=i

(−Fji + Fij) + Ii − F0i.(3.1)

In this paper the flows Fij will be assumed to be linearly dependent on qj :

Fij = fijqj , i = 0, . . . , n, j = 1, . . . , n, i 6= j,

in which fij are called the fractional transfer coefficients. In general, fij are functionsof q and time t. If fij is independent of q, the system is a linear system. In thispaper it is assumed that fij is also independent of the time t; i.e., the system is atime-invariant linear system. Using this, (3.1) can be written as

q = Fq + I,

where q =(q1 · · · qn

)T ∈ Rn+, F = (fij) ∈ Rn×n, with fii = −(f0i +∑j 6=i fji)

and fij constant for i 6= j, and I denotes the inflow from outside the system. Sinceqi ≥ 0 and Ii ≥ 0, this system is easily seen to be a positive linear system, if theoutput is taken as

y = Cq, y ∈ Rk, C ∈ Rk×n+ ,

where y denotes the vector of the observations. Note that the output is not the outflowof the compartmental system. The outflow, which is sometimes also called excretion,represents the flow of material leaving the system. The outputs of an experiment are

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measurements and usually differ from the material outflows. On the other hand, theterms inflow and input can be used interchangeably.

Another property of compartmental systems is that the total flow out of a com-partment over any time interval cannot be larger than the amount that was initiallypresent plus the amount that flowed into the compartment during that interval. To-gether with the constraints on positive linear systems, this comes down to

1. fij ≥ 0 for all i, j ∈ Zn, i 6= j,

2. − fjj ≥n∑

i=1,i 6=jfij ≥ 0 for all j ∈ Zn.

A matrix F satisfying conditions 1 and 2 above is said to be a compartmental matrix.There is an extensive amount of literature on compartmental systems. See, for exam-ple, [1, 7, 8, 9]. Condition 2 states that all column sums of F are less than or equalto zero.

Below some properties of compartmental matrices from the literature will bediscussed that are needed in this paper. References are [5, 6, 9, 15].

DEFINITION 3.1. A matrix A ∈ Rn×n is said to be reducible if there exists apermutation matrix P ∈ Rn×n such that

PAPT =(U 0Q R

),

with U and R square matrices. A is said to be irreducible if A is not reducible.Let F ∈ Rn×n be a compartmental matrix. Then it follows from [2, Theo-

rem 6.4.6] that σ(F ) ⊆ {λ ∈ C | Re(λ) < 0 or λ = 0}. Since a system x = Fx isasymptotically stable if and only if σ(F ) ⊆ {λ ∈ C | Re(λ) < 0}, a compartmentalmatrix is asymptotically stable if and only if 0 /∈ σ(F ). In the rest of this subsectioncompartmental matrices with zero eigenvalues are characterized.

PROPOSITION 3.2 (adapted from [15, Theorem III]). Let F ∈ Rn×n be an irre-ducible compartmental matrix. Then 0 ∈ σ(F ) if and only if

∑ni=1 fij = 0 for all

j ∈ ZnDEFINITION 3.3. Consider an n-compartmental system. A trap is a compartment

or a set of compartments from which there are no transfers or flows to the environmentnor to compartments that are not in that set. A trap is said to be simple if it does notstrictly contain a trap.

In the physical literature traps are usually referred to as sinks.Let S be a linear compartmental system consisting of the compartments C1,

C2, . . . , Cn and let qj be the amount of material in Cj . Let T ⊆ S be a subsys-tem of S. Renumbering the compartments, assume T consists of the compartmentsCm, Cm+1, . . . , Cn, for m ≤ n. Let F ∈ Rn×n be the compartmental matrix cor-responding to S, consistent with this renumbering. Then T is a trap if and onlyif

fij = 0 for all (i, j) such that j = m,m+ 1, . . . , n, i = 0, 1, . . . ,m− 1.(3.2)

The following two theorems are due to Fife [5].THEOREM 3.4. S contains a trap if and only if one of the following conditions

holds:1. for all j ∈ Zn

n∑i=1

fij = 0;

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2. there exists a permutation matrix P ∈ Rn×n such that

PFPT =(U 0Q R

),

with U , R square matrices and the sum of entries of every column of R is zero.THEOREM 3.5. S contains a trap if and only if 0 ∈ σ(F ).In response to Fife [5], Foster and Jacquez [6] derived the following result. See

also Theorems 1 and 2 together with their proofs in [9].THEOREM 3.6. Let S be a compartmental system with system matrix F .

1. Zero is an eigenvalue of F of multiplicity m ∈ Z+ if and only if S containsm simple traps.

2. Assume zero is an eigenvalue of F of multiplicity m ∈ Z+. Then there existsa partition of S into a disjoint union of subsystems

S = S1 ∪ S2 ∪ · · · ∪ Sp

such that Si receives no input from Si+1, . . . , Sp, i = 1, . . . , p−1, and Sp−m+1, . . . , Spare traps. Relative to this partition the system matrix is given by

PFPT = F =

F11 0 0 0 · · · 0...

. . . 0Fp−m,1 Fp−m,p−m 0Fp−m+1,1 Fp−m+1,p−m Fp−m+1,p−m+1

...... 0

. . . 0Fp1 · · · Fp,p−m 0 0 Fpp

,

where Fii is irreducible for all i ∈ Zp and zero is an eigenvalue of Fii of multiplicity1 for i = p − m + 1, . . . , p, and the sum of entries of every column of Fii, i =p−m+ 1, . . . , p, is zero.

An additional consequence of this theorem is that if zero is an eigenvalue of acompartmental matrix of (algebraic) multiplicity m, the geometric multiplicity is alsom, so there are always m independent eigenvalues for the eigenvalue zero.

3.2. Positive linear observers. In this subsection conditions for the existenceof a positive linear observer for continuous-time linear compartmental systems willbe derived. Consider a compartmental matrix F ∈ Rn×n. If F ∈ Rn×n is a com-partmental matrix and K ∈ Rn×k+ , C ∈ Rk×n+ are such that F − KC is a Metzlermatrix, then F −KC is also a compartmental matrix, since condition 1 in section 3.1is satisfied because F −KC is a Metzler matrix and condition 2 becomes

n∑i=1

(F −KC)ij =n∑i=1

fij − (KC)ij ≤n∑i=1

fij ≤ 0.

Therefore, for the special class of compartmental matrices, the problem to be solvedis the following.

Problem 3.7. Formulate necessary and sufficient conditions on a compartmentalmatrix F ∈ Rn×n and a positive matrix C ∈ Rk×n+ such that there exists a K ∈ Rn×k+ ,K 6= 0, with

1. F −KC a compartmental matrix;2. σ(F −KC) ⊆ {λ ∈ C | Re(λ) < 0}.

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To solve this problem, the notions of positive modifiability and positive detectabil-ity will be defined.

DEFINITION 3.8. Let F ∈ Rn×n be a compartmental matrix and C ∈ Rk×n+ . Thematrix pair (F,C) is said to be positively modifiable if there exists a K ∈ Rn×k+ suchthat KC 6= 0 and F−KC is a compartmental matrix. This implies that σ(F−KC) ⊆{λ ∈ C | Re(λ) < 0 or λ = 0}. (F,C) is said to be positively detectable if there existsa K ∈ Rn×k+ such that KC 6= 0 and F−KC is an asymptotically stable compartmentalmatrix. This implies that σ(F −KC) ⊆ {λ ∈ C | Re(λ) < 0}.

Note that solving Problem 3.7 is equivalent to checking positive detectability.To solve Problem 3.7, positive modifiability will be used, for which the followingcharacterization can be given.

PROPOSITION 3.9. Let F ∈ Rn×n be a compartmental matrix and C ∈ Rk×n+ .The matrix pair (F,C) is positively modifiable if and only if there exists an i ∈ Znand an r ∈ Zk such that the rth row in C is nonzero and

{for all j 6= i with crj 6= 0, also fij 6= 0}.(3.3)

Remark 3.10. In terms of compartments, Proposition 3.9 can be interpretedas follows: an output can be seen as a strictly positive linear combination of oneor more compartments. These compartments contribute to this output. For positivemodifiability there should exist an output such that all the compartments contributingto this output have a direct flow to one and the same compartment. A compartmentalsystem with system matrix F can be represented by a unique directed graph; see,for example, [4] or [8, Chapter 3]. Every compartment is represented by a vertexand there is a directed arc from xi to xj if and only if fji > 0. If compartmenti contributes to output j, i.e., cji 6= 0, this will be represented by a dashed arc.The claim in Proposition 3.9 is equivalent to saying that the graph of F shouldcontain a subgraph of the form given in Figure 3.2(a) if the considered output hasone contributing compartment or, for example, Figure 3.2(b) if the considered outputhas three contributing compartments.

Proof. (⇒) Assume (F,C) is positively modifiable, so a K ∈ Rn×k+ can be foundsuch that KC 6= 0 and F − KC is a compartmental matrix. KC ∈ Rn×n+ , sinceK ∈ Rn×k+ and C ∈ Rk×n+ . Therefore there exist i, s ∈ Zn such that 0 < (KC)is =∑kt=1 kitcts. Hence there exists an r ∈ Zk such that kircrs > 0, which implies kir > 0

and crs > 0. From this it follows that the rth row in C is nonzero. Next, the followingholds for j ∈ Zn, j 6= i, since F −KC is a compartmental matrix,

0 ≤ (F −KC)ij = fij −k∑t=1

kitctj .(3.4)

Suppose crj 6= 0; i.e., crj > 0. Since kir > 0, this implies∑kt=1 kitctj ≥ kircrj > 0.

Then it follows from (3.4) that fij > 0. So there exist i ∈ Zn, r ∈ Zk such that rowr in C is nonzero and for all j 6= i with crj 6= 0, also, fij 6= 0.

(⇐) Assume there exist i ∈ Zn, r ∈ Zk such that row r in C is nonzero, and forall j 6= i with crj 6= 0, fij 6= 0 also. Since row r in C is nonzero and C ∈ Rk×n+ ,there exists either an s ∈ Zn \ {i} such that crs > 0, or crs = 0 for all s ∈ Zn \ {i}and cri > 0. Assume first that there exists an s ∈ Zn \ {i} such that crs > 0. Byassumption, this implies fis > 0, and in general, for all v ∈ Zn \ {i}, crv > 0 impliesfiv > 0. Now take

0 < kir < minv∈Zn\{i},crv>0

fivcrv

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(a) (b)

xi

yr

xi

xj

yr

xq

yr

xi

xj

xq

xs

xi xj

yr

xq

FIG. 3.2. Subgraphs.

and all other entries of K equal to zero. Then K ∈ Rn×k+ and

(F −KC)iw = fiw − kircrw ={fiw − kircrw ∈ (0, fiw), if crw > 0, w 6= i,fiw ≥ 0, if crw = 0, w 6= i;

(F −KC)hw = fhw ≥ 0 for h 6= i, w 6= h;

(F −KC)ii = fii − kircri ≤ fii ≤ −n∑

q=1,q 6=ifqi ≤ −

n∑q=1,q 6=i

(F −KC)qi;

(F −KC)hh = fhh ≤ −n∑

q=1,q 6=hfqh ≤ −(fih − kircrh)−

∑q=1,q 6=h,q 6=i

fqh

= −∑

q=1,q 6=h(F −KC)qh for h 6= i.

It follows that K given above satisfies KC 6= 0, and F −KC satisfies the conditionsfor a compartmental matrix.

Now assume crs = 0 for all s ∈ Zn \ {i} and cri > 0. Take kir > 0, any positiveconstant, and all other entries of K equal to zero. Then K ∈ Rn×k+ ,

(F −KC)hw = fhw ≥ 0 for h ∈ Zn, w 6= h;

(F −KC)hh = fhh ≤ −n∑

q=1,q 6=hfqh = −

∑q=1,q 6=h

(F −KC)qh for h 6= i;

(F −KC)ii = fii − kircri < fii ≤ −n∑

q=1,q 6=ifqi ≤ −

n∑q=1,q 6=i

(F −KC)qi.

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Again it follows that K given above satisfies KC 6= 0, and F − KC satisfies theconditions for a compartmental matrix.

A matrix K ∈ Rn×k+ such that KC 6= 0 and F −KC is a compartmental matrixcan be found by the following algorithm.

ALGORITHM 3.11. Consider F ∈ Rn×n, C ∈ Rk×n+ . Define the sets

Rc = {(i, r) ∈ Zn×Zk | row r of C is nonzero and (3.3) holds for (i, r)}and

T(i,r) = {j ∈ Zn \ {i} | crj 6= 0}.

Form the matrix K ∈ Rn×k+ as follows.1. For every pair (i, r) ∈ Rc with T(i,r) 6= ∅, take

0 ≤ kir < minj∈T(i,r)

fijcrj

.

2. For every pair (i, r) /∈ Rc with T(i,r) 6= ∅, take kir = 0.3. For every pair (i, r) ∈ Zn × Zk with T(i,r) = ∅, take any positive constant

kir ≥ 0.Of course, for KC to be nonzero, at least one kir, for a pair (i, r) ∈ Rc, should

be strictly positive. It follows from Proposition 3.9 that the set Rc is nonempty ifand only if the pair (F,C) is positively modifiable, and if this is the case, K can bechosen in such a way that KC 6= 0.

Before presenting the main theorem of this section, the following proposition isstated.

PROPOSITION 3.12. Let F ∈ Rn×n be an irreducible compartmental matrix. As-sume C ∈ Rk×n+ and K ∈ Rn×k+ are such that KC 6= 0 and F−KC is a compartmentalmatrix. Then F −KC is asymptotically stable.

Proof. Let

K =

K1...Kn

, with Ki ∈ R1×k+ , C =

(C1 · · · Cn

), with Ci ∈ Rk×1

+ .

From K ∈ Rn×k+ and C ∈ Rk×n+ it follows that (F −KC)rs ≤ frs for all r, s ∈ Zn.First, assume F − KC is irreducible. Since KC 6= 0, there exist i, j ∈ Zn such

that (KC)ij > 0, or equivalently, (F −KC)ij < fij . It follows thatn∑q=1

(F −KC)qj <n∑q=1

fqj ≤ 0.

With Proposition 3.2 this implies that zero is not an eigenvalue of F−KC, so F−KCis asymptotically stable.

Now, assume F − KC is reducible. Suppose zero is an eigenvalue of F − KC.Without loss of generality it may be assumed that

F −KC =(U 0Q R

),

where U ∈ Rr×r, with 1 ≤ r < n, R ∈ R(n−r)×(n−r), and the sum of entries of everycolumn of R is zero (see Theorem 3.4). For all j = 1, . . . , n− r,

n−r∑i=1

Rij =n∑i=1

(F −KC)i,r+j =n∑i=1

(fi,r+j −KiCr+j).

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Since F is irreducible, there exists a t ∈ Zr such that ft,r+j > 0, and because ft,r+j −KtCr+j = (F −KC)t,r+j = 0, KtCr+j = ft,r+j > 0. This implies

n−r∑i=1

Rij =n∑i=1

(fi,r+j −KiCr+j) <n∑i=1

fi,r+j ≤ 0.

This contradicts the requirements on R. It follows that zero is not an eigenvalue ofF −KC, so F −KC is asymptotically stable.

From Proposition 3.12 it follows that for an irreducible compartmental matrix F ∈Rn×n and a positive matrix C ∈ Rk×n+ , positive modifiability of (F,C) is equivalentto positive detectability of (F,C).

Consider a linear compartmental system S. Assume S contains m ≥ 0 traps. Ifm ≥ 1, then S can be partitioned as in Theorem 3.6, with system matrix

F =

F11 0 0 0 · · · 0...

. . . 0Fp−m,1 Fp−m,p−m 0Fp−m+1,1 Fp−m+1,p−m Fp−m+1,p−m+1

...... 0

. . . 0Fp1 · · · Fp,p−m 0 0 Fpp

∈ Rn×n,(3.5)

in which Fii is irreducible for all i ∈ Zp and the sum of entries of every column of thesquare matrices Fp−m+1,p−m+1, . . . , Fpp equals zero. Note that

1. 0 /∈ σ(Fii) for i = 1, . . . , p−m;2. 0 ∈ σ(Fii) with multiplicity 1 for i = p−m+ 1, . . . , p.

Consider C ∈ Rk×n+ , K ∈ Rn×k+ , and decompose them to conform to the partition in(3.5):

C =(C1 · · · Cp

), K =

K1...Kp

.(3.6)

Now the main theorem of this subsection can be stated. It solves Problem 3.7.THEOREM 3.13. Consider a linear compartmental system S, as defined above,

with m ≥ 0 traps.1. If m = 0, then F − KC is asymptotically stable for all K ∈ Rn×k+ with

F −KC a compartmental matrix. Moreover, (F,C) is positively detectable if and onlyif (F,C) is positively modifiable.

2. For m ≥ 1, let F , K, and C be partitioned as in (3.5) and (3.6). (F,C) ispositively detectable if and only if (Fii, Ci) is positively modifiable for all i = p−m+1, . . . , p.

Proof. 1. Since S contains no traps, it follows from Theorem 3.5 that 0 /∈ σ(F ),so F itself is asymptotically stable. Let K ∈ Rn×k+ be such that F − KC is acompartmental matrix. Suppose 0 ∈ σ(F − KC). Because 0 /∈ σ(F ), KC 6= 0 andwith Theorem 3.4 it follows that either

n∑q=1

(F −KC)qj = 0 for all j ∈ Zn(3.7)

or there exists a permutation matrix P ∈ Rn×n such that

P (F −KC)PT =(U 0Q R

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where U ∈ Rr×r, with 1 ≤ r < n, and R ∈ R(n−r)×(n−r), with

n−r∑q=1

Rqj = 0 for all j = 1, . . . , n− r.(3.8)

Since there exist s, t ∈ Zn such that (KC)st > 0, the equation

n∑q=1

(F −KC)qt <n∑q=1

Fqt ≤ 0

contradicts (3.7). For the other possibility, assume without loss of generality thatP = I. Since F contains no traps, the last n − r columns of F − KC cannot beidentical to the last n− r columns of F , so there exist s ∈ Zn and t ∈ Zn−r such that(F −KC)s,r+t < Fs,r+t, from which it follows that

n−r∑q=1

Rqt =n∑q=1

(F −KC)q,r+t <n∑q=1

Fq,r+t ≤ 0,

which contradicts (3.8). It follows that 0 /∈ σ(F −KC) for all K ∈ Rn×k+ such thatF −KC is a compartmental matrix. By the definition of positive modifiability andpositive detectability, the second statement in 1 follows.

2. The blocks of F − KC are Fij − KiCj , for i, j ∈ Zp. Consider Fii − KiCifor i = p −m + 1, . . . , p. There exists a positive matrix Ki such that KiCi 6= 0 andFii −KiCi is a compartmental matrix if and only if (Fii, Ci) is positively modifiable,by definition.

(⇐) Assume (Fii, Ci) is positively modifiable for all i = p − m + 1, . . . , p. Leti ∈ {p−m + 1, . . . , p}. Since Fii is irreducible, it follows from Proposition 3.12 thatif (Fii, Ci) is positively modifiable, i.e., KiCi 6= 0 and Fii −KiCi is a compartmentalmatrix for some positive matrix Ki, then 0 /∈ σ(Fii −KiCi). For j ∈ {1, . . . , p−m},0 /∈ σ(Fjj), which implies by 1 above that 0 /∈ σ(Fjj −KjCj) for all positive matricesKj such that Fjj −KjCj are compartmental matrices. Hence 0 /∈ σ(F −KC), and itfollows that (F,C) is positively detectable.

(⇒) Assume (F,C) is positively detectable; i.e., there exists a K ∈ Rn×k+ suchthat KC 6= 0, F − KC is a compartmental matrix, and 0 /∈ σ(F − KC). Then0 /∈ σ(Fii−KiCi) for all i ∈ Zp. In particular, 0 /∈ σ(Fii−KiCi) for i = p−m+1, . . . , p.But 0 ∈ σ(Fii), which implies Fii − KiCi 6= Fii; i.e., KiCi 6= 0. Since Fii − KiCiis also a compartmental matrix, it follows that (Fii, Ci) is positively modifiable fori = p−m+ 1, . . . , p.

To construct a positive linear observer, the following algorithm can be used.ALGORITHM 3.14. Consider a linear compartmental system S, with system matrix

F ∈ Rn×n and C ∈ Rk×n+ . Assume S contains m ≥ 0 traps.1. Write F and C in the forms (3.5) and (3.6), and decompose a matrix K ∈

Rn×k+ accordingly.2. With Proposition 3.9 check positive modifiability of (Fii, Ci) for every i =

p−m+ 1, . . . , p.3. Execute Algorithm 3.11 for every pair (Fii, Ci), i = 1, . . . , p.4. If (Fii, Ci) is positively modifiable for every i = p − m + 1, . . . , p, step 3

provides a positive linear observer. Otherwise (F,C) is not positively detectable.Note that in step 3 it is not necessary to have KiCi 6= 0 for i = 1, . . . , p − m,

whereas it is necessary for i = p−m+ 1, . . . , p.

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To illustrate the theory, this section will be concluded with an example.Example 3.15. Consider a continuous-time compartmental system with matrices

F =(−2 01 0

), C =

(0 1

).

The second compartment turns out to be a trap, and F has the form (3.5), with m = 1and p = 2. Since (F22, C2) = (0, 1) satisfies the conditions stated in Proposition 3.9,(F22, C2) is positively modifiable. Hence with Theorem 3.13, (F,C) is positivelydetectable, so there exist k1, k2 ∈ R+ such that

F −KC =(−2 −k11 −k2

)is an asymptotically stable compartmental matrix. Indeed, this can be achieved bychoosing k1 = 0 and k2 > 0. Note that the eigenvalues of F − KC cannot bearbitrarily located in the complex plane, because of the necessary condition k1 = 0.One eigenvalue, −2, cannot be moved. The other eigenvalue can be placed, but onlyon the real negative axis. The larger k2, the deeper this latter eigenvalue is placedin the left-half complex plane, but this makes the observer very sensitive to possibleobservation noise. The problem of choosing a suitable k2 has not been solved yet.Because of the restriction k1 = 0, the theory for linear optimal observers, as describedin, for example, [10], cannot be used.

4. Discrete time. In this section conditions for the existence of a positive linearobserver for discrete-time linear compartmental systems will be derived. Most of theresults are closely related to the continuous-time case. Again, first, some theory oncompartmental systems will be presented.

4.1. Discrete-time compartmental systems. In this subsection discrete-timecompartmental systems will be considered. For that purpose it is assumed that trans-fer of material occurs at discrete times t1, t2, . . . , or a continuous-time system is sam-pled at discrete times, in which case the state at time tk has been changed into thestate at time tk+1. What happens in between will not be considered explicitly. There-fore, this can also be seen as if a transfer has occurred at time tk+1. The discretetimes will be assumed to be equally spaced to obtain a time-invariant system. Letthis space be the unit time, so tk+1 = tk + 1.

Let qi(t) be the amount of material in the ith compartment at time t. The amounttransferred from the jth to the ith compartment between time t and time t + 1 isGij(t). This transferred material will be assumed to be linearly dependent on qj ; i.e.,Gij(t) = gijqj(t). The state at time t+ 1 will be given by

qi(t+ 1) =∑j 6=i

gijqj(t) + Ii(t) + giiqi(t),

where giiqi(t) is the amount of material that was in compartment i at time t and isstill (or again) in compartment i at time t+ 1. This amount giiqi(t) is equal to qi(t)minus the amount that left compartment j:

giiqi(t) = qi(t)− goiqi(t)−n∑

j=1,j 6=igjiqi(t) =

1− goi −n∑

j=1,j 6=igji

qi(t).

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Hence define

gii = 1− goi −n∑

j=1,j 6=igji.

The total outflow of a compartment at time t+ 1 cannot be larger than the amountthat was present at time t if the inflow from outside is assumed to be zero. Togetherwith the constraints on positive linear systems (in discrete time), this comes down to

1. gij ≥ 0 for all i, j ∈ Zn;

2.n∑i=1

gij ≤ 1 for all j ∈ Zn.

A matrix G satisfying conditions 1 and 2 above is said to be a compartmental matrix(in the discrete-time case). Condition 2 states that all column sums of G = (gij) ∈Rn×n+ are less than or equal to one.

Below properties of compartmental matrices in discrete time will be discussed,analogous to the continuous-time case. In the rest of this subsection, G refers to adiscrete-time compartmental matrix, whereas F refers to a continuous-time compart-mental matrix.

Let G ∈ Rn×n+ be a compartmental matrix. Then σ(G) ⊆ {λ ∈ C | |λ| ≤ 1}, sincethe sum of entries of every column of G is less than or equal to one; see [12, Section 6.2]or [2, Chapter 2]. Because a system x(t + 1) = Gx(t) is asymptotically stable if andonly if σ(G) ⊆ {λ ∈ C | |λ| < 1}, it follows from the Perron–Frobenius theorem (see[12]) that a compartmental system is asymptotically stable if and only if the spectralradius ρ(G) 6= 1, which is equivalent to 1 /∈ σ(G). Analogously to the continuous-timecase, compartmental matrices having spectral radius one are characterized.

PROPOSITION 4.1. Let G ∈ Rn×n+ be an irreducible compartmental matrix. Thenρ(G) = 1 if and only if

∑ni=1 gij = 1 for all j ∈ Zn.

Proof. This follows from [2, Theorem 2.2.35].A trap in an n-compartmental system is defined in the same way as for continuous-

time systems; see Definition 3.3. As in the continuous-time case, let C1, C2, . . . , Cn bethe compartments of a linear compartmental system S. After renumbering, let T ⊆ Sconsist of the compartments Cm, Cm+1, . . . , Cn, for m ≤ n. Then T is a trap if andonly if

gij = 0 for all (i, j) such that j = m,m+ 1, . . . , n, i = 0, 1, . . . ,m− 1,(4.1)

where G = (gij) ∈ Rn×n+ is the compartmental matrix corresponding to S. ConsiderF = G− I. Since

1. fij = gij ≥ 0, for i, j ∈ Zn, i 6= j;

2.n∑j=1

fji = gii − 1 +n∑

j=1,j 6=igji =

n∑j=1

gji − 1 ≤ 0.

The matrix F is a continuous-time compartmental matrix. Assume F is the systemmatrix for a continuous-time compartmental system SF and let TF ⊆ SF consist ofthe last n−m+ 1 compartments Cm, . . . , Cn.

PROPOSITION 4.2. Consider T ⊆ S and TF ⊆ SF defined above. Then T is a(simple) trap if and only if TF is a (simple) trap.

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Proof. T is a trap if and only if (4.1) holds, which is equivalent togij = 0 for all (i, j) such that j = m,m+ 1, . . . , n, i = 1, 2, . . . ,m− 1,andgjj = 1−

∑ni=1,i 6=j gij for all j = m,m+ 1, . . . , n,

(4.2)

since g0j = 0. Because fij = gij for i 6= j and fjj = gjj − 1, (4.2) is equivalent tofij = 0 for all (i, j) such that j = m,m+ 1, . . . , n, i = 1, 2, . . . ,m− 1,andfjj = −

∑ni=1,i 6=j fij for all j = m,m+ 1, . . . , n,

which is, because f0j = −∑ni=1 fij , equivalent to (3.2); i.e., TF is a trap. In the same

way it can be proved that T is a simple trap if and only if TF is a simple trap.Using Proposition 4.2, the following theorems, analogous to Theorems 3.4, 3.5,

and 3.6, can be proved.THEOREM 4.3. S contains a trap if and only if one of the following conditions

holds.1. For all j ∈ Zn

n∑i=1

gij = 1;

2. There exists a permutation matrix P ∈ Rn×n such that

PGPT =(U1 0Q1 R1

),

with U1, R1 square matrices and the sum of entries of every column of R1 being one.Proof. Since

n∑i=1

fij =

(n∑i=1

gij

)− 1 and

PFPT = P (G− I)PT = PGPT − I =(U1 − I 0Q1 R1 − I

)=:(U 0Q R

),

in which the sum of entries of every column of R = R1 − I is equal to the sum ofentries of every column of R1 minus one, it follows that the conditions stated in thetheorem are equivalent to the conditions stated in Theorem 3.4. The theorem nowfollows using Proposition 4.2.

THEOREM 4.4. S contains a trap if and only if 1 ∈ σ(G).Proof. The following statements are equivalent: (i) 0 ∈ σ(F ); (ii) det(F ) = 0;

(iii) det(G − I) = 0; (iv) 1 ∈ σ(G); and (v) ρ(G) = 1. The last equivalence relationfollows from the Perron–Frobenius theorem. With Proposition 4.2, the theorem isproved.

THEOREM 4.5. Let S be a compartmental system with system matrix G.1. One is an eigenvalue of G of multiplicity m ∈ Z+ if and only if S contains

m simple traps.2. Assume one is an eigenvalue of G of multiplicity m ∈ Z+. Then there exists

a partition of S into a disjoint union of subsystems

S = S1 ∪ S2 ∪ · · · ∪ Sp

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such that Si receives no input from Si+1, . . . , Sp, i = 1, . . . , p−1, and Sp−m+1, . . . , Spare traps. Relative to this partition the system matrix is given by

PGPT = G =

G11 0 0 0 · · · 0...

. . . 0Gp−m,1 Gp−m,p−m 0Gp−m+1,1 Gp−m+1,p−m Gp−m+1,p−m+1

...... 0

. . . 0Gp1 · · · Gp,p−m 0 0 Gpp

,

where Gii is irreducible for all i ∈ Zp and one is an eigenvalue of Gii of multiplicityone for i = p − m + 1, . . . , p, and the sum of entries of every column of Gii, i =p−m+ 1, . . . , p, is one.

Proof. 1. The following statements are equivalent:(i) one is an eigenvalue of G of multiplicity m ∈ Z+;(ii) det(G− λI) = (λ− 1)mp(λ) with p(1) 6= 0;(iii) det(F − λI) = λmp1(λ) with p1(0) = p(1) 6= 0;(iv) zero is an eigenvalue of F of multiplicity m ∈ Z+.

The equivalence between the second and third statements follows from det(F −λI) =det(G − I − λI) = det(G − (λ + 1)I) = ((λ + 1) − 1)mp(λ + 1) = λmp1(λ) withp1(λ) = p(λ + 1). Now statement 1 follows from Proposition 4.2 and statement 1 ofTheorem 3.6.

2. Consider the following statements.a. one is an eigenvalue of G of multiplicity m ∈ Z+;b. zero is an eigenvalue of F of multiplicity m ∈ Z+;c. statement 2 in Theorem 3.6;d. statement 2 in Theorem 4.5.From 1 it follows that a ⇔ b, and Theorem 3.6 provides b ⇒ c. Noting that

PGPT = PFPT + I, Gij = Fij for i 6= j, and Gii = Fii + I, where the sum of entriesof every column of Gii is equal to the sum of entries of every column of Fii plus 1;the implication c ⇒ d follows from the statements of the proof of part 1 for m = 1.This completes the proof of part 2.

4.2. Positive linear observers. In this subsection conditions for the existenceof a positive linear observer for discrete-time linear compartmental systems will bederived. Consider a compartmental matrix G ∈ Rn×n. If G ∈ Rn×n+ is a compartmen-tal matrix and K ∈ Rn×k+ , C ∈ Rk×n+ are such that G −KC ∈ Rn×n+ , then G −KCis also a compartmental matrix, since condition 1 in section 4.1 is satisfied becauseG−KC ∈ Rn×n+ and condition 2 becomes

n∑i=1

(G−KC)ij =n∑i=1

gij − (KC)ij ≤n∑i=1

gij ≤ 1.

The problem for the special class of compartmental matrices is stated below.Problem 4.6. Formulate necessary and sufficient conditions on a compartmental

matrix G ∈ Rn×n+ and a positive matrix C ∈ Rk×n+ such that there exists a K ∈ Rn×k+ ,K 6= 0, with

1. G−KC ∈ Rn×n+ being a compartmental matrix;2. σ(G−KC) ⊆ {λ ∈ C | |λ| < 1}.

Note that 2 is equivalent to ρ(G−KC) < 1, under the assumption that 1 holds.

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(b)(a)

yr

xi

yr

xixi

xj

xq

yr

xj

xi

yr

xq

xi

yr

xj xq

xs

FIG. 4.1. Subgraphs.

The notions of positive modifiability and positive detectability are defined anal-ogously to the continuous-time case.

DEFINITION 4.7. Let G ∈ Rn×n+ be a compartmental matrix and C ∈ Rk×n+ . Thematrix pair (G,C) is said to be positively modifiable if there exists a K ∈ Rn×k+ suchthat KC 6= 0 and G−KC is a compartmental matrix. This implies that σ(G−KC) ⊆{λ ∈ C | |λ| ≤ 1}. The matrix pair (G,C) is said to be positively detectable if thereexists a K ∈ Rn×k+ such that KC 6= 0 and G − KC is an asymptotically stablecompartmental matrix. This implies that σ(G−KC) ⊆ {λ ∈ C | |λ| < 1}.

For the discrete-time case, the condition for positive modifiability is somewhatdifferent, because the diagonal elements of G also play a role.

PROPOSITION 4.8. Let G ∈ Rn×n+ be a compartmental matrix and C ∈ Rk×n+ .(G,C) is positively modifiable if and only if there exists an i ∈ Zn and an r ∈ Zp suchthat the rth row in C is nonzero and

{for all j ∈ Zn with crj > 0, also gij > 0}.(4.3)

Remark 4.9. The interpretation of Proposition 4.8 is that for positive modifiabilitythere should exist an output such that all the compartments contributing to thisoutput have a direct flow to one and the same compartment. Note that in contrastto the continuous-time case, if this last mentioned compartment is a compartmentthat contributes to the output, also is “flow” to itself needed. This means thatsome of the material in this compartment is still in this compartment one time stepahead. Defining a graph for system matrix G as in Remark 3.10, then the condition inProposition 4.8 says that the graph of G should contain a subgraph of the form shownin Figure 4.1. In this case, “flow” to itself is represented by a loop, which occurs ifgii > 0.

Proof. (⇒) Assume (G,C) is positively modifiable, so a K ∈ Rn×k+ can be found,such that KC 6= 0 and G − KC is a compartmental matrix. KC ∈ Rn×n+ , since

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K ∈ Rn×k+ and C ∈ Rk×n+ . Therefore, there exist i, s ∈ Zn such that 0 < (KC)is =∑kt=1 kitcts. Hence there exists an r ∈ Zk such that kircrs > 0, which implies kir > 0

and crs > 0. From this it follows that the rth row in C is nonzero. Next,

0 ≤ (G−KC)ij = gij −k∑t=1

kitctj(4.4)

holds for all j ∈ Zn, since G−KC is a compartmental matrix. Suppose crj > 0. Sincekir > 0, this implies

∑kt=1 kitctj ≥ kircrj > 0. From (4.4) it follows that gij > 0. So

there exist i ∈ Zn, r ∈ Zk such that row r in C is nonzero, and for all j ∈ Zn withcrj > 0, gij > 0 also.

(⇐) Assume there exist i ∈ Zn, r ∈ Zk such that row r in C is nonzero, and forall j ∈ Zn with crj > 0, gij > 0 also. Since row r in C is nonzero, there exists ans ∈ Zn such that crs > 0. This implies gis > 0, and in general, for all v ∈ Zn, crv > 0implies giv > 0. Now take

0 < kir < minv∈Zn,crv>0

givcrv

and all other entries of K equal to zero. Then K ∈ Rn×k+ and

(G−KC)iw = giw − kircrw ={giw − kircrw ∈ (0, giw) if crw > 0,giw ≥ 0 if crw = 0;

(G−KC)hw = ghw ≥ 0 for h 6= i.

It follows that K given above satisfies KC 6= 0, and G−KC satisfies the conditionsfor a compartmental matrix.

A matrix K ∈ Rn×k+ such that KC 6= 0 and G−KC is a compartmental matrixcan be found by the following algorithm.

ALGORITHM 4.10. Consider G ∈ Rn×n+ , C ∈ Rk×n+ . Define the sets

Rd = {(i, r) ∈ Zn×Zk | row r of C is nonzero and (4.3) holds for (i, r)}

and

D(i,r) = {j ∈ Zn | crj 6= 0}.

Form the matrix K ∈ Rn×k+ as follows.1. For every pair (i, r) ∈ Rd, take

0 ≤ kir < minj∈D(i,r)

gijcrj

.

2. For every pair (i, r) /∈ Rd, take kir = 0.Of course, for KC to be nonzero, at least one kir, for a pair (i, r) ∈ Rd, should be

strictly positive. It follows from Proposition 4.8 that the set Rd is nonempty if andonly if the pair (G,C) is positively modifiable, and if this is the case, K can be chosenin such a way that KC 6= 0. Analogous to the continuous-time case, the followingresults can be stated.

PROPOSITION 4.11. Let G ∈ Rn×n+ be an irreducible compartmental matrix. As-sume C ∈ Rk×n+ and K ∈ Rn×k+ are such that KC 6= 0 and G−KC is a compartmentalmatrix. Then G−KC is asymptotically stable.

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Consider a linear compartmental system S. Assume S contains m ≥ 0 traps. Ifm ≥ 1, then S can be partitioned as in Theorem 4.5, with system matrix

G =

G11 0 0 0 · · · 0...

. . . 0Gp−m,1 Gp−m,p−m 0Gp−m+1,1 Gp−m+1,p−m Gp−m+1,p−m+1

...... 0

. . . 0Gp1 · · · Gp,p−m 0 0 Gpp

∈ Rn×n+ ,(4.5)

in which Gii is irreducible for all i ∈ Zp and the sum of entries of every column of thesquare matrices Gp−m+1,p−m+1, . . . , Gpp equals one. Note that

1. 1 /∈ σ(Gii) for i = 1, . . . , p−m;2. 1 ∈ σ(Gii) with multiplicity 1 for i = p−m+ 1, . . . , p.

Consider C ∈ Rk×n+ , K ∈ Rn×k+ , and decompose them to conform to the partition in(4.5):

C =(C1 · · · Cp

), K =

K1...Kp

.(4.6)

The main theorem of this subsection, solving Problem 4.6, is stated below.THEOREM 4.12. Consider a linear compartmental system S, as defined above,

with m ≥ 0 traps.1. If m = 0, then G − KC is asymptotically stable for all K ∈ Rn×k+ with

G − KC a compartmental matrix. Moreover, (G,C) is positively detectable if andonly if (G,C) is positively modifiable.

2. For m ≥ 1, let G, K, and C be partitioned as in (4.5) and (4.6). (G,C) ispositively detectable if and only if (Gii, Ci) is positively modifiable for all i = p−m+1, . . . , p.

Proofs of Proposition 4.11 and Theorem 4.12. These proofs are analogous to theproofs of Proposition 3.12 and Theorem 3.13, using Proposition 4.1, Theorem 4.3, andTheorem 4.4, respectively, instead of Proposition 3.2, Theorem 3.4, and Theorem 3.5,respectively, and changing F into G and the appropriate zeros into ones. Details areleft for the reader.

To construct a positive linear observer, the following algorithm can be used.ALGORITHM 4.13. Consider a linear compartmental system S, with system matrix

G ∈ Rn×n+ and C ∈ Rk×n+ . Assume S contains m ≥ 0 traps.1. Write G and C in the forms (4.5) and (4.6), and decompose a matrix K ∈

Rn×k+ accordingly.2. With Proposition 4.8 check positive modifiability of (Gii, Ci) for every i =

p−m+ 1, . . . , p.3. Execute Algorithm 4.10 for every pair (Gii, Ci), i = 1, . . . , p.4. If (Gii, Ci) is positively modifiable for every i = p − m + 1, . . . , p, step 3

provides a positive linear observer. Otherwise (G,C) is not positively detectable.Note that in step 3 it is not necessary to have KiCi 6= 0 for i = 1, . . . , p − m,

whereas it is necessary for i = p−m+ 1, . . . , p.

5. Concluding remarks. Positive linear observers for linear compartmentalsystems have been considered. Conditions on the system matrices A and C have been

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derived for the existence of positive linear observers, i.e., linear observers that providepositive estimates of the state in case the estimate of the initial state and the inputis positive. As has been shown in the example in section 3.2, the problem of findingan optimal positive linear observer is also worthwhile to be studied. By an optimalpositive linear observer we mean on the one hand one with a “large” gain K, buton the other hand one that is not too sensitive to possible observation noise. Thisproblem remains to be investigated.

In linear system theory, the dual of the observer problem is the stabilizationproblem by linear state feedback; see, for example, [3, 10]. Of course, duals of theresults in this chapter can be derived. But these results will have no physical meaningsince a stabilization problem by linear state feedback would be to design for a positivelinear system

x(t) = Ax(t) +Bu(t)

a positive linear control law

u(t) = Fx(t) + v(t),

with v(t) ∈ Rm+ a new input, such that the closed loop system

x(t) = (A+BF )x(t) +Bv(t)

is asymptotically stable. For physical reasons, this control law should produce apositive input u, given positive state x and positive input v, so F ∈ Rm×n+ . ThereforeA + BF ≥ A, whereas A − KC ≤ A. So new results for this problem have to befound, and they are definitely not dual to the results in this chapter.

REFERENCES

[1] D. H. ANDERSON, Compartmental Modeling and Tracer Kinetics, Lecture Notes in Biomathe-matic 50, Springer-Verlag, Berlin, 1983.

[2] A. BERMAN AND R. J. PLEMMONS, Nonnegative Matrices in the Mathematical Sciences, Com-puter Science and Applied Mathematics, Academic Press, New York, 1979.

[3] F. M. CALLIER AND C. A. DESOER, Linear System Theory, Springer Texts in Electrical Engi-neering, Springer-Verlag, New York, 1991.

[4] E. J. DAVISON, Connectability and structural controllability of composite systems, Automatica,13 (1977), pp. 109–123.

[5] D. FIFE, Which compartmental systems contain traps?, Math. Biosciences, 14 (1972), pp.311–315.

[6] D. M. FOSTER AND J. A. JACQUEZ, Multiple zeros for eigenvalues and the multiplicity of trapsof a linear compartmental system, Math. Biosciences, 26 (1975), pp. 89–97.

[7] K. GODFREY, Compartmental Models and Their Applications, Academic Press, London, 1983.[8] J. A. JACQUEZ, Compartmental Analysis in Biology and Medicine, The University of Michigan

Press, Ann Arbor, MI, 1985.[9] J. A. JACQUEZ AND C. P. SIMON, Qualitative theory of compartmental systems, SIAM Rev.,

35 (1993), pp. 43–79.[10] H. KWAKERNAAK AND R. SIVAN, Linear Optimal Control Theory, John Wiley, New York,

1972.[11] D. G. LUENBERGER, An introduction to observers, IEEE Trans. Automat. Control, 16 (1966),

pp. 596–602.[12] D. G. LUENBERGER, Introduction to Dynamic Systems: Theory, Models, and Applications,

John Wiley, New York, 1979.[13] E. D. SONTAG, Nonlinear regulation: The piecewise linear approach, IEEE Trans. Automat.

Control, 26 (1981), pp. 346–358.[14] E. D. SONTAG, Remarks on piecwise-linear algebra, Pacific J. Math., 98 (1982), pp. 183–201.[15] O. TAUSSKY, A recurring theorem on determinants, Amer. Math. Monthly, 56 (1949), pp.

672–676.

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