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Fig. 1. Level curves of Lyapunov functions corresponding to the conditions (1.2), (1.3) and (1.4) in Proposition 1: If Aξ < 0, then V (x) = maxi(xi/ξi) is a Lyapunov function with rectangular level curves. If T , then T is a linear z A < 0 V (x) = z x x3 Lyapunov function. Finally if AT P + PA ≺ 0 and P ≻ 0, then V (x)= xT Px is a quadratic Lyapunov function for the system x˙ = Ax. Fig. 2. A graph of an interconnected system. In Example 1 the interpretation is a transportation network and each arrow indicates a transportation link. In Example 2 the interpretation is instead a vehicle formation and each arrow Hurwitz if all eigenvalues have negative real part. It is Schur indicates the use of a distance measurement. if all eigenvalues are strictly inside the unit circle. Finally, the matrix is said to be Metzler if all off-diagonal elements are T T nonnegative. The notation RH represents the set of rational gives (A P + P A)ξ = A z + PAξ < 0 so the symmetric ∞ T ✷ functions with real coefficients and without poles in the closed matrix A P + P A is Hurwitz and (1.4) follows. right half plane. The set of n m matrices with elements in Example 1. Linear transportation network. Consider a RH is denoted RHn×m. × dynamical system interconnected according to the graph il- ∞ ∞ lustrated in Figure 2:

x˙ 1 −1 − ℓ31 ℓ12 0 0 x1 III. DISTRIBUTED STABILITY VERIFICATION x˙   0 −ℓ − ℓ ℓ 0  x  2 = 12 32 23 2 The following well known characterizations of stability will x˙ 3  ℓ31 ℓ32 −ℓ23 − ℓ43 ℓ34  x3 x˙ 4  0 0 ℓ43 −4 − ℓ34 x4 be used extensively: (1) Proposition 1: Given a Metzler matrix A Rn×n, the The model could for example be used to describe a transporta- following statements are equivalent: ∈ tion network connecting four buffers. The states x1, x2, x3, x4 (1.1) The matrix A is Hurwitz. represent the contents of the buffers and the parameter ℓ Rn ij (1.2) There exists a ξ such that ξ > 0 and Aξ < 0. determines the rate of transfer from buffer j to buffer i. Such ∈Rn T (1.3) There exists a z such that z > 0 and z A< 0. transfer is necessary to stabilize the content of the second and (1.4) There exists a diagonal∈ matrix P 0 such that ≻ third buffer. AT P + P A 0. Notice that the dynamics has the form x˙ = Ax where A ≺ (1.5) The matrix A−1 exists and has nonnegative entries. is a Metzler matrix provided that every ℓij is nonnegative. − Hence, by Proposition 1, stability is equivalent to existence of Moreover, if ξ = (ξ1,...,ξn) and z = (z1,...,zn) satisfy numbers ξ1,...,ξ4 > 0 such that the conditions of (1.2) and (1.3) respectively, then P = −1 − ℓ31 ℓ12 0 0 ξ1 0 diag(z1/ξ1,...,zn/ξn) satisfies the conditions of (1.4).  0 −ℓ − ℓ ℓ 0  ξ  0 12 32 23 2 < Remark 1. Each of the conditions (1.2), (1.3) and (1.4)  ℓ31 ℓ32 −ℓ23 − ℓ43 ℓ34  ξ3 0  0 0 ℓ −4 − ℓ  ξ  0 corresponds to a Lyapunov function of a specific form. See 43 34 4 Figure 1. Given these numbers, stability can be verified by a distributed test where the first buffer verifies the first inequality, the second Remark 2. One of the main observations of this paper is that buffer verifies the second and so on. In particular, the relevant verification and synthesis of positive control systems can be test for each buffer only involves parameter values at the local done with methods that scale linearly with the number of node and the neighboring nodes, so a global model is not interconnections. For stability, this claim follows directly from needed anywhere. ✷ Proposition 1: Given ξ, verification of the inequality Aξ < 0 requires a number of scalar additions and multiplications that Example 2. Vehicle formation (or distributed Kalman is directly proportional to the number of nonzero elements in filter). Another system structure, which can be viewed as a the matrix A. In fact, the search for a feasible ξ also scales dual of the previous one, is the following: linearly, since integration of the differential equation ξ˙ = Aξ x˙ 1 = x1 + ℓ13(x3 x1) with ξ(0) = ξ0 for an arbitrary ξ0 > 0 generates a feasible − − x˙ = ℓ (x x )+ ℓ (x x ) ξ(t) in finite time provided that A is Metzler and Hurwitz.  2 21 1 − 2 23 3 − 2 (2) x˙ 3 = ℓ32(x2 x3)+ ℓ34(x4 x3) Proof of Proposition 1. The equivalence between (1.1), (1.2),  − − x˙ 4 = 4x4 + ℓ43(x3 x4) (1.4) and (1.5) is the equivalence between the statements G20, − −  I27, H24 and N38 in [2, Theorem 6.2.3]. The equivalence be- This model could for example be used to describe a formation tween (1.1) and (1.3) is obtained by applying the equivalence of four vehicles. The parameters ℓij represent position adjust- between (1.1) and (1.2) to the transpose of A. Moreover, if ments based on distance measurements between the vehicles. ξ = (ξ ,...,ξ ) and z = (z ,...,z ) satisfy the conditions of The terms x and 4x reflect that the first and fourth 1 n 1 n − 1 − 4 (1.2) and (1.3) respectively, then P = diag(z1/ξ1,...,zn/ξn) vehicle can maintain stable positions on their own, but the 3

second and third vehicle rely on the distance measurements for Proof. It is well known that g 2−ind = maxω G(iω) 2−ind stabilization. Again, stability can be verified by a distributed for general linear time-invariantk k systems. Whenkg(t) k0, the test where the first vehicle verifies the first inequality, the maximum must be attained at ω =0 since ≥ second vehicle verifies the second inequality and so on. ✷ ∞ G(iω)w g(t)e−iωt dt w A discrete time counterpart to Proposition 1 is given next: | |≤ 0 · | | Proposition 2: For B Rn×n, the following statements are Z ∞ + G equivalent: ∈ = g (t)dt w = (0) w 0 · | | | | (2.1) The matrix B is Schur stable. Z for every w Cm. This completes the proof for p = 2. For (2.2) There is a ξ Rn such that ξ > 0 and Bξ <ξ. p =1, the fact∈ follows from the calculations (2.3) There exists a∈z Rn such that z > 0 and BT z

(4.1) The matrix A is Hurwitz and g ∞−ind <γ. Proof. By Theorem 3, the inequality g 1−ind < γ can k k equivalently be written D CA−1B k k <γ or (4.2) There exists ξ Rn such that k − k1−ind ∈ + −1 T A B ξ 0 (D CA B) 1 <γ1. (11) < . (4) − CD 1 γ1      Assume that (5.1) holds. By Proposition 1 there exists z > 0 Moreover, if ξ satisfies (4), then ξ < x(t) < ξ for all T −T T − such that z A< 0. Define p = z A C . Then p z > 0. solutions to the equation x˙ = Ax + Bw with x(0) = 0 and Moreover − ≥ w 1. k k∞ ≤ Proof of Theorem 4. A is Metzler, so eAt 0 and the AT p + CT = AT z < 0 ≥ assumptions of Theorem 3 hold. Hence g ∞−ind < γ can −1 k k If is sufficiently small, we also get T T 1 1 so equivalently be written D CA B ∞−ind <γ or z B p + D < γ k − k (5.2) follows. (D CA−1B)1 <γ1. (5) − Conversely, suppose that (5.2) holds. Then A is Hurwitz by Assume that (4.2) holds. Then A is Hurwitz by Proposition 1. Proposition 1. Consider any solutions to Multiplying the inequality Aξ + B1 < 0 with the non- positive matrix CA−1 from the left gives Cξ +CA−1B1 0. x˙ = Ax + Bw x(0) = 0 Subtracting this from the inequality Cξ + D1 <γ1 gives≥ (5), y˙ = Ay + B w y(0) = 0. | | so (4.1) follows. Conversely, suppose that (4.1) and therefore (5) holds. By A is Metzler, so x(t) y(t) for all t 0. Multiplying the transpose of (9) by| (y,| ≤w ) from the right≥ gives Proposition 1 there exists x > 0 such that Ax < 0. Define | | ξ = x A−1B. Then ξ x> 0. Moreover − ≥ pT y˙ + Cy + D w γ w . Aξ + B = Ax < 0 | |≥ | | Integrating of t and using that x(t) y(t) gives (10). Then If x is sufficiently small, we also get Cξ + D1 <γ1 so (4.2) (5.1) follows as t and the| proof|≤ is complete. ✷ follows. → ∞ To prove the last statement, suppose that ξ satisfies (4) and A discrete time counterpart of Theorem 4 and Theorem 5 define x, y and z by is given without proof: Theorem 6: Given matrices A,B,C,D 0, let y˙ = Ay + u y(0) = ξ (6) ≥ − D t =0 x˙ = Ax + Bw x(0) = 0 (7) g(t)= CAt−1B t =1, 2,... z˙ = Az + v z(0) = ξ (8)  where w 1, u = Aξ and v = Aξ. Then the solutions Then the following two statements are equivalent: k k∞ ≤ − of (6) and (8) are constantly equal to ξ and ξ respectively. (6.1) The matrix A is Schur and g ∞−ind <γ. Moreover, the inequalities − k k Rn (6.2) There exists ξ + such that u Bw v ∈ ≤ ≤ A B ξ ξ follow from (4). Together with the assumption that A is < . (12) CD 1 γ1 Metzler, gives that y(t) x(t) z(t) for all t. This completes      ≤ ≤ ✷ the proof. If ξ satisfies (12), then ξ < x(t) < ξ for all solutions to Theorem 5: Suppose that g(t) = CeAtB + Dδ(t) where the equation x(t +1) =−Ax(t)+ Bw(t) with x(0) = 0 and Rn×n Rn×m Rr×n A is Metzler and B + , C + , D w ∞ 1. Rr×∈m ∈ ∈ ∈ k k ≤ + . Then the following statements are equivalent: The following two statements are also equivalent: (5.1) The matrix A is Hurwitz and g 1−ind <γ. (6.3) The matrix A is Schur and g <γ. k k k k1−ind Rn (5.2) There exists p + such that (6.4) There exists p Rn such that ∈ ∈ + T A B p 0 T < . (9) A B p p CD 1 γ1 < . (13)       CD 1 γ1 Moreover, if p satisfies (9), then all solutions to the equation       x˙ = Ax + Bw with x(0) = 0 satisfy Moreover, if p satisfies (13), then all solutions to the equation t t x(t +1)= Ax(t)+ Bw(t) with x(0) = 0 satisfy pT x(t) + Cx + Dw dτ γ w dτ (10) t t | | 0 | | ≤ 0 | | Z Z pT x(t) + Cx(τ)+ Dw(τ) γ w(τ) with equality only if w is identically zero. | | | |≤ | | τ=0 τ=0 Remark 3. The first part of Theorem 4 and Theorem 5 X X previously appeared in [5]. with equality only if w is identically zero. 5

V. DISTRIBUTED CONTROL SYNTHESIS BY LINEAR −1 PROGRAMMING s Equipped with scalable analysis methods for stability and x˙ x performance, we are now ready to consider synthesis of controllers by distributed optimization. We will start by re- A B E visiting an example of section III. z C D G w Example 3. Consider again the transportation network (1), this F H 0  time with the flow parameters ℓ31 = 2, ℓ34 = 1 and ℓ43 = 2   fixed: y u x˙ 1 3 ℓ12 0 0 x1 x˙ −0 ℓ ℓ ℓ 0 x L 2 = 12 32 23 2 (14) x˙   2 − ℓ − ℓ 2 1  x  3 32 − 23 − 3 x˙ 4  0 0 2 5 x4    −          Fig. 3. Theorem 7 shows how to determine the L that We will ask the question how to find the remaining parameters minimizes the L∞-induced gain from w to z. Theorem 8 shows how to ℓ12, ℓ23 and ℓ32 in the interval [0, 1] such that the closed minimize the L1-induced gain. If w and z are scalar, both gains are equal loop system (14) becomes stable. According to Proposition 1, to the the standard H∞-norm. Extension to the case of system matrix with non-zero lower right corner is straightforward, but omitted from this paper. stability is equivalent to existence of ξ1,...,ξ4 > 0 such that 3 ℓ 0 0 ξ − 12 1 0 ℓ12 ℓ32 ℓ23 0 ξ2 Remark 4. If the diagonal elements of are restricted to R  − −    < 0 D + 2 ℓ32 ℓ23 2 1 ξ3 instead of [0, 1], then the condition F ξ + H1 µ is replaced − − ≥  0 0 2 5 ξ4 by F ξ + H1 0.  −    ≥ At first sight, this looks like a difficult problem   due to multi- Remark 5. When the matrices have a sparsity pattern corre- plications between the two categories of parameters. However, sponding to a graph, each row of the vector inequalities in (7.2) a closer look suggests the introduction of new variables: can be verified separately to get a distributed performance test. µ := ℓ ξ , µ := ℓ ξ and µ := ℓ ξ . The problem 12 12 2 32 32 2 23 23 3 Also finding a solution to the linear programming problem then reduces to linear programming: Find ξ , ξ , ξ , ξ > 0 1 2 3 4 can be done with distributed methods, where each node in the and µ ,µ ,µ 0 such that 12 32 23 ≥ graph runs a local algorithm involving only local variables and 30 0 0 ξ1 1 0 0 information exchange only with its neighbors. For example, − µ12 000 0 ξ2 1 1 1 given a stable Metzler matrix A, consider the problem to find a     + − −  µ32 < 0 2 0 2 1 ξ3 0 1 1   stability certificate ξ > 0 satisfying Aξ < 0. This can be done − − µ23  0 0 2 5 ξ   0 0 0   −   4     in a distributed way by simulating the system using Euler’s µ ξ µ ξ µ  ξ  method until the state is close to a dominating eigenvector of 12 ≤ 2 32 ≤ 2 23 ≤ 3 the A. Then it must satisfy the conditions on ξ. with the solution (ξ1, ξ2, ξ3, ξ4) = (0.5, 0.5, 1.69, 0.87) and (µ12,µ32,µ23) = (0.5, 0.5, 0). The corresponding stabilizing Remark 6. It is interesting to compare our results with the anal- gains can then be computed as ysis and synthesis methods proposed by Tanaka and Langbort

ℓ12 = µ12/ξ2 =1 ℓ32 = µ32/ξ2 =1 ℓ23 = µ23/ξ3 =0 in [24] and Briat in [5]. Our mathematical treatment has much in common with theirs. However, none of them is discussing ✷ scalable design, nor verification, of distributed controllers. The idea can be generalized into the following theorem: Moreover, our “static output feedback” expression A+ELF is Theorem 7: Let be the set of m m diagonal matrices D × significantly more general than the “state feedback” expression with entries in [0, 1]. Suppose that A + ELF is Metzler and A + BL used in both those references. This gives us a C + GLF 0, B + ELH 0, D + GLH 0 for all L . ≥ ≥ ≥ ∈ D higher degree of flexibility, particularly in the specification of Let gL(t) be the impulse response of distributed controllers. On the other hand, their parametrization (C + GLF )[sI (A + ELF )]−1(B + ELH)+ D + GLH has the advantage that the Metzler property of the closed loop − system matrix can be enforced as a constraint in the synthesis If F 0, then the following two conditions are equivalent: ≥ procedure, rather than being verified a priori for all L . (7.1) There exists L with A + ELF is Hurwitz and ∈ D ∈ D gL ∞−ind <γ. Proof. Suppose (7.1) holds. Then, according to Theorem 4, (7.2)k Therek exist ξ Rn , µ Rm with there exists ξ Rn such that ∈ + ∈ + ∈ + 1 Aξ + B + Eµ < 0 A + ELF B + ELH ξ 0 < . (15) Cξ + D1 + Gµ<γ1 C + GLF D + GLH 1 γ1      F ξ + H1 µ ≥ Setting µ = LF ξ +LH1 gives (7.2). Conversely, suppose that Moreover, if ξ,µ satisfy (7.2), then (7.1) holds for every L (7.2) holds. Choose L to get µ = LF ξ + LH1. Then such that µ = LF ξ + LH1. (15) holds and (7.1) follows∈ D by Theorem 4. ✷ 6

2 2 2 Theorem 7 was inspired by the transportation network in Example 3, where non-negativity of F is natural assumption. However, this condition would fail in a vehicle formation problem, where control is based on distance measurements. 1 3 4 1 3 4 1 3 4 T T T B = 1 1 1 1 B = 10 10 1 1 B = 1 1 10 10 For such problems, the following dual formulation is useful:       Theorem 8: Let be the set of m m diagonal matrices D × with entries in [0, 1]. Suppose that A + ELF is Metzler and Fig. 4. Illustration of optimal gains for disturbance rejection in a vehicle T C + GLF 0, B + ELH 0, D + GLH 0 for all L . formation. When B = (1 1 1 1) , all four vehicles face unit disturbances and ≥ ≥ ≥ ∈ D the optimal L = diag{0, 1, 1, 0, 1, 0} illustrated by arrows in the left diagram Let gL(t) be the impulse response of gives γ = 4.125. Apparently, the first vehicle should ignore the distance to the third vehicle, while the third vehicle should ignore the second vehicle and the −1 (C + GLF )[sI (A + ELF )] (B + ELH)+ D + GLH fourth should ignore the third. The middle diagram illustrates a situation where − the disturbances on vehicle 1 and 2 are ten times bigger. Then the minimal If the matrices B,D and E have nonnegative coefficients, then value γ = 15.562 is attained with L = diag{1, 1, 1, 0, 1, 0}, so the first the following two conditions are equivalent: vehicle should use distance measurements to the third. The converse situation in the right diagram gives γ = 12.750 for L = diag{0, 1, 0, 1, 1, 0}. (8.1) There exists L with A + ELF is Hurwitz and ∈ D gL 1−ind <γ. (8.2)k Therek exist p Rn , q Rm with ∈ + ∈ + First we need to define a notion of positivity for input-output models. One option would be to work with non-negative AT p + CT 1 + F T q< 0 impulse responses like in Theorem 3. However, to verify for T T 1 T 1 B p + D + H q<γ a given rational transfer function that the impulse response is ET p + GT 1 q non-negativehas proved to be NP-hard! See [3] for the discrete ≥ time problem and [1] for continuous time. Instead we will use Moreover, if satisfy (8.2), then (8.1) holds for every p, q L the following definition. such that q = LET p + LGT 1. G RHm×n is called positively dominated if every matrix ✷ ∈ ∞ Proof. The proof is analogous to the proof of Theorem 7. entry satisfies Gjk(iω) Gjk(0) for all ω R. The set | | ≤ DHm×n ∈ Example 4. Disturbance rejection in vehicle formation. of all such matrices is denoted ∞ . The essential scalar Consider the vehicle formation model frequency inequality can be tested by semi-definite program- ming, since b(iω)/a(iω) b(0)/a(0) holds for ω R if x˙ = x + ℓ (x x )+ w | | ≤ 2 2 2 ∈2 1 − 1 13 3 − 1 and only if the polynomial a(iω) b(0) b(iω) a(0) can x˙ = ℓ (x x )+ ℓ (x x )+ w be written as a sum of squares.| | − | |  2 21 1 2 23 3 2 (16) − − Some properties of positively dominated transfer functions x˙ 3 = ℓ32(x2 x3)+ ℓ34(x4 x3)+ w  − − x˙ = 4x + ℓ (x x )+ w follow immediately: 4 4 43 3 4 G H DHn×n GH DHn×n  − − Proposition 9: Let , ∞ . Then ∞ where is an external disturbance acting on the vehicles. G H DHn×n ∈ R ∈ G w and a +b ∞ when a,b +. Moreover ∞ = Our problem is to find feedback gains gains ℓij [0, 1] that G(0) . ∈ ∈ k k stabilize the formation and minimize the gain from∈ w to x. k k The problem can be solved by applying Theorem 8 with The following property is also fundamental: G DHn×n G −1 DHn×n Theorem 10: Let ∞ . Then (I ) ∞ A = diag{−1, 0, 0, −4} C = 1 1 1 1 if and only if G(0) is Schur.∈ − ∈ 100000  Proof. That (I G)−1 is stable and positively dominated 011000 −G −1 G E = D = 0 implies that [I (0)] exists and is nonnegative, so (0)  000110  must be Schur− according to Proposition 2. On the other hand, 000001   if G is Schur we may choose R and with   (0) ξ + ǫ > 0 G ∈ Cn L = diag{ℓ13,ℓ21,ℓ23,ℓ32,ℓ34,ℓ43} (0)ξ < (1 ǫ)ξ. Then for every z with 0 < z < ξ and s C with− Re s 0 we have ∈ | | −1 0 1 0 0 ∈ ≥ 1 −1 0 0 0 G(s)tz G(0)t z < (1 ǫ)t z for t =1, 2, 3,...  0 −1 1 0   0  | |≤ | | − | | F = H = 0 1 −1 0 0 ∞ G t     Hence k=0 (s) z is convergent and bounded above by  0 0 −1 1   0  ∞     G(0)t z = [I G(0)]−1 z . The sum of the series  0 0 1 −1   0  k=0     solves theP equation| | −G ∞| |G t , so therefore     [I (s)] k=0 (s) z = z Solutions for three different cases are illustrated in Figure 4. P∞ G t −G −1 G −1 k=0 (s) z = [I (s)] z. This proves (I ) is ✷ stable and positively− dominatedP and the proof is complete.− ✷ P Theorem 8 has the following counterpart for positively VI. POSITIVELY DOMINATED SYSTEMS dominated systems, as illustrated in Figure 5. So far, the emphasis has been on state space models. Theorem 11: Let be the set of m m diagonal matrices D B DH×n×k C DHl×n However, for many applications input-output models are more with entries in [0, 1], while ∞ , ∞ , D DHl×k E DHn×m ∈ F RHm×n∈ natural as a starting point. In this section, we will therefore ∞ , ∞ and ∞ . Suppose A +∈ELF DHn×∈n for all L . ∈ extend the main ideas of the previous sections to such models. ∈ ∞ ∈ D 7

Similarly to Example 3, we write this on matrix form as x X = (A + ELF)X + BW A B E The transfer matrices B, E and A + ELF are positively z C D 0 w   dominated for all provided that . F 0 0 L di ki + j ℓij Hence Theorem 11 can∈ D then be applied to find≥ the optimal   P y u spring constants. Notice that ℓij and ℓji must be optimized separately, even though by symmetry they must be equal at L optimum. ✷

Fig. 5. Theorem 11 shows how to determine the diagonal matrix L that VII. SCALABLE VERIFICATION OF THE LYAPUNOV minimizes the L1-induced gain from w to z. Extension to the case of a matrix with all non-zero blocks is straightforward, but omitted from this paper. The INEQUALITY same is true for the transpose version corresponding to Theorem 7. In the preceding sections we have derived scalable condi- tions for verification of stability and optimality, using gen- Then the following two conditions are equivalent: eralizations of the linear inequalities in (1.2) and (1.3) of A E F −1 DHn×n Proposition 1. To address multi-variable systems using linear (11.1) There is L with (I L ) ∞ ∈ D −−1 − ∈ and C(I A ELF) B + D 1−ind <γ. programming, the natural performance measures have been k − −Rn Rm k input-output gains with signals measured L -norm or L - (11.2) There exist p +, q + with 1 ∞ ∈ ∈ norm. A(0)T p + C(0)T 1 + F(0)T q

2 lem on the left is generally not a convex program, since the s + dis + ki + ℓij Xi(s) matrices Mk may be indefinite. However, the maximization on  j  2 2 X the left is concave in (x1,...,xn) [17]. This is because every = ℓ X (s) + (ℓ ℓ )X (s) + W (s). product xixj is the geometric mean of two such variables, ij j ij − ij i i j hence concave [4, p. 74]. X   8

Remark 8. The second statement of Proposition 13 is important (14.1) For ω [0, ] is is true that ∈ ∞ for scalability, since the condition X X has a natural ∗ ∈ (iωI A)−1B (iωI A)−1B decomposition and only entries of X that correspond to non- − Q − 0 I I  zero entries of Mk need to be taken into account.     T Proof of Proposition 13. Every x satisfying the constraints on A−1B A−1B the left hand side of (17) corresponds to a matrix T (14.2) − Q − 0. X = xx I I  satisfying the constraints on the right hand side. This shows     that the right hand side of (17) is at least as big as the left. (14.3) There exists a diagonal P 0 such that On the other hand, let X = (xij ) be a positive definite  T matrix. In particular, the diagonal elements x11,...,xnn are A P + PA PB Q + T 0 non-negative and xij √xiixjj . Let x = (√x11,..., √xnn). B P 0  T ≤   Then the matrix xx has the same diagonal elements as X, but Rn Rm (14.4) Thereexist x, p +, u + with Ax+Bu 0, has off-diagonal elements √xiixjj instead of xij . The fact that ∈ ∈ ≤ xxT has off-diagonal elements at least as big as those of X, x AT Q + T p 0 together with the assumption that the matrices Mk are Metzler, u B ≤ gives T for . This shows     x Mkx trace(MkX) k = 1,...,K Moreover, if all inequalities are replaced by strict ones, then that the left hand≥ side of (17) is at least as big as the right. the equivalences hold even without the stabilizability assump- Nothing changes if is not positive definite but X, so X X tion. the second statement is also proved. ∈ 0 1 For the last statement, note that the conditions Remark 9. For A = 1, B =0, Q = 1 0 , condition (14.1) holds, but not (14.3).− This demonstrates that the stabilizability trace(MkX) bk are linear in X, so strong duality holds [22, Theorem≥ 28.2] and the right hand side of (17) has of ( A, B) is essential. K − a finite maximum if and only if M0 + k=1 τkMk 0 for Remark 10. Our statement of the KYP lemma for continuous  ✷ some τ1,...,τK 0. and discrete time positive systems extends earlier versions ≥ P Proof of Theorem 12. Let be the set of indices (k,l) of of [24], [19] in several respects: Non-strict inequality, more non-zero off-diagonal entriesE in M. Define general Q and a fourth equivalent condition in terms of linear programming rather than semi-definite programming. X = X Rn×n :[e e ]T X[e e ] 0 for all (k,l) E ∈ k l k l  ∈E Proof. One at a time, we will prove the implications (14.1) n ⇒ where e1,...,en are the unit vectors in R . If M is negative (14.2) (14.3) (14.1) and (14.2) (14.4). Putting ω =0 semi-definite, then immediately⇒ gives⇒ (14.2) from (14.1).⇔ −1 xT Mx MX Assume that (14.2) holds. The matrix A is nonnegative, 0 = max = max trace( ) T |x|≤1 X∈XE x x Rn R−m so Q 0 for all x +, w + with T w w ≤ ∈ ∈ = min max trace(MX)+ trace Nkl[ek el] X[ek el]     × Nkl0 X∈Rn n −1 (k,l)∈E x A Bw (18) X   ≤− −1 T The inequality (18) follows (by multiplication with A = min max trace M + [ek el]Nkl[ek el] X × − Nkl0 X∈Rn n    from the left) from the constraint 0 Ax + Bw, which can (k,lX)∈E ≤    also be written 0 Aix + Biw for i = 1,...,n, where Ai R2×2 ≤ where Nkl for every k and l. In particular, and denote the :th rows of and respectively. For ∈ Bi i A B there exists a choice of the matrices Nkl that makes M + non-negative x and w, this is equivalent to T (k,l)∈E [ek el]Nkl[ek el] = 0. This completes the proof. ✷ 0 x (A x + B w) i =1,...,n (19) ≤ i i i P T Hence (14.2) implies x Q x 0 for x Rn , w Rm w w ≤ ∈ + ∈ + VIII. THE KYP LEMMA FOR POSITIVE SYSTEMS satisfying (19). Proposition  13  will next be used to verify Input-output gain is certainly not the only way to quantify existence of τ ,...,τ 0 such that the quadratic form 1 n ≥ the performance of a linear time-invariant system. A more T general class of specifications known as Integral Quadratic x x σ(x, w)= Q + τixi(Aix + Biw) Constraints [18] can be tested using the Kalman-Yakubovich- w w " # " # i Popov lemma. It is therefore of interest to see that the corre- X is negative semi-definite. However, the application of Propo- sponding result of [24] for positive systems can be generalized sition 13 requires existence of a positive definite such that the following way: X all diagonal elements of Theorem 14: Let A Rn×n be Metzler and Hurwitz, while n×m ∈ B R and the pair ( A, B) is stabilizable. Suppose that I ∈ + − A B X all entries of Q R(n+m)×(n+m) are nonnegative, except for 0 ∈   the last m diagonal elements. Then the following statements   are positive. The pair ( A, B) is stabilizable, so there exists are equivalent: K that make all eigenvalues− of A+BK unstable and therefore 9

(A + BK)Z + Z(A + BK)T = I has a symmetric positive For strict inequalities, the proofs that (14.2) (14.4) and definite solution Z. Hence the desired X can be constructed (14.3) (14.1) (14.2) remain the same. Assuming⇔ that as (14.2) holds⇒ with⇒ strict inequality, we get

ZZKT ∗ X = A−1B A−1B KZ − (Q + ǫI) − 0  ∗  I I      where the lower right corner is chosen big enough to make X 0. for some scalar ǫ > 0. Hence, there exists a diagonal P 0 ≻  Define P = diag(τ ,...,τ ) 0. Then σ being negative such that 1 n  definite means that AT P + PA PB Q + ǫI + 0 AT P + PA PB BT P 0  Q + 0   BT P 0    Adding a small multiple of the identity to P gives P 0 such so (14.3) follows. that ≻ Assume that (14.3) holds. Integrating σ(x(t), w(t)) over AT P + PA PB time gives Q + 0 BT P 0 ≺ T   ∞ x x 0 Q + xT P (Ax + Bw) dt so also (14.3) holds with strict inequality. Hence the proof is ≥  w w  Z0 " # " # complete. ✷   For square integrable solutions to x˙ = Ax + Bw, x(0) = 0 An analogous discrete time result is stated here and proved we get in the appendix: Rn×n Rn×m Theorem 15: Let A + be Schur, while B + T ∈ ∈ ∞ x x d and the pair (A, B) is anti-stabilizable. Suppose that all 0 Q + (xT P x/2) dt R(n+m)×(n+m) ≥  w w dt  entries of Q are nonnegative, except for Z0 " # " # the last m diagonal∈ elements. Then the following statements  T  are equivalent: ∞ x(t) x(t) = Q dt 0 " w(t) # " w(t) # Z (15.1) For ω [0, ] is is true that which in frequency domain implies (14.1). Hence (14.1) ∈ ∞ ∗ (14.2) (14.3) (14.1). ⇒ (eiωI A)−1B (eiωI A)−1B ⇒ ⇒ − Q − 0 Assuming again (14.2) gives, by Proposition 1, existence of I I      u Rm such that ∈ + −1 ∗ −1 T (I A) B (I A) B A−1B A−1B (15.2) − Q − 0. − Q − u 0 I I  I I ≤        ! Setting x = A−1u gives x Rn and (15.3) There exists a diagonal P 0 such that − ∈ +  T T T A−1 A−1B x A P A P A PB − Q 0 Q + T − T 0 0 I u ≥ B PA B PB   −      n m T T T (15.4) There are x, p R , u R with x Ax + Bu, due to the sign structure of Q. Let p q be the column ∈ + ∈ + ≥ T A B T on the left hand side. Multiplying with  from the x A I 0 I Q + −T p 0 u B ≤ left gives       AT 0 p x Moreover, if all inequalities are taken to be strict, then the T = Q equivalences hold even without the anti-stabilizability assump- B I q − u      tion. and (14.4) follows. Proof. The theorem can be proved in analogy with the proof of Finally, suppose that (14.4) holds. Then x A−1Bu. Theorem 14. Alternatively, it can be derived from Theorem 14 Multiplying the main inequality from the≥ − left with T −T using a bilinear transformation in the following way: B A I gives iω 1+iω − Instead of e , one can parametrize the unit circle as 1−iω . T T  A−1B x A−1B A−1B Hence (15.1) is equivalent to saying that 0 − Q = − Q − u ≥ I u I I ∗        ! x x Q 0 u u ≤ and (14.2) follows.     10

i1 i4 and the external currents are determined by Kirchoff’s law v1 v4 i (t)= i (t) i (t) 1 − 41 − 21 i (t)= i (t) i (t) i (t)  2 21 − 32 − 42 (21) i3(t)= i32(t) i3 i2  v3 v2  i4(t)= i41(t)+ i42(t)  Fig. 6. Illustration of a dynamic power transmission network with inductive The generation and consumption of power is subject to con- transmission lines studied in Example 6. straints of the form 1 T ik(t)vk(t)dt p (22) 1+iω k for all solutions (ω,x,u) to the equation I A x = T 0 ≤ 1−iω − Z Bu. Alternatively, introducing   If k is a generator node, then pk > 0 indicates production capacity. Similarly, for loads p < 0 represents power demand. A = (A I)(A + I)−1 k − Transmission lines have capacity constraints of the form B = 2(A + I)−1B T b 1 2 x = x + Ax + Bu vk(t) vj (t) dt ckj (23) b T 0 | − | ≤ (A + I)−1 (A + I)−1B Z S = b 0 − I Finally, the voltages are non-negative and subject to magnitude   bounds Q = ST QS T 2 1 2 2 the condition can be re-written as the statement that vk vk(t) dt vk (24) b ∗ ≤ T 0 ≤ x x Z Q 0 We are now interested to minimize the resistive power losses u u ≤     in the network subject to the given constraints: for all solutions (ω, x, ub) tob theb equation (iωI A)x = Bu. − 1 4 T According to Theorem 14, this is equivalent to validity of Minimize T k=1 0 ik(t)vk(t)dt the inequality for ω b=b 0, i.e. (15.2). It is also equivalentb b b to subject to (20)P (24)R existence of a diagonal P 0 such that  −  Using the theory above, our goal is to prove that minimal T  A P + P A P B losses can be attained with constant voltages and currents. Q + T 0 " B P 0 #  With line currents being states and voltage differences being b b b bA + I B inputs, this is a problem of the form Multiplying by b from the right and its transpose ∗ 0 I T x x from the left, the matrix inequality (after trivial manipulations) Maximize Q0 dt 0 "u# "u# becomes  subject to xR˙(t)= Ax(t)+ Bu(t) (25) T T  A P A P A PB  ∗ Q +2 T − T 0  T x x B PA B PB  and Qk dt qk, k =1,...,m   0 "u# "u# ≤  Replacing 2P by P gives equivalence to (15.3). Also by  R n  Rn×n Rn×m Theorem 14, it is equivalent to existence of R , where A is Metzler and Hurwitz, B + x, p + ∈ ∈ Rm ∈ and all entries of Q R(n+m)×(n+m) are nonnegative u + such that k ∈ except possibly for the last∈ m diagonal elements. To bring b x AT the problem on a form where Theorem 14 can be applied, we b Q + p 0 u BT ≤ will apply relaxation in two different ways: The inequalities   " # b b are handled using Lagrange relaxation and the time interval is Left multiplication byb −T and substitution S b (x, u)= S(x, u) extended to [0, ]. This brings the problem to the dual form gives equivalence to (15.4). ✷ ∞ Minimize m τ q subject to τ 0 such that As an application of the equivalence between (14.1)b and − k=1 k k k ≥ (14.2), we consider an example devoted to optimal power flow ∗  ∞ x P x (Q0 + k τkQk) dt 0 in an electrical network, a time-varying version of a problem  0 "u# "u# ≤  considered in[13]: forR all solutions toPx˙(t)= Ax(t)+ Bu(t). Example 6. Optimal power flow in an electrical network.  Consider a power transmission network as in Figure 6. The In frequency domain, this is written as current from node j to node k is governed by the voltage m Minimize k=1 τkqk subject to τk 0 and difference vj vk according to the differential equation ≥ − ∗  (iωI − A)P−1B (iωI − A)−1B dijk  (Q0 + τkQk)  0 Ljk = Rjkijk + vj (t) vk(t) (20)  I k I dt − − " # " # P   11

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X. ACKNOWLEDGMENTS

The author is grateful for suggestions and comments by numerous colleagues, including the anonymous reviewers. In particular, Theorem 3 was updated based on suggestions by Andrej Ghulchak and Figure 4 was drawn by Enrico Lovis- ari. The work has been supported by the Swedish Research Council through the Linnaeus Center LCCC.