R := 0 d + 1 θ. We obtain that the A-module E is defined by. generators which satisfy. A-linear relations. We do not print the large outputs of Endo.

Transcription

1 > restart: > with(oremodules): > with(oremorphisms); > with(linalg): We consider the differential time-delay model of a stirred tank studied in H. Kwakernaak, R. Sivan, Linear Optimal Control Systems, Wiley-Interscience, 972. The system matrix is defined by: > A:=DefineOreAlgebra(diff=d,t,dual_shift=delta,s,polynom=t,s, > comm=theta,c,c,c2,v): > R:=matrix(2,4,d+/2/theta,,-,-,,d+/theta,-(c-c)/V*delta, > -(c2-c)/v*delta); d + R := d + θ (c c) δ (c2 c) δ Let us consider the A = Q(c, c, c 2,, θ)d, δ-module M = A 4 /(A 2 R) finitely presented by R. We compute the A-module structure of the endomorphism ring E = end A (M) of M: > Endo:=MorphismsConstCoeff(R,R,A): We obtain that the A-module E is defined by > nops(endo); generators which satisfy 8 > rowdim(endo2); A-linear relations. We do not print the large outputs of Endo. 4 Let us now search for idempotents of E defined by two idempotent matrices P A 4 4 and Q A 2 2, i.e., P and Q satisfy the relations R P = Q R, P 2 = P and Q 2 = Q: > Idem:=IdempotentsMatConstCoeff(R,Endo,A,,alpha): > nops(idem); 2 We obtain 2 different matrices P satisfying the previous relations. Let us consider the first one where we have set to zero the two arbitrary constants: > P:=subs(c8=,c2=,evalm(Idem,)); Q:=Factorize(Mult(R,P,A),R,A); P := c c c 2 c Q := c c 2 c c 2 c c c c 2 c2 c c c 2 As we have P 2 = P and Q 2 = Q, we know that the A-modules ker A (.P ), im A (.P ) = ker A (.(I 4 P )), ker A (.Q) and im A (.Q) = ker A (.(I 2 Q)) are projective, and thus, free by the Quillen-Suslin theorem. Let us compute bases of those free modules: > U:=SyzygyModule(P,A): U2:=SyzygyModule(evalm(-P),A): > U:=stackmatrix(U,U2);

2 U := c c c 2 c > X:=SyzygyModule(Q,A): X2:=SyzygyModule(evalm(-Q),A): > X:=stackmatrix(X,X2); X := We then know that the matrix U GL 4 (A) is such that the matrix R is equivalent to the following block-diagonal matrix S = R U : > S:=Mult(R,LeftInverse(U,A),A); d + S := d + θ We note that the second entry of the matrix S is invertible over A. Hence, we can use an elementary row operation to reduce the first row. In order to do that, we introduce the following unimodular matrix: > X:=evalm(,,,,-,d+/(2*theta),,,,,,,,,,); d + X := Then, the matrix S X has the form: > Mult(S,X,A); d + θ Hence, if we denote by Y = X U GL 4 (A) defined by > Y:=Mult(LeftInverse(X,A),U,A); d + Y := c c c 2 c then, the matrix R is equivalent to the following simple block-diagonal matrix V R Y : > Mult(V,R,LeftInverse(Y,A),A); d + θ Therefore, we have M = A 2 /(A ( )) A 2 /(A (d + /θ δ/ )), i.e.: M = A A 2 /(A (d + /θ / )). 2

3 If F denotes an A-module (e.g., F = C (R)), then we obtain that the linear differential time-delay system ker F (R.) is equivalent to the linear system ker F (S.), i.e.: ζ =, ζ 2 F, ζ3 (t) ζ 3 (t)/θ + ζ 4 (t h)/ =. Let us study the A-module structure of the endomorphism ring E = end A (M) of M: > ext:=exti(involution(endo2,a),a,): ext; As we have ext A (N, A) =, where N = A 2 /(A 4 R T ), because the previous matrix is the identity matrix (see F. Chyzak, A. Quadrat, D. Robertz, OreModules: A symbolic package for the study of multidimensional linear systems, in the book Applications of Time-Delay Systems, J. Chiasson and J. - J. Loiseau (Eds.), Lecture Notes in Control and Information Sciences (LNCIS) 352, Springer, ), we obtain that the A-module E is torsion-free. Let us check whether or not the A-module E is reflexive: > ext2:=exti(involution(endo2,a),a,2); δ d θ + δ d θ + δ d θ + δ d θ + As the previous matrix is not the identity matrix, we obtain that ext 2 A (N, A), a fact proving that the A-module E is not reflexive. Hence, the A-module E is torsion-free but not free. We proved that M = A N, where N = A 2 /(A (d + /θ δ/ )). Hence, we get: E = end A (M) = end A (A) hom A (N, A) hom A (A, N) end A (N). We have end A (A) = A and hom A (A, N) = N. Let us compute the A-modules hom A (N, A) and end A (N). In order to do that, we introduce the two matrices Z = and T = (d + /θ δ/ ): > Z:=evalm(); Z := > T:=submatrix(S,2..2,3..4); T := d + θ Let us check that we have hom A (A, N) = N by computing the A-module hom A (A, N): > E:=MorphismsConstCoeff(Z,T,A); E :=,, d θ + δ θ We obtain that the A-module hom A (A, N) is defined by two generators f and f 2 satisfying the A-linear relation θ d f θ δ f 2 =, which is precisely the definition of the A-module N. 3

4 Let us now compute the A-module hom A (N, A): > E2:=MorphismsConstCoeff(T,Z,A); E 2 := δ θ d θ + We obtain that A-module hom A (N, A) is defined by one generator which does not satisfy any A-linear relation, i.e., it is not a torsion element. Hence, we obtain hom A (N, A) = A. We note that we have T E2 =, which is consistent with the fact that hom A (N, A) = ker A (T.). Let us compute the A-module end A (N): > E3:=MorphismsConstCoeff(T,T,A): > E3; δ θ d θ +,, The A-module end A (N) is defined by two generators g and g 2 = id N which satisfy the relation: > E32; V d θ + Hence, we obtain that g = (θ d + ) id M, i.e., the A-module end A (N) is generated by id M, which proves that end A (N) = A. We finally obtain: E = A N A A = A 3 N. We can check the previous result by studying the A-module E. One way to do that is to find an injective A-morphism ϕ : N E such that coker ϕ = A 3. Using OreMorphisms, we can try to handle the corresponding computations. We first compute the A-module hom A (N, E): > Morph:=MorphismsConstCoeff(T,Endo2,A): If we consider the matrix P = Morph6 A 2 8 defining the element h of hom A (N, E) and compute a matrix Q A 8 satisfying T P = Q Endo2, > Y:=Morph6; Z:=Factorize(Mult(T,Y,A),Endo2,A); Y := ( ) ( ) θ c2 c c 2 c + c 2 c c θ c2 c c 2 c + c 2 c c θ ( c + c 2 ) ( θ 2 c2 c + c 2 + c ) 2 2 ( θ 2 c c + c 2 + c ) 2 then we can compute the A-module ker ϕ: > K:=KerMorphism(T,Endo2,Y,Z,A); Z := K :=, d θ + δ θ, d + θ, θ As the first matrix K is, we obtain that ϕ hom A (N, E) is injective. Let us now compute the A-module coker ϕ: > Coker:=CokerMorphism(T,Endo2,Y,Z,A): Let us compute its rank: 4

5 > OreRank(Coker,A); We finally need to check whether or not the A-module coker ϕ is free: > Ext:=Exti(Involution(Coker,A),A,): Ext; We obtain that the A-module coker ϕ is torsion-free. Moreover, we have coker ϕ = A 8 Ext3. > map(factor,leftinverse(ext3,a)); 2 (c 2 c ) 2 θ c 2 c 2 ( c +c 2) 2 θ c c 2 2 θ (c 2 c ) 3 (c 2 c ) 2 (c c ) c c 2 (c 2 c ) 2 (c c ) (c 2 c ) 2 (c c ) (c 2 c ) 2 c c 2 c c 2 (c c 2)(c c ) As the matrix Ext3 admits a left-inverse over A, we obtain that coker ϕ = A 3, which proves that we have the split exact sequence of A-modules N ϕ E A 3, a fact implying that E = N A 3 and proves the result. In all the previous computations, we have assumed that we were in the generic situation, i.e., the constants c, c and c 2 are pairwise different. As the module properties of M are known to depend on the system parameters (see F. Chyzak, A. Quadrat, D. Robertz, OreModules project, for the precise details), we let the reader handle the different non-generic situations. 5