# Differential Transcendency in the Theory of Linear Differential Systems with Constant Coefficients

Authors: Branko Malešević; Dragana Todorić; Ivana Jovović; Sonja Telebaković
Journal: ISRN Mathematical Analysis
 (2012)

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## Abstract

We consider a reduction of a nonhomogeneous linear system of first-order operator equations to a totally reduced system. Obtained results are applied to Cauchy problem for linear differential systems with constant coefficients and to the question of differential transcendency.


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## Body

## 1. Introduction

Linear systems with constant coefficients are considered in various fields  (see [1–5]). In our paper [5] we use the rational canonical form and a certain sum of principal minors to reduce a linear system of first-order operator equations with constant coefficients to an equivalent, so called partially reduced, system. In this paper we obtain more general results regarding sums of principal minors and a new type of reduction. The obtained formulae of reduction allow some new considerations in connection with Cauchy problem for linear differential systems with constant coefficients and in connection with the differential transcendency of the solution coordinates.


## 2. Notation

Let us recall some notation. LetK be a field, and let B∈Kn×n be an n-square matrix. We denote by(δk)
δk=δk(B) the sum of its principal minors of order k(1≤k≤n),(δki)
δki=δki(B) the sum of its principal minors of order k containing ith column (1≤i,k≤n).Letv1,…,vn be elements of K. We write Bi(v1,…,vn)∈Kn×n for the matrix obtained by substituting v→=[v1…vn]T in place of ith column of B. Furthermore, it is convenient to use
(2.1)δki(B;v→)=δki(B;v1,…,vn)=δki(Bi(v1,…,vn)),
and the corresponding vector
(2.2)δ→k(B;v→)=[δk1(B;v→)…δkn(B;v→)]T.
The characteristic polynomial ΔB(λ) of the matrix B∈Kn×n has the following form:
(2.3)ΔB(λ)=det⁡(λI-B)=λn+d1λn-1+⋯+dn-1λ+dn,
where dk=(-1)kδk(B), 1≤k≤n; see [6, page 78].Denote byB¯(λ) the adjoint matrix of λI-B, and let B0,B1,…,Bn-2,Bn-1 be n-square matrices over K determined by
(2.4)B¯(λ)=adj(λI-B)=λn-1B0+λn-2B1+⋯+λBn-2+Bn-1.Recall that(λI-B)B¯(λ)=ΔB(λ)I=B¯(λ)(λI-B).The recurrenceB0=I;Bk=Bk-1B+dkIfor1≤k<n follows from equation B¯(λ)(λI-B)=ΔB(λ)I; see [6, page 91].


## 3. Some Results about Sums of Principal Minors

In this section we give two results about sums of principal minors.Theorem 3.1.
ForB∈Kn×n and v→=[v1…vn]T∈Kn×1, one has:
(3.1)δki(B;∑j=1nb1jvj,…,∑j=1nbnjvj)+δk+1i(B;v1,…,vn)=δk(B)vi.Remark 3.2.
The previous result can be described by(3.2)δki(B;Bv→)+δk+1i(B;v→)=δk(B)vi,1≤i≤n,
or simply by the following vector equation:
(3.3)δ→k(B;Bv→)+δ→k+1(B;v→)=δk(B)v→.Proof of Theorem3.1.
Lete→s∈Kn×1 denote the column whose only nonzero entry is 1 in sth position. We also write B↓s for sth column of the matrix B and [B]s^ for a square matrix of order n-1 obtained from B by deleting its sth column and row. According to the notation used in  (2.1), let [Bi(B↓s)]s^ stand for the matrix of order n-1 obtained from B by substituting sth column B↓s in place of ith column, and then by deleting sth column and sth row of the new matrix. By applying linearity of δ→k(B;v→) with respect to v→, we have
(3.4)δki(B;Bv→)+δk+1i(B;v→)=δki(B;∑s=1nvsB↓s)+δk+1i(B;∑s=1nvse→s)=∑s=1nvsδki(B;B↓s)+∑s=1nvsδk+1i(B;e→s)=∑s=1nvs(δki(B;B↓s)+δk+1i(B;e→s))=vi(δki(B;B↓i)+δk+1i(B;e→i))+∑s=1s≠invs(δki(B;B↓s)+δk+1i(B;e→s)).
First, we compute vi(δki(B;B↓i)+δk+1i(B;e→i))=vi(δki(B)+δk([B]i^))=viδk(B). Then, it remains to show that ∑s=1,s≠invs(δki(B;B↓s)+δk+1i(B;e→s))=0.
It suffices to prove that each term in the sum is zero, that is,(3.5)δki(B;B↓s)+δk+1i(B;e→s)=0,fors≠i.
Suppose that s≠i. We now consider minors in the sum δki(B;B↓s). All of them containing sth column are equal to zero, so we deduce
(3.6)δki(B;B↓s)=δki(Bi(B↓s))=δki([Bi(B↓s)]s^).
If s≠i, then each minor in the sum δk+1i(B;e→s) necessarily contains sth row and ith column. By interchanging ith and sth column, we multiply each minor by -1. We now proceed by expanding these minors along sth column to get -1 times the corresponding kth order principal minors of matrix Bi(B↓s) which do not include sth column. Hence, δk+1i(B;e→s)=-δki([Bi(B↓s)]s^), and the proof is complete.In the following theorem, we give some interesting correspondence between the coefficientsBk of the matrix polynomial B¯(λ)=adj(λI-B) and sums of principal minors δ→k+1(B;v→), 0≤k<n.Theorem 3.3.
Given an arbitrary column[v1…vn]T∈Kn×1, it holds
(3.7)Bkv→=Bk[v1v2⋮vn]=(-1)k[δk+11(B;v1,…,vn)δk+12(B;v1,…,vn)⋮δk+1n(B;v1,…,vn)]=(-1)kδ→k+1(B;v→).Proof.
The proof proceeds by induction onk. It is being obvious for k=0. Assume, as induction hypothesis (IH), that the statement is true for k-1. Multiplying the right side of the equation Bk=Bk-1B+dkI, by the vector v→, we obtain that
(3.8)Bkv→=Bk-1(Bv→)+dkv→=(IH)⁡(-1)k-1δ→k(B;Bv→)+dkv→=(-1)k-1(δ→k(B;Bv→)-δkv→)=(5)⁡(-1)kδ→k+1(B;v→).Remark 3.4.
Theorem3.3 seems to have an independent application. Taking v→=e→j,1≤j≤n, we prove formulae (8)–(10) given in  [7].


## 4. Formulae of Total Reduction

We can now obtain a new type of reduction for the linear systems with constant coefficients from [5] applying results of previous section. For the sake of completeness, we introduce some definition.LetK be a field, V a vector space over field K, and let A:V→V be a linear operator. We consider a linear system of first-order A-operator equations with constant coefficients in unknowns xi, 1≤i≤n, is
(4.1)A(x1)=b11x1+b12x2+⋯+b1nxn+φ1A(x2)=b21x1+b22x2+⋯+b2nxn+φ2⋮A(xn)=bn1x1+bn2x2+⋯+bnnxn+φn
for bij∈K and φi∈V. We say that B=[bij]i,j=1n∈Kn×n is the system matrix, and φ→=[φ1…φn]T∈Vn×1 is the free column.Letx→=[x1…xn]T be a column of unknowns and A→:Vn×1→Vn×1 be a vector operator defined componentwise A→(x→)=[A(x1)…A(xn)]T. Then system (4.1) can be written in the following vector form:
(4.2)A→(x→)=Bx→+φ→.
Any column v→∈Vn×1 which satisfies the previous system is its solution.Powers of operatorA are defined as usual Ai=Ai-1∘A assuming that A0=I:V→V is the identity operator. By nth order linear A-operator equation with constant coefficients, in unknown x, we mean
(4.3)An(x)+b1An-1(x)+⋯+bnI(x)=φ,
where b1,…,bn∈K are coefficients and φ∈V. Any vector v∈V which satisfies (4.3) is its solution.The following theorem separates variables of the initial system.Theorem 4.1.
Assume that the linear system of first-orderA-operator equations is given in the form  (4.2), A→(x→)=Bx→+φ→, and that matrices B0,…,Bn-1 are coefficients of the matrix polynomial B¯(λ)=adj(λI-B). Then it holds the following:
(4.4)(ΔB(A→))(x→)=∑k=1nBk-1A→n-k(φ→).Proof.
LetLB:Vn×1→Vn×1 be a linear operator defined by LB(x→)=Bx→. Replacing λI by A→ in the equation ΔB(λ)I=B¯(λ)(λI-B), we obtain that
(4.5)ΔB(A→)=B¯(A→)∘(A→-LB),henceΔB(A→)(x→)=B¯(A→)((A→-LB)(x→))=B¯(A→)(A→(x→)-Bx→)=(8)⁡B¯(A→)(φ→)=∑k=1nBk-1A→n-k(φ→).The next theorem is an operator generalization of Cramer's rule.Theorem 4.2 (the theorem of total reduction-vector form).
Linear system of first-orderA-operator equations  (4.2) can be reduced to the system of nth order A-operator equations
(4.6)(ΔB(A→))(x→)=∑k=1n(-1)k-1δ→k(B;A→n-1(φ→)).Proof.
It is an immediate consequence of Theorems4.1 and 3.3 as follows:
(4.7)ΔB(A→)(x→)=(10)⁡∑k=1nBk-1A→n-k(φ→)=(6)⁡∑k=1n(-1)k-1δ→k(B;A→n-k(φ→)).We can now rephrase the previous theorem as follows.Theorem 4.3 (the theorem of total reduction).
Linear system of first-orderA-operator equations (4.1) implies the system, which consists of nth order A-operator equations as follows:
(4.8)(ΔB(A))(x1)=∑k=1n(-1)k-1δk1(B;A→n-k(φ→))⋮(ΔB(A))(xi)=∑k=1n(-1)k-1δki(B;A→n-k(φ→))⋮(ΔB(A))(xn)=∑k=1n(-1)k-1δkn(B;A→n-k(φ→)).Remark 4.4.
System (4.8) has separated variables, and it is called totally reduced. The obtained system is suitable for applications since it does not require a change of base. This system consists of nth order linear A-operator equations which differ only in the variables and in the nonhomogeneous terms.
Transformations of the linear systems of operator equations into independent equations are important in applied mathematics [1]. In the following two sections: we apply our theorem of total reduction to the specific linear operators A.


## 5. Cauchy Problem

Let us assume thatA is a differential operator on the vector space of real functions and that system (4.1) has initial conditions xi(t0)=ci, for 1≤i≤n. Then the Cauchy problem for system (4.1) has a unique solution. Using form (4.2), we obtain additional n-1 initial conditions of ith equation in system (4.8). Consider
(5.1)(Aj(xi))(t0)=[Bjx→(t0)]i+∑k=0j-1[Bj-1-kA→k(φ→)(t0)]i,1≤j≤n-1,
where []i denotes ith coordinate. Then each equation in system (4.8) has a unique solution under given conditions and additional conditions (5.1), and these solutions form a unique solution to system (4.1). Therefore, formulae (4.8) can be used for solving systems of differential equations.It is worth pointing out that the above method can be also extended to systems of difference equations.


## 6. Differential Transcendency

Now suppose thatV is the vector space of meromorphic functions over the complex field C and that A is a differential operator, A(x)=(d/dz)(x). Let us consider system  (4.1) under these assumptions.Recall that a functionx∈V is differentially algebraic if it satisfies a differential algebraic equation with coefficients from C; otherwise, it is differentially transcendental (see [2–4, 8–10]).Let us consider nonhomogenous linear differential equation ofnth order in the form  (4.3), where b1,…,bn∈C are constants and φ∈V. If x is differentially transcendental then ΔB(A)(x) is also a differentially transcendental function. On the other hand, if φ is differentially transcendental, then, based on Theorem 2.8. from [10], the solution x of (4.3) is a differentially transcendental function. Therefore, we obtain the equivalence.Theorem 6.1.
Letx be a solution of (4.3), and then x is a differentially transcendental function if and only if φ is a differentially transcendental function.We also have the following statement about differential transcendency.Theorem 6.2.
Letφj be the only differentially transcendental component of the free column φ→. Then for any solution x→ of system (4.2), the corresponding entry xj is also a differentially transcendental function.Proof.
The sum∑k=1n(-1)k-1δkj(B;A→n-k(φ→)) must be a differentially transcendental function. The previous theorem applied to the following equation:(6.1)ΔB(A)(xj)=∑k=1n(-1)k-1δkj(B;A→n-k(φ→))
implies that xj is a differentially transcendental function too.Let us consider system  (4.1), and let φ1 be the only differentially transcendental component of the free column φ→. Then, the coordinate x1 is a differentially transcendental function too. Whether the other coordinates xk are differentially algebraic depends on the system matrix B. From the formulae of total reduction and Theorem 6.1. we obtain the following statement.Theorem 6.3.
Letφ1 be the only differentially transcendental component of the free column φ→ of system  (4.1). Then the coordinate xk, k≠1, of the solution x→, is differentially algebraic if and only if in the sum ∑j=1n(-1)j-1δjk(B;A→n-j(φ→)) appears no function An-j(φ1)(j=1,…,n).Example 6.4.
Let us consider system  (4.1) in the form (4.8) in dimensions n=2 and n=3 with φ1 as the only differentially transcendental component. The function x1 is differentially transcendental. For n=2 the function x2 is differentially algebraic if and only if b21=0. For n=3 the function x2 is differentially algebraic if and only if b21=0∧b31·b23=0 and the function x3 is differentially algebraic iff b31=0∧b31·b22=0.Let us emphasize that if we consider two or more differentially transcendental components of the free columnφ→, then the differential transcendency of the solution coordinates also depends on some kind of their differential independence (see e.g., [8]).



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