ON THE PERIODIC SOLUTIONS OF LINEAR HOMOGENEOUS SYSTEMS OF DIFFERENTIAL EQUATIONS

Given a fundamental matrix ϕ(x) of an n-th order system of linear homogeneous differential equations Y′=A(x)Y, a necessary and sufficient condition for the existence of a k-dimensional (k≤n) periodic sub-space (of period T) of the solution space of the above system is obtained in terms of the rank of the scalar matrix ϕ(t)−ϕ(0).

The purpose of this note is to deduce a necessary and sufficient condition for the existence of periodic sub-spaces (of period T) of the solution space S of n the system (1.1) and to show that the existence and dimensions of these periodic sub-spaces depend not on any prior assumption about the periodicity (of period T) of the elements a..(x) of the coefficient matrix A(x) of the system (I.i) (that is, all the elements a..(x) of A(x) need not be periodic of period T), but precisely on the rank of the scalar matrix (T) (0). (1.3) 2. MAIN RESULTS.
The condition (1.3) is stated more explicitly in the following theorem: THEOREM.Let k be a non-negative integer, 0 < k < n.There exists a k-dimen- sional sub-space S k of the solution space Sn of the linear homogeneous system (i.i)such that each member of S k is periodic of period T and no member of Sn S k is periodic of period T if and only if the rank of the scalar matrix (T) (0) is

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The above theorem can also be phrased is terms of the eigen values of the sca- -i far matrix (0)(T) as follows.
COROLLARY.Let k be a non-negative integer, 0 < k < n.There exists a k-di- mensional sub-space S k of the solution space Sn of the linear homogeneous system (i.i)such that each member of S k is periodic of period T and no member of S S k n is periodic of period T if and only if i is an eigen value of the scalar matrix -I(0)(T) of multiplicity k.
PROOF OF THE THEOREM.Let k be a non-negative integer, 0 -< k -< n and rank of (T) (0) is n-k.Then the dimension of the kernel of (T) (0) is k.Hence there exists k linearly independent vectors v. col Let fi(x) (x)vi, i 1,2,...,k.The linear independence of the vectors vl, v2,... v k implies the linear independence of the k solution vectors fl(x)' f2(x) fk(x) of the system (i.i).Also, fi(T) fi(0) ((T) (0))v i 0, i 1,2 k, (2.2) implies by the uniqueness of the solutions of initial value problem that fi(x+T) fi(x) 0 for all x, i 1,2 k.Hence each solution vector f'l(x)' i 1,2 k, is periodic of period T. Let S k be the k-dimensional periodic (of period T) sub-space of Sn generated by fl(x)"'''fk(x)" We need to show that no member of Sn S k is periodic of period T. Let gl(x) g Sn S k.Then gl(x) is non- trivial and fl(x), f2(x), 'fk(x)' gl(x) are k+l linearly independent members of S Let n g!(x) (X)Vk+l, where Vk+1 col(Ck+ 1 iCk+l 2,...,Ck+l n ).
If possible, let gl(x) be periodic of period T. That is gl(x + T) gl(x) 0 for all x.
Then gl(r) gl(0) 0. (2.3) Since any set of linearly independent members of S form a part of a basis of n Sn, let fl(x),f2(x),...,fk (x)'gl (x)'g2 (x)'''''gn-k(x) be a basis of Sn and gi(x) (X)Vk+i, i 2,3 ,n-k, where Vk+i col(Ck+ i 1 Ck+i 2 Ck+i n i 2,3, n-k.The linear independence of the basis vectors fl(x)' f2(x)"''fk(x)'gl(x) 'gn-k(X) implies that the But, by actual multiplication and using (2.2) and ( 2.3), we see that the first k+l column vectors of (#(T) #( 0))C are zero-vectors and hence the rank of ((T) @(O))C is at most n-k-l.Therefore, from (2.4)   n-k rank of ((T) #( 0)) rank of ((T) #(0))C < n-k-i implying a contradiction.Hence gl(x) cannot be periodic of period T. That is no member of S S k is periodic of period T.
n Conversely, suppose that S k be a k-dimensional sub-space of Sn such that every member of S k is periodic of period T and no member of Sn S k is periodic of period T. We need to show that the rank of (T) (0) is n-k.
Let H(x) be the fundamental matrix of the linear system (i.i)whose column vectors are