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    MATH 623 Ordinary Differential Equations Fall 2010Instructor: Georgi Medvedev

    5 Lecture 5. Linear systems and Floquet theory.

    5.1 Linear systems

    Consider the homogeneous system of ODEs

    x= A(t)x and (5.1)

    and inhomogeneous systems of differential equations

    x= A(t)x + h(t), x Rn, (5.2)

    whereA(t) Rnn is a continuous matrix andh(t) Rn is a continuous vector.

    Definition 5.1. Suppose{x(1), x(2), . . . , x(n)}is a set of solutions of (5.1). Let

    X(t) = col (x(1), x(2), . . . , x(n)) Rnn

    such thatdetX(t0)= 0for somet0 R. ThenX(t)is called fundamental matrix solution of (5.1).

    Definition 5.2. Fundamental matrix solution of (5.1),X(t), is called a principal matrix solution of (5.1) attimet0 R ifX(t0) =I.

    Lemma 5.3. The rank ofX(t)does not depend ont.

    Proof. Let x(1), x(2), . . . , x(r) be solutions of (5.1). Then

    x(t) =c1x(1)(t) + c2x

    (2)(t) + + crx(r)(t) (5.3)

    solves (5.1). Ifx(t0) = 0for somet0 R then

    c1x(1)(t) + c2x

    (2)(t) + + crx(r)(t) = 0t R, (5.4)

    by uniqueness of solutions of the intial value problems for (5.1). Therefore,x(1)(t), x(2)(t), . . . , x(r)(t)are

    linearly independent for everyt iff they are linearly independent for some t0.

    We will need the following properties of solutions of (5.1) and (5.2).

    Lemma 5.4.

    1. IfX(t) is a fundamental matrix solution then X(t)X1(t0) is the principal matrix solution at timet0.

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    2. The variation of constants formula:

    x(t) =X(t)X1(t0

    )x(t0

    ) + t

    0X(t)X1(s)h(s)ds. (5.5)

    3. The Liouvilles formula:

    detX(t) = detX(t0)exp{

    tt0

    trA(s)}. (5.6)

    Proof.

    1. follows by verification.

    2. By a change of variable in (5.2)

    x(t) =X(t)y(t), (5.7)we have

    Xy+ Xy= AX y+ h. (5.8)

    PlugX=AXin (5.2) to obtainy= X1(t)h(t). (5.9)

    The statement in 2) follows after integrating (5.9) and using (5.7).

    3. Use product rule to verify the Leibniz formula for determinants:

    (det X) =n

    i=1

    det

    X1,1 X1,2 X1,n...

    ......

    Xi,1 Xi,2 X

    i,n

    ......

    ...

    Xn,1 Xn,2 Xn,n

    . (5.10)

    Since X=AX,

    Xi,k =n

    j=1

    ai,jXj,k, i, k {1, . . . , n},

    and

    ( Xi,1, . . . , Xi,n) =n

    j=1

    ai,j(Xj,1, . . . , X j,n), i {1, . . . , n}. (5.11)

    Thus,

    (det X) =n

    i=1

    det

    X1,1 X1,2 X1,n...

    ......n

    j=1 a1,jXj,1n

    j=1 a2,jXj,2 n

    j=1 an,jXj,n...

    ......

    Xn,1 Xn,2 Xn,n

    . (5.12)

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    Subtract from rowi on the rhs of (5.13) the linear combination:

    n

    j=1j=i

    ai,j(Xj,1, . . . , X j,n)

    to obtain (5.13)

    (det X) =n

    i=1

    det

    X1,1 X1,2 X1,n...

    ......

    ai,iXi,1 ai,iXi,2 ai,iXi,n...

    ......

    Xn,1 Xn,2 Xn,n

    = trA(t)det X(t) (5.13)

    The statement in 3) follows by integrating (5.13).

    Example 5.5. Consider

    x= A(t)x, A(t) =

    1 4(cos 2t)2 2 + 2 sin 4t2 + 2 sin 4t 1 4(sin2t)2

    . (5.14)

    From the characteristic equation

    2 + 2 + 1 = 0 (5.15)

    one finds that the eigenvalues ofA(t)are negative1,2 = 1. Nonetheless, the equilibrium at the origin is

    unstable, because

    x(t) =et

    sin2tcos2t

    (5.16)

    solves (5.14).

    5.2 Linear systems with periodic coefficients

    Consider

    x= A(t)x, x Rn, (5.17)

    whereA(t)is continuous periodic matrix with the least period >0,A(t + ) =A(t) tR

    .

    Theorem 5.6. A fundamental matrix solution of (5.17) can be written as

    X(t) =P(t)etB, (5.18)

    whereP(t)is periodic andB (in general complex) is a constant(n n)matrix.

    Remark5.7. Equation (5.18) implies that the stability of the equilibrium at the origin is deteremined by the

    spectrum ofB .

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    Proof. Note thatX(t) =X(t + )is a fundamental matrix solution of (5.17):

    X(t) =A(t + )X(t + ) =A(t)X(t).

    Thus, for some constant nonsingular matrix C Rnn,

    X(t + ) =X(t)C. (5.19)

    Set

    C=eB (5.20)

    and define

    P(t) =X(t)etB (5.21)

    ThenP is periodic, because

    P(t + ) =X(t + )eBetB =X(t)etB =P(t).

    Remark5.8. When Chas negative eigenvalues, Bas defined in (5.20) is complex. However, if one redefines

    C=X(t)1X(t + 2)thenB can be chosen real.

    Definition 5.9. MatrixC in (5.19) is called the monodromy matrix. The eigenvalues

    1, 2, . . . , n

    ofCare called characteristic multipliers. Complexk is called a characteristic exponent if

    k =ek.

    Theorem 5.10.

    12 . . . n = exp{

    0

    trA(t)dt}, (5.22)

    ni=1

    i = 1

    0

    trA(t)dtmod 21i (5.23)

    Proof. Let X(t)be a principal matrix solution. Then by Liouvilles formula,

    det X(t) = exp{ 0

    trA(s)ds}. (5.24)

    On the other hand

    det X(t) = det P(t)etB.

    Thus,

    det P()eB = exp{

    0

    trA(s)ds}.

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    But,

    det P() = det P(0) = det X(0) = 1.

    Finally,12 . . . n = det P()e

    B = exp{

    0

    trA(s)ds}.

    This shows (5.22). Equation (5.23) follows from (5.22).

    For2Dflows Theorem it allows to compute the characteristic multipliers. To show this, consider

    x= f(x), x R2, (5.25)

    where as usualf(x) = (f1(x), f2(x))T is a smooth vector field. Supposex = u(t)is a periodic solution of(5.25) with period1. The variational equation is obtained by linearizing the flow near O = {x= u(t) : t

    S1}, i.e., pluggingx = u(t) + (t)in (5.25) and keeping the linear term:

    = A(t), A(t) =Df(u(t)) . (5.26)

    We are interested in stability of the trivial solution = 0 of the variational equation (5.25). By differentiatingboth sides of

    u= F(u(t)) ,

    we find that= u(t)solves (5.26). Therefore, one can choose the fundamental matrix solution as

    X(t) = (u(t)).

    Thus,

    X(t + ) = (u(t + )) = (u(t)) =

    1 0

    . (5.27)

    From (5.27) conclude that one multiplier cahrachteristic multiplier is1 = 1. Theorem 5.2 then yields thesecond multiplier

    2 = exp{

    0

    trDf(u(s))ds}= exp{

    0

    divf(u(s))ds}.

    Therefore, the sign of0 divf(u(s))dsdetermines the stability of the periodic orbit of a 2Dflow.

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