(1984), Periodic - MATHfe.math.kobe-u.ac.jp/FE/FE_pdf_with_bookmark/FE21... · Funkcialaj Ekvacioj, 27 (1984), 25-48 Perturbation Method for Linear Periodic Systems I By SAITO Tosiya

Funkcialaj Ekvacioj, 27 (1984), 25-48

Perturbation Method for LinearPeriodic Systems I

By

Tosiya SAITO and Ichiro TSUKAMOTO

(Keio University, Japan)

§1. Let there be given a linear system

(1) $¥frac{dx}{dt}=A(t, ¥epsilon)x$, $X¥in C^{n}$ , $t$ $¥in R$, $¥epsilon¥in C$.

Here $A(t, ¥epsilon)$ is an n-by-n matrix with complex-valued entries which are continuousand periodic in $t$ with period $T$ and holomorphic in $¥epsilon$ for $t¥in R$ and $|¥epsilon|¥leqq K$. Then$A(t, ¥epsilon)$ can be expressed as

(2) $¥mathrm{A}(t, ¥epsilon)=¥sum_{0}^{¥infty}A_{k}(t)¥epsilon^{k}$, $A_{k}(t+T)=A_{k}(t)$ .

Let $X(t, ¥epsilon)$ be a fundamental matrix of (1) such that

$X(0, ¥epsilon)=E$

where $E$ denotes an $¥mathrm{n}$-dimensional unit matrix. Then since $X(t_{ 6},)$ is holomorphicin $¥epsilon$ at $¥epsilon=0$, it can be represented as

(3) $X(t, ¥epsilon)=¥sum_{0}^{¥infty}X_{k}(t)¥epsilon^{k}$.

Therefore usual perturbation method can be applied and we can calculate$X_{k}(t)$ , $k=1,2$, $¥ldots$ , whenever $X_{0}(t)$ can be obtained explicitly by solving an unper-turbed system

$¥frac{dx}{dt}=A(t, 0)x=A_{0}(t)x$ .

However expression such as (3) of $X(t, ¥epsilon)$ is not a convenient one in discussingthe asymptotic properties of solutions of (1). For example, consider a linear system

$¥frac{dx_{1}}{dt}=X_{2}$ , $¥frac{dx_{2}}{dt}=-¥epsilon^{2}x_{l}$ .

This system can easily be solved and we get

26 T. SAITO and I. TSUKAMOTO

$X(t, ¥epsilon)=¥left(¥begin{array}{ll}¥mathrm{c}¥mathrm{o}¥mathrm{s}¥epsilon t & ¥mathrm{s}¥mathrm{i}¥mathrm{n}¥epsilon t/¥epsilon¥¥-¥epsilon ¥mathrm{s}¥mathrm{i}¥mathrm{n}¥epsilon t & ¥mathrm{c}¥mathrm{o}¥mathrm{s}¥epsilon t¥end{array}¥right)$

which shows that every solution is periodic with period 2 $¥pi/¥epsilon$ , while the perturbationmethod gives us

$X(t, ¥epsilon)=¥left(¥begin{array}{ll}1 & t¥¥0 & 1¥end{array}¥right)$ $-¥epsilon^{2}[_{i}^{¥frac{t^{2}}{2^{1}}}$ $¥frac¥frac{3t^{3}t^{2}¥dagger}{2^{1}}1¥rfloor|+¥epsilon^{4}[[¥frac{4t^{4}t^{3}!}{3¥dagger}¥frac|$ $¥frac¥frac{5t^{5}t^{4}¥dagger}{4¥dagger}]--$ .

Periodicity of solutions can hardly be read from such an expression.On the other hand, it is well known that $X(t, ¥epsilon)$ admits following Floquet

representation:

$X(t, ¥epsilon)=F(t, ¥epsilon)e^{t¥Lambda(¥epsilon)}$

where $F(t+T, ¥epsilon)=F(t, ¥epsilon)$ and $¥Lambda(¥epsilon)$ is a constant matrix. In discussing periodicity,boundedness or stability properties of solutions, this representation is most useful.Especially the eignevalues of $¥Lambda(¥epsilon)$ , which are usually called characteristic exponentsof (1), will give us extremely important information regarding the asymptoticbehavior of solutions. So it is very desirable to find a perturbation method whichallows us to calculate $F_{-}(t, ¥epsilon)$ and $¥Lambda(¥epsilon)$ . The object of this paper is to develop onesuch perturbation method which is valid under fairly general assumptions.

§2. We begin with following general consideration.Suppose a periodic system

(4) $¥frac{dx}{dt}=A(t)x$ , $¥mathrm{A}(t+T)=A(t)$ ,

is given where $A(t)$ is assumed to be continuously differentiable. Let $X(t)$ be anarbitrary fundamental matrix of (4).

Instead of Floquet representation

$X(t)=F(t)e^{t¥Lambda}$, $F(t+T)=F(t)$

we propose here a representation of the following form:

(5) $X(t)=V(t)W(t)$,

where $¥nabla(t+T)=V(t)$ and

$W(t)=¥left(¥begin{array}{ll}W_{1}(t) & ¥mathrm{C}¥¥0 & W_{r}(t)¥end{array}¥right)$ ,

Perturbation Method 27

$W_{k}(t)=e^{r_{k}c}$ $¥left¥{¥begin{array}{lll}w_{1k}(t). & w_{2k}(t.)¥cdot¥cdot & .w_{n_{k}k^{¥prime}}(t)¥¥0 & & ¥vdots¥¥ & & w_{2k}(t)¥¥ & & w_{1k}(t)¥end{array}¥right¥}$ $n_{1}+-+n_{r}=n$.

Here $¥lambda_{k}$ are constants and $w_{jk}(t)$ are polynomials of $t$ whose coefficients are allperiodic functions of $t$ (with period $T$). Let us call (5) a quasi-Floquet representa-tion of $X(t)$ .

Quasi-Floquet representation of a fundamental matrix is not unique. Howeverif we can obtain one such representation of $X(t)$ , we can immediately see that $¥lambda_{k}$ arecharacteristic exponents of (4). Moreover it is not difficult to construct Floquetrepresentation of $X(t)$ from this.

§3. To simplify the description, we introduce following definitions.

De finition 1. Let $C$ be an n-by-n matrix and its Jordan’s normal form be

$¥left(¥begin{array}{ll}J_{1} & 0¥¥0 & .J_{¥gamma}¥end{array}¥right)$ , $J_{k}=¥left(¥begin{array}{lll}¥lambda_{k}. & 1. & 0¥¥ & & .1¥¥0 & & ¥lambda_{k}¥end{array}¥right)$, $k=1$,?, $r$ .

Then $C$ is said to be nondegenerate if

$¥lambda_{i}¥neq¥lambda_{j}$ for $i¥neq j$,

or, equivalently, if the rank of $C-¥lambda_{k}E$ is equal to n?l for every $¥dot{¥Lambda}_{k}$ .

Deffiition 2. Let $M$ be a monodromy matrix of a fundamental matrix $X(t)$ of(4), i.e.

$X(t+T)=X(t)M$.

Then the system (4) is called a nondegenerate system if $M$ is nondegsnerate.

Let us consider a matrix differential equation

(6) $¥frac{dU}{dt}=A(t)U-$ $X(t)$

along with the system (4). As can easily be verified by direct calculation, generalsolution of (6) is given by

(7) $U(t)=X(t)CX^{-1}(t)$

where $X(t)$ is a fundamental matrix of (4), and $C$ is an arbitrary constant matrix.Let $M$ be a monodromy matrix of $X(t)$ . Then


$U(t+T)=X(t+T)CX^{-1}(t+T)$

$=X(t)MCM^{-1}X(t)$ .

Therefore we have

$U(t+T)=U(t)$

if and only if

$MC=CM$.

A periodic solution of (6)

$U(t)=X(t)¥Gamma x^{-1}(t)$ , $M¥Gamma=¥Gamma M$,

is called a nondegenerate periodic solution if $¥Gamma$ is nondegenerate. Then we canprove the following:

Theorem 1. Suppose that the system (4) is nondegenerate and that a nondegene-rate periodic solution of (6) has been obtained, then we can construct a quasi-Floquetrepresentation of the fundamental matrix of (A).

Proof. Let

(8) $U(t)=X(t)¥Gamma X^{-1}(t)$ , $M¥Gamma=¥Gamma M$,

be a nondegenerate periodic solution of (6). We suppose that $X(t)$ is so chosen thatits monodromy matrix $M$ is reduced to its Jordan’s normal form, namely

(9) $M=¥left(¥begin{array}{ll}M_{1} & 0¥¥0 & ¥dot{M}_{r}¥end{array}¥right)$ , $M_{k}=¥underline{¥left(¥begin{array}{lll}¥mu_{k}. & 1. & 0¥¥ & & .1¥¥0 & & .¥mu_{k}¥end{array}¥right)}-¥}n_{k}$ rows.

$n_{k}$ columns

Since the system (4) is assumed to be nondegenerate,

$¥mu_{i}¥neq¥mu_{j}$ , if $i¥neq j$.

Then, since $M¥Gamma=¥Gamma M$ and $¥Gamma$ is also nondegenerate, $¥Gamma$ must be of the followingform:

(10) $¥Gamma=¥left(¥begin{array}{ll}¥Gamma_{1} & 0¥¥0 & ¥dot{¥Gamma}_{r}¥end{array}¥right)$, $¥Gamma_{k}=[_{0}^{¥gamma_{1k}}¥gamma_{2k}.-¥gamma_{1¥mathrm{k}}¥gamma_{2k}...¥rfloor¥gamma_{n_{k}k}]$ ,

$r_{1i}¥neq r_{1j}$ if $i¥neq.¥dot{f}$.


Notice that $¥gamma_{1k}$ , the eigenvalues of $¥Gamma$ , are also the eigenvalues of $U$, and hence canbe calculated explicitly.

Let $J$ be the Jordan’s normal form of $¥Gamma$ , then obviously we have

(11) $J=¥left(¥begin{array}{ll}J_{1} & 0¥¥0 & .J_{r}¥end{array}¥right)$ , $J_{k}=¥left(¥begin{array}{lll}¥gamma_{1k}. & 1. & 0¥¥ & & .1¥¥0 & & .¥gamma_{1k}¥end{array}¥right)¥}n_{k}$ rows.

$¥overline{n_{k}¥mathrm{c}¥mathrm{o}1¥mathrm{u}¥mathrm{m}¥mathrm{n}¥mathrm{s}}$

Since $U(t)$ is equivalent to $¥Gamma$ , its Jordan’ $¥mathrm{s}$ normal form is also $J$ and we can find anonsingular matrix $V(t)$ such that

(12) $V^{-1}(t)U(t)V(t)=J$, $V(t+T)=V(t)$ .

Obviously $V(t)$ can be chosen so that it is continuosly differentiable. Then from (8)and (12), we have

$V^{-1}X¥Gamma X^{-1}V=J$.

So if we put

(13) $V^{-1}X=W$,

we get

$W¥Gamma W^{-1}=J$.

Comparing (10) and (11), we immediately see that $W(t)$ has the following form:

$¥mathrm{r}^{w_{1k}(t)}¥cdot.w_{2k}(t.)-w_{n_{h^{k}}}.(t)]$

(14) $W(t)=¥left(¥begin{array}{ll}W_{1}(t). & 0¥¥0 & ¥dot{W}_{¥gamma}(t)¥end{array}¥right)$ ,$W_{k}(t)=1|$ 0

.$¥dot{w}_{2k}w_{1k}.(t)(t)¥rfloor|$

and from (12) and (13) we have

(15) $X(t)=V(t)W(t)$ , $V(t+T)=V(t)$ .

We shall show that $W(t)$ can be calculated explicitly and (15) gives a quasi-Floquetrepresentation of $X(t)$ .

Substituting (15) into (6), we obtain a following differential equation

(16) $¥frac{dW}{dt}=V^{-1}(t)(A(t)V(t)-¥dot{V}(t))W$.

Let us put


$V^{-1}(t)(A(t)V(t)-¥dot{V}(t))=P(t)$ .

Then obviously we have

$P(t+T)=P(t)$ .

Since $V(t)$ can be calculated from $U(t)$ , $P(t)$ is a known matrix whenever $U(t)$ isknown.

Also, since

$P(t)=¥dot{W}W^{-1}$

and $W(t)$ is of the form (14), $P(t)$ must have a similar form, that is:

(17) $P(t)=¥left(¥begin{array}{ll}P_{1}(t). & 0¥¥0 & .P_{r}(t)¥end{array}¥right)$ $P_{k}(t)=¥left¥{¥begin{array}{lllll} & P_{1k}. & P_{2k} & ¥cdots & P_{n_{k}k}¥¥ & & & & ¥vdots¥¥¥mathrm{l} & & & & .P_{1k}P_{2k}¥end{array}¥right¥}$.

Therefore, from (14), (16) and (17), we get

$¥frac{dW_{k}}{dt}=P_{k}(t)W_{k}$ , $k=1$,?, $r$ ,

or, written explicitly,

(18) $¥dot{w}_{1k}=p_{1k}w_{1k},¥dot{w}_{2k}=p_{1k}w_{2k}+p_{2k}w_{1k}$ ,?, $¥dot{w}_{n_{k}k}=p_{1k}w_{nk}h+-+p_{n}k^{k}w_{1k}$ ,

$k=1$,?, $r$ .

$W_{k}$ can be calculated by solving (18).From the first equation of (18), we obtain

$ w_{1k}(t)=¥exp¥int^{t}p_{1k}(¥tau)d¥tau$.

Hence if we denote by $¥lambda_{k}$ the constant term in the Fourier expansion of $p_{1k}(t)$ , $w_{1k}(t)$

can be written as

$w_{1k}(t)=e^{¥lambda_{k}t}q_{1k}(t)$ , $q_{1k}(t+T)=q_{1k}(t)$ .

Substituting this into the second equation, we get

$ w_{2k}(t)=w_{1k}(t)¥int^{t}w_{1k}^{-1}(¥tau)p_{2k}(¥tau)w_{1k}(¥tau)d¥tau$

$=e^{¥lambda_{k}t}q_{1k}(t)¥int^{t}p_{2k}(¥tau)d¥tau$

$=e^{¥lambda_{k}t}q_{2k}(t)$ ,


where $q_{2k}(t)$ is a polynomial (of the first degree) of $t$ whose coefficients are periodicfunctions of $t$ .

Proceeding in this way, we can show that

$w_{jk}(t)=e^{¥lambda_{k}t}q_{jk}(t)$, $j=1$,?, $n_{k}$

wher $q_{jk}(t)$ are polynomials of $t$ whose coefficients are periodic functions of $t$ .

Therefore

$X(t)=V(t)W(t)$

is a quasi-Floquet representation of $X(t)$ and Theorem 1 is proved.

§4. If we want to solve the periodic linear system (4) by applying the resultof Theorem 1, we have first to confirm that (4) is a nondegenerate system and nextto obtain a nondegenerate periodic solution of (6). This is usually impossible.Therefore Theorem 1 will not work as a tool to solve periodic linear systems ingeneral. However it works as a perturbation method as we shall show from now on.

So let us return to $¥mathrm{t}¥mathrm{n}¥mathrm{e}$ linear system (1) mentioned at the beginning of §1,namely

$¥frac{dx}{dt}=A(t, ¥epsilon)x$ ,

$A(t, ¥epsilon)=¥sum_{0}^{¥infty}A_{k}(t)¥epsilon^{k}$ , $A_{k}(t+T)=A_{k}(t)$ .

As a starting point of the perturbation method, we assume that:(a) an unperturbed system

(19) $¥frac{dx}{dt}=A(t, 0)x=A_{0}(t)x$

can be solved and is a nondegenerate system.

Let $X(t, ¥epsilon)$ be a fundamental matrix of (1) such that

$X(0, ¥epsilon)=E$,

and $M(¥epsilon)$ be its monodromy matrix. Then $X(t, 0)$ is a fundamental matrix of (19)and $M(0)$ is a corresponding monodromy matrix. By asumption (a), $M(0)$ is anondegenerate matrix.

Since

$X(T, ¥epsilon)=X(0, ¥epsilon)M(¥epsilon)=M(¥epsilon)$.

and $X(t, ¥epsilon)$ is holomorphic in $¥epsilon$ for any $t$ , $M(¥epsilon)$ is also holomorphic in $¥epsilon$ at $¥epsilon=0$ . Soif we denote by $¥mu_{1}(¥epsilon)$ , $¥mu_{2}(¥epsilon)$,?, the eigenvalues of $M(¥epsilon)$ , each $¥mu_{k}(¥epsilon)$ is a continuous


function of $¥epsilon$ near $¥epsilon=0$ . So if we put

$P_{k}(¥epsilon)=M(¥epsilon)-¥mu_{k}(¥epsilon)E$ ,

$P_{k}(¥epsilon)$ is also continuous in $¥epsilon$ . Since $M(0)$ is assumed to be nondegenerate,

rank $(P_{k}(0))=n-l$ .

Therefore

rank $(P_{k}(¥epsilon))¥geqq n-l$

for sufficiently small $¥epsilon$ . However, as $¥mu_{k}(¥epsilon)$ is an eignevalue of $M(¥epsilon)$ , we should have

rank $(P_{k}(¥epsilon))¥leqq n-l$ .

Hence we have

rank $(P_{k}(¥epsilon))=$ rank $(M(¥epsilon)-¥mu_{k}(¥epsilon)E)=¥mathrm{n}-1$

for each $k$ which shows that $M(¥epsilon)$ is nondegenerate. Consequently if $M(0)$ is non-degenerate as we have assumed above, $M(¥epsilon)$ is also nondegenerate if $|¥epsilon|$ is smallenough.

Following our scheme developed in §3, we consider a matrix differentialequation.

(20) $¥frac{dU}{dt}=A(t, ¥epsilon)U-U_{t}4(t, ¥epsilon)$ .

If a nondegenerate periodic solution of (20) can be found, then we can construct aquasi-Floquet representation of $X(t, ¥epsilon)$ by Theorem 1.

Let 1 $(¥epsilon)$ be a nondegenerate matrix which is holomorphic in $¥epsilon$ at $¥epsilon=0$ , and

(21) $M(¥epsilon)¥Gamma(¥epsilon)=¥Gamma(¥epsilon)M(¥epsilon)$ .

Then

(22) $U(t_{ 6},)=X(t, ¥epsilon)¥Gamma(¥epsilon)X^{-1}(t, ¥epsilon)$

is a nondegenerate periodic solution of (20). Existence ofsuch $¥Gamma(¥epsilon)$ is evident because$M(¥epsilon)$ itself has these properties. Since $X(t, ¥epsilon)$ and $¥Gamma(¥epsilon)$ are both holomorphic in $¥epsilon$ ,$U(t, ¥epsilon)$ is also holomorphic in $¥epsilon$ and hence can be expressed as

(23) $U(t, ¥epsilon)=¥sum_{0}^{¥infty}U_{k}(t)¥epsilon^{k}$, $U_{k}(¥mathrm{r}+T)=U_{h}(t)$ ,

in the neighbourhood of $¥epsilon=0$ . Substituting (23) into (20) and comparing the termsof the same degree in $¥epsilon$ on both sides, we obtain


$¥dot{U}_{0}=A_{0}U_{0}-U_{0}A_{0}$,$¥dot{U}_{1}=A_{0}U_{1}-U_{1}A_{0}+A_{1}U_{0}-U_{0}A_{1}$ ,

(24)

$¥dot{U}_{k}=A_{0}U_{k}-U_{k}A_{0}+¥sum_{m=1}^{k}(A_{m}U_{k-m}-U_{k-m}A_{m})$ ,

.

Obviously $U_{0}(t)$ is given by

$U_{0}(t)=X(t, 0)¥Gamma(0)X^{-1}(t, 0)$.

As we have assumed that the unperturbed system (19) can be solved (assumption$(¥mathrm{a}))$ , $X(t, 0)$ may be supposed to be known. As $¥Gamma(0)$ , we may take

$¥Gamma(0)=M(0)$

because the existence of $¥Gamma(¥epsilon)$ with $¥Gamma(0)=M(0)$ is assured (at least $M(¥epsilon)$ itself is such).Hence, by writing

$X(t, 0)=X_{0}(t)$

for simplicity, we have

$U_{0}(t)=X_{0}(t)M(0)X_{0}^{-1}(t)$ .

Suppose that $U_{0}$ , $U_{1}$,?, $U_{k-1}$ have been determined, then $U_{k}(t)$ is given by

$¥dot{U}_{k}=A_{0}U_{k}-U_{k}A_{0}+F_{k}(t)$ .

(25)$F_{k}(t)=¥sum_{m=1}^{k}(A_{m}(t)U_{k-m}(t)-U_{k-m}(t)A_{m}(t))$ .

To solve this differential equation, we put

$U_{k}=X_{0}(t)Z_{k}X_{0}^{-1}(t)$ .

Inserting this into (25), we get

$¥dot{Z}_{k}=X_{0}^{-1}(t)F_{k}(t)X_{0}(t)$ .

Consequently

$U_{k}(t)=X_{0}(t)[¥int_{0}^{t}X_{0}^{-1}(¥tau)F_{k}(¥tau)X_{0}(¥tau)d_{T}+C_{k}]X_{0}^{-1}(t)$ ,

where $C_{k}$ is a constant matrix. Since $U_{k}(t)$ should be so determined that

$U_{k}(t+T)=U_{k}(t)$ ,


we should have

$U_{k}(T)=X_{0}(T)[¥int_{0}^{T}X_{0}^{-1}(¥tau)F_{k}(¥tau)X_{0}(¥tau)d_{T}+C_{k}]X_{0}^{-1}(T)$

(26) $=U_{k}(0)$

$=X_{0}(0)C_{k}X_{0}^{-1}(0)$ .

As $X(t, ¥epsilon)$ has been so chosen that

$X(0, ¥epsilon)=E$

and

$X(T, ¥epsilon)=M(¥epsilon)$ ,

we have

$X_{0}(0)=X(0,0)=E$,

$X_{0}(T)=X(T, 0)=M(0)$,

So (26) will become

$M(0)[¥int_{0}^{T}X_{0}^{-1}(¥tau)F_{k}(¥tau)X_{0}(¥tau)d_{T}+C_{k}]M^{-1}(0)=C_{k}$,

which is equivalent to

(27) $ M(0)^{-1}C_{k}M(0)-C_{k}=¥int_{0}^{T}X_{0}^{-1}(¥tau)F_{k}(¥tau)X_{0}(¥tau)d¥tau$.

If $C_{k}$ could be determined uniquely from (27), then our perturbation method wouldbe complete. Unfortunately this is not the case. Indeed if a matrix $C_{k}$ satisfies$¥backslash ^{¥gamma}.27)$ , then $C_{k}+D$ will also satisfy (27) if

$M(0)D=DM(0)$.

Therefore there exist infinitely many ways of choosing $C_{k}$ , and an arbitrary choiceof $C_{k}$ will not assure the convergence of

$¥sum_{0}^{¥infty}U_{k}(t)¥epsilon^{k}$ .

Thus there still remains a problem: “What should be the correct choice of $C_{k}?$”

In our next paper, we shall solve this problem when $A_{0}(t)$ is a constant matrix.

§5. As a special case of our theory, we consider here a single equation of thesecond order:

Perturba’ion Method 35

$¥frac{d^{2}x}{dt^{2}}+p_{1}(t)¥frac{dx}{dt}+p_{2}(t)x=0$ ,

$p_{1}(t+T)=p_{1}(t)$ , $p_{2}(t+T)=p_{2}(t)$.

In this case, calculation of $X(t)$ from $U(t)$ becomes much simpler and also theperturbation method can be carried out without any difficulty.

As is well known, a transformation

$x=¥exp(-¥frac{1}{2}¥int p_{1}dt)y$

will reduce the equation to

$¥frac{d^{2}y}{dt^{2}}=(¥frac{1}{2}I_{1}+¥frac{1}{4}p_{1}^{2}-p_{2})y$.

Therefore we restrict ourselves to the equation

(28) $¥frac{d^{2}x}{dt^{2}}=p(t)x$, $p(t+T)=p(t)$,

which is usually known as Hill’s equation. Here we assume that $p(t)$ is continuouslydifferentiate. Putting

$A(t)=¥left(¥begin{array}{ll}0 & 1¥¥p(t) & 0¥end{array}¥right)$,

(28) can be written as

(29) $¥frac{d}{dt}¥left(¥begin{array}{l}x¥¥¥dot{X}¥end{array}¥right)=A(t)¥left(¥begin{array}{l}X¥¥¥dot{X}¥end{array}¥right)$ .

We assume that the linear system (29) is nondegenerate.Following our scheme developed in preceding sections, we look for a non-

degenerate periodic solution of a matrix differential equation

(30) $¥frac{dU}{dt}=AU-UA$ .

If we put

$U=¥left(¥begin{array}{ll}u_{11} & u_{12}¥¥u_{21} & u_{22}¥end{array}¥right)$ ,$uu_{12)}22$ ,

then (30) means that


$¥frac{du_{11}}{dt}=u_{21}-pu_{12}$ , $¥frac{du_{12}}{dt}=u_{22}-u_{11}$ ,

(31)$¥frac{du_{21}}{dt}=p(u_{11}-u_{22})$, $¥frac{du_{22}}{dt}=pu_{12}-u_{21}$ .

From (31), we immediately get

$u_{11}+u_{22}=c$ , $u_{11}=¥frac{1}{2}(c-¥dot{u}_{12})$ , $u_{21}=pu_{12}-¥frac{1}{2}¥ddot{u}_{12}$

where $c$ is an arbitrary constant. Also it is not difficult to see that $u_{12}$ satisfiesfollowing linear equation of the third order:

(32) $¥ddot{¥dot{u}}=4p¥dot{u}+2¥dot{p}u$ .

Therefore a periodic solution of (30) has a following form:

(33) $U=¥left¥{¥begin{array}{l}¥frac{c-¥dot{u}}{2}¥¥¥mathrm{l}^{pu-¥frac{¥ddot{u}}{2}}¥end{array}¥right.$ $¥frac{c+¥dot{u}}{2}u]¥rfloor|$

where $c$ is an arbitrary constant and $u=u(t)$ is any periodic solution of (32).If $U$ is degenerate, it must be of the form

$X(t)¥left(¥begin{array}{ll}U & 0¥¥0 & U¥end{array}¥right)X^{-1}(t)=¥left(¥begin{array}{ll}U & 0¥¥0 & U¥end{array}¥right)$ .

Hence the degeneracy of $U$ implies $u¥equiv 0$ . Therefore if $u(t)$ is any nontrivial periodicsolution of (32), then (33) is a nondegenerate periodic solution of (30).

Suppose that such $U(t)$ has been obtained and let $¥gamma_{1}$ and $¥gamma_{2}$ be its eigenvalues.Since they are the roots of the quadratic equation

$|_{pu-¥frac{¥ddot{u}}{2}}^{¥frac{c-¥dot{u}}{2}-¥gamma}$ $¥frac{c+¥dot{u}}{2}u-¥gamma|=r^{2}-c¥mathcal{T}+¥frac{c^{2}-¥dot{u}^{2}}{4}-pu^{2}+¥frac{u¥ddot{u}}{2}=0$,

we have

$r_{1}=¥frac{1}{2}(c+¥vee¥overline{¥ddot{u}^{2}-2u¥ddot{u}+4pu^{2}})$ , $r_{2}=¥frac{1}{2}(c-¥sqrt{¥dot{u}^{2}-2u¥ddot{u}+4pu^{2}})$ .

Expression in the square root looks like a function of $t$ at first sight. Howeverit is actually a constant because


$¥frac{d}{dt}(¥dot{u}^{2}-2u¥ddot{u}+4pu^{2})=2u(-¥ddot{¥dot{u}}+4p¥dot{u}+2¥dot{p}u)=0$.

Let us put

$¥vee¥overline{¥ddot{u}^{2}-2u¥ddot{u}+4pu^{2}}=¥beta$ ,

and write

(34) $r_{1}=¥frac{1}{2}(c+¥beta)$ , $¥gamma_{2}=¥frac{1}{2}(c-¥beta)$ .

To calculate the solutions of (28) from $U(t)$ , we have to consider following twocases separately.

The case when $¥beta¥neq 0$ . In this case, $r_{1}¥neq r_{2}$ and hence the Jordan’s normal formof $U$ is

$J=¥left(¥begin{array}{ll}¥gamma_{1} & 0¥¥0 & ¥gamma_{2}¥end{array}¥right)$

and by choosing a suitable fundamental matrix

$X(t)=¥left(¥begin{array}{ll}X_{1} & x_{2}¥¥¥dot{X}_{1} & ¥dot{X}_{2}¥end{array}¥right)$

of (2), $U$ can be written as

$U=XJX^{-1}$ .

From this we get

$U¥left(¥begin{array}{l}x_{1}¥¥¥dot{X}_{1}¥end{array}¥right)=r_{1}¥left(¥begin{array}{l}x_{1}¥¥¥dot{X}_{1}¥end{array}¥right)$ , $U¥left(¥begin{array}{l}x_{2}¥¥¥dot{X}_{2}¥end{array}¥right)=r_{2}¥left(¥begin{array}{l}X_{2}¥¥¥dot{X}_{2}¥end{array}¥right)$ ,

which shows that

$¥left(¥begin{array}{l}x_{1}¥¥¥dot{X}_{1}¥end{array}¥right)$ and $¥left(¥begin{array}{l}x_{2}¥¥¥dot{X}_{2}¥end{array}¥right)$

are eigenvectors of $U$ corresponding to eigenvalues $¥gamma_{1}$ and $¥gamma_{2}$ respectively.On the other hand, by (33) and (34), we can easily show that

$U¥left(¥begin{array}{l}1¥¥¥frac{¥dot{u}}{2u}+¥frac{¥beta}{2u}¥end{array}¥right)=r_{1}¥left(¥begin{array}{l}1¥¥¥frac{¥dot{u}}{2u}+¥frac{¥beta}{2u}¥end{array}¥right)$ , $U¥left(¥begin{array}{l}1¥¥¥frac{¥dot{u}}{2u}-¥frac{¥beta}{2u}¥end{array}¥right)=r_{2}¥left(¥begin{array}{l}1¥¥¥frac{¥dot{u}}{2u}-¥frac{¥beta}{2u}¥end{array}¥right)$ .

Since each eigenvalue is simple, this fact implies that


$¥left(¥begin{array}{l}X_{1}¥¥¥dot{X}_{1}¥end{array}¥right)=x_{1}¥left(¥begin{array}{l}1¥¥¥frac{¥dot{u}}{2u}+¥frac{¥beta}{2u}¥end{array}¥right)$ , $¥left(¥begin{array}{l}x_{2}¥¥¥dot{X}_{2}¥end{array}¥right)=x_{2}¥left(¥begin{array}{l}1¥¥¥frac{¥dot{u}}{2u}-¥frac{¥beta}{2u}¥end{array}¥right)$ .

Thus we are led to differential equations

$¥dot{x}_{1}=(¥frac{¥dot{u}}{2u}+¥frac{¥beta}{2u})X_{1}$ , $¥dot{x}_{2}=(_{¥frac{¥dot{u}}{2u}}-¥frac{¥beta}{2u})x_{2}$ ,

which can be easily solved and we obtain

(35) $x_{1}=u^{1/2}¥exp(¥int¥frac{¥beta}{2u}dt)$ , $x_{2}=u^{1/2}¥exp(-¥int_{¥_}^{¥beta}2udt)$ .

The case when $¥beta=0$ . In this case $r_{1}=r_{2}=c/2$ and the Jordan’s normal form of$U$ is

$J=¥left(¥begin{array}{ll}c/2 & 1¥¥0 & c/2¥end{array}¥right)$

if we recall that $U$ is nondegenerate. If we choose a fundamental matrix

$X=¥left(¥begin{array}{ll}x_{1} & x_{2}¥¥¥dot{X}_{1} & ¥dot{X}_{2}¥end{array}¥right)$

so that we have

$U=XJX^{-1}$,

then we get

$U¥left(¥begin{array}{l}X_{1}¥¥¥dot{X}_{1}¥end{array}¥right)=¥frac{c}{2}¥left(¥begin{array}{l}X_{1}¥¥¥dot{X}_{1}¥end{array}¥right)$ .

On the other hand, we can also show that

$U¥left(¥begin{array}{l}1¥¥¥dot{u}/2u¥end{array}¥right)=¥frac{c}{2}¥left(¥begin{array}{l}1¥¥¥dot{u}/2u¥end{array}¥right)$ .

Hence we have that

$¥left(¥begin{array}{l}X_{1}¥¥¥dot{X}_{1}¥end{array}¥right)=x_{1}¥left(¥begin{array}{l}1¥¥¥dot{u}/2u¥end{array}¥right)$ ,

or, what is the same thing, that

$¥dot{u}$

$¥dot{x}_{1}=_{¥overline{2u}}x_{1}$ ,


and we obtain

$x_{1}=u^{1/2}$ .

Since the $¥mathrm{t}¥mathrm{r}¥mathrm{a}¥mathrm{c}¥mathrm{e}$ of the matrix $A(t)$ in (29) is equal to zero, $|X(t)|$ must be a constant.So, without loss of generality, we may put

$|X(t)|=x_{1}¥dot{x}_{2}-¥dot{x}_{1}x_{2}=1$ .

Regarding this as a differential equation satisfied by $x_{2}$ , we immediately get

$x_{2}=x_{1}¥int¥frac{dt}{X_{1}^{2}}=u^{1/2}¥int¥frac{dt}{u}$ .

Thus, in this case, we have

(36) $x_{1}=u^{1/2}$ , $x_{2}=u^{1/2}¥int¥frac{dt}{u}$ .

§6. Now let us apply our perturbation method to the equation

$¥frac{d^{2}x}{dt^{2}}=p(t, ¥epsilon)x$,(37)

$p(t, ¥epsilon)=p_{0}(t)+¥epsilon p_{1}(t)+¥epsilon^{2}p_{2}(t)+-$ , $p_{k}(t+T)=p_{k}(t)$.

As a startpoint of the perturbation, we have to assume that the unperturbedequation

$¥frac{d^{2}x}{dt^{2}}=p_{0}(t)x$,

or equivalently, the system

(38) $¥frac{d}{dt}¥left(¥begin{array}{l}X¥¥¥dot{X}¥end{array}¥right)=¥left(¥begin{array}{ll}0 & 1¥¥p_{0}(t) & 0¥end{array}¥right)¥left(¥begin{array}{l}x¥¥¥dot{X}¥end{array}¥right)$

is solvable and nondegenerate. Let $X_{0}(t)$ be its fundamental matrix and $M_{0}$ be thecorresponding monodromy matrix.

According to the general theory developed in§4, a matrix differential equation

$¥frac{dU}{dt}=A(t, ¥epsilon)U-UA(t, ¥epsilon)$ ,

$A(t, ¥epsilon)=¥left(¥begin{array}{ll}0 & 1¥¥p(t,¥epsilon) & 0¥end{array}¥right)$

has a nondegenerate periodic solution $U(t, ¥epsilon)$ which can be expressed as a convergentpower series in $¥epsilon$ in the following way:


$U(t, ¥epsilon)=U_{0}(t)+¥epsilon U_{1}(t)+¥epsilon^{2}U_{2}(t)+$ ?, $U_{k}(t+T)=U_{k}(t)$,(39)

$U_{0}(t)=X_{0}(t)M_{0}X_{0}^{-1}(t)$ .

If we recall the discussions given in §5, such $U(t, ¥epsilon)$ is of the form:

$¥mathrm{r}¥frac{c-¥dot{u}(t,¥epsilon)}{2}$ $u(t, ¥epsilon)$ $1$

$[p(t, ¥epsilon)-¥frac{1}{2}$ u(t, $¥epsilon$) $¥frac{c+¥dot{u}(t,¥epsilon)}{2}¥rfloor$

and $u(t, ¥epsilon)$ is a nontrivial periodic solution of

(40) $¥ddot{¥dot{u}}=4p(t, ¥epsilon)¥dot{u}+2¥dot{p}(t, ¥epsilon)u$.

Since $U(t, ¥epsilon)$ is holomorphic in $¥epsilon$ at $¥epsilon=0$ , $u(t, ¥epsilon)$ should also be holomorphic in $¥epsilon$

and hence it can be expressed as

$u(t, ¥epsilon)=u_{0}(t)+¥epsilon u_{1}(t)+¥epsilon^{2}u_{2}(t)+$ ?,

(41)$u_{k}(t+T)=u_{k}(t)$ .

Here, by the second formula of (39), $u_{0}(t)$ is an element of

$X_{0}(t)M_{0}X_{0}^{-1}(t)$

situated in the first row and in the second column. Using such $u(t, ¥epsilon)$ , the solutionof (37) is expressed either by (35) or by (36) of§5 according as $¥beta^{2}=u^{2}-2u¥ddot{u}+4pu^{2}$

$¥neq 0$ or not.Now let us put

$¥omega(¥epsilon)=¥frac{1}{T}¥int_{0}^{T}u(¥tau, ¥epsilon)d¥tau$, $¥omega_{k}=¥frac{1}{T}¥int_{0}^{T}u_{h}(¥tau)d¥tau$ .

Then $¥omega(¥epsilon)$ is holomorphic in $¥epsilon$ and can be expressed as

$¥omega(¥epsilon)=¥omega_{0}+¥epsilon¥omega_{1}+¥epsilon^{2}¥omega_{2}+-$ .

If $¥omega_{0}¥neq 0$ , then

$v(t, ¥epsilon)=¥frac{u(t,¥epsilon)}{¥omega(¥epsilon)}$

is holomorphic in $¥epsilon$ and can be expressed as

$v(t, ¥epsilon)=v_{0}(t)+¥epsilon v_{1}(t)+¥epsilon^{2}v_{2}(t)+-$ .

Since


$¥frac{1}{T}¥int_{0}^{T}v(¥tau, ¥epsilon)d¥tau=1$ ,

we have

$¥frac{1}{T}¥int_{0}^{T}v_{0}(¥tau)d¥tau=1$ , $¥int_{0}^{T}v_{k}(¥tau)d¥tau=0$ , $k=1,2$,?.

It is obvious that $v(t, ¥epsilon)$ is also a solution of (40) and we can use $v(t, ¥epsilon)$ in place of$u(t, ¥epsilon)$ in the expression (35) and (36). Therefore we may assume from the outsetthat

(42) $¥frac{1}{T}¥int_{0}^{T}u_{0}(t)dt=1$ , $¥int_{0}^{T}u_{k}(t)dt=0$ , $k=1,2$, $¥cdot$ .

in solving the equation (40) by perturbation method.Substituting (41) into (40), we get the following system of differential equations;

$¥ddot{¥dot{u}}_{k}-4p_{0}¥dot{u}_{k}-2¥dot{p}_{0}u_{k}=4¥sum_{m=0}^{k-1}p_{k-m}u_{m}+2¥sum_{m=0}^{k-1}¥dot{p}_{k-m}u_{m}$

(43)$k=1,2$,? $¥cdot$

(43) can be solved successively if we know the basis of solutions of a homo-geneous equation

(44) $¥ddot{¥dot{u}}-4p_{0}¥dot{u}-2¥dot{p}_{0}u=0$ .

Let $x_{1}(t)$ and $x_{2}(t)$ form a basis of solutions of the unperturbed equation

$¥ddot{x}=p_{0}(t)x$

which we have assumed to be solvable. Then, as direct calculation shows,

$¥{x_{1}(t)¥}^{2}$ , $x_{1}(t)x_{2}(t)$ , $¥{x_{2}(t)¥}^{2}$

form the basis of the solutions of (44). Hence we can solve (43) and determine $u_{lt}$

successively. If all the $u_{k}$ can be determined uniquely by conditions

(45) $u_{k}(t+T)=u_{k}(t)$ , $¥int_{0}^{T}u_{k}(t)dt=0$ , $k=1,2$,?,

then our perturbation procedure will be completed.

§7. Hereafter we consider the case when $p_{0}(t)$ is a constant and shall showthat, in this case, $u_{k}$ can be determined uniquely by (45) and hence our perturbationmethod is carried out successfully.

Let us put


$p_{0}(t)=¥lambda^{2}$ .

Then we have to treat following two cases separately.

Case I. $¥lambda¥neq 0$ . In this case, the unperturbed equation

$¥frac{d^{2}x}{dt^{2}}=¥lambda^{2}x$

has two linearly independent solutions

$x_{1}=e^{¥lambda t}$ , $x_{2}=e^{-¥lambda t}$ .

So the fundamental matrix $X_{0}(t)$ is

$¥left(¥begin{array}{ll}x_{1} & x_{2}¥¥¥dot{X}_{1} & ¥dot{X}_{2}¥end{array}¥right)=¥left(¥begin{array}{ll}e^{¥lambda t} & e^{-¥lambda t}¥¥¥lambda e^{¥lambda t} & -¥lambda e^{-¥lambda t}¥end{array}¥right)$,

and the corresponding monodromy matrix is

$M_{0}=¥left(¥begin{array}{ll}e^{lT} & 0¥¥0 & e^{-¥lambda T}¥end{array}¥right)$.

For $M_{0}$ to be nondegenerate, we have to assume that

(46) $¥lambda T¥neq n¥pi i$

where $n$ is an arbitrary integer.Then

$U_{0}(t)=X_{0}(t)M_{0}X_{0}^{-1}(t)=¥frac{1}{2}¥left(¥begin{array}{ll}e^{¥lambda T}+e^{-¥lambda T} & (1/¥lambda)(e^{¥lambda T}-e^{-¥lambda T})¥¥¥lambda(e^{¥lambda T}-e^{-¥lambda T}) & e^{¥lambda T}+e^{-¥lambda T}¥end{array}¥right)$

which shows that $u_{0}(t)$ is a constant. Since we choose $u_{0}$ so that the first formulaof (42):

$¥frac{1}{T}¥int_{0}^{T}u_{0}d¥tau=1$ ,

is satisfied, we have

$u_{0}=1$ .

The equation (43) will become

$¥ddot{¥dot{u}}_{k}-4¥lambda^{2}¥dot{u}_{k}=f_{k}(t)$ , $f_{k}(t)=4¥sum_{m=0}^{k-1}p_{k-m}(t)¥dot{u}_{m}(t)+2¥sum_{m=0}^{h-1}¥dot{p}_{k-m}(t)u_{m}(t)$ ,(47)

$k=1,2$,?.


As a basis of solutions of

$¥ddot{¥dot{u}}-4¥lambda^{2}¥dot{u}=0$,

we choose

$v_{1}=e^{2¥lambda t}$, $v_{2}=1$ , $v_{3}=e^{-2¥lambda t}$ .

Then since

$|_{¥ddot{v}_{1}}^{U}¥dot{v}_{1}1$$¥ddot{v}_{2}¥dot{v}_{2}v_{2}$

$¥dot{v}_{3}¥ddot{v}_{3}v_{3}|=-16¥lambda^{3}$ ,

the general solution of (47) is given by

$u_{k}(t)=-¥frac{1}{16¥lambda^{3}}[v_{1}(t)¥int(v_{2}(t)¥dot{v}_{3}(t)-¥dot{v}_{2}(t)v_{3}(t))f_{k}(t)dt$

$+v_{2}(t)¥int(v_{3}(t)¥dot{v}_{1}(t)-¥dot{v}_{3}(t)v_{1}(t))f_{k}(t)dt$

(48)$+v_{3}(t)¥int(v_{1}(t)¥dot{v}_{2}(t)-¥dot{v}_{1}(t)v_{2}(t))f_{k}(t)dt]$

$=-¥frac{1}{16¥lambda^{3}}[e^{2¥lambda t}¥int(-2¥lambda e^{-2¥lambda t})f_{k}(t)dt+¥int 4¥lambda f_{k}(t)dt+e^{-2¥lambda t}¥int(-2¥lambda e^{2¥lambda t})f_{k}(t)dt]$ .

Suppose that $u_{m}(t)$ $(m=0,1, --, k-l)$ have been determined uniquely so as tosatisfy

$u_{m}(t+T)=u_{m}(t)$ , $¥int_{0}^{T}u_{m}(t)dt=0$, $m=1,2$,?, $k-l$ .

Then $f_{k}(t)$ is a periodic function of $t$ with period $T$. So if we assume that $p_{j}(t)$ areall continuously differentiable, $f_{k}(t)$ can be expressed by a uniformly convergentFourier series such as

$f_{k}(t)=¥sum_{m=-¥infty}^{¥infty}c_{k}(m)¥exp(¥frac{2¥pi imt}{T})$ .

Substituting it into (48), we obtain

$u_{k}(t)=¥frac{1}{8¥lambda^{2}}[¥sum¥frac{Tc_{k}(m)}{2¥pi im-2¥lambda T}¥exp(¥frac{2¥pi imt}{T})+c_{1}e^{2¥lambda t}-2c_{k}(0)t$

? $¥sum_{m¥neq 0}¥frac{Tc_{k}(m)}{¥pi im}¥exp(¥frac{2¥pi imt}{T})+c_{2}+¥sum¥frac{Tc_{k}(m)}{2¥pi im+2¥lambda T}¥exp(¥frac{2¥pi imt}{T})+c_{3}e^{-2¥lambda t}]$ .


$¥mathrm{S}¥mathrm{i}¥mathrm{n}¥dot{¥mathrm{c}}¥mathrm{e}$ the existence of a periodic solution of (43) is already assured, the secular term$-2c_{k}(0)t$ will never appear if $u_{0}$ , $u_{1}$,?, $u_{k-1}$ have been determined correctly. Sowe may suppose that $c_{k}(0)=0$ is satisfied automatically. Denominators of Fouriercoefficients will never vanish because of the nondegeneracy condition (46). Perio-dicity of $u_{k}(t)$ will imply

$c_{1}=c_{3}=0$

also because of (46). The second condition of (42):

$¥int_{0}^{T}u_{k}(t)dt=0$

will require

$c_{2}=0$ .

Consequently $u_{k}(t)$ is uniquely determined in a form

(49) $u_{k}(t)=-¥frac{T^{3}}{8}¥sum_{m¥neq 0}¥frac{c_{k}(m)}{¥pi im(¥pi^{2}m^{2}+¥lambda^{2}T^{2})}¥exp(¥frac{2¥pi imt}{T})$ .

Case $¥mathrm{I}¥mathrm{I}$ . $¥lambda=0$ . The unperturbed equation

$¥frac{d^{2}x}{dt^{2}}=0$

has two linearly independent solutions

$x_{1}=1$ , $x_{2}=t$ .

So the fundamental matrix $X_{0}(t)$ is

$¥left(¥begin{array}{ll}x_{1} & x_{2}¥¥¥dot{X}_{1} & ¥dot{X}_{2}¥end{array}¥right)=¥left(¥begin{array}{ll}1 & t¥¥0 & 1¥end{array}¥right)$ ,

and the corresponding monodromy matrix is

$M_{0}=¥left(¥begin{array}{ll}1 & T¥¥0 & 1¥end{array}¥right)$ ,

which is clearly nondegenerate. Then

$U_{0}(t)=X_{0}(t)M_{0}X_{0}^{-1}(t)=¥left(¥begin{array}{ll}1 & T¥¥0 & 1¥end{array}¥right)$

which shows that $u_{0}(t)$ is a constant. Hence, from the condition

Perturbation $f¥nu_{A}^{¥prime}e^{t}$hod 45

$¥frac{1}{T}¥int_{0}^{T}u_{0}dt=1$ ,

we should put

$u_{0}=1$ .

The equation (43) will become

$¥ddot{¥dot{u}}_{k}=f_{k}(t)$ , $f_{k}(t)=4¥sum_{m=0}^{k-1}p_{k-m}(t)¥dot{u}_{m}(t)+2¥sum_{m=0}^{k-1}¥dot{p}_{k-m}(t)u_{m}(t)$ ,

$k=1,2$, $¥cdots$ .

As a basis of solutions of

$¥ddot{¥dot{u}}=0$ ,

we choose

$v_{1}=1$ , $v_{2}=t$, $v_{3}=t^{2}$ .

Calculation similar to the case I will give

(50) $u_{k}(t)=-¥frac{T^{3}}{8¥pi^{3}i}¥sum_{m¥neq 0}¥frac{c_{k}(m)}{m^{3}}¥exp(¥frac{2¥pi imt}{T})$ ,

where

$f_{k}(t)=¥sum c_{k}(m)¥exp(¥frac{2¥pi imt}{T})$ .

§8. Let us apply our result to Mathieu’s equation

$¥frac{d^{2}x}{dt^{2}}=(¥lambda^{2}+2¥epsilon¥cos t)x$ .

In this case

$p_{0}=¥lambda^{2}$ , $p_{1}=2¥cos t=e^{it}+e^{-it}$ , $p_{2}=p_{3}=-=0$.

The case when $¥lambda¥neq 0$ . We first assume that

ni$¥lambda¥neq_{¥overline{2}}$ , $n=0$, $¥pm 1$ , $¥pm 2$ , $¥cdots$ ,

to guarantee the nondegeneracy of unperturbed equation. Our perturbation methodgives us


$u_{0}=1$ ,

$u_{1}=-¥frac{2}{1+4¥lambda^{2}}(e^{it}+e^{-it})$.

$u_{2}=¥frac{3}{¥mathfrak{x}^{2(1+¥lambda^{2})(1+4¥lambda^{2})}¥mathrm{E}}(e^{2it}+e^{-2it})$ ,

.

Next we have to calculate

$¥beta=¥sqrt{¥dot{u}^{2}-2u¥ddot{u}+4pu^{2}}$.

Since $¥beta$ is a constant, we have only to calculate the value of

$¥dot{u}^{2}-2u¥ddot{u}+4pu^{2}$

for some particular value of $t$ . So we put $t=¥pi/2$ and get

$¥beta^{2}=4¥lambda^{2}(1-¥frac{2(1+12¥lambda^{2})}{¥lambda^{2}(1+4¥lambda^{2})^{2}}¥epsilon^{2}+¥cdots)$.

Hence

$¥beta=2¥lambda(1-¥frac{1+12¥lambda^{2}}{¥lambda^{2}(1+4¥lambda^{2})^{2}}¥epsilon^{2}+-)$.

Then a straightforward calculation will give

$¥frac{¥beta}{2u}=¥lambda[1+¥frac{4¥epsilon}{1+4¥lambda^{2}}¥cos t+¥epsilon^{2}(-¥frac{1}{¥lambda^{2}(1+4¥lambda^{2})}+¥frac{5-4¥lambda^{2}}{(1+¥lambda^{2})(1+4¥lambda^{2})^{2}}¥cos 2t)+-]$ ,

$u^{1/2}=1-¥frac{2¥epsilon}{1+4¥lambda^{2}}¥cos t+¥epsilon^{2}(-¥frac{1}{(1+4¥lambda^{2})^{2}}+¥frac{1+10¥lambda^{2}}{2(1+¥lambda^{2})(1+4¥lambda^{2})^{2}}¥cos 2t)+-$.

Inserting these expressions into (35), we finally obtain

$x_{1}(t)=¥phi(t)¥exp(¥lambda-¥frac{¥epsilon^{2}}{¥lambda(1+4¥lambda^{2})}+-)t$ , $x_{2}(t)=x_{1}(-t)$,

$¥phi(t)=1-¥epsilon(_{¥frac{2}{1+4¥lambda^{2}}¥cos t-}¥frac{4¥lambda}{1+4¥lambda^{2}}¥sin t)$

$+¥epsilon^{2}(_{¥frac{8¥lambda^{2}+1}{(1+4¥lambda^{2})^{2}}-¥frac{16¥lambda^{4}+6¥lambda^{2}-1}{2(1+¥lambda^{2})(1+4¥lambda^{2})^{2}}¥cos 2t-}¥frac{12¥lambda^{3}+3¥lambda}{2(1+¥lambda^{2})(1+4¥lambda^{2})^{2}}¥sin 2t)+¥cdots$ .

The case when $¥lambda=0$ . Following our perturbation method, we have

$u_{0}=1$ ,

$u_{1}=-2(e^{it}+e^{-it})$ ,

Perturbation Method

$u_{2}=¥frac{3}{2}(e^{2it}+e^{-2it})$,

$u_{3}=-¥frac{5}{9}(e^{3it}+e^{-3it})-9(e^{it}+e^{-it})$ ,

$u_{4}=¥frac{35}{288}(e^{4it}+e^{-4it})+¥frac{67}{9}(e^{2it}+e^{-2it})$

$¥beta=2¥sqrt{2}i¥epsilon(1+¥frac{5}{6}¥epsilon^{2}+-)$ .

Hence

$¥frac{¥beta}{2u}=¥sqrt{2}i¥epsilon+4¥sqrt{2}i¥epsilon^{2}¥cos t+¥epsilon^{3}$($¥frac{53}{6}¥sqrt{2}i+5¥sqrt{2}$icos $ 2t)+¥cdots$ ,

$u^{1/2}=1-2¥epsilon¥cos t+¥epsilon^{2}(-1+¥frac{1}{2}¥cos 2t)-¥epsilon^{3}(¥frac{1}{8}¥cos 3t+¥frac{21}{2}¥mathrm{c}¥mathrm{o}¥mathrm{s}¥mathrm{t})+-$ .

Inserting there expressions into (35), we obtain

$x_{1}(t)=¥phi(t)¥exp¥sqrt{2}i(¥epsilon+¥frac{53}{6}¥epsilon^{3}+¥cdots)t$, $x_{2}(t)=x_{1}(-t)$ ,

$¥phi(t)=1-2¥epsilon¥cos t+¥epsilon^{2}(-1+¥frac{1}{2}¥cos 2t+4¥sqrt{2}$ isin $t)$

$+¥epsilon^{3}(-¥frac{21}{2}¥cos t-¥frac{1}{8}¥cos 3t-¥frac{9¥sqrt{2}}{4}i¥sin 2t)+-$ .

If we want to get the solution in real form, we have only to take

$¥frac{1}{2}(x_{1}(t)+x_{2}(t))$ , $¥frac{1}{2i}(x_{1}(t)-x_{2}(t))$

as a basis. This will give

$¥cos¥sqrt{2}$ ($¥epsilon+¥frac{53}{6}¥epsilon^{3}+$ ?) $t¥cdot[1-2¥epsilon¥cos t+¥epsilon^{2}(-1+¥cos 2t)$

$+¥epsilon^{3}(-¥frac{21}{2}¥mathrm{c}¥mathrm{o}vt-¥frac{1}{8}¥cos 3t)+¥cdots]+¥sin¥sqrt{2}(¥epsilon+¥frac{53}{6}¥epsilon^{3}+$? $)t$

$-4¥sqrt{2}¥epsilon^{2}¥sin t+¥frac{9}{4}¥sqrt{2}¥epsilon^{3}¥sin 2t+-$

and


$¥sin¥sqrt{2}(¥epsilon+¥frac{53}{6}¥epsilon^{3}+-)t[1-2¥epsilon¥cos t+¥epsilon^{2}(-1+¥cos 2t)$

$+¥epsilon^{3}(-¥frac{21}{2}¥cos t-¥frac{1}{8}¥cos 3t)+-]-¥cos¥sqrt{2}(¥epsilon+¥frac{53}{6}¥epsilon^{3}+-)t$

$-4¥sqrt{2}¥epsilon^{2}¥sin t+¥frac{9¥sqrt{2}}{2}¥epsilon^{3}¥sin 2t+-$ .

Remark. The underlying idea of this paper was already suggested by one ofthe authors and Y. Sibuya concerning the study of an irregular singular point ofthe second order linear differential equation in complex domain [1], [2].

Acknowledgment. The authors express their gratitude to the referee for hishelpful suggestion for the improvement of the paper.

References

[1] Saito, T., On a singular point of a second order linear differential equation containinga parameter, Funkcial. Ekvac., 5 (1963), 1-29.

[2] Sibuya, Y., Perturbation at an irregular singular point, Japan-U.S. seminar on ordinaryand functional equations. Lecture Notes in Math., Springer, 243 (1971), 148-168.

nuna adreso:Department of MathematicsFaculty of Science and TechnologyKeio UniversityHiyoshi, Kohoku-ku, Yokohama 223Japan

(Ricevita la 1-an de februaro, 1983)(Reviziita la 9-an de septembro, 1983)

Documents

(1984), Periodic - MATHfe.math.kobe-u.ac.jp/FE/FE_pdf_with_bookmark/FE21... · Funkcialaj Ekvacioj, 27 (1984), 25-48 Perturbation Method for Linear Periodic Systems I By SAITO Tosiya