ISSN 0012-2661, Differential Equations, 2012, Vol. 48, No. 6, pp. 809–819. c© Pleiades Publishing, Ltd., 2012.Original Russian Text c© A.A. Pokutnyi, 2012, published in Differentsial’nye Uravneniya, 2012, Vol. 48, No. 6, pp. 803–813.

PARTIAL DIFFERENTIAL EQUATIONS

Bounded Solutions of Linear and Weakly NonlinearDifferential Equations in a Banach Space

with an Unbounded Operator in the Linear Part

A. A. PokutnyiInstitute for Mathematics, National Academy of Sciences, Kiev, Ukraine

Received September 10, 2010

Abstract—For linear and weakly linear differential equations in a Banach space, we obtainnecessary and sufficient conditions for the existence of bounded solutions on the entire real lineunder the assumption that the homogeneous equation dx/dt = A(t)x(t) admits an exponentialdichotomy on the half-lines.

DOI: 10.1134/S0012266112060055

Numerous papers deal with problems of the existence of bounded solutions of linear and nonlineardifferential equations in Banach spaces. Of them, we note the papers [1, 2], where such problemswere considered in finite-dimensional spaces, and the monographs [3–7]. The papers [8, 9] (see thebibliography therein) and [10] deal with the Fredholm property of the operator corresponding toa differential equation. The normal solvability of a differential operator was considered in [11].

STATEMENT OF THE PROBLEM AND BASIC NOTIONS

Consider the homogeneous differential equation

dx

dt= A(t)x, t ∈ J, (1)

in a Banach space B, where, for each t ∈ J ⊂ R, the operator A(t) is closed with dense domainD(A(t)) = D ⊂ B independent of t. Throughout the following, by L(B) we denote the space of alllinear bounded operators mapping the space B into itself.

If the Cauchy problem generated by Eq. (1) and the initial condition x(s, s, x0) = x0 ∈ D isuniformly well posed [12, p. 237], then one can define a linear operator T (t, s) : D → B on J fort ≥ s by the rule

T (t, s)x0 = x(t, s, x0).

The family of operators thus defined is a strongly continuous operator [9, 12] by virtue of[12, pp. 237–239]. In this case, one says that T (t, s) is the family of evolution operators asso-ciated with Eq. (1).

The operator T (t, s) is a bounded linear operator for fixed t ≥ s, and since the set D is densein B, we find that it can be extended to the entire space B by continuity, which is assumed inforthcoming considerations. The extension of the family of evolution operators to the entire spaceis denoted in the same way.

Throughout the following, we use the notion of exponential dichotomy in the sense of [13, p. 245of the Russian translation]. It is of special interest to analyze the exponential dichotomy on thehalf-lines R

−s = (−∞, s] and R

+s = [s,∞). [In this case, the projection-valued functions defined on

half-lines will be denoted by P−(t) for all t ≥ s and P+(t) for all t ≤ s with constants M1, α1 andM2, α2, respectively.]

809

810 POKUTNYI

The present paper dealt with the derivation of necessary and sufficient conditions for the exis-tence of weak [9] bounded solutions of the inhomogeneous equation

dx

dt= A(t)x(t) + f(t), t ∈ J, (2)

with f ∈ BC(J,B) = {f : J → B; the function f is continuous and bounded}. Here theboundedness is treated in the sense that |||f ||| = supt∈J ‖f(t)‖ < ∞.

Throughout the following, the generalized inverse operator [14, p. 138] of an operator L isdenoted by L−.

Auxiliary Result (Linear Case)

The main result of this section can be stated as follows.

Theorem 1. Let {T (t, s)| t ≥ s ∈ R} be the family of strongly continuous evolution operatorsassociated with Eq. (1). Suppose that the following conditions are satisfied.

1. The operator T (t, s) admits exponential dichotomy on the half-lines R+0 and R

−0 with projec-

tion-valued operator-functions P+(t) and P−(t), respectively.2. The operator D = P+(0) − (I − P−(0)) is generalized-invertible.Then the following assertions hold.1. There exist weak solutions of Eq. (2) bounded on the entire line if and only if the vector

function f ∈ BC(R,B) satisfies the condition

+∞∫

−∞

H(t)f(t) dt = 0, (3)

where H(t) = PN(D∗)P−(0)T (0, t).2. Under condition (3), the weak solutions of Eq. (2) bounded on the entire line have the form

x0(t, c) = T (t, 0)P+(0)PN(D)c + (G[f ])(t, 0) ∀c ∈ B, (4)

where

(G[f ])(t, s) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

t∫s

T (t, τ)P+(τ)f(τ) dτ −+∞∫t

T (t, τ)(I − P+(τ))f(τ) dτ

+ T (t, s)P+(s)D−[ ∞∫

s

T (s, τ)(I − P+(τ))f(τ) dτ

+s∫

−∞T (s, τ)P−(τ)f(τ) dτ

]for t ≥ s

t∫−∞

T (t, τ)P−(τ)f(τ) dτ −s∫t

T (t, τ)(I − P−(τ))f(τ) dτ

+ T (t, s)(I − P−(s))D−[ ∞∫

s

T (s, τ)(I − P+(τ))f(τ) dτ

+s∫

−∞T (s, τ)P−(τ)f(τ) dτ

]for s ≥ t

is the generalized Green operator of the problem on the bounded , on the entire line, solutions

(G[f ])(0+, 0) − (G[f ])(0−, 0) = −+∞∫

−∞

H(t)f(t) dt; L(G[f ])(t, 0) = f(t), t ∈ R,

DIFFERENTIAL EQUATIONS Vol. 48 No. 6 2012

BOUNDED SOLUTIONS OF LINEAR AND WEAKLY NONLINEAR . . . 811

and(Lx)(t) =

dx

dt− A(t)x(t);

D− is the generalized inverse to the operator D; PN(D) = I −D−D and PN(D∗) = I −DD− are theprojections [14] onto the kernel and cokernel of the operator D.

Remark 1. A similar theorem holds for the case in which the family of evolution operatorsT (t, s) admits exponential dichotomy on the half-lines R

+s and R

−s . In the present paper, we consider

the special case in which s = 0.Proof of Theorem 1. The solutions of Eq. (2) bounded on the half-lines R

+0 and R

−0 have the

form

x(t, ξ1, ξ2)

=

⎧⎪⎪⎨⎪⎪⎩

T (t, 0)P+(0)ξ1 −+∞∫t

T (t, τ)(I − P+(τ))f(τ) dτ +t∫0

T (t, τ)P+(τ)f(τ) dτ for t ≥ 0

T (t, 0)(I − P−(0))ξ2 +t∫

−∞T (t, τ)P−(τ)f(τ) dτ −

0∫t

T (t, τ)(I − P−(τ))f(τ) dτ for t ≤ 0.(5)

Indeed, we have

‖T (t, 0)P+(0)ξ1‖ ≤ ‖T (t, 0)P+(0)‖‖ξ1‖ ≤ M1e−α1t‖ξ1‖

and∂(T (t, 0)P+(0)ξ1)

∂t= A(t)T (t, 0)P+(0)ξ1.

Therefore, the expression T (t, 0)P+(0)ξ1 provides all bounded solutions of the homogeneousequation (1) on R

+0 .

Now let us prove the boundedness of a certain integral occurring in the representation (5) :∥∥∥∥∥

+∞∫

t

T (t, τ)(I − P+(τ))f(τ) dτ

∥∥∥∥∥ ≤+∞∫

t

‖T (t, τ)(I − P+(τ))‖‖f(τ) ‖dτ

≤ supt∈R

‖f(t)‖+∞∫

t

M1eα1(t−τ) dτ =

M1

α1

|||f ||| < ∞.

The boundedness of the remaining integrals can be verified in a similar way. By using the repre-sentation

∂

∂t

(−

+∞∫

t

T (t, τ)(I − P+(τ))f(τ) dτ +

t∫

0

T (t, τ)P+(τ)f(τ) dτ

)

= T (t, t)(I − P+(t))f(t) − A(t)

+∞∫

t

T (t, τ)(I − P+(τ))f(τ) dτ

+ T (t, t)P+(t)f(t) + A(t)

t∫

0

T (t, τ)P+(τ)f(τ) dτ

= f(t) + A(t)

{−

+∞∫

t

T (t, τ)(I − P+(τ))f(τ) dτ +

t∫

0

T (t, τ)P+(τ)f(τ) dτ

},

one can show that relation (5) indeed defines all bounded solutions of Eq. (2) on the half-lines.

DIFFERENTIAL EQUATIONS Vol. 48 No. 6 2012

812 POKUTNYI

Relation (5) defines weak bounded solutions on the entire line if and only if

x(0+, ξ1, ξ2) = x(0−, ξ1, ξ2),

which is equivalent to the solvability of the operator equation

P+(0)ξ1 −+∞∫

0

T (0, τ)(I − P+(τ))f(τ) dτ = (I − P−(0))ξ2 +

0∫

−∞

T (0, τ)P−(τ)f(τ) dτ. (6)

If ξ1 and ξ2 are solutions of Eq. (6), then, by substituting them into relation (5), we obtain a weakbounded solution of Eq. (2) on the entire line. Let us show that if the assumptions of the theoremhold, then the set of weak bounded solutions of Eq. (2) on the entire line can be sought in the form

x(t, ξ) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

T (t, 0)P+(0)ξ −+∞∫t

T (t, τ)(I − P+(τ))f(τ)dτ

+t∫0

T (t, τ)P+(τ)f(τ)dτ for t ≥ 0

T (t, 0)(I − P−(0))ξ +t∫

−∞T (t, τ)P−(τ)f(τ)dτ

−0∫t

T (t, τ)(I − P−(τ))f(τ)dτ for t ≤ 0;

(7)

i.e., the cardinalities of the sets of weak bounded solutions (5) and (7) coincide, and the elementsξ1 and ξ2 can be chosen to be equal, ξ = ξ1 = ξ2. Obviously, each bounded solution of the form (7)lies in the set of weak bounded solutions (5) under the solvability condition.

We rewrite Eq. (6) in the form

P+(0)ξ1 = (I − P−(0))ξ2 + g, (8)

g =

0∫

−∞

T (0, τ)P−(τ)f(τ) dτ +

+∞∫

0

T (0, τ)(I − P+(τ))f(τ) dτ.

Since P 2+(0) = P+(0), we have P 2

+(0)ξ1 = P+(0)ξ1 for each element ξ1. By using this relationin (8), we obtain

P+(0)((I − P−(0))ξ2 + g) = (I − P−(0))ξ2 + g,

orP+(0)(I − P−(0))ξ2 − (I − P−(0))ξ2 = g − P+(0)g.

Now, by using the relation (I − P−(0))2 = I − P−(0), we obtain the equation

P+(0)(I − P−(0))ξ2 − (I − P−(0))2ξ2 = g − P+(0)g,

which is equivalent to the equation

D(I − P−(0))ξ2 = (I − P+(0))g. (9)

By the assumptions of the theorem, the operator D is normally solvable (and even generalized-invertible); therefore [15, p. 227], a necessary and sufficient condition for the solvability of Eq. (9)is the following: PN(D∗)(I − P−(0))g = 0. (By [14, p. 139], PN(D∗) can be found from the relationPN(D∗) = I −DD−.) Since PN(D∗)D = 0 [which follows, for example, from the relation PN(D∗)D =

DIFFERENTIAL EQUATIONS Vol. 48 No. 6 2012

BOUNDED SOLUTIONS OF LINEAR AND WEAKLY NONLINEAR . . . 813

(I −DD−)D = D−DD−D = D−D = 0], or, which is the same, PN(D∗)(P+(0)− (I −P−(0))) = 0,we have PN(D∗)(I − P+(0)) = PN(D∗)P−(0). Hence we find PN(D∗)g :

PN(D∗)

0∫

−∞

T (0, τ)P−(τ)f(τ) dτ + PN(D∗)

+∞∫

0

T (0, τ)(I − P+(τ))f(τ) dτ

= PN(D∗)

(P 2−(0)

0∫

−∞

T (0, τ)f(τ) dτ + (I − P+(0))2+∞∫

0

T (0, τ)f(τ) dτ

)

= PN(D∗)P−(0)

( 0∫

−∞

P−(0)T (0, τ)f(τ) dτ +

+∞∫

0

(I − P+(0))T (0, τ)f(τ) dτ

)

= PN(D∗)P−(0)g = PN(D∗)(I − P+(0))g = 0.

The condition PN(D∗)g = 0 is necessary and sufficient for the solvability of the equation Dξ = g,or the equation Pξ = (I − Q)ξ + g, which coincides with Eq. (8) if we set ξ = ξ1 = ξ2.

We have thereby shown that the set of bounded solutions (5) is a subset of the set of boundedsolutions (7); therefore, they coincide. By virtue of the preceding argument, we find that thecondition for the existence of bounded solutions of Eq. (2) on the entire line is equivalent tothe solvability of the operator equation

Dξ =

∞∫

0

T (0, τ)(I − P+(τ))f(τ) dτ +

0∫

−∞

T (0, τ)P−(τ)f(τ) dτ. (10)

Since D is a generalized-invertible operator, it follows that Eq. (10) has solutions if and only if

PN(D∗)

{ ∞∫

0

T (0, τ)(I − P+(τ))f(τ) dτ +

0∫

−∞

T (0, τ)P−(τ)f(τ) dτ

}= 0.

Under this condition, Eq. (10) has the set of solutions

ξ = D−

( ∞∫

0

T (0, τ)(I − P+(τ))f(τ) dτ +

0∫

−∞

T (0, τ)P−(τ)f(τ) dτ

)+ PN(D)c,

where c is an arbitrary element of the Banach space B. By substituting these solutions into (5),we obtain the representation (4).

Let us prove the property of the generalized Green operator concerning the jump at zero withthe use of the notation

(G[f ])(0 + 0) − (G[f ])(0 − 0)

= −∞∫

0

T (0, τ)(I − P+(τ))f(τ) dτ + P+(0)D−g −0∫

−∞

T (0, τ)P−(τ)f(τ) dτ − (I − P−(0))D−g

= −g + P+(0)D−g − D−g + P−(0)D−g = (P+(0) − I + P−(0))D−g − g

= DD−g − g = −(I − DD−)g = −PN(D∗)g = −+∞∫

−∞

H(t)f(t) dt.

The second property can be verified by a straightforward substitution of the Green operator intoEq. (2). The proof of the theorem is complete.

DIFFERENTIAL EQUATIONS Vol. 48 No. 6 2012

814 POKUTNYI

To clarify the relationship of the proved assertion with the results in [9], we state and prove oneauxiliary assertion.

Lemma. If the projections P+(0) and P−(0) commute ([P+(0), P−(0)] = 0), then the operatorD = P+(0) − (I − P−(0)) always has a generalized inverse, which coincides with the operator D.

Proof. Obviously, D is a bounded operator. First, let us find the operator

D2 = (P+(0) − I + P−(0))(P+(0) − I + P−(0))= P 2

+(0) − P+(0) + P−(0)P+(0) − P+(0) + I − P−(0) + P+(0)P−(0) − P−(0) + P 2+(0)

= 2P+(0)P−(0) − P+(0) − P−(0) + I.

ThenD3 = D2D = (2P+(0)P−(0) − P+(0) − P−(0) + I)(P+(0) − I + P−(0))

= 2P+(0)P−(0)P+(0) − P+(0)2 − P−(0)P+(0) + P+(0) − 2P+(0)P−(0)+ P+(0) + P−(0) − I + 2P+(0)P−(0)2 − P+(0)P−(0) − P−(0)2 + P−(0)

= −2P+(0)P−(0) + 2P+(0)P−(0) + P+(0) + P−(0) − I = D.

It follows from the resulting relation that DDD = D; therefore, D is a generalized invertibleoperator, and D = D−. The proof of the lemma is complete.

Remark 2. From the above-proved lemma and Theorem 1, we obtain, as a corollary, the resultsin [9], which were obtained under the assumption that the projections P+(0) and P−(0) commute.

Since each Fredholm operator is generalized-invertible, it follows that the case considered in [8]satisfies the assumptions of Theorem 1 as well.

MAIN RESULT (NONLINEAR CASE)

In the Banach space B, consider the differential equation

dx

dt= A(t)x(t) + εZ(x, t, ε) + f(t). (11)

We seek a bounded solution x(t, ε) of Eq. (11) that becomes one of the solutions x(t, 0) = x0(t, c)of the generating equation for ε = 0.

To find a necessary condition on the operator function Z(x, t, ε), we impose the joint constraints

Z(· , · , ·) ∈ BC(R,B) × C[0, ε0] × C[‖x − x0‖ ≤ q],

where q is some positive constant.Let us show that this problem can be solved with the use of the operator “equation for generating

constants”1

F (c) =

+∞∫

−∞

H(t)Z(x0(t, c), t, 0) dt = 0. (12)

Theorem 2 (necessary condition). Let the homogeneous equation (1) admit exponential di-chotomy on the half-lines R

+0 and R

−0 with projection-valued operator functions P+(t) and P−(t),

respectively , and let the nonlinear equation (11) have a bounded solution x(· , ε) that becomes oneof the solutions of the generating equation (1) with constant c = c0, x(t, 0) = x0(t, c0) for ε = 0.Then this constant should satisfy the equation for generating constants (12).

1 In the case of a classical periodic problem, Eq. (12) is an equation for generating amplitudes.

DIFFERENTIAL EQUATIONS Vol. 48 No. 6 2012

BOUNDED SOLUTIONS OF LINEAR AND WEAKLY NONLINEAR . . . 815

Proof. If Eq. (11) has bounded solutions x(t, ε), then, by Theorem 1, the following solvabilitycondition should be satisfied:

+∞∫

−∞

H(t){f(t) + εZ(x(t, ε), t, ε)} dt = 0. (13)

By using condition (4), we find that condition (13) is equivalent to the following:

+∞∫

−∞

H(t)Z(x(t, ε), t, ε) dt = 0.

Since x(t, ε) → x0(t, c0) as ε → 0, we finally [by using the continuity of the operator functionZ(x, t, ε)] have

F (c0) =

+∞∫

−∞

H(t)Z(x0(t, c0), t, 0) dt = 0.

The proof of the theorem is complete.To find a sufficient condition for the existence of bounded solutions of Eq. (11), we additionally

assume that the operator function Z(x, t, ε) is strongly differentiable in a neighborhood of thegenerating solution (Z(·, t, ε) ∈ C1[‖x − x0‖ ≤ q]).

This problem can be solved with the use of the operator

B0 =

+∞∫

−∞

H(t)A1(t)T (t, 0)P+(0)PN(D) dt : B → B,

where A1(t) = Z1(v, t, ε)|v=x0 , ε=0 (the Frechet derivative).

Theorem 3 (sufficient condition). Suppose that the homogeneous equation (1) admits exponen-tial dichotomy on the half-lines R

+0 and R

−0 with projection-valued operator functions P+(t) and

P−(t), respectively , and let Eq. (2) have bounded solutions of the form (5). In addition, let theoperator B0 satisfy the following conditions.

1. The operator B0 is generalized-invertible.2. PN(B∗

0 )PN(D∗)P−(0) = 0.Then for an arbitrary element c = c0 ∈ B satisfying the equation for generating constants (12),

there exists at least one weak bounded solution of Eq. (11). This solution can be found with the useof the iterative process

yk+1(t, ε) = εG[Z(x0(τ, c0) + yk, τ, ε)](t, 0),

ck = −B−0

+∞∫

−∞

H(τ){A1(τ)yk(τ, ε) + R(yk(τ, ε), τ, ε)} dτ,

yk+1(t, ε) = T (t, 0)P+(0)PN(D)ck + yk+1(t, 0, ε),xk(t, ε) = x0(t, c0) + yk(t, ε), k = 0, 1, 2, . . . , y0(t, ε) = 0, x(t, ε) = lim

k→∞xk(t, ε).

Proof. By the change of variables x(t, ε) = x0(t, c0)+y(t, ε), where the constant c0 satisfies (12),we reduce Eq. (11) to the equation for y,

dy

dt= A(t)y(t) + εZ(x0(t, c0) + y(t, ε), t, ε). (14)

DIFFERENTIAL EQUATIONS Vol. 48 No. 6 2012

816 POKUTNYI

We need to find a bounded solution y(t, ε) : y(· , ε) ∈ BC1(R,B), y(t, ·) ∈ C[0, ε0], y(t, 0) = 0.Obviously, the solvability of Eq. (14) is equivalent to the solvability of Eq. (11). We write out thiscondition for y,

+∞∫

−∞

H(t)Z(x0(t, c0) + y(t, ε), t, ε) dt = 0. (15)

If condition (15) is satisfied, then the set of bounded solutions of Eq. (14) has the form

y(t, ε) = T (t, 0)P+(0)PN(D)c + y(t, ε),

wherey(t, ε) = εG[Z(x0 + y, τ, ε)](t, 0).

Since the operator Z(x, t, ε) is Frechet differentiable in a neighborhood of the generating solution,we find that it admits the representation

Z(x0(t, c0) + y(t, ε), t, ε) = Z(x0(t, c0), t, 0) + A1(t)y(t, ε) + R(y(t, ε), t, ε),

where A1(t) = Z(1)(v, t, ε)|v=x0 , ε=0 and the higher-order terms R(y, t, ε) satisfy the relations

R(0, t, 0) = 0, R(1)x (0, t, 0) = 0.

Then condition (15) can be represented in the form

+∞∫

−∞

H(t){Z(x0(t, c0), t, 0) + A1(t){T (t, 0)P+(0)PN(D)c + y(t, ε)}} dt

+

+∞∫

−∞

H(t)R(y(t, ε), t, ε) dt = 0, (16)

or in the form of the operator equation

B0c = −+∞∫

−∞

H(t){A1(t)y(t, ε) + R(y(t, ε), t, ε)} dt. (17)

By Theorem 1, the operator equation (17) is solvable if and only if

PN(B∗0 )

+∞∫

−∞

H(t){A1(t)y(t, ε) + R(y(t, ε), t, ε)} dt = 0,

which is true by virtue of assumption 2

(PN(B∗0 )H(t) = PN(B∗

0 )PN(D∗)P−(0)T (0, t) = 0).

Then the element c can be chosen in the form

c = −B+0

+∞∫

−∞

H(t){A1(t)y(t, ε) + R(y(t, ε), t, ε)} dt.

DIFFERENTIAL EQUATIONS Vol. 48 No. 6 2012

BOUNDED SOLUTIONS OF LINEAR AND WEAKLY NONLINEAR . . . 817

Therefore, we obtain the operator system

y(t, ε) = T (t, 0)P+(0)PN(D)c + y(t, ε),

c = −B−0

+∞∫

−∞

H(t){A1(t)y(t, ε) + R(y(t, ε), t, ε)} dt,

y(t, ε) = εG[Z(x0 + y, τ, ε)](t, 0).

(18)

Let us introduce the auxiliary vector u = (y, c, y)T ∈ B × B × B, which belongs to the Cartesianproduct B3 (T stands for the transposition), and the auxiliary operator

L1g = −B−0

+∞∫

−∞

H(t)A1(t)g(t) dt.

Then we rewrite the operator system (18) in the form

u =

⎡⎢⎣

0 T (t, 0)P+(0)PN(D) I

0 0 L1

0 0 0

⎤⎥⎦ u +

⎛⎜⎜⎜⎝

0

−B−0

+∞∫−∞

H(t)R(y, t, ε) dt

εG[Z(x0 + y, τ, ε)](t, 0)

⎞⎟⎟⎟⎠ .

In turn, the operator system is equivalent to the following:

Lu = g, (19)

where

L =

⎡⎢⎣

I −T (t, 0)P+(0)PN(D) −I

0 I −L1

0 0 I

⎤⎥⎦ , g =

⎛⎜⎜⎜⎝

0

−B−0

+∞∫−∞

H(t)R(y, t, ε) dt

εG[Z(x0 + y, τ, ε)](t, 0)

⎞⎟⎟⎟⎠ .

The operator L has a bounded inverse L−1. Indeed, the operator L−1 can be represented in theclosed form

L−1 =

⎡⎢⎣

I T (t, 0)P+(0)PN(D) T (t, 0)P+(0)PN(D)L1 + I

0 I L1

0 0 I

⎤⎥⎦ .

By a straightforward verification, one can show that the operator thus defined satisfies therelation LL−1 = L−1L = I.

Let us show that this relation indeed defines a bounded operator. We need to prove the existenceof a constant c1 > 0 such that the inequality ‖L−1u‖B3 ≤ c1‖u‖B3 is satisfied for all u ∈ B3. Thisinequality is equivalent [16, p. 47] to the following inequality with a constant c2 > 0 :

|||L−1(y, c, y)t|||B3 ≤ c2(|||y|||B + |||c|||B + |||y|||B),

L−1(y, c, y)t = (y + T (t, 0)P+(0)PN(D)c + T (t, 0)P+(0)PN(D)L1y + y, c + L1y, y)T

for all y, c, y ∈ B. Let us prove the boundedness of the norm of each component of the last vectorin the Banach space B :

|||y + T (· , 0)P+(0)PN(D)c + T (· , 0)P+(0)PN(D)L1y + y|||B≤ |||y|||B + |||T (· , 0)P+(0)PN(D)|||B|||c|||B + |||T (· , 0)P+(0)PN(D)L1y|||B + |||y|||B≤ |||y|||B + c1|||c|||B + c2|||y|||B.

DIFFERENTIAL EQUATIONS Vol. 48 No. 6 2012

818 POKUTNYI

In a similar way, we obtain the estimate

|||c + L1y|||B ≤ |||c|||B + |||L1|||B|||y|||B ≤ |||c|||B + c3|||y|||B.

Therefore,|||L−1(y, c, y)T|||B3 ≤ |||y|||B + (c1 + 1)|||c|||B + (1 + c2 + c3)|||y|||B

≤ c4(|||y|||B + |||c|||B + |||y|||B),

where c4 = max{1, 1 + c1, 1 + c2 + c3}. Consequently, the proof of the boundedness of the operatorL−1 is complete.

Then the operator system (18) acquires the form

u = L−1g = L−1S(ε)u,

where the operator S(ε) is not necessarily linear. By changing the parameter ε and by using theboundedness of the operator L−1, one can ensure that L−1S(ε) is a contraction operator. It followsfrom the contraction mapping principle [16, p. 318] that the operator system (18) has a uniquefixed point, which provides a bounded solution of Eq. (11). The subsequent argument is obvious.

RELATIONSHIP BETWEEN NECESSARY AND SUFFICIENT CONDITIONS

First, we prove the following assertion.

Corollary. Let a functional F (c) have the Frechet derivative F (1)(c) for each element c0 ofthe Banach space B satisfying the equation for generating constants (12). If F (1)(c) has a boundedinverse, then Eq. (11) has a unique bounded solution on the entire line for each c0.

Proof. The representation

F (1)(c)[h] =

+∞∫

−∞

H(t)Z(1)(v, t, ε)|v=x0 , ε=0[x(1)0 (t, s, c)[h]] dt

follows from the theorem on the composition of differentiable mappings in the Banach space[17, p. 145]. Let us find the derivative of the solution x

(1)0 (t, c) with respect to t. Since

x0(t, c) = T (t, 0)P+(0)PN(D)c + (G[f ])(t, 0),

we have [17, p. 161]

x(1)0 (t, c)[h] =

∂x0(t, c + αh)∂α

∣∣∣∣α=0

=∂

∂α(T (t, 0)P+(0)PN(D)c + αT (t, 0)P+(0)PN(D)h + (G[f ])(t, 0))|α=0

=∂

∂α(T (t, 0)P+(0)PN(D)c)|α=0 +

∂

∂α(αT (t, 0)P+(0)PN(D)h)|α=0 +

∂

∂α(G[f ])(t, 0)|α=0

= T (t, 0)P+(0)PN(D)h, Z(1)(v, t, ε)|v=x0 , ε=0 = A1(t).

Therefore, we finally obtain

F (1)(c)[h] =

+∞∫

−∞

H(t)A1(t)T (t, 0)P+(0)PN(D) dt[h] = B0[h].

By virtue of the invertibility of the operator F (1)(c), the operator B0 is invertible as well. Conse-quently, Eq. (12) has a unique solution, and then Eq. (11) has a unique bounded solution on theentire line as well.

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BOUNDED SOLUTIONS OF LINEAR AND WEAKLY NONLINEAR . . . 819

Remark 3. If assumptions of the corollary are satisfied, then it follows from its proof that theoperators B0 and F (1)(c0) are equal. Since the operator F (1)(c) is invertible, it follows that assump-tions 1 and 2 of Theorem 3 are necessarily satisfied for the operator B0. In this case, Eq. (11) hasa unique bounded solution for each c0 ∈ B. Therefore, the invertibility condition for the operatorF (1)(c) relates the necessary and sufficient conditions. In the finite-dimensional case, the conditionof invertibility of the operator F (1)(c) is equivalent to the condition of simplicity of the root c0 ofthe equation for generating amplitudes [5].

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