2005-Oujda International Conference on Nonlinear Analysis. Electronic Journal of Differential Equations, Conference 14, 2006, pp. 191–205. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu ftp ejde.math.txstate.edu (login: ftp)

NON-AUTONOMOUS INHOMOGENEOUS BOUNDARY CAUCHY PROBLEMS

MOHAMMED FILALI, BELHADJ KARIM

Abstract. In this paper we prove existence and uniqueness of classical solutions for the non-autonomous inhomogeneous Cauchy problem

\frac{d}{dt} u(t) = A(t)u(t) + f(t), 0 ≤ s ≤ t ≤ T,

L(t)u(t) = Φ(t)u(t) + g(t), 0 ≤ s ≤ t ≤ T,

u(s) = x.

The solution to this problem is obtained by a variation of constants formula.

1. Introduction

Consider the boundary Cauchy problem

\begin{matrix}\frac{d}{dt}u(t) = A(t)u(t), 0 ≤ s ≤ t ≤ T, \\ L(t)u(t) = Φ(t)u(t), 0 ≤ s ≤ t ≤ T, \\ u(s) = x. \end{matrix}  (1.1)

In the autonomous case (A(t) = A, L(t) = L), the Cauchy problem (1.1) was studied by Greiner [3]. The author used the perturbation of domains of infinitesimal generators to study the homogeneous boundary Cauchy problem. He has also showed the existence of classical solution of (1.1) via a variation of constants formula. In the non-autonomous case, Kellerman [5] and Lan [6] showed the existence of an evolution family (U(t, s))_{0≤s≤t≤T} which provides classical solutions of homogeneous boundary Cauchy problems. Filali and Moussi [2] showed the existence and uniqueness of classical solutions to the problem

\begin{matrix}\frac{d}{dt}u(t) = A(t)u(t), 0 ≤ s ≤ t ≤ T, \\ L(t)u(t) = Φ(t)u(t) + g(t), 0 ≤ s ≤ t ≤ T, \\ u(s) = x. \end{matrix}  (1.2)

2000 Mathematics Subject Classification. 34G10, 47D06.

Key words and phrases. Boundary Cauchy problem; evolution families; solution;

well posedness; variation of constants formula.

\begin{matrix}c \\ ◯ \end{matrix}2006 Texas State University - San Marcos.

Published ??, 2006.

In this paper, we prove existence and uniqueness of classical solutions to the problem

\begin{matrix}\frac{d}{dt}u(t) = A(t)u(t) + f(t), 0 ≤ s ≤ t ≤ T, \\ L(t)u(t) = Φ(t)u(t) + g(t), 0 ≤ s ≤ t ≤ T, \\ u(s) = x. \end{matrix}  (1.3)

Our technique consists on transforming (1.3) into an ordinary Cauchy problem and giving an equivalence between the two problems. The solution is explicitly given by a variation of constants formula.

2. Evolution Family

Definition 2.1. A family of bounded linear operators (U(t, s))_{0≤s≤t≤T} on X is an evolution family if

(a) U(t, r)U(r, s) = U(t, s) and U(t, t) = Id for all 0 ≤ s ≤ r ≤ t ≤ T; and

(b) the mapping (t, s) → U(t, s)x is continuous on △, for all x ∈ X with

△ = {(t, s) ∈ ℝ_{+}^{2} : 0 ≤ s ≤ t ≤ T }.

Definition 2.2. A family of linear (unbounded) operators (A(t))_{0≤t≤T} on a Banach space X is a stable family if there are constants M ≥ 1, ω ∈ ℝ such that ]ω, +∞[⊂ ρ(A(t)) for all 0 ≤ t ≤ T and

∥ ∏_{i=1}^{m} R(λ, A(t_{i}))∥ ≤ M \frac{1}{(λ − ω)^{m}}

for λ > ω and any finite sequence 0 ≤ t_{1} ≤ t_{2} ≤ ⋯ ≤ t_{m} ≤ T.

Let D, X and Y be Banach spaces, D densely and continuously embedded in X. Consider families of operators A(t) ∈ L(D, X), L(t) ∈ L(D, Y), Φ(t) ∈ L(X, Y) for 0 ≤ t ≤ T. In this section, we use the operator matrices method to prove the existence of classical solutions for the non-autonomous inhomogeneous boundary Cauchy problem (1.3). We use the following theorem due to Tanaka [9].

Theorem 2.3. Let (A(t))_{0≤t≤T} be a stable family of linear operators on a Banach space X such that

(a) the domain D = (D(A(t), ∥.∥_{D}) is a Banach space independent of t,

(b) the mapping t → A(t)x is continuously differentiable in X for every x ∈ D. Then there is an evolution family (U(t, s))_{0≤s≤t≤T} on \overline{D}. Moreover, we have the following properties: (1) U(t, s)D(s) ⊂ D(t) for all 0 ≤ s ≤ t ≤ T, where

D(r) = \{x ∈ D : A(r)x ∈ \overline{D}\}, 0 ≤ r ≤ T;

(2) the mapping t → U(t, s)x is continuously differentiable in X on [s, T] and

\frac{d}{dt}U(t, s)x = A(t)U(t, s)x

for all x ∈ D(s) and t ∈ [0, T].

We will assume that the following hypotheses:

(H1) The mapping t → A(t)x is continuously differentiable for all x ∈ D.

(H2) The family (A_{0}(t))_{0≤t≤T} , A_{0}(t) = A(t)/ ker L(t) the restriction of A(t) to ker L(t), is stable, with M_{0} and ω_{0} constants of stability.

(H3) The operator L(t) is surjective for every t ∈ [0, T] and the mapping t →

L(t)x is continuously differentiable for all x ∈ D. (H4) The mapping t → Φ(t)x is continuously differentiable for all x ∈ X. (H5) There exist constants γ > 0 and ω_{1} ∈ ℝ such that

∥L(t)x∥_{Y} ≥ \frac{λ − ω_{1}}{γ} ∥x∥_{X}  (2.1)

for x ∈ ker(λI − A(t)), ω_{1} < λ and t ∈ [0, T]. Note that under the above hypotheses, Lan [6] has showed that A_{0}(t) generates

an evolution family (U(t, s))_{0≤s≤t≤T} such that:

(a) U(t, r)U(r, s) = U(t, s) and U(t, t) = Id_{X} for all 0 ≤ s ≤ t ≤ T;

(b) (t, s) → U(t, s)x is continuously differentiable on Δ for all x ∈ X with Δ = {(t, s) ∈ ℝ_{+}^{2} : 0 ≤ s ≤ t ≤ T };

(c) there exists constants M_{0} ≥ 1 and ω_{0} ∈ ℝ such that ∥U(t, s)∥ ≤ M_{0}e^{ω_{0}(t−s)}. The following results with will be used in this article.

Lemma 2.4 ([3]). For t ∈ [0, T] and λ ∈ ρ(A_{0}(t)), following properties are satisfied: (1) D = D(A_{0}(t)) ⊕ ker(λI − A(t))

(2) L(t)/ ker(λI − A(t)) is an isomorphism from ker(λI − A(t)) onto Y

(3) t ↦ L_{λ,t} := (L(t)/ ker(λI − A(t)))^{−1} is strongly continuously differentiable.

As a consequence of this lemma, we have L(t)L_{λ,t} = Id_{Y}, L_{λ,t}L(t) and (I − L_{λ,t}L(t)) are the projections from D onto ker(λI − A(t)) and D(A_{0}(t)) .

3. The Homogeneous Problem

In this section, we consider the Cauchy problem (1.1). A function u : [s, T] → X is called classical solution if it is continuously differentiable, u(t) ∈ D for all 0 ≤ s ≤ t ≤ T and u satisfies (1.1).

We now introduce the Banach spaces Z = X × Y, Z_{0} = X × {0} ⊂ Z and we consider the projection of Z onto X: p_{1}(x, y) = x. Let M(t) be the matrix-valued operator defined on Z by

M(t) = (\begin{matrix}A(t) & 0 \\ −L(t) + Φ(t) & 0 \end{matrix}) = l(t) + ϕ(t),

where

l(t) = (\begin{matrix}A(t) & 0 \\ −L(t) & 0 \end{matrix}) , ϕ(t) = (\begin{matrix}0 & 0 \\ Φ(t) & 0 \end{matrix}) ,

and D(M(t)) = D × {0}.

Now, we consider the Cauchy problem

\begin{matrix}\frac{d}{dt}u(t) = M(t)u(t), 0 ≤ s ≤ t ≤ T, \\ u(s) = (x, 0). \end{matrix}  (3.1)

We start by proving the following lemma.

Lemma 3.1. Assume that hypothesis (H1)–(H5) hold. Then, the family of operators (M(t))_{0≤t≤T} is stable.

Remark 3.2. Since L_{λ,t}L(t) is the projection from D onto ker(λI − A(t)) and x − L_{λ,t}L(t)x ∈ D(A_{0}(t)), we have

R(λ, A_{0}(t))((λI − A(t))x) + L_{λ,t}L(t)x
= R(λ, A_{0}(t))((λI − A(t))(x − L_{λ,t}L(t)x) + L_{λ,t}L(t)x

and

R(λ, A_{0}(t))((λI − A(t))x) + L_{λ,t}L(t)x = x.  (3.2)

Proof of Lemma 3.1. Since M(t) is a perturbation of l(t) by a linear bounded operator on E, hence, in view of the perturbation result [7, Theorem 5.2.3], it is sufficient to show the stability of l(t). For λ > ω_{0} and λ ≠ 0, let

R(λ) = (\begin{matrix}R(λ, A_{0}(t)) & L_{λ,t} \\ 0 & 0 \end{matrix}) .

We have D(l(t)) = D × {0} and

(λI − l(t)) (\begin{matrix}x \\ 0 \end{matrix}) = (\begin{matrix}(λI − A(t))x \\ L(t)x \end{matrix})

for (x

0) ∈ D × {0}. By Remark 3.2, we obtain

R(λ)(λI − l(t)) (\begin{matrix}x \\ 0 \end{matrix}) = (\begin{matrix}R(λ, A_{0}(t))((λI − A(t))x) + L_{λ,t}L(t)x \\ 0 \end{matrix}) .

So that

R(λ)(λI − l(t)) (\begin{matrix}x \\ 0 \end{matrix}) = (\begin{matrix}x \\ 0 \end{matrix}) .  (3.3)

On the other hand, for (x, y) ∈ X × Y, we have

(λI − l(t))R(λ) (\begin{matrix}x \\ y \end{matrix}) = (\begin{matrix}λI − A(t) & 0 \\ L(t) & λ \end{matrix}) (\begin{matrix}R(λ, A_{0}(t))x + L_{λ,t}y \\ 0 \end{matrix}) = (\begin{matrix}x \\ y \end{matrix}) (3.4)

from (3.3) and (3.4), we obtain that the resolvent of l(t) is given by

R(λ, l(t)) = (\begin{matrix}R(λ, A_{0}(t)) & L_{λ,t} \\ 0 & 0 \end{matrix}) .  (3.5)

By a direct computation, we obtain

∏_{i=1}^{m} R(λ, l(t_{i})) = (\begin{matrix}∏_{i=1}^{m} R(λ, A_{0}(t_{i})) & ∏_{i=1}^{m−1} R(λ, A_{0}(t_{i})L_{λ,t_{m}} \\ 0 & 0 \end{matrix})

for a finite sequence 0 ≤ t_{1} ≤ t_{2} ≤ ⋯ ≤ t_{m} ≤ T and we have

∏_{i=1}^{m} R(λ, l(t_{i})) (\begin{matrix}x \\ y \end{matrix}) = (\begin{matrix}∏_{i=1}^{m} R(λ, A_{0}(t))x + ∏_{i=1}^{m−1} R(λ, A_{0}(t))L_{λ,t_{m}}y \\ 0 \end{matrix}) .

From hypothesis (H5), we conclude that ∥L_{λ,t}∥ ≤ \frac{γ}{(λ−ω)} for all t ∈ [0, T] and λ > ω and by using (H2), we obtain

\begin{aligned}∥ ∏_{i=1}^{m} R(λ, l(t_{i})) (\begin{matrix}x \\ y \end{matrix}) ∥ & ≤ ∥ ∏_{i=1}^{m} R(λ, A_{0}(t))x∥ + ∥ ∏_{i=1}^{m−1} R(λ, A_{0}(t))L_{λ,t_{m}}y∥ \\ & ≤ \frac{M}{(λ − ω_{0})^{m}} ∥x∥ + \frac{γM}{(λ − ω_{0})^{m−1}} \frac{1}{λ − ω_{1}} ∥y∥. \end{aligned} (3.6)

For ω_{2} = max(ω_{0}, ω_{1}), we have

∥ ∏_{i=1}^{m} R(λ, l(t_{i}) (\begin{matrix}x \\ y \end{matrix}) ∥ ≤ \frac{M′}{(λ − ω_{2})^{m}} (∥x∥ + ∥y∥),

where M′ = max(M, Mγ). On E = X × Y equipped with the norm ∥(x, y)∥_{1} = ∥x∥ + ∥y∥, we have:

∥ ∏_{i=1}^{m} R(λ, l(t_{i}) (\begin{matrix}x \\ y \end{matrix}) ∥ ≤ \frac{M′}{(λ − ω_{2})^{m}} (∥(x, y)∥_{1}).

□

In the following proposition we give the equivalence between the boundary problem (1.1) and the Cauchy problem (3.1).

Proposition 3.3. Let (x, 0) ∈ D × {0}.

(1) If the function t → U(t) = (u_{1}(t), 0) is a classical solution of (3.1) with an initial value (x, 0) then t → u_{1}(t) is a classical solution of (1.1) with the initial value x.

(2) Let u be a classical solution of (1.1) with the initial value x. Then the function t → U(t) = (u(t), 0) is a classical solution of (3.1) with the initial value (x, 0).

Proof. (1) Since U(t) = (u_{1}(t), 0) is a classical solution of (3.1), u_{1} is continuously differentiable on [s, T] and u_{1}(t) ∈ D. Moreover,

\frac{d}{dt}U(t) = (\begin{matrix}\frac{d}{dt}u_{1}(t) \\ 0 \end{matrix}) = M(t)U(t) and U(s) = (\begin{matrix}x \\ 0 \end{matrix}) .  (3.7)

Therefore,

\begin{matrix}\frac{d}{dt}u_{1}(t) = A(t)u_{1}(t), 0 ≤ s ≤ t ≤ T, \\ L(t)u_{1}(t) = Φ(t)u_{1}(t), 0 ≤ s ≤ t ≤ T, \\ u_{1}(s) = x. \end{matrix}  (3.8)

This implies that u_{1} is a classical solution of (1.1).

(2) Let u is a classical solution of (1.1), then u is continuously differentiable, u(t) ∈ D for t ≥ s and

\frac{d}{dt}u(t) = A(t)u(t), 0 ≤ s ≤ t ≤ T,

L(t)u(t) = Φ(t)u(t), 0 ≤ s ≤ t ≤ T,

u(s) = x.

Hence

(\begin{matrix}\frac{d}{dt}u(t) \\ 0 \end{matrix}) = (\begin{matrix}A(t) & 0 \\ −L(t) + Φ(t) & 0 \end{matrix}) (\begin{matrix}u(t) \\ 0 \end{matrix}) ,

with (u(s), 0) = (x, 0). This implies that U(t) = (u(t), 0) is a classical solution of (3.2) with the initial value (x, 0). □

The above proposition allows us to get the aim of this section by showing the well-posedness of the Cauchy problem (1.1).

Theorem 3.4. Assume that the hypotheses (H1)–(H5) hold. Then for every x ∈ D, such that −L(s)x + Φ(s)x = 0, the problem (1.1) has a unique classical solution.

Moreover, u is given by t → p_{1}(U(t, s) (\begin{matrix}x \\ 0 \end{matrix}), where U(t, s) is the evolution family

generated by (M(t)_{0≤t≤T}).

Proof. For the Cauchy problem (3.1), we have the following:

(1) D(M(t)) = D × {0} is independent of t.

(2) t → M(t) (\begin{matrix}x \\ 0 \end{matrix}) is continuously differentiable for (x, 0) ∈ D × \{0\}.
(3) The family (M(t))_{0≤t≤T} is stable.

Then the family M(t) satisfies all conditions of Theorem 2.3. Thus, there exist an evolution family (U(t, s))_{0≤s≤t} generated by the family (M(t))_{0≤t≤T} such that

(a) U(t, t) = Id_{X×{0}},

(b) U(t, r)U(r, s) = U(t, s), 0 ≤ s ≤ r ≤ t ≤ T,

(c) (t, s) → U(t, s) is strongly continuous,

(d) the function t → U(t, s) (\begin{matrix}x \\ 0 \end{matrix}) is continuously differentiable in X × \{0\} on

[s, T], and satisfies

\frac{d}{dt}U(t, s) (\begin{matrix}x \\ 0 \end{matrix}) = M(t)U(t, s) (\begin{matrix}x \\ 0 \end{matrix}) for (\begin{matrix}x \\ 0 \end{matrix}) ∈ D(s),

and

U(t, s)D(s) ⊂ D(t), for all 0 ≤ s ≤ t ≤ T,  (3.9)

where

\begin{aligned}D(s) & = \{(\begin{matrix}x \\ 0 \end{matrix}) ∈ D × \{0\} : M(s) (\begin{matrix}x \\ 0 \end{matrix}) ∈ X × \{0\}\} \\ & = ker (L(s) − Φ(s)) × \{0\}. \end{aligned}  (3.10)

Let U(t, s)(x, 0) = (u_{1}(t), 0). We have

(\begin{matrix}\frac{d}{dt}u_{1}(t) \\ 0 \end{matrix}) = M(t) (\begin{matrix}u_{1}(t) \\ 0 \end{matrix}) ,

and for u(t) = (u_{1}(t), 0), we have \frac{d}{dt}u(t) = M(t)u(t), with u(s) = (x, 0), thus u(t) = (u_{1}(t), 0) is a classical solution of (3.1) and from Proposition 3.3, we have u_{1} is a classical solution of (1.1) and

u_{1}(t) = p_{1}(U(t, s) (\begin{matrix}x \\ 0 \end{matrix}) ).  (3.11)
□

4. First Inhomogeneous Problem

In this section, we consider the inhomogeneous Cauchy problem

\begin{matrix}\frac{d}{dt}u(t) = A(t)u(t) + f(t), 0 ≤ s ≤ t ≤ T, \\ L(t)u(t) = Φ(t)u(t), 0 ≤ s ≤ t ≤ T, \\ u(s) = x. \end{matrix}  (4.1)

A function u : [s, T] → X is called classical solution if it is continuously differentiable, u(t) ∈ D, t ≥ s and u satisfies (4.1).

Consider the Banach space E = X×Y ×C^{1}([0, T], X), T > 0, where C^{1}([0, T], X) is the space of continuously differentiable functions from [0, T] into X equipped with the norm ∥f∥ = ∥f∥_{∞} + ∥f′∥_{∞}, for f ∈ C^{1}([0, T], X). Let B(t) be the operator matrices defined on E by

B(t) = (\begin{matrix}A(t) & 0 & δ_{t} \\ −L(t) + Φ(t) & 0 & 0 \\ 0 & 0 & 0 \end{matrix})  (4.2)

with D(B(t)) = D × {0} × C^{1}([0, T], X). Where δ_{t} : C^{1}([0, T], X) → X is the Dirac function concentrated at the point t with δ_{t}(f) = f(t). To the family B(t) we associate the homogeneous Cauchy problem

\begin{matrix}\frac{d}{dt}u(t) = B(t)u(t), 0 ≤ s ≤ t ≤ T, \\ u(s) = (x, 0, f). \end{matrix}  (4.3)

with (x, 0, f) ∈ D × {0} × C^{1}([0, T].

Lemma 4.1. Assume that hypothesis (H1)–(H5) hold. Then the family operators (B(t))_{0≤t≤T} is stable.

Proof. For t ∈ [0, T], we write the operator B(t) as B(t) = l(t) + ϕ(t), with

l(t) = (\begin{matrix}A(t) & 0 & 0 \\ −L(t) & 0 & 0 \\ 0 & 0 & 0 \end{matrix}) and ϕ(t) = (\begin{matrix}0 & 0 & δ_{t} \\ Φ(t) & 0 & 0 \\ 0 & 0 & 0 \end{matrix}) .

We must show that l(t) is stable and that

R(λ, l(t)) = (\begin{matrix}R(λ, A_{0}(t)) & L_{λ,t} & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1/λ \end{matrix}) .  (4.4)

For λ > ω_{0}, λ ≠ 0, and t ∈ [0, T], let

R(λ) = (\begin{matrix}R(λ, A_{0}(t)) & L_{λ,t} & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1/λ \end{matrix}) .

For (x, y, f) ∈ X × Y × C^{1}([0, T], X), we have

(\begin{matrix}R(λ, A_{0}(t) & L_{λ,t} & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1/λ \end{matrix}) (\begin{matrix}x \\ y \\ f \end{matrix}) = (R(λ, A_{0}(t))x + L_{λ,t}y ; 0 ; \frac{f}{λ}) ,

by the Remark 3.2, we obtain

(λI − l(t))R(λ) (\begin{matrix}x \\ y \\ f \end{matrix}) = (\begin{matrix}(λI − A(t))[R(λ, A_{0}(t))x + L_{λ,t}y] \\ L(t)[R(λ, A_{0}(t))x + L_{λ,t}y] \\ f \end{matrix}) = (\begin{matrix}x \\ y \\ f \end{matrix}) . (4.5)

On the other hand, for (x, 0, f) ∈ D × {0} × C^{1}([0, T], X), we have

(λI − l(t)) (\begin{matrix}x \\ 0 \\ f \end{matrix}) = (\begin{matrix}(λI − A(t))x \\ L(t)x \\ λf \end{matrix}) ,

and

R(λ)(λI − l(t)) (\begin{matrix}x \\ 0 \\ f \end{matrix}) = (\begin{matrix}R(λ, A_{0}(t))((λI − A(t))x) + L_{λ,t}L(t)x \\ 0 \\ f \end{matrix}) .

From Remark 3.2, we have

R(λ)(λI − l(t)) (\begin{matrix}x \\ 0 \\ f \end{matrix}) = (\begin{matrix}x \\ 0 \\ f \end{matrix}) .  (4.6)

From (4.5) and (4.6), we obtain that the resolvent of l(t) is given by

R(λ, l(t)) = (\begin{matrix}R(λ, A_{0}(t)) & L_{λ,t} & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1/λ \end{matrix}) .

By recurrence we can obtain

∏_{i=1}^{m} R(λ, l(t_{i})) = (\begin{matrix}∏_{i=1}^{m} R(λ, A_{0}(t_{i})) & ∏_{i=1}^{m−1} R(λ, A_{0}(t_{i}))L_{λ,t_{m}} & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1/λ^{m} \end{matrix}) .

For a finite sequence 0 ≤ t_{1} ≤ t_{2} ⋯ ≤ t_{m} ≤ T and for (x, y, f) ∈ E, we have

∏_{i=1}^{m} R(λ, l(t_{i})) (\begin{matrix}x \\ y \\ f \end{matrix}) = (\begin{matrix}∏_{i=1}^{m} R(λ, A_{0}(t_{i}))x + ∏_{i=1}^{m−1} R(λ, A_{0}(t_{i}))L_{λ,t_{m}}y \\ 0 \\ f/λ^{m} \end{matrix}) .

Using (H5), we obtain

∥ ∏_{i=1}^{m} R(λ, l(t_{i})) (\begin{matrix}x \\ y \\ f \end{matrix}) ∥ ≤ ∥ ∏_{i=1}^{m} R(λ, A_{0}(t_{i}))x + ∏_{i=1}^{m−1} R(λ, A_{0}(t_{i}))L_{λ,t_{m}}y∥ + \frac{∥f∥}{λ^{m}}

≤ \frac{M}{(λ − ω_{0})^{m}} ∥x∥ + \frac{M}{(λ − ω_{0})^{m−1}} \frac{γ}{λ − ω_{1}} ∥y∥ + \frac{∥f∥}{λ^{m}}.

Define ω_{2} = max(0, ω_{0}, ω_{1}). Then

∥ ∏_{i=1}^{m} R(λ, l(t_{i})) (\begin{matrix}x \\ y \\ f \end{matrix}) ∥ ≤ \frac{M′}{(λ − ω_{2})^{m}} (∥x∥ + ∥y∥ + ∥f∥),

where M′ = max(M, Mγ) and

∥ ∏_{i=1}^{m} R(λ, l(t_{i}))∥ ≤ \frac{M′}{(λ − ω_{2})^{m}} .  (4.7)

This inequality shows that the family l(t) is stable and by using [7, Theorem 5.2.3], the family B(t) is stable. □

Proposition 4.2. Let (x, 0, f) ∈ D × {0} × C^{1}([0, T], X).

(1) If the function t → u(t) = (u_{1}(t), 0, u_{2}(t)) is a classical solution of (4.3) with an initial value (x, 0, f) then t → u_{1}(t) is a classical solution of (4.1) with the initial value x.

(2) Let u is a classical solution of (4.1) with the initial value x .Then, the function t → U(t) = (u(t), 0, f) is a classical solution of (4.3) with the initial value (x, 0, f).

Proof. (1) If u(t) = (u_{1}(t), 0, u_{2}(t)) is a classical solution of (4.3), then u_{1} is continuously differentiable on [s, T], u_{1} ∈ D and we have

\frac{d}{dt}u(t) = (\begin{matrix}\frac{d}{dt}u_{1}(t) \\ 0 \\ \frac{d}{dt}u_{2}(t) \end{matrix}) = B(t)u(t),

which implies

(\begin{matrix}\frac{d}{dt}u_{1}(t) \\ 0 \\ \frac{d}{dt}u_{2}(t) \end{matrix}) = (\begin{matrix}A(t) & 0 & δ_{t} \\ −L(t) + Φ(t) & 0 & 0 \\ 0 & 0 & 0 \end{matrix}) (\begin{matrix}u_{1}(t) \\ 0 \\ u_{2}(t) \end{matrix}) ,

and

(\begin{matrix}\frac{d}{dt}u_{1}(t) \\ 0 \\ \frac{d}{dt}u_{2}(t) \end{matrix}) = (\begin{matrix}A(t)u_{1}(t) + δ_{t}u_{2}(t) \\ −L(t)u_{1}(t) + Φ(t)u_{1}(t) \\ 0 \end{matrix}) ,

with

u(s) = (\begin{matrix}u_{1}(s) \\ 0 \\ u_{2}(s) \end{matrix}) = (\begin{matrix}x \\ 0 \\ f \end{matrix}) .

One has \frac{d}{dt}u_{2}(t) = 0. This implies u_{2}(t) = u_{2}(s) = f; therefore, δ_{t}u_{2}(t) = δ_{t}f = f(t) and we have

\frac{d}{dt}u_{1}(t) = A(t)u_{1}(t) + f(t), 0 ≤ s ≤ t ≤ T,

L(t)u_{1}(t) = Φ(t)u_{1}(t), 0 ≤ s ≤ t ≤ T,

u_{1}(s) = x.

Therefore, u_{1} is a classical solution of (4.1) with the initial value x.

(2) If u is a classical solution of (4.1), then u is continuously differentiable, u(t) ∈ D and

\frac{d}{dt}u(t) = A(t)u(t) + f(t), 0 ≤ s ≤ t ≤ T,

L(t)u(t) = Φ(t)u(t), 0 ≤ s ≤ t ≤ T,

u(s) = x.

Moreover,

(\begin{matrix}\frac{d}{dt}u(t) \\ 0 \\ 0 \end{matrix}) = (\begin{matrix}A(t) & 0 & δ_{t} \\ −L(t) + Φ(t) & 0 & 0 \\ 0 & 0 & 0 \end{matrix}) (\begin{matrix}u(t) \\ 0 \\ f \end{matrix}) .

With u(s) = x, U(t) = (u(t), 0, f) is continuously differentiable, U(t) ∈ D(B(t)) = D × {0} × C^{1}([0, T], X) then it is a classical solution of (4.3) with the initial value (x, 0, f). □

Theorem 4.3. Let f ∈ C^{1}([0, T], X). Assume that the hypothesis (H1)–(H5) hold. Then for all x ∈ D, such that −L(s)x + Φ(s)x = 0, problem (4.1) has a unique classical solution solution u. Moreover, u is given by

u(t) = U_{Φ}(t, s)x + ∫_{s}^{t} U_{Φ}(t, s)f(r)dr,  (4.8)

where U_{Φ}(t, s) is an evolution family solution of the problem (3.1)

Proof. Consider the problem

\frac{d}{dt}u(t) = B(t)u(t), 0 ≤ s ≤ t ≤ T,

u(s) = (x, 0, f).

We have showed that (B(t))_{0≤t≤T} is a stable family and the function t → B(t)y is continuously differentiable, for all y ∈ D(B(t)) = D × {0} × C^{1}([0, T], X) and that D(B(t)) is independent of t. Then there exist an evolution system U(t, s) on X × {0} × C^{1}([0, T], X) such that

U(t, s) (\begin{matrix}x \\ 0 \\ f \end{matrix}) = (\begin{matrix}u_{1}(t) \\ 0 \\ u_{2}(t) \end{matrix}) = u(t)

is a classical solution of (4.3) and from the Proposition 4.2, u_{1} is a classical solution of (4.1), for (x, 0, f) ∈ ker(L(s) − Φ(s)) × {0} × C^{1}([0, T], X). Let v(r) = U_{Φ}(t, r)u_{1}(r). Then v is differentiable and

\frac{d}{dr} v(r) = −U_{Φ}(t, r)A_{Φ}(r)u_{1}(r) + U_{Φ}(t, r)[A_{Φ}(r)u_{1}(r) + f(r)],

where A_{Φ}(t) = A(t)/ ker(L(t) − Φ(t)); therefore,

\frac{d}{dr} v(r) = U_{Φ}(t, r)f(r).  (4.9)

Integrating (4.9) from s to t, we obtain

u_{1}(t) = U_{Φ}(t, s)x + ∫_{s}^{t} U_{Φ}(t, r)f(r)dr,

which completes the proof. □

5. Second Inhomogeneous Problem

In this section, we consider the Inhomogeneous Cauchy problem

\begin{matrix}\frac{d}{dt}u(t) = A(t)u(t) + f(t), 0 ≤ s ≤ t ≤ T, \\ L(t)u(t) = Φ(t)u(t) + g(t), 0 ≤ s ≤ t ≤ T, \\ u(s) = x. \end{matrix}  (5.1)

A function u : [s, T] → X is a classical solution if it is continuously differentiable, u(t) ∈ D, for all t ≥ s and u satisfies (5.1).

Consider the Banach space E = X × Y × C^{1}([0, T], X) × C^{1}([0, T], Y), where C^{1}([0, T], X) and C^{1}([0, T], Y) are equipped with the norm ∥f∥ = ∥f∥_{∞} + ∥f′∥_{∞} for f in C^{1}([0, T], X) or in C^{1}([0, T], Y). Consider the operator matrices

B(t) = (\begin{matrix}A(t) & 0 & δ_{t} & 0 \\ −L(t) + Φ(t) & 0 & 0 & \overline{δ_{t}} \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}) ,  (5.2)

with

D(B(t)) = D × {0} × C^{1}([0, T], X) × C^{1}([0, T], Y)

where δ_{t} : C^{1}([0, T], X) → X such that δ_{t}(f) = f(t) and \overline{δ_{t}} : C^{1}([0, T], Y) → Y such that \overline{δ_{t}}(g) = g(t). To the family B(t) ,we associate the homogeneous Cauchy problem

\begin{matrix}\frac{d}{dt}u(t) = B(t)u(t), 0 ≤ s ≤ t ≤ T, \\ u(s) = (x, 0, f, g) \end{matrix}  (5.3)

for (x, 0, f, g) ∈ D × {0} × C^{1}([0, T], X) × C^{1}([0, T], Y) = D_{1}.

Lemma 5.1. Assume that the hypothesis (H1)–(H5) hold. Then the family operators B(t) is stable.

Proof. For t ∈ [0, T], we write the B(t) defined in (5.2) as B(t) = l(t) + ϕ(t), where

l(t) = (\begin{matrix}A(t) & 0 & 0 & 0 \\ −L(t) & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}) and ϕ(t) = (\begin{matrix}0 & 0 & δ_{t} & 0 \\ Φ(t) & 0 & 0 & \overline{δ_{t}} \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}) ,

we must show that the family l(t) is stable. Let

R(λ) = (\begin{matrix}R(λ, A_{0}(t)) & L_{λ,t} & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1/λ & 0 \\ 0 & 0 & 0 & 1/λ \end{matrix}) .

For λ > ω_{0}, λ ≠ 0 and t ∈ [0, T] we show that R(λ, l(t)) = R(λ). For (x, y, f, g) ∈ X × Y × C^{1}([0, T], X) × C^{1}([0, T], Y), we have

R(λ) (\begin{matrix}x \\ y \\ f \\ g \end{matrix}) = (\begin{matrix}R(λ, A_{0}(t))x + L_{λ,t}y \\ 0 \\ f/λ \\ g/λ \end{matrix}) ,  (5.4)

by the Remark 3.2 and with the same proof as Lemma 4.1 we obtain

(λI − l(t))R(λ) (\begin{matrix}x \\ y \\ f \\ g \end{matrix}) = (\begin{matrix}x \\ y \\ f \\ g \end{matrix}) .  (5.5)

On the other hand, for (x, 0, f, g) ∈ D ×{0}×C^{1}([0, T], X)×C^{1}([0, T], Y), we have

(λI − l(t)) (\begin{matrix}x \\ 0 \\ f \\ g \end{matrix}) = (\begin{matrix}(λI − A(t))x \\ L(t)x \\ λf \\ λg \end{matrix}) ,

and

R(λ)(λI − l(t)) (\begin{matrix}x \\ 0 \\ f \\ g \end{matrix}) = (\begin{matrix}R(λ, A_{0}(t))((λI − A(t))x) + L_{λ,t}L(t)x \\ 0 \\ f \\ g \end{matrix}) = (\begin{matrix}x \\ 0 \\ f \\ g \end{matrix}) , (5.6)

then from (5.5), (5.6) and Remark 3.2, we have R(λ) = R(λI, l(t)). By recurrence we obtain

∏_{i=1}^{m} R(λ, l(t_{i})) = (\begin{matrix}∏_{i=1}^{m} R(λ, A_{0}(t_{i})) & ∏_{i=1}^{m−1} R(λ, A_{0}(t_{i}))L_{λ,t_{m}} & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1/λ^{m} & 0 \\ 0 & 0 & 0 & 1/λ^{m} \end{matrix}) ,

for a finite sequence 0 ≤ t_{1} ≤ t_{2} ≤ ⋯ ≤ t_{m} ≤ T.

Now on the space X × Y × C^{1}([0, T], X) × C^{1}([0, T], Y),we consider the norm

∥(x, y, f, g)∥ = (∥x∥ + ∥y∥ + ∥f∥ + ∥g∥).  (5.7)

For (x, y, f, g) ∈ X × Y × C^{1}([0, T], X) × C^{1}([0, T], Y), we have

∥ ∏_{i=1}^{m} R(λ, l(t_{i})) (\begin{matrix}x \\ y \\ f \\ g \end{matrix}) ∥ ≤ \frac{M}{(λ − ω_{0})^{m}} ∥x∥ + \frac{Mγ}{(λ − ω_{0})^{m−1}} \frac{1}{λ − ω_{1}} ∥y∥ + \frac{∥f∥}{λ^{m}} + \frac{∥g∥}{λ^{m}}

≤ \frac{M′}{(λ − ω_{2})^{m}} (∥x∥ + ∥y∥ + ∥f∥ + ∥g∥),

where ω_{2} = max(0, ω_{0}, ω_{1}) and M′ = max(M, Mγ). Since B(t) is a perturbation of l(t), by a linear operator ϕ(t) on E; hence, in view of perturbation result [7, Theorem 5.2.3], B(t) is stable. □

Proposition 5.2. Let (x, 0, f, g) ∈ D × {0} × C^{1}([0, T], X) × C^{1}([0, T], Y)

(1) If the function t → u(t) = (u_{1}(t), 0, u_{2}(t), u_{3}(t)) is a classical solution of (5.3) with an initial value (x, 0, f, g) then t → u_{1}(t) is a classical solution of (5.1) with the initial value x.

(2) Let u is a classical solution of (5.1) with the initial value x. Then, the function t → U(t) = (u(t), 0, f, g) is a classical solution of (5.3) with the initial value (x, 0, f, g).

Proof. (1) If u(t) = (u_{1}(t), 0, u_{2}(t), u_{3}(t)) is a classical solution of (5.3), then u_{1} is continuously differentiable on [s, T] and we have

(\begin{matrix}\frac{d}{dt}u_{1}(t) \\ 0 \\ \frac{d}{dt}u_{2}(t) \\ \frac{d}{dt}u_{3}(t) \end{matrix}) = (\begin{matrix}A(t) & 0 & δ_{t} & 0 \\ −L(t) + Φ(t) & 0 & 0 & \overline{δ_{t}} \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}) (\begin{matrix}u_{1}(t) \\ 0 \\ u_{2}(t) \\ u_{3}(t) \end{matrix}) .

This implies

\frac{d}{dt}u_{1}(t) = A(t)u_{1}(t) + δ_{t}u_{2}(t), 0 ≤ s ≤ t ≤ T,

L(t)u_{1}(t) = Φ(t)u_{1}(t) + \overline{δ_{t}}u_{3}(t), 0 ≤ s ≤ t ≤ T,

\frac{d}{dt}u_{2}(t) = 0,

\frac{d}{dt}u_{3}(t) = 0.

One has \frac{d}{dt}u_{3}(t) = 0 which implies u_{3}(t) = u_{3}(s) = g and L(t)u_{1}(t) = Φ(t)u_{1}(t) + g(t). Also \frac{d}{dt}u_{2}(t) = 0 implies u_{2}(t) = u_{2}(s) = f and \frac{d}{dt}u_{1}(t) = A(t)u_{1}(t) + f(t).

Then

\frac{d}{dt}u_{1}(t) = A(t)u_{1}(t) + f(t), 0 ≤ s ≤ t ≤ T,

L(t)u_{1}(t) = Φ(t)u_{1}(t) + g(t), 0 ≤ s ≤ t ≤ T,

u_{1}(s) = x.

Thus u_{1} is a classical solution of (5.1) with the initial value x.

(2) Let u is a classical solution of (5.1). This implies that u is continuously differentiable and u(t) ∈ D × {0} × C^{1}([0, T], X) × C^{1}([0, T], Y). Moreover,

\frac{d}{dt}u(t) = A(t)u(t) + f(t), 0 ≤ s ≤ t ≤ T

L(t)u(t) = Φ(t)u(t) + g(t), 0 ≤ s ≤ t ≤ T

u(s) = x.

This implies

(\begin{matrix}\frac{d}{dt}u(t) \\ 0 \\ 0 \\ 0 \end{matrix}) = (\begin{matrix}A(t) & 0 & δ_{t} & 0 \\ −L(t) + Φ(t) & 0 & 0 & \overline{δ_{t}} \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}) (\begin{matrix}u(t) \\ 0 \\ f \\ g \end{matrix}) ,

with u(s) = x. Then U(t) = (u(t), 0, f, g) is continuously differentiable, U(t) ∈ D × {0} × C^{1}([0, T], X) × C^{1}([0, T], Y), for all t ∈ [s, T] and U(t) is a classical solution of (5.1) with the initial value (x, 0, f, g). □

Theorem 5.3. Let f ∈ C^{1}([0, T], X) and g ∈ C^{1}([0, T], Y). Assume that the hypothesis (H1)–(H5) hold. Then for every x ∈ D such that −L(s)x+Φ(s)x+g(s) = 0, problem (5.1) has a unique classical solution.

Proof. Consider the homogenous Cauchy problem

\frac{d}{dt}u(t) = B(t)u(t), 0 ≤ s ≤ t ≤ T,

u(s) = (x, 0, f, g).

By Lemma 5.1, B(t) is a stable family and the function t → B(t)y is continuously differentiable for all y ∈ D_{1} = D(B(t)) independent of t. Then there exist an evolution family U(t, s) on X × {0} × C^{1}([0, T], X) × C^{1}([0, T], Y) such that

U(t, s) (\begin{matrix}x \\ 0 \\ f \\ g \end{matrix}) = (\begin{matrix}u_{1}(t) \\ 0 \\ u_{2}(t) \\ u_{3}(t) \end{matrix}) = u(t)

is a classical solution of (5.3) and from the Proposition 5.2, u_{1} is a classical solution of (5.1). The uniqueness of u_{1} comes from the uniqueness of the solution of (5.3) and Proposition 5.2. □

Theorem 5.4. Let f ∈ C^{1}([0, T], X) and g ∈ C^{1}([0, T], Y). If u is a classical solution of (5.1) then u is given by the variation of constants formula

u(t) = U(t, s)(I−L_{λ,s}L(s))x+g(t, u(t))+∫_{s}^{t} U(t, r)[λg(r, u(r))−g(r, u(r))′+f(r)]dr,

(5.8) where U(t, s) is the evolution family generated by A_{0}(t) and

g(t, u(t)) = L_{λ,t}(Φ(t)u(t) + g(t)).

Proof. Let now u be a classical solution of (5.1). Take

u_{2}(t) = L_{λ,t}L(t)u(t) and u_{1}(t) = (I − L_{λ,t}L(t))u(t).

Then the functions

u_{2}(t) = g(t, u(t)) = L_{λ,t}(Φ(t)u(t) + g(t)) and u_{1}(t)

are differentiable. Since u_{2}(t) ∈ ker(λI − A(t)), we have A(t)u_{2}(t) = λu_{2}(t) and

\frac{d}{dt}u_{1}(t) = \frac{d}{dt}u(t) − \frac{d}{dt}u_{2}(t)
= A(t)u(t) − (g(t, u(t)))′ + f(t)
= A(t)(u_{1}(t) + u_{2}(t)) + f(t) − (g(t, u(t)))′
= A(t)u_{1}(t) + λ(g(t, u(t)) + f(t) − (g(t, u(t)))′.

When we define h(t) := λg(t, u(t)) + f(t) − (g(t, u(t)))′, we get

u_{1}(t) = U(t, s)u_{1}(s) + ∫_{s}^{t} U(t, r)h(r)dr.  (5.9)

By replacing u_{1}(s) by (I − L_{λ,s}L(s))x, we obtain

u_{1}(t) = U(t, s)(I − L_{λ,s}L(s))x + ∫_{s}^{t} U(t, r)h(r)dr,  (5.10)

it follows that

u(t) = u(t, s)(I − L_{λ,s}L(s))x + g(t, u(t))
+ ∫_{s}^{t} u(t, r)[λg(r, u(r)) − (g(r, u(r)))′ + f(r)]dr,

which completes the proof. □

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Mohammed Filali Département de Mathématiques et Informatique, Faculté des Sciences, Université Mohammed 1er, Oujda, Maroc

E-mail address: filali@sciences.univ-oujda.ac.ma

Belhadj Karim Département de Mathématiques et Informatique, Faculté des Sciences, Université Mohammed 1er, Oujda, Maroc

E-mail address, B. Karim: karim@sciences.univ-oujda.ac.ma