Generalized solutions to boundary value problems for quasilinear hyperbolic systems of partial differential-functional equations

POLONICI MATHEMATICI LVII.2 (1992)

Generalized solutions to boundary value problems for quasilinear hyperbolic systems

of partial differential-functional equations

by Tomasz Cz lapi´ nski (Gda´ nsk)

Abstract. Generalized solutions to quasilinear hyperbolic systems in the second canonical form are investigated. A theorem on existence, uniqueness and continuous dependence upon the boundary data is given. The proof is based on the methods due to L. Cesari and P. Bassanini for systems which are not functional.

1. Introduction. We denote by C(X, Y ) the set of all continuous functions from X to Y , where X, Y are any metric spaces. Let a

> 0, B = [−b

, 0] × [−b, b], where b

∈ R

, b = (b

, . . . , b

) ∈ R

, R

= [0, ∞).

For z : [−b

, a

] × R

→ R

and (x, y) = (x, y

, . . . , y

) ∈ [0, a

] × R

, define z

: B → R

by z

(t, s) = z(x + t, y + s), (t, s) ∈ B. Put Ω = [0, a

] × R

× C(B, R

), and denote by M (m, r) the set of all real m × r matrices.

Let

A : Ω → M (m, m) , A = [A

], i, j = 1, . . . , m,

% : Ω → M (m, r), % = [%

], i = 1, . . . , m, j = 1, . . . , r, f : Ω → M (m, 1), f = [f

, . . . , f

]

,

be given functions of the variables (x, y, w), y=(y

, . . . , y

), w=(w

, . . . , w

), and T the transpose symbol. Note that if (x, y) ∈ [0, a

] × R

is fixed then A(x, y, ·), %(x, y, ·), f (x, y, ·) are operators on the function space C(B, R

).

Furthermore, let

B

: R

→ M (m, m), B

= [B

], i, j = 1, . . . , m, ψ : R

→ R

, ψ = (ψ

, . . . , ψ

) ,

where l=1, . . . , N , m≤N , be given functions of the variable y=(y

, . . . , y

).

1991 Mathematics Subject Classification: 35L50, 35D05.

Key words and phrases: differential-functional system, second canonical form, gener- alized solutions.

We consider the following hyperbolic differential-functional system in the second canonical form:

(1)

m

X

j=1

A

(x, y, z

) h

D

z

(x, y) +

r

X

k=1

%

(x, y, z

)D

z

(x, y) i

= f

(x, y, z

), i = 1, . . . , m , with the boundary data on N arbitrary (not necessarily distinct) hyper- planes x = a

, a

∈ [0, a

], l = 1, . . . , N ,

(2)

N

X

l=1

B

(y)z(a

, y) = ψ(y), y ∈ R

.

A function z ∈ C(I

, R

), I

= [0, a]×R

, where a ∈ (0, a

], is a solution of (1), (2) if z has partial derivatives D

z, D

z, k = 1, . . . , r, a.e. on I

, satisfies (1) a.e. on I

and (2) for all y ∈ R

.

If b

= 0 and b = 0 then the differential-functional system (1) reduces to a differential system in the second canonical form. Generalized (a.e.) solutions of systems of that form have been investigated in a large number of papers. We mention here the papers of P. Bassanini [1]–[3] and L. Ce- sari [5], [6]. Quasilinear hyperbolic systems with a retarded argument have been studied by Z. Kamont and J. Turo [12]–[15]. In [19]–[22] J. Turo has also considered the existence and uniqueness of generalized solutions for differential-functional hyperbolic systems with operators of Volterra type.

In that case the given functions are superpositions of functions defined on subsets of a finite-dimensional Euclidean space with operators of Volterra type. The main assumptions for the operators have been formulated in the form of linear inequalities given on some function space. Classical solutions to nonlinear equations with the same model of functional dependence have been investigated in [11], [18].

In this paper we consider a new model of differential-functional systems

which has been used in [9] for initial value problems. For each (x, y) ∈

[−b

, a

] × R

and for any z : [−b

, a

] × R

→ R

we denote by z

the

translation of the restriction of z to B. In other words, the graph of z

is

the graph of z : [x − b

, x] × [y − b, y + b] → R

shifted to B. Since the given

functions are now operators on z

we no longer need the assumption that

the right-hand side of the system is the superposition of some function and

an operator of Volterra type. This simple model is well known for ordi-

nary differential-functional equations (see [10], [16]). It is also very general

since systems of differential equations with retarded argument ([12]–[15]),

differential-integral systems ([17]) and differential-functional systems with

operators of Volterra type ([19]–[22]) can be obtained from (1) by special-

izing the given functions. Our formulation of the problem enables us to get

new examples of differential-functional systems which cannot be obtained from the results cited above. More detailed comparison between the model of functional dependence proposed in this paper and that used in [19]–[22]

is presented in [7], [8]. The method which we use is based on the Banach fixed point theorem in a product space and it is close to that used in [14].

2. Assumptions and notations. Let kηk

= max

|η

| in R

and kU k

= max

P

j=1

|U

| in M (m, k). For short we write kηk, kU k.

For w ∈ C(B, R

) put kwk

= sup{kw(t, s)k : (t, s) ∈ B}. Let C

(B, R

) denote the set of all functions w ∈ C(B, R

) such that

kwk

= sup{[|t − t| + kw − wk]

kw(t, s) − w(t, s)k : (t, s), (t, s) ∈ B} < ∞ . We equip this set with the norm kwk

= kwk

+ kwk

, w ∈ C

(B, R

).

Furthermore, for any q ∈ R

we set C(B, R

, q) = {w ∈ C(B, R

) : kwk

≤ q}, C

(B, R

, q) = {w ∈ C

(B, R

) : kwk

≤ q}.

Assumption H

. 1

A ∈ C(Ω, M (m, m)) and there is a constant ν > 0 such that det A(x, y, w) ≥ ν for any (x, y, w) ∈ Ω.

2

There are nondecreasing functions n, l : R

→ R

such that for any q ∈ R

we have

(i) kA(x, y, w)k ≤ n(q), (x, y, w) ∈ [0, a

] × R

× C(B, R

, q),

(ii) kA(x, y, w) − A(x, y, w)k ≤ l(q)[|x − x| + ky − yk + kw − wk

], (x, y, w), (x, y, w) ∈ [0, a

] × R

× C

(B, R

, q).

R e m a r k 1. Assumption H

yields that A

(x, y, w) exists for any (x, y, w) ∈ Ω. Furthermore, there are nondecreasing n

, l

: R

→ R

such that condition 2

of Assumption H

holds with A, n, l replaced by A

, n

, l

, respectively.

Let θ denote the set of all functions l : [0, a

] × R

→ R

such that l(·, q) : [0, a

] → R

is measurable for any q ∈ R

, and l(x, ·) : R

→ R

is nondecreasing for a.e. x ∈ [0, a

].

Assumption H

. 1

%(·, y, w) : [0, a

] → M (m, r) is measurable for all (y, w) ∈ R

× C(B, R

).

2

%(x, ·) : R

× C(B, R

) → M (m, r) is continuous for a.e. x ∈ [0, a

].

3

There are a nondecreasing function n

: R

→ R

and l

∈ θ such that for any q ∈ R

we have

(i) k%(x, y, w)k ≤ n

(q), (x, y, w) ∈ [0, a

] × R

× C(B, R

, q),

(ii) k%(x, y, w)−%(x, y, w)k≤l

(x, q)[ky − yk+kw − wk

] for all (y, w), (y, w) ∈ R

× C

(B, R

, q) and for a.e. x ∈ [0, a

].

Assumption H

. 1

f (·, y, w) : [0, a

] → M (m, 1) is measurable for all (y, w) ∈ R

× C(B, R

).

2

f (x, ·) : R

× C(B, R

) → M (m, 1) is continuous for a.e. x ∈ [0, a

].

3

There are a nondecreasing function n

: R

→ R

and l

∈ θ such that for any q ∈ R

we have

(i) kf (x, y, w)k ≤ n

(q), (x, y, w) ∈ [0, a

] × R

× C(B, R

, q),

(ii) kf (x, y, w) − f (x, y, w)k ≤ l

(x, q)[ky − yk + kw − wk

] for all (y, w), (y, w) ∈ R

× C

(B, R

, q) and for a.e. x ∈ [0, a

].

For (x, y, w) ∈ Ω, l = 1, . . . , N , we put

(3) A(x, y, w) = E + e A(x, y, w), A

(x, y, w) = E + A(x, y, w), B

(y) = E

+ e B

(y), E = [δ

], E

= [δ

δ

], i, j = 1, . . . , m , and

σ

= sup n X

l=1

k e B

(y)k : y ∈ R

o , σ

= sup{k e A(x, y, w)k : (x, y, w) ∈ Ω} , σ

= sup{kA(x, y, w)k : (x, y, w) ∈ Ω} .

Assumption H

. 1

ψ ∈ C(R

, R

) and there are constants Γ, Λ ∈ R

such that for all y, y ∈ R

we have

kψ(y)k ≤ Γ, kψ(y) − ψ(y)k ≤ Λky − yk .

2

B

∈ C(R

, M (m, m)), det B

(y) 6= 0 for all l = 1, . . . , N , y ∈ R

, and there is a constant G ∈ R

such that

N

X

l=1

kB

(y) − B

(y)k ≤ Gky − yk, y, y ∈ R

. 3

(σ

+ σ

)(1 + σ

) < 1.

We denote by B

(a, Q, Q

, Q

), where a ∈ (0, a

], Q, Q

, Q

∈ R

, the set of all functions z ∈ C(I

, R

) such that for all (x, y), (x, y) ∈ I

we have

(i) kz(x, y)k ≤ Q,

(ii) kz(x, y) − z(x, y)k ≤ Q

|x − x| + Q

ky − yk.

Note that B

(a, Q, Q

, Q

) is a closed subset of the Banach space of all continuous and bounded vector functions z : I

→ R

with the norm kzk

= sup{kz(x, y)k : (x, y) ∈ I

}.

For a ∈ (0, a

] define the constants L

(Q, Q

, Q

) = R

0

l

(t, Q + Q

+ Q

) dt, i = 1, 2. In the sequel we write L

for short.

Let a ∈ (0, a

] be so small that

(4) L

(1 + Q

) < 1 ,

and let Q, Q

, Q

∈ R

, p ∈ (0, 1). Then we denote by B

(a, Q, Q

, Q

, p)

the set of all functions h ∈ C(∆

, M (m, r)), ∆

= [0, a] × [0, a] × R

, such

that for all (x, x, y), (ξ, x, y), (ξ, x, y) ∈ ∆

, i = 1, . . . , m, we have

(i) h

(x, x, y) = 0,

(ii) kh

(ξ, x, y) − h

(ξ, x, y)k ≤ n

(Q)|ξ − ξ| + γ

n

(Q)|x − x| + pky − yk, where γ

= [1 − L

(1 + Q

)]

, h

= (h

, . . . , h

). It is easily seen that B

(a, Q, Q

, Q

, p) is a closed subset of the Banach space of all continuous and bounded matrix functions h : ∆

→ M (m, r) with the norm khk

= max

sup{kh

(ξ, x, y)k : (ξ, x, y) ∈ ∆

}. Indeed, for all (ξ, x, y) ∈ ∆

, i = 1, . . . , m, we have

kh

(ξ, x, y)k = kh

(ξ, x, y) − h

(x, x, y)k ≤ n

(Q)|ξ − x| ≤ n

(Q)a , which yields khk

≤ n

(Q)a.

For any h ∈ B

(a, Q, Q

, Q

, p) let g : ∆

→ M (m, r) be defined by (5) g

(ξ, x, y) = y + h

(ξ, x, y), (ξ, x, y) ∈ ∆

, i = 1, . . . , m ,

where g

= (g

, . . . , g

), h

= (h

, . . . , h

). Then for all (x, x, y), (ξ, x, y), (ξ, x, y) ∈ ∆

, i = 1, . . . , m, we have

(i) g

(x, x, y) = y,

(ii) kg

(ξ, x, y)−g

(ξ, x, y)k ≤ n

(Q)|ξ−ξ|+γ

n

(Q)|x−x|+(1+p)ky−yk.

3. The operators V

, V

and their properties. Let h·, ·i be a scalar product in R

, let g

(t, x, y) be defined by (5) and set

z

(t, x, y) = (z

(t, g

(t, x, y)), . . . , z

(t, g

(t, x, y))) , A

(t, x, y)

= (A

(t, g

(t, x, y), z

), . . . , A

(t, g

(t, x, y), z

)) . We consider the operators Z = V

[z, h], H = V

[z, h] defined for z ∈ B

(a, Q, Q

, Q

), h ∈ B

(a, Q, Q

, Q

, p) as follows:

Z(x, y) = A

(x, y, z

)(∆

(x, y) + ∆

(x, y) + ∆

(x, y)) ,

∆

(x, y) = (∆

(x, y), . . . , ∆

(x, y)), k = 1, 2, 3,

∆

(x, y) = ψ

(g

(a

, x, y)) +

x

R

ai

f

(t, g

(t, x, y), z

) dt,

∆

(x, y) =

x

R

ai

hD

A

(t, x, y), z

(t, x, y)i dt (6)

+ hA

(a

, x, y), z

(a

, x, y)i ,

∆

(x, y) = −

N

X

l=1

hB

(g

(a

, x, y)), z(a

, g

(a

, x, y))i,

i = 1, . . . , m, (x, y) ∈ I

;

and

(7)

H(ξ, x, y) = [H

(ξ, x, y)], i = 1, . . . , m, j = 1, . . . , r, H

(ξ, x, y) =

ξ

R

x

%

(t, g

(t, x, y), z

) dt,

i=1, . . . , m, (ξ, x, y) ∈ ∆

. R e m a r k 2. By (3) we may simultaneously replace D

A

, A

, B

in the above equalities by D

A e

, e A

, e B

, respectively.

Lemma 1. Suppose that Assumptions H

–H

hold and that (8) (1 + σ

)Γ + (1 + σ

)(σ

+ σ

)Q < Q,

(9) l

(Q)(1 + Q

)[Γ + (σ

+ σ

)Q] + (1 + σ

)[n

(Q)(Λ + GQ) + l(Q)Q(2 + n

(Q) + 2Q

+ Q

n

(Q))]

+ n

(Q)(1 + σ

)(σ

+ σ

)Q

< Q

, (10) l

(Q)(1+Q

)[Γ +(σ

+σ

)Q]+(1+σ

)[(1+p)(Λ+GQ)+l(Q)Q(1+Q

)]

+ (1 + p)(1 + σ

)(σ

+ σ

)Q

< Q

. Then for a∈(0, a

] sufficiently small the operator V

maps B

(a, Q, Q

, Q

)

× B

(a, Q, Q

, Q

, p) into B

(a, Q, Q

, Q

).

P r o o f. Let Z=V

[z, h], z∈B

(a, Q, Q

, Q

), h ∈ B

(a, Q, Q

, Q

, p).

By the estimates k∆

(x, y)k ≤ max

1≤i≤m

n

|ψ

(g

(a

, x, y))|

+   

x

R

ai

f

(t, g

(t, x, y), z

) dt    o

≤ Γ +

x

R

ai

n

(Q) dt ≤ Γ + n

(Q)a,

k∆

(x, y)k ≤ max

1≤i≤m

n  

x

R

ai

hD

A e

(t, x, y), z

(t, x, y)i dt    + |h e A

(a

, x, y), z

(a

, x, y)i|

o

≤

x

R

ai

l(Q)(1 + n

(Q) + Q

+ Q

n

(Q))Q dt + σ

Q

≤ l(Q)(1 + n

(Q) + Q

+ Q

n

(Q))Qa + σ

Q, k∆

(x, y)k ≤ max

1≤i≤m

   −

N

X

l=1

h e B

(g

(a

, x, y)), z(a

, g

(a

, x, y))i  

 ≤ σ

Q ,

we obtain kZ(x, y)k

≤ (1 + σ

)[Γ + n

(Q)a + l(Q)(1 + n

(Q) + Q

+ Q

n

(Q))Qa + σ

Q + σ

Q] , (x, y) ∈ I

. By (8) we can choose a so small that

(11) (1 + σ

)[Γ + (σ

+ σ

)Q]

+(1 + σ

)[n

(Q) + l(Q)(1 + n

(Q) + Q

+ Q

n

(Q))Q]a ≤ Q , and thus for (x, y) ∈ I

we obtain

(12) kZ(x, y)k ≤ Q .

For any (x, y), (x, y) ∈ I

, we have

Z(x, y) − Z(x, y) = α

+ α

+ α

+ α

, where

α

= (A

(x, y, z

) − A

(x, y, z

))[∆

(x, y) + ∆

(x, y) + ∆

(x, y)], α

= A

(x, y, z

)[∆

(x, y) − ∆

(x, y)], i = 1, 2, 3 .

We can estimate the above terms as follows:

kα

k ≤ [l

(Q)(1 + Q

)|x − x| + l

(Q)(1 + Q

)ky − yk]

× [k∆

(x, y)k + k∆

(x, y)k + k∆

(x, y)k]

≤ [Γ + n

(Q)a + l(Q)(1 + n

(Q) + Q

+ Q

n

(Q))Qa + σ

Q + σ

Q]l

(Q)(1 + Q

)|x − x|

+ [Γ + n

(Q)a + l(Q)(1 + n

(Q) + Q

+ Q

n

(Q))Qa + σ

Q + σ

Q]l

(Q)(1 + Q

)ky − yk,

kα

k ≤ (1 + σ

) max

1≤i≤m

n

|ψ

(g

(a

, x, y)) − ψ

(g

(a

, x, y))|

+   

x

R

ai

[f

(t, g

(t, x, y), z

)

− f

(t, g

(t, x, y), z

)] dt    +

x

R

x

f

(t, g

(t, x, y), z

) dt    o

≤ (1 + σ

)[Λγ

+ L

(1 + Q

)γ

+ n

(Q)a]|x − x|

+ (1 + σ

)[Λ(1 + p) + L

(1 + Q

)(1 + p)]ky − yk, kα

k ≤ (1 + σ

) max

1≤i≤m

n  

x

R

ai

h e A

(t, x, y) − e A

(t, x, y), D

z

(t, x, y)i dt

+   

x

R

ai

hD

A e

(t, x, y), z

(t, x, y) − z

(t, x, y)i dt    +

x

R

x

hD

A e

(t, x, y), z

(t, x, y)i dt    + |h e A

(x, x, y) − e A

(x, x, y), z

(x, x, y)i|

+ |h e A

(a

, x, y), z

(a

, x, y) − z

(a

, x, y)i| o

≤ (1 + σ

){l(Q)(1 + Q

)γ

(Q

+ Q

n

(Q))a + l(Q)(1 + n

(Q) + Q

+ Q

n

(Q))Q

γ

a + l(Q)(1 + n

(Q) + Q

+ Q

n

(Q))Q + l(Q)(1 + Q

)Q + σ

Q

γ

}|x − x|

+ (1 + σ

){l(Q)(1 + Q

)(1 + p)(Q

+ Q

n

(Q))a + l(Q)(1 + n

(Q) + Q

+ Q

n

(Q))Q

(1 + p)a + l(Q)(1 + Q

)Q + σ

Q

(1 + p)}ky − yk, kα

k ≤ (1 + σ

)

× max

1≤i≤m

n  

N

X

l=1

h e B

(g

(a

, x, y)) − e B

(g

(a

, x, y)), z(a

, g

(a

, x, y))i    +

N

X

l=1

h e B

(g

(a

, x, y)), z(a

, g

(a

, x, y)) − z(a

, g

(a

, x, y))i    o

≤ (1 + σ

)γ

[GQ + σ

Q

]|x − x| + (1 + σ

)(1 + p)[GQ + σ

Q

]ky − yk . In estimating α

we have used integration by parts. From the above in- equalities we obtain

kZ(x, y) − Z(x, y)k ≤ W

|x − x| + W

ky − yk , where

W

= l

(Q)(1 + Q

)[Γ + (σ

+ σ

)]

+ (1 + σ

)[γ

(Λ + GQ) + l(Q)Q(2 + n

(Q) + 2Q

+ Q

n

(Q))]

+ γ

(1 + σ

)(σ

+ σ

)Q

+ W

, W

= l

(Q)(1 + Q

)[Γ + (σ

+ σ

)]

+ (1 + σ

)[(1 + p)(Λ + GQ) + l(Q)Q(1 + Q

)]

+ (1 + p)(1 + σ

)(σ

+ σ

)Q

+ W

,

and W

≥ 0, W

≥ 0 are some constants such that lim

W

= lim

W

= 0. By (9), (10) we can choose a so small that

(13) W

≤ Q

, W

≤ Q

.

Hence, for any (x, y), (x, y) ∈ I

we get

(14) kZ(x, y) − Z(x, y)k ≤ Q

|x − x| + Q

ky − yk .

Finally, if a ∈ (0, a

] is so small that inequalities (4), (11), (13) hold, then by (12), (14) we have Z ∈ B

(a, Q, Q

, Q

). This completes the proof.

Lemma 2. Suppose that Assumption H

holds, Q, Q

, Q

∈ R

, p ∈ (0, 1) and a ∈ (0, a

] is so small that

(15) L

(1 + p)(1 + Q) ≤ p .

Then the operator V

maps B

(a, Q, Q

, Q

) × B

(a, Q, Q

, Q

, p) into B

(a, Q, Q

, Q

).

P r o o f. Let H=V

[z, h], z∈B

(a, Q, Q

, Q

), h∈B

(a, Q, Q

, Q

, p). It is easy to see that for any (ξ, x, y), (ξ, x, y), (ξ, x, y) ∈ ∆

, i = 1, . . . , m, we have

H

(x, x, y) = 0, kH

(ξ, x, y) − H

(ξ, x, y)k ≤ n

(Q)|ξ − ξ|, kH

(ξ, x, y) − H

(ξ, x, y)k

≤   

ξ

R

x

k%

(t, g

(t, x, y), z

) − %

(t, g

(t, x, y), z

)k dt   

≤   

ξ

R

x

l

(t, P + Q)(1 + Q)kg

(t, x, y) − g

(t, x, y)k dt   

≤ L

(1 + Q)(1 + p)ky − yk ≤ pky − yk

by (15). Note that (15) implies (4), and thus for any (ξ, x, y), (ξ, x, y) ∈ ∆

, i = 1, . . . , m, we have

kH

(ξ, x, y) − H

(ξ, x, y)k ≤   

x

R

x

k%

(t, g

(t, x, y), z

)k dt    +

ξ

R

x

k%

(t, g

(t, x, y), z

) − %

(t, g

(t, x, y), z

)k dt   

≤ [n

(Q) + L

(1 + Q)γ

]|x − x| = γ

|x − x| .

From the above inequalities we derive H ∈ B

(a, Q, Q

, Q

, p). This is our claim.

For a ∈ (0, a

] define constants

E

= l

(Q)[Γ + n

(Q)a + l(Q)(1 + n

(Q) + Q

+ Q

n

(Q))Qa + σ

Q + σ

Q] + (1 + σ

)[L

+ l(Q)(Q

+ Q

n

(Q))a

+ l(Q)(1 + n

(Q) + Q

+ Q

n

(Q))a + l(Q)Q + σ

+ σ

],

E

= (1 + σ

)[Λ + L

(1 + Q

) + l(Q)(1 + Q

)(Q

+ Q

n

(Q))a + l(Q)Q

(1 + n

(Q) + Q

+ Q

n

(Q))a + σ

Q

+ GQ + σ

Q

] . Lemma 3. Let Assumptions H

–H

hold , Q, Q

, Q

∈ R

, p ∈ (0, 1) and a ∈ (0, a

] be so small that inequality (4) holds. Then for any z, z

∈ B

(a, Q, Q

, Q

), h, h

∈ B

(a, Q, Q

, Q

, p), we have

kV

[z, h] − V

[z

, h

]k

≤ E

kz − z

k

+ E

kh − h

k

, (16)

kV

[z, h] − V

[z

, h

]k

≤ L

kz − z

k

+ L

(1 + Q)kh − h

k

. (17)

P r o o f. Let z, z

∈ B

(a, Q, Q

, Q

) and h, h

∈ B

(a, Q, Q

, Q

, p). For any (x, y) ∈ I

we have

V

[z, h](x, y) − V

[z

, h

](x, y) = β

+ β

+ β

+ β

, where

β

= (A

(x, y, z

) − A

(x, y, z

))[∆

(x, y) + ∆

(x, y) + ∆

(x, y)], β

= A

(x, y, z

)[∆

(x, y) − ∆

(x, y)], i = 1, 2, 3,

and formulas for ∆

, ∆

, ∆

arise from (6) by replacing z, h by z

, h

, re- spectively. Analogously to Lemma 1 we have

kβ

k ≤ l

(Q)[Γ + n

(Q)a + l(Q)(1 + n

(Q)

+ Q

+ Q

n

(Q))Qa + σ

Q + σ

Q]kz − z

k

,

kβ

k ≤ (1 + σ

)L

kz − z

k

+ (1 + σ

)[Λ + L

(1 + Q

)]kh − h

k

, kβ

k ≤ (1 + σ

)[l(Q)(Q

+ Q

n

(Q))a

+ l(Q)(1 + n

(Q) + Q

+ Q

n

(Q))a + l(Q)Q + σ

]kz − z

k

+ (1 + σ

)[l(Q)(1 + Q

)(Q

+ Q

n(Q))a

+ l(Q)(1 + n

(Q) + Q

+ Q

n

(Q))a + σ

Q]kh − h

k

, kβ

k ≤ (1 + σ

)σ

kz − z

k

+ (1 + σ

)[GQ + σ

Q

]kh − h

k

.

From the above estimates we get (16). In a similar way we derive (17), which completes the proof.

4. The main theorem

Theorem. Suppose that Assumptions H

–H

, conditions (8)–(10) and the inequality

(18) l

(Q)[Γ + Q(σ

+ σ

)] + (1 + σ

)l(Q)Q + (1 + σ

)(σ

+ σ

) < k , are satisfied , where k ∈ (0, 1) is a constant such that k > (σ

+ σ

)(1 + σ

).

Then for a ∈ (0, a

] sufficiently small and for any system of numbers a

∈

[0, a], l = 1, . . . , N , there is a function z ∈ B

(a, Q, Q

, Q

) which is a

unique solution of the problem (1), (2) in the class B

(a, Q, Q

, Q

). Fur- thermore, if z, z

are solutions of (1), (2) with functions ψ, ψ

respectively then

(19) kz − z

k

≤ (1 − k)

(1 + σ

)kψ − ψ

k

, where kψ − ψ

k

= sup{kψ(y) − ψ

(y)k : y ∈ R

}.

P r o o f. Let V = (V

, V

), where V

, V

are defined by (6), (7).

From Lemmas 1–3 it follows that

V : B

(a, Q, Q

, Q

) × B

(a, Q, Q

, Q

, p)

→ B

(a, Q, Q

, Q

) × B

(a, Q, Q

, Q

, p) is continuous provided that a is so small that (11), (13) and (15) hold. By (18) we may additionally assume that a is small enough that

(20) E

< k, L

(1 + Q

) < k, E

L

< (k − E

)(k − L

(1 + Q

)) . Let α, β > 0 satisfy L

(k − E

)

≤ α/β ≤ (k − L

(1 + Q

))E

. Analogously to [4] we define in B

(a, Q, Q

, Q

) × B

(a, Q, Q

, Q

, p) the following weighted norm:

(21) kwk

= αkzk

+ βkhk

, w = (z, h) ,

where z ∈ B

(a, Q, Q

, Q

), h ∈ B

(a, Q, Q

, Q

, p). It is easy to check (cf. [4]) that V is a contraction with constant k with respect to this norm.

Thus V has a unique fixed point w = V w, w = (z, h) ∈ B

(a, Q, Q

, Q

) × B

(a, Q, Q

, Q

, p). Let us prove that z is a solution of (1), (2). From (6) by integration by parts we obtain

(22) 0 = A

(x, y, z

)[∆

(x, y) + e ∆

(x, y) + ∆

(x, y)], (x, y) ∈ I

, where ∆

, ∆

are defined by (6) with z = z, h = h and e ∆

(x, y) = ( e ∆

(x, y), . . . , e ∆

(x, y)) is defined by

∆ e

(x, y) = −

x

R

ai

hA

(t, x, y), D

z

(t, x, y)i dt, i = 1, . . . , m . Multiplying (22) by A(x, y, z

) we obtain

∆

(x, y) + e ∆

(x, y) + ∆

(x, y) = 0, (x, y) ∈ I

, i = 1, . . . , m , which for x = a

yields that z satisfies the boundary condition (2). Thus (22) reduces to

x

R

ai

[f

(t, g

(t, x, y), z

) − hA

(t, x, y), D

z

(t, x, y)i] dt = 0 ,

(x, y) ∈ I

, i = 1, . . . , m, where g is defined by (5) for h = h. By the same

considerations as in [6] (in particular by using the group property for g) we

see that z satisfies (1) a.e. in I

. By the reverse considerations and by the uniqueness of the fixed point of V we conclude that z is a unique solution of (1), (2).

It remains to prove (19). Let w = (z, h) = V w, w

= (z

, h

) = V w

be the fixed points of V = (V

, V

) defined by (6), (7) with functions ψ, ψ

respectively. We then have

kz − z

k

≤ (1 + σ

)kψ − ψ

k

+ E

kz − z

k

+ E

kh − h

k

, kh − h

k

≤ L

kz − z

k

+ L

(1 + Q

)kh − h

k

.

From this we obtain (1 − E

− E

L

(1 − L

(1 + Q

))

)kz − z

k

≤ (1 + σ

)kψ − ψ

k

. By (20) we have 1 − k ≤ 1 − E

− E

L

(1 − L

(1 + Q

))

, and from the last inequality we derive (19). This ends the proof.

5. Examples. Put b Ω = [0, a

] × R

× R

× R

.

Assumption H

. 1

A ∈ C( b b Ω, M (m, m)) and there is a constant ν > 0 b such that det b A(x, y, u, v) ≥ b ν for any (x, y, u, v) ∈ b Ω.

2

There are constants n, l ∈ R

such that for all (x, y, u, v), (x, y, u, v) ∈ Ω we have b

k b A(x, y, u, v)k ≤ n ,

k b A(x, y, u, v) − b A(x, y, u, v)k ≤ l[|x − x| + ky − yk + ku − uk + kv − vk] . 3

%(·, y, u, v) : [0, a b

] → M (m, r), b f (·, y, u, v) : [0, a

] → M (m, 1) are measurable for any (y, u, v) ∈ R

× R

× R

and %(x, ·) : R b

× R

× R

→ M (m, r), b f (x, ·) : R

× R

× R

→ M (m, 1) are continuous for a.e.

x ∈ [0, a

].

4

There are Lebesgue integrable functions b l

, b l

: [0, a

] → R

and con- stants n

, n

∈ R

such that for a.e. x ∈ [0, a

] and for all (y, u, v), (y, u, v) ∈ R

× R

× R

we have

k %(x, y, u, v)k ≤ n b

, k b f (x, y, u, v)k ≤ n

,

k %(x, y, u, v) − b %(x, y, u, v)k ≤ l b

(x)[ky − yk + ku − uk + kv − vk] , k b f (x, y, u, v) − b f (x, y, u, v)k ≤ l

(x)[ky − yk + ku − uk + kv − vk] . Let α, β, γ : [0, a

] × R

→ R

, where α = (α

, α

, . . . , α

), β = (β

, β

, . . . , β

), γ = (γ

, γ

, . . . , γ

), be given functions such that −b

≤ α

(x, y), β

(x, y), γ

(x, y) ≤ x for all (x, y) ∈ [0, a

] × R

. Suppose that α is Lipschitz continuous on [0, a

] × R

and that β, γ are Lipschitz continuous with respect to the second variable on R

. Then, if we define

A(x, y, w) = b A(x, y, w(0, 0), w(α(x, y) − (x, y))) ,

%(x, y, w) = %(x, y, w(0, 0), w(β(x, y) − (x, y))) , b

f (x, y, w) = b f (x, y, w(0, 0), w(γ(x, y) − (x, y))) ,

and if Assumption H

holds then the functions A, %, f satisfy Assumptions H

–H

respectively. Thus the hyperbolic system with retarded argument (23)

m

X

j=1

A b

(x, y, z(x, y), z(α(x, y))) h

D

z

(x, y)

+

r

X

k=1

%(x, y, z(x, y), z(β(x, y)))D b

z

(x, y) i

= b f (x, y, z(x, y), z(γ(x, y))), i = 1, . . . ., m, is a particular case of (1).

The differential-integral system (24)

m

X

j=1

A b



x, y, z(x, y),

˜ α(x,y)

R

α(x,y)

z(t, s)K(t, s, x, y) dt ds h

D

z

(x, y)

+

r

X

k=1

% b



x, y, z(x, y),

β(x,y)˜

R

β(x,y)

z(t, s)K

(t, s, x, y) dt ds



D

z

(x, y) i

= b f



x, y, z(x, y),

˜ γ(x,y)

R

γ(x,y)

z(t, s)K

(t, s, x, y) dt ds 

, i = 1, . . . , m ,

where α, α, β, e e β, γ, e γ : [0, a

] × R

→ R

and K, K

, K

: [0, a

] × R

× [0, a

] × R

→ M (m, m), is also a particular case of (1). In this case if we make suitable assumptions on α, α, β, e e β, γ, e γ, K, K

, K

and if Assumption H

holds then A(x, y, w) defined by

A(x, y, w) = b A 

x, y, w(0, 0),

˜ α(x,y)

R

α(x,y)

w(t − x, s − y)K(t, s, x, y) dt ds  satisfies Assumption H

, and %, f similarly defined satisfy Assumptions H

, H

respectively.

Our last example is the system (25)

m

X

j=1

A b

(x, y, z(x, y), z(α(x, y, z

))) h

D

z

(x, y)

+

r

X

k=1

% b

(x, y, z(x, y), z(β(x, y, z

)))D

z

(x, y) i

= b f

(x, y, z(x, y), z(γ(x, y, z

))) ,

i = 1, . . . , m, α, β, γ : Ω → R

.

Lemma 4. Suppose that 1

Assumption H

holds;

2

α(·, y, w), β(·, y, w), γ(·, y, w) : [0, a

] → R

are measurable for any (y, w) ∈ R

× C(B, R

), and α(x, ·), β(x, ·), γ(x, ·) : R

× C(B, R

) → R

are continuous for a.e. x ∈ [0, a

];

3

for any (x, y, w) ∈ Ω we have

α(x, y, w) ∈ B, β(x, y, w) ∈ B, γ(x, y, w) ∈ B ;

4

there is a nondecreasing function D : R

→ R

such that for all q ∈ R

, (x, y, w), (x, y, w) ∈ [0, a

] × R

× C

(B, R

, q) we have

kα(x, y, w) − α(x, y, w)k ≤ D(q)[|x − x| + ky − yk + kw − wk

] ; 5

there are nondecreasing functions D

, D

: R

→ R

such that for all q ∈ R

and (y, w), (y, w) ∈ R

× C

(B, R

, q) and for a.e. x ∈ [0, a

] we have

kβ(x, y, w) − β(x, y, w)k ≤ D

(q)[ky − yk + kw − wk

] , kγ(x, y, w) − γ(x, y, w)k ≤ D

(q)[ky − yk + kw − wk

] . Then the functions A, %, f defined by

A(x, y, w) = b A(x, y, w(0, 0), w(α(x, y, w) − (x, y))) ,

%(x, y, w) = %(x, y, w(0, 0), w(β(x, y, w) − (x, y))) , b f (x, y, w) = b f (x, y, w(0, 0), w(γ(x, y, w) − (x, y))) , satisfy Assumptions H

, H

, H

respectively.

R e m a r k 3. Systems (23), (24) can also be obtained from the theory of differential-functional equations with operators of Volterra type [19]–[22].

System (25) cannot be obtained from that theory.

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250.

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INSTITUTE OF MATHEMATICS UNIVERSITY OF GDA ´NSK WITA STWOSZA 57 80-952 GDA ´NSK, POLAND

Re¸cu par la R´edaction le 4.11.1991 R´evis´e le 15.2.1992