POLONICI MATHEMATICI LXVII.3 (1997)
On existence and uniqueness of solutions
of nonlocal problems for hyperbolic differential-functional equations in two independent variables
by Tomasz Cz lapi´ nski (Gda´ nsk)
Abstract. We seek for classical solutions to hyperbolic nonlinear partial differential- functional equations of the second order. We give two theorems on existence and unique- ness for problems with nonlocal conditions in bounded and unbounded domains.
1. Introduction. If X, Y are any metric spaces then we denote by C (X; Y ) the set of all continuous functions from X to Y . Let R
+= [0, ∞), a
0, b
0∈ R
+and a > 0, b > 0. We put Ω = [0, a] × [0, b], e Ω = [−a
0, a] × [−b
0, b], Ω
0= [−a
0, a] × [−b
0, b] \ (0, a] × (0, b] and B = [−a
0, 0] × [−b
0, 0].
If z : e Ω → R then for any (x, y) ∈ Ω we define a function z
(x,y): B → R by z
(x,y)(s, t) = z(x + s, y + t), (s, t) ∈ B.
In this paper we will deal with the following nonlinear hyperbolic differ- ential-functional equation:
(1) D
xyu(x, y) = f (x, y, u
(x,y), D
xu(x, y), D
yu(x, y)), (x, y) ∈ Ω, where f : Ω × C(B; R) × R
2→ R, together with the nonlocal conditions
u
(x,0)+ X
n i=1(h
i)
(x,0)u
(x,bi)= φ
(x,0), x ∈ [0, a], (2)
u
(0,y)+ X
m j=1(g
j)
(0,y)u
(aj,y)= φ
(0,y), y ∈ [0, b], (3)
where φ : Ω
0→ R, h
i: [−a
0, a] × [−b
0, 0] → R, i = 1, . . . , m, g
j: [−a
0, 0] × [−b
0, b ] → R, j = 1, . . . , m, and a
j, j = 1, . . . , m, b
i, i = 1, . . . , n, are finite numbers such that 0 < a
1< . . . < a
m≤ a, 0 < b
1< . . . < b
n≤ b. The
1991 Mathematics Subject Classification: 35L10, 35L20, 35R10.
Key words and phrases : differential-functional equations, nonlinear hyperbolic prob- lems, nonlocal conditions.
[205]
nonlocal conditions (2), (3) may also be written in the form (2
′) u(x, y) +
X
n i=1h
i(x, y)u(x, b
i+ y) = φ(x, y),
(x, y) ∈ [−a
0, a] × [−b
0, 0], (3
′) u(x, y) +
X
m j=1g
j(x, y)u(a
j+ x, y) = φ(x, y),
(x, y) ∈ [−a
0, 0] × [−b
0, b].
R e m a r k 1. Note that the domains on which the nonlocal conditions (2
′), (3
′) are considered overlap. To ensure compatibility we need addi- tional assumptions on the functions h
i, g
j. In the sequel we will assume that h
i(x, y) = g
j(x, y) = 0 for (x, y) ∈ B, i = 1, . . . , n, j = 1, . . . , m.
If a
0= 0, b
0= 0 then the differential-functional problem (1)–(3) reduces to a differential nonlocal problem. A problem of this type, in which g
j= 0, j = 1, . . . , m, was considered by L. Byszewski [2]. The nonlocal conditions were introduced for the first time by J. Chabrowski [3] in the study of linear parabolic problems. Conditions of this type can be applied in the theory of elasticity with better results than the initial or Darboux conditions.
Nonlinear differential problems of parabolic type with nonlocal inequalities together with their physical interpretation were considered by L. Byszewski [1]. His results concerning nonlocal problems are generalizations of those given in [4], [7] since in case h
i= g
j= 0, i = 1, . . . , n, j = 1, . . . , m, the nonlocal conditions reduce to the classical Darboux conditions (see also [6], [8]).
In this paper we consider nonlocal problems for the differential-functional equation (1), i.e. for the equation in which the right hand side is an operator on the function space C(B; R) with respect to the third variable. The Dar- boux problem for nonlinear hyperbolic differential-functional equations in a Banach space was studied in [5] with the use of the Kuratowski α-measure of noncompactness. In this paper we give a theorem on existence and unique- ness of classical solutions of the problem (1)–(3). We also give an analogous theorem for a problem on an unbounded domain, with Ω, [0, a] and [0, b]
replaced by R
2+, R
+and R
+, respectively. The proof is based on the Banach fixed-point theorem and it is close to that given in [2].
2. Theorems. By C
1(Ω; R) we denote the space of all functions u ∈
C (Ω; R) which have continuous derivatives D
xu, D
yu on Ω. Let U be the
set of all functions u ∈ C( e Ω ; R) which are C
1on Ω and such that the mixed
derivative D
xyu exists and is continuous on Ω, and let k · k denote the usual
supremum norm in C(B; R).
Theorem 1. Suppose that
(i) f ∈ C(Ω × C(B, R) × R
2; R) and there is a nonnegative constant L such that for all (x, y) ∈ Ω, z, z ∈ C(B; R), p, p, q, q ∈ R, we have
(4) |f (x, y, z, p, q) − f (x, y, z, p, q)| ≤ L{kz − zk + |p − p| + |q − q|};
(ii) φ ∈ C(Ω
0; R) and φ(0, ·) ∈ C
1([0, a]; R), φ(·, 0) ∈ C
1([0, b]; R);
(iii) h
i∈ C
1([−a
0, a] × [−b
0, 0]; R), i = 1, . . . , n, g
j∈ C
1([−a
0, 0] × [−b
0, b ]; R), j = 1, . . . , m, and
(5) h
i(x, y) = 0, g
j(x, y) = 0, (x, y) ∈ B, i = 1, . . . , n, j = 1, . . . , m;
(iv) K, M are nonnegative constants such that for all (x, y) ∈ Ω, i = 1, . . . , n, j = 1, . . . , m, we have
(6) |h
i(x, y)| ≤ N, |D
xh
i(x, y)| ≤ N,
|g
j(x, y)| ≤ M, |D
yg
j(x, y)| ≤ M ; (v) there is a positive constant C such that
(7) L(2C + 1)
C
2+ 2nN e
Cbn+ 2mM e
Cam< 1.
Then there is a unique solution of the problem (1)–(3) in the class U . P r o o f. Let b U be the set of all functions u ∈ C( e Ω ; R) which are C
1on Ω. In b U we introduce the norm
kuk
C= kuk
C,0+ kuk
C,1+ kuk
C,2, where
kuk
C,0= sup
(x,y)∈ eΩ
e
−C(x+y)|u(x, y)|, kuk
C,1= sup
(x,y)∈Ω
e
−C(x+y)|D
xu(x, y)|, kuk
C,2= sup
(x,y)∈Ω
e
−C(x+y)|D
yu(x, y)|.
We define the following operator on b U:
(8) (T u)(x, y) = φ(x, 0) + φ(0, y) − φ(0, 0)
− X
n i=1h
i(x, 0)u(x, b
i) − X
m j=1g
j(0, y)u(a
j, y)
+
x
\
0 y
\
0
f (s, t, u
(s,t), D
xu(s, t), D
yu(s, t)) dt ds,
(x, y) ∈ Ω,
(9) (T u)(x, y) = φ(x, y) − X
n i=1h
i(x, y)u(x, b
i+ y),
(x, y) ∈ [−a
0, a] × [−b
0, 0), (10) (T u)(x, y) = φ(x, y) −
X
m j=1g
j(x, y)u(a
j+ x, y),
(x, y) ∈ [−a
0, 0) × [−b
0, b].
The above definition is correct even though the domains in (9) and (10) overlap. Indeed, if (x, y) ∈ B then by (5) we see that (T u)(x, y) = φ(x, y) in (9) as well as in (10).
It is easy to see that if u ∈ b U then T u is continuous on e Ω \ Ω and C
1on Ω. The continuity of T u on e Ω, i.e. the continuity on {0}×[0, b]∪[0, a]×{0}, follows immediately from (5). Therefore, T maps b U into itself.
If u ∈ U is a solution of the problem (1)–(3) then integrating (1) on [0, x] × [0, y] and making use of (2), (3) we find that u is a fixed point of T . Conversely, if u ∈ b U is a fixed point of T then from (8) we see that u has a continuous derivative D
xyu on Ω, and that equation (1) holds true.
Morover, (2), (3) follow immediately from (9), (10). Thus we will seek for a fixed point of the operator T .
If u, u ∈ b U then by (4), (6), (8)–(10), we have
|(T u)(x, y) − (T u)(x, y)|
≤ L
x
\
0 y
\
0
[ku
(s,t)− u
(s,t)k
+ |D
xu(s, t) − D
xu(s, t)| + |D
yu(s, t) − D
yu(s, t)|] dt ds +
X
n i=1|h
i(x, 0)| · |u(x, b
i) − u(x, b
i)| + X
m j=1|g
j(0, y)| · |u(a
j, y) − u(a
j, y)|
≤ L
x
\
0 y
\
0
[ku − uk
C,0+ ku − uk
C,1+ ku − uk
C,2]e
C(s+t)dt ds
+ X
ni=1
N e
C(bi+x+y)ku − uk
C,0+ X
m j=1M e
C(aj+x+y)ku − uk
C,0≤ e
C(x+y)×
L
C
2ku − uk
C+ [nN e
Cbn+ mM e
Cam]ku − uk
C,0for (x, y) ∈ Ω,
|(T u)(x, y) − (T u)(x, y)|
≤ X
n i=1|h
i(x, y)| · |u(x, b
i+ y) − u(x, b
i+ y)|
≤ X
n i=1N e
C(bi+x+y)ku − uk
C,0≤ e
C(x+y)nN e
Cbnku − uk
C,0for (x, y) ∈ [−a
0, a] × [−b
0, 0], and
|(T u)(x, y) − (T u)(x, y)|
≤ X
m j=1|g
j(x, y)| · |u(a
j+ x, y) − u(a
j+ x, y)|
≤ X
m j=1M e
C(aj+x+y)ku − uk
C,0≤ e
C(x+y)mM e
Camku − uk
C,0for (x, y) ∈ [−a
0, 0] × [−b
0, b].
From the above estimates we get (11) kT u − T uk
C,0≤ L
C
2ku − uk
C+ [nN e
Cbn+ mM e
Cam]ku − uk
C,0. In the same manner from (4), (6), (8), and from the estimates
|D
x(T u)(x, y) − D
x(T u)(x, y)|
≤ L
y
\
0
[ku
(x,t)− u
(x,t)k
+ |D
xu(x, t) − D
xu(x, t)| + |D
yu(x, t) − D
yu(x, t)|] dt +
X
n i=1|D
xh
i(x, 0)| · |u(x, b
i) − u(x, b
i)|
+ X
n i=1|h
i(x, 0)| · |D
xu(x, b
i) − D
xu(x, b
i)|
and
|D
y(T u)(x, y) − D
y(T u)(x, y)|
≤ L
x
\
0
[ku
(s,y)− u
(s,y)k
+ |D
xu(s, y) − D
xu(s, y)| + |D
yu(s, y) − D
yu(s, y)|] ds
+ X
m j=1|D
yg
j(0, y)| · |u(a
j, y) − u(a
j, y)|
+ X
m j=1|g
j(0, y)| · |D
yu(a
i, y) − D
yu(a
i, y)|, which hold true for (x, y) ∈ Ω, we get
kT u − T uk
C,1≤ L
C ku − uk
C+ nN e
Cbnku − uk
C,0(12)
+ nN e
Cbnku − uk
C,1, kT u − T uk
C,2≤ L
C ku − uk
C+ mM e
Camku − uk
C,0(13)
+ mM e
Camku − uk
C,2. Therefore, by (11)–(13), we get
kT u − T uk
C≤
L(2C + 1)
C
2+ 2nN e
Cbn+ 2mM e
Canku − uk
C. This together with (7) yields that T is a contraction. By the Banach fixed point theorem we conclude that T has a unique fixed point in b U, which completes the proof of Theorem 1.
R e m a r k 2. The proof of Theorem 1 still goes when the condition (iii) is replaced by: For all i = 1, . . . , n, j = 1, . . . , m, we have
(a) h
i∈ C([−a
0, a] × [−b
0, 0]; R), g
j∈ C([−a
0, 0] × [−b
0, b ]; R);
(b) h
iis C
1on [0, a] × [−b
0, 0] and g
jis C
1on [−a
0, 0] × [0, b];
(c) h
i(x, y) = g
j(x, y) = 0, for (x, y) ∈ B.
Now, we consider the problem (1)–(3) with Ω, [0, a] and [0, b] replaced by R
2+, R
+and R
+, respectively. Consequently, in the definitions of the spaces U , b U , of the operator T , and of the norm k · k
C, we replace Ω, e Ω and Ω
0by R
2+, [−a
0, ∞) × [−b
0, ∞) and [−a
0, ∞) × [−b
0, ∞) \ (0, ∞)
2, respectively.
Theorem 2. Suppose that
(i) f ∈ C(R
2+× C(B; R) × R
2; R) and there is a nonnegative constant L such that for all (x, y) ∈ R
+, z, z ∈ C(B, R), p, p, q, q ∈ R, we have (14) |f (x, y, z, p, q) − f (x, y, z, p, q)| ≤ L{kz − zk + |p − p| + |q − q|};
(ii) φ ∈ C([−a
0, ∞) × [−b
0, ∞) \ (0, ∞)
2; R) and φ(0, ·) ∈ C
1(R
+; R), φ(·, 0) ∈ C
1(R
+, R);
(iii) h
i∈ C
1([−a
0, ∞) × [−b
0, 0]; R), i = 1, . . . , n, g
j∈ C
1([−a
0, 0] × [−b
0, ∞); R), j = 1, . . . , m, and
h
i(x, y) = 0, g
j(x, y) = 0, (x, y) ∈ B, i = 1, . . . , n, j = 1, . . . , m;
(iv) there are nonnegative constants M , N such that for all (x, y) ∈ R
2+, i = 1, . . . , n, j = 1, . . . , m, we have
(15) |h
i(x, y)| ≤ N, |D
xh
i(x, y)| ≤ N,
|g
j(x, y)| ≤ M, |D
yg
j(x, y)| ≤ M ; (v) there is a positive constant C such that
L(2C + 1)
C
2+ 2nN e
Cbn+ 2mM e
Cam< 1;
(vi) there are nonnegative constants K
1, K
2such that
|φ(x, 0)| ≤ K
1e
Cx, |D
xφ(x, 0)| ≤ K
1e
Cxfor x ∈ R
+|φ(0, y)| ≤ K
1e
Cy, |D
yφ(0, y)| ≤ K
1e
Cyfor y ∈ R
+, (16)
|f (x, y, 0, 0, 0)| ≤ K
2e
C(x+y)for (x, y) ∈ R
+(17)
Then there is a unique solution of the problem (1)–(3) in the class of func- tions u ∈ U such that kuk < ∞.
P r o o f. Let e U be the space of functions u ∈ b U such that kuk
C< ∞. We first prove that the operator T defined by (8)–(10) maps e U into itself. By the same argument as in the proof of Theorem 1, if u ∈ e U then T u ∈ b U . It remains to show that kT uk
C< ∞. Note that
(18) (T u)(x, y)
= φ(x, 0) + φ(0, y) − φ(0, 0)
− X
n i=1h
i(x, 0)u(x, b
i) − X
m j=1g
j(0, y)u(a
j, y)
+
x
\
0 y
\
0
[f (s, t, u
(s,t), D
xu(s, t), D
yu(s, t)) − f (s, t, 0, 0, 0)] dt ds
+
x
\
0 y
\
0
f (s, t, 0, 0, 0) dt ds for (x, y) ∈ R
2+, and hence, by (14)–(17), we get
|(T u)(x, y)| ≤ |φ(x, 0)| + |φ(0, y)| + |φ(0, 0)|
+ X
ni=1
|h
i(x, 0)| · |u(x, b
i)| + X
m j=1|g
j(0, y)| · |u(a
j, y)|
+ L
x
\
0 y
\
0
[ku
(s,t)k + |D
xu(s, t)| + |D
yu(s, t)|] dt ds
+
x
\
0 y
\
0
|f (s, t, 0, 0, 0)| dt ds
≤ K
1e
Cx+ K
1e
Cy+ K
1+ X
n i=1N e
C(bi+x)kuk
C,0+ X
m j=1M e
C(aj+y)kuk
C,0+ L
x
\
0 y
\
0
[kuk
C,0+ kuk
C,1+ kuk
C,2]e
C(s+t)dt ds +
x
\
0 y
\
0
K
2e
C(s+t)dt ds
≤ 3K
1e
C(x+y)+ [nN e
Cbn+ mM e
Cam]kuk
C,0e
C(x+y)+ 1
C
2[Lkuk
C+ K
2]e
C(x+y)for (x, y) ∈ R
2+.
Since the above estimate also holds for (x, y) ∈ [−a
0, ∞)×[−b
0, ∞)\(0, ∞)
2we see that
(19) kT uk
C,0≤ 3K
1+ [nN e
Cbn+ mM e
Cam]kuk
C,0+ 1
C
2[Lkuk
C+ K
2].
Analogously, by (14)–(18), and by the estimates
|D
x(T u)(x, y)| ≤ |D
xφ(x, 0)|
+ X
n i=1|D
xh
i(x, 0)| · |u(x, b
i)| + X
n i=1|h
i(x, 0)| · |D
xu(x, b
i)|
+
y
\
0
|f (x, t, u
(x,t), D
xu(x, t), D
yu(x, t)) − f (x, t, 0, 0, 0)| dt
+
y
\
0
|f (x, t, 0, 0, 0)| dt
≤ K
1e
Cx+ X
ni=1
N e
C(bi+x)[kuk
C,0+ kuk
C,1]
+ L
y
\
0
[kuk
C,0+ kuk
C,1+ kuk
C,2]e
C(x+t)dt
+
y
\
0
K
2e
C(x+t)dt
≤ K
1e
C(x+y)+ nN e
Cbn[kuk
C,0+ kuk
C,1]e
C(x+y)+ 1
C
Lkuk
C+ K
2e
C(x+y)and
|D
y(T u)(x, y)| ≤ |D
yφ(0, y)|
+ X
m j=1|D
yg
j(0, y)| · |u(a
j, y)| + X
m j=1|g
j(0, y)| · |D
yu(a
j, y)|
+
x
\
0
|f (s, y, u
(s,y), D
xu(s, y), D
yu(s, y)) − f (x, t, 0, 0, 0)| ds
+
x
\
0
|f (s, y, 0, 0, 0)| ds
≤ K
1e
Cy+ X
m j=1M e
C(aj+y)[kuk
C,0+ kuk
C,2]
+ L
x
\
0
[kuk
C,0+ kuk
C,1+ kuk
C,2]e
C(s+y)ds +
x
\
0
