POLONICI MATHEMATICI LXXII.2 (1999)
Nonlocal problems for first order functional partial differential equations
by Jan Turo (Gda´ nsk)
Abstract. Local existence of generalized solutions to nonlocal problems for nonlinear functional partial differential equations of first order is investigated. The proof is based on the bicharacteristics and successive approximations methods.
1. Introduction. Let B = [−b
0, 0]×[−b, b], I
t= [0, t] × R
nand E
t= [−b
0, t] × R
n, where t ∈ [0, a), a > 0, b = (b
1, . . . , b
n) ∈ R
n+, R
+= [0, ∞), and b
0∈ R
+. For any function z : E
a→ R with a > 0 and a fixed (x, y) = (x, y
1, . . . , y
n) ∈ I
awe define the function z
(x,y): B → R by z
(x,y)(t, s) = z(x + t, y + s), (t, s) ∈ B. We denote by C
0(B, R) the set of all continuous real functions on B.
We assume that f : I
a× C
0(B, R) × R
n→ R, and h
l, ϕ : E
0= [−b
0] × R
n→ R, l = 1, . . . , r, are given functions.
We consider the nonlinear functional partial differential equation (1) D
xz(x, y) = f (x, y, z
(x,y), D
yz(x, y)),
with nonlocal condition (2) z(x, y) +
X
r l=1h
l(x, y)z(a
k+ x, y) = ϕ(x, y), (x, y) ∈ E
0,
where D
yz = (D
y1z, . . . , D
ynz), and a
l, i = 1, . . . , r, are numbers such that 0 < a
1< . . . < a
r≤ a.
The nonlocal condition (2) may also be written in the form
(3) z
(0,y)+
X
r l=1(h
l)
(0,y)z
(al,y)= ϕ, y ∈ R
n.
1991 Mathematics Subject Classification: 35L50, 35R10.
Key words and phrases: nonlocal problem, differential-functional equation, generalized solution.
[99]
In this paper we extend the method of quasilinearization to problem (1), (2), for which we construct two uniformly convergent sequences: the first converges to a solution of the problem, the second to its derivative.
The technique essentially involves the method of bicharacteristics.
We consider generalized solutions of problem (1), (2). More precisely, a function u : E
c→ R, where 0 < c ≤ a, is a solution of (1), (2) provided
(i) u ∈ C(E
c, R) and D
yu(x, y) exists for (x, y) ∈ I
c;
(ii) the function u(·, y) : [0, c] → R is absolutely continuous on [0, c] for each y ∈ R
n;
(iii) for each y ∈ R
nequation (1) is satisfied for almost all x ∈ [0, c], and condition (2) holds.
For r = n, b
0= 0 and h
lij= h
ijδ
li(δ
liis the Kronecker symbol) the nonlocal boundary condition (2) reduces to the nonlocal condition “` a la Cesari” [1]–[4], [11], [12], [20], [26], [27]. If h
lij= δ
liδ
ijthen (2) reduces to the Nicoletti condition [21], [23]. Furthermore, if all a
l= 0, l = 1, . . . , r, then we get the usual Cauchy condition.
Classical solutions to nonlocal problems were considered for parabolic equations in [8], [9], [13], and for hyperbolic equations in [6], [7], [10], [15].
Generalized solutions to nonlinear first order partial differential equa- tions have been investigated in a large number of papers by various au- thors. Theorems on existence, uniqueness and continuous dependence upon Cauchy or boundary data for quasilinear systems were given in [1]–[3], [11], [12], [16], [17], [22]. Quasilinear differential-integral systems and systems with a retarded argument were considered in [4], [19], [20]. Nonlinear differ- ential equations were studied in [14]. Generalized solutions of quasilinear or nonlinear equations with operators of the Volterra type were investigated in [24]–[27]. Generalized solutions of (1) with the Cauchy condition were considered in [5].
Equation (1) contains as particular cases the differential equations with a retarded argument, differential-integral equations and differential-functional equations with operators of the Volterra type (see Section 6).
2. Preliminaries and assumptions. For y ∈ R
nwe write kyk = P
ni=1
|y
i|. Let k · k
0denote the supremum norm in C
0(B, R). We denote by C
0,L(B, R) the class of all functions w ∈ C
0(B, R) such that
kwk
L= sup{|w(t, s) − w(t, s)| · ks − sk
−1: (t, s), (t, s) ∈ B} < ∞.
For w ∈ C
0,L(B, R) we put kwk
0,L= kwk
0+ kwk
L.
Let C
1(B, R) be the set of all continuous functions w : B → R such that
the derivatives D
sw = (D
s1w, . . . , D
snw) exist and D
sw ∈ C
0(B, R
n). For
w ∈ C
1(B, R) we put kwk
1= kwk
0+ max{kD
sw(t, s)k : (t, s) ∈ B}.
We denote by C
1,L(B, R) the set of all functions w ∈ C
1(B, R) such that D
sw ∈ C
0,L(B, R). For w ∈ C
1,L(B, R) we write kwk
1,L= kwk
1+ kD
swk
L.
Let Θ be the class of all functions γ : [0, a]×R
+→ R
+such that γ(·, t) ∈ L([0, a], R
+) for every t ∈ R
+, and γ(s, ·) : R
+→ R
+is nondecreasing for almost all s ∈ [0, a], where L([0, a], R
+) is the class of all integrable functions on [0, a].
We put Ω
(i)= I
a× C
i(B, R) × R
nfor i = 1, 2, and Ω
(1,L)= I
a× C
1,L(B, R) × R
n.
Given S = (S
0, S
1, S
2) ∈ R
3+, we denote by C
1,L[S] the set of all func- tions ϕ : E
0→ R such that
(i) ϕ ∈ C(E
0, R) and D
yϕ(x, y) exists for (x, y) ∈ E
0; (ii) |ϕ(x, y)| ≤ S
0, kD
yϕ(x, y)k ≤ S
1on E
0, and
kD
yϕ(x, y) − D
yϕ(x, y)k ≤ S
2ky − yk for (x, y), (x, y) ∈ E
0.
Given S = (S
0, S
1, S
2) ∈ R
3+, S
0, S
1∈ [0, 1/2), we denote by C
1,L[S]
the set of all functions h
l: E
0→ R such that
(i) h
l∈ C(E
0, R) and D
yh
l(x, y) exists for (x, y) ∈ E
0, l = 1, . . . , r;
(ii) P
rl=1
|h
l(x, y)| ≤ S
0, P
rl=1
kD
yh
l(x, y)k ≤ S
1on E
0, and X
rl=1
kD
yh
l(x, y) − D
yh
l(x, y)k ≤ S
2ky − yk for (x, y), (x, y) ∈ E
0.
Let ϕ ∈ C
1,L[S], h
l∈ C
1,L[S], l = 1, . . . , r, be given and let 0 < c ≤ a, Q = (Q
0, Q
1, Q
2), λ = (λ
0, λ
1), where Q
i∈ R
+, i = 0, 1, 2, and λ
j∈ L([0, a], R
+), j = 0, 1. We denote by C
1,L[c, ϕ, h, Q, λ] the set of all functions z ∈ C(E
c, R) such that
(i) z satisfies condition (2) on E
0; (ii) D
yz(x, y) exists for (x, y) ∈ I
c;
(iii) |z(x, y)| ≤ Q
0and kD
yz(x, y)k ≤ Q
1for (x, y) ∈ I
c; (iv) for (x, y), (x, y), (x, y) ∈ I
cwe have
|z(x, y) − z(x, y)| ≤
x\
x
λ
0(t) dt ,
kD
yz(x, y) − D
yz(x, y)k ≤
x\
x
λ
1(t) dt
+ Q
2ky − yk.
We denote by C
0,L[c, Q, λ
1] the class of all functions v ∈ C(I
c, R
n) such
that
(i) kv(x, y)k ≤ Q
1for (x, y) ∈ I
c; (ii) kv(x, y) − v(x, y)k ≤ |
Tx
x
λ
1(t) dt| + Q
2ky − yk for (x, y), (x, y) ∈ I
c. Let CL(B, R) denote the set of all continuous linear operators defined on C(B, R) and taking values in R. The norm in CL(B, R) is denoted by k · k
CL(B,R).
Assumption H
1. 1
◦The function f (·, y, w, q) : [0, a] → R is measurable for every (y, w, q) ∈ R
n×C
0(B, R)×R
nand f (x, ·) : R
n×C
0(B, R)×R
n→ R is continuous for a.e. x ∈ [0, a];
2
◦there is β ∈ Θ such that |f (x, y, w, q)| ≤ β(x, kwk
0) on Ω
(0);
3
◦for every (x, y, w, q) ∈ Ω
(1), D
yf (x, y, w, q) ∈ R
nand D
wf (x, y, w, q)
∈ CL(B, R) exist;
4
◦there is α ∈ Θ such that
kD
yf (x, y, w, q)k ≤ α(x, kwk
1), kD
wf (x, y, w, q)k
CL(B,R)≤ α(x, kwk
1), on Ω
(1);
5
◦there is γ ∈ Θ such that
(4) kD
yf (x, y, w, q) − D
yf (x, y, w + h, q)k
≤ γ(x, kwk
1,L)[ky − yk + khk
1+ kq − qk], and
kD
wf (x, y, w, q) − D
wf (x, y, w + h, q)k
CL(B,R)≤ γ(x, kwk
1,L)[ky − yk + khk
1+ kq − qk], for (x, y, w, q) ∈ Ω
(1,L), y, q ∈ R
n, h ∈ C
1(B, R).
Assumption H
2. 1
◦D
qf exists for every (x, y, w, q) ∈ Ω
(1)and D
qf (·, y, w, q) ∈ L([0, a], R
n) for (y, w, q) ∈ R
n× C
1(B, R) × R
n;
2
◦for (x, y, w, q) ∈ Ω
(1)we have the estimate kD
qf (x, y, w, q)k ≤ α(x, kwk
1),
3
◦for (x, y, w, q) ∈ Ω
(1,L), y, q ∈ R
nand h ∈ C
1(B, R) we have the estimate
kD
qf (x, y, w, q) − D
qf (x, y, w + h, q)k
≤ γ(x, kwk
1,L)[ky − yk + khk
1+ kq − qk], Remark 1. We prove the theorem on the existence of solutions and we give an estimate of the domain of solutions. For simplicity of the estimates of the constant c we have assumed the same estimate for the derivatives D
yf, D
wf, D
qf , and we have also assumed the Lipschitz condition for these derivatives with the same constant.
Remark 2. It is important in our considerations that the Lipschitz condi-
tions for the given functions are local with respect to the functional variable.
Consider the function D
yf only and the simplest assumptions on it.
Suppose that there is a constant eγ such that
(5) kD
yf (x, y, w, q) − D
yf (x, y, w + h, q)k ≤ eγ[ky − yk + khk
1+ kq − qk].
Of course, our results are true if we assume (5) instead of (4).
Now we show that there is a class of nonlinear equations satisfying (4) and not satisfying (5).
Consider the equation with a deviated argument
(6) D
xz(x, y) = e f (x, y, z(λ(x), ψ(x, y)), D
yz(x, y)),
where e f : I
a× R × R
n→ R, λ ∈ C([0, a], R), ψ ∈ C(I
a, R
n), −b
0≤ λ(x) − x ≤ 0, −b ≤ ψ(x, y) − y ≤ b, (x, y) ∈ I
a, and the derivative D
yψ exists on I
a. We get equation (6) by putting in (1)
f (x, y, w, q) = e f (x, y, w(λ(x) − x, ψ(x, y) − y), q).
Suppose that there are constants α
1, γ
1, δ
0, δ
1∈ R
+such that kD
yf (x, y, w, q) − D e
yf (x, y, w, q)k ≤ γ e
1[ky − yk + |w − w| + kq − qk], kD
wf (x, y, w, q) − D e
wf (x, y, w, q)k ≤ γ e
1[ky − yk + |w − w| + kq − qk],
kD
wf (x, y, w, q)k ≤ α e
1, kD
yψ(x, y)k ≤ δ
0, kD
yψ(x, y) − D
yψ(x, y)k ≤ δ
1ky − yk.
Observe that in Ω
(1)we have
D
yif (x, y, w, q) = D
yif (x, y, w(λ(x) − x, ψ(x, y) − y), q) e + D
wf (x, y, w(λ(x) − x, ψ(x, y) − y), q) e
· h X
nj=1
D
yjw(λ(x) − x, ψ(x, y) − y)(D
yiψ
j(x, y) − 1) i . Then it is easy to see that
kD
yf (x, y, w, q) − D
yf (x, y, w + h, q)k
≤ {γ
1[1 + (1 + δ
0)kwk
1]
2+ α
1[(1 + δ
0)
2kwk
1,L+ δ
1kwk
1]}
× [ky − yk + khk
1+ kq − qk].
Thus condition (4) is satisfied. We see at once that the function D
yf does not satisfy the global Lipschitz condition (5). Similar considerations apply to D
wf, D
qf .
3. Bicharacteristics. Suppose that h
l∈ C
1,L[S], l = 1, . . . , r, ϕ ∈ C
1,L[S], z ∈ C
1,L[c, ϕ, h, Q, λ] and u ∈ C
0,L[c, Q, λ
1]. We consider the Cauchy problem
(7) η
′(t) = −D
qf (t, η(t), z
(t,η(t)), u(t, η(t))), η(x) = y,
and denote by g[z, u](·, x, y) its solution.
By using the Gronwall inequality, we can prove, similarly to [5], [28], the following.
Lemma 1. Let ϕ, ϕ ∈ C
1,L[S], h
l, h
l∈ C
1,L[S], z ∈ C
1,L[c, ϕ, h, Q, λ], z ∈ C
1,L[c, ϕ, h, Q, λ], u, u ∈ C
0,L[c, Q, λ
1] be given, and let Assumption H
2be satisfied. Then the solutions g[z, u](·; x, y) and g[z, u](·, x, y) of (7) exist on [0, c], they are unique and we have the estimates
kg[z, u](t; x, y) − g[z, u](t; x, y)k ≤ G(x) h
x\
x
α(s, Q
′) ds
+ ky − yk i
and
kg[z, u](t; x, y) − g[z, u](t; x, y)k
≤ G(x)
t
\
x
γ(s, Q
∗) ds [kz − zk
C(Es,R)+ kD
yz − D
yzk
C(Es,Rn)+ ku − uk
C(Is,Rn)] , where
G(x) = exp h Q
x\
0
γ(s, Q
∗) ds i ,
and Q
′= Q
0+ Q
1, Q = 1 + Q
1+ 2Q
2, Q
∗= Q
0+ Q
1+ Q
2.
For given ϕ ∈ C
1,L[S], h
l∈ C
1,L[S], z ∈ C
1,L[c, ϕ, h, Q, λ] and u ∈ C
0,L[c, Q, λ
1], we define
F [z, u](x, y) = ϕ(0, g(0; x, y)) − X
r l=1h
l(0, g(0; x, y))z(a
l, g(0; x, y))
+
x
\
0
h
f (P [z, u](s; x, y))
− X
n k=1D
qkf (P [z, u](s; x, y))u
k(s, g(s; x, y)) i ds, and
T
i[z, u](x, y) = D
yiϕ(0, g(0; x, y)) − X
r l=1D
yih
l(0, g(0; x, y))z(a
l, g(0; x, y))
− X
rl=1
h
l(0, g(0; x, y))u(a
l, g(0; x, y))
+
x\
0
[D
yif (P [z, u](s; x, y))
+ D
wf (P [z, u](s; x, y))(u
i)
(s,g(s;x,y))] ds,
where
P [z, u](τ ; x, y) = (τ, g[z, u](τ ; x, y), z
(τ,g[z,u](τ ;x,y)), u(τ, g[z, u](τ ; x, y))), and i = 1, . . . , n.
4. Successive approximations. Now we consider the system of inte- gral-functional equations which are generated by (1), (2):
(8)
z(x, y) = F [z, u](x, y), u(x, y) = T [z, u](x, y), (x, y) ∈ I
a, z(x, y) +
X
r l=1h
l(x, y)z(a
k+ x, y) = ϕ(x, y), (x, y) ∈ E
0,
g(t; x, y) = y +
x\
t
D
qf (P [z, u](s; x, y)) ds, where T = (T
1, . . . , T
n).
Remark 3. The functional integral system (8) is obtained in the follow- ing way. By introducing an additional unknown function u = D
yz in (1) we consider the linearization of this equation with respect to u, i.e.
(9) D
xz(x, y) = f (V ) + X
n j=1D
qjf (V )(D
yjz(x, y) − u
j(x, y)),
where V = (x, y, z
(x,y), u(x, y)). From (1) we get the following differential system for the unknown function u:
(10) D
xu
i(x, y) = D
yif (V ) + D
wf (V )(D
yiz)
(x,y)+ X
n j=1D
qjf (V )u
i(x, y), i = 1, . . . , n. Finally, we put D
yz = u in (10).
Considering (9), (10) along the bicharacteristic g[z, u](·; x, y), we obtain d
dt z(t, g[z, u](t; x, y)) = f (P [z, u](t; x, y))
− X
n j=1D
qjf (P [z, u](t; x, y))u
j(t, g[z, u](t; x, y)), d
dt u
i(t, g[z, u](t; x, y)) = D
yif (P [z, u](t; x, y))
+ D
wf (P [z, u](t; x, y))(u
i)
(t,g[z,u](t;x,y)).
Now we construct successive approximations. Suppose that ϕ ∈ C
1,L[S], h
l∈ C
1,L[S], l = 1, . . . , r, and that Assumptions H
1–H
2are satisfied. We define the sequences {z
(m)} and {u
(m)} in the following way:
(11) z
(0)is an arbitrary element of C
1,L[c, ϕ, h, Q, λ],
u
(0)= D
yz
(0)on I
c.
Evidently, u
(0)∈ C
0,L[c, Q, λ
1]. Now, if (z
(m), u
(m)) ∈ C
1,L[c, ϕ, h, Q, λ] × C
0,L[c, Q, λ
1] are known functions, then u
(m+1)is a solution of the equation
(12) u = T
(m)[u],
and
(13) z
(m+1)= F
(m)[z
(m), u
(m+1)], where T
(m)= (T
1(m), . . . , T
n(m)) is defined by
T
i(m)[z, u](x, y) = D
yiϕ(0, g(0; x, y))
− X
r l=1D
yih
l(0, g(0; x, y))z
(m)(a
l, g(0; x, y))
− X
r l=1h
l(0, g(0; x, y))u
(m)(a
l, g(0; x, y))
+
x\
0
[D
yif (P [z
(m), u](s; x, y))
+ D
wf (P [z
(m), u](s; x, y))(u
(m)i)
(s,g(s;x,y))] ds, i = 1, . . . , n. We denote by g the solution of
(14) g(t; x, y) = y +
x\
t
D
qf (P [z
(m), u](τ ; x, y)) dτ, and for simplicity we ignore the dependence of g on z
(m)and u.
Note that the equations u = T
(m)[z
(m), u] and (12) are not identical.
Lemma 2. If Assumptions H
1–H
2are satisfied and if h
l∈ C
1,L[S], ϕ ∈ C
1,L[S], Q
i> S
i, i = 0, 1, 2, then there is c ∈ (0, a] such that for any m ≥ 0 we have
(a
m) z
(m)∈ C
1,L[c, ϕ, h, Q, λ], u
(m)∈ C
0,L[c, Q, λ
1];
(b
m) D
yz
(m)(x, y) = u
(m)(x, y) in I
c.
P r o o f. We first suppose that c ∈ (0, a] is a small constant such that G(c) h
S
0+
c
\
0
α(s, Q
′) ds i
< 1, h
1 − S
0−
c
\
0
α(s, Q
′) ds i
−1h
S
1+ S
1Q
0+
c
\
0
α(s, Q
′) ds i
≤ Q
1, n
1 − G(c) h S
0+
c
\
0
α(s, Q
′) ds io
−1P (c) ≤ Q
2, M (c) < 1,
(1 − S
0)
−1n S
0+
c
\
0
[β(s, Q
0) + Q
1α(s, Q
′)] ds o
≤ Q
0,
where
P (x) = G(x) n
S
2+ S
1Q
1+ S
2Q
0+ S
1Q
1+ Q(1 + Q
1)
x
\
0
γ(s, Q
∗) ds o ,
M (x) = n
G(c)Q
2h S
0+
x\
0
α(s, Q
′) ds i
+ P (x) o
exp h
x\0
γ(s, Q
∗) ds i
+ S
0+ (1 + Q
1)
x\
0
γ(s, Q
∗) ds.
Put
λ
1(x) = n
G(c)Q
2h S
0+
c
\
0
α(s, Q
′) ds i
+ P (c) + 1 + Q
1o
α(x, Q
′), λ
0(x) = β(x, Q
0) + n
Q
1+ G(c) h
S
1+ S
1Q
0+ S
0Q
1+ Q
c\
0
[α(s, Q
′) + Q
1γ(s, Q
∗)] ds io
α(x, Q
′).
We prove (a
m) and (b
m) by induction. It follows from the definitions of z
(0)and u
(0)that (a
0) and (b
0) are satisfied. Suppose now that (a
m) and (b
m) hold. First we prove that
(15) T
(m): C
0,L[c, Q, λ
1] → C
0,L[c, Q, λ
1].
Indeed, it follows from Assumptions H
1–H
2and Lemma 1 that for u ∈ C
0,L[c, Q, λ
1] and (x, y), (x, y) ∈ I
c, we have
(16) kT
(m)[u](x, y)k ≤ S
1+S
1Q
0+S
0Q
1+(1+Q
1)
x\
0
α(s, Q
′) ds ≤ Q
1, and
kT
(m)[u](x, y) − T
(m)[u](x, y)k
≤ kD
yϕ(0, g(0; x, y)) − D
yϕ(0, g(0; x, y))k +
X
r l=1kD
yh
l(0, g(0; x, y))k · |z
(m)(a
l, g(0; x, y)) − z
(m)(a
l, g(0; x, y))|
+ X
rl=1
kD
yh
l(0, g(0; x, y)) − D
yh
l(0, g(0; x, y))k · |z
(m)(a
l, g(0; x, y))|
+ X
rl=1
X
n k=1|h
l(0, g(0; x, y))| · |u
(m)k(a
l, g(0; x, y)) − u
(m)k(a
l, g(0; x, y))|
+ X
rl=1
X
n k=1|h
l(0, g(0; x, y)) − h
l(0, g(0; x, y))| · |u
(m)k(a
l, g(0; x, y))|
+
x\
0
kD
yf (P [z
(m), u](s; x, y)) − D
yf (P [z
(m), u](s; x, y))k ds
+
x
\
0
kD
yf (P [z
(m), u](s; x, y))k ds
+
x\
x
X
n k=1D
wf (P [z
(m), u](s; x, y))(u
(m)k)
(s,g(s;x,y))ds
+
x\
0
X
n k=1|D
wf (P [z
(m), u](s; x, y))(u
(m)k)
(s,g(s;x,y))− D
wf (P [z
(m), u](s; x, y))(u
(m)k)
(s,g(s;x,y))| ds
≤
x\
x
λ
1(t) dt
+ Q
2ky − yk.
By the above inequality and by (16) we get (15).
From Assumptions H
1–H
2, applying Lemma 1, we obtain kT
(m)[u](x, y) − T
(m)[u](x, y)k
≤ kD
yϕ(0, g(0; x, y)) − D
yϕ(0, g(0; x, y))k +
X
r l=1kD
yh
l(0, g(0; x, y))k · |z
(m)(a
l, g(0; x, y)) − z
(m)(a
l, g(0; x, y))|
+ X
rl=1
kD
yh
l(0, g(0; x, y)) − D
yh
l(0, g(0; x, y))k · |z
(m)(a
l, g(0; x, y))|
+ X
rl=1
X
n k=1|h
l(0, g(0; x, y))| · |u
(m)k(a
l, g(0; x, y)) − u
(m)k(a
l, g(0; x, y))|
+ X
rl=1
X
n k=1|h
l(0, g(0; x, y))| · |u
(m)k(a
l, g(0; x, y)) − u
(m)k(a
l, g(0; x, y))|
+ X
rl=1
X
n k=1|h
l(0, g(0; x, y)) − h
l(0, g(0; x, y))| · |u
(m)k(a
l, g(0; x, y))|
+
x\
0
kD
yf (P [z
(m), u](s; x, y)) − D
yf (P [z
(m), u](s; x, y))k ds
+
x\
0
X
n k=1|D
wf (P [z
(m), u](s; x, y))
− D
wf (P [z
(m), u](s; x, y))(u
(m)k)
(s,g(s;x,y))| ds
≤ M (c)ku − uk
C(Ic,Rn),
where g is the solution of equation (14) with u instead of u on the right-hand
side.
Hence, using the Banach fixed point theorem we conclude that there is exactly one u
(m+1)∈ C
0,L[c, Q, λ
1] satisfying (12).
Now we prove that the function z
(m+1)given by (13) satisfies (b
m+1).
Put
∆(x, y, y) = z
(m+1)(x, y) − z
(m+1)(x, y) − X
n k=1u
(m+1)k(x, y)(y
k− y
k).
Similarly to [5], [18] we can prove the estimate
|∆(x, y, y)| ≤ (S
2+ S
2Q
0+ S
0Q
2)kg(0; x, y) − g(0; x, y)k
2+
x\
0
kg(s; x, y) − g(s; x, y)k
2[Q(1 + Q
1+ Q
2)γ(s, Q
∗) + Q
2α
1(s, Q
′)] ds.
It follows from Lemma 1 that
kg(s; x, y) − g(s; x, y)k ≤ G(c)ky − yk.
Therefore there is C
∗> 0 such that
|∆(x, y, y)| ≤ C
∗ky − yk
2, (x, y), (x, y) ∈ I
c. Consequently,
D
yz
(m+1)(x, y) = u
(m+1)(x, y), (x, y) ∈ I
c.
It is easily shown that z
(m+1)∈ C
1,L[c, ϕ, h, Q, λ]. The proof of Lemma 2 is complete.
5. Uniform convergence. We can now formulate the result on the convergence of the sequence of successive approximations.
Lemma 3. If Assumptions H
1–H
2are satisfied and if h
l∈ C
1,L[S], ϕ ∈ C
1,L[S], Q
i> S
i, i = 0, 1, 2, then there is c ∈ (0, a] such that the sequences {z
(m)} and {u
(m)} are uniformly convergent on I
c.
P r o o f. Suppose that c ∈ (0, a] is a small constant such that h S
1+ S
0+
c\
0
H
1(s) ds i
exp n
c\0
H
0(s) ds o +
c
\
0
H e
1(s) ds < 1, where
H
0(x) = n
1 + Q
1+ G(c)Q
2h S
0+
c
\
0
α(s, Q
′) ds i
+ P (x) o
γ(x, Q
∗), H
1(x) = H
0(x) + α(x, Q
′),
H e
i(x) = K
0(x) + α(x, Q
′) + K
1(c)H
i(x), i = 1, 2,
K
0(x) = γ(x, Q
∗) n
Q
2+ G(c)
S
1+ S
0Q
1+ S
2Q
0+ Q
c
\
0
[α(s, Q
′) + Q
1γ(s, Q
∗)] ds o ,
K
1(x) = exp h
x\0
H
0(s) ds i
x\0
[K
0(s) + 2α(s, Q
′)] ds.
For t ∈ [0, c] and m ≥ 1 we put
Z
(m)(t) = sup{|z
(m)(x, y) − z
(m−1)(x, y)| : (x, y) ∈ I
t}, U
(m)(t) = sup{ku
(m)(x, y) − u
(m−1)(x, y)k : (x, y) ∈ I
t}.
According to (12) and Assumptions H
1, H
2, on I
cwe have
|u
(m+1)i(x, y) − u
(m)i(x, y)| ≤
x\
0
H
0(s)U
(m+1)(s) ds
+
x\
0
[H
0(s)Z
(m)(s) + H
1(s)U
(m)] ds + S
1Z
(m)(x) + S
0U
(m)(x).
Then we obtain the following integral inequality on [0, c]:
U
(m+1)(x) ≤
x\
0
H
0(s)U
(m+1)(s) ds +
x\
0
[H
0(s)Z
(m)(s) + H
1(s)U
(m)] ds + S
1Z
(m)(x) + S
0U
(m)(x).
The above estimate and the Gronwall lemma imply U
(m+1)(x) ≤ n
x\0
[H
0(s)Z
(m)(s) + H
1(s)U
(m)] ds (17)
+ S
1Z
(m)(x) + S
0U
(m)(x) o
exp n
x\0
H
0(s) ds o . Using (13) and Assumptions H
1, H
2we can prove that
Z
(m+1)(x) ≤ S
0Z
(m)(x) (18)
+
x\
0
[ e H
0(s)Z
(m)(s) + e H
1(s)U
(m)(s)] ds, x ∈ [0, c].
From (17), (18) we conclude that there exists δ ∈ (0, 1) such that
(19) U
(m+1)(x)+Z
(m+1)(x) ≤ δ[U
(m)(x)+Z
(m)(x)], x ∈ [0, c], m ≥ 0.
From (11) and from Assumptions H
1, H
2we have Z
(1)(x) ≤
x
\
0
[β(s, Q
0) + e Q
1α(s, Q
′)] ds,
U
(1)(x) ≤ (1 + e Q
1)
x\
0
α(s, Q
′) ds.
Finally, the assertion of Lemma 3 follows from (19).
6. Theorem and particular cases. We are now able to state the main result of the paper.
Theorem . Let Assumptions H
1–H
2hold , and let Q
i> S
i, i = 0, 1, 2.
Then there is c ∈ (0, a] such that for any numbers a
l, 0 ≤ a
l≤ c, l = 1, . . . , r, and functions ϕ ∈ C
1,L[S], h
l∈ C
1,L[S], there exists a solution z : E
c→ R of problem (1), (2).
P r o o f. By Lemma 3 there exist z ∈ C
1,L[c, ϕ, h, Q, λ] and u ∈ C
0,L[c, Q, λ
1] such that {z
(m)} converges to z and {u
(m)} converges to u uniformly on I
c. Furthermore, D
yz exists on I
cand D
yz = u. Thus we obtain
z(x, y) = ϕ(0, g(0; x, y)) − X
r l=1h
l(0, g(0; x, y))z(a
l, g(0; x, y)) (20)
+
x\
0
h
f (P [z, D
yz](s; x, y))
− X
n k=1D
qkf (P [z, D
yz](s; x, y))D
ykz(s; x, y) i ds, where
g(t; x, y) = y +
x\
t
D
qf (P [z, D
yz](τ ; x, y)) dτ.
Now we prove that z is a solution of (1). By taking y = g(x; 0, η) in (20), we get
(21) z(x, g(x; 0, η))
= ϕ(0, η) − X
rl=1
h
l(0, η)z(a
l, η)
+
x\
0
h
f (τ, g(x; 0, η), z
(τ,g(x;0,η)), D
yz(τ, g(x; 0, η)))
− X
n k=1D
qkf (τ, g(x; 0, η), z
(τ,g(x;0,η)), D
yz(τ, g(x; 0, η)))D
ykz(τ, g(x; 0, η)) i
dτ,
since the function g satisfies the group property. By differentiation of (21) with respect to x, and by putting again y = g(x; 0, η), we find that z satisfies (1). It follows immediately that z satisfies (2). The proof of the Theorem is complete.
Now we list examples of equations which can be derived from (1).
1) Suppose that f : I
a× R × R
n→ R, λ : [0, a] → R, ψ : I
a→ R
n, are given. Assume that
−b
0≤ λ(x) − x ≤ 0, −b ≤ ψ(x, y) − y ≤ b, (x, y) ∈ I
a. We consider the operator f defined by
f (x, y, w, q) = f (x, y, w[λ(x) − x, ψ(x, y) − y], q), (x, y, w, q) ∈ Ω
(0). In this case equation (1) reduces to the following differential equation with a retarded argument:
D
xz(x, y) = f (x, y, z(λ(x), ψ(x, y)), D
yz(x, y)), z(x, y) = ϕ(x, y), (x, y) ∈ E
0.
2) Suppose that f : I
a× R × R
n→ R is given function. Let f (x, y, w, q) = f
x, y,
\
B
w(t, s) dt ds, q . Then equation (1) reduces to the differential-integral equation
D
xz(x, y) = f x, y,
0
\
−b0
b
\
−b