POLONICI MATHEMATICI LXIV.1 (1996)
Convergence results for unbounded solutions of first order non-linear differential-functional equations
by Henryk Leszczy´ nski (Gda´ nsk)
Abstract. We consider the Cauchy problem in an unbounded region for equations of the type either D
tz(t, x) = f (t, x, z(t, x), z
(t,x), D
xz(t, x)) or D
tz(t, x) = f (t, x, z(t, x), z, D
xz(t, x)). We prove convergence of their difference analogues by means of recurrence inequalities in some wide classes of unbounded functions.
Introduction. Basic uniqueness results for first order differential equa- tions were proved by Szarski [8], and then generalized by Kamont [3], Besala [1] and others. Let us mention [6] where the case of differential-functional equations was treated.
Uniqueness, existence and convergence results for parabolic equations require some assumptions on the class of solutions, namely one ought to assume that the solutions and their derivatives grow at most as exp ckxk
2(see [4]). The convergence of difference schemes was proved first locally, next in the unbounded case for differential problems [2], and finally for differential-functional systems using a special type of difference operators [7], and with general difference analogues consistent with the differential- functional problem [5].
We extend general methods of proving convergence by means of dif- ference inequalities described in [8] and in the references mentioned there.
Working in wide functional classes (see [6]), we prove recurrence estimates in a way similar to that used for parabolic equations. We deal simultaneously with two main types of functional dependence: first, with the variable z
(t,x)as an extension of retardations and integrations over a rectangular bounded left-side neighbourhood of the point (t, x), and secondly, with z appearing as variable in a function of the Volterra type. These two quite general models of functional dependence coincide in classes of bounded solutions; however, if we investigate unbounded functions, then two slightly different sets of as-
1991 Mathematics Subject Classification: Primary 35A35; Secondary 35B35.
Key words and phrases : error estimates, recurrence inequalities, difference scheme.
[1]
sumptions imply uniqueness and convergence as shown in [6]. Finally, we remark that our results, formulated for one equation, can easily be proved for weakly coupled systems.
1. Basic notations and formulation of the first differential-func- tional problem. Let E
0= [−τ
0, 0] × R
n, E = [0, a] × R
nand D = [−τ
0, 0] × [−τ, τ ], where τ
0∈ R
+, a > 0 and τ = (τ
1, . . . , τ
n) ∈ R
n+. If z ∈ C(E
0∪ E, R) and (t, x) ∈ E, then z
(t,x): D → R is defined by z
(t,x)(t, x) = z(t + t, x + x) for (t, x) ∈ D.
A function H is of class H iff H ∈ C(E
0∪ E, (0, ∞)), the functions H
|E0and H
|Eare continuously differentiable, and (i) H(·, x) is non-decreasing for every x ∈ R
n,
(ii) x
iD
xiH(t, x) ≥ 0 for i = 1, . . . , n and (t, x) ∈ E
0∪ E, where x = (x
1, . . . , x
n).
If H ∈ H, then a function ε is said to be of class E
Hiff ε ∈ C(E
0∪ E, (0, ∞)), ε(t, x) ≤ ε(t, x) for kxk ≥ kxk, 0 ≤ t, t ≤ a, and the function e H defined by e H(t, x) = H(t, x)ε(t, x) for (t, x) ∈ E
0∪ E belongs to H.
A function z is of class C
H(resp. C
H,ε) iff z ∈ C(E
0∪ E, R) and
|z(t, x)|/H(t, x) (resp. |z(t, x)|/ e H (t, x), where e H(t, x) = H(t, x)ε(t, x)) is bounded.
The classes C
Hand C
H,εare equipped with the seminorms k · k
H(t) defined by
(1.1) kzk
H(t) = sup{|z(t, x)|/H(t, x) | (t, x) ∈ E
0∪ E, t ≤ t}
for (t, x) ∈ (0, a] and z ∈ C
H. The space C(D, R) is equipped with the maximum norm k · k
D. Denote by Ω
(0)and Ω
H(1)the sets
Ω
(0)= E × R × C(D, R) × R
nand Ω
H(1)= E × R × C
H× R
n. If L ∈ C(E
0∪ E, R
+) and L
1, L
2∈ R
+, then a function f is of class Lip(Ω
(0); L, L
1, L
2) iff f ∈ C(Ω
(0), R) and
(1.2) |f (t, x, p, w, q) − f (t, x, p, w, q)|
≤ L
1|p − p| + L(t, x)kw − wk
D+ L
2kq − qk for all (t, x, p, w, q), (t, x, p, w, q) ∈ Ω
(0).
Now, let f ∈ C(Ω
(0), R). We consider the differential-functional equation (1.3) D
tz(t, x) = f (t, x, z(t, x), z
(t,x), D
xz(t, x)),
where D
xz(t, x) = (D
x1z(t, x), . . . , D
xnz(t, x)), with the initial condition (1.4) z(t, x) = φ(t, x), (t, x) ∈ E
0,
where φ ∈ C(E
0, R).
We assume throughout the paper that the Cauchy problem (1.3), (1.4) has a unique solution of class C
Hdefined on E
0∪ E (see [6]).
Let Γ : R
+→ R
+and κ, ψ : [−τ
0, a] → (0, ∞) be continuously dif- ferentiable functions such that κ
′(t) ≥ 0 and ψ
′(t) ≥ 0 for t ∈ [0, a];
κ
′(t) = ψ
′(t) = 0 for t ∈ [−τ
0, 0]; and Γ
′(t) ≥ 0 for t ∈ R
+. We define
(1.5) H(t, x) = Γ (ψ(t) p
1 + kxk
2) for (t, x) ∈ E
0∪ E, and
(1.6) L(t, x) = p/Γ (κ(t) p
1 + kxk
2)
for (t, x) ∈ E, where p ∈ R
+. Some conditions, assumed in [6], on the functions Γ, κ, ψ imply uniqueness for problem (1.3), (1.4) as well as for the Cauchy problem with another type of functional dependence. The properties of these functions assumed in the present paper are very close to those in [6].
Example. Let us list a few examples of functions Γ appearing in (1.5), (1.6):
(i) Γ (t) = t
kfor t ∈ R
+and k > 0.
(ii) Γ (t) = exp t
mfor t ∈ R and m ∈ N.
(iii) Γ (t) = e
l(t) for t ∈ R
+and l ∈ N, where e
0(t) = t and e
l+1(t) = exp(e
l(t)) for l = 0, 1, . . . and t ∈ R.
(iv) Γ (t) = Γ
T(t) = exp(
Tt
0
Γ (r + 1) dr) for t ∈ R, where e e Γ (r) = (r − n)e
n+1(n + 1) + (n + 1 − r)e
n(n) for r ∈ [n, n + 1] and n = 0, 1, . . . This example shows that Γ can grow faster than all e
lfor l = 0, 1, . . . It is possible to construct still faster growing functions:
(v) Γ (t) = Γ
T,i(t) for t ∈ R
+and i = 0, 1, . . . , where Γ
T,0(t) = Γ
T(t) and Γ
T,k+1(t) = exp(Γ
T,k(t)) for k = 0, 1, . . . ; and next:
(vi) Γ (t) = Γ
TT(t) = exp(
Tt
0
Γ (r + 1) dr) for t ∈ R e
+, where e Γ (r) = (r − l)Γ
T,l+1(l + 1) + (l + 1 − r)Γ
T,l(l) for r ∈ [l, l + 1] and l = 0, 1, . . . , and so on.
(vii) If ξ ∈ C(R, R) and
ω(t) = exp(− max{0, (1 − t)
−1})
∞\1
e
−ss
−3/2(s − 1)
−1/2ds
−1for t ∈ R, then
Γ (t) = C
0exp
∞\−∞
s∈[−r−1,r+1]
max |ξ(s)|ω( p
1 + t
2− r) dr
, t ∈ R
+,
where C
0= 1 + max{|ξ(r)| | r ∈ [−1, 1]}, is differentiable and satisfies
Γ (|t|) ≥ |ξ(t)| for t ∈ R.
3. Formulation of the difference problem and consistency lem- mas. Let h = (h
0, h
′) ∈ (0, ∞)
1+n, where h
′= (h
1, . . . , h
n), and I
dbe a non-empty subset of
h = (h
0, h
′) ∈ R
1+n+h
′= (h
1, . . . , h
n); h
i∈ (0, h
i], i = 1, . . . , n;
h
0N
0= τ
0for some N
0∈ {0, 1, . . .}
. Let (t, x)
(η)= (t
(η), x
(η)), where t
(η)= h
0η
0and x
(η)= (h
1η
1, . . . , h
nη
n) for η = (η
0, η
′) ∈ Z
1+nand η = (η
1, . . . , η
n). For h = (h
0, h
′) ∈ I
dthere is a natural constant N
∗and N
1, . . . , N
n∈ Z
+such that N
∗h
0≤ a < (N
∗+ 1)h
0and τ
i≤ N
ih
i< τ
i+ h
ifor i = 1, . . . , n.
Let h = (h
0, h
′) ∈ I
d. Then we define
E
0,h= {(t, x)
(η)| η = (η
0, η
′) ∈ Z
1+n, η
0∈ {−N
0, . . . , 0}}, E
h= {(t, x)
(η)| η = (η
0, η
′) ∈ Z
1+n, η
0∈ {0, . . . , N
∗− 1}}, E b
h= {(t, x)
(η)| η = (η
0, η
′) ∈ Z
1+n, η
0∈ {−N
0, . . . , N
∗}}, D
h=
(t, x)
(η)η = (η
0, η
′) ∈ Z
1+n, η
0∈ {−N
0, . . . , 0};
η
′= (η
1, . . . , η
n); |η
i| ≤ N
i, i = 1, . . . , n
.
Let z
hbe a function defined on b E
h. Set z
h(η)= z
h(t
(η), x
(η)) for (t, x)
(η)∈ b E
h. If (t, x)
(η)∈ E
hand z
his a function defined on b E
h, then the function (z
h)
(η): D
h→ R is defined by
(z
h)
(η)((t, x)
(η)) = z
h((t, x)
(η+η)) for (t, x)
(η)∈ D
h.
Denote by F( b E
h, R) the set of all functions from b E
hto R. If z ∈ C(E
0∪E, R), then z
h∈ F( b E
h, R) denotes the restriction of z to the mesh.
Let λ ∈ {1, 2, . . .} and S
λ=
s = (s
1, . . . , s
n) ∈ Z
n| |s
i| ≤ λ, i = 1, . . . , n}.
We define the difference operators A, ∆
0and ∆ = (∆
1, . . . , ∆
n) by
(2.1)
Az
h(η)= X
s∈Sλ
a
(η)sz
h(η0,η′+s), (t, x)
(η)∈ E
h,
∆
0z
h(η)= h
−10(z
h(η0+1,η′)− Az
(η)h), (t, x)
(η)∈ E
h,
∆
lz
h(η)= h
−1lX
s∈Sλ
b
(η)s,lz
h(η0,η′+s), (t, x)
(η)∈ E
h, l = 1, . . . , n,
where a
(η)sand b
(η)s,lare real coefficients.
Assumption H
1. Suppose that the discrete operators A and ∆ defined by (2.1) satisfy the following conditions
(i) P
s∈Sλ
a
(η)s= 1, P
s∈Sλ
a
(η)ss
l= 0 for l = 1, . . . , n and (t, x)
(η)∈ E
h.
(ii) P
s∈Sλ
b
(η)s,l= 0, P
s∈Sλ
b
(η)s,ls
j= δ
ljfor j, l = 1, . . . , n and (t, x)
(η)∈ E
h.
(iii) there is c ∈ R
+such that |a
(η)s|, |b
(η)s,l| ≤ c for s ∈ S
λ, (t, x)
(η)∈ E
h, l = 1, . . . , n and h ∈ I
d.
(iv) there are constants c
0, c
1∈ (0, ∞) such that c
0h
0≤ h
l≤ c
1h
0for l = 1, . . . , n.
This assumption is necessary to prove that the discrete operators ∆
lfor l = 0, 1, . . . , n approximate the differential operators D
tand D
xlfor l = 1, . . . , n.
The operator [·]
h: F( b E
h, R) → C(E
0∪ E, R) is defined by (2.2) [z
h]
h(t, x) =
X
0 r=−λX
s∈Sλ
p
(η)r,s(t, x)z
(ηh0+r,η′+s)for z
h: b E
h→ R, t
(η−1)< t ≤ t
(η)and x
(η−1)< x ≤ x
(η), where (t, x)
(η−1), (t, x)
(η)∈ b E
hand 1 = (1, . . . , 1) ∈ Z
1+n. If (t, x)
(η0+r,η′+s)6∈ b E
h, then z
h(η0+r,η′+s)means the same as z
(−Nh 0,η′+s).
Assumption H
2. Suppose that the functions p
(η)r,s: R
1+n→ R in formula (2.2) are bounded and satisfy
(2.3)
X
0 r=−λX
s∈Sλ
p
(η)r,s(t, x) = 1 for t
(η−1)< t ≤ t
(η)and x
(η−1)< x ≤ x
(η), and (2.4) p = sup b
X
0 r=−λX
s∈Sλ
|p
(η)r,s(t, x)|
t
(η−1)< t ≤ t
(η), x
(η−1)< x ≤ x
(η), h ∈ I
d,
(t, x)
(η−1), (t, x)
(η)∈ b E
h.
< ∞ Now, we define difference schemes which correspond to the differential- functional problem (1.3), (1.4). Let f ∈ C
H(Ω
(0)). Then we consider the following difference-functional problem (cf. (1.3)):
(2.5) ∆
0z
h(η)= f ((t, x)
(η), z
h(η), ([z
h]
h)
(η), ∆z
(η)h), (t, x) ∈ E
h, with the initial condition
(2.6) z
h(η)= φ
h(η), (t, x) ∈ E
0,h.
In the literature the function f is often replaced by a function f
hwhich is defined by use of a finite Taylor expansion of f . This way one can obtain better difference approximations to the differential problem.
Let H ∈ H and ε ∈ E
H. A function z
his of class F
H(h)(resp. of class
F
H,ε(h)) iff there is a function z ∈ C
H(resp. z ∈ C
H,ε) such that z
h= z
h,
where h ∈ I
d.
Now, we prove an auxiliary lemma on consistency.
Lemma 2.1. Suppose that H ∈ H and Assumptions H
1and H
2are sat- isfied. Let f ∈ C
H(Ω
(0)) with constants L
0, L
1, L
2∈ R
+and L ∈ C(E, R
+).
Let u ∈ C(E
0∪ E, R) be a solution of (1.3), (1.4) such that u, D
tu, D
xlu ∈ C
Hfor l = 1, . . . , n, and there is L
u∈ R
+such that
(2.7) |D
xlu(t, x) − D
xlu(t, x)| + |D
tu(t, x) − D
tu(t, x)|
≤ L
u|t − t| + X
n j=1|x
j− x
j|
max{H(t, x), H(t, x)}
for (t, x), (t, x) ∈ E and l = 1, . . . , n. Then
(2.8) |∆
0u
(η)h− f ((t, x)
(η), u
(η)h, ([u
h]
h)
(η), ∆z
h(η))| ≤ µ
(η)h, where
µ
(η)h= h
0L
uH
h(η0+1,η′)+ h
−10baλ
2X
nl=1
h
l 2L
uH
h(η0,|η′|+λ′)(2.9)
+ L((t, x)
(η))H
h(η0,|η′|+λ′+1′)p b
×
h
0λkD
tuk
H(a) + (λ + 1) X
nl=1
h
lkD
xluk
H(a)
+ L
2k(h
−11, . . . , h
−1n)kbb X
nl=1
h
l2λ
2L
uH
h(η0,|η′|+λ′), where λ
′= (λ, . . . , λ) ∈ Z
n, 1
′= (1, . . . , 1) ∈ Z
n, and
(2.10) ba = sup
η
X
s∈Sλ\{0′}
|a
(η)s|, bb = sup
l,η
X
s∈Sλ\{0′}
|b
(η)s,l|.
P r o o f. First, using the mean value theorem we have (2.11) |D
tu((t, x)
(η)) − ∆
0u
(η)h|
= D
tu((t, x)
(η)) − h
−10n
u
(η)h+ h
0D
tu((t, x)
(η)+ θ
0(η)(h))
− X
s∈Sλ
a
(η)su
(η)h+
X
n l=1h
ls
lD
xlu((t, x)
(η)+ θ
s(η)(h)) o for (t, x)
(η)∈ E
h, where
(2.12) θ
0(η)(h) ∈ (0, h
0) × {0
′},
θ
s(η)(h) ∈ {0} × [−λh
1, λh
1] × . . . × [−λh
n, λh
n]
for s ∈ S
λ.
From (2.7), (2.11) and Assumption H
1we obtain (2.13) |D
tu((t, x)
(η)) − ∆
0u
(η)h|
≤ L
uh
0H
h(η0+1,η′)+ h
−10baλ
2X
nl=1
h
l 2L
uH
h(η0,|η′|+λ′), where ba is defined by (2.10). In a similar way we obtain
(2.14) |D
xlu((t, x)
(η)) − ∆
lu
(η)h|
= D
xlu((t, x)
(η)) − h
−1lX
s∈Sλ
b
(η)s,lu
(η)h+
X
n l=1h
ls
lD
xlu((t, x)
(η)+ θ
(η)s(h))
for (t, x)
(η)∈ E
hand l = 1, . . . , n, and thus (2.15) |D
xlu((t, x)
(η)) − ∆
lu
(η)h| ≤ h
−1lbb X
nl=1
h
l2λ
2L
uH
h(η0,|η′|+λ′)for (t, x)
(η)∈ E
hand l = 1, . . . , n. If (t, x) ∈ E
0∪ E and t
(η−1)< t ≤ t
(η), x
(η−1)< x ≤ x
(η), where (t, x)
(η−1), (t, x)
(η)∈ b E
h, then
(2.16) |[u
h]
h(t, x) − u(t, x)|
= X
0 r=−λX
s∈Sλ
p
(η)r,s(t, x)
u(t, x) + (t
(η0+r,η′+s)− t)D
tu(θ
(η)r,s(t, x, h))
+ X
n l=1(x
(η0+r,η′+s)− x
l)D
xlu(θ
r,s(η)(t, x, h))
− u(t, x)
≤ b p
kD
tuk
H(a) λh
0+ X
n l=1h
l(λ + 1)kD
xluk
H(a)
H
h(η0,|η′|+λ′),
where b p is defined by (2.4) and θ
r,s(η)(t, x, h) is an intermediate point between (t, x)
(η0+r,η′+s)and (t, x).
Let (t, x) ∈ E. From (2.16) we have (2.17) k([u
h]
h)
(t,x)− u
(t,x)k
D= max
(t,x)∈D
k[u
h]
h(t + t, x + x) − u(t + t, x + x)k
≤ b p
kD
tuk
H(a)λh
0+ X
nl=1
h
l(λ + 1)kD
xluk
H(a)
H
(e
η0,|e
η′|+λ′)h
,
where (e η
l− 1)h
l< x
l≤ e η
lh
lfor l = 0, . . . , n. Condition (2.8) follows from
(2.13), (2.15) and (2.17). This finishes the proof.
3. Convergence theorem. In this section we will prove that natural assumptions imply the convergence of the difference scheme (2.5), (2.6).
Theorem 3.1. Suppose that
1) H ∈ H is given by (1.5), and L is defined by (1.6), where κ(t) >
ψ(t) > 0 for t ∈ [−τ
0, a], ψ, κ are increasing on [0, a], and f ∈ C
H(Ω
(0)) with L
1, L
2∈ R
+,
2) u ∈ C(E
0∪E, R) is a solution of (1.3), (1.4) such that u, D
tu, D
xlu ∈ C
Hfor l = 1, . . . , n, and there is L
u∈ R
+such that condition (2.7) is satisfied ,
3) v
h∈ F( b E
h, R) is a solution of (2.5), (2.6), 4) the following monotonicity condition is satisfied:
(3.1) a
(η)s+ h
0X
nl=1
h
−1lb
(η)s,lD
qlf (P
(η)) ≥ 0
for s ∈ S
λ, (t, x)
(η)∈ E
hand P
(η)= ((t, x)
(η), p, w, q) ∈ Ω
(0)H, 5) there is M
φ∈ R
+such that
(3.2) |φ
(η)h− φ
(η)h| ≤ h
0M
φH
h(η), (t, x)
(η)∈ E
0,h. Then
(3.3) |v
h(η)− u
(η)h| ≤ h
0Ψ (t
(η))H
h(η), (t, x)
(η)∈ b E
h, where Ψ : [−τ
0, a] → R satisfies
(3.4) Ψ (t) ≥ M
φ, t ∈ [−τ
0, 0], (3.5) Ψ
′(t) − L
u− B
0(h) − L
1Ψ (t)
− B
1(h, ψ(t))Γ (ψ(t) p
1 + (r
1(t) + kτ k + kh
′k(λ + 1))
2)
× (Γ (Ψ (t)e θ)Γ (κ(t)e θ))
−1≥ 0 for t ∈ [0, a] and h ∈ I
d, where e θ = p
1 + (r
0(t, h))
2, and B
0and B
1are defined by
B
0(h) = λ
2L
uh
−10X
nl=1
h
l 2(ba + bbL
2h
0k(h
−11, . . . , h
−1n)k), (3.6)
B
1(h, p) = p p b
p + λkD
tuk
H(a) + (λ + 1) X
n l=1h
lh
−10kD
xluk
H(a) (3.7)
for p ∈ R
+, and
(3.8)
r
0(t, h) = 1 + p
1 + (λkh
′k/h
0)
2ψ
′(t)/ψ(t) λkh
′k/h
0, r
1(t) = ψ(t)(kτ k + kh
′k(λ + 1))
κ(t) − ψ(t) . Moreover , we have
(3.9) Ψ
′(t) − L
1− L
u+ (Ψ (t) − h
0L
u)ψ
′(t)Γ
′(ψ(t))/Γ (ψ(t))
− B
0(h)Γ (ψ(t) p
1 + (kh
′kλ/h
0+ r
0(t, h))
2)/Γ (t)
− B
1(h, Ψ (t))Γ ψ(t)
q
1 + (r
0(t, h) + kτ k + kh
′k(1 + λ))
2× (Γ (κ(t))Γ (ψ(t)))
−1≥ 0
for t ∈ [0, a] with Ψ (t) so large that Ψ (t) − h
0L
u≥ 0.
P r o o f. Let w
(η)h= u
(η)h− v
h(η)for (t, x)
(η)∈ b E
h. From (2.5) we obtain the recurrence equality
w
h(η0+1,η′)= Aw
(η)h(3.10)
+ h
0(f ((t, x)
(η), u
(η)h, ([u
h]
h)
(η), ∆u
(η)h)
− f ((t, x)
(η), u
(η)h, ([u
h]
h)
(η), ∆v
h(η))) + h
0(f ((t, x)
(η), u
(η)h, ([u
h]
h)
(η), ∆v
(η)h)
− f ((t, x)
(η), v
h(η), ([v
h]
h)
(η), ∆v
h(η)))
+ h
0(∆
0u
(η)h− f ((t, x)
(η), u
(η)h, ([u
h]
h)
(η), ∆u
(η)h)) for (t, x)
(η)∈ E
h. Using the mean value theorem, the Lipschitz condition for f , and Lemma 2.1, we obtain the recurrence estimate
|w
h(η0+1,η′)| ≤ X
s∈Sλ
|w
(ηh0,η′+s)| a
(η)s+ h
0X
n l=1h
−1lb
(η)s,lD
qlf (P
(η)) (3.11)
+ h
0L
1|w
(η)h| + h
0L((t, x)
(η)) k([w
h]
h)
(η)k
D+ h
0µ
(η)hfor (t, x)
(η)∈ E
h. Now, using (3.1) and Assumption H
1, we easily obtain from (3.11) the following inequality which is much easier to analyse:
|w
h(η0+1,η′)| ≤ max
s∈Sλ
|w
(ηh0,η′+s)| + h
0L
1|w
(η)h| (3.12)
+ h
0L((t, x)
(η))k([w
h]
h)
(η)k
D+ h
0µ
(η)hfor (t, x)
(η)∈ E
h. If we prove that the function W
h(η)= h
0Ψ (t
(η))H
h(η),
(t, x)
(η)∈ b E
h, satisfies a comparison inequality with respect to (3.12), then
(3.3) will be established. Thus, in order to finish the proof of our theorem it is enough to prove the following
Lemma 3.1. If the assumptions of Theorem 3.1 are satisfied, then W
h(η0+1,η′)≥ W
h(η0,|η′|+λ′)+ h
0L
1W
h(η)(3.13)
+ h
0L((t, x)
(η))k([W
h]
h)
(η)k
D+ h
0µ
(η)hfor (t, x)
(η)∈ E
h, and
(3.14) W
h(η)≥ h
0M
φH
h(η), (t, x)
(η)∈ E
0,h.
P r o o f. Condition (3.14) follows immediately from (3.4) and (3.2). Con- dition (3.13) is a consequence of
(3.15) (Ψ (t + h
0) − h
0L
u)Γ (ψ(t + h
0) p 1 + r
2)
≥ (Ψ (t) + h
0B
0(h))Γ (ψ(t) p
1 + (r + kh
′kλ)
2) + h
0B
1(h, Ψ (t))Γ (ψ(t) p
1 + (r + kτ k + kh
′k(λ + 1))
2)
× (Γ (ψ(t) p
1 + r
2))
−1for t ∈ [0, a − h
0] and r = kxk ∈ R
+, where B
0(h) and B
1(h, Ψ (t)) are given by (3.6) and (3.7). This implication follows from (3.9).
If r is greater than r
0(t, h) given by (3.8), then (3.16) ψ(t + h
0) p
1 + r
2≥ ψ(t) p
1 + (r + kh
′kλ)
2, and (3.15) follows from
Ξ(θ) := Ψ (t + θ) − θL
u− Ψ (t) − θB
0(h) − θL
1Ψ (t) (3.17)
− θB
1(h, Ψ (t))Γ (ψ(t) p
1 + (r
1(t) + kτ k + kh
′k(λ + 1))
2)
× (Γ (ψ(t)e θ)Γ (κ(t)e θ))
−1≥ 0,
where θ ∈ [0, h
0], t ∈ [0, a − h
0] and e θ is the same as in (3.5). Now, (3.17) holds true because Ξ(0) = 0 and Ξ
′(θ) ≥ 0 for θ ∈ [0, h
0] as we have (3.5).
For r ≤ r
0(x
0, h) and t ∈ [0, a − h
0], formula (3.15) is a consequence of the inequality Ξ
1(θ; t, r) ≥ 0 for θ ∈ [0, h
0], where
(3.18) Ξ
1(θ; t, r)
:= (Ψ (t + θ) − θL
u)Γ (ψ(t + θ) p
1 + r
2) − θL
1Γ (ψ(t) p 1 + r
2)
− (Ψ (t) + θB
0(h))Γ (ψ(t) p
1 + (r + θλkh
′k/h
0)
2)
− θB
1(h, Ψ (t))Γ ψ(t)
q
1 + (r
0(t, h) + kτ k + kh
′k(λ + 1))
2Γ (κ(t))
for θ ∈ [0, h
0] and t ∈ [0, a − h
0]. From (3.18) we find a lower estimate of Ξ
1′(θ; t, h):
(3.19) Ξ
1′(θ; t, h)
≥ (Ψ
′(t + θ) − L
u)Γ (ψ(t + θ) p 1 + r
2) + (Ψ (t + θ) − θL
u)Γ
′(ψ(t + θ) p
1 + r
2)ψ
′(t + θ) p 1 + r
2− L
1Γ (ψ(t) p
1 + r
2) − B
0(h)Γ (ψ(t) p
1 + (r + kh
′kλ)
2)
− (Ψ (t) + θB
0(h))Γ
′(ψ(t) p
1 + (r + λkH
′k)
2)ψ
′(t)
− B
1(h, Ψ (t))Γ ψ(t)
q
1 + (r
0(t, h) + kτ k + h
′k(λ + 1))
2Γ (κ(t)).
From (3.19) and (3.9) we obtain Ξ
′(θ; t, h) ≥ 0, and (3.18) implies Ξ(0; t, h) = 0. Therefore, Ξ(h
0; t, h) ≥ 0 for t ∈ [0, a − h
0]. This completes the proof.
R e m a r k. Condition 1) of Theorem 3.1 can be much weaker, namely L might be defined by
L(t, x) = Γ (ψ(t) p
1 + kxk
2)/Γ (κ(t) p
1 + kxk
2), (t, x) ∈ E.
In this case the function Ψ satisfies stronger conditions than (3.4), (3.5) and (3.9).
If Γ (t) ≥ const e
2(t), then L might be defined by L(t, x) = (Γ (ψ(t) p
1 + kxk
2))
ν/Γ (κ(t) p
1 + kxk
2), (t, x) ∈ E, where ν ≥ 0, and the proof works for a sufficiently large function Ψ .
4. Convergence result for another functional dependence. If L, L
1, L
2∈ R
+, then f is of class Lip(Ω
(1)H; L, L
1, L
2) iff f ∈ C(Ω
H(1), R) and (4.1) |f (t, x, p, w, q) − f (t, x, p, w, q)|
≤ L
1|p − p| + LH(t, x)kw − wk
H(t) + L
2kq − qk for all (t, x, p, w, q), (t, x, p, w, q) ∈ Ω
(1)H. For f ∈ C(Ω
H(1), R) we consider the Cauchy problem for the equation
(4.2) D
tz(t, x) = f (t, x, z(t, x), z, D
xz(t, x)).
We assume that the Cauchy problem (4.2), (1.4) has a unique solution of class C
Hdefined on E
0∪ E.
Let f ∈ C
H(Ω
H(1)). Then the difference analogue of (4.2) reads
(4.3) ∆
0z
(η)h= f ((t, x)
(η), z
h(η), [z
h]
h, ∆z
h(η)), (t, x)
(η)∈ E
h.
Problem (4.3) is also considered with initial condition (2.6).
Lemma 4.1. Suppose that Assumptions H
1and H
2are satisfied and H ∈ H, ε ∈ E
H. Let f ∈ C
H(Ω
H(1)) with constants L, L
0, L
1, L
2∈ R
+. Let u ∈ C(E
0∪ E, R) be a solution of (4.2), (1.4) such that u, D
tu, D
xlu ∈ C
H,εfor l = 1, . . . , n, and there is L
u∈ R
+such that
(4.4) |D
xl(t, x) − D
xlu(t, x)| + |D
tu(t, x) − D
tu(t, x)|
≤ L
u|t − t| + X
n j=1|x
j− x
j|
max{H(t, x)ε(t, x), H(t, x)ε(t, x)}
for (t, x), (t, x) ∈ E and l = 1, . . . , n. Assume also that for h ∈ I
dand s ∈ S
λ+1there is R(h) ∈ R
+such that lim sup{R(h) | h ∈ I
d} < ∞, and (4.5)
X
n l=1x
lD
xl( e H(t + h
0, x + (x
(s)))/H(t, x)) ≤ 0 for kxk ≥ R(h) and (t + h
0, x) ∈ E, where
(4.6) H(t, x) = H(t, x)ε(t, x), e (t, x) ∈ E.
Then
(4.7) |∆
0u
(η)h− f ((t, x)
(η), u
(η)h, [u
h]
h, ∆u
(η)h)| ≤ µ
(η)h, where
µ
(η)h= h
0L
uH e
h(η0+1,η′)+ h
−10baλ
2X
nl=1
h
l 2L
uH e
h(η0,|η′|+λ′)(4.8)
+ L e H
h(η)p b
h
0λkD
tuk
H(a) + (λ + 1) X
nl=1
h
lkD
xluk
H(a)
× sup
η
sup
x(η−1)<x≤x(η)
H e
h(η0,|η′|+λ′)H(t, x) + L
2k(h
−11, . . . , h
−1n)kbb X
nl=1
h
l 2λ
2L
uH e
h(η0,|η′|+λ′),
where (t, x)
(η)∈ E
h, and e H is defined by (4.6).
R e m a r k. Condition (4.5) is satisfied if, for example, H is defined by (1.5), and
(4.9) ε(t, x) = Γ (ξ(t) p
1 + kxk
2)/Γ (ψ(t) p
1 + kxk
2),
where ξ ∈ C
1([−τ
0, a], R
+) is increasing, and 0 < ξ(t) < ψ(t) for t ∈ [−τ
0, a].
We should only assume that h
0< (ψ(t) − ξ(t))/ξ
′(a) for t ∈ [−τ
0, a].
P r o o f o f L e m m a 4.1. Estimates (2.13)–(2.16) remain true if we replace H by e H defined by (4.6). Thus, in order to obtain (4.7) with µ
hdefined by (4.8), it is sufficient to estimate LH
h(η)k[u
h]
h− uk
H(a) by a suit- ably chosen expression basing on formula (2.16) with H replaced by e H. The double supremum appearing in (4.8) is finite because (4.5) guarantees that H e
h(η0,|η′|+λ′)/H(t, x) is bounded by
(4.10) M
h= sup
H(t + h e
0, x + (x
(s)))/H(t, x)
kxk ≤ R(h), (t + h
0, x) ∈ E, s ∈ S
λ+1
, where h ∈ I
d. The rest works as in the proof of Lemma 2.1.
Theorem 4.1. Suppose that
1) Assumptions H
1and H
2are satisfied ,
2) H ∈ H and ε ∈ E
Hare given by (1.5) and (4.9), respectively, where ξ ∈ C([−τ
0, a], R
+) has positive derivative on [0, a]; 0 < ξ(t) < ψ(t) for t ∈ [−τ
0, a]; e H ∈ H is defined by (4.6), and f ∈ C
H(Ω
H(1)) with constants L, L
1, L
2,
3) u ∈ C(E
0∪E, R) is a solution of (4.2), (1.4) such that u, D
tu, D
xlu ∈ C
H,εfor l = 1, . . . , n, and there exists L
u∈ R
+such that condition (4.4) is satisfied , and the constants M
hdefined by (4.10) for h ∈ I
dare such that lim sup{M
h| h ∈ I
d} ≤ M ,
4) v
h∈ F
H,ε(h)is a solution of (4.3), (2.6),
5) the monotonicity condition (3.1) is satisfied for s ∈ S
λ, (t, x)
(η)∈ E
hand P
(η)= ((t, x)
(η), p, z, q) ∈ Ω
H(1), 6) there is M
φ∈ R such that
(4.11) |φ
(η)h− φ
h(η)| ≤ h
0M
φH e
h(η), (t, x)
(η)∈ E
h, 7) h
0is so small that ψ(t) − ξ(t) − h
0ξ
′(t) > 0 for t ∈ [0, a].
Then
(4.12) |v
h(η)− u
(η)h| ≤ h
0Ψ (t
(η)) e H
h(η), (t, x)
(η)∈ b E
h, where Ψ : [−τ
0, a] → R
+satisfies inequality (3.4) and
(4.13) Ψ
′(θ) − L
u− L
1− e B
0(h)
− e B(h, Ψ (θ))Γ (ξ(θ) p
1 + (r
1(t))
2)/Γ (ψ(0)) ≥ 0
for θ ∈ [0, a], where e B
0(h), r
1(t) and e B(h, p) for p ∈ R
+are defined by
(4.14)
B e
0(h) = λ
2L
uX
nl=1
h
l/h
0 2(ba + L
2bbk(h
0h
−11, . . . , h
0h
−1n)k),
B(h, p) = Lb e p
λkD
tuk
H(a) + (λ + 1) X
nl=1
h
lh
−10kD
xluk
H(a) + p , r
1(t) = (λ + 1)kh
′k(1 + ψ(t)(ψ(t) − ξ(t) − h
0ξ
′(t))
−1),
where θ, t ∈ [0, a], h ∈ I
d, and
(4.15) (Ψ
′(θ) − L
u)Γ (ξ(θ)) + Γ
′(ξ(θ))ξ
′(θ)(Ψ (θ) − h
0L
u)
− Γ (ξ(θ) p
1 + (r
0(t))
2)
× (L
1Ψ (θ) + e B(h, Ψ (t))Γ (ξ(θ) p
1 + (r
1(θ))
2)/Γ (ψ(0)))
− (Ψ (θ) + h
0B e
0(h))
× Γ
′(ξ(θ) p
1 + (r
0(θ) + λkh
′k)
2)ξ(θ) λkh
′k/h
0− e B
0(h)Γ (ξ(θ) p
1 + (r
0(θ) + λkh
′k)
2) ≥ 0
for θ ∈ [0, a], h ∈ I
dwith h
0so small that Ψ (θ) − h
0L
u≥ 0 on [−τ
0, a].
P r o o f. Let w
h(η)= u
(η)h− v
h(η)for (t, x)
(η)∈ b E
h. From (4.11) it fol- lows that (4.12) is satisfied for (t, x)
(η)∈ E
0,h. Subtracting the recurrence expressions of v
(ηh0+1,η′)from u
(ηh0+1,η′)leads to the recurrence error esti- mates similar to (3.12), as in the proof of Theorem 3.1 (compare (3.10) and (3.11)):
|w
h(η+1,η′)| ≤ max
s∈Sλ