Anna Szafrańska
Implicit difference methods for infinite systems of hyperbolic functional differential equations
Abstract. The paper deal with classical solutions of initial boundary value pro- blems for infinite systems of nonlinear differential functional equations. Two types of difference schemes are constructed. First we show that solutions of our differential problem can be approximated by solutions of infinite difference functional schemes.
In the second part of the paper we proof that solutions of finite difference systems approximate the solutions of aur differential problem.
We give a complete convergence analysis for both types of difference methods. We adopt nonlinear estimates of the Perron type for given functions with respect to the functional variable. The proof of the stability is based on the comparison technique.
Numerical examples are presented.
2000 Mathematics Subject Classification: 65M12, 65M15, 35R10 .
Key words and phrases: initial boundary value problems, difference functional equ- ations, difference methods, stability and convergence, interpolating operators, nonli- near estimates of the Perron type.
1. Introduction. For any metric spaces X and Y , by C(X, Y ) we denote the class of all continuous functions from X into Y . We use vectorial inequalities, assu- ming that the same inequalities hold between their corresponding components. Let N and Z be the sets of natural numbers and integers respectively.
Let a > 0, d
0∈ R
+, b = (b
1, . . . , b
n) ∈ R
n+and d = (d
1, . . . , d
n) ∈ R
n+where R
+= [0, +∞). Suppose that κ, 0 ¬ κ ¬ n, is a fixed integer. For each x = (x
1, . . . , x
n) ∈ R
nwe write x = (x
0, x
00) where x
0= (x
1, . . . , x
κ), x
00= (x
κ+1, . . . , x
n). If κ = n we have x
0= x, if κ = 0 then x
00= x. We define the sets
E = [0, a] × [−b
0, b
0) × (−b
00, b
00], D = [−d
0, 0] × [0, d
0] × [−d
00, 0].
Let c = (c
1, . . . , c
n) = b + d and
E
0= [−d
0, 0] × [−b
0, c
0] × [−c
00, b
00],
∂
0E = ((0, a] × [−b
0, c
0] × [−c
00, b
00]) \ E, B = E
0∪ E ∪ ∂
0E.
Let l
∞be the class of all sequences p = {p
k}
k∈N, p
k∈ R such that kpk
∞= sup{|p
k| : k ∈ N} < ∞.
Suppose that z : B → l
∞and (t, x) ∈ [0, a] × [−b, b]. We define the function z
(t,x): D → l
∞as follows
z
(t,x)(s, y) = z(t + s, x + y), (s, y) ∈ D.
The function z
(t,x)is the restriction of z to the set [t−d
0, t] ×[x
0, x
0+d
0]×[x
00−d
00, x
00] and this restriction is shifted to the set D. If w ∈ C(D, l
∞) then we write
kwk
D= max {kw(t, x)k
∞: (t, x) ∈ D}.
We set Ω = E × C(D, l
∞) × R
nand assume that the functions
f = {f
k}
k∈N, f : Ω → l
∞, and ϕ = {ϕ
k}
k∈N, ϕ : E
0∪ ∂
0E → l
∞, are given. For a function z : B → l
∞, z = {z
k}
k∈N, and for a point (t, x) ∈ E we write
∂
tz(t, x) = n
∂
tz
k(t, x) o
k∈N
, f (t, x, z
(t,x), ∂
xz(t, x)) = n
f
k(t, x, z
(t,x), ∂
xz
k(t, x) o
k∈N
where ∂
xz
k= (∂
x1z
k, . . . , ∂
xnz
k). We consider the initial boundary problem (1) ∂
tz(t, x) = f (t, x, z
(t,x), ∂
xz(t, x))
(2) z(t, x) = ϕ(t, x) for (t, x) ∈ E
0∪ ∂
0E
We call a function v : B → l
∞, v = {v
k}
k∈N, a solution of problem (1), (2) if (i) v ∈ C(B, l
∞) and v is of class C
1on E,
(ii) v satisfies (1) on E and the condition (2) holds.
We deal with the numerical methods for (1), (2). Classical difference methods for partial differential or functional differential equations consist in replacing partial derivatives by difference expressions. Then the original problem is transformed into difference or difference functional equations. The main question in these conside- rations is to find a finite difference approximation which is stable. The method of difference inequalities or simple theorems on recurrent inequalities for functions of one variable are used in the investigations of the stability of difference schemes.
In recent years, a number of papers concerning above subjects have been publi- shed. It is not our purpose to show full bibliography relating to a difference methods for hyperbolic differential functional problems. We need to mention a few papers.
They are [1], [3], [4], [7].
Theorem on the local existence of generalized solutions of the infinite systems of nonlinear differential functional problems can be found in [2]. A general uniqu- eness result with nonlinear estimates of the Perron type for infinite systems with initial boundary value conditions is obtained in the paper [6]. The results concerning uniqueness of the solutions of infinite systems can be found also in [8], [9].
The paper is organized as follows. Section 2 deals with infinite system of func- tional difference equations generated by (1), (2). Constructed difference method is implicit with respect to the time variable. The theorems on the existence and uni- queness of considering method are proved. In Section 3 we consider the systems of functional differential equations which are finite. Components of the unknown func- tion z = {z
k}
k∈Nwith k > µ where µ ∈ N is fixed, are replaced by the respective components of an extension of the initial boundary function. For such differential problem we construct implicit difference scheme. The solutions of above difference method approximate the solutions of differential problem (1), (2) if the step of the mesh is tending to zero and if the number of equations used in this scheme is increasing to infinity. We proved that considering implicit difference method is co- nvergent. The last section of the paper contains results of the numerical examples.
2. Infinite systems of difference equations. We formulate a difference problem corresponding to (1), (2). Let us denote by F (X, Y ) the class of all functions defined on X and taking values in Y , where X and Y are arbitrary sets. For x ,¯x ∈ R
nx = (x
1, . . . , x
n), ¯x = (¯x
1, . . . , ¯ x
n), we write x ¯x = (x
1¯x
1, . . . , x
n¯x
n). We define a mesh on the set B in the following way. Let (h
0, h
0), h
0= (h
1, . . . , h
n), stand for steps of the mesh. For h = (h
0, h
0) and (r, m) ∈ Z
1+n, m = (m
1, . . . , m
n), we define nodal points as follows
t
(r)= rh
0, x
(m)= m h
0, x
(m)= (x
(m1 1), . . . , x
(mn n)).
By H we will denote the set of all h = (h
0, h
0) such that there are N = (N
1, . . . , N
n), N ∈ N
n, N
0∈ N with the properties: N h
0= d, N
0h
0= d
0. Let K ∈ N be defined by the relations Kh
0¬ a < (K + 1)h
0. We define the sets
R
h1+n= {(t
(r), x
(m)) : (r, m) ∈ Z
1+n},
E
h= E ∩ R
1+nh, E
h.0= E
0∩ R
1+nh, ∂
0E
h= ∂
0E ∩ R
1+nhand
B
h= E
0.h∪ E
h∪ ∂
0E
h, I
h= {t
(r): 0 ¬ r ¬ K}.
Write
B
h.r= B
h∩
[−d
0, t
(r)] × R
n, 0 ¬ r ¬ K, and
E
h0= {(t
(r), x
(m)) ∈ E
h: 0 ¬ r ¬ K − 1}.
For functions z : B
h→ l
∞and w : D
h→ l
∞we write
z
(r,m)= z(t
(r), x
(m)) on B
hand w
(r,m)= w(t
(r), x
(m)) on D
h.
For the above z and for a point (t
(r), x
(m)) ∈ E
hwe define the function z
[r,m]: D
h→ l
∞by the formula
z
[r,m](τ, y) = z(t
(r)+ τ, x
(m)+ y), (τ, y) ∈ D
h. The function z
[r,m]is the restriction of z to the set
([t
(r)− d
0] × [x
(m0), x
(m0)+ d
0] × [x
(m00)− d
00, x
(m00)]) ∩ R
1+nhand the restriction is shifted to the set D
h. For a function w : D
h→ l
∞we write kwk
Dh= max {kw
(r,m)k
∞: (t
(r), x
(m)) ∈ D
h}.
The equation (1) contain the functional variable z
(t,x)which is the element of the space C(D, l
∞). Therefore we need an interpolating operator T
h: F (D
h, l
∞) → C(D
h, l
∞) and the following assumptions on T
h.
Assumption H[T
h]. Suppose that the operator T
h: F (D
h, l
∞) → F (D, l
∞) satisfies the conditions
1) if w, e w ∈ F (D
h, l
∞) then T
h[w], T
h[ e w] ∈ C(D, l
∞) and (3) kT
h[w] − T
h[ e w] k
D¬ kw − e w k
Dh,
2) if w : D → l
∞is of class C
1then there is γ : H → R
+such that (4) k T
h[w
h] − w k
D¬ γ(h) and lim
h→0
γ(h) = 0, where w
his the restriction of w to the set D
h.
Remark 2.1 It is easy to see that the interpolating operator considered in [5]
(Chapter V) satisfies the Assumption H[T
h].
Let e
j= (0, . . . , 0, 1, 0, . . . , 0) ∈ R
n, 1 ¬ j ¬ n, where 1 is the j-th coordinate.
For w : B
h→ R and (t
(r), x
(m)) ∈ E
h0we define the difference operators δ
0, δ = (δ
1, . . . , δ
n) in the following way:
(5) δ
0w
(r,m)= 1
h
0w
(r+1,m)− w
(r,m)(6) δ
jw
(r+1,m)= 1
h
jw
(r+1,m+ej)− w
(r+1,m), 1 ¬ j ¬ κ,
(7) δ
jw
(r+1,m)= 1 h
jw
(r+1,m)− w
(r+1,m−ej), κ + 1 ¬ j ¬ n,
where 1 ¬ κ ¬ n is fixed natural number. Note that δw
(r+1,m)is given by (7) if κ = 0 and δw
(r+1,m)is defined by (6) for κ = n. Suppose that
ϕ
h= {ϕ
h.k}
k∈N, ϕ
h: E
h.0∪ ∂
0E
h→ l
∞is given function. For z : B
h→ l
∞and for a point (t
(r), x
(m)) ∈ E
h0we define δ
0z
(r,m)= n
δ
0z
(r,m)ko
k∈N
, F
h[z]
(r,m)= {F
h.k[z]
(r,m)}
k∈Nand
F
h.k[z]
(r,m)= f(t
(r), x
(m), T
hz
[r,m], δz
(r+1,m)), k ∈ N.
We will approximate solutions of problem (1), (2) by means of solutions of the problem
(8) δ
0z
(r,m)= F
h[z]
(r,m),
(9) z
(r,m)= ϕ
(r,m)hon E
h.0∪ ∂
0E
hwhere z = {z
k}
k∈N. We first prove that there exists exactly one solution u
h: B
h→ l
∞of problem (8), (9). Let
S = [ −b
0, b
0) × (b
00, b
00], S
∗= [−b
0, c
0] × [−c
00, b
00] and
R
nh= {x
(m): m ∈ Z
n} where h ∈ H. Lets consider the sets
S
h= S ∩ R
nh, S
h∗∩ R
nh, ∂
0S
h= S
h∗\ S
h. For each x
(m)∈ S
hwe put
∆
(m)= {x
(m+ej): 1 ¬ j ¬ κ} ∪ {x
(m−ej): κ + 1 ¬ j ¬ n}.
Assumption H[f]. Suppose that the function f : Ω → l
∞, f = {f
k}
k∈N, of the variables (t, x, w, q) is such that
1) for k ∈ N and each P = (t, x, w, q) ∈ Ω there exists partial derivatives
∂
qf
k(P ) =
∂
q1f
k(P ), . . . , ∂
qnf
k(P ) and ∂
qf
k(t, x, w, ·) ∈ C(R
n, R
n),
2) for each P ∈ Ω and for k ∈ N the estimates
∂
qjf
k(P ) 0 for 1 ¬ j ¬ κ, ∂
qjf
k(P ) ¬ 0 for κ + 1 ¬ j ¬ n are satisfied.
Lemma 2.1 Suppose that assumption H[f] is satisfied and ϕ
h: E
h.0∪ ∂
0E
h→ l
∞then there exists exactly one solution u
h: B
h→ l
∞of (8), (9).
Proof It follows from (9) that u
his defined on E
h.0. Suppose that 0 ¬ r < K is fixed and that u
h.kis defined on B
h.rfor k ∈ N. Assume now that k ∈ N is fixed.
Consider the problem
(10) z
k(r+1,m)= u
(r,m)h.k+ h
0f
k(t
(r), x
(m), (u
h)
[r,m], δu
(r+1,m)k)
(11) u
(r+1,m)h.k= ϕ
(r+1,m)h.kfor x
(m)∈ ∂
0S
h.
Assume now that the numbers u
h.k(t
(r+1), y) where y ∈ ∆
(m)are known. Write ψ
k(τ) = u
(r,m)h.k+ h
0f
k(t
(r), x
(m), (u
h)
[r,m], Q
(r+1,m)k(τ)),
where
Q
(r+1,m)k(τ) =
1
h
1(u
(r+1,m+eh.k 1)− τ), . . . , 1 h
κ(u
(r+1,m+eh.k κ)− τ), 1
h
κ+1(τ − u
(r+1,m−eh.k κ+1)), . . . , 1
h
n(τ − u
(r+1,m−eh.k n))
, Then ψ
k: R → l
∞is of class C
1and
ψ
0k(τ) = −h
0X
n j=11
h
j|∂
qjf
k(t
(r), x
(m), (u
h)
[r,m], Q
(r+1,m)k(τ))| ¬ 0
for τ ∈ R. Then equation τ = ψ
k(τ) has exactly one solution and consequently the number u
(r+1,m)h.kcan be calculated. Since u
(r+1,m)h.kis given for x
(m)∈ ∂
0S
hit follows that there exists exactly one solution u
(r+1,m)h.kof (10), (11) for x
(m)∈ S
h. Then u
h.kis defined on B
h.r+1. Then by induction the solution exists and it is unique on B
h.
■Now we formulate the main theorem of this section. We will proof convergence of the method (8), (9). We will need the following assumptions.
Assumption H[f, σ]. Suppose that
1) the function σ : [0, a] × R
+→ R
+satisfies the following conditions:
(i) σ(t, ·) : R
+→ R
+is continuous and nondecreasing for each t ∈ [0, a], (ii) the maximal solution of the Cauchy problem
(12) w
0(t) = σ(t, w(t)), w(0) = 0, is e w(t) = 0 for t ∈ [0, a],
2) the estimate
(13) kf(t, x, w, q) − f(t, x, ¯ w, q) k
∞¬ σ(t, kw − ¯ w k
D)
is satisfied on Ω.
Theorem 2.2 Suppose that Assumptions H[f], H[f, σ] are satisfied and
1) the function v : B → l
∞, v = {v
k}
k∈N, is a solution of (1), (2) and is of class C
1,
2) h ∈ H and the function u
h: B
h→ l
∞, u
h= {u
h.k}
k∈Nis a solution of (8), (9) and there is α
0: H → R
+such that
kϕ
(r,m)− ϕ
(r,m)hk
∞¬ α
0(h) on E
h.0∪ ∂
0E
hand lim
h→0
α
0(h) = 0, 3) the operator T
h: F (D
h, l
∞) → C(D, l
∞) satisfies the Assumption H[T
h].
Then there exists a function α : H → R
+such that (14) kv
(r,m)− u
(r,m)hk
∞¬ α(h) on E
hand lim
h→0
α(h) = 0.
Proof Put ev
h= {v
h.k}
k∈N, where v
h.kis the restriction of v
kto the set B
h, k ∈ N.
Let the function Γ
h: E
h0→ l
∞, Γ
h= {Γ
h.k}
k∈N, be defined by (15) δ
0ev
(r,m)h= F
h[ev
h]
(r,m)+ Γ
(r,m)hon E
h0.
We conclude from the assumption 3) that there exists a function eγ : H → R
+such that
kΓ
(r,m)hk
∞¬ eγ(h) on E
h0and lim
h→0
eγ(h) = 0.
Adapting the mean-value theorem to the difference
f
k(t
(r), x
(m), (T
hu
h)
[r,m], δ ev
h.k(r+1,m)) − f
k(t
(r), x
(m), (T
hu
h)
[r,m], δu
(r+1,m)h.k)
= X
n i=1∂
qif
k(P
ki)δ(ev
h.k− u
h.k)
(r+1,m), k ∈ N,
where P
ki∈ Ω, k ∈ N, are intermediate points. Then using the definitions of difference operator δ
0and δ we get
(ev
h.k− u
h.k)
(r+1,m)1 + h
0X
n i=11
h
i|∂
qif
k(P
ki)|
= (ev
h.k− u
h.k)
(r,m)+h
0X
κ i=1∂
qif
k(P
ki) 1
h
i(ev
h.k− u
h.k)
(r+1,m+ei)−h
0X
n i=κ+1∂
qif
k(P
ki) 1
h
i(ev
h.k− u
h.k)
(r+1,m−ei)+h
0h f
k(t
(r), x
(m), ( ev
h)
(t(r),x(m)), δ ev
h.k(r+1,m))
−f
k(t
(r), x
(m), (T
hu
h)
[r,m], δ ev
h.k(r+1,m)) i
+ h
0Γ
(r,m)h.k, k ∈ N.
We define
ε
(r)h.k= max {|(ev
h.k− u
h.k)
(ν,m)| : (t
(nu), x
(m)) ∈ B
h.r}, η
h(r)= sup {ε
(r)h.k: k ∈ N}.
According to Assumption H[T
h] we have the estimate
k(ev
h)
(t(r),x(m))− (T
hu
h)
[r,m]k
D¬ γ(h) + η
(r)h. Then for η
(r)hthe following inequalities
(16) η
(r+1)h¬ η
h(r)+ h
0σ(t, γ(h) + η
h(r)) + h
0eγ(h), η
(0)h¬ α
0(h)
hold. Let us consider the Cauchy problem
(17) w
0(t) = σ(t, γ(h) + w(t)) + eγ(h),
(18) w(0) = α
0(h).
It follows from condition 1)-(ii) of Assumption H[f, σ] that there exists the maximal solution e w
hof the problem (17), (18) and e w
his defined on [0, a]. Moreover
h
lim
→0w e
h(t) = 0 uniformly on [0, a].
It is easily seen that e w
hsatisfies the recurrent inequality e
w
h(r+1) e w
(r)h+ h
0σ(t
(r), γ(h) + w e
(r)h) + h
0eγ(h), 0 ¬ r ¬ K − 1 and it follows from (18) that the inequality
e
w
(r)h¬ α
0(h), −K
0¬ r ¬ 0 is satisfied. By the above relations and (16) we have
η
(r)h¬ e w
(r)h, 0 ¬ r ¬ K.
Thus we get (14) for α(h) = e w
h(a). This completes the proof.
■3. Finite systems of difference equations. We take into consideration the problem (1), (2). We construct auxiliary differential functional problem. Let the function φ : B → l
∞, φ = {φ
k}
k∈N, be such that φ(t, x) = ϕ(t, x), (t, x) ∈ E
0∪∂
0E.
Fix µ ∈ N. For the function w = (w
1, . . . , w
µ) : D → R
µor w : D → l
∞, w = {w
k}
k∈N, and for ¯ w : D → l
∞, ¯ w = { ¯ w
k}
k∈N, we define
[w, ¯ w]
µ= { e w
k}
k∈Nwhere w e
k=
( w
kfor 1 ¬ k ¬ µ,
w ¯
kfor k > µ.
Consider now the differential functional problem
(19) ∂
tz
k(t, x) = f
k(t, x, [z
(t,x), φ
(t,x)]
µ, ∂
xz
k(t, x)), 1 ¬ k ¬ µ,
(20) z
k(t, x) = ϕ
k(t, x), 1 ¬ k ¬ µ, where z = (z
1, . . . , z
µ). We will need the following assumptions.
Assumption H[f, ϕ]. Suppose that the function f : Ω → l
∞satisfies Assump- tion H[f, σ] and
1) ϕ ∈ C(E
0∪ ∂
0E, l
∞) is such that there exists φ ∈ C(B, l
∞), φ = {φ
k}
k∈N, with the properties
(i) φ(t, x) = ϕ(t, x) for (t, x) ∈ E
0∪ ∂
0E,
(ii) for each k ∈ N the function φ
k: Ω → l
∞is of class C
1,
2) for every k ∈ N there exist C
k∈ R
+such that for (t, x) ∈ E we have
|∂
tφ
k(t, x) − f
k(t, x, φ
(t,x), ∂
xφ
k(t, x))| ¬ C
kand lim
k→∞C
k= 0.
Lemma 3.1 Suppose that Assumptions H[f] and H[f, ϕ] are satisfied and the func- tion v : B → l
∞, v = {v
k}
k∈N, is a solution of (1), (2) then for each k ∈ N there exists β
k∈ C([0, a], R
+) such that
(21) |v
k(t, x) − φ
k(t, x)| ¬ β
k(t), (t, x) ∈ E, and lim
k→∞β
k(t) = 0 uniformly on [0, a].
Proof Let the function ev : B → l
∞, ev = {ev
k}
k∈Nbe defined by ev(t, x) = v(t, x) − φ(t, x), (t, x) ∈ B. Then from condition 1) of Assumption H[f, ϕ] we get that ev(t, x) = 0 on E
0∪ ∂
0E. We have
∂
tev
k(t, x) = f
k(t, x, ev
(t,x)+ φ
(t,x), ∂
xev
k(t, x) + ∂
xφ
k(t, x)) − ∂
tφ
k(t, x), k ∈ N.
Form above we get
|∂
tev
k(t, x) − X
n i=1∂
qif
k(P
ki)∂
xiev
k(t, x)|
¬ |f
k(t, x, ev
(t,x)+ φ
(t,x), ∂
xφ
k(t, x)) − f
k(t, x, φ
(t,x), ∂
xφ
k(t, x))|
+|f
k(t, x, φ
(t,x), ∂
xφ
k(t, x)) − ∂
tφ
k(t, x)| ¬ σ(t, kevk
D) + C
k, k ∈ N where P
ki∈ Ω, k ∈ N are intermediate points. Let the function β
kbe the solution of problem
w
0(t) = σ(t, w(t)) + C
k, w(0) = 0.
It follows from the comparison theorem for infinite systems of hyperbolic functional differential equations (see [6]) that
|ev
k(t, x)| ¬ β
k(t) on E.
Since lim
k→∞C
k= 0, from the Assumption H[f, σ] we have
k
lim
→∞β
k(t) = 0 uniformly on [0, a].
Then the assertion (22) holds.
■Lemma 3.2 Suppose that Assumptions H[f] and H[f, ϕ] are satisfied and 1) v : B → ∞, v = {v
k}
k∈N, is a solution of (1), (2),
2) for every µ ∈ N the function v
[µ]= (v
[µ]1, . . . , v
µ[µ]) : B → R
µ, is a solution of (19), (20).
Then for every µ ∈ N there exists the continuous function β
[µ]: [0, a] → R
+such that
(22) |v
k(t, x) − v
k[µ](t, x)| ¬ β
[µ](t), (t, x) ∈ E, where 1 ¬ k ¬ µ and lim
µ→∞β
[µ](t) = 0 uniformly on [0, a].
Proof We define the function u
[µ]= (u
[µ]1, . . . , u
[µ]) : B → R
µas u
[µ]k(t, x) = v
[µ]k(t, x) − v
k(t, x), 1 ¬ k ¬ µ. Then from condition 1) of Assumption H[f, ϕ] we get that u
[µ]k(t, x) = 0 on E
0∪ ∂
0E. We have
∂
tu
[µ]k(t, x) = ∂
t(v
[µ]k(t, x) − v
k(t, x))
= f
k(t, x, [v
[µ](t,x), φ
(t,x)]
k, ∂
xv
[µ]k(t, x)) − f
k(t, x, v
(t,x), ∂
xv
k(t, x)), 1 ¬ k ¬ µ.
Based on Assumption H[f, σ] and Lemma (3.1) we have the estimations
|∂
tu
[µ]k(t, x) − X
n i=1∂
qif
k(P
ki)∂
xiu
[µ]k(t, x)|
= |f
k(t, x, [v
[µ](t,x), φ
(t,x)]
k, ∂
xv
k(t, x)) − f
k(t, x, v
(t,x), ∂
xv
k(t, x))|
¬ |f
k(t, x, [v
[µ](t,x), φ
(t,x)]
k, ∂
xv
k(t, x)) − f
k(t, x, [v
(t,x), φ
(t,x)]
k, ∂
xv
k(t, x))|
+|f
k(t, x, [v
(t,x), φ
(t,x)]
k, ∂
xv
k(t, x)) − f
k(t, x, v
(t,x), ∂
xv
k(t, x))|
¬ σ(t, ku
[µ]k
D.µ) + σ(t, e β
µ(t)), where
ku
[µ]k
D.µ= max {|u
[µ]k(τ, y)| : (τ, y) ∈ D, 1 ¬ k ¬ µ},
β e
µ(t) = sup {β
k(t) : k > µ},
and P
ki∈ Ω are intermediate points, 1 ¬ k ¬ µ. Put ε
µ= σ(a, e β
µ(a)). From the comparison theorem for infinite systems of hyperbolic functional differential equations we have
|u
[µ]k(t, x)| ¬ β
[µ](t) on E where β
[µ]is a solution of the problem
w
0(t) = σ(t, w(t)) + ε
µ, w(0) = 0.
Since lim
µ→∞ε
µ= 0, then based on Assumption H[f, σ] we have lim
µ→∞β
[µ](t) =
0 uniformly on [0, a]. This completes the proof.
■Assume now that
ϕ
h= (ϕ
h.1, . . . , ϕ
h.µ) : E
h.0∪ ∂
0E
h→ R
µis given function. The difference operators δ
0, δ = (δ
1, . . . , δ
n) are given by (5)-(7).
We will approximate solutions of (1) ,(2) by the solutions of the difference functional problem
(23) δ
0z
k(r,m)= f
k(t
(r), x
(m), [T
hz
[r,m], φ
(t(r),x(m))]
k, δz
(r+1,m)k), 1 ¬ k ¬ µ,
(24) z
k(r,m)= ϕ
(r,m)h.kon E
h.0∪ ∂
0E
h, 1 ¬ k ¬ µ.
Remark 3.3 Existence and uniqueness of solutions of the problem (23), (24) we proof in a similarly way as in the case of infinite difference functional schemes (8), (9).
We proof the convergence of the above implicit difference method.
Theorem 3.4 Suppose that the Assumptions H[f], H[f, ϕ] are satisfied and 1) the function v : B → l
∞, v = {v
k}
k∈N, is a solution of (1), (2),
2) for every µ ∈ N the function v
[µ]= (v
[µ]1, . . . , v
µ[µ]) : B → R
µis a solution of (19), (20),
3) for every µ ∈ N and h ∈ H the function u
[µ]h= (u
[µ]h.1, . . . , u
[µ]h.µ) : B
h→ R
µis a solution of (23) ,(24),
4) for every µ ∈ N there is α
[µ]0: H → R
+such that
|ϕ
(r,m)h.k− ϕ
k(t
(r), x
(m))| ¬ α
[µ]0(h) on E
h.0∪ ∂
0E
h, 1 ¬ k ¬ µ,
and lim
h→0α
0(h) = 0.
Then for every µ ∈ N there exists γ
[µ]: H → R
+and λ
µ∈ R
+such that (25) |v
k(t
(r), x
(m)) − (u
[µ]h.k)
(r,m)| ¬ γ
[µ](h) + λ
µ, 1 ¬ k ¬ µ, and lim
h→0γ
[µ](h) = 0, lim
µ→∞λ
µ= 0.
Proof Let us fix µ ∈ N. We have
|v
k(t
(r), x
(m)) − (u
[µ]h.k)
(r,m)|
¬ |v
k(t
(r), x
(m)) − v
[µ]k(t
(r), x
(m))| + |v
[µ]k(t
(r), x
(m)) − (u
[µ]h.k)
(r,m)|, 1 ¬ k ¬ µ.
Form Lemma (3.2) we get the estimation
|v
k(t
(r), x
(m)) − v
k[µ](t
(r), x
(m))| ¬ β
[µ](t
(r)), 1 ¬ k ¬ µ.
In the similarly way as in the proof of Theorem (2.2) we can show that
|v
[µ]k(t
(r), x
(m)) − (u
[µ]h.k)
(r,m)| ¬ (ω
h[µ])
(r), 1 ¬ k ¬ µ, where ω
[µ]his a solution of the problem
w
0(t) = σ(t, w(t)) + eγ
[µ](h), w(0) = α
[µ]0(h),
where lim
h→0eγ
[µ](h) = 0. Therefore with γ
[µ](h) = ω
[µ]h(a) and λ
µ= β
[µ](a) we get
the assertion of Theorem.
■4. Numerical example. For n = 2 we put
(26) B = {(t, x, y) : t ∈ [0, a], x ∈ [−1, 1], y ∈ [−1, 1]}.
Consider the system of differential functional equations
(27) ∂
tz
k(t, x, y) = z
k(t, x, y)z
k+2(t, x, y) + ∂
xz
k(t, x, y) − ∂
yz
k(t, x, y) + cos (∂
xz
k(t, x, y) + ∂
yz
k(t, x, y) + 2t(y − x)z
k(t, x, y))
−2t(x + y)z
k(t, x, y) − k
2e
2t(x2−y2)− 1 + (x
2− y
2)z
k(t, x, y) with the initial boundary condition
(28) z
k(0, x, y) = k, x ∈ [−1, 1], y ∈ [−1, 1], z
k(t, 1, y) = ke
t(1−y2), t ∈ [0, a], y ∈ [−1, 1], z
k(t, x, −1) = ke
t(x2−1), t ∈ [0, a], x ∈ [−1, 1].
where 1 ¬ k ¬ 4. The exact solution of this problem is known. It is v
k(t, x, y) =
ke
t(x2−y2). Put h = (h
0, h
1, h
2) stand for the steps of the mesh on B.
Let us denote by z
k.h: B
h→ l
∞, 1 ¬ k ¬ 4, the solution of the explicit difference problem corresponding to (27), (28) and u
k.h: B
h→ l
∞, 1 ¬ k ¬ 4, the solution of the implicit difference problem. Write
(29) η
k.h(r)= 1 (2N + 1)
2X
N m1=−NX
N m2=−N|z
(r,mk.h 1,m2)− v
(r,mk 1,m2)|,
(30) η e
(r)k.h= 1 (2N + 1)
2X
N m1=−NX
N m2=−N|u
(r,mk.h 1,m2)− v
k(r,m1,m2)|,
where N ∈ N is defined by the condition (Nh
1, N h
2) = (1, 1) and 1 ¬ k ¬ 4. The numbers η
k.h(r), eη
k.h(r)are the arithmetical mean of the errors with fixed t
(r). The values of the functions η
k.h, eη
k.hare listed in the table. We write ”x”for η
k.h> 100.
Table of errors η
h, eη
hwith h = (0.05, 0.005, 0.005)
η
he η
ht = 0.2 z
1: 0.0000038 0.0000038 z
2: 0.0000086 0.0000086 z
3: 0.0000099 0.0000096 z
4: 0.0000146 0.0000137 t = 0.3 z
1: 0.0000045 0.0000082 z
2: 0.0000074 0.0000211 z
3: 0.0004797 0.0000179 z
4: 0.0010949 0.0000277 t = 0.4 z
1: 0.0523412 0.0000171 z
2: 0.3810366 0.0000553 z
3: 0.9636259 0.0000305 z
4: 2.1082880 0.0000498
t = 0.5 z
1: x 0.0000395
z
2: x 0.0001658
z
3: x 0.0000490
z
4: x 0.0000848
The results given in the table are consistent with our theoretical results. Our experiments have the following property. The explicit method for steps h = (0.05, 0.005, 0.005) which fail to satisfy condition (CFL)
1 − h
0X
2 j=11
h
j|∂
qjf (P ) | 0,
is not stable. The implicit difference method is stable aside from steps of the mesh.
References
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Gdańsk University of Technology
Gabriela Narutowicza 11-12, Gdańsk, Poland E-mail: annak@mif.pg.gda.pl
(Received: 10.05.2009)