Asymptotic behavior of solutions to partial difference equations of parabolic type
YU SAKAMAKI
Abstract
We establish stabilization results for the solutions to partial difference equations of parabolic type. We impose either the Dirichlet boundary condition or the Robin boundary condition.
Keywords: parabolic partial difference equation, the Dirichlet boundary condition, the Robin boundary condition, stabilization, large-time limit.
MSC2010: 39A14 (primary), 39A06 (secondary)
1 Introduction and main results
In this paper we establish stabilization results for the solutions to partial difference equations of parabolic type. We shall impose either the Dirichlet boundary condition or the Robin boundary condition. Let us formulate our main results. For m, n ∈
Zwith m ≤ n, let m, n = { j ∈
Z| m ≤ j ≤ n } , which we call a discrete interval. Let also m, ∞ = { j ∈
Z| m ≤ j } for m ∈
Z. For(n, m) ∈
Z2, we set
N (n, m) = { (n − 1, m), (n + 1, m), (n, m − 1), (n, m + 1) } .
We call N (n, m) the neighborhood of the point (n, m). Let E be a finite subset of
Z2. Define E
int= { (n, m) ∈
Z2| (n, m) ∈ E and N (n, m) ⊂ E } ,
which we call the interior of E. We suppose that E
int̸= ϕ. Suppose also that E satisfies the condition
E =
∪(n,m)∈Eint
[N (n, m) ∪ { (n, m) } ]. (1.1)
Let ∂E = E \ E
int; we call it the boundary of E. Put D = (E × 1, ∞ ) ∪ (E
int× { 0 } ), D
0= E
int× 1, ∞ and S = ∂E × 1, ∞ . Consider the operator
(Lz)(n, m, l) =k
1(n, m, l)z(n − 1, m, l) + k
2(n, m, l)z(n + 1, m, l)
+ k
3(n, m, l)z(n, m − 1, l) + k
4(n, m, l)z(n, m + 1, l) − k
5(n, m, l)z(n, m, l)
− { z(n, m, l) − z(n, m, l − 1) } ,
(n, m, l) ∈ D
0, where z(n, m, l) is a real-valued function on D and the coefficients k
j(n, m, l), 1 ≤ j ≤ 5, are real-valued functions on D
0. We suppose that L is parabolic in the following sense:
(H.1) k
j(n, m, l) ≥ 0 on D
0for 1 ≤ j ≤ 4.
We also suppose that (H.2) k
5(n, m, l) ≥
∑4j=1
k
j(n, m, l) on D
0.
We introduce a terminology. Let (n, m), (n
′, m
′) ∈
Z2. We say that (n, m) is adjacent to (n
′, m
′) if | n − n
′| + | m − m
′| = 1. For (n, m) ∈ ∂E, we designate by q(n, m) the number of the points (n
′, m
′) ∈ E
intwhich are adjacent to (n, m). Let P
1, ..., P
q(n,m)be the points in E
intwhich are adjacent to (n, m). By N
n,miwe denote the vector from (n, m) to P
ifor 1 ≤ i ≤ q(n, m), i.e.
N
n,mi= P
i− (n, m). We call them the inward normal vectors at (n, m). We also consider the operator
(Bz)(n, m, l) =
q(n,m)∑
i=1
B
i(n, m, l) { z((n, m) + N
n,mi, l) − z(n, m, l) } , (1.2) (n, m, l) ∈ S. Given a real-valued function f (n, m, l) on D
0and real-valued function h(n, m, l) on S, consider the system
{
(Lz)(n, m, l) = f (n, m, l) for (n, m, l) ∈ D
0,
z(n, m, l) = h(n, m, l) for (n, m, l) ∈ S. (1.3) Note that we impose the Dirichlet boundary condition in (1.3). For a real-valued function f (n, m, l) on D
0and real-valued functions g(n, m, l) and h(n, m, l) on S, we also consider the system
{
(Lz)(n, m, l) = f(n, m, l) for (n, m, l) ∈ D
0,
(Bz)(n, m, l) + g(n, m, l)z(n, m, l) = h(n, m, l) for (n, m, l) ∈ S. (1.4) Note that we impose the Robin boundary condition in (1.4). Put Λ = { (1, 2), (2, 1), (3, 4), (4, 3) } . We impose the following hypotheses:
(H.3) There is a pair (I, J ) ∈ Λ such that there exists s > 0 and M ≥ 0 for which k
I≥ s on D
0and k
J≤ M on D
0;
(H.4) There is a positive number K such that B
i(n, m, l) < K for (n, m, l) ∈ S and 1 ≤ i ≤ q(n, m);
(H.5) B
i(n, m, l) ≥ 0 for (n, m, l) ∈ S and 1 ≤ i ≤ q(n, m);
(H.6) There is a positive constant µ
1such that g(n, m, l) < − µ
1on S.
This paper possesses four main theorems. The first and second one read as follows.
Theorem 1.
Adopt the assumptions (H.1), (H.2) and (H.3). Let z be a solution of (1.3). If f (n, m, l) → 0 as l → ∞ for (n, m) ∈ E
intand h(n, m, l) → 0 as l → ∞ for (n, m) ∈ ∂E, then
l
lim
→∞z(n, m, l) = 0 f or each (n, m) ∈ E.
Theorem 2.
Adopt the hypotheses (H.1), (H.2), (H.3), (H.4), (H.5) and (H.6). Let z be a solution of (1.4). If f (n, m, l) → 0 as l → ∞ for (n, m) ∈ E
intand h(n, m, l) → 0 as l → ∞ for (n, m) ∈ ∂E, then
l
lim
→∞z(n, m, l) = 0 f or each (n, m) ∈ E.
Next, we consider the situation where the coefficients in (1.3) and those in (1.4) converge as l → ∞ . We impose the following hypotheses:
(H.7) For (n, m) ∈ E
intand 1 ≤ j ≤ 5, the limit d
j(n, m) ≡ lim
l→∞k
j(n, m, l) exists in
R. Itholds that d
j(n, m) > 0 on E
intfor 1 ≤ j ≤ 4;
(H.8) The limit f
0(n, m) ≡ lim
l→∞f (n, m, l) exists in
Rfor each (n, m) ∈ E
int. The limit h
0(n, m) ≡ lim
l→∞h(n, m, l) exists in
Rfor each (n, m) ∈ ∂E;
(H.9) The limit g
0(n, m) ≡ lim
l→∞g(n, m, l) exists in
Rfor each (n, m) ∈ ∂E. The limit B ˜
i(n, m) ≡ lim
l→∞B
i(n, m, l) exists in
Rfor each (n, m) ∈ ∂E. It holds that ˜ B
i(n, m) > 0 on ∂E for 1 ≤ i ≤ q(n, m).
We introduce the operators
(L
0w)(n, m) =d
1(n, m)w(n − 1, m) + d
2(n, m)w(n + 1, m)
+ d
3(n, m)w(n, m − 1) + d
4(n, m)w(n, m + 1) − d
5(n, m)w(n, m), ( ˜ Bw)(n, m) =
q(n,m)∑
i=1
B ˜
i(n, m) { w((n, m) + N
n,mi) − w(n, m) } ,
where w(n, m) is a real-valued function on E. Consider the elliptic boundary value problems
{(L
0w)(n, m) = f
0(n, m) for (n, m) ∈ E
int,
w(n, m) = h
0(n, m) for (n, m) ∈ ∂E, (1.5)
{
(L
0w)(n, m) = f
0(n, m) for (n, m) ∈ E
int,
( ˜ Bw)(n, m) + g
0(n, m)w(n, m) = h
0(n, m) for (n, m) ∈ ∂E. (1.6) The system (1.6) possesses a unique solution; see Lemma 4. It follows from (H.2) and (H.7) that d
5(n, m) ≥
∑4j=1
d
j(n, m). Using this and (H.7), we obtain the following implication: If u is a real-valued function on E such that L
0u ≥ 0 on E
intand u ≤ 0 on ∂E, then u ≤ 0 on E.
The proof of this fact is similar to Lemma 3. This, combined with the argument in the proof of Lemma 4, we infer that the system (1.5) also possesses a unique solution. The third and fourth main theorems are stated as follows.
Theorem 3.
Adopt the assumptions (H.1), (H.2), (H.7) and (H.8). Let z be a solution of (1.3).
By w we denote the solution of (1.5). Then, we have
l
lim
→∞z(n, m, l) = w(n, m) f or each (n, m) ∈ E.
Theorem 4.
Suppose that conditions (H.1), (H.2), (H.6), (H.7), (H.8) and (H.9) are fulfilled.
Let z be a solution of (1.4). We designate the solution of (1.6) by w. Then, we have
l
lim
→∞z(n, m, l) = w(n, m) f or each (n, m) ∈ E.
We address the background of our work here. Partial difference equations have been studied
by numerous authors; we refer to [1, 10] and the references therein for a thorough review. It
is worth mentioning that I. G. Petrovskii established the existence of solutions to the initial
boundary value problems for the heat equation by using the discrete heat equation; see [7,
Section 42]. Though the stabilization theory of parabolic differential equations is a significant
topic, its discrete version was overlooked; this motivates our current study. Our work here is closely related with the stability theory of parabolic difference equations. The supposition which we impose on the coefficients is weaker than those in the literature of the stability theory; cf. [1, Chapter 6]. Our formulation of the discrete boundary operators appears to be new. Theorems 1, 2, 3 and 4 are discrete analogues of [4, Chapter 6, Theorems 1, 4, 2 and 5], respectively (see also [5]).
Partial difference equations play an important role in numerical analysis and simulations; we refer to [2, 9] for this connection. Our results can also be utilized in the detection of a failure in numerical simulations. For the basic theory of difference equations, we refer to [3, 6, 8].
2 Auxiliary lemmas
We prove the following implications.
Lemma 1.
(Comparison Theorem I) Suppose that (H.1), (H.2), (H.5) and (H.6) hold. Let v and w be real-valued functions on E × 0, ∞ . If v > w on E × { 0 } , Lv ≤ 0 on D
0, Lw ≥ 0 in D
0and
(Bv)(n, m, l) + g(n, m, l)v(n, m, l) < (Bw)(n, m, l) + g(n, m, l)w(n, m, l) on S, then we have v > w on E × 0, ∞ .
Proof. Put z = v − w. Pick a T ∈
N, arbitrarily. Setα = max{j ∈ 0, T | z(n, m, l) > 0 for all (n, m, l) ∈ E × 0, j}.
We note that α exists, since v > w on E × { 0 } by assumption. It suffices to show that α = T . Seeking a contradiction, we suppose that α ≤ T − 1. Let (n
0, m
0) ∈ E be such that
z(n
0, m
0, α + 1) = min
(n,m)∈E
z(n, m, α + 1). (2.1)
It follows from the definition of α that z(n
0, m
0, α + 1) ≤ 0. Let us prove that (n
0, m
0) ∈ E
intby contradiction. Suppose that (n
0, m
0) ∈ ∂E. Then, we get
(Bz)(n
0, m
0, α + 1) < −g(n
0, m
0, α + 1)z(n
0, m
0, α + 1) ≤ 0
from (H.6) and z(n
0, m
0, α + 1) ≤ 0. However, (Bz)(n
0, m
0, α + 1) ≥ 0 by (H.5), which is a contradiction. Therefore, (n
0, m
0) ∈ E
int. We have
(Lz)(n
0, m
0, α + 1) =k
1(n
0, m
0, α + 1) { z(n
0− 1, m
0, α + 1) − z(n
0, m
0, α + 1) } + k
2(n
0, m
0, α + 1) { z(n
0+ 1, m
0, α + 1) − z(n
0, m
0, α + 1) } + k
3(n
0, m
0, α + 1){z(n
0, m
0− 1, α + 1) − z(n
0, m
0, α + 1)}
+ k
4(n
0, m
0, α + 1) { z(n
0, m
0+ 1, α + 1) − z(n
0, m
0, α + 1) } +
∑4 j=1
k
j(n
0, m
0, α + 1) − k
5(n
0, m
0, α + 1)
z(n
0, m
0, α + 1) + {z(n
0, m
0, α) − z(n
0, m
0, α + 1)}
≡ I
1+ I
2+ I
3+ I
4+ I
5+ I
6.
(2.2)
We have I
6> 0. It follows from (2.1) that
z(n
0− 1, m
0, α + 1) − z(n
0, m
0, α + 1) ≥ 0,
so that I
1≥ 0. Similarly, we have I
2≥ 0, I
3≥ 0 and I
4≥ 0. Since z(n
0, m
0, α + 1) ≤ 0, we infer from (H.2) that I
5≥ 0. So, we obtain (Lz)(n
0, m
0, α + 1) > 0. Because
(Lv)(n
0, m
0, α + 1) ≤ 0 and (Lw)(n
0, m
0, α + 1) ≥ 0
by assumption, we have (Lz)(n
0, m
0, α + 1) ≤ 0, which is a contradiction. Thereby, α = T .
Lemma 2.(Comparison Theorem II) Suppose that (H.1) and (H.2) hold. Let v and w be real- valued functions on D. If v > w on (E
int× { 0 } ) ∪ S, Lv ≤ 0 on D
0and Lw ≥ 0 in D
0, then we have v > w on D.
Proof. Though the proof is somewhat similar to the previous one, we provide it for the sake of completeness. Put z = v − w. Pick a T ∈
N, arbitrarily. Setα = max { j ∈ 0, T | z(n, m, l) > 0 for all (n, m, l) ∈ E
int× 0, j } .
We note that α exists, since v > w on E
int× { 0 } by assumption. It suffices to show that α = T . Seeking a contradiction, we suppose that α ≤ T − 1. Let (n
0, m
0) ∈ E
intbe such that
z(n
0, m
0, α + 1) = min
(n,m)∈Eint
z(n, m, α + 1). (2.3)
It follows from the definition of α that z(n
0, m
0, α+ 1) ≤ 0. We employ the decomposition (2.2).
We have I
6> 0. In the case where (n
0− 1, m
0) ∈ E
int, we have z(n
0− 1, m
0, α + 1) − z(n
0, m
0, α + 1) ≥ 0
from (2.3). In the case where (n
0− 1, m
0) ̸∈ E
int, we have (n
0− 1, m
0, α + 1) ∈ S, so that z(n
0− 1, m
0, α + 1) − z(n
0, m
0, α + 1) > 0.
Thus, I
1≥ 0 in any case. Similarly, we have I
2≥ 0, I
3≥ 0 and I
4≥ 0. Since z(n
0, m
0, α+1) ≤ 0, we infer from (H.2) that I
5≥ 0. So, we obtain (Lz)(n
0, m
0, α + 1) > 0. Because
(Lv)(n
0, m
0, α + 1) ≤ 0 and (Lw)(n
0, m
0, α + 1) ≥ 0
by assumption, we have (Lz)(n
0, m
0, α + 1) ≤ 0, which is a contradiction. Thereby, α = T .
Remark 1.Our Lemmas 1 and 2 are discrete analogues of [4, Chapter 2, Theorems 17 and 16], respectively. Because of a discrete character of our problem, we need the condition (H.6), which is not employed in [4, Chapter 3, Theorem 17]. (Note that β in [4, Chapter 2, (6.5)] is any function.)
Lemma 3.
(Weak maximum principle) Suppose that the conditions (H.6), (H.7) and (H.9) hold.
If u is a real-valued function on E such that L
0u ≥ 0 on E
intand Bu ˜ + g
0u ≥ 0 on ∂E, then
u ≤ 0 on E.
Proof. Put
M = max
(n,m)∈E
u(n, m).
Seeking a contradiction, we suppose that M > 0. Let (n
0, m
0) ∈ E be such that u(n
0, m
0) = M . We prove that (n
0, m
0) ∈ E
intby contradiction. Suppose that (n
0, m
0) ∈ ∂E. Then, it follows from the definition of (n
0, m
0), (H.6) and (H.9) that
( ˜ Bu)(n
0, m
0) + g
0(n
0, m
0)u(n
0, m
0) < 0.
However, ˜ Bu + g
0u ≥ 0 on ∂E by assumption, which is a contradiction. Hence, we have (n
0, m
0) ∈ E
int. By assumption
(L
0u)(n
0, m
0)
=d
1(n
0, m
0) { u(n
0− 1, m
0) − u(n
0, m
0) } + d
2(n
0, m
0) { u(n
0+ 1, m
0) − u(n
0, m
0) } + d
3(n
0, m
0) { u(n
0, m
0− 1) − u(n
0, m
0) } + d
4(n
0, m
0) { u(n
0, m
0+ 1) − u(n
0, m
0) }
−
d
5(n
0, m
0) −
∑4 j=1
d
j(n
0, m
0)
u(n
0, m
0)
≥ 0.
This, together with (H.7) and the fact that u(n
0, m
0) = M , implies that u(n
0, m
0) = u(n, m) for every (n, m) ∈ N (n
0, m
0). Since there exists k ∈
Nsuch that (n
0− j, m
0) ∈ E
intfor 0 ≤ j ≤ k − 1 and (n
0− k, m
0) ∈ ∂E, we have u(n
0− k, m
0) = u(n
0, m
0) = M by induction.
This is a contradiction.
Remark 2.
In contrast to the maximum principle in [1, Theorem 50], our Lemma 3 requires no connectivity condition on E
int.
Lemma 4.
Adopt the assumptions in Lemma 3. The system (1.6) possesses a unique solution.
Proof. Let F (E) be the set of real-valued functions on E. For w ∈ F (E), we define T w ∈ F (E) by the formula
(T w)(n, m) =
{(L
0w)(n, m) for (n, m) ∈ E
int,
( ˜ Bw)(n, m) + g
0(n, m)w(n, m) for (n, m) ∈ ∂E.
Since T : F (E) → F (E) is linear, we have
dim Ran T = ♯E − dim Ker T ,
where ♯E stands for the number of the elements of E. It follows from Lemma 3 that Ker T = { 0 } , whence Ran T = F(E).
3 Proofs of Theorems
Proof of Theorem 1. We proceed by an argument parallel to that employed in the proof of
[4, Chapter 6, Theorem 1]. Recall the assumption (H.3). We first consider the case where
(I, J ) = (2, 1). Consider the function φ(n) ≡ (1 + λ)
R− (1 + λ)
n, where R is any positive number
satisfying R > n for all (n, m, l) ∈ D and λ is a positive constant for which M − (1 + λ)s < 0.
We note that φ(n) is always positive. Put
δ = λ { (1 + λ)s − M } min
(n,m)∈Eint
(1 + λ)
n−1. We have δ > 0. It follows from (H.2) and (H.3) that
Lφ(n)
=λ(1 + λ)
n−1{ k
1(n, m, l) − (1 + λ)k
2(n, m, l) } −
k
5(n, m, l) −
∑4 j=1
k
j(n, m, l)
φ(n)
≤ λ(1 + λ)
n−1{ k
1(n, m, l) − (1 + λ)k
2(n, m, l) }
≤ λ(1 + λ)
n−1{ M − (1 + λ)s }
≤ − δ.
(3.1)
We set
δ
0= min
(n,m)∈E
φ(n), δ
1= max
(n,m)∈E
φ(n).
Pick an ϵ > 0, arbitrarily. Then, there exists σ ∈
Nsuch that
| f (n, m, l) | < ϵ for (n, m, l) ∈ E
int× σ, ∞ (3.2) and
| h(n, m, l) | < ϵ for (n, m, l) ∈ ∂E × σ, ∞ . (3.3) Consider the function
ψ(n, m, l) ≡ ϵ φ(n)
δ + ϵ φ(n) δ
0+ A φ(n) δ
0(1 − γ)
l−σ, (n, m, l) ∈ E × σ, ∞ , (3.4) where 0 < γ < 1 and A ≥ 0 are constants which we specify later. Using Lφ ≤ − δ, we obtain
Lψ ≤ − ϵ − ϵ δ δ
0− A δ
δ
0(1 − γ)
l−σ− A φ(n)
δ
0{ (1 − γ )
l−σ− (1 − γ)
l−1−σ}
= − ϵ − ϵ δ δ
0− A δ
δ
0(1 − γ)
l−σ+ A φ(n)
δ
0γ(1 − γ)
l−1−σ≤ − ϵ − ϵ δ δ
0− A δ
δ
0(1 − γ)
l−σ+ A δ
1δ
0γ(1 − γ)
l−1−σ. If we take
γ = δ δ
1+ δ , then we have 0 < γ < 1 and δ(1 − γ ) = γδ
1, so that
Lψ ≤ − ϵ on E × σ, ∞ . (3.5)
We have
ψ(n, m, l) > ϵ φ(n)
δ
0≥ ϵ for (n, m, l) ∈ ∂E × σ + 1, ∞ , (3.6) ψ(n, m, σ) > A φ(n)
δ
0≥ A in (n, m) ∈ E. (3.7)
Taking
A = max
(n,m)∈E
| z(n, m, σ) |
and using the inequalities (3.2), (3.3), (3.5), (3.6) and (3.7), we get ψ ± z > 0 on (E × { σ } ) ∪ (∂E × σ + 1, ∞ ), L(ψ ± z) < 0 on E
int× σ + 1, ∞ .
Hence, we can apply Lemma 2 to ψ ± z and 0 on E × σ, ∞ . The result is | z(n, m, l) | < ψ(n, m, l) on E × σ, ∞ . Therefore,
| z(n, m, l) | < ψ(n, m, l) ≤ A
1ϵ + A
2(1 − γ )
l−σfor (n, m, l) ∈ E
int× σ + 1, ∞ , where
A
1= δ
1δ + δ
1δ
0and A
2= A δ
1δ
0.
Thus,
| z(n, m, l) | < 2A
1ϵ for (n, m) ∈ E
intand l ≥ σ + max
{
log
AA1ϵ2
log(1 − γ) , 0
}.
So, we have the assertion.
Likewise, we obtain the conclusion in the case where (I, J) ∈ {(1, 2), (3, 4), (4, 3)}.
Proof of Theorem 3. Set u(n, m, l) = z(n, m, l) − w(n, m). Since z is a solution of (1.3) and w is a solution of (1.5), it follows that
{
Lu = f (n, m, l) − f
0(n, m) − (L − L
0)w on D
0, u(n, m, l) = h(n, m, l) − h
0(n, m) on S.
We have
(L − L
0)w = { k
1(n, m, l) − d
1(n, m) } w(n − 1, m) + { k
2(n, m, l) − d
2(n, m) } w(n + 1, m) + { k
3(n, m, l) − d
3(n, m) } w(n, m − 1) + { k
4(n, m, l) − d
4(n, m) } w(n, m + 1)
− { k
5(n, m, l) − d
5(n, m) } w(n, m).
Using (H.7), we obtain
((L − L
0)w)(n, m, l) → 0 as l → ∞ for (n, m) ∈ E
int.
This, combined with (H.8) and Theorem 1, implies that u(n, m, l) → 0 as l → ∞ for each (n, m) ∈ E
int, i.e. z(n, m, l) → w(n, m) as l → ∞ .
Proof of Theorem 2. Though the proof of Theorem 2 is somewhat similar to that of Theorem 1, we give it for the sake of completeness. We first consider the case where (I, J) = (2, 1). Consider the function φ(n) ≡ (1 + λ)
R− (1 + λ)
n, where R is any positive number satisfying R > n for all (n, m, l) ∈ D and λ is a positive constant for which M − (1 + λ)s < 0. We note that φ(n) is always positive. For any function H(l) > 0,
(B(Hφ))(n, m, l) + g(n, m, l)H(l)φ(n) < H(l) { Kλ(1 + λ)
n−1− µ
1φ(n) }
by (H.4) and (H.6). We take R such that Kλ(1 + λ)
n−1− µ
1{ (1 + λ)
R− (1 + λ)
n} < − µ
2on S for some positive constant µ
2. Then,
(B (Hφ))(n, m, l) + g(n, m, l)H(l)φ(n) < −H(l)µ
2on S. (3.8) Put
δ = λ{(1 + λ)s − M } min
(n,m)∈Eint
(1 + λ)
n−1.
We have δ > 0. It follows from (H.2) and (H.3) that Lφ(n) ≤ −δ (cf. (3.1)). We set δ
0= min
(n,m)∈E
φ(n), δ
1= max
(n,m)∈E
φ(n).
Pick an ϵ > 0, arbitrarily. Then, there exists σ ∈
Nsuch that
| f (n, m, l) | < ϵ for (n, m, l) ∈ E
int× σ, ∞ (3.9) and
| h(n, m, l) | < ϵ for (n, m, l) ∈ ∂E × σ, ∞ . (3.10) Consider the function
ψ(n, m, l) ≡ ϵ φ(n)
δ + ϵ φ(n)
µ
2+ A φ(n)
δ
0(1 − γ)
l−σ, (n, m, l) ∈ E × σ, ∞ ,
where γ = δ/(δ
1+δ) and A = max
(n,m)∈E|z(n, m, σ)|. Note that this function is slightly different from (3.4). As in the derivation of (3.5), we obtain
Lψ ≤ − ϵ on E × σ, ∞ . (3.11)
By (3.8),
(Bψ)(n, m, l) + g(n, m, l)ψ(n, m, l) < − ϵ for (n, m, l) ∈ ∂E × σ + 1, ∞ . (3.12) We have
ψ(n, m, σ) > A φ(n)
δ
0≥ A in (n, m) ∈ E. (3.13)
Using the inequalities (3.9)–(3.13), we get
ψ ± z > 0 on (E × {σ}), L(ψ ± z) < 0 on E
int× σ + 1, ∞ , B(ψ ± z) + g(ψ ± z) < 0 on ∂E × σ + 1, ∞ .
Hence, we can apply Lemma 1 to ψ ± z and 0 on E × σ, ∞ . The result is | z(n, m, l) | < ψ(n, m, l) on E × σ, ∞ . Therefore, we have the assertion as in the discussion at the end of the proof of Theorem 1.
Likewise, we obtain the conclusion in the case where (I, J) ∈ { (1, 2), (3, 4), (4, 3) } .
Proof of Theorem 4. Set u(n, m, l) = z(n, m, l) − w(n, m). Since z is a solution of (1.4) and w is a solution of (1.6), it follows that
{
Lu = f(n, m, l) − f
0(n, m) − (L − L
0)w on D
0,
Bu + g(n, m, l)u = h(n, m, l) − h
0(n, m) − (B − B)w ˜ − (g(n, m, l) − g
0(n, m))w on S.
We have
(L − L
0)w = { k
1(n, m, l) − d
1(n, m) } w(n − 1, m) + { k
2(n, m, l) − d
2(n, m) } w(n + 1, m) + { k
3(n, m, l) − d
3(n, m) } w(n, m − 1) + { k
4(n, m, l) − d
4(n, m) } w(n, m + 1)
− { k
5(n, m, l) − d
5(n, m) } w(n, m).
Using (H.7), we obtain
((L − L
0)w)(n, m, l) → 0 as l → ∞ for (n, m) ∈ E
int. By (H.9),
((B − B)w)(n, m, l) ˜ → 0 as l → ∞ for (n, m) ∈ ∂E and
(g(n, m, l) − g
0(n, m))w(n, m) → 0 as l → ∞ for (n, m) ∈ ∂E.
This, combined with (H.8) and Theorem 2, implies that u(n, m, l) → 0 as l → ∞ for each (n, m) ∈ E, i.e. z(n, m, l) → w(n, m) as l → ∞ .
4 Remarks
4.1
Needless to say, the conditions (H.1), (H.2), (H.3) and (H.7) are satisfied in the case where L is a discrete heat operator, that is,
(Lz)(n, m, l) = K { z(n − 1, m, l)+z(n + 1, m, l) + z(n, m − 1, l) + z(n, m + 1, l) − 4z(n, m, l) }
− { z(n, m, l) − z(n, m, l − 1) } , where K is a positive constant.
4.2
One can state Theorems 1, 2, 3 and 4 in the setting where the dimension of the base E is arbitrary.
4.3
One may suspect that (1.1) is a restrictive condition. However, it is not so. In order to see this, let E be any finite subset of
Z2. Set
E
′=
∪(n,m)∈Eint
[N (n, m) ∪ { (n, m) } ].
Consider the initial boundary value problem
Lz = f on E
int× 1, ∞ , z = g on (E \ E
int) × 1, ∞ , z(n, m, 0) = h(n, m) on E.
Suppose that k
j> 0 on E
int× 1, ∞ for 1 ≤ j ≤ 4. Then, the system above admits a unique solu- tion. The values of the solution z on E
int× 1, ∞ depend only on those of g on (E
′\ E
int) × 1, ∞ and those of h on E
int.
4.4
Consider the homogeneous initial boundary value problem
Lz = 0 on E
int× 1, ∞ , z = 0 on ∂E × 1, ∞ ,
z(n, m, 0) = h(n, m) on E
int,
(4.1)
where E is a subset of
Z2satisfying E
int̸= ∅ and (1.1). The argument in the proof of Theorem
1 gives the following
Assertion 1.
Suppose that (H.1), (H.2), and (H.3) are satisfied. Then, there exist constants A > 0 and 0 < σ < 1 depending only on E, s, and M such that if z is a solution of (4.1), then
| z(n, m, l) | ≤ Aσ
lmax
Eint
| h | , (n, m, l) ∈ D.
This general stability result is not derived in [1].
Appendix
Our purpose here is to observe that the boundary operator B from (1.2) is a discrete analogue of the conormal derivative operator. We achieve this by four steps.
A.1
First, we take notice of the continuous case. Let Ω be a bounded domain in
R2with a smooth boundary. For x ∈ ∂Ω, we denote by N (x) = (N
1(x), N
2(x)) the inward unit normal vector to ∂Ω at x. Consider the operator of the form
H =
∑2 i,j=1
a
ij(x) ∂
2∂x
i∂x
j.
We suppose that the coefficients a
ijbelong to C
1(Ω). For u ∈ C
2(Ω) and v ∈ C
1(Ω), we have
∫
Ω
Hu · vdx = −
∫
∂Ω
Cu · vdS −
∑2 i,j=1
∫
Ω
∂u
∂x
i(x) ∂
∂x
j(a
ij(x)v(x))dx, (A.1) where
C =
∑2 i,j=1
a
ij(x)N
j(x) ∂
∂x
i.
The operator C is called the conormal derivative operator associated with H (cf. [4, Chapter 5, Section 2]).
A.2
Our operator L
0is obtained by discretizing the elliptic operator of the form H
′= a
11(x) ∂
2∂x
21+ a
22(x) ∂
2∂x
22+ b
1(x) ∂
∂x
1+ b
2(x) ∂
∂x
2+ c(x).
A.3
(summation by parts formula) Let k, l ∈
Zand k ≤ l. For a function u on k − 1, l and a function v on k, l, we have
∑l n=k
{ u(n) − u(n − 1) } v(n) = u(l)v(l) − u(k − 1)v(k) −
l−1
∑
n=k
u(n) { v(n + 1) − v(n) } . (A.2)
A.4We define
(T
1u)(n, m) = u(n, m) − u(n − 1, m),
(S
1u)(n, m) = u(n, m) − u(n + 1, m),
(T
2u)(n, m) = u(n, m) − u(n, m − 1),
(S
2u)(n, m) = u(n, m) − u(n, m + 1).
Having been suggested by
A.2, we consider the operator of the form(L
′u)(n, m) = − a
11(n, m)(T
1S
1u)(n, m) − a
22(n, m)(T
2S
2u)(n, m),
where (n, m) ∈ E
intand u is a function on E. We suppose that L
′is elliptic, i.e. a
11≥ 0 and a
22≥ 0 on E
int. Now, we perform a computation similar to (A.1). For m ∈
Z, we define(E
int)
m= { n ∈
Z| (n, m) ∈ E
int} , (E
int)
m= { n ∈
Z| (m, n) ∈ E
int} .
Put F
1= {m ∈
Z| (E
int)
m̸= ∅} and F
2= {m ∈
Z| (E
int)
m̸= ∅}. For each m ∈ F
1, there is a finite sequence of discrete intervals I
lm= a
ml, b
ml, l = 1, 2, ..., j
m, such that a
ml+1− b
ml≥ 2 for l = 1, 2, ..., j
m− 1 and (E
int)
m=
∪jml=1
I
lm. For each m ∈ F
2, there is a finite sequence of discrete intervals J
lm= c
ml, d
ml, l = 1, 2, ..., s
msuch that d
ml+1− c
ml≥ 2 for l = 1, 2, ..., s
m− 1 and (E
int)
m=
∪smj=1
J
lm. Using formula (A.2) we obtain, for real-valued functions u, v on E,
∑
(n,m)∈Eint
(L
′u)(n, m)v(n, m)
= −
∑m∈F1
jm
∑
l=1 bml
∑
n=aml
a
11(n, m) { (S
1u)(n, m) − (S
1u)(n − 1, m) } v(n, m)
−
∑m∈F2
sm
∑
l=1 dml
∑
n=cml
a
22(m, n) { (S
2u)(m, n) − (S
2u)(m, n − 1) } v(m, n)
= −
∑m∈F1
jm
∑
l=1
[a
11(b
ml, m) { u(b
ml, m) − u(b
ml+ 1, m) } v(b
ml, m)
+ a
11(a
ml, m){u(a
ml, m) − u(a
ml− 1, m)}v(a
ml, m)]
−
∑m∈F2
sm
∑
l=1
[a
22(m, d
ml) { u(m, d
ml) − u(m, d
ml+ 1) } v(m, d
ml)
+ a
22(m, c
ml) { u(m, c
ml) − u(m, c
ml− 1) } v(m, c
ml)]
−
∑m∈F1
jm
∑
l=1 b∑ml −1 n=aml
(S
1u)(n, m)(S
1(a
11v))(n, m)
−
∑m∈F2
sm
∑
l=1 d∑ml −1 n=cml
(S
2v)(m, n)(S
2(a
22v))(m, n)
≡ − I
1− I
2− I
3− I
4.
Arranging I
1+ I
2, we obtain the boundary operator of the form (1.2) satisfying (H.5).
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Sui Sun Cheng, Partial Difference Equations, Taylor & Francis, London and New York, 2003.
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G. E. Forsytheand
W. R. Wasow, Finite-Difference Methods for Partial Differential
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T. Fort, Partial Linear Difference Equations, Amer. Math. Monthly
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A. Friedman, Convergence of Solutions of Parabolic Equations to a Steady State, J. Math.
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YU SAKAMAKI