Hindawi Publishing Corporation
Boundary Value Problems
Volume 2007, Article ID 38230, 12 pages
doi:10.1155/2007/38230
Research Article
Existence of Positive Solutions for Boundary Value
Problems of Nonlinear Functional Difference Equation
with p-Laplacian Operator
S. J. Yang, B. Shi, and D. C. Zhang
Received 18 March 2007; Accepted 23 May 2007
Recommended by Raul Manasevich
The existence of positive solutions for boundary value problems of nonlinear functional
difference equations with p-Laplacian operator is investigated. Sufficient conditions are
obtained for the existence of at least one positive solution and two positive solutions.
Copyright © 2007 S. J. Yang et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and
reproduction in any medium, provided the original work is properly cited.
1. Introduction
In recent years, boundary value problems of differential and difference equations have
been studied widely and there are many excellent results (see Erbe and Wang [1], Grimm
and Schmitt [2], Gustafson and Schmitt [3], Weng and Jiang [4], Weng and Tian [5],
Wong [6], and Yang et al. [7]). Weng and Guo [8] considered two-point boundary value
problem of a nonlinear functional difference equation with p-Laplacian operator
ΔΦ p Δx(t) + r(t) f xt = 0,
x0 = ϕ ∈ C ,
+
t ∈ [0,T],
Δx(T + 1) = 0,
(1.1)
where Φ p (u) = |u| p−2 u, p > 1, φ(0) = 0, C + = {ϕ | ϕ ∈ C, ϕ(k) ≥ 0, k ∈ [−τ,0]}.
Ntouyas et al. [9] investigated the existence of solutions of a boundary value problem
for functional differential equations
x (t) = f t,xt ,x (t) ,
t ∈ [0,T],
α0 x0 − α1 x (0) = φ,
β0 x(T) + β1 x (T) = A,
(1.2)
2
Boundary Value Problems
where f : [0,T] × Cr × Rn → Rn is a continuous function, ϕ ∈ Cr , A ∈ Rn , Cr =
C([−r,0], Rn ).
Let
R+ = {x | x ∈ R, x ≥ 0 },
[a,b] = {a,...,b},
[a,b) = {a,...,b − 1},
[a, ∞) = {a,a + 1,... }
(1.3)
for a,b ∈ N and a < b. For τ,T ∈ N and 0 τ < T, we define
Cτ = ϕ | ϕ : [−τ,0] −→ R ,
C+τ = ϕ ∈ Cτ | ϕ(ϑ) ≥ 0, ϑ ∈ [−τ,0] .
(1.4)
Then Cτ and C+τ are both Banach spaces endowed with the max-norm
ϕτ = max ϕ(k).
(1.5)
k∈[−τ,0]
For any real function x defined on the interval [−τ,T] and any t ∈ [0,T], we denote by
xt an element of Cτ defined by xt (k) = x(t + k), k ∈ [−τ,0].
In this paper, we consider the following nonlinear difference boundary value problems:
ΔΦ p Δx(t) + r(t) f x(t),xt = 0,
α0 x0 − α1 Δx(0) = h,
t ∈ [1,T],
t ∈ [−τ,0],
(1.6)
β0 x(T + 1) + β1 Δx(T + 1) = A,
where Φ p (u) = |u| p−2 u, p > 1, q > 1 are positive constants satisfying 1/ p + 1/q = 1, Δx(t) =
x(t + 1) − x(t), f : R × Cτ → R is a continuous function, h ∈ C+τ and h(t) ≥ h(0) ≥ 0,
t ∈ [−τ,0], A ∈ R+ , α0 , α1 , β0 < β1 are nonnegative real constants such that
α0 β0 T + α0 β1 + α1 β0 = 0.
(1.7)
At this point, it is necessary to make some remarks on the first boundary condition in
(1.6). This condition is a generalization of the classical condition
α0 x(0) − α1 Δx(0) = c
(1.8)
from ordinary difference equations. Here this condition connects the history x0 with the
single value Δx(0). This is suggested by the well posedness of the BVP (1.6), since the
function f depends on the terms xt and x(t).
The case α0 = 0 must be treated separately, since in this case, the BVP (1.6) is not well
posed. Indeed, if α0 = 0, the first boundary condition yields
−α1 Δx(0) = h,
(1.9)
S. J. Yang et al. 3
where now h must be a constant in R and α1 = 0, because of (1.7). In this case, we consider
the next boundary conditions instead of the two boundary conditions in (1.6):
x0 = x(0),
−α1 Δx(0) = h,
(1.10)
β0 x(T) + β1 Δx(T + 1) = A.
As usual, a sequence {u(−τ),...,u(T + 2)} is said to be a positive solution of BVP (1.6)
if it satisfies (1.6) with u(k) > 0 for k ∈ {1,...,T + 1}.
We will need the following well-known lemma (See Guo [10]).
Lemma 1.1. Assume that X is a Banach space and K ⊂ X is a cone in X. Ω1 , Ω2 are two
open sets in X with 0 ∈ Ω1 ⊂ Ω2 . Furthermore, assume that Ψ : K ∩ (Ω2 \ Ω1 ) → K is a
completely continuous operator and satisfies one of the following two conditions:
(1) Ψx x for x ∈ K ∩ ∂Ω1 , Ψx ≥ x for x ∈ K ∩ ∂Ω2 ;
(2) Ψx x for x ∈ K ∩ ∂Ω2 , Ψx ≥ x for x ∈ K ∩ ∂Ω1 .
Then Ψ has a fixed point in K ∩ (Ω2 \ Ω1 ).
2. Main results
Suppose that x(t) is a solution of BVP (1.6).
If h(0) = 0, then
(i) if α0 = 0, β1 = 0,
⎧ t −1
T
⎪
⎪
⎪
⎪
Φq
r(n) f x(n),xn
⎪
⎪
⎪
n =m
⎪
m=0
⎪
⎪
⎪
⎪
α1
⎪
⎪
⎪
x(1)
⎪
⎨ α0 + α1
x(t) = ⎪
⎪
α1 Δx(0) + h(t)
⎪
⎪
⎪
⎪
⎪
α0
⎪
⎪
⎪
⎪
⎪
⎪
⎪
β1 − β 0
1
⎪
⎪
⎩ A+
x(T + 1)
β1
β1
if t ∈ [1,T + 1],
if t = 0,
(2.1)
if t ∈ [−τ,0),
if t = T + 2;
(ii) if α0 = 0, β1 = 0,
⎧
T
t
−1
⎪
⎪
⎪
⎪
Φq
r(n) f x(n),xn
⎪
⎪
⎪
⎪
n =m
m=0
⎪
⎪
⎪
⎪
α1
⎪
⎪
⎪
x(1)
⎨
x(t) = ⎪
α0 + α1
⎪
⎪
α1 Δx(0) + h(t)
⎪
⎪
⎪
⎪
⎪
⎪
α0
⎪
⎪
⎪1
⎪
⎪
⎪
⎩ A
β0
if t ∈ [1,T],
if t = 0,
(2.2)
if t ∈ [−τ,0),
if t = T + 1;
4
Boundary Value Problems
(iii) if α0 = 0, β1 = 0,
⎧
T
t
−1
⎪
⎪
⎪
⎪
Φ
r(n)
f
x(n),x
⎪
q
n
⎪
⎪
⎪m=0
n =m
⎪
⎪
⎨
x(t) = ⎪x(1) + 1 h
⎪
α1
⎪
⎪
⎪
⎪
⎪
⎪
β1 − β 0
1
⎪
⎪
x(T + 1)
⎩ A+
β1
β1
if t ∈ [1,T + 1],
if t ∈ [−τ,0],
(2.3)
if t = T + 2;
(iv) if α0 = 0, β1 = 0,
⎧
T
t
−1
⎪
⎪
⎪
⎪
Φq
r(n) f x(n),xn
⎪
⎪
⎪
⎪
n =m
⎪m=0
⎪
⎨
x(t) = ⎪x(1) + 1 h
⎪
⎪
α1
⎪
⎪
⎪
⎪
1
⎪
⎪ A
⎪
⎩
β0
if t ∈ [1,T],
if t ∈ [−τ,0],
(2.4)
if t = T + 2.
We only prove (i), the proofs of (ii)–(iv) are similar and we will omit them.
Assume that f ≡ 0, then BVP (1.6) may be rewritten as
ΔΦ p Δx(t) = 0,
α0 x0 − α1 Δx(0) = h,
t ∈ [1,T],
t ∈ [−τ,0],
(2.5)
β0 x(T + 1) + β1 Δx(T + 1) = A.
Assume that x(t) is a solution of system (2.5), then
⎧
⎪
⎪
0
⎪
⎪
⎪
⎪
⎪
⎪
⎨1
if t ∈ [0,T + 1],
x(t) = ⎪ α h(t) if t ∈ [−τ,0),
0
⎪
⎪
⎪
⎪
1
⎪
⎪
⎪
⎩ A
β1
(2.6)
if t = T + 2.
Assume that x(t) is a solution of BVP (1.6). Let u(t) = x(t) − x(t). Then for t ∈ [1,T + 1],
we have u(t) ≡ x(t), and
⎧ t −1
T
⎪
⎪
⎪
⎪
Φq
r(n) f u(n) + x(n),un + xn
⎪
⎪
⎪
⎪
n =m
⎪m=0
⎪
⎨
u(t) = ⎪
α1
u(1)
⎪
α0 + α1
⎪
⎪
⎪
⎪
⎪ β 1 − β0
⎪
⎪
⎪
u(T + 1)
⎩
β1
if t ∈ [1,T + 1],
if t ∈ [−τ,0],
if t = T + 2.
(2.7)
S. J. Yang et al. 5
Let
u =
max u(t),
t ∈[−τ,T+2]
E = y | y : [−τ,T + 2] −→ R ,
K = y | y ∈ E : y(t) =
α1
y(1) for t ∈ [−τ,0],
α0 + α1
(2.8)
β1 − β0
y for t ∈ [1,T + 2] .
β1 (T + 1)
y(t) ≥
Then E is a Banach space endowed with norm · and K is a cone in E.
For y ∈ K, we have y(t) = (α1 /(α0 + α1 ))y(1) for t ∈ [−τ,0]. So,
y =
max y(t) = max y(t).
t ∈[−τ,T+2]
(2.9)
t ∈[1,T+2]
Define an operator Ψ : K → E,
⎧t−1
T
⎪
Φ
r(n)
f
y(n)
+
x(n),
y
+
x
⎪
q
n
n
=
0
=
m
m
n
⎪
⎪
⎪
⎪
⎪
⎪
⎨ α1
if t ∈ [1,T + 1],
Ψy(t) = ⎪ α0 + α1 Ψy(1)
if t ∈ [−τ,0],
⎪
⎪
⎪
⎪
⎪
β 1 − β0
⎪
⎪
⎩
Ψy(T + 1)
(2.10)
if t = T + 2.
β1
Then we may transform our existence problem of BVP (1.6) into a fixed point problem
of the operator (2.10).
By (2.10), we have
Ψy = (Ψy)(T + 1) =
(T + 1)Φq
T
m=0
T
n =0
Φq
T
n =m
r(n) f y(n) + x(n), yn + xn
r(n) f y(n) + x(n), yn + xn
(2.11)
.
Lemma 2.1. Ψ(K) ⊂ K.
Proof. If t ∈ [−τ,0], then Ψy(t) = (α1 /(α0 + α1 ))Ψy(1).
If t ∈ [1,T + 1], then by (2.10) and (2.11), we have
Ψy(t) ≥ Φq
T
n =0
r(n) f y(n) + x(n), yn + xn
β1 − β0
1
Ψy ≥
Ψy .
≥
T +1
β1 (T + 1)
(2.12)
6
Boundary Value Problems
If t = T + 2, then
Ψy(T + 2) =
β1 − β0
β1 − β0
Ψy(T + 1) ≥
Ψy .
β1
β1 (T + 1)
(2.13)
So, by the definition of K, we have Ψ(K) ⊂ K.
Lemma 2.2. Ψ : K → K is completely continuous.
Proof. Notice that yn + xn = (y(n − τ) + x(n − τ),..., y(n) + x(n)). So f : Rτ+2 → R. Then
by [10, Theorem 2.6, page 33], f is completely continuous. Hence, Ψ is completely con
tinuous.
In this paper, we always assume that
(H1 ) Tn=τ+1 r(n) > 0,
(H2 ) f : R+ × C+τ → R+
hold.
Then we have the following main results.
Theorem 2.3. Assume that (H1 ), (H2 ) hold. Then BVP (1.6) has at least one positive solution if the following conditions are satisfied:
(H3 ) there exist ρ1 > 0, such that if ϕ ρ1 + ρ0 , then
f ϕ(n),ϕn bρ1
p −1
;
(2.14)
(H4 ) there exists ρ2 > ρ1 + 2, such that if ϕ ≥ ρ2 , then
f ϕ(n),ϕn ≥ Bρ2
p −1
(2.15)
or
(H5 ) there exists 0 < r1 < ρ1 , such that if ϕ ≥ r1 , then
f ϕ(n),ϕn ≥ Br1
p −1
;
(2.16)
(H6 ) there exists R1 > ρ2 , such that if ϕ R1 + ρ0 , then
f ϕ(n),ϕn BR1
p −1
,
(2.17)
where
ρ0 =
h τ
α0
,
b=
1
(T + 1)Φq
T
n=0 r(n)
,
B=
Φq
1
T
n=0 r(n)
.
(2.18)
Theorem 2.4. Assume that (H1 ), (H2 ) hold. Then BVP (1.6) has at least one positive solution if one of the following conditions is satisfied:
p −1
p −1
(H7 ) limsupϕn τ →0 (f (ϕ(n),ϕn)/ ϕn τ )<m p−1 , liminf ϕn τ →∞ (f (ϕ(n),ϕn)/ ϕn τ )>
M p−1 , h(ϑ) = 0,ϑ ∈ [−τ,0];
p −1
p −1
(H8 ) liminf ϕn τ →0 (f (ϕ(n),ϕn)/ ϕn τ )>M p−1 , limsupϕn τ →∞ (f (ϕ(n),ϕn)/ ϕn τ )<
p
−1
m ,
S. J. Yang et al. 7
where
m=
1
(T + 1)Φq
T
n=0 r(n)
β1 (T + 1)
.
M=
T
β1 − β0 Φ q
n=τ+1 r(n)
,
(2.19)
Theorem 2.5. Assume that (H1 ), (H2 ) hold. Then BVP (1.6) has at least two positive solutions if the conditions (H3 )–(H5 ) or (H3 ), (H4 ), and (H6 ) hold.
Theorem 2.6. Assume that (H1 ), (H2 ) hold. Then BVP (1.6) has at least three positive
solutions if the conditions (H3 )–(H6 ) hold.
3. Proofs of the theorems
Proof of Theorem 2.3. Assume that (H3 ) and (H4 ) hold.
For every y ∈ K ∩ ∂Ωρ1 , y = ρ1 , y + x y + x ρ1 + ρ0 , then by (2.10) and
(H3 ),
Ψy =
T
m=0
T
m=0
Φq
n =m
Φq
T
T
n =m
r(n) f y(n) + x(n), yn + xn
r(n) bρ1
p −1
bρ1 (T + 1)Φq
T
n =0
(3.1)
r(n) = ρ1 = y .
For every y ∈ K ∩ ∂Ωρ2 , y = ρ2 , y + x = max{ρ2 ,ρ0 } ≥ ρ2 , then by (2.10) and
(H4 ),
Ψy =
T
m=0
T
≥
m=0
Φq
n =m
Φq
T
T
n =0
r(n) f y(n) + x(n), yn + xn
n =m
≥ Bρ2 Φq
T
r(n) Bρ2
p −1
(3.2)
r(n) = ρ2 = y .
So by (3.1), (3.2) and Lemma 1.1, there exists one positive fixed point y1 of operator Ψ
with y1 ∈ K ∩ (Ωρ2 \ Ωρ1 ).
Assume that (H5 ) and (H6 ) hold. Similar to the above proof, we have that for every
y ∈ K ∩ ∂Ωr1 ,
Ψy ≥ y ,
(3.3)
Ψy y .
(3.4)
and for every y ∈ K ∩ ∂ΩR1 ,
So by (3.3) and (3.4), there exists one positive fixed point y2 of operator Ψ with y2 ∈
K ∩ (ΩR1 \ Ωr1 ). Consequently, x1 = y1 + x or x2 = y2 + x is a positive solution of BVP
(1.6).
8
Boundary Value Problems
Proof of Theorem 2.4. Assume that (H7 ) holds. By h(ϑ) = 0, ϑ ∈ [−τ,0], we have x(n) = 0
for n ∈ [−τ,T + 1].
From
f ϕ(n),ϕn
< m p −1 ,
limsup p−1
ϕn ϕn τ →0
τ
(3.5)
there exists a constant ρ1 > 0, such that for ϕn τ < ρ1 ,
p −1
f ϕ(n),ϕn mϕn τ
.
(3.6)
Let Ωρ = { y ∈ K | y < ρ}.
For every y ∈ K ∩ ∂Ωρ1 , yn τ y ρ1 , then by (2.10) and (3.6),
Ψy =
T
m=0
T
m=0
Φq
T
n =m
Φq
T
n =m
r(n) f y(n), yn
T
m=0
Φq
T
n =m
r(n)m
p −1 r(n)m p−1 y p−1 m(T + 1) y Φq
T
n =0
p −1
yn τ
(3.7)
r(n) = y .
Furthermore, by
f ϕ(n),ϕn
liminf
ϕn τ →∞
p −1
ϕn > M p −1 ,
(3.8)
τ
there exists a positive constant ρ2 > ρ1 , such that for ϕn τ ≥ ((β1 − β0 )/β1 (T + 1))ρ2 ,
p −1
f ϕ(n),ϕn ≥ M ϕn τ
.
(3.9)
For y ∈ K, we have y(t) ≥ ((β1 − β0 )/β1 (T + 1)) y for t ∈ [1,T + 2]. So, if n ∈ [τ +
1,T + 1], then
yn ≥ β1 − β0 y = β1 − β0 ρ2 .
τ
β1 (T + 1)
(3.10)
β1 (T + 1)
For y ∈ K ∩ ∂Ωρ2 , by (2.10) and (3.9),
Ψy =
≥
T
m=0
Φq
T
m=τ+1
=
T
n =m
Φq
r(n) f y(n), yn
T
n =m
≥
m=τ+1
p −1
r(n) M yn τ
M β1 − β0
y Φ q
β1 (T + 1)
T
n=τ+1
T
Φq
≥ Φq
T
n =m
T
r(n) f y(n), yn
M β1 − β0
r(n)
y
β1 (T + 1)
n=τ+1
p −1 r(n) = y .
(3.11)
S. J. Yang et al. 9
So, by (3.7), (3.11), and Lemma 1.1, there exists a positive fixed point y3 of operator
Ψ with y3 ∈ K ∩ (Ωρ2 \ Ωρ1 ), such that
0 < ρ1 y ρ2 .
(3.12)
Assume that (H8 ) holds. From
f ϕ(n),ϕn
> M p −1 ,
liminf p−1
ϕn τ
ϕn τ →0
(3.13)
there exists a constant ρ1 > 0, such that for ϕn τ < ρ1 ,
p −1
f ϕ(n),ϕn ≥ M ϕn τ
.
(3.14)
For every y ∈ K ∩ ∂Ωρ1 , yn τ y ρ1 , then by (2.10), (3.10), and (3.14),
Ψy =
≥
T
m=0
Φq
T
n =m
T
m=τ+1
Φq
r(n) f y(n) + x(n), yn + xn
T
n =m
r(n) M y τ
M β1 − β0
≥
y Φ q
β1 (T + 1)
p −1
≥
T
≥
m=τ+1
T
m=τ+1
Φq
Φq
T
T
n =m
r(n) f y(n), yn
M β1 − β0
r(n)
y
β1 (T + 1)
n =m
p −1 T
n=τ+1
r(n) = y .
(3.15)
Furthermore, by
f ϕ(n),ϕn
< m p −1 ,
limsup p−1
ϕn τ
ϕn τ →∞
(3.16)
there exists a positive constant N > max{ρ1 , hτ }, such that for ϕn τ ≥ N,
p −1
f ϕ(n),ϕn mϕn τ
.
(3.17)
Let
ρ2 = N + 2
h τ
α0
+ m−1 max m ρ2 +
h τ
α0
, Φq max f ϕ(n),ϕn : ϕn τ ρ2 +
h τ
α0
.
(3.18)
10
Boundary Value Problems
For y ∈ K ∩ ∂Ωρ2 , by (2.10), (3.17),
Ψy =
T
m=0
Φq
T
n =m
(T + 1)Φq
r(n) f y(n) + x(n), yn + xn
T
n =0
(T + 1)Φq
r(n) f y(n) + x(n), yn + xn
(T + 1)Φq
n =0
+
yn τ >N+hτ /α0
T
r(n) f y(n) + x(n), yn + xn
yn τ N+hτ /α0
r(n)
h τ
h τ
, Φq max f ϕ(n),ϕn : ϕn τ ρ2 +
× max m ρ2 +
α0
ρ2 = y .
α0
(3.19)
So, by (3.15), (3.19), and Lemma 1.1, there exists a positive fixed point y4 of operator
Ψ with y4 ∈ K ∩ (Ωρ2 \ Ωρ1 ), such that
0 < ρ1 y ρ2 .
(3.20)
Hence, x3 (t) = y3 (t) + x(t) or x4 (t) = y4 (t) + x(t) is a positive solution of BVP (1.6).
If h(0) = 0, then by the transformation
z=x−
h(0)
,
α0
(3.21)
the BVP (1.6) is reduced to the following BVP:
h(0)
h(0)
,zt +
= 0, t ∈ [1,T]
α0
α0
α0 z0 − α1 Δz(0) = h = h − h(0), t ∈ [−τ,0]
ΔΦ p Δz(t) + r(t) f z(t) +
β0 x(T + 1) + β1 Δx(T + 1) = A +
(3.22)
β0 h(0)
,
α0
where obviously h(0) = 0.
Similar to the above proof, we can prove that BVP (3.22) has at least one positive
solution. Consequently, BVP (1.6) has at least one positive solution.
Proof of Theorem 2.5. By (3.1)–(3.3) and Lemma 1.1, or by (3.1), (3.2), (3.4), and
Lemma 1.1, it is easy to see that BVP (1.6) has two positive solutions.
S. J. Yang et al.
11
Proof of Theorem 2.6. By (3.1)–(3.4) and Lemma 1.1, it is easy to see that BVP (1.6) has
three positive solutions.
4. An example
Consider BVP
ΔΦ3/2 Δx(t) + t f x(t),xt = 0,
x0 − Δx(0) = h,
t ∈ [1,4],
t ∈ [−2,0],
(4.1)
Δx(5) = 1,
where h(t) = −t, for (ϕ(t),ϕt ) ∈ R+ × C+τ ,
⎧
⎪
10−2 ,
⎪
⎪
⎪
⎪
⎪
⎪
⎪ 44 × 10−4
⎪
⎪
⎪
(s − 3)2 + 10−2 ,
⎪
⎪
49
⎨
f ϕ(t),ϕt = ⎪7956 × 10−4 (s − 8),
⎪
⎪
⎪
⎪
⎪
⎪
10−2 100 − 19(s − 52)2 ,
⎪
⎪
⎪
⎪
⎪
⎪
⎩1,
0 < s 3,
3 < s 8,
8 < s 9,
(4.2)
9 < s 52,
52 < s,
where s = ϕ.
In BVP (4.1), p = 3/2, q = 3, T = 4, τ = 2, r(t) = t, α0 = 1, α1 = 1, β0 = 0, β1 = 1,
A = 1, ρ0 = 2, b = 0.02, B = 0.1.
Let r1 = 1, ρ1 = 6, ρ2 = 9, R1 = 50. Then by simple computation, we can show that
⎧ p −1
⎪
= 0.01
if s ≥ r1 = 1,
⎪
⎪≥ Br1 ⎪
p −1
⎪
⎨ bρ1
= 1.44 × 10−2 if s ρ1 + ρ0 = 8,
f ϕ(t),ϕt ⎪ p−1
⎪
≥ Bρ2
= 0.81
if s ≥ ρ2 = 9,
⎪
⎪
⎪ p −1
⎩
Br1
=1
if s R1 + ρ0 = 52,
⎧
⎪
⎪
if t ∈ [0,T + 1],
⎪
⎨0
x(t) = ⎪−t
⎪
⎪
⎩1
(4.3)
if t ∈ [−τ,0),
if t = T + 2.
By Theorem 2.6, BVP (4.1) has three positive solutions
x1 = y1 + x,
x2 = y2 + x,
x3 = y3 + x,
(4.4)
with
y1 ∈ K ∩ Ωρ1 \ Ωr1 ,
y2 ∈ K ∩ Ωρ2 \ Ωρ1 ,
y3 ∈ K ∩ ΩR1 \ Ωρ2 .
(4.5)
Acknowledgment
This work is supported by Distinguished Expert Science Foundation of Naval Aeronautical Engineering Institute.
12
Boundary Value Problems
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S. J. Yang: Institute of Applied Mathematics, Naval Aeronautical Engineering Institute,
Yantai 264001, Shandong, China
Email address: yangshujie@163.com
B. Shi: Institute of Applied Mathematics, Naval Aeronautical Engineering Institute,
Yantai 264001, Shandong, China
Email address: baoshi781@sohu.com
D. C. Zhang: Institute of Applied Mathematics, Naval Aeronautical Engineering Institute,
Yantai 264001, Shandong, China
Email address: dczhang1967@tom.com