- Email: [email protected]
On discrete boundary value problems arising in transport phenomena
On Discrete Boundary Value Problems Arising in Transport Phenomena Peter Y. H. Pang and Ravi P. Agarwal Department of Mathematics National University of Sirzgapore Kent Ridge, Singapore OS11
Transmitted
by V. Lakshmikantham
ABSTRACT
For the discrete boundary value problems arising in transport processes we provide comparison results. These results are used to develop monotone iterative methods for the construction of the maximal and minimal solutions in a sector. The advantage of this technique is that the successive approximations are the solutions of the initial and terminal value problems. Numerical illustration showing the sharpness as well the importance of the obtained results is also included.
1.
INTRODUCTION Let T = {to, t,, . . . , t,v} denote
a set of increasing
time instances;
x: T +
R’” and xk = (xl,..., x”)(t,); y:T + R” and yk = (y’,..., ym)(tk); f:T X R” X R” --) R” andj-(k, x’k, yk) stands for f(tk, xk, yk); g: T X R” X R” + R” and g(k, xk, yk) stands for g(tk, xk, yk); LY and p are given vectors in R” and R”, respectively. Let A denote the forward difference operator, i.e., Ax k = Xktl - xk. In this paper, we shall consider the following discrete system
AXk=fck,
xk> Yk)
(1.1) APPLIED MATHEMATZCS AND COMPUTATION 0 Elsevier Science Inc., 1994 655 Avenue of the Americas, New York, NY 10010
60:193-214
(1994)
193
0096-3003/94/$7.00
194 together
P. Y. H. PANG AND R. P. AGARWAL with the boundary
conditions x0
=
a,
YN =
The motivation to study the problem (l.lI, the natural discrete analog of the continuous
P.
(1.2)
(1.2) is due to the fact that it is boundary value problem
x’ =f(t, x, y) -
y’ = g(t, x, y)
x(O) = a, which occurs in transport
theory,
y(l)
= P
and has been the subject
investigations, e.g., [5, 7, 8, 10, 11, 141. The problem (l.l), includes the two-point focal boundary value problem A%(k) Ah(O)
= Ai,
(1.3) matter of several (1.2) in particular
=f(k,u(k),Au(k),...,b”-‘u(k)) o
(lfp
which has been extensively analyzed in [2, 3, and the references therein]. Discretizations of the potential equation lead to boundary value problems of the form (l.l), (1.2) with f and g 1’mear, e.g., [l]. Further, several numerical methods to solve the boundary value problem (l.l), (1.2) have been proposed in [l] and [4]. The plan of this paper is as follows: In Section 2 for the problem cl.]), (1.2) we shall define upper and lower solutions (u, u), (w, z>, and a U-L pair of upper and lower solutions. Our main result here provides sufficient conditions so that for the U-L pair of upper and lower solutions the inequality (u, u) 2 (w, Z> holds. Examples which dwell upon the restrictiveness and the nontriviality of these conditions are illustrated. An analog of this result for the continuous boundary value problem (1.3) can be modeled rather easily. In Section 3, we prove the existence of a U-L pair of upper and lower solutions of the problem (l.l), (1.2). I n S ec t ion 4, we shall construct the maximal and minimal solutions of (l.l), (1.2) m . a sector (defined by the upper and the lower solutions) by developing an iterative scheme which converges monotonically. The advantage of this method is that the successive approximations are the solutions of the initial value problems rather than the boundary value
195
Discrete Boundary Value Problems
problems. Two numerical examples which demonstrate the monotone convergence are also included. Since monotone iterative methods have superiority over ordinary iterative methods, monotone schemes similar to the one we consider in Section 4 for ordinary (in particular for (I.3)), partial as well as difference equations have been analyzed in [6-131. Finally, in Section 5, we address the questions problem (1.11, (1.2). 2.
COMPARISON
related
to the
uniqueness
of the
solutions
of the
RESULTS
We begin with the following:
DEFINITION 2.1. A function (u, v>: T --f R”‘” (1.2) if and only if
is called an upper solution
of &I),
Similarly,
(2.2) tith
a lower solution (w, z): T -+ Rntm all inequalities reversed.
CONDITION C,. exist nonnegative
(1.2) satisfies
(2.Q
For (t, X, y), (t, x’, y), (t, X, y’) E T X R” X R” there cj, KI; i, j = 1,. . . , n, 1 = 1,. . . , WI such that
constants
f’(t, x’, provided
of (l.l),
y)
-f’(t,
x, y) > q
*‘j - d)
xtp = x P, p = 1, . . . , n, p # j, x’j > xj; and
0
?m, x, y’)
--f”(t,
x, y) > Kg
y”
-
y”)
if y” = yr, r = 1,. . . , m, r # I, y’” < y’.
CONDITION C, in each component
The function
g(t, X, y> is monotonically
nondecreasing
of x, and for (t, X, y), (t, X, y’) E T x R” x Rm there
P. Y. H. PANG AND R. P. AGARWAL
196 exist nonnegative < 1, and
constants
P,‘; r,s = 1, . . . , m such that for each r, Cy= r c,’
0 > g’(t, x, y’) - g’(t, provided
ylq = yq, q = l,...,
DEFINITION 2.2.
Under
x, y> > K( y’” - y”)
m, 9 # s, y’” < y’.
the conditions
C, and C,
a pair of upper
and
lower solutions (u, u), (w, z) is called a U-L pair if for each i = 1,. . . ,n and each t, E T,
where
v - = min(v, 0), and - denotes
the usual dot product
of vectors.
REMARK2.1. It is obvious that if u 2 w and v > z, then (u, II), (w, z) is a U-L pair. The main result of this section asserts that the converse is true as well.
THEOREM 2.1. Let the conditions C, and C, be satisfied, and let (u, v>, (w, z) be a U-L pair of upper and lower solutions, then (u, v) > (w, z).
PROOF. We need to prove the following: (Qr) There does not exist any t, E T such that uk > wk and v; < .zL for any 1 < r < m. Suppose not, and suppose vk - zk. Then,
II; - z: is a (negative)
in view of condition
4+1 > UI,
-w; +fi(k,u,,vk)
-fi(k,wk,
-w; +f”(ku,,vk)
-fi(kw,,v,)
+fi(kw,,v,) >
uf,- w; + Li’(Uk
> 0.
minimal
component
C,, for each 1 < i < n it follows that
-fi(kw,,
zk)
zk)
- wk) + K”. (vk - zk)
of
197
Discrete Boundary Value Problems And, using the condition
Q
C, we find that
v; -
2; -g’(k,uk,v,)
vk) + g’(k,w,,
- g’(kq,
<
v; - .z; -
Q
0.
+gj(kwk,vd
P’.
Q)
(Ok - zk)-
However, then by an inductive argument, we arrive at vi; < .zi for some which contradicts the boundary conditions. 1676m, (Q2) There does not exist any t, E T such that uk 2 wk and u:+ 1 < wL+ 1 for any 1 < i < n. Indeed,
o>u:+l
if not then
-u; - (4+1 - 4)
z=fVw,,v,)
-fi(k,w
=f’(k,uli, ok)
-fi(k,w,,v,)
Q) +fi(k,wk,ok)
-fi(kwk>
Q).
Since by the conditions C,, .f”(k,uk,v,>-f “(lc,u+,vk> z LJ * (IA, - wk> > 0 we must have f ‘(k, wk, v,> -f ‘(k, wk, zk) < 0. But, then again condition C, implies that u[ < z( for some 1 < p Q n. However, this contradicts (91). Next from the boundary conditions ZL,,> wO. Thus, from (Q1> and (Q2> it is clear that oug> .q,, and ur 2 wr. The result now follows by an inductive cl argument.
From the proof of the above theorem it is clear that if for a REMARK 2.2. U-L pair u0 > w,,, uN > zN then (u, w> > (w, z). However, q, = wO, oN = .zN need not imply that (u, V> = (w, .z>.
P. Y. H. PANG AND R. P. AGARWAL
198
To illustrate the restrictiveness and the nontriviality of U-L pairs, we shall consider the following autonomous scaler system (n = m = 1)
Ax, = axk + bYk
-Ayk
=
x()=a,
cxk
ON=
+
(2.3)
dyk
(2.4
P.
EXAMPLE 2.1. Let in the above system (2.3), (2.4) the constants are a = 1, d = $, b = c > 0 and (Y = 0, p = 1 so that the conditions C, and C, with L,\ = 1, K: = 1 and Pi = 1 are satisfied. We shall show that the construction of a U-L pair satisfying ua > wa, + > xN becomes difficult as b increases. In fact, we shall compute the largest time interval, i.e., the maximal N on which the U-L pair exists. By Remark 2.2 such a U-L pair must satisfy (u, u) > (w, z). For this, from (2.3) we have
Uk+l
2t‘,
>
+
bVk 2 274. + bz, >
2
2(uk - wk) + b(V, - zk)
wk+l
which imply that
uk+
vk+l
1 -
-
wk+l
zk+l
’<
-b(u,
- wk) + ;(t$
- zk).
In the following we let & = uk - wk and vk = 2)k - zk. Clearly, positive on T. (i) For b = $, th e ab ove inequalities imply that
5 and 7) are
However, this contradicts the positivity of 5 and 77. Hence, a U-L pair cannot exist for N > 1. The construction of a U-L pair for N = 1 is trivial.
199
Discrete Boundary Value Problems
(ii> Similarly,
for
b = +, the above inequalities
As in (i) this contradicts exist for N > 3. (iii) Once again, for
which contradicts for
imply that
the positivity of 5 and v. Hence,
b = A, the above inequalities
the positivity of 5 and 7. Hence,
N > 4. However,
a U-L
pair
a U-L pair cannot
imply that
a U-L pair cannot exist
on T = {to, t,, t,, t,, t4} can
easily
be
obtained.
EXAMPLE 2.2. Let in the system (2.3) the constants a, b, c, d are nonnegative and satisfy the condition (a + l)(l - d) = -bc. Since such a system can be written as
Yk+l the above condition means (singular) transformation.
= -mk
+
t1 -
that the time
In this case, also following as in Example is necessary that
vk<(a--++)
Thus, the singularity assumption that a U-L pair cannot exist.
advance
map is a noninvertible
2.1, we find that for a U-L pair it
k-l[ -c&l
ek < (u - d + 2)k-‘[(u
d)yk
+ (1 - 47701
+ l)&,
+ bq&
which implies that (1 - d) < 0 concludes
P. Y. H. PANG AND R. P. AGARWAL
200
EXAMPLE 2.3. Once again let in the system (2.3) the constants a, are nonnegative and b = c. It is clear that the solution of the system (u
Uk+l=
+
=
_cuk
wk+l
=
(U
wk+l
=
-czL’k
vk+l
l)u,
+
+
bv,
+
(1 - d)ck
l)wk
+
b, c, d
bZk
+
(1
-
d)?,
together with the initial conditions u0 = wg = (Y and the assumed vO, z0 so that v. 2 z,, and vN > p > zh, hold, should be an upper and lower solution of (2.3), (2.4). In the above we let s1 = b, d, = 1 - d, &+i = (a + 1>6, + bd,, c?k+l
= -c&k f (1 - d>&. It is clear that the above construction U-L pair provided
actually yields a
dk > 0, 1 < k < N. Further,
(i> & > 0 is equivalent
to b < 1 - d; in particular,
if d = a, then we
must have b < f for a U-L pair to be defined on T = {to, t,, ts} (refer to Example 2.1 (i)); (ii) Za > 0 is e q uiva 1ent to (a - d + 2)b2 + (1 - d>‘b < (1 - dj3; in particular, if a = 1 and d = i, then we must have b < 26 - l/20 for a U-L pair to be defined on T = {t,, t,, t,, ts} (refer to Example 2.1 (ii)). In fact, in this case with cr = 0, p = 1 it is easy to construct a table (Table 1); (iii) zd > 0 is equivalent to the following cubic inequality in b b3(1
- d) - b2[( a + 1)2 + (u + l)(l
- b(u + 1)(1 in particular,
- d) + (1 - d)‘]
- d)’ + (1 - d)” > 0;
if a = 1 and d = $ then we must have b < 0.071
pair to be defined
on T = {t,, t,, t,, t,, t4} (refer
3.
OF
EXISTENCE
U-L
PAIRS
to Example
IN BOUNDED
SETS
Let /_L,Y, 4, $ be fixed vectors and the system (1.1) is defined
1cE [ p, VI, y E [ 4, $1. Consider
for a U-L
2.1 (iii)).
for t E T,
the systems
(3.1)
201
Discrete Boundary Value Problems TABLE 1 k 0 1 2 3
xk
0.0000000 0.1052632 0.2631579 0.5515789
0 1 1 1
0.1052632 0.5263158 0.2526316 0.1000000
vk
0.1100000 0.5500000 0.2640000 0.1045000
Uk
Yk
+ + + +
+ + + +
2 1 1 1
0.0000000 0.1100000 0.2750000 0.5764000
wk
+ + + +
2 1 1 1
+ + + +
0 1 1 1
+ + + +
1 1 1 1
zk
- 0.1000000 - 0.1600000 -0.3590000 -0.7059000
+ + + +
1 1 1 1
0.4000000 0.2100000 0.1210000 0.9640000
and
Awk =f(k>w,>
+>
Azk = g(k
$)
-
Since in view of the conditions
Pu,
C, and C, for (w,,
zk), (u,,
uk) E [CL, V] X
0 < k < N we have f(k, uk, rci) >f(k, uk, ok), g(k, Y, ok) > g(k, uk> ok), f(k, wk> 4) z,> and g(k, /A x,> 6 g(k, wk, z,>. The solutions (u, v), (w, z) of (3.11, (3.2) are upper and lower solutions of (1.1), (1.2) and 14,
#I,
uk+
1 -wk+l
=tik
-
zL;k
-[g(k
+f(k,u,,
+> -f(k,‘-%,
v> zk) - g(k,
/J>
+)
zk)].
It is easy to see that if u0 - w0 > 0, then uk - wk > 0 for all k. On the other hand, if there exists some s and k such that vi - zi < 0 (which we may assume
to be the minimum
of such negative
components),
then
by
P. Y. H. PANG AND R. P. AGARWAL conditions
C, and C, we have
<
v; - z; - (Vk - zk)-
But, by an inductive argument this leads to a contradiction to the boundary condition at N. Thus, we have v - z 2 0. Therefore, we can conclude that if solutions of (3.1) and (3.2) exist, then they must form a U-L pair for the system (1.1). The existence
problem
is discussed
in the following:
THEOREM 3.1. In the above setting, suppose that there exist a natural number N and fixed vectors IL”, too E [ p, u] c R” and vO, .q, E [$, I)] c R” such that
for all 0 < k < N domain
[P,
1. Th en, the system (1.11, (1.2) has a U-L pair in the
~1 X [4, $1.
PROOF. The above conditions clearly specify solutions (3.1) and (3.2) which we have shown must form a U-L pair.
REMARK 3.1.
Obviously,
in the
above
theorem
of the
systems 0
the conditions
(ii>-(v)
impose restrictions on the growth of the functions f and g with respect to their arguments. To illustrate this for the linear autonomous system (2.3) we
Discrete Boundq
Value Problems
see that these conditions
respectively
can be written as
qku, + [(cl + l)i-l
+
p
203
(u +
+
@<
Ok
=
(1 - dyv,
(b <
zk
=
(1 -
-
[(l - ,)i-l
d)kz, - [(l -,y
+ (1 - ,y
+ .** +1]cv
+
+ ...
(1 - ,)i-2
Q *;
for all 0 < k < N. In the case (Y > 0, these conditions, especially the first one, lead to stringent conditions on the constant b, as we have already experienced in Examples 2.1-2.3. However, the following example illustrates the existence of a U-L pair in a bounded domain on T with an arbitrary N.
EXAMPLE3.1.
Consider
the scalar system
AX, = 1 + xk - exp( -zjk)
(3.3) with boundary conditions
X”
Following
Theorem
=
0,
yN = 1.
(3.4
3.1, we consider UO
=w,=O
1+
exp( -vk)
Uk+l
= uk
+
‘k-cl
=
vk
-
exp( -k)(uk
+ &,)
zk+l
=
zk
-
eXp( -k)(w,
+ &).
uk
-
204
P. Y. H. PANG AND
Then,
R. P. AGARWAL
it is easy to see that
OGU,,
w,<2k--1,
so that we have the bounds p = 0, v = 2N in the notation of Theorem 3.1. Using these, it is not difficult to establish the bounds C#J= 0, I,!I= 2N+ ’ for u and z. Hence, the problem (3.31, (3.4) satisfies the required boundedness conditions of Theorem 3.1, and therefore it has a U-L pair within the bounds. For N = 10 we can take v,, = 3600 and z,, = 2 to obtain the following U-L pair (Table 2). 4.
MONOTONE
ITERATIVE
METHOD
Here, assuming the existence of upper and lower solutions, where the upper solution is greater than or equal to the lower solution on T, we shall develop a monotone iterative scheme for the boundary value problem (LI), (1.2). This scheme converges to the maximal and minimal solutions of (1.11, (1.2) in the sector defined by the upper and lower solutions. the following conditions which will be used later.
For this, we list
CONDITION D,. There exist upper and lower solutions (1.11, (1.2) satisfying (u, u> > (w, .z>.
(u, zj), (w, z> of
TABLE2 k
uk
0
0.0000000 + 0 0.1000000 + 1
1 2 3 4 5 6 7 a 9 10
0.3000000 + 0.7000000 + 0.1500000 + 0.3100000 + 0.6300000 + 0.1270000 + 0.2550000 + 0.5110000 + 0.1023000 +
w
'k
1 1 2 2 2 3 3 3 4
0.3600000 + 0.7760000 + 0.2565542 + 0.1006105 + 0.4712395 + 0.2793718 + 0.2094340 + 0.1837920 + 0.1743706 + 0.1709062 + 0.1696319 +
4 3 3 3 2 2 2 2 2 2 2
k
0.0000000
+ 0
0.0000000 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000
+ + + + + + +
0.0000000
+ 0
0
0 0 0 0 0 0
0.0000000 + 0 0.0000000 + 0
0.2000000 + 0.1000000 + 0.8160603 + 0.7608394 + 0.7418995 + 0.7351053 + 0.7326287 + 0.7317207 + 0.7313871 + 0.7312645 + 0.7312193 +
1 1 0 0 0 0 0 0 0 0 0
Discrete Boundu y Value Problems
CONDITIONV, .
For
each
205 fixed
I. G k g N,
1 < i < n and (t, x, y),
(t, x’, y) E T X R” x R”, there exists a constant 0 Q M Q 1, such f’(k, x, y) is monotone nondecreasing in each component of y, and
f”(k x, y) -fj(k, whenever
- 2)
wk < x’ < x < uk.
CONDITIOND3.
For each
(t, x, y’) E T x R” x R”, component of x, and
fixed
1 $ k < N, 1 < s < m and
x, y> is monotone
g”(k,
g”fk, x* y) whenever
x‘, y) 23 -M(d
that
nondecreasing
x, y’) > -M( ys -
g”(k,
zk < y’ < y < vk, where
(t , x,
y),
in each
y’“)
hl is the same constant
as in condition
&-
Now let the conditions D, - D, be satisfied, and (5,~): T + R”‘” be given such that (w, z) < ( 5,~) < (u, 0). Consider the following linear boundary value problem
x0
=
@,
YN =
P.
Clearly, (4.1) (4.2) h as a unique solution. We define an operator functions (5, n) by
where (x, y) is the unique solution of (4.1) to define the monotone iterative scheme.
LEMMA4.1.
(4-Q A on such
(4.2). We shall use this operator
The hollowing hold:
(i) If p: T -+ RP sati.&s 0 GM < 1, then pk 0.
Ap,
> - Mpk, with p0 a 0, for some constant
P. Y. H. PANG AND R. P. AGARWAL
206
Zf p: T + RP satisfies Apk < Mp,
(ii)
with pN > 0, for some constant
0 d M < 1, then p > 0.
PROOF. The proof is obvious.
LEMMA4.2. (i)
the unique
Assume that the conditions solution
(u, u); (ii) the operator functions
( 5,7j),
(The conclusion
PROOF.
n
D, - D, hold. Then,
(x, y) of (4.11, (4.2) satisfies (w, z> < (x, y> <
A is monotone
( 5’, 77’) satisfying
in the following (w, z> < (5,771
sense: < (5’,
for each pair of 77’) < (u, 01,
(i) implies that (w, z> < A(w, Z) and A(u, 0) < (u, u).)
To prove (i), we let
p’ = x - w and
p” = u - X. Then,
pS > 0 and
Ap; = Axk - Awk > f( k, &, vk) - M(x,
-
>f(k,
- 5k) -f(k,w,,
&, .zk) - M(x~
> -M(
tk - q)
6,)
-f(k,
wk7 %) G)
- M(xk - &)
= -MP;, also Ap;
z f(k,uk,vk) 2
-M(u,
= -Mp;l.
-f(k> - &)
~5, S)
+ M(x,
+ M(x,
- &)
- &)
pb>
207
Discrete Boundary Value Problems
Thus, by Lemma 4.1 it follows that u > x > W. Similarly, to prove that v 2 I_J> z, we consider p’ = y - x and p” = v - y. Then, we have ph , pk > 0, and
Ap; = Ayk - Azk
also
Ap; < -g(k,u,,uk)
< M(Vk - vk) -
-
f"ftYk
qk)
- MtYk -
+g(k&>qk) vk)
= Mpk”. Again, Lemma 4.1 gives the desired result. To prove (ii>, we let (x, y> = Act, 71, (x’, y’) = Act’, and p” = y’ - y. Then,
pb = pi
77’h P’ = X’ - x,
= 0, and
= Axk - Ax,
Ap;
=f(k,
c$, 7;)
- M( x; -
af(k,
,$,r)k)
-f(k>
> =
-M(
5; -
&)
6;)
‘$k>
-f(k>
vk)
- M(+
-
-
M(?k
6;)
‘6k>vk) -
6%)
+ M(r,
-
+
txk
+
M(xk
-
tk)
-
Sk)
‘tk)
-MP;,
also Ap;
<
-g(k,
&‘, vi)
+ M( y; -
<
-g(k,
s$k,
+
<
M( 7: -
= Mp;.
7$)
qk)
g(k,
+ M( y; -
tkv
vi) qk)
Vi)
+ g(k, +
M(?di
- M( yk -
Sk>
vk)
-
Mt
-
7;)
-
M(Yk
%)
Yk
-
vk)
-
77k)
208
P. Y. H. PANG AND R. P. AGARWAL
Thus, from Lemma 4.1 it follows that (x’, y’) 2 (x, y).
THEOREM 4.3. Let the conditions D, - D, be sntisfied. Then, the sequences {( &), y(J))} and {( xCjj, yv,)} defined by the iterative schemes ( x(j+l),
y(j+‘))
= A (,(I),
y(j));
j
=
0,
1,
. . .
and
where (x(“), y”‘) = (u, v) and (xCgj, y& = (w, z) converge monotonicaZly to a maximal and a minimal solution of the boundary value problem (Ll), (1.2).
PROOF. {Cx(j)>
Y(j))}
By Lemma
4.2 it is clear that the sequences
are well defined
(ZL‘, 2) < (U,V)
...
>
< (&,
and satisfy the monotonicity y(j))
G (,(j+l),
**. a (‘(,j)‘Y(,))
a (“(j+l)’
y(j+r))
Y(j+l))
{(&,
y(j))}
requirements,
and i.e.,
G . . . G (U,v)
a
"*
2
C",')'
Thus, these sequences converge to the functions, say, (X, Y > and (x, y>, respectively. It is clear that (X, Y > and (x, y) are the solutions of the boundary value problem (l.l), (1.2). Th erefore, it remains to show that these functions are the maximal and minimal solutions of (l.l), (1.2) in the sector [u, w] x [v, z]. For this, we assume that (5, v) is a solution of (l.l), (1.2) in this sector and ( xCj _ , ), yC,j Ij > < (5,$ =G(x(j-‘), y (j-l)). Let p’ = 5 - xCij, p” = x(j) - 5, (T’ = 77 - yv, and c#’ = y(j) - 77. Then, & = p’; = cr.&+ cri = 0, and
AP~=f(k, 5ka~k)-f(‘, aff(k,
tcj-l),k,
+ M( 2
=
-“(
- MP;
)
Y(j-l),k
x(j), k - ‘(jtk -
"(j-l),k
Y(j-1j.k)+ M(X(j1.k-~(j-~),t)
X(j-l),k>
-fCk,
‘(jPI),k’Y(j-I),k)
l), k 1
>
+
M(
‘(j),
k - ‘(,j-
11,k >
Discrete
Boundary
Value Problems
209
also
Thus, Lemma 4.2 is applicaand similarly, A oi < Mu: and Au[ < -MU:. ble and it follows that (xcjj, yv,) < (5,~) < (x(j), y(j)). Now an inductive argument easily gives that (x, y> < (5,~) < (X, Y). This completes the n proof of our theorem. To demonstrate the effectiveness of the above iterative procedure, we once again consider the previous examples. For the linear problem in Example 2.1 (ii) with N = 3, the numerical results are presented in Table 3. Next, we consider the problem (3.3), (3.4) with N = 10. Some iterates showing the monotone as well as rapid convergence are listed in Table 4. 5.
UNIQUENESS
For the problem general, uniqueness
CONSIDERATIONS (l.l), (1.2) with (Y = 0 and p = 0 we shall show that, in does not hold. For this, once again we consider the scalar
linear autonomous system (2.3). We shall show that this problem for certain choices of the constants a, b, c, d, has an infinite number of solutions. However, in the generic case (in a sense to be explained later), uniqueness does hold. To show this it suffices to consider the system (2.3) together with the initial conditions x0 = 0, y0 = y, where y is to be determined so that the boundary condition yN = 0 is satisfied. It is easy to see that yN can be expressed in terms of a, 0, c, d and y as
YN =
P(a, b, c, Or,
where P is a polynomial of degree N in four variables. Thus, yN = 0 can be achieved in two ways: (i) if (a, b, c, d) IS a root of P, then any choice of y will give a solution of the problem. In this case, the system has infinitely many solutions; or (ii) if (a, b, c, d) IS not a root of P, which is the generic case, the only solution is given by choosing Y = 0, in which case the only solution is the zero solution.
3
Y(j),3
YY’
‘(j).
Y(j). 2 XY’
YP
‘(jh 2
Y(j), 1 XSp
“WI yp
,$j)
Y(J). 0
Yb’
‘(j).o
Xlr”
j
0.6955472 + 1
0.4036000 + 1
0.2936498 + 1
0.1000000 + 1
0.1000000 + 1
0.1000000 + 1
0.1000000 + 1
0.5764000 + 1
+ 1
0.2109172 + 1
0.5655978 + 1
0.1246000 + 1
-0.5459000
0.2547161 + 1
0.2595000 + 1
0.2692603 + 1
0.2750000 + 1
0.1478378 + 1
0.3908912 + 1
0.2136000 + 1
+ 1
0.5329860 + 1
0.5455000 + 1
-0.1990000
0.6093120 + 0
+ 0
-0.6000000
0.1075835 + 1
0.1070903 + 2
0.1095500 + 2
0.1100000 + 1
0.0000000 + 0
0.0000000 + 0
0.0000000 + 0
0.1000000 + 1
0.1000000 + 1
0.4674909 + 1
0.5558348 + 1
0.2400198 + 1
0.2532720 + 1
0.2268578 + 1
0.2649955 + 1
0.4869069 + 1
0.5283145 + 1
0.9149346 + 0
0.1059603 + 1
0.9443742 + 1
0.1058116 + 2
0.0000000 + 0
10
0.0000000 + 0
5
0.0000000 + 0
1
TABLE 3
0.1000000 + 1
0.1000000 + 1
0.5439459 + 1
0.5519659 + 1
0.2514823 + 1
0.2526899 + 1
0.2598609 + 1
0.2633250 + 1
0.5227271 + 1
0.5264977 + 1
0.1040121 + 1
0.1053266 + 1
0.1042788 + 2
0.1053131 + 2
0.0000000 + 0
0.0000000 + 0
20
0.1000000 + 1
0.1000000 + 1
0.5508844 + 1
0.5516143 + 1
0.2525270 + 1
0.2526369 + 1
0.2629218 + 1
0.2631731 + 1
0.5259893 + 1
0.5263324 + 1
0.1051493 + 1
0.1052689 + 1
0.1051736 + 2
0.1052677 + 2
0.0000000 + 0
0.0000000 + 0
30
0.1000000 + 1
0.1000000 + 1
0.5515599 + 1
0.5515799 + 1
0.2526287 + 1
0.2526317 + 1
0.2631497 + 1
0.2631583 + 1
0.5263068 + 1
0.5263163 + 1
0.1052600 + 1
0.1052633 + 1
0.1052607 + 2
0.1052633 + 2
0.0000000 + 0
0.0000000 + 0
45
0
2
8
6
6
4
4
10
y(J).
10
Yid’
Vj),
Yls”
“W.
LX&f’
Y(j).
yp)
“(jh
xp
Y(j),
#)
X(I).
x,’
Y(j).2
YP
‘(j),
xp
Y(j),0
Yb'
‘(j).
j
0.2940340 + 1
0.2928195 + 1
0.2595371 + 1
0.1479592 + 2
0.1402962 + 2
0.1919088 + 1
0.1679520 + 1
0.6189307 + 2
0.5405985 + 2
0.1417965 + 1
0.1248537 + 1
0.2508922 + 3
0.1772042 + 3
0.1147249 + 1
0.1064565 + 1
0.1010207 + 4
0.4890514 + 3
0.1000000 + 1
0.1000000 + 1
0.2308766 + 2
0.1084841 + 1
0.1500000 + 2
0.2587341 + 1
0.2468175 + 1
0.1010680 + 1
0.6300000 + 2
0.3631678 + 1
0.1458892 + 1
0.1001410 + 1
0.2550000 + 3
0.4669954 + 1
0.1152585 + 1
0.1000168 + 1
0.1023000 + 4
0.5707412 + 1
0.1000000 + 1
0.1000000 + 1
0.6212954 + 1
0.2268781 + 1
0.1496785 + 1
0.1365834 + 3
0.1966193 + 4
0.2994087 + 1
0.0000000 + 0
0.0000000 + 0
0.0000000 + 0
0.3000000 + 1
o.ooooooo +0
o.ooooooo +0
0.0000000 + 0
0.2910048 + 1 0.1471711+ 2
0.1471720 + 2
0.1471858 + 2
0.1472089 + 2
0.1908217 + 1 0.6137408 + 2 0.6137408 + 2 0.1411663 + 1
0.6137483 + 2 0.6137401 + 2 0.1411676 + 1
0.1000000 + 1
0.1000000 + 1
0.9785051 + 3
0.9950454 + 3
0.1142324 + 1
0.1144925 + 1
0.2463916 + 3
0.2479718 + 3
0.1000000 + 1
0.1000000 + 1
0.9926620 + 3
0.9931036 + 3
0.1144548 + 1
0.1144615 + 1
0.2476996 + 3
0.2477645 + 3
0.1411736 + 1 0.1411594 + 1
0.1407849 + 1
0.6137158 + 2
0.6138226 + 2
0.1412287 + 1
0.6124956 + 2
0.6139945 + 2
0.9929156 + 3 0.9929154 + 3
0.9929099 + 3
0.1000000 + 1
0.1000000 + 1 0.1000000 + 1
0.1144587 + 1 0.9929579 + 3
0.1000000 + 1
0.1144587 + 1 0.1144586 + 1
0.2477246 + 3
0.2477246 + 3
0.1144594 + 1
0.2477240 + 3
0.2477305 + 3
0.1411663 + 1
0.1908218 + 1 0.1908215 + 1
0.1908126 + 1
0.1903239 + 1
0.1411661 + 1
0.1471711 + 2
0.1908238 + 1
0.1908361 + 1
0.1471687 + 2
0.1909064 + 1
0.1470579 + 2
0.1471710 + 2
0.2910311 + 1
0.2910048 + 1
0.2911301 + 1
0.2910078 + 1
E -
0.2981337 + 1
0.2910046 + 1
s F * 0’ n% 2
0.2981337 + 1
0.2981337 + 1
0.8033249 + 1
:
ii
ii
Is 3 t
0.2981347 + 1
0.8033056 + 1
0.8033256 + 1
0.2909919 + 1
0.2981319 + 1
0.2981548 + 1
0.8027928 + 1
0.8037378 + 1
o.ooooooo +0
o.ooooooo +0
0.0000000 + 0 0.8160162 + 1
o.ooQoooo + 0
0.0000000 + 0
o.ooooooo +0
30
20
15
0.2903144 + 1
0.2980609 + 1
0.2982054 + 1
0.7906032 + 1
0.1206182 + 2
10
5
1
TABLE 4
212
P. Y. H. PANG AND R. P. AGARWAL
We will illustrate
EXAMPLE 5.1.
this result by an example.
It is easy to see that the system AX k+l = 6x,
with the boundary respectively
conditions
+ :yk
x,, = y3 = 0 has upper
and lower solutions
as follows: U
0 = 0,
u1 = 0.25,
v. =
WO
= 0,
20
2,
u2 = 1.875,
01 = 1,
w1 = 0.125,
= 1,
us = 13.18359
v2 = 0.46875, w, = 0.9375,
zr = 0.5,
z2 = 0.234375,
v3 = 0 ws = 6.591797 z3 = 0.
And these solutions are respectively the maximal and minimal solutions in the sector they define. Further, between these maximal and minimal solutions, it is clear that there are infinitely many solutions of this problem.
REMARK 5.1. We note that in the continuous analogue of the problem (1.11, (1.2) an additional Lipschitz condition is not sufficient for the uniqueness of the solutions, contrary to the assertion in [7, their Theorem 21. For this, we furnish the following counter example: I
x
=Y
-y’=”
x(0)
= y
(1 ;
= 0
for which in the sector [0, sin t] X [O, cos t], (O,O> and (sin t, cos t> are the minimal and maximal solutions. Further, for this problem for each 0 < c < 1, (c sin t, c cos t) is also a solution in this sector.
Discrete
Boundary
213
Value Problems
REFERENCES
1 R. P. Agarwal, scheme Mach. 2
The
method
for higher Sot.
Ser.
R. P. Agarwal,
order
of upper,
A 47:153-170 Da@-ence
lower
hyperbolic
partial
solutions
and monotone
differential
equations,
iterative
J.
Austral.
(1989).
Equations
Marcel
and Inequalities,
Dekker,
New York,
1992. 3
R. P. Agarwal
and B. S. La&
tion, maximum Math. 4
E.
principles,
and
R.
difference
equations
6:345-355
(1970).
Kalaba,
R. Bellman,
Time
Appl. 43~232-240
G. S. Ladde,
dependent
Lakshmikantham
boundary 9:35-41 8
11 12
order discrete
14
B.
for
vector-matrix
/, Optim. Theory
and invariant
and A. S. Vatsala,
G.
Pitman,
Appl.
imbedding,
Monotone Boston,
Monotone
Pachpatte,
arising
in a Banach
in transport
J.
Iterative
Tech-
1985.
method
processes,
Monotone
space,
Appl.
C. V. Pao, Successive Anal.
Q. Sheng
Appl.
for
nonlinear
Math.
Comput.
boundary
Math. Comput.
for solving a two-point
value problems
Modelling
16:101-112
boundary
describing
9:165-174
(1981).
for first and (1992).
value problem,
J.
(1975).
approximations
and R. P. Aganval,
Periodic
PanAmerican
and R. P. Agarwaf, Periodic
equations
(to appear). An Introduction
for a time dependent
scattering
process,
J.
(1975).
Q. Sheng
G. M. Wing,
for equations
Math. Comput.
method
Appl. 49:545-560 equations,
methods
systems,
Phys. 16:1134-1138
differential 13
and
method
conditions,
processes
Equations,
and A. S. Vatsala,
phenomena
C. V. Pao, An iterative
Math.
func-
Comput.
(1973).
P. Y. H. Pang and R. P. Agarwaf, Periodic
Math.
Green’s
value problems,
(1981).
B. G. Pachpatte
second 10
scattering
Differential
value problems
transport 9
interpolation,
numerical
boundary
V. Lakshmikantham,
niques for Nonlinear V.
polynomial
and boundary
A one-sweep
with two-point
Math.
Anal.
bounds
(1993).
Appl. 25:3-39
Angel
Discrete:
error
solutions
of higher order hyperbolic
J. Math. 2:1-22 solutions
to Transport
partial
(1992).
of higher order partial difference
Theory,
Wiley,
New York, 1962.
Transmitted
by V. Lakshmikantham
ABSTRACT
For the discrete boundary value problems arising in transport processes we provide comparison results. These results are used to develop monotone iterative methods for the construction of the maximal and minimal solutions in a sector. The advantage of this technique is that the successive approximations are the solutions of the initial and terminal value problems. Numerical illustration showing the sharpness as well the importance of the obtained results is also included.
1.
INTRODUCTION Let T = {to, t,, . . . , t,v} denote
a set of increasing
time instances;
x: T +
R’” and xk = (xl,..., x”)(t,); y:T + R” and yk = (y’,..., ym)(tk); f:T X R” X R” --) R” andj-(k, x’k, yk) stands for f(tk, xk, yk); g: T X R” X R” + R” and g(k, xk, yk) stands for g(tk, xk, yk); LY and p are given vectors in R” and R”, respectively. Let A denote the forward difference operator, i.e., Ax k = Xktl - xk. In this paper, we shall consider the following discrete system
AXk=fck,
xk> Yk)
(1.1) APPLIED MATHEMATZCS AND COMPUTATION 0 Elsevier Science Inc., 1994 655 Avenue of the Americas, New York, NY 10010
60:193-214
(1994)
193
0096-3003/94/$7.00
194 together
P. Y. H. PANG AND R. P. AGARWAL with the boundary
conditions x0
=
a,
YN =
The motivation to study the problem (l.lI, the natural discrete analog of the continuous
P.
(1.2)
(1.2) is due to the fact that it is boundary value problem
x’ =f(t, x, y) -
y’ = g(t, x, y)
x(O) = a, which occurs in transport
theory,
y(l)
= P
and has been the subject
investigations, e.g., [5, 7, 8, 10, 11, 141. The problem (l.l), includes the two-point focal boundary value problem A%(k) Ah(O)
= Ai,
(1.3) matter of several (1.2) in particular
=f(k,u(k),Au(k),...,b”-‘u(k)) o
(lfp
which has been extensively analyzed in [2, 3, and the references therein]. Discretizations of the potential equation lead to boundary value problems of the form (l.l), (1.2) with f and g 1’mear, e.g., [l]. Further, several numerical methods to solve the boundary value problem (l.l), (1.2) have been proposed in [l] and [4]. The plan of this paper is as follows: In Section 2 for the problem cl.]), (1.2) we shall define upper and lower solutions (u, u), (w, z>, and a U-L pair of upper and lower solutions. Our main result here provides sufficient conditions so that for the U-L pair of upper and lower solutions the inequality (u, u) 2 (w, Z> holds. Examples which dwell upon the restrictiveness and the nontriviality of these conditions are illustrated. An analog of this result for the continuous boundary value problem (1.3) can be modeled rather easily. In Section 3, we prove the existence of a U-L pair of upper and lower solutions of the problem (l.l), (1.2). I n S ec t ion 4, we shall construct the maximal and minimal solutions of (l.l), (1.2) m . a sector (defined by the upper and the lower solutions) by developing an iterative scheme which converges monotonically. The advantage of this method is that the successive approximations are the solutions of the initial value problems rather than the boundary value
195
Discrete Boundary Value Problems
problems. Two numerical examples which demonstrate the monotone convergence are also included. Since monotone iterative methods have superiority over ordinary iterative methods, monotone schemes similar to the one we consider in Section 4 for ordinary (in particular for (I.3)), partial as well as difference equations have been analyzed in [6-131. Finally, in Section 5, we address the questions problem (1.11, (1.2). 2.
COMPARISON
related
to the
uniqueness
of the
solutions
of the
RESULTS
We begin with the following:
DEFINITION 2.1. A function (u, v>: T --f R”‘” (1.2) if and only if
is called an upper solution
of &I),
Similarly,
(2.2) tith
a lower solution (w, z): T -+ Rntm all inequalities reversed.
CONDITION C,. exist nonnegative
(1.2) satisfies
(2.Q
For (t, X, y), (t, x’, y), (t, X, y’) E T X R” X R” there cj, KI; i, j = 1,. . . , n, 1 = 1,. . . , WI such that
constants
f’(t, x’, provided
of (l.l),
y)
-f’(t,
x, y) > q
*‘j - d)
xtp = x P, p = 1, . . . , n, p # j, x’j > xj; and
0
?m, x, y’)
--f”(t,
x, y) > Kg
y”
-
y”)
if y” = yr, r = 1,. . . , m, r # I, y’” < y’.
CONDITION C, in each component
The function
g(t, X, y> is monotonically
nondecreasing
of x, and for (t, X, y), (t, X, y’) E T x R” x Rm there
P. Y. H. PANG AND R. P. AGARWAL
196 exist nonnegative < 1, and
constants
P,‘; r,s = 1, . . . , m such that for each r, Cy= r c,’
0 > g’(t, x, y’) - g’(t, provided
ylq = yq, q = l,...,
DEFINITION 2.2.
Under
x, y> > K( y’” - y”)
m, 9 # s, y’” < y’.
the conditions
C, and C,
a pair of upper
and
lower solutions (u, u), (w, z) is called a U-L pair if for each i = 1,. . . ,n and each t, E T,
where
v - = min(v, 0), and - denotes
the usual dot product
of vectors.
REMARK2.1. It is obvious that if u 2 w and v > z, then (u, II), (w, z) is a U-L pair. The main result of this section asserts that the converse is true as well.
THEOREM 2.1. Let the conditions C, and C, be satisfied, and let (u, v>, (w, z) be a U-L pair of upper and lower solutions, then (u, v) > (w, z).
PROOF. We need to prove the following: (Qr) There does not exist any t, E T such that uk > wk and v; < .zL for any 1 < r < m. Suppose not, and suppose vk - zk. Then,
II; - z: is a (negative)
in view of condition
4+1 > UI,
-w; +fi(k,u,,vk)
-fi(k,wk,
-w; +f”(ku,,vk)
-fi(kw,,v,)
+fi(kw,,v,) >
uf,- w; + Li’(Uk
> 0.
minimal
component
C,, for each 1 < i < n it follows that
-fi(kw,,
zk)
zk)
- wk) + K”. (vk - zk)
of
197
Discrete Boundary Value Problems And, using the condition
Q
C, we find that
v; -
2; -g’(k,uk,v,)
vk) + g’(k,w,,
- g’(kq,
<
v; - .z; -
Q
0.
+gj(kwk,vd
P’.
Q)
(Ok - zk)-
However, then by an inductive argument, we arrive at vi; < .zi for some which contradicts the boundary conditions. 1676m, (Q2) There does not exist any t, E T such that uk 2 wk and u:+ 1 < wL+ 1 for any 1 < i < n. Indeed,
o>u:+l
if not then
-u; - (4+1 - 4)
z=fVw,,v,)
-fi(k,w
=f’(k,uli, ok)
-fi(k,w,,v,)
Q) +fi(k,wk,ok)
-fi(kwk>
Q).
Since by the conditions C,, .f”(k,uk,v,>-f “(lc,u+,vk> z LJ * (IA, - wk> > 0 we must have f ‘(k, wk, v,> -f ‘(k, wk, zk) < 0. But, then again condition C, implies that u[ < z( for some 1 < p Q n. However, this contradicts (91). Next from the boundary conditions ZL,,> wO. Thus, from (Q1> and (Q2> it is clear that oug> .q,, and ur 2 wr. The result now follows by an inductive cl argument.
From the proof of the above theorem it is clear that if for a REMARK 2.2. U-L pair u0 > w,,, uN > zN then (u, w> > (w, z). However, q, = wO, oN = .zN need not imply that (u, V> = (w, .z>.
P. Y. H. PANG AND R. P. AGARWAL
198
To illustrate the restrictiveness and the nontriviality of U-L pairs, we shall consider the following autonomous scaler system (n = m = 1)
Ax, = axk + bYk
-Ayk
=
x()=a,
cxk
ON=
+
(2.3)
dyk
(2.4
P.
EXAMPLE 2.1. Let in the above system (2.3), (2.4) the constants are a = 1, d = $, b = c > 0 and (Y = 0, p = 1 so that the conditions C, and C, with L,\ = 1, K: = 1 and Pi = 1 are satisfied. We shall show that the construction of a U-L pair satisfying ua > wa, + > xN becomes difficult as b increases. In fact, we shall compute the largest time interval, i.e., the maximal N on which the U-L pair exists. By Remark 2.2 such a U-L pair must satisfy (u, u) > (w, z). For this, from (2.3) we have
Uk+l
2t‘,
>
+
bVk 2 274. + bz, >
2
2(uk - wk) + b(V, - zk)
wk+l
which imply that
uk+
vk+l
1 -
-
wk+l
zk+l
’<
-b(u,
- wk) + ;(t$
- zk).
In the following we let & = uk - wk and vk = 2)k - zk. Clearly, positive on T. (i) For b = $, th e ab ove inequalities imply that
5 and 7) are
However, this contradicts the positivity of 5 and 77. Hence, a U-L pair cannot exist for N > 1. The construction of a U-L pair for N = 1 is trivial.
199
Discrete Boundary Value Problems
(ii> Similarly,
for
b = +, the above inequalities
As in (i) this contradicts exist for N > 3. (iii) Once again, for
which contradicts for
imply that
the positivity of 5 and v. Hence,
b = A, the above inequalities
the positivity of 5 and 7. Hence,
N > 4. However,
a U-L
pair
a U-L pair cannot
imply that
a U-L pair cannot exist
on T = {to, t,, t,, t,, t4} can
easily
be
obtained.
EXAMPLE 2.2. Let in the system (2.3) the constants a, b, c, d are nonnegative and satisfy the condition (a + l)(l - d) = -bc. Since such a system can be written as
Yk+l the above condition means (singular) transformation.
= -mk
+
t1 -
that the time
In this case, also following as in Example is necessary that
vk<(a--++)
Thus, the singularity assumption that a U-L pair cannot exist.
advance
map is a noninvertible
2.1, we find that for a U-L pair it
k-l[ -c&l
ek < (u - d + 2)k-‘[(u
d)yk
+ (1 - 47701
+ l)&,
+ bq&
which implies that (1 - d) < 0 concludes
P. Y. H. PANG AND R. P. AGARWAL
200
EXAMPLE 2.3. Once again let in the system (2.3) the constants a, are nonnegative and b = c. It is clear that the solution of the system (u
Uk+l=
+
=
_cuk
wk+l
=
(U
wk+l
=
-czL’k
vk+l
l)u,
+
+
bv,
+
(1 - d)ck
l)wk
+
b, c, d
bZk
+
(1
-
d)?,
together with the initial conditions u0 = wg = (Y and the assumed vO, z0 so that v. 2 z,, and vN > p > zh, hold, should be an upper and lower solution of (2.3), (2.4). In the above we let s1 = b, d, = 1 - d, &+i = (a + 1>6, + bd,, c?k+l
= -c&k f (1 - d>&. It is clear that the above construction U-L pair provided
actually yields a
dk > 0, 1 < k < N. Further,
(i> & > 0 is equivalent
to b < 1 - d; in particular,
if d = a, then we
must have b < f for a U-L pair to be defined on T = {to, t,, ts} (refer to Example 2.1 (i)); (ii) Za > 0 is e q uiva 1ent to (a - d + 2)b2 + (1 - d>‘b < (1 - dj3; in particular, if a = 1 and d = i, then we must have b < 26 - l/20 for a U-L pair to be defined on T = {t,, t,, t,, ts} (refer to Example 2.1 (ii)). In fact, in this case with cr = 0, p = 1 it is easy to construct a table (Table 1); (iii) zd > 0 is equivalent to the following cubic inequality in b b3(1
- d) - b2[( a + 1)2 + (u + l)(l
- b(u + 1)(1 in particular,
- d) + (1 - d)‘]
- d)’ + (1 - d)” > 0;
if a = 1 and d = $ then we must have b < 0.071
pair to be defined
on T = {t,, t,, t,, t,, t4} (refer
3.
OF
EXISTENCE
U-L
PAIRS
to Example
IN BOUNDED
SETS
Let /_L,Y, 4, $ be fixed vectors and the system (1.1) is defined
1cE [ p, VI, y E [ 4, $1. Consider
for a U-L
2.1 (iii)).
for t E T,
the systems
(3.1)
201
Discrete Boundary Value Problems TABLE 1 k 0 1 2 3
xk
0.0000000 0.1052632 0.2631579 0.5515789
0 1 1 1
0.1052632 0.5263158 0.2526316 0.1000000
vk
0.1100000 0.5500000 0.2640000 0.1045000
Uk
Yk
+ + + +
+ + + +
2 1 1 1
0.0000000 0.1100000 0.2750000 0.5764000
wk
+ + + +
2 1 1 1
+ + + +
0 1 1 1
+ + + +
1 1 1 1
zk
- 0.1000000 - 0.1600000 -0.3590000 -0.7059000
+ + + +
1 1 1 1
0.4000000 0.2100000 0.1210000 0.9640000
and
Awk =f(k>w,>
+>
Azk = g(k
$)
-
Since in view of the conditions
Pu,
C, and C, for (w,,
zk), (u,,
uk) E [CL, V] X
0 < k < N we have f(k, uk, rci) >f(k, uk, ok), g(k, Y, ok) > g(k, uk> ok), f(k, wk> 4)
#I,
uk+
1 -wk+l
=tik
-
zL;k
-[g(k
+f(k,u,,
+> -f(k,‘-%,
v> zk) - g(k,
/J>
+)
zk)].
It is easy to see that if u0 - w0 > 0, then uk - wk > 0 for all k. On the other hand, if there exists some s and k such that vi - zi < 0 (which we may assume
to be the minimum
of such negative
components),
then
by
P. Y. H. PANG AND R. P. AGARWAL conditions
C, and C, we have
<
v; - z; - (Vk - zk)-
But, by an inductive argument this leads to a contradiction to the boundary condition at N. Thus, we have v - z 2 0. Therefore, we can conclude that if solutions of (3.1) and (3.2) exist, then they must form a U-L pair for the system (1.1). The existence
problem
is discussed
in the following:
THEOREM 3.1. In the above setting, suppose that there exist a natural number N and fixed vectors IL”, too E [ p, u] c R” and vO, .q, E [$, I)] c R” such that
for all 0 < k < N domain
[P,
1. Th en, the system (1.11, (1.2) has a U-L pair in the
~1 X [4, $1.
PROOF. The above conditions clearly specify solutions (3.1) and (3.2) which we have shown must form a U-L pair.
REMARK 3.1.
Obviously,
in the
above
theorem
of the
systems 0
the conditions
(ii>-(v)
impose restrictions on the growth of the functions f and g with respect to their arguments. To illustrate this for the linear autonomous system (2.3) we
Discrete Boundq
Value Problems
see that these conditions
respectively
can be written as
qku, + [(cl + l)i-l
+
p
203
(u +
+
@<
Ok
=
(1 - dyv,
(b <
zk
=
(1 -
-
[(l - ,)i-l
d)kz, - [(l -,y
+ (1 - ,y
+ .** +1]cv
+
+ ...
(1 - ,)i-2
Q *;
for all 0 < k < N. In the case (Y > 0, these conditions, especially the first one, lead to stringent conditions on the constant b, as we have already experienced in Examples 2.1-2.3. However, the following example illustrates the existence of a U-L pair in a bounded domain on T with an arbitrary N.
EXAMPLE3.1.
Consider
the scalar system
AX, = 1 + xk - exp( -zjk)
(3.3) with boundary conditions
X”
Following
Theorem
=
0,
yN = 1.
(3.4
3.1, we consider UO
=w,=O
1+
exp( -vk)
Uk+l
= uk
+
‘k-cl
=
vk
-
exp( -k)(uk
+ &,)
zk+l
=
zk
-
eXp( -k)(w,
+ &).
uk
-
204
P. Y. H. PANG AND
Then,
R. P. AGARWAL
it is easy to see that
OGU,,
w,<2k--1,
so that we have the bounds p = 0, v = 2N in the notation of Theorem 3.1. Using these, it is not difficult to establish the bounds C#J= 0, I,!I= 2N+ ’ for u and z. Hence, the problem (3.31, (3.4) satisfies the required boundedness conditions of Theorem 3.1, and therefore it has a U-L pair within the bounds. For N = 10 we can take v,, = 3600 and z,, = 2 to obtain the following U-L pair (Table 2). 4.
MONOTONE
ITERATIVE
METHOD
Here, assuming the existence of upper and lower solutions, where the upper solution is greater than or equal to the lower solution on T, we shall develop a monotone iterative scheme for the boundary value problem (LI), (1.2). This scheme converges to the maximal and minimal solutions of (1.11, (1.2) in the sector defined by the upper and lower solutions. the following conditions which will be used later.
For this, we list
CONDITION D,. There exist upper and lower solutions (1.11, (1.2) satisfying (u, u> > (w, .z>.
(u, zj), (w, z> of
TABLE2 k
uk
0
0.0000000 + 0 0.1000000 + 1
1 2 3 4 5 6 7 a 9 10
0.3000000 + 0.7000000 + 0.1500000 + 0.3100000 + 0.6300000 + 0.1270000 + 0.2550000 + 0.5110000 + 0.1023000 +
w
'k
1 1 2 2 2 3 3 3 4
0.3600000 + 0.7760000 + 0.2565542 + 0.1006105 + 0.4712395 + 0.2793718 + 0.2094340 + 0.1837920 + 0.1743706 + 0.1709062 + 0.1696319 +
4 3 3 3 2 2 2 2 2 2 2
k
0.0000000
+ 0
0.0000000 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000
+ + + + + + +
0.0000000
+ 0
0
0 0 0 0 0 0
0.0000000 + 0 0.0000000 + 0
0.2000000 + 0.1000000 + 0.8160603 + 0.7608394 + 0.7418995 + 0.7351053 + 0.7326287 + 0.7317207 + 0.7313871 + 0.7312645 + 0.7312193 +
1 1 0 0 0 0 0 0 0 0 0
Discrete Boundu y Value Problems
CONDITIONV, .
For
each
205 fixed
I. G k g N,
1 < i < n and (t, x, y),
(t, x’, y) E T X R” x R”, there exists a constant 0 Q M Q 1, such f’(k, x, y) is monotone nondecreasing in each component of y, and
f”(k x, y) -fj(k, whenever
- 2)
wk < x’ < x < uk.
CONDITIOND3.
For each
(t, x, y’) E T x R” x R”, component of x, and
fixed
1 $ k < N, 1 < s < m and
x, y> is monotone
g”(k,
g”fk, x* y) whenever
x‘, y) 23 -M(d
that
nondecreasing
x, y’) > -M( ys -
g”(k,
zk < y’ < y < vk, where
(t , x,
y),
in each
y’“)
hl is the same constant
as in condition
&-
Now let the conditions D, - D, be satisfied, and (5,~): T + R”‘” be given such that (w, z) < ( 5,~) < (u, 0). Consider the following linear boundary value problem
x0
=
@,
YN =
P.
Clearly, (4.1) (4.2) h as a unique solution. We define an operator functions (5, n) by
where (x, y) is the unique solution of (4.1) to define the monotone iterative scheme.
LEMMA4.1.
(4-Q A on such
(4.2). We shall use this operator
The hollowing hold:
(i) If p: T -+ RP sati.&s 0 GM < 1, then pk 0.
Ap,
> - Mpk, with p0 a 0, for some constant
P. Y. H. PANG AND R. P. AGARWAL
206
Zf p: T + RP satisfies Apk < Mp,
(ii)
with pN > 0, for some constant
0 d M < 1, then p > 0.
PROOF. The proof is obvious.
LEMMA4.2. (i)
the unique
Assume that the conditions solution
(u, u); (ii) the operator functions
( 5,7j),
(The conclusion
PROOF.
n
D, - D, hold. Then,
(x, y) of (4.11, (4.2) satisfies (w, z> < (x, y> <
A is monotone
( 5’, 77’) satisfying
in the following (w, z> < (5,771
sense: < (5’,
for each pair of 77’) < (u, 01,
(i) implies that (w, z> < A(w, Z) and A(u, 0) < (u, u).)
To prove (i), we let
p’ = x - w and
p” = u - X. Then,
pS > 0 and
Ap; = Axk - Awk > f( k, &, vk) - M(x,
-
>f(k,
- 5k) -f(k,w,,
&, .zk) - M(x~
> -M(
tk - q)
6,)
-f(k,
wk7 %) G)
- M(xk - &)
= -MP;, also Ap;
z f(k,uk,vk) 2
-M(u,
= -Mp;l.
-f(k> - &)
~5, S)
+ M(x,
+ M(x,
- &)
- &)
pb>
207
Discrete Boundary Value Problems
Thus, by Lemma 4.1 it follows that u > x > W. Similarly, to prove that v 2 I_J> z, we consider p’ = y - x and p” = v - y. Then, we have ph , pk > 0, and
Ap; = Ayk - Azk
also
Ap; < -g(k,u,,uk)
< M(Vk - vk) -
-
f"ftYk
qk)
- MtYk -
+g(k&>qk) vk)
= Mpk”. Again, Lemma 4.1 gives the desired result. To prove (ii>, we let (x, y> = Act, 71, (x’, y’) = Act’, and p” = y’ - y. Then,
pb = pi
77’h P’ = X’ - x,
= 0, and
= Axk - Ax,
Ap;
=f(k,
c$, 7;)
- M( x; -
af(k,
,$,r)k)
-f(k>
> =
-M(
5; -
&)
6;)
‘$k>
-f(k>
vk)
- M(+
-
-
M(?k
6;)
‘6k>vk) -
6%)
+ M(r,
-
+
txk
+
M(xk
-
tk)
-
Sk)
‘tk)
-MP;,
also Ap;
<
-g(k,
&‘, vi)
+ M( y; -
<
-g(k,
s$k,
+
<
M( 7: -
= Mp;.
7$)
qk)
g(k,
+ M( y; -
tkv
vi) qk)
Vi)
+ g(k, +
M(?di
- M( yk -
Sk>
vk)
-
Mt
-
7;)
-
M(Yk
%)
Yk
-
vk)
-
77k)
208
P. Y. H. PANG AND R. P. AGARWAL
Thus, from Lemma 4.1 it follows that (x’, y’) 2 (x, y).
THEOREM 4.3. Let the conditions D, - D, be sntisfied. Then, the sequences {( &), y(J))} and {( xCjj, yv,)} defined by the iterative schemes ( x(j+l),
y(j+‘))
= A (,(I),
y(j));
j
=
0,
1,
. . .
and
where (x(“), y”‘) = (u, v) and (xCgj, y& = (w, z) converge monotonicaZly to a maximal and a minimal solution of the boundary value problem (Ll), (1.2).
PROOF. {Cx(j)>
Y(j))}
By Lemma
4.2 it is clear that the sequences
are well defined
(ZL‘, 2) < (U,V)
...
>
< (&,
and satisfy the monotonicity y(j))
G (,(j+l),
**. a (‘(,j)‘Y(,))
a (“(j+l)’
y(j+r))
Y(j+l))
{(&,
y(j))}
requirements,
and i.e.,
G . . . G (U,v)
a
"*
2
C",')'
Thus, these sequences converge to the functions, say, (X, Y > and (x, y>, respectively. It is clear that (X, Y > and (x, y) are the solutions of the boundary value problem (l.l), (1.2). Th erefore, it remains to show that these functions are the maximal and minimal solutions of (l.l), (1.2) in the sector [u, w] x [v, z]. For this, we assume that (5, v) is a solution of (l.l), (1.2) in this sector and ( xCj _ , ), yC,j Ij > < (5,$ =G(x(j-‘), y (j-l)). Let p’ = 5 - xCij, p” = x(j) - 5, (T’ = 77 - yv, and c#’ = y(j) - 77. Then, & = p’; = cr.&+ cri = 0, and
AP~=f(k, 5ka~k)-f(‘, aff(k,
tcj-l),k,
+ M( 2
=
-“(
- MP;
)
Y(j-l),k
x(j), k - ‘(jtk -
"(j-l),k
Y(j-1j.k)+ M(X(j1.k-~(j-~),t)
X(j-l),k>
-fCk,
‘(jPI),k’Y(j-I),k)
l), k 1
>
+
M(
‘(j),
k - ‘(,j-
11,k >
Discrete
Boundary
Value Problems
209
also
Thus, Lemma 4.2 is applicaand similarly, A oi < Mu: and Au[ < -MU:. ble and it follows that (xcjj, yv,) < (5,~) < (x(j), y(j)). Now an inductive argument easily gives that (x, y> < (5,~) < (X, Y). This completes the n proof of our theorem. To demonstrate the effectiveness of the above iterative procedure, we once again consider the previous examples. For the linear problem in Example 2.1 (ii) with N = 3, the numerical results are presented in Table 3. Next, we consider the problem (3.3), (3.4) with N = 10. Some iterates showing the monotone as well as rapid convergence are listed in Table 4. 5.
UNIQUENESS
For the problem general, uniqueness
CONSIDERATIONS (l.l), (1.2) with (Y = 0 and p = 0 we shall show that, in does not hold. For this, once again we consider the scalar
linear autonomous system (2.3). We shall show that this problem for certain choices of the constants a, b, c, d, has an infinite number of solutions. However, in the generic case (in a sense to be explained later), uniqueness does hold. To show this it suffices to consider the system (2.3) together with the initial conditions x0 = 0, y0 = y, where y is to be determined so that the boundary condition yN = 0 is satisfied. It is easy to see that yN can be expressed in terms of a, 0, c, d and y as
YN =
P(a, b, c, Or,
where P is a polynomial of degree N in four variables. Thus, yN = 0 can be achieved in two ways: (i) if (a, b, c, d) IS a root of P, then any choice of y will give a solution of the problem. In this case, the system has infinitely many solutions; or (ii) if (a, b, c, d) IS not a root of P, which is the generic case, the only solution is given by choosing Y = 0, in which case the only solution is the zero solution.
3
Y(j),3
YY’
‘(j).
Y(j). 2 XY’
YP
‘(jh 2
Y(j), 1 XSp
“WI yp
,$j)
Y(J). 0
Yb’
‘(j).o
Xlr”
j
0.6955472 + 1
0.4036000 + 1
0.2936498 + 1
0.1000000 + 1
0.1000000 + 1
0.1000000 + 1
0.1000000 + 1
0.5764000 + 1
+ 1
0.2109172 + 1
0.5655978 + 1
0.1246000 + 1
-0.5459000
0.2547161 + 1
0.2595000 + 1
0.2692603 + 1
0.2750000 + 1
0.1478378 + 1
0.3908912 + 1
0.2136000 + 1
+ 1
0.5329860 + 1
0.5455000 + 1
-0.1990000
0.6093120 + 0
+ 0
-0.6000000
0.1075835 + 1
0.1070903 + 2
0.1095500 + 2
0.1100000 + 1
0.0000000 + 0
0.0000000 + 0
0.0000000 + 0
0.1000000 + 1
0.1000000 + 1
0.4674909 + 1
0.5558348 + 1
0.2400198 + 1
0.2532720 + 1
0.2268578 + 1
0.2649955 + 1
0.4869069 + 1
0.5283145 + 1
0.9149346 + 0
0.1059603 + 1
0.9443742 + 1
0.1058116 + 2
0.0000000 + 0
10
0.0000000 + 0
5
0.0000000 + 0
1
TABLE 3
0.1000000 + 1
0.1000000 + 1
0.5439459 + 1
0.5519659 + 1
0.2514823 + 1
0.2526899 + 1
0.2598609 + 1
0.2633250 + 1
0.5227271 + 1
0.5264977 + 1
0.1040121 + 1
0.1053266 + 1
0.1042788 + 2
0.1053131 + 2
0.0000000 + 0
0.0000000 + 0
20
0.1000000 + 1
0.1000000 + 1
0.5508844 + 1
0.5516143 + 1
0.2525270 + 1
0.2526369 + 1
0.2629218 + 1
0.2631731 + 1
0.5259893 + 1
0.5263324 + 1
0.1051493 + 1
0.1052689 + 1
0.1051736 + 2
0.1052677 + 2
0.0000000 + 0
0.0000000 + 0
30
0.1000000 + 1
0.1000000 + 1
0.5515599 + 1
0.5515799 + 1
0.2526287 + 1
0.2526317 + 1
0.2631497 + 1
0.2631583 + 1
0.5263068 + 1
0.5263163 + 1
0.1052600 + 1
0.1052633 + 1
0.1052607 + 2
0.1052633 + 2
0.0000000 + 0
0.0000000 + 0
45
0
2
8
6
6
4
4
10
y(J).
10
Yid’
Vj),
Yls”
“W.
LX&f’
Y(j).
yp)
“(jh
xp
Y(j),
#)
X(I).
x,’
Y(j).2
YP
‘(j),
xp
Y(j),0
Yb'
‘(j).
j
0.2940340 + 1
0.2928195 + 1
0.2595371 + 1
0.1479592 + 2
0.1402962 + 2
0.1919088 + 1
0.1679520 + 1
0.6189307 + 2
0.5405985 + 2
0.1417965 + 1
0.1248537 + 1
0.2508922 + 3
0.1772042 + 3
0.1147249 + 1
0.1064565 + 1
0.1010207 + 4
0.4890514 + 3
0.1000000 + 1
0.1000000 + 1
0.2308766 + 2
0.1084841 + 1
0.1500000 + 2
0.2587341 + 1
0.2468175 + 1
0.1010680 + 1
0.6300000 + 2
0.3631678 + 1
0.1458892 + 1
0.1001410 + 1
0.2550000 + 3
0.4669954 + 1
0.1152585 + 1
0.1000168 + 1
0.1023000 + 4
0.5707412 + 1
0.1000000 + 1
0.1000000 + 1
0.6212954 + 1
0.2268781 + 1
0.1496785 + 1
0.1365834 + 3
0.1966193 + 4
0.2994087 + 1
0.0000000 + 0
0.0000000 + 0
0.0000000 + 0
0.3000000 + 1
o.ooooooo +0
o.ooooooo +0
0.0000000 + 0
0.2910048 + 1 0.1471711+ 2
0.1471720 + 2
0.1471858 + 2
0.1472089 + 2
0.1908217 + 1 0.6137408 + 2 0.6137408 + 2 0.1411663 + 1
0.6137483 + 2 0.6137401 + 2 0.1411676 + 1
0.1000000 + 1
0.1000000 + 1
0.9785051 + 3
0.9950454 + 3
0.1142324 + 1
0.1144925 + 1
0.2463916 + 3
0.2479718 + 3
0.1000000 + 1
0.1000000 + 1
0.9926620 + 3
0.9931036 + 3
0.1144548 + 1
0.1144615 + 1
0.2476996 + 3
0.2477645 + 3
0.1411736 + 1 0.1411594 + 1
0.1407849 + 1
0.6137158 + 2
0.6138226 + 2
0.1412287 + 1
0.6124956 + 2
0.6139945 + 2
0.9929156 + 3 0.9929154 + 3
0.9929099 + 3
0.1000000 + 1
0.1000000 + 1 0.1000000 + 1
0.1144587 + 1 0.9929579 + 3
0.1000000 + 1
0.1144587 + 1 0.1144586 + 1
0.2477246 + 3
0.2477246 + 3
0.1144594 + 1
0.2477240 + 3
0.2477305 + 3
0.1411663 + 1
0.1908218 + 1 0.1908215 + 1
0.1908126 + 1
0.1903239 + 1
0.1411661 + 1
0.1471711 + 2
0.1908238 + 1
0.1908361 + 1
0.1471687 + 2
0.1909064 + 1
0.1470579 + 2
0.1471710 + 2
0.2910311 + 1
0.2910048 + 1
0.2911301 + 1
0.2910078 + 1
E -
0.2981337 + 1
0.2910046 + 1
s F * 0’ n% 2
0.2981337 + 1
0.2981337 + 1
0.8033249 + 1
:
ii
ii
Is 3 t
0.2981347 + 1
0.8033056 + 1
0.8033256 + 1
0.2909919 + 1
0.2981319 + 1
0.2981548 + 1
0.8027928 + 1
0.8037378 + 1
o.ooooooo +0
o.ooooooo +0
0.0000000 + 0 0.8160162 + 1
o.ooQoooo + 0
0.0000000 + 0
o.ooooooo +0
30
20
15
0.2903144 + 1
0.2980609 + 1
0.2982054 + 1
0.7906032 + 1
0.1206182 + 2
10
5
1
TABLE 4
212
P. Y. H. PANG AND R. P. AGARWAL
We will illustrate
EXAMPLE 5.1.
this result by an example.
It is easy to see that the system AX k+l = 6x,
with the boundary respectively
conditions
+ :yk
x,, = y3 = 0 has upper
and lower solutions
as follows: U
0 = 0,
u1 = 0.25,
v. =
WO
= 0,
20
2,
u2 = 1.875,
01 = 1,
w1 = 0.125,
= 1,
us = 13.18359
v2 = 0.46875, w, = 0.9375,
zr = 0.5,
z2 = 0.234375,
v3 = 0 ws = 6.591797 z3 = 0.
And these solutions are respectively the maximal and minimal solutions in the sector they define. Further, between these maximal and minimal solutions, it is clear that there are infinitely many solutions of this problem.
REMARK 5.1. We note that in the continuous analogue of the problem (1.11, (1.2) an additional Lipschitz condition is not sufficient for the uniqueness of the solutions, contrary to the assertion in [7, their Theorem 21. For this, we furnish the following counter example: I
x
=Y
-y’=”
x(0)
= y
(1 ;
= 0
for which in the sector [0, sin t] X [O, cos t], (O,O> and (sin t, cos t> are the minimal and maximal solutions. Further, for this problem for each 0 < c < 1, (c sin t, c cos t) is also a solution in this sector.
Discrete
Boundary
213
Value Problems
REFERENCES
1 R. P. Agarwal, scheme Mach. 2
The
method
for higher Sot.
Ser.
R. P. Agarwal,
order
of upper,
A 47:153-170 Da@-ence
lower
hyperbolic
partial
solutions
and monotone
differential
equations,
iterative
J.
Austral.
(1989).
Equations
Marcel
and Inequalities,
Dekker,
New York,
1992. 3
R. P. Agarwal
and B. S. La&
tion, maximum Math. 4
E.
principles,
and
R.
difference
equations
6:345-355
(1970).
Kalaba,
R. Bellman,
Time
Appl. 43~232-240
G. S. Ladde,
dependent
Lakshmikantham
boundary 9:35-41 8
11 12
order discrete
14
B.
for
vector-matrix
/, Optim. Theory
and invariant
and A. S. Vatsala,
G.
Pitman,
Appl.
imbedding,
Monotone Boston,
Monotone
Pachpatte,
arising
in a Banach
in transport
J.
Iterative
Tech-
1985.
method
processes,
Monotone
space,
Appl.
C. V. Pao, Successive Anal.
Q. Sheng
Appl.
for
nonlinear
Math.
Comput.
boundary
Math. Comput.
for solving a two-point
value problems
Modelling
16:101-112
boundary
describing
9:165-174
(1981).
for first and (1992).
value problem,
J.
(1975).
approximations
and R. P. Aganval,
Periodic
PanAmerican
and R. P. Agarwaf, Periodic
equations
(to appear). An Introduction
for a time dependent
scattering
process,
J.
(1975).
Q. Sheng
G. M. Wing,
for equations
Math. Comput.
method
Appl. 49:545-560 equations,
methods
systems,
Phys. 16:1134-1138
differential 13
and
method
conditions,
processes
Equations,
and A. S. Vatsala,
phenomena
C. V. Pao, An iterative
Math.
func-
Comput.
(1973).
P. Y. H. Pang and R. P. Agarwaf, Periodic
Math.
Green’s
value problems,
(1981).
B. G. Pachpatte
second 10
scattering
Differential
value problems
transport 9
interpolation,
numerical
boundary
V. Lakshmikantham,
niques for Nonlinear V.
polynomial
and boundary
A one-sweep
with two-point
Math.
Anal.
bounds
(1993).
Appl. 25:3-39
Angel
Discrete:
error
solutions
of higher order hyperbolic
J. Math. 2:1-22 solutions
to Transport
partial
(1992).
of higher order partial difference
Theory,
Wiley,
New York, 1962.