Internat. J. Math. & Math. Sci.
VOL. 13 NO.
(1990) 159-164
159
ITERATIVE SOLUTION OF NEGATIVE EXPONENT EMDEN-FOWLER PROBLEMS
C.D. LUNING
Department of Mathematics
Sam Houston State University
IIunt sville, TX 77340
W.L. PERRY
Department of Mathematics
Texas A & M University
College Station, TX 77843
(Received June 21, 1988)
ABSTRACT.
A monotone iterative technique for proving existence of a positive solution
pair (,y) of y"(x) +X
shown.
a(x)y(x)
<
0, 0
x
<
I,
y(0+)
y(l-)
u < -I is
0,
The method is computationally effective and an example is given.
KEY WORDS AND PHRASES.
Monotone iterations, singular Emden-Fowler problems.
34B15.
1980 AMS SUBJECT CLASSIFICATION CODE.
INTRODUCTION.
In this paper a monotone Iteratlve technique for finding a positive solution
pair (k,y) of
/’(x) +
y(O+)
is shown.
and
a(x)
y(1-)
y(x)
<
0, 0
x
<
(I.I)
O,
We require a(x) to be non-negative and continuous on [0,I] with x -I a(x)
(l-x) -I a(x)
being bounded on (0,1).
and y’(0+) and y’(l-) are
finite.
Under these conditions, a solution exists
was
This
shown
by Tallaferro
methods in that paper were not constructive in nature.
sequence which is
paper.
proved to converge
In other work the authors [2]
to
a
positive
[I];
Development
solution
is
have obtained results for tJ
o
however
the
an iteratlve
this
our
aim
in
>-
I.
Negative
C.D. LUNING AND W.L. PERRY
160
problems
Emden-Fowler
exponent
have
some
significant
catalysis and fluid mechanics (Lunlng and Perry
in
applications
permeable
[3], [4]).
THE MONOTONE ITERATION.
2.
The iteratlve sequence is defined as follows:
(1-x)
(1-)x
K(x,)
0
<
x
x
(
1,
Let
Uo(X)
x,
and
n
u (x)
n
f
0
K(x,) a()u n-I () d
(I-O a()
0
n-- 1,2
....
(2.1)
Un_
1,2,3,..., on
(x), n
Moreover, since K(x,) is the Green’s
The conditions on a(x) guarantee the existence of u
<
0
<
x
and that u’(l-) is finite for all n.
u"
function for
O, u(O)
+
u"(X)n
u(1)
O, the sequences
" (x)
n a(X)Un-I
Un(l-)
u (0+)
n
{Un(X)} {n
satisfy
0
(2.2)
0
u’ (0+)
n
The sequences
LEMMA.
0
<
x
<
2,
{Un(X)} {n
I,
For n
are actually alternating monotone.
U2n_l(X) < U2n+l(X) < U2n(X) < UZn_Z(X)
2n < 2n+2 < 2n+l < 2n-l"
one value of x e
(0,I) such that
Moreover, for any 0
Muir(x)
B(x)
u
<
M
<
there is at most
O, where the pair (i,J) can be
(2n-2,2n-l), (2n,2n-l), (2n-2,2n), or (2n+l,2n-l).
Before proving the lemnm, we note that the result implies that {u
,o (l-)}n=
is a
bounded sequence.
PROOF OF THE LEMMA.
(u0
u
l)’’(x)
so that u
0
> Ul,
u
)’la(x)
0
<
x
First we compare u
<
I.
(x)
>
Let
O, 0
<
fO, l(X)
x
<
M
and u
-
0
1,
I.
We have
(Uo-U I)’(0+)’= (Uo-UI)(0)
u0(x) Ul(X) where 0 < M <
O,
I.
Now
fO,l ’’(x) -Ul"(x) )’I a(x) u((x) > O, 0 < x < 1, fO,l(O) O, fO,l’(O) M-1/-1 <
Thus
fo ,I (x) can cross the x axis at most once interior to (0,I). Since u0 > Ul, the
definition of
k leads to < I"
O.
ITERATIVE SOLUTION OF NEGATIVE EXPONENT EMDEN-FOWLER PROBLEMS
In a completely similar way it is shown that u 0
M-l/Bu0(x)-u2(x) has
f0,2(x)
(0,I) such that
12n+2 < 2n+!
O, 0
<
x
<
l,
To show that
For the induction
>
()
(U2n-U2n+2)
-I/u2n-u2n+l
(M-I/U2n-U2n+?)(x)_ O.
2n+3 < 12n+l"
and
>
<
0, 0
x
<
l,
O, and there is at
(x)
(O,l) such that (M
there is at most one x
most one x
(x)
and
(0,I).
at most one zero interior to
(U2n-U2n+l)
hypothesis we assume
3 < ll’
> ’*2’
161
This implies that
(U2n+2-U2n+l)(X) >
0, 0
<
<
x
I, we
proceed as follows
(u
2 n+ 2
-u
X2n+ 2 a
’2n+
2n+
-n+la(x)Un(X)U2n+l (x)
12n+2
n+l
X2n+2/X2n+l <
Since
(x)]
2n- ( x)-u n+
(U2n+2-U2n+l)
we conclude that
(x)
(0,1) such that
is shown that there is at most one x e
>
O, 0
X2n+3 > )’2n+2" One then proceeds in the same manner
(U2n+3-U2n+l) (x), (M-1/U2n+3-U2n+l) (x), (U2n+2-U2n+3) (x),
(U2n+2-U2n+4)
the ’s.
<
1.
Similarly it
O.
to consider
(M-1/U2n+2-U2n+3)
(x) along with the definition of
(x),
Xn to order
This completes the induction proof of the lemma.
< k2n < 12n+2 < A2n+l < A2n-l’
, llmn
Because 0
2n
llmn
Uo(X),
(M-1/laU2n+2-U2n+4)
(x),
x
(M-1/U2n+2-U2n+l) (x)
Therefore
and
<
0
<
x
<
" 2n_l}
sequences {u
<
A
<
A such that
< U2n+l(X) < U2n(X) < U2n_2(x) <
Since 0
-2n+l
I, the
there exists 0
and
{U2n}
are
uniformly
bounded
monotone
sequence s.
We can also conclude that the sequences are equicontinuous because
By Ascoli’s lemma there exist functions
uniformly bounded.
0
< (x)
[O,l].
4
(x),
0
<
x
<
I, and
limn
and
(R)U2n-I
5,
llmn + -U2n
{lu n ’(x) i}
C[O,I] such that
-u uniformly on
Using the dominated convergence theorem and (2.1) we have
(x)
f K(x,)a()51()
d,
0
(x)
(2.3)
X f K(x,)a()tl()
0
d
or, equivalently,
a"(x) +
a(x)[tl(x)
O,
0
<
x
<
is
C.D. LUNING AND W.L. PERRY
162
"(x)
.
+
G(o)
a(x)(x)
G(o)
fi(1)
0
G(1)
o.
Now we have two possibilities.
some
subinterval
(0,I).
of
l,(o)
G(1)
Either
x
c
o
,
(2.4)
1,
-
(x) on (0,I) or (x) < fi(x) on
the pair (k,y) -(,) is a
and
Otherwise we proceed as follows.
Now let the constant C e (0,1) be chosen so that
-(cG")
<
(x)
then
If
solution to the problem (I.I).
<
O,
a(x)G"
< c 3, a(x)
CI-
.
Since
we have
Then we have
,"
c-P(x) (c[,)"
,
c,(o):cG()
o.
and a subsolutlon
In this way we have a supersolut[on
a(x)
-y"(x)
y(O)
.
yU(x),
y(1)
(cG)"
a(x)
0
<
x
<
C
for the problem
I,
(2.5)
0
p-I
Since x
a(x) and (l-x)-I a(x) are bounded, the function
@ )
Therefore there
f(x,u)= ka(x)u has @f/@u bounded, 0 < x < I, (x) < u < x).
with
exists m
>
0 such that
f(x,)
for all x e
[0,I] and
f(x,n)
,
>
-m(-l)
n e [,@] with
>
n (where one sided limits are used
at 0
and I).
The foregoing conditions are sufficient for the iteration schemes (see Amann
page 648).
[5],
ITERATIVE SOLUTION OF NEGATIVE EXPONENT EMDEN-FOWLER PROBLEMS
-Vk+ l(x) +Vk+ l(x)
Vk+ 1(0)
v
0
,
Vk+ l(0)
or
Vk+ 1(l)
in
,taxi.at qol,tion
Iteratlo
scheme
(2.6).
0
,) of the problem
positive solution pair (\,y) exists for problem (1.[)
(?.I).
Moreover the ei,envalue
the
If
)y
(that is if
iteration scheme (2.[)
i teratlve scheme,
in
(if v
[,].
and is obtainable by monotone iteration.
the
(2.6)
0
@) or a
0
Thus we have shown t-hat
by
Vk(X)
Vo
to converge to a minimal (if
(2,5), the solution lying
a(x) vkU(x) +
i
163
functiou
y
s
not
is determined
determined
by
the
is found by the appllca{ion of the second
To implement the method numerically, special care must be
taken in all calculations near the endpolnts.
3.
EXAL LE.
Calculations were performed for the problem
y"() + k 3(i_x)3
0.
y(1)
y(0)
y- 2 ()
=0
The iterative method was applied using the integral formulation, using Simpson’s rule
away from the endpoints.
"-
’10
and y
Xll "(ulO + u II)/2;
7 173,
"pinched together".
The sequences were alternating monotone and we found
7 2213
that
maxlu
Because
is,
we
-u
<
10
considered
that
.0003 we used Athe
Iterative
(0
+
kl
sequences
)/2
had
The calculated solution was
"-
7.1975
0
0
.2
1.3
.4
.5
where y(x) is symmetric about x
I/.
I181
I".2002
.222
.2292
Vntqueness of solution follows from a result
of "raliaferro [I].
4.
NOTE.
Our conditions on a(x)
existence res,It.
if
f
0
t(l-t)a(t)dt
are
more
For a(x) continuous
< -Ill.
restrictive
e,n
than those
for a nonconstructive
(0,I) a solution exists if and only
164
C.D. LUNING AND W.L. PERRY
REFERENCES
I.
TALIAFERRO, S.., A nontlnear singular boundary value problem Nonlinear
Aalysi 3 (1979), 897-904.
2.
LUNING, C.D. and PERRY, W.L., Positive solutions of negative exponent
generalized Emden-Fowler boundary value problems, qIAM J. ,Math. Analysis 12
3.
(198[), 874-879.
PERRY, W.L., A monotone terative technique for solution of
reaction-diffusion problems
4.
(1984), 353-357.
LUNING, C.D. and PERRY, .L.,
in permeable
tteratlve
catalysts.
J.
p order
(p
<
0)
Comput. Chemistry 5
method for solution of a boundary
value problem in non-Newtonlan fluid flow, J. of Non-Newtonlan Fluid Mech. 15
5.
(1984), 145-154.
AMANN, H., Fixed point equations aud nonlinear elgenvalue problems
Banach spaces, SIAM Review. 18 (1976), 620-70q.
in ordered