Volume 2011, Article ID 483057,10pages doi:10.1155/2011/483057
Research Article
Iterative Solutions of Singular Boundary Value
Problems of Third-Order Differential Equation
Peiguo Zhang
Department of Elementary Education, Heze University, Heze 274000, Shandong, China
Correspondence should be addressed to Peiguo Zhang,[email protected]
Received 19 January 2011; Revised 20 February 2011; Accepted 6 March 2011
Academic Editor: Kanishka Perera
Copyrightq2011 Peiguo Zhang. 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.
By using the cone theory and the Banach contraction mapping principle, the existence and uniqueness results are established for singular third-order boundary value problems. The theorems obtained are very general and complement previous known results.
1. Introduction
Third-order differential equations arise in a variety of different areas of applied mathematics and physics, such as the deflection of a curved beam having a constant or varying cross section, three-layer beam, electromagnetic waves, or gravity-driven flows1. Recently, third-order boundary value problems have been studied extensively in the literaturesee, e.g.,2– 13, and their references. In this paper, we consider the following third-order boundary value problem:
ut ft, ut 0, t∈0,1,
u0 u0 0, u1 αuη, 1.1
whereft, x∈C0,1×−∞,∞,−∞,∞, 0< η <1.
ft, x is separable by using cone expansion-compression fixed point theorem. In 3, the singular third-order three-point BVP1.1is considered under some conditions concerning the first eigenvalues corresponding to the relevant linear operators, where 1 < α < 1/η,
ft, x is separable and is not necessary to be nonnegative, and the existence results of nontrivial solutions and positive solutions are given by means of the topological degree theory. Motivated by the above works, we consider the singular third-order three-point BVP
1.1. Here, we give the unique solution of BVP1.1under the conditions thatαη /1 and
ft, x is mixed nonmonotone in x and does not need to be separable by using the cone theory and the Banach contraction mapping principle.
2. Preliminaries
LetJ 0,1,I 0,1. By2, Lemma 2.1, we have thatxis a solution of1.1if and only if
xt 1
0
Gt, sfs, xsds, t∈I, 2.1
where
Gt, s 1
21−αη ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
1−αηs2−2ts−1−αt2s, s≤min η, t,
αη−1t2 1−αt2s, t≤s≤η,
1−αηs2−2tss−αηt2, η≤s≤t,
s−1t2, max η, t≤s.
2.2
It is shown in 2 that Gt, s is the Green’s function to −u 0, u0 u0 0, and
u1 αuη. Let
ht, s 1
s1−s|Gt, s|,
I1t 1
0
ht, sds,
In1t 1
0
ht, sInsds, n1,2, . . .,
rG lim
n→ ∞
sup
t∈J
Int −1/n
.
2.3
It is easy to see thatrG>0.
Lemma 2.1Guo14,15. P is generating if and only if there exists a constantτ > 0such that every elementx ∈CIcan be represented in the formxy−z, wherey, z∈Pandy ≤τx,
3. Singular Third-Order Boundary Value Problem
This section discusses singular third-order boundary value problem1.1.
LetP {x ∈ CI | xt ≥ 0, ∀t ∈ 0,1}. Obviously, P is a normal solid cone of Banach spaceCI; by16, Lemma 2.1.2, we have thatPis a generating cone inCI.
Theorem 3.1. Suppose thatft, x gt, x, x, and there exist two positive linear bounded operators
B:CI → CIandC:CI → CIwithrBC< rGsuch that for anyt∈I,x1, x2, y1, y2∈ CI,x1≥x2,y1 ≤y2, we have
−Bx1−x2−C
y2−y1
≤t1−tgt, x1, y1
−t1−tgt, x2, y2
≤Bx1−x2 C
y2−y1
,
3.1
and there existsx0, y0∈CI, such that
1
0
t1−tgt, x0t, y0t
dtcan converge toσ∈R. 3.2
Then1.1has a unique solutionx∗inCI. And moreover, for anyx0 ∈CI, the iterative sequence
xnt 1
0
Gt, sft, xn−1sds n1,2, . . . 3.3
converges tox∗ n → ∞.
Remark 3.2. Recently, in the study of BVP1.1, almost all the papers have supposed that the Green’s functionGt, sis nonnegative. However, the scope ofαis not limited to 1< α <1/η
inTheorem 3.1, so, we do not need to suppose thatGt, sis nonnegative.
Remark 3.3. The functionf inTheorem 3.1is not monotone or convex; the conclusions and the proof used in this paper are different from the known papers in essence.
Proof. It is easy to see that, for anyt∈J,ht, scan be divided into finite partitioned monotone and bounded function on0,1, and then by3.2, we have
1
0
Gt, sgs, x0s, y0s
dsconverges toσt∈R. 3.4
For anyx, y∈CI, letu|x0||x|,v−|y0| − |y|, thenu≥x0, v≤y0, by3.1, we have
−Bu−x0t−C
y0−v
t≤t1−tgt, ut, vt−t1−tgt, x0t, y0t
≤Bu−x0t C
y0−v
Hence
t1−tgt, ut, vt−t1−tgt, x0t, y0t≤ Bu−x0tC
y0−v
t ≤ Bu−x0Cy0−v.
3.6
Following the former inequality, we can easily have
1
0
Gt, sgs, us, vs−gs, x0s, y0s
dsconverges to some elementσ1t∈R;
3.7
thus
1
0
Gt, sgs, us, vsd 1
0
Gt, sgs, x0s, y0s
ds
1
0
Gt, sgs, us, vs−gs, x0s, y0s
dsis converged.
3.8
Similarly, byu≥x, v≤yand01Gt, sgs, us, vsdsbeing converged, we have that
1
0
Gt, sgs, xs, ysdsconverges to some elementσ2t∈R. 3.9
Define the operatorA:CI×CI → CIby
Ax, yt 1
0
Gt, sgs, xs, ysds, ∀t∈I. 3.10
Thenxis the solution of BVP1.1if and only ifxAx, x. Let
Sxt 1
0
ht, sBxsds, Tyt 1
0
By3.1and3.10, for anyx1, x2, y1, y2∈CI,x1≥x2,y1 ≤y2, we have
−Sx1−x2−T
y2−y1
≤Ax1, y1
−Ax2, y2
≤Sx1−x2 T
y2−y1
, 3.12
STxt 1
0
ht, sBCxsds,
STn1xt 1
0
ht, sBCSTnxsds
BCn1In1t, n1,2, . . . , STn ≤ BCnsup
t∈J
Int,
rST ≤ rBC rG <1,
3.13
so we can choose anα, which satisfies limn→ ∞STn1/nrST< α <1, and so there exists a positive integern0such that
STn< αn<1, n≥n0. 3.14
SincePis a generating cone inCI, fromLemma 2.1, there existsτ >0 such that every elementx∈CIcan be represented in
xy−z, y, z∈P, y ≤τx, z ≤τx. 3.15
This implies
−yz≤x≤yz. 3.16
Let
x0 inf{u |u∈P,−u≤x≤u}. 3.17
By3.16, we know thatx0 is well defined for anyx∈CI. It is easy to verify that
· 0is a norm inCI. By3.15–3.17, we get
x0≤ yz ≤2τx, ∀x∈CI. 3.18
On the other hand, for anyu∈Pwhich satisfies−u≤x≤u, we haveθ≤xu≤2u. Thusx ≤ xu −u ≤2N1u, whereNdenotes the normal constant ofP. Since
uis arbitrary, we have
It follows from3.18and3.19that the norms · 0and · are equivalent. Now, for anyx, y∈CIandu∈Pwhich satisfies−u≤x−y≤u, let
u1 1 2
xy−u, u2 1 2
x−yu, u3 1 2
−xyu; 3.20
thenx≥u1,y≥u1,x−u1u2,y−u1u3, andu2u3u. It follows from3.12that
−Su2≤Ax, x−Au1, x≤Su2, 3.21 −Su3−Tu2≤A
y, u1
−Au1, x≤Su2Tu3, 3.22
−Tu3 ≤A
y, u1
−Ay, y≤Tu3. 3.23
Subtracting3.22from3.21 3.23, we obtain
−STu≤Ax, x−Ay, y≤STu. 3.24
LetAx Ax, x; then we have
−STu≤Ax−Ay≤STu. 3.25
AsSandT are both positive linear bounded operators, so,ST is a positive linear bounded operator, and thereforeSTu∈P. Hence, by mathematical induction, it is easy to know that for natural numbern0in3.14, we have
−STn0u≤An0x−An0y≤STn0u, STn0u∈P. 3.26
SinceSTn0u∈P, we see that
An0x−An0y
0≤ ST
n0u, 3.27
which implies by virtue of the arbitrariness ofuthat
An0x−An0y
0≤ ST
n0x−y 0 ≤α
n0x−y
0. 3.28
Example 3.4. In this paper, the results apply to a very wide range of functions, we are following only one example to illustrate.
Consider the following singular third-order boundary value problem:
ut k1tm1
1−ttant
u2t k2tm22 t
0
ps, us
ttan1−sds, t∈J,
u0 u0 0, u1 αuη,
3.29
wherek1, m1, k2, m2 ∈Rand there existsM≥0, such that for anyt∈I,y1, y2∈CI,y1≤y2, we have
−My2−y1
t≤pt, y1t
−pt, y2t
≤My2−y1
t. 3.30
ApplyingTheorem 3.1, we can find that3.29has a unique solutionx∗t∈C2Iprovided Nmax{|m1|,|k1m1|}< rG. And moreover, for anyw0∈CI, the iterative sequence
wnt 1
0 Gt, s
k1sm1 stan1−s
w2
n−1s k2sm22 s
0
pτ, wn−1τ
1−τtans dτ
ds,
n1,2, . . .
3.31
converges tox∗ n → ∞. To see that, we put
gt, xt, yt k1tm1
1−ttant
x2t k2tm22 t
0
ps, ys ttan1−sds,
Bxt Nxt, Cyt M t
0
ysds.
3.32
Then3.1is satisfied for anyt∈I,x1, x2, y1, y2∈CI,x1≥x2, andy1≤y2. In fact, ifx1t x2t, then
t1−tgt, x1t, y1t
−t1−tgt, x2t, y2t
≤k1tm1
x2
1t k2tm2 2−k
1tm1
x2
2t k2tm2 2
t
0
ps, y1s
−ps, y2s ds t 0
ps, y1s
−ps, y2s ds ≤ t 0
My2s−y1s
ds
Bx1−x2t C
y2−y1
t.
Ifx1t> x2t, then
t1−tgt, x1t, y1t
−t1−tgt, x2t, y2t
≤k1tm1
x21t k2tm22−k1tm1
x22t k2tm22
t
0
ps, y1s
−ps, y2s
ds
k1tm1
x21t−x22t
x2
1t k2tm2 2x2
2t k2tm2 2
t
0
ps, y1s
−ps, y2s
ds
≤k1tm1|x1t| − |x2t| t
0
My2s−y1s
ds
≤ |k1tm1|x1t−x2t M t
0
y2s−y1s
ds
≤Bx1−x2t C
y2−y1
t.
3.34
Similarly,
t1−tgt, x1t, y1t
−t1−tgt, x2t, y2t
≥ −Bx1−x2t−C
y2−y1
t. 3.35
Next, for anyt∈I, by3.30and3.32, we get
Tut ≤Mtuc. 3.36
Then, from3.32and3.36, we have
T2ut≤M t
0
Tusds≤Muc t
0
s ds M 2t2
2! uc, ∀t∈I, 3.37
so it is easy to know by induction, for anyn, we get
Tnut ≤ M
ntn
n! uc, ∀t∈I; 3.38
thus
Tnumax
t∈I T
nut ≤ Mntn
so
rT lim
n→ ∞T
n1/n0. 3.40
then we get
rBC≤rB rC N0< rG. 3.41
Letx0 y01; then 1
0
t1−tgt, x0t, y0t
dtis converged. 3.42
Thus all conditions inTheorem 3.1are satisfied.
Acknowledgment
The author is grateful to the referees for valuable suggestions and comments.
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