- Research Article
- Open access
- Published:
Existence and Uniqueness of Solutions for Higher-Order Three-Point Boundary Value Problems
Boundary Value Problems volume 2009, Article number: 362983 (2009)
Abstract
We are concerned with the higher-order nonlinear three-point boundary value problems: with the three point boundary conditions
;
where
is continuous,
are continuous, and
are arbitrary given constants. The existence and uniqueness results are obtained by using the method of upper and lower solutions together with Leray-Schauder degree theory. We give two examples to demonstrate our result.
1. Introduction
Higher-order boundary value problems were discussed in many papers in recent years; for instance, see [1–22] and references therein. However, most of all the boundary conditions in the above-mentioned references are for two-point boundary conditions [2–11, 14, 17–22], and three-point boundary conditions are rarely seen [1, 12, 13, 16, 18]. Furthermore works for nonlinear three point boundary conditions are quite rare in literatures.
The purpose of this article is to study the existence and uniqueness of solutions for higher order nonlinear three point boundary value problem
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ1_HTML.gif)
with nonlinear three point boundary conditions
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ2_HTML.gif)
where ,
is a continuous function,
are continuous functions, and
are arbitrary given constants. The tools we mainly used are the method of upper and lower solutions and Leray-Schauder degree theory.
Note that for the cases of or
in the boundary conditions (1.2), our theorems hold also true. However, for brevity we exclude such cases in this paper.
2. Preliminary
In this section, we present some definitions and lemmas that are needed to our main results.
Definition 2.1.
are called lower and upper solutions of BVP (1.1), (1.2), respectively, if
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ3_HTML.gif)
Definition 2.2.
Let be a subset of
. We say that
satisfies the Nagumo condition on
if there exists a continuous function
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ4_HTML.gif)
Lemma 2.3 (see [10]).
Let be a continuous function satisfying the Nagumo condition on
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ5_HTML.gif)
where are continuous functions such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ6_HTML.gif)
Then there exists a constant (depending only on
and
such that every solution
of (1.1) with
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ7_HTML.gif)
satisfies
Lemma 2.4.
Let be a continuous function. Then boundary value problem
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ8_HTML.gif)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ9_HTML.gif)
has only the trivial solution.
Proof.
Suppose that is a nontrivial solution of BVP (2.6), (2.7). Then there exists
such that
or
. We may assume
. There exists
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ10_HTML.gif)
Then ,
. From (2.6) we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ11_HTML.gif)
which is a contradiction. Hence BVP (2.6), (2.7) has only the trivial solution.
3. Main Results
We may now formulate and prove our main results on the existence and uniqueness of solutions for -order three point boundary value problem (1.1), (1.2).
Theorem 3.1.
Assume that
(i)there exist lower and upper solutions of BVP (1.1), (1.2), respectively, such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ12_HTML.gif)
(ii) is continuous on
,
is nonincreasing in
on
, and
is nonincreasing in
on
and satisfies the Nagumo condition on
, where
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ13_HTML.gif)
(iii) is continuous on
, and
is nonincreasing in
and nondecreasing in
on
;
(iv) is continuous on
, and nonincreasing in
and nondecreasing in
on
Then BVP (1.1), (1.2) has at least one solution such that for each
,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ14_HTML.gif)
Proof.
For each define
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ15_HTML.gif)
where ,
.
For , we consider the auxiliary equation
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ16_HTML.gif)
where is given by the Nagumo condition, with the boundary conditions
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ17_HTML.gif)
Then we can choose a constant such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ18_HTML.gif)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ19_HTML.gif)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ20_HTML.gif)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ21_HTML.gif)
In the following, we will complete the proof in four steps.
Step 1.
Show that every solution of BVP (3.5), (3.6) satisfies
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ22_HTML.gif)
independently of .
Suppose that the estimate is not true. Then there exists
such that
or
. We may assume
. There exists
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ23_HTML.gif)
There are three cases to consider.
Case 1 ().
In this case, and
. For
, by (3.8), we get the following contradiction:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ24_HTML.gif)
and for , we have the following contradiction:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ25_HTML.gif)
Case 2 ().
In this case,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ26_HTML.gif)
and . For
, by (3.6) we have the following contradiction:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ27_HTML.gif)
For , by (3.9) and condition (iii) we can get the following contradiction:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ28_HTML.gif)
Case 3 ().
In this case,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ29_HTML.gif)
and . For
, by (3.6) we have the following contradiction:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ30_HTML.gif)
For , by (3.10) and condition (iv) we can get the following contradiction:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ31_HTML.gif)
By (3.6), the estimates
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ32_HTML.gif)
are obtained by integration.
Step 2.
Show that there exists such that every solution
of BVP (3.5), (3.6) satisfies
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ33_HTML.gif)
independently of .
Let
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ34_HTML.gif)
and define the function as follows:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ35_HTML.gif)
In the following, we show that satisfies the Nagumo condition on
, independently of
. In fact, since
satisfies the Nagumo condition on
, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ36_HTML.gif)
Furthermore, we obtain
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ37_HTML.gif)
Thus, satisfies the Nagumo condition on
, independently of
. Let
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ38_HTML.gif)
By Step 1 and Lemma 2.3, there exists such that
for
. Since
and
do not depend on
, the estimate
on
is also independent of
.
Step 3.
Show that for , BVP (3.5), (3.6) has at least one solution
.
Define the operators as follows:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ39_HTML.gif)
by
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ40_HTML.gif)
by
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ41_HTML.gif)
with
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ42_HTML.gif)
Since is compact, we have the following compact operator:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ43_HTML.gif)
defined by
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ44_HTML.gif)
Consider the set
By Steps 1 and 2, the degree is well defined for every
and by homotopy invariance, we get
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ45_HTML.gif)
Since the equation has only the trivial solution from Lemma 2.4, by the degree theory we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ46_HTML.gif)
Hence, the equation has at least one solution. That is, the boundary value problem
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ47_HTML.gif)
with the boundary conditions
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ48_HTML.gif)
has at least one solution in
.
Step 4.
Show that is a solution of BVP (1.1), (1.2).
In fact, the solution of BVP (3.36), (3.37) will be a solution of BVP (1.1), (1.2), if it satisfies
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ49_HTML.gif)
By contradiction, suppose that there exists such that
. There exists
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ50_HTML.gif)
Now there are three cases to consider.
Case 1 ().
In this case, since on
, we have
and
. By conditions (i) and (ii), we get the following contradiction:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ51_HTML.gif)
Case 2 ().
In this case, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ52_HTML.gif)
and . By (3.37) and conditions (i) and (iii) we can get the following contradiction:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ53_HTML.gif)
Case 3 ().
In this case, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ54_HTML.gif)
and . By (3.37) and conditions (i) and (iv) we can get the following contradiction:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ55_HTML.gif)
Similarly, we can show that on
. Hence
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ56_HTML.gif)
Also, by boundary condition (3.37) and condition (i), we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ57_HTML.gif)
Therefore by integration we have for each ,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ58_HTML.gif)
that is,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ59_HTML.gif)
Hence is a solution of BVP (1.1), (1.2) and satisfies (3.3).
Now we give a uniqueness theorem by assuming additionally the differentiability for functions ,
and
, and a kind of estimating condition in Theorem 3.1.
Theorem 3.2.
Assume that
(i)there exist lower and upper solutions of BVP (1.1), (1.2), respectively, such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ60_HTML.gif)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_IEq142_HTML.gif)
(ii) and its first-order partial derivatives in
are continuous on
,
on
,  
on
and satisfy the Nagumo condition on
(iii) is continuous on
and continuously partially differentiable on
, and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ61_HTML.gif)
(iv) is continuous on
and continuously partially differentiable on
, and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ62_HTML.gif)
(v)there exists a function such that
on
and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ63_HTML.gif)
Then BVP (1.1), (1.2) has a unique solution satisfying (3.3).
Proof.
The existence of a solution for BVP (1.1), (1.2) satisfying (3.3) follows from Theorem 3.1.
Now, we prove the uniqueness of solution for BVP (1.1), (1.2). To do this, we let and
are any two solutions of BVP (1.1), (1.2) satisfying (3.3). Let
. It is easy to show that
is a solution of the following boundary value problem
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ64_HTML.gif)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ65_HTML.gif)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ66_HTML.gif)
where for each ,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ67_HTML.gif)
By conditions (ii), (iii), and (iv), we have that and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ68_HTML.gif)
Now suppose that there exists such that
. Without loss of generality assume
, and let
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ69_HTML.gif)
It is easy to see that by condition (v), hence
. Let
. We have that
,
on
, and there exists a point
such that
. Furthermore
. In fact, if
, then
. By condition (v) and (3.55) we can easily show that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ70_HTML.gif)
In particular
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ71_HTML.gif)
Hence
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ72_HTML.gif)
which contradicts to (3.54). Thus . Similarly we can show that
. Consequently
.
Now, there are two cases to consider, that is
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ73_HTML.gif)
If , then by (3.59) we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ74_HTML.gif)
Thus, by (3.53) and condition (v) we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ75_HTML.gif)
Consequently, by Taylor's theorem there exists such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ76_HTML.gif)
which is a contradiction.
A similar contradiction can be obtained if . Hence
on
. By (3.55), we obtain
on
. This completes the proof of the theorem.
Next we give two examples to demonstrate the application of Theorem 3.2.
Example 3.3.
Consider the following third-order three point BVP:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ77_HTML.gif)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ78_HTML.gif)
Let
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ79_HTML.gif)
Choose and
. It is easy to check that
, and
are lower and upper solutions of BVP (3.66), (3.67) respectively, and all the assumptions in Theorem 3.2 are satisfied. Therefore by Theorem 3.2 BVP (3.66), (3.67) has a unique solution
satisfying
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ80_HTML.gif)
Example 3.4.
Consider the following fourth-order three point BVP:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ81_HTML.gif)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ82_HTML.gif)
Let
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ83_HTML.gif)
Choose and
. It is easy to check that
, and
are lower and upper solutions of BVP (3.70), (3.71), respectively, and all the assumptions in Theorem 3.2 are satisfied. Therefore by Theorem 3.2 BVP (3.70), (3.71) has a unique solution
satisfying
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F362983/MediaObjects/13661_2009_Article_842_Equ84_HTML.gif)
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Pei, M., Chang, S. Existence and Uniqueness of Solutions for Higher-Order Three-Point Boundary Value Problems. Bound Value Probl 2009, 362983 (2009). https://doi.org/10.1155/2009/362983
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DOI: https://doi.org/10.1155/2009/362983