Nonlinear fourth order boundary value problem
© Yardımcı and Uğurlu; licensee Springer. 2014
Received: 24 March 2014
Accepted: 23 July 2014
Published: 25 September 2014
In this paper we consider a nonlinear boundary value problem generated by a fourth order differential equation on the semi-infinite interval in which the lim-4 case holds for fourth order differential expression at infinity. Using the well-known Banach and Schauder fixed point theorems we prove the existence and uniqueness theorems for the nonlinear boundary value problem.
MSC: 34A34, 34B15, 34B16, 34G20.
is in a squarely integrable space on , where λ is a complex parameter. This result follows from the convergence of the corresponding nested circles. These circles either converge to a circle or a point. In the primary case, two linearly independent solutions of (9) and any combinations of them belong to the squarely integrable space and (9) is said to be the limit-circle case. Otherwise (9) is said to be of limit-point case. These Weyl’s results have been generalized to the fourth order case as well as 2n th order case by Everitt – (further see ). Moreover, we should note that limit-point/circle classifications does not depend on the spectral parameter λ. Using these ideas we generalize the results of (5), (6) to the fourth order case as given in (10), (13)-(16). Using Banach and Schauder fixed point theorems we establish the existence and uniqueness theorems for the singular fourth order nonlinear boundary value problem (10), (13)-(16) in the lim-4 case.
2 Nonlinear problem
where and .
Let denote the Hilbert space consisting of all real-valued functions y such that with the inner product and the norm .
where and . Green’s formula implies that for two functions , the limit exists and is finite.
where is the Kronecker delta and . This means that for arbitrary , the values , , and exist and are finite.
where and () are defined as above and and are some real numbers.
3 Green’s function
where is defined by (20). Note that since the lim-4 case holds for , one finds that is a Hilbert-Schmidt kernel.
Therefore solving (23) in is equivalent to find the fixed points of ℒ.
Now we recall the well-known fixed point theorem.
Banach fixed point theorem
Let B be a Banach space and S a nonempty closed subset of B. Assumeis a contraction, i.e., there is a λ, , such thatfor all u, v in S. Then A has a unique fixed point in S.
Therefore ℒ is a contraction mapping. Hence from the Banach fixed point theorem the proof is completed. □
This implies that . Since K is a closed subset of , the Banach fixed point theorem can be applied to obtain a unique solution of (21) in K. This completes the proof. □
4 Fixed points on Banach space
Nonlinear boundary value problems may have solutions without uniqueness. To show that the boundary value problem (10), (13)-(16) have solutions may be without uniqueness, we recall the following well-known theorems.
Schauder fixed point theorem
Let B be a Banach space and K a nonempty bounded, convex, and closed subset of B. Assumeis a completely continuous operator. Then L has at least one fixed point in K provided that.
A setis relatively compact if and only if for every, K is bounded, there exists asuch thatfor alland allwith, there exists a numbersuch thatfor all.
Now we can state the following theorem.
It is clear that K is bounded, convex, and closed. Further one can see that ℒ maps K into itself. Hence the proof of Theorem 4.2 will be completed with the next lemma. □
Hence we have obtained for and : there exists a such that implies for . This implies that ℒ is continuous.
for all . Therefore is bounded in .
for all and all with .
This implies that for there exists () such that . Consequently is a relatively compact set in . This completes the proof of Lemma 4.3 and therefore Theorem 4.2. □
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