Abstract
We consider the following complementary Lidstone boundary value problem
where λ > 0. The values of λ are characterized so that the boundary value problem has a positive solution. Moreover, we derive explicit intervals of λ such that for any λ in the interval, the existence of a positive solution of the boundary value problem is guaranteed. Some examples are also included to illustrate the results obtained. Note that the nonlinear term F depends on y' and this derivative dependence is seldom investigated in the literature.
AMS Subject Classification: 34B15.
Keywords:
eigenvalues; positive solutions; complementary Lidstone boundary value problems1 Introduction
In this article, we shall consider the complementary Lidstone boundary value problem
where m ≥ 1, λ > 0, and F is continuous at least in the interior of the domain of interest. Note that the nonlinear term F involves a derivative of the dependent variablethis is seldom studied in the literature and most research articles on boundary value problems consider nonlinear terms that involve y only.
We are interested in the existence of a positive solution of (1.1). By a positive solution y of (1.1), we mean a nontrivial y ∈ C^{(2m+1)}(0, 1) satisfying (1.1) and y(t) ≥ 0 for t ∈ (0, 1). If, for a particular λ the boundary value problem (1.1) has a positive solution y, then λ is called an eigenvalue and y is a corresponding eigenfunction of (1.1). We shall denote the set of eigenvalues of (1.1) by E, i.e.,
The focus of this article is eigenvalue problem, as such we shall characterize the values of λ so that the boundary value problem (1.1) has a positive solution. To be specific, we shall establish criteria for E to contain an interval, and for E to be an interval (which may either be bounded or unbounded). In addition explicit subintervals of E are derived.
The complementary Lidstone interpolation and boundary value problems are very recently introduced in [1], and studied by Agarwal et. al. [2,3] where they consider an odd order ((2m+ 1)th order) differential equation together with boundary data at the odd order derivatives
The boundary conditions (1.2) are known as complementary Lidstone boundary conditions, they naturally complement the Lidstone boundary conditions [47] which involve even order derivatives. To be precise, the Lidstone boundary value problem comprises an even order (2mth order) differential equation and the Lidstone boundary conditions
There is a vast literature on Lidstone interpolation and boundary value problems. The Lidstone interpolation has a long history from 1929 when Lidstone [8] introduced a generalization of Taylor's series that approximates a given function in the neighborhood of two points instead of one. Further characterization can be found in the study of [916]. More research on Lidstone interpolation as well as Lidstone spline is seen in [1,1723]. On the other hand, the Lidstone boundary value problems and several of its particular cases have been the subject matter of numerous investigations, see [4,18,2437] and the references cited therein. It is noted that in most of these studies the nonlinear terms considered do not involve derivatives of the dependent variable, only a handful of articles [30,31,34,35] tackle nonlinear terms that involve even order derivatives. In the present study, our study of the complementary Lidstone boundary value problem (1.1) where F depends on a derivative certainly extends and complements the rich literature on boundary value problems and in particular on Lidstone boundary value problems.
The plan of the article is as follows. In Section 2, we shall state a fixed point theorem due to Krasnosel'skii [38], and develop some inequalities for certain Green's function which are needed later. The characterization of the set E is presented in Section 3. Finally, in Section 4, we establish explicit subintervals of E.
2 Preliminaries
Theorem 2.1. [38] Let B be a Banach space, and let C(⊂ B) be a cone. Assume Ω_{1}, Ω_{2 }are open subsets of B with , and let be a completely continuous operator such that, either
(a) ∥Sy∥ ≤ ∥y∥, y ∈ C∩∂Ω_{1}, and ∥Sy∥ ≥ ∥y∥, y ∈ C ∩ ∂Ω_{2}, or
(b) ∥Sy∥ ≥ ∥y∥, y ∈ C∩∂Ω_{1}, and ∥Sy∥ ≤ ∥y∥, y ∈ C ∩ ∂Ω_{2}.
Then, S has a fixed point in .
To tackle the complementary Lidstone boundary value problem (1.1), let us review certain attributes of the Lidstone boundary value problem. Let g_{m}(t, s) be the Green's function of the Lidstone boundary value problem
The Green's function g_{m}(t, s) can be expressed as [4,5]
where
Further, it is known that
We also have the inequality
The following two lemmas give the upper and lower bounds of g_{m}(t, s), they play an important role in subsequent development.
Lemma 2.1. For (t, s) ∈ [0, 1] × [0, 1], we have
Proof. For (t, s) ∈ [0, 1] × [0, 1], it is clear from (2.3) that
Using (2.7), (2.4), and (2.5) in (2.2) yields for (t, s) ∈ [0, 1] × [0, 1],
By induction, we can show that
Now (2.6) is immediate by applying (2.9) to (2.8).
Lemma 2.2. Let be given. For (t, s) ∈ [δ, 1δ] × [0, 1], we have
Proof. For (t, s) ∈ [δ, 1δ] × [0, 1], from (2.3) we find
Then, using (2.11), (2.4), and (2.5) in (2.2), we get for (t, s) ∈ [δ, 1  δ ] × [0, 1],
which, in view of (2.9), gives (2.10) immediately.
Remark 2.1. The bounds in Lemmas 2.1 and 2.2 are sharper than those given in the literature [4,5,35,37].
3 Eigenvalues of (1.1)
To tackle (1.1) we first consider the initial value problem
whose solution is simply
Taking into account (3.1) and (3.2), the complementary Lidstone boundary value problem (1.1) reduces to the Lidstone boundary value problem
If (3.3) has a positive solution x*, then by virtue of (3.2), is a positive solution of (1.1). Hence, the existence of a positive solution of the complementary Lidstone boundary value problem (1.1) follows from the existence of a positive solution of the Lidstone boundary value problem (3.3). It is clear that an eigenvalue of (3.3) is also an eigenvalue of (1.1), thus
With the lemmas developed in Section 2 and a technique to handle the nonlinear term F, we shall study the eigenvalue problem (1.1) via (3.3).
For easy reference, we list below the conditions that are used later. In these conditions, f, α, and β are continuous functions with f : (0, ∞) × (0, ∞) → (0, ∞) and α, β : (0, 1) → [0, ∞).
(A1) f is nondecreasing in each of its arguments, i.e., for u, u_{1}, u_{2}, v, v_{1}, v_{2 }∈ (0, ∞) with u_{1 }≤ u_{2 }and v_{1 }≤ v_{2}, we have
(A2) for t ∈ (0, 1) and u, v ∈ (0, ∞),
(A3) α(t) is not identically zero on any nondegenerate subinterval of (0, 1) and there exists a_{0 }∈ (0, 1] such that α(t) ≥ a_{0}β(t) for all t ∈ (0, 1);
(A5) for t ∈ (0, 1) and u, u_{1}, u_{2}, v, v_{1}, v_{2 }∈ (0, ∞) with u_{1 }≤ u_{2 }and v_{1 }≤ v_{2}, we have
We shall consider the Banach space B = C[0, 1] equipped with the norm
For a given , let the cone C_{δ }be defined by
where (a_{0 }is defined in (A3)). Further, let
Let the operator S : C_{δ }→ B be defined by
To obtain a positive solution of (3.3), we shall seek a fixed point of the operator S in the cone C_{δ}.
Further, we define the operators U, V : C_{δ }→ B by
and
If (A2) holds, then
Lemma 3.1. Let (A1)(A4) hold. Then, the operator S is compact on the cone C_{δ}.
Proof. Let us consider the case when α(t) is unbounded in a deleted right neighborhood of 0 and also in a deleted left neighborhood of 1. Clearly, β(t) is also unbounded near 0 and 1.
For n ∈ {1, 2, 3, ...}, let α_{n}, β_{n }: [0, 1] → [0, ∞) be defined by
and
Also, we define the operators U_{n}, V_{n }: C_{δ }→ B by
and
It is standard that for each n, both U_{n }and V_{n }are compact operators on C_{δ}. Let M > 0 and x ∈ C_{δ}(M). For t ∈ [0, 1], we get
By the monotonicity of f (see (A1)), we have
Coupling with Lemma 2.1, it follows that
The integrability of β(t) sin πt (see (A4)) ensures that V_{n }converges uniformly to V on C_{δ}(M). Hence, V is compact on C_{δ}. By a similar argument, we see that U_{n }converges uniformly to U on C_{δ}(M) and therefore U is also compact on C_{δ}. It follows immediately from inequality (3.5) that the operator S is compact on C_{δ}.
Remark 3.1. From the proof of Lemma 3.1, we see that if the functions α and β are continuous on the close interval [0, 1], then the conditions (A1) and (A4) are not needed in Lemma 3.1.
The first result shows that E contains an interval.
Theorem 3.1. Let (A1)(A4) hold. Then, there exists c > 0 such that the interval (0, c] ⊆ E.
Proof. Let M > 0 be given. Define
Let λ ∈ (0, c]. We shall prove that S(C_{δ}(M)) ⊆ C_{δ}(M). Let x ∈ C_{δ }(M). First, we shall show that Sx ∈ C_{δ}. It is clear from (3.5) that
Further, from (3.5) and Lemma 2.1 we get
which leads to
Now, applying (3.5), Lemma 2.2, (A3) and (3.9) successively, we find for t ∈ [δ, 1δ],
Therefore,
Inequalities (3.8) and (3.10) imply that Sx ∈ C_{δ}.
Next, we shall verify that ∥Sx∥ ≤ M. For this, an application of (3.5), Lemma 2.1, (3.6) and (3.7) provides
or equivalently
Hence, S(C_{δ}(M)) ⊆ C_{δ}(M). Also, the standard arguments yield that S is completely continuous. By Schauder fixed point theorem, S has a fixed point in C_{δ}(M). Clearly, this fixed point is a positive solution of (3.3) and therefore λ is an eigenvalue of (3.3). Since λ ∈ (0, c] is arbitrary, it follows immediately that the interval (0, c] ⊆ E.
Remark 3.2. From the proof of Theorem 3.1, we see that (A2) and (A3) lead to S : C_{δ }→ C_{δ}.
Theorem 3.2. Let (A1)(A5) hold. Suppose that λ* ∈ E, for any λ ∈ (0, λ*), we have λ ∈ E, i.e., (0, λ*] ⊆ E.
Proof. Let x* be the eigenfunction corresponding to the eigenvalue λ*. Thus, we have
Define
Let λ ∈ (0, λ*) and x ∈ K*. Using (A5), we get
where the last equality follows from (3.11). This immediately implies that the operator S maps K* into K*. Moreover, the operator S is continuous and completely continuous. Schauder's fixed point theorem guarantees that S has a fixed point in K*, which is a positive solution of (3.3). Hence, λ is an eigenvalue, i.e., λ ∈ E.
The following result shows that E is an interval.
Corollary 3.1. Let (A1)(A5) hold. If , then E is an interval.
Proof. Suppose E is not an interval. Then, there exist and with τ ∉ E. However, this is not possible as Theorem 3.2 guarantees that τ ∈ E. Hence, E is an interval.
The following two results give the upper and lower bounds of an eigenvalue in terms of some parameters of the corresponding eigenfunction.
Theorem 3.3. Let (A1) and (A2) hold. Assume that m is odd. Let λ be an eigenvalue of (3.3) and x ∈ C_{δ }be a corresponding eigenfunction. If x^{(i) }(0) = b_{i}, i = 1, 3, ..., 2m  1, where b_{2m1 }> 0, then λ satisfies
where
and
Proof. For n ∈ {1, 2, 3, ...}, we define f_{n }= f * ω_{n}, where ω_{n }is a standard mollifier [25] such that f_{n }is Lipschitz and converges uniformly to f.
For a fixed n, let λ_{n }be an eigenvalue and x_{n}(t), with be a corresponding eigenfunction of the following boundary value problem
where F_{n }converges uniformly to F, and for u, v ∈ (0, ∞),
(see the proof of Lemma 3.1 for the definitions of α_{n}(t) and β_{n}(t)).
It is clear that x_{n}(t) is the unique solution of the initial value problem (3.13),
First, we shall establish an upper bound for x_{n}. Since
we have is nonincreasing and hence
In view of the initial conditions (3.16) and also (3.17), we find
Next, an application of (3.18) gives
By repeating the process, we get
By the monotonicity of f_{n}, we have
and
Coupling with (3.13) and (3.15), it follows that
Once again, using the initial conditions (3.16), repeated integration of (3.20) from 0 to t provides
where
and
In order to satisfy the boundary conditions , from inequality (3.21) it is necessary that
This readily implies
where
and
From (3.20) it is observed (by using the initial conditions (3.16) and repeated integration) that is a uniformly bounded sequence on [0, 1]. Thus, there exists a subsequence, which can be relabeled as , that converges uniformly (in fact, in C^{(2m1)}norm) to some x on [0, 1]. We note that each x_{n}(t) can be expressed as
Since is a bounded sequence (from (3.22)), there is a subsequence, which can be relabeled as , that converges to some λ. Then, letting n → ∞ in (3.23) yields
This means that x(t) is an eigenfunction of (3.3) corresponding to the eigenvalue λ. Further, x^{(i)}(0) = b_{i}, i = 1, 3, ..., 2m  1 and inequality (3.12) follows from (3.22) immediately.
Theorem 3.4. Let (A1)(A4) hold. Let λ be an eigenvalue of (3.3) and x ∈ C_{δ }be a corresponding eigenfunction. Further, let ∥x∥ = p. Then,
and
where t_{1 }is any number in (0, 1) such that x(t_{1}) ≠ 0.
Proof. Let t_{0 }∈ [0, 1] be such that
Then, using (3.5), Lemma 2.1 and the monotonicity of f, we find
which gives (3.24) readily.
Next, we employ (3.5), the monotonicity of f and the fact that min_{t∈[δ, 1δ] }x(t) ≥ γp to get
from which (3.25) is immediate.
The following result gives the criteria for E to be a bounded/unbounded interval.
Theorem 3.5. Define
(a) Let (A1)(A5) hold. If f ∈ W_{B}, then E = (0, c) or (0, c] for some c ∈ (0, ∞).
(b) Let (A1)(A5) hold. If f ∈ W_{0}, then E = (0, c] for some c ∈ (0, ∞).
(c) Let (A1)(A4) hold. If f ∈ W_{∞}, then E = (0, ∞).
Proof. (a) This is immediate from (3.25) and Corollary 3.1.
(b) Since W_{0 }⊆ W_{B}, it follows from Case (a) that E = (0, c) or (0, c] for some c ∈ (0, ∞).
In particular,
Let be a monotonically increasing sequence in E which converges to c, and let be a corresponding sequence of eigenfunctions in the context of (3.3). Further, let p_{n }= ∥x_{n}∥. Then, (3.25) together with f ∈ W_{0 }implies that no subsequence of can diverge to infinity. Thus, there exists R > 0 such that p_{n }≤ R for all n. So is uniformly bounded. This implies that there is a subsequence of , relabeled as the original sequence, which converges uniformly to some x, where x(t) ≥ 0 for t ∈ [0, 1]. Clearly, we have Sx_{n }= x_{n}, i.e.,
Since x_{n }converges to x and λ_{n }converges to c, letting n → ∞ in (3.26) yields
Hence, c is an eigenvalue with corresponding eigenfunction x, i.e., c = sup E ∈ E. This completes the proof for Case (b).
(c) Let λ > 0 be fixed. Choose ε > 0 so that
By definition, if f ∈ W_{∞}, then there exists M = M(ε) > 0 such that
We shall prove that S(C_{δ}(M)) ⊆ C_{δ}(M). Let x ∈ C_{δ }(M). As in the proof of Theorem 3.1, we have (3.8) and (3.10) and so Sx ∈ C_{δ}. Thus, it remains to show that ∥Sx∥ ≤ M. Using (3.5), Lemma 2.1, (3.6), (3.28), and (3.27), we find for t ∈ [0, 1],
It follows that ∥Sx∥ ≤ M and hence S(C_{δ}(M)) ⊆ C_{δ}(M). Also, S is continuous and completely continuous. Schauder's fixed point theorem guarantees that S has a fixed point in C_{δ}(M). Clearly, this fixed point is a positive solution of (3.3) and therefore λ is an eigenvalue of (3.3). Since λ > 0 is arbitrary, we have proved that E = (0, ∞).
Example 3.1. Consider the complementary Lidstone boundary value problem
where λ > 0 and q ≥ 0.
Here, m = 2 and
Clearly, F(t, u, v) is nondecreasing in u and v, thus (A5) is satisfied.
Choose
and
We see that (A1)(A4) are satisfied.
Case 1. 0 ≤ q < 1. Clearly f ∈ W_{∞}. It follows from Theorem 3.5(c) that the set E = (0, ∞). As an example, when λ = 24, the boundary value problem (3.29) has a positive solution given by .
Case 2. q = 1. Here f ∈ W_{B}. By Theorem 3.5(a) the set E is an open or a halfclosed interval. Further, from Case 1 and Theorem 3.2 we note that E contains the interval (0, 24].
Case 3. q > 1. Clearly f ∈ W_{0}. By Theorem 3.5(b) the set E is a halfclosed interval. Again, as in Case 2 we note that (0, 24] ⊆ E.
4 Eigenvalue intervals
In this section, we shall establish explicit subintervals of E. Here, the functions α and β in (A2)(A4) are assumed to be continuous on the closed interval [0, 1]. Hence, noting Remark 3.1, we shall not require conditions (A1) and (A4) to show the compactness of the operator S. For the function f in (A2), we define
Theorem 4.1. Let (A2)(A4) hold. Then, λ ∈ E if λ satisfies
Proof. We shall use Theorem 2.1. Let λ satisfy (4.2) and let ε > 0 be such that
First, we pick p > 0 so that
Let x ∈ C_{δ }be such that ∥x∥ = p. Note that for s ∈ [0, 1],
Then, using (3.5), Lemma 2.1, (4.4) and (4.3) successively, we find for t ∈ [0, 1],
Hence,
If we set Ω_{1 }= {x ∈ B ∥x∥ < p}, then (4.5) holds for x ∈ C_{δ }∩ ∂ Ω_{1}.
Next, let q > 0 be such that
Let x ∈ C_{δ }be such that
It is clear that
Then, an application of (3.5), (4.7), and (4.6) gives for t ∈ [0, 1],
Taking supremum both sides and using (4.3) then provides (see (4.1) for the definition of t*)
Therefore, if we set Ω_{2 }= {x ∈ B ∥x∥ < q_{0}}, then for x ∈ C_{δ }∩ ∂ Ω_{2 }we have
Now that we have obtained (4.5) and (4.8), it follows from Remark 3.2 and Theorem 2.1 that S has a fixed point such that p ≤ ∥x∥ ≤ q_{0}. Obviously, this x is a positive solution of (3.3) and hence λ ∈ E. □
Theorem 4.2. Let (A2)(A4) hold. Then, λ ∈ E if λ satisfies
Proof. We shall apply Theorem 2.1 again. Let λ satisfy (4.9) and let ε > 0 be such that
First, we choose r > 0 so that
Let x ∈ C_{δ }be such that ∥x∥ = r. Then, on using (3.5), (4.11), and (4.10) successively, we have for t ∈ [0, 1],
Taking supremum both sides and using (4.10) then yields (see (4.1) for the definition of )
Hence, if we set Ω_{1 }= {y ∈ B ∥x∥ < r}, then (4.8) holds for x ∈ C_{δ }∩ ∂ Ω_{1}.
Next, pick w > 0 such that
We shall consider two cases  when f is bounded and when f is unbounded.
Case 1. Suppose that f is bounded. Then, there exists some M > 0 such that
Let x ∈ C_{δ }be such that
From (3.5), Lemma 2.1 and (4.13), it is clear for t ∈ [0, 1] that
Hence, (4.5) holds.
Case 2. Suppose that f is unbounded. Then, there exists w_{0 }> max {r + 1, w} such that
Let x ∈ C_{δ }be such that ∥x∥ = w_{0}. Then, applying (3.5), Lemma 2.1, (4.14), (4.12), and (4.10) successively gives for t ∈ [0, 1],
Thus, (4.5) follows immediately.
In both Cases 1 and 2, if we set Ω_{2 }= {x ∈ B ∥x∥ < w_{0}}, then (4.5) holds for x ∈ C_{δ }∩ ∂Ω_{2}.
Now that we have obtained (4.8) and (4.5), it follows from Remark 3.2 and Theorem 2.1 that S has a fixed point such that r ≤ ∥x∥ ≤ w_{0}. It is clear that this x is a positive solution of (3.3) and hence λ ∈ E.
Remark 4.1. In (4.2) and (4.9), although t* and can be computed from (4.1), we can circumvent the computation by giving further bounds. Indeed, applying Lemma 2.2 we find
and
The following corollary is immediate from Theorems 4.1, 4.2 and Remark 4.1.
Corollary 4.1. Let (A2)(A4) hold. Then,
and
Remark 4.2. If f is superlinear (i.e., and ) or sublinear (i.e., and ), then we conclude from Corollary 4.1 that E = (0, ∞), i.e., the boundary value problem (3.3) (or (1.1)) has a positive solution for any λ > 0.
Example 4.1. Consider the complementary Lidstone boundary value problem
where λ, a, b, c > 0 and r ≤ 1.
Here, m = 2. It is clear that (A2)(A4) are satisfied with
Case 1. r < 1. It is clear that f is sublinear. Therefore, by Remark 4.2 the boundary value problem (4.15) has a positive solution for any λ > 0. In fact, we note that when λ = 120, (4.15) has a positive solution given by .
Case 2. r = 1, a = b = 0.5, c = 10. Here, and . It follows from Corollary 4.1 that
Once again we note that when λ = 120 ∈ (0,528.99), the corresponding eigenfunction is given by .
Competing interests
The authors declare that they have no competing interests.
Authors' contributions
All authors read and approved the final manuscript.
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