Non-conjugate boundary value problem of a third order differential equation

This paper is devoted to prove the existence of the optimal interval where the Green’s function is negative definite. The left and right endpoints of the interval are found. Then, a new principle of comparison of a third-order differential equation is established. As an application of our results, the solvability of a non-conjugate boundary value problem is discussed.


Introduction
The third-order differential equation has attracted considerable attention because of its wide applications in the deflection of a curved beam having a constant or varying cross section, in the three layer beam, in electromagnetic waves and gravity-driven flows [4].Many authors have used a great number of theories and methods to deal with the third-order differential equation, such as the approaches based on differential inequality [5,6] disconjugacy theory [1,8], fixed point method [9], variational method [10], topological degree theory [7], the upper and lower solutions method [2].These techniques can be interconnected and have proved to be very strong and fruitful.The Green's function plays an important role in the solvability of differential equations.We can transform the differential equation into integral equations by utilizing Green's function.
Using of the spectral theory for self-adjoint operators, P. J. Torres [9] studied the secondorder differential equations with periodic boundary value problem and obtained the condition under which the Green's function had constant sign.R. Ma [8] discussed a class of the third-order differential equations (Non-self-adjoint operator) with conjugate boundary value condition and got the symmetric optimal interval in the neighborhood of origin 0 by utilizing 2 Preliminaries Definition 2.1 ([1]).Let p k ∈ C [a, b] for k = 1, . . ., n.A linear differential equation of order n Ly ≡ y (n) + p 1 (t)y (n−1) + • • • + p n (t)y = 0 (2.1) is said to be non-conjugate on an interval [a, b] if every nontrivial solution has less than n zeros on [a, b], multiple zeros being accounted according to their multiplicity.
(1) In the case m = 0, the corresponding fundamental system of solutions to the homogeneous differential equation are To get a particular solution, let then we have the system of linear inhomogeneous equations The solutions are Thus the particular solution is given by the corresponding general solution of the differential equation is f (s) ds. (2.5) By the boundary condition u(0) = u ′ (0) = u ′ (1) = 0, we have , . Hence (2.6) (2) .
By Taylor's series expansion at the origin we have For 0 ≤ t ≤ s ≤ 1, the result can be proved by the same method as employed above.The Green's function is (2.7) Remark 2.4.(1) This result just coincides with [3] in the case m = 0.
(2) In the case 0 In the following discussion about the sign of the Green's function, we make an appointment for t, s ∈ (0, 1) from Lemma 3.1 to Theorem 3.7.

Existence of the optimal interval
In this section, we prove the existence of the optimal interval on which the Green's function will have constant sign.Let g(m), h(m) be as in Lemma 2.3, M = m 3 for parameter M. By Lemma 3.1 we can easily get the following. then (2) In the case m > 0, F ′′′ t,s (0) Let us define By Lemma 3.3, we know that T 0 = φ, T 1 = φ.
Proof.In fact, let ∈ T 1 , which implies m 2 is the upper bound of the set T 1 . Let It is easy to verify that T 2 = φ, Lemma 3.6.Let m > 0, λ (λ ≈ 3.01674) be the smallest positive zero of the equation g(m) = 0 and µ (µ ≈ 4.23321) be the smallest positive zero of the equation h(m) = 0, then Proof.
(1) The proof will be given in two steps.
Proof.It is easy to verify the assertion by Lemma 3.5 and Lemma 3.6.

Right and left endpoints of the optimal interval
In this section, we set out to find the right and left endpoints of the optimal interval.
Note: In fact, we just need to inspect whether the maximum value of F t,s (m) is 0, namely to calculate every place of the three-dimensional range where the ternary function F t,s (m) will possibly get maximum value.So our task is to check 9 edges and the stagnation points on 5 faces and the interior of the right prism region.
Proof.(See Figure 4.1.)For the edges and the face By the properties of functions h(m) and g(m), we know that The face Using the properties of functions h(m) and g(m), we have For the edges and the face For the edge Since λ is the smallest positive solution of the function g(m), we have

The face
We have verified 4 edges for the 3-dimensional area, and it remains to calculate the stagnation points inside the face.Let Obviously, {(m, s) | m = 0, s ∈ [0, 1]} solve the system of nonlinear equations.While there may exist other solutions.Combining (4.1) with (4.2), we can get corresponding curves of the system in Figure 4.2, which clearly shows there exists no other solution.Hence, all the solutions of the system are {(m, s) | m = 0, s ∈ [0, 1]}.When m = 0, F t,s (0) ≡ 0; thus, on the face ABB ′ A ′ , F t,s (m) ≤ 0.
Combining (4.3), (4.4), with (4.5), we know that {(m, t, s) | m = 0, ∀ 0 ≤ s ≤ t ≤ 1} or {(m, t, s) | ∀m ∈ [0, λ] , t = s = 0} are the solutions of the system.While there may be other solutions.But in Figure 4.3 it is obvious that there exists no other solution for the system.So we obtain that all the solutions of the system are {(m, t, s) In summary, when ∀m ∈ [0, λ], ∀ 0 ≤ s ≤ t ≤ 1, the ternary function F t,s (m) ≤ 0, namely G(t, s) ≤ 0 (∀ 0 ≤ s ≤ t ≤ 1).Moreover, in the case 0 ≤ t ≤ s ≤ 1, from the expression of the Green's function, we can get G(t, s) ≤ 0. Theorem 4.2.The smallest positive solution of the equation : g(m) = 0 is just the right endpoint of the optimal interval where the Green's function will be negative definite.
Proof.From Lemma 3.6, we know that the right endpoint value of the optimal interval couldn't be larger than λ.Combining it with Lemma 4.1, we have G(t, s) ≤ 0 for arbitrary m ∈ [0, λ].This completes the proof.
Proof.(See Figure 4.4) Following the same manner of Lemma 4.1, it is easy to prove that on the edges and the faces AB, OB, AO, ABO,OO  lead to F t,s (m) | t=1 < 0. As a result, the Green's function will not be negative definite when 0 ≤ s ≤ t ≤ 1.This completes the proof of assertion (1).
For the edge 2 (−µ)s 2 > 0, and then it is a unary function with respect to s.Noting that J m (0) = J m (1) ≡ 0, we only need to compute the stagnation points with respect to s ∈ [0, 1] .Additionally, Non-conjugate BVP of a third order ODE 13 Namely, to seek out the stagnation points of the system (4.7) on [0, 1] cos By calculating, we get s * .= 0.564746, and J(−µ, s * ) .= 2012.74≫ 0. So on the edge A ′ B ′ ,we  have F t,s (−µ) ≥ 0. When m = 0, F t,s (0) = J(0, s) ≡ 0, and we have verified 4 edges on the face ABB ′ A ′ .In the following step we will compute the stagnation points inside this face, since It is easy to know that {(m, s) | m = 0, s ∈ [0, 1]} solves the system (4.9), while there may be other solutions.It can be seen from the Figure 4.7 that there exists no other solution for the system (4.9).When m = 0, F t,s (0) ≡ 0. So on the face ABB ′ A ′ we have F t,s (m) ≥ 0.
In the case t = 1, it is just the edge A ′ B ′ , hence J(−µ, s) ≥ 0. When s = 0, F t,s (−µ) ≡ 0. If s = 1, then t = 1, thus F t,s (−µ) ≡ 0. It only remains to verify the stagnation points inside the face O ′ A ′ B ′ , so let It is evident to know that from (4.3), (4.4) {(t, s) | t = s = 0} solves the system (4.10), while there may be other solutions.It can be seen from the Figure 4.8 that there exists no other solution for the system (4.10),so all the solution of the system (4.10) is {(t, s) | t = s = 0}.When t = s = 0, F(0, 0, −µ) ≡ 0. Thus on the face O ′ A ′ B ′ we can get F t,s (m) ≥ 0.
Finally, we will compute the stagnation points inside the 3-dimensional region Combining (4.3), (4.4), with (4.5), we have ] , t = s = 0} solve the system of nonlinear equations, while there may be other solutions.It can be seen from Figure 4.9 that there exists no other solution for the system (4.11), and we will find F t,s (m) ≡ 0 in both cases.
All in all, we know that the ternary function F(t, s, m) ≥ 0 always holds in the 3  where Apparently, {(m, t) | m = 0, t ∈ [0, 1]} solve the system (5.2), while there may exist other solutions.In Figure 5.1, by computing we obtain t * ≈ 0.4274818, m * ≈ −1.47531, and 1] it is easy to prove that if m > 0, then w(m) > 0. When m = λ, we get a unary function with respect to t.So, let By calculating, we gain ∂χ(m,t) ∂t t=0 = √ 3me 3m 2 g(m), and the stagnation point t * = 0, hence χ(m, 0) ≡ 0. It only remains to verify the stagnation point inside the rectangle region.Therefore, we also use the system (5.2).

Application
In this section we will utilize Theorem 5.