Electronic Journal of Qualitative Theory of Differential Equations

By using variational methods and maximum principles we discuss the existence, uniqueness and multiplicity of solutions for a semilinear sixth-order ODE. The main difference between our work and other related papers is that we treat a general case and we do not impose sign restrictions on the nonlinearity f or on its potential F.


Introduction
In this paper, we study the existence and uniqueness of solutions of the following boundary value problem u (6) + Au (4) + Bu ′′ − C(x)u + f (x, u) = 0 in Ω u = u ′′ = u (4) = 0 on ∂Ω, (1.1) where A, B are some given constants, C(x) is a given function, f is a continuous function on [0, L] × IR and Ω = (0, L). The treatment of (1.1) is motivated by the study of stationary solutions (which leads to sixth-order ODEs) of the sixth-order parabolic differential equation arising in the formation of spatial periodic patterns in bistable systems and is also a model for describing the behaviour of phase fronts in materials that are undergoing a transition between the liquid and the solid state. The case f (u) = u − u 3 was treated by Gardner and Jones [13] as well as by Caginalp and Fife [7].
We also note that the deformation of the equilibrium state of an elastic circular ring segment with its two ends simply supported can be described by a boundary value of sixth-order (see [1]): u (6) + 2u (4) + u ′′ = f (x, u) in Ω = (0, 1) Boundary value problems of sixth-order also arise in sandwich beam deflection under transverse shear [2].
It is worth mentioning the new paper of Bonanno and Livrea [4], where the problem −u (6) + Au (4) is treated. The authors prove the existence of infinitely many solutions to problem (1.6) under different assumptions on A, B, C and by requiring an oscillation on f (x, ·) at infinity. More precisely if i). F(x, t) ≥ 0 for every (x, t) ∈ ([0, 5/12]  Using variational methods we present here some new existence results (Section 3.1). The main difference between our work and the above mentioned papers is that we treat a general case and we do not impose sign restrictions on f or F. We note that we cover nonlinearities that are not treated elsewhere, e.g., the cases f (u) = ln(|u| + 1) + |u| |u|+1 + u and f (x, u) = a(x) cos(u n + C)u n−1 , where a is a bounded function, C is a constant and n is a natural number. We see that these cases are not covered in [14] since the assumption (H1) in [14], i.e. (1.4) is not satisfied. In particular, since (2.10) holds (here A = 2, B = 1, C = 0, L = 1), our results apply to (1.3).
We obtain our main existence results under the restriction where K 1 , K 2 , r > 0. In Section 3.2 we will briefly present some uniqueness results for the corresponding nonhomogeneous linear equation.
The last section is devoted to a multiplicity result. As we mentioned above, the available multiplicity results (see [20,Theorem 3] and [9,Theorem B]) are stated under the restriction F ≤ 0. Here we strengthen relation (1.7), more precisely we impose ∀ (x, s) ∈ Ω × IR, (1.8) where K, K 1 , K 2 > 0, 0 < r < 2, p > 2 and obtain for sufficiently large L a multiplicity result that holds without a sign restriction on F. We also note that the multiplicity result holds if (1.4) is satisfied without the sign restriction on F.
We note that if f is a continuous function on [0, L] × IR, then a weak solution is a classical solution (for a proof see [20]).
The problem (1.1) has a variational structure and the weak solutions in the space H(Ω) can be found as critical points of the functional which is Fréchet differentiable and its Fréchet derivative is given by Throughout the paper C denotes a universal positive constant depending on the indicated quantities, unless otherwise specified.
The following results will be useful. .
where L represents the length of Ω. The next key result is more general version for bounded domains of the result presented in [20], Lemma 5 and will be used to handle the existence in the case r > 2 as well the multiplicity result.
Then there exists a constant k 1 such that Then there exists a constant k 2 > 0 such that The inequality (2.8) also holds if c).
Then there exists a constant k 3 > 0 such that Proof. a). We borrow some ideas from the paper of Bonheure (see [5,Lemma 5]). It is easy to see that for any real α Hence for any α the quantity For arbitrary ε > 0 we have by Lemma 2. 2 Choosing ε sufficiently small, using that Q A We first note that if one of the inequalities (2.5), (2.8) or (2.11) holds for u ∈ H 3 (IR), then it follows that the inequalities are also true for u ∈ H(Ω).
Indeed, for u ∈ H(Ω), we have Here * stands for one of the expressions in the inequalities (2.5), (2.8) or (2.11) that is inside the square brackets. We now prove the required inequalities for u ∈ H 3 (IR). We note that the proof of (2.5) under the conditions (2.6) is similar to the proof of (2.8) under the hypothesis (2.7) and hence is omitted.
To prove inequality (2.8) we see that for all ξ ∈ IR Hence As a consequence, we get Letû(ξ) be the Fourier transform of u(x) ∈ H 3 (IR). By Parseval's identity and (2.15) we get which is the desired result.
If (2.9) holds then we can achieve the proof in a similar way by showing that . .
where C is a positive constant depending only on the indicated quantities and S is the best constant in the imbedding H 3 (Ω) ⊂ C 0 (Ω).
Proof. a). Follows from inequality (2.1). b). By the Sobolev imbedding and Lemma 2.2 we get

Existence
We split the study of existence into three cases: Then the boundary value problem (1.1) has at least one solution.
Proof. The result is a consequence of the Weierstrass theorem, which tells us that if the functional J is coercive and weakly lower semicontinuous on H(Ω), then J has a global minimum. We first establish that J(u) is coercive.
If we choose now ε > 0 sufficiently small we get that J(u) is coercive on H(Ω).
We now show that J(u) is weakly lower semicontinuous on the reflexive space H(Ω).
Since A ≤ 0 and B ≥ 0 we get that is sequentially weakly continuous.
Therefore, J(u) is weakly lower semicontinuous by the result in [3, Criterion 6.1.3, p. 30], and the proof follows.
The next lemma ensures that the solution we have found is nontrivial.
Suppose that the following condition holds: Then there exists e ∈ H(Ω) such that J(e) < 0.
We note that by the first part of relation (3.3) and by the fact that where N is a positive constant.
Hence by the dominated convergence theorem and (3.3) which is the desired result. Hence there exists e = sφ ∈ H(Ω) such that J(e) < 0.
Our first main existence result reads. Case r = 2 Lemma 3.5. Suppose that F satisfies . If in addition we assume that Similarly, we get the corresponding existence result in the case r = 2.
Theorem 3.6. Suppose that F satisfies If in addition we assume that Case r > 2, K 2 = 0 The existence for the case r > 2 will be treated differently. We shall see that J(u) has a mountain-pass structure and the nontrivial critical points of J(u) will be found by using the Mountain-Pass theorem of Brézis and Nirenberg.
The following two lemmas show when J(u) has a mountain-pass structure.
Here ∥u∥ i denotes one of the following norms when A ≤ 0, B, C ≥ 0 or when one of the relations (2.4) or (2.6) is satisfied; when one of the relations (2.7) or (2.9) is satisfied; when the relation (2.10) is satisfied.
Proof. By virtue of Lemma 2.3 we see that H(Ω) endowed with one of the scalar products (u, v) i , i = 1, 2, 3, becomes a Hilbert space.
We give the proof in the case when (2.4) is satisfied. The cases when relations (2.7), (2.9) or (2.10) hold can be treated similarly.
We note that (2.5) reads Since r > 2, we can choose q > 1 such that r = 1 + q.
The following celebrated result is useful. Theorem 3.9 (Mountain Pass Theorem [6]). Let E be a real Banach space with its dual E * and suppose that J ∈ C 1 (E, R) satisfies for some constants µ < η, ρ > 0 and e ∈ E with ∥e∥ > ρ. Let λ ≥ η be characterized by Then there exists a sequence {u n } ⊂ E such that J(u n ) → λ ≥ η and ∥J ′ (u n )∥ E * → 0, as n → ∞.
By (3.20) and (3.21) − By Young's inequality we get which is similar to the first inequality in (3.17). Now the proof follows exactly as the proof of Lemma 3.10. Proof. By Lemma 3.10 there is a bounded Cerami type sequence {u n }. Hence we can extract a subsequence, still denoted {u n }, such that u n ⇀ u 0 weakly in H(Ω), u n → u 0 strongly in C 2 (Ω).
We can now conclude the existence result in the case r > 2, K 2 = 0.
Theorem 3.14. Let F satisfy where K 1 > 0, r > 2. Suppose that one of the conditions of Lemma 2.3 is satisfied and that (3.10) holds.

If the condition (3.3) is satisfied, then problem (1.1) has a nontrivial solution in H(Ω).
We end this section by giving the following examples as an application of the results.

Example 3.15.
We see that the theory presented includes the typical example where b is a bounded function which is either strictly positive or strictly negative in Ω (no sign changing is allowed). For the sake of simplicity we take p even. We can check that If b > 0 then we can choose 2 < θ < p and see that the left hand side of (3.25) becomes negative and hence (3.25) is satisfied. Due to the negativity of the left hand side of (3.25) for s ≤ 0 it is also obvious that (3.21) is satisfied.
We can argue similarly if b < 0 by choosing θ > p. Also since (2.10) holds with A = 2, B = 1, C = 0, L = 1 we get by Theorem 3.14 that the boundary value problem that describes the deformation of the equilibrium state of an elastic circular ring segment with its two ends simply supported (see [1]) has at least one nontrivial solution.
Example 3.16. We consider the following function g(x, s) = a(x) cos(s n + C)s n−1 , where a is a bounded function, C is a constant and n is a natural number.
Since the potential of g is and satisfies the requirements of Lemma 3.1, we get that problem (1.1) with f replaced by g has a nontrivial solution in H(Ω) if A ≤ 0, B, C ≥ 0, C ∈ C 0 (Ω).
Proof. By the proof of Lemma 3.1, J(u) can be represented as the sum J(u) = J 1 (u) + J 2 (u), where J 1 (u) is convex and Condition (3.27) assures that the function s → F(x, s) is strictly convex and hence J 2 (u) is strictly convex. The last statement implies that J(u) is strictly convex and the uniqueness follows.
The next uniqueness result is a consequence of the following one dimensional generalized maximum principle (for results concerning the generalized maximum principle see [18, p. 73]) and collects several author's uniqueness results in the case when the coefficients A, B, C are nonconstant or have arbitrary sign and f = f (x). holds.
Then, there exists a function w > 0 in Ω, w ∈ C ∞ (Ω) such that u/w satisfies a generalized maximum principle in Ω, i.e., there exists a constant k ∈ IR such that u/w ≡ k in Ω or u/w does not attain a nonnegative maximum in Ω.
Proof. The proof follows directly from [11, Theorem 2.1] (which holds for all dimensions n ≥ 1).
The interested reader may consult the paper [16] for a different kind of one dimensional maximum principle for sixth order operators. The authors prove (Theorem 3.1) the positivity of the solution u that satisfies a sixth order differential inequality assuming that u, u ′ are positive on the boundary of the domain Ω = (a, b) and (in particular) u ′′′ (a) ≤ 0, u ′′′ (b) ≥ 0.
Theorem 3.20. The boundary value problem

29)
has at most one solution if one of the following conditions is satisfied (here g i , i = 1, 2, 3 are arbitrary constants) 1). sup Here A < −1, B > 0 are constants and C > 0 in Ω is a function.
2). Suppose that the functions A, C satisfy −A = C > 0 in Ω and that the function B satisfies 3). Suppose that the functions A < 0, B, C > 0 in Ω satisfy and also (3.31) holds.

33)
where the functions A < 0, C > 0 satisfy A + C + 1 < 0 in Ω and 1/( Proof. 1). The proof uses the P-function method introduced by L. E. Payne [17]. Many results concerning the P-function method and its applications can be found in the book [19]. We give the proof when (3.30) holds. We define u = u 1 − u 2 , where u 1 and u 2 are solutions of (3.29). Then u satisfies (3.29), where f = 0 and with zero boundary data u = u ′′ = u (4) = 0 on ∂Ω.
According to [11], Lemma 3.1, i), the function P = (−Au (4) satisfies the inequality Hence by Theorem 3.19 there exists w > 0 in Ω such that P/w satisfies a generalized maximum principle in Ω, i.e., either there exists a constant k ∈ IR such that If (3.34) holds then since the function P/w is smooth (3.34) holds in Ω. By the zero boundary conditions we have P = 0 on ∂Ω, i.e., k = 0. It follows that P = 0 in Ω. Since P is a sum of squares multiplied by positive constants, P = 0 in Ω implies u ≡ 0 in Ω. Hence u 1 = u 2 in Ω.
Alternatively, if (3.35) holds, then by the zero boundary conditions. It follows that i.e., P = 0 in Ω. Using the same arguments as above, we get u ≡ 0 in Ω, i.e., u 1 = u 2 in Ω.

Multiplicity
Finally, we present a multiplicity result for (1.1) that is based on the result presented in [20,Theorem 3], Lemma 2.3 and the next result.
Lemma 3.21. Let A, B, C be real constants such that C > 0. The polynomial has exactly one positive zero ξ 0 if either holds. Moreover, for L > ξ 0 we have P(L) > 0.
The proof is a direct consequence of the result presented in [8,Lemma 4.3].
To prove the multiplicity result we need the following Definition 3.22. Let X be a Banach space and J ∈ C 1 (X, IR). We say that J satisfies a Palais-Smale condition if any sequence {u n } n in X for which J(u n ) is bounded and J ′ (u n ) → 0 as n → ∞, has a convergent subsequence.
Theorem 3.23 (Clark [10]). Let X be a Banach space, J ∈ C 1 (X, IR) be even, bounded below and satisfy the Palais-Smale condition. Suppose that J(0) = 0 and there is a set Y ⊂ X such that Y is homeomorphic to S m−1 by an odd map and sup Y J < 0. Then J possesses at least m distinct pairs of critical points.
Our multiplicity result reads Proof. We first note that if relation (1.4) holds, then by [14,Lemma 7], we get that J(u) is bounded from below and satisfies the Palais-Smale condition for any real constants A, B, C. If one of the relations (3.41) or (3.42) is assumed, in view of Lemma 2.3 and the structure condition (1.8) we can use a similar argument that was used in the proof of Lemma 3.1 to show that J(u) is bounded from below on H(Ω) by a negative constant. Since f is continuous on IR 2 we can follow [20], proof of Theorem 3, to get that J(u) satisfies the Palais-Smale condition.
We now use the same techniques as in [20,Theorem 3] and prove the case when (3.40) holds. We can treat similarly the other cases.
Hence by Lemma 3.21 we see that for L > ξ 0 Q(L) < 0 and by choosing ρ sufficiently small, we get J(v) ≤ ∥v∥ 2 m L 4 Q(L) + C(K, m, p, L)∥v∥ p−2 m < 0, for any v ∈ Y. Now the proof follows from Clark's theorem.