Aequationes Math. 74 (2007) 111–118 0001-9054/07/010111-8 DOI 10.1007/s00010-006-2855-5
c Birkh¨ ° auser Verlag, Basel, 2007
Aequationes Mathematicae
Multiplicity theorems for discrete boundary value problems Francesca Faraci and Antonio Iannizzotto
Summary. In this paper we study a boundary value problem for difference equations in a variational framework. We establish the existence of at least three solutions for a perturbed problem using a recent result by Ricceri, in which the non-convexity of a superlevel of a suitable functional is assumed. Mathematics Subject Classification (2000). 39A10, 47J30. Keywords. Discrete boundary value problem, multiple solutions, convexity.
1. Introduction In this paper we deal with boundary value problems for difference equations of the type ( −∆2 y(k − 1) = λf (k, y(k) + y0 (k)), k ∈ [1, T ] (P ) y(0) = y(T + 1) = 0 where T is a positive integer (T > 1), y0 is a real-valued function defined in the discrete interval [1, T ] = {1, .., T } (vanishing at 0 and at T + 1), λ is a real parameter, f : [1, T ] × R → R is a continuous function and the difference operators involved are defined, for all y : [1, T ] → R, by putting ∆y(k) = y(k + 1) − y(k), ∆2 y(k) = y(k + 2) − 2y(k + 1) + y(k). Our aim is to establish, under very general assumptions, the existence of some y0 and λ such that problem (P ) admits at least three solutions y. We follow a variational approach, applying the critical point theory. Such a method was already employed by Guo and Yu in [3], [4], where periodic solutions are found for nonlinear difference equations with superlinear or subquadratic nonlinearities. We also mention the papers of Henderson and Lauer [5] and Reid and Yin [6], where boundary value problems are studied and an interval of eigenvalues, assuring the existence of a positive solution, is found (for an exhaustive discussion of difference equations, we refer to [1]).
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We set the problem in the variational framework described in [2], where Agarwal, Perera and O’Regan studied the problem ( −∆2 y(k − 1) = f (k, y(k)), k ∈ [1, T ] (Q) y(0) = y(T + 1) = 0. In [2], as the authors are interested in positive solutions, it is assumed that f (k, 0) ≥ 0 for all k ∈ [1, T ]; moreover, the classical critical point methods employed (namely the mountain pass lemma) require the assumption f (k, x) > λ1 for all k ∈ [1, T ], x where λ1 > 0 is the smallest eigenvalue of the linear problem ( −∆2 y(k − 1) = λy(k), k ∈ [1, T ] lim inf
x→+∞
(1)
y(0) = y(T + 1) = 0.
Under such hypotheses, problem (Q) admits at least two non-negative (and nonzero) solutions. Problem (P ) is of a completely different nature (see Example 11). The parameter λ and the function y0 for which (P ) has at least three solutions remain undetermined, as to balance the generality of our hypotheses. Our main assumption is the non-convexity of a superlevel of the integral functional associated to the nonlinearity f , defined on a suitable Hilbert space. Such a ‘geometric’ condition allows us to apply some recent multiplicity results established by Ricceri in [7], which, for the finite-dimensional case, can be stated as follows. Theorem 1 ([7], Theorem 2). Let X be a real Hilbert space with dim X < ∞, J ∈ C 1 (X) be a nonconstant functional, r ∈ ] inf X J, supX J[ such that the set J −1 ([r, +∞[) is not convex. Assume that J(y) ≤ 0. 2 kyk kyk→+∞ lim sup
(2)
Then there exist y0 ∈ J −1 (] − ∞, r[) and λ > 0 such that the equation y = λJ ′ (y) + y0
(E)
has at least three solutions in X. Theorem 2 ([7], Theorem 4). Let X, J be as in Theorem 1, r ∈ ] inf X J, supX J[ such that the set J −1 (r) is not convex. Assume that lim
kyk→+∞
J(y) = 0. kyk2
(3)
Then there exist y0 ∈ X \ J −1 (r) and λ ∈ R such that equation (E) has at least three solutions in X.
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The paper is divided into three sections: Section 2 is devoted to our main results, and in Section 3 we discuss the existence of positive solutions.
2. Results We are going to seek the solutions of problem (P ) in the T -dimensional real Hilbert space H = {y : [0, T + 1] → R : y(0) = y(T + 1) = 0}, endowed with the inner product (y, z) =
T +1 X
∆y(k − 1)∆z(k − 1) = −
T X
∆2 y(k − 1)z(k)
k=1
k=1
T
(hence H is linearly homeomorphic to R , endowed with the Euclidean norm). Let f be as in Section 1; then we define Z x f (k, ξ)dξ, F (k, x) = 0
and the functional J : H → R by
J(y) =
T X
F (k, y(k)).
k=1
We introduce now a ∈ R ∪ {−∞}, b ∈ R ∪ {+∞} defined by a := inf J = H
T X
k=1
inf F (k, x),
x∈R
b := sup J = H
Our first result can be stated as follows.
T X
k=1
sup F (k, x). x∈R
Theorem 3. Let f : [1, T ] × R → R be a continuous function, r ∈]a, b[. Assume that J −1 ([r, +∞[) is not convex and that for all k ∈ [1, T ] F (k, x) ≤ 0. |x|2 |x|→+∞ lim sup
(4)
Then there exist y0 ∈ J −1 (] − ∞, r[) and λ > 0 such that problem (P ) has at least three solutions. Proof. We are going to apply Theorem 1: let us check that all the hypotheses are satisfied. First of all, it is easily seen that J is continuously Gˆateaux-differentiable with derivative given by (J ′ (y), z) =
T X
k=1
f (k, y(k))z(k).
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Condition (2) follows from assumption (4) (actually, they are equivalent). Indeed, note that (due to the equivalence of the norms in H) there exists c > 0 such that for all y ∈ H v u T uX y(k)2 . kyk ≥ ct k=1
By (4), for every ε > 0, there exists a δ > 0 such that for all k ∈ [1, T ], x ∈ R with |x| > δ, one has F (k, x) ε < c2 . x2 T Let y ∈ H satisfy s kyk ≥
Then
J(y) = kyk2
X
|y(k)|>δ
T ε
max
k∈[1,T ],|x|≤δ
F (k, x).
X F (k, y(k)) + 2 kyk
|y(k)|≤δ
F (k, y(k)) < ε, kyk2
that is, (2) is satisfied. By Theorem 1, equation (E) admits at least three solutions for some y0 ∈ H, λ > 0. We observe that problem (P ) and equation (E) are equivalent in the following sense: y ∈ H is a solution of (P ) if and only if −
T X
[∆2 y(k − 1) + λf (k, y(k) + y0 (k))]z(k) = (y − λJ ′ (y + y0 ), z) = 0
k=1
for all z ∈ H, that is, y + y0 is a solution of (E). Thus, we obtain three pairwise distinct solutions of (P ). ¤ Theorem 4. Let f : [1, T ] × R → R be a continuous function, r ∈]a, b[ such that J −1 (r) is not convex and that, for all k ∈ [1, T ], F (k, x) = 0. |x|2 |x|→+∞ lim
(5)
Then there exist y0 ∈ H \ J −1 (r) and λ ∈ R such that (P ) has at least three solutions. Proof. Arguing as above, it is easily seen that condition (3) is equivalent to assumption (5); then, by applying Theorem 2, we obtain, for suitable y0 ∈ H, λ ∈ R, at least three solutions of (E), i.e. three solutions of (P ). ¤ Remark 5. Let f (k, ·) be a decreasing function for all k ∈ [1, T ] and assume condition (5). Then F (k, ·) is strictly concave for all k ∈ [1, T ]. For all r ∈ ]a, b[,
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choose y, z ∈ J −1 ([r, +∞[), y 6= z and µ ∈ ]0, 1[ such that J(µy + (1 − µ)z) =
T X
F (k, µy(k) + (1 − µ)z(k))
k=1
>
T X £
µF (k, y(k)) + (1 − µ)F (k, z(k))
k=1
= µJ(y) + (1 − µ)J(z) ≥ r.
¤
Thus, J −1 ([r, +∞[) is convex, while J −1 (r) is not. Theorem 3 cannot be applied to such a function, while Theorem 4 yields the existence of y0 ∈ H \ J −1 (r) and λ ∈ R such that (P ) has at least three solutions. In order to get more precise information about y0 and λ, we apply Theorem 3 to −f (in fact J −1 (] − ∞, r]) is not convex: otherwise J −1 (r) should be convex), finding that problem (P ) has at least three solutions for some y0 ∈ J −1 (]r, +∞[) and λ < 0. Remark 6. We wish to show how a homogeneity condition can be applied to get a multiplicity result for a problem not involving the parameter λ. Let β 6= 1 be a real number, and suppose that, for all k ∈ [1, T ], the function f (k, ·) is positively β-homogeneous, that is, for all x ∈ R, µ > 0 we have f (k, µx) = µβ f (k, x). Then there exists z0 ∈ H such that the problem ( −∆2 z(k − 1) = f (k, z(k) + z0 (k)), k ∈ [1, T ]
(R)
z(0) = z(T + 1) = 0
admits at least three solutions. Indeed, by Theorem 3, there exist y0 ∈ H, λ > 0 such that problem (P ) has at least three solutions: let y ∈ H be one of these. We 1 1 set z0 = λ β−1 y0 , and observe that z = λ β−1 y is a solution of (R): thus, the latter has at least three solutions. In the next corollary we use a coercivity assumption in order to assure that the functional J has a non-convex superlevel set: Corollary 7. Let f : [1, T ] × R → R be a continuous function, satisfying (4) and such that lim sup F (h, x) = +∞ (6) |x|→+∞
for some h ∈ [1, T ]. Then for all r ∈]a, +∞[ there exist y0 ∈ J −1 (] − ∞, r[) and λ > 0 such that (P ) has at least three solutions. Proof. First of all we notice that in this case b = +∞. We prove now that J −1 ([r, +∞[) is not convex for any r ∈]a, +∞[. In fact, let y ∈ J −1 (] − ∞, r[); by
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(6) we find ρ > 0 such that F (h, y(h) ± ρ) ≥ r −
X
F (k, y(k)).
k6=h
We define y1 , y2 ∈ H by putting y1 (k) = y(k) − ρδhk ,
y2 (k) = y(k) + ρδhk
(where δhk is the Kronecker symbol), and we get y1 , y2 ∈ J −1 ([r, +∞[), while y = 12 (y1 + y2 ) does not lie in this set. An application of Theorem 3 completes the proof. ¤ We give next an example of a nonlinearity satisfying the assumptions of Corollary 7. Example 8. Let q1 , . . . , qT ∈ ]1, 2[. Then for all r > 0 there exist y0 ∈ H with T X
|y0 (k)|qk < r
k=1
and λ > 0 such that the problem ( −∆2 y(k − 1) = λqk (y(k) + y0 (k))|y(k) + y0 (k)|qk −2 , k ∈ [1, T ] y(0) = y(T + 1) = 0
has at least three solutions (notice that here F (k, x) = |x|qk is coercive on R for all k ∈ [1, T ]). ¤
3. Positive solutions As in [2], we pay special attention to positive solutions of problem (P ). Assume that f is non-negative: in this case, we observe that a=
T X
k=1
lim F (k, x),
x→−∞
b=
T X
k=1
so a ≤ 0, b ≥ 0. The following result holds.
lim F (k, x),
x→+∞
Theorem 9. Let f : [1, T ] × R → [0, +∞[ be a continuous function, r ∈]a, b[. Assume that J −1 ([r, +∞[) is not convex and for all k ∈ [1, T ] lim
x→+∞
F (k, x) = 0. |x|2
(7)
Then there exist y0 ∈ J −1 (] − ∞, r[) and λ > 0 such that (P ) has at least three non-negative solutions.
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Proof. It is easily seen that condition (4) is equivalent to (7). Then applying Theorem 3 we obtain three solutions of (P ) for suitable y0 ∈ J −1 (] − ∞, r[) and λ > 0. Note that every solution y of (P ) is non-negative: define y − ∈ H by putting − y (k) = max{0, −y(k)}; since y + y0 is a solution of (E), we get 0 = (y − λJ ′ (y + y0 ), y − ) = (y, y − ) − λ
T X
f (k, y(k) + y0 (k))y − (k)
k=1
≤ (y, y − ) ≤ −ky − k2 , so y − = 0, that is, y is non-negative.
¤
Remark 10. As the nonlinearity is supposed to be non-negative, if y is a solution of (P ), either y = 0 or y(k) > 0 for all k ∈ [1, T ] (see [2], Lemma 2.1). Thus, from the three solutions assured by Theorem 9, at least two are strictly positive. Example 11. Consider the following problem of the type (Q): k −∆2 y(k − 1) = , k ∈ [1, T ] 1 + (ky(k))2 y(0) = y(T + 1) = 0.
(Q0 )
For T = 2, by elementary arguments it is shown that such a problem has only one solution, which is positive by the above considerations. Obviously, the results of [2] cannot be applied here, for the lack of the main hypothesis: in fact lim inf
x→+∞
k =0 x(1 + (kx)2 )
for all k ∈ [1, 2],
so condition (1) is not satisfied. In order to get multiple solutions (for any T > 1), one can modify (Q0 ) by introducing λ and y0 and study the (P )-type problem k −∆2 y(k − 1) = λ , k ∈ [1, T ] 1 + k 2 (y(k) + y0 (k))2 (P0 ) y(0) = y(T + 1) = 0. i π h From Theorem 9 it follows that, for all r ∈ −T , 0 , there exist y0 ∈ H with 2 T X
arctan(ky0 (k)) < r
k=1
and λ > 0 such that problem (P0 ) has at least three positive solutions. Let us check the hypotheses of Theorem 9: condition (7) is obviously satisfied, since the function x → F (k, x) = arctan(kx) is bounded for all k.
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Moreover, J −1 ([r, +∞[) is not convex: in fact, our functional reads as J(y) =
T X
arctan(ky(k)),
k=1
so, given x < 0 such that
r r < arctan x < , T −2 T x define y ∈ H by putting y(k) = for all k ∈ [1, T ], hence J(y) < r. For all ρ > 0, k define y1,ρ , y2,ρ ∈ H by ρ ρ y1,ρ (k) = y(k) + ρδ1k − δ2k , y2,ρ (k) = y(k) − ρδ1k + δ2k . 2 2 We have lim J(y1,ρ ) = lim J(y2,ρ ) = (T − 2) arctan x > r. ρ→+∞
ρ→+∞
So, for ρ big enough we get y1,ρ , y2,ρ ∈ J −1 ([r, +∞[), while y = 12 (y1,ρ + y2,ρ ) does not belong to this set. Thus, J −1 ([r, +∞[) is not convex.
References [1] R. P. Agarwal, Difference equations and inequalities, Monogr. Textbooks Pure Appl. Math. Vol. 228, Marcel Dekker Inc., New York, 2000. [2] R. P. Agarwal, K. Perera and D. O’Regan, Multiple positive solutions of singular and nonsingular discrete problems via variational methods, Nonlinear Anal. 58 (2004), 69–73. [3] Z. Guo and J. Yu, Existence of periodic and subharmonic solutions for second-order superlinear difference equations, Sci. China Ser. A 46 (2003), 506–515. [4] Z. Guo and J. Yu, The existence of periodic and subharmonic solutions of subquadratic second order difference equations, J. London Math. Soc. 68 (2003) 419–430. [5] J. Henderson and S. Lauer, Existence of a positive solution for an nth order boundary value problem for nonlinear difference equations, Abstr. Appl. Anal. 2 (1997), 271–279. [6] D. Reid and W. Yin, Existence of solutions of some functional difference equations, Comm. Appl. Nonlinear Anal. 6 (1999), 51–59. [7] B. Ricceri, A general multiplicity theorem for certain nonlinear equations in Hilbert spaces, Proc. Amer. Math. Soc. 133 (2005), 3255–3261. Francesca Faraci and Antonio Iannizzotto Department of Mathematics and Computer Science University of Catania Viale A. Doria 6 I–95125 Catania Italy e-mail:
[email protected] [email protected] Manuscript received: December 5, 2005 and, in final form, May 24, 2006.
