Bull. Iran. Math. Soc. https://doi.org/10.1007/s41980-018-0025-x ORIGINAL PAPER
Variational Approaches for a p-Laplacian Boundary-Value Problem with Impulsive Effects Shapour Heidarkhani1 · Shahin Moradi2 · Giuseppe Caristi3
Received: 20 October 2017 / Revised: 25 January 2018 / Accepted: 22 February 2018 © Iranian Mathematical Society 2018
Abstract In this paper, a p-Laplacian boundary-value problem with impulsive effects is considered. The existence of at least one non-trivial weak solution and at least three non-negative weak solutions via variational methods and critical point theory is obtained. Some recent results are extended and improved. Some examples are presented to demonstrate the application of our main results. Keywords Weak solution · p-Laplacian boundary-value problem · Impulsive effect · Variational methods AMS Subject Classification 34B15 · 34B37 · 58E30
B
Shapour Heidarkhani
[email protected] Shahin Moradi
[email protected] Giuseppe Caristi
[email protected]
1
Department of Mathematics, Faculty of sciences, Razi University, 67149 Kermanshah, Iran
2
Young Researchers and Elite Club, Kermanshah Branch, Islamic Azad University, Kermanshah, Iran
3
Department of Economics, University of Messina, 98122 Messina, Italy
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1 Introduction In this work, we are interested in ensuring the existence of at least one weak solution and three weak solutions for the following nonlinear problem:
−(ρ(x) p (u (x))) + s(x) p (u (x)) = λ f (x, u(x)), α1 u (a + ) − α2 u(a) = 0,
x = x j , a.e. x ∈ (a, b), β1 u (b− ) + β2 u(b) = 0
(1.1)
with the impulsive conditions (ρ(x j ) p (u (x j ))) = I j (u(x j )),
j = 1, 2, . . . , l
(1.2)
where a, b ∈ R with a < b, p > 1, p (t) = |t| p−2 t, ρ, s ∈ L ∞ ([a, b]), α1 , α2 , β1 , β2 are positive constants, f : [a, b] × R → R is an L 1 -Carathéodory function, x0 = a < x1 < x2 < · · · < xl < xl+1 = b, − + − (ρ(x j ) p (u (x j ))) = ρ(x + j ) p (u (x j )) − ρ(x j ) p (u (x j ))
where z(y + ) and z(y − ) denote the right and left limits of z(y) at y, respectively, I j : R → R for j = 1, . . . , l are continuous and λ is a positive parameter. We assume that ρ0 := essinf x∈[a,b] ρ(x) > 0, s0 := essinf x∈[a,b] s(x) > 0, ρ(a + ) = ρ(a) > 0, ρ(b− ) = ρ(b) > 0. We refer to the impulsive problem (1.1)–(1.2) as (IP). The theory of impulsive differential equations has undergone rapid development over the years and played a very important role in modern applied mathematical models of real processes arising in phenomena studied in physics, population dynamics, chemical technology and economics. For the background, theory and applications of impulsive differential equations, we refer the interested readers to [4,13,26,27,30]. In recent years, problems of impulsive differential equations have been studied by a number of authors; for some recent work, we refer the reader to [2,3,10,11,14,16,18,33–37] for advances and references of this area. For instance, Tain and Ge in [34] using variational methods have investigated the existence of positive solutions to a second-order Sturm–Liouville boundary-value problem with impulsive effects, similar to (1.1), while in [35] they have studied the existence of positive solutions to linear and nonlinear Sturm–Liouville impulsive problems using variational methods. In fact, they have generalized the results of [34]. Zhang and Dai in [36] using critical point theory have obtained the existence of solutions for the following class of nonlinear impulsive problems with Dirichlet boundary conditions ⎧ ⎨ −u (t) + g(t)u(t) = f (t, u(t)), − u (t j ) = u (t + j ) − u (t j ) = Ii (u(ti )), ⎩ αu (a) − βu(0) = u(T ) = 0
a.e. t ∈ [0, T ], i = 1, 2, . . . , p,
where T > 0, f : [0, T ] × R → R is continuous, g ∈ L ∞ [0, T ], p is a positive − integer, 0 = t0 < t1 < t2 < · · · < t p < t p+1 = T , u (t j ) = u (t + j ) − u (t j ) =
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limt→t + u (t)−limt→t − u (t), I j : R → R are continuous. In particular, in [3] Bai and j j Dai, employing a three-critical-point theorem by Ricceri [32], have ensured the existence of at least three solutions for the problem (IP). Shi et al. [33] using variational methods and critical point theorems have obtained the existence of infinitely many solutions for the problem (IP), in the case λ = 1. In [10], Bonanno et al., employing variational methods and critical point theory, have studied a second-order impulsive differential equation with Dirichlet boundary conditions, depending on two real parameters. In fact, they showed that an appropriate growth condition of the nonlinear term, under small perturbations of impulsive terms, ensures the existence of three solutions, while in [11] applying variational methods, they have established multiplicity results to obtain the existence of infinitely many solutions for the same problem. In [16] using variational methods the existence of weak solutions for a perturbed p-Laplacian boundary-value problem with impulsive effects, similar to (1.1), was discussed. More precisely, the existence of an exactly determined open interval of positive parameters for which the problem admits infinitely many weak solutions was established. We also refer the reader to [5,17,22,28,38,39] in which existence results for impulsive boundary-value problems have been established. Motivated by the above facts, in the present paper, first using Ricceri’s variational principle [31] which we recall in the Sect. 3 (Theorem 3.1), we study the existence of at least one non-trivial weak solution for the problem (IP) under an asymptotical behaviour of the nonlinear datum at zero, see Theorem 3.2. Example 3.4 illustrates Theorem 3.2. We give some remarks on our results. In Theorem 3.8 we present an application of Theorem 3.2. As a special case of Theorem 3.8, we obtain Theorem 3.9 considering the case p = 2. We present Example 3.10 in which the hypotheses of Theorem 3.9 are fulfilled. Next, using two kinds of three-critical-point theorems by Bonanno and Candito [7] which we recall in Sect. 4 (Theorems 4.1 and 4.2), we establish the existence of at least three non-negative solutions for the problem (IP), see Theorems 4.3 and 4.4. In particular, we require that there is a growth of the antiderivative of f which is greater than quadratic in a suitable interval (see, for instance, condition (B2 ) of Theorem 4.5), and which is less than quadratic in the following suitable interval (see, for instance, condition (B2 ) of Theorem 4.5). We present Examples 4.6 and 4.7 in which the hypotheses of Theorems 4.3 and 4.5 are fulfilled, respectively. Theorems 4.8 and 4.9 are consequence of Theorems 4.3 and 4.5, respectively. Compared to the previous results, we give some new assumptions to obtain the existence of at least one non-trivial weak solution and at least three non-negative weak solutions of the problem (IP). We give exact intervals of the parameter λ. Recent related works are generalized. The paper is organized as follows. In Sect. 2, we recall some basic definitions and our main tool, while Sects. 3 and 4 are devoted to our abstract results. For a through on the subject, we also refer the reader to [6].
2 Preliminaries Let X := W 1, p ([a, b]) equipped with the norm
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b
u :=
1/ p
b
ρ(x)|u (x)| dx + p
p
s(x)|u(x)| dx
a
a
which is equivalent to the usual one. The following lemma is useful for proving our main results. Lemma 2.1 [34, Lemma 2.6] Let u ∈ X . Then u∞ = max |u(x)| ≤ Mu
(2.1)
x∈[a,b]
where M =2
1/q
max
1 (b
1/ p − a)1/ p s0
,
(b − a)1/ p
1 1 + = 1. p q
,
1/ p ρ0
By a classical solution of problem (IP), we mean a function u ∈ {u(x) ∈ X : ρ(x) p (u )(.) ∈ W 1,∞ (x j , x j+1 ), j = 0, 1, . . . , l} such that u satisfies (1.1)–(1.2). We say that a function u ∈ X is a weak solution of problem (IP) if
b
b
ρ(x) p (u (x))v (x)dx +
a
+ ρ(b) p
s(x) p (u(x))v(x)dx + ρ(a) p
a
α2 u(a) v(a) α1
b l β2 u(b) v(b) + I j (u(x j ))v(x j ) − λ f (x, u(x))v(x)dx = 0 β1 a j=1
for every v ∈ X . For the sake of convenience, in the sequel, we define Mp C1 = p
p−1
ρ(a)α2
p−1
α1
p−1
+
ρ(b)β2
p−1
β1
and C2 =
1+ Mp . p
For the given constants δ1 , δ2 , η1 and η2 put α1 1 δ1 δ2 β1 1 + δ1 + δ2 (b − a) K 1 := (b − a) + + −1 , η1 η2 α2 β2 η1 η2 a+ b−a b− b−a b η1 η2 ρ(x)dx + |K 1 | p ρ(x)dx + |δ2 | p ρ(x)dx, K 2 := |δ1 | p a
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α1 b−a α1 b−a β1 β1 |δ1 |, |δ2 |, |δ1 |, + + |δ2 | , α2 η1 α2 η2 β2 β2 b p K 4 := (C1 + C2 ) K 2 + K 3 s(x)dx , a α1 b−a b−a α1 + δ1 , − a , h 2 (x) = K 1 x − a − + h 1 (x) = δ1 x + α2 η1 η1 α2 β1 h 3 (x) = δ2 x − −b , β2
K 3 = max
and K
F
a+ b−a η 1
:=
F(x, h 1 (x))dx +
a
b− b−a η 2
a+ b−a η
F(x, h 2 (x))dx +
1
b b− b−a η
F(x, h 3 (x))dx.
2
Set F(x, t) =
t
f (x, ξ )dξ for all (x, t) ∈ [a, b] × R.
0
We need the following proposition in the proof our main result. Proposition 2.2 [18, Proposition 2.4] Let T : X → X ∗ be the operator defined by T (u)h = a
b
ρ(x) p (u (x))h (x)dx
+
b
s(x) p (u(x))h(x)dx + ρ(a) p
a
+ ρ(b) p
α2 u(a) h(a) α1
l β2 u(b) h(b) + I j (u(x j ))h(x j ) β1 j=1
for every u, h ∈ X . Then T admits a continuous inverse on X ∗ . In this paper, we assume throughout, and without further mention, that the following condition holds: (H ) 0≤
l
u(x j )
j=1 0
I j (t)dt ≤
1 p
max |u(t j )| p
j∈{1,...,l}
and I j (0) = 0 for j = 1, . . . , l.
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3 Main Results: One Solution We shall prove the existence of at least one non-trivial weak solution to the problem (IP) applying Ricceri’s variational principle [31, Theorem 2.1] (see also [12, Theorem 2.1]). Theorem 3.1 Let X be a reflexive real Banach space, let , : X −→ R be two Gâteaux differentiable functionals such that is sequentially weakly lower semicontinuous, strongly continuous and coercive in X and is sequentially weakly upper semicontinuous in X . Let Iλ be the functional defined as Iλ := − λ, λ ∈ R, and for every r > inf X , let ϕ be the function defined as ϕ(r ) :=
inf
supu∈−1 (−∞,r ) (u) − (u) r − (u)
u∈−1 (−∞,r )
.
1 Then, for every r > inf X and every λ ∈ 0, ϕ(r ) , the restriction of the functional Iλ to −1 (−∞, r ) admits a global minimum, which is a critical point (precisely a local minimum) of Iλ in X . We refer the interested reader to the papers [1,19,24,29] in which Theorem 3.1 has been successfully employed to the existence of at least one non-trivial solution for some boundary-value problems. We formulate our main results on the existence of at least one weak solution for the problem (IP). Theorem 3.2 Assume that sup b
γ >0 a
γp sup|t|≤γ F(x, t)dx
> pM p
(D F )
and there are non-empty open sets D ⊆ (a, b) and B ⊂ D of positive Lebesgue measure such that lim sup
ess inf x∈B F(x, ξ ) = +∞ |ξ | p
(3.1)
lim inf
ess inf x∈D F(x, ξ ) > −∞. |ξ | p
(3.2)
ξ →0+
and
ξ →0+
Then, for each
γp 1 sup b λ ∈ = 0, p pM γ >0 a sup|t|≤γ F(x, t)dx
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the problem (IP) possesses at least one non-trivial weak solution u λ ∈ X. Moreover, one has lim u λ = 0
λ→0+
and the real function l u λ (x j ) p−1 p−1 ρ(a)α2 ρ(b)β2 1 p p I j (t)dt + |u (a)| + |u λ (b)| p λ → u λ + λ p−1 p−1 p 0 pα pβ j=1 1 1 b − λ F(x, u λ (x))dx a
is negative and strictly decreasing in . Proof To apply Theorem 3.1 to our problem, we introduce the functionals , : X → R for each u ∈ X , as follows: 1 u p + p l
(u) =
u(x j )
p−1
I j (t)dt +
j=1 0
ρ(a)α2
p−1
|u(a)| p +
p−1 pα1
ρ(b)β2
p−1 pβ1
|u(b)| p (3.3)
and
b
(u) =
F(x, u(x))dx,
(3.4)
a
and put Iλ (u) = (u) − λ(u) for every u ∈ X . We now show that the functionals and satisfy the required conditions. It is well known that is a differentiable functional whose differential at the point u ∈ X is
(u)(v) =
b
f (x, u(x))v(x)dx,
a
for every v ∈ X, as well as, is sequentially weakly upper semicontinuous. Moreover, is continuously differentiable and whose differential at the point u ∈ X is
(u)v =
b
b
ρ(x) p (u (x))v (x)dx + s(x) p (u(x))v(x)dx a a α2 u(a) β2 u(b) v(b) v(a) + ρ(b) p + ρ(a) p α1 β1
+
l
I j (u(x j ))v(x j )
j=1
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for every v ∈ X . Furthermore, is sequentially weakly lower semicontinuous. Indeed, let for fixed u ∈ X , assume u n → u weakly in X as n → +∞. The continuity and convexity of u p imply u p is sequentially weakly lower semicontinuous, which combining with the continuity of I j for j = 1, . . . , l yields that ⎛
1 lim ⎝ u n p + n→+∞ p l
u n (x j )
p−1
I j (t)dt +
ρ(a)α2
p−1
pα1
j=1 0
1 u p + p l
≥
u(x j )
p−1
|u n (a)| p +
p−1
I j (t)dt +
ρ(a)α2
j=1 0
p−1 pα1
ρ(b)β2
p−1
pβ1
⎞ |u n (b)| p ⎠
p−1
|u(a)| p +
ρ(b)β2
p−1 pβ1
|u(b)| p ,
namely lim inf (u n ) ≥ (u) n→+∞
which means is sequentially weakly lower semicontinuous. Furthermore, using the condition (H ), we have l
0≤
u(t j )
I j (s)ds ≤
j=1 0
Mp u p , p
(3.5)
and this in conjunction with the second inequality in (3.5) ensures 1 u p ≤ (u) ≤ (C1 + C2 )u p p
(3.6)
for all u ∈ X . The first inequality in (3.6) follows limu→+∞ (u) = +∞, namely is coercive. Clearly, the weak solutions of problem (IP) are exactly the solutions of the equation Iλ (u)v = 0. Using the condition (D F ), there exists γ¯ > 0 such that b a
Putting r =
γ¯ p pM p ,
γ¯ p sup|t|≤γ¯ F(x, t)dx
> pM p .
from (2.1) we have p
|u(x)| p ≤ u∞ ≤ M p u p ≤ pM p (u) < pM p r = γ¯ p for all x ∈ (a, b). Hence, −1 (−∞, r ) = {u ∈ X : (u) < r } ⊆ {u ∈ X : u∞ ≤ γ¯ } , and this ensures (u) ≤
123
(3.7)
sup u∈−1 (−∞,r ) a
b
F(x, u(x))dx ≤
b
sup F(x, t)dx
a |t|≤γ¯
(3.8)
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for every u ∈ X such that (u) < r . Then sup (u) ≤
b
sup F(x, t)dx.
a |t|≤γ¯
(u)
By simple calculations and from the definition of ϕ(r ), since 0 ∈ −1 (−∞, r ) and (0) = (0) = 0, one has ϕ(r ) =
inf
(supu∈−1 (−∞,r ) (u)) − (u) r − (u)
u∈−1 (−∞,r )
b
≤ pM p
a
sup|t|≤γ¯ F(x, t)dx γ¯ p
≤
supu∈−1 (−∞,r ) (u) r
.
Hence, setting λ∗ =
γp 1 sup b p pM γ >0 a sup|t|≤γ F(x, t)dx
1 Theorem 3.1 ensures that for every λ ∈ (0, λ∗ ) ⊆ (0, ϕ(r ) ), the functional Iλ admits −1 at least one critical point (local minima) u λ ∈ (−∞, r ). We will show that the function u λ cannot be trivial. Let us prove that
lim sup u→0+
(u) = +∞. (u)
(3.9)
Owing to the assumptions (3.1) and (3.2), we can consider a sequence {ξn } ⊂ R+ converging to zero and two constants σ, κ (with σ > 0) such that ess inf x∈B F(x, ξn ) = +∞ n→+∞ |ξn | p lim
and ess inf F(x, ξ ) ≥ κ|ξ | p x∈D
for every ξ ∈ [0, σ ]. We consider a set G ⊂ B of positive measure and a function v ∈ X such that (k1 ) v(x) ∈ [0, 1] for every x ∈ (a, b), (k2 ) v(x) = 1 for every x ∈ G, (k3 ) v(x) = 0 for every x ∈ (a, b)\D. Hence, fix M > 0 and consider a real positive number η with η meas(G) + κ D\G |v(x)| p dx M< . (C1 + C2 )v p
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Then, there is n 0 ∈ N such that ξn < σ and ess inf F(x, ξn ) ≥ η|ξn | p x∈B
for every n > n 0 . Now, for every n > n 0 , by considering the properties of the function v (that is 0 ≤ ξn v(x) < σ for n large enough), we have (ξn v) G F(x, ξn )dx + D\G F(x, ξn v(x))dx = (ξn v) (ξ v) n η meas(G) + κ D\G |v(x)| p dx > > M. (C1 + C2 )v p Since M could be arbitrarily large, it yields lim
n→∞
(ξn v) = +∞, (ξn v)
from which (3.9) clearly follows. Hence, there exists a sequence {wn } ⊂ X strongly converging to zero such that, for n large enough, wn ∈ −1 (−∞, r ) and Iλ (wn ) = (wn ) − λ(wn ) < 0. Since u λ is a global minimum of the restriction of Iλ to −1 (−∞, r ), we obtain Iλ (u λ ) < 0;
(3.10)
hence, u λ is not trivial. From (3.10) we easily observe that the map (0, λ∗ ) λ → Iλ (u λ )
(3.11)
is negative. Also, one has lim u λ = 0.
λ→0+
Indeed, bearing in mind that is coercive and for every λ ∈ (0, λ∗ ) the solution u λ ∈ −1 (−∞, r ), one has that there exists a positive constant L such that u λ ≤ L for every λ ∈ (0, λ∗ ). We can easily see that there exists positive constant N such that
a
b
f (x, u λ (x))u λ (x)dx ≤ N u λ ≤ N L
(3.12)
for every λ ∈ (0, λ∗ ). Since u λ is a critical point of Iλ , we have Iλ (u λ )(v) = 0 for every v ∈ X and every λ ∈ (0, λ∗ ). In particular, Iλ (u λ )(u λ ) = 0, that is, (u λ )(u λ ) = λ
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a
b
f (x, u λ (x))u λ (x)dx
(3.13)
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for every λ ∈ (0, λ∗ ). Since 0 ≤ u λ p ≤ (u λ )(u λ ), from (3.13), we have 0 ≤ u λ p ≤ λ
a
b
f (x, u λ (x))u λ (x)dx
(3.14)
for any λ ∈ (0, λ∗ ). Letting λ → 0+ , (3.14) together with (3.12), yields lim u λ = 0.
λ→0+
Then, we have obviously the desired conclusion. Finally, we have to show that the map λ → Iλ (u λ ) is strictly decreasing in (0, λ∗ ). For our goal, we see that for any u ∈ X , one has
(u) Iλ (u) = λ − (u) . λ
(3.15)
Now, let us consider 0 < λ1 < λ2 < λ∗ and let u λi be the global minimum of the functional Iλi restricted to (−∞, r ) for i = 1, 2. Also, set m λi =
(v) (u λi ) − (u λi ) = inf − (v) λi λi v∈−1 (−∞,r )
for every i = 1, 2. Clearly, (3.11) in conjunction with (3.15) and the positivity of λ implies that m λi < 0 for i = 1, 2.
(3.16)
m λ2 ≤ m λ1 ,
(3.17)
Moreover,
due to the fact that 0 < λ1 < λ2 . Then, by (3.15)–(3.17) and again by the fact that 0 < λ1 < λ2 , we get that Iλ2 (u λ2 ) = λ2 m λ2 ≤ λ2 m λ1 < λ1 m λ1 = Iλ1 (u λ1 ), so that the map λ → Iλ (u λ ) is strictly decreasing in λ ∈ (0, λ∗ ). The arbitrariness of λ < λ∗ shows that λ → Iλ (u λ ) is strictly decreasing in (0, λ∗ ). The proof is complete.
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Remark 3.3 If f is non-negative then the weak solution ensured in Theorem 3.2 is positive. Indeed, let u 0 be a (non-trivial) weak solution of the problem (IP). Arguing by a contradiction, assume that the set A = x ∈ (a, b) : u 0 (x) < 0 is non-empty and of positive measure. Put v(x) ¯ = min{0, u 0 (x)} for all x ∈ (a, b). Clearly, v¯ ∈ X and one has
b
a
ρ(x) p (u 0 (x))v¯ (x)dx +
+ ρ(a) p − λ
b
s(x) p (u 0 (x))v(x)dx ¯
a
l α2 u 0 (a) β2 u 0 (b) v(a) ¯ + ρ(b) p v(b) ¯ + I j (u 0 (x j ))v(x ¯ j) α1 β1 j=1
b
f (x, u 0 (x))v(x)dx ¯ = 0.
a
Using the fact that u 0 also is a weak solution of (IP) and by choosing v¯ = u 0 and since f is non-negative, for λ > 0 we have
ρ(x) p (u 0 (x))u 0 (x)dx + s(x) p (u 0 (x))u 0 (x)dx A A α2 u 0 (a) β2 u 0 (b) u 0 (b) u 0 (a) + ρ(b) p + ρ(a) p α1 β1 + I j (u 0 (x j ))u 0 (x j ) = λ f (x, u 0 (x))u 0 (x)dx ≤ 0.
uw1, p (A) ≤
A
A
Hence, that is, u 0 w1, p (A) = 0 which contradicts with this fact that u 0 is a non-trivial weak solution. Hence, our claim is proved. Here we present an example in which the hypotheses of Theorem 3.2 are fulfilled. Example 3.4 Let a = 0, b = 1 and p = 4. Consider the following problem: ⎧ ⎨ −(ρ(x)4 (u (x))) + s(x)4 (u (x)) = λ f (u(x)), (ρ(x1 )4 (u (x1 ))) = I1 (u(x1 )), ⎩ + u (0 ) − u(0) = 0, u (1− ) + u(1) = 0
x = x1 , a.e. x ∈ (0, 1),
(3.18) where ρ(x) = x 2 + 16 for every x ∈ [0, 1], s(x) = 8e x + 16 for every x ∈ [0, 1], I1 (ξ ) = ξ 3 for every ξ ∈ R and f (t) =
1 3 t 4t + e + 2 sin(t) cos(t) 104
for every t ∈ R. By the expression of f , we have F(t) =
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1 4 t 2 t + e + sin (t) − 1 104
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for every t ∈ R. A simple calculation gives ρ0 = 16, s0 = 24 and M = sup
γ >0
γ4 sup|t|≤γ F(t)
1 √ 4 . 2
Since
= 104 > 2 = 4M 4 ,
we observe that all assumptions of Theorem 3.2 are fulfilled. Hence, Theorem 3.2 4 concludes that for each λ ∈ (0, 102 ) the problem (3.18) possesses at least one nontrivial weak solution u λ ∈ W 1,4 ([0, 1]). Moreover, one has lim u λ = 0
λ→0+
and the real function 1 λ → u λ 4 + 4
0
u λ (x1 )
17 I1 (ξ )dξ + 4|u λ (0)| + |u λ (1)|4 − λ 4
4
0
1
F(u λ (x))dx
4 is negative and strictly decreasing in 0, 102 . Now, we give some remarks of our results. Remark 3.5 In Theorem 3.2 we searched for the critical points of the functional Iλ naturally associated with the problem (IP). We note that, generally Iλ can be unbounded from the below equation in X . Indeed, for example, if we take f (ξ ) = 1+|ξ |γ − p ξ p−1 for every ξ ∈ R with γ > p, for any fixed u ∈ X \{0} and ι ∈ R, we obtain Iλ (ιu) = (ιu) − λ
b
F(ιu(x))dx a
≤ ι p (C1 + C2 )u p − λιu L 1 − λ
ιγ γ u L γ → −∞ γ
as ι → +∞. Hence, we cannot use direct minimization to find critical points of the functional Iλ . Remark 3.6 For fixed γ¯ > 0 let b Theorem 3.2 holds with u λ ∞
γ¯ p
−
a sup|t|≤γ¯ F(x, t)dx ≤ γ¯ .
> pM p . Then the result of
Remark 3.7 We observe that Theorem 3.2 is a bifurcation result in the sense that the pair (0, 0) belongs to the closure of the set {(u λ , λ) ∈ X × (0, +∞) : u λ isanon − trivialweaksolutionof (I P)} in X × R. Indeed, by Theorem 3.2 we have that u λ → 0 as λ → 0.
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Hence, there exist two sequences {u j } in X and {λ j } in R+ (here u j = u λ j ) such that λ j → 0+ and u j → 0, as j → +∞. Moreover, we emphasis that due to the fact that the map (0, λ∗ ) λ → Iλ (u λ ) is strictly decreasing, for every λ1 , λ2 ∈ (0, λ∗ ), with λ1 = λ2 , the weak solutions u λ1 and u λ2 ensured by Theorem 3.2 are different. When f does not depend upon x, we obtain the following autonomous version of Theorem 3.2. Theorem 3.8 Let f : R → R be a non-negative continuous function. Put F(ξ ) = ξ 0 f (t)dt for all ξ ∈ R. Assume that lim
ξ →0+
F(ξ ) = +∞. |ξ | p
Then, for each
γp 1 sup λ ∈ = 0, pM(b − a) γ >0 F(γ )
the problem ⎧ ⎨ −(ρ(x) p (u (x))) + s(x) p (u (x)) = λ f (u(x)), (ρ(x j ) p (u (x j ))) = I j (u(x j )), ⎩ α1 u (a + ) − α2 u(a) = 0, β1 u (b− ) + β2 u(b) = 0
x = x j , a.e. x ∈ (a, b), j = 1, 2, . . . , l, (3.19)
possesses at least one non-trivial weak solution u λ ∈ X such that lim u λ = 0
λ→0+
and the real function l u λ (x j ) p−1 p−1 ρ(a)α2 ρ(b)β2 1 p p I j (t)dt + |u λ (a)| + |u λ (b)| p λ → u λ + p−1 p−1 p 0 pα pβ j=1 1 1 b − λ F(u λ (x))dx a
is negative and strictly decreasing in .
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Now, we point out the following version of Theorem 3.2, in the case p = 2. Theorem 3.9 Assume that sup b
γ >0 a
γ2 sup|t|≤γ F(x, t)dx
> 2M 2
and there are a non-empty open set D ⊆ (a, b) and B ⊂ D of positive Lebesgue measure such that lim sup
ess inf x∈B F(x, ξ ) = +∞ |ξ |2
lim inf
ess inf x∈D F(x, ξ ) > −∞. |ξ |2
ξ →0+
and ξ →0+
Then, for each
1 γ2 λ ∈ = 0, sup 2M 2 γ >0 b sup|t|≤γ F(x, t)dx
a
the problem
−(ρ(x)u (x)) + s(x)u (x) = λ f (x, u(x)), α1 u (a + ) − α2 u(a) = 0, β1 u (b− ) + β2 u(b) = 0
x = x j , a.e. x ∈ (a, b),
with the impulsive conditions (ρ(x j )u (x j )) = I j (u(x j )),
j = 1, 2, . . . , l
possesses at least one non-trivial weak solution u λ ∈ X. Moreover, one has lim u λ = 0
λ→0+
and the real function l u λ (x j ) 1 ρ(a)α2 ρ(b)β2 2 λ → u λ + I j (t)dt + |u λ (a)|2 + |u λ (b)|2 2 2α 2β 1 1 j=1 0 b − λ F(x, u λ (x))dx a
is negative and strictly decreasing in .
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We present an example in which the hypotheses of Theorem 3.9 are satisfied. Example 3.10 Consider the problem ⎧ ⎨ −(ρ(x)u (x)) + s(x)u (x) = λ f (u(x)), (ρ(x1 )u (x1 )) = I1 (u(x1 )), ⎩ α1 u (0+ ) − α2 u(0) = 0, β1 u (1− ) + β2 u(1) = 0
x = x1 , a.e. x ∈ (0, 1),
(3.20) where ρ(x) = x 4 + 4 for every x ∈ [0, 1], s(x) = 4e x + 4 for every x ∈ [0, 1], I1 (ξ ) = ξ for every ξ ∈ R and 1 f (t) = 2 10
2t +
2t 1 + t2
for every t ∈ R. By the expression of f , we have F(t) =
1 2 2 t + ln(1 + t ) 102
for every t ∈ R. By simple calculations, we obtain ρ0 = 4, s0 = 8 and M = Since sup
γ2
γ >0 sup|t|≤γ F(t)
√1 . 2
= 102 > 1 = 2M 2 ,
we observe that all assumptions of Theorem 3.9 are fulfilled. Hence, Theorem 3.9 implies that for each λ ∈ (0, 102 ) the problem (3.20) possesses at least one non-trivial weak solution u λ ∈ W 1,2 ([0, 1]). Moreover, one has lim u λ = 0
λ→0+
and the real function 1 λ → u λ 2 + 2
u λ (x1 ) 0
5 I1 (ξ )dξ + 2|u λ (0)| + |u λ (1)|2 − λ 2
2
0
1
F(u λ (x))dx
is negative and strictly decreasing in (0, 102 ).
4 Main Results: Three Solutions Now, let us give the main tools, two kinds of three-critical-point theorems due Bonanno and Candito, we will use in this section to discuss the existence of multiple solutions for the problem (IP).
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Theorem 4.1 [7, Corollary 3.1] Let X be a reflexive real Banach space, : X −→ R be a convex, coercive and continuously Gâteaux differentiable functional whose derivative admits a continuous inverse on X ∗ , : X −→ R be a continuously Gâteaux differentiable functional whose derivative is compact, such that (a1 ) inf X = (0) = (0) = 0. Assume that there are two positive constants r1 , r2 and v ∈ X, with 2r1 < (v) < such that supu∈−1 (−∞,r1 ) (u)
r2 2,
2 (v) ; r1 3 (v) supu∈−1 (−∞,r2 ) (u) 1 (v) (a3 ) < ; r2 3 (v) (a4 ) for each
(a2 )
λ ∈ r 1 ,r2 :=
<
r2 r1 3 (v) 2 , min , 2 (v) supu∈−1 (−∞,r1 ) (u) supu∈−1 (−∞,r2 ) (u)
and for every u 1 , u 2 ∈ X which are local minima for the functional − λ and such that (u 1 ) ≥ 0 and (u 2 ) ≥ 0, one has inf (su 1 + (1 − s)u 2 ) ≥ 0.
s∈[0,1]
Then, for each λ ∈ r 1 ,r2 , the functional − λ has at least three distinct critical points which lie in −1 (−∞, r2 ). Theorem 4.1 has been successfully used to ensure the existence of at least three solutions for boundary-value problems in the papers [8,15]. Let X be a non-empty set and , : X → R be two functions. For all r, r1 , r2 > inf X , r2 > r1 , r3 > 0, we define ϕ(r ) :=
inf
(supu∈−1 (]−∞,r [) (u)) − (u)
r − (u) (v) − (u) , β(r1 , r2 ) := inf sup u∈−1 (]−∞,r1 [) v∈−1 ([r1 ,r2 [) (v) − (u) γ (r2 , r3 ) :=
u∈−1 (]−∞,r [)
,
supu∈−1 (]−∞,r2 +r3 [) (u)
, r3 α(r1 , r2 , r3 ) := max{ϕ(r1 ), ϕ(r2 ), γ (r2 , r3 )}. Theorem 4.2 [7, Theorem 3.3] Let X be a reflexive real Banach space, : X → R be a convex, coercive and continuously Gâteaux differentiable functional whose Gâteaux derivative admits a continuous inverse on X ∗ , : X → R be a continuously Gâteaux differentiable functional whose Gâteaux derivative is compact such that
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(b1 ) inf X = (0) = (0) = 0; (b2 ) for every u 1 , u 2 ∈ X such that (u 1 ) ≥ 0 and (u 2 ) ≥ 0, one has inf (su 1 + (1 − s)u 2 ) ≥ 0.
s∈[0,1]
Assume that there are three positive constants r1 , r2 , r3 with r1 < r2 , such that (b3 ) ϕ(r1 ) < β(r1 , r2 ); (b4 ) ϕ(r2 ) < β(r1 , r2 ); (b5 ) γ (r2 , r3 ) < β(r1 , r2 ). Then, for each λ ∈] β(r11,r2 ) , α(r1 ,r12 ,r3 ) [ the functional − λ admits three distinct critical points u 1 , u 2 , u 3 such that u 1 ∈ −1 (] − ∞, r1 [), u 2 ∈ −1 ([r1 , r2 [) and u 3 ∈ −1 (] − ∞, r2 + r3 [). We refer the interested reader to the papers [9,20,21,23,25] in which Theorem 4.2 has been successfully employed to obtain the existence of at least three solutions for boundary-value problems. For our first goal, let us fix positive constants θ1 , θ2 , η1 and η2 and constants δ1 and δ2 such that
p p θ1 θ2 3 pM p K 4 , b , < min b 2 KF sup F(x, t)dx 2 sup F(x, t)dx |t|≤θ |t|≤θ a a 1 2 and take
p p θ1 θ2 3 K4 1 λ ∈ := , b . , min b 2 K F pM p sup F(x, t)dx 2 sup F(x, t)dx |t|≤θ |t|≤θ a a 1 2
Theorem 4.3 Let f be a non-negative function. Assume that there exist constants δ1 and δ2 , and θ1 , θ2 , η1 and η2 with δ12 + δ22 = 0, η1 + η2 < η1 η2 , positive constants √ θ1 < M p K22 and M p 2 pK 4 < θ2 such that (A1 ) b max
a
sup|t|≤θ1 F(x, t)dx p θ1
,
2
b a
sup|t|≤θ2 F(x, t)dx
p θ2
<
2 KF . 3 pM p K 4
Then, for each λ ∈ the problem (IP) possesses at least three distinct non-negative weak solutions u i for i = 1, 2, 3, such that 0 ≤ u i (x) < θ2 , ∀ x ∈ [a, b], (i = 1, 2, 3). Proof To apply Theorem 4.1 to our problem. Take and as in the proof of Theorem 3.2. Furthermore, Proposition 2.2 gives that admits a continuous inverse on
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X ∗ . We observe that the regularity the assumptions of Theorem 4.1 on and are fulfilled. Put p
r1 =
p
θ1 θ and r2 = 2 p . pM p pM
Define w by setting ⎧ ⎪ ⎨ h 1 (x), w(x) = h 2 (x), ⎪ ⎩ h (x), 3
x ∈ [a, a + b−a η1 ), x ∈ [a + b−a η1 , b − b−a x ∈ (a + η2 , b].
b−a η2 ],
(4.1)
It is easy to see that w ∈ X and, in particular, in view of
b
ρ(x)|w (x)| p dx = K 2 and 0 ≤
a
b
a
p
s(x)|w(x)| p dx ≤ K 3
b
s(x)dx, a
we have
1 p
K 2 ≤ w ≤
K2 +
p K3
1p
b
,
s(x)dx a
from (3.6), we have K2 ≤ (w) ≤ K 4 . p
(4.2)
Taking (3.8) into account, we have −1 (−∞, r1 ) ⊆ {u ∈ X ; u∞ ≤ θ1 } , and it follows that sup u∈−1 (−∞,r1 )
(u) =
From conditions θ1 < M have
sup u∈−1 (−∞,r1 ) a
p
K2 2
and M
supu∈−1 (−∞,r1 ) (u) r1
b
=
√ p
F(x, u(x))dx ≤
b
sup F(x, t)dx.
a |t|≤θ1
2 pK 4 < θ2 , we have 2r1 < (w) <
supu∈−1 (−∞,r1 ) b
≤
a
b a
r2 . We 2
F(x, u(x))dx
r1 sup|t|≤θ1 F(x, t)dx p
θ1 pM p
<
1 λ
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b 2 a F(x, w(x))dx < 3 K4 2 KF 2 (w) ≤ ≤ 3 K4 3 (w)
(4.3)
and 2 supu∈−1 (−∞,r2 ) (u) r2
=
2 supu∈−1 (−∞,r2 ) b
≤2
a
b a
F(x, u(x))dx
r2 sup|t|≤θ2 F(x, t)dx
<
p θ2
b
pM p
1 λ
2 a F(x, w(x))dx 3 K4 2 KF 2 (w) . ≤ ≤ 3 K4 3 (w)
<
(4.4)
Therefore, (a2 ) and (a3 ) of Theorem 4.1 are satisfied. Finally, we prove that − λ satisfies the assumption (a4 ) of Theorem 4.1. For this, let u 1 and u 2 be two local minima for − λ. Then u 1 and u 2 are critical points for − λ, and so, they are solutions for the problem (IP). Since f is a non-negative function, then, we observe u 1 (x) ≥ 0 and u 2 (x) ≥ 0 for every x ∈ [a, b]. Thus, it follows that λ f (x, su 1 + (1 − s)u 2 ) ≥ 0 for all s ∈ [0, 1], and consequently, (su 1 + (1 − s)u 2 ) ≥ 0, for every s ∈ [0, 1]. From Theorem 4.1, for every ⎧ ⎫⎡ ⎤ r2 ⎪ ⎪ ⎨ ⎬ r1 ⎢ ⎥ 3 (w) 2 , min , λ∈⎦ ⎣, ⎪ ⎪ 2 (w) sup (u) sup (u) ⎩ ⎭ u∈−1 (−∞,r1 )
u∈−1 (−∞,r2 )
the functional − λ has at least three distinct critical points. Then, taking into account the fact that the weak solutions of problem (IP) are exactly critical points of the functional − λ we have the desired conclusion. Now, we present a variant of Theorem 4.3. Theorem 4.4 Let f be a non-negative function. Assume that there exist constants δ1 and δ2 , and positive constants θ√1 , θ2 , θ3 , η1 and η2 with δ12 + δ22 = 0, η1 + η2 < η1 η2 , √ p θ3 > θ2 , θ1 < M K 2 and M p pK 4 < θ2 such that (B1 ) b a
sup|t|≤θ1 F(x, t)dx
b a
sup|t|≤θ2 F(x, t)dx
, p p θ1 θ2 b 1 K F − a sup|t|≤θ1 F(x, t)dx . < pM p K4
max
123
b ,
a
sup|t|≤θ3 F(x, t)dx p
p
θ3 − θ2
Bull. Iran. Math. Soc.
Then, for every p θ1 1 , , min λ∈ b b p K F − a sup|t|≤θ1 F(x, t)dx pM a sup|t|≤θ1 F(x, t)dx
p p p θ2 θ3 − θ2 , b b a sup|t|≤θ2 F(x, t)dx a sup|t|≤θ3 F(x, t)dx
K4
the problem (IP) possesses at least three non-negative weak solutions u 1 , u 2 , and u 3 such that max |u 1 (x)| < θ1 , max |u 2 (x)| < θ2 and
x∈[a,b]
x∈[a,b]
max |u 3 (x)| < θ3 .
x∈[a,b]
Proof To apply Theorem 4.2 to our problem. Let and be as given in (3.3) and (3.4), respectively. These functionals satisfy the required hypotheses of Theorem 4.2. θ
p
θ
p
p
θ −θ
p
Put r1 : pM1 p , r2 := pM2 p , r3 := 3pM p2 and w as given in (4.1). From the conditions √ √ θ3 > θ2 , θ1 < M p K 2 and M p pK 4 < θ2 we get r3 > 0 and r1 < (w) < r2 . By the same arguments as given in the proof of Theorem 4.3, we have sup (u) ≤
(u)
b
sup F(x, t)dx,
a |t|≤θ1
sup (u) ≤
(u)
b
sup F(x, t)dx
a |t|≤θ2
and sup
(u)
(u) ≤
b
sup F(x, t)dx.
a |t|≤θ3
Therefore, since 0 ∈ −1 (− ∞, r1 ) and (0) = (0) = 0, one has ϕ(r1 ) = ≤
a
r1 sup|t|≤θ1 F(x, t)dx p
θ1 pM p
sup ϕ(r2 ) ≤
r1 − (u)
1)
supu∈−1 (−∞,r1 ) (u) b
≤
(supu∈−1 (−∞,r1 ) (u)) − (u)
inf
u∈−1 (−∞,r
u∈−1 (−∞,r2 )
,
(u)
r2
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b ≤
a
sup|t|≤θ2 F(x, t)dx p
θ2 pM p
and sup γ (r2 , r3 ) ≤
u∈−1 (−∞,r2 +r3 )
b ≤
a
(u)
r3 sup|t|≤θ3 F(x, t)dx p
p
θ3 −θ2 pM p
.
For each u ∈ −1 (− ∞, r1 ) one has β(r1 , r2 ) ≥ ≥
KF − KF
b a
sup|t|≤θ1 F(x, t)dx
(w) − (u) b − a sup|t|≤θ1 F(x, t)dx K4
.
Due to (B1 ) we get α(r1 , r2 , r3 ) < β(r1 , r2 ). By the same arguments as given in the proof of Theorem 4.3 we observe that the functional − λ satisfies the assumption (b2 ) of Theorem 4.2. Hence, Theorem 4.2 implies that for every
p θ1 1 λ∈ , , min b b p K F − a sup|t|≤θ1 F(x, t)dx pM a sup|t|≤θ1 F(x, t)dx
p p p θ2 θ3 − θ2 , b b a sup|t|≤θ2 F(x, t)dx a sup|t|≤θ3 F(x, t)dx K4
the functional Iλ has three critical points u i , i = 1, 2, 3, in X such that (u 1 ) < r1 , (u 2 ) < r2 and (u 3 ) < r2 + r3 , that is, max |u 1 (x)| < θ1 , max |u 2 (x)| < θ2 and
x∈[a,b]
x∈[a,b]
max |u 3 (x)| < θ3 .
x∈[a,b]
Then, taking into account the fact that the weak solutions of the problem (IP) are exactly critical points of the functional Iλ we have the desired conclusion. Now, we deduce the following straightforward consequence of Theorem 4.4.
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Theorem 4.5 Let f be a non-negative function. Assume that there exist constants δ1 and δ2 , and positive constants θ1 , θ4 , η1 and η2 with δ12 + δ22 = 0, η1 + η2 < η1 η2 , √ √ p p θ1 < M K 2 and M 2 pK 4 < θ4 such that (B2 ) b
sup|t|≤θ1 F(x, t)dx
a
max
p θ1
,
2
b a
sup|t|≤θ2 F(x, t)dx
<
p θ4
1 KF . 2 pM p K 4
Then, for every
λ∈ =
p p θ1 θ4 2K 4 1 , b , , min b K F pM p a sup|t|≤θ1 F(x, t)dx 2 a sup|t|≤θ4 F(x, t)dx
the problem (IP) possesses at least three non-negative weak solutions u 1 , u 2 and u 3 such that 1 max |u 2 (x)| < √ θ4 and p x∈[a,b] 2
max |u 1 (x)| < θ1 ,
x∈[a,b]
Proof Choose θ2 = b a
1 √ p θ4 2
sup|t|≤θ2 F(x, t)dx p θ2
max |u 3 (x)| < θ4 .
x∈[a,b]
and θ3 = θ4 . So, from (B2 ) one has b
2
a
= <
sup|t|≤
θ4 √ p2
F(x, t)dx ≤
p θ4
2
b a
sup|t|≤θ4 F(x, t)dx p
θ4
1 KF 2 pM p K 4
(4.5)
and b a
sup|t|≤θ3 F(x, t)dx p θ3
p − θ2
=
2
b a
sup|t|≤θ4 F(x, t)dx p θ4
<
1 KF . 2 pM p K 4
(4.6)
Moreover, using (B2 ) we have 1 KF − pM p >
1 pM p
b
a
sup|t|≤θ1 F(x, t)dx K4
KF K4
−
KF
1 2 K4
⎛ ⎞ b sup F(x, t)dx 1 ⎝KF |t|≤θ 1 ⎠ > − a p θ1 pM p K 4 pM p
=
KF
1 . 2 pM p K 4
Hence, from (B2 ), (4.5) and (4.6), it is easy to see that the assumption (B1 ) of Theorem 4.4 is satisfied, and it follows the conclusion.
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We now present the following examples to illustrate Theorems 4.3 and 4.5, respectively. Example 4.6 Consider the problem ⎧ ⎨ −(ρ(x)3 (u (x))) + s(x)3 (u (x)) = λ f (u(x)) (ρ(x1 )3 (u (x1 ))) = I1 (u(x1 )), ⎩ α1 u (0+ ) − α2 u(0) = 0, β1 u (1− ) + β2 u(1) = 0
x = x1 , a.e. x ∈ (0, 1), (4.7)
where ρ(x) = x 2 + 32, s(x) = e x + 32 for every x ∈ [0, 1], I1 (ξ ) = ξ 2 for every ξ ∈ R and f (t) =
5t 5 , 5 , t2
if t < 1, if t ≥ 1.
By the expression of f , we have F(t) =
t 5, 6 − 5t ,
if t < 1, if t ≥ 1.
Choose η1 = η2 = 4, θ1 = 10−6 , θ2 = 103 , δ1 = −1 and δ2 = 1. Direct calculations 65 9 5 37 1555 125 , C2 = 24 , K 1 = 0, K 2 = 1555 give C1 = 24 96 , K 3 = 4 and K 4 = 12 ( 96 + 64 (31 + e)). We clearly observe that all assumptions of Theorem 4.3 are satisfied. Then, it follows that for each λ ∈ :=
8 × 109 3K 4 , , 8 + 40 ln(2) − 20 ln(5) 18 − 153 10
the problem (4.7) possesses at least three distinct non-negative weak solutions u i , i = 1, 2, 3, such that 0 < u i (x) < 103 for all x ∈ [0, 1], i = 1, 2, 3. Example 4.7 Let p = 4, a = 0 and b = 1. We consider the problem ⎧ ⎨ −(ρ(x)4 (u (x))) + s(x)4 (u (x)) = λ f (u(x)) (ρ(x1 )4 (u (x1 ))) = I1 (u(x1 )), ⎩ α1 u (0+ ) − α2 u(0) = 0, β1 u (1− ) + β2 u(1) = 0
x = x1 , a.e. x ∈ (0, 1), (4.8)
where ρ(x) = x 2 + 128, s(x) = x 4 + 128 for every x ∈ [0, 1], I1 (ξ ) = ξ 3 for every ξ ∈ R and ⎧ 6 ⎨ 7t , f (t) = 7t, ⎩ 448 , t2
123
if t < 1, if 1 ≤ t ≤ 4, if t > 4.
Bull. Iran. Math. Soc.
By the expression of f , we have ⎧ 7 ⎨t , F(t) = 27 t 2 − 25 , ⎩ 331 448 2 − t ,
if t < 1, if 1 ≤ t ≤ 4, if t > 4.
Take η1 = η2 = 4, θ1 = 10−8 , θ4 = 103 , δ1 = −1 and δ2 = 1. By simple 17 6163 5 calculations, we obtain C1 = 257 64 , C 2 = 64 , K 1 = 0, K 2 = 96 , K 3 = 4 and 137 6163 80125 K 4 = 32 ( 96 + 256 ). Then all conditions in Theorem 4.5 are fulfilled. Therefore, for each
48K 4 16 × 1012 , λ∈ , 59 662 − 1792 3 10
the problem (4.8) possesses at least three distinct non-negative weak solutions u i (i = 1, 2, 3) such that u i ∞ < 103 , i = 1, 2, 3. Here, we present a simple consequence of Theorem 4.3 in the case f does not depend upon x. Theorem 4.8 Let f : R → R be a non-negative continuous function, and put F(t) = t 0 f (ξ )dξ forall t ∈ R. Assume that lim
t→0+
F(t) =0 t3
and 106 F(102 ) <
1 4
0
F(x + 1)dx +
1 2F
5 4
+
1 3 4
F(2 − x)dx .
9M 3 K 4
Then, for every ⎡
⎤ ⎥3 λ∈⎦ 2
106
1 4
0
K4 ⎢ , ⎣, 6M 3 F(102 ) 1 1 5 F(x + 1)dx + 2 F 4 + 3 F(2 − x)dx 4
the problem ⎧ ⎨ −(ρ(x)3 (u (x))) + s(x)3 (u (x)) = λ f (u(x)) 3 (u (x j ))) = I j (u(x j )), ⎩ α1 u (0+ ) − α2 u(0) = 0, β1 u (1− ) + β2 u(1) = 0
x = x j , a.e. x ∈ (0, 1), j = 1, 2, . . . , l,
possesses at least three distinct non-negative solutions.
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Proof Our aim is to employ Theorem 4.3 by choosing η1 = η2 = 4, θ2 = 102 , 1 1 , 1/3 }, C1 = δ1 = −1 and δ2 = 1. Direct calculations give M = 22/3 max{ 1/3
M 3 (ρ(0)+ρ(1)) , 3
K4 =
1
3 4 C2 = 1+M 3 , K 1 = 0, K 2 = 0 ρ(x)dx 1 (C1 + C2 )(K 2 + 125 64 0 s(x)dx). We have
3 K4 3 = F 2K 2
+
1 3 4
s0
ρ0
ρ(x)dx, K 3 =
5 4
and
K4 1 F(x + 1)dx + 21 F 45 + 3 F(2 − x)dx
1 4
0
4
and p
θ2 pM p
2 sup|t|≤θ2 F(t)
=
106 . 6M 3 F(102 )
Moreover, we have lim
t→0+
F(t) = 0. t3
Then, there exists a positive constant θ1 < M F(θ1 ) 2 < 3 9M 3 θ1
1 4
0
3
K2 2
F(x + 1)dx + 21 F
such that
5 4
K4
+
1 3 4
F(2 − x)dx
.
and θ13 106 > . F(θ1 ) 2F(102 ) Finally, we easily observe that all assumptions of Theorem 4.3 are satisfied, and it follows the result. We end this paper by giving the following consequence of Theorem 4.5. 4.9 Let f : R → R be a continuous and non-zero function, and put F(t) = Theorem t f (ξ )dξ forall t ∈ R. Assume that f (t) ≥ 0 for all t ∈ [0, +∞) such that 0 (B3 ) limt→0+
F(t) tp
= limt→+∞
F(t) tp
= 0.
Then, for every λ > λ where λ = inf
123
2K 4 F : h (x) > 0 forall x ∈ [a, b], K > 0, i = 1, 2, 3 i KF
Bull. Iran. Math. Soc.
where
a+ b−a η 1
K F :=
F(h 1 (x))dx +
a
b− b−a η 2
F(h 2 (x))dx +
a+ b−a η 1
b
b− b−a η
F(h 3 (x))dx,
2
the problem (3.19) possesses at least two non-trivial weak solutions.
Proof Fix λ > λ∗ , and let h i (x) > 0 for all x ∈ [a, b] for i = 1, 2, 3, K F > 0 and λ > 2KF4 . From (B3 ) there is θ1 > 0 such that K
θ1 < M
) p
K 2 and
sup|t|≤θ1 F(t) p θ1
<
1 λpM p (b
− a)
and there is θ4 > 0 such that M
) p
2 pK 4 < θ4 and
sup|t|≤θ4 F(t) p
θ4
Therefore, Theorem 4.3 ensures the conclusion.
<
1 . 2λpM p (b − a)
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