Abstract
This paper investigates the existence of solutions for weighted p(r)Laplacian impulsive system mixed type boundary value problems. The proof of our main result is based upon Gaines and Mawhin's coincidence degree theory. Moreover, we obtain the existence of nonnegative solutions.
Keywords:
Weighted p(r)Laplacian; impulsive system; coincidence degree1 Introduction
In this paper, we mainly consider the existence of solutions for the weighted p(r)Laplacian system
where u: [0, T] → ℝ^{N}, with the following impulsive boundary conditions
where p ∈ C ([0, T], ℝ) and p(r) > 1, Δ_{p(r) }u:= (w(r) u'^{p(r)2 }u'(r))' is called weighted p(r)Laplacian; 0 < r_{1 }< r_{2 }< ⋯ < r_{k }< T; A_{i}, B_{i }∈ C(ℝ^{N }× ℝ^{N}, ℝ^{N}); a, b, c, d ∈ [0, +∞), ad + bc > 0.
Throughout the paper, o(1) means functions which uniformly convergent to 0 (as n → +∞); for any v ∈ ℝ^{N}, v^{j }will denote the jth component of v; the inner product in ℝ^{N }will be denoted by 〈·,·〉; · will denote the absolute value and the Euclidean norm on ℝ^{N}. Denote J = [0, T], J' = [0, T]\{r_{0}, r_{1},..., r_{k+1}}, J_{0 }= [r_{0}, r_{1}], J_{i }= (r_{i}, r_{i+1}], i = 1, ..., k, where r_{0 }= 0, r_{k+1 }= T. Denote the interior of J_{i}, i = 0, 1,..., k. Let PC(J, ℝ^{N}) = {x: J → ℝ^{N } x ∈ C(J_{i}, ℝ^{N}), i = 0, 1,..., k, and exists for i = 1,..., k}; w ∈ PC(J, ℝ) satisfies 0 < w(r), ∀r ∈ J', and ; , and exists for i = 0, 1,..., k}. For any x = (x^{1},..., x^{N}) ∈ PC(J, ℝ^{N}), denote x^{i}_{0 }= sup_{r∈J' }x^{i}(r). Obviously, PC(J, ℝ^{N}) is a Banach space with the norm , PC^{1}(J, ℝ^{N}) is a Banach space with the norm . In the following, PC(J, ℝ^{N}) and PC^{1}(J, ℝ^{N}) will be simply denoted by PC and PC^{1}, respectively. Let L^{1 }= L^{1}(J, ℝ^{N}) with the norm , ∀x ∈ L^{1}, where . We will denote
The study of differential equations and variational problems with nonstandard p(r)growth conditions is a new and interesting topic. It arises from nonlinear elasticity theory, electrorheological fluids, image processing, etc. (see [14]). Many results have been obtained on this problems, for example [125]. If p(r) ≡ p (a constant), (1) is the wellknown pLaplacian system. If p(r) is a general function, Δ_{p(r) }represents a nonhomogeneity and possesses more nonlinearity, thus Δ_{p(r) }is more complicated than Δ_{p}; for example, if Ω ⊂ ℝ^{N }is a bounded domain, the Rayleigh quotient
is zero in general, and only under some special conditions λ_{p(·) }> 0 (see [8,1719]), but the property of λ_{p }> 0 is very important in the study of pLaplacian problems.
Impulsive differential equations have been studied extensively in recent years. Such equations arise in many applications such as spacecraft control, impact mechanics, chemical engineering and inspection process in operations research (see [2628] and the references therein). It is interesting to note that p(r)Laplacian impulsive boundary problems are about comparatively new applications like ecological competition, respiratory dynamics and vaccination strategies. On the Laplacian impulsive differential equation boundary value problems, there are many results (see [2937]). There are many methods to deal with this problem, e.g., subsupersolution method, fixed point theorem, monotone iterative method and coincidence degree. Because of the nonlinearity of Δ_{p}, results on the existence of solutions for pLaplacian impulsive differential equation boundary value problems are rare (see [38,39]). On the Laplacian (p(x) ≡ 2) impulsive differential equations mixed type boundary value problems, we refer to [30,32,34].
In [39], Tian and Ge have studied nonlinear IBVP
where Φ_{p}(x) = x^{p2 }x, p > 1, ρ, s ∈ L^{∞ }[a, b] with essin f_{[a, b] }ρ > 0, and essin f_{[a,b] }s > 0, 0 < ρ(a), p(b) <∞, σ_{1 }≤ 0, σ_{2 }≥ 0, α, β, γ, σ > 0, a = t_{0 }< t_{1 }< ⋯ < t_{l }< t_{l+1 }= b, I_{i }∈ C([0, +∞), [0, ∞)), i = 1,..., l, f ∈ C ([a, b] × [0, +∞), [0, ∞)), f(·, 0) is nontrivial. By using variational methods, the existence of at least two positive solutions was obtained.
In [24,25], the present author investigates the existence of solutions of p(r)Laplacian impulsive differential equation (13) with periodiclike or multipoint boundary value conditions.
In this paper, we consider the existence of solutions for the weighted p(r)Laplacian impulsive differential system mixed type boundary value condition problems, when p(r) is a general function. The proof of our main result is based upon Gaines and Mawhin's coincidence degree theory. Since the nonlinear term f in (5) is independent on the firstorder derivative, and the impulsive conditions are simpler than (2), our main results partly generalized the results of [30,32,34,39]. Since the mixed type boundary value problems are different from periodiclike or multipoint boundary value conditions, and this paper gives two kinds of mixed type boundary value conditions (linear and nonlinear), our discussions are different from [24,25] and have more difficulties. Moreover, we obtain the existence of nonnegative solutions. This paper was motivated by [2426,38,40].
Let N ≥ 1, the function f: J × ℝ^{N }× ℝ^{N }→ ℝ^{N }is assumed to be Caratheodory; by this, we mean:
(i) for almost every t ∈ J, the function f(t, ·, ·) is continuous;
(ii) for each (x, y) ∈ ℝ^{N }× ℝ^{N}, the function f(·, x, y) is measurable on J;
(iii) for each R > 0, there is a α_{R }∈ L^{1 }(J, ℝ), such that, for almost every t ∈ J and every (x, y) ∈ ℝ^{N }× ℝ^{N }with x ≤ R, y ≤ R, one has
We say a function u: J → ℝ^{N }is a solution of (1) if u ∈ PC^{1 }with w(·) u'^{p(·)2 }u'(·) absolutely continuous on every , i = 0, 1,..., k, which satisfies (1) a.e. on J.
This paper is divided into three sections; in the second section, we present some preliminary. Finally, in the third section, we give the existence of solutions and nonnegative solutions of system (1)(4).
2 Preliminary
Let X and Y be two Banach spaces and L: D(L) ⊂ X → Y be a linear operator, where D(L) denotes the domain of L. L will be a Fredholm operator of index 0, i.e., ImL is closed in Y and the linear spaces KerL and coImL have the same dimension which is finite. We define X_{1 }= KerL and Y_{1 }= coImL, so we have the decompositions X = X_{1 }⊕ coKerL and Y = Y_{1 }⊕ ImL. Now, we have the linear isomorphism Λ: X_{1 }→ Y_{1 }and the continuous linear projectors P: X → X_{1 }and Q: Y → Y_{1 }with KerQ = ImL and ImP = X_{1}.
Let Ω be an open bounded subset of X with Ω ∩ D(L) ≠ ∅. Operator be a continuous operator. In order to define the coincidence degree of (L, S) in Ω, as in [40,41], denoted by d(L  S, Ω), we assume that
It is easy to see that the operator , M = (L + ΛP)^{1 }(S + ΛP) is well defined, and
If M is continuous and compact, then S is called Lcompact, and the LeraySchauder degree of I_{X } M (where I_{X }is the identity mapping of X) is well defined in Ω, and we will denote it by d_{LS }(I_{X } M, Ω, 0). This number is independent of the choice of P, Q and Λ (up to a sign) and we can define
Definition 2.1. (see [40,41]) The coincidence degree of (L, S) in Ω, denoted by d(L  S, Ω), is defined as d(L  S, Ω) = d_{LS }(I_{X } M, Ω, 0).
There are many papers about coincidence degree and its applications (see [4043]).
Proposition 2.2. (see [40]) (i) (Existence property). If d(L  S, Ω) ≠ 0, then there exists x ∈ Ω such that Lx = Sx.
(ii) (Homotopy invariant property). If is continuous, Lcompact and Lx ≠ H(x, λ) for all x ∈ ∂Ω and λ ∈ [0, 1], then d(L  H (·, λ), Ω) is independent of λ.
The effect of small perturbations is negligible, as is proved in the next Proposition (see [41] Theorem III.3, page 24).
Proposition 2.3. Assume that Lx ≠ Sx for each x ∈ ∂Ω. If S_{ε }is such that sup_{x∈∂Ω}S_{ε}x_{Y }is sufficiently small, then Lx ≠ Sx + S_{ε}x for all x ∈ ∂Ω and d(L  S  S_{ε}, Ω) = d(L  S, Ω).
For any (r, x) ∈ (J × ℝ^{N}), denote φ_{p(r)}(x) = x^{p(r)2}x. Obviously, φ has the following properties
Proposition 2.4 (see [41]) φ is a continuous function and satisfies
(i) For any r ∈ [0, T], φ_{p(r)}(·) is strictly monotone, i.e.,
(ii) There exists a function η: [0, +∞) → [0, +∞), η(s) → +∞ as s → +∞, such that
It is well known that φ_{p(r)}(·) is a homeomorphism from ℝ^{N }to ℝ^{N }for any fixed r ∈ J. Denote
It is clear that is continuous and sends bounded sets to bounded sets, and where . Let X = {(x_{1}, x_{2})  x_{1 }∈ PC, x_{2 }∈ PC} with the norm (x_{1}, x_{2})_{X }=  x_{1}_{0 }+ x_{2}_{0}, Y = L^{1 }× L^{1 }× ℝ^{2(k + 1)N}, and we define the norm on Y as
where y_{1}, y_{2 }∈ L^{1}, z_{m }∈ ℝ^{N}, m = 1,..., 2(k + 1), then X and Y are Banach spaces.
Define L: D(L) ⊂ X → Y and S: X → Y as the following
where
Obviously, the problem (1)(4) can be written as Lx = Sx, where L: X → Y is a linear operator, S: X → Y is a nonlinear operator, and X and Y are Banach spaces.
Since
we have dimKerL = dim(Y/ImL) = 2N < +∞ is even and ImL is closed in Y, then L is a Fredholm operator of index zero. Define
at the same time the projectors P: X → X and Q: Y → Y satisfy
Since ImQ is isomorphic to KerL, there exists an isomorphism Λ: KerL → ImQ. It is easy to see that L _{D(L)∩KerP }: D(L) ∩ KerP → ImL is invertible. We denote the inverse of that mapping by K_{p}, then K_{p }: ImL → D(L) ∩ KerP as
then
Proposition 2.5 (i) K_{p}(·) is continuous;
(ii) K_{p }(I  Q)S is continuous and compact.
Proof. (i) It is easy to see that K_{p}(·) is continuous. Moreover, the operator sends equiintegrable set of L^{1 }to relatively compact set of PC.
(ii) It is easy to see that K_{p}(I  Q)Sx ∈ X, ∀x ∈ X. Since and f is Caratheodory, it is easy to check that S is a continuous operator from X to Y, and the operators (x_{1}, x_{2}) → φ_{q(r) }((w(r))^{1 }x_{2}) and (x_{1}, x_{2}) → f (r, x_{1}, φ_{q(r) }((w(r))^{1 }x_{2})) both send bounded sets of X to equiintegrable set of L^{1}. Obviously, A_{i}, B_{i }and QS are compact continuous. Since f is Caratheodory, by using the AscoliArzela theorem, we can show that the operator is continuous and compact. This completes the proof.
Denote
where A_{i}, B_{i }are defined in (6), i = 1,..., k.
Consider
Define as M(·, ·) = (L + ΛP)^{1 }(S(·, ·) + ΛP), then
Since (I  Q)S(·, 0) = 0 and K_{p }(0) = 0, we have
It is easy to see that all the solutions of Lx = S(x, 0) belong to KerL, then
Notice that P _{KerL }= I_{KerL}, then
Proposition 2.6 (continuation theorem) (see [40]). Suppose that L is a Fredholm operator of index zero and S is Lcompact on , where Ω is an open bounded subset of X. If the following conditions are satisfied,
(i) for each λ ∈ (0, 1), every solution x of
is such that x ∉ ∂Ω;
(ii) QS(x, 0) ≠ 0 for x ∈ ∂Ω ∩ KerL and d_{B}(Λ^{1 }QS(·,0), Ω ∩ KerL, 0) ≠ 0, then the operator equation Lx = S(x, 1) has one solution lying in .
The importance of the above result is that it gives sufficient conditions for being able to calculate the coincidence degree as the Brouwer degree (denoted with d_{B}) of a related finite dimensional mapping. It is known that the degree of finite dimensional mappings is easier to calculate. The idea of the proof is the use of the homotopy of the problem Lx = S(x, 1) with the finite dimensional one Lx = S(x, 0).
Let us now consider the following simple impulsive problem
where J' = [0, T]\{r_{0}, r_{1}, ..., r_{k+1}}, a_{i}, b_{i }∈ ℝ^{N}; g ∈ L^{1}.
If u is a solution of (7), then we have
Denote ρ_{0 }= w(0)_{φp(0) }(u'(0)). Obviously, ρ_{0 }is dependent on g, a_{i}, b_{i}. Define F: L^{1 }→ PC as
By (8), we have
If a ≠ 0, then the boundary condition implies that
The boundary condition implies that
Denote H = L^{1 }× ℝ^{2kN }with the norm
then H is a Banach space. For fixed h ∈ H, we denote
Lemma 2.7 The mapping Θ_{h}(·) has the following properties
(i) For any fixed h ∈ H, the equation
has a unique solution ρ(h) ∈ ℝ^{N}.
(ii) The mapping ρ: H → ℝ^{N}, defined in (i), is continuous and sends bounded sets to bounded sets. Moreover, , where h = (g, a_{i}, b_{i}) ∈ H, , the notation p^{# }means .
Proof. (i) From Proposition 2.4, it is immediate that
and hence, if (10) has a solution, then it is unique.
Let . Since and F(g) ∈ PC, if ρ > R_{0}, it is easy to see that there exists a j_{0 }such that, the j_{0}th component of ρ satisfies
Obviously,
then
and
By (11) and (12), the j_{0}th component of keeps the same sign of on J and
Combining (13) and (14), the j_{0}th component of satisfies
From the definition , we have , then , and
Without loss of generality, we may assume that , then we have
Therefore, the j_{0}th component of keeps the same sign of . Since the j_{0}th component of keeps the same sign of , a, b, c, d ∈ [0, +∞) and ad + bc > 0, we can easily see that the j_{0}th component of Θ_{h}(ρ) keeps the same sign of , and thus
Let us consider the equation
According to the above discussion, all the solutions of (15) belong to b(R_{0 }+ 1) = {x ∈ ℝ^{N} x < R_{0 }+ 1}. So, we have
It means the existence of solutions of Θ_{h}(ρ) = 0.
In this way, we define a mapping ρ(h): H → ℝ^{N}, which satisfies
(ii) By the proof of (i), we also obtain ρ sends bounded set to bounded set, and
It only remains to prove the continuity of ρ. Let {u_{n}} is a convergent sequence in H and u_{n }→ u, as n → +∞. Since {ρ(u_{n})} is a bounded sequence, it contains a convergent subsequence satisfies as j → +∞. Since Θ_{h}(ρ) consists of continuous functions, and
Letting j → +∞, we have
from (i) we get ρ_{* }= ρ(u), it means that ρ is continuous.
This completes the proof.
If a = 0, the boundary condition implies that
Since ad + bc > 0, we have c > 0. Thus,
the boundary condition implies that
Denote G: H → ℝ^{N }as
It is easy to see that
Lemma 2.8 The function G(·) is continuous and sends bounded sets to bounded sets. Moreover, , where , the notation p* means .
3 Main results and proofs
In this section, we will apply coincidence degree to deal with the existence of solutions for (1)(4). In the following, we always use C and C_{i }to denote positive constants, if it cannot lead to confusion.
Theorem 3.1 Assume that Ω is an open bounded set in X such that the following conditions hold.
(1^{0}) For each λ ∈ (0, 1) the problem
has no solution on ∂Ω.
(2^{0}) (0, 0) ∈ Ω.
Then, problem (1)(4) has a solution u satisfies , where v = w(r)φ_{p(r)}(u'(r)), ∀r ∈ J'.
Proof. Let us consider the following operator equation
It is easy to see that x = (x_{1}, x_{2}) is a solution of Lx = S(x, 1) if and only if x_{1}(r) is a solution of (1)(4) and , ∀r ∈ J'.
According to Proposition 2.5, we can conclude that S(·, ·) is Lcompact from X × [0, 1] to Y. We assume that for λ = 1, (16) does not have a solution on ∂Ω, otherwise we complete the proof. Now from hypothesis (1^{0}), it follows that (16) has no solutions for (x, λ) ∈ ∂Ω × (0, 1]. For λ = 0, (17) is equivalent to Lx = S(x, 0), namely the following usual problem
The problem (??) is a usual differential equation. Hence,
where c_{1}, c_{2 }∈ ℝ^{N }are constants. The boundary value condition of (??) holds,
Since (ad + bc) > 0, we have
which together with hypothesis (2^{0}), implies that (0, 0)∈ Ω. Thus, we have proved that (16) has no solution on ∂Ω × [0, 1]. It means that the coincidence degree d[L  S(·, λ), Ω] is well defined for each λ ∈ [0, 1]. From the homotopy invariant property of that degree, we have
Now, it is clear that the following problem
is equivalent to problem (1)(4), and (18) tells us that problem (19) will have a solution if we can show that
Since by hypothesis (2^{0}), this last degree
where ω_{*}(c_{1}, c_{2}) = (ac_{1 } bφ_{q(0)}(c_{2}), cc_{1 }+ dc_{2}). This completes the proof.
Our next theorem is a consequence of Theorem 3.1. Denote
Theorem 3.2 Assume that the following conditions hold
(1^{0}) a > 0;
(2^{0}) lim_{u + v → +∞ }(f(r, u, v)/(u + v)^{β(r) 1}) = 0, for r ∈ J uniformly, where β(r) ∈ C(J, ℝ), and 1<β ^{ }≤ β ^{+ }< p ^{};
(3^{0}) when u + v is large enough, where ;
(4^{0}) when u + v is large enough, where 0 ≤ ε < β ^{+ } 1.
Then, problem (1)(4) has at least one solution.
Proof. Now, we consider the following operator equation
For any λ ∈ (0, 1], x = (x_{1}, x_{2}) = (u, v) is a solution of (20) if and only if and u(r) is a solution of the following
We claim that all the solutions of (21) are uniformly bounded for λ ∈ (0, 1]. In fact, if it is false, we can find a sequence (u_{n}, λ_{n}) of solutions for (21), such that u_{n}_{1 }> 1 and u_{n}_{1 }→ +∞ when n → +∞, λ_{n }∈ (0, 1]. Since (u_{n}, λ_{n}) are solutions of (21), we have
By computation, we have
Denote
We claim that
If it is false, without loss of generality, we may assume that
then for any n = 1, 2, ..., there is a j_{n }∈ {1, ..., N} such that the j_{n}th component of ρ_{n }satisfies
Thus, when n is large enough, the j_{n}th component of Γ_{n}(r) keeps the same sign as and satisfies
When n is large enough, we can conclude that the j_{n}th component of F{φ_{q(r)}[Γ_{n}(r)]} (T) keeps the same sign as and satisfies
Since
from (22) and (24), we can see that keeps the same sign as , when n is large enough.
But the boundary value conditions (4) mean that
It is a contradiction. Thus (23) is valid. Therefore,
It means that
From (22), (23) and (25), for any r ∈ J, we have
then
From (25) and (26), we get that all the solutions of (20) are uniformly bounded for any λ ∈ (0, 1].
When λ = 0, if (x_{1}, x_{2}) is a solution of (20), then (x_{1}, x_{2}) is a solution of the following usual equation
we have
Thus, there exists a large enough R_{0 }> 0 such that all the solutions of (20) belong to B(R_{0}) = {x ∈ X   x _{X }< R_{0}}. Thus, (20) has no solution on ∂B (R_{0}). From theorem 3.1, we obtain that (1)(4) has at least one solution. This completes the proof.
Theorem 3.3 Assume that the following conditions hold
(1^{0}) a = 0;
(2^{0}) lim_{u + v → +∞ }f(r, u, v)/(u + v)^{ε }= 0 for r ∈ J uniformly, where 0 ≤ ε min(1, p^{ } 1);
(3^{0}) when u + v is large enough, where 0 < θ < 1;
(4^{0}) when u + v is large enough, where 0 ≤ ε < min(1, p^{ } 1).
Then, problem (1)(4) has at least one solution.
Proof Now, we consider the following operator equation
If (x_{1}, x_{2}) is a solution of (27) when λ = 0, then (x_{1}, x_{2}) is a solution of the following usual equation
Then, we have
For any λ ∈ (0, 1], x = (x_{1}, x_{2}) = (u, v) is a solution of (27) if and only if and u(r) is a solution of the following
We only need to prove that all the solutions of (28) are uniformly bounded for λ ∈ (0, 1].
In fact, if it is false, we can find a sequence (u_{n}, λ_{n}) of solutions for (28), such that u_{n}_{1 }> 1 and u_{n}_{1 }→ +∞ when n → +∞. Since (u_{n}, λ_{n}) are solutions of (28), we have
From conditions (2^{0}) and (4^{0}), we have
Thus,
Denote
By solving u_{n}(r), we have
From condition (3^{0}), we have
The boundary value condition implies
From (31) and conditions (2^{0}), (3^{0}) and (4^{0}), we have
From (30) and (32), we have
From (29) and (33), we can conclude that {u_{n}_{1}} is uniformly bounded for λ ∈ (0, 1]. This completes the proof.
Now, let us consider the following mixed type boundary value condition
Theorem 3.4 Assume that the following conditions hold
(1^{0}) , for r ∈ J uniformly, where β(r) ∈ C(J, ℝ), and 1 < β ^{ }≤ β ^{+}< p^{};
(2^{0}) when u + v is large enough, where ;
(3^{0}) when u + v is large enough, where 0 ≤ ε < β ^{+ } 1.
Then, problem (1) with (2), (3) and (34) has at least one solution.
Proof. It is similar to the proof of Theorem 3.2 and Theorem 3.3, we omit it.
Denote
where f_{* }(r, u, v) is Caratheodory.
Let us consider
Theorem 3.5 Under the conditions of Theorem 3.2, Theorem 3.3 or Theorem 3.4, then (35) with (2), (3) and (4) or (34) has at least a solution when δ is small enough.
Proof We only need to prove the existence of solutions under the conditions of Theorem 3.2, the rest is similar. If δ = 0, the proof of Theorem 3.2 means that all the solutions of (35) with (2), (3) and (4) are bounded and belong to U(R_{0}) = {u ∈ PC^{1} u_{1 }< R_{0}}. Define S_{δ }: X → Y as
Since f_{* }(r, u, v) is a Caratheodory function, we have S_{δ}x  S_{0}x_{Y }→ 0 as δ → 0, for uniformly. According to Proposition 2.3, we get the existence of solutions.
In the following, we will consider the existence of nonnegative solutions. For any x = (x^{1}, ..., x^{N}) ∈ ℝ^{N}, the notation x ≥ 0 means x^{i }≥ 0 for any i = 1, ..., N.
Theorem 3.6 We assume
(i) f(r, u, v) ≤ 0, ∀(r, u, v) ∈ J × ℝ^{N }× ℝ^{N};
(ii) for any i = 1, ..., k, B_{i}(u, v) ≤ 0,∀(u, v) ∈ ℝ^{N }× ℝ^{N};
(iii) for any i = 1, ..., k, j = 1, ..., N, ∀(u, v) ∈ ℝ^{N }× ℝ^{N}.
Then, the solution u in Theorem 3.2, Theorem 3.3 or Theorem 3.4 is nonnegative.
Proof We only need to prove that the solution u in Theorem 3.2 is nonnegative, and the rest is similar. Denote
where
Similar to (8) and (9), we have
where
and ρ is the solution (unique) of
Denote
From (i), (ii) and (36), we can see that Φ(r) is decreasing, namely
We claim that
If it is false, then there exists some j_{0 }∈ {1, ..., N}, such that the j_{0}th component of ρ satisfies
Combining (i), (ii), (iii) and (40), we can see that the j_{0}th component of D is negative. It is a contradiction to (37). Thus, (39) is valid. So, we have
We claim that
If it is false. Then, there exists some j_{1 }∈ {1,..., N}, such that the j_{1}th component of Φ(T) satisfies
From (38) and (42), we have
Combining (i), (ii), (iii) and (42), we can see that the j_{1}th component of D is positive. It is a contradiction to (37). Thus, (41) is valid.
If c > 0. We have
Since Φ(r) is decreasing, Φ(0) = ρ ≥ 0 and Φ(T) ≤ 0, for any j = 1, ..., N, there exists ξ_{j }∈ J such that
Combining condition (iii), we can conclude that u^{j}(r) is increasing on [0, ξ_{j}], and u^{j}(r) is decreasing on (ξ_{j}, T]. Notice that u(0) ≥ 0 and u(T) ≥ 0, then we have u(r) ≥ 0,∀r ∈ [0, T].
If c = 0, boundary condition (4) means that Φ(T) = 0. Since Φ(r) is decreasing, we get that Φ(r) ≥ 0. Combining condition (iii), we can conclude that u(r) is increasing on J, namely u(t_{2})  u(t_{1}) ≥ 0,∀t_{2}, t_{1 }∈ J, t_{2 }> t_{1}. Notice that u(0) ≥ 0, then we have u(r) ≥ 0,∀t ∈ J. This completes the proof.
Corollary 3.7 We assume
(i) f(r, u, v) ≤ 0,∀(r, u, v) ∈ J × ℝ^{N }× ℝ^{N }with u ≥ 0;
(ii) for any i = 1, ..., k, B_{i}(u, v) ≤ 0,∀(u, v) ∈ ℝ^{N }× ℝ^{N }with u ≥ 0;
(iii) for any i = 1, ..., k, j = 1, ..., N, ,∀(u, v) ∈ ℝ^{N }× ℝ^{N }with u ≥ 0.
Then, we have
(1_{0}) Under the conditions of Theorem 3.2 or Theorem 3.3, (1)(4) has a nonnegative solution.
(2_{0}) Under the conditions of Theorem 3.4, (1) with (2), (3) and (34) has a nonnegative solution.
Proof We only need to prove that (1)(4) has a nonnegative solution under the conditions of Theorem 3.2, and the rest is similar. Define
where
Denote
then satisfies Caratheodory condition, and for any (r, u, v) ∈ J × ℝ^{N }× ℝ^{N}.
For any i = 1, ..., k, we denote
then and are continuous and satisfy
Obviously, we have
(2^{0})' , for r ∈ J uniformly, where β(r) ∈ C(J, ℝ), and1 < β ^{ }≤ β ^{+}< p^{ };
(3^{0})' when u + v is large enough, where ;
(4^{0})' when u + v is large enough, where 0 ≤ ε < β ^{+ } 1.
Let us consider
From Theorem 3.2 and Theorem 3.6, we can see that (43) has a nonnegative solution u. Since u ≥ 0, we have ϕ(u) = u, and then
Thus, u is a nonnegative solution of (1)(4). This completes the proof.
4 Examples
Example 4.1. Consider the following problem
where p(r) = 5 + cos 3r, , 0 ≤ g(r) ∈ L^{1}, e_{0}, e_{1 }∈ ℝ^{N}, w(r) = 3 + sin r, σ is a nonnegative parameter.
Obviously, is Caratheodory, q(r) ≤ 3.5 < 4 ≤ p (r) ≤ 6, then the conditions of Theorem 3.5 are satisfied, then (P_{1}) has a solution when δ > 0 is small enough. Moreover, when σ = 0, the conditions of Corollary 3.7 are satisfied, then (P_{1}) has a nonnegative solution.
Example 4.2. Consider the following problem
where p(r) = 5 + cos 3r, , 0 ≤ g(r) ∈ L^{1}, e_{0}, e_{1 }∈ ℝ^{N}, w(r) = 3+sin r, σ is a nonnegative parameter.
Obviously, is Caratheodory, 1 < q(r) < 2 < 4 ≤ p (r) ≤ 6, then conditions of Theorem 3.5 are satisfied, then (P_{2}) has a solution when δ > 0 is small enough. Moreover, when σ = 0, the conditions of Corollary 3.7 are satisfied, and (P_{2}) has a nonnegative solution.
Competing interests
The authors declare that they have no competing interests.
Authors' contributions
All authors typed, read and approved the final manuscript.
Acknowledgements
The authors would like to thank the referees very much for their helpful comments and suggestions. This study was partly supported by the National Science Foundation of China (10701066 & 10926075 & 10971087) and China Postdoctoral Science Foundation funded project (20090460969) and the Natural Science Foundation of Henan Education Committee (200875565).
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