Boundary value problems for systems of second-order functional differential equations

Svatoslav Staněk^{∗}^{†}

Department of Mathematical Analysis, Faculty of Science, Palacký University, Tomkova 40, 779 00 Olomouc, Czech Republic e–mail: stanek@risc.upol.cz

Abstract. Systems of second-order functional differential equations (x′(t)+L(x)(t))′ = F (x)(t) together with nonlinear functional boundary conditions are considered. Here L : C^{1}([0, T]; ℝ^{n}) → C^{0}([0, T]; ℝ^{n}) and F : C^{1}([0, T]; ℝ^{n}) → L_{1}([0, T]; ℝ^{n}) are continuous operators. Existence results are proved by the Leray-Schauder degree and the Borsuk antipodal theorem for α-condensing operators. Examples demonstrate the optimality of conditions.

Key words and phrases. Functional boundary value problem, existence, Leray-Schauder degree, Borsuk theorem, α-condensing operator.

1991 Mathematics Subject Classification. 34K10, 34B15

1 Introduction, notation

Let J = [0, T] be a compact interval, n ∈ ℕ. For a ∈ ℝ^{n}, a = (a_{1}, …, a_{n}), we set |a| = max{|a_{1}|, …, |a_{n}|}. For any x : J → ℝ^{n} (n ≥ 2) we write x(t) = (x_{1}(t), …, x_{n}(t)) and ∫_{a}^{b} x(t) dt = (∫_{a}^{b} x_{1}(t) dt, …, ∫_{a}^{b} x_{n}(t) dt) for 0 ≤ a < b ≤ T.

From now on, C^{0}(J; ℝ), C^{0}(J; ℝ^{n}), C^{1}(J; ℝ^{n}), C^{0}(J; ℝ^{n}) × ℝ^{n} × ℝ^{n}, L_{1}(J; ℝ) and L_{1}(J; ℝ^{n}) denote the Banach spaces with the norms ∥x∥_{0} = max{|x(t)| : t ∈ J}, ∥x∥ = max{∥x_{1}∥_{0}, …, ∥x_{n}∥_{0}}, ∥x∥_{1} = max{∥x∥, ∥x′∥}, ∥(x, a, b)∥_{∗} = ∥x∥ + |a| + |b|, ∥x∥_{L_{1}}^{0} = ∫_{0}^{T} |x(t)| dt and ∥x∥_{L_{1}} = max{∥x_{1}∥_{L_{1}}^{0}, …, ∥x_{n}∥_{L_{1}}^{0}}, respectively. K(J × [0, ∞); [0, ∞)) denotes the set of all functions ω : J × [0, ∞) → [0, ∞) which are integrable on J in the first variable, nondecreasing on [0, ∞) in the

second variable and lim_{ϱ→∞} \frac{1}{ϱ} ∫_{0}^{T} ω(t, ϱ) dt = 0.

∗This work was supported by grant no. 201/98/0318 of the Grant Agency of Czech Republic and by the Council of Czech Government J14/98:153100011

†This paper is in final form and no version of it will be submitted for publication elsewhere

Denote by A_{0} the set of all functionals α : C^{0}(J; ℝ) → ℝ which are

a) continuous, Im(α)=ℝ, and

b) increasing (i.e. x, y ∈ C^{0}(J; ℝ), x(t) < y(t) for t ∈ J ⇒ α(x) < α(y)).

Here Im(α) stands for the range of α. If k is an increasing homeomorphism on ℝ and 0 ≤ a < b ≤ T, then the following functionals

max{k(x(t)) : a ≤ t ≤ b}, min{k(x(t)) : a ≤ t ≤ b}, ∫_{a}^{b} k(x(t)) dt

belong to the set A_{0}. Next examples of functionals belonging to the set A_{0} can be found for example in [2], [3].

Let A = \underbrace{A_{0} × … × A_{0}}_{n}. For each x ∈ C^{0}(J; ℝ^{n}), x(t) = (x_{1}(t), …, x_{n}(t)) and

φ ∈ A, φ = (φ_{1}, …, φ_{n}), we define φ(x) by

φ(x) = (φ_{1}(x_{1}), …, φ_{n}(x_{n})).  (1)

Let L : C^{1}(J; ℝ^{n}) → C^{0}(J; ℝ^{n}), F : C^{1}(J; ℝ^{n}) → L_{1}(J; ℝ^{n}) be continuous operators, L = (L_{1}, …, L_{n}), F = (F_{1}, …, F_{n}). Consider the functional boundary value problem (BVP for short)

(x′(t) + L(x)(t))′ = F (x)(t),  (2)
φ(x) = A, ψ(x′) = B.  (3)

Here φ, ψ ∈ A, φ = (φ_{1}, …, φ_{n}), ψ = (ψ_{1}, …, ψ_{n}) and A, B ∈ ℝ^{n}, A = (A_{1}, …, A_{n}), B = (B_{1}, …, B_{n}).

A function x ∈ C^{1}(J; ℝ^{n}) is said to be a solution of BVP (2), (3) if the vector function x′(t)+L(x)(t) is absolutely continuous on J, (2) is satisfied for a.e. t ∈ J and x satisfies the boundary conditions (3).

The aim of this paper is to state sufficient conditions for the existence results of BVP (2), (3). These results are proved by the Leray-Schauder degree and the Borsuk theorem for α-condensing operators (see e.g. [1]). In our case α-condensing operators have the form U + V, where U is a compact operator and V is a strict contraction. We recall that this paper is a continuation of the previous paper by the author [3], where the scalar BVP

(x′(t) + L_{1}(x′)(t))′ = F_{1}(x)(t),
φ_{1}(x) = 0, ψ_{1}(x′) = 0

was considered. Here L_{1} : C^{0}(J; ℝ) → C^{0}(J; ℝ), F_{1} : C^{1}(J; ℝ) → L_{1}(J; ℝ) are continuous operators and φ_{1}, ψ_{1} ∈ A_{0} satisfy φ_{1}(0) = 0 = ψ_{1}(0).

We assume throughout the paper that the continuous operators L and F in (2) satisfy the following assumptions:

(H_{1}) There exists k ∈ [0, \frac{1}{2μ}), μ = max\{1, T \}, such that

∥L(x) − L(y)∥ ≤ k∥x − y∥_{1} for x, y ∈ C^{1}(J; ℝ^{n}),

(H_{2}) There exists ω ∈ K(J × [0, ∞); [0, ∞)) such that

|F (x)(t)| ≤ ω(t, ∥x∥_{1})
for a.e. t ∈ J and each x ∈ C^{1}(J; ℝ^{n}).

Remark 1. If assumption (H_{1}) is satisfied then

∥L(x)∥ ≤ k∥x∥_{1} + ∥L(0)∥ for x ∈ C^{1}(J; ℝ^{n}).

Example 1. Let w ∈ C^{0}(J × ℝ^{n} × ℝ^{n} × ℝ^{n} × ℝ^{n}; ℝ^{n}), χ, ϕ ∈ C^{0}(J; J) and

|w(t, r_{1}, u_{1}, v_{1}, z_{1}) − w(t, r_{2}, u_{2}, v_{2}, z_{2})|
≤ k max{|r_{1} − r_{2}|, |u_{1} − u_{2}|, |v_{1} − v_{2}|, |z_{1} − z_{2}|}

for t ∈ J and r_{i}, u_{i}, v_{i}, z_{i} ∈ ℝ^{n} (i = 1, 2), where k ∈ [0, \frac{1}{2μ}). Then the Nemytskii operator L : C^{1}(J; ℝ^{n}) → C^{0}(J; ℝ^{n}),

L(x)(t) = w(t, x(t), x(χ(t)), x′(t), x′(ϕ(t)))

satisfies assumption (H_{1}).

Example 2. Let f : J × ℝ^{n} × ℝ^{n} → ℝ^{n} satisfy the Carathéodory conditions on J × ℝ^{n} × ℝ^{n} and

|f(t, u, v)| ≤ ω(t, max{|u|, |v|})

for a.e. t ∈ J and each u, v ∈ ℝ^{n}, where ω ∈ K(J × [0, ∞); [0, ∞)). Then the Nemytskii operator F : C^{1}(J; ℝ^{n}) → L_{1}(J; ℝ^{n}),

F (x)(t) = f(t, x(t), x′(t))

satisfies assumption (H_{2}).

The existence results for BVP (2), (3) are given in Sec. 3. Here the optimality of our assumptions (H_{1}) and (H_{2}) is studied as well. We shall show that k ∈ [0, \frac{1}{2}) can not be replaced be the weaker assumption k ∈ [0, \frac{1}{2}] in (H_{1}) provided T ≤ 1 (see Example 4), and if k > \frac{1}{2μ} in (H_{2}) then there exists unsolvable BVP of the type (2), (3) provided T > 1 (see Example 5). Example 6 shows that the

condition lim_{ϱ→∞} \frac{1}{ϱ} ∫_{0}^{T} ω(t, ϱ) dt = 0 which appears for ω in (H_{2}) can not be replaced

by lim sup_{ϱ→∞} \frac{1}{ϱ} ∫_{0}^{T} ω(t, ϱ) dt < ∞.

2 Auxiliary results

For each α ∈ A_{0}, we define the function p_{α} ∈ C^{0}(ℝ; ℝ) by

p_{α}(c) = α(c). ^{1}

Then p_{α} is increasing on ℝ and maps ℝ onto itself. Hence there exists the inverse function p_{α}^{−1} : ℝ → ℝ to p_{α}.

From now on, m_{γC} ∈ ℝ is defined for each γ ∈ A, γ = (γ_{1}, …, γ_{n}) and C ∈ ℝ^{n}, C = (C_{1}, …, C_{n}), by

m_{γC} = max{|p_{γ_{i}}^{−1}(C_{i})| : i = 1, …, n}.  (4)

Lemma 1. Let γ ∈ A, A ∈ ℝ^{n} and let γ(x) = A for some x ∈ C^{0}(J; ℝ^{n}). Then there exists ξ ∈ ℝ^{n} such that

(x_{1}(ξ_{1}), …, x_{n}(ξ_{n})) = (p_{γ_{1}}^{−1}(A_{1}), …, p_{γ_{n}}^{−1}(A_{n})).

Proof. Fix j ∈ {1, …, n}. If x_{j}(t) > p_{γ_{j}}^{−1}(A_{j}) (resp. x_{j}(t) < p_{γ_{j}}^{−1}(A_{j})) on J, then γ_{j}(x_{j}) > γ_{j}(p_{γ_{j}}^{−1}(A_{j})) = A_{j} (resp. γ_{j}(x_{j}) < γ_{j}(p_{γ_{j}}^{−1}(A_{j})) = A_{j}), contrary to γ_{j}(A_{j}) = A_{j}. Hence there exists ξ_{j} ∈ ℝ such that x_{j}(ξ_{j}) = p_{γ_{j}}^{−1}(A_{j}). □

Define the operators

Π : C^{0}(J; ℝ^{n}) × ℝ^{n} → C^{1}(J; ℝ^{n}), P : C^{0}(J; ℝ^{n}) × ℝ^{n} → C^{0}(J; ℝ^{n}),

Q : C^{0}(J; ℝ^{n}) × ℝ^{n} → L_{1}(J; ℝ^{n})

by the formulas

Π(x, a)(t) = ∫_{0}^{t} x(s) ds + a,  (5)
P (x, a)(t) = L(Π(x, a))(t)  (6)

and

Q(x, a)(t) = F (Π(x, a))(t).  (7)

Here L and F are the operators in (2).

Consider BVP

x(t) = a + λ( − P (x, b)(t) + ∫_{0}^{t} Q(x, b)(s) ds),  (8)_{(λ,a,b)}

φ(∫_{0}^{t} x(s) ds + b) = A,  (9)_{b}
ψ(x) = B  (10)

1We identify the set of all constant scalar functions on J with ℝ.

depending on the parameters λ, a, b, (λ, a, b) ∈ [0, 1] × ℝ^{n} × ℝ^{n}. Here φ, ψ ∈ A and A, B ∈ ℝ^{n}.

We say that x ∈ C^{0}(J; ℝ^{n}) is a solution of BVP (8)_{(λ,a,b)}, (9)_{b}, (10) for some (λ, a, b) ∈ [0, 1] × ℝ^{n} × ℝ^{n} if (8)_{(λ,a,b)} is satisfied for t ∈ J and x(t) satisfies the boundary conditions (9)_{b}, (10).

Lemma 2. (A priori bounds). Let assumptions (H_{1}) and (H_{2}) be satisfied. Let x(t) be a solution of BVP (8)_{(λ,a,b)}, (9)_{b}, (10) for some (λ, a, b) ∈ [0, 1]×ℝ^{n} ×ℝ^{n}. Then

∥x∥ < S, |a| < (1 − kμ)S, |b| < m_{φA} + ST,

where S is a positive constant such that

\frac{m_{ψB} + 2km_{φA} + 2∥L(0)∥}{u} + \frac{1}{u} ∫_{0}^{T} ω(t, m_{φA} + μu) dt < 1 − 2kμ  (11)

for u ∈ [S, ∞) and m_{φA}, m_{ψB} are given by (4).

Proof. By Lemma 1 (cf. (9)_{b} and (10)), there exist ξ, ν ∈ ℝ^{n} such that

∫_{0}^{ξ_{i}} x_{i}(s) ds + b_{i} = p_{φ_{i}}^{−1}(A_{i}), x_{i}(ν_{i}) = p_{ψ_{i}}^{−1}(B_{i}), i = 1, …, n.  (12)

Then (cf. (8)_{(λ,a,b)})

p_{ψ_{i}}^{−1}(B_{i}) = a_{i} + λ(−P_{i}(x, b)(ν_{i}) + ∫_{0}^{ν_{i}} Q_{i}(x, b)(s) ds),  (13)

and consequently (for i = 1, …, n)

x_{i}(t) = p_{ψ_{i}}^{−1}(B_{i}) + λ(P_{i}(x, b)(ν_{i}) − P_{i}(x, b)(t) + ∫_{ν_{i}}^{t} Q_{i}(x, b)(s) ds).

Hence (cf. (4), (H_{1}), (H_{2}) and Remark 1)

|x_{i}(t)| ≤ m_{ψB} + 2k∥Π(x, b)∥_{1} + 2∥L(0)∥ + ∫_{0}^{T} ω(t, ∥Π(x, b)∥_{1}) dt (14)

for t ∈ J and i = 1, …, n. Since (cf. (5) and (12))

∥Π(x, b)∥ = ∥(∫_{0}^{t} x_{1}(s) ds + b_{1}, …, ∫_{0}^{t} x_{n}(s) ds + b_{n})∥
= ∥(∫_{ξ_{1}}^{t} x_{1}(s) ds + p_{φ_{1}}^{−1}(A_{1}), …, ∫_{ξ_{n}}^{t} x_{n}(s) ds + p_{φ_{n}}^{−1}(A_{n}))∥  (15)
= max{∥ ∫_{ξ_{i}}^{t} x_{i}(s) ds + p_{φ_{i}}^{−1}(A_{i})∥_{0} : i = 1, …, n} ≤ m_{φA} + T ∥x∥,

we have

∥Π(x, b)∥_{1} ≤ max{m_{φA} + T ∥x∥, ∥x∥} ≤ m_{φA} + μ∥x∥.  (16)

Then (cf. (14)-(16))

∥x∥ ≤ m_{ψB} + 2k(m_{φA} + μ∥x∥) + 2∥L(0)∥ + ∫_{0}^{T} ω(t, m_{φA} + μ∥x∥) dt. (17)

Set

q(u) = \frac{m_{ψB} + 2km_{φA} + 2∥L(0)∥}{u} + \frac{1}{u} ∫_{0}^{T} ω(t, m_{φA} + μu) dt

for u ∈ (0, ∞). Then lim_{u→∞} q(u) = 0. Whence there exists S > 0 such that q(u) < 1 − 2kμ for u ≥ S, and so (cf. (17))

∥x∥ < S.

Therefore (cf. (12), (13) and (15))

|b_{i}| = |p_{φ_{i}}^{−1}(A_{i}) − ∫_{0}^{ξ_{i}} x_{i}(s) ds| < m_{φA} + ST,

|a_{i}| = |p_{ψ_{i}}^{−1}(B_{i}) + λ(P_{i}(x, b)(ν_{i}) − ∫_{0}^{ν_{i}} Q_{i}(x, b)(s) ds)|
≤ m_{ψB} + k∥Π(x, b)∥_{1} + ∥L(0)∥ + ∫_{0}^{T} ω(t, ∥Π(x, b)∥_{1}) dt
≤ m_{ψB} + k(m_{φA} + μS) + ∥L(0)∥ + ∫_{0}^{T} ω(t, m_{φA} + μS) dt
< kμS + (1 − 2kμ)S = (1 − kμ)S

for i = 1, …, n, and consequently

|a| < (1 − kμ)S, |b| < m_{φA} + ST.  □

Lemma 3. Let assumption (H_{2}) be satisfied, φ, ψ ∈ A, A, B ∈ ℝ^{n} and S > 0 be a constant such that (11) is satisfied for u ≥ S. Set

\begin{matrix}Ω = \{(x, a, b) : (x, a, b) ∈ C^{0}(J; ℝ^{n}) × ℝ^{n} × ℝ^{n}, \\ ∥x∥ < S, |a| < S, |b| < m_{φA} + ST \} \end{matrix}  (18)

and let Γ : Ω̄ → C^{0}(J; ℝ^{n}) × ℝ^{n} × ℝ^{n} be given by

Γ(x, a, b) = (a, a + φ(∫_{0}^{t} x(s) ds + b) − A, b + ψ(x) − B).  (19)

Then

D(I − Γ, Ω, 0) ≠ 0,  (20)

where “D” denotes the Leray-Schauder degree and I is the identity operator on C^{0}(J; ℝ^{n}) × ℝ^{n} × ℝ^{n}.

Proof. Let U : [0, 1] × Ω̄ → C^{0}(J; ℝ^{n}) × ℝ^{n} × ℝ^{n},

U(λ, x, a, b) = (a, a + φ(∫_{0}^{t} x(s) ds + b) − (1 − λ)φ(− ∫_{0}^{t} x(s) ds − b) − λA,

b + ψ(x) − (1 − λ)ψ(−x) − λB).

By the theory of homotopy and the Borsuk antipodal theorem, to prove (20) it is sufficient to show that

(j) U(0, ·) is an odd operator,

(jj) U is a compact operator, and

(jjj) U(λ, x, a, b) ≠ (x, a, b) for (λ, x, a, b) ∈ [0, 1] × ∂Ω.

Since

U(0, −x, −a, −b) = (−a, −a + φ(− ∫_{0}^{t} x(s) ds − b) − φ(∫_{0}^{t} x(s) ds + b),

−b + ψ(−x) − ψ(x)) = −U(0, x, a, b)

for (x, a, b) ∈ Ω̄, U is an odd operator.

The compactness of U follows from the properties of φ, ψ and applying the Bolzano-Weierstrass theorem.

Assume that U(λ_{0}, x_{0}, a_{0}, b_{0}) = (x_{0}, a_{0}, b_{0}) for some (λ_{0}, x_{0}, a_{0}, b_{0}) ∈ [0, 1]×∂Ω, a_{0} = (a_{01}, …, a_{0n}), b_{0} = (b_{01}, …, b_{0n}). Then

x_{0}(t) = a_{0}, t ∈ J,  (21)
φ(a_{0}t + b_{0}) = (1 − λ_{0})φ(−a_{0}t − b_{0}) + λ_{0}A,  (22)
ψ(a_{0}) = (1 − λ_{0})ψ(−a_{0}) + λ_{0}B,  (23)

and consequently (cf. (22) and (23))

φ_{i}(a_{0i}t + b_{0i}) = (1 − λ_{0})φ_{i}(−a_{0i}t − b_{0i}) + λ_{0}A_{i},  (24)
ψ_{i}(a_{0i}) = (1 − λ_{0})ψ_{i}(−a_{0i}) + λ_{0}B_{i}  (25)

for i = 1, …, n. Fix i ∈ {1, …, n}. If a_{0i} > 0 then ψ_{i}(−a_{0i}) < ψ_{i}(a_{0i}), and so (cf. (25)) ψ_{i}(a_{0i}) ≤ (1 − λ_{0})ψ_{i}(a_{0i}) + λ_{0}B_{i}. Therefore

λ_{0}ψ_{i}(a_{0i}) ≤ λ_{0}B_{i}.  (26)

For λ_{0} = 0 we obtain (cf. (25)) ψ_{i}(a_{0i}) = ψ_{i}(−a_{0i}), a contradiction. Let λ_{0} ∈ (0, 1]. Then (cf. (26)) ψ_{i}(a_{0i}) ≤ B_{i} and

0 < a_{0i} ≤ p_{ψ_{i}}^{−1}(B_{i}) ≤ m_{ψB}.  (27)

If a_{0i} < 0 then ψ_{i}(a_{0i}) < ψ_{i}(−a_{0i}) and (cf. (25)) ψ_{i}(a_{0i}) ≥ (1 − λ_{0})ψ_{i}(a_{0i}) + λ_{0}B_{i}. Hence

λ_{0}ψ_{i}(a_{0i}) ≥ λ_{0}B_{i}.  (28)

For λ_{0} = 0 we obtain (cf. (25)) ψ_{i}(a_{0i}) = ψ_{i}(−a_{0i}), which is impossible. Let λ_{0} ∈ (0, 1]. Then (cf. (28))

0 > a_{0i} ≥ p_{ψ_{i}}^{−1}(B_{i}) ≥ −m_{ψB}.  (29)

From (27) and (29) we deduce

|a_{0i}| ≤ m_{ψB}.  (30)

Assume that a_{0i}t + b_{0i} > 0 for t ∈ J. Then φ_{i}(−a_{0i}t − b_{0i}) < φ_{i}(a_{0i}t + b_{0i}), and so (cf. (24)) λ_{0} ≠ 0 and φ_{i}(a_{0i}t + b_{0i}) ≤ (1 − λ_{0})φ_{i}(a_{0i}t + b_{0i}) + λ_{0}A_{i}. Hence

φ_{i}(a_{0i}t + b_{0i}) ≤ A_{i}.

If a_{0i}t + b_{0i} > p_{φ_{i}}^{−1}(A_{i}) for t ∈ J then A_{i} ≥ φ_{i}(a_{0i}t + b_{0i}) > φ_{i}(p_{φ_{i}}^{−1}(A_{i})) = A_{i}, a contradiction. Thus there is ξ_{i} ∈ J such that

0 < a_{0i}ξ_{i} + b_{0i} ≤ p_{φ_{i}}^{−1}(A_{i}) ≤ m_{φA}.  (31)

Let a_{0i}t + b_{0i} < 0 for t ∈ J. Then φ_{i}(a_{0i}t + b_{0i}) < φ_{i}(−a_{0i}t − b_{0i}) and (24) implies that λ_{0} ≠ 0 and φ_{i}(−a_{0i}t − b_{0i}) ≤ A_{i}. If −a_{0i}t − b_{0i} > p_{φ_{i}}^{−1}(A_{i}) for t ∈ J then A_{i} ≥ φ_{i}(−a_{0i}t−b_{0i}) > φ_{i}(p_{φ_{i}}^{−1}(A_{i})) = A_{i}, a contradiction. Hence there exists ν_{i} ∈ J such that

0 < −a_{0i}ν_{i} − b_{0i} ≤ p_{φ_{i}}^{−1}(A_{i}) ≤ m_{φA}.  (32)

We have proved that there exists τ_{i} ∈ J such that (cf. (31) and (32))

|a_{0i}τ_{i} + b_{0i}| ≤ m_{φA},

and consequently (cf. (30))

|b_{0i}| ≤ |a_{0i}τ_{i} + b_{0i}| + |a_{0i}τ_{i}| ≤ m_{φA} + T m_{ψB}.  (33)

Since (cf. (11)) m_{ψB} < (1 − kμ)S ≤ S, it follows that (cf. (21), (30) and (33))

∥x_{0}∥ < S, |a| < S, |b| < m_{φA} + ST,

contrary to (x_{0}, a_{0}, b_{0}) ∈ ∂Ω. □

3 Existence results, examples

The main result of this paper is given in the following theorem.

Theorem 1. Let assumptions (H_{1}) and (H_{2}) be satisfied. Then for each φ, ψ ∈ A and A, B ∈ ℝ^{n}, BVP (2), (3) has a solution.

Proof. Fix φ, ψ ∈ A and A, B ∈ ℝ^{n}. Let S be a positive constant such that (11) is satisfied for u ≥ S and Ω ⊂ C^{0}(J; ℝ^{n}) × ℝ^{n} × ℝ^{n} be defined by (18). Let U, V : Ω̄ → C^{0}(J; ℝ^{n}) × ℝ^{n} × ℝ^{n},

U(x, a, b) = (a + ∫_{0}^{t} Q(x, b)(s) ds, a + φ(∫_{0}^{t} x(s) ds + b) − A, b + ψ(x) − B),

V (x, a, b) = (−P (x, b)(t), 0, 0)

and let W, Z : [0, 1] × Ω̄ → C^{0}(J; ℝ^{n}) × ℝ^{n} × ℝ^{n},

W (λ, x, a, b) = (a + λ ∫_{0}^{t} Q(x, b)(s) ds, a + φ(∫_{0}^{t} x(s) ds + b) − A, b + ψ(x) − B),

Z(λ, x, a, b) = λV (x, a, b).

Then W (0, ·)+Z(0, ·) = Γ(·) and W (1, ·)+Z(1, ·) = U(·)+V (·), where Γ is defined by (19). By Lemma 3, D(I − W (0, ·) − Z(0, ·), Ω, 0) ≠ 0, and consequently, by the theory of homotopy (see e.g. [1]), to show that

D(I − U − V, Ω, 0) ≠ 0  (34)

it suffices to prove:

(i) W is a compact operator,

(ii) there exists m ∈ [0, 1) such that

∥Z(λ, x, a, b) − Z(λ, y, c, d)∥_{∗} ≤ m∥(x, a, b) − (y, c, d)∥_{∗}

for λ ∈ [0, 1] and (x, a, b), (y, c, d) ∈ Ω̄,

(iii) W (λ, x, a, b) + Z(λ, x, a, b) ≠ (x, a, b) for (λ, x, a, b) ∈ [0, 1] × ∂Ω.

The continuity of W follows from that of Q, φ and ψ. We claim that W ([0, 1]× Ω̄) is a relatively compact subset of the Banach space C^{0}(J; ℝ^{n}) × ℝ^{n} × ℝ^{n}.

Indeed, let {(λ_{j}, x_{j}, a_{j}, b_{j})} ⊂ [0, 1] × Ω̄, x_{j} = (x_{j1}, …, x_{jn}), a_{j} = (a_{j1}, …, a_{jn}), b_{j} = (b_{j1}, …, b_{jn}) (j ∈ ℕ). Then (cf. (7), (H_{2}) and (18))

|a_{ji} + λ ∫_{0}^{t} Q_{i}(x_{j}, b_{j})(s) ds| ≤ |a_{ji}| + ∫_{0}^{T} |Q_{i}(x_{j}, b_{j})(s)| ds
< S + ∫_{0}^{T} ω(t, ∥Π(x_{j}, b_{j})∥_{1}) dt ≤ S + ∫_{0}^{T} ω(t, μ∥x_{j}∥ + |b_{j}|) dt
≤ S + ∫_{0}^{T} ω(t, m_{φA} + S(μ + T)) dt,

| ∫_{t_{1}}^{t_{2}} Q_{i}(x_{j}, b_{j})(s) ds| ≤ | ∫_{t_{1}}^{t_{2}} ω(t, m_{φA} + S(μ + T)) dt|,

|a_{ji} + φ_{i}(∫_{0}^{t} x_{ji}(s) ds + b_{ji}) − A_{i}|
< S + max{|p_{φ_{i}}(−m_{φA} − 2ST)|, |p_{φ_{i}}(m_{φA} + 2ST)|} + |A|

and

|b_{ji} + ψ_{i}(x_{ji}) − B_{i}| < m_{φA} + ST + max{|p_{ψ_{i}}(−S)|, |p_{ψ_{i}}(S)|} + |B|

for t, t_{1}, t_{2} ∈ J, i = 1, …, n and j ∈ ℕ. Therefore there exists a convergent subsequence of {W (λ_{j}, x_{j}, a_{j}, b_{j})} by the Arzelà-Ascoli theorem and the Bolzano-Weierstrass theorem. Hence W is a compact operator.

Let (λ, x, a, b), (λ, y, c, d) ∈ [0, 1] × Ω̄. Then (cf. (H_{1}) and (6))

∥Z(λ, x, a, b) − Z(λ, y, c, d)∥_{∗} ≤ ∥P (x, b) − P (y, d)∥ = ∥L(Π(x, b)) − L(Π(y, d))∥

≤ k∥Π(x, b) − Π(y, d)∥_{1} = k max{∥Π(x, b) − Π(y, d)∥, ∥x − y∥}
≤ k max{∥x − y∥T + |b − d|, ∥x − y∥}
≤ kμ(∥x − y∥ + |b − d|) ≤ kμ∥(x, a, b) − (y, c, d)∥_{∗}.

Hence (ii) holds with m = kμ < \frac{1}{2}.

Suppose (iii) was false. Then we could find (λ_{0}, x_{0}, a_{0}, b_{0}) ∈ [0, 1] × ∂Ω such that

W (λ_{0}, x_{0}, a_{0}, b_{0}) + Z(λ_{0}, x_{0}, a_{0}, b_{0}) = (x_{0}, a_{0}, b_{0}).

Then

x_{0}(t) = a_{0} + λ_{0}(−P (x_{0}, b_{0})(t) + ∫_{0}^{t} Q(x_{0}, b_{0})(s) ds) for t ∈ J,

φ(∫_{0}^{t} x_{0}(s) ds + b_{0}) = A, ψ(x_{0}) = B,

and consequently x_{0}(t) is a solution of BVP (8)_{(λ_{0},a_{0},b_{0})}, (9)_{b_{0}}, (10). By Lemma 2, ∥x_{0}∥ < S, |a_{0}| < (1−kμ)S ≤ S and |b_{0}| < m_{φA}+ST, contrary to (x_{0}, a_{0}, b_{0}) ∈ ∂Ω.

We have proved (34). Therefore there exists a fixed point of the operator U + V, say (u, a, b). It follows that

u(t) = a − P (u, b)(t) + ∫_{0}^{t} Q(u, b)(s) ds for t ∈ J,  (35)

φ(∫_{0}^{t} u(s) ds + b) = A, ψ(u) = B.  (36)

Set x(t) = ∫_{0}^{t} u(s) ds + b, t ∈ J. Then (cf. (5)-(7), (35) and (36))

x′(t) = a − L(x)(t) + ∫_{0}^{t} F (x)(s) ds for t ∈ J,
φ(x) = A, ψ(x′) = B,

and we see that x(t) is a solution of BVP (2), (3). □

Example 3. Let w_{ji} ∈ C^{0}(J; ℝ), α_{i}, β_{i}, γ_{i}, δ_{i} ∈ C^{0}(J; J) for j = 1, 2, …, 9 and i = 1, 2. Define L_{i} : C^{1}(J; ℝ^{2}) → C^{0}(J; ℝ) (i = 1, 2) by

L_{i}(x)(t) = w_{1i}(t)x_{1}(t) + w_{2i}(t)x_{2}(t) + w_{3i}(t)x_{1}(α_{i}(t)) + w_{4i}(t)x_{2}(β_{i}(t))
+ w_{5i}(t)x′_{1}(t) + w_{6i}(t)x′_{2}(t) + w_{7i}(t)x′_{1}(γ_{i}(t)) + w_{8i}(t)x′_{2}(δ_{i}(t)) + w_{9i}(t).

Let F_{i} : C^{1}(J; ℝ^{2}) → L_{1}(J; ℝ) (i = 1, 2) be continuous operators such that

|F_{i}(x)(t)| ≤ ω̃(t, ∥x∥_{1})

for a.e. t ∈ J and each x ∈ C^{1}(J; ℝ^{2}), where ω̃ ∈ K(J × [0, ∞); [0, ∞)). Consider BVP

\begin{matrix}(x′_{1}(t) + L_{1}(x)(t))′ = F_{1}(x)(t), \\ (x′_{2}(t) + L_{2}(x)(t))′ = F_{2}(x)(t), \end{matrix}  (37)
φ_{1}(x_{1}) = A_{1}, φ_{2}(x_{2}) = A_{2}, ψ_{1}(x′_{1}) = B_{1}, ψ_{2}(x′_{2}) = B_{2}.  (38)

By Theorem 1, for each φ_{i}, ψ_{i} ∈ A_{0} and A_{i}, B_{i} ∈ ℝ (i = 1, 2), BVP (37), (38) has

a solution provided ∑_{j=1}^{8} ∥w_{ji}∥_{0} < \frac{1}{2μ} for i = 1, 2.

Next Example 4 shows that for T ≤ 1 the condition k ∈ [0, \frac{1}{2}) in (H_{1}) is optimal and can not be replaced by k ∈ [0, \frac{1}{2}]. In the case of T > 1 we will show (see Example 5) that for each k > \frac{1}{2T} in (H_{1}) there exists an unsolvable BVP of the type (2), (3) satisfying (H_{2}).

Example 4. Let T ≤ 1. Consider BVP

\begin{matrix}(x′_{1}(t) + α(t)(x′_{1}(T) + x′_{2}(T)))′ = 1, \\ (x′_{2}(t) + α(t)(x′_{1}(T) + x′_{2}(T)))′ = 1, \end{matrix}  (39)

\begin{matrix}φ_{1}(x_{1}) = A_{1}, min\{x′_{1}(t) : t ∈ J\} = 0, \\ φ_{2}(x_{2}) = A_{2}, min\{x′_{2}(t) : t ∈ J\} = 0, \end{matrix}  (40)

where α ∈ C^{0}(J; ℝ), ∥α∥_{0} = \frac{1}{4}, α(0) = \frac{1}{4}, α(T) = −\frac{1}{4}, φ_{1}, φ_{2} ∈ A_{0} and A_{1}, A_{2} ∈ ℝ.

Let L_{i} : C^{1}(J; ℝ^{2}) → C^{0}(J; ℝ), L_{i}(x)(t) = α(t)(x′_{1}(T) + x′_{2}(T)) (i = 1, 2). Then

∥L_{i}(x) − L_{i}(y)∥_{0} ≤ ∥α∥_{0}(|x′_{1}(T) − y′_{1}(T)| + |x′_{2}(T) − y′_{2}(T)|)

≤ \frac{1}{4}(∥x′_{1} − y′_{1}∥_{0} + ∥x′_{2} − y′_{2}∥_{0}) ≤ \frac{1}{2}∥x′ − y′∥ ≤ \frac{1}{2}∥x − y∥_{1},

and so ∥L(x) − L(y)∥ ≤ \frac{1}{2}∥x − y∥_{1} for x, y ∈ C^{1}(J; ℝ^{2}) where L = (L_{1}, L_{2}). BVP (39), (40) satisfies (H_{2}) with ω(t, ϱ) = 1 but in (H_{1}) we have k = \frac{1}{2} (= \frac{1}{2μ}).

Assume that u(t) = (u_{1}(t), u_{2}(t)) is a solution of BVP (39), (40). Then u′_{1} = u′_{2}. Indeed, since (u′_{1}(t)−u′_{2}(t))′ = 0 for t ∈ J there exists c ∈ ℝ such that u′_{1}(t) = u′_{2}(t) + c on J. From min{u′_{1}(t) : t ∈ J} = min{u′_{2}(t) : t ∈ J} = 0 we deduce that u′_{1}(ν) = 0, u′_{2}(τ) = 0 for some ν, τ ∈ J, and so 0 = u′_{1}(ν) = u′_{2}(ν) + c ≥ c. If c < 0 then 0 ≤ u′_{1}(τ) = c, a contradiction. Hence c = 0 and then

(u′_{1}(t) + 2α(t)u′_{1}(T))′ = 1 for t ∈ J.

Using the equality u′_{1}(ν) = 0 we have

u′_{1}(t) = 2(α(ν) − α(t))u′_{1}(T) + t − ν for t ∈ J.  (41)

If ν = 0 then (cf. (41) with t = T) u′_{1}(T) = u′_{1}(T) + T, which is impossible. Assume ν ∈ (0, T]. Then (cf. (41) with t = 0)

u′_{1}(0) = 2(α(ν) − \frac{1}{4})u′_{1}(T) − ν ≤ −ν

contrary to u′_{1}(t) ≥ 0 for t ∈ J. It follows that BVP (39), (40) is unsolvable.

Example 5. Let T > 1 and ε > 1. Consider BVP

\begin{matrix}(x′_{1}(t) + α(t)(x_{1}(T) + x_{2}(T)))′ = 1, \\ (x′_{2}(t) + α(t)(x_{1}(T) + x_{2}(T)))′ = 1, \end{matrix}  (42)

min{x_{i}(t) : t ∈ J} = 0, min{x′_{i}(t) : t ∈ J} = 0, i = 1, 2,  (43)

where α ∈ C^{0}(J; ℝ), ∥α∥_{0} = \frac{ε}{4T}, ∫_{0}^{T} α(s) ds = −\frac{1}{4}, α(0) = \frac{1}{4T}, α(T) = −\frac{ε}{4T} and α(t) ≤ \frac{1}{4T} for t ∈ J.

Let L_{i} : C^{1}(J; ℝ^{2}) → C^{0}(J; ℝ), (L_{i}x)(t) = α(t)(x_{1}(T) + x_{2}(T)) (i = 1, 2). Then

∥L_{i}(x) − L_{i}(y)∥_{0} ≤ ∥α∥_{0}(|x_{1}(T) − y_{1}(T)| + |x_{2}(T) − y_{2}(T)|)

≤ \frac{ε}{4T} (∥x_{1} − y_{1}∥_{0} + ∥x_{2} − y_{2}∥_{0}) ≤ \frac{ε}{2T} ∥x − y∥ ≤ \frac{ε}{2T} ∥x − y∥_{1},

and so ∥Lx − Ly∥ ≤ \frac{ε}{2T} ∥x − y∥_{1} for x, y ∈ C^{1}(J; ℝ^{2}) where L = (L_{1}, L_{2}). Hence BVP (42), (43) satisfies (H_{2}) with ω(t, ϱ) = 1 but in (H_{1}) we have k = \frac{ε}{2T} (> \frac{1}{2μ}).

Assume that u(t) = (u_{1}(t), u_{2}(t)) is a solution of BVP (42), (43). Applying the same procedure as in Example 4, it is obvious that u_{1} = u_{2}. Hence

(u′_{1}(t) + 2α(t)u_{1}(T))′ = 1 for t ∈ J,

and since min{u_{1}(t) : t ∈ J} = 0 and min{u′_{1}(t) : t ∈ J} = 0 we have u_{1}(t) ≥ 0, u′_{1}(t) ≥ 0 on J and u′_{1}(ν) = 0 for some ν ∈ J. Therefore

u′_{1}(t) = 2(α(ν) − α(t))u_{1}(T) + t − ν for t ∈ J.  (44)

Assume ν = 0. Then

u′_{1}(t) = 2(\frac{1}{4T} − α(t))u_{1}(T) + t ≥ t,

and so u_{1}(t) is increasing on J and min{u_{1}(t) : t ∈ J} = 0 implies u_{1}(0) = 0. Hence

u_{1}(t) = 2(\frac{t}{4T} − ∫_{0}^{t} α(s) ds)u_{1}(T) + \frac{t^{2}}{2} for t ∈ J

and

u_{1}(T) = 2(\frac{1}{4} − ∫_{0}^{T} α(s) ds)u_{1}(T) + \frac{T^{2}}{2} = u_{1}(T) + \frac{T^{2}}{2} ,

which is impossible.

Let ν ∈ (0, T]. Then (cf. (44))

u′_{1}(0) = 2(α(ν) − \frac{1}{4T} )u_{1}(T) − ν ≤ −ν,

contrary to min{u′_{1}(t) : t ∈ J} = 0. We have proved that BVP (42), (43) is unsolvable.

The following example demonstrates that the condition lim_{ϱ→∞} ∫_{0}^{T} ω(t, ϱ) dt = 0

in (H_{2}) can not be replaced by lim sup_{ϱ→∞} ∫_{0}^{T} ω(t, ϱ) dt < ∞.

Example 6. Consider BVP

x′′_{1}(t) = 1 + \frac{2}{T^{2}} ∥x∥_{1}, x′′_{2}(t) = 1 + \sqrt{∥x∥},  (45)
min{x_{1}(t) : t ∈ J} = 0, φ_{1}(x_{2}) = A, min{x′_{1}(t) : t ∈ J} = 0, φ_{2}(x′_{2}) = B, (46)

where φ_{1}, φ_{2} ∈ A_{0} and A, B ∈ ℝ. Assume that BVP (45), (46) is solvable and let u(t) = (u_{1}(t), u_{2}(t)) be its solution. Then u′′_{1}(t) ≥ 1 on J and the equality min{u′_{1}(t) : t ∈ J} = 0 implies u′_{1}(0) = 0. Hence

u′_{1}(t) = (1 + \frac{2}{T^{2}} ∥u∥_{1})t for t ∈ J,  (47)

and consequently u_{1}(t) is increasing on J. From min{u_{1}(t) : t ∈ J} = 0 we deduce that u_{1}(0) = 0 and then (cf. (47))

u_{1}(t) = \frac{1}{2}(1 + \frac{2}{T^{2}} ∥u∥_{1})t^{2} for t ∈ J.

Therefore

∥u_{1}∥_{0} = \frac{T^{2}}{2} + ∥u∥_{1} ≥ \frac{T^{2}}{2} + ∥u_{1}∥_{0}.

which is impossible. Hence BVP (45), (46) is unsolvable.

We note that for (45) the inequality |F (x)(t)| ≤ ω(t, ∥x∥_{1}) in (H_{2}) is optimal

with respect to the function ω for ω(t, ϱ) = 1 + max \{\frac{2}{T^{2}} ϱ, \sqrt{ϱ}\} and we see that

lim_{ϱ→∞} \frac{1}{ϱ} ∫_{0}^{T} ω(t, ϱ) dt = \frac{2}{T}.

Reference

[1] Deimling K., Nonlinear Functional Analysis. Springer Berlin Heidelberg, 1985.

[2] Staněk S., Multiple solutions for some functional boundary value problems. Nonlin. Anal. 32(1998), 427–438.

[3] Staněk S., Functional boundary value problems for second order functional differential equations of the neutral type. Glasnik Matematički (to apear).