Sixth Mississippi State Conference on Differential Equations and Computational Simulations, Electronic Journal of Differential Equations, Conference 15 (2007), pp. 127–139. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu ftp ejde.math.txstate.edu (login: ftp)

A NON-RESONANT GENERALIZED MULTI-POINT BOUNDARY-VALUE PROBLEM OF DIRICHELET TYPE INVOLVING A P-LAPLACIAN TYPE OPERATOR

CHAITAN P. GUPTA

Abstract. We study the existence of solutions for the generalized multi-point boundary-value problem

(ϕ(x′))′ = f(t, x, x′) + e 0 < t < 1,

x(0) = ∑_{i=1}^{m−2} a_{i}x(ξ_{i}), x(1) = ∑_{j=1}^{n−2} b_{j}x(τ_{j}),

in the non-resonance case. Our methods consist in using topological degree and some a priori estimates.

1. Introduction

Let ϕ be an odd increasing homeomorphism from ℝ onto ℝ satisfying ϕ(0) = 0, f : [0, 1] × ℝ × ℝ → ℝ be a function satisfying Carathéodory conditions and e : [0, 1] → ℝ be a function in L^{1}[0, 1]. Let ξ_{i},τ_{j} ∈ (0, 1), a_{i}, b_{j} ∈ ℝ, i = 1, 2, …, m−2, j = 1, 2, …, n − 2, 0 < ξ_{1} < ξ_{2} < ⋯ < ξ_{m−2} < 1, 0 < τ_{1} < τ_{2} < ⋯ < τ_{n−2} < 1 be given. We study the problem of existence of solutions for the generalized multi-point boundary-value problem

\begin{matrix}(ϕ(x′))′ = f(t, x, x′) + e, 0 < t < 1, \\ x(0) = ∑_{i=1}^{m−2} a_{i}x(ξ_{i}), x(1) = ∑_{j=1}^{n−2} b_{j}x(τ_{j}), \end{matrix}  (1.1)

in the non-resonance case. We say that this problem is non-resonant if the associated problem:

\begin{matrix}(ϕ(x′))′ = 0, 0 < t < 1, \\ x(0) = ∑_{i=1}^{m−2} a_{i}x(ξ_{i}), x(1) = ∑_{j=1}^{n−2} b_{j}x(τ_{j}), \end{matrix}  (1.2)

2000 Mathematics Subject Classification. 34B10, 34B15, 34L30, 34L90.

Key words and phrases. Generalized multi-point boundary value problems; non-resonance;

p-Laplace type operator; a priori estimates; topological degree.

\begin{matrix}c \\ ◯ \end{matrix}2007 Texas State University - San Marcos.

Published February 28, 2007.

has the trivial solution as its only solution. This is the case, (see Proposition 2.1 below), if

( ∑_{i=1}^{m−2} a_{i}ξ_{i})(1 − ∑_{j=1}^{n−2} b_{j}) ≠ (1 − ∑_{i=1}^{m−2} a_{i})( ∑_{j=1}^{n−2} b_{j}τ_{j} − 1).

This problem was studied by Gupta, Ntouyas, and Tsamatos in [20] and by the author in [16] when the homeomorphism ϕ from ℝ onto ℝ is the identity homeomorphism, i.e for second order ordinary differential equations. The study of multi-point boundary value problems for second order ordinary differential equations was initiated by Il’in and Moiseev in [22, 23] motivated by the works of Bitsadze and Samarskii on nonlocal linear elliptic boundary value problems, [2, 3, 4] and has been the subject of many papers, see for example, [5, 6, 11, 12, 13, 14, 15, 17, 18, 19, 21, 24, 27, 28]. More recently multipoint boundary value problems involving a p-Lalacian type operator or the more general operator −(ϕ(x′))′ has been studied in [1, 7, 8, 9, 10, 25] to mention a few.

We present in Section 2 some a priori estimates for functions x(t) that satisfy the boundary conditions in (1.1). Our a priori estimates are sharper versions of the corresponding estimates in [16] and explicitly utilize the non-resonance condition for the boundary value problem (1.1). In section 3, we present an existence theorem for the boundary value problem (1.1) using degree theory.

2. A Priori Estimates

We shall assume throughout that ϕ is an odd increasing homeomorphism from ℝ onto ℝ satisfying ϕ(0) = 0. We shall also assume that the homeomorphism ϕ satisfies the following conditions:

(a) For any constant M > 0,

lim sup_{z→∞} \frac{ϕ(Mz)}{ϕ(z)} ≡ α(M) < ∞.  (2.1)

(b) For any σ, 0 ≤ σ < 1,

α̃(σ) ≡ lim sup_{z→∞} \frac{ϕ(σz)}{ϕ(z)} < 1.  (2.2)

Proposition 2.1. The boundary-value problem (1.2) has only the trivial solution if and only if

( ∑_{i=1}^{m−2} a_{i}ξ_{i})(1 − ∑_{j=1}^{n−2} b_{j}) ≠ (1 − ∑_{i=1}^{m−2} a_{i})( ∑_{j=1}^{n−2} b_{j}τ_{j} − 1).  (2.3)

Proof. It is obvious that x(t) = At + B, t ∈ [0, 1], A, B ∈ ℝ, is a general solution for the differential equation

(ϕ(x′))′ = 0, 0 < t < 1,

in (1.2). If, now, x(t) = At + B, t ∈ [0, 1], A, B ∈ ℝ, is a solution to the boundary value problem (1.2) then we must have

B = ∑_{i=1}^{m−2} a_{i}(Aξ_{i} + B), A + B = ∑_{j=1}^{n−2} b_{j}(Aτ_{j} + B).

In other words A, B must satisfy the system of equations

\begin{matrix}A( ∑_{i=1}^{m−2} a_{i}ξ_{i}) + B( ∑_{i=1}^{m−2} a_{i} − 1) = 0, \\ A(∑_{j=1}^{n−2} b_{j}τ_{j} − 1) + B(∑_{j=1}^{n−2} b_{j} − 1) = 0. \end{matrix}  (2.4)

Now, the system of equations (2.4) has A = 0, B = 0 as the only solution if and only if

det (\begin{matrix}∑_{i=1}^{m−2} a_{i}ξ_{i} & ∑_{i=1}^{m−2} a_{i} − 1 \\ ∑_{j=1}^{n−2} b_{j}τ_{j} − 1 & ∑_{j=1}^{n−2} b_{j} − 1 \end{matrix}) ≠ 0,

or

( ∑_{i=1}^{m−2} a_{i}ξ_{i})( ∑_{j=1}^{n−2} b_{j} − 1) − ( ∑_{i=1}^{m−2} a_{i} − 1)( ∑_{j=1}^{n−2} b_{j}τ_{j} − 1) ≠ 0.  (2.5)

It is now obvious that (2.5) is equivalent to (2.3). Hence the boundary value problem (1.2) has only the trivial solution if and only if the condition (2.3) holds. This completes the proof of the Proposition. □

We shall assume in the following that ξ_{i}, τ_{j} ∈ (0, 1), a_{i}, b_{j} ∈ ℝ, i = 1, 2, …, m−2, j = 1, 2, …, n − 2, 0 < ξ_{1} < ξ_{2} < ⋯ < ξ_{m−2} < 1, 0 < τ_{1} < τ_{2} < ⋯ < τ_{n−2} < 1 satisfy the condition (2.3). We observe that when condition (2.3) holds at least one of 1 − ∑_{i=1}^{m−2} a_{i}, 1 − ∑_{j=1}^{n−2} b_{j} is non-zero. Now, for a ∈ R, we set a^{+} = max(a, 0), a^{−} = max(−a, 0) so that a = a^{+} − a^{−} and |a| = a^{+} + a^{−}. Next, in case 1 − ∑_{i=1}^{m−2} a_{i} ≠ 0, we notice that

σ_{1} ≡ min \{\frac{∑_{i=1}^{m−2} a_{i}^{+}}{1 + ∑_{i=1}^{m−2} a_{i}^{−}} , \frac{1 + ∑_{i=1}^{m−2} a_{i}^{−}}{∑_{i=1}^{m−2} a_{i}^{+}} \} ∈ [0, 1)

is well-defined. Similarly, if 1 − ∑_{j=1}^{n−2} b_{j} ≠ 0, we see that

σ_{2} ≡ min\{\frac{∑_{j=1}^{n−2} b_{j}^{+}}{1 + ∑_{j=1}^{n−2} b_{j}^{−}} , \frac{1 + ∑_{j=1}^{n−2} b_{j}^{−}}{∑_{j=1}^{n−2} b_{j}^{+}} \} ∈ [0, 1)

is well-defined. Accordingly, let us define

σ_{1} ≡ \begin{cases}min\{\frac{∑_{i=1}^{m−2} a_{i}^{+}}{1+∑_{i=1}^{m−2} a_{i}^{−}} , \frac{1+∑_{i=1}^{m−2} a_{i}^{−}}{∑_{i=1}^{m−2} a_{i}^{+}} \} ∈ [0, 1) & if 1 − ∑_{i=1}^{m−2} a_{i} ≠ 0, \\ 1 & if 1 − ∑_{i=1}^{m−2} a_{i} = 0; \end{cases}  (2.6)

and

σ_{2} ≡ \begin{cases}min\{\frac{∑_{j=1}^{n−2} b_{j}^{+}}{1+∑_{j=1}^{n−2} b_{j}^{−}} , \frac{1+∑_{j=1}^{n−2} b_{j}^{−}}{∑_{j=1}^{n−2} b_{j}^{+}} \} ∈ [0, 1) & if 1 − ∑_{j=1}^{n−2} b_{j} ≠ 0, \\ 1 & if 1 − ∑_{j=1}^{n−2} b_{j} = 0. \end{cases}  (2.7)

The a priori estimate obtained in the following proposition is a sharpening of the a priori estimate of Lemma 2 of [16]. We repeat the details given in Lemma 2 of [16] for the sake of completeness.

Proposition 2.2. Let ξ_{i},τ_{j} ∈ (0, 1), a_{i}, b_{j} ∈ ℝ, i = 1, 2, …, m−2, j = 1, 2, …, n− 2, 0 < ξ_{1} < ξ_{2} < ⋯ < ξ_{m−2} < 1, 0 < τ_{1} < τ_{2} < ⋯ < τ_{n−2} < 1, with (∑_{i=1}^{m−2} a_{i}ξ_{i})(1 − ∑_{j=1}^{n−2} b_{j}) ≠ (1 − ∑_{i=1}^{m−2} a_{i})(∑_{j=1}^{n−2} b_{j}τ_{j} − 1) be given. Also let the function x(t) be such that x(t), x′(t) be absolutely continuous on [0, 1] and x(0) = ∑_{i=1}^{m−2} a_{i}x(ξ_{i}), x(1) = ∑_{j=1}^{n−2} b_{j}x(τ_{j}). Then

∥x∥_{∞} ≤ M∥x′∥_{∞},  (2.8)

where

M = min \{\frac{1}{| ∑_{i=1}^{m−2} a_{i}|}( ∑_{i=1}^{m−2} |a_{i}|λ_{i} + \frac{∑_{i=1}^{m−2} |a_{i}ξ_{i}|}{|1 − ∑_{i=1}^{m−2} a_{i}|}),

\frac{1}{| ∑_{j=1}^{n−2} b_{j}|}(∑_{j=1}^{n−2} |b_{j}|μ_{j} + \frac{∑_{j=1}^{n−2} |b_{j}(1 − τ_{j})|}{|1 − ∑_{j=1}^{n−2} b_{j}|}), 1 + \frac{∑_{i=1}^{m−2} |a_{i}ξ_{i}|}{|1 − ∑_{i=1}^{m−2} a_{i}|},

1 + \frac{∑_{j=1}^{n−2} |b_{j}(1 − τ_{j})|}{|1 − ∑_{j=1}^{n−2} b_{j}|}, \frac{1}{1 − σ_{1}} , \frac{1}{1 − σ_{2}} \}

with λ_{i} = max(ξ_{i}, 1 − ξ_{i}) for i = 1, 2 …, m − 2, μ_{j} = max(τ_{j}, 1 − τ_{j}) for j = 1, 2, …, n − 2, σ_{1} as defined in (2.6) and σ_{2} as defined in (2.7).

Proof. We first observe that at least one of (1−∑_{i=1}^{m−2} a_{i}), (1−∑_{j=1}^{n−2} b_{j}) is non-zero, in view of our assumption

( ∑_{i=1}^{m−2} a_{i}ξ_{i})(1 − ∑_{j=1}^{n−2} b_{j}) ≠ (1 − ∑_{i=1}^{m−2} a_{i})( ∑_{j=1}^{n−2} b_{j}τ_{j} − 1).

Accordingly, M < ∞. Next, we see from x(ξ_{i}) − x(0) = ∫_{0}^{ξ_{i}} x′(s)ds for i = 1, 2, …, m − 2 and the assumption that x(0) = ∑_{i=1}^{m−2} a_{i}x(ξ_{i}), that

(1 − ∑_{i=1}^{m−2} a_{i})x(0) = ∑_{i=1}^{m−2} a_{i} ∫_{0}^{ξ_{i}} x′(s)ds.

It then follows that

|x(0)| ≤ \frac{∑_{i=1}^{m−2} |a_{i}ξ_{i}|}{|1 − ∑_{i=1}^{m−2} a_{i}|}∥x′∥_{∞}.  (2.9)

Also, since x(t) = x(ξ_{i}) + ∫_{ξ_{i}}^{t} x′(s)ds, we see that

( ∑_{i=1}^{m−2} a_{i})x(t) = ∑_{i=1}^{m−2} a_{i}x(ξ_{i}) + ∑_{i=1}^{m−2} a_{i} ∫_{ξ_{i}}^{t} x′(s)ds = x(0) + ∑_{i=1}^{m−2} a_{i} ∫_{ξ_{i}}^{t} x′(s)ds.

We, now, use (2.9) to get

| ∑_{i=1}^{m−2} a_{i}||x(t)| ≤ |x(0)| + ∑_{i=1}^{m−2} |a_{i}|| ∫_{ξ_{i}}^{t} x′(s)ds|

≤ (\frac{∑_{i=1}^{m−2} |a_{i}ξ_{i}|}{|1 − ∑_{i=1}^{m−2} a_{i}|} + ∑_{i=1}^{m−2} λ_{i}|a_{i}|)∥x′∥_{∞}.

It is now immediate that

∥x∥_{∞} ≤ \frac{1}{| ∑_{i=1}^{m−2} a_{i}|}(\frac{∑_{i=1}^{m−2} |a_{i}ξ_{i}|}{|1 − ∑_{i=1}^{m−2} a_{i}|} + ∑_{i=1}^{m−2} λ_{i}|a_{i}|)∥x′∥_{∞}.  (2.10)

Similarly, starting from x(1) − x(τ_{j}) = ∫_{τ_{j}}^{1} x′(s)ds and proceeding, as above, we obtain the estimate

∥x∥_{∞} ≤ \frac{1}{| ∑_{j=1}^{n−2} b_{j}|}(\frac{∑_{j=1}^{n−2} |b_{j}(1 − τ_{j})|}{|1 − ∑_{j=1}^{n−2} b_{j}|} + ∑_{j=1}^{n−2} μ_{j}|b_{j}|)∥x′∥_{∞}.  (2.11)

If we next use the equation x(t) = x(0) + ∫_{0}^{t} x′(s)ds and the estimate (2.9) we obtain

∥x∥_{∞} ≤ (\frac{∑_{i=1}^{m−2} |a_{i}ξ_{i}|}{|1 − ∑_{i=1}^{m−2} a_{i}|} + 1)∥x′∥_{∞}.  (2.12)

Similarly, starting from the equation x(t) = x(1) − ∫_{t}^{1} x′(s)ds, we obtain the estimate

∥x∥_{∞} ≤ (\frac{∑_{j=1}^{n−2} |b_{j}(1 − τ_{j})|}{|1 − ∑_{j=1}^{n−2} b_{j}|} + 1)∥x′∥_{∞}.  (2.13)

Next, since x(0) = ∑_{i=1}^{m−2} a_{i}x(ξ_{i}) we see that

x(0) + ∑_{i=1}^{m−2} a_{i}^{−}x(ξ_{i}) = ∑_{i=1}^{m−2} a_{i}^{+}x(ξ_{i}).

It follows that there must exist χ_{1}, χ_{2} in [0, 1] such that

(1 + ∑_{i=1}^{m−2} a_{i}^{−})x(χ_{1}) = ( ∑_{i=1}^{m−2} a_{i}^{+})x(χ_{2}).  (2.14)

If, now, one of x(χ_{1}), x(χ_{2}) is zero, we see using one of the two equations

x(t) = x(χ_{k}) + ∫_{τ_{k}}^{t} x′(s)ds, k = 1, 2; t ∈ [0, 1]  (2.15)

that

∥x∥_{∞} ≤ ∥x′∥_{∞}.  (2.16)

If both x(χ_{1}), x(χ_{2}) are non-zero and 1 − ∑_{i=1}^{m−2} a_{i} ≠ 0, so that 1 + ∑_{i=1}^{m−2} a_{i}^{−} ≠ ∑_{i=1}^{m−2} a_{i}^{+}, it is easy to see from (2.14) that x(χ_{1}) ≠ x(χ_{2}). It then follows easily from (2.14) and (2.15) that

∥x∥_{∞} ≤ \frac{1}{1 − σ_{1}} ∥x′∥_{∞},  (2.17)

where

σ_{1} = min\{\frac{∑_{i=1}^{m−2} a_{i}^{+}}{1 + ∑_{i=1}^{m−2} a_{i}^{−}} , \frac{1 + ∑_{i=1}^{m−2} a_{i}^{−}}{∑_{i=1}^{m−2} a_{i}^{+}} \} ∈ [0, 1).

Similarly, we see from x(1) = ∑_{j=1}^{n−2} b_{j}x(τ_{j}) that either (2.16) holds or

∥x∥_{∞} ≤ \frac{1}{1 − σ_{2}} ∥x′∥_{∞},  (2.18)

where

σ_{2} = min\{\frac{∑_{j=1}^{n−2} b_{j}^{+}}{1 + ∑_{j=1}^{n−2} b_{j}^{−}} , \frac{1 + ∑_{j=1}^{n−2} b_{j}^{−}}{∑_{j=1}^{n−2} b_{j}^{+}} \} ∈ [0, 1).

The proposition is now immediate from (2.10), (2.11), (2.12), (2.13), (2.16), (2.17) and (2.18) and the definitions of σ_{1}, σ_{2} as given in (2.6), (2.7). □

The following lemma is needed in the next proposition.

Lemma 2.3. Let us set

\begin{aligned}A & = [(1 − ∑_{i=1}^{m−2} a_{i})(1 − ∑_{j=1}^{n−2} b_{j})]^{+} + ∑_{j=1}^{n−2}[b_{j}(1 − τ_{j})(1 − ∑_{i=1}^{m−2} a_{i})]^{+} \\ & + ∑_{i=1}^{m−2}[a_{i}ξ_{i}(1 − ∑_{j=1}^{n−2} b_{j})]^{+} \end{aligned}  (2.19)

and

\begin{aligned}B & = [(1 − ∑_{i=1}^{m−2} a_{i})(1 − ∑_{j=1}^{n−2} b_{j})]^{−} + ∑_{j=1}^{n−2}[b_{j}(1 − τ_{j})(1 − ∑_{i=1}^{m−2} a_{i})]^{−} \\ & + ∑_{i=1}^{m−2}[a_{i}ξ_{i}(1 − ∑_{j=1}^{n−2} b_{j})]^{−}. \end{aligned}  (2.20)

Then A ≠ B, when the non-resonance assumption (2.3) holds.

Proof. We note that

A − B

= (1 − ∑_{i=1}^{m−2} a_{i})(1 − ∑_{j=1}^{n−2} b_{j}) + ∑_{j=1}^{n−2} b_{j}(1 − τ_{j})(1 − ∑_{i=1}^{m−2} a_{i}) + ∑_{i=1}^{m−2} a_{i}ξ_{i}(1 − ∑_{j=1}^{n−2} b_{j})

= 1 − ∑_{i=1}^{m−2} a_{i} − (1 − ∑_{i=1}^{m−2} a_{i})(∑_{j=1}^{n−2} b_{j}) + (∑_{j=1}^{n−2} b_{j})(1 − ∑_{i=1}^{m−2} a_{i})

− (1 − ∑_{i=1}^{m−2} a_{i})(∑_{j=1}^{n−2} b_{j}τ_{j}) + ( ∑_{i=1}^{m−2} a_{i}ξ_{i})(1 − ∑_{j=1}^{n−2} b_{j})

= 1 − ∑_{i=1}^{m−2} a_{i} − (1 − ∑_{i=1}^{m−2} a_{i})(∑_{j=1}^{n−2} b_{j}τ_{j}) + ( ∑_{i=1}^{m−2} a_{i}ξ_{i})(1 − ∑_{j=1}^{n−2} b_{j})

= ( ∑_{i=1}^{m−2} a_{i}ξ_{i})(1 − ∑_{j=1}^{n−2} b_{j}) − (1 − ∑_{i=1}^{m−2} a_{i})(∑_{j=1}^{n−2} b_{j}τ_{j} − 1) ≠ 0,

in view of the non-resonance assumption (2.3). Hence A ≠ B. This completes the proof of the lemma. □

Let us define

σ^{∗} = min\{\frac{A}{B} , \frac{B}{A} \} ∈ [0, 1),  (2.21)

where A, B are as defined in Lemma 2.3. Accordingly, we see that

α̃(σ^{∗}) = lim sup_{z→∞} \frac{ϕ(σ^{∗}z)}{ϕ(z)} < 1,

in view of our assumption (2.2). Let ε > 0 be such that α̃(σ^{∗}) + ε < 1 and the constant C_{ε} be such that

ϕ(σ^{∗}z) ≤ (α̃(σ^{∗}) + ε)ϕ(z) + C_{ε}, for every z ∈ ℝ.  (2.22)

Proposition 2.4. Let ξ_{i},τ_{j} ∈ (0, 1), a_{i}, b_{j} ∈ ℝ, i = 1, 2, …, m−2, j = 1, 2, …, n− 2, 0 < ξ_{1} < ξ_{2} < ⋯ < ξ_{m−2} < 1, 0 < τ_{1} < τ_{2} < ⋯ < τ_{n−2} < 1, with (∑_{i=1}^{m−2} a_{i}ξ_{i})(1 − ∑_{j=1}^{n−2} b_{j}) ≠ (1 − ∑_{i=1}^{m−2} a_{i})(∑_{j=1}^{n−2} b_{j}τ_{j} − 1) be given. Also let the function x(t) be such that x(t), x′(t) be absolutely continuous on [0, 1] with (ϕ(x′))′ ∈ L^{1}(0, 1) and x(0) = ∑_{i=1}^{m−2} a_{i}x(ξ_{i}), x(1) = ∑_{j=1}^{n−2} b_{j}x(τ_{j}). Then

∥ϕ(x′)∥_{∞} ≤ \frac{1}{1 − α̃(σ^{∗}) − ε}∥(ϕ(x′))′∥_{L^{1}(0,1)} + \frac{C_{ε}}{1 − α̃(σ^{∗}) − ε},  (2.23)

where ε and C_{ε} are as in (2.22).

Proof. For i = 1, 2, …, m − 2 we see using mean value theorem that there exist χ_{i} in [0, 1] such that

x(ξ_{i}) − x(0) = ξ_{i}x′(χ_{i}).

It then follows using x(0) = ∑_{i=1}^{m−2} a_{i}x(ξ_{i}) that

(1 − ∑_{i=1}^{m−2} a_{i})x(0) = ∑_{i=1}^{m−2} a_{i}ξ_{i}x′(χ_{i}).  (2.24)

Again, for j = 1, 2, …, n − 2 we see using mean value theorem that there exist λ_{j} in [0, 1] such that

x(1) − x(τ_{j}) = (1 − τ_{j})x′(λ_{j}),

and we see using x(1) = ∑_{j=1}^{n−2} b_{j}x(τ_{j}) that

(∑_{j=1}^{n−2} b_{j} − 1)x(1) = ∑_{j=1}^{n−2} b_{j}(1 − τ_{j})x′(λ_{j}).  (2.25)

Also, we see that there exists a λ ∈ [0, 1] such that

x(1) − x(0) = x′(λ).  (2.26)

Now, we see from equations (2.24), (2.25), (2.26) that

(1 − ∑_{i=1}^{m−2} a_{i})(∑_{j=1}^{n−2} b_{j} − 1)x′(λ)

= (1 − ∑_{i=1}^{m−2} a_{i})(∑_{j=1}^{n−2} b_{j} − 1)(x(1) − x(0))

= (1 − ∑_{i=1}^{m−2} a_{i})(∑_{j=1}^{n−2} b_{j}(1 − τ_{j})x′(λ_{j})) − (∑_{j=1}^{n−2} b_{j} − 1)( ∑_{i=1}^{m−2} a_{i}ξ_{i}x′(χ_{i})).

It follows that

(1 − ∑_{i=1}^{m−2} a_{i})(1 − ∑_{j=1}^{n−2} b_{j})x′(λ) + ∑_{j=1}^{n−2} b_{j}(1 − τ_{j})(1 − ∑_{i=1}^{m−2} a_{i})x′(λ_{j})

+ ∑_{i=1}^{m−2} a_{i}ξ_{i}(1 − ∑_{j=1}^{n−2} b_{j})x′(χ_{i}) = 0.

Using, next, the intermediate value theorem we see that there exist υ_{1}, υ_{2} in [0, 1] such that

Ax′(υ_{1}) − Bx′(υ_{2}) = 0,  (2.27)

where A, B are as defined in (2.19), (2.20). Suppose, now, one of x′(υ_{1}), x′(υ_{2}) is zero. We then see from one of the following equations

ϕ(x′(t)) = ϕ(x′(υ_{k})) + ∫_{υ_{k}}^{t}(ϕ(x′))′(s)ds, k = 1, 2; t ∈ [0, 1]  (2.28)

that

∥ϕ(x′)∥_{∞} ≤ ∥(ϕ(x′))′∥_{L^{1}(0,1)}.  (2.29)

Let us, next, suppose that both x′(υ_{1}), x′(υ_{2}) are non-zero. Since, now, A ≠ B, in view of Lemma 2.3 we see from equation (2.27) that x′(υ_{1}) ≠ x′(υ_{2}). We now use the equations

ϕ(x′(t)) = ϕ(x′(υ_{1})) + ∫_{υ_{k}}^{t}(ϕ(x′))′(s)ds = ϕ(\frac{B}{A} x′(υ_{2})) + ∫_{υ_{k}}^{t}(ϕ(x′))′(s)ds,

ϕ(x′(t)) = ϕ(x′(υ_{2})) + ∫_{υ_{k}}^{t}(ϕ(x′))′(s)ds = ϕ(\frac{A}{B} x′(υ_{1})) + ∫_{υ_{k}}^{t}(ϕ(x′))′(s)ds,

along with the definition of σ^{∗}, as given in (2.21), (2.22) and the estimate (2.29) to obtain the estimate (2.23). This completes the proof of the proposition. □

3. Existence Theorem

Let ϕ be an odd increasing homeomorphism from ℝ onto ℝ satisfying ϕ(0) = 0, f : [0, 1] × ℝ × ℝ → ℝ be a function satisfying Carathéodory conditions and e : [0, 1] → ℝ be a function in L^{1}[0, 1]. Let ξ_{i}, τ_{j} ∈ (0, 1), a_{i}, b_{j} ∈ ℝ, i = 1, 2, …, m−2, j = 1, 2, …, n − 2, 0 < ξ_{1} < ξ_{2} < ⋯ < ξ_{m−2} < 1, 0 < τ_{1} < τ_{2} < ⋯ < τ_{n−2} < 1 with (∑_{i=1}^{m−2} a_{i}ξ_{i})(1 − ∑_{j=1}^{n−2} b_{j}) ≠ (1 − ∑_{i=1}^{m−2} a_{i})(∑_{j=1}^{n−2} b_{j}τ_{j} − 1).

Theorem 3.1. Let f : [0, 1] × ℝ × ℝ → ℝ be a function satisfying Carathéodory’s conditions such that there exist non-negative functions d_{1}(t), d_{2}(t), and r(t) in L^{1}(0, 1) such that

|f(t, u, v)| ≤ d_{1}(t)ϕ(|u|) + d_{2}(t)ϕ(|v|) + r(t),

for a. e. t ∈ [0, 1] and all u, v ∈ ℝ. Suppose, further,

α(M)∥d_{1}∥_{L^{1}(0,1)} + ∥d_{2}∥_{L^{1}(0,1)} < 1 − α̃(σ^{∗})  (3.1)

where M is as defined in Proposition 2.2, α(M) is as defined in (2.1), σ^{∗} and α̃(σ^{∗}) are as defined in (2.21), (2.22). Then, for every given function e(t) ∈ L^{1}[0, 1], the boundary value problem (1.1) has at least one solution x(t) ∈ C^{1}[0, 1].

Proof. We consider the family of boundary-value problems

\begin{matrix}(ϕ(x′))′ = λf(t, x, x′) + λe, 0 < t < 1, λ ∈ [0, 1] \\ x(0) = ∑_{i=1}^{m−2} a_{i}x(ξ_{i}), x(1) = ∑_{j=1}^{n−2} b_{j}x(τ_{j}). \end{matrix}  (3.2)

Also, we define an operator Ψ : C^{1}[0, 1] × [0, 1] → C^{1}[0, 1] by setting for (x, λ) ∈ C^{1}[0, 1] × [0, 1]

\begin{aligned}Ψ(x, λ)(t) & = x(0) + ∫_{0}^{t} ϕ^{−1}(ϕ(x′(0)) + λ ∫_{0}^{s}(f(τ, x(τ), x′(τ)) + e(τ))dτ)ds \\ & + (x(0) − ∑_{i=1}^{m−2} a_{i}x(ξ_{i})) + t(x(1) − ∑_{j=1}^{n−2} b_{j}x(τ_{j})) \end{aligned} (3.3)

Let us, suppose that x(t) ∈ C^{1}[0, 1] is a solution to the operator equation, for some λ ∈ [0, 1],

\begin{aligned}x & = Ψ(x, λ) \\ & = x(0) + ∫_{0}^{t} ϕ^{−1}(ϕ(x′(0)) + λ ∫_{0}^{s}(f(τ, x(τ), x′(τ)) + e(τ))dτ)ds \\ & + (x(0) − ∑_{i=1}^{m−2} a_{i}x(ξ_{i})) + t(x(1) − ∑_{j=1}^{n−2} b_{j}x(τ_{j})) \end{aligned}  (3.4)

Evaluating this equation at t = 0 we see that x(t) satisfies the boundary condition

x(0) = ∑_{i=1}^{m−2} a_{i}x(ξ_{i}).

Next, we differentiate the equation (3.4) with respect to t to get

x′(t) = ϕ^{−1}(ϕ(x′(0)) + λ ∫_{0}^{t}(f(τ, x(τ), x′(τ)) + e(τ))dτ) + x(1) − ∑_{j=1}^{n−2} b_{j}x(τ_{j}). (3.5)

Evaluating, now, the equation (3.5) at t = 0 we see that x(t) satisfies the boundary condition

x(1) = ∑_{j=1}^{n−2} b_{j}x(τ_{j}),

and on differentiating the equation (3.5) with respect to t we get

(ϕ(x′))′ = λf(t, x, x′) + λe, 0 < t < 1, λ ∈ [0, 1].

Thus we see that if x(t) ∈ C^{1}[0, 1] is a solution to the operator equation x = Ψ(x, λ) for some λ ∈ [0, 1] then x(t) is a solution to the boundary value problems (3.2) for the corresponding λ ∈ [0, 1]. Conversely, it is easy to see that if x(t) ∈ C^{1}[0, 1] is a solution to the boundary value problems (3.2) for some λ ∈ [0, 1] then x(t) ∈ C^{1}[0, 1] is a solution to the operator equation x = Ψ(x, λ) for the corresponding λ ∈ [0, 1].

Next, it is easy to show, following standard arguments, that Ψ : C^{1}[0, 1]×[0, 1] → C^{1}[0, 1] is a completely continuous operator.

We shall next show that there is a constant R > 0, independent of λ ∈ [0, 1], such that if x(t) ∈ C^{1}[0, 1] is a solution to (3.4), equivalently to the boundary value problems (3.2), for some λ ∈ [0, 1] then ∥x∥_{C^{1}[0,1]} < R.

We note first that if x(t) ∈ C^{1}[0, 1] satisfies

x = Ψ(x, 0),  (3.6)

then x(t) = 0 for all t ∈ [0, 1]. Indeed, from the definition of Ψ or from the boundary value problem (3.2), it follows that x(t) = x(0) + x′(0)t. It then follows from the two boundary conditions in (3.2) and the non-resonance assumption (2.3) that x(0) = x′(0) = 0, implying x(t) = 0 for all t ∈ [0, 1].

We shall assume, in the following, that λ ∈ (0, 1]. We shall also assume that σ^{∗}, as defined in (2.21) is positive, since the proof for the case σ^{∗} = 0 is simpler. Let us choose ε > 0 such that α̃(σ^{∗}) + ε < 1 and

(α(M) + ε)∥d_{1}∥_{L^{1}(0,1)} + ∥d_{2}∥_{L^{1}(0,1)} < 1 − α̃(σ^{∗}) − ε,  (3.7)

which is possible to do, in view of our assumption (3.1). Here M is as defined in Propostion 2.2 and α(M) is as defined in (2.1) so that for the ε > 0, chosen above, there exists a constant C_{ε}^{1} > 0 such that

ϕ(Mz) ≤ (α(M) + ε)ϕ(z) + C_{ε}^{1}, for every z ∈ ℝ.  (3.8)

Also, from Proposition 2.4 we see that there is a constant C_{ε}^{2} > 0, for the chosen ε > 0, such that

ϕ(∥x′∥_{∞}) ≤ \frac{1}{1 − α̃(σ^{∗}) − ε}∥(ϕ(x′))′∥_{L^{1}(0,1)} + C_{ε}^{2}.  (3.9)

We, now, see from the equation in (3.2), using our assumptions on the function f, Proposition 2.2, and estimates (3.8), (3.9) that

∥(ϕ(x′))′∥_{L^{1}(0,1)}

≤ ϕ(∥x∥_{∞})∥d_{1}∥_{L^{1}(0,1)} + ϕ(∥x′∥_{∞})∥d_{2}∥_{L^{1}(0,1)} + ∥r∥_{L^{1}(0,1)} + ∥e∥_{L^{1}(0,1)}

≤ ϕ(M∥x′∥_{∞})∥d_{1}∥_{L^{1}(0,1)} + ϕ(∥x′∥_{∞})∥d_{2}∥_{L^{1}(0,1)} + ∥r∥_{L^{1}(0,1)} + ∥e∥_{L^{1}(0,1)}

≤ ((α(M) + ε)∥d_{1}∥_{L^{1}(0,1)} + ∥d_{2}∥_{L^{1}(0,1)})ϕ(∥x′∥_{∞}) + ∥r∥_{L^{1}(0,1)} + ∥e∥_{L^{1}(0,1)}

+ C_{ε}^{1}∥d_{1}∥_{L^{1}(0,1)}

≤ \frac{(α(M) + ε)∥d_{1}∥_{L^{1}(0,1)} + ∥d_{2}∥_{L^{1}(0,1)}}{1 − α̃(σ^{∗}) − ε} ∥(ϕ(x′))′∥_{L^{1}(0,1)} + Cε,

where Cε = ∥r∥_{L^{1}(0,1)} + ∥e∥_{L^{1}(0,1)} + C_{ε}^{1}∥d_{1}∥_{L^{1}(0,1)} + C_{ε}^{2}[(α(M) + ε)∥d_{1}∥_{L^{1}(0,1)} + ∥d_{2}∥_{L^{1}(0,1)}]. It, now, follows from (3.7) that there exists a constant R_{0}, independent of λ ∈ [0, 1], such that if x(t) ∈ C^{1}[0, 1] is a solution to the boundary value problems (3.2) for some λ ∈ [0, 1] then

∥(ϕ(x′))′∥_{L^{1}(0,1)} ≤ R_{0}.

This combined with (3.9) and (2.8) give that there exists a constant R > 0 such that

∥x∥_{C^{1}[0,1]} < R.

This then implies that deg_{LS}(I −Ψ(·, λ), B(0, R), 0) is well-defined for all λ ∈ [0, 1], where B(0, R) is the ball with center 0 and radius R in C^{1}[0, R].

Let, now, X denote the two-dimensional subspace of C^{1}[0, 1] given by

X = {A + Bt | for A, B ∈ ℝ}.  (3.10)

Let us define the isomorphism i : ℝ^{2} → X by

i (\begin{matrix}A \\ B \end{matrix}) = i(\begin{matrix}A \\ B \end{matrix}) ∈ X, for (\begin{matrix}A \\ B \end{matrix}) ∈ ℝ^{2},  (3.11)

where

i(\begin{matrix}A \\ B \end{matrix})(t) = A + Bt, for t ∈ [0, 1].  (3.12)

Also, we define a 2 × 2 matrix

𝔸 = (\begin{matrix}−(1 − ∑_{i=1}^{m−2} a_{i}) & ∑_{i=1}^{m−2} a_{i}ξ_{i} \\ −(1 − ∑_{j=1}^{n−2} b_{j}) & −(1 − ∑_{j=1}^{n−2} b_{j}τ_{j}) \end{matrix}) .  (3.13)

We note that

det 𝔸 = (1 − ∑_{i=1}^{m−2} a_{i})(1 − ∑_{j=1}^{n−2} b_{j}τ_{j}) + ( ∑_{i=1}^{m−2} a_{i}ξ_{i})(1 − ∑_{j=1}^{n−2} b_{j}) ≠ 0,

in view of the non-resonance assumption (2.3).

Next, we define a function G : ℝ^{2} → ℝ^{2} by setting

\begin{aligned}G (\begin{matrix}A \\ B \end{matrix}) & = 𝔸 · (\begin{matrix}A \\ B \end{matrix}) \\ & = (\begin{matrix}−A(1 − ∑_{i=1}^{m−2} a_{i}) + B(∑_{i=1}^{m−2} a_{i}ξ_{i}) \\ −A(1 − ∑_{j=1}^{n−2} b_{j}) − B(1 − ∑_{j=1}^{n−2} b_{j}τ_{j}) \end{matrix}) for (\begin{matrix}A \\ B \end{matrix}) ∈ ℝ^{2}. \end{aligned} (3.14)

We note that for v(t) = A + Bt ∈ X we have

(I − Ψ(·, 0))(v) = i_{G(\begin{matrix}A \\ B \end{matrix})}

and it follows that

G = i^{−1} ∘ ((I − Ψ(·, 0))|_{X} ∘ i.

Now, we see from the homotopy invariance property of the Leray-Schauder degree that

deg_{LS}(I − Ψ(·, 1), B(0, R), 0) = deg_{LS}(I − Ψ(·, 0), B(0, R), 0)
= deg_{B}(I − Ψ(·, 0)|_{X}, X ∩ B(0, R), 0)
= deg_{B}(G, 𝔹(0, R), 0),

where 𝔹(0, R) denotes the ball of radius R in ℝ^{2} with center at the origin. Finally, we have that

deg_{B}(G, 𝔹(0, R), 0) = \begin{cases}1, & if det 𝔸 > 0 \\ −1, & if det 𝔸 < 0. \end{cases}

Accordingly, we see from the non-resonance assumption (2.3) i.e.

det 𝔸 = (1 − ∑_{i=1}^{m−2} a_{i})(1 − ∑_{j=1}^{n−2} b_{j}τ_{j}) + ( ∑_{i=1}^{m−2} a_{i}ξ_{i})(1 − ∑_{j=1}^{n−2} b_{j}) ≠ 0

that deg_{LS}(I − Ψ(·, 1), B(0, R), 0) ≠ 0 and there is x(t) ∈ B(0, R) ⊂ C^{1}[0, 1] that satisfies

x = Ψ(x, 1),

equivalently x(t) is a solution to the boundary value (1.1). This completes the proof of the theorem. □

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Chaitan P. Gupta Department of Mathematics, 084, University of Nevada, Reno, Reno, NV 89557, USA

E-mail address: gupta@unr.edu