A fixed point theorem for weakly inward A-proper maps and application to a Picard boundary value problem

Abstract

A fixed point theorem for weakly inward A-proper maps defined on cones in Banach spaces is established using a fixed point index for such maps. The result generalizes a theorem in Deimling (Nonlinear Functional Analysis, 1985) for weakly inward maps defined on a cone in $\mathbb{R}^{n}$. We then apply the theorem to a Picard boundary value problem and obtain the existence of a positive solution.

1 Introduction

The purpose of this paper is to establish a fixed point theorem for weakly inward A-proper maps defined on cones in Banach spaces that generalizes a result in Deimling [1], p.254, for weakly inward maps defined on a cone in $\mathbb{R}^{n}$. We use the fixed point index for weakly inward A-proper maps introduced by Lan and Webb [2] to obtain our new result. As an application, we obtain a positive solution to the Picard boundary value problem

$$- x''(t) = f\bigl(t,x(t),x'(t),x''(t) \bigr),\quad \mbox{where } x(0) = x(1) = 0, $$

under suitable conditions on f. This problem has been extensively studied; in particular, we refer to [3], where the concept of $P_{\gamma}$-compact maps and quasinormal cones is used, [4], where the problem is formulated as a semilinear equation, [5], where f is allowed to take negative values, and [6], where positive solutions for three-point boundary value problems are obtained. As mentioned in [5], in [3] and [4], examples were provided with conflicting hypotheses; our theorem will allow a different approach, which corrects the hypotheses of the analogous examples.

2 Preliminaries

Let X be a Banach space, $X_{n} \subset X$ a sequence of oriented finite-dimensional subspaces, and $P_{n}:X \to X_{n}$ a sequence of continuous linear projections such that $P_{n} x \to x$ for each $x \in X$.

Then X is called a Banach space with projection scheme $\Gamma = \{ X_{n},P_{n} \}$.

A map $f:\operatorname{dom} f \subset X \to X$ is said to be A-proper with respect to Γ if $P_{n}f:X_{n} \to X_{n}$ is continuous for each n and for any bounded sequence $\{ x_{n_{j}}|x_{n_{j}} \in X_{n_{j}} \}$ such that $f_{n_{j}}(x_{n_{j}}) \to y$, there exists a subsequence $\{ x_{n_{j_{k}}} \}$ such that $x_{n_{j_{k}}} \to x$ and $f(x) = y$.

A closed convex set K in a Banach space X is called a cone if $\lambda K \subset K$ for all $\lambda \ge 0$ and $K \cap \{ - K \} = 0$.

Let $K \subset X$ be a closed convex set. For each $x \in K$, the set $I_{K}(x) = \{ x + c(z - x):z \in K, c \ge 0 \}$ is called the inward set of x with respect to K. A map $f:K \to X$ is called inward (respectively, weakly inward) if for all $x \in K$, $f(x) \in I_{K}(x)$ ($f(x) \in \bar{I}_{K}(x)$).

A map $f:\overline{\Omega}_{K} \to X$ is said to be inward (respectively, weakly inward) on $\overline{\Omega}_{K}$ relative to K if $f(x) \in I_{K}(x)$ (respectively, $f(x) \in \bar{I}_{K}(x)$) for $x \in \overline{\Omega}_{K}$, where $\Omega \subset X$ is open and bounded with $\Omega_{K} = \Omega \cap K \ne \emptyset$.

For the definition and properties of the Lan-Webb fixed point index, see [2].

3 An existence theorem for weakly inward A-proper maps

Theorem 3.1

Let K be a closed convex set, and $f:K \to X$ be weakly inward on K, where $I - f$ is A-proper. Suppose that

(a):

$f(x)\not\le x$ on $\Vert x \Vert = r$, and

(b):

there exists $\rho \in (0, r)$ such that $\lambda x \not\le f (x)$ for $\Vert x \Vert = \rho$ and $\lambda > 1$.

Then f has a fixed point in $\{ x \in K:\rho < \Vert x \Vert < r\}$.

Proof

Let $B_{r} = \{ x \in X:\Vert x \Vert < r\}$, $B_{r_{K}} = B_{r} \cap K$, $B_{\rho} = \{ x \in X:\Vert x \Vert < \rho \}$, and $B_{\rho_{K}} = B_{\rho} \cap K$. We show that $i_{K}(f,B_{r_{K}}) = \{ 0\}$ and $i_{K}(f,B_{\rho_{K}}) = \{ 1\}$, so that by the additivity property of the index $i_{K}(f,B_{r_{K}}\backslash B_{\rho_{K}}) = i_{K}(f,B_{r_{K}}) - i_{K}(f,B_{\rho_{K}}) = \{ 0\} - \{ 1\} = \{ - 1\} \ne \{ 0\}$, which implies the existence of a fixed point $x \in K$ such that $\rho < \Vert x \Vert < r$.

To show that $i_{K}(f,B_{r_{K}}) = \{ 0\}$, suppose instead that $i_{K}(f,B_{r_{K}}) \ne \{ 0\}$. Then we choose an a with $\Vert f(x) \Vert \le a$ on $\overline{B}_{r_{K}}$ and an $e \in K$ with $\Vert e \Vert > r + a$. Define the weakly inward A-proper homotopy $H(x,t) = f(x) + te$. Now if $H(x,t) = x$ for some $(x,t) \in \partial B_{r_{K}} \times [0, 1]$, then $f(x) + te = x$, so that $x \in K$ and $x - f(x) = te \in K$ so $f (x) \leq x$, which contradicts (a). Thus, H is an admissible homotopy, and $i_{K}(H(x, 1),B_{r_{K}}) = i_{K}(f,B_{r_{K}}) \ne \{ 0\}$. Then there exists $x \in B_{r_{K}}$ with $f(x) + e = x$, so that $\Vert e \Vert = \Vert x - f(x) \Vert \le \Vert x \Vert + \Vert f(x) \Vert \le r + a$, which contradicts $\Vert e \Vert > r + a$. Hence, $i_{K}(f,B_{r_{K}}) = \{ 0\}$.

Now we show that $i_{K}(f,B_{\rho_{K}}) = \{ 1\}$. Define the weakly inward A-proper homotopy $H(x,t) = tf(x)$.

If $H(x,t) = x$ for some $(x,t) \in \partial B_{\rho_{K}} \times [0, 1]$, then $t \ne 0$ (this would give $0 = x$ on $\partial B_{r_{K}}$) and $tf(x) = x$ and $x \in K$, so that $f(x) = \frac{1}{t} x \ge x$, which contradicts (b).

Thus $H(x,t) \ne x$ on $\partial B_{\rho_{K}} \times [0, 1]$.

By the homotopy property of the index, $i_{K}(H(x,0),B_{\rho_{K}}) = i_{K}(H(x,1),B_{\rho_{K}}) = i_{K}(f,B_{\rho_{K}}) = \{ 1\}$.

Consequently, $i_{K}(f,B_{r_{K}}\backslash B_{\rho_{K}}) = i_{K}(f,B_{r_{K}}) - i_{K}(f,B_{\rho_{K}}) = \{ 0\} - \{ 1\} = \{ - 1\}$.

Since the index is not 0, the existence property implies that there exists a fixed point $x \in K$ such that

$$\rho < \Vert x \Vert < r. $$

 □

Remark 3.1

The conclusion of Theorem 3.1 is valid if condition (a) holds for $\Vert x \Vert = \rho$ and condition (b) holds for $\Vert x \Vert = r$, that is,

(a):

$f(x) \not\le x$ on $\Vert x \Vert = \rho$, and

(b):

$\lambda x \not\le f (x)$ for $\Vert x \Vert = r$ and $\lambda > 1$.

We shall use these conditions in the following application.

4 Application

We formulate the Picard boundary value problem

$$ - x''(t) = f\bigl(t,x(t),x'(t),x''(t) \bigr),\quad \mbox{where } x(0) = x(1) = 0 $$

(1)

as a fixed point equation of the operator $T:\overline{K}_{r} \to K$, $K_{r} = \{ x \in K:\Vert x \Vert < r \}$,

$$Ty(t) = f\biggl(t,L^{ - 1}y,\frac{d}{dt}\bigl(L^{ - 1}y \bigr), - y\biggr), $$

where $L:X \to Y$ is defined by $Lx = - x''(t)$. Observe that (1) is equivalent to $y = Ty$.

Let $X = \{ x \in C^{2}[0, 1]:x(0) = x(1) = 0\}$, $Y = C[0, 1]$, and $K = \{ y \in C[1, 0]:y(t) \ge 0\}$ with norms $\Vert x \Vert _{X} = \max \{ \Vert x \Vert _{Y},\Vert x' \Vert _{Y},\Vert x'' \Vert _{Y}\}$ and $\Vert x \Vert _{Y} = \max_{t \in [0,1]}\{ \vert x(t) \vert \}$. Then L is a linear bounded isometric homeomorphism.

Theorem 4.1

Under the above assumptions, suppose also that

(a′):

there exist $r > 0$ and $k \in (0, 1)$ such that $f:[0, 1] \times [0, r] \times [ - r, r] \times R^{ -} \to R^{ +}$ is continuous with $\vert f(t,p,q,s_{1}) - f(t,p,q,s_{2}) \vert \le k\vert s_{1} - s_{2} \vert$ for $t \in [0, 1]$, $p \in [0, r]$, $q \in [ - r, r]$, $s_{1},s_{2} \in R^{ -}$;

(b′):

$f(t,p,q,s) < r$ for every $t \in [0, 1]$, $p \in [0, r]$, $q \in [ - r, r]$, $s = - r$;

(c′):

there are $\rho \in (0, r)$, $t_{0} \in [0, 1]$ such that $f(t_{0},p,q,s) > \rho$ for $p \in [0, \rho ]$, $q \in [ - \rho, \rho ]$, $s = - \rho$.

Then there exists a positive solution $x \in K$ to equation (1) with $\rho < \Vert x \Vert _{X} < r$.

Proof

Since T maps K to K, T is weakly inward. Condition (a′) implies that T is $(\beta_{K})k$-ball contractive, where $\beta_{K}$ is the ball measure of noncompactness associated with K, and thus $\lambda I - T$ is A-proper with respect to the projection scheme $\Gamma = \{ X_{n},P_{n}\}$ for every $\lambda \ge \gamma$, $\gamma \in (k, 1)$ (cf. [3]). To verify the remaining hypotheses of Remark 3.1, we first show that (b′) implies (b). Let r be as in (b′) and $y \in K$ such that $\Vert y \Vert _{Y} = r$. Then there exists $x \in L^{ - 1}(K)$ such that $Lx = y$ and $\Vert x \Vert _{X} = \Vert y \Vert _{Y} = \Vert x'' \Vert _{Y}$, so that $r = \Vert x'' \Vert _{Y} = \Vert x \Vert _{X}$ and there exists $t_{0} \in [0, 1]$ such that $y(t_{0}) = r$. Now since $y = Lx$ for some $x \in L^{ - 1}(K)$, we have that $x(t) \in [0, r]$, $x'(t) \in [ - r, r]$ for all $t \in [0, 1]$ and $r = - x''(t_{0})$. Then if $Ty = \lambda y$ for some $\lambda > 1$ and $y \in K$ with $\Vert y \Vert _{Y} = r$, we would have $f(t,x(t),x'(t),x''(t)) = \lambda y(t)$ for all $t \in [0, 1]$, including $t_{0}$, but then this implies $\lambda r < r$, a contradiction. So (b) holds.

To show that (c′) implies (a) of Remark 3.1, let $x \in K$ with $\Vert x \Vert _{X} = \rho$. Then $\Vert Lx \Vert _{Y} = \Vert - x'' \Vert _{Y} = \rho$, and there exists $t_{1} \in [0, 1]$ such that $- x''(t_{1}) = \rho$ or $x''(t_{1}) = - \rho$. So we have for $t \in [0, 1]$ that $x(t) \in [0, \rho ]$, $x'(t_{1}) \in [ - \rho, \rho ]$, and $x''(t_{1}) = - \rho$. By (c′) we have $Ty(t_{1}) = f(t_{1},x(t_{1}),x'(t_{1}),x''(t_{1})) > \rho$, and so (a) is satisfied.

Thus, there exists a solution to equation (1) with $x \in K$ and $\rho < \Vert x \Vert < r$. □

Example 4.1

The function $f(t,x,x',x'') = 1 + \frac{3}{4}\sin x''$ with $r = \frac{3\pi}{2}$ and $\rho = \frac{\pi}{2}$ shows that the class of maps that satisfy the conditions of Theorem 4.1 is nonempty.