Sex differences in behavioral, cognitive and voluntary ethanol-intake effects in Dexamethasone-induced depression-like state in Wistar rat

1.   Introduction

In this paper, we are concerned with the uniqueness and multiplicity of positive solutions for the quasilinear two-point boundary value problem

where $\lambda > 0$ is a parameter, $f: \mathbb{R} \rightarrow [0, \infty)$. This is the one-dimensional version of the Dirichlet problem associated with the prescribed mean curvature equation in Minkowski space

The study of spacelike submanifolds of codimension one in the flat Minkowski space $\mathbb{L}^{N+1}$ with prescribed mean extrinsic curvature can lead to the type of problems (1.2), where

endowed with the Lorentzian metric

and $(x, t)$ is the canonical coordinate in $\mathbb{R}^{N+1}$ (see ^[1]). Problem (1.1) is also related to classical relativity ^[2], the Born-Infeld model in the theory of nonlinear electrodynamics ^[3,4] and cosmology ^[5,6].

In 1934, in order to combine Maxwell's equation with the quantum theory, Born ^[7] proposed a modified energy function

where $a$ is a constant of the dimensions $r_0^2/e$ with $e =$ charge, $r_0 =$ radius of the electron, $H$ is the magnetic field and $E$ is the electric field. Then, it is easy to verify that $\lim_{a\to 0} L(a) = L = \frac{1}{2}(H^2-E^2)$. Here $L$ is the Lagrangian of Maxwell's equation. When $H = 0$ and $a = 1$, the Poisson equation of $L(a)$ is

where $\rho$ denotes the charge density and $\nabla u = E$. This is closely related to the first equation of problem (1.2). See ^[7,8] for more details.

The existence/ multiplicity of positive solutions for this kind of problems have been extensively studied by using the method of lower and upper solutions, variational methods, fixed point theorem in cones as well as the bifurcation theory, see ^{[9,10,11,12,13,14]} and the references therein.

Recently, the authors of ^[9] obtained some existence results for positive radial solutions of problem (1.2) without parameter $\lambda$. In another paper, they ^[10] established some nonexistence, existence and multiplicity results for positive radial solutions of problem (1.2) with $\lambda f(|x|, s) = \lambda \mu(|x|)s^q$, where $q > 1, \mu:[0, \infty)\to \mathbb{R}$ is continuous, strictly positive on $(0, \infty)$. In ^[11], under some assumptions on $f$, the authors proved that problem (1.2) has infinitely many radial solutions. In ^[12], Coelho et al. discussed the existence of either one, or two, or three, or infinitely many positive solutions of the quasilinear two-point boundary value problem (1.1).

However, as far as we know, there are few works on the uniqueness of positive solutions of problem (1.1). Very recently, Zhang and Feng ^[13] discussed the exactness of positive solutions (1.1) under some special cases of $f$. For example, $f(u) = u^p (p > 0), \ u^p+u^q(0 < p < q\leq1)$.

Let us briefly recall their main results.

Theorem A. Assume that $f(u) = u^p$ with $p < 1$, then for any $\lambda > 0$ (1.1) has exactly one positive solution.

Theorem B. Assume that $f(u) = u^p+u^q$ with $0 < p < q\leq 1$, then for any $\lambda > 0$ (1.1) has exactly one positive solution.

Motivated on these studies, we are interested in investigating a general class of $f$, and we point out $f(u) = u^p$ with $0 < p < 1$, $f(u) = u^p+u^q$ with $0 < p < q\leq 1$ as the special case. In addition, we consider the case where $f$ has $n$ zeros and obtain the existences of arbitrarily many positive solutions of the problem (1.1).

Throughout this paper, we assume that

(H1): $f: \mathbb{R} \rightarrow [0, \infty)$ is continuous function, and $f$ is not identical to zero.

If $\Omega$ is an open bounded subset in a Banach space $E$ and $T: \bar\Omega\to E$ is compact, with $0\not\in (I-T)(\partial \Omega)$, then $i(T, \Omega, E)$ is the fixed point index of $T$ on $\Omega$ with respect to $E$. For the definition and properties of the fixed point index we refer the reader to e.g., ^[15].

The rest of the paper is organized as follows. In section 2, we give some preliminary results. In section 3, under some assumptions on $f$, we prove the uniqueness of positive solutions for problem (1.1). In section 4, we prove the existence of arbitrarily many positive solutions for problem (1.1).

2.   Preliminaries

Below, let $X$ be the Banach space $C[0, 1]$ endowed with the norm $||v|| = \sup_{t\in[0, 1]} \, |v(t)|$. Let $P = \{v\in X: v(t)\ \text{is decreasing in}\ t, v(1) = 0, \text{and}\ v(t)\geq ||v||(1-t)\ \text{for all} \ t\in [0, 1]\}$, then $P$ is a cone in $X$, and the corresponding open ball of center 0 and radius $r > 0$ will be denoted by $B_r$.

By an argument similar to that of Lemma 2.4 of ^[14], we have the following

Lemma 2.1. Let $(\lambda, u)$ be a positive solution of the problem (1.1) with $||u|| = \rho_1$ and $\lambda > 0$. Let $x_0\in (-1, 1)$ be such that $u'(x_0) = 0.$ Then

(ⅰ) $x_0 = 0$;

(ⅱ) $x_0$ is the unique point on which $u$ attains its maximum;

(ⅲ) $u'(t) > 0, t\in (-1, 0)$ and $u'(t) < 0, t\in (0, 1)$.

By a solution of problem (1.1), we understand that it is a function which belongs to $C^1[0, 1]$ with $||u'|| < 1$, such that $u'/\sqrt{1-u'^2}$ is differentiable and problem (1.1) is satisfied.

By Lemma 2.1, $(\lambda, u)$ is a positive solution of (1.1) if and only if $(\lambda, u)$ is a positive solution of the mixed boundary value problem

Now we treat positive classical solutions of (2.1). The following well-known result of the fixed point index is crucial in our arguments.

Lemma 2.2. ^[15,16,17] Let $E$ be a Banach space and $K$ a cone in $E$. For $r > 0$, define $K_r = K \cap B_r$. Assume that $T:\bar{K}_r \to K$ is completely continuous such that $Tx\neq x$ for $x\in \partial K_r = \{x\in K: \|x\| = r\}$.

(ⅰ) If $\|Tx\|\geq \|x\|$ for $x\in \partial K_r$, then $i(T, K_r, K)$ = 0.

(ⅱ) If $\|Tx\|\leq \|x\|$ for $x\in \partial K_r$, then $i(T, K_r, K)$ = 1.

Now we define a nonlinear operator $T_\lambda$ on $P\cap B_1$ as follows:

where $\phi(s) = s/\sqrt{1-s^2}$.

We point out that $u$ is a positive solution of problem (2.1) if $u\in P\cap B_1$ is a fixed point of the nonlinear operator $T_\lambda$.

Next, we give an important Lemma which will be used later.

Lemma 2.3. Let $v\in X$ with $v(t) \geq 0$ for $t\in [0, 1]$. If $v$ concave on $[0, 1]$, then

In particular, for any pair $0 < \alpha < \beta < 1$ we have

Furthermore, if $v(0) = \|v\|$, then we have

Proof. It is an immediate consequence of the fact that $u$ is concave down in $[0, 1]$.

Lemma 2.4. $T_\lambda (P\cap B_1)\subset P$ and the map $T_\lambda: P\cap B_\rho\to P$ for all $\rho\in (0, 1)$ is completely continuous.

Proof. From the condition (H1) one can easily see that $T_\lambda v\in X$ with $T_\lambda v(1) = 0$, and for $v\in P$, a simple computation shows that

So $T_\lambda v(t)$ is decreasing on [0, 1], which implies that

Moreover, it is easy to see that $(T_\lambda v)'$ decreasing in $[0, 1]$, then it follows from Lemma 2.3 that

Therefore, we obtain that $T_\lambda (P\cap B_1)\subset P.$ Moreover, the operator $T_\lambda$ is compact on $P\cap \bar B_\rho$ is an immediate consequence of [9,Lemma 2].

Simple computations leads lead to the following lemma.

Lemma 2.5. Let $\phi(s): = s/\sqrt{1-s^2}$. Then $\phi^{-1}(s) = s/\sqrt{1+s^2}$ and

In particular, for $0 < s_1\leq 1$ we have

3.   Uniqueness results

In this section, we are going to study the positive solution of problem (1.1). Results to be proved in this section are true for any positive parameter $\lambda$. So, we may assume $\lambda = 1$ for simplicity and therefore consider

and the corresponding operator

defined on $P\cap B_1$.

Now, we present the main results of this section as follows.

Theorem 3.1. Assume (H1), and $f$ is increasing, such that for any $u > 0$ and $t\in (0, 1)$ there always exists some $\xi > 1$ such that

Then problem (3.1) can have at most one positive solution.

Remark 3.1. Since $f$ is increasing, condition (3.3) implies that $f(u) > 0$ when $u > 0$, and can be replaced by the following simpler condition:

Remark 3.2. To compare with the Theorem 3.1 and Theorem 3.4 in ^[13], ours in this section cover a very broad class of functions of $f$. For example, $f(u) = \sum_{i = 1}^{i = m}u^{p_i}$ with $0 < p_i < 1$ or $f(u) = 1+\sqrt{u}$.

Theorem 3.1 is proved via a sequence of Lemmas. But we need to have a definition first.

Definition 3.1. ^[16,17] Let $K$ be a cone in real Banach space $Y$ and $\leq$ be the partial ordering defined by $K$. Let $A :K\to K$ and $u_0 > \theta$, where $\theta$ denotes the zero element $Y$.

(a) for any $x > \theta$, there exist $\alpha, \ \beta > 0$ such that

(b) for any $\alpha u_0\leq x\leq \beta u_0$ and $t\in (0, 1)$, there exists some $\eta > 0$ such that

Then $A$ is called $u_0$-sublinear

Lemma 3.2. Suppose that $f$ satisfies the conditions of Theorem 3.1. Then the operator $T_1$, defined by 3.2, is $u_0$-sublinear with $u_0 = 1-t$.

Proof. We divide the proof into two steps.

Step 1. We check condition (a) of Definition 3.1. First, we show that for any $u > 0$ from $P\cap B_1$, there exist $\alpha, \beta > 0$ such that

Let $M = \max_{t\in [0, 1]}\{f(u(t))\}$. It follows from Lemma 2.5, we have

So, we may take $\beta = M$.

Notice that $(T_1 u)(t)$ is strictly decreasing in $t$ and vanishes at $t = 1$. Choose any $c\in (0, 1)$ and set

Then for all $t\in [c, 1]$, we have

Since $(T_1u)(t)\geq (T_1u)(c) = \frac{m}{\sqrt{1+m^2}}(1-c)$ for all $t\in [0, c]$, we have

for all $t\in [0, 1]$. So, we may choose $\alpha = \frac{m}{\sqrt{1+m^2}}(1-c)$ and (3.4) is proved.

Step 2. We check condition (b) of Definition 3.1. We need to show that for any $\alpha u_0\leq u\leq \beta u_0$ and $\xi\in (0, 1)$, there exists some $\eta > 0$ such that

To this end, due to the conditions satisfied by $f$, there exists an $\eta > 0$ such that

It follows from Lemma 2.5 that

Thus we have

Therefore, the proof is complete.

Proof of Theorem 3.1. Let $f$ be from Theorem 3.1. From Lemma 3.2, we know that the operator $T_1$ is increasing and $u_0$-sublinear with $u_0 = 1-t.$

Next, we show that $T_1$ can have at most one positive fixed point. Assume to the contrary that there exist two positive fixed points $u$ and $x^\ast$. Now, we claim that there exists $c > 0$ such that $u\geq c x^\ast$. By (3.2), we know that

So, $x^*(t)$ is strictly decreasing in $t$ and vanishes at $t = 1$. There exist $\sigma_1, \sigma_2\in (0, \infty)$ such that

Using the same method, there exist $\delta_1, \delta_2\in (0, \infty)$ such that

Therefore, $\forall t\in [0, 1]$

So, we can choose $c = \frac{\delta_1}{\sigma_2}$. Set $c^\ast = \sup \{t|u\geq t x^\ast\}$. We claim that

If not, then there exists some $\xi > 0$ such that $T_1(c^\ast x^\ast)\geq (1+\xi)c^\ast T_1 x^\ast$. This combine the fact that $T_1$ is increasing imply that

which is a contraction since $(1+\xi)c^\ast > c^\ast$. So, $u\geq x^\ast$ and similarly we can show that $x^\ast\geq u.$ Now with the above lemmas in place, Theorem 3.1 follows readily. Therefore, the proof is complete.

Example 3.1. Let us consider the problem

By using Theorem 3.1, we know that the problem (3.5) has a unique positive solution, this is just the conclusion of [13, Theorem 3.1].

4.   Multiplicity solutions

In this section, we shall investigate the positive solutions of problem (1.1). Furthermore, we are going to examine how the number of zeros of $f$ may have a huge impact on the number of solutions of (1.1). Here is the main result of this section.

Theorem 4.1. Assume (H1), and there exist two sequences of positive numbers $a_i$ and $b_i$, satisfying

and such that $f(a_i) = 0$ and $f(b_i) = 0$, and $f(u) > 0$ on $(a_i, b_i)$, for all $i = 1, \cdots, n\ (n\in \mathbb{N} = \{1, 2, \cdots\}).$ Then there exists $\lambda_0$ such that for all $\lambda\geq \lambda_0$ the problem (1.1) has $n$ distinct positive solutions, $u_1, u_2, \cdots, u_n$, such that

for each $i = 1, \cdots, n.$

Remark 4.1. Note that $f$ will satisfy the conditions in the theorem if there exist numbers $a_n > a_{n-1} > \cdots > a_1 > a_0 = 0$ such that $f(a_i) = 0$ for $i = 1, \cdots, n$, and $f(u) > 0$ for $a_{i-1} < u < a_i, \ i = 1, \cdots, n.$

Now we define a nonlinear operator $T_\lambda^i, \ i = 1, \cdots n$ as follows:

In order to prove the main result, we define $f_i$ by

For any $0 < r < 1$, define $\mathcal{O}_{r}$ by

Note that $\partial\mathcal{O}_{r} = \{u\in P:\|u\| = r\}.$ Similar to $T_1$, it is easy to verify that $T^i_\lambda$ is compact on $P\cap B_r$ for all $r\in (0, 1)$.

In order to prove the theorem, we need the following two Lemmas.

Lemma 4.2. Suppose that $f$ satisfies the conditions of Theorem 4.1. If $v\in P\cap B_1$ is a positive fixed point of (4.1), i.e. $T_\lambda^i v = v$, then $v$ is a solution of (2.1) such that

Proof. Notice that $v$ satisfies

If on the contrary that $\sup_{t\in [0, 1]}v(t) = v(0) > b_i$, then there exists a $t_0\in (0, 1)$ such that $v(t) > b_i$ for $t\in [0, t_0)$ and $v(t_0) = b_i$. It follows that

Thus, $-\Big(\frac{v'}{\sqrt{1-v'^2}}\Big)$ is constant on $[0, t_0]$. Since $v'(0) = 0,$ it follows that $v'(t) = 0$ for $t\in [0, t_0]$. Consequently, $v(t)$ is a constant on $[0, t_0]$. This is a contradiction. Therefore, $\sup_{t\in [0, 1]}v(t)\leq b_i$. On the other hand, since $f(v)\equiv f_i(v)$ for $0\leq v\leq b_i$, $v$ is a solution of (2.1).

Choose any $\zeta$ such that

Lemma 4.3. Let the conditions of Theorem 4.1 be satisfied. For any $i\in \{1, \cdots, n\}$, there exists $r_i$ such that $[\zeta r_i, r_i]\subset(a_i, b_i)$. Furthermore, for any $v\in \partial \mathcal{O}_{r_i}$ we have

where $\omega_{r_i} = \min_{\zeta r_i\leq v\leq r_i}\{f_i(v)\} > 0.$

Proof. Based on the choice of $\zeta$, the existence of $r_i$ with $[\zeta r_i, r_i]\subset(a_i, b_i)$ is obvious. Notice that for any $v\in P$ we have that $v(t)\geq v(0)(1-t)$ for all $t\in [0, 1]$. In particular, we have $\zeta v(0)\leq v(t)\leq v(0)$ for all $t\in [0, 1-\zeta]$. Let $v\in \partial \mathcal{O}_{r_i}$, then $f_i(v(t))\geq \omega_{r_i}$ for $t\in [0, 1-\zeta]$. It follows that

Proof of Theorem 4.1. Define $\lambda_0$ by

For each $i = 1, \cdots, n$ and $\lambda > \lambda_0$, by Lemma 4.3 we infer that

On the other hand, for each fixed $\lambda > \lambda_0$, since $f_i(v)$ is bounded, this combine the fact that the rage of $\phi^{-1}$ is $(-1, 1)$ imply that there is an $R_i > r_i$ such that

It follows from Lemma 2.2 that

and hence

Thus, $T_\lambda^i$ has a fixed point $v_i^*$ in $\mathcal{O}_{R_i}\backslash \bar{\mathcal{O}}_{r_i}$. Lemma 4.2 implies that the fixed point $v_i^*$ is a solution of (2.1) such that

Consequently, (2.1) has $n$ positive solutions, $v_1, v_2, \cdots, v_n$ for each $\lambda > \lambda_0$, such that

Therefore, the proof is complete.

Remark 4.1. It is interesting to compare our main result(Theorem 4.1) to Candito et al. ^[18]. In ^[18], the authors showed the existence of three classical solutions of the one dimensional prescribed mean curvature equation in Euclid space by using a variational approach.

Example 4.1. Let us consider the problem

where $g(u) = \sin8\pi u+1$.

Let $a_1 = \frac{3}{16}, \ a_2 = \frac{7}{16}, \ a_3 = \frac{11}{16}, \ a_4 = \frac{15}{16}$, we can easily check that $g(a_i) = 0$, and $g(u) > 0$ for $a_{i} < u < a_{i+1}, \ i = 1, 2, 3$.

From Theorem 4.1, the problem (4.2) has at least three positive solutions $u_1, u_2, u_3$ with $a_j < \sup_{r\in [0, 1]}u_j(r)\leq a_{j+1}, \ j = 1, 2, 3$ provided that $\lambda$ is large enough.

Acknowledgments

The authors are very grateful to the anonymous referees for their valuable suggestions.

Conflict of interest

The authors declare that they have no conflicts of interest.