Two Positive Solutions for a Nonlinear Neumann Problem Involving the Discrete p-Laplacian

Abstract

This paper is devoted to study of existence of at least two positive solutions for a nonlinear Neumann boundary value problem involving the discrete p-Laplacian.

1 Introduction

In this paper, we investigate the existence of two positive solutions for the following nonlinear discrete Neumann boundary value problem

$$\begin{aligned} \left\{ \begin{array}{l} -\varDelta (\phi _p(\varDelta u(k-1)))+q(k)\phi _p(u(k))= \lambda f(k,u(k)) ,\quad k \in [1,N], \\ {\varDelta u(0)=\varDelta u(N)=0, } \quad \quad \qquad \qquad \qquad \\ \end{array} \right. \qquad {N_{\lambda , \underline{f}}} \end{aligned}$$

where $\lambda$ is a positive parameter, N is a fixed positive integer, $[0,N+1]$ is the discrete interval $\{0,...,N+1\}$, $\phi _p(s):=|s|^{p-2}s$, $1<p<+\infty$ and for all $k\in [0,N+1]$, $q(k)>0$, $\varDelta u(k):=u(k+1)-u(k)$ denotes the forward difference operator and is a continuous function.

The theory of difference equations employs numerical analysis, fixed point methods, upper an lower solutions methods (see, for instance, [3, 5, 7, 23]). The variational approach represents an important advance as it allows to prove multiplicity results, usually, under a suitable condition on the nonlinearities, see [1, 2, 7,8,9,10,11, 14,15,16,17,18,19,20,21,22, 24, 25].

In the present paper, we study the problem ($N_{\lambda , \underline{f}}$) following a variational approach, based on a recent result of Bonanno and D’Aguì (see [6]), that assures the existence of at least two non trivial critical points for a certain class of functionals defined on infinite-dimensional Banach space. This theorem is obtained by combining a local minimum result given in [13], together with the Ambrosetti-Rabinowitz theorem (see [4]). In the application of the mountain pass theorem, to prove the Palais-Smale condition of the energy functional associated to the nonlinear differential problems, the Ambrosetti-Rabinowitz condition is requested on the nonlinear term, in particular this means that the nonlinear term has to be more than p-superlinear at infinity.

In this paper, exploiting that the variational framework of the problem ($N_{\lambda , \underline{f}}$) is defined in a finite-dimensional space, we prove that the p-superlinearity at infinity of the primitive on the nonlinearity is enough to prove the Palais-Smale condition. For a complete overview on variational methods on finite Banach spaces and discrete problems, see [12]. We obtain, here, Theorem 2, which gived the existence of two positive solutions, by requiring an algebraic condition on the nonlinearity (we mean (6) in 2).

The paper is so organized: Sect. 2, contains basic definitions and main results on difference equations and some critical point tools, in addition, Lemma 2 is given in order to prove the Palais-Smale condition of the functional associated to problem ($N_{\lambda , \underline{f}}$). Section 3 is devoted to our main result. In particular, our main theorem allows us to obtain two positive solutions with only one hypothesis on the primitive of the nonlinear term f without any asymptotic behaviour at zero. Moreover, a consequence (Corollary 1) (requiring the p-superlinearity at infinity and the p-sublinearity at zero on the primitive of f) of our main result is presented in order to show the applicability of our results.

2 Mathematical Background

In the $N+2$-dimensional Banach space

we consider the norm

$$ \Vert u\Vert :=\left( \displaystyle { \sum _{k=1}^{N+1}}|\varDelta u(k-1)|^p+ \displaystyle { \sum _{k=1}^{N}}q(k)|u(k)|^p\right) ^{1/p} \quad \forall u\in X. $$

Moreover, we will use also the equivalent norm

$$ \Vert u\Vert _\infty :=\displaystyle \max _{k\in [0,N+1]}|u(k)|, \quad \forall u\in X. $$

For our purpose, it will be useful the following inequality

$$\begin{aligned} \displaystyle \Vert u\Vert _\infty \le \Vert u\Vert {q}^{-1/p}, \quad \forall u\in X, \quad \text {where} \quad q:=\displaystyle {\min _{k\in [1, N]}q_k}. \end{aligned}$$

(1)

Moreover, we mention the classical Hölder norm on X.

$$\begin{aligned} \Vert u\Vert _p = \left( \sum _{k=0}^{N+1} |u(k)|^p \right) ^{\frac{1}{p}}. \end{aligned}$$

We observe that being X a finite dimensional Banach space, all norms defined on it are equivalent and in particular, there exist two positive constants $L_1$ and $L_2$ such that

$$\begin{aligned} L_1 \Vert u\Vert _p \le \Vert u\Vert \le L_2 \Vert u\Vert _p. \end{aligned}$$

(2)

To describe the variational framework of problem ($N_{\lambda , \underline{f}}$), we introduce the following two functions

$$\begin{aligned} \varPhi (u):=\frac{\Vert u\Vert ^{p}}{p} \quad \text {and} \quad \varPsi (u):=\sum _{k=1}^{N}F(k,u(k)),\quad \forall u\in X, \end{aligned}$$

(3)

where $F(k,t):=\int _{0}^{t}f(k,\xi )d \xi$ for every . Clearly, $\varPhi$ and $\varPsi$ are two functionals of class whose Gâteaux derivatives at the point $u \in X$ are given by

$$\varPhi '(u)(v) = \sum _{k=1}^{N+1}\phi _p\left( \varDelta u\left( k-1\right) \right) \varDelta v\left( k-1\right) + q(k) \left| u\left( k\right) \right| ^{p-2}u\left( k\right) v\left( k\right) ,$$

and

$$\varPsi '(u)(v) =\sum _{k=1}^{N} f\left( k,u\left( k\right) \right) v(k),$$

for all u, $v \in X$. Taking into account that

$$ {\displaystyle -\sum _{k=1}^{N}\varDelta (\phi _p(\varDelta u(k-1)))v(k)}={\displaystyle \sum _{k=1}^{N+1}\phi _p(\varDelta u(k-1))\varDelta v(k-1)}, \quad \forall \ u\, \ v, \in X, $$

it is easy to verify, see also [25], that

Lemma 1

A vector $u\in X$ is a solution of problem ($N_{\lambda , \underline{f}}$) if and only if u is a critical point of the function $I_{\lambda }=\varPhi -\lambda \varPsi$.

Let $(X, \Vert \cdot \Vert )$ be a Banach space and let . We say that I satisfies the Palais-Smale condition (in short (PS)-condition), if any sequence such that

• 1. is bounded,

• 2. converges to 0 in $X^*$,

admits a subsequence which is convergent in X.

Here, we recall the abstract result established in [6], on the existence of two non-zero critical points.

Theorem 1

Let X be a real Banach space and let $\varPhi$, be two functionals of class $C^1$ such that $\displaystyle \inf _X \varPhi = \varPhi (0) = \varPsi (0) = 0$. Assume that there are and $\tilde{u} \in X$, with $0< \varPhi (\tilde{u}) < r$, such that

(4)

and, for each

the functional $I_{\lambda } = \varPhi - \lambda \varPsi$ satisfies the (PS)-condition and it is unbounded from below.

Then, for each $\lambda \in \varLambda$, the functional $I_{\lambda }$ admits at least two non-zero critical points $u_{\lambda ,1}$, $u_{\lambda ,2}$ such that $I(u_{\lambda ,1})< 0 < I(u_{\lambda ,2})$.

Here and in the sequel we suppose $f(k,0) \ge 0$ for all $k \in [1, N]$. We assume that $f(k,x) = f(k,0)$ for all $x < 0$ and for all $k \in [1, N]$. Put

$$\begin{aligned} L_{\infty }(k):=\liminf _{s\rightarrow +\infty }\dfrac{F(k,s)}{s^p}, \quad L_{\infty }:= \min _{k\in [1,N]} L_{\infty }(k). \end{aligned}$$

We give the following lemma.

Lemma 2

If $L_{\infty } > 0$ then $I_{\lambda }$ satisfies (PS)-condition and it is unbounded from below for all $\lambda \in \left]\dfrac{L_2^p}{pL_{\infty }}, +\infty \right[$, where $L_2$ is given in (2).

Proof

Since $L_{\infty } > 0$ we put $\lambda >\dfrac{L_2^p}{pL_{\infty }}$ and l such that $L_{\infty }> l >\dfrac{L_2^p}{p\lambda }$. Let $\{u_n\}$ be a sequence such that $\displaystyle \lim _{n\rightarrow +\infty } I_{\lambda }(u_n) = c$ and $\displaystyle \lim _{n\rightarrow +\infty } I'_{\lambda }(u_n) = 0$. Put $u^+_n=\max \{u_n, 0\}$ and $u^-_n= \max \{-u_n, 0\}$ for all . We have that $\{u^-_n\}$ is bounded. In fact, one has

$$\begin{aligned} \left| \varDelta u^-_n (k - 1)\right| ^p \le - \phi _p \left( \varDelta u_n (k - 1)\right) \varDelta u^-_n (k -1), \end{aligned}$$

for all $k \in [1, N + 1]$, and

$$\begin{aligned} q(k)\left| u^-_n (k)\right| ^p = - q(k)\left| u_n (k)\right| ^{p-2}u_n (k) u^-_n (k), \end{aligned}$$

for all $k \in [1, N + 1]$.

So we have,

$$\begin{aligned}&\sum _{k=1}^{N+1} \left( \left| \varDelta u^-_n (k - 1)\right| ^p +q(k)\left| u^-_n (k)\right| ^p \right) \\&\le - \sum _{k=1}^{N+1} \left( \phi _p \left( \varDelta u_n (k - 1)\right) \varDelta u^-_n (k -1)+q(k)\left| u_n (k)\right| ^{p-2}u_n (k) u^-_n (k)\right) . \end{aligned}$$

So,

$$\begin{aligned} \Vert u_n^-\Vert ^p= & {} \sum _{k=1}^{N+1} \left( \left| \varDelta u^-_n (k - 1)\right| ^p +q(k)\left| u^-_n (k)\right| ^p \right) \\\le & {} - \sum _{k=1}^{N+1} \left( \phi _p \left( \varDelta u_n (k - 1)\right) \varDelta u^-_n (k -1)+q(k)\left| u_n (k)\right| ^{p-2}u_n (k) u^-_n (k)\right) \\= & {} -\varPhi '(u_n)(u_n^-). \end{aligned}$$

By definition of $u_n^-$ and taking into account that $f(k,x) = f(k,0)$ for all $x < 0$ and for all $k \in [1, N]$, we have

$$\begin{aligned} \varPsi '(u_n)(u_n^-)= \sum _{k=1}^{N} f\left( k,u_n(k)\right) u_n^-(k) \ge 0. \end{aligned}$$

So, we get

$$\begin{aligned} \Vert u_n^-\Vert ^p \le -\varPhi '(u_n)(u_n^-)\le -\varPhi '(u_n)(u_n^-) + \lambda \varPsi '(u_n)(u_n^-), \end{aligned}$$

that is

$$\begin{aligned} \Vert u_n^-\Vert ^p \le -I'_{\lambda }(u_n)(u_n^-), \end{aligned}$$

(5)

for all . Now, from $\displaystyle \lim _{n\rightarrow +\infty } I'_{\lambda }(u_n) = 0$, one has $\displaystyle \lim _{n\rightarrow +\infty } \dfrac{I'_{\lambda }(u_n)(u_n^-)}{\Vert u_n^-\Vert } = 0$, for which, taking (5) into account, gives $\displaystyle \lim _{n\rightarrow +\infty } \Vert u_n^-\Vert = 0$. So, we obtain the claim. And, there is $M > 0$ such that $\Vert u_n^-\Vert \le M$, $\Vert u_n^-\Vert _p \le \dfrac{M}{L_1}=L$, $0\le u_n^-(k)\le L$ for all $k \in [1, N]$ for all .

At this point, by contradiction argument, assume that $\{u_n\}$ is unbounded (that is, $\{u^+_n\}$ is unbounded).

From $\displaystyle \liminf _{s\rightarrow +\infty }\dfrac{F(k,s)}{s^p} = L_{\infty }(k)\ge L_{\infty }>l$ there is $\delta _k > 0$ such that $F(k,s) > ls^p$ for all $s > \delta _k$. Moreover,

$$\begin{aligned} F(k,s)\ge & {} \min _{s\in [-L,\delta _k]} F(k,s) \ge ls^p - l\left( \max \{\delta _k, L\}\right) ^p + \min _{s\in [-L,\delta _k]}F(k,s) \\\ge & {} ls^p- \max \{l\left( \max {\delta _k, L}\right) ^p- \min _{s\in [-L,\delta _k]}F(k,s),0\} = ls^p - Q(k) \end{aligned}$$

for all $s \in [-L, \delta _k]$. Hence, $F(k,s) \ge ls^p - Q(k)$ for all $s \ge -L$. It follows that $F\left( k,u_n(k)\right) \ge l\left( u_n(k)\right) ^p - Q(k)$ for all and for all $k \in [1, N]$, $\displaystyle \sum _{k=1}^{N} F(k,u_n(k)) \ge \sum _{k=1}^{N} \left[ l\left( u_n(k)\right) ^p - Q(k)\right] =l\Vert u_n\Vert _p^p - \displaystyle \sum _{k=1}^{N} Q(k)= l\Vert u_n\Vert _p^p -\overline{Q}$, that is,

$$\begin{aligned} \varPsi (u_n)\ge l\Vert u_n\Vert _p^p -\overline{Q}, \end{aligned}$$

for all . Therefore, one has

$$I_{\lambda }(u_n)=\varPhi (u_n) - \lambda \varPsi (u_n)= \dfrac{1}{p}\Vert u_n\Vert ^p-\lambda \varPsi (u_n) \le \dfrac{L_2^p}{p}\Vert u_n\Vert _p^p-\lambda l\Vert u_n\Vert _p^p + \lambda \overline{Q},$$

that is

$$I_{\lambda }(u_n)\le \left( \dfrac{L_2^p}{p}- \lambda l\right) \Vert u_n\Vert _p^p + \lambda \overline{Q},$$

for all . Since $\Vert u_n\Vert _p\rightarrow + \infty$ and $\dfrac{L_2^p}{p}- \lambda l <0$, one has $\displaystyle \lim _{n\rightarrow +\infty } I_{\lambda }(u_n) = -\infty$ and this is absurd. Hence, $I_{\lambda }$ satisfies (PS)-condition.

Finally, we get that $I_{\lambda }$ is unbounded from below. Let $\{u_n\}$ be such that $\{u^-_n\}$ is bounded and $\{u^+_n\}$ is unbounded. As before, we obtain $\varPsi (u_n) \ge l\Vert u_n\Vert _p^p -\overline{Q}$, for all and, consequently, $I_{\lambda }(u_n)\le \left( \dfrac{L_2^p}{p}- \lambda l\right) \Vert u_n\Vert _p^p + \lambda \overline{Q}$, for all . Hence, $\displaystyle \lim _{n\rightarrow +\infty } I_{\lambda }(u_n) = -\infty$ and the proof is complete.

3 Main Results

In this section, we present the main existence result of our paper. We start putting

$$\begin{aligned} Q= \sum _{k=1}^{N} q(k). \end{aligned}$$

Theorem 2

Let be a continuous function such that $f(k,0)\ge 0$ for all $k\in [1,N]$, and $f(k,0)\ne 0$ for some $k\in [1,N]$. Assume also that there exist two positive constants c and d with $d<c$ such that

$$\begin{aligned} \dfrac{\displaystyle \sum _{k=1}^{N} \max _{\left| \xi \right| \le c}F(k,\xi )}{c^p} < q\min \left\{ \dfrac{1}{Q}\dfrac{\displaystyle \sum _{k=1}^{N} F(k,d)}{d^{p}},\dfrac{L_{\infty }}{L^p_2}\right\} . \end{aligned}$$

(6)

Then, for each $\lambda \in \bar{\varLambda }$ with

the problem ($N_{\lambda , \underline{f}}$) admits at least two positive solutions.

Proof

We consider the functionals $\varPhi$ and $\varPsi$ given in (3). $\varPhi$ and $\varPsi$ satisfy all regularity assumptions requested in Theorem 1, moreover we have that any critical point in X of the functional $I_{\lambda }$ is exactly a solution of problem ($N_{\lambda , \underline{f}}$). Furthermore, $\displaystyle \inf _S \varPhi = \varPhi (0) = \varPsi (0) = 0$. In order to prove our result, we need to verify condition (4) of Theorem 1. Fix $\lambda \in \bar{\varLambda }$, from (6) one has that $L_{\infty } > 0$ and $\bar{\varLambda }$ is non-degenerate. From Lemma 2, the functional $I_{\lambda }$ satisfies the (PS)-condition for each $\lambda > \dfrac{L_2^p}{pL_{\infty }}$, and it is unbounded from below. Now, put $r= \dfrac{qc^p}{p}$, an condier ; so such a u satisfies

$$\dfrac{1}{p}\Vert u\Vert ^p\le r,$$

so

$$\Vert u\Vert \le (pr)^{\frac{1}{p}}.$$

One has

$$\left| u\right| \le \dfrac{1}{q^{\frac{1}{p}}} \Vert u\Vert \le {\left( \dfrac{pr}{q}\right) }^{\frac{1}{p}} =c.$$

So,

$$\begin{aligned} \varPsi (u) = \sum _{k=1}^{N} F(k,u(k)) \le \sum _{k=1}^{N}\max _{|\xi |\le c}F(k,\xi ), \end{aligned}$$

for all $u \in X$ such that .

Hence,

(7)

Now, let be be such that $\tilde{u}(k) = d$ for all $k \in [0,N+1]$. Clearly, $\tilde{u} \in X$ and it holds

$$\begin{aligned} \varPhi (\tilde{u}) = \dfrac{Q d^p}{p}, \end{aligned}$$

(8)

and so, we have

$$\begin{aligned} \dfrac{\varPsi (\tilde{u})}{\varPhi (\tilde{u})}=\dfrac{p}{Q}\dfrac{\displaystyle \sum _{k=1}^{N} F(k,d)}{d^{p}}. \end{aligned}$$

(9)

Therefore, from (7), (9) and assumption (6) one has

Moreover, taking into account that $0< d < c$ and again by (6), we have that

$$\begin{aligned} 0< d < \left( \dfrac{q}{Q}\right) ^{\frac{1}{p}}c. \end{aligned}$$

(10)

Indeed, by contradiction, if we suppose that $d \ge \left( \dfrac{q}{Q}\right) ^{\frac{1}{p}}c$, we have

$$\dfrac{\displaystyle \sum _{k=1}^{N}\max _{|\xi | \le c}F(k,\xi )}{c^p}\ge \dfrac{\displaystyle \sum _{k=1}^{N} F(k,d)}{c^p}\ge \dfrac{q}{Q}\dfrac{\displaystyle \sum _{k=1}^{N} F(k,d)}{d^{p}},$$

which contradicts (6). Hence by (8) and (10) we get $0< \varPhi (\tilde{u}) < r$.

So, finally we obtain that $I_{\lambda }$ admits at least two non-zero critical points and then, for all $\lambda \in \bar{\varLambda } \subset \varLambda$, these are non zero solutions of ($N_{\lambda , \underline{f}}$).

Since we are interested to obtain a positive solution for problem ($N_{\lambda , \underline{f}}$), we adopt the following truncation on the functions f(k, s),

$$ f^+(k,s)=\left\{ \begin{array}{ll} f(k,s), &{} \hbox {if }s\ge 0; \\ f(k,0), &{} \hbox {if }s<0. \end{array} \right. $$

Fixed $\lambda \in \varLambda _c^+$. Working with the truncations $f^+(k,s)$, since we have that $f(k(0,s)\ne 0$ for some $k\in [1,N]$, let u a non trivial solution guaranteed in the first part of the proof, now, to prove the u is nonnegative, we exploit the u is a critical point of the energy functional $I_\lambda =\varPhi -\lambda \varPsi$ associated to problem $(N_{\lambda , f^{+}})$. In other words, we have that $u\in X$ satisfies the following condition

$$\begin{aligned} {\displaystyle \sum _{k=1}^{N+1}\phi _p(\varDelta u(k-1))\varDelta v(k-1)}+\displaystyle \sum _{k=1}^{N}q(k)\phi _p(u(k))v(k)=\displaystyle \sum _{k=1}^{N}f^+(k,u(k))v(k), \ \forall u, v \in X. \end{aligned}$$

(11)

From this, taking as test function $v=-u^-$, it is a simple computation to prove that $\Vert u^-\Vert =0$, that is u is nonnegative. Moreover, arguing by contradiction, we show that u is also a positive solution of problem ($N_{\lambda , \underline{f}}$). Suppose that $u(k)=0$ for some $k\in [1,N]$. Being u a solution of problem ($N_{\lambda , \underline{f}}$) we have

$$ \phi _p(\varDelta u(k-1))-\phi _p(\varDelta u(k))=f(k,0)\ge 0, $$

which implies that

$$ 0\ge -|u(k-1)|^{p-2}u(k-1)-|u(k+1)|^{p-2}u(k+1)\ge 0. $$

So, we have that $u(k-1)=u(k+1)=0$. Hence, iterating this process, we get that $u(k)=0$ for every $k\in [1,N]$, which contradicts that u is nontrivial and this completes the proof.

Now, we present a particular case of Theorem 2.

Corollary 1

Assume that f is a continuous function such that $f(k,0)>0$ for all $k\in [0,N]$ and

$$\begin{aligned} \limsup _{t \rightarrow 0^+} \dfrac{F(k,t)}{t^{p}} = +\infty , \end{aligned}$$

(12)

and

$$\begin{aligned} \lim _{t \rightarrow +\infty } \dfrac{F(k,t)}{t^{p}} = +\infty , \end{aligned}$$

for all $k\in [0,N]$, and put $\lambda ^*= \dfrac{q}{p} \displaystyle \sup _{c > 0}\dfrac{c^{p}}{\displaystyle \sum _{k=1}^{N} \max _{|\xi |\le c}F(k,\xi )}$.

Then, for each , the problem ($N_{\lambda , \underline{f}}$) admits at least two positive solutions.

Proof

First, note that $L_{\infty }= + \infty$. Then, fix and $c > 0$ such that

$$\lambda < \dfrac{q}{p} \displaystyle \dfrac{c^{p}}{\displaystyle \sum _{k=1}^{N} \max _{|\xi |\le c}F(k,\xi )}.$$

From (12) we have

$$\begin{aligned} \limsup _{t \rightarrow 0^+} \dfrac{\displaystyle \sum _{k=1}^{N} F(k,t)}{t^{p}} = +\infty , \end{aligned}$$

then there is $d>0$ with $d<c$ such that $\dfrac{p}{Q}\dfrac{\displaystyle \sum _{k=1}^{N} F(k,d)}{d^{p}}>\dfrac{1}{\lambda }$. Hence, Theorem 2 ensures the conclusion.