Eigenvalue of boundary value problem for nonlinear singular third-order q-difference equations

Abstract

In this paper, we establish the existence of positive solutions of a boundary value problem for nonlinear singular third-order q-difference equations $D_q^{3}u(t)+\lambda a(t)f(u(t))=0$, $t\in I_q$, $u(0)=0$, $D_{q}u(0)=0$, $\alpha D_{q}u(1)+\beta D_q^{2}u(1)=0$, by using Krasnoselskii’s fixed-point theorem on a cone, where λ is a positive parameter. Finally, we give an example to demonstrate the use of the main result of this paper. The conclusions in this paper essentially extend and improve known results.

1 Introduction

The q-difference equations initiated in the beginning of the 20th century [1–4], is a very interesting field in difference equations. In the last few decades, it has evolved into a multidisciplinary subject and plays an important role in several fields of physics, such as cosmic strings and black holes [5], conformal quantum mechanics [6], and nuclear and high-energy physics [7]. For some recent work on q-difference equations, we refer the reader to [8–12]. However, the theory of boundary value problems (BVPs) for nonlinear q-difference equations is still in an early stage and many aspects of this theory need to be explored. To the best of our knowledge, for the BVPs of nonlinear third-order q-difference equations, a few works were done, see [13, 14] and the references therein.

Recently, in [15], El-Shahed has studied the existence of positive solutions for the following nonlinear singular third-order BVP:

$$\{\begin{array}{l}{u}^{‴}(t)+\lambda a(t)f(u(t))=0,0\le t\le 1,\\ u(0)=u^{\prime }(0)=0,\alpha u^{\prime }(1)+\beta u^{″}(1)=0,\end{array}$$

by Krasnoselskii’s fixed-point theorem on a cone.

More recently, in [13] Ahmad has studied the existence of positive solutions for the following nonlinear BVP of third-order q-difference equations:

$$\{\begin{array}{l}{D}_q^{3}u(t)=f(t,u(t)),0\le t\le 1,\\ u(0)=0,D_{q}u(0)=0,u(1)=0,\end{array}$$

by Leray-Schauder degree theory and some standard fixed-point theorems.

Motivated by the work above, in this paper, we will study the following BVP of nonlinear singular third-order q-difference equations:

$$\{\begin{array}{l}{D}_q^{3}u(t)+\lambda a(t)f(u(t))=0,t\in I_q,\\ u(0)=0,D_{q}u(0)=0,\alpha D_{q}u(1)+\beta D_q^{2}u(1)=0,\end{array}$$

(1.1)

where $\lambda >0$ is a positive parameter, $a:(0,1)\to [0,\mathrm{\infty })$ is continuous and $0<{\int }_0^{1}a(t)d_{q}t<\mathrm{\infty }$, f is a continuous function, $I_q=\{q^n:n\in N\}\cup \{0,1\}$, $q\in (0,1)$ is a fixed constant, and $\alpha ,\beta \ge 0$, $\alpha +\beta >0$.

Obviously, when $q\to 1^{-}$, BVP (1.1) reduces to the standard BVP in [15].

Throughout this paper, we always suppose the following conditions to hold:

(C₁) $f\in C([0,1],[0,+\mathrm{\infty }))$;

(C₂) $\alpha ,\beta \ge 0$, $\alpha +\beta >0$ and $\frac{\alpha -\beta }{\alpha +\beta }\le q$.

2 Preliminary results

In this section, firstly, let us recall some basic concepts of q-calculus [16, 17].

Definition 2.1 For $0<q<1$, we define the q-derivative of a real-value function f as

$$D_{q}f(t)=\frac{f(t)-f(qt)}{(1-q)t},t\in I_q-\{0\},D_{q}f(0)=\underset{t\to 0}{lim}{D}_{q}f(t).$$

Note that ${lim}_{q\to 1^{-}}{D}_{q}f(t)=f^{\prime }(t)$.

Definition 2.2 The higher-order q-derivatives are defined inductively as

$$D_q^{0}f(t)=f(t),D_q^{n}f(t)=D_q{D}_q^{n-1}f(t),n\in N.$$

For example, $D_q(t^k)={[k]}_q{t}^{k-1}$, where k is a positive integer and the bracket ${[k]}_q=(q^k-1)/(q-1)$. In particular, $D_q(t^2)=(1+q)t$.

Definition 2.3 The q-integral of a function f defined in the interval $[a,b]$ is given by

$${\int }_a^{x}f(t)d_{q}t:=\sum _{n=0}^{\mathrm{\infty }}x(1-q)q^{n}f\left(x{q}^n\right)-af\left(a{q}^n\right),x\in [a,b],$$

and for $a=0$, we denote

$$I_{q}f(x)={\int }_0^{x}f(t)d_{q}t=\sum _{n=0}^{\mathrm{\infty }}x(1-q)q^{n}f\left(x{q}^n\right),$$

then

$${\int }_a^{b}f(t)d_{q}t={\int }_0^{b}f(t)d_{q}t-{\int }_0^{a}f(t)d_{q}t.$$

Similarly, we have

$$I_q^{0}f(t)=f(t),I_q^{n}f(t)=I_q{I}_q^{n-1}f(t),n\in N.$$

Observe that

$$D_q{I}_{q}f(x)=f(x),$$

and if f is continuous at $x=0$, then $I_q{D}_{q}f(x)=f(x)-f(0)$.

In q-calculus, the product rule and integration by parts formula are

$$D_q(gh)(t)=D_{q}g(t)h(t)+g(qt)D_{q}h(t),$$

(2.1)

$${\int }_0^{x}f(t)D_{q}g(t)d_{q}t={[f(t)g(t)]}_0^x-{\int }_0^x{D}_{q}f(t)g(qt)d_{q}t.$$

(2.2)

Remark 2.1 In the limit $q\to 1^{-}$, the above results correspond to their counterparts in standard calculus.

Definition 2.4 Let E be a real Banach space. A nonempty closed convex set $P\subset E$ is called a cone if it satisfies the following two conditions:

• (i) $x\in P$, $\lambda \ge 0$ implies $\lambda x\in P$;

• (ii) $x\in P$, $-x\in P$ implies $x=0$.

Theorem 2.1 (Krasnoselskii) [18]

Let E be a Banach space and let $K\in E$ be a cone in E. Assume that ${\mathrm{\Omega }}_1$ and ${\mathrm{\Omega }}_2$ are open subsets of E with $0\in {\mathrm{\Omega }}_1$ and ${\overline{\mathrm{\Omega }}}_1\subset {\mathrm{\Omega }}_2$. Let $T:K\cap ({\overline{\mathrm{\Omega }}}_2\mathrm{\setminus }{\mathrm{\Omega }}_1)\to K$ be a completely continuous operator. In addition, suppose either

(H₁) $\parallel Tu\parallel \le \parallel u\parallel$, $\mathrm{\forall }u\in K\cap \partial {\mathrm{\Omega }}_1$ and $\parallel Tu\parallel \ge \parallel u\parallel$, $\mathrm{\forall }u\in K\cap \partial {\mathrm{\Omega }}_2$ or

(H₂) $\parallel Tu\parallel \le \parallel u\parallel$, $\mathrm{\forall }u\in K\cap \partial {\mathrm{\Omega }}_2$ and $\parallel Tu\parallel \ge \parallel u\parallel$, $\mathrm{\forall }u\in K\cap \partial {\mathrm{\Omega }}_1$

holds. Then T has a fixed point in $K\cap ({\overline{\mathrm{\Omega }}}_2\mathrm{\setminus }{\mathrm{\Omega }}_1)$.

Lemma 2.1 Let $y\in C[0,1]$, then the BVP

$$\{\begin{array}{l}{D}_q^{3}u(t)+y(t)=0,t\in I_q,\\ u(0)=0,D_{q}u(0)=0,\alpha D_{q}u(1)+\beta D_q^{2}u(1)=0,\end{array}$$

(2.3)

has a unique solution

$$u(t)={\int }_0^{1}G(t,s;q)y(s)d_{q}s,$$

where

$$G(t,s;q)=\frac{1}{(1+q)(\alpha +\beta )}\{\begin{array}{ll}\alpha t^2(1-qs)+\beta t^2-(t-qs)(t-q^{2}s)(\alpha +\beta ),& 0\le s\le t\le 1,\\ \alpha t^2(1-qs)+\beta t^2,& 0\le t\le s\le 1.\end{array}$$

Proof Integrate the q-difference equation from 0 to t, we get

$$D_q^{2}u(t)=-{\int }_0^{t}y(s)d_{q}s+a_2.$$

(2.4)

Integrate (2.4) from 0 to t, and change the order of integration, we have

$$D_{q}u(t)=-{\int }_0^t(t-qs)y(s)d_{q}s+a_{2}t+a_1.$$

(2.5)

Integrating (2.5) from 0 to t, and changing the order of integration, we obtain

$$u(t)=-{\int }_0^t(\frac{t^2+q^3{s}^2}{1+q}-qts)y(s)d_{q}s+\frac{a_2}{1+q}{t}^2+a_{1}t+a_0,$$

(2.6)

where $a_2$, $a_1$, $a_0$ are arbitrary constants. Using the boundary conditions $u(0)=0$, $D_{q}u(0)=0$, $\alpha D_{q}u(1)+\beta D_q^{2}u(1)=0$ in (2.6), we find that $a_0=a_1=0$, and

$$a_2=\frac{1}{\alpha +\beta }(\alpha {\int }_0^1(1-qs)y(s)d_{q}s+\beta {\int }_0^{1}y(s)d_{q}s).$$

Substituting the values of $a_2$, $a_1$, and $a_0$ in (2.6), we obtain

$$\begin{array}{rcl}u(t)& =& -{\int }_0^t(\frac{t^2+q^3{s}^2}{1+q}-qts)y(s)d_{q}s\\ +\frac{t^2}{(1+q)(\alpha +\beta )}(\alpha {\int }_0^1(1-qs)y(s)d_{q}s+\beta {\int }_0^{1}y(s)d_{q}s)\\ =& {\int }_0^{1}G(t,s;q)y(s)d_{q}s,\end{array}$$

where

$$G(t,s;q)=\frac{1}{(1+q)(\alpha +\beta )}\{\begin{array}{ll}\alpha t^2(1-qs)+\beta t^2-(t-qs)(t-q^{2}s)(\alpha +\beta ),& 0\le s\le t\le 1,\\ \alpha t^2(1-qs)+\beta t^2,& 0\le t\le s\le 1.\end{array}$$

This completes the proof. □

Remark 2.2 For $q\to 1$, equation (2.6) takes the form

$$u(t)=-\frac{1}{2}{\int }_0^t{(t-s)}^{2}y(s)d_{q}s+\frac{a_2}{2}{t}^2+a_{1}t+a_0,$$

which is the solution of a classical third-order ordinary differential equation $u^{‴}(t)+y(t)=0$ and the associated form of Green’s function for the classical case is

$$G(t,s)=\frac{1}{2(\alpha +\beta )}\{\begin{array}{ll}\alpha t^2(1-s)+\beta t^2-{(t-s)}^2(\alpha +\beta ),& 0\le s\le t\le 1,\\ \alpha t^2(1-s)+\beta t^2,& 0\le t\le s\le 1.\end{array}$$

It is obvious that, when (C₂) holds, $G(t,s;q)\ge 0$, and $G(t,s;q)\le G(1,s;q)$, $0\le t,s\le 1$.

Lemma 2.2 Let (C₂) hold, then $G(t,s;q)\ge g(t)G(1,s;q)$ for $0\le t,s\le 1$, where $g(t)=\frac{4\beta }{5(\alpha +\beta )}{t}^2$.

Proof If $t\le s$, then

$$\begin{array}{rcl}\frac{G(t,s;q)}{G(1,s;q)}& =& \frac{\frac{\alpha t^2(1-qs)}{(1+q)(\alpha +\beta )}+\frac{\beta t^2}{(1+q)(\alpha +\beta )}}{\frac{\alpha (1-qs)}{(1+q)(\alpha +\beta )}+\frac{\beta }{(1+q)(\alpha +\beta )}}=\frac{t^2-\frac{\alpha qs}{\alpha +\beta }{t}^2}{1-\frac{\alpha qs}{\alpha +\beta }}\\ \ge & t^2-\frac{\alpha }{\alpha +\beta }{t}^2=\frac{\beta }{\alpha +\beta }{t}^2\ge \frac{4\beta }{5(\alpha +\beta )}{t}^2.\end{array}$$

If $t\ge s$, then

$$\begin{array}{rcl}\frac{G(t,s;q)}{G(1,s;q)}& =& \frac{\frac{\alpha t^2(1-qs)}{(1+q)(\alpha +\beta )}+\frac{\beta t^2}{(1+q)(\alpha +\beta )}-\frac{t^2+q^3{s}^2}{1+q}+qts}{\frac{\alpha (1-qs)}{(1+q)(\alpha +\beta )}+\frac{\beta }{(1+q)(\alpha +\beta )}-\frac{1+q^3{s}^2}{1+q}+qs}=\frac{-\frac{\alpha qs}{\alpha +\beta }{t}^2-q^3{s}^2+(1+q)qts}{-\frac{\alpha qs}{\alpha +\beta }-q^3{s}^2+(1+q)qs}\\ \ge & \frac{(1+q)q{t}^2-\frac{\alpha qs}{\alpha +\beta }{t}^2-q^3{s}^2}{(1+q)qs-q^3{s}^2}=\frac{(1+q)t^2-\frac{\alpha s}{\alpha +\beta }{t}^2-q^2{s}^2}{(1+q)s-q^2{s}^2}\\ \ge & \frac{t^2-\frac{\alpha }{\alpha +\beta }{t}^2}{1+q-q^2}\ge \frac{4\beta }{5(\alpha +\beta )}{t}^2.\end{array}$$

The proof is complete. □

We consider the Banach space $C_q=C(I_q,R)$ equipped with standard norm $\parallel u\parallel =sup\{|u(t)|,t\in I_q\}$, $u\in C_q$. Define a cone P by

$$P=\{u\in C_q|u(t)\ge 0,u(t)\ge g(t)\parallel u\parallel ,t\in I_q\}.$$

It is easy to see that if $u\in P$, then $\parallel u\parallel =u(1)$.

Define an integral operator $T:P\to C_q$ by

$$Tu(t)=\lambda {\int }_0^{1}G(t,s;q)a(s)f(u(s))d_{q}s,t\in I_q,u\in P.$$

(2.7)

Obviously, T is well defined and $u\in P$ is a solution of BVP (1.1) if and only if u is a fixed point of T.

Remark 2.3 By Lemma 2.2, we obtain, for $u\in P$, $Tu(t)\ge 0$ on $I_q$ and

$$\begin{array}{rcl}Tu(t)& =& \lambda {\int }_0^{1}G(t,s;q)a(s)f(u(s))d_{q}s\ge \lambda g(t){\int }_0^{1}G(1,s;q)a(s)f(u(s))d_{q}s\\ \ge & \lambda g(t)\underset{t\in I_q}{sup}{\int }_0^{1}G(t,s;q)a(s)f(u(s))d_{q}s=g(t)\parallel Tu\parallel .\end{array}$$

Thus $T(P)\subset P$.

We adopt the following assumption:

(C₃) $a(t)\in C((0,1),R^{+})$ may be singular at $t=0,1$, $0<{\int }_0^{1}a(t)d_{q}t<+\mathrm{\infty }$, and $0<{\int }_0^{1}G(1,s;q)a(t)d_{q}t<+\mathrm{\infty }$.

Lemma 2.3 Assume (C₁), (C₂), and (C₃) hold, then $T:P\to P$ is completely continuous.

Proof Define the functions $a_n(t)$ for $n\ge 2$ by

$$a_n(t)=\{\begin{array}{ll}inf\{a(t),a(\frac{1}{n})\},& 0\le t\le \frac{1}{n},\\ a(t),& \frac{1}{n}\le t\le 1-\frac{1}{n},\\ inf\{a(t),a(1-\frac{1}{n})\},& 1-\frac{1}{n}\le t\le 1.\end{array}$$

Next, for $n\ge 2$, we define the operator $T_n:P\to P$ by

$$T_{n}u(t)=\lambda {\int }_0^{1}G(t,s;q)a_n(s)f(u(s))d_{q}s,t\in I_q,u\in P.$$

Obviously, $T_n$ is completely continuous on P for any $n\ge 2$ by an application of the Ascoli-Arzelá theorem. Denote $B_K=\{u\in P:\parallel u\parallel \le K\}$. Then $T_n$ converges uniformly to T as $n\to \mathrm{\infty }$. In fact, for any $t\in I_q$, for each fixed $K>0$ and $u\in B_K$, from (C₁), we obtain

$$\begin{array}{rcl}|T_{n}u(t)-Tu(t)|& =& |\lambda {\int }_0^{1}G(t,s;q)[a(s)-a_n(s)]f(u(s))d_{q}s|\\ \le & \lambda {\int }_0^{\frac{1}{n}}G(1,s;q)|a(s)-a_n(s)|f(u(s))d_{q}s\\ +\lambda {\int }_{\frac{1}{n}}^{1-\frac{1}{n}}G(1,s;q)|a(s)-a_n(s)|f(u(s))d_{q}s\\ +\lambda {\int }_{1-\frac{1}{n}}^{1}G(1,s;q)|a(s)-a_n(s)|f(u(s))d_{q}s\to 0(n\to \mathrm{\infty }),\end{array}$$

where we have used the fact that $G(t,s;q)\ge 0$, and $G(t,s;q)\le G(1,s;q)$, $0\le t,s\le 1$. Hence, $T_n$ converges uniformly to T as $n\to \mathrm{\infty }$, and therefore T is completely continuous also. This completes the proof. □

3 Main results

In this section, we will apply Krasnoselskii’s fixed-point theorem to the eigenvalue problem (1.1). First, we define some important constants:

$$\begin{array}{c}{A}_q={\int }_0^{1}G(1,s;q)a(s)g(s)d_{q}s,B_q={\int }_0^{1}G(1,s;q)a(s)d_{q}s,\hfill \\ F_0=\underset{u\to 0^{+}}{lim}sup\frac{f(u)}{u},f_0=\underset{u\to 0^{+}}{lim}inf\frac{f(u)}{u},\hfill \\ F_{\mathrm{\infty }}=\underset{u\to +\mathrm{\infty }}{lim}sup\frac{f(u)}{u},f_{\mathrm{\infty }}=\underset{u\to +\mathrm{\infty }}{lim}inf\frac{f(u)}{u}.\hfill \end{array}$$

Here we assume that $\frac{1}{A_q{f}_{\mathrm{\infty }}}=0$ if $f_{\mathrm{\infty }}=\mathrm{\infty }$ and $\frac{1}{B_q{F}_0}=\mathrm{\infty }$ if $F_0=0$ and $\frac{1}{A_q{f}_0}=0$ if $f_0=\mathrm{\infty }$ and $\frac{1}{B_q{F}_{\mathrm{\infty }}}=\mathrm{\infty }$ if $F_{\mathrm{\infty }}=0$.

The main result of this paper is the following.

Theorem 3.1 Suppose that (C₁), (C₂) and (C₃) hold and $A_q{f}_{\mathrm{\infty }}>B_q{F}_0$. Then for each $\lambda \in (\frac{1}{A_q{f}_{\mathrm{\infty }}},\frac{1}{B_q{F}_0})$, BVP (1.1) has at least one positive solution.

Proof By the definition of $F_0$, we see that there exists an $l_1>0$, such that $f(u)\le (F_0+\epsilon )u$ for $0\le u\le l_1$. If $u\in P$ with $\parallel u\parallel =l_1$, we have

$$\parallel Tu\parallel =Tu(1)=\lambda {\int }_0^{1}G(1,s;q)a(s)f(u(s))d_{q}s\le \lambda (F_0+\epsilon )\parallel u\parallel B_q.$$

Choose $\epsilon >0$ sufficiently small such that $\lambda (F_0+\epsilon )B_q\le 1$. Then we obtain $\parallel Tu\parallel \le \parallel u\parallel$. Thus if we let ${\mathrm{\Omega }}_1=\{u\in C_q|\parallel u\parallel <l_1\}$, then $\parallel Tu\parallel \le \parallel u\parallel$ for $u\in P\cap \partial {\mathrm{\Omega }}_1$.

From the definition of $f_{\mathrm{\infty }}$, we see that there exist an $l_3>0$ and $l_3>l_1$, such that $f(u)\ge (f_{\mathrm{\infty }}-\epsilon )u$ for $u>l_2$. Let $l_2>l_3$, if $u\in P$ with $\parallel u\parallel =l_2$ we have

$$\begin{array}{rcl}\parallel Tu\parallel & =& Tu(1)=\lambda {\int }_0^{1}G(1,s;q)a(s)f(u(s))d_{q}s\\ \ge & \lambda {\int }_0^{1}G(1,s;q)a(s)g(s)f(u(s))d_{q}s\ge \lambda (f_{\mathrm{\infty }}-\epsilon )\parallel u\parallel A_q.\end{array}$$

Choose $\epsilon >0$ sufficiently small such that $\lambda (f_{\mathrm{\infty }}-\epsilon )A_q\ge 1$. Then we have $\parallel Tu\parallel \ge \parallel u\parallel$. Let ${\mathrm{\Omega }}_2=\{u\in C_q|\parallel u\parallel <l_2\}$, then ${\mathrm{\Omega }}_1\subset {\overline{\mathrm{\Omega }}}_2$ and $\parallel Tu\parallel \ge \parallel u\parallel$ for $u\in P\cap \partial {\mathrm{\Omega }}_2$.

Condition (H₁) of Krasnoselskii’s fixed-point theorem is satisfied. Hence, by Theorem 2.1, the result of Theorem 3.1 holds. This completes the proof of Theorem 3.1. □

Theorem 3.2 Suppose that (C₁), (C₂) and (C₃) hold and $A_q{f}_0>B_q{F}_{\mathrm{\infty }}$. Then for each $\lambda \in (\frac{1}{A_q{f}_0},\frac{1}{B_q{F}_{\mathrm{\infty }}})$, BVP (1.1) has at least one positive solution.

Proof It is similar to the proof of Theorem 3.1. □

Theorem 3.3 Suppose that (C₁), (C₂) and (C₃) hold and $\lambda B_{q}f(u)<u$ for $u\in (0,+\mathrm{\infty })$. Then BVP (1.1) has no positive solution.

Proof Assume to the contrary that u is a positive solution of BVP (1.1). Then

$$\begin{array}{rcl}u(1)& =& \lambda {\int }_0^{1}G(1,s;q)a(s)f(u(s))d_{q}s<\frac{1}{B_q}{\int }_0^{1}G(1,s;q)a(s)u(s)d_{q}s\\ \le & \frac{u(1)}{B_q}{\int }_0^{1}G(1,s;q)a(s)d_{q}s=u(1).\end{array}$$

This is a contradiction and completes the proof. □

Theorem 3.4 Suppose that (C₁), (C₂) and (C₃) hold and $\lambda A_{q}f(u)>u$ for $u\in (0,+\mathrm{\infty })$. Then BVP (1.1) has no positive solution.

Proof It is similar to the proof of Theorem 3.3. □

4 Example

Consider the following BVP:

$$\{\begin{array}{l}{D}_{\frac{1}{2}}^{3}u(t)+\lambda t^{-\frac{1}{2}}\frac{10u^2+u}{u+1}(5+sinu)=0,t\in I_q,\\ u(0)=0,D_{\frac{1}{2}}u(0)=0,D_{\frac{1}{2}}u(1)+3D_{\frac{1}{2}}^{2}u(1)=0.\end{array}$$

(4.1)

Then $F_0=6$, $f_0=4$, $F_{\mathrm{\infty }}=60$, $f_{\mathrm{\infty }}=40$, and $4u\le f(u)\le 60u$. By direct calculations, we obtain $A_q=0.110963$ and $B_q=0.271661$. From Theorem 3.1 we see that if $\lambda \in (0.225299,0.613510)$ then the problem (4.1) has a positive solution. From Theorem 3.3 we see that if $\lambda <0.061351$ then the problem (4.1) has no positive solution. By Theorem 3.4 we see that if $\lambda >2.252986$ then the problem (4.1) has no positive solution.