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A fixed point operator for systems of vector p-Laplacian with singular weights
- Xianghui Xu1,
- Inbo Sim2 and
- Yong-Hoon Lee1Email author
- Received: 11 January 2016
- Accepted: 30 March 2016
- Published: 5 April 2016
Abstract
In this paper, after establishing a fixed point operator for a strongly coupled vector p-Laplacian with a singular and sign-changing weight function, which may not be integrable, we investigate the existence for the Dirichlet boundary value problems of strongly coupled vector p-Laplacian systems with a nonlinear term consisting of Hadamard product. The proofs are mainly based on topological degree arguments and the global continuation theorem.
Keywords
- p-Laplacian system
- sign-changing weight
- existence
- nontrivial solution
MSC
- 34B16
- 34B18
1 Introduction
We are concerned with the existence of nontrivial solutions for strongly coupled nonlinear differential systems of the form where \(p>1\), \(\Psi_{p}: \mathbb{R}^{N} \to\mathbb{R}^{N}\) is defined by \(\Psi _{p}(x)=|x|^{p-2}x\), \(\lambda>0\) is a parameter, \(h(t)=(h_{1}(t),\ldots ,h_{N}(t))\) with \(h_{i} :(0,1) \to\mathbb{R}\), and \(f(u)=(f_{1}(u),\ldots,f_{N}(u))\) with continuous \(f_{i}: \mathbb{R}^{N}\to\mathbb{R}\). Here we denote \(x \cdot y = (x_{1}y_{1}, x_{2}y_{2}, \ldots, x_{N}y_{N})\) the Hadamard product of x and y in \(\mathbb{R}^{N}\). Thus, problem (\(P_{\lambda}\)) can be rewritten as
$$ (P_{\lambda})\quad \textstyle\begin{cases} -\Psi_{p}(u')'= \lambda h(t)\cdot f(u), \quad t\in(0,1),\\ u(0)= 0 = u(1), \end{cases} $$
$$ \textstyle\begin{cases} -(|u'(t)|^{p-2}u'_{1}(t))'= \lambda h_{1}(t)f_{1}(u),\\ \vdots\\ -(|u'(t)|^{p-2}u'_{N}(t))'= \lambda h_{N}(t)f_{N}(u),\quad t\in(0,1),\\ u_{i}(0)=0=u_{i}(1), \quad i=1,\ldots,N. \end{cases} $$
Throughout the paper, we denote by \(|\cdot|\) the absolute value on \(\mathbb{R}\) or the Euclidean norm on \(\mathbb{R}^{N}\) and by \(\langle\cdot,\cdot\rangle\) the inner product on \(\mathbb{R}^{N}\) and define \(\varphi_{p}: \mathbb{R} \to\mathbb{R}\) by \(\varphi _{p}(s)=|s|^{p-2}s\). For a weight function h, we assume that \(h_{i} \in \mathcal{H}\), where
$$\mathcal{H}= \biggl\{ g\in L^{1}_{\mathrm{loc}}\bigl((0,1),\mathbb{R} \bigr) \Bigm| \int_{0}^{\frac{1}{2}}\varphi_{p}^{-1} \biggl( \int_{s}^{\frac{1}{2}} \bigl|g(\tau)\bigr|\,d\tau \biggr)\,ds+ \int_{\frac{1}{2}}^{1} \varphi_{p}^{-1} \biggl( \int_{\frac{1}{2}}^{s} \bigl|g(\tau)\bigr|\,d\tau \biggr)\,ds< \infty \biggr\} . $$
It is well known that \(L^{1} (0,1) \subsetneqq\mathcal{H}\). Thus, a function in \(\mathcal{H}\) may have stronger singularity at the boundary than a function in \(L^{1}(0,1)\) (see examples in Section 4). If \(h_{i} \in\mathcal{H}\) for all \(i=1,2,\ldots,N\), then \(|h| \in \mathcal{H}\). In this sense, we shall denote \(h \in\mathcal{H}\) whenever \(h_{i} \in\mathcal{H}\) for all \(i=1,2,\ldots,N\).
Scalar equations or systems of p-Laplacian-like problem (\(P_{\lambda}\)) appear in various applications, which describe reaction-diffusion systems, nonlinear elasticity, glaciology, population biology, combustion theory, and non-Newtonian fluids (see [1–4]). The study on the existence of solutions for p-Laplacian scalar equations or systems or more generalized Laplacian systems has attracted much attention recently (see [5–18] and the references therein).
Among their general setup, a solution operator for nonlinear p-Laplacian systems was introduced in the pioneering works of Manásevich and Mawhin [19, 20]. They applied the solution operator to study the existence of solutions for systems of strongly coupled vector p-Laplacian-like operators with \(L^{1}\)-Carathéodory nonlinear perturbations.
We see that the \(L^{1}\)-Carathéodory condition in problem (\(P_{\lambda}\)) corresponds to the condition \(h \in L^{1}((0,1),\mathbb{R}^{N})\). As a generalization of the \(L^{1}\)-Carathéodory condition, it is interesting to consider the case \(h \in\mathcal{H}\). Since our problem involves systems of strongly coupled differential operators and the weight function h may change sign, related studies are not known yet, as far as the authors know. Recently, for a scalar equation of (\(P_{\lambda}\)), Sim and Lee [21] established a new solution operator and proved an existence result by the global continuation theorem.
Thus, the goal of this paper is to get an existence result for (\(P_{\lambda}\)) where the differential operator is related to strongly coupled vector p-Laplacian and the weight function has stronger singularity at the boundary than \(L^{1}\) and sign-changing. The novelty of the paper is providing a new solution operator, which is the most generalized so far.
This paper is organized as follows. In Section 2, we derive a solution operator for problem (W)+(D) with \(g \in\mathcal{H}\). In Section 3, we prove the compactness of the solution operator for (\(P_{\lambda}\)) with \(\lambda=1\). In Section 4, we show the existence of solutions and give some illustrative examples, which satisfy all assumptions in the paper and are not given in other studies.
2 A fixed point operator
In this section, we construct a solution operator for a strongly coupled vector p-Laplacian. Let us consider a problem of the form where \(g\in\mathcal{H}\). Since g may not be in \(L^{1}((0,1),\mathbb{R}^{N})\), the solution of (W)+(D) may not be in \(C^{1}([0,1],\mathbb{R}^{N})\). For an example of a simple scalar case, take \(g(t) = (p-1) t^{-1}|1+ \ln t|^{p-2}\), \(p>2\); then \(g \notin L^{1}(0,1)\), but \(g\in\mathcal{H}\), and the solution u is given by \(u(t) = -t \ln t\), which is not in \(C^{1}[0,1]\).
$$\textstyle\begin{array}{@{}lll} &\displaystyle(W)&\quad {-}\Psi_{p}\bigl(w'\bigr)'= g(t),\quad t\in(0,1),\\ &\displaystyle(D)&\quad w(0)=0=w(1), \end{array} $$
So by a solution to this problem we mean a function \(w \in C([0,1],\mathbb{R}^{N}) \cap C^{1}((0,1),\mathbb{R}^{N})\) with \(\Psi_{p}(w')\) absolutely continuous that satisfies equations (W)+(D).
We first give some remarks for calculations later on.
Remark 2.1
From the definition of \(\Psi_{p}\) and \(\varphi_{p}\) we get, for any \(x, y\in\mathbb{R}^{N}\), where
$$\bigl|\Psi_{p}^{-1}(x+y)\bigr|\leq\varphi_{p}^{-1} \bigl(|x|+|y| \bigr)\leq C_{p} \bigl(\varphi_{p}^{-1}\bigl(|x|\bigr)+ \varphi_{p}^{-1}\bigl(|y|\bigr) \bigr), $$
$$\begin{aligned} C_{p}= \textstyle\begin{cases} 1, & p>2,\\ 2^{\frac{2-p}{p-1}},& 1< p\leq2. \end{cases}\displaystyle \end{aligned}$$
Remark 2.2
By the homogeneity of \(\varphi_{p}^{-1}\) we can deduce that if \(h \in \mathcal{H}\), then \(\alpha\cdot h \in\mathcal{H}\) for all \(\alpha\in C([0,1],\mathbb{R}^{N})\).
Let w be a solution of (W)+(D). Then integrating both sides of \((W)\) on the intervals \([s,\frac{1}{2}]\) and \([\frac{1}{2},s]\) for \(s\in(0,\frac {1}{2}]\) and \(s\in[\frac{1}{2},1)\), respectively, we find that (W)+(D) is equivalent to where \(a=\Psi_{p}(w'(\frac{1}{2}))\). Applying Remark 2.1 with \(x=a\) and \(y=\int_{s}^{\frac{1}{2}}g(\tau)\,d\tau\), we get Since \(g \in\mathcal{H}\), we know that Thus, we may integrate both sides of (2.1) on the interval \([0,t]\) for \(t\in[0,\frac{1}{2}]\) and on the interval \([t,1]\) for \(t\in[\frac{1}{2},1]\), and we get We need to check that \(w(\frac{1}{2}^{-})=w(\frac{1}{2}^{+})\). For \(a \in\mathbb{R}^{N}\), define Then the function \(G_{g}: \mathbb{R}^{N}\rightarrow\mathbb{R}^{N}\) is well defined. If \(G_{g}\) has a unique zero, then \(w(\frac{1}{2}^{-})=w(\frac{1}{2}^{+})\). For this, we give the following lemma.
$$ \textstyle\begin{cases} w'(s)=\Psi_{p}^{-1} (a+\int_{s}^{\frac{1}{2}}g(\tau)\,d\tau ), \qquad w(0)=0, \quad s\in(0,\frac{1}{2}],\\ w'(s)=\Psi_{p}^{-1} (a-\int_{\frac{1}{2}}^{s}g(\tau)\,d\tau ), \qquad w(1)=0, \quad s\in[\frac{1}{2},1), \end{cases} $$
(2.1)
$$\begin{aligned} \biggl\vert \Psi_{p}^{-1} \biggl(a+ \int_{s}^{\frac{1}{2}}g(\tau)\,d\tau \biggr)\biggr\vert \leq& \varphi_{p}^{-1} \biggl(|a|+ \int_{s}^{\frac{1}{2}}\bigl|g(\tau)\bigr|\,d\tau \biggr) \\ \leq& C_{p}\varphi_{p}^{-1}\bigl(|a|\bigr)+ C_{p}\varphi_{p}^{-1} \biggl( \int_{s}^{\frac {1}{2}}\bigl|g(\tau)\bigr|\,d\tau \biggr). \end{aligned}$$
$$\Psi_{p}^{-1} \biggl(a+ \int_{s}^{\frac{1}{2}}g(\tau)\,d\tau \biggr) \in L^{1} \biggl(\biggl(0,\frac{1}{2}\biggr]\biggr), \qquad \Psi_{p}^{-1} \biggl(a- \int_{\frac{1}{2}}^{s}g(\tau)\,d\tau \biggr) \in L^{1}\biggl(\biggl[\frac{1}{2},1\biggr)\biggr). $$
$$\begin{aligned} w(t)= \textstyle\begin{cases} \int_{0}^{t}\Psi_{p}^{-1} (a+\int_{s}^{\frac{1}{2}}g(\tau)\,d\tau )\,ds, & t\in[0,\frac{1}{2}],\\ \int_{t}^{1}\Psi_{p}^{-1} (-a+\int_{\frac{1}{2}}^{s}g(\tau)\,d\tau )\,ds, & t\in[\frac{1}{2},1]. \end{cases}\displaystyle \end{aligned}$$
$$ G_{g} (a)= \int_{0}^{\frac{1}{2}}\Psi_{p}^{-1} \biggl(a+ \int_{s}^{\frac {1}{2}}g(\tau)\,d\tau \biggr)\,ds- \int_{\frac{1}{2}}^{1}\Psi_{p}^{-1} \biggl(-a+ \int_{\frac{1}{2}}^{s}g(\tau )\,d\tau \biggr)\,ds. $$
(2.2)
Lemma 2.3
For given \(g \in\mathcal{H}\), the function \(G_{g}\) defined in (2.2) has a unique zero \(a=a(g)\) in \(\mathbb{R}^{N}\).
Proof
I. Existence. We claim that there exists \(r>0\) such that \(\langle G_{g} (a),a \rangle>0\) for all \(a \in\partial B_{r}(0)\subset\mathbb{R}^{N}\). If the claim is valid, then we consider the homotopy By the claim, for any \(a\in\partial B_{r}(0)\), \(\lambda\in[0,1]\). Taking \(\Omega=B_{r}(0)\), we see that the Brouwer degree \(d_{B}(h(\lambda,a), \Omega,0)\) is well defined, and by the homotopy invariance property we get since \(0\in\Omega\). This completes the proof of the existence of a zero of \(G_{g}\). We now prove the claim. For convenience, we denote Then it suffices to show that there exists \(r>0\) such that \(\langle H_{g} (a), a\rangle> 0\) and \(\langle W_{g} (a), a \rangle< 0\) for all \(a \in\partial B_{r}(0)\subset \mathbb{R}^{N}\). Indeed, we have where \(\delta\in(0,\frac{1}{2})\) will be determined later. Since \(g\in \mathcal{H}\), both integrations are well defined, and we denote We first consider \(H_{1,\delta}\). Since applying Remark 2.1, we obtain where \(p^{\ast}=\frac{p}{p-1}\). Thus, we get Now we consider \(H_{2,\delta}\). Since \(\langle\Psi_{p}(x),x\rangle =|x|^{p}\), \(x \in\mathbb{R}^{N}\), we see that Moreover, for \(s\in[\delta,\frac{1}{2}]\), \(|\int_{s}^{\frac{1}{2}}g(\tau )\,d\tau|\leq\int_{\delta}^{\frac{1}{2}}|g(\tau)|\,d\tau<\infty\); thus, denoting \(\int_{\delta}^{\frac{1}{2}}|g(\tau)|\,d\tau\triangleq M_{\delta}\), we obtain Since \(p^{*} >1\) and for \(s\in[\delta,\frac{1}{2}]\), taking \(|a|\) large enough to satisfy \(|a|-M_{\delta}>0\), we get Combining (2.3) and (2.4), we get that Since \(g\in\mathcal{H}\), we have that \(\varphi_{p}^{-1} (\int _{s}^{\frac{1}{2}}|g(\tau)|\,d\tau )\in L^{1}(0,\delta]\). Choosing \(\delta>0\) sufficiently small and \(|a|=r\) sufficiently large, we can make the right-hand side of (2.5) strictly greater than 0. This implies that there exists \(r>0\) such that \(\langle H_{g}(a),a \rangle>0\) for all \(a \in\partial B_{r}(0)\). Applying a similar argument, we can show that \(\langle W_{g} (a), -a \rangle> 0\) for all \(a \in\partial B_{r}(0)\). Therefore, we conclude that there exists \(r>0\) such that \(\langle G_{g}(a),a \rangle>0\) for all \(a \in\partial B_{r}(0)\), and the claim is proved.
$$h(\lambda,a)=\lambda a+(1-\lambda)G_{g} (a) \quad\mbox{for } \lambda \in[0,1]. $$
$$ \bigl\langle h(\lambda,a),a\bigr\rangle =\lambda\langle a,a\rangle+(1- \lambda )\bigl\langle G_{g} (a),a\bigr\rangle >0 $$
$$d_{B}\bigl(G_{g}(\cdot),\Omega,0\bigr)=d_{B} \bigl(h(0,a),\Omega,0\bigr)=d_{B}\bigl(h(1,a),\Omega ,0 \bigr)=d_{B}(id,\Omega,0)=1 $$
$$\begin{aligned}& H_{g}(a)\triangleq \int_{0}^{\frac{1}{2}}\Psi_{p}^{-1} \biggl(a+ \int_{s}^{\frac {1}{2}}g(\tau)\,d\tau \biggr)\,ds, \qquad W_{g} (a) \triangleq \int_{\frac {1}{2}}^{1}\Psi_{p}^{-1} \biggl(-a+ \int_{\frac{1}{2}}^{s}g(\tau)\,d\tau \biggr)\,ds. \end{aligned}$$
$$\begin{aligned} \bigl\langle H_{g}(a), a\bigr\rangle &= \int_{0}^{\frac{1}{2}}\biggl\langle \Psi_{p}^{-1} \biggl(a+ \int_{s}^{\frac {1}{2}}g(\tau)\,d\tau \biggr), a\biggr\rangle \,ds \\ & = \int_{0}^{\delta}\biggl\langle \Psi_{p}^{-1} \biggl(a+ \int_{s}^{\frac {1}{2}}g(\tau)\,d\tau \biggr), a\biggr\rangle \,ds+ \int_{\delta}^{\frac {1}{2}}\biggl\langle \Psi_{p}^{-1} \biggl(a+ \int_{s}^{\frac{1}{2}}g(\tau)\,d\tau \biggr), a\biggr\rangle \,ds, \end{aligned}$$
$$\begin{aligned}& H_{1,\delta} \triangleq \int_{0}^{\delta}\biggl\langle \Psi_{p}^{-1} \biggl(a+ \int_{s}^{\frac{1}{2}}g(\tau)\,d\tau \biggr), a\biggr\rangle \,ds, \\& H_{2,\delta} \triangleq \int_{\delta}^{\frac{1}{2}}\biggl\langle \Psi _{p}^{-1} \biggl(a+ \int_{s}^{\frac{1}{2}}g(\tau)\,d\tau \biggr), a\biggr\rangle \,ds. \end{aligned}$$
$$\biggl| \int_{s}^{\frac{1}{2}}g(\tau)\,d\tau\biggr|\leq \int_{s}^{\frac{1}{2}}\bigl|g(\tau )\bigr|\,d\tau, $$
$$\begin{aligned} |H_{1,\delta}| &\leq \int_{0}^{\delta} \biggl|\biggl\langle \Psi_{p}^{-1}\biggl(a+ \int_{s}^{\frac {1}{2}}g(\tau)\,d\tau\biggr),a\biggr\rangle \biggr|\,ds \leq \int_{0}^{\delta} \biggl\vert \Psi_{p}^{-1} \biggl(a+ \int_{s}^{\frac {1}{2}}g(\tau)\,d\tau \biggr)\biggr\vert |a|\,ds \\ & \leq \int_{0}^{\delta} \varphi_{p}^{-1} \biggl(|a|+\biggl|\int_{s}^{\frac {1}{2}}g(\tau)\,d\tau\biggr| \biggr) |a|\,ds \leq \int_{0}^{\delta} \varphi_{p}^{-1} \biggl(|a|+ \int_{s}^{\frac {1}{2}}\bigl|g(\tau)\bigr|\,d\tau \biggr) |a|\,ds \\ &\leq \int_{0}^{\delta} C_{p} \biggl( \varphi_{p}^{-1}\bigl(|a|\bigr)+\varphi_{p}^{-1} \biggl( \int_{s}^{\frac{1}{2}}\bigl|g(\tau)\bigr|\,d\tau \biggr) \biggr)|a|\,ds \\ & = C_{p} \delta|a|^{p^{\ast}} + C_{p} \biggl[ \int_{0}^{\delta}\varphi _{p}^{-1} \biggl( \int_{s}^{\frac{1}{2}}\bigl|g(\tau)\bigr|\,d\tau \biggr)\,ds\biggr] |a|, \end{aligned}$$
$$\begin{aligned} H_{1,\delta}&\geq -C_{p} \delta|a|^{p^{\ast}} - C_{p} \biggl[ \int_{0}^{\delta }\varphi_{p}^{-1} \biggl( \int_{s}^{\frac{1}{2}}\bigl|g(\tau)\bigr|\,d\tau \biggr)\,ds\biggr] |a| \\ & = |a|^{p^{\ast}} \biggl[-C_{p} \delta- C_{p} \biggl[ \int_{0}^{\delta}\varphi _{p}^{-1} \biggl( \int_{s}^{\frac{1}{2}}\bigl|g(\tau)\bigr|\,d\tau \biggr)\,ds\biggr] \frac {1}{|a|^{p^{\ast}-1}} \biggr]. \end{aligned}$$
(2.3)
$$\bigl\langle \Psi_{p}^{-1}(x),x\bigr\rangle =\bigl| \Psi_{p}^{-1}(x)\bigr|^{p}=|x|^{(p^{\ast }-1)p}=|x|^{p^{\ast}}. $$
$$\begin{aligned} H_{2,\delta}={}& \int_{\delta}^{\frac{1}{2}}\biggl\langle \Psi_{p}^{-1} \biggl(a+ \int_{s}^{\frac{1}{2}}g(\tau)\,d\tau \biggr),a + \int_{s}^{\frac {1}{2}}g(\tau)\,d\tau\biggr\rangle \,ds \\ &{}- \int_{\delta}^{\frac{1}{2}}\biggl\langle \Psi_{p}^{-1} \biggl(a+ \int_{s}^{\frac{1}{2}}g(\tau)\,d\tau \biggr), \int_{s}^{\frac {1}{2}}g(\tau)\,d\tau\biggr\rangle \,ds \\ \geq{}& \int_{\delta}^{\frac{1}{2}}\biggl|a+ \int_{s}^{\frac{1}{2}}g(\tau)\,d\tau \biggr|^{p^{\ast}}\,ds-M_{\delta} \int_{\delta}^{\frac{1}{2}}\biggl|\Psi_{p}^{-1} \biggl(a+ \int_{s}^{\frac{1}{2}}g(\tau)\,d\tau \biggr)\biggr|\,ds. \end{aligned}$$
$$\biggl|a+ \int_{s}^{\frac{1}{2}}g(\tau)\,d\tau\biggr|\ge|a|-\biggl| \int_{s}^{\frac {1}{2}}g(\tau)\,d\tau\biggr|\ge|a|-M_{\delta} $$
$$\begin{aligned} H_{2,\delta} &\ge \int_{\delta}^{\frac{1}{2}}\bigl(|a|-M_{\delta}\bigr)^{p^{\ast }}\,ds-M_{\delta} \int_{\delta}^{\frac{1}{2}}\bigl(|a|+M_{\delta}\bigr)^{p^{\ast}-1}\,ds \\ & =\biggl(\frac{1}{2}-\delta\biggr) \bigl(|a|-M_{\delta}\bigr)^{p^{\ast}}-\frac{M_{\delta }}{2}\bigl(|a|+M_{\delta}\bigr)^{p^{\ast}-1} \\ & = |a|^{p^{\ast}} \biggl[\biggl(\frac{1}{2}-\delta\biggr) \biggl(1- \frac{M_{\delta }}{|a|}\biggr)^{p^{\ast}}-\frac{M_{\delta}}{2} \biggl(1+ \frac{M_{\delta}}{|a|}\biggr)^{p^{\ast}-1}\frac{1}{|a|} \biggr]. \end{aligned}$$
(2.4)
$$\begin{aligned} \bigl\langle H_{g}(a),a\bigr\rangle \geq{}&|a|^{p^{\ast}} \biggl[\biggl(\frac{1}{2}-\delta \biggr) \biggl(1-\frac{M_{\delta}}{|a|}\biggr)^{p^{\ast}} - \frac{M_{\delta}}{2}\cdot \biggl(1+\frac{M_{\delta}}{|a|}\biggr)^{p^{\ast}-1}\cdot\frac{1}{|a|} \\ &{} -C_{p}\delta -C_{p} \int_{0}^{\delta}\varphi_{p}^{-1} \biggl( \int_{s}^{\frac{1}{2}}\bigl|g(\tau)\bigr|\,d\tau \biggr)\,ds\cdot \frac {1}{|a|^{p^{\ast}-1}} \biggr]. \end{aligned}$$
(2.5)
II. Uniqueness. Suppose that \(a_{1}\) and \(a_{2}\) are two distinct zeros of \(G_{g}\). Then On the contrary, Therefore, we get since \(\langle\Psi_{p}^{-1}(x)-\Psi_{p}^{-1}(y),x-y\rangle>0\) for all \(x,y\in\mathbb{R}^{N}\), \(x\neq y\). This contradiction completes the proof of uniqueness. □
$$\bigl\langle G_{g}(a_{1})-G_{g}(a_{2}),a_{1}-a_{2} \bigr\rangle =0. $$
$$\begin{aligned} & \bigl\langle G_{g}(a_{1})-G_{g}(a_{2}),a_{1}-a_{2} \bigr\rangle \\ & \quad= \bigl\langle H_{g}(a_{1})-H_{g}(a_{2}),a_{1}-a_{2} \bigr\rangle + \bigl\langle W(a_{2})-W(a_{1}),a_{1}-a_{2} \bigr\rangle \\ &\quad = \int_{0}^{\frac{1}{2}}\biggl\langle \Psi_{p}^{-1} \biggl(a_{1}+ \int _{s}^{\frac{1}{2}}g(\tau)\,d\tau \biggr)- \Psi_{p}^{-1} \biggl(a_{2}+ \int_{s}^{\frac{1}{2}}g(\tau)\,d\tau \biggr),a_{1}-a_{2} \biggr\rangle \,ds \\ & \qquad{} + \int_{\frac{1}{2}}^{1}\biggl\langle \Psi_{p}^{-1} \biggl(-a_{2}+ \int _{\frac{1}{2}}^{s}g(\tau)\,d\tau \biggr)- \Psi_{p}^{-1} \biggl(-a_{1}+ \int_{\frac{1}{2}}^{s}g(\tau)\,d\tau \biggr),a_{1}-a_{2} \biggr\rangle \,ds. \end{aligned}$$
$$\begin{aligned} & \bigl\langle G_{g}(a_{1})-G_{g}(a_{2}),a_{1}-a_{2} \bigr\rangle \\ &\quad = \int_{0}^{\frac{1}{2}}\biggl\langle \Psi_{p}^{-1} \biggl(a_{1}+ \int _{s}^{\frac{1}{2}}g(\tau)\,d\tau \biggr)- \Psi_{p}^{-1} \biggl(a_{2}+ \int_{s}^{\frac{1}{2}}g(\tau)\,d\tau \biggr), \\ & \qquad{} \biggl( a_{1}+ \int_{s}^{\frac{1}{2}}g(\tau )\,d\tau \biggr)- \biggl(a_{2}+ \int_{s}^{\frac{1}{2}}g(\tau)\,d\tau \biggr)\biggr\rangle \,ds \\ &\qquad{} + \int_{\frac{1}{2}}^{1}\biggl\langle \Psi_{p}^{-1} \biggl(-a_{2}+ \int _{\frac{1}{2}}^{s}g(\tau)\,d\tau \biggr)- \Psi_{p}^{-1} \biggl(-a_{1}+ \int_{\frac{1}{2}}^{s}g(\tau)\,d\tau \biggr), \\ & \qquad{} \biggl( -a_{2}+ \int_{\frac{1}{2}}^{s}g(\tau )\,d\tau \biggr) - \biggl(-a_{1}+ \int_{\frac{1}{2}}^{s}g(\tau)\,d\tau \biggr)\biggr\rangle \,ds>0 \end{aligned}$$
Lemma 2.3 implies that if \(g\in\mathcal{H}\), then the solution w of (W)+(D) can be represented by where \(a(g)\in\mathbb{R}^{N}\) satisfies
$$ w(t)= \textstyle\begin{cases} \int_{0}^{t}\Psi_{p}^{-1} (a(g)+\int_{s}^{\frac{1}{2}}g(\tau)\,d\tau )\,ds, & t\in[0,\frac{1}{2}],\\ \int_{t}^{1}\Psi_{p}^{-1} (-a(g)+\int_{\frac{1}{2}}^{s}g(\tau)\,d\tau )\,ds, & t\in[\frac{1}{2},1], \end{cases} $$
(2.6)
$$ \int_{0}^{\frac{1}{2}}\Psi_{p}^{-1} \biggl(a(g)+ \int_{s}^{\frac{1}{2}}g(\tau )\,d\tau \biggr)\,ds= \int_{\frac{1}{2}}^{1}\Psi_{p}^{-1} \biggl(-a(g)+ \int_{\frac {1}{2}}^{s}g(\tau)\,d\tau \biggr)\,ds. $$
(2.7)
We note that \(a(g)\) is determined uniquely up to g, and from this uniqueness property the following corollary is obvious.
Corollary 2.4
Let
\(g\in\mathcal{H}\), Then, as a function of
g, a
is homogeneous, that is,
$$a(\lambda g)=\lambda a(g) \quad\textit{for all } \lambda\in\mathbb{R}. $$
On the other hand, it is not hard to see that the function w defined in (2.6) satisfies \(w\in C([0,1],\mathbb{R}^{N})\cap C^{1}((0,1),\mathbb{R}^{N})\), \(\Psi_{p}(w')\) is absolutely continuous on \((0,1)\), and w satisfies (W)+(D). Therefore, we conclude that if \(g\in \mathcal{H}\), then w is a solution of (W)+(D) if and only if w satisfies (2.6).
3 Compactness of the fixed point operator
Consider a nonlinear problem of the form where \(h \in\mathcal{H}\) and \(f \in C(\mathbb{R}^{N},\mathbb{R}^{N})\). We note that, by Remark 2.2, \(h\cdot f(u)\in\mathcal{H}\). Let us apply the solution representation for (W)+(D) given in (2.6) replacing g with \(h\cdot f(u)\). Then we may rewrite problem (P) equivalently as where \(T: C([0,1],\mathbb{R}^{N}) \to C([0,1],\mathbb{R}^{N})\) is defined by
$$ (P)\quad\textstyle\begin{cases} -\Psi_{p}(u')'=h(t)\cdot f(u), \quad t\in(0,1),\\ u(0)= 0 = u(1), \end{cases} $$
$$u=T(u), $$
$$\begin{aligned} T(u) (t) = \textstyle\begin{cases} \int_{0}^{t} \Psi_{p}^{-1} (a( h\cdot f(u))+ \int_{s}^{\frac{1}{2}} h(\tau)\cdot f(u(\tau))\,d\tau )\,ds, & t\in[0,\frac{1}{2}],\\ \int_{t}^{1} \Psi_{p}^{-1} (-a( h\cdot f(u))+ \int_{\frac{1}{2}}^{s} h(\tau)\cdot f(u(\tau))\,d\tau )\,ds, & t\in[\frac{1}{2},1]. \end{cases}\displaystyle \end{aligned}$$
In this section, we prove that the solution operator T is completely continuous. For this, we need two lemmas about the properties of \(a( h\cdot f(u))\). Since h and f are fixed, we regard \(a( h\cdot f(u))\) as a function of \(u\in C([0,1],\mathbb{R}^{N})\).
Lemma 3.1
The function a sends bounded sets in \(C([0,1],\mathbb{R}^{N})\) into bounded sets in \(\mathbb{R}^{N}\).
Proof
Assume that a sequence \(\{u_{n}\}\) is bounded in \(C([0,1],\mathbb {R}^{N})\). Let us denote \(a_{n}\triangleq a(h\cdot f(u_{n}))\) and \(G_{n} \triangleq G_{h\cdot f(u_{n})}\). Suppose that \(\{a_{n}\}\) is unbounded in \(\mathbb{R}^{N}\). Then there exists a subsequence \(\{a_{n_{k}}\}\) such that \(|a_{n_{k}}|\rightarrow \infty\) as \(k\rightarrow\infty\). Since each \(a_{n_{k}}\) is a zero of \(G_{n_{k}}\), we see that \(\langle G_{n_{k}}(a_{n_{k}}), a_{n_{k}}\rangle=0\) for all k. On the other hand, by the same calculation as in the proof of Lemma 2.3 we obtain where \(M=\sup_{k\in\mathbb{N}} \|f(u_{n_{k}})\|_{\infty}\) and \(H_{\delta} = \int_{\delta}^{\frac{1}{2}}|h(\tau)|\,d\tau\). Since \(|a_{n_{k}}|\rightarrow\infty\) as \(k\rightarrow\infty\), we may choose sufficiently large k and then \(\delta>0\) small enough to satisfy \(\langle H_{n_{k}} (a_{n_{k}}),a_{n_{k}} \rangle>0\). Applying a similar argument for \(W_{n_{k}}\), we conclude that \(\langle G_{n_{k}} (a_{n_{k}}),a_{n_{k}} \rangle>0\) for sufficiently large k, and this contradiction completes the proof. □
$$\begin{aligned} \bigl\langle H_{n_{k}} (a_{n_{k}}),a_{n_{k}} \bigr\rangle \geq{}& |a_{n_{k}}|^{p^{\ast}}\biggl[\biggl(\frac{1}{2}- \delta\biggr) \biggl(1-\frac{MH_{\delta }}{|a_{n_{k}}|}\biggr)^{p^{\ast}} - \frac{MH_{\delta}}{2}\cdot\biggl(1+\frac {MH_{\delta}}{|a_{n_{k}}|}\biggr)^{p^{\ast}-1}\cdot \frac{1}{|a_{n_{k}}|} \\ &{}-C_{p}\delta-C_{p}\varphi_{p}^{-1}(M) \int _{0}^{\delta}\varphi_{p}^{-1} \biggl( \int_{s}^{\frac{1}{2}}\bigl|h(\tau)\bigr|\,d\tau \biggr)\,ds\cdot \frac {1}{|a_{n_{k}}|^{p^{\ast}-1}}\biggr], \end{aligned}$$
Remark 3.2
If B is a bounded set in \(C([0,1],\mathbb{R}^{N})\), then \(\{a(h\cdot v) | v\in B\}\) is also bounded in \(\mathbb{R}^{N}\). The proof is similar to that of Lemma 3.1 by replacing M with \(\sup_{v\in B}\|v\|_{\infty}\).
Lemma 3.3
The function \(a: C([0,1],\mathbb{R}^{N})\rightarrow\mathbb{R}^{N}\) is continuous.
Proof
Assume that \(u_{n} \to u\) in \(C([0,1],\mathbb{R}^{N})\). Then for the continuity of a, we need to show that \(a(h\cdot f(u_{n}))\rightarrow a(h\cdot f(u))\) in \(\mathbb{R}^{N}\) as \(n\rightarrow\infty\). Denote again \(a_{n}\triangleq a(h\cdot f(u_{n}))\). We know that \(\{a_{n}\}\) is bounded in \(\mathbb{R}^{N}\) by Lemma 3.1; thus, it has a convergent subsequence \(\{a_{n_{k}}\}\), which converges to, say, \(\hat{a}\in\mathbb{R}^{N}\). We first claim that Indeed, let us take \(K=\sup_{n\in\mathbb{N}} |a_{n}|\), \(M=\sup_{n\in \mathbb{N}} \|f(u_{n})\|_{\infty}\) and fix \(s\in(0, {\frac{1}{2}}]\). Then we get for all \(\tau\in[s, {\frac{1}{2}}]\). Moreover, \(h_{i} \in L^{1}_{\mathrm{loc}} (0,1)\) implies \(|h|\in L^{1} [s, {\frac{1}{2}}]\). Thus, by the continuity of \(\Psi _{p}^{-1}\) and applying the Lebesgue dominated convergence theorem componentwise, we get Similarly, for \(k\in\mathbb{N}\), where \(A=C_{p}\varphi_{p}^{-1}(K)\) and \(B=C_{p}\varphi_{p}^{-1}(M)\). Since \(h\in\mathcal{H}\), the right-hand side of the last inequality is in \(L^{1} (0, {\frac{1}{2}}]\). Thus, applying the Lebesgue dominated convergence theorem componentwise again, we have By the same argument, for fixed \(s\in[{\frac{1}{2}}, 1)\), we also get Moreover, by the definition of \(a_{n_{k}}\) given in (2.7), we know that This implies that both limits in (3.2) and (3.3) are the same, and thus (3.1) is valid. Equation (3.1) implies that \(\hat{a}=a(h\cdot f(u))\) by the uniqueness of â. So we conclude that \(\lim_{k\to\infty} a_{n_{k}} ( = a(h\cdot f(u_{n_{k}})) ) = a(h\cdot f(u))\) in \(\mathbb{R}^{N}\). It is not hard to see by the standard subsequence argument that \(\lim_{n\to\infty} a_{n} ( = a(h\cdot f(u_{n})) ) = a(h\cdot f(u))\), and the proof is done. □
$$\begin{aligned} & \int_{0}^{\frac{1}{2}}\Psi_{p}^{-1} \biggl(\hat{a}+ \int_{s}^{\frac {1}{2}}h(\tau)\cdot f\bigl(u(\tau)\bigr)\,d\tau \biggr)\,ds \\ &\quad = \int_{\frac{1}{2}}^{1}\Psi_{p}^{-1} \biggl(-\hat{a}+ \int_{\frac {1}{2}}^{s}h(\tau)\cdot f\bigl(u(\tau)\bigr)\,d\tau \biggr)\,ds. \end{aligned}$$
(3.1)
$$\bigl|h(\tau)\cdot f\bigl(u_{n_{k}}(\tau)\bigr)\bigr|\le M\bigl|h(\tau)\bigr| $$
$$\lim_{k\to\infty} \Psi_{p}^{-1} \biggl(a_{n_{k}}+ \int_{s}^{\frac {1}{2}}h(\tau)\cdot f\bigl(u_{n_{k}}( \tau)\bigr)\,d\tau \biggr) =\Psi_{p}^{-1} \biggl(\hat{a}+ \int_{s}^{\frac{1}{2}}h(\tau)\cdot f\bigl(u(\tau )\bigr)\,d\tau \biggr). $$
$$\biggl|\Psi_{p}^{-1} \biggl(a_{n_{k}}+ \int_{s}^{\frac{1}{2}}h(\tau)\cdot f\bigl(u_{n_{k}}( \tau)\bigr)\,d\tau \biggr) \biggr| \leq A+B \varphi_{p}^{-1} \biggl( \int_{s}^{\frac{1}{2}}\bigl|h(\tau)\bigr|\,d\tau \biggr), $$
$$\begin{aligned} &\lim_{k\to\infty} \int_{0}^{\frac{1}{2}}\Psi_{p}^{-1} \biggl(a_{n_{k}}+ \int_{s}^{\frac{1}{2}}h(\tau)\cdot f\bigl(u_{n_{k}}( \tau)\bigr)\,d\tau \biggr)\,ds \\ &\quad = \int_{0}^{\frac{1}{2}}\Psi_{p}^{-1} \biggl(\hat{a}+ \int_{s}^{\frac {1}{2}}h(\tau)\cdot f\bigl(u(\tau)\bigr)\,d\tau \biggr)\,ds. \end{aligned}$$
(3.2)
$$ \begin{aligned}[b] &\lim_{k\to\infty} \int_{\frac{1}{2}}^{1}\Psi_{p}^{-1} \biggl(-a_{n_{k}}+ \int_{\frac{1}{2}}^{s}h(\tau)\cdot f\bigl(u_{n_{k}}( \tau)\bigr)\,d\tau \biggr)\,ds \\ &\quad= \int_{\frac{1}{2}}^{1}\Psi_{p}^{-1} \biggl(-\hat{a}+ \int_{\frac {1}{2}}^{s}h(\tau)\cdot f\bigl(u(\tau)\bigr)\,d\tau \biggr)\,ds. \end{aligned} $$
(3.3)
$$\begin{aligned} & \int_{0}^{\frac{1}{2}}\Psi_{p}^{-1} \biggl(a_{n_{k}}+ \int_{s}^{\frac {1}{2}}h(\tau)\cdot f\bigl(u_{n_{k}}( \tau)\bigr)\,d\tau \biggr)\,ds \\ & \quad= \int_{\frac{1}{2}}^{1}\Psi_{p}^{-1} \biggl(-a_{n_{k}}+ \int_{\frac {1}{2}}^{s}h(\tau)\cdot f\bigl(u_{n_{k}}( \tau)\bigr)\,d\tau \biggr)\,ds. \end{aligned}$$
Remark 3.4
If \(v_{n} \in C([0,1],\mathbb{R}^{N})\) with \(v_{n}\to v\) as \(n\to\infty\), then \(a(h\cdot v_{n})\to a(h\cdot v)\) as \(n\to\infty\). In particular, if \(v=0\), then \(a(h\cdot v_{n})\to0 \) as \(n\to\infty\). The proof is similar to that of Lemma 3.3 by replacing M with \(\sup_{v\in B}\|v\|_{\infty}\).
Lemma 3.5
The operator \(T: C([0,1],\mathbb{R}^{N})\rightarrow C([0,1],\mathbb{R}^{N})\) is completely continuous.
Proof
The continuity of T is easily verified mainly by Lemma 3.1 and the Lebesgue dominated convergence theorem. Let B be a bounded subset of \(C([0,1],\mathbb{R}^{N})\). Then by the Arzelà-Ascoli theorem, it suffices to show that \(T(B)\) is uniformly bounded and equicontinuous. Take \(M_{B}=\sup_{u\in B} \|f(u)\|_{\infty}\), \(K_{B}=\sup_{u\in B} |a(h\cdot f(u))|\), and denote \(a_{u}\triangleq a(h\cdot f(u))\). Then, for \(t \in(0,\frac{1}{2}]\), Since \(h\in\mathcal{H}\), we see that the last bound is independent of \(u\in B\) and \(t \in(0,\frac{1}{2}]\). The bound on the interval \([\frac{1}{2},1)\) can be obtained similarly, and thus \(T(B)\) is uniformly bounded.
$$\begin{aligned} \bigl|T(u) (t)\bigr| \leq& \int_{0}^{t} \biggl|\Psi_{p}^{-1} \biggl(a_{u}+ \int_{s}^{\frac {1}{2}}h(\tau)\cdot f\bigl(u(\tau)\bigr)\,d\tau \biggr) \biggr|\,ds \\ \leq& \int_{0}^{t}\varphi_{p}^{-1} \biggl(K_{B}+M_{B} \int_{s}^{\frac {1}{2}}\bigl|h(\tau)\bigr|\,d\tau \biggr)\,ds \\ \leq& \frac{1}{2}C_{p}\varphi_{p}^{-1}(K_{B})+C_{p} \varphi _{p}^{-1}(M_{B}) \int_{0}^{\frac{1}{2}}\varphi_{p}^{-1} \biggl( \int_{s}^{\frac{1}{2}}\bigl|h(\tau)\bigr|\,d\tau \biggr)\,ds. \end{aligned}$$
To show the equicontinuity of \(T(B)\), let \(t_{1}, t_{2}\in[0,1]\) with \(t_{1}< t_{2}\).
Case 1. \(t_{1},t_{2}\in[0,\frac{1}{2} ]\) or \(t_{1},t_{2}\in[\frac{1}{2}, 1 ]\). We have The bound is independent of \(u\in B\) and \(\varphi_{p}^{-1} (\int _{s}^{\frac{1}{2}} |h(\tau)|\,d\tau )\in L^{1} (0,\frac{1}{2} ]\) since \(h \in\mathcal {H}\); thus, we see that the bound converges to 0 as \(|t_{1}-t_{2}| \to0\). The case of \(t_{1},t_{2}\in[\frac{1}{2},1]\) can be similarly proved.
$$\begin{aligned} & \bigl|T(u) (t_{1})-T(u) (t_{2}) \bigr| \\ &\quad\leq \int_{t_{1}}^{t_{2}} \biggl|\Psi_{p}^{-1} \biggl(a_{u}+ \int_{s}^{\frac {1}{2}}h(\tau)\cdot f\bigl(u(\tau)\bigr)\,d\tau \biggr) \biggr|\,ds \\ &\quad \leq C_{p} \varphi_{p}^{-1}(K_{B}) (t_{2}-t_{1})+C_{p} \varphi _{p}^{-1}(M_{B}) \int_{t_{1}}^{t_{2}}\varphi_{p}^{-1} \biggl( \int_{s}^{\frac{1}{2}} \bigl|h(\tau)\bigr|\,d\tau \biggr)\,ds. \end{aligned}$$
Case 2. \(0< t_{1}\leq\frac{1}{2}< t_{2}<1\). Since \(t_{1}\) and \(t_{2}\) can be considered sufficiently close, without loss of generality, we assume that \(\frac{1}{4}\leq t_{1}\leq\frac {1}{2}< t_{2}\leq\frac{3}{4}\). Then, by the definition of T, and
$$\begin{aligned} T(u) (t_{1}) ={}& \int_{0}^{t_{1}}\Psi_{p}^{-1} \biggl(a_{u}+ \int_{s}^{\frac {1}{2}}h(\tau)\cdot f\bigl(u(\tau)\bigr)\,d\tau \biggr)\,ds \\ ={}& \int_{0}^{\frac{1}{2}}\Psi_{p}^{-1} \biggl(a_{u}+ \int_{s}^{\frac {1}{2}}h(\tau)\cdot f\bigl(u(\tau)\bigr)\,d\tau \biggr)\,ds \\ &{}- \int_{t_{1}}^{\frac{1}{2}}\Psi_{p}^{-1} \biggl(a_{u}+ \int_{s}^{\frac {1}{2}}h(\tau)\cdot f\bigl(u(\tau)\bigr)\,d\tau \biggr)\,ds \end{aligned}$$
$$\begin{aligned} T(u) (t_{2}) ={}& \int_{t_{2}}^{1}\Psi_{p}^{-1} \biggl(-a_{u}+ \int_{\frac {1}{2}}^{s}h(\tau)\cdot f\bigl(u(\tau)\bigr)\,d\tau \biggr)\,ds \\ = {}& \int_{\frac{1}{2}}^{1}\Psi_{p}^{-1} \biggl(-a_{u}+ \int_{\frac {1}{2}}^{s}h(\tau)\cdot f\bigl(u(\tau)\bigr)\,d\tau \biggr)\,ds \\ &{}- \int_{\frac{1}{2}}^{t_{2}}\Psi_{p}^{-1} \biggl(-a_{u}+ \int_{\frac {1}{2}}^{s}h(\tau)\cdot f\bigl(u(\tau)\bigr)\,d\tau \biggr)\,ds. \end{aligned}$$
Since, by the definition of \(a_{u}\), we get
$$\begin{aligned} &\int_{0}^{\frac{1}{2}}\Psi_{p}^{-1} \biggl(a_{u}+ \int_{s}^{\frac {1}{2}}h(\tau)\cdot f\bigl(u(\tau)\bigr)\,d\tau \biggr)\,ds \\ &\quad= \int_{\frac{1}{2}}^{1}\Psi_{p}^{-1} \biggl(-a_{u}+ \int_{\frac {1}{2}}^{s}h(\tau)\cdot f\bigl(u(\tau)\bigr)\,d\tau \biggr)\,ds, \end{aligned}$$
$$\begin{aligned} & \bigl|T(u) (t_{1})-T(u) (t_{2})\bigr| \\ &\quad= \biggl| \int_{\frac{1}{2}}^{t_{2}}\Psi_{p}^{-1} \biggl(-a_{u}+ \int_{\frac {1}{2}}^{s}h(\tau)\cdot f\bigl(u(\tau)\bigr)\,d\tau \biggr)\,ds\\ &\qquad{} - \int_{t_{1}}^{\frac{1}{2}}\Psi_{p}^{-1} \biggl(a_{u}+ \int_{s}^{\frac {1}{2}}h(\tau)\cdot f\bigl(u(\tau)\bigr)\,d\tau \biggr)\,ds \biggr| \\ &\quad\leq \int_{\frac{1}{2}}^{t_{2}}\varphi_{p}^{-1} \biggl(K_{B}+M_{B} \int _{\frac{1}{2}}^{s}\bigl|h(\tau)\bigr|\,d\tau \biggr)\,ds + \int_{t_{1}}^{\frac{1}{2}} \varphi_{p}^{-1} \biggl(K_{B}+M_{B} \int _{s}^{\frac{1}{2}}\bigl|h(\tau)\bigr|\,d\tau \biggr)\,ds \\ &\quad\leq \int_{\frac{1}{2}}^{t_{2}}\varphi_{p}^{-1} \biggl(K_{B}+M_{B} \int _{\frac{1}{2}}^{\frac{3}{4}}\bigl|h(\tau)\bigr|\,d\tau \biggr)\,ds + \int_{t_{1}}^{\frac{1}{2}} \varphi_{p}^{-1} \biggl(K_{B}+M_{B} \int_{\frac {1}{4}}^{\frac{1}{2}}\bigl|h(\tau)\bigr|\,d\tau \biggr)\,ds. \end{aligned}$$
Thus, using Remark 2.1, we obtain Since the coefficient at \(t_{2} -t_{1}\) is a constant independent on \(u\in B\), the proof of the equicontinuity of \(T(B)\) is complete. □
$$\begin{aligned} & \bigl|T(u) (t_{1})-T(u) (t_{2})\bigr| \\ &\quad\leq C_{p} \int_{\frac{1}{2}}^{t_{2}} \varphi_{p}^{-1}(K_{B})\,ds +C_{p} \int_{\frac{1}{2}}^{t_{2}} \varphi_{p}^{-1}(M_{B}) \varphi _{p}^{-1} \biggl( \int_{\frac{1}{2}}^{\frac{3}{4}}\bigl|h(\tau)\bigr|\,d\tau \biggr)\,ds \\ &\qquad{}+C_{p} \int_{t_{1}}^{\frac{1}{2}} \varphi_{p}^{-1}(K_{B})\,ds +C_{p} \int_{t_{1}}^{\frac{1}{2}} \varphi_{p}^{-1}(M_{B}) \varphi _{p}^{-1} \biggl( \int_{\frac{1}{4}}^{\frac{1}{2}}\bigl|h(\tau)\bigr|\,d\tau \biggr)\,ds \\ & \quad\leq \biggl[ C_{p}\varphi_{p}^{-1}(K_{B})+C_{p} \varphi_{p}^{-1}(M_{B}) \varphi_{p}^{-1} \biggl( \int_{\frac{1}{4}}^{\frac{3}{4}}\bigl|h(\tau)\bigr|\,d\tau \biggr) \biggr] (t_{2}-t_{1}). \end{aligned}$$
4 Applications
In this section, we apply the solution operator obtained in Section 2 and use the compactness of the operator in Section 3 to show the existence of nontrivial solutions for the problem
$$ (P_{\lambda})\quad\textstyle\begin{cases} -\Psi_{p}(u')'= \lambda h(t) \cdot f(u), \quad t\in(0,1), \\ u(0)= 0 = u(1). \end{cases} $$
For this, we first give one assumption on f.
- (F)
\(f_{i}(0,\ldots,0)>0\) and \(\lim_{|s|\to\infty} f_{i}(s)/|s|^{p-1}=0\) for \(s\in\mathbb{R}^{N}\), \(i=1,\ldots,N\).
Let X be a Banach space, and \(G : \mathbb{R} \times X \to X\) be completely continuous with \(G(0,u)=0\). Consider Denote by \(\mathcal{S}\) the set of solutions of (4.1), \(\mathbb{R}_{+}=[0,\infty)\), and \(\mathbb{R}_{-}=(-\infty,0]\). As the basic tool for the proof of our main theorem, we introduce the following theorem known as the global continuation theorem.
$$ u= G(\lambda,u). $$
(4.1)
Theorem 4.1
([22])
Let X be a Banach space, and \(G : \mathbb{R} \times X \to X\) be continuous and compact with \(G(0,u)=0\). Then \(\mathcal{S}\) contains a pair of unbounded components \(\mathcal{C}^{+}\) and \(\mathcal{C}^{-}\) in \(\mathbb{R}_{+} \times X\) and \(\mathbb{R}_{-} \times X\), respectively, and \(\mathcal{C}^{+} \cap\mathcal{C}^{-} = \{(0,0)\}\).
For our fitting, let us take \(X = C([0,1],\mathbb{R}^{N})\). Then the usual norm for X to be a Banach space is defined by \(\|u\|_{\infty}= \sum_{i=1}^{N} \|u_{i}\|_{\infty}\). In this paper, for the convenience of computation, we establish an equivalent norm, which is defined by Indeed, it is easy to see that
$$\|u\|_{X} = \max_{0\le t \le1} \bigl|\bigl(u_{1}(t), \ldots, u_{N}(t)\bigr)\bigr| = \max_{0\le t \le1} \bigl(u_{1}^{2}(t)+ \cdots+ u_{N}^{2}(t) \bigr)^{1/2}. $$
$$\|u\|_{X} \le\|u\|_{\infty}\le N \|u\|_{X}. $$
We are ready to state our main existence theorem.
Theorem 4.2
Assume that \(h \in\mathcal{H}\) and that (F) holds. Then (\(P_{\lambda}\)) has at least one nontrivial solution for all \(\lambda>0\).
We know that to solve (\(P_{\lambda}\)) is equivalent to solve where \(G: (0,\infty) \times X \to X\) is defined by By Remark 2.2 and Lemma 3.5 we can easily show that G is continuous and compact with \(G(0,u)=0\). Since Theorem 4.1 guarantees an unbounded continuum \(\mathcal{C}^{+}\), if we provide the a priori boundedness of solutions for (\(P_{\lambda}\)), then the unbounded continuum allows the existence of solutions for all \(\lambda>0\).
$$u= G(\lambda,u), $$
$$\begin{aligned} G(\lambda,u) (t) = \textstyle\begin{cases} \int_{0}^{t} \Psi_{p}^{-1} (a(\lambda h\cdot f(u))+ \int_{s}^{\frac{1}{2}} \lambda h(\tau)\cdot f(u(\tau))\,d\tau )\,ds, & t\in[0,\frac{1}{2}],\\ \int_{t}^{1} \Psi_{p}^{-1} (-a(\lambda h\cdot f(u)) + \int_{\frac{1}{2}}^{s} \lambda h(\tau)\cdot f(u(\tau))\,d\tau )\,ds, & t \in[\frac{1}{2},1]. \end{cases}\displaystyle \end{aligned}$$
Lemma 4.3
Assume that \(h \in\mathcal{H}\) and that f satisfies (F). Let any \(\Lambda>0\) be given, and let \((\lambda,u)\) be a solution for (\(P_{\lambda}\)) with \(\lambda\in(0,\Lambda]\). Then there exists a constant \(C(\Lambda)>0\), depending only on Λ, such that \(\|u\|_{X} \le C(\Lambda)\).
Proof
Assume that there exists a sequence \((\lambda_{n},u_{n})\in(0,\Lambda] \times X\) such that, for any \(n\in\mathbb{N}\), with \(\|u_{n}\|_{X} \to\infty\) as \(n\to\infty\).
$$u_{n}=G(\lambda_{n}, u_{n}) $$
By using Remark 2.1 with \(x=a(\lambda_{n} h\cdot f(u_{n}))\), \(y=\int_{s}^{\frac{1}{2}}\lambda_{n} h(\tau)\cdot f(u_{n}(\tau))\,d\tau\) and the homogeneity of \(\varphi_{p}^{-1}\) and a we can estimate the solution \(u_{n}\) as follows: for all \(t\in[0,\frac{1}{2}]\). By the homogeneity of a again, we get By (F), for any \(\varepsilon>0\), there exists \(l_{\epsilon}>0\) such that for all \(s\in\mathbb{R}^{N}\) with \(|s|\ge l_{\epsilon}\), Since \(f_{i}\) is continuous on \(\{s \in\mathbb{R}^{N} \mid |s| \le l_{\epsilon}\}\), there exists a constant \(M_{\epsilon}>0\) such that on \(\{s \in\mathbb{R}^{N} \mid |s| \le l_{\epsilon}\}\) for \(i=1,\ldots,N\). Thus, we have Since \(\|u_{n}\|_{X} \to\infty\) as \(n\to\infty\), there exists \(n_{\epsilon}\in\mathbb{N}\) such that for any \(n\geq n_{\epsilon}\), we have that is, Using (4.2), we get that, for any \(n\geq n_{\epsilon}\) and \(t\in[0,1/2]\), and Take Then B is a bounded subset in X. Thus, by Remark 3.2 we see that the set \(\{a(h\cdot v) \mid v\in B\}\) is bounded in \(\mathbb{R}^{N}\). Moreover, by (4.3) and Remark 3.4 we may choose a constant \(C_{\epsilon}=C_{\epsilon}(\epsilon N)>0\) satisfying \(C_{\epsilon}\to0\) as \(\epsilon\to0\) such that Therefore, for \(t\in[0,\frac{1}{2}]\), we obtain By similar arguments, for \(t\in[\frac{1}{2},1]\), we obtain Denoting \(C_{h} \triangleq\max\{ \int_{0}^{\frac{1}{2}}\varphi_{p}^{-1}(\int_{s}^{\frac{1}{2}} |h(\tau)|\,d\tau )\,ds,\int_{\frac{1}{2}}^{1}\varphi_{p}^{-1}(\int_{\frac{1}{2}}^{s} |h(\tau )|\,d\tau)\,ds\} \), we can choose \(\epsilon>0\) small enough such that Consequently, combining (4.4) and (4.5), we obtain, for \(t\in[0,1]\), This implies that which contradicts and this completes the proof. □
$$\begin{aligned} \bigl|u_{n}(t)\bigr| =& \biggl| \int_{0}^{t}\Psi_{p}^{-1} \biggl(a\bigl(\lambda_{n} h\cdot f(u_{n})\bigr)+ \int_{s}^{\frac{1}{2}} \lambda_{n} h(\tau)\cdot f \bigl(u_{n}(\tau)\bigr)\,d\tau \biggr)\,ds\biggr| \\ \le& \int_{0}^{t}\biggl|\Psi_{p}^{-1} \biggl(a\bigl(\lambda_{n} h\cdot f(u_{n})\bigr)+ \int_{s}^{\frac{1}{2}} \lambda_{n} h(\tau)\cdot f \bigl(u_{n}(\tau)\bigr)\,d\tau \biggr)\biggr|\,ds \\ \le& \int_{0}^{t}\varphi_{p}^{-1} \biggl(\bigl|a\bigl(\lambda_{n} h\cdot f(u_{n})\bigr)\bigr|+ \biggl| \int_{s}^{\frac{1}{2}} \lambda_{n} h(\tau)\cdot f \bigl(u_{n}(\tau)\bigr)\,d\tau\biggr| \biggr)\,ds \\ \le& \varphi_{p}^{-1}(\lambda_{n}) \int_{0}^{\frac{1}{2}}\varphi _{p}^{-1} \biggl(\bigl|a\bigl(h\cdot f(u_{n})\bigr)\bigr|+ \biggl| \int_{s}^{\frac{1}{2}} h(\tau)\cdot f\bigl(u_{n}( \tau)\bigr)\,d\tau\biggr| \biggr)\,ds \\ \le& \varphi_{p}^{-1}(\Lambda) \int_{0}^{\frac{1}{2}}\varphi_{p}^{-1} \biggl(\frac{|a(h\cdot f(u_{n}))|}{\|u_{n}\|_{X}^{p-1}}+ \frac{|\int_{s}^{\frac{1}{2}} h(\tau)\cdot f(u_{n}(\tau))\,d\tau|}{\|u_{n}\|_{X}^{p-1}} \biggr)\,ds\|u_{n} \|_{X} \end{aligned}$$
$$\bigl|u_{n}(t)\bigr|\le \varphi_{p}^{-1}(\Lambda) \int_{0}^{\frac{1}{2}}\varphi _{p}^{-1} \biggl(\biggl|a\biggl(h\cdot\frac{f(u_{n})}{\|u_{n}\|_{X}^{p-1}}\biggr)\biggr|+ \int_{s}^{\frac{1}{2}} \bigl|h(\tau)\bigr| \frac{|f(u_{n}(\tau))|}{\|u_{n}\|_{X}^{p-1}}\,d\tau \biggr)\,ds\|u_{n}\|_{X}. $$
$$ \bigl|f_{i}(s)\bigr| \le\varepsilon|s|^{p-1} \quad \mbox{for } i=1,\ldots,N. $$
$$ \bigl|f_{i}(s)\bigr| \le M_{\epsilon}$$
$$ \bigl|f_{i}(s)\bigr| \le\varepsilon|s|^{p-1}+M_{\epsilon}\quad\mbox{for all } s \in \mathbb{R}^{N} , i=1,\ldots,N. $$
(4.2)
$$\|u_{n}\|_{X}\geq\biggl(\frac{M_{\epsilon}}{\epsilon} \biggr)^{\frac{1}{p-1}}, $$
$$\frac{1}{\|u_{n}\|_{X}^{p-1}}\leq\frac{\epsilon}{M_{\epsilon}}. $$
$$\frac{|f_{i}(u_{n}(t))|}{\|u_{n}\|_{X}^{p-1}}\leq\epsilon\cdot\frac {|u_{n}(t)|^{p-1}}{\|u_{n}\|_{X}^{p-1}}+\frac{M_{\epsilon}}{\|u_{n}\| _{X}^{p-1}}\leq \epsilon+M_{\epsilon}\cdot\frac{\epsilon}{M_{\epsilon}}=2\epsilon $$
$$ \frac{\|f(u_{n})\|_{X}}{\|u_{n}\|_{X}^{p-1}}\leq\frac{\|f(u_{n})\|_{\infty}}{\| u_{n}\|_{X}^{p-1}}=\frac{\sum_{i=1}^{N}\|f_{i} (u_{n})\|_{\infty}}{\|u_{n}\| _{X}^{p-1}}\leq N \cdot2\epsilon=2\epsilon N. $$
(4.3)
$$B=\biggl\{ \frac{f(u_{n})}{\|u_{n}\|_{X}^{p-1}}\biggr\} _{n\geq n_{\epsilon}}. $$
$$\biggl|a\biggl(h\cdot\frac{f(u_{n})}{\|u_{n}\|_{X}^{p-1}}\biggr)\biggr| \leq C_{\epsilon}\quad\mbox{for any } n\geq n_{\epsilon}. $$
$$\begin{aligned} \bigl|u_{n}(t)\bigr| \leq{}&\biggl[\varphi_{p}^{-1}( \Lambda) \int_{0}^{\frac{1}{2}}\varphi _{p}^{-1} \biggl(C_{\epsilon}+2\epsilon \int_{s}^{\frac{1}{2}} \bigl|h(\tau)\bigr|\,d\tau \biggr)\,ds\biggr] \|u_{n}\|_{X} \\ \leq{}& \biggl[\frac{1}{2}\varphi_{p}^{-1}( \Lambda)C_{p} \varphi_{p}^{-1}(C_{\epsilon}) \\ &{}+\varphi_{p}^{-1}(\Lambda)C_{p} \varphi_{p}^{-1}(2\epsilon) \int _{0}^{\frac{1}{2}}\varphi_{p}^{-1} \biggl( \int_{s}^{\frac{1}{2}} \bigl|h(\tau)\bigr|\,d\tau \biggr)\,ds\biggr] \|u_{n}\|_{X}. \end{aligned}$$
(4.4)
$$\begin{aligned} \bigl|u_{n}(t)\bigr| \leq{}&\biggl[\varphi_{p}^{-1}( \Lambda) \int_{\frac{1}{2}}^{1}\varphi _{p}^{-1} \biggl(C_{\epsilon}+2\epsilon \int_{\frac{1}{2}}^{s} \bigl|h(\tau)\bigr|\,d\tau \biggr)\,ds\biggr] \|u_{n}\|_{X} \\ \leq{}& \biggl[\frac{1}{2}\varphi_{p}^{-1}( \Lambda)C_{p} \varphi_{p}^{-1}(C_{\epsilon}) \\ &{}+\varphi_{p}^{-1}(\Lambda)C_{p} \varphi_{p}^{-1}(2\epsilon) \int_{\frac {1}{2}}^{1}\varphi_{p}^{-1} \biggl( \int_{\frac{1}{2}}^{s} \bigl|h(\tau)\bigr|\,d\tau\biggr)\,ds\biggr]\| u_{n}\|_{X}. \end{aligned}$$
(4.5)
$$\frac{1}{2}\varphi_{p}^{-1}(\Lambda)C_{p} \varphi_{p}^{-1}(C_{\epsilon}) +\varphi_{p}^{-1}( \Lambda)C_{p} \varphi_{p}^{-1}(2\epsilon) C_{h} \leq\frac{1}{2}. $$
$$\bigl|u_{n}(t)\bigr|\leq\frac{1}{2}\|u_{n}\|_{X}. $$
$$\|u_{n}\|_{X}\leq0 \quad\mbox{for } n\geq n_{\epsilon}, $$
$$\|u_{n}\|_{X}\geq\biggl(\frac{M_{\epsilon}}{\epsilon} \biggr)^{\frac{1}{p-1}}>0 \quad\mbox{for } n\geq n_{\epsilon}$$
Example 1
Consider the following p-Laplacian system: where \(\mathbf{u}=(u,v)\), \(\lambda>0\) is a parameter, and \(h(t)=(h_{1}(t),h_{2}(t))\) is given by We note that \(h\in L^{1} _{\mathrm{loc}}\) but \(h_{1}\notin L^{1}\). We now show that \(h\in\mathcal{H}\). Indeed, Since \(1<\alpha<p\), we have \(\frac{1}{\alpha-1}s^{-(\alpha-1)}>0\) for \(s\in(0,1)\) and In addition, since \(h_{1}\) and \(h_{2}\) are constants on \((\frac{1}{2},1)\) and \((0,1)\), respectively, by Remark 2.1 we get \(h\in\mathcal{H}\).
$$ (E_{1})\quad\textstyle\begin{cases} -(|\mathbf{u}|^{p-2}u')'= \lambda h_{1}(t)[ (u^{2}+v^{2})^{\frac {p-1}{4}}+1],\\ -(|\mathbf{u}|^{p-2}v')'= \lambda h_{2}(t)e^{-v^{2}}[1+(u^{2})^{\frac {p-1}{3}}],\quad t\in(0,1),\\ u(0)=v(0)= 0 = u(1)=v(1), \end{cases} $$
$$\begin{aligned}& h_{1}(t)= \textstyle\begin{cases} t^{-\alpha}, & t\in(0,\frac{1}{2}],\\ -1, & t\in(\frac{1}{2},1), 1< \alpha< p, \end{cases}\displaystyle \qquad h_{2}(t)=-1, \quad t\in(0,1). \end{aligned}$$
$$\begin{aligned} \int_{s}^{\frac{1}{2}}\tau^{-\alpha}\,d\tau = & - \frac{1}{\alpha-1}\tau^{-(\alpha-1)} \bigg|_{s}^{\frac {1}{2}}=- \frac{1}{\alpha-1} \biggl[\biggl(\frac{1}{2}\biggr)^{-(\alpha-1)}-s^{-(\alpha-1)} \biggr] \\ = & \frac{1}{\alpha-1} \bigl[s^{-(\alpha-1)}-2^{\alpha-1}\bigr]\le \frac{1}{\alpha-1}s^{-(\alpha-1)}. \end{aligned}$$
$$\begin{aligned} \int_{0}^{\frac{1}{2}}\varphi_{p}^{-1} \biggl( \int_{s}^{\frac{1}{2}}\tau ^{-\alpha}\,d\tau \biggr)\,ds \le& \int_{0}^{\frac{1}{2}}\varphi_{p}^{-1} \biggl(\frac{1}{\alpha -1}s^{-(\alpha-1)} \biggr)\,ds = \int_{0}^{\frac{1}{2}} \biggl(\frac{s^{-(\alpha-1)}}{\alpha-1} \biggr)^{\frac{1}{p-1}}\,ds \\ = & \frac{p-1}{(\alpha-1)^{\frac{1}{p-1}}(p-\alpha)}s^{\frac{p-\alpha }{p-1}} \bigg|_{0}^{\frac{1}{2}} < \infty. \end{aligned}$$
Next, we need to check that both \(f_{1}(u,v)=(u^{2}+v^{2})^{\frac {p-1}{4}}+1\) and \(f_{2}(u,v)= e^{-v^{2}}[1+(u^{2})^{\frac{p-1}{3}}]\) satisfy assumption (F). In fact, \(f_{1}(0,0)=f_{2}(0,0)=1>0\), and that is, \(\lim_{|(u,v)|\to\infty}\frac{f_{2}(u,v)}{|(u,v)|^{p-1}}=0\). Consequently, by Theorem 4.2 we see that problem (\(E_{1}\)) has at least one nontrivial solution for all \(\lambda>0\).
$$\begin{aligned}& \begin{aligned}[b] \lim_{|(u,v)|\to\infty}\frac{f_{1}(u,v)}{|(u,v)|^{p-1}} &= \lim _{|(u,v)|\to\infty}\frac{(u^{2}+v^{2})^{\frac {p-1}{4}}+1}{(u^{2}+v^{2})^{\frac{p-1}{2}}} \\ &= \lim_{|(u,v)|\to\infty} \biggl(\frac{1}{(u^{2}+v^{2})^{\frac {p-1}{4}}}+\frac{1}{(u^{2}+v^{2})^{\frac{p-1}{2}}} \biggr)=0, \end{aligned} \\& \begin{aligned}[b] 0&\leq\lim_{|(u,v)|\to\infty}\frac{f_{2}(u,v)}{|(u,v)|^{p-1}} =\lim _{|(u,v)|\to\infty}\frac{e^{-v^{2}}[1+(u^{2})^{\frac {p-1}{3}}]}{(u^{2}+v^{2})^{\frac{p-1}{2}}} \\ &\leq \lim_{|(u,v)|\to\infty} \biggl(\frac {1}{e^{v^{2}}(u^{2}+v^{2})^{\frac{p-1}{2}}}+\frac {1}{e^{v^{2}}(u^{2}+v^{2})^{\frac{p-1}{6}}} \biggr)=0. \end{aligned} \end{aligned}$$
Example 2
Consider the following p-Laplacian system with \(p=6\): where \(\mathbf{u}=(u,v)\), \(\lambda>0\) is a parameter, and \(h(t)=(h_{1}(t),h_{2}(t))\) is given by and By similar arguments as in Example 1, we can easily check that \(h\in \mathcal{H}\) and \(f_{1}\), \(f_{2}\) satisfy assumption (F). Consequently, by Theorem 4.2 we see that problem (\(E_{2}\)) has at least one nontrivial solution for all \(\lambda>0\).
$$ (E_{2})\quad\textstyle\begin{cases} -(|\mathbf{u}|^{4}u')'= \lambda h_{1}(t)[ 1- (u^{2}+v^{2})^{\frac {5}{3}}],\\ -(|\mathbf{u}|^{4}v')'= \lambda h_{2}(t)[2-e^{-(u^{2}+v^{4})}], \quad t\in (0,1),\\ u(0)=v(0)= 0 = u(1)=v(1), \end{cases} $$
$$ h_{1}(t)= \textstyle\begin{cases} t^{-2}, & t\in(0,\frac{1}{2}],\\ -1,& t\in(\frac{1}{2},1), \end{cases} $$
$$ h_{2}(t)= \textstyle\begin{cases} t^{-4},& t\in(0,\frac{1}{2}],\\ 1,& t\in(\frac{1}{2},1). \end{cases} $$
Example 3
Consider the following p-Laplacian system: where \(\mathbf{u}=(u_{1},\ldots,u_{N})\), \(\lambda>0\) is a parameter, \(h(t)=(h_{1}(t),\ldots,h_{N}(t))\) is defined by and We note that each \(h_{i}\) is not in \(L^{1}(0,1)\), \(h_{i}(\frac {1}{2})=4^{\alpha}-4^{p}<0\) for \(1<\alpha<p\), and \(h:(0,1)\to\mathbb {R}^{N}\) is locally integrable. By similar arguments as in Example 1, we can easily check that \(h\in \mathcal{H}\).
$$ (E_{3})\quad\textstyle\begin{cases} -(|\mathbf{u}|^{p-2}u_{1}')'= \lambda h_{1}(t)\ln ((u_{1}^{2}+\cdots +u_{N}^{2})^{\frac{1}{2}}+2 ),\\ \vdots\\ -(|\mathbf{u}|^{p-2}u_{N}')'= \lambda h_{N}(t)\ln ((u_{1}^{2}+\cdots +u_{N}^{2})^{\frac{1}{2}}+N+1 ),\quad t\in(0,1),\\ u_{i}(0)=0=u_{i}(1), \quad i=1,\ldots,N, \end{cases} $$
$$ h_{i}(t)=\frac{1}{t^{\alpha}(1-t)^{\alpha}}-4^{p}, \quad t\in(0,1), 1< \alpha < p, i=1,\ldots,N, $$
$$f_{i}(u_{1},\ldots,u_{N})=\ln \bigl( \bigl(u_{1}^{2}+\cdots+u_{N}^{2} \bigr)^{\frac {1}{2}}+i+1 \bigr),\quad i=1,\ldots,N. $$
Next, let us check (F) for \(f_{i}(u_{1},\ldots,u_{N})=\ln ((u_{1}^{2}+\cdots+u_{N}^{2})^{\frac{1}{2}}+i+1 )\). In fact, \(f_{i}(0,\ldots,0)= \ln(i+1)>0\), and setting \(x :=(u_{1}^{2}+\cdots +u_{N}^{2})^{\frac{1}{2}}\), we have that is, \(\lim_{|(u_{1},\ldots,u_{N})|\to\infty}\frac{f_{i}(u_{1},\ldots ,u_{N})}{|(u_{1},\ldots,u_{N})|^{p-1}}=0\) for \(i=1,\ldots,N\). Consequently, by Theorem 4.2 we see that problem (\(E_{3}\)) has at least one nontrivial solution for all \(\lambda>0\).
$$\begin{aligned} 0 \leq&\lim_{|(u_{1},\ldots,u_{N})|\to\infty}\frac{f_{i}(u_{1},\ldots ,u_{N})}{|(u_{1},\ldots,u_{N})|^{p-1}} = \lim _{|(u_{1},\ldots,u_{N})|\to\infty}\frac{\ln ((u_{1}^{2}+\cdots +u_{N}^{2})^{\frac{1}{2}}+i+1 )}{(u_{1}^{2}+\cdots+u_{N}^{2})^{\frac {p-1}{2}}} \\ =& \lim_{x\to+\infty}\frac{\ln(x+i+1)}{x^{p-1}} \\ =& \lim_{x\to+\infty} \frac{1}{x+i+1}\cdot\frac{1}{(p-1)x^{p-2}} \\ \leq& \lim_{x\to+\infty} \frac{1}{(p-1)x^{p-1}}=0, \end{aligned}$$
Declarations
Acknowledgements
The second author was supported by the National Research Foundation of Korea, Grant funded by the Korea Government (MEST) (NRF2012R1A1A2000739). The third author was supported by the National Research Foundation of Korea, Grant funded by the Korea Government (MEST) (NRF2014R1A1A2056339).
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
Authors’ Affiliations
(1)
Department of Mathematics, Pusan National University, Busan, 609-735, Republic of Korea
(2)
Department of Mathematics, University of Ulsan, Ulsan, 680-749, Republic of Korea
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