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
$$ u= G(\lambda,u). $$
(4.1)
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.
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
$$\|u\|_{X} = \max_{0\le t \le1} \bigl|\bigl(u_{1}(t), \ldots, u_{N}(t)\bigr)\bigr| = \max_{0\le t \le1} \bigl(u_{1}^{2}(t)+ \cdots+ u_{N}^{2}(t) \bigr)^{1/2}. $$
Indeed, it is easy to see that
$$\|u\|_{X} \le\|u\|_{\infty}\le N \|u\|_{X}. $$
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
$$\begin{aligned} G(\lambda,u) (t) = \textstyle\begin{cases} \int_{0}^{t} \Psi_{p}^{-1} (a(\lambda h\cdot f(u))+ \int_{s}^{\frac{1}{2}} \lambda h(\tau)\cdot f(u(\tau))\,d\tau )\,ds, & t\in[0,\frac{1}{2}],\\ \int_{t}^{1} \Psi_{p}^{-1} (-a(\lambda h\cdot f(u)) + \int_{\frac{1}{2}}^{s} \lambda h(\tau)\cdot f(u(\tau))\,d\tau )\,ds, & t \in[\frac{1}{2},1]. \end{cases}\displaystyle \end{aligned}$$
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 \(\lambda>0\).
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 \(n\in\mathbb{N}\),
$$u_{n}=G(\lambda_{n}, u_{n}) $$
with \(\|u_{n}\|_{X} \to\infty\) as \(n\to\infty\).
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:
$$\begin{aligned} \bigl|u_{n}(t)\bigr| =& \biggl| \int_{0}^{t}\Psi_{p}^{-1} \biggl(a\bigl(\lambda_{n} h\cdot f(u_{n})\bigr)+ \int_{s}^{\frac{1}{2}} \lambda_{n} h(\tau)\cdot f \bigl(u_{n}(\tau)\bigr)\,d\tau \biggr)\,ds\biggr| \\ \le& \int_{0}^{t}\biggl|\Psi_{p}^{-1} \biggl(a\bigl(\lambda_{n} h\cdot f(u_{n})\bigr)+ \int_{s}^{\frac{1}{2}} \lambda_{n} h(\tau)\cdot f \bigl(u_{n}(\tau)\bigr)\,d\tau \biggr)\biggr|\,ds \\ \le& \int_{0}^{t}\varphi_{p}^{-1} \biggl(\bigl|a\bigl(\lambda_{n} h\cdot f(u_{n})\bigr)\bigr|+ \biggl| \int_{s}^{\frac{1}{2}} \lambda_{n} h(\tau)\cdot f \bigl(u_{n}(\tau)\bigr)\,d\tau\biggr| \biggr)\,ds \\ \le& \varphi_{p}^{-1}(\lambda_{n}) \int_{0}^{\frac{1}{2}}\varphi _{p}^{-1} \biggl(\bigl|a\bigl(h\cdot f(u_{n})\bigr)\bigr|+ \biggl| \int_{s}^{\frac{1}{2}} h(\tau)\cdot f\bigl(u_{n}( \tau)\bigr)\,d\tau\biggr| \biggr)\,ds \\ \le& \varphi_{p}^{-1}(\Lambda) \int_{0}^{\frac{1}{2}}\varphi_{p}^{-1} \biggl(\frac{|a(h\cdot f(u_{n}))|}{\|u_{n}\|_{X}^{p-1}}+ \frac{|\int_{s}^{\frac{1}{2}} h(\tau)\cdot f(u_{n}(\tau))\,d\tau|}{\|u_{n}\|_{X}^{p-1}} \biggr)\,ds\|u_{n} \|_{X} \end{aligned}$$
for all \(t\in[0,\frac{1}{2}]\). By the homogeneity of a again, we get
$$\bigl|u_{n}(t)\bigr|\le \varphi_{p}^{-1}(\Lambda) \int_{0}^{\frac{1}{2}}\varphi _{p}^{-1} \biggl(\biggl|a\biggl(h\cdot\frac{f(u_{n})}{\|u_{n}\|_{X}^{p-1}}\biggr)\biggr|+ \int_{s}^{\frac{1}{2}} \bigl|h(\tau)\bigr| \frac{|f(u_{n}(\tau))|}{\|u_{n}\|_{X}^{p-1}}\,d\tau \biggr)\,ds\|u_{n}\|_{X}. $$
By (F), for any \(\varepsilon>0\), there exists \(l_{\epsilon}>0\) such that for all \(s\in\mathbb{R}^{N}\) with \(|s|\ge l_{\epsilon}\),
$$ \bigl|f_{i}(s)\bigr| \le\varepsilon|s|^{p-1} \quad \mbox{for } i=1,\ldots,N. $$
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
$$ \bigl|f_{i}(s)\bigr| \le M_{\epsilon}$$
on \(\{s \in\mathbb{R}^{N} \mid |s| \le l_{\epsilon}\}\) for \(i=1,\ldots,N\). Thus, we have
$$ \bigl|f_{i}(s)\bigr| \le\varepsilon|s|^{p-1}+M_{\epsilon}\quad\mbox{for all } s \in \mathbb{R}^{N} , i=1,\ldots,N. $$
(4.2)
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
$$\|u_{n}\|_{X}\geq\biggl(\frac{M_{\epsilon}}{\epsilon} \biggr)^{\frac{1}{p-1}}, $$
that is,
$$\frac{1}{\|u_{n}\|_{X}^{p-1}}\leq\frac{\epsilon}{M_{\epsilon}}. $$
Using (4.2), we get that, for any \(n\geq n_{\epsilon}\) and \(t\in[0,1/2]\),
$$\frac{|f_{i}(u_{n}(t))|}{\|u_{n}\|_{X}^{p-1}}\leq\epsilon\cdot\frac {|u_{n}(t)|^{p-1}}{\|u_{n}\|_{X}^{p-1}}+\frac{M_{\epsilon}}{\|u_{n}\| _{X}^{p-1}}\leq \epsilon+M_{\epsilon}\cdot\frac{\epsilon}{M_{\epsilon}}=2\epsilon $$
and
$$ \frac{\|f(u_{n})\|_{X}}{\|u_{n}\|_{X}^{p-1}}\leq\frac{\|f(u_{n})\|_{\infty}}{\| u_{n}\|_{X}^{p-1}}=\frac{\sum_{i=1}^{N}\|f_{i} (u_{n})\|_{\infty}}{\|u_{n}\| _{X}^{p-1}}\leq N \cdot2\epsilon=2\epsilon N. $$
(4.3)
Take
$$B=\biggl\{ \frac{f(u_{n})}{\|u_{n}\|_{X}^{p-1}}\biggr\} _{n\geq n_{\epsilon}}. $$
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
$$\biggl|a\biggl(h\cdot\frac{f(u_{n})}{\|u_{n}\|_{X}^{p-1}}\biggr)\biggr| \leq C_{\epsilon}\quad\mbox{for any } n\geq n_{\epsilon}. $$
Therefore, for \(t\in[0,\frac{1}{2}]\), we obtain
$$\begin{aligned} \bigl|u_{n}(t)\bigr| \leq{}&\biggl[\varphi_{p}^{-1}( \Lambda) \int_{0}^{\frac{1}{2}}\varphi _{p}^{-1} \biggl(C_{\epsilon}+2\epsilon \int_{s}^{\frac{1}{2}} \bigl|h(\tau)\bigr|\,d\tau \biggr)\,ds\biggr] \|u_{n}\|_{X} \\ \leq{}& \biggl[\frac{1}{2}\varphi_{p}^{-1}( \Lambda)C_{p} \varphi_{p}^{-1}(C_{\epsilon}) \\ &{}+\varphi_{p}^{-1}(\Lambda)C_{p} \varphi_{p}^{-1}(2\epsilon) \int _{0}^{\frac{1}{2}}\varphi_{p}^{-1} \biggl( \int_{s}^{\frac{1}{2}} \bigl|h(\tau)\bigr|\,d\tau \biggr)\,ds\biggr] \|u_{n}\|_{X}. \end{aligned}$$
(4.4)
By similar arguments, for \(t\in[\frac{1}{2},1]\), we obtain
$$\begin{aligned} \bigl|u_{n}(t)\bigr| \leq{}&\biggl[\varphi_{p}^{-1}( \Lambda) \int_{\frac{1}{2}}^{1}\varphi _{p}^{-1} \biggl(C_{\epsilon}+2\epsilon \int_{\frac{1}{2}}^{s} \bigl|h(\tau)\bigr|\,d\tau \biggr)\,ds\biggr] \|u_{n}\|_{X} \\ \leq{}& \biggl[\frac{1}{2}\varphi_{p}^{-1}( \Lambda)C_{p} \varphi_{p}^{-1}(C_{\epsilon}) \\ &{}+\varphi_{p}^{-1}(\Lambda)C_{p} \varphi_{p}^{-1}(2\epsilon) \int_{\frac {1}{2}}^{1}\varphi_{p}^{-1} \biggl( \int_{\frac{1}{2}}^{s} \bigl|h(\tau)\bigr|\,d\tau\biggr)\,ds\biggr]\| u_{n}\|_{X}. \end{aligned}$$
(4.5)
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
$$\frac{1}{2}\varphi_{p}^{-1}(\Lambda)C_{p} \varphi_{p}^{-1}(C_{\epsilon}) +\varphi_{p}^{-1}( \Lambda)C_{p} \varphi_{p}^{-1}(2\epsilon) C_{h} \leq\frac{1}{2}. $$
Consequently, combining (4.4) and (4.5), we obtain, for \(t\in[0,1]\),
$$\bigl|u_{n}(t)\bigr|\leq\frac{1}{2}\|u_{n}\|_{X}. $$
This implies that
$$\|u_{n}\|_{X}\leq0 \quad\mbox{for } n\geq n_{\epsilon}, $$
which contradicts
$$\|u_{n}\|_{X}\geq\biggl(\frac{M_{\epsilon}}{\epsilon} \biggr)^{\frac{1}{p-1}}>0 \quad\mbox{for } n\geq n_{\epsilon}$$
and this completes the proof. □
Example 1
Consider the following p-Laplacian system:
$$ (E_{1})\quad\textstyle\begin{cases} -(|\mathbf{u}|^{p-2}u')'= \lambda h_{1}(t)[ (u^{2}+v^{2})^{\frac {p-1}{4}}+1],\\ -(|\mathbf{u}|^{p-2}v')'= \lambda h_{2}(t)e^{-v^{2}}[1+(u^{2})^{\frac {p-1}{3}}],\quad t\in(0,1),\\ u(0)=v(0)= 0 = u(1)=v(1), \end{cases} $$
where \(\mathbf{u}=(u,v)\), \(\lambda>0\) is a parameter, and \(h(t)=(h_{1}(t),h_{2}(t))\) is given by
$$\begin{aligned}& h_{1}(t)= \textstyle\begin{cases} t^{-\alpha}, & t\in(0,\frac{1}{2}],\\ -1, & t\in(\frac{1}{2},1), 1< \alpha< p, \end{cases}\displaystyle \qquad h_{2}(t)=-1, \quad t\in(0,1). \end{aligned}$$
We note that \(h\in L^{1} _{\mathrm{loc}}\) but \(h_{1}\notin L^{1}\). We now show that \(h\in\mathcal{H}\). Indeed,
$$\begin{aligned} \int_{s}^{\frac{1}{2}}\tau^{-\alpha}\,d\tau = & - \frac{1}{\alpha-1}\tau^{-(\alpha-1)} \bigg|_{s}^{\frac {1}{2}}=- \frac{1}{\alpha-1} \biggl[\biggl(\frac{1}{2}\biggr)^{-(\alpha-1)}-s^{-(\alpha-1)} \biggr] \\ = & \frac{1}{\alpha-1} \bigl[s^{-(\alpha-1)}-2^{\alpha-1}\bigr]\le \frac{1}{\alpha-1}s^{-(\alpha-1)}. \end{aligned}$$
Since \(1<\alpha<p\), we have \(\frac{1}{\alpha-1}s^{-(\alpha-1)}>0\) for \(s\in(0,1)\) and
$$\begin{aligned} \int_{0}^{\frac{1}{2}}\varphi_{p}^{-1} \biggl( \int_{s}^{\frac{1}{2}}\tau ^{-\alpha}\,d\tau \biggr)\,ds \le& \int_{0}^{\frac{1}{2}}\varphi_{p}^{-1} \biggl(\frac{1}{\alpha -1}s^{-(\alpha-1)} \biggr)\,ds = \int_{0}^{\frac{1}{2}} \biggl(\frac{s^{-(\alpha-1)}}{\alpha-1} \biggr)^{\frac{1}{p-1}}\,ds \\ = & \frac{p-1}{(\alpha-1)^{\frac{1}{p-1}}(p-\alpha)}s^{\frac{p-\alpha }{p-1}} \bigg|_{0}^{\frac{1}{2}} < \infty. \end{aligned}$$
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 \(h\in\mathcal{H}\).
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
$$\begin{aligned}& \begin{aligned}[b] \lim_{|(u,v)|\to\infty}\frac{f_{1}(u,v)}{|(u,v)|^{p-1}} &= \lim _{|(u,v)|\to\infty}\frac{(u^{2}+v^{2})^{\frac {p-1}{4}}+1}{(u^{2}+v^{2})^{\frac{p-1}{2}}} \\ &= \lim_{|(u,v)|\to\infty} \biggl(\frac{1}{(u^{2}+v^{2})^{\frac {p-1}{4}}}+\frac{1}{(u^{2}+v^{2})^{\frac{p-1}{2}}} \biggr)=0, \end{aligned} \\& \begin{aligned}[b] 0&\leq\lim_{|(u,v)|\to\infty}\frac{f_{2}(u,v)}{|(u,v)|^{p-1}} =\lim _{|(u,v)|\to\infty}\frac{e^{-v^{2}}[1+(u^{2})^{\frac {p-1}{3}}]}{(u^{2}+v^{2})^{\frac{p-1}{2}}} \\ &\leq \lim_{|(u,v)|\to\infty} \biggl(\frac {1}{e^{v^{2}}(u^{2}+v^{2})^{\frac{p-1}{2}}}+\frac {1}{e^{v^{2}}(u^{2}+v^{2})^{\frac{p-1}{6}}} \biggr)=0. \end{aligned} \end{aligned}$$
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 \(\lambda>0\).
Example 2
Consider the following p-Laplacian system with \(p=6\):
$$ (E_{2})\quad\textstyle\begin{cases} -(|\mathbf{u}|^{4}u')'= \lambda h_{1}(t)[ 1- (u^{2}+v^{2})^{\frac {5}{3}}],\\ -(|\mathbf{u}|^{4}v')'= \lambda h_{2}(t)[2-e^{-(u^{2}+v^{4})}], \quad t\in (0,1),\\ u(0)=v(0)= 0 = u(1)=v(1), \end{cases} $$
where \(\mathbf{u}=(u,v)\), \(\lambda>0\) is a parameter, and \(h(t)=(h_{1}(t),h_{2}(t))\) is given by
$$ h_{1}(t)= \textstyle\begin{cases} t^{-2}, & t\in(0,\frac{1}{2}],\\ -1,& t\in(\frac{1}{2},1), \end{cases} $$
and
$$ h_{2}(t)= \textstyle\begin{cases} t^{-4},& t\in(0,\frac{1}{2}],\\ 1,& t\in(\frac{1}{2},1). \end{cases} $$
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 \(\lambda>0\).
Example 3
Consider the following p-Laplacian system:
$$ (E_{3})\quad\textstyle\begin{cases} -(|\mathbf{u}|^{p-2}u_{1}')'= \lambda h_{1}(t)\ln ((u_{1}^{2}+\cdots +u_{N}^{2})^{\frac{1}{2}}+2 ),\\ \vdots\\ -(|\mathbf{u}|^{p-2}u_{N}')'= \lambda h_{N}(t)\ln ((u_{1}^{2}+\cdots +u_{N}^{2})^{\frac{1}{2}}+N+1 ),\quad t\in(0,1),\\ u_{i}(0)=0=u_{i}(1), \quad i=1,\ldots,N, \end{cases} $$
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
$$ h_{i}(t)=\frac{1}{t^{\alpha}(1-t)^{\alpha}}-4^{p}, \quad t\in(0,1), 1< \alpha < p, i=1,\ldots,N, $$
and
$$f_{i}(u_{1},\ldots,u_{N})=\ln \bigl( \bigl(u_{1}^{2}+\cdots+u_{N}^{2} \bigr)^{\frac {1}{2}}+i+1 \bigr),\quad i=1,\ldots,N. $$
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 \(h\in \mathcal{H}\).
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
$$\begin{aligned} 0 \leq&\lim_{|(u_{1},\ldots,u_{N})|\to\infty}\frac{f_{i}(u_{1},\ldots ,u_{N})}{|(u_{1},\ldots,u_{N})|^{p-1}} = \lim _{|(u_{1},\ldots,u_{N})|\to\infty}\frac{\ln ((u_{1}^{2}+\cdots +u_{N}^{2})^{\frac{1}{2}}+i+1 )}{(u_{1}^{2}+\cdots+u_{N}^{2})^{\frac {p-1}{2}}} \\ =& \lim_{x\to+\infty}\frac{\ln(x+i+1)}{x^{p-1}} \\ =& \lim_{x\to+\infty} \frac{1}{x+i+1}\cdot\frac{1}{(p-1)x^{p-2}} \\ \leq& \lim_{x\to+\infty} \frac{1}{(p-1)x^{p-1}}=0, \end{aligned}$$
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 \(\lambda>0\).