Consider the following $p$-Laplacian problem

$ \begin{equation}\label{sep} \left\{ \begin{array}{ll} -\text{div}\left( \vert \nabla u\vert^{p-2}\nabla u\right)=\lambda f(u)\, \, &\text{in}\, \, \Omega, \\ u=0 &\text{on}\, \, \partial\Omega, \end{array} \right. \end{equation} $

(1.1)

where $\lambda$ is a nonnegative parameter, $\Omega$ is a bounded domain of $\mathbb{R}^N$ with smooth boundary $\partial\Omega$, $p\in(1, +\infty)$ and $f:[0, +\infty)\rightarrow[0, +\infty)$ is some given continuous nonlinearity. We also assume that $f(s)>0$ for $s>0$ and there exist $f_0$, $f_\infty\in[0, +\infty]$ such that

$ \begin{equation} f_0=\lim\limits_{s\rightarrow 0^+}\frac{f(s)}{\varphi_p(s)}, \, \, f_{\infty}=\lim\limits_{s\rightarrow +\infty}\frac{f(s)}{\varphi_p(s)}, \nonumber \end{equation} $

where $\varphi_p(s)=\vert s\vert^{p-2}s$.

If $f_0$, $f_\infty\in(0, +\infty)$ with $f_0\neq f_\infty$, it follows from Theorem 5.1-5.2 in [1] that problem (1.1) has at least one positive solution for any $\lambda\in\left(\min\left\{\lambda_1/f_0, \lambda_1/f_\infty\right\}, \max\left\{\lambda_1/f_0, \lambda_1/f_\infty\right\}\right)$, where $\lambda_1$ is the first eigenvalue of problem (1.1) with $f(s)=\varphi_p(s)$. Here we study the cases of $f_0\not\in(0, +\infty)$ or $f_\infty\not\in(0, +\infty)$. If $f$ is superlinear, we also require that $f$ satisfies the following subcritical growth condition

$ \begin{equation} \lim\limits_{s\rightarrow +\infty}\frac{f(s)}{s^{q-1}}=C\nonumber \end{equation} $

for some $q\in\left(p, p_*\right)$ and positive constant $C$, where

$ \begin{equation} p_*=\left\{ \begin{array}{ll} \frac{(N-1)p}{N-p}\, \, &\text{if}\, \, p<N, \\ +\infty &\text{if}\, \, p\geq N \end{array} \right.\nonumber \end{equation} $

is the Serrin's exponent (see [2]).

Our main result is the following theorem.

Theorem 1.1 (a) If $f_0\in(0, +\infty)$ and $f_\infty=0$, problem (1.1) has at least one positive solution for every $\lambda\in\left(\lambda_{1}/f_0, +\infty\right)$.

(b) If $f_0\in(0, +\infty)$ and $f_\infty=+\infty$, problem (1.1) has at least one positive solution for every $\lambda\in\left(0, \lambda_{1}/f_0\right)$.

(c) If $f_0=0$ and $f_\infty\in(0, +\infty)$, problem (1.1) has at least one positive solution for every $\lambda\in\left(\lambda_{1}/f_\infty, +\infty\right)$.

(d) If $f_0=f_\infty=0$, there exists $\lambda_*>0$ such that problem (1.1) has at least two positive solutions for any $\lambda\in\left(\lambda_*, +\infty\right)$.

(e) If $f_0=0$ and $f_\infty=+\infty$, problem (1.1) has at least one positive solution for any $\lambda\in (0, +\infty)$.

(f) If $f_0=+\infty$ and $f_\infty=0$, problem (1.1) has at least one positive solution for any $\lambda\in\left(0, +\infty\right)$.

(g) If $f_0=+\infty$ and $f_\infty\in(0, +\infty)$, problem (1.1) has at least one positive solution for any $\lambda\in\left(0, \lambda_1/f_\infty\right)$.

(h) If $f_0=f_\infty=\infty$, there exists $\lambda^*>0$ such that problem (1.1) has at least two positive solutions for any $\lambda\in\left(0, \lambda^*\right)$.

We first have the following two nonexistence results.

Lemma 2.1 Assume that there exists a positive constant $\rho>0$ such that

$ \begin{equation} f(s)/\varphi_p(s)\geq \rho\nonumber \end{equation} $

for any $s>0$. Then there exists $\xi_*>0$ such that problem (1.1) has no positive solution for any $\lambda\in\left(\xi_*, +\infty\right)$.

Proof By contradiction, assume that $u_n$ $(n=1, 2, \cdots)$ are positive solutions of problem (1.1) with $\lambda=\lambda_n$ $(n=1, 2, \cdots)$ such that $\lambda_n\rightarrow+\infty$ as $n\rightarrow+\infty$. Then we have that $\lambda_n f\left(u_n\right)/\varphi_p\left(u_n\right)>\lambda_1$ for $n$ large enough. By Theorem 2.6 of [3], we know that $u_n$ must change sign in $\Omega$ for $n$ large enough, which is a contradiction.

Lemma 2.2 Assume that there exists a positive constant $\varrho>0$ such that

$ \begin{equation} f(s)/\varphi_p(s)\leq\varrho\nonumber \end{equation} $

for any $s>0$. Then there exists $\eta_*>0$ such that problem (1.1) has no positive solution for any $\lambda\in\left(0, \eta_*\right)$.

Proof Suppose, on the contrary, that there exists one positive solution $u$. Then we have that

$ \begin{eqnarray} \lambda_1\int_\Omega\vert u\vert^p\, dx\leq \int_\Omega\vert \nabla u\vert^p\, dx=\lambda \int_\Omega \frac{f\left(u\right)}{\varphi_p\left(u\right)} u^p\, dx \leq \lambda \varrho \int_\Omega u^p\, dx, \nonumber \end{eqnarray} $

which implies that $\lambda\geq \lambda_1/\varrho$.

Let

$ \begin{equation} E=\left\{u\in C^{1}(\overline{\Omega}):u=0\, \, \text{on}\, \, \partial\Omega\right\} \nonumber \end{equation} $

with the usual norm

$ \begin{equation} \Vert u\Vert=\max\limits_{\overline{\Omega}}\vert u\vert+\max\limits_{\overline{\Omega}}\left\vert \nabla u\right\vert.\nonumber \end{equation} $

Set

$ \begin{equation} \mathbb{P}:=\left\{u\in E:u> 0\, \text{ in }\, \, \Omega\, \,\text{and }\, \,\frac{\partial u}{\partial \omega}<0\, \, \text{on}\, \, \partial\Omega\right\}, \nonumber \end{equation} $

where $\omega$ is the outward pointing normal to $\partial\Omega$.

Proof of Theorem 1.1 (a) From Lemma 5.4 of [1], there exists a continuum $\mathscr{C}$ of nontrivial solutions of problem (1.1) emanating from $\left(\lambda_{1}/f_0, 0\right)$ such that $\mathscr{C}\subset (\mathbb{R}\times \mathbb{P})\cup\left\{\left(\lambda_1/f_0, 0\right)\right\}$, meets $\infty$ in $\mathbb{R}\times E$. It suffices to show that $\mathscr{C}$ joins $\left(\lambda_1/f_0, 0\right)$ to $\left(+\infty, +\infty\right)$. Lemma 2.2 implies that $\lambda>0$ on $\mathscr{C}$ and $\lambda=0$ is not the blow up point of $\mathscr{C}$.

We claim that $\mathscr{C}$ is unbounded in the direction of $E$. Suppose, by contradiction, that $\mathscr{C}$ is bounded in the direction of $E$. So there exist $\left(\lambda_n, u_n\right) \in \mathscr{C}$ and a positive constant $M$ such that $\lambda_n\rightarrow+\infty$ as $n\rightarrow+\infty$ and $\left\Vert u_n\right\Vert\leq M$ for any $n\in \mathbb{N}$. It follows that $f\left(u_n\right)/u_n\geq\delta$ for some positive constant $\delta$ and all $n\in \mathbb{N}$. Lemma 2.1 implies that $u_n\equiv 0$ for $n$ large enough, which is a contradiction. Lemma 5.1 of [4] implies that the unique blow up point of $\mathscr{C}$ is $\lambda=+\infty$. Now the desired conclusion can be got immediately from the global structure of $\mathscr{C}$.

(b) It is enough to show that $\mathscr{C}$ joins $\left(\lambda_1/f_0, 0\right)$ to $\left(0, +\infty\right)$. Lemma 2.1 implies that $\mathscr{C}$ is bounded in the direction of $\lambda$. By virtue of Lemma 5.1 of [4], we know that $(0, +\infty)$ is the unique blow up point of $\mathscr{C}$.

(c) If $(\lambda, u)$ is any solution of (1.1) with $\Vert u\Vert\not\equiv 0$, dividing (1.1) by $\Vert u\Vert^{2(p-1)}$ and setting $w=u/\Vert u\Vert^2$ yield

$ \begin{equation}\label{C4} \left\{ \begin{array}{ll} -\text{div}\left( \left\vert \nabla w\right\vert^{p-2}\nabla w\right)=\lambda \frac{f(u)}{\Vert u\Vert^{2(p-1)}}\, \, &\text{in}\, \, \Omega, \\ w=0 &\text{on}\, \, \partial\Omega. \end{array} \right. \end{equation} $

(2.1)

Define

$ \begin{eqnarray} \widetilde{f}(w)=\left\{ \begin{array}{lll} \Vert w\Vert^{2(p-1)}f\left(\frac{w}{\Vert w\Vert^2}\right)\, \, &\text{if}\, w\neq 0, \\ 0 &\text{if}\, \, w=0. \end{array} \right.\nonumber \end{eqnarray} $

Then (2.1) is equivalent to

$ \begin{equation} \left\{ \begin{array}{ll} -\text{div}\left( \left\vert \nabla w\right\vert^{p-2}\nabla w\right)=\lambda \widetilde{f}(w)\, \, &\text{in}\, \, \Omega, \\ w=0 &\text{on}\, \, \partial\Omega. \end{array} \right.\nonumber \end{equation} $

By doing some simple calculations, we can show that $\widetilde{f}_0=f_\infty$ and $\widetilde{f}_\infty=f_0$. Applying the conclusion of (a) and the inversion $w\rightarrow w/\Vert w\Vert^2=u$, we obtain the desired conclusion.

(d) Define

$ \begin{equation} f^n(s)=\left\{ \begin{array}{ll} \varphi_p(s)/n, \, \, & s\in\left[0, 1/n\right], \\ \left(f\left(2/n\right)-1/n^{p}\right)ns+2/n^{p}-f\left(2/n\right), &s\in\left(1/n, 2/n\right), \\ f(s), &s\in\left[2/n, +\infty\right) \end{array} \right.\nonumber \end{equation} $

and consider the following problem

$ \begin{equation} \left\{ \begin{array}{ll} -\text{div}\left( \left\vert \nabla u\right\vert^{p-2}\nabla u\right)=\lambda f^n(u)\, \, &\text{in}\, \, \Omega, \\ u=0 &\text{on}\, \, \partial\Omega. \end{array} \right.\nonumber \end{equation} $

It is easy to see that $\lim_{n\rightarrow+\infty}f^n(s)=f(s)$, $f_0^n=1/n$ and $f_\infty^n=f_\infty=0$. From the conclusion of (a), we obtain a sequence of unbounded continua $\mathscr{C}_n$ emanating from $\left(n\lambda_1, 0\right)$ and joining to $\left(+\infty, +\infty\right):=z_*$. Let $\mathscr{C}=\mathop {\lim \sup }\limits_{n \to + \infty } \mathscr{C}_n$. For any $(\lambda, u)\in \mathscr{C}$, the definition of superior limit (see [5]) shows that there exists a sequence $\left(\lambda_n, u_n\right)\in \mathscr{C}_n$ such that $\left(\lambda_n, u_n\right)\rightarrow (\lambda, u)$ as $n\rightarrow+\infty$. Then a continuity argument shows that $u$ is a solution of problem (1.1).

From Proposition 2 of [4], for each $\epsilon> 0$ there exists an $N_0$ such that for every $n > N_0$, $\mathscr{C}_n\subset V_\epsilon\left(\mathscr{C}\right)$ with $V_\epsilon\left(\mathscr{C}\right)$ denoting the $\epsilon$-neighborhood of $\mathscr{C}$. It follows that $ \left(n\lambda_1, +\infty\right)\subseteq \text{Proj}\left(\mathscr{C}_n\right)\subseteq \text{Proj}\left(V_\epsilon\left(\mathscr{C}\right)\right), $ where $\text{Proj}\left(\mathscr{C}_n\right)$ denotes the projection of $\mathscr{C}_n$ on $\mathbb{R}$. So we have that $\left(n\lambda_1+\epsilon, +\infty\right)\subseteq \text{Proj}\left(\mathscr{C}\right)$ for any $n > N_0$. Hence, we have $\mathscr{C}\setminus\{\infty\}\neq\emptyset$.

Let

$ \begin{equation} S_1=\left\{(+\infty, u):0<\Vert u\Vert<+\infty\right\}.\nonumber \end{equation} $

For any fixed $n\in \mathbb{N}$, we claim that $\mathscr{C}_n\cap S_1=\emptyset$. Otherwise, there exists a sequence $\left(\lambda_m, u_m\right)\in \mathscr{C}_n$ such that $\left(\lambda_m, u_m\right)\rightarrow\left(+\infty, u_*\right)\in S_1$ with $\left \Vert u_*\right\Vert<+\infty$. It follows that $\left\Vert u_m\right\Vert\leq M_n$ for some constant $M_n>0$. It implies that $f^n\left(u_m\right)/u_m\geq\delta_n$ for some positive constant $\delta_n$ and all $m\in \mathbb{N}$. Lemma 2.1 implies that $u_m\equiv 0$ for $m$ large enough, which contradicts the fact of $\left \Vert v_*\right\Vert>0$. It follows that $\left(\cup_{n=1}^{+\infty}\mathscr{C}_n\right)\cap S_1=\cup_{n=1}^{+\infty}\left(\mathscr{C}_n\cap S_1\right)=\emptyset$. Since ${\mathscr{C}}\subseteq \left(\cup_{n=1}^{+\infty}\mathscr{C}_n\right)$, one has that ${\mathscr{C}}\cap S_1=\emptyset$. Furthermore, set

$ \begin{equation} S_2:=\left\{(\lambda, +\infty):0\leq\lambda<+\infty\right\}.\nonumber \end{equation} $

For any fixed $n\in \mathbb{N}$, by $f_\infty=0$ and an argument similar to that of (a), we have that $\mathscr{C}_n \cap S_2=\emptyset$. Reasoning as the above, we have that $\mathscr{C} \cap S_2=\emptyset$. Hence, $\mathscr{C}\cap \left(S_1\cup S_2\right)=\emptyset$. Taking $z^*=(+\infty, 0)$, we have $z^*\in \mathop {\lim \inf }\limits_{n \to + \infty } \mathscr{C}_n$ with $\left\Vert z^*\right\Vert_{\mathbb{R}\times E}=+\infty$. Therefore, we obtain that $\mathscr{C}\cap\{\infty\}=\left\{z_*, z^*\right\}$. Clearly, $\left(\cup_{n=1}^{+\infty} \mathscr{C}_n\right)\cap \overline{B}_R$ is pre-compact. So Lemma 3.1 of [6] implies that $\mathscr{C}$ is connected. By an argument similar to that of Theorem 1.3 of [4], we can show that $\mathscr{C}\cap([0, +\infty)\times\{0\})=\emptyset$. Now the desired conclusion can be deduced from the global structure of $\mathscr{C}$.

(e) By an argument similar to that of (d), in view of the conclusion of (b), we can get the desired conclusion.

(f) By an argument similar to that of (c) and the conclusion of (e), we can prove it.

(g) By an argument similar to that of (c) and the conclusion of (b), we can obtain it.

(h) Define

$ \begin{eqnarray} f^n(s)=\left\{ \begin{array}{lll} n\varphi_p(s), \, \, \quad\quad\quad\quad \quad\quad\quad\quad\, \, \quad & s\in\left[0, 1/n\right], \\ n\left(f\left(2/n\right)-1/n^{p-2}\right)\left(s-1/n\right)+1/n^{p-2}, \, \, & s\in \left(1/n, 2/n\right), \\ f(s), \quad\quad\quad\quad\quad\, \quad\quad\quad\quad\quad\, & s\in\left[2/n, +\infty\right). \end{array} \right.\nonumber \end{eqnarray} $

Clearly, we have that $\mathop {\lim }\limits_{n \to + \infty } f^n(s)=f(s)$, $f_0^n=n$ and $f_\infty^n=f_\infty=+\infty$. The conclusion of (b) implies that there exists a sequence unbounded continua $\mathscr{C}_n$ emanating from $\left(\lambda_1/n, 0\right)$ and joining to $\left(0, +\infty\right)$. Taking $z^*=\left(0, 0\right)$, then $z^*\in \mathop {\lim \inf }\limits_{n \to + \infty } \mathscr{C}_n$. Lemma 2.1 of [4] implies that $\mathscr{C}=\mathop {\lim \sup }\limits_{n \to + \infty } \mathscr{C}_n$ is unbounded and connected such that $z^*\in \mathscr{C}$ and $\left(0, +\infty\right)\in \mathscr{C}$. We claim that $\mathscr{C}\cap((0, +\infty]\times\{0\})=\emptyset$. If there exists a sequence $\left\{\left(\lambda_n, u_n\right)\right\}$ with $u_n\in \mathbb{P}$ such that $\underset{n\rightarrow +\infty}{\lim}\lambda_n=\mu>0$ and $\underset{n\rightarrow +\infty}{\lim}\left\Vert u_n\right\Vert=0$ as $n\rightarrow+\infty$. $f_0=+\infty$ implies that $ \lambda_n\frac{f\left(u_n\right)}{u_n^{p-1}}>\lambda_1$ for $n$ large enough. By Theorem 2.6 of [3], we know that $u_n$ must change its sign for $n$ large enough, which is a contradiction.