On a class of one-dimensional superlinear semipositone $(p, q)$ -Laplacian problem

1.   Introduction

In this paper, we investigate positive solutions for the one-dimensional BVP

where $\varepsilon \geq 0, \ \phi _{\varepsilon }(s) = |s|^{p-2}s+\varepsilon |s|^{q-2}s, p > q > 1, a, b, c, d\ $ are nonnegative constants with $ac+ad+bc >$ $0$, $f:(0, 1)\times \lbrack 0, \infty)\rightarrow \mathbb{R},$ $g:(0, \infty)\rightarrow \lbrack 0, \infty), \ $ and $\lambda$ is a nonnegative parameter.

For $\varepsilon = 0, \ -(\phi _{\varepsilon }(u^{\prime }))^{\prime }$ is the usual $p$-Laplacian while for $\varepsilon > 0,$ the operator is referred to as the $(p, q)$ -Laplacian. We are focusing on the case when $f$ ($\cdot, u)\ $ is $p\,$-superlinear, and $g\ $ is allowed to exhibit semipositone structure i.e., $-g(0^{+})\in \lbrack -\infty, 0).$ For a rich literature on semipositone problems and their applications, see ^[9]. Using Amann's Fixed Point Theorem, we shall establish here the existence of a positive solution to (1.1) for $\lambda \geq 0$ small when $f(\cdot, u)$ is $p$-superlinear at $0$ and $\infty,$ and the superlinearity is involved with the first eigenvalue of the p-Laplacian operator when $\varepsilon = 0.\ $ Our result in the p-Laplacian case improves previous ones in ^{[3,4,8,10,12]} (see Remark 1.1 below), while producing a new existence criteria in the (p, q)-Laplacian case. We refer to ^[6,7,13] and the references therein for related existence results to (1.1) in the superlinear/sublinear cases when $\varepsilon = 0.$

Let $\lambda _{1}$ be the principal eigenvalue of $-(\phi _{0}(u^{\prime }))^{\prime }$ on $(0, 1)$ with Sturm-Liouville boundary condition in (1.1), (see ^[2,5]).

We consider the following hypotheses:

(A1) $g:(0, \infty)\rightarrow \lbrack 0, \infty)$ is continuous, non-increasing, and integrable near $0.$

(A2) $f:(0, 1)\times \lbrack 0, \infty)\rightarrow \mathbb{R}$ is a Carath éodory function, and there exists $\gamma \in L^{1}(0, 1)$ such that

for a.e. $t\in (0, 1).$

(A3) $\ \sup\limits_{z\in (0, c)}|f(t, z)|$ is integrable on $(0, 1)$ for all $c > 0.$

(A4)There exists a number $\sigma > 0$ such that

for $z\in (0, \sigma]$ and a.e. $t\in (0, 1)$, and in addition $f(t, z)\not\equiv \lambda _{1}z^{p-1}$ on any subinterval of $[0, \sigma]$ if $\varepsilon = 0.$

(A5) $\lim\limits_{z\rightarrow \infty }\frac{f(t, z)}{z^{p-1}} = \infty$ if $\varepsilon > 0,$ and $\lim\limits_{z\rightarrow \infty }\inf \frac{f(t, z)}{ z^{p-1}} > \lambda _{1}$ if $\varepsilon = 0, \ $ where the limits are uniform for a.e. $t\in (0, 1).$

Let $p(t) = \min (t, 1-t).\ $ By a positive solution of (1.1), we mean a function $u\in C^{1}[0, 1]$ with $\inf\limits_{(0, 1)}\frac{u}{p} > 0$ and satisfying (1.1).

Our main result is

Theorem 1.1. Let (A1)–(A5) hold. Then there exists a number $\lambda _{0} > 0$ such that (1.1) has a positive solution for $0\leq \lambda < \lambda _{0}.$

Remark 1.1. (i) When $\varepsilon = 0,$ the existence of a positive solution to (1.1) was established in ^[3,4], where $g\equiv 0$ with Sturm-Liouville condition in ^[3], and $g(u) = u^{-\delta }, \delta \in (0, 1)$ with Dirichlet boundary condition in ^[4], under the assumption

The results in ^[3,4] provided extensions of the work in ^[8,10,12]. Note that our condition (A4) allows the case $\limsup\limits_{z\rightarrow 0^{+}}\frac{f(t, z)}{z^{p-1}} = \lambda _{1}.$

(ii) In the case $bd = 0,$ the proof of Theorem 1.1 shows that when $\varepsilon = 0, \ $ (A4) can be replaced by the weaker condition $f(t, z)\leq \lambda _{1}z^{p-1}$ and $f(t, z)\not\equiv \lambda _{1}z^{p-1}$ on $[0, \sigma]$ for a.e. $t\in (0, 1).$

Example 1.1. Let $\delta, \nu \in (0, 1)$ with $\delta +\nu < 1,$ and $r > p-1.\ $ By Theorem 1.1, the following problems have a positive solution for $\lambda \geq 0$ small:

(i)

where $\varepsilon > 0$ and $r > s > p-1.\ $ Indeed, here

i.e., (A4) holds. Since

for $z\in (0, \infty), \ $ (A2) holds. Clearly (A1), (A3), and (A5) are satisfied.

(ii)

where $\alpha \in (0, r-p+1).\ $ Note that (A4) with $\varepsilon = 0$ is equivalent to

on $[0, \sigma]\ $ and $\lambda _{1}(1-e^{-z^{\alpha }})\not\equiv z^{r-(p-1)}$ on any subinterval of $[0, \sigma]$ for some $\sigma > 0.$ This is true since $\lim\limits_{z\rightarrow 0^{+}}\frac{1-e^{-z^{\alpha }}}{z^{r-p+1}} = \infty.$ Clearly the remaining conditions are satisfied.

Note that $\limsup\limits_{z\rightarrow 0^{+}}\frac{f(t, z) }{z^{p-1}} = \lambda _{1}$ in both examples.

2.   Preliminaries

Let $0\leq \alpha < \beta \leq 1.\ $ In what follows, $\gamma \in L^{1}(\alpha, \beta)$ with $\gamma \geq 0$ and we shall denote the norm in $L^{q}(\alpha, \beta)$ and $C^{1}[\alpha, \beta]$ by $||\cdot ||_{q}$ and $|\cdot |_{1}$ respectively.

Lemma 2.1. Let $u, v\in C^{1}[\alpha, \beta]\ $ satisfy

Then $u\geq v$ on $[\alpha, \beta].$

Proof. Suppose $u(t_{0}) < v(t_{0})$ for some $t_{0}\in (\alpha, \beta).$ Let $I = (\alpha _{0}, \beta _{0})\subset (\alpha, \beta)$ be the largest open interval containing $t_{0}$ such that $u < v$ on $I.$ Then $u(\alpha _{0}) = v(\alpha _{0})$ if $\alpha _{0} > \alpha$ and $u(\beta _{0}) = v(\beta _{0})$ if $\beta _{0} < \beta.$ Multiplying the inequation in (2.1) by $\ u-v$ and integrating on I gives

since $\gamma \geq 0$ and $-(\phi _{\varepsilon }(u^{\prime })-\phi _{\varepsilon }(v^{\prime })(u-v)|_{\alpha _{0}}^{\beta _{0}}\geq 0$ $\ $ in view of the boundary conditions at $\alpha, \beta.$ Since $\phi _{\varepsilon }$ is increasing, it follows that $u^{\prime } = v^{\prime }$ on $I$ and hence $u = v+\sigma$ on $I,$ where $\sigma$ is a negative constant. If $\alpha _{0} > \alpha$ or $\beta _{0} < \beta$ then $\sigma = 0,$ a contradiction. On the other hand, if $\alpha _{0} = \alpha$ and $\beta _{0} = \beta$ then the boundary conditions in (2.1) gives $a\sigma, c\sigma \geq 0$ and thus $a = c = 0,$ a contradiction and hence the result follows. □

Lemma 2.2. Let $k\in L^{1}(\alpha, \beta)$. Then the problem

has a unique solution $z\equiv T_{\varepsilon }k\in C^{1}[\alpha, \beta]$ with

where the constant $K$ is independent of $k, \alpha, \beta, z, \varepsilon.\ $ In addition, the map $T_{\varepsilon }:L^{1}(\alpha, \beta)\rightarrow C[\alpha, \beta]$ is completely continuous.

Proof. Suppose first that $\gamma \equiv 0.$

By integrating, we see that the solution of (2.2) is given by

where the constants $C, C_{1}$ satisfy

Note that (2.5) has a unique solution $(C, C_{1})\;$ since if $a = 0$ then $C = 0$ and

while if $a > 0$ then $C_{1} = -\frac{b}{a}\phi _{\varepsilon }^{-1}(C),$ where $C$ is the unique solution of

Indeed, $g_{\varepsilon }(C) > 0$ for $C > ||k||_{1}$ and $g_{\varepsilon }(C) < 0$ for $C < -||k||_{1}.$ Thus (2.7) has a unique solution $C$ with $|C|\leq ||k||_{1}$ since $g_{\varepsilon }$ is continuous and increasing.

Using the inequality (see Proposition A(ii) in Appendix)

for $m\geq 1$, $x\geq 0,$ and (2.4)–(2.6), we get

for all $t\in \lbrack \alpha, \beta], \ $ where $c_{0} = (d/c+1)$ if $a = 0$, $c_{0} = b/a$ if $a > 0,$ from which (2.3) follows.

Next, we consider the general case $\gamma \in L^{1}(\alpha, \beta)$ with $\gamma \geq 0.$ In view of the above, there exist $z_{1}, z_{2}\in C^{1}[\alpha, \beta]$ satisfying

with Sturm-Liouville boundary conditions.

By Lemma 2.1, $z_{1}\leq 0\leq z_{2}$ on $(\alpha, \beta),$ which implies

and

i.e., $(z_{1}, z_{2})$ is a pair of sub- and supersolution of (2.2) with $z_{1}\leq z_{2}$ on $(\alpha, \beta).$ Thus (2.2) has a solution $z\in C^{1}[\alpha, \beta]$ with $z_{1}\leq z\leq z_{2}$ on $(\alpha, \beta).$ The solution is unique due to Lemma 2.1.

Since

and $||z||_{\infty }\leq \max (||z_{1}||_{\infty }, ||z_{2}||_{\infty })\leq K\phi _{\varepsilon }^{-1}(||k||_{1})$ in view of (2.3) when $\gamma = 0$, it follows that

where $K_{1} = \max (K, 1)$ and $K_{2} = 1+K_{1}^{p-1}||\gamma ||_{1}.\ $ Here we have used Proposition A(iii) in Appendix. Consequently, it is

where we have used Proposition A(ii) in Appendix. Thus (2.3) holds. Next, we verify that $T_{\varepsilon }$ is continuous. Let $(k_{n})\subset L^{1}(\alpha, \beta)$ and $k\in L^{1}(\alpha, \beta)\ $ be such that $||k_{n}-k||_{1}\rightarrow 0.$ Let $u_{n} = T_{\varepsilon }k_{n}$ and $u = T_{\varepsilon }k.$

Multiplying the equation

by $u_{n}-u$ and integrating between $\alpha$ and $\beta,$ we obtain

where $c_{n} = -(\phi _{\varepsilon }(u_{n}^{\prime })-\phi _{\varepsilon }(u^{\prime })(u_{n}-u)|_{\alpha }^{\beta }\geq 0.\ $ By [11,Lemma 30],

for all $x, y\in \mathbb{R}$ with $|x|+|y|\leq 2M, \ $ where $c_{0}$ $> 0$ is a constant depending only on $p\ $ and $M.\ $ Applying (2.9) with $x = u_{n}^{\prime }, y = u^{\prime }$ and note that $|u_{n}|_{1}+|u|_{1}\leq 2M,$ where $M = K\max \left(\phi ^{-1}(||k_{n}||_{1}), \phi ^{-1}(||k||_{1}\right)),$ we obtain from (2.8) that

If $b = 0$ then $(u_{n}-u)(\alpha) = 0$ and the Mean Value Theorem implies that

for $t\in \lbrack \alpha, \beta].\ $ Hence $||u_{n}-u||_{\infty }\rightarrow 0$ as $n\rightarrow \infty$ in view of (2.10).

If $b > 0$ then $u_{n}^{\prime }(\alpha) = \frac{a}{b}u_{n}(\alpha), u^{\prime }(\alpha) = \frac{a}{b}u(\alpha),$ and since (2.9) gives

it follows from the Mean Value Theorem and (2.10) that

as $n\rightarrow \infty.$ Hence $T_{\varepsilon }$ is continuous. Since $(u_{n})$ is bounded in $C^{1}[\alpha, \beta],$ $T_{\varepsilon }$ is completely continuous, which completes the proof. □

Lemma 2.3. Let $k\in L^{1}(0, 1)\ $ with $k\geq 0, \ $ and $u\in C^{1}[0, 1]$ satisfy

Then there exist constants $\kappa, C > 0$ independent of $u, k, \varepsilon$ such that if $||u||_{\infty }\geq C\phi _{\varepsilon }^{-1}(||k||_{1})$ then

for $t\in \lbrack 0, 1].$

Proof. Let $v\in C^{1}[0, 1]$ satisfy

By Lemma 2.2, $|v|_{1}\leq K\phi _{\varepsilon }^{-1}(||k||_{1}),$ where $K$ is independent of $k$. $\ $ By Lemma 2.1, $u\geq v$ on $[0, 1].\ $ Suppose $||u||_{\infty } > K\phi _{\varepsilon }^{-1}\left(||k||_{1}\right),$ and $||u||_{\infty } = |u(\tau)|$ for some $\tau \in \lbrack 0, 1].$ Then $u(\tau) > 0$ because if $u(\tau)\leq 0$ then $||u||_{\infty } = -u(\tau)\leq -v(\tau)\leq K\phi _{\varepsilon }^{-1}\left(||k||_{1}\right),$ a contradiction. In what follows, we may increase $K$ without mentioning if needed.

Suppose first that $\tau \in (0, 1).\ $ Let $z\in C^{1}[0, \tau]$ satisfying

Note that $z_{0}$ is a subsolution of (2.11) and $z_{0}+||u||_{\infty }$ is a supersolution of (2.11), where $z_{0}$ satisfies

from which the existence of $z$ follows. By Lemma 2.1, $\ u\geq z\geq v\geq -K\phi _{\varepsilon }^{-1}(||k||_{1})$ on $[0, \tau]$. $\ $ Define $z_{1}(t) = z(t)+K\phi _{\varepsilon }^{-1}(||k||_{1}).$ Then $z_{1}\geq 0$ on $[0, 1]$ and

where $K_{1} = K$ if $b = 0$ and $K_{1} = K(1+a/b)$ if $b > 0.\ $ Indeed, if $b = 0$ then $z(0) = v(0) = 0$ and so $\ z_{1}^{\prime }(0) = z^{\prime }(0)\geq v^{\prime }(0)\geq -K\phi _{\varepsilon }^{-1}(||k||_{1}),$ while if $b > 0$ then $z_{1}^{\prime }(0) = (a/b)z(0)\geq -K(a/b)\phi _{\varepsilon }^{-1}(||k||_{1}).\ $

Since $z\leq z_{1}$ on $(0, \tau)\ $ and $\ z_{1}^{\prime }(0)+K_{1}\phi _{\varepsilon }^{-1}(||k||_{1})\geq 0, \ $ it follows upon integrating the equation

that

Applying $\phi _{\varepsilon }$ on both sides of (2.12) and using the inequality (see Proposition A(i) in Appendix)

where $M = 2^{\max (p-2, 0)}, \ $ we obtain

By Gronwall's inequality,

for $t\in \lbrack 0, \tau].$ In particular when $t = \tau,$ we obtain

where $K_{2} = (2M)^{-1}e^{-M||\gamma ||_{1}}.\ $ Since $\phi _{\varepsilon }(x)+\phi _{\varepsilon }(y)\leq 2\phi _{\varepsilon }(x+y)$ for $x, y\geq 0,$ this implies

where $K_{3} = K_{2}/2 < 1$ and $K_{4} = K_{3}^{\frac{1}{q-1}}, \ $ provided that $\phi _{\varepsilon }(||u||_{\infty })\geq ||k||_{1}/K_{2}$ which is true if $||u||_{\infty } \geq (1/K_{2})^{\frac{1}{q-1}}\phi _{\varepsilon }^{-1}(||k||_{1})$. Consequently,

which implies

where $K_{5} = K_{4}/2, \ $ provided that $||u||_{\infty }\geq \frac{2(K+K_{1})}{ K_{4}}\phi ^{-1}(||k||_{1}).\ $ Since

it follows that

for $t\in \lbrack 0, \tau].\ $ If $b = 0$ then $z(0) = 0$ and (2.13) becomes $z^{\prime }(0)\geq K_{5}||u||_{\infty },$ from which (2.14) implies

where $K_{6} = 2^{1-q}K_{5},$ provided that $\phi _{\varepsilon }(K_{5}||u||_{\infty })\geq 2K^{\frac{1}{p-1}}||\gamma ||_{1}||k||_{1}$ which is true if $||u||_{\infty }\geq K_{5}^{-1}\left(2K^{\frac{1}{p-1} }||\gamma ||_{1}\right) ^{\frac{1}{q-1}}\phi _{\varepsilon }^{-1}(||k||_{1})$. Consequently,

which implies upon integrating that

If $b > 0$ then $z^{\prime }(0) = (a/b)z(0)$ and (2.13) becomes

Since $z^{\prime }(0)\geq 0,$ (2.14) gives

where $\tilde{K} = \left(K^{\frac{1}{p-1}}||\gamma ||_{1}\right) ^{\frac{1}{ q-1}}.$ This, together with (2.16), implies

Hence

where $K_{7} = \frac{K_{5}b}{2(a+b)},$ provided that $||u||_{\infty }\geq \frac{2\tilde{K}(a+b)}{K_{5}b}\phi _{\varepsilon }^{-1}(||k||_{1}).$ Combining (2.15) and (2.17), we obtain

where $\kappa _{0} = \min (K_{6}, K_{7}).$ Next, let $w\in C^{1}[\tau, 1]$ be the solution of

Then $u\geq w$ on $[\tau, 1],$ and using similar arguments as above, we obtain

where $\kappa _{1} > 0$ is a constant independent of $k,$ provided that $||u||_{\infty } > C\phi _{\varepsilon }^{-1}(||k||)$ for some large constant $C$ independent of $u.$

Combining (2.18) and (2.19), we see that Lemma 2.3 holds with $\kappa = \min (\kappa _{0}, \kappa _{1}).\ $ If $\tau = 0$ then (2.19) holds on $[0, 1],$ and if $\tau = 1$ then (2.17) holds on $[0, 1],$ which completes the proof. □

3.   Proof of the main result

Let $E = C[0, 1]$ be with the usual sup-norm.

Proof of Theorem 1.1. Let $C, \kappa$ be given by Lemma 2.3 and define $\sigma _{0} = \kappa \sigma, h(t) = g(\sigma _{0}p(t)).\ $ For $v\in E,$ $g(\max (v, \sigma _{0}p))\in L^{1}(0, 1)$ by (A1), $\ $ and $0\leq f(t, |v|)+\gamma (t)\phi _{\varepsilon }(|v|)\in L^{1}(0, 1)$ by (A2) and (A3). Let $\lambda \geq 0$ be small so that $C\phi _{\varepsilon }^{-1}\left(\lambda ||h||_{1}\right) < \sigma.$ Then the problem

has a unique solution $u = A_{\varepsilon }v\in C^{1}[0, 1]$ in view of Lemma 2.2. Since the operator $S:E\rightarrow L^{1}(0, 1)$ defined by $(Sv)(t) = -\lambda g(\max (v, \sigma _{0}p))+f(t, |v|)+\gamma (t)|v|^{p-1}$ is continuous, it follows from Lemma 2.2 that $A_{\varepsilon }:E\rightarrow E\ $ is completely continuous. We shall verify that

(i) $u = \theta A_{\varepsilon }u, \ \theta \in (0, 1]\Longrightarrow ||u||_{\infty }\neq \sigma.$

Let $u\in E$ satisfy $u = \theta A_{\varepsilon }u$ for some $\theta \in (0, 1]$ with $||u||_{\infty } = \sigma.$

Suppose $\varepsilon > 0$. Then $\ $ $\ $

on $(0, 1),$ which implies upon multiplying by $\theta ^{p-1}$ that

Since $||u||_{\infty } > C\phi _{\varepsilon }^{-1}\left(\lambda ||h||_{1}\right)$, Lemma 2.3 gives

for $t\in (0, 1)\ $ (recall that $\kappa \sigma = \sigma _{0}).\ $ Hence it follows from (3.1) and (A4) that

on $(0, 1).\ $ Multiplying (3.2) by $u$ and integrating gives

Since $au(0)-bu^{\prime }(0) = 0 = cu(1)+du^{\prime }(1)$ and $\varepsilon > 0,$ this implies

Consequently,

Since $\lambda _{1}$ is characterized by the Raleigh formula

where $V = \{u\in C^{1}[0, 1]:au(0)-bu^{\prime }(0) = 0 = cu(1)+du^{\prime }(1)\},$ we get a contradiction. $\ $ Thus (i) holds.

Next, suppose $\varepsilon = 0.$ Then the $<$ inequality in (3.3) is replaced by $\leq \,$ which together with (3.4) imply

i.e., $u$ is an eigenfunction corresponding to $\lambda _{1}.$ Hence (3.2) gives

from which it follows that $f(t, u) = \lambda _{1}u^{p-1}$ for a.e. $t\in (0, 1).$ Since $||u||_{\infty } = \sigma,$ we get a contradiction with (A4) with $\varepsilon = 0.$ If $bd = 0,$ then $u(0) = 0$ or $u(1) = 0,$ and since $||u||_{\infty } = \sigma,$ we have $u[0, 1] = [0, \sigma]$, we get a contradiction if $f(t, z)\not\equiv \lambda _{1}z^{p-1}$ on $[0, \sigma]\ $ for a.e. $t\in (0, 1).\ $ Thus (i) holds.

Next, we verify that

(ii) There exists a constant $R > \sigma$ such that $u = A_{\varepsilon }u+\xi, \ \xi \geq 0\Longrightarrow ||u||_{\infty }\neq R.$

Let $u\in E$ satisfy $u = A_{\varepsilon }u+\xi$ for some $\xi \geq 0.$ Then $u$ satisfies

on $(0, 1), \ $ which implies

on $(0, 1).\ $ Note that

Suppose $||u||_{\infty } = R > \sigma.\ $ Then Lemma 2.3 gives

for $t\in (0, 1).\ $ Using (3.8) in (3.5), we get

Suppose $\varepsilon > 0$ and let $M > 0.\ $ Since $\lim\limits_{z\rightarrow \infty }\frac{f(t, z)-\lambda g(z)}{\phi _{\varepsilon }(z)} = \infty$ by (A1) and (A5), there exists a positive constant $L$ such that

for a.e. $t\in (0, 1)$ and $z > L.$ By (3.8),

for $R$ large, from which (3.9) and (3.10) imply

Since $u(1/4)$ and $u(3/4)$ are positive, the comparison principle gives $u\geq \tilde{u}$ on $[1/4, 3/4],$ where $\tilde{u}$ is the solution of

Let $||\tilde{u}||_{\infty } = \tilde{u}(\tau)$ for some $\tau \in (1/4, 3/4).$ If $\tau \leq 1/2$ then we have

while if $\tau > 1/2$,

Hence using Proposition A(iii) we see that in either case,

i.e., $||u||_{\infty }\geq \frac{\kappa (M/8)^{\frac{1}{p-1}}||u||_{\infty } }{32},$ a contradiction if $M$ is large enough, which proves (ii).

Suppose next that $\varepsilon = 0.\ $ Since $\liminf\limits_{z\rightarrow \infty }\frac{f(t, z)-\lambda g(z)}{z^{p-1}} > \lambda _{1}$ uniformly for a.e. $t\in (0, 1), \ $ there exist positive constants $L_{0}, \tilde{\lambda}\ $ with $\tilde{\lambda} > \lambda _{1}$ such that

for a.e. $t\in (0, 1)$ and all $z\geq L_{0}.$ For $\delta _{1}\in (0, 1/2),$ let $\lambda _{1, \delta _{1}}$ be the first eigenvalue of the problem

where $\delta _{2} = 1-\delta _{1}.$ By the continuity of the first eigenvalue with respect to the domain, $\lambda _{1, \delta _{1}}\rightarrow \lambda _{1}$ as $\delta _{1}\rightarrow 0.$ Hence there exits $\delta > 0$ such that $\lambda _{1, \delta _{1}} < \tilde{\lambda}\ $ for $\delta _{1}\leq \delta.\ \ $ Let $\ \delta _{1} = \delta /2$, $\delta _{2} = 1-\delta /2,$ and $\mu \in (\lambda _{1, \delta _{1}}, \tilde{\lambda}).\ $ By decreasing $\delta$ if necessary, we have from (3.7) that

where $\bar{u} = u+1.\ $ By (3.8),

for $t\in \lbrack \delta /4, 1-\delta /4]$ for $R$ large. It follows from (3.9), (3.11) and (3.14) that

By (3.6) and (3.15),

for a.e. $t\in (0, 1),$ where $\gamma _{L}(t) = \lambda h(t)+\gamma (t)\phi _{0}(L)\geq 0.\ $ We claim that the eigenvalue problem

has a positive solution, provided that $R$ is large enough.

Let $\psi _{1}$ be the positive solution of (3.12) with $||\psi _{1}||_{\infty } = 1.$ Then clearly $\psi _{1}$ is a subsolution of (3.17). Since (3.14) implies

for $R$ large and $\frac{\kappa R\delta /4}{1+\kappa R\delta /4}\rightarrow 1$ as $R\rightarrow \infty$ $,$ it follows from (3.15) that

for $R$ large.

Case 1. $a, c > 0.\ $ Then $\bar{u}$ is a supersolution of (3.17) in view of (3.13) and (3.18).

Case 2. $ac = 0.$ If $a = 0$ then (3.7) gives $u^{\prime }(0) = 0$. Combining (3.14)–(3.16), we deduce that for $R$ large,

i.e., $u^{\prime }(\delta _{1}) < 0.\ $ Similarly if $c=0$ then $u^{\prime }(1)=0,$ and

i.e., $u^{\prime }(\delta _{2}) > 0.$ Since $a\bar{u}(\delta _{1})-b\bar{u} ^{\prime }(\delta _{1}) > 0$ and $c\bar{u}(\delta _{2})+d\bar{u}^{\prime }(\delta _{2}) > 0,$ it follows from (3.18) that $\bar{u}$ is a supersolution of (3.17).

Since $\psi _{1}\leq 1\leq \bar{u}$ on $[\delta _{1}, \delta _{2}],$ the existence of a solution $v$ to (3.17) with $\psi _{1}\leq v\leq \bar{u}$ on $(\delta _{1}, \delta _{2})\ $ follows, which is a contradiction. Thus (ii) holds. By Amann's fixed point theorem [1,Theorem 12.3], $A_{\varepsilon }$ has a fixed point $u\in E$ with $||u||_{\infty } > \sigma.$ Using $\xi = 0$ in (ii) and (3.8), we obtain $u(t)\geq \sigma _{0}p(t)$ for $t\in \lbrack 0, 1]$ i.e., $g(\max (u, \sigma _{0}p(t))) = g(u)$ and therefore $u$ is a positive solution of (1.1), which completes the proof. □

Use of AI tools declaration

The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

Conflict of interest

The authors declare no conflict of interest.

Appendix A

We provide here some inequalities regarding the operator $\phi _{\varepsilon }.$

Proposition A.

(i) $\phi _{\varepsilon }(x+y)\leq M(\phi _{\varepsilon }(x)+\phi _{\varepsilon }(y))$ for $x, y\geq 0,$ where $M = 2^{\max (p-2, 0)}.$

(ii) $\phi _{\varepsilon }^{-1}(mx)\leq m^{\frac{1}{q-1}}\phi _{\varepsilon }^{-1}(x)$ for $m\geq 1, x\geq 0.$

(iii) $\phi _{\varepsilon }(cx)\leq c^{p-1}\phi _{\varepsilon }(x)$ for $c\geq 1, x\geq 0.$

Proof. (i) Let $x, y\geq 0.\ $ Since the function $z^{r}$ is convex on $[0, \infty)$ for $r\geq 1,$

i.e.,

On the other hand if $0 < r < 1,$ we have

Hence for $r > 0,$

which implies

i.e., (i) holds.

(ii) Let $z\geq 0$ and $c\geq 1.$ We claim that

Indeed,

i.e., (A.1) holds. Let $m\geq 1, x\geq 0.$ Then by using (A.1) with $c = m^{\frac{1}{ q-1}}$ and $z = \phi _{\varepsilon }^{-1}(x),$ we obtain

i.e., (ii) holds.

(iii) Let $c\geq 1$ and $x\geq 0.$ Then

i.e., (iii) holds. □