Existence of positive solutions for eigenvalue problem of nonlinear fractional differential equations

Abstract

In this article, by using the fixed point theorem, existence of positive solutions for eigenvalue problem of nonlinear fractional differential equations

$$\left\{\begin{array}{c}\hfill D_{0+}^{\alpha }u\left(t\right)+\lambda a\left(t\right)f\left(t,u\left(t\right)\right)=0,0<t<1,\hfill \\ \hfill u\left(0\right)=u\left(1\right)=0\hfill \end{array}\right.$$

is considered, where 1 < α < 2 is a real number, $D_{0+}^{\alpha }$ is the standard Riemann-Liouville derivate, λ is a positive parameter and a(t) ∈ C([0, 1], [0, ∞)), f(t, u) ∈ C([0, 1] × [0, ∞), [0, ∞)).

MSC(2010): 34B18.

1 Introduction

Fractional differential equations have been of great interest recently. It is caused both by the intensive development of the theory of fractional calculus itself and by the applications of such constructions in various sciences such as physics, mechanics, chemistry, engineering, etc. For details, see [1–6] and references therein.

Recently, many results were obtained dealing with the existence and multiplicity of solutions of nonlinear fractional differential equations by the use of techniques of nonlinear analysis, see [7–22] and the reference therein. Bai and Lu [7] studied the existence of positive solutions of nonlinear fractional differential equation

$$\left\{\begin{array}{c}\hfill D_{0+}^{\alpha }u\left(t\right)+f\left(t,u\left(t\right)\right)=0,0<t<1,\hfill \\ \hfill u\left(0\right)=u\left(1\right)=0,\hfill \end{array}\right.$$

(1.1)

where 1 < α≤ 2 is a real number, $D_{0+}^{\alpha }$ is the standard Riemann-Liouville differentiation, and f : [0, 1] × [0, ∞) → [0, ∞) is continuous. They derived the corresponding Green function and obtained some properties as follows.

Proposition 1 The Green function G(t, s) satisfies the following conditions:

(R1) G(t, s) ∈ C([0,1] × [0,1]), and G(t, s) > 0 for t, s ∈ (0, 1);

(R2) There exists a positive function γ ∈ C(0, 1) such that

$$\underset{\frac{1}{4}\le t\le \frac{3}{4}}{\text{min}}G\left(t,s\right)\ge \gamma \left(s\right)\underset{0\le t\le 1}{\text{max}}G\left(t,s\right)\ge \gamma \left(s\right)G\left(s,s\right),s\in \left(0,1\right),$$

(1.2)

where

$$G\left(t,s\right)=\left\{\begin{array}{c}\hfill \frac{{\left[t\left(1-s\right)\right]}^{\alpha -1}-{\left(t-s\right)}^{\alpha -1}}{\Gamma \left(\alpha \right)},0\le s\le t\le 1,\hfill \\ \hfill \frac{{\left[t\left(1-s\right)\right]}^{\alpha -1}}{\Gamma \left(\alpha \right)},0\le t\le s\le 1.\hfill \end{array}\right.$$

(1.3)

It is well known that the cone plays a very important ^... role in applying the Green function in research area. In [7], the authors cannot acquire a positive constant taken instead of the role of positive function γ(s) with 1 < α < 2 in (1.2). In [9], Jiang and Yuan obtained some new properties of the Green function and established a new cone. The results can be stated as follows.

Proposition 2 The Green function G(t, s) defined by (1.3) has the following properties: G(t, s) = G(1 - s, 1 - t) and

$$\frac{\alpha -1}{\Gamma \left(\alpha \right)}{t}^{\alpha -1}\left(1-t\right){\left(1-s\right)}^{\alpha -1}s\le G\left(t,s\right)\le \frac{1}{\Gamma \left(\alpha \right)}{t}^{\alpha -1}\left(1-t\right){\left(1-s\right)}^{\alpha -2}\forall t,s\in \left(0,1\right).$$

(1.4)

Proposition 3 The function G*(t, s):= t^2-αG(t, s) has the following properties:

$$q\left(t\right)\Phi \left(s\right)\le G^{*}\left(t,s\right)\le \Phi \left(s\right)\forall t,s\in \left[0,1\right].$$

(1.5)

where q(t) = (α - 1)t(1 - t),$\Phi \left(s\right)=\frac{1}{\Gamma \left(\alpha \right)}s{\left(1-s\right)}^{\alpha -1}$.

The purpose of this article is to establish the existence of positive solutions for eigenvalue problem of nonlinear fractional differential equations

$$\left\{\begin{array}{c}\hfill D_{0+}^{\alpha }u\left(t\right)+\lambda a\left(t\right)f\left(t,u\left(t\right)\right)=0,0<t<1,\hfill \\ \hfill u\left(0\right)=u\left(1\right)=0\hfill \end{array}\right.$$

(1.6)

where 1 < α < 2 is a real number, $D_{0+}^{\alpha }$ is the standard Riemann-Liouville derivate, λ is a positive parameter and a(t) ∈ C([0,1], [0, ∞)), f(t, u) ∈ C([0,1] × [0, ∞), [0, ∞)).

2 The preliminary lemmas

For the convenience of the reader, we present here the necessary definitions from fractional calculus theory. These definitions can be found in the recent literature.

Definition 2.1 The fractional integral of order α > 0 of a function y : (0, ∞) → R is given by

$$I_{0+}^{\alpha }y\left(t\right)=\frac{1}{\Gamma \left(\alpha \right)}{\int }_0^t{\left(t-s\right)}^{\alpha -1}y\left(s\right)ds$$

provided the right side is pointwise defined on (0, ∞).

Definition 2.2 The fractional derivative of order α > 0 of a function y : (0, ∞) → R is given by

$$D_{0+}^{\alpha }y\left(t\right)=\frac{1}{\Gamma \left(n-\alpha \right)}{\left(\frac{d}{dt}\right)}^n{\int }_0^t\frac{y\left(s\right)}{{\left(t-s\right)}^{\alpha -n+1}}ds$$

where n = [α] + 1, provided the right side is pointwise defined on (0, ∞).

Lemma 2.1 Let α > 0. If we assume u ∈ C(0, 1) ∩ L(0, 1), then the fractional differential equation

$$D_{0+}^{\alpha }u\left(t\right)=0$$

has u(t) = C₁t^α-1 + C₂t^α-2 + ... + C _N t^α-N , C _i ∈ R, i = 1, 2, ..., N, where N is the smallest integer greater than or equal to α, as unique solutions.

Lemma 2.2[7] Assume that u ∈ C(0, 1) ∩ L(0, 1) with a fractional derivative of order α > 0 that belongs to u ∈ C(0, 1) ∩ L(0, 1). Then

$$I_{0+}^{\alpha }{D}_{0+}^{\alpha }u\left(t\right)=u\left(t\right)+C_1{t}^{\alpha -1}+C_2{t}^{\alpha -2}+\cdots +C_N{t}^{\alpha -N}$$

for some C _i ∈ R, i = 1, 2, . . . , N.

Lemma 2.3[7] Given y ∈ C[0,1] and 1 < α ≤ 2, the unique solution of

$$\left\{\begin{array}{c}\hfill D_{0+}^{\alpha }u\left(t\right)+y\left(t\right)=0,0<t<1,\hfill \\ \hfill u\left(0\right)=u\left(1\right)=0,\hfill \end{array}\right.$$

(2.1)

is

$$u\left(t\right)={\int }_0^{1}G\left(t,s\right)y\left(s\right)ds,$$

where

$$G\left(t,s\right)=\left\{\begin{array}{c}\hfill \frac{{\left[t\left(1-s\right)\right]}^{\alpha -1}-{\left(t-s\right)}^{\alpha -1}}{\Gamma \left(\alpha \right)},0\le s\le t\le 1,\hfill \\ \hfill \frac{{\left[t\left(1-s\right)\right]}^{\alpha -1}}{\Gamma \left(\alpha \right)},0\le t\le s\le 1.\hfill \end{array}\right.$$

(2.2)

Lemma 2.4[10] Let K be a cone in Banach space E. Suppose that $T:{\overline{K}}_r\to K$ is a completely continuous operator.

• (i) If there exists u ₀ ∈ K \ {θ} such that u - Tu ≠ μu ₀ for any u ∈ ∂K _r and μ ≥ 0, then i(T, K _r, K) = 0.

• (ii) If Tu ≠ μu for any u ∈ ∂K _r and μ ≥ 1, then i(T, K _r, K) = 1.

Lemma 2.5[8] Let P be a cone in Banach space X. Suppose that T : P → P is a completely continuous operator. If there exists a bounded open set Ω(P) such that each solution of

$$u=\sigma Tu,u\in P,\sigma \in \left[0,1\right]$$

satisfies u ∈ Ω(P), then the fixed point index i(T, Ω(P), P) = 1.

3 The main results

Let E = C[0,1] be endowed with the ordering u ≤ v if u(t) ≤ v(t) for all t ∈ [0,1], and the maximum norm, $∥u∥=\underset{0\le t\le 1}{\text{max}}\left|u\left(t\right)\right|$. Define the cone P ⊂ E by P = {u ∈ E|u(t) ≥ 0}, and

$$K=\left\{u\in P|u\left(t\right)\ge q\left(t\right)|\left|u\right||\right\}.$$

where q(t) is defined by (1.5).

It is easy to see that P and K are cones in E. For any 0 < r < R < +∞, let K _r = {u ∈ K| ||u|| < r}, ∂K _r = {u ∈ K| ||u|| = r}, ${\overline{K}}_r=\left\{u\in K\left|\right|\left|u\right||\le r\right\}$and${\overline{K}}_R\backslash K_r=\left\{u\in K|r\le |\left|u\right||\le R\right\}$.

For convenience, we introduce the following notations

$$\begin{array}{c}{g}_0=\underset{u\to 0^{+}}{\text{lim}}\frac{g\left(u\right)}{u},g_{\infty }=\underset{u\to +\infty }{\text{lim}}\frac{g\left(u\right)}{u},\\ {\int }_{\frac{1}{4}}^{\frac{3}{4}}{G}^{*}\left(\tau ,s\right)a\left(s\right)q_1\left(s\right)ds=\underset{\frac{1}{4}\le t\le \frac{3}{4}}{\text{max}}{\int }_{\frac{1}{4}}^{\frac{3}{4}}{G}^{*}\left(t,s\right)a\left(s\right)q_1\left(s\right)ds.\end{array}$$

We assume the following conditions hold throughout the article:

(H1) a(t) ∈ C([0, 1], [0, ∞)), a(t) ≢ 0;

(H2) f(t, u) ∈ C([0, 1] × [0, ∞), [0, ∞)), and there exist g ∈ C([0, +∞), [0, +∞)), q_1,q₂ ∈ C((0, 1), (0, +∞)) such that

$$q_1\left(t\right)g\left(u\right)\le f\left(t,t^{\alpha -2}u\right)\le q_2\left(t\right)g\left(u\right),t\in \left(0,1\right),u\in \left[0,+\infty \right),$$

where ${\int }_0^1{q}_i\left(s\right)ds<+\infty ,i=1,2$.

By similar arguments to Lemma 4.1 and Theorem 1.3 of [9], we obtain the following result.

Lemma 3.1 Assume that (H1)(H2) hold. Let

$$Tu\left(t\right):=\lambda {\int }_0^1{G}^{*}\left(t,s\right)a\left(s\right)f\left(s,s^{\alpha -2}u\left(s\right)\right)ds,$$

(3.1)

then T : K → K is completely continuous. Moreover, if u is a fixed point of T in ${\overline{K}}_R\backslash K_r$, then y = t^α-2u is a positive solution of BVP (1.6).

By similar arguments to Theorems 2 and 3 of [18], we obtain the following result.

Theorem 3.1 Assume that (H1)(H2) hold. Then, for each λ satisfying

$$\frac{16}{\left[\left(\alpha -1\right){\int }_{\frac{1}{4}}^{\frac{3}{4}}{G}^{*}\left(\tau ,s\right)a\left(s\right)q_1\left(s\right)ds\right]g_{\infty }}<\lambda <\frac{1}{\left({\int }_0^1\Phi \left(s\right)a\left(s\right)q_2\left(s\right)ds\right)g_0},$$

(3.2)

there exists at least one positive solution of BVP (1.6) in K.

Theorem 3.2 Assume that (H1)(H2) hold. Then, for each λ satisfying

$$\frac{16}{\left[\left(\alpha -1\right){\int }_{\frac{1}{4}}^{\frac{3}{4}}{G}^{*}\left(\tau ,s\right)a\left(s\right)q_1\left(s\right)ds\right]g_0}<\lambda <\frac{1}{\left({\int }_0^1\Phi \left(s\right)a\left(s\right)q_2\left(s\right)ds\right)g_{\infty }},$$

(3.3)

there exists at least one positive solution of BVP (1.6) in K.

Set

$$L_{1}u\left(t\right):={\int }_0^1{G}^{*}\left(t,s\right)a\left(s\right)q_1\left(s\right)u\left(s\right)ds,$$

(3.4)

$$L_{2}u\left(t\right):={\int }_0^1{G}^{*}\left(t,s\right)a\left(s\right)q_2\left(s\right)u\left(s\right)ds.$$

(3.5)

It is clear that L₁, L₂ is a completely continuous linear operator and L₁(P) ⊂ K, L₂(P) ⊂ K. By virtue of the Krein-Rutman theorem and Proposition 3, we have the following lemma.

Lemma 3.2 Assume that (H1)(H2) hold. Then the spectral radius r(L₁) > 0, r(L₂) > 0 and L₁, L₂, respectively, has a positive eigenfunction φ₁, φ₂ corresponding to its first eigenvalue λ₁ = (r(L₁))^-1, λ₂ = (r(L₁))^-1, that is φ₁ = λ₁L₁φ₁, φ₂ = λ₂L₂φ₂.

Obviously, L₁φ₁ ∈ K, L₂φ₂ ∈ K, so φ₁ ∈ K, φ₂ ∈ K.

In the following, we will obtain some existence results under some conditions concerning the first eigenvalue with respect to linear operator L₁, L₂.

Assume that

(H3)$\frac{{\lambda }_1}{g_0}<\lambda <\frac{{\lambda }_2}{g_{\infty }};$

(H3)$\frac{{\lambda }_1}{g_{\infty }}<\lambda <\frac{{\lambda }_2}{g_0}.$

Theorem 3.3 Assume (H1)(H2)(H3) hold. Then the BVP (1.6) has at least one positive solution.

Proof. First, by Lemma 3.1, we know that T : K → K is completely continuous.

By (H3), we have $g_0>\frac{{\lambda }_1}{\lambda }$, there exists r₁> 0 such that

$$g\left(u\right)\ge \frac{{\lambda }_1}{\lambda }u,0<u\le r_1.$$

Then for $u\in \partial K_{r_1}$, we have

$$\begin{array}{ll}\hfill \left(Tu\right)\left(t\right)& =\lambda {\int }_0^1{G}^{*}\left(t,s\right)a\left(s\right)f\left(s,s^{\alpha -2}u\left(s\right)\right)ds\\ \ge \lambda {\int }_0^1{G}^{*}\left(t,s\right)a\left(s\right)q_1\left(s\right)g\left(u\left(s\right)\right)ds\\ \ge {\lambda }_1{\int }_0^1{G}^{*}\left(t,s\right)a\left(s\right)q_1\left(s\right)u\left(s\right)ds={\lambda }_1\left(L_{1}u\right)\left(t\right).\end{array}$$

By Lemma 3.2, φ₁ = λ₁L₁φ₁. We may suppose that T has no fixed points on $\partial K_{r_1}$ (otherwise, the proof is finished). Now we show that

$$u-Tu\ne \mu {\phi }_1,u\in \partial K_{r1},\mu \ge 0.$$

(3.6)

Assume by contradicts that there exist $u\in \partial K_{r_1}$ and μ₁ ≥ 0 such that u₁ - Tu₁ = μ₁φ₁, then μ₁> 0 and u₁ = Tu₁ + μ₁φ₁ ≥ μ₁φ₁. Let $\bar{\mu }=\text{sup}\left\{\mu |u_1\ge \mu {\phi }_1\right\}$, then $\bar{\mu }\ge {\mu }_1,u_1\ge \bar{\mu }{\phi }_1$ and $T{u}_1\ge {\lambda }_1\bar{\mu }{L}_1{\phi }_1=\bar{\mu }{\phi }_1$. Thus,

$$u_1=T{u}_1+{\mu }_1{\phi }_1\ge \bar{\mu }{\phi }_1+{\mu }_1{\phi }_1=\left(\bar{\mu }+{\mu }_1\right){\phi }_1,$$

which contradicts the definition of $\bar{\mu }$ So (3.6) is true and by lemma 2.4 we have

$$i\left(T,K_{r_1},K\right)=0.$$

(3.7)

On the other hand, by (H3), we have $g_{\infty }<\frac{\lambda 2}{\lambda }$, there exists 0 < σ < 1 and r₂> r₁> 0 such that

$$g\left(u\right)\le \sigma \frac{{\lambda }_2}{\lambda }u,u\ge r_2.$$

Let Lφ = σλ₂L₂φ, φ ∈ C[0,1]. Then L : C[0,1] → C[0,1] is a bounded linear operator and L(K) ⊂ K. Denote

$$M=\left(\underset{0\le t,s\le 1}{\text{max}}{G}^{*}\left(t,s\right)\right)\underset{\phi \in \partial K_{r_2}}{\text{sup}}{\int }_0^1\lambda a\left(s\right)q_2\left(s\right)g\left(\phi \left(s\right)\right)ds.$$

It is clear that M < +∞ Let

$$W=\left\{\phi \in K|\phi =\mu T\phi ,0\le \mu \le 1\right\}.$$

In the following, we prove that W is bounded.

For any φ ∈ W , set $\bar{\phi }\left(t\right)=\text{min}\left\{\phi \left(t\right),r_2\right\}$ and denote E(φ) = {t ∈ [0,1]|φ(t) > r₂}, then

$$\begin{array}{ll}\hfill \phi \left(t\right)& =\mu \left(T\phi \right)\left(t\right)\le \left(T\phi \right)\left(t\right)=\lambda {\int }_0^1{G}^{*}\left(t,s\right)a\left(s\right)f\left(s,s^{\alpha -2}\phi \left(s\right)\right)ds\\ \le \lambda {\int }_0^1{G}^{*}\left(t,s\right)a\left(s\right)q_2\left(s\right)g\left(\phi \left(s\right)\right)ds\\ =\lambda {\int }_{E\left(\phi \right)}{G}^{*}\left(t,s\right)a\left(s\right)q_2\left(s\right)g\left(\phi \left(s\right)\right)ds+\lambda {\int }_{\left[0,1\right]\backslash E\left(\phi \right)}{G}^{*}\left(t,s\right)a\left(s\right)q_2\left(s\right)g\left(\bar{\phi }\left(s\right)\right)ds\\ \le \sigma {\lambda }_2{\int }_0^1{G}^{*}\left(t,s\right)a\left(s\right)q_2\left(s\right)\phi \left(s\right)ds+\lambda {\int }_0^1{G}^{*}\left(t,s\right)a\left(s\right)q_2\left(s\right)g\left(\bar{\phi }\left(s\right)\right)ds\\ \le \left(L\phi \right)\left(t\right)+M,t\in \left[0,1\right].\end{array}$$

Thus ((I - L)φ)(t) ≤ M, t ∈ [0,1]: Since λ₂ is the first eigenvalue of L₂ and 0 < σ < 1, the first eigenvalue of L, (r(L))^-1> 1. Therefore, the inverse operator (I - L)^-1 exists and

$${\left(I-L\right)}^{-1}=I+L+L^2+\cdot \cdot \cdot +L^n\cdot \cdot \cdot .$$

It follows from L(K) ⊂ K that (I - L)^-1(K) ⊂ K. So we have φ(t) ≤ (I - L)^-1M, t ∈ [0,1] and W is bounded.

Choose r₃> max{r₂, || (I - L)^-1M||}. Then by lemma 2.5, we have

$$i\left(T,K_{r_3},K\right)=1.$$

(3.8)

By (3.7) and (3.8), one has

$$i\left(T,K_{r_3}\backslash {\bar{K}}_{r_1},K\right)=i\left(T,K_{r_3},K\right)-i\left(T,K_{r_1},K\right)=1.$$

Then T has at least one fixed point on $K_{r_3}\backslash {\bar{K}}_{r_1}$ By Lemma 3.1, this means that problem (1.6) has at least one positive solution. The proof is complete.

Theorem 3.4 Assume (H1)(H2)(H4) hold. Then the BVP (1.6) has at least one positive solution.

Theorem 3.5 Assume (H1)(H2) hold and $\frac{{\lambda }_1}{\lambda }<g_0\le \infty ,\frac{{\lambda }_1}{\lambda }<g_{\infty }\le \infty$. Moreover, the following condition holds:

(H5) there exists a p > 0 such that 0 ≤ u ≤ p implies g(u) < M₁p, where $M_1={\left[\lambda {\int }_0^1\Phi \left(s\right)a\left(s\right)q_2\left(s\right)ds\right]}^{-1}$.

Then the BVP (1.6) has at least two positive solution.

Theorem 3.6 Assume (H1)(H2) hold and $0\le g_0<\frac{{\lambda }_2}{\lambda }$, $0\le g_{\infty }<\frac{{\lambda }_2}{\lambda }$. Moreover, the following condition holds:

(H6) there exists a p > 0 such that $\frac{p\left(\alpha -1\right)}{16}\le u\le p$ implies g(u) > M₂p, where $M_2={[\lambda {\displaystyle {\int }_{\frac{1}{4}}^{\frac{3}{4}}{G}^{\ast }(\frac{1}{2},s)a(s)q_1(s)ds}]}^{-1}$. Then the BVP (1.6) has at least two positive solution.

Remark 3.1 Theorems 3.3, 3.4 extend and improve Theorem 1.5 in [9], Theorems 3.5 and 3.6 extend and improve Theorems 1.3 and 1.4 in [9], respectively.

4 Example

Consider the boundary value problem

$$\left\{\begin{array}{c}\hfill D_{0+}^{\alpha }u\left(t\right)+\lambda \mu \left(u^a\left(t\right)+u^b\left(t\right)\right)=0,0<t<1,0<a<1<b<\frac{1}{2-\alpha },1<\alpha <2,\hfill \\ \hfill u\left(0\right)=u\left(1\right)=0.\hfill \end{array}\right.$$

(3.9)

Then (3.9) have at least two positive solutions u₁ and u₂, for each 0 < μ < μ*, where μ* is some positive constant.

Proof. We will apply Theorem 3.5. To this end we take a(t) ≡ 1, f(t, u) = μ(u^a (t) + u^b (t)), then f(t, t^α-2y) = μ(t^a(α-2)y^a (t) + t^b(α-2)yb (t)).

Let q₁(t) = t^a(α-2), q₂(t) = t^b(α-2) and g(y) = μ(y^a + y^b ), then q₁(t)g(y) ≤ f(t, t^α-2y) ≤ q₂(t)g(y) and q₁, q₂ ∈ L¹(0, 1), g ∈ C([0, +∞), [0, +∞)). Thus (H1)(H2)is satisfied, and it is to see that $\frac{{\lambda }_1}{\lambda }<g_0\le \infty$,$\frac{{\lambda }_1}{\lambda }<g_{\infty }\le \infty$.

Since

$$\begin{array}{ll}\hfill \frac{1}{\Gamma \left(\alpha \right)}{\int }_0^{1}s{\left(1-s\right)}^{\alpha -1}{q}_2\left(s\right)ds& =\frac{1}{\Gamma \left(\alpha \right)}{\int }_0^{1}s{\left(1-s\right)}^{\alpha -1}{s}^{b\left(\alpha -2\right)}ds\\ =\frac{1}{\Gamma \left(\alpha \right)}{\int }_0^1{s}^{\left(2+b\left(\alpha -2\right)\right)-1}{\left(1-s\right)}^{\alpha -1}ds\\ =\frac{1}{\Gamma \left(\alpha \right)}\frac{\Gamma \left(2+b\left(\alpha -2\right)\right)\Gamma \left(\alpha \right)}{\Gamma \left(2+b\left(\alpha -2\right)+\alpha \right)}\\ =\frac{\Gamma \left(2+b\left(\alpha -2\right)\right)}{\Gamma \left(2+b\left(\alpha -2\right)+\alpha \right)},\end{array}$$

then $M_1={\left(\lambda \frac{\Gamma \left(2+b\left(\alpha -2\right)\right)}{\Gamma \left(2+b\left(\alpha -2\right)+\alpha \right)}\right)}^{-1}$.

Let

$${\mu }^{*}=F\left(p\right)=\underset{y>0}{\text{sup}}\left\{F\left(y\right):F\left(y\right)=M_1\frac{y}{y^a+y^b}\right\}=F\left({\left(\frac{1-a}{b-1}\right)}^{\frac{1}{b-a}}\right),$$

where $p={\left(\frac{1-a}{b-1}\right)}^{\frac{1}{b-a}}$.

Then, for μ < μ*, we have

$$g\left(y\right)=\mu \left(y^a+y^b\right)\le \mu \left(p^a+p^b\right)<M_{1}p,0\le y\le p,$$

thus (H5) holds. Therefore, (3.9) have at least two positive solutions u₁ and u₂, for each 0 < μ < μ*.