Positive Solutions of the One-Dimensional 𝑝-Laplacian with Nonlinearity Defined on a Finite Interval

Abstract

We use the quadrature method to show the existence and multiplicity of positive solutions of the boundary value problems involving one-dimensional 𝑝-Laplacian (|𝑢^′(𝑡)|^𝑝−2𝑢^′(𝑡))^′+𝜆𝑓(𝑢(𝑡))=0, 𝑡∈(0,1), 𝑢(0)=𝑢(1)=0, where 𝑝∈(1,2], 𝜆∈(0,∞) is a parameter, 𝑓∈𝐶¹([0,𝑟),[0,∞)) for some constant 𝑟>0, 𝑓(𝑠)>0 in (0,𝑟), and lim_{𝑠→𝑟−}(𝑟−𝑠)^𝑝−1𝑓(𝑠)=+∞.

1. Introduction and the Main Results

Let 𝑎∶[0,1]→[0,∞) be continuous and 𝑎(𝑡)≢0 on any subset of [0,1], and let 𝑓∶[0,∞)→ℝ be a continuous function. Wang [1] proved the existence of positive solutions of nonlinear boundary value problems 𝑢^′′(𝑡)(𝑡)𝑓(𝑢(𝑡(),𝑢(0)(1)+𝑎))=0,𝑡∈0,1=𝑢=0, (1) under the following assumptions: 𝑓₀∶=lim_𝑠→0𝑓(𝑠)𝑠=0,𝑓_∞∶=lim_𝑠→+∞𝑓(𝑠)𝑠=∞. (2) Since then, the existence and multiplicity of positive solutions of (1) and its generalized forms have been extensively studied via the fixed point theorem in cones. For example, Ge [2] showed a series of results on the existence and multiplicity of solutions of nonlinear ordinary differential equation of second order/higher order subjected with diverse boundary conditions via topological degree and fixed point theorem in cones; Wang [3] use fixed point theorem in cones to study the existence of positive solutions for the one dimensional p-Laplacian. For other recent results along this line, see [4–11] and the bibliographies in [2]. For the special case 𝑎(𝑡)≡1, beautiful results have been obtained via the quadrature method; see Fink et al. [12], Brown and Budin [13], Addou and Wang [14], Cheng and Shao [15], Karátson and Simon [16], and the references therein.

The nonlinearity 𝑓(𝑢) that appeared in the above previously papers is assumed to be well defined in [0,∞) or (−∞,∞). Of course, natural question is what would happen if 𝑓(𝑢) is only well defined in a finite interval [0,𝑟), where 𝑟 is a positive constant; that is, what would happen if (2) is replaced with the following limit 𝑓_𝑟∶𝑓_𝑟∶=lim_{𝑠→𝑟−}𝑓(𝑠)𝑠^𝑝−1=+∞. (3)

It is worth remarking that the fixed point theorem in cones method in [1–3] cannot be used to deal with the existence of positive solutions of the problem 𝑢^′(𝑡)^𝑝−2𝑢^′(𝑡)^′(𝑢(𝑡(),𝑢(0)(1)+𝜆𝑓))=0,𝑡∈0,1=𝑢=0, (4) under the restriction (3) any more since the appearance of singularity of 𝑓 at 𝑟. The purpose of this paper is to use the quadrature method to show the existence and multiplicity of positive solutions of (4), in which 𝜆∈(0,∞) is a parameter, and 𝑓 satisfies the following assumptions: (H1) 𝑓∈𝐶¹([0,𝑟),[0,∞)); (H2) 𝑓(𝑠)>0 in (0,𝑟); (H3) lim_{𝑠→𝑟−}(𝑟−𝑠)^𝑝−1𝑓(𝑠)=+∞.

Let 𝑓₀∶=lim_𝑠→0⁺(𝑓(𝑠)/𝑠^𝑝−1), 𝐹(𝑠)=∫^𝑠₀𝑓(𝑣)𝑑𝑣.

The main result of the paper is the following.

Theorem 1. Let 𝑝∈(1,2], and let (H1), (H2), and (H3) hold. Then,(i)if 𝑓₀=+∞, then there exist 𝜆_∗>0, such that (4) has at least two positive solutions for 𝜆∈(0,𝜆_∗), has at least one positive solution for 𝜆=𝜆_∗, and has no positive solution for 𝜆>𝜆_∗;(ii)if 𝑓₀=0, then (4) has at least one positive solution for 𝜆∈(0,∞);(iii)if 𝑓₀∈(0,∞), then (4), has at least one positive solution for 𝜆∈(0,𝜋^𝑝_𝑝/𝑓₀), where 𝜋_𝑝∶=2^{(𝑝−1)1/𝑝}₀11−𝑠^𝑝/(𝑝−1)^1/𝑝𝑑𝑠. (5)

The proof of our main result is motivated by Laetsch [17] in which the existence and multiplicity of positive solutions of (4) with 𝑝=2 were studied via the quadrature method. Since then, there are plenty of research papers on the study of exact multiplicity of positive solutions of the 𝑝-Laplacian problem with general 𝑝>1 and some more special nonlinearities; see [18, 19] and the references therein. To find the exact number of positive solutions, the nonlinearity 𝑓 needs to satisfy some restrictive conditions, such as the monotonic condition or convex condition …. Our conditions (H1)–(H3) are not strong enough to guarantee the problem exact number of positive solutions.

The rest of the paper is arranged as follows. In Section 2, we state and prove some preliminary results. Finally in Section 3, we give the proof of Theorem 1.

2. Preliminaries

To prove our main results, we will use the uniqueness results due to Reichel and Walter [20] on the initial value problem 𝑢^′(𝑡)^𝑝−2𝑢^′(𝑡)^′(𝑢(𝑡𝑢(𝑎)+𝜆𝑓))=0,=𝑏,𝑢^′(𝑎)=𝑑, (6) where 𝑎∈[0,1], and 𝑏,𝑑∈ℝ.

Lemma 2. Let (H1) and (H2) hold. Then,(a)(6) with 𝑑≠0 has a unique local solution, and, the extension 𝑢(𝑡) remains unique as long as 𝑢^′(𝑡)≠0;(b)(6) with 𝑏∈(0,𝑟) and 𝑑=0 has a unique local solution.(c)(6) with 𝑏=0 and 𝑑=0 has a unique local solution 𝑢≡0.

Proof . (a) It is an immediate consequence of Reichel and Walter [20, Theorem 2].
(b) (H1) implies that f is local Lipschitz continuous. Combining this with the fact that 𝑓(𝑏)≠0 and using [20, (iii) and (v) in the case (𝛽) of Theorem 4], it follows that (6) with 𝑏∈(0,𝑟) and 𝑑=0 has a unique solution in some neighborhood of 𝑎.
(c) Define 𝑓(𝑠)=𝑓(𝑠)(),0≤𝑠<𝑟,−𝑓−𝑠,0≤−𝑟<𝑠<0, (7) and consider the auxiliary problem 𝑢^′(𝑡)^𝑝−2𝑢^′(𝑡)^′𝑓(𝑢(𝑡𝑢(𝑎)+𝜆))=0,=𝑢^′(𝑎)=0. (8) It follows from [20, (i) in the case (𝛿) of Theorem 4] that (8) has a unique local solution 𝑢≡0 in some neighborhood of 𝑎, and consequently, (6) with 𝑏=𝑑=0 has a unique local solution 𝑢≡0 in some neighborhood of 𝑎.

Lemma 3. Let (H1) and (H2) hold. Let (𝜆,𝑢) be a solution of 𝑢^′(𝑡)^𝑝−2𝑢^′(𝑡)^′(𝑢(𝑡(),𝑢(0)(1)+𝜆𝑓))=0,𝑡∈0,1=𝑢=0, (9) with ‖𝑢‖_∞=𝜌<𝑟. Let 𝑥₀∈(0,1) be such that 𝑢(𝑥₀)=‖𝑢‖_∞. Then, 𝑢(𝑡)>0 on (0,1), 𝑢^′(𝑡)>0 on (0,𝑥₀), and 𝑢^′(𝑡)<0 on (𝑥₀,1), and 𝑢𝑥₀−𝑡=𝑢𝑥₀+𝑡,𝑡∈0,min𝑥₀,1−𝑥₀. (10)

Proof . Since 𝑓(𝑠)≥0 for 𝑠∈[0,𝜌], it follows from (9) that 𝑢^′(𝑡)^𝑝−2𝑢^′(𝑡)^′≤0,𝑡∈[0,1]. (11) This, together with the fact that 𝑢^′(𝑥₀)=0, implies that 𝑢 is nondecreasing in [0,𝑥₀], and 𝑢 is nonincreasing in [𝑥₀,1].
We claim that 𝑢(𝑡)>0,𝑡∈(0,1). (12)
In fact, suppose on the contrary that there exists 𝜏∈(0,𝑥₀) such that 𝑢(𝜏)=0; then, 𝜏=max𝜏∈0,𝑥₀∣𝑢(𝜏)=0 (13) is well defined. Moreover, 𝑢(𝑡)≡0,𝑡∈(0,𝜏]. (14) By Lemma 2 (c), 𝑢(𝑡)≡0,𝑡∈0,𝜏+𝜂 (15) for some 𝜂>0. However, this contradicts the definition of 𝜏, see (13). Therefore, 𝑢(𝑡)>0 in (0,𝑥₀]. Similarly, we may show that 𝑢(𝑡)>0 in [𝑥₀,1).
Notice that (H2), (12), and (9) yield that 𝑢^′(𝑡)>0,𝑡∈0,𝑥₀,𝑢^′(𝑡)<0,𝑡∈𝑥₀,1. (16)
Now, since 𝑓 is independent of 𝑡, both 𝑢(𝑥₀−𝑡) and 𝑢(𝑥₀+𝑡) satisfy the initial value problem 𝑤^′(𝑡)^𝑝−2𝑤^′(𝑡)^′(𝑤(𝑡𝑤(0)𝑥+𝜆𝑓))=0,=𝑢₀,𝑤^′(0)=0. (17) From (16) and Lemma 2(a), it follows that both 𝑢(𝑥₀−𝑡) and 𝑢(𝑥₀+𝑡) can be uniquely extended to [0,min{𝑥₀,1−𝑥₀}]. Thus, we have from Lemma 2(b) that (17) has a unique solution 𝑢≡0 on [0,min{𝑥₀,1−𝑥₀}], and, accordingly, (10) is true.

Lemma 4. Let (H1) and (H2) hold. Assume that (𝜆,𝑢) is a positive solution of the problem (9) with ‖𝑢‖_∞=𝜌<𝑟 and 𝜆>0. Let 𝑥₀∈(0,1) be such that 𝑢(𝑥₀)=‖𝑢‖_∞. Then,(a)𝑥₀=1/2;(b)𝑥₀ is the unique point on which 𝑢 attains its maximum;(c)𝑢^′(𝑡)>0,𝑡∈(0,1/2).

Proof . (a) Suppose on the contrary that 𝑥₀≠1/2, and say that 𝑥₀∈(1/2,1); then, 2𝑥₀−1∈(0,1), and 0=𝑢(1)=𝑢2𝑥₀−1. (18) However, this contradicts (12). Therefore, 𝑥₀=1/2.
Also (b) and (c) can be easily deduced from (16).

3. Proof of the Main Result

To prove Theorem 1, we need the following quadrature method.

Lemma 5. For any 𝜌<𝑟, there exists a unique 𝜆>0 such that 𝑢^′(𝑡)^𝑝−2𝑢^′(𝑡)^′(𝑢(𝑡(),𝑢(0)(1)+𝜆𝑓))=0,𝑡∈0,1=𝑢=0, (19) has a positive solution (𝜆,𝑢) with ‖𝑢‖=𝜌. Moreover, 𝜌→𝜆(𝜌) is a continuous function on [0,𝑟).

Proof . By Lemma 4, (𝜆,𝑢) is a positive solution of (19), if and only if (𝜆,𝑢) is a positive solution of 𝑢^′(𝑡)^𝑝−2𝑢^′(𝑡)^′+𝜆𝑓(𝑢(𝑡))=0,𝑡∈(0,1),𝑢(0)=𝑢^′12=0. (20)(21) Suppose that (𝜆,𝑢) is a solution of (20), with ‖𝑢‖=𝜌. Then, 𝑢^′(𝑡)^𝑝𝑝=𝜆𝐹𝜌(𝑢(𝑡1𝑝−1−𝐹)),𝑡∈0,2, (22) and so 𝑡𝑝𝜆𝑝−1^1/𝑝=₀^𝑢(𝑡)𝐹𝜌−𝐹(𝑠)^−1/𝑝1𝑑𝑠,𝑡∈0,2. (23) Putting 𝑡=1/2, we obtain 𝜆^1/𝑝=2𝑝−1𝑝^1/𝑝𝜌₀𝐹𝜌−𝐹(𝑠)^−1/𝑝𝑑𝑠. (24) Hence, 𝜆 (if exists) is uniquely determined by 𝜌.
If 𝜌<𝑟, we define 𝜆(𝜌) by (24) and 𝑢(𝑡) by (23); it is straightforward to verify that 𝑢 is twice differentiable, 𝑢 satisfies (21), 𝑢>0 in (0,1), and 𝑢(1/2)=𝜌. The continuity of 𝜆(⋅) is implied by (24), and this completes the proof.

Lemma 6. Let (H3) hold, and let 𝑝∈(1,2). Then lim_{𝜌→𝑟−}𝜆𝜌=0. (25)

Proof . By (H3), there are positive numbers 𝑅<𝑟 and 𝑘 such that 𝑓(𝑠)≥𝑘^𝑝(𝑟−𝑠)^𝑝−1,𝑅≤𝑠<𝑟. (26) Thus, if 𝑅<𝜌<𝑟, (24) implies that 𝜆𝜌^1/𝑝=2𝑝−1𝑝^1/𝑝𝜌₀1𝐹𝜌−𝐹(𝑠)^1/𝑝𝑑𝑠=2𝑝−1𝑝^1/𝑝×^𝑅₀1𝐹𝜌−𝐹(𝑠)^1/𝑝+𝑑𝑠_𝜌𝑅1𝐹𝜌−𝐹(𝑠)^1/𝑝𝑑𝑠=2𝑝−1𝑝^1/𝑝×^𝑅₀1∫^𝜌_𝑠𝑓(𝑤)𝑑𝑤^1/𝑝+𝑑𝑠_𝜌𝑅1∫^𝜌_𝑠𝑓(𝑤)𝑑𝑤^1/𝑝≤2𝑑𝑠𝑘𝑝−1𝑝^1/𝑝2−𝑝^1/𝑝×^𝑅₀1(𝑟−𝑠)^2−𝑝−𝑟−𝜌^{2−𝑝1/𝑝}+𝑑𝑠_𝜌𝑅1(𝑟−𝑠)^2−𝑝−𝑟−𝜌^{2−𝑝1/𝑝}𝑑𝑠by()≤226𝑘𝑝−1𝑝^1/𝑝2−𝑝^1/𝑝×𝑅𝜉𝜌^−1/𝑝+12−𝑝₀^𝜉(𝜌)𝑥^−1/𝑝×𝑥+𝑟−𝜌^{2−𝑝(𝑝−1)/(2−𝑝)}𝑑𝑥⟶0, (27) as 𝜌→𝑟⁻, where 𝜉𝜌∶=(𝑟−𝑅)^2−𝑝−𝑟−𝜌^2−𝑝,𝑥=∶(𝑟−𝑠)^2−𝑝−𝑟−𝜌^2−𝑝. (28) If 𝜌→𝑟⁻, then 𝜉𝜌∶=(𝑟−𝑅)^2−𝑝−𝑟−𝜌^2−𝑝⟶(𝑟−𝑅)^2−𝑝,12−𝑝₀^𝜉(𝜌)𝑥^−1/𝑝𝑥+𝑟−𝜌^{2−𝑝(𝑝−1)/(2−𝑝)}⟶1𝑑𝑥2−𝑝^{(𝑟−𝑅)2−𝑝}₀𝑥^{(𝑝2−2)/𝑝(2−𝑝)}=𝑝𝑑𝑥2()𝑝−1𝑟−𝑅^{2(𝑝−1)/𝑝}. (29)

Lemma 7. Let (H1) and (H2) hold, and let 𝑝=2. Then, lim_{𝜌→𝑟−}𝜆𝜌=0. (30)

Proof . There are positive numbers 𝑅<𝑟 and 𝑘 such that 𝑓(𝑠)≥𝑘²𝑟−𝑠,𝑅≤𝑠<𝑟. (31) Thus, if 𝑅<𝜌<𝑟, (24) implies that 12𝜆𝜌^1/2=^𝜌₀1𝐹𝜌(𝑠)=−𝐹𝑑𝑠_𝑅0𝑑𝑠𝐹𝜌(𝑠)+−𝐹^𝜌𝑅𝑑𝑠𝐹𝜌(𝑠)=−𝐹_𝑅0𝑑𝑠∫^𝜌_𝑠𝑓(𝑤)+𝑑𝑤^𝜌𝑅𝑑𝑠∫^𝜌_𝑠𝑓(𝑤)𝑑𝑤≤^𝑅₀𝑘²ln𝑟−𝑅𝑟−𝜌^−1/2+𝑑𝑠_𝜌𝑅𝑘²ln𝑟−𝑠𝑟−𝜌^−1/2𝑑𝑠by()≤131𝑘𝑅𝜉𝜌^−1/2+×𝑟−𝜌_0𝜉(𝜌)𝑥^−1/2𝑒^𝑥𝑑𝑥⟶0, (32) as 𝜌→𝑟⁻, where 𝜉𝜌∶=ln𝑟−𝑅𝑟−𝜌,𝑥=∶ln𝑟−𝑠.𝑟−𝜌 (33)

Lemma 8. Let (H1), (H2), and (H3) hold, and assume that 𝑝∈(1,+∞). Then,(a)if 𝑓₀=∞, then lim_𝜌→0⁺𝜆(𝜌)=0;(b)if 𝑓₀=0, then lim_𝜌→0⁺𝜆(𝜌)=+∞;(c)if 𝑓₀∈(0,∞), then lim_𝜌→0⁺𝜆(𝜌)=𝜋^𝑝_𝑝/𝑓₀, where 𝜋_𝑝∶=2^{(𝑝−1)1/𝑝}₀11−(𝑠^𝑝/(𝑝−1))^1/𝑝𝑑𝑠. (34)

Proof . (a) If 𝑓₀=∞, then for any positive constant 𝑘, there exists 𝑅∈(0,𝑟) such that 𝑓(𝑠)≥𝑘^𝑝𝑠^𝑝−1,0<𝑠<𝑅. (35) Thus, if 0<𝜌<𝑅, (24) implies that 𝜆𝜌^1/𝑝=2𝑝−1𝑝^1/𝑝𝜌₀1𝐹𝜌−𝐹(𝑠)^1/𝑝𝑑𝑠=2𝑝−1𝑝^1/𝑝𝜌₀1∫^𝜌_𝑠𝑓(𝑣)𝑑𝑣^1/𝑝𝑑𝑠≤2𝑝−1𝑝^1/𝑝𝑝^1/𝑝𝑘^𝜌₀1𝜌^𝑝−𝑠^𝑝1/𝑝𝑑𝑠by(35)=2𝑝−1^1/𝑝𝑘¹₀1(1−𝑠^𝑝)^1/𝑝=2𝑑𝑠𝑘^{(𝑝−1)1/𝑝}₀11−𝑠^𝑝/(𝑝−1)^1/𝑝=1𝑑𝑠𝑘𝜋_𝑝, (36) which implies that lim_𝜌→0⁺𝜆(𝜌)=0.
(b) If 𝑓₀=0, then for any 𝜀>0, there exists 𝛿∈(0,𝑟) such that 𝑓(𝑠)≤𝜀^𝑝𝑠^𝑝−1,0<𝑠<𝛿. (37) Thus, if 0<𝜌<𝛿, (24) implies that 𝜆𝜌^1/𝑝=2𝑝−1𝑝^1/𝑝𝜌₀1𝐹𝜌−𝐹(𝑠)^1/𝑝𝑑𝑠=2𝑝−1𝑝^1/𝑝𝜌₀1∫^𝜌_𝑠𝑓(𝑣)𝑑𝑣^1/𝑝𝑑𝑠≥2𝑝−1𝑝^1/𝑝𝑝^1/𝑝𝜀^𝜌₀1𝜌^𝑝−𝑠^𝑝1/𝑝𝑑𝑠by(37)=2𝑝−1^1/𝑝𝜀¹₀1(1−𝑠^𝑝)^1/𝑝=2𝑑𝑠𝜀^{(𝑝−1)1/𝑝}₀11−𝑠^𝑝/(𝑝−1)^1/𝑝=1𝑑𝑠𝜀𝜋_𝑝, (38)
which implies that lim_𝜌→0⁺𝜆(𝜌)=+∞.
(c) If 𝑓₀∈(0,∞), then for any 𝜀∈(0,𝑓₀/2), there exists 𝛿∈(0,𝑟) such that 𝑓₀𝑓(𝑠)−𝜀≤𝑠^𝑝−1≤𝑓₀+𝜀,0<𝑠<𝛿. (39) Thus, if 0<𝜌<𝛿, (24) and the second part of (39) imply that 𝜆𝜌^1/𝑝=2𝑝−1𝑝^1/𝑝𝜌₀1𝐹𝜌−𝐹(𝑠)^1/𝑝𝑑𝑠=2𝑝−1𝑝^1/𝑝𝜌₀1∫^𝜌_𝑠𝑓(𝑣)𝑑𝑣^1/𝑝𝑑𝑠≥2𝑝−1𝑝^1/𝑝𝑝𝑓₀+𝜀^1/𝑝𝜌₀1𝜌^𝑝−𝑠^𝑝1/𝑝𝑑𝑠=2𝑝−1𝑓₀+𝜀^1/𝑝1₀1(1−𝑠^𝑝)^1/𝑝=2𝑑𝑠𝑓₀+𝜀^{1/𝑝(𝑝−1)1/𝑝0}11−𝑠^𝑝/(𝑝−1)^1/𝑝=1𝑑𝑠𝑓₀+𝜀^1/𝑝𝜋_𝑝, (40)
which implies that lim_𝜌→0+𝜆𝜌≥𝜋^𝑝_𝑝𝑓₀. (41)
Similarly, 𝜆𝜌^1/𝑝=2𝑝−1𝑝^1/𝑝𝜌₀1𝐹𝜌−𝐹(𝑠)^1/𝑝𝑑𝑠=2𝑝−1𝑝^1/𝑝𝜌₀1∫^𝜌_𝑠𝑓(𝑣)𝑑𝑣^1/𝑝𝑑𝑠≥2𝑝−1𝑝^1/𝑝𝑝𝑓₀−𝜀^1/𝑝𝜌₀1𝜌^𝑝−𝑠^𝑝1/𝑝𝑑𝑠=2𝑝−1𝑓₀−𝜀^1/𝑝1₀1(1−𝑠^𝑝)^1/𝑝=2𝑑𝑠𝑓₀−𝜀^{1/𝑝(𝑝−1)1/𝑝0}11−𝑠^𝑝/(𝑝−1)^1/𝑝=1𝑑𝑠𝑓₀−𝜀^1/𝑝𝜋_𝑝, (42)
which implies that if 0<𝜌<𝛿, (24) and the first part of (39) imply that lim_𝜌→0+𝜆𝜌≤𝜋^𝑝_𝑝𝑓₀. (43) Combining (41) and (43), it follows that lim_𝜌→0+𝜆𝜌=𝜋^𝑝_𝑝𝑓₀. (44)

Proof of Theorem 1. (i) It is from Lemma 8(a) that lim_𝜌→0+𝜆𝜌=0. (45) From Lemma 6 or Lemma 7, we have that lim_{𝜌→𝑟−}𝜆𝜌=0. (46) Combining this with (45) and using the facts that 𝜆(𝜌)>0 and 𝜆(𝜌) is continuous, it concludes that there exists 𝜆_∗>0 such that (4) has at least two positive solutions for 𝜆∈(0,𝜆_∗), has at least one positive solution for 𝜆=𝜆_∗, and has no positive solution for 𝜆>𝜆_∗.
(ii) It is an immediate consequence of Lemma 6, and Lemma 7, and Lemma 8(b).
(iii) It is an immediate consequence of Lemma 6, and Lemma 7, and Lemma 8(c).

Acknowledgments

The authors are very grateful to the anonymous referees for their valuable suggestions. This work is supported by the NSFC (no.11061030), SRFDP (no.20126203110004), and Gansu Provincial National Science Foundation of China (no. 1208RJZA258).