Existence of Sign-Changing Solutions to Equations Involving the One-Dimensional p-Laplacian RuyunMa and

u 󸀠 (t)) 󸀠 + λf(u(t)) = 0, 0 < t < 1, and u(0) = u(1) = 0, where p > 1, λ > 0, f ∈ C1(R;R), f(s)s > 0, and s ̸ = 0. We show the existence of sign-changing solutions under the assumptions f ∞ = lim |s|→∞ (f (s) /s p−1 ) = +∞ and f 0 = lim |s|→0 (f(s)/s p−1 ) ∈ [0,∞]. We also show that (P) has exactly one solution having specified nodal properties for λ ∈ (0, λ∗) for some λ∗ ∈ (0,∞). Our main results are based on quadrature method.


Introduction and Main Results
Existence and multiplicity of positive solutions of nonlinear second order boundary value problem 1/(1 − (  /( − 1))) 1/ .The main results of this paper are the following.
Remark 5.It is worth noticing that Lee and Sim [7] studied the nonautonomous cases (2), (3) and obtained the existence of positive solutions with  ∞ = ∞,  0 ∈ (0, ∞).They gave no information about the sign-changing solutions.In Theorem 1, we show the existence of solutions having specified nodal properties.
Remark 6. Very little is known in the available literature even in the special case  = 2.We establish uniqueness results in this paper; see Theorems 1 and 2.
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 proofs of Theorems 1, 2, and 3.
The proof of the above theorem follows by carefully extending the arguments used in [15,Theorem 2.2] for second order differential equation to the case of one-dimensional -Laplacian.
Using the same argument, with obvious changes, we may deduce the following.

The Proofs of the Main Results
Proof of Theorem 1. First, we consider  = 2 + 1.

2(
− 1  ) To this end, we have from (H1) that, for any  ∈ N, there exists If  < −  , it follows from (28) and (29) that we have that where It follows from the fact that  is sufficiently large and (30) that Therefore, from ( 27) and (32), we have that lim ℎ → ∞  2 (ℎ) = 0.

2(
− 1  ) Since  0 = 0, then, for any  > 0, there exists  ∈ (0, ∞) such that Thus, if − <  < 0, from (54), we have that From the fact that  is small and combining (52) and (55), we get that lim By analyzing 1 2 (ℎ) defined in (20) instead of  2 (ℎ) in the proof of the above, we have the same result.The proof of  * is similar to the proof of Theorem 1.We omit it here.Finally, if  = 2, then it clearly follows by analyzing  2−1 (ℎ) defined in (21) instead of  2 (ℎ) in the proof of the case  = 2 + 1.
It follows from the quadrature method that a solution with 2 interior zeros exists if for  > 0 there exists ℎ ∈ (0, ∞) such that () 1/ =  2 (ℎ).We prove this by proving The proof of (A2) is the same as the proof of (A) of Theorem 1, so we omit it here; we are only to prove (B2).