Nodal Solutions of the p-Laplacian with Sign-Changing Weight

Theorem 1. Let (H₁), ($$), and ($$) hold. Assume that, for some k ∈ ℕ, either

Then (1) has two solutions $$ and $$ such that $$ has exactly k − 1 zeros in (0,1) and is positive near 0 and $$ has exactly k − 1 zeros in (0,1) and is negative near 0.

Lemma 2 (see [18], [19].)Let (H₂) hold. Then

• (i)

  the set of all eigenvalues of the problem (13) is two infinite sequences of simple eigenvalues as follows:

  $$ ()

• (ii)

  for k ∈ ℕ and ν ∈ {+, −}, Ker$$ is a space of E with dimensional 1;

• (iii)

  the eigenfunction corresponding to $$ has exactly k − 1 simple zeros in (0,1).

Remark 3. Using the Gronwall inequality, we can easily show that all zeros of eigenfunction corresponding to eigenvalue $$ are simple.

It is very known that $$ is completely continuous in C¹[0,1]. Thus, the Leray-Schauder degree $$ is well-defined for arbitrary r-ball B_r(0) and $$ and ν ∈ {+, −}.

Lemma 4. For r > 0, we have

Proof. We divide the proof into two cases.

Case  1. λ ≥ 0. Since $$ is compact and linear, by [20, Theorem 8.10] and Lemma 2 (ii) with p = 2,

where m(λ) is the sum of algebraic multiplicity of the eigenvalues μ of (13) satisfying μ⁻¹λ > 1.

If $$, then there are no such μ at all; then

If $$ for some k ∈ ℕ, then

This together with Lemma 2 (ii) implies the following:

Case  2. λ < 0. In this case, we consider a new sign-changing eigenvalue problem as follows

where λ = −λ, $$. It is easy to check that Thus, we may use the result obtained in Case 1 to deduce the desired result.

Proposition 5. The eigenvalue function $$ is continuous.

Proof. We only show that $$ is continuous since the case of $$ is similar. In the following proof, we will shorten $$ to μ₁. From the variational characterization of μ₁(p), it follows that

Let $$ be a sequence in (1, +∞) convergent to p > 1. We will show that

To do this, let $$. Then, from (24),

On applying the Dominated Convergence Theorem, we find that

Relation (27), the fact that u is arbitrary and (24) yield

Thus, to prove (25), it suffices to show that

Let $$ be a subsequence of $$ such that $$.

Let us fix ɛ₀ > 0 so that p − ɛ₀ > 1 and, for each 0 < ɛ < ɛ₀, $$ is compactly embedded into L^p+ɛ(0,1). For k ∈ ℕ, let us choose $$ such that

For 0 < ɛ < ɛ₀, there exists k₀ ∈ ℕ such that p − ɛ < p_k < p + ɛ for any k ≥ k₀. Thus, for k ≥ k₀, (30) and Hölder′s inequality imply that This shows that $$ is a bounded sequence in $$. Passing to a subsequence if necessary, we can assume that u_k⇀u in $$ and hence that u_k → u in L^p+ɛ(0,1). Furthermore, u ∈ L^p(0,1) and u_k → u in $$ for k ≥ k₀. It follows that as k → +∞. It is clear that

Thus,

Similarly, we can also obtain that

where m⁺(t) = max {m(t), 0} and m⁻(t) = −min {m(t), 0}. Therefore,

We note that (30) and (31) imply that

for all k ∈ ℕ. Thus, letting k go to +∞ in (38) and using (37), we find that

On the other hand, since u_k⇀u in $$, from (32) we obtain that

Now, letting ɛ → 0⁺ and applying Fatou′s Lemma, we find that

Hence, u ∈ W^1,p(0,1); here W^1,p(0,1) denotes the radially symmetric subspace of W^1,p(0,1). We claim that actually $$. Indeed, we know that $$ for each 0 < ɛ < ɛ₀. For $$, it is easy to see that Then, letting ɛ → 0⁺, we obtain that where p^′ = p/(p − 1). Since ϕ is arbitrary, from Proposition IX-18 of [21], we find that $$, as desired.

Finally, combining (39) and (41), we obtain that

This and the variational characterization of μ₁(p) imply (29) and hence (25). This concludes the proof of the lemma.

Lemma 6. For fixed 2 ≤ k ∈ ℕ and ν ∈ {+, −}, $$ as a function of p ∈ (1, +∞) is continuous.

Proof. Let $$ be an eigenfunction corresponding to $$. By Lemma 2 and Remark 3, we know that u has exactly k − 1 simple zeros in I; that is, there exist c_k,1, …, c_k,k−1 ∈ I such that u(c_k,1) = ⋯ = u(c_k,k−1) = 0. For convenience, we set c_k,0 = 0, c_k,k = 1, and J_i = (c_k,i−1, c_k,i) for i = 1, …, k. Let $$ denote the first positive or negative eigenvalue of the restriction of problem (13) on J_i for i = 1, …, k. Lemma 3 of [18] follows that $$ for i = 1, …, k. Using a similar proof to Proposition 5, we can show that $$ is continuous with respect to p for i = 1, …, k. Therefore, $$ is also continuous with respect to p.

Lemma 7. (i) Let $$ be the sequence of positive eigenvalues of (13). Let λ be a constant with $$ for all k ∈ ℕ. Then, for arbitrary r > 0,

where β is the number of eigenvalues $$ of problem (13) less than λ.

(ii) Let $$ be the sequence of negative eigenvalues of (13). Consider $$, k ∈ ℕ; then

where β is the number of eigenvalues $$ of problem (25) larger than λ.

Proof. We will only prove the case $$ since the proof for the other cases is similar. We also only give the proof for the case p > 2. Proof for the case 1 < p < 2 is similar. Assume that $$ for some k ∈ ℕ. Since the eigenvalues depend continuously on p, there exists a continuous function χ : [2, p] → ℝ and q ∈ [2, p] such that $$ and λ = χ(p). Define

It is easy to show that Φ(q, u) is a compact perturbation of the identity such that, for all u ≠ 0, by definition of χ(q), Φ(q, u) ≠ 0, for all q ∈ [2, p]. Hence, the invariance of the degree under homotopology and the classical result for p = 2 imply

Lemma 8 (see [17].)Let X be a Banach space. Let F : ℝ × X → X be completely continuous such that F(λ, 0) = 0 for all λ ∈ ℝ. Suppose that there exist constants ρ, η ∈ ℝ, with ρ < η, such that (ρ, 0) and (η, 0) are not bifurcation points for the equation

Furthermore, assume that where B_r(0) = {u ∈ X : ∥u∥ < r} is an isolating neighborhood of the trivial solution for both constants ρ and η. Let and let 𝒞 be the component of 𝒮 containing [ρ, η]×{0}. Then, either

• (i)

  𝒞 is unbounded in ℝ × X or

• (ii)

  𝒞∩[(ℝ∖[ρ, η]) × {0}] ≠ ∅.

Lemma 9. Assume that (H₁), (H₂), and (H₃) hold. Then, for fixed p > 1 and for fixed σ ∈ {+, −}, each $$ is a bifurcation point of (12) and the associated bifurcation branch $$ satisfies the following;

• (1)

  $$ is unbounded in E;

• (2)

  $$, where $$ is the set of function $$ which has exact k − 1 simple zeros in (0,1), and σu is positive near 0.

Lemma 10. Let (H₂) hold. Let I = [a, b] be such that I ⊂ I₊ and

Let g_n : [0,1]→(0, +∞) be such that

Let y_n ∈ E be a solution of the equation

Then, the number of zeros of y_n|_I goes to infinity as n → +∞.

Proof. After taking a subsequence if necessary, we may assume that

as j → +∞. It is easy to check that the distance between any two consecutive zeros of any nontrivial solution of the equation goes to zero as j → +∞. Using this with [21, Lemma 2.5], it follows the desired results.

Theorem 11. Let (H₁), (H₂), (H₃), and (H₄) hold. Assume that, for some k ∈ ℕ, either

or

Then, (4) has two solutions $$ and $$ such that $$ has exactly k − 1 zeros in (0,1) and is positive near 0 and $$ has exactly k − 1 zeros in (0,1) and is negative near 0.

Proof. We only prove the case of γ > 0. The case of γ < 0 is similar. Consider the problem

Considering the results of Lemma 9, we have that, for each integer k ≥ 1, σ ∈ {+, −}, there exists a continuum $$ of solutions of (62) joining $$ to infinity in $$. Moreover, $$.

It is clear that any solution of (62) of the form (1, u) yields a solution u of (4). We will show that $$ crosses the hyperplane {1} × E in ℝ × E. To this end, it will be enough to show that $$ joins $$ to $$. Let $$ satisfy

We note that η_n > 0 for all n ∈ ℕ since (0,0) is the only solution of (62) for λ = 0 and $$.

Case  1. $$. In this case, we only need to show that

We divide the proof into two steps.

Step  1. We show that, if there exists a constant number M > 0 such that

for n ∈ ℕ large enough, then $$ joins $$ to $$.

In this case, it follows that

Let ξ ∈ C(ℝ) be such that

Then, Let Then, $$ is nondecreasing and

We divide the equation

by ∥y_n∥ and set $$. Since $$ is bounded in E, after taking a subsequence if necessary, we have $$ for some $$ and $$ in Y with $$. Moreover, from (70) and the fact that $$ is nondecreasing, we have since

By the continuity and compactness of G_p, it follows that

where $$, again choosing a subsequence and relabeling if necessary.

We claim that

Suppose on the contrary that $$. Since $$ is a solution of (74) and all zeros of $$ in [0,1] are simple, it follows that $$ for some h ∈ ℕ and ι ∈ {+, −}.

By the openness of $$, we have that there exists a neighborhood $$ such that

which contradicts the facts that $$ in E and $$. Therefore, $$. Moreover, by Lemma 2, $$, so that Therefore, $$ joins  $$ to $$.

Step  2. We show that there exists a constant M such that μ_n ∈ (0, M] for n ∈ ℕ large enough.

On the contrary, we suppose that

Since $$, it follows that

Let

be the zeros of y_n in [0,1]. Then, after taking a subsequence if necessary, Notice that Lemma 10 and the fact that y_n has exactly k − 1 simple zeros in [0,1] yield which implies that

However, this contradicts (H₂): 0 < meas   I⁻ < 1.

Case  2. $$. In this case, we have that

Assume that $$ is such that

If η_n → +∞, then we are done!

If there exists M > 0, such that, for n ∈ ℕ sufficiently large,

Applying the same method used in Step 1 of Case 1, after taking a subsequence and relabeling if necessary, it follows that

Thus, $$ joins $$ to $$.