A Multiplicity Result for Quasilinear Problems with Nonlinear Boundary Conditions in Bounded Domains

Proposition 1.1. Assume (𝒜₁), then for every $$,

defines a norm in W^1,p(Ω) which is equivalent to the usual norm of W^1,p(Ω), that is,

Proof. See [2].

Theorem 1.2. Assume (𝒜₁), (𝒜₂), (ℬ₁), and (ℬ₂). Then, for every $$, problem (1.2) admits at least one positive solution u ∈ W^1,p(Ω)∩L^∞(Ω).

Theorem 1.3. Assume (𝒜₁), (𝒜₂), (ℬ₁), (ℬ₂), and (ℬ₃). Then, problem (1.2) has at least one positive solution u ∈ W^1,p(Ω)∩L^∞(Ω) for $$.

Theorem 1.4. Assume (𝒜₁), (𝒜₂), (ℬ₁), (ℬ₂), and (ℬ₃). Then, there exists $$ such that problem (1.2) admits at least two distinct positive weak solutions in W^1,p(Ω)∩L^∞(Ω), whenever $$.

Corollary 1.5. Assume ∫_Ωb(x)dx < 0. Then, the problem

has a positive solution.

Lemma 2.1. Let u ∈ X − {0} and t > 0. Then, $$ if and only if $$.

Proof. The result is an immediate consequence of the fact that

Lemma 2.2. Suppose that u₀ is a critical point of J_λ, where A_λ(u₀) and B(u₀) have the same sign. Then, $$.

Proof. Let u₀ ∈ X, B(u₀) ≠ 0, then

for all v ∈ X. This completes the proof.

Lemma 2.3. Suppose that H is a well-defined functional on X and u₀ is a minimizer of J_λ on

for some c ≠ 0. Then,$$.

Proof. If u₀ is a local minimizer of J_λ on S, then u₀ is a solution of the optimization problem:

where γ(u) = H(u) − c. Hence, by the theory of the Lagrange multipliers, there exists μ ∈ R such that $$. Thus, Note that the functional J_λ is 0-homogeneous and the Gateaux derivative of J_λ at the point v ∈ X, B(v) ≠ 0, in direction v, is zero, that is, $$. It then follows that μ = 0 due to u₀ ∈ S. Hence, the proof is complete.

Proof of Theorem 1.2. Suppose that (𝒜₁), (𝒜₂), (ℬ₁), (ℬ₂) satisfy. It follows from variational characterization of μ(α, λ) that A_λ(u) ≥ 0 for all u ∈ X and $$. Moreover, we have

Hence, for using Lemma 2.3, it is sufficient to consider the case A_λ(u) = c ≠ 0, for example c = 1. In this case, we have From the necessary condition for the existence of t(u), we have B(u) > 0. It follows that we must consider a critical point with B(u) > 0.

The functional J_λ(u) is nonnegative and so bounded below, hence we can look for positive local minimizer for J_λ(u) on X.

Let us consider variational problem:

Note that for $$, this set is not empty, and from Lemma 2.3, the solution of this problem is a minimizer of J_λ(u) on X.

Suppose {u_n} is the maximizing sequence of this problem. It follows from the equivalent property in Proposition 1.1, {u_n} is bounded and so we may assume that u_n⇀u₀ in X. Since X may be compactly embedded in L^p(Ω), L^γ(Ω), and L^p(∂Ω), we have u_n → u₀ in L^p(Ω), L^γ(Ω) and L^p(∂Ω). Hence,

Moreover, we have Here, the weak lower semicontinuity of the equivalent norm was used.

Assume that A_λ(u₀) < 1. By using the map:

we have L(1) < 1 and lim _t→∞L(t) = ∞ and so L(t₀) = 1 for some t₀ > 1, that is, A_λ(t₀u₀) = 1. So we derive

which is a contradiction. Hence, A_λ(u₀) = 1 and so u₀ is a maximizer. As Lemma 2.2, the result would follow by considering u₁ = t(u₀)u₀ ∈ X. Then, u₁ is a weak solution of (1.2) and u₁ ≥ 0 in Ω. Now, following the bootstrap argument (used, e.g., in [9]), we prove u₁ ∈ L^∞(Ω). Then, we can apply the Harnack inequality due to Trudinger [5] in order to get u₁ > 0 in Ω (cf. [9]).

Proof of Theorem 1.3. If $$, It is easy to see that the set {u ∈ X; A_λ(u) = 1} is unbounded and we are forced to require an additional assumption (ℬ₃). In this case, again we are looking for a maximizing of the problem:

Suppose {u_n} is a maximizing sequence of this problem. First, we investigate the case when {u_n} is unbounded. Then, we may assume without loss of generality that ∥u_n∥→∞. So, we obtain where v_n = u_n/∥u_n∥. Thus, we have Since ∥v_n∥ = 1, we may assume that v_n⇀v₀ in X. Again, since X may be compactly embedded in L^p(Ω), L^γ(Ω) and L^p(∂Ω), we have So, And, therefore, v₀≢0. Returning to $$, we have also Due to simplicity of $$, there exists c ≠ 0 such that $$. Therefore, we have that implies $$, which contradicts (ℬ₃). Therefore, {u_n} is bounded and so u_n⇀u₀ in X and u_n → u₀ in L^p(Ω), L^γ(Ω) and L^p(∂Ω). Hence, $$, and so u₀ ≠ 0.

Moreover, we obtain $$. Here, the variational characteristic of $$ and the weak lower semicontinuity of the norm were used. Let us now consider the case $$. Again, we obtain $$ for some c ≠ 0, and so $$ which contradicts (ℬ₃).

Now we prove that $$. Suppose otherwise, then by using L(t) in (2.20), we obtain some t₀ > 1 such that $$. By direct calculation, we get a contradiction like (4.1). This yields that u₀ is a nonnegative solution of problem (1.2). Using the same ideas as the proof of Theorem 1.2, we have proved Theorem 1.3.

Lemma 4.1. Under the assumption of Theorem 1.4, there exists a maximizer u₁ of the problem:

whenever $$, for some δ > 0. Moreover, u₁ ∈ X∩L^∞(Ω) is a positive weak solution of problem (1.2).

Proof. First note that using L(t) in (2.20), it is easy to see that u₁ is a maximizer of problem (4.2) if and only if u₁ is a maximizer of the problem:

Now suppose that the result is false. Then, there exists a sequence {δ_k} such that δ_k → 0 and problem (4.1) has no solution for $$. For simplicity, we use $$. Let $$ be a maximizing sequence of this problem, that is, We prove that if $$ be bounded or unbounded, we arrive at a contradiction and so the lemma is proved.

The first case: $$ is bounded. Thus, $$ in X for some $$ and $$ in L^p(Ω), L^γ(Ω), and L^p(∂Ω). Hence, $$ and the weak lower semicontinuity of the norm gives $$. Therefore, $$ is a solution of problem (4.5), which is a contradiction.

The case $$ is unbounded. Then we may assume that $$. Let $$. Then, $$ in X for some $$ and $$ in L^p(Ω), L^γ(Ω) and L^p(∂Ω). This implies that

Therefore, $$. Furthermore, $$ and so $$. Thus, we arrive at which implies It is a direct consequent of the comactly embedding of X in L^p(Ω), L^γ(Ω) and L^p(∂Ω). Now, we pass to the limit for k → ∞. Then, $$ and since $$ is a bounded sequence, we may assume that $$ in X for some v₀ ∈ X and $$ in L^p(Ω), L^γ(Ω), and L^p(∂Ω).

It follows from (4.10) that

and from the variational characteristic of $$ and (4.6) that which contradicts (ℬ₃).

Hence, for some δ > 0, problem (4.5) has at least one nonnegative solution u₁ for any $$.

Lemma 4.2. Under the assumption of Theorem 1.4, there exists ɛ > 0 such that for $$, problem (1.1) has a nonnegative solution u₂ satisfying A_λ(u₂) < 0.

Proof. First note that using the auxiliary function L(t) = |t|^γB(u) and the assumption (ℬ₃), it is easy to see that $$ for some t₁ < 1. Thus, the set {u ∈ X; B(u) = −1} is nonempty. Also, for this t₁, we have

for $$.

Again, we assume that the result is not true. Then, there exists ɛ_k → 0 such that for $$, problem (1.1) has no solution.

Let $$ be a minimizing sequence of this problem, that is,$$

Assume that $$ is bounded. Using similar argument as in the proof of Lemma 4.1, we find a solution of problem (1.1) which is a contradiction. Let us assume $$ is bounded. Again, we consider

$$ is bounded, and so $$ in X for some $$. Thus,$$.

Letting n → ∞, we arrive at $$. It follows easily that $$, and so $$. As might be expected, we arrive at a contradiction with assumption (ℬ₃), by using similar steps as in the proof of Theorem 1.3. This completes the proof.

Proof of Theorem 1.4. Let η = min {ɛ, δ}, $$ and w₁ = t(u₁)u₁ and w₂ = t(u₂)u₂, where t(u_i) is defined by (2.16).

It easy to see that A_λ(w₁) > 0 and A_λ(w₂) ≤ 0, and so w₁≢w₂. Thus, w₁ and w₂ are two distinct non negative weak solutions to (1.2). Other properties of w₁ and w₂ follow in the same way as in Section 2.