Bifurcation analysis for a modified quasilinear equation with negative exponent

Proposition 3.1

Assume that the changing of variables h is defined section 2. Then the map F : ℝ × Cϕα+ → 𝓒_ϕα is well defined and analytic.

Proof

We split the proof in three steps.

1. h(ω) ∈ Cϕα+ for any ω ∈ Cϕα+ . We just consider the case 0 < α < 1, because the case α ≥ 1 is similar. Since ϕ_α = φ₁, it follows from the properties of h(t) and ∥φ₁∥_L^∞(Ω) = 1, that there exist positive constants C₁, C₂ depend on ω such that

   0<C1≤h(ω)φ1(x)≤ωφ1(x)≤C2<∞,∀x∈Ω.

   Then h(ω)^−α ∈ Cϕα−α+ (Ω) for ω ∈ Cϕα+ . By the fact that λ > 0, 0 < a(x) ∈ 𝓒(Ω) and h^′(ω) ∈ 𝓒(Ω), we conclude that λ a(x)h(ω)^−αh^′(ω) ∈ Cϕα−α+ (Ω).

2. For any ω ∈ Cϕα−α (Ω), by similar ideas made in the proofs of Proposition 2.3 in [5], we are able to obtain that ω ↦ (−Δ)⁻¹ω ∈ 𝓒_ϕα(Ω) is a linear continuous map and hence analytic.

3. Due to α > 0, 1 < β ≤ 22^* − 1 and the properties of h(t), it is easy to see that (−Δ)⁻¹b(x)h(ω)^βh^′(ω) ∈ 𝓒_ϕα(Ω) for any ω ∈ Cϕα+ .

Hence, by the above three steps, we can obtain the result.□

Now, to show the existence of the analytic global path of solutions to F in ℝ⁺ × Cϕα+ (Ω), we consider the following problem:

where k ≥ 0 and g(x) is a local Hölder continuous function in Ω.

To begin with, we have the following comparison principle.

Lemma 3.2

Assume that there exist u and v satisfying the following inequalities in the weak sense,

then there holds u ≤ v in Ω.

Proof

Arguing by contradiction, assume that Ω₀ := {x ∈ Ω: u(x) > v(x)} ≠ ∅. For fixed ϵ > 0, let us define

and

By pointwise limit, we get

where χ_Ω0 represent the characteristic function of Ω₀.

By taking derivatives, we get

and

in Ω₀. From the above argument, we have that φ_ϵ, ψ_ϵ ∈ Hloc1 (Ω) ∩ 𝓒₀(Ω). So, by density arguments, we are able to test the first and second inequalities in (3.2) against φ_ϵ and ψ_ϵ, respectively, to obtain

Now, set W1:=∇ln⁡uϵ=∇uuϵ and W2:=∇ln⁡vϵ=∇vvϵ in Ω₀, it follows from (3.3) and Lemma 4.2 of [23], that

On the other hand, since a(x) > 0 and h(t)^−αh^′(t) is decreasing for t > 0 and α > 0, we have

Besides, we are able to obtain

This, together with (3.4) and (3.5), implies

which is a contradiction. Thus, Ω₀ = ∅, and hence the proof is completed.□

Lemma 3.3

There exists a unique weak solution w ∈ W01,q (Ω) ∩ 𝓒₀(Ω) for some q > 1 to problem (3.1). Furthermore, w ∈ [cϕ_α, ϕ] for c > 0 small enough if there exists ϕ ∈ Cϕα+ (Ω) which is a super-solution of (3.1). In particular, w ∈ Cϕα+ (Ω).

Proof

Given k > 0, 0 ≤ g(x) ∈ L^∞(Ω), we consider the following approximated problem:

It is easy to check that for c > 0 small enough, ψ_ϵ=(c1+α2φ1+ε1+α2)21+α−ε is a sub-solution of (3.6) for ε > 0. The unique positive solution ψ_ε ∈ H01 (Ω) of

is a super-solution of (3.6). Indeed, it follows from the monotonicity of h(t)^−αh^′(t) that

By comparison principle, it is obvious that ψ^ϵ < ψ_ε. Then we obtain a solution w_ε ∈ [ψ^ϵ, ψ_ε] to (3.6) by standard arguments, and uniquely by the non-increasing nature of the right hand side in (3.6). Thus, we can infer that w_ε is Hölder continuous on Ω through elliptic regularity. In addition, w_ε > 0 in Ω by maximum principle.

Now we prove that w_ε is monotone as ε → 0⁺ by a comparison argument: let 0 < ε^′ < ε, then we have

On the other hand, assume x0=arg⁡minΩ¯(wε′−wε)∈Ω, and (w_ε^′ − w_ε)(x₀) ≤ 0, then it follows that

which is a contradiction with the last equation. Thus we have w_ε^′ > w_ε in Ω if 0 < ε^′ < ε. Therefore, we obtain that

and satisfies in the sense of distributions of (3.1).

Furthermore, w ∈ W01,q (Ω) for some q > 1. Indeed, from g(x) ∈ L^∞(Ω) and (3.7), it is easy to get that g(x)+λ a(x)h(w)^−αh^′(w) ∈ L¹(Ω, d(x, ∂Ω)^s) for some s < 1. Then, it follows from Theorems 3 and 4 of [15], that w ∈ W01,q (Ω) for some q > 1. Using the comparison principle again, we can derive that w is the unique weak solution of (3.1).

Now, assume that (3.1) has a super-solution ϕ ∈ Cϕα+ (Ω). It is clear that ϕ is also a super-solution of (3.6). Thus, for c > 0 small enough, we get cϕ_α ≤ ϕ. Hence ψ_ε can be replaced by ϕ and repeat the above argument to get a solution w ∈ Cϕα+ (Ω). Thus we complete the proof.□

Since 0 < a(x) ∈ 𝓒(Ω) and h(t)^−αh^′(t), t > 0, is decreasing, then (h(t)^−αh^′(t))^′ ≤ 0. From the idea of [5], let m(x) := −a(x)(h(ω)^−αh^′(ω))^′, then we have m(x) ≥ 0. Now, we consider the following problem

We have the following result.

Lemma 3.4

Assume that m(x) is defined as above, 0 ≤ m(x) ≤ m₁d(x, ∂Ω)⁻² for some positive constant m₁. Then, for given z ∈ 𝓒_ϕα(Ω), there exists a unique v ∈ 𝓒_ϕα(Ω) solves −Δ v + m(x)v = m(x)z in Ω. Furthermore, ∥v∥Cϕα(Ω)≤C∥mz∥Cϕα−α(Ω) for some constant C > 0 independent of z.

To this end, we need the next Lemma.

Lemma 3.5

If m(x) is defined as above, then there exists positive constant m₁ such that m(x) ≤ m₁d(x, ∂Ω)⁻².

Proof

Since a(x) ∈ 𝓒(Ω) be bounded, it suffices to prove that there exists a positive constant C such that −(h(ω)^−αh^′(ω))^′ ≤ Cd(x, ∂Ω)⁻² for all ω > 0. In the following process, the C represents different positive constant. A direct calculation shows that

Note that h(ω)^−α−1 ∼ d(x, ∂Ω)⁻² near ∂Ω. Besides this, by Lemma 2.1-(3) and −(6), we have (h^′(ω))² ≤ C and [h(ω)h^′(ω)]² ≤ C for all ω > 0. Now, it is obvious that −(h(ω)^−αh^′(ω))^′ ≤ Cd(x, ∂Ω)⁻² for all ω > 0. Consequently, one can choose a positive constant m₁ such that m(x) ≤ m₁d(x, ∂Ω)⁻².□

Corollary 3.6

Let g(x) ∈ Cϕα−α (Ω), m(x) ∈ 𝓒(Ω) and 0 ≤ m(x) ≤ m₁ d(x, ∂Ω)⁻² for some positive constant m₁. If v ∈ 𝓒_ϕα(Ω) ∩ 𝓒²(Ω) is a classical solution of

then ∥v∥Cϕα(Ω)≤c∥g∥Cϕα−α(Ω) for c > 0 independent of g.

Lemma 3.7

The map ∂_ωF(λ, ω) is Fredholm with index 0 for all (λ, ω) ∈ ℝ⁺ × Cϕα+ (Ω). %for b⁺ ≠ 0.

Proof

Rewrite ∂_ωF(λ, ω) = I + A_λ + B, where A_λv = −λ(−Δ)⁻¹ a(x)(h(ω)^−αh^′(ω))^′v, Bv = −(−Δ)⁻¹b(x)(h(ω)^βh^′(ω))^′ v.

Applying Lemma 3.4 with m(x) = −λ a(x)(h(ω)^−αh^′(ω))^′, which turns out that I + A_λ is invertible on 𝓒_ϕα(Ω). On the other hand, B is compact on 𝓒_ϕα(Ω). Thus, ∂_ωF(λ, ω) is Fredholm with index 0.□

Lemma 3.8

It holds 0 < Λ < ∞ and (2.1) admits a minimal solution ω_λ ∈ Cϕα+ (Ω) for all 0 < λ < Λ with b⁺(x) ≠ 0.

Proof

Let ϕ_λ=cλ11+αϕα, then we can check that ϕ_λ is a sub-solution of (2.1) for all λ > 0 if c > 0 is chosen small enough. Next, we find a super-solution to (2.1). Consider the following problem

Since the conditions (g1) and (g2) of Theorem 2.2 in [10] are fulfilled by the nonlinear perturbation of (3.8), we conclude that (3.8) admits a unique solution ψ_λ ∈ Cϕα+ (Ω) for any α > 0. Let ν solves

Define ϕ_λ := ψ_λ + Mν. For an appropriate constant M > 0 and some λ > 0, ϕ_λ is a super-solution of (2.1) for 0 < λ < λ. Indeed, combining the monotonicity of h(t)^−αh^′(t) with the properties of h(t), we have

if λ + M > λ + b(x)h(ϕ_λ)^βh^′(ϕ_λ) sufficiently large. This also implies λ > 0. Note that we can obtain ϕ_λ ∈ Cϕα+ (Ω), then there holds liminfλ→0+infx∈Ωϕ¯λλ11+αϕα>0. This implies for all 0 < λ < λ, there holds ϕ_λ < ϕ_λ in Ω if choose c > 0 small enough.

Consider the following monotone iterative scheme for all λ ∈ (0, λ),

with ω₀ = ϕ_λ and k = k(λ) > 0 large enough such that b(x)h(t)^βh^′(t) + kt is non-decreasing on [0, ∥ϕ_λ∥_L^∞(Ω)]. From the comparison principle and Lemma 3.3, we can get the existence of ω_n, and ϕ_λ < ω_n < ϕ_λ. It is easy to check that ϕ_λ, ϕ_λ are respectively sub and super-solution to (3.10). Furthermore, due to the monotonicity of the left side of (3.10), it follows that the monotonicity of the iterates, that is ω_n ≥ ω_n−1. By Ascoli-Arzela Theorem, there exists ω_λ such that ω_n → ω_λ in Cϕα+ (Ω) as n → ∞ and ϕ_λ ≤ ω_λ ≤ ϕ_λ. That is, ω_λ is a minimal solution of (2.1) for 0 < λ < λ.

Now, we set

From the above argument, we have Λ > 0. We claim that Λ < ∞. In fact, taking φ₁ as the test function in (2.1), we obtain

If we chose a λ > 0, large enough if necessary, such that λ a(x)h(t)^−αh^′(t) + b(x)h(t)^βh^′(t) > 2λ₁t for all t > 0, which is a contradiction with (3.11). Thus Λ < ∞ must holds. Now, we can further get the existence of minimal solution ω_λ ∈ Cϕα+ (Ω) for (2.1) with any 0 < λ < Λ. In fact, taking ϕ_λ=cλ11+αϕα as a sub-solution, and ϕ_λ^′, solves problem (2.1)_λ^′, for appropriate λ < λ^′ < Λ as a super-solution of (2.1). Then by the similar proceeding above, we can conclude that there exists a minimal solution ω_λ ∈ Cϕα+ (Ω) for all 0 < λ < Λ, which completes the proof.□

Remark 3.9

Suppose that there exists M₀ > 0 such that

if further choose λ₀ small enough such that sup0<λ<λ0∥ωλ∥C(Ω¯)<M0. Then ω_λ is the unique solution in (0, λ₀) × {ω ∈ C0+ (Ω) : ∥ω∥_𝓒(Ω) < M₀}.

Indeed, suppose ω̃_λ is another solution and satisfies ∥ω̃_λ∥_𝓒(Ω) < M₀ with λ < λ₀. Let ψ_λ = ω_λ − ω̃_λ, then ψ_λ solves

where ξ_λ lies between ω_λ and ω̃_λ. It is easy to see λ a(x)(h(ξ_λ)^−αh^′(ξ_λ))^′ + b(x)(h(ξ_λ)^βh^′(ξ_λ))^′ ≤ 0 and hence ψ_λ ≡ 0.

Lemma 3.10

Let ω ∈ 𝓒²(Ω) ∩ Cϕα+ (Ω), λ > 0. If ∂_ωF(λ, ω)φ = 0 for some φ ∈ 𝓒²(Ω) ∩ 𝓒_ϕα(Ω), then φ ∈ H01 (Ω) ∩ 𝓒_φ1(Ω) and is a H¹-weak solution for −Δφ − [λ a(x)(h(ω)^−αh^′(ω))^′ + b(x)(h(ω)^βh^′(ω))^′]φ = 0. Inversely, if φ ∈ H01 (Ω) is a non-negative H¹-weak solution for −Δφ − [λ a(x)(h(ω)^−αh^′(ω))^′ + b(x)(h(ω)^βh^′(ω))^′]φ = θφ for some θ ∈ ℝ, then φ ∈ 𝓒²(Ω) ∩ 𝓒_φ1(Ω).