A Hopf Bifurcation in a Three-Component Reaction-Diffusion System with a Chemoattraction

Lemma 1. The functions G(·, η):(0, ∞) → X, $$, J(·, η):(0, ∞) → X, C(·) : W → ℂ, and f : W → X × ℝ are continuously differentiable with derivatives given by

Theorem 2. Suppose that (1/2) − a(1 − s₀) < μ/(1 + μ) and C^′(γ(η); a(α(η) − s₀)) + $$ for all η > 0. Then (18) has at least one equilibrium solution (0,0, 0, η^*) for κ < κ_c, where κ_c is a solution of

The linearization of f at the stationary solution (0,0, 0, η^*) is

where $$ − $$. The pair (0,0, 0, η^*) corresponds to a unique steady state (v^*, p^*, w^*, η^*) of (10) for σ ≠ 0 with v^*(x) = g(x, η^*), $$, and w^*(x) = j(x, η^*) − s₀.

Proof. From the system of (22), we have u^* = 0, z^* = 0, and q^* = 0. In order to show existence of η^*, we define

Then Since γ^′(η) < 0 and α^′(η) < 0 for all η > 0, Γ(η, κ) = 0 is solvable with η^* if Γ(0, κ) > 0, Γ(∞, κ) < 0, and (∂/∂η)Γ(η, κ) < 0, which means that C(γ(0); a(α(0) − s₀)) > 0, C(γ(∞); a(α(∞) − s₀))   + $$, and $$.

Let  κ_c  be a solution of

Then Γ(∞, κ) < Γ(∞, κ_c) < Γ(0, κ_c) with Γ(∞, κ_c) = 0. Hence, η^* exists for κ < κ_c.

The formula for Df(0,0, 0, η^*) follows from the relation C^′(1/2) = 4, and the corresponding steady state (v^*, p^*, w^*, η^*) for (10) is obtained by using Theorem  2.1 in [19].

Definition 3. Under the assumptions of Theorem 2, define (for 1 ≥ θ > 3/4) the linear operator B from $$ to $$ by

We then define (0,0, 0, η^*) to be a Hopf point for (18) if and only if there exist an ϵ₀ > 0 and a C¹-curve (Y_C denotes the complexification of the real space Y) of eigendata for $$ with

• (i)

  $$, $$,

• (ii)

  λ(τ^*) = iβ with β > 0,

• (iii)

  Re  (λ) ≠ 0 for all λ in the spectrum of $$,

• (iv)

  Reλ^′(τ^*) ≠ 0 (transversality),

where τ = 1/σ.

Theorem 4. Suppose that (1/2) − a(1 − s₀)<(μ/(1 + μ))  and C^′(γ(η); a(α(η) − s₀)) + $$ for all η > 0. Assume that $$. Additionally, suppose that the operator $$ has a unique pair {±iβ}, β > 0 of purely imaginary eigenvalues for some τ^* > 0. Then, (0,0, 0, η^*, τ^*) is a Hopf point for (18).

Proof. We assume, without loss of generality, that β > 0, and Φ^* is the (normalized) eigenfunction of $$ with eigenvalue iβ. We have to show that (Φ^*, iβ) can be extended to a C¹-curve τ ↦ (Φ(τ), λ(τ)) of eigendata for $$ with Re(λ^′(τ^*)) ≠ 0.

For this, let Φ^* = (ψ₀,  z₀,  q₀,  η₀) ∈ D(A) × D(A) × D(A₀) × ℝ. First, we note that η₀ ≠ 0. Otherwise, by (31), (A + iβ)ψ₀ = μ  iβ  η₀  G(·, η^*) = 0 and $$, which is not possible because A is symmetric. So, without loss of generality, let η₀   = 1. Then E(ψ₀,  z₀,  q₀,  iβ,  τ^*) = 0 by (31), where

The equation E(u,  z,  q,  λ,  τ) = 0 is equivalent to λ being an eigenvalue of $$ with eigenfunction (u, z, q, 1). We will apply the implicit function theorem to E. For this, we check that E is of C¹-class and that

is an isomorphism. In addition, the mapping is a compact perturbation of the mapping which is invertible. Thus, D_(u,z,q,λ)E(ψ₀, z₀, q₀, iβ, τ^*) is a Fredholm operator of index  0. Therefore, in order to verify (33), it suffices to show that the system of equations

which is equivalent to

necessarily implies that $$, $$,   $$, and $$. If we define ϕ : = ψ₀ − μG(·, η^*), $$, and ρ = q₀ − J(·, η^*),  then (37) becomes On the other hand, since E(ψ₀,  z₀,  q₀,  iβ,  τ^*) = 0,  ϕ,  ξ and ρ are solutions to the equations, we have

Multiplying (39) and (43) by ϕ and (38) and (42) by ξ and subtracting one from the other, we obtain

Multiplying (38) by $$, (39) by $$, and (40) by $$ and adding the resultants to each, we have Multiplying (42) by $$, (43) by $$, and (44) by $$ and adding the resultants to each, we obtain From (45), we get and thus (47) implies that

Now, multiplying (38) by $$, (42) by $$, and (40) by $$ and adding the resultants to each, we have

From (50), we get Multiplying (42) by $$ and (43) by $$, we get and applying (46) to the above equation, we have Now, multiplying (38) by $$ and (42) by $$ and subtracting the resultants to each other, we obtain Applying (54) to the above equation, we have and thus (52) implies that Since d₂ − (κd₁/η^*) > 0 and a₁ > 0, we have $$ and $$, and so, $$ and $$.

Theorem 5. Under the same condition as in Theorem 4, (0,0, 0, η^*, τ^*) satisfies the transversality condition. Hence, this is a Hopf point for (18).

Proof. By the implicit differentiation of E(ψ₀(τ), z₀(τ), q₀(τ), λ(τ),  τ) = 0, we find

This means that the functions $$,   $$,   $$, and $$ satisfy the equations

By letting ϕ : = ψ₀  −  μG(·, η^*),   $$, and ρ = q₀ − J(·, η^*) as before, we obtain Multiplying (60) by $$, (61) by$$, and (62) by $$ and adding the resultants to each, we obtain From (49) and (63), the above equation implies that

Multiplying (60) by$$, (61) by $$ and (62) by $$ and adding the resultants to each, we have

From (65), we have and the real part is Now, multiplying (60) by $$ and (61) by $$ and applying (54) to resultants, we obtain and thus (68) implies that which is positive since d₂ − (κd₁/η^*) > 0 and a₁ > 0. We have Reλ^′(τ^*) > 0 for β > 0, and thus, by the Hopf-bifurcation theorem in [19], there exists a family of periodic solutions which bifurcates from the stationary solution as τ passes τ^*.

Lemma 6. Suppose that d₂ − (κd₁/η^*) > 0. Let G_β, $$, and J_β be Green functions of the differential operators A + iβ, A + i/β and A₀ + iβ satisfying (42), (43), and (44), respectively. Then, $$ and Re(J_β(η^*, η^*)) are strictly decreasing in β ∈ ℝ⁺ with

Moreover, $$ and Im (J_β(η^*, η^*)) < 0 for β > 0.

Proof. First, we have (A + iβ) ⁻¹ = (A − iβ)(A² + β²) ⁻¹. So, if L(β): = Re(A + iβ) ⁻¹, then L(β) = A(A² + β²) ⁻¹. Moreover, L(β) → A⁻¹ as β → 0 and L(β) → 0 as β → ∞, which results in the corresponding limiting behavior for Re(G_β(η^*, η^*)).

To show that $$ is strictly increasing, we define $$. Then (in the weak sense initially)

As a result, h(β) ∈ D(A) _C and h : ℝ⁺ → D(A) _C is differentiable with ih(β) + (A + iβ)h^′(β) = $$, and therefore Thus, we get It follows that Since (d₂   − (κd₁/η^*)) > 0, we have $$ for β > 0.

In order to show $$, from (72), we have

which implies that −β(d₂ − (κd₁/η^*))Im h(β)(η^*) = ∥Ah(β)∥² > 0. Since (d₂ − (κd₁/η^*)) > 0, we have Im h(β)(η^*) < 0 for  β > 0.

Let k(β)(x): = J_β(x, η^*) − J(x, η^*). Then we have (∂/∂β)(Re(J_β(η^*, η^*))) < 0 and Im J_β(η^*, η^*) < 0 for β > 0.

Theorem 7. Under the same condition as in Theorem 4, for a unique critical point τ^* > 0, there exists a unique, purely imaginary eigenvalue λ = iβ of (31) with β > 0.

Proof. We only need to show that the function (u, z, q, β, τ) ↦ E(u, z, q, iβ, τ) has a unique zero with β > 0 and τ > 0. This means solving the system of (31) with λ = iβ,  u = v − μG(·, η^*),  $$, and $$,

The real and imaginary parts of the above equation are given by Since $$ by Lemma 6, there is a critical point τ^*, provided the existence of β. We now define Using Lemma 6, we have T^′(β) > 0 for β > 0 and $$ + $$ if $$. Moreover, for d₂ > (κd₁/η^*) and a₁ > 0. Hence, there exists a unique β > 0.

Theorem 8. Suppose that (1/2) − a(1 − s₀)<(μ/(1 + μ)) and C^′(γ(η);  a(α(η) − s₀)) + $$ for all η > 0. Then (18) and (10) have at least one stationary solution (u^*, z^*, q^*, η^*), where u^* = z^* = q^* = 0, and (v^*, p^*, w^*, η^*) where v^*(x) = g(x, η^*), $$ and w^*(x) = j(x, η^*) − s₀, for all τ and for κ < κ_c, respectively, where κ_c is a solution of

Assume that $$. Then there exists a unique τ^* such that the linearization $$ has a purely imaginary pair of eigenvalues. The point (0,0, 0, η^*, τ^*) is then a Hopf point for (18), and there exists a C⁰-curve of nontrivial periodic orbits for (18) and (10), bifurcating from (0,   0,  0,   η^*,   τ^*) and (v^*, z^*, w^*, η^*, τ^*), respectively.