Hopf bifurcations in Lengyel–Epstein reaction–diffusion model with discrete time delay

Abstract

We investigate bifurcations of the Lengyel–Epstein reaction–diffusion model involving time delay under the Neumann boundary conditions. Choosing the delay parameter as a bifurcation parameter, we show that Hopf bifurcation occurs. We also determine two properties of the Hopf bifurcation, namely direction and stability, by applying the normal form theory and the center manifold reduction for partial functional differential equations.

1 Introduction

In this paper, we consider the following reaction–diffusion model with time delay under the Neumann boundary conditions

(1)

where $\varOmega$ is a bounded domain in ${\mathbf {R}}^{\mathbf {m}},$ $m\ge 1$, with smooth boundaries $\partial \varOmega$, $\overrightarrow{n}$ is the unit outer normal to $\partial \varOmega$ and $u_{0},v_{0}\in C^{2}(\varOmega )\cap C( \overline{\varOmega }).$ When there is no time delay, system (1) reduces to the Lengyel–Epstein reaction–diffusion model based on the chlorite–iodide–malonic acid chemical (CIMA) reaction (see Turing [1], Lengyel et al. [2, 3], De Kepper et al. [4], Epstein et al. [5] and the references therein). In the model, $u(x,t)$ and $v(x,t)$ denote chemical concentrations of the activator iodide and the inhibitor chlorite, respectively. $a>0$ and $b>0$ are parameters related to the feed concentrations, and $\sigma >0$ is a rescaling parameter depending on the concentration of the starch. Here, the positive constants $d_{1}$ and $d_{2}$ are diffusion coefficients of the activator and the inhibitor, respectively, and $\tau$ is delay parameter.

Studies on stability analysis of reaction–diffusion systems have attracted very much interest in mathematical biology, medicine, ecology, economics and so on (see, e.g., [5–10] and [11]). In the last two decades, some mathematical investigations were conducted for the Lengyel–Epstein reaction–diffusion system with $\tau =0$ in system (1) (see, e.g., [2, 3, 12–18]). For example, Yi et al. [13] and Rovinsky and Menzinger [17] derived the conditions on the parameters at which the spatial homogenous equilibrium solution and the spatial homogenous periodic solution became Turing unstable, and performed a Hopf bifurcation analysis for both ODE and PDE models. Du and Wang [12] investigated the existence of multiple spatially nonhomogenous periodic solutions when all parameters of the system are spatially homogenous in the one-dimensional case. Ni and Tang [16] obtained a priori bound of solutions, nonexistence of nonconstant steady states for small effective diffusion rate, and existence of nonconstant steady states for large effective diffusion rate which partially verify Turing stability for the CIMA reaction. Jang et al. [15] studied the limiting behavior of the steady state solutions by using a shadow system approach, and the global bifurcation of the nonconstant equilibriums for the one-dimensional case.

The ODE model corresponding to system (1) is

$$\begin{aligned} \left\{ \begin{array}{l} \frac{du(t)}{dt}=a-u(t)-4\frac{u(t)v(t-\tau )}{1+u(t)^{2}} \\ \frac{dv(t)}{dt}=\sigma b\left( u(t)-\frac{u(t)v(t-\tau )}{1+u(t)^{2}} \right) . \end{array} \right. \end{aligned}$$

(2)

Recently, Çelik and Merdan [19] studied bifurcation of system (2 ). They gave a detailed Hopf bifurcation analysis by choosing the delay parameter $\tau$ as a bifurcation parameter and analyzed (locally) stability of the equilibrium point of system (2). Their work illustrates that Hopf bifurcation occurs when the delay parameter passes through a sequence of critical values, namely $(\tau _{n})_{n=1}^{\infty }$. They also determined the direction of the Hopf bifurcation and the stability of bifurcating periodic solution through the normal form theory and the center manifold reduction for functional differential equations. They also performed some numerical simulations to support analytical results.

In order to analyze the dynamical behavior of models that depend on the past history of systems, it is often necessary to incorporate time delays into models. The aim of this work is to study the delay effect on the Lengyel–Epstein reaction–diffusion model with the Neumann boundary conditions. We include a time delay into $v(x,t)$ to discuss dynamical behavior of the model from a mathematical point of view. Through the setting in system (1), we give a detailed Hopf bifurcation analysis. We first investigate the existence of a periodic solution. Later, we determine the direction of the solution by using Poincaré normal form and the center manifold reduction for partial functional differential equations.

The paper is organized as follows. Section 2 gives a detail Hopf bifurcation analysis of system (1). Section 3 presents some of the properties of the Hopf bifurcation. Finally, the paper ends with some concluding remarks and discussion.

2 Existence of Hopf bifurcations

In this section, we study the Hopf bifurcation of system (1) on a spatial domain. For simplicity, we chose the spatial domain as $\varOmega =(0,\pi )\subset \mathbb {R}$, but all calculations can be extended for the higher dimensions. In this case, one-dimensional- delayed reaction–diffusion Lengyel–Epstein model can be written as follows

(3)

System (3) has a unique equilibrium point $(u^{*},v^{*})=(\alpha ,1+\alpha ^{2})$ where $\alpha =\frac{a}{5}$. First, we get the following system by shifting the equilibrium point $(u^{*},v^{*})$ to the origin

(4)

where

$$\begin{aligned} f(u,v,\tau )&:= \frac{4\alpha (3{-}\alpha ^{2})}{\left( 1{+}\alpha ^{2}\right) ^{2}}u^{2}{+}\frac{4(\alpha ^{2}{-}1)}{\left( 1{+}\alpha ^{2}\right) ^{2}}uv+h.o.t., \nonumber \\ \end{aligned}$$

(5)

$$\begin{aligned} g(u,v,\tau )&:= \frac{\sigma b}{4}f(u,v,\tau ), \end{aligned}$$

(6)

in which the term $h.o.t.$ denotes the higher-order terms. Second, we find the characteristic equation of system (4). Let the linear operator $\varDelta$ be defined by $\varDelta :=$diag$\left\{ \frac{\partial ^{2}}{\partial x^{2}},\frac{\partial ^{2}}{\partial x^{2}}\right\}$ and $U(t):=(u(t),v(t))^{T}=(u(\cdot ,t),v(\cdot ,t))^{T}$. With this notation, system (4) can be rewritten as an abstract ordinary differential equation in the Banach Space $C=C\left( [-\tau ,0],X\right)$ where

$$\begin{aligned} X=\left\{ (u,v):u,v\in W^{2,2}(0,\pi );\,\frac{du}{dx}=\frac{dv}{dx}=0, \,x=0,\pi \right\} \end{aligned}$$

as follows

$$\begin{aligned} \frac{d}{dt}U(t)=d\varDelta U(t)+L(U_{t})+h.o.t., \end{aligned}$$

(7)

where $d=(d_{1},d_{2})^{T}$, $U_{t}(\theta )=U(t+\theta ),$ $-\tau \le \theta \le 0$, $L:C\rightarrow X$. Here, $L$ is defined by

$$\begin{aligned} L(\varphi )=\left( \begin{array}{cc} \frac{3\alpha ^{2}-5}{1+\alpha ^{2}}\varphi _{1}(0) &{} \frac{-4\alpha }{1+\alpha ^{2}} \varphi _{2}(-\tau ) \\ \frac{2\sigma b\alpha ^{2}}{1+\alpha ^{2}}\varphi _{1}(0) &{} \frac{-\sigma b\alpha }{1+\alpha ^{2}}\varphi _{2}(-\tau ) \end{array} \right) _{2x2} \end{aligned}$$

(8)

for $\varphi (\theta )=U_{t}(\theta ),$ $\varphi =(\varphi _{1,}\varphi _{2})^{T}\in C$. The characteristic equation of (7) is equivalent to

$$\begin{aligned} \lambda y-d\varDelta y-L(e^{\lambda }y)=0, \end{aligned}$$

(9)

where $y\in dom(\varDelta )$ and $y\ne 0$, $dom(\varDelta )\subset X.$ From properties of the Laplacian operator defined on a bounded domain, the operator $\varDelta$ has eigenvalues $-n^{2},$ $n\in \mathbb {N} _{0}=\{0,1,2,...\}$. The corresponding eigenfunctions for each $n$ are

$$\begin{aligned} \beta _{n}^{1}=\left( \begin{array}{c} \gamma _{n} \\ 0 \end{array} \right) ,\, \beta _{n}^{2}=\left( \begin{array}{c} 0 \\ \gamma _{n} \end{array} \right) ,\, \gamma _{n}=\cos (nx). \end{aligned}$$

It is easy to see that $\left( \beta _{n}^{1},\beta _{n}^{2}\right) _{n=0}^{\infty }$ forms a basis for the phase space $X$. Therefore, any arbitrary $y$ in $X$ can be written as a Fourier series in the following form

$$\begin{aligned} y=\underset{n=0}{\overset{\infty }{\sum }}Y_{n}^{T}\left( \begin{array}{c} \beta _{n}^{1} \\ \beta _{n}^{2} \end{array} \right) ,\, Y_{n}^{T}=\left( \begin{array}{c} <y,\beta _{n}^{1}> \\ <y,\beta _{n}^{2}> \end{array} \right) . \end{aligned}$$

(10)

One can show that

$$\begin{aligned} L\left( \varphi ^{T}\Bigg ( \begin{array}{c} \beta _{n}^{1} \\ \beta _{n}^{2} \end{array} \Bigg ) \right) =L(\varphi )^{T}\left( \begin{array}{c} \beta _{n}^{1} \\ \beta _{n}^{2} \end{array} \right) ,\, n\in \mathbb {N} _{0}. \end{aligned}$$

(11)

From (10) and (11), (9) is equivalent to

$$\begin{aligned} \underset{n=0}{\overset{\infty }{\sum }}Y_{n}^{T}&\left[ \left( \lambda I_{2}+dn^{2}I_{2}\right) \nonumber \right. \\&\left. -\left( \begin{array}{cc} \frac{3\alpha ^{2}-5}{1+\alpha ^{2}} &{} \frac{-4\alpha }{1+\alpha ^{2}}e^{-\lambda \tau } \\ \frac{2\sigma b\alpha ^{2}}{1+\alpha ^{2}} &{} \frac{-\sigma b\alpha }{ 1+\alpha ^{2}}e^{-\lambda \tau } \end{array} \right) \right] \left( \begin{array}{c} \beta _{n}^{1} \\ \beta _{n}^{2} \end{array} \right) =0, \end{aligned}$$

(12)

where $I_{2}$ is the 2$\times$2 identity matrix here. Notice that the sum in (12) is zero if and only if the determinant of the matrix in bracket is zero, i.e., $\det (\lambda I_{2}-J)=0$, where

$$\begin{aligned} J=\left( \begin{array}{cc} -d_{1}n^{2}+\frac{3\alpha ^{2}-5}{1+\alpha ^{2}} &{} -\frac{4\alpha }{1+\alpha ^{2}}e^{-\lambda \tau } \\ \frac{2\sigma b\alpha ^{2}}{1+\alpha ^{2}} &{} -d_{2}n^{2}-\frac{\sigma b\alpha }{1+\alpha ^{2}}e^{-\lambda \tau } \end{array} \right) . \end{aligned}$$

(13)

Hence, we conclude that the characteristic equation of system (4) is

$$\begin{aligned} \lambda ^{2}+A\lambda +Be^{-\lambda \tau }+C\lambda e^{-\lambda \tau }+D=0, \end{aligned}$$

(14)

where

$$\begin{aligned} A&= \left( d_{1}+d_{2}\right) n^{2}-m,\, B=5k+kd_{1}n^{2},\, C=k,\,\nonumber \\ D&= d_{1}d_{2}n^{4}-md_{2}n^{2} \end{aligned}$$

(15)

in which

$$\begin{aligned} m=\frac{3\alpha ^{2}-5}{1+\alpha ^{2}},\, k=\frac{\sigma b\alpha }{ 1+\alpha ^{2}},\, n\in \mathbb {N} _{0}=\{0,1,2,...\}. \end{aligned}$$

(16)

We also conclude that system (4) is equivalent to the following system of delay differential equations

$$\begin{aligned} \frac{du}{dt}&= \left( -d_{1}n^{2}+\frac{3\alpha ^{2}-5}{1+\alpha ^{2}} \right) u(t)\nonumber \\&\quad +\left( \frac{-4\alpha }{1+\alpha ^{2}}\right) v(t-\tau )+f(u,v,\tau ), \\ \frac{dv}{dt}&= \left( \frac{2\sigma b\alpha ^{2}}{1+\alpha ^{2}}\right) u(t)-d_{2}n^{2}v(t)\nonumber \\&\quad +\left( \frac{-\sigma b\alpha }{1+\alpha ^{2}}\right) v(t-\tau )+g(u,v,\tau ), \end{aligned}$$

where $f$ and $g$ are defined by (5) and (6), respectively. Now, we can apply the general Hopf bifurcation theorem [20] to this system. We will state the main theorem of this work after two lemmas below.

Lemma 1

The characteristic Eq. (14) has a pair of pure imaginary roots $\lambda =\pm i\omega ,$ $\omega >0,$ if either of two conditions below holds:

• 1. $(X_{n}^{2}-4Y_{n}=0)$ and $(X_{n}<0),$

• 2. $(X_{n}^{2}-4Y_{n}>0)$ and

  • (a) $(X_{n}=0$ and $Y_{n}<0)$ or

  • (b) $(X_{n}>0$ and $Y_{n}<0)$ or

  • (c) $(X_{n}<0$ and $Y_{n}<0)$ or

  • (d) $(X_{n}<0$ and $Y_{n}=0)$ or

  • (e) $(X_{n}<0$ and $Y_{n}>0),$

where $X_{n}=(A^{2}-C^{2}-2D)$ and $Y_{n}=\left( D^{2}-B^{2}\right)$.

Proof

Assume that $\lambda =i\omega$, $\omega \in R$ and $\omega >0$, is a solution of (14). First, substituting it into the characteristic Eq.  (14) and then separating its real and imaginary parts by utilizing Euler’s formula, one gets the following two equations in $\omega$

$$\begin{aligned} \omega ^{2}-D&= B\cos (\omega \tau )+C\omega \sin (\omega \tau ), \\ A\omega&= B\sin (\omega \tau )-C\omega \cos (\omega \tau ). \end{aligned}$$

Second, squaring each side of these equations and then adding them up, one can obtain the following equation

$$\begin{aligned} \omega ^{4}+(A^{2}-C^{2}-2D)\omega ^{2}+D^{2}-B^{2}=0. \end{aligned}$$

(17)

Its roots are given by

$$\begin{aligned} \omega ^{2}=\frac{-(A^{2}-C^{2}-2D)\pm \sqrt{(A^{2}-C^{2}-2D)^{2}-4\left( D^{2}-B^{2}\right) }}{2}. \end{aligned}$$

(18)

Since $X_{n}=(A^{2}-C^{2}-2D)$ and $Y_{n}=\left( D^{2}-B^{2}\right)$, one can now write (18) as follows

$$\begin{aligned} \omega ^{2}=\frac{-X_{n}\pm \sqrt{X_{n}^{2}-4Y_{n}}}{2}. \end{aligned}$$

(19)

Notice that for each $n\in \mathbb {N} _{0}=\{0,1,2,...\}$, we have a different value of $\omega$ since $A$, $B$ and $D$ depend on $n$ (see (15)). Therefore, for each $n\in \mathbb {N} _{0},$ let us denote this value by $\omega _{n}$, i.e., $\omega _{n}^{2}:=\omega ^{2}$. Our goal is to get a strictly positive real $\omega _{n}$. Analyzing the quantity in the radical in (19) yields the following results:

• 1. If $X_{n}^{2}-4Y_{n}<0,$ then $\omega _{n}^{2}\in \mathbb {C}$ so that there is no real root.

• 2. If $X_{n}^{2}-4Y_{n}=0,$ then $\omega _{n_{1,2}}=\pm \sqrt{\frac{ \left( -X_{n}\right) }{2}}$. Thus,

  • (a) $X_{n}>0\Longrightarrow \omega _{n_{1,2}}\in \mathbb {C},$

  • (b) $X_{n}=0\Longrightarrow \omega _{n_{1,2}}=0$ and

  • (c) $X_{n}<0\Longrightarrow$ there is only one positive real root, namely $\omega _{n}=\sqrt{\frac{\left( -X_{n}\right) }{2}}$, where $n\in \mathbb {N} _{0}$.

• 3. If $X_{n}^{2}-4Y_{n}>0,$ then

  • (a) $X_{n}=0$ and $Y_{n}<0\Longrightarrow$ there is only one positive real root $\omega _{n}^{2}=\sqrt{-Y_{n}},$ where $n\in \mathbb {N}_{0},$

  • (b) $X_{n}>0$ and $Y_{n}>0\Longrightarrow X_{n}>\sqrt{X_{n}^{2}-4Y_{n}} \Longrightarrow \omega _{n}^{2}<0$ so that there is no real root,

  • (c) $X_{n}>0$ and $Y_{n}=0\Longrightarrow \omega _{n}^{2}=\frac{-X_{n}\pm \sqrt{X_{n}^{2}}}{2}\Longrightarrow \omega _{n}^{2}=-X_{n}$ or $\omega _{n}^{2}=0$ so that there is no positive real root,

  • (d) $X_{n}>0$ and $Y_{n}<0\Longrightarrow$ there is only one positive real root, namely $\omega _{n}=\sqrt{\frac{-X_{n}+\sqrt{X_{n}^{2}-4Y_{n}}}{2}},$ where $n\in \mathbb {N} _{0},$

  • (e) $X_{n}<0$ and $Y_{n}<0\Longrightarrow$ there is only one positive real root which is $\omega _{n}=\sqrt{\frac{-X_{n}+\sqrt{X_{n}^{2}-4Y_{n}}}{2}},$ where $n\in \mathbb {N} _{0},$

  • (f) $X_{n}<0$ and $Y_{n}=0\Longrightarrow \omega _{n}^{2}=\frac{-X_{n}\pm \sqrt{X_{n}^{2}}}{2}\Longrightarrow \omega _{n}^{2}=-X_{n}$ or $\omega _{n}^{2}=0\Longrightarrow$ there is only one positive real root, namely $\omega _{n}^{2}=-X_{n},$ where $n\in \mathbb {N}_{0},$

  • (g) $X_{n}<0$ and $Y_{n}>0\Longrightarrow$ there are two positive real roots which are $\omega _{n_{1}}=\sqrt{\frac{-X_{n}+\sqrt{X_{n}^{2}-4Y_{n}}}{2}}$ and $\omega _{n_{2}}=\sqrt{\frac{-X_{n}-\sqrt{X_{n}^{2}-4Y_{n}}}{2}},$ where $n\in \mathbb {N} _{0}.$

We conclude from the analysis above that there exists only one positive real $\omega _{n}$ for 2c, 3a, 3d, 3e and 3f, while there exist two different positive real $\omega _{n}$ values for 3g. This completes the proof. $\square$

Lemma 1 basically underlines that the characteristic Eq. (14) has a pair of complex conjugate eigenvalues of the form $\lambda (\tau )=\gamma (\tau )\pm i\omega (\tau ),$ and there are some critical values, namely $\tau _{n},$ of the bifurcation parameter $\tau$ at which $\gamma (\tau _{n})=0$ and $\omega (\tau _{n})=\omega _{n}$ for each $n\in \mathbb {N} _{0}.$ Next, we determine these critical values $\tau _{n}$. To do this, we substitute $\lambda (\tau _{n})=i\omega (\tau _{n})=i\omega _{n}$ into (14), separate real and imaginary parts utilizing Euler’s formula and obtain the following two equations in $\omega _{n}$ and $\tau _{n}:$

$$\begin{aligned} \omega _{n}^{2}-D&= B\cos (\omega _{n}\tau _{n})+C\omega _{n}\sin (\omega _{n}\tau _{n}), \\ A\omega _{n}&= B\sin (\omega _{n}\tau _{n})-C\omega _{n}\cos (\omega _{n}\tau _{n}). \end{aligned}$$

Solving these equations for $\tau _{n}$, one has the following

$$\begin{aligned} \tau _{n}=\frac{1}{\omega _{n}}\arctan \left( \frac{C\omega _{n}^{3}+(AB-CD)\omega _{n}}{(B-AC)\omega _{n}^{2}-BD}\right) . \end{aligned}$$

(20)

On the other hand, since $\tan x$ is a periodic function with period $\pi ,$ the critical values have the following form for each $n$ and $k:$

$$\begin{aligned} \tau _{n,k}=\frac{1}{\omega _{n}}\arctan \left( \frac{C\omega _{n}{}^{3}+(AB-CD)\omega _{n}}{(B-AC)\omega _{n}^{2}-BD}\right) +\frac{k\pi }{\omega _{n}}, \end{aligned}$$

(21)

where $n,k\in \mathbb {N} _{0}.$ Note that $\gamma (\tau _{n})=\gamma (\tau _{n,k})=0$ and $\omega (\tau _{n})=\omega (\tau _{n,k})=\omega _{n}.$ Note also that for each $n^{*}\in \mathbb {N} _{0}$, we uniquely determine $\tau _{n^{*},k}$ such that $\lambda (\tau _{n^{*},k})=i\omega _{n^{*}}.$ This underlines that all other roots of the characteristic Eq. (14) have nonzero real parts at $\tau =\tau _{n^{*},k}$.

We now check whether the transversality condition holds. The following lemma gives the required conditions under which it holds.

Lemma 2

The transversality condition holds, i.e.,

$$\begin{aligned} \left. \frac{d(Re\lambda ) }{d\tau }\right| _{\tau =\tau _{n,k}}\ne 0, \end{aligned}$$

where $n,k\in \mathbb {N} _{0},$ if both $X_{n}^{2}-4Y_{n}>0$ and one of the following conditions is satisfied:

• 1. $X_{n}=0$ and $Y_{n}<0,$

• 2. $X_{n}>0$ and $Y_{n}<0$ and $\left( d_{1}\right) ^{2}>\left( d_{2}\right) ^{2}$,

• 3. $X_{n}<0$ and $Y_{n}=0$,

• 4. $X_{n}<0$ and $Y_{n}<0$,

• 5. $X_{n}<0$ and $Y_{n}>0$ and

  • (a) $X_{n}^{2}-(2+2\sqrt{5})Y_{n}>0$ or

  • (b) $X_{n}B^{2}C^{2}+B^{4}C^{4}+Y_{n}<0$,

where $X_{n}=(A^{2}-C^{2}-2D)$ and $Y_{n}=\left( D^{2}-B^{2}\right)$.

Proof

Differentiating the characteristic Eq. (14) with respect to $\tau$, we get the following equation

$$\begin{aligned} \frac{d\lambda }{d\tau }=\frac{B\lambda +C\lambda ^{2}}{2\lambda e^{\lambda \tau }+Ae^{\lambda \tau }-B\tau +C-C\lambda \tau }. \end{aligned}$$

(22)

Substituting $\tau =\tau _{n,k}$ in the equation above yields

$$\begin{aligned} \left. \frac{d\lambda }{d\tau }\right| _{\tau =\tau _{n,k}}=\frac{ Bi\omega _{n}+C\left( i\omega _{n}\right) ^{2}}{2i\omega _{n}e^{i\omega _{n}\tau _{n,k}}+Ae^{i\omega _{n}\tau _{n,k}}-B\tau _{n,k}+C-Ci\omega _{n}\tau _{n,k}}. \end{aligned}$$

(23)

Since $\left. \frac{d\lambda }{d\tau }\right| _{\tau =\tau _{n,k}}=\left. \frac{d\gamma }{d\tau }\right| _{\tau =\tau _{n,k}}+i\left. \frac{d\omega }{d\tau }\right| _{\tau =\tau _{n,k}}$, we can find the equation of $\left. \frac{d\gamma }{d\tau }\right| _{\tau =\tau _{n,k}}$ explicitly from (23). Notice that

$$\begin{aligned} Re\left( \left. \frac{d\lambda }{d\tau }\right| _{\tau =\tau _{n,k}}\right) \ne 0 \Longleftrightarrow Re\left( \left. \frac{d\lambda }{d\tau } \right| _{\tau =\tau _{n,k}}\right) ^{-1}\ne 0.\nonumber \\ \end{aligned}$$

(24)

From (23), we obtain that

$$\begin{aligned}&Re\left( \left. \frac{d\lambda }{d\tau }\right| _{\tau =\tau _{n,k}}\right) ^{-1} \!\!{=}\frac{(C^{2}+2DC^{2}-AC^{2}+2B)\omega _{n}^{2}}{(B^{2}+C^{2}\omega _{n}^{2})} \nonumber \\&\quad +\frac{2C^{2}(B^{2}-D^{2})+B^{2}(A^{2}-C^{2}-2D)}{(B^{2}+C^{2}\omega _{n}^{2})}. \end{aligned}$$

(25)

Then, by substituting $\omega _{n}$ values which are obtained from 2c, 3a, 3d, 3e, 3f and 3g in the proof of Lemma 1 into (25), we can check whether transversality condition holds. We conclude that $\left. \frac{d\gamma }{d\tau }\right| _{\tau =\tau _{n,k}}\ne 0$ for some of these $\omega _{n}$ values together with the following conditions. These conditions and the corresponding $\omega _{n}$ values are as follows:

• 1. If $X_{n}=0$, $Y_{n}<0$ and $X_{n}^{2}-4Y_{n}>0$, then $\omega _{n}=\sqrt{-Y_{n}}$,

• 2. If $X_{n}>0$, $Y_{n}<0$ and $X_{n}^{2}-4Y_{n}>0$ and $\left( d_{1}\right) ^{2}>\left( d_{2}\right) ^{2}$, then $\omega _{n}=\sqrt{\frac{-X_{n}+\sqrt{X_{n}^{2}-4Y_{n} }}{2}}$,

• 3. If $X_{n}<0$, $Y_{n}=0$ and $X_{n}^{2}-4Y_{n}>0$, then $\omega _{n}=-X_{n}$,

• 4. If $X_{n}<0$, $Y_{n}<0$ and $X_{n}^{2}-4Y_{n}>0$, then $\omega _{n}=\sqrt{\frac{-X_{n}+\sqrt{ X_{n}^{2}-4Y_{n}}}{2}}$,

• 5. If $X_{n}<0$, $Y_{n}>0$ and $X_{n}^{2}-4Y_{n}>0$ and

  • (a) $X_{n}^{2}-(2+2\sqrt{5})Y_{n}{>}0$, then $\omega _{n}{=}\sqrt{\frac{-X_{n}{+} \sqrt{X_{n}^{2}-4Y_{n}}}{2}}$ or,

  • (b) $X_{n}B^{2}C^{2}+B^{4}C^{4}+Y_{n}<0$, then $\omega _{n}=\sqrt{\frac{ -X_{n}-\sqrt{X_{n}^{2}-4Y_{n}}}{2}}$.

This completes the proof. $\square$

Thus, using the Hopf bifurcation theorem [20] together with Lemma 1 and Lemma 2, one can be able to show that for each of these cases obtained above, system (4) undergoes a Hopf bifurcation at $(u^{*},v^{*})$ as $\tau$ passing through $\tau _{n,k}$ ($n,k\in \mathbb {N} _{0})$ and possesses a family of real-valued periodic solutions at these values. These results are summarized in the following theorem.

Theorem 1

Let $X_{n}=(A^{2}-C^{2}-2D)$, $Y_{n}=\left( D^{2}-B^{2}\right)$ and $X_{n}^{2}-4Y_{n}>0$. If one of the following conditions holds

• 1. $X_{n}=0$ and $Y_{n}<0$,

• 2. $X_{n}>0$ and $Y_{n}<0$ and $\left( d_{1}\right) ^{2}>\left( d_{2}\right) ^{2}$,

• 3. $X_{n}<0$ and $Y_{n}=0$,

• 4. $X_{n}<0$ and $Y_{n}<0$,

• 5. $X_{n}<0$ and $Y_{n}>0$ and

  • (a) $X_{n}^{2}-(2+2\sqrt{5})Y_{n}>0$ or,

  • (b) $X_{n}B^{2}C^{2}+B^{4}C^{4}+Y_{n}<0,$

then system (3) undergoes a Hopf bifurcation at $(u^{*},v^{*})$ as $\tau$ passing through $\tau _{n,k}$ and possesses a family of real-valued periodic solutions when $\lambda (\tau )$ crosses the imaginary axis at $\tau =\tau _{n,k}$.

3 Direction and stability of the Hopf bifurcation

In this section, we determine some of the properties of Hopf bifurcation by applying the normal form theory and the center manifold reduction for partial functional differential equations.

Remember that the system whose equilibrium is shifted to the origin is

$$\begin{aligned} \left\{ \begin{array}{l} u_{t}(x,t)=d_{1}u_{xx}(x,t)+\left( \frac{3\alpha ^{2}-5}{1+\alpha ^{2}} \right) u(x,t)\\ +\left( \frac{-4\alpha }{1+\alpha ^{2}}\right) v(x,t-\tau )+f(u,v,\tau ), \\ v_{t}(x,t)=d_{2}v_{xx}(x,t)+\left( \frac{2\sigma b\alpha ^{2}}{1+\alpha ^{2}} \right) u(x,t)\\ +\left( \frac{-\sigma b\alpha }{1+\alpha ^{2}}\right) v(x,t-\tau )+g(u,v,\tau ), \end{array} \right. \end{aligned}$$

(26)

where the functions $f$ and $g$ have the forms in (5) and (6), respectively. In order to determine the direction and the stability of the Hopf bifurcation, we consider the following system which is equivalent to (26)

$$\begin{aligned} \left\{ \begin{array}{l} \frac{du}{dt}=\left( -d_{1}n^{2}+\frac{3\alpha ^{2}-5}{1+\alpha ^{2}} \right) u(t)\\ \quad +\left( \frac{-4\alpha }{1+\alpha ^{2}}\right) v(t-\tau )+f(u,v,\tau ), \\ \frac{dv}{dt}=\frac{2\sigma b\alpha ^{2}}{1+\alpha ^{2}} u(t)-d_{2}n^{2}v(t)\\ \quad +\left( \frac{-\sigma b\alpha }{1+\alpha ^{2}}\right) v(t-\tau )+g(u,v,\tau ), \end{array} \right. \end{aligned}$$

(27)

where $u(t)=u(.,t)$ and $v(t)=v(.,t)$, so we can continue our analysis with system (27). Let $\phi (\theta )=\left( \begin{array}{c} \phi _{1}(\theta )\\ \phi _{2}(\theta ) \end{array} \right) \in C^{1}\left[ -\tau ,0\right]$ and $L_{n}:C^{1}\left[ -\tau ,0 \right] \rightarrow \mathbb {R} ^{2}$. Now we define $L_{n}$ and $F$ as follows

$$\begin{aligned} L_{n}\left( \phi (\theta )\right)&= \left[ \begin{array}{cc} \frac{3\alpha ^{2}-5}{1+\alpha ^{2}}-d_{1}n^{2} &{} 0 \\ \frac{2\sigma b\alpha ^{2}}{1+\alpha ^{2}} &{} -d_{2}n^{2} \end{array} \right] \left( \begin{array}{c} \phi _{1}(0) \\ \phi _{2}(0) \end{array} \right) \\&+\left[ \begin{array}{cc} 0 &{} \frac{-4\alpha }{1+\alpha ^{2}} \\ 0 &{} \frac{-\sigma b\alpha }{1+\alpha ^{2}} \end{array} \right] \left( \begin{array}{c} \phi _{1}(-\tau ) \\ \phi _{2}(-\tau ) \end{array} \right) , \\ F\left( \phi (\theta )\right)&= \left( \begin{array}{c} f(\phi (\theta )) \\ g(\phi (\theta )) \end{array} \right) , \end{aligned}$$

where $f,$ $g:C^{1}\left[ -1,0\right] \rightarrow \mathbb {R}$

$$\begin{aligned} f(\phi (\theta ))&= \frac{4\alpha \left( 3-\alpha ^{2}\right) }{\left( 1+\alpha ^{2}\right) ^{2}}\phi _{1}(0)^{2}\\&\quad +\frac{4\left( \alpha ^{2}-1\right) }{\left( 1+\alpha ^{2}\right) ^{2}}\phi _{1}(0)\phi _{2}(-\tau )+h.o.t., \\ g(\phi (\theta ))&= \frac{\sigma b}{4}f(\phi (\theta )). \end{aligned}$$

Let $U(t)=$ $\left( \begin{array}{c} u(t) \\ v(t) \end{array} \right)$ and $U_{t}$ be two notations such that $U_{t}=U(t+\theta ),$ $\theta \in \left[ -\tau ,0\right]$ so that system (27) turns into

$$\begin{aligned} \frac{\partial U}{\partial t}=L_{n}U_{t}+F(U_{t}). \end{aligned}$$

(28)

By now taking $t=\tau s$ and $\mu =\tau -\tau _{n}$, the new scaled system whose bifurcation value is shifted to $0$ can be written as follows

$$\begin{aligned} \frac{\partial U}{\partial s}=(\tau _{n}+\mu )L_{n}U_{s}+(\tau _{n}+\mu )F(U_{s}), \end{aligned}$$

(29)

where $U_{s}=U(s+\theta ),$ $\theta \in \left[ -1,0\right] .$

Let

$$\begin{aligned}&\!\!\!L_{n_{\mu }}\left( \phi (\theta )\right) \nonumber \\&\!\!\!\quad =(\tau _{n}+\mu )\left( \begin{array}{c} \left[ \begin{array}{cc} \frac{3\alpha ^{2}-5}{1+\alpha ^{2}}-d_{1}n^{2} &{} 0 \\ \frac{2\sigma b\alpha ^{2}}{1+\alpha ^{2}} &{} -d_{2}n^{2} \end{array} \right] \left( \begin{array}{c} \phi _{1}(0) \\ \phi _{2}(0) \end{array} \right) \\ +\left[ \begin{array}{cc} 0 &{} \frac{-4\alpha }{1+\alpha ^{2}} \\ 0 &{} \frac{-\sigma b\alpha }{1+\alpha ^{2}} \end{array} \right] \left( \begin{array}{c} \phi _{1}(-1) \\ \phi _{2}(-1) \end{array} \right) \end{array} \right) \nonumber \\ \end{aligned}$$

(30)

and

$$\begin{aligned} \widetilde{F}(\phi (\theta ))=(\tau _{n}+\mu )F(U_{s}). \end{aligned}$$

(31)

For convenience, we continue our calculations by taking $s=t$ and $\widetilde{F}(\phi (\theta ))=F(\phi (\theta ))$ in the rest of the paper. We rewrite system (29) in the following form

$$\begin{aligned} \frac{\partial U}{\partial t}=L_{n_{\mu }}U_{t}+F(U,\mu ). \end{aligned}$$

(32)

Notice that system (32) has two different unknown functions, namely $U(x,t)$ and $U_{t}=U(x,t+\theta )$. Applying the Riesz Representation Theorem yields that there exists a matrix valued function $\eta (\cdot ,\mu )$ where $\eta (\cdot ,\mu ):\left[ -1,0\right] \rightarrow \mathbb {R} ^{2}$ and $\phi \in C^{1}\left[ -1,0\right]$ so that

$$\begin{aligned} L_{n_{\mu }}\left( \phi (\theta )\right) =\int _{-1}^{0}d\eta (\theta ,\mu )\phi (\theta ). \end{aligned}$$

Let us choose $d\eta (\theta ,\mu )$ as follows

$$\begin{aligned}&\!\!\!d\eta (\theta ,\mu )\\&=(\tau _{n}+\mu )\left( \begin{array}{c} \left[ \begin{array}{cc} \frac{3\alpha ^{2}-5}{1+\alpha ^{2}}-d_{1}n^{2} &{} 0 \\ \frac{2\sigma b\alpha ^{2}}{1+\alpha ^{2}} &{} -d_{2}n^{2} \end{array} \right] \delta \left( \theta \right) \\ +\left[ \begin{array}{cc} 0 &{} \frac{-4\alpha }{1+\alpha ^{2}} \\ 0 &{} \frac{-\sigma b\alpha }{1+\alpha ^{2}} \end{array} \right] \delta \left( \theta +1\right) \end{array} \right) d\theta , \end{aligned}$$

where $\delta \left( \theta \right)$ is the Dirac delta function here. Using them, we define the operators $A(\mu )\phi$ and $R(\mu )\phi$ as follows

$$\begin{aligned} A(\mu )\phi {=}\left\{ \begin{array}{ll} \frac{d\phi (\theta )}{d\theta } &{} ,\theta {\in } \left[ {-}1,0\right) \\ \int _{-1}^{0}d\eta (\theta ,\mu )\phi (\theta )=L_{n_{\mu }}\left( \phi \right) &{} ,\theta =0 \end{array} \right. \end{aligned}$$

(33)

and

$$\begin{aligned} R(\mu )\phi =\left\{ \begin{array}{ll} 0 &{} ,\theta \in \left[ -1,0\right) \\ F(\theta )&{} ,\theta =0 \end{array} \right. . \end{aligned}$$

(34)

Now we can state system (32) as follows

$$\begin{aligned} \frac{\partial U_{t}}{\partial t}=A(\mu )U_{t}+R(\mu )U_{t} \end{aligned}$$

(35)

which involves only one unknown function. In order to construct center manifold coordinates, we need to define an inner product. For $\psi ,\phi \in C\left[ -1,0\right]$, one can define it as follows

$$\begin{aligned}&<\psi ,\phi >=\overline{\psi }(0)\cdot \phi (0)\nonumber \\&\quad -\int _{\theta =-r}^{0} \int _{\xi =0}^{\theta }\overline{\psi }^{T}(\xi -\theta )d\eta (\theta ,\mu )\phi (\xi )d\xi . \end{aligned}$$

(36)

Let $q\left( \theta \right)$ be an eigenvector of $A(0)$ corresponding to $\lambda (0)=i\omega _{n}$ and $q^{*}\left( s\right)$ be an eigenvector of $A^{*}(0)$ associated with $\overline{\lambda }(0)=-i\omega _{n}$ satisfying

$$\begin{aligned}&<q^{*}\left( s\right) ,q\left( \theta \right) >=1\text { and } <q^{*}\left( s\right) ,\overline{q}\left( \theta \right) >=0, \nonumber \\ \end{aligned}$$

(37)

$$\begin{aligned}&A(0)q\left( \theta \right) =i\omega _{n}q\left( \theta \right) \text { and } A^{*}(0)q\left( s\right) \nonumber \\&\quad =-i\omega _{n}q^{*}\left( s\right) , \end{aligned}$$

(38)

where $A^{*}(\mu )$ is adjoint operator of $A(\mu )$ defined as

$$\begin{aligned} A^{*}(\mu )\phi = \left\{ \begin{array}{ll} -\frac{d\phi (s)}{ds}&{},s\in \left[ -1,0\right) \\ \int _{-1}^{0}d\eta ^{T}(s,\mu )\phi (-s)&{},s=0. \end{array} \right. \end{aligned}$$

First, we determine $q\left( \theta \right)$ from $A(0)q\left( \theta \right) =i\omega _{n}q\left( \theta \right)$ in (38). It will be done in two cases as follows:

Case A1: If $\theta \in \left[ -1,0\right)$, then, by (33),

$$\begin{aligned} A(0)q\left( \theta \right) =\frac{dq(\theta )}{d\theta }=i\omega _{n}q\left( \theta \right) \end{aligned}$$

(39)

so that we obtain that $q\left( \theta \right) =\left( {\begin{array}{c}1\\ c\end{array}}\right) e^{i\omega _{n}\theta }$ from ( 39) where $c$ will be determined in Case A2.

Case A2: When $\theta =0$, utilizing (33) we have

$$\begin{aligned} A(0)q\left( \theta \right)&= \int _{-1}^{0}d\eta (\theta ,\mu )q(\theta ) \\&= \tau _{n}\left[ \begin{array}{cc} m-d_{1}n^{2} &{} 0 \\ f &{} -d_{2}n^{2} \end{array} \right] \int _{-1}^{0}\delta \left( \theta \right) q(\theta )d\theta \\&+\tau _{n}\left[ \begin{array}{cc} 0 &{} g \\ 0 &{} -k \end{array} \right] \int _{-1}^{0}\delta \left( \theta +1\right) q(\theta )d\theta \\&= \tau _{n}\left[ \begin{array}{cc} m-d_{1}n^{2} &{} 0 \\ f &{} -d_{2}n^{2} \end{array} \right] q(0)\\&+\tau _{n}\left[ \begin{array}{cc} 0 &{} g \\ 0 &{} -k \end{array} \right] q(-1) \\&= \left[ \begin{array}{c} \tau _{n}\left( m-d_{1}n^{2}\right) +\tau _{n}ge^{-i\omega _{n}}c \\ \tau _{n}f-\tau _{n}d_{2}n^{2}c+\tau _{n}\left( -k\right) e^{-i\omega _{n}}c \end{array} \right] \\&= i\omega _{n}q\left( 0\right) =i\omega _{n}\left( {\begin{array}{c}1\\ c\end{array}}\right) e^{i\omega _{n}0}= \left( {\begin{array}{c}i\omega _{n}\\ i\omega _{n}c\end{array}}\right) , \end{aligned}$$

where

$$\begin{aligned} m=\frac{3\alpha ^{2}-5}{1+\alpha ^{2}},f=\frac{2\sigma b\alpha ^{2}}{ 1+\alpha ^{2}},g=\frac{-4\alpha }{1+\alpha ^{2}},k=\frac{\sigma b\alpha }{ 1+\alpha ^{2}}. \end{aligned}$$

(40)

From the calculations above, one obtains $c$ as follows

$$\begin{aligned} c=\left( \frac{i\omega _{n}-\tau _{n}\left( m-d_{1}n^{2}\right) }{\tau _{n}ge^{-i\omega _{n}}}\right) . \end{aligned}$$

(41)

Second, we determine $q^{*}\left( s\right)$ from $A(0)q^{*}\left( s\right) =-i\omega _{n}q^{*}\left( s\right)$ in (38). Once again, it will be done in two cases as follows:

Case B1: If $\theta \in \left[ -1,0\right) ,$ then, by (33), one has

$$\begin{aligned} A^{*}(0)q^{*}\left( s\right) =-\frac{dq^{*}(s)}{ds}=-i\omega _{n}q^{*}\left( s\right) \end{aligned}$$

(42)

so that one obtains that $q^{*}\left( s\right) =E\left( {\begin{array}{c}c^{*}\\ 1\end{array}}\right) e^{i\omega _{n}\theta }$. The constant $c^{*}$ will be calculated below.

Case B2: When $\theta =0$, we have (see (33))

$$\begin{aligned}&A^{*}(0)q^{*}\left( s\right) =\int _{-1}^{0}d\eta ^{T}(s,\mu )\phi (-s) \\&= \tau _{n}\left[ \begin{array}{cc} m-d_{1}n^{2} &{} f \\ 0 &{} -d_{2}n^{2} \end{array} \right] \int _{-1}^{0}\delta \left( s\right) q^{*}(-s)ds \\&+\tau _{n}\left[ \begin{array}{cc} 0 &{} 0 \\ g &{} -k \end{array} \right] \int _{-1}^{0}\delta \left( s+1\right) q^{*}(-s)ds \\&= \tau _{n}\left[ \begin{array}{cc} m-d_{1}n^{2} &{} f \\ 0 &{} -d_{2}n^{2} \end{array} \right] q^{*}(0)+\tau _{n}\left[ \begin{array}{cc} 0 &{} 0 \\ g &{} -k \end{array} \right] q^{*}(1) \\&= E\left[ \begin{array}{c} \tau _{n}(m-d_{1}n^{2})c^{*}+\tau _{n}f \\ \tau _{n}e^{i\omega _{n}}gc^{*}-\tau _{n}d_{2}n^{2}+\tau _{n}e^{i\omega _{n}}(-k) \end{array} \right] \\&= -i\omega _{n}q^{*}\left( 0\right) =-i\omega _{n}E\left( {\begin{array}{c}c^{*}\\ 1\end{array}}\right) =E\left( \begin{array}{c} -i\omega _{n}c^{*} \\ -i\omega _{n} \end{array} \right) . \end{aligned}$$

These calculations yield that $c^{*}$ has the following form

$$\begin{aligned} c^{*}=\left( \frac{-\tau _{n}f}{\tau _{n}(m-d_{1}n^{2})+i\omega _{n}} \right) . \end{aligned}$$

(43)

These two eigenvectors must satisfy the properties given in (37). Since $\lambda (\mu )$ is a simple eigenvalue, one can show that $<q^{*}\left( s\right) ,\overline{q}\left( \theta \right) >=0$ (see [20] and [21]). Let us now choose $E$ such that $<q^{*}\left( s\right) ,q\left( \theta \right) >=1.$ By the definition of inner product (see (36)), one has

$$\begin{aligned}&< q^{*}\left( s\right) ,q\left( \theta \right) >=\overline{q^{*}} (0)\cdot q(0)\\&-\int _{\theta =-r}^{0}\int _{\xi =0}^{\theta }\overline{q^{*} }^{T}(\xi -\theta )d\eta (\theta ,\mu )q(\xi )d\xi \\&= \overline{E}\left( \overline{c^{*}}+c\right) -\overline{E}\left( \begin{array}{cc} \overline{c^{*}}&1 \end{array} \right) \left( \int _{-1}^{0}d\eta (\theta ,0)e^{i\omega _{n}\theta }\theta \right) \nonumber \\&\times \left( \begin{array}{c} 1 \\ c \end{array} \right) . \end{aligned}$$

First, we calculate the integral on the right-hand side of the latter equation as follows

$$\begin{aligned}&\left( \int _{-1}^{0}d\eta (\theta ,0)e^{i\omega _{n}\theta }\theta \right) \\&\quad =\int _{-1}^{0}\tau _{n}\left( \begin{array}{c} \left[ \begin{array}{cc} m-d_{1}n^{2} &{} 0 \\ f &{} -d_{2}n^{2} \end{array} \right] \delta \left( \theta \right) \\ +\left[ \begin{array}{cc} 0 &{} g \\ 0 &{} -k \end{array} \right] \delta \left( \theta +1\right) \end{array} \right) e^{i\omega _{n}\theta }\theta d\theta \\&\quad =\left[ \begin{array}{cc} 0 &{} -\tau _{n}ge^{-i\omega _{n}} \\ 0 &{} \tau ke^{-i\omega _{n}} \end{array} \right] . \end{aligned}$$

Second, we substitute the result into the equation above to determine $\overline{E}$

$$\begin{aligned}&< q^{*}\left( s\right) ,q\left( \theta \right) >=\overline{E}\left( \overline{c^{*}}+c\right) \\&-\overline{E}\left( \begin{array}{cc} \overline{c^{*}}&1 \end{array} \right) \left[ \begin{array}{cc} 0 &{} -\tau _{n}ge^{-i\omega _{n}} \\ 0 &{} \tau _{n}ke^{-i\omega _{n}} \end{array} \right] \left( \begin{array}{c} 1 \\ c \end{array} \right) \\&= \overline{E}\left( \overline{c^{*}}+c+\tau _{n}ge^{-i\omega _{n}}c \overline{c^{*}}-\tau _{n}ke^{-i\omega _{n}}c\right) . \end{aligned}$$

Finally, since $<q^{*}\left( s\right) ,q\left( \theta \right) >=1$, we obtain $\overline{E}$ as follows

$$\begin{aligned} \overline{E}=\frac{1}{\left( \overline{c^{*}}+c+\tau _{n}ge^{-i\omega _{n}}c\overline{c^{*}}-\tau _{n}ke^{-i\omega _{n}}c\right) }. \end{aligned}$$

Next, we define center manifold coordinates by using these eigenvectors. Let $X$ denote domain of the operator $L_{n_{\mu }}$ (see (3)). We decompose $X=X^{C}+X^{S}$ with $X^{C}:=\left\{ \left. zq+\overline{z}\overline{q} \right| z\in \mathbb {C} \right\}$, $X^{S}:=\left\{ \left. w\in X\right| <q^{*},w>=0\right\}$. For any $U=\left( \begin{array}{c} u \\ v \end{array} \right) \in X$, there exists $z\in \mathbb {C}$ and $w=\left( \begin{array}{c} w_{1} \\ w_{2} \end{array} \right) \in X^{s}$ such that

$$\begin{aligned} U=\left( \begin{array}{c} u \\ v \end{array} \right) =zq+\overline{z}\overline{q}+\left( \begin{array}{c} w_{1} \\ w_{2} \end{array} \right) . \end{aligned}$$

(44)

Thus, at $\mu =0$, system (35) is reduced to the following system in $(z,w)$-coordinates

$$\begin{aligned} \frac{\partial z}{\partial t}&= i\omega _{n}z+<q^{*},F_{0}>=i\omega _{n}z+g(z,\overline{z}) \\ \frac{\partial w}{\partial t}&= A(0)w+H(z,\overline{z},\theta ), \end{aligned}$$

where

$$\begin{aligned} F_{0}:=F(zq+\overline{z}\overline{q}+w,0),\,<q^{*},F_{0}>= \overline{q^{*}}(0)\cdot F_{0}, \\ H(z,\overline{z},\theta )=F_{0}-<q^{*},F_{0}>q-<\overline{q^{*}} ,F_{0}>\overline{q}. \end{aligned}$$

From (31), we have

$$\begin{aligned}&\!\!\!F_{0} =F(zq+\overline{z}\overline{q}+w,0)=F(U_{t},0) \nonumber \\&\!\!\!\quad =\tau _{n}\left( \begin{array}{c} \frac{4\alpha \left( 3-\alpha ^{2}\right) }{\left( 1+\alpha ^{2}\right) ^{2}} u_{t}(0)^{2}+\frac{4\left( \alpha ^{2}-1\right) }{\left( 1+\alpha ^{2}\right) ^{2}}u_{t}(0)v_{t}(-1)+h.o.t. \\ \frac{\alpha \sigma b\left( 3-\alpha ^{2}\right) }{\left( 1+\alpha ^{2}\right) ^{2}}u_{t}(0)^{2}+\frac{\sigma b\left( \alpha ^{2}-1\right) }{ \left( 1+\alpha ^{2}\right) ^{2}}u_{t}(0)v_{t}(-1)+h.o.t. \end{array} \right) . \nonumber \\ \end{aligned}$$

(45)

Representing $w$ as $w(z,\overline{z})=\sum \frac{1}{i!j!}w_{ij}(z)^{i}( \overline{z})^{j}$ and using (44), we get

$$\begin{aligned} U_{t}(\theta )&= zq(\theta )+\overline{z}\overline{q}(\theta )+w_{20}(\theta ) \frac{z^{2}}{2}+w_{11}(\theta )z\overline{z}\nonumber \\&+\,w_{02}(\theta )\frac{\left( \overline{z}\right) ^{2}}{2}+h.o.t. \end{aligned}$$

(46)

To get $u_{t}(0)$, we put $\theta =0$ in (46) that leads to the following equation

$$\begin{aligned} u_{t}(0)&= z+\overline{z}+w_{20_{1}}(0)\frac{z^{2}}{2}+w_{11_{1}}(0)z\overline{ z}\\&+\,w_{02_{1}}(0)\frac{\left( \overline{z}\right) ^{2}}{2}+h.o.t. \end{aligned}$$

Similarly, $v_{t}(-1)$ can be obtained by plugging in $\theta =-1$ into (46), so we have the following

$$\begin{aligned} v_{t}(-1)&= z+\overline{z}+w_{20_{2}}(-1)\frac{z^{2}}{2}+w_{11_{2}}(-1)z \overline{z}\nonumber \\&+\,w_{02_{2}}(-1)\frac{\left( \overline{z}\right) ^{2}}{2}+h.o.t. \end{aligned}$$

Substituting now $u_{t}(0)$ and $v_{t}(-1)$ into (45), one obtains $F_{0}$ as follows

$$\begin{aligned} F_{0}&= F(zq+\overline{z}\overline{q}+w,0) \nonumber \\&= \left( \begin{array}{c} F_{0_{1}} \\ F_{0_{2}} \end{array} \right) \\&= \left( \begin{array}{c} K_{20}z^{2}+K_{11}z\overline{z}+K_{02}\left( \overline{z}\right) ^{2}+K_{21}z^{2}\overline{z} \\ \frac{\sigma b}{4}\left( K_{20}z^{2}+K_{11}z\overline{z}+K_{02}\left( \overline{z}\right) ^{2}+K_{21}z^{2}\overline{z}\right) \end{array} \right) , \nonumber \end{aligned}$$

(47)

where

$$\begin{aligned} K_{20}&= \tau _{n}\left( p+rce^{-i\omega _{n}}\right) ,K_{02}=\tau _{n}\left( p+r\overset{-}{c}e^{i\omega _{n}}\right) , \\ K_{11}&= \tau _{n}\left( 2p+r\overset{-}{c}e^{i\omega _{n}}+rce^{-i\omega _{n}}\right) \\ K_{21}&= \tau _{n}\left( \begin{array}{l} 2pw_{11_{1}}(0)+pw_{20_{1}}(0)+2rw_{11_{2}}(-1) \\ +rw_{20_{2}}(-1)+r\overset{-}{c}e^{i\omega _{n}}w_{20_{1}}(0)+2rce^{-i\omega _{n}}w_{11_{1}}(0) \end{array} \right) \end{aligned}$$

and

$$\begin{aligned} p=\frac{4\alpha \left( 3-\alpha ^{2}\right) }{\left( 1+\alpha ^{2}\right) ^{2}},\, r=\frac{4\left( \alpha ^{2}-1\right) }{\left( 1+\alpha ^{2}\right) ^{2}}. \end{aligned}$$

(48)

Since $g(z,\overline{z})=\overline{q^{*}}(0)\cdot F_{0}(z,\overline{z} )=\sum \frac{1}{i!j!}g_{ij}(z)^{i}(\overline{z})^{j}$, we have

$$\begin{aligned} \sum \frac{1}{i!j!}g_{ij}(z)^{i}(\overline{z})^{j}&= \overline{q^{*}} (0)\cdot F_{0}(z,\overline{z}) \nonumber \\&= \overline{E}\left[ \begin{array}{cc} \overline{c^{*}}&1 \end{array} \right] \left[ \begin{array}{c} F_{0_{1}} \\ \frac{\sigma b}{4}F_{0_{1}} \end{array} \right] \\&= \overline{E}\overline{c^{*}}F_{0_{1}}+\overline{E}\frac{\sigma b}{4} F_{0_{1}}. \nonumber \end{aligned}$$

(49)

In order to determine the stability and the direction of the Hopf bifurcation, we need to find the Liapunov coefficient ([20, 31]) that is given by the following formula

$$\begin{aligned} c_{1}(\tau _{n})=\frac{i}{2\omega _{n}}\left( g_{20}g_{11}-2\left| g_{11}\right| ^{2}-\frac{1}{3}\left| g_{02}\right| ^{2}\right) + \frac{g_{21}}{2}, \end{aligned}$$

(50)

where from (49)

$$\begin{aligned} g_{20}&= 2\overline{E}\left( \overline{c^{*}}\!+\!\frac{\sigma b}{4} \right) K_{20}, \, g_{11}\!=\!\overline{E}\left( \overline{c^{*}}\!+\!\frac{\sigma b}{4}\right) K_{11},\nonumber \\ g_{02}&= 2\overline{E}\left( \overline{c^{*}}\!+\!\frac{\sigma b}{4} \right) K_{02}, \, g_{21}\!=\!\overline{E}\left( \overline{c^{*}}\!+\!\frac{\sigma b}{4}\right) K_{21}.\nonumber \\ \end{aligned}$$

(51)

To calculate $g_{21}$, we first need to find $w_{20}$ and $w_{11}$. We have

$$\begin{aligned} \frac{\partial w}{\partial t}=A(0)w+H(z,\overline{z},\theta ), \end{aligned}$$

where

$$\begin{aligned} H(z,\overline{z},\theta )=F_{0}-<q^{*},F_{0}>q-<\overline{q^{*}} ,F_{0}>\overline{q}. \end{aligned}$$

(52)

On the other hand, we also have

$$\begin{aligned}&\frac{\partial w}{\partial t} =\frac{\partial w}{\partial z}\frac{\partial z}{\partial t}+\frac{\partial w}{\partial \overset{-}{z}}\frac{\partial \overline{z}}{\partial t}, \nonumber \\&w(z,\overline{z},\theta )=\sum \frac{1}{i!j!}w_{ij}(\theta )(z)^{i}( \overline{z})^{j}, \\&H(z,\overline{z},\theta ) =\sum \frac{1}{i!j!}H_{ij}(\theta )(z)^{i}( \overline{z})^{j}. \nonumber \end{aligned}$$

(53)

Substituting (52) into (53) yields the following equalities

$$\begin{aligned} H_{20}&= (2i\omega _{n}-A)w_{20}, \nonumber \\ H_{11}&= (-A(0))w_{11}, \\ w_{02}&= \overline{w}_{20}. \nonumber \end{aligned}$$

(54)

First, we find $w_{20}$. From (52), $H_{20}$ equals to

$$\begin{aligned} H_{20}(\theta )=F_{0}-g_{20}q(\theta )-\overline{g}_{02}\overline{q}(\theta ). \end{aligned}$$

(55)

We analyze the right-hand side of the latter equation with respect to the position of $\theta$ as follows

Case C1: If $\theta \in \left[ -1,0\right)$, then using (33) we can write (55) as follows

$$\begin{aligned} H_{20}(\theta )=2i\omega _{n}w_{20}(\theta )-\frac{dw_{20}(\theta )}{d\theta }. \end{aligned}$$

(56)

Combining (54) and (56), one obtains the following differential equation

$$\begin{aligned} \frac{dw_{20}(\theta )}{d\theta }-2i\omega _{n}w_{20}(\theta )=g_{20}q(\theta )+\overline{g}_{02}\overline{q}(\theta ). \end{aligned}$$

(57)

Its solution is

$$\begin{aligned} w_{20}(\theta )&= -\frac{1}{i\omega _{n}}q(0)e^{i\omega _{n}\theta }g_{20}\nonumber \\&- \frac{1}{3i\omega _{n}}\overline{q}(0)e^{-i\omega _{n}\theta }\overline{g} _{02}+Se^{2i\omega _{n}\theta }. \end{aligned}$$

(58)

Case C2: If $\theta =0,$ then (55) becomes

$$\begin{aligned} H_{20}(0)=2K_{20}\left( \begin{array}{c} 1 \\ \frac{\sigma b}{4} \end{array} \right) -g_{20}q(0)-\overline{g}_{02}\overline{q}(0). \end{aligned}$$

(59)

Both (54) and (59) give us

$$\begin{aligned} A(0)w_{20}(0)&= 2i\omega _{n}w_{20}(0)+g_{20}q(0)\nonumber \\&+\frac{1}{3}\overline{g}_{02} \overline{q}(0)-2K_{20}\left( \begin{array}{c} 1 \\ \frac{\sigma b}{4} \end{array} \right) . \end{aligned}$$

(60)

From definition of the operator $A(0)$ (see (33)), we get

$$\begin{aligned} A(0)w_{20}(0)&= \int _{-1}^{0}d\eta (\theta ,0)w_{20}(\theta ) \nonumber \\&= -g_{20}q(0)+\frac{1}{3}\overline{g}_{02}\overline{q}(0)\nonumber \\&+S\int _{-1}^{0}d \eta (\theta ,0)e^{2i\omega _{n}\theta }, \end{aligned}$$

(61)

so that (60) and (61) yield the following equation

$$\begin{aligned} A(0)w_{20}(0)&= -g_{20}q(0)+\frac{1}{3}\overline{g}_{02}\overline{q} (0)\nonumber \\&+S\int _{-1}^{0}d\eta (\theta ,0)e^{2i\omega _{n}\theta } \nonumber \\&= 2i\omega _{n}w_{20}(0)+g_{20}q(0)\nonumber \\&+\overline{g}_{02}\overline{q} (0)-2K_{20}\left( \begin{array}{c} 1 \\ \frac{\sigma b}{4} \end{array} \right) . \end{aligned}$$

(62)

From Case C1, we have a formula for $w_{20}(\theta )$, namely (58). By substituting $w_{20}(0)$ (see (58)) into (62), we obtain

$$\begin{aligned} A(0)w_{20}(0)&= -g_{20}q(0)+\frac{1}{3}\overline{g}_{02}\overline{q} (0)\nonumber \\&+S\int _{-1}^{0}d\eta (\theta ,0)e^{2i\omega _{n}\theta } \\&= -g_{20}q(0)+\frac{1}{3}\overline{g}_{02}\overline{q}(0)\nonumber \\&+\,2i\omega _{n}S-2K_{20}\left( \begin{array}{c} 1 \\ \frac{\sigma b}{4} \end{array} \right) . \end{aligned}$$

Thus, we get the following equality that will give us $S$

$$\begin{aligned} S\left( 2i\omega _{n}-\int _{-1}^{0}d\eta (\theta ,0)e^{2i\omega _{n}\theta }\right) =2K_{20}\left( \begin{array}{c} 1 \\ \frac{\sigma b}{4} \end{array} \right) . \end{aligned}$$

(63)

Evaluating the integral above, one obtains the following:

$$\begin{aligned}&\int _{-1}^{0}d\eta (\theta ,0)e^{2i\omega _{n}\theta }\\&\quad =\int _{-1}^{0}\tau _{n}\left( \begin{array}{c} \left[ \begin{array}{cc} m-d_{1}n^{2} &{} 0 \\ f &{} -d_{2}n^{2} \end{array} \right] \delta \left( \theta \right) \\ +\left[ \begin{array}{cc} 0 &{} g \\ 0 &{} -k \end{array} \right] \delta \left( \theta +1\right) \end{array} \right) e^{2i\omega _{n}\theta }d\theta \\&\quad =\tau _{n}\left[ \begin{array}{cc} m-d_{1}n^{2} &{} ge^{-2i\omega _{n}} \\ f &{} -d_{2}n^{2}-ke^{-2i\omega _{n}} \end{array} \right] . \end{aligned}$$

Hence, $S$ is equal to

$$\begin{aligned}&{\small S=}\left( \left[ \begin{array}{cc} 2i\omega _{n}\!-\!\tau _{n}\left( m-d_{1}n^{2}\right) &{} \tau _{n}ge^{-2i\omega _{n}} \\ \tau _{n}f &{} 2i\omega _{n}\!+\tau _{n}d_{2}n^{2}\!+\tau _{n}ke^{-2i\omega _{n}} \end{array} \right] \right) ^{-1}\\&\quad \quad {\small 2K}_{20}\left( \begin{array}{c} 1 \\ \frac{\sigma b}{4} \end{array} \right) . \end{aligned}$$

Similarly, we will find $w_{11}$ so that, from (52), $H_{11}$ will be equal to

$$\begin{aligned} H_{11}(\theta )=F_{0}-g_{11}q(\theta )-\overline{g}_{11}\overline{q}(\theta ). \end{aligned}$$

(64)

To do this, we consider two cases as follows.

Case D1: If $\theta \in \left[ -1,0\right) ,$ then because of definition of the operator $A(\theta )$ (see (33)) the equality (54) becomes

$$\begin{aligned} H_{20}(\theta )=-\frac{dw_{11}(\theta )}{d\theta }. \end{aligned}$$

(65)

Both (54) and (65) give us

$$\begin{aligned} \frac{dw_{11}(\theta )}{d\theta }=g_{11}q(\theta )+\overline{g}_{11} \overline{q}(\theta ) \end{aligned}$$

(66)

so that we have

$$\begin{aligned} w_{11}(\theta )=\frac{1}{i\omega _{n}}q(0)e^{i\omega _{n}\theta }g_{11}- \frac{1}{i\omega _{n}}\overline{q}(0)e^{-i\omega _{n}\theta }\overline{g} _{11}+G, \end{aligned}$$

(67)

where $G$ will be determined in Case D2.

Case D2: If $\theta =0,$ then from (64) we have

$$\begin{aligned} H_{11}(0)=g_{11}q(0)+\overline{g}_{11}\overline{q}(0)-K_{11}\left( \begin{array}{c} 1 \\ \frac{\sigma b}{4} \end{array} \right) . \end{aligned}$$

(68)

(54) and (68) give us

$$\begin{aligned} A(0)w_{11}(0)=g_{11}q(0)+\overline{g}_{11}\overline{q}(0)-K_{11}\left( \begin{array}{c} 1 \\ \frac{\sigma b}{4} \end{array} \right) . \end{aligned}$$

(69)

From (33), we get

$$\begin{aligned} A(0)w_{11}(0)&= \int _{-1}^{0}d\eta (\theta ,0)w_{11}(\theta ) \nonumber \\&= g_{11}q(0)+\overline{g}_{11}\overline{q}(0)+G\int _{-1}^{0}d\eta (\theta ,0) \nonumber \\ \end{aligned}$$

(70)

so that (67) and (70) yield

$$\begin{aligned} A(0)w_{11}(0)&= g_{11}q(0)+\overline{g}_{11}\overline{q}(0)+G\int _{-1}^{0}d \eta (\theta ,0) \nonumber \\&= g_{11}q(0)+\overline{g}_{11}\overline{q}(0)-K_{11}\left( \begin{array}{c} 1 \\ \frac{\sigma b}{4} \end{array} \right) . \nonumber \\ \end{aligned}$$

(71)

Equating the right-hand sides of the latter equation, one obtains the following identity that is used to determine $G$

$$\begin{aligned} G\left( \int _{-1}^{0}d\eta (\theta ,0)\right) =K_{11}\left( \begin{array}{c} 1 \\ \frac{\sigma b}{4} \end{array} \right) . \end{aligned}$$

(72)

First, we calculate the integral above as follows

$$\begin{aligned} \int _{-1}^{0}d\eta (\theta ,0)&= \int _{-1}^{0}\tau _{n}\left( \begin{array}{c} \left[ \begin{array}{cc} m-d_{1}n^{2} &{} 0 \\ f &{} -d_{2}n^{2} \end{array} \right] \delta \left( \theta \right) \\ +\left[ \begin{array}{cc} 0 &{} g \\ 0 &{} -k \end{array} \right] \delta \left( \theta +1\right) \end{array} \right) d\theta \\&= \tau _{n}\left[ \begin{array}{cc} m-d_{1}n^{2} &{} g \\ f &{} \left( -k-d_{2}n^{2}\right) \end{array} \right] . \end{aligned}$$

So, we have

$$\begin{aligned} G\tau _{n}\left[ \begin{array}{cc} m-d_{1}n^{2} &{} g \\ f &{} \left( -k-d_{2}n^{2}\right) \end{array} \right] =K_{11}\left( \begin{array}{c} 1 \\ \frac{\sigma b}{4} \end{array} \right) \end{aligned}$$

so that $G$ is equal to

$$\begin{aligned} G=\left( \tau _{n}\left[ \begin{array}{cc} m-d_{1}n^{2} &{} g \\ f &{} \left( -k-d_{2}n^{2}\right) \end{array} \right] \right) ^{-1}K_{11}\left( \begin{array}{c} 1 \\ \frac{\sigma b}{4} \end{array} \right) . \end{aligned}$$

(73)

In order to determine direction of the bifurcation, we also need to know sign of the Liapunov coefficient. It can be determined by using following formula

$$\begin{aligned} Re(c_{1}(\tau _{n}))&= Re\left( \frac{g_{21}}{2}\right) -\frac{1}{2\omega _{n}} \left( Re(g_{20})Im(g_{11})\right. \nonumber \\&\left. +\,Im(g_{20})Re(g_{11})\right) . \end{aligned}$$

(74)

From the analysis above and the general Hopf bifurcation theorem (see [12], page 479 and references therein), we can deduce the following results.

Theorem 2

If $\frac{1}{\alpha ^{^{\prime }}(0)}Re(c_{1}(\tau _{n}))<0$ $(\frac{1}{ \alpha ^{^{\prime }}(0)}Re(c_{1}(\tau _{n}))$ $>0)$, then the bifurcation is supercritical (subcritical). In addition, if other eigenvalues of $L_{n}$ have negative real parts, then the bifurcating periodic solution is stable (unstable) if $Re(c_{1}(\tau _{n,k}))<0$ ( $Re(c_{1}(\tau _{n,k}))>0$).

4 Analysis of spatially homogeneous Lengyel–Epstein system with delay

When $n=0$ in (12), the characteristic equation (14) becomes

$$\begin{aligned} \lambda ^{2}+A\lambda +Be^{-\lambda \tau }+C\lambda e^{-\lambda \tau }+D=0, \end{aligned}$$

(75)

where

$$\begin{aligned} A=-m,\, B=5k,\, C=k \end{aligned}$$

and $m$ and $k$ are given in (40). By substituting $\lambda =i\omega _{0},$ $\omega _{0}>0$, into (75) we get the following equation

$$\begin{aligned} \omega _{0}^{4}+(A^{2}-C^{2})\omega _{0}^{2}-B^{2}=0. \end{aligned}$$

(76)

From (76), we have

$$\begin{aligned} \omega _{0}^{2}=\frac{-(A^{2}-C^{2})\pm \sqrt{(A^{2}-C^{2})^{2}-4\left( -B^{2}\right) }}{2}. \end{aligned}$$

(77)

Let $X_{0}=(A^{2}-C^{2})$ and $Y_{0}=-B^{2}$. So, (18) can be written as follows

$$\begin{aligned} \omega _{0}^{2}=\frac{-X_{0}\pm \sqrt{X_{0}^{2}-4Y_{0}}}{2}. \end{aligned}$$

(78)

Our aim is to get at least one $\omega _{0}\in \mathbb {R} ^{+}$. By analyzing (78), we get the following results:

• 1. If $X_{0}^{2}-4Y_{0}<0,$ then $\omega _{0}^{2}\in \mathbb {C}$. Hence, there is no real root.

• 2. If $X_{0}^{2}-4Y_{0}=0,$ then $B=0$ and $A^{2}=C^{2}\Longrightarrow m=k=b=\sigma =0$ so that there is no positive real root.

• 3. If $X_{0}^{2}-4Y_{0}>0$, then there are three possibilities as follows:

  • (a) $X_{0}<0\Longrightarrow$ there is only one positive real root which is

    $$\begin{aligned} \omega _{0}=\sqrt{\frac{-X_{0}+\sqrt{X_{0}^{2}-4Y_{0}}}{2}}. \end{aligned}$$

    (79)

  • (b) $X_{0}=0\Longrightarrow$ there is only one positive real root, namely

    $$\begin{aligned} \omega _{0}=\sqrt{\frac{\sqrt{-4Y_{0}}}{2}}=\sqrt{B}=\sqrt{5k}. \end{aligned}$$

    (80)

  • (c) $X_{0}>0\Longrightarrow$ there is only one positive real root which is

    $$\begin{aligned} \omega _{0}=\sqrt{\frac{-X_{0}+\sqrt{X_{0}^{2}-4Y_{0}}}{2}}. \end{aligned}$$

    (81)

In conclusion, we have only one positive real $\omega _{0}$ for $3a$, $3b$ and $3c$.

Finally, we need to check the transversality condition. From (23), we can obtain $Re\left( \left. \frac{d\alpha }{d\tau }\right| _{\tau =\tau _{0}}\right) ^{-1}$ as follows

$$\begin{aligned}&Re\left( \left. \frac{d\alpha }{d\tau }\right| _{\tau =\tau _{0}}\right) ^{-1}\nonumber \\&=\frac{1}{\left( B^{2}+C^{2}\omega _{0}^{2}\right) ^{2}} \left( \begin{array}{c} 2C^{2}\omega _{0}^{4} \\ +\left( A^{2}C^{2}+2B^{2}-C^{4}\right) \omega _{0}^{2} \\ +A^{2}B^{2}-B^{2}C^{2} \end{array} \right) , \nonumber \\ \end{aligned}$$

(82)

where (see (76))

$$\begin{aligned} \omega _{0}^{4}=B^{2}-(A^{2}-C^{2})\omega _{0}^{2}. \end{aligned}$$

(83)

Plugging in (83) into (82) yields

$$\begin{aligned}&Re\left( \left. \frac{d\alpha }{d\tau }\right| _{\tau =\tau _{0}}\right) ^{-1} \nonumber \\&= \frac{1}{\left( B^{2}+C^{2}\omega _{0}^{2}\right) ^{2}} \left( \begin{array}{c} \left( C^{4}-A^{2}C^{2}+2B^{2}\right) \omega _{0}^{2} \\ +A^{2}B^{2}+B^{2}C^{2} \end{array} \right) \nonumber \\&= \frac{1}{\left( 25+\omega _{0}^{2}\right) ^{2}k^{4}}\left( \begin{array}{c} k^{4}\omega _{0}^{2}+\left( 50-m^{2}\right) k^{2}\omega _{0}^{2} \\ +25k^{2}m^{2}+25k^{4} \end{array} \right) , \end{aligned}$$

(84)

and

$$\begin{aligned} \left( 50-m^{2}\right) =\frac{25+130\alpha ^{2}+41\alpha ^{4}}{1+2\alpha ^{2}+\alpha ^{4}}>0. \end{aligned}$$

(85)

Hence, we have $Re\left( \left. \frac{d\alpha }{d\tau }\right| _{\tau =\tau _{0}}\right) ^{-1}>0$ which means $Re\left( \left. \frac{ d\alpha }{d\tau }\right| _{\tau =\tau _{0}}\right) >0.$ It shows that the transversality condition holds.

By Theorem 2, when $n=0$, the Hopf bifurcation occurs at $\mu =0$ for each $\omega _{0}$ which we found above, and system (32) possesses a family of real-valued periodic solutions bifurcating from the equilibrium point $(0,0)$ at $\mu =0$.

Next, we find the direction of this bifurcation and analyze the stability of periodic solutions. For these goals, we need to know the sign of the Liapunov coefficient which is given by (74) where

$$\begin{aligned}&Re(c_{1}(\tau _{n}))=Re(g_{21}/2)\\&-\frac{1}{2\omega _{n}} \left( Re(g_{20})Im(g_{11})+Im(g_{20})Re(g_{11})\right) . \end{aligned}$$

So, we need to compute $g_{20}, g_{11}$ and $g_{21}$ for$\ n=0.$

From (51), we have

$$\begin{aligned} g_{20}=2\overline{E}\left( \overline{c^{*}}+\frac{\sigma b}{4}\right) K_{20}, \end{aligned}$$

where

$$\begin{aligned} \overline{E}&= \frac{1}{\left( \overline{c^{*}}+c+\tau _{0}ge^{-i\omega _{0}}c\overline{c^{*}}-\tau _{0}ke^{-i\omega _{0}}c\right) }, \\ c^{*}&= \left( \frac{-\tau _{0}f}{\tau _{0}m+i\omega _{0}}\right) , \,\ c=\left( \frac{i\omega _{0}-\tau _{0}m}{\tau _{0}ge^{-i\omega _{0}}} \right) , \,\ \\&K_{20} =\tau _{0}\left( p+rce^{-i\omega _{0}}\right) \end{aligned}$$

in which $m$, $f$, $g$, $k$, $p$ and $r$ are given in (40) and (48). Similarly, the coefficient $g_{11}$ has the following form

$$\begin{aligned} g_{11}=\overline{E}\left( \overline{c^{*}}+\frac{\sigma b}{4}\right) K_{11,} \end{aligned}$$

where

$$\begin{aligned} \overline{E}&= \frac{1}{\left( \overline{c^{*}}+c+\tau _{0}ge^{-i\omega _{0}}c\overline{c^{*}}-\tau _{0}ke^{-i\omega _{0}}c\right) }, \\ c^{*}&= \left( \frac{-\tau _{0}f}{\tau _{0}m+i\omega _{0}}\right) , \,\ c=\left( \frac{i\omega _{0}-\tau _{0}m}{\tau _{0}ge^{-i\omega _{0}} }\right) , \,\ \\&K_{11} =\tau _{0}\left( 2p+r\overline{c}e^{i\omega _{0}}+rce^{-i\omega _{0}}\right) . \end{aligned}$$

From (51), we also have

$$\begin{aligned} g_{21}=\overline{E}(\overline{c^{*}}+\frac{\sigma b}{4})K_{21}, \end{aligned}$$

where

$$\begin{aligned} \overline{E}&= \frac{1}{\left( \overline{c^{*}}+c+\tau _{0}ge^{-i\omega _{0}}c\overline{c^{*}}-\tau _{0}ke^{-i\omega _{0}}c\right) }, \\ c^{*}&= \left( \frac{-\tau _{0}f}{\tau _{0}m+i\omega _{0}}\right) , \,\ c=\left( \frac{i\omega _{0}-\tau _{0}m}{\tau _{0}ge^{-i\omega _{0}} }\right) , \\ K_{21}&= \tau _{0}\left( \begin{array}{c} 2pw_{11_{1}}(0)+pw_{20_{1}}(0)+2rw_{11_{2}}(-1) \\ +rw_{20_{2}}(-1) +r\overline{c}e^{i\omega _{0}}w_{20_{1}}(0) \\ +2rce^{-i\omega _{0}}w_{11_{1}}(0) \end{array} \right) \end{aligned}$$

and

$$\begin{aligned} w_{20}(\theta )&= -\frac{1}{i\omega _{0}}q(0)e^{i\omega _{0}\theta }g_{20}\\&-\frac{1}{3i\omega _{0}}\overline{q}(0)e^{-i\omega _{0}\theta }\overline{g} _{02}+Se^{2i\omega _{0}\theta }, \end{aligned}$$

where

$$\begin{aligned}&S=\left( \left[ \begin{array}{cc} 2i\omega _{0}-\tau _{0}m &{} \tau _{0}ge^{-2i\omega _{0}} \\ \tau _{0}f &{} 2i\omega _{0}+\tau _{0}ke^{-2i\omega _{0}} \end{array} \right] \right) ^{-1}\\&\quad \times 2K_{20}\left( \begin{array}{c} 1 \\ \frac{\sigma b}{4} \end{array} \right) , \end{aligned}$$

and

$$\begin{aligned} w_{11}(\theta )=\frac{1}{i\omega _{0}}q(0)e^{i\omega _{0}\theta }g_{11}- \frac{1}{i\omega _{0}}\overline{q}(0)e^{-i\omega _{0}\theta }\overline{g} _{11}+G, \end{aligned}$$

where

$$\begin{aligned} G=\left( \tau _{0}\left[ \begin{array}{cc} m &{} g \\ f &{} -k \end{array} \right] \right) ^{-1}K_{11}\left( \begin{array}{c} 1 \\ \frac{\sigma b}{4} \end{array} \right) . \end{aligned}$$

Now repeating a similar calculation as in the former section, we can determine the direction of the Hopf bifurcation with respect to Theorem 2.

5 Conclusion

Former studies show that delay parameter plays an important role on the stability analysis of positive equilibrium points of a dynamical system (see [6, 7, 9, 19–28] and references therein). In addition, diffusion-driven instability, which is also known as turing instability in the literature, has been studied extensively in the last two decades (see [2, 3, 8, 10–17, 29, 30] and references therein).

In this paper, we study the delay effect on the Lengyel–Epstein reaction–diffusion model with the Neumann boundary conditions. First, we investigated the necessary conditions at which Hopf bifurcation occurs by choosing the delay parameter $\tau$ as a bifurcation parameter. Using Poincaré normal form and the center manifold reduction for partial functional differential equations, the formulas that determine the direction of bifurcation and the stability of periodic solutions are obtained. We showed that when the bifurcation parameter $\tau$ passes through a critical bifurcation value $\tau _{n,k}$ $(n,k=0,1,2,\ldots )$, stability of the positive equilibrium point of system (2) changes from stable to unstable or vice versa, and Hopf bifurcation occurs at these critical values when the associated characteristic equation has only one pure imaginary root. Moreover, if the characteristic equation has two different pure imaginary roots, then not only Hopf bifurcation occurs but also stability of equilibrium point switches.

This work differs from [19] since system (1) is a PDE model and involves both diffusion term and delay while the model in [19] is an ODE model and only consists of delay. This work also differs from [13] since it consists of Hopf bifurcation analysis in the Lengyel–Epstein reaction–diffusion model involving delay while the other studied that of the model without delay.