From [1] we know that the diffusion can influence the stability of the homogeneous solutions. Turing instability occurs when the homogeneous solutions, which should be stable for system (5), become unstable due to diffusions. In this section, we focus on the full reaction–diffusion model (
3
)–(
4
) to obtain the parametric ranges in which the homogeneous solutions undergo Turing instability. The analysis of Turing instabilities for the equilibrium
\((a_{*},h_{*})\) and the periodic solution bifurcated from Hopf bifurcation is carried out respectively in Sects. 3.1 and 3.2.
Turing instability of the equilibrium for the full reaction–diffusion model
In this section, we follow the standard treatment for this type of problems as in [9]. First, we assume that one of conditions (H1) and (H2) holds and study system (3) with the Neumann boundary conditions (4) in the Banach space \(\mathbb {H}^{2}((0,\pi))\times\mathbb {H}^{2}((0,\pi))\), where \(\mathbb {H}^{2}((0,\pi ))=\{w(\cdot)\mid\frac{\partial^{i}w}{\partial x^{i}}(\cdot)\in\mathbb {L}^{2}((0,\pi)),i=0,1,2\}\). Obviously, \((a_{*},h_{*})\) is a steady state for system (3)–(4).
Let \(a=u_{1}+a_{*},h=u_{2}+h_{*}\). The linearized system of (3) at the equilibrium \((a_{*},h_{*})\) is
\begin{array}{rl}\left(\begin{array}{c}\frac{\partial {u}_{1}}{\partial t}\\ \frac{\partial {u}_{2}}{\partial t}\end{array}\right)& =\left(\begin{array}{cc}\mu +r{\nu}^{r}{a}_{\ast}^{1+r{r}^{2}}+{\partial}_{xx}& r{\nu}^{1+r}{a}_{\ast}^{{r}^{2}}\\ r{a}_{\ast}^{1+r}& \nu +d{\partial}_{xx}\end{array}\right)\left(\begin{array}{c}{u}_{1}\\ {u}_{2}\end{array}\right)\\ & :=L(c)\left(\begin{array}{c}{u}_{1}\\ {u}_{2}\end{array}\right).\end{array}
(10)
Here the boundary conditions are:
$$ \frac{\partial u_{1}}{\partial x}(0,t)=\frac{\partial u_{1}}{\partial x}(\pi ,t)=0, \qquad \frac{\partial u_{2}}{\partial x}(0,t)= \frac{\partial u_{2}}{\partial x}(\pi,t)=0. $$
(11)
Considering the boundary conditions, the solution \((u_{1},u_{2})\in\mathbb {H}^{2}((0,\pi))\times\mathbb {H}^{2}((0,\pi))\) of system (10) can formally be written as
\left(\begin{array}{c}{u}_{1}(x,t)\\ {u}_{2}(x,t)\end{array}\right)=\sum _{k=0}^{\mathrm{\infty}}\left(\begin{array}{c}{A}_{k}\\ {H}_{k}\end{array}\right){e}^{{\lambda}_{k}t}coskx,
(12)
where \(\lambda_{k} \in\mathbb {C}\) is the temporal spectrum, k is the wave number, and \(A_{k},H_{k} \in\mathbb{R}\) for \(k=0,1,2 ,\ldots\) . Substituting (12) into system (10), we have
\sum _{k=0}^{\mathrm{\infty}}\left(\begin{array}{c}{A}_{k}\\ {H}_{k}\end{array}\right){\lambda}_{k}{e}^{{\lambda}_{k}t}coskx=\sum _{k=0}^{\mathrm{\infty}}{J}_{k}(c)\left(\begin{array}{c}{A}_{k}\\ {H}_{k}\end{array}\right){e}^{{\lambda}_{k}t}coskx,
where
{J}_{k}(c)=\left(\begin{array}{cc}\mu +r{\nu}^{r}{a}_{\ast}^{1+r{r}^{2}}{k}^{2}& r{\nu}^{1+r}{a}_{\ast}^{{r}^{2}}\\ r{a}_{\ast}^{1+r}& \nu d{k}^{2}\end{array}\right).
(13)
Equating the like powers of k, we have
({\lambda}_{k}I{J}_{k}(c))\left(\begin{array}{c}{A}_{k}\\ {H}_{k}\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right),\phantom{\rule{1em}{0ex}}k=0,1,2,\dots .
(14)
Equation (14) has a nonzero solution \((A_{k},H_{k})^{T}\) if and only if
$$ \operatorname{Det} \bigl(\lambda_{k} IJ_{k}(c) \bigr)=0 $$
(15)
for some \(k=0,1,2,\ldots \) . Namely,
$$ \lambda_{k}^{2}\operatorname{Tr} \bigl(J_{k}(c) \bigr) \lambda_{k}+\operatorname{Det} \bigl(J_{k}(c) \bigr)=0, $$
(16)
where
$$\operatorname{Tr} \bigl(J_{k}(c) \bigr)=\mu\nu+r \nu^{r}a_{*}^{1+rr^{2}}(1+d)k^{2} $$
and
$$\operatorname{Det} \bigl(J_{k}(c) \bigr)= \bigl(k^{2}+\mu \bigr)\nu+(r1)r\nu ^{1+r}a_{*}^{1+rr^{2}}+dk^{2} \bigl[k^{2} \bigl(r\nu^{r}a_{*}^{1+rr^{2}}\mu \bigr) \bigr]. $$
We will be concerned with the requirement \(\operatorname{Re}(\lambda_{k})>0\) on the solutions of (16).
For any wave number k, there exists a pair of temporal eigenvalues \(\lambda_{k}^{\pm}= \frac{\operatorname{Tr}(J_{k}(c))\pm\sqrt{\operatorname {Tr}^{2}(J_{k}(c))4\operatorname{Det}(J_{k}(c))}}{2}\). If (H1) or (H2) holds, then, for \(k=0\), \(\operatorname{Tr}(J_{k}(c))=\operatorname{Tr}J(c)<0\) and \(\operatorname {Det}(J_{k}(c))=\operatorname{Det}J(c)>0\), which implies that the real parts of the related temporal spectra \(\lambda_{0}^{\pm}\) are negative. Since \(\operatorname{Tr}(J_{k}(c))=\operatorname{Tr}J(c)(1+d)k^{2} < 0\) for \(k=1,2,\ldots \) , we need further consider the sign of \(\operatorname{Det}(J_{k}(c))\). If \(r\nu^{r}a_{*}^{1+rr^{2}}\mu\leq1\), then we have \(a_{*} \geq(\frac{r\nu ^{r}}{\mu+1})^{\frac{1}{1r+r^{2}}}\) since \(1+rr^{2} <0\). Denoting \(\bar {a}= (\frac{r\nu^{r}}{\mu+1})^{\frac{1}{1r+r^{2}}}\), from Lemma 2.1 and Proposition 2.1 we have \(\phi(\bar{a}) \geq0\), that is, \(c \geq\frac{\bar{a}}{r}[\mu(r1)1]\). Therefore \(k^{2}(r\nu ^{r}a_{*}^{1+rr^{2}}\mu)\geq0\), which means that \(\operatorname{Det}(J_{k}(c)) >0\). Moreover, if \(m^{2} < r\nu^{r}a_{*}^{1+rr^{2}}\mu\leq(m+1)^{2}, m \in \mathbb {N}^{+}\), then we have \([\frac{r\nu^{r}}{\mu+(m+1)^{2}}]^{\frac {1}{1r+r^{2}}} \leq a_{*} < (\frac{r\nu^{r}}{\mu+m^{2}})^{\frac{1}{1r+r^{2}}}\) since \(1+rr^{2} <0\). Denoting \(a_{m+1}=[\frac{r\nu^{r}}{\mu +(m+1)^{2}}]^{\frac{1}{1r+r^{2}}}\) and \(a_{m}=(\frac{r\nu^{r}}{\mu+m^{2}})^{\frac {1}{1r+r^{2}}}\), from Lemma 2.1 and Proposition 2.1 we have \(\phi(a_{m})<0\leq\phi(a_{m+1})\), that is, \(\frac{a_{m+1}}{r}[\mu (r1)(m+1)^{2}] \leq c < \frac{a_{m}}{r}[\mu(r1)m^{2}]\). In this case, for any \(k=m+1,m+2,\ldots \) , we have \(k^{2}(r\nu^{r}a_{*}^{1+rr^{2}}\mu)\geq0\), which indicates that \(\operatorname{Det}(J_{k}(c)) >0\), whereas for any \(k=1,2,\ldots,m\), we have \(k^{2}(r\nu^{r}a_{*}^{1+rr^{2}}\mu)<0\). Further computations reveal that if \(0< d < \min_{1\leq k \leq m}\frac{(k^{2}+\mu )\nu+(r1)r\nu^{1+r}a_{*}^{1+rr^{2}}}{k^{2}(r\nu^{r}a_{*}^{1+rr^{2}}\mu k^{2})}\), then \(\operatorname{Det}(J_{k}(c))>0\).
The above discussion implies that, for any spatial spectrum \(k=1,2,\ldots \) , the real parts of the corresponding temporal spectrum \(\lambda_{k}^{\pm}\) are negative. This means the real parts of the solutions of (16) are negative for all \(k=0,1,2,\ldots \) , so that \((a_{*},h_{*})\) is asymptotically stable for (3). According to [28], because of the sectorial of L, \((a_{*},h_{*})\) is also locally uniformly stable, and therefore there is no Turing pattern for system (3) under the above conditions. However, if \(d > \min_{1\leq k \leq m}\frac{(k^{2}+\mu)\nu+(r1)r\nu ^{1+r}a_{*}^{1+rr^{2}}}{k^{2}(r\nu^{r}a_{*}^{1+rr^{2}}\muk^{2})}\), then at least one of \(\operatorname{Det}(J_{1}(c)), \operatorname{Det} (J_{2}(c)), \ldots, \operatorname{Det}(J_{m}(c))\) is negative. Thus the equilibrium is unstable for (3). In this case, Turing patterns will occur. Summarizing the discussion, we have the following:
Theorem 3.1
Let (H1) or (H2) hold. Denote
$$\hat{d}=\min_{1\leq k \leq m}\frac{(k^{2}+\mu)\nu+(r1)r\nu ^{1+r}a_{*}^{1+rr^{2}}}{k^{2}(r\nu^{r}a_{*}^{1+rr^{2}}\muk^{2})}, \quad m \in\mathbb {N}^{+}. $$
Then the equilibrium
\((a_{*},h_{*})\)
of system (3) persists the stability if one of the following conditions holds:

(H4)
\(c\geq\frac{\bar{a}}{r}[\mu(r1)1]\),

(H5)
\(A_{m+1}\leq c < A_{m},0<d < \hat{d}\).
It is unstable if

(H6)
\(A_{m+1}\leq c < A_{m}, d>\hat{d}\),
where
\(\bar{a}=(\frac{r\nu^{r}}{\mu+1})^{\frac{1}{1r+r^{2}}}\), \(A_{j}=\frac {a_{j}}{r}[\mu(r1)j^{2}]\), \(a_{j}=(\frac{r\nu^{r}}{\mu+j^{2}})^{\frac {1}{1r+r^{2}}}, j=m, m+1, m \in\mathbb {N}^{+}\).
Remark 3.1
It is worth pointing out that in Theorem 3.1 if \(d=\hat{d}\), then from the discussion it follows that there exists at least one \(k, k=1,2,\ldots,m\), \(m \in\mathbb {N}^{+}\), such that the linearized system of (3) at \((a_{*},h_{*})\) has a zero eigenvalue. At this time, the stability of \((a_{*},h_{*})\) for system (3) cannot be determined by the linearized system.
Turing instability of the limit cycle for the full reaction–diffusion model
From Theorem 2.1 note that if \(0<\nu<(r1)\mu\), then there is a limit cycle bifurcated from \((a_{*},h_{*})\) for c sufficiently close to \(c_{h}\). We investigate the stability of the limit cycle obtained in Theorem 2.1. Throughout this section, we assume condition (H3), so that the limit cycle is stable to homogeneous perturbation.
Let \(a=u_{1}+a_{*}, h=u_{2}+h_{*}, c=c_{h}\), \(U=(u_{1},u_{2})^{T}\), \(V=(v_{1},v_{2})^{T}\), and \(W=(w_{1},w_{2})\). Then system (3)–(4) can be rewritten as [29, 30]
$$ \textstyle\begin{cases} \frac{\partial U}{\partial t}=[J(c_{h})+D ( {\scriptsize\begin{matrix}{} \partial_{xx}& 0\\ 0 & \partial_{xx} \end{matrix}} ) ] U+\tilde{F}(U,c_{h}), \\ \frac{\partial U}{\partial x}(0,t)=\frac{\partial U}{\partial x}(\pi ,t)=(0,0)^{T}, \end{cases} $$
(17)
where
J({c}_{h})=\left(\begin{array}{cc}\nu & r{\nu}^{r+1}{a}_{0}^{{r}^{2}}\\ r{a}_{0}^{r1}& \nu \end{array}\right),\phantom{\rule{2em}{0ex}}D=\left(\begin{array}{cc}1& 0\\ 0& d\end{array}\right),
\(\tilde{F}(U,c_{h})=(f_{2}(u_{1},u_{2},c_{h}),g_{2}(u_{1},u_{2},c_{h}))^{T}\), and \(f_{2}\) and \(g_{2}\) are defined in (9). We write \(\tilde{F}(U,c_{h})=\frac{1}{2}Q(U,U)+\frac {1}{6}C(U,U,U)+O(\lvert U \rvert^{4})\), where \(Q(U,V)=(Q_{1}(U,V),Q_{2}(U,V))^{T}\), \(C(U,V,W)=(C_{1}(U,V,W),C_{2}(U,V,W))^{T}\) with
$$\begin{aligned} &Q_{1}(U,V)=f_{2uu}u_{1}v_{1}+f_{2uv}u_{1}v_{2}+f_{2vu}u_{2}v_{1}+f_{2vv}u_{2}v_{2} \\ &\phantom{Q_{1}(U,V)}=(\mu+\nu) \bigl[(r1)a_{0}^{1}u_{1}v_{1}r \nu a_{0}^{r}(u_{2}v_{1}+u_{1}v_{2}) \\ &\phantom{Q_{1}(U,V)=}{}+(r+1)\nu^{2}a_{0}^{12r}u_{2}v_{2} \bigr], \\ &Q_{2}(U,V)=g_{2uu}u_{1}v_{1}+g_{2uv}u_{1}v_{2}+g_{2vu}u_{2}v_{1}+g_{2vv}u_{2}v_{2} \\ &\phantom{Q_{2}(U,V)}=(r1)ra_{0}^{r2}u_{1}v_{1}, \\ &C_{1}(U,V,W)=f_{2uuu}u_{1}v_{1}w_{1}+f_{2uuv}u_{1}v_{1}w_{2}+f_{2uvu}u_{1}v_{2}w_{1}+f_{2vuu}u_{2}v_{1}w_{1} \\ &\phantom{C_{1}(U,V,W)=}{}+f_{2uvv}u_{1}v_{2}w_{2}+f_{2vuv}u_{2}v_{1}w_{2}+f_{2vvu}u_{2}v_{2}w_{1}+f_{2vvv}u_{2}v_{2}w_{2} \\ &\phantom{C_{1}(U,V,W)}=(\mu+\nu) \bigl[(r2) (r1)a_{0}^{2}u_{1}v_{1}w_{1}+(1r)r \nu a_{0}^{1r}(u_{1}v_{1}w_{2} \\ &\phantom{C_{1}(U,V,W)=}{}+u_{1}v_{2}w_{1}+u_{2}v_{1}w_{1})+(1+r)r \nu^{2} a_{0}^{2r}(u_{1}v_{2}w_{2}+u_{2}v_{1}w_{2} \\ &\phantom{C_{1}(U,V,W)=}{}+u_{2}v_{2}w_{1})(r+1) (2+r)\nu^{3} a_{0}^{13r}u_{2}v_{2}w_{2} \bigr], \\ &C_{2}(U,V,W)=g_{2uuu}u_{1}v_{1}w_{1}+g_{2uuv}u_{1}v_{1}w_{2}+g_{2uvu}u_{1}v_{2}w_{1}+g_{2vuu}u_{2}v_{1}w_{1} \\ &\phantom{C_{2}(U,V,W)=}{}+g_{2uvv}u_{1}v_{2}w_{2}+g_{2vuv}u_{2}v_{1}w_{2}+g_{2vvu}u_{2}v_{2}w_{1}+g_{2vvv}u_{2}v_{2}w_{2} \\ &\phantom{C_{2}(U,V,W)}=(r2) (r1)ra_{0}^{r3}u_{1}v_{1}w_{1}, \end{aligned}$$
and \(U,V,W \in\mathbb{H}^{2}((0,\pi))\times\mathbb{H}^{2}((0,\pi))\).
For \(c=c_{h}\), the linear operator L defined in (10) is
LU=[J({c}_{h})+D\left(\begin{array}{cc}{\partial}_{xx}& 0\\ 0& {\partial}_{xx}\end{array}\right)]U,
and the corresponding adjoint operator, denoted by \(L^{*} \), is
{L}^{\ast}U=[{J}^{\ast}({c}_{h})+\left(\begin{array}{cc}{\partial}_{xx}& 0\\ 0& {\partial}_{xx}\end{array}\right)]U,
where
J({c}_{h})=\left(\begin{array}{cc}\nu & r{\nu}^{r+1}{a}_{0}^{{r}^{2}}\\ r{a}_{0}^{r1}& \nu \end{array}\right),\phantom{\rule{2em}{0ex}}{J}^{\ast}({c}_{h})=\left(\begin{array}{cc}\nu & r{a}_{0}^{r1}\\ r{\nu}^{r+1}{a}_{0}^{{r}^{2}}& \nu \end{array}\right).
The inner product in \(\mathbb{H}^{2}((0,\pi))\times\mathbb{H}^{2}((0,\pi ))\) is given by \(\langle U,V\rangle=\frac{1}{\pi}\int_{0}^{\pi}\overline {U}^{T} V\,\mathrm{d}x\) for \(U,V \in\mathbb{H}^{2}((0,\pi))\times\mathbb {H}^{2}((0,\pi))\). Note that \(\langle L^{*}U,V\rangle= \langle U,LV\rangle \). The linearized system of (17) evaluated at
\((0,0)\) is
$$ \frac{\partial U}{\partial t}=LU $$
(18)
with the Neumann boundary conditions
$$ \frac{\partial U}{\partial x}(0,t)=\frac{\partial U}{\partial x}(\pi,t)=(0,0)^{T}. $$
(19)
System (18) with boundary conditions (19) has a solution, which can be formally written as
U=\sum _{k=0}^{\mathrm{\infty}}\left(\begin{array}{c}{a}_{k}\\ {h}_{k}\end{array}\right){e}^{\lambda (k)t}coskx,
(20)
where \(\lambda(k) \in\mathbb{C}\) are the temporal eigenvalues, k is the wave number, and \(a_{k},h_{k} \in\mathbb{R}\) for \(k=0,1,2,\ldots\) . Substituting (20) into (18) and comparing the like terms about k, we have
(\lambda (k)I{L}_{k})\left(\begin{array}{c}{a}_{k}\\ {h}_{k}\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right),\phantom{\rule{1em}{0ex}}k=0,1,2,\dots ,
(21)
where
{L}_{k}=\left(\begin{array}{cc}\nu {k}^{2}& r{\nu}^{r+1}{a}_{0}^{{r}^{2}}\\ r{a}_{0}^{r1}& \nu d{k}^{2}\end{array}\right).
For some k, a sufficient and necessary condition for (21) to have a nonzero solution \((a_{k},h_{k})^{T}\) is \(\operatorname{Det}(\lambda(k) IL_{k})=0\), that is,
$$ \lambda(k)^{2}\operatorname{Tr}(L_{k}) \lambda(k)+\operatorname{Det}(L_{k})=0, \quad k=0,1,2,\ldots, $$
(22)
where \(\operatorname{Tr}(L_{k})=(1+d)k^{2}\) and \(\operatorname{Det}(L_{k})=k^{2}\nu+\nu (r\mu\nu+r\nu)+k^{2}(k^{2}\nu)d\). We are interested in the solutions \(\lambda(k)\) such that \(\operatorname{Re}(\lambda(k))>0\).
With \(0<\nu<(r1)\mu\) and \(c=c_{h}\), we have \(\operatorname{Tr}(L_{0})=0\), \(\operatorname{Det}(L_{0})=\nu(r\mu\nu+r\nu)>0\), and \(\operatorname{Tr}(L_{k})<0\) for \(k=1,2,\ldots \) . This means that, for \(k=0\), the real parts of the eigenvalues of L are zero. We then have to do the center manifold reduction.
First of all, for \(k=1,2,\ldots \) , if \(0<\nu\leq1\), then \(\operatorname {Det}(L_{k})>0\). Furthermore, if \(m^{2}<\nu\leq(m+1)^{2}\) and \(0< d<\bar {d}\), where \(\bar{d}=\min_{1 \leq k\leq m} \frac{k^{2}\nu+\nu(r\mu\nu +r\nu)}{k^{2}(k^{2}\nu)}, m \in\mathbb{N}^{+}\), then we have \(\operatorname {Det}(L_{k})>0\) for \(k=1,2,\ldots \) , and if \(m^{2}<\nu\leq(m+1)^{2}\) and \(d>\bar{d}\), then at least one of \(\operatorname{Det}(L_{1}),\operatorname {Det}(L_{2}),\ldots,\operatorname{Det}(L_{m})\) is negative.
Letting \(\xi=(\nu(\mu+\nu)a_{0}^{1r},\nui\omega_{0})^{T}\) and \(\xi^{*}=\frac {1}{2\omega_{0}} (\frac{\omega_{0}+i\nu}{r\nu^{r+1}}a_{0}^{r^{2}},i )^{T}\), we have \(L\xi=i\omega_{0}\xi\), \(L^{*}\xi^{*}=i\omega_{0}\xi^{*}\), \(\langle\xi ^{*},\bar{\xi}\rangle=0\), and \(\langle\xi^{*},\xi\rangle=1\). Let \(U=z\xi +\bar{z}\bar{\xi}+w\), \(z=\langle\xi^{*},U\rangle\), \(w=(w_{1},w_{2})^{T}\). Then
$$ \textstyle\begin{cases} u=\nu(\mu+\nu)a_{0}^{1r}(z+ \bar{z})+w_{1}, \\ v=(\nui\omega_{0})z+(\nu+i\omega_{0}) \bar{z}+w_{2}. \end{cases} $$
In \((z,w)\) coordinates, system (17) is
$$ \textstyle\begin{cases} \frac{dz}{dt}=i\omega_{0}z+ \langle\xi^{*},\tilde{f} \rangle, \\ \frac{dw}{dt}=Lw+\tilde{H}(z,\bar{z},w), \end{cases} $$
(23)
where \(\tilde{f}=\tilde{F}(z\xi+\bar{z}\bar{\xi}+w,c_{h})\) and \(\tilde {H}(z,\bar{z},w)=\tilde{f}\langle\xi^{*},\tilde{f}\rangle\xi\langle \bar{\xi}^{*},\tilde{f}\rangle\bar{\xi}\). From further calculation we have
$$\begin{aligned} &\tilde{F}(z\xi+\bar{z}\bar{\xi}+w,c_{h})= \frac{1}{2}Q(\xi,\xi)z^{2}+Q(\xi ,\bar{\xi})z\bar{z}+ \frac{1}{2}Q(\bar{\xi},\bar{\xi})\bar{z}^{2}+O \bigl( \vert z \vert ^{3}, \vert z \vert \cdot \vert w \vert , \vert w \vert ^{2} \bigr), \\ &\bigl\langle \xi^{*},\tilde{f} \bigr\rangle =\frac{1}{2} \bigl\langle \xi^{*},Q( \xi,\xi ) \bigr\rangle z^{2}+ \bigl\langle \xi^{*},Q(\xi,\bar{ \xi}) \bigr\rangle z\bar{z}+\frac {1}{2} \bigl\langle \xi^{*},Q(\bar{\xi},\bar{\xi}) \bigr\rangle \bar{z}^{2} \\ &\phantom{\bigl\langle \xi^{*},\tilde{f} \bigr\rangle =}{}+O \bigl( \vert z \vert ^{3}, \vert z \vert \cdot \vert w \vert , \vert w \vert ^{2} \bigr), \\ &\bigl\langle \bar{\xi}^{*},\tilde{f} \bigr\rangle =\frac{1}{2} \bigl\langle \bar{\xi }^{*},Q(\xi,\xi) \bigr\rangle z^{2}+ \bigl\langle \bar{\xi}^{*},Q( \xi, \bar{\xi}) \bigr\rangle z\bar{z}+\frac{1}{2} \bigl\langle \bar{ \xi}^{*},Q(\bar{ \xi},\bar{\xi}) \bigr\rangle \bar{z}^{2} \\ &\phantom{\bigl\langle \bar{\xi}^{*},\tilde{f} \bigr\rangle =}{}+O \bigl( \vert z \vert ^{3}, \vert z \vert \cdot \vert w \vert , \vert w \vert ^{2} \bigr). \end{aligned}$$
So \(\tilde{H}(z,\bar{z},w)=\frac{1}{2}z^{2}\tilde{H}_{20}+z\bar{z}\tilde {H}_{11}+\frac{1}{2}\bar{z}^{2}\tilde{H}_{02}+O(\vert z\vert^{3},\vert z \vert\cdot\vert w \vert,\vert w \vert^{2})\), where
$$\begin{aligned} &\tilde{H}_{20}=Q(\xi,\xi) \bigl\langle \xi^{*},Q(\xi,\xi) \bigr\rangle \xi \bigl\langle \bar{\xi}^{*},Q(\xi,\xi) \bigr\rangle \bar{\xi} \\ &\phantom{\tilde{H}_{20}}= \bigl(Q_{1}(\xi,\xi),Q_{2}(\xi,\xi) \bigr)^{T} \frac{1}{2\omega_{0}} \biggl[\frac{\omega_{0}i\nu }{\nu(\mu+\nu)}a_{0}^{r1}Q_{1}( \xi,\xi)+iQ_{2}(\xi,\xi) \biggr]\xi \\ &\phantom{\tilde{H}_{20}=}{}\frac{1}{2\omega_{0}} \biggl[\frac{\omega_{0}+i\nu}{\nu(\mu+\nu )}a_{0}^{r1}Q_{1}( \xi,\xi)iQ_{2}(\xi,\xi) \biggr]\bar{\xi} \\ &\phantom{\tilde{H}_{20}}=(0,0)^{T}, \\ &\tilde{H}_{11}=Q(\xi,\bar{\xi}) \bigl\langle \xi^{*},Q(\xi,\bar{\xi}) \bigr\rangle \xi \bigl\langle \bar{\xi}^{*},Q(\xi,\bar{\xi}) \bigr\rangle \bar{\xi} \\ &\phantom{\tilde{H}_{11}}= \bigl(Q_{1}(\xi,\bar{\xi}),Q_{2}(\xi,\bar{\xi}) \bigr)^{T}\frac{1}{2\omega_{0}} \biggl[\frac {\omega_{0}i\nu}{\nu(\mu+\nu)}a_{0}^{r1}Q_{1}( \xi,\bar{\xi})+iQ_{2}(\xi,\bar {\xi}) \biggr]\xi \\ &\phantom{\tilde{H}_{11}=}{}\frac{1}{2\omega_{0}} \biggl[\frac{\omega_{0}+i\nu}{\nu(\mu+\nu )}a_{0}^{r1}Q_{1}( \xi,\bar{\xi})iQ_{2}(\xi,\bar{\xi}) \biggr]\bar{\xi} \\ &\phantom{\tilde{H}_{11}}=(0,0)^{T}, \\ &\tilde{H}_{02}=Q(\bar{\xi},\bar{\xi}) \bigl\langle \xi^{*},Q(\bar{\xi}, \bar {\xi}) \bigr\rangle \xi \bigl\langle \bar{\xi}^{*},Q(\bar{\xi},\bar{\xi}) \bigr\rangle \bar {\xi} \\ &\phantom{\tilde{H}_{02}}= \bigl(Q_{1}(\bar{\xi},\bar{\xi}),Q_{2}(\bar{\xi},\bar{ \xi}) \bigr)^{T}\frac {1}{2\omega_{0}} \biggl[\frac{\omega_{0}i\nu}{\nu(\mu+\nu)}a_{0}^{r1}Q_{1}( \bar{\xi },\bar{\xi})+iQ_{2}(\bar{\xi},\bar{\xi}) \biggr]\xi \\ &\phantom{\tilde{H}_{02}=}{}\frac{1}{2\omega_{0}} \biggl[\frac{\omega_{0}+i\nu}{\nu(\mu+\nu )}a_{0}^{r1}Q_{1}( \bar{\xi},\bar{\xi})iQ_{2}(\bar{\xi},\bar{\xi}) \biggr]\bar{\xi } \\ &\phantom{\tilde{H}_{02}}=(0,0)^{T}. \end{aligned}$$
Then we have \(\tilde{H}(z,\bar{z},w)=O(\vert z\vert^{3},\vert z \vert \cdot\vert w \vert,\vert w \vert^{2})\). System (23) has a center manifold of the form \(w=\frac {w_{20}}{2}z^{2}+w_{11}z\bar{z}+\frac{w_{02}}{2}\bar{z}^{2}+O(\vert z\vert^{3})\). With \(Lw+\tilde{H}(z,\bar{z},w)=\frac{dw}{dt}=\frac{\partial w}{\partial z}\frac{dz}{dt}+\frac{\partial w}{\partial\bar{z}}\frac {d\bar{z}}{dt}\), we obtain \(w_{20}=[2i\omega_{0}L]^{1}\tilde {H}_{20}=(0,0)^{T}\), \(w_{11}=L^{1}\tilde{H}_{11}=(0,0)^{T}\), \(w_{02}=[2i\omega_{0}L]^{1}\tilde{H}_{02}=(0,0)^{T}\). Therefore, on the center manifold, the reaction–diffusion system is
$$ \dot{z}=i\omega_{0}z+ \bigl\langle \xi^{*},\tilde{f} \bigr\rangle =i\omega_{0}z+\sum_{2\leq i+j \leq3} \frac{g_{ij}}{i!j!} z^{i}\bar{z}^{j}+O \bigl( \bigl\vert z^{4} \bigr\vert \bigr), $$
(24)
where
$$\begin{aligned} &g_{20}= \bigl\langle \xi^{*},Q(\xi,\xi) \bigr\rangle ,\qquad g_{11}= \bigl\langle \xi^{*},Q(\xi,\bar {\xi}) \bigr\rangle ,\\ & g_{02}= \bigl\langle \xi^{*},Q(\bar{\xi},\bar{\xi}) \bigr\rangle ,\qquad g_{21}= \bigl\langle \xi^{*},C( \xi,\xi,\bar{\xi}) \bigr\rangle . \end{aligned}$$
We rewrite (17) into the Poincaré normal form in a neighborhood of \(c_{h}\):
$$ \frac{\mathrm{d}z}{\mathrm{d}t}= \bigl(\alpha(c)+i\omega(c) \bigr)z+z\sum _{j=1}^{M}\delta_{j}(c) (z \bar{z})^{j}, $$
(25)
where the variable \(z \in\mathbb{C}\), \(M\geq1\), and the coefficients \(\delta_{j}(c)\) are complex. Direct computation shows that
$$\delta_{1}(c)=\frac{g_{20}g_{11}[3\alpha(c)+i\omega(c)]}{2[\alpha ^{2}(c)+\omega^{2}(c)]}+\frac{ \vert g_{11} \vert ^{2}}{\alpha(c)+i\omega (c)}+ \frac{ \vert g_{02} \vert ^{2}}{2[\alpha(c)+3i\omega(c)]}+\frac{g_{21}}{2}, $$
and thus \(\operatorname{Re}(\delta_{1}(c_{h}))=\operatorname{Re}[\frac {g_{20}g_{11}}{2\omega_{0}}i+\frac{g_{21}}{2}]\). Since \(\alpha(c_{h})=0 \), \(\omega(c_{h})=\omega_{0}>0\), and \(0<\nu<(r1)\mu\), we then have
$$\operatorname{Re} \bigl[\delta_{1}(c_{h}) \bigr]= \frac{[(r1)\mu\nu]\nu^{3}(\mu+\nu )^{2}[(1+r^{2})\mu+(r1)\nu]}{4\omega_{0}^{2}a_{0}^{2r}}< 0. $$
Therefore, the supercritical Hopf bifurcation occurs at \(c=c_{h}\). The following theorem is a summary of the preceding analysis.
Theorem 3.2
Let (H3) hold, and let
$$\bar{d}=\min_{1\leq k\leq m}\frac{k^{2}\nu+\nu(r\mu\nu+r\nu )}{k^{2}(k^{2}\nu)},\quad m\in\mathbb {N}^{+}. $$
Then the spatially homogenous periodic solution for (3) is stable if either (H7) or (H8) holds and is unstable if (H9) holds, where

(H7)
\(0<\nu\leq1 \),

(H8)
\(m^{2} <\nu\leq(m+1)^{2}\), \(0< d< \bar{d}\), and

(H9)
\(m^{2} <\nu\leq(m+1)^{2}\), \(d>\bar{d}\).
Remark 3.2
Theorem 3.1 and 3.2 show that system (3) subjected to (4) undergoes Turing instability under either of assumptions (H1) and (H6), or (H2) and (H6), or (H3) and (H9), which is responsible for the patterns of stripe and spot types. We will see these patterns numerically in the next section.