Let {X}_{1}, X be two Banach spaces, {X}_{1}\subset X be a dense and compact inclusion. Consider the following nonlinear evolution equations
\{\begin{array}{c}\frac{du}{dt}=Lu+Gu,\hfill \\ u(0)=\phi ,\hfill \end{array}
(4.1)
where L:{X}_{1}\to X is bounded linear operator, G:{X}_{1}\to X is {C}^{r} (r\ge 1) mapping.
Definition 4.1 [14]
Let \mathrm{\Omega}\subset X be a bounded open set, we say that (4.1) is Lyapunov stable in Ω, if the solution of (4.1) u(t,\phi )\in \mathrm{\Omega}, \mathrm{\forall}t\ge 0, for any initial point \phi \in \mathrm{\Omega}.
Definition 4.2 [14]
We say (4.1) is asymptotically Lyapunov stable in Ω, or {u}_{0} is an asymptotically stable equilibrium point of (4.1) in Ω, if there exists a steady solution {u}_{0}\in \mathrm{\Omega}, satisfying
\underset{t\to \mathrm{\infty}}{lim}{\parallel u(t,\phi ){u}_{0}\parallel}_{X}=0\phantom{\rule{1em}{0ex}}\text{for any}\phi \in \mathrm{\Omega}.
Lemma 4.3 [14]
Let L:{X}_{1}\to x be a sectorial operator, G:{X}_{\alpha}\to X be a {C}^{r} (r\ge 1) mapping for certain 0\le \alpha <1, v be a steady solution of (4.1). If the spectral \beta (\lambda ) of L+DG(v) satisfy Re\beta (\lambda )<\epsilon, for certain \epsilon >0, then v is a locally asymptotically stable equilibrium point of (4.1), and it decays exponentially. Namely, there exist M>0, \sigma >0, \delta >0, for the solution of (4.1) u(t,\phi ), {\parallel u(t,\phi )v\parallel}_{X}\le M{e}^{\sigma t} (\mathrm{\forall}t\ge 0) hold true, as \parallel \phi v\parallel <\delta.
Let
\delta =\frac{3{r}_{2}^{2}2{r}_{2}1}{{r}_{2}^{2}({r}_{2}1)({r}_{2}+3)}>0,\phantom{\rule{1em}{0ex}}\text{as}{r}_{2}{r}_{1}=1,
and as above
\mu =\frac{{\mathrm{\Omega}}_{2}}{{\mathrm{\Omega}}_{1}},\phantom{\rule{2em}{0ex}}\eta =\frac{{r}_{1}}{{r}_{2}}.
Theorem 4.4 If \mu \le \delta, or {\eta}^{2}\le \mu, then ({u}_{z},{u}_{r},{u}_{\theta})=0 is a locally asymptotically stable equilibrium point of (3.4), and it decays exponentially.
Theorem 4.5 If \delta <\mu <{\eta}^{2} (\mu \ne 0), we have the following conclusions for (3.4):

(1)
Equation (3.4) happens with a continuous transition at (u,\lambda )=(0,\sqrt{{T}_{c}}), namely there is an attractor bifurcation, and it bifurcates exactly into two singular points {v}_{i} (i=1,2), which attract two open subsets of U separately. U is the neighborhood of u=0.

(2)
The two singular points {v}_{i} (i=1,2) can be written by:
{v}_{1,2}(\lambda )=\pm {{\beta}_{{n}_{1}1}/\sigma }^{\frac{1}{2}}{\psi}_{{n}_{1}1}+o\left({{\beta}_{{n}_{1}1}/\sigma }^{\frac{1}{2}}\right),
where \sigma =\frac{1}{4{\beta}_{2{n}_{1}2}}{a}_{1}^{4}{b}_{1}^{2}, {\beta}_{{n}_{1}1} is the first eigenvalue.
Remark 4.6 From the point of view of mathematics, Theorem 4.4 explains that there exists an open set \mathrm{\Omega}\subset {H}_{1}, which guarantees that if the initial point {u}_{0}\in \mathrm{\Omega}, then the solution of (3.4) satisfies {\parallel u(t,{u}_{0})\parallel}_{H}\to 0 (t\to \mathrm{\infty}).
Remark 4.7 From the point of view of physics, Theorem 4.4 explains that in the widegap case, the Couette flow of the Taylor problem is metastable. Namely, if the initial perturbation is in a certain range, the disturbed fluid will become the Couette flow in a short time. But if the initial perturbation is beyond that range above, the disturbed fluid will become another steady flow. The explanation is the same as that for Theorem 4.5.
Remark 4.8 Theorems 4.44.5 give the entire results of stability of the Taylor problem in the widegap case; see Figure 1. {\mathrm{\Sigma}}_{1} represents the unstable area, {\mathrm{\Sigma}}_{2} represents the stable area, {T}_{0}\ne 0 and
{T}_{0}=\underset{(n,l)}{min}\frac{{\alpha}^{3}{\nu}^{2}{(1{\eta}^{2})}^{2}}{4{a}^{2}{\eta}^{2}(\frac{1}{{r}_{0}^{2}}{\eta}^{2})}.
Remark 4.9 [15]
In the narrowgap case, let {r}_{1}=3.55\phantom{\rule{0.3em}{0ex}}\text{cm} and {r}_{2}=4.035\phantom{\rule{0.3em}{0ex}}\text{cm}, the results can be seen in Figure 2. {\mathrm{\Sigma}}_{1} represents the unstable area, {\mathrm{\Sigma}}_{2} represents the stable area.
Remark 4.10 By Figure 1 and Figure 2, we can conclude that the results of the stability of the Taylor problem in the widegap case, obtained by using the averaging method, are essentially the same as those of the Taylor problem in the narrowgap case.
Proof of Theorem 4.4 Obviously, A and B are linear operators. According to [16], A is a homeomorphism, then it is a sectorial operator. According to the Sobolev compact embedding theorem [17] and the Leray projection, P is bounded, B is compact. So,
{L}_{\lambda}=A+\lambda B:{H}_{1}\to H
is a sectorial operator, and also a linear completely continuous field.
Now,
\begin{array}{rcl}{\int}_{\mathrm{\Omega}}(u\cdot \mathrm{\nabla})u\cdot v+{u}^{2}\cdot v\phantom{\rule{0.2em}{0ex}}dx& \le & {\parallel u\parallel}_{{C}^{0}}{\parallel \mathrm{\nabla}u\parallel}_{{L}^{2}}{\parallel v\parallel}_{{L}^{2}}+{\parallel u\parallel}_{{C}^{0}}{\parallel u\parallel}_{{L}^{2}}{\parallel v\parallel}_{{L}^{2}}\\ =& {\parallel u\parallel}_{{C}^{0}}{\parallel u\parallel}_{{W}^{1,2}}{\parallel v\parallel}_{{L}^{2}},\end{array}
we have
{\parallel Gu\parallel}_{H}\le {\parallel u\parallel}_{{C}^{0}}{\parallel u\parallel}_{{W}^{1,2}}.
According to [18], Theorem 7.2.2, the fraction spaces of {L}_{\lambda},{H}_{\alpha}=D({L}_{\lambda}^{\alpha}), satisfy
{\parallel u\parallel}_{{C}^{0}}+{\parallel u\parallel}_{{W}^{1,2}}\le C{\parallel u\parallel}_{{H}_{\alpha}},
where \alpha >\frac{3}{4}. Then
G:{H}_{\alpha}\to H\phantom{\rule{1em}{0ex}}(\frac{3}{4}<\alpha <1)\text{is a bounded analytic mapping}.
Consider the eigenvalue problem:
{L}_{\lambda}\psi =\beta (\lambda )\psi ,\phantom{\rule{1em}{0ex}}\psi =({u}_{z},{u}_{r},{u}_{\theta})\in {H}_{1}.
By (3.3), the abstract form can be referred to the following eigenvalue equations in {H}_{1}:
\mathrm{\Delta}{u}_{z}\frac{\partial p}{\partial z}=\beta (\lambda ){u}_{z},
(4.2)
\mathrm{\Delta}{u}_{r}\frac{\partial p}{\partial r}+\lambda (\frac{1}{{r}_{0}^{2}}\kappa ){u}_{\theta}=\beta (\lambda ){u}_{r},
(4.3)
\mathrm{\Delta}{u}_{\theta}+\lambda \kappa {u}_{r}=\beta (\lambda ){u}_{\theta},
(4.4)
\frac{\partial {u}_{r}}{\partial r}+\frac{\partial {u}_{z}}{\partial z}=0,
(4.5)
\frac{\partial {u}_{z}}{\partial r}=0,\phantom{\rule{2em}{0ex}}{u}_{r}={u}_{\theta}=0,\phantom{\rule{1em}{0ex}}\text{at}r=1,{r}_{2},
(4.6)
{u}_{z}=0,\phantom{\rule{2em}{0ex}}\frac{\partial {u}_{r}}{\partial z}=\frac{\partial {u}_{\theta}}{\partial z}=0,\phantom{\rule{1em}{0ex}}\text{at}z=0,L.
(4.7)
According to [5], we take the separation of variables as follows:
\begin{array}{r}{u}_{z}=\frac{dh(z)}{dt}\frac{dR(r)}{dr},\\ {u}_{r}={a}^{2}h(z)R(r),\\ {u}_{\theta}=h(z)\phi (r).\end{array}
(4.8)
We utilize (4.8) in (4.5) and (4.7) to obtain:
\{\begin{array}{c}{h}^{\u2033}{R}^{\prime}+{a}^{2}h{R}^{\prime}=0,\hfill \\ {h}^{\prime}(0)={h}^{\prime}(L)=0\hfill \end{array}\phantom{\rule{1em}{0ex}}\Rightarrow \phantom{\rule{1em}{0ex}}\{\begin{array}{c}h=cosaz,\hfill \\ a=\frac{n\pi}{L},\phantom{\rule{1em}{0ex}}n=1,2,\dots .\hfill \end{array}
(4.9)
We utilize (4.8), (4.9) in (4.4) to obtain:
{h}^{\u2033}\phi +h{\phi}^{\u2033}+\lambda \kappa {a}^{2}hR=\beta (\lambda )h\phi
that equals to
(\frac{{d}^{2}}{d{r}^{2}}{a}^{2})\phi =\lambda \kappa {a}^{2}R+\beta (\lambda )\phi .
(4.10)
We utilize (4.8), (4.9) in (4.2), then differentiate with respect to r,
{h}^{\u2034}{R}^{\u2033}+{h}^{\prime}{R}^{\prime \prime \prime \prime}\frac{{\partial}^{2}p}{\partial r\partial z}=\beta (\lambda ){h}^{\prime}{R}^{\u2033},
and utilize (4.8), (4.9) in (4.3), then differentiate with respect to z,
{a}^{2}{h}^{\u2034}R+{a}^{2}{h}^{\prime}{R}^{\u2033}+\lambda (\frac{1}{{r}_{0}^{2}}\kappa ){h}^{\prime}\phi \frac{{\partial}^{2}p}{\partial z\partial r}=\beta (\lambda ){a}^{2}{h}^{\prime}R,
the two equations above subtract to obtain:
{(\frac{{d}^{2}}{d{r}^{2}}{a}^{2})}^{2}R=\lambda (\frac{1}{{r}_{0}^{2}}\kappa )\phi +\beta (\lambda )(\frac{{d}^{2}}{d{r}^{2}}{a}^{2})R.
(4.11)
Utilizing (4.8) in (4.6), we have
\phi (1)=\phi ({r}_{2})=R(1)=R({r}_{2})={R}^{\u2033}(1)={R}^{\u2033}({r}_{2})=0,
then let
R=sinl\pi \left(\frac{r1}{{r}_{2}1}\right),\phantom{\rule{2em}{0ex}}\phi ={C}_{nl}sinl\pi \left(\frac{r1}{{r}_{2}1}\right),\phantom{\rule{1em}{0ex}}l=1,2,\dots .
Combine (4.10), (4.11) to obtain:
\{\begin{array}{c}\alpha {C}_{nl}={a}^{2}\lambda \kappa {C}_{nl}\beta (\lambda ),\hfill \\ {\alpha}^{2}={C}_{nl}(\frac{1}{{r}_{0}^{2}}\kappa )\lambda \alpha \beta (\lambda ),\hfill \end{array}\phantom{\rule{1em}{0ex}}\alpha ={a}^{2}+{\left(\frac{l\pi}{{r}_{2}1}\right)}^{2},
(4.12)
therefore, we get
\{\begin{array}{c}{C}_{nl}=\frac{2\alpha \lambda \kappa {a}^{2}}{\pm \sqrt{\mathrm{\Delta}}},\hfill \\ {\beta}_{nl}=\alpha \pm \frac{\sqrt{\mathrm{\Delta}}}{2\alpha},\hfill \end{array}
(4.13)
where
\mathrm{\Delta}=4\alpha {a}^{2}{\lambda}^{2}\kappa (\frac{1}{{r}_{0}^{2}}\kappa ).
The corresponding eigenvectors of (4.2)(4.7) are as follows:
{\psi}_{nl}=\{\begin{array}{c}\frac{nl{\pi}^{2}}{L({r}_{2}1)}sin\frac{n\pi z}{L}cos\frac{l\pi (r1)}{({r}_{2}1)},\hfill \\ {(\frac{n\pi}{L})}^{2}cos\frac{n\pi z}{L}sin\frac{l\pi (r1)}{({r}_{2}1)},\hfill \\ {C}_{nl}cos\frac{n\pi z}{L}sin\frac{l\pi (r1)}{({r}_{2}1)}.\hfill \end{array}
(4.14)
Now, as (n,l)\ne (0,0),
\begin{array}{r}{L}_{\lambda}{\psi}_{nl}={\beta}_{nl}(\lambda ){\psi}_{nl},\\ {E}_{1}=\{{\psi}_{nl}{\psi}_{nl}\text{satisfies}(\text{4.14}),(n,l)\in N\times N\},\\ {B}_{1}=\{{\beta}_{nl}(\lambda ){\beta}_{nl}(\lambda )\text{satisfies}(\text{4.13}),(n,l)\in N\times N\},\end{array}
as n=0, from (4.2)(4.8),
{L}_{\lambda}{\psi}_{0l}={\beta}_{0l}(\lambda ){\psi}_{0l},
where
\begin{array}{r}{\psi}_{0l}\in {E}_{2}=\left\{{\psi}_{0l}\right{\psi}_{0l}={(0,0,sin\frac{l\pi (r1)}{({r}_{2}1)})}^{T},l=1,2,\dots \},\\ {\beta}_{0l}\in {B}_{2}=\left\{{\beta}_{0l}\right{\beta}_{0l}={\left(\frac{l\pi}{{r}_{2}1}\right)}^{2},l=1,2,\dots \},\end{array}
as l=0, {\psi}_{n0}=0.
Now, according to the Fourier expansion, {B}_{1}\cup {B}_{2} are all the eigenvalues of (4.2)(4.7), the corresponding eigenvectors {E}_{1}\cup {E}_{2} form a complete basis in {H}_{1}.
In view of the condition \mu <\delta, by simple calculation, we know that \kappa >0, \kappa {r}_{0}^{2}1>0, then
thus,
Re{\beta}_{nl}<\epsilon \phantom{\rule{1em}{0ex}}(0<\epsilon <\alpha ),
and
{\beta}_{0l}={\left(\frac{l\pi}{{r}_{2}1}\right)}^{2}<0.
Now, by Lemma 4.3, the proof of Theorem 4.4 is completed. □
Proof of Theorem 4.5 Utilizing the method in Theorem 4.4, we have the following results for the eigenvalue problem of {L}_{\lambda}^{\ast}, here {L}_{\lambda}^{\ast} is adjoint operator of {L}_{\lambda}.
As (n,l)\ne (0,0),
\{\begin{array}{c}{C}_{nl}^{\ast}=\frac{2\alpha (\frac{1}{{r}_{0}^{2}}\kappa ){\alpha}^{2}\lambda}{\pm \sqrt{\mathrm{\Delta}}},\hfill \\ {\beta}_{nl}(\lambda )=\alpha \pm \frac{\sqrt{\mathrm{\Delta}}}{2\alpha},\hfill \end{array}
(4.15)
where
\mathrm{\Delta}=4\alpha {a}^{2}{\lambda}^{2}\kappa (\frac{1}{{r}_{0}^{2}}\kappa ),
the corresponding eigenvectors are as follows:
{\psi}_{nl}^{\ast}=\{\begin{array}{c}\frac{nl{\pi}^{2}}{L({r}_{2}1)}sin\frac{n\pi z}{L}cos\frac{l\pi (r1)}{({r}_{2}1)},\hfill \\ {(\frac{n\pi}{L})}^{2}cos\frac{n\pi z}{L}sin\frac{l\pi (r1)}{({r}_{2}1)},\hfill \\ {C}_{nl}^{\ast}cos\frac{n\pi z}{L}sin\frac{l\pi (r1)}{({r}_{2}1)},\hfill \end{array}
(4.16)
satisfying
\begin{array}{c}{E}_{1}^{\ast}=\{{\psi}_{nl}^{\ast}{\psi}_{nl}^{\ast}\text{satisfies}(\text{4.16}),(n,l)\in N\times N\},\hfill \\ {B}_{1}=\{{\beta}_{nl}(\lambda ){\beta}_{nl}(\lambda )\text{satisfies}(\text{4.15}),(n,l)\in N\times N\}.\hfill \end{array}
As n=0,
{L}_{\lambda}^{\ast}{\psi}_{0l}^{\ast}={\beta}_{0l}(\lambda ){\psi}_{0l}^{\ast},
where
\begin{array}{c}{\psi}_{0l}^{\ast}\in {E}_{2}^{\ast}=\left\{{\psi}_{0l}^{\ast}\right{\psi}_{0l}^{\ast}={(0,0,sin\frac{l\pi (r1)}{({r}_{2}1)})}^{T},l=1,2,\dots \},\hfill \\ {\beta}_{0l}\in {B}_{2}=\left\{{\beta}_{0l}\right{\beta}_{0l}={\left(\frac{l\pi}{{r}_{2}1}\right)}^{2},l=1,2,\dots \}.\hfill \end{array}
As l=0, {\psi}_{n0}^{\ast}=0.
Consider equation {\beta}_{nl}(\lambda )=0, by (4.15), we have
\begin{array}{c}\alpha {(\alpha +\frac{1}{{r}_{0}^{2}})}^{2}{\lambda}^{2}{a}^{2}\kappa \left(\frac{1}{{r}_{0}^{2}\kappa}\right)=0,\hfill \\ {\lambda}^{2}=\frac{{\alpha}^{3}}{{a}^{2}\kappa (\frac{1}{{r}_{0}^{2}}\kappa )}.\hfill \end{array}
Now, we have the critical Taylor number
{T}_{c}=\underset{(n,l)\in {N}^{2}}{min}\frac{{\alpha}^{3}}{{a}^{2}\kappa (\frac{1}{{r}_{0}^{2}}\kappa )},
(4.17)
through sample calculation, as (n,l)=(\frac{L}{\sqrt{2}({r}_{2}1)},1), \frac{{\alpha}^{3}}{{a}^{2}\kappa (\frac{1}{{r}_{0}^{2}}\kappa )} reaches its minimal value. We assume that n+\frac{1}{2}\ne \frac{L}{\sqrt{2}({r}_{2}1)} for \mathrm{\forall}n>0, then there exists only one couple ({n}_{1},l), making \frac{{\alpha}^{3}}{{a}^{2}\kappa (\frac{1}{{r}_{0}^{2}}\kappa )} reach the minimal value.
Therefore, we have
\begin{array}{r}{\beta}_{{n}_{1}l}(\lambda )=\{\begin{array}{cc}<0,\hfill & \text{if}\lambda {T}_{c},\hfill \\ =0,\hfill & \text{if}\lambda ={T}_{c},\hfill \\ 0,\hfill & \text{if}\lambda {T}_{c},\hfill \end{array}\\ {\beta}_{nl}({T}_{c})0,\phantom{\rule{1em}{0ex}}\mathrm{\forall}(n,l)\ne ({n}_{1},1).\end{array}
(4.18)
The reduced form of (3.4) into its center manifold in H is:
\frac{dx}{dt}={\beta}_{{n}_{1}1}x+\frac{1}{{\u3008{\psi}_{{n}_{1}1},{\psi}_{{n}_{1}1}^{\ast}\u3009}_{H}}{\u3008G(\psi ,\psi ),{\psi}_{{n}_{1}1}^{\ast}\u3009}_{H},
(4.19)
where \psi \in {H}_{1} can be written by
\psi =x{\psi}_{{n}_{1}1}+\mathrm{\Phi},
(4.20)
Φ is a center manifold function. G(u,v) is a bilinear operator, satisfying
G(u,v)=P((\tilde{u}\cdot \mathrm{\nabla}){v}_{z},(\tilde{u}\cdot \mathrm{\nabla}){v}_{r}\frac{{u}_{\theta}{v}_{\theta}}{{r}_{0}},(\tilde{u}\cdot \mathrm{\nabla}){v}_{\theta}+\frac{{u}_{\theta}{v}_{r}}{{r}_{0}}).
(4.21)
Through direct calculation, we have
{\u3008G({\psi}_{{n}_{1}1},{\psi}_{{n}_{1}1}),{\psi}_{J}^{\ast}\u3009}_{H}=\{\begin{array}{cc}0,\hfill & {\psi}_{J}^{\ast}\ne {\psi}_{2{n}_{1}2}^{\ast},\hfill \\ {a}_{1}^{2}{b}_{1}{c}_{11}{c}_{11}^{\ast}L({r}_{2}1),\hfill & {\psi}_{J}^{\ast}={\psi}_{2{n}_{1}2}^{\ast},\hfill \end{array}
(4.22)
where,
{a}_{1}=\frac{{n}_{1}\pi}{L},\phantom{\rule{2em}{0ex}}{b}_{1}=\frac{\pi}{{r}_{2}1},\phantom{\rule{2em}{0ex}}{c}_{11}={C}_{{n}_{1}1},\phantom{\rule{2em}{0ex}}{c}_{11}^{\ast}={C}_{{n}_{1}1}^{\ast}.
Here, we use the mark as follows:
o(2)=o\left({x}^{2}\right)+O\left({\beta}_{{n}_{1}l}(\lambda ){x}^{2}\right),\phantom{\rule{2em}{0ex}}o(3)=o\left({x}^{3}\right)+O\left({\beta}_{{n}_{1}l}(\lambda ){x}^{3}\right).
Then the center manifold function can be written by
\mathrm{\Phi}=\frac{{x}^{2}{\u3008G({\psi}_{{n}_{1}1},{\psi}_{{n}_{1}1}),{\psi}_{2{n}_{1}2}^{\ast}\u3009}_{H}}{{\beta}_{2{n}_{1}2}{\u3008{\psi}_{2{n}_{1}2},{\psi}_{2{n}_{1}2}^{\ast}\u3009}_{H}}{\psi}_{2{n}_{1}2}+o(2).
(4.23)
Direct computation yields that
\begin{array}{r}{\u3008G({\psi}_{{n}_{1}1},{\psi}_{2{n}_{1}2}),{\psi}_{{n}_{1}1}^{\ast}\u3009}_{H}=2{a}_{1}^{2}{b}_{1}{c}_{11}{c}_{11}^{\ast}L({r}_{2}1),\\ {\u3008G({\psi}_{2{n}_{1}2},{\psi}_{{n}_{1}1}),{\psi}_{{n}_{1}1}^{\ast}\u3009}_{H}={a}_{1}^{2}{b}_{1}{c}_{11}{c}_{11}^{\ast}L({r}_{2}1),\\ {\u3008{\psi}_{{n}_{1}1},{\psi}_{{n}_{1}1}^{\ast}\u3009}_{H}=\frac{1}{2}{c}_{11}{c}_{11}^{\ast}L({r}_{2}1),\phantom{\rule{2em}{0ex}}{\u3008{\psi}_{2{n}_{1}2},{\psi}_{2{n}_{1}2}^{\ast}\u3009}_{H}=8{c}_{11}{c}_{11}^{\ast}L({r}_{2}1).\end{array}
(4.24)
Finally, we utilize (4.20), (4.22)(4.24) in (4.19) to obtain
\frac{dx}{dt}={\beta}_{{n}_{1}1}x\sigma {x}^{3}+o(3),
(4.25)
where
\sigma =\frac{1}{4{\beta}_{2{n}_{1}2}{a}_{1}^{4}{b}_{1}^{2}}>0.
Now, according to Theorem 6.9 in [14], the proof of Theorem 4.5 is completed. □