Hopf bifurcation

In the previous section, we considered a system which becomes unstable as a real eigenvalue changes its sign. The other possibility for an instability is when a pair of complex conjugate eigenvalues crosses the imaginary axis. Again, we shall consider a system of the form  

$$\dot u=A(\lambda)u+f(u,\lambda),$$ (114)

where the nonlinearity f has the form  

$$f(u,\lambda)=N_2(u,u;\lambda)+N_3(u,u,u;\lambda)+O(\Vert u\Vert^4).$$ (115)

We now assume that A(0) has a pair of simple purely imaginary eigenvalues $\pm i\omega_0$, while the rest of the spectrum is in the left half plane.

In an analogous fashion as before, we denote by $\sigma(\lambda)$ the eigenvalue which passes through $i\omega_0$ as $\lambda$ passes through 0, and we denote by $a(\lambda)$ the eigenvector and $b(\lambda)$ is the adjoint eigenvector:  

$$A(\lambda)a(\lambda)=\sigma(\lambda)a(\lambda),\quad (b(\lambda),(A(\lambda) -\sigma(\lambda))u)=0.$$ (116)

We normalize the eigenvectors such that $(b(\lambda), a(\lambda))=1$. The analogue of assumption (3.25) is  

$${d\over d\lambda}({\rm Re}\,\sigma(\lambda))={\rm Re}(b(0),A'(0)a(0))\gt.$$ (117)

We introduce the projection operator $P(\lambda)$ as follows:  

$$P(\lambda)u=u-(b(\lambda),u)a(\lambda)-(\overline{b(\lambda)},u)\overline{a( \lambda)}.$$ (118)

We note for the following that  

$$(b(\lambda),\overline{a(\lambda)})=(b(\lambda),(A(\lambda)-\... ...(\lambda)}\over \overline{\sigma(\lambda)}-\sigma(\lambda)})=0,$$ (119)

and consequently

$$(b(\lambda),P(\lambda)u)=0.$$ (120)

The analogue of the decomposition (3.30) is  

$$u=za(\lambda)+\bar z\overline{a(\lambda)}+y,$$ (121)

where $y=P(\lambda)u$ and $(b(\lambda),u)=z$. The scalar z is complex. The analogue of (3.31) is

   $$\begin{eqnarray} \dot z&=&\sigma(\lambda)z+g(z,y,\lambda),\nonumber\ \dot y&=&B(\lambda)y+h(z,y,\lambda).\end{eqnarray}$$

Here we have

   $$\begin{eqnarray} g(z,y,\lambda)&=&(b(\lambda),f(za(\lambda)+\bar z\overline{a(\l... ...ambda)}+y, \lambda),\nonumber\ B(\lambda)&=&P(\lambda)A(\lambda).\end{eqnarray}$$

As before, we use the center manifold reduction. We need the center manifold up to quadratic order. The result is  

$$y=\phi(z,\lambda)=\psi(\lambda)z^2+\overline{\psi(\lambda)}\bar z^2+ 2\chi(\lambda)z\bar z+O(\vert z\vert^3).$$ (122)

Here

   $$\begin{eqnarray} \psi(\lambda)&=&-(B(\lambda)-2\sigma(\lambda))^{-1}P(\lambda)N_... ...)})^{-1} P(\lambda)N_2(a(\lambda), \overline{a(\lambda)};\lambda).\end{eqnarray}$$

We can use (3.66) in the first equation of (3.64), and obtain the following equation for z, up to cubic order:

   $$\begin{eqnarray} \dot z&=&\sigma(\lambda)z+\alpha_1(\lambda)z^2+\alpha_2(\lambda... ...)\vert z\vert^2\bar z +\beta_4(\lambda)\bar z^3+O(\vert z\vert^4).\end{eqnarray}$$

The coefficients in this equation are

   $$\begin{eqnarray} \alpha_1(\lambda)&=&(b(\lambda),N_2(a(\lambda),a(\lambda);\lamb... ...ine{a(\lambda)},\overline{\psi(\lambda)};\lambda)), \nonumber\ &&\end{eqnarray}$$

The definition of N₃ for unequal arguments is analogous to that for N₂ in the previous section:

   $$\begin{eqnarray} N_3(u,v,w;\lambda)&=& {1\over 24}(N_3(u+v+w,u+v+w,u+v+w;\lambda... ...a)\nonumber\ &+&N_3(-u-v+w,-u-v+w,-u-v+w;\lambda)).\nonumber\ &&\end{eqnarray}$$

We can simplify (3.68) further by a transformation of variables which removes ``nonresonant" terms. The simplified equation is known as the Birkhoff normal form. We need to explain what is meant by nonresonant. We are interested in solutions for which $\lambda$ and |z| are small. The leading contribution to (3.68) is therefore the one arising from linearizing with respect to z and setting $\lambda=0$. In that limit, the equation becomes $\dot z=i\omega_0 z$, and the solution is proportional to $\exp(i\omega_0t)$. We now divide the nonlinear contributions into ``resonant" ones which formally have the same time dependence (for instance |z|²z) and nonresonant ones which have a different dependence (e.g. if z is proportional to $\exp(i\omega_0t)$, then z² is proportional to $\exp(2i\omega_0t)$. We see that the only resonant term in (3.68) is the one proportional to |z|²z, all other terms will be transformed away. We show how to transform away the quadratic terms. For this purpose, we set

$$w=z+\gamma_1(\lambda)z^2+\gamma_2(\lambda)z\bar z+\gamma_3(\lambda)\bar z^2 +O(\vert z\vert^3),$$ (123)

which for small |z| can be inverted to yield

$$z=w-\gamma_1(\lambda)w^2-\gamma_2(\lambda)w\bar w-\gamma_3(\lambda)\bar w^2 +O(\vert w\vert^3),$$ (124)

We conclude that

   $$\begin{eqnarray} \dot w&=&\dot z+2\gamma_1(\lambda)z\dot z+\gamma_2(\lambda)(\do... ...a(\lambda)-\sigma(\lambda)) +\alpha_3(\lambda))+O(\vert w\vert^3).\end{eqnarray}$$

The quadratic terms disappear if we choose the $\gamma_i(\lambda)$ such that

   $$\begin{eqnarray} \gamma_1(\lambda)\sigma(\lambda)+\alpha_1(\lambda)&=&0,\nonumbe... ...bda)(2\bar\sigma(\lambda)-\sigma(\lambda))+\alpha_3(\lambda) &=&0.\end{eqnarray}$$

By including cubic terms in the transformation, we can also transform away the nonresonant cubic terms. For the resonant term, this does not work because the analogue of (3.74) would involve the term

$$\gamma_i(\lambda)(\sigma(\lambda)+\bar\sigma(\lambda)),$$ (125)

and $\sigma(0)+\bar\sigma(0)=0$, so we cannot solve for $\gamma_i(\lambda)$.We shall not carry out the tedious details of the transformation; this is best left to symbolic manipulation packages.

The Birkhoff normal form of (3.68) takes the form  

$$\dot w=\sigma(\lambda)w+\beta(\lambda)\vert w\vert^2w+O(\vert w\vert^5),$$ (126)

we note that there are no resonant fourth order terms, hence the remainder O(|w|⁵). If we truncate the equation at the cubic order, then we can easily find an explicit periodic solution  

$$w(t)=R\exp(i\omega t),$$ (127)

where

   $$\begin{eqnarray} {\rm Re}(\sigma(\lambda))+R^2\,{\rm Re}(\beta(\lambda))&=&0,\no... ... {\rm Im}(\sigma(\lambda))+R^2\,{\rm Im}(\beta(\lambda))&=&\omega.\end{eqnarray}$$

If higher order corrections are taken into account, then this explicit solution arises as the first term in an asymptotic approximation, i.e. there is a bifurcating family of periodic solutions, parametrized by their amplitude R such that

   $$\begin{eqnarray} \lambda&=&-R^2{{\rm Re}(\beta(0))\over{\rm Re}(\sigma'(0))}+O(R... ...)){\rm Re}(\beta(0)) \over {\rm Re}(\sigma'(0))})+O(R^4).\nonumber\end{eqnarray}$$

Since we assumed that ${\rm Re}(\sigma'(0))\gt$, we have a supercritical bifurcation if ${\rm Re}(\beta(0))<0$ and a subcritical bifurcation if ${\rm Re}(\beta(0))\gt$.

To analyze the stability of the bifurcating periodic solution, it suffices to study the cubic truncation. In the equation  

$$\dot w=\sigma(\lambda)w+\beta(\lambda)\vert w\vert^2w,$$ (128)

we set $w=(R+v)\exp(i\omega t)$, where R and $\omega$ are as above, and we linearize with respect to v. The resulting linearized equation is

$$\dot v+i\omega v=\sigma(\lambda)v+2\beta(\lambda)R^2v+\beta(\lambda)R^2\bar v.$$ (129)

Keeping in mind that $\sigma+R^2\beta=i\omega$, we can simplify this to  

$$\dot v=\beta(\lambda)R^2(v+\bar v).$$ (130)

This system has eigenvalues $2{\rm Re}(\beta(\lambda))R^2$and . The zero eigenvalue results from the neutral direction which allows for a phase shift in the periodic solution. The stability is determined by the other eigenvalue. We find that the branch of periodic solutions is stable if it is supercritical and unstable if it is subcritical.