2

Consider the equation $$ u_t=u_{xx}+\cos u-1+\mu~~\qquad (1) $$

Ignoring the term $u_{xx}$, i.e. considering the quation $$ u_t=\cos u-1+\mu\quad (2) $$

for $0<\mu<2$, we have two equilibria $u_1$ and $u_2$ on $[0,2\pi)$. As far as I see, one is stable and the other is unstable.

I have mainly three questions:

(0) Are the two equilibria I mentioned for the case $0<\mu<2$ also all equilibria for equation (1) on $[0,2\pi)$, i.e. when adding the term $u_{xx}$ again?

(1) What happens for $\mu=0$ and $\mu=2$? As far as I see, we then have one equilibrium on$[0,2\pi)$ and this is a saddle in both cases?

(2) For $\mu\notin [0,2]$ we have no equilibria. So, does this mean that at $\mu=0$ and $\mu=2$ we have a saddle-node bifurcation resp. a saddle-node on a limit cycle bifurcation when considering the equation on $S^1=\mathbb{R}/2\pi n$?

mathfemi
  • 2,631
  • For first equation, your $mu$ should be in $[-2,0]$. However, your $mu$ should be in $[-1,1]$. – xpaul Sep 05 '16 at 16:18
  • Shouldn't you add boundary conditions. E.g., periodic ones $\lim_{x\rightarrow2\pi} u(x) = u(0)$? By the way, what is the space $u$ should live in? For spatial constant $u$ you get $u_{xx}=0$ which gives the same effect as ignoring $u_{xx}$. So you have equilibria for the case $0<\mu<2$. – Tobias Sep 06 '16 at 09:24
  • @Tobias To be honest, I am not sure where $u$ is defined on and where it lives in. The task just was "determine all equilibria in $[0,2\pi)$." What do you think this means for the domain of $u$ and the space it lives in? – mathfemi Sep 06 '16 at 10:10
  • Additional equilibria of (1) require $u''=1-\mu-\cos(u)$. Multiplication of $u'$ and integration gives $\frac12((u')^2-(u'0)^2)=(1-\mu)(u-u_0)+(\sin(u_0)-\sin(u))$ where $u_0:=u(x_0),u'_0:=u'(x_0)$. After resolution for $u'$ you obtain $\int{u_0}^u \frac{du}{\sqrt{u'_0 + 2(1-\mu)(u-u_0) + 2(\sin(u_0)-\sin(u))}} = x-x_0$ with the help of the separation of variables. The integral exists at least for some neighborhood of $u_0$ if $u_0'\neq 0$. So if the domain of the solution candidates $u$ is small enough you have other equilibria. (Hopefully, there are no fatal errors in the calc.) – Tobias Sep 06 '16 at 13:54

0 Answers0