I am thinking about a geometric problem which boils down to the following parabolic equation: Suppose $u=u(r,t)$, $r$ is defined on $[0,1]$ and $t>0$ $$\begin{cases}\displaystyle \frac{\partial u}{\partial t}=(\ln u)''+\frac{1}{r}(\ln u)'-4u(k\sqrt{u(1)}-1)\\u'(0)=0\\u'(1)=2u(1)(k\sqrt{u(1)}-1)\end{cases}$$ Here $'$ means differentiating respect to $r$. $u(1)$ is $u(1,t)$ for short. $k$ is some nice positive constant fixed. You can assume it is very small.

Obviously this is a nonlocal quasilinear equation with nonlinear neumann boundary condition. We always assume that the initial data $u(\cdot,0)>0$. And it has a particular solution which is $u\equiv \frac{1}{k^2}$. A nice property of this equation is $\int_0^1 ru(r,t)dr$ is unchanged along the flow.

I am very interested in the long time existence of this equation. Suppose we have very nice initial data which is positive

(1)Will $u$ exists forever?

(2)If this is true, will $u$ converge to some solution?

(3) What about the stability of this particular solution?

(4)All the above question is linked to examine the behavior of $u(1)$. If one can control $u(1)$, then it is done. Can anyone construct one solution $u$ such that $u(1,t)\to \infty$ when $t\to T$ where $T$ can be finite or infinite? Or $u(1,t)\to 0$ when $t\to T$? Conunterexample is welcomed, because I know every few about this equation.

Thank you for any observation in advance.

  • $\begingroup$ Isn't the particular solution $1/k^2$, not $1/k$? $\endgroup$ – Nate Eldredge Jan 16 '15 at 4:58
  • $\begingroup$ @NateEldredge Sorry, that is a typo. Thank you. $\endgroup$ – Slm2004 Jan 16 '15 at 14:29
  • $\begingroup$ Can you rewrite the final term? I can't tell if its $u$ evaluated at some quantity, or if its $u(t,r)$ times that weird quantity.. $\endgroup$ – Otis Chodosh Jan 16 '15 at 15:51
  • $\begingroup$ @OtisChodosh it is $4u(t,r)\cdot(k\sqrt{u(1,t)}-1)$, and boundary condition is $u'(1,t)=2u(1,t)\cdot(k\sqrt{u(1,t)}-1)$. I omit $t$ in the expression $\endgroup$ – Slm2004 Jan 17 '15 at 13:40

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