# Exponential boundedness for PDE

Suppose we have $\kappa$ bounded and the following equation:

$${\partial^2\kappa\over\partial t\partial\theta}=\kappa^2{\partial^3\kappa\over\partial\theta^3}+2\kappa{\partial\kappa\over\partial\theta}{\partial^2\kappa\over\partial\theta^2}+3\kappa^2{\partial\kappa\over\partial\theta}.$$

It is claimed in the paper where I found this that this implies that ${\partial\kappa\over\partial\theta}$ grows at most exponentially as can be seen by considering the PDE satisfied by $e^{\alpha t}{\partial\kappa\over\partial\theta}$ and choosing $\alpha$ so that the maximum principle can be used.

Since I don't know much about PDE and parabolic equations theory, the explanation above is unclear to me. Could anyone give me some insight about this argument or any reference where I might find PDE theory that allows me to understand it better?

Thank you in advance.

## 1 Answer

$\newcommand{\pd}{\frac{\partial#1}{\partial#2}}$ $\newcommand{\pdk}{\frac{\partial^#3#1}{\partial#2^#3}}$ $\newcommand{\pdd}{\frac{\partial^2#1}{\partial#2\ \partial #3}}$ Let $Q(\theta,t) = e^{\alpha t} \pd\kappa\theta$ and compute the time derivative:

$$\pd Q t = \alpha e^{\alpha t}\pd \kappa \theta+e^{\alpha t} \pdd\kappa\theta t.$$

Substituting in the given expression for $\pdd \kappa \theta t$:

$$\pd Q t = e^{\alpha t}\left(\alpha\pd \kappa \theta + \kappa^2 \pdk \kappa \theta 3 + 2 \kappa \pd \kappa \theta \pdk \kappa \theta 2 + 3 \kappa^2 \pd \kappa \theta \right).$$

We want to put this as much in terms of $Q$ (rather than $\kappa$) as is possible. We already know $\pd \kappa \theta = e^{-\alpha t} Q$, and differentiating this gives us expressions for the higher derivatives of $\kappa$. Substituting these all in to $\pd Q t$ we get a parabolic equation for $Q$:

$$\pd Q t = \kappa^2 \pdk Q \theta 2 + 2 e^{-\alpha t} \kappa Q \pd Q \theta + (\alpha + 3 \kappa^2) Q.$$

Since $\kappa$ is bounded we can choose $\alpha$ so that $\alpha + 3\kappa^2 \le 0$, so the maximum principle tells you that $Q$ achieves its maximum on the parabolic boundary: that is, either at the initial time or on the spatial boundary. I'm guessing from the notation that the problem is periodic in $\theta$ so that there is no spatial boundary, in which case you get

$$\pd \kappa \theta \le e^{-\alpha t} \sup_{t=0} Q = e^{-\alpha t} \sup_{t=0} \pd \kappa\theta.$$

• Thanks, for the answer. I think the second term in the last equation of ${\partial Q\over\partial t}$ should be $2\kappa\left({\partial Q\over\partial\theta}\right)^2$, right? I guess this does not affect the application of the max. principle – Edu Jun 15 '16 at 13:28
• @EduardoMSánchez: Actually I think it should be $2 e^{-\alpha t} \kappa Q \partial_\theta Q$. As you say, it doesn't really matter for the purposes of applying the maximum principle - the presence of the $\partial_\theta Q$ in that term means it won't give you any trouble. – Anthony Carapetis Jun 15 '16 at 13:40
• Yes, okay, $2e^{-\alpha t}\kappa Q\partial_{\theta}Q$, which can be written as $2\kappa(\partial_{\theta}Q)^2$. And yes btw, the function $\kappa$ is $2\pi$-periodic in $\theta$! Thanks for your answers, got it! – Edu Jun 15 '16 at 13:44