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.
 A: $\newcommand{\pd}[2]{\frac{\partial#1}{\partial#2}}$
$\newcommand{\pdk}[3]{\frac{\partial^#3#1}{\partial#2^#3}}$
$\newcommand{\pdd}[3]{\frac{\partial^2#1}{\partial#2\ \partial #3}}$
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
$$ \pd \kappa \theta \le e^{-\alpha t} \sup_{t=0} Q = e^{-\alpha t} \sup_{t=0} \pd \kappa\theta. $$
