Show $v(x,t) \in L^2([0,T];H^2(\mathbb{R}))$ when $v(x,t)$ is a transformation of a $L^2([0,T];H^2(\mathbb{R}))$ function

Question

Context: I am reading a paper on Long-Time Asymptotics of the thin film equations, in which the authors consider the strong solutions of the thin film equation in 1-D and transform them using a time-dependent scaling into a solution for a Fokker-Planck type equation. I am meant to translate the properties similarly.

So, given a strong solution, $u(x,t) \in L^2([0,T];H^2(\mathbb{R}))$, we take the functions $\alpha(t), \beta(t)$ which are injective and continuous, such that we define: $$v(x,t) = \alpha(t)u(\alpha(t)x, \beta(t)).$$

At this point, I am struggling to understand how to justify that $v(x,t)$ is also in $L^2([0,T];H^2(\mathbb{R}))$. What must $\alpha(t), \beta(t)$ satisfy? Or just $\alpha(t)$, since it is the only one "outside" which I would need to control for when taking the norm?

In my case, $\alpha(t)=e^t$ and $\beta(t)=\frac{e^{5t}-1}{5}$.

Answer 1

First, you probably want to check that $v(x,t)$ is in $L^2([0,\log(5T+1)];H^2(\mathbb{R}))$.

Notice that $\alpha(t)$ is bounded and $\beta(t)$ is bijective and bi-Lipschitz. This is the only information that we will actually use. Now, let us compute the norm of $v$ using the chain rule and the change of variables theorem.

\begin{align*} \|v\| &= \int_0^{\log(5T+1)}\left(\int_{\mathbb R^n}|v|^2+|Dv|^2+|D^2v|^2\mathrm{d}x\right)^{1/2}\mathrm{d}t\\ &= \int_0^{\log(5T+1)}\left(\int_{\mathbb R^n}\big(\alpha(t)^2|u|^2+\alpha(t)^4|Du|^2+\alpha(t)^6|D^2u|^2\big) (\alpha(t)x,\beta(t))\mathrm{d}x\right)^{1/2}\mathrm{d}t\\ &= \int_0^{\log(5T+1)}\left(\int_{\mathbb R^n}\big(|u|^2+\alpha(t)^2|Du|^2+\alpha(t)^4|D^2u|^2\big) (x,\beta(t))\mathrm{d}x\right)^{1/2}\mathrm{d}t\\ &\leq \int_0^{\log(5T+1)}\max\{1,\alpha(t)^2)\}\|u(\cdot,\beta(t))\|_{H^2(\mathbb R^n)}\mathrm{d}t = \int_0^{T}\frac{\alpha(t)^2}{\beta'(\beta^{-1}(t))}\|u(\cdot,t)\|_{H^2(\mathbb R^n)}\mathrm{d}t\\ &\leq C\|u\|. \end{align*}

In the last step we used that $\beta'(s) \geq 1$ for all nonnegative $s$, and that $\alpha(t) \leq e^T$ in our domain.