I'm considering a problem about time varying domain in Chen Wenxiong and Li Congming 's study on $-\Delta u=\exp u$ in $\mathbb R^2$ and $\int\_{\mathbb R^2} \exp u(x) \, d x< +\infty$.


LEMMA 1.1 (Ding). If $u$ is a solution of $-\Delta u=\exp u$ in $\mathbb R^2$ and $\int\_{\mathbb R^2} \exp u(x) \, d x<$ $+\infty$, then $\int\_{\mathbb R^2} \exp u(x) \, d x \geqslant 8 \pi$.


Proof. For $-\infty<t<\infty$ let $\Omega\_t=\{x \mid u(x)>t\}$; one can obtain
$$
\begin{aligned}
\int\_{\Omega\_t} \exp u(x) \, d x & =-\int\_{\Omega\_t} \Delta u=\int\_{\partial \Omega\_t}|\nabla u| \,d s \\
-\frac{d}{d t}\left|\Omega\_t\right| & =\int\_{\partial \Omega\_t} \frac{d s}{|\nabla u|} .
\end{aligned}
$$
By the Schwarz inequality and the isoperimetric inequality,
$$
\int\_{\partial \Omega\_t} \frac{d s}{|\nabla u|} \cdot \int\_{\partial \Omega\_t}|\nabla u| \geqslant\left|\partial \Omega\_t\right|^2 \geqslant 4 \pi\left|\Omega\_t\right| .
$$
Hence
$$
-\left(\frac{d}{d t}\left|\Omega\_t\right|\right) \cdot \int\_{\Omega\_t} \exp u(x) \, d x \geqslant 4 \pi\left|\Omega\_t\right|
$$
and so
$$
\frac{d}{d t}\left(\int\_{\Omega\_t} \exp u(x)\, d x\right)^2=2 \exp t \cdot\left(\frac{d}{d t}\left|\Omega\_t\right|\right) \cdot \int\_{\Omega\_t} \exp u(x)\, d x \leqslant-8 \pi \exp t \cdot\left|\Omega\_t\right| .
$$
Integrating from $-\infty$ to $\infty$ gives
$$\tag{1}
-\left(\int\_{\mathbb R^2} \exp u(x) \,d x\right)^2 \leqslant-8 \pi \int\_{\mathbb R^2} \exp u(x) \,d x
$$
which implies $\int\_{\mathbb R^2} \exp u(x) \,d x \geqslant 8 \pi$ as desired.


**The proof details before (1) is presented by the discussions between me and other users in [Differentiation in time varying domain](https://math.stackexchange.com/questions/4528507/differentiation-in-time-varying-domain/4677803#4677803), I'm confused about the last step, how they integrate from $-\infty$ to $\infty$?**


In detail, for the LHS, how to compute the limit of $\lim\_{t \rightarrow +\infty} \int\_{\Omega\_t} \exp u(x) \, d x$ and $\lim\_{t \rightarrow -\infty} \int\_{\Omega\_t} \exp u(x)\, d x$, I think $\lim\_{t \rightarrow -\infty} \int\_{\Omega\_t} \exp u(x)\, d x$ must be $\int\_{\mathbb R^2} \exp u(x) \,d x$ but I don't know how to deal with $\lim\_{t \rightarrow +\infty} \int\_{\Omega\_t} \exp u(x) \,d x$, from the fact that $\lim\_{t \rightarrow -\infty} \int\_{\Omega\_t} \exp u(x) \,d x = \int\_{\mathbb R^2} \exp u(x) \,d x$ which is bounded and $\frac{d}{d t}\left(\int\_{\Omega\_t} \exp u(x) \,d x\right)^2=2 \exp t \cdot\left(\frac{d}{d t}\left|\Omega\_t\right|\right) \cdot \int\_{\Omega\_t} \exp u(x) \,d x \leqslant 0 $ we know that when $t \rightarrow +\infty$, $\lim\_{t \rightarrow +\infty} \int\_{\Omega\_t} \exp u(x) \, d x$ is also bounded, but based on the proof, it seems that it is 0.


For the RHS, how to integrate $\exp t \cdot\left|\Omega\_t\right|$ from $-\infty$ to $\infty$? I try to use Reynolds transport theorem
\begin{align}
& \frac{\mathrm{d}}{\mathrm{d} t} \int\_{\Omega(t)} \mathbb{A}(\boldsymbol{x}, t)\, \mathrm{d}^n \boldsymbol{x} \\[8pt]
= {} & \int\_{\partial \Omega(t)}(\boldsymbol{V}(\boldsymbol{x}, t) \cdot \boldsymbol{n}(\boldsymbol{x}, t)) \mathbb{A}(\boldsymbol{x}, t) \,\mathrm{d}^{n-1} \boldsymbol{x}+\int\_{\Omega(t)} \partial\_t \mathbb{A}(\boldsymbol{x}, t) \,\mathrm{d}^n \boldsymbol{x}
\end{align}
I set $A\_1=e^u$ and $A\_2=e^t$, because on $\partial \Omega(t)$ we have $u=t$ so we have
$$\exp t \cdot\left|\Omega\_t\right|=
\int\_{\Omega t} e^t \,d x =\frac{d}{d t} \int\_{\Omega t} e^t \,d x - \frac{d}{d t}\int\_{\Omega t} e^u \,d x 
$$
Now we have to face another question which is $\lim\_{t \rightarrow +\infty} \exp t \cdot\left|\Omega\_t\right|$ and $\lim\_{t \rightarrow -\infty} \exp t \cdot\left|\Omega\_t\right|$



Time varying domain in Chen Wenxiong and Li Congming 's study on $-\Delta u=\exp u$ in $\mathbb R^2$ and $\int_{\mathbb R^2}\exp u(x) \, dx< +\infty$
I think your question is just basic real analysis, not at the research level, you should ask this question on stack exchange instead of overflow.


Since $\int\_{\mathbb{R}^2} e^u \, dx <\infty$, for every $\epsilon>0$, there exists $\delta>0$ such that if $|A|<\delta$, then $\int\_A e^u \, dx <\epsilon$. By Chebyshev's Inequality, it is easy to see that $|\Omega\_t|\to 0$ as $t\to +\infty$.
So $\lim\_{t\to 0} \int\_{\Omega\_t} e^u \, dx=0$. On the other hand, by the description of the $L^1$ norm in terms of the distribution function, we have
$$\int\_{\mathbb{R}^2} e^u \, dx =\int\_0^\infty |\Omega\_{\ln s}| \, ds 
=\int\_{-\infty}^{+\infty}e^t|\Omega\_{t}| \, dt.
$$
In the last step, we use the formula of change of variables by setting $s=e^t.$


