Differentiation in time varying domain I am stuck at two steps in the following PDE . So it's let $u$ be solution of $-\Delta u= \exp u$ in $\mathbb{R^2}$ and $\int_{\mathbb{R^2}}\exp(u(x))dx<\infty$ given ; then for $-\infty<t<\infty$ we defined $\Omega_t=\{x| u(x)>t\}$ then we have
$\int_{\Omega_t}\exp u(x)dx= -\int_{\Omega_t}\Delta u=\int_{\partial\Omega_t}|\nabla u|dS...(1)$
In this line I am getting $-\int_{\Omega_t}\Delta u= \int_{\partial\Omega_t}\nabla u.n dS$ now unit normal is $\frac{\nabla u}{|\nabla u|}$ replacing this I am getting equation (1) with a minus sign in the right. Where I am doing it wrong?
In the very next line it's written that $-\frac{d}{dt}|\Omega_t|=\int_{\partial\Omega_t}\frac{ds}{|\nabla u|}$ ($|\Omega_t|$ means measure ). But I am not getting how this is coming what I was doing is $ -\frac{d}{dt}|\Omega_t|=-\frac{d}{dt}\int_{\Omega_t}\chi_{\Omega_t}dt=-\frac{d}{dt}\int_{\partial\Omega_t} (v.n)f dt$ is this This is using this link https://en.wikipedia.org/wiki/Reynolds_transport_theorem  but I don't know how the RHS i.e $\frac{ds}{|\nabla u|}$ is coming?.
Then again it uses
$\frac{d}{dt}(\int_{\Omega_t}\exp (u(x)) dx)^2=2\exp t.\frac{d}{dt}|\Omega_t|\int_{\Omega_t}\exp u(x)dx$. This also how it's coming I am not getting. It would be very helpful if someone can tell me how to get these
 A: Not a complete answer.
This is a hard question. I can see why you struggled with it.

Problem. Let $\phi:\Bbb R^n\to\Bbb R$ be a solution of $-\Delta \phi=\exp\phi$ and let $\Omega(t)=\{\boldsymbol x\in\Bbb R^n : \phi(\boldsymbol x)>t\}$. Let $F(t)=\int_{\Omega(t)}\exp\phi(\boldsymbol x)\mathrm d^n \boldsymbol x$ and $M(t)=\mu^n(\Omega(t))$ where $\mu^n$ is the $n$ dimensional Lesbesgue measure.
Show:

*

*$F(t)=\int_{\partial\Omega(t)}|\nabla\phi (\boldsymbol x)|\mathrm d^{n-1} \boldsymbol x$


*$M'(t)=-\int_{\partial\Omega(t)}\frac{1}{|\nabla\phi(\boldsymbol x)|}\mathrm d^{n-1}\boldsymbol x$


*Show $\frac{\mathrm d}{\mathrm dt}\big(F(t)\big)^2=2\exp(t) M'(t)F(t)$

Part 1.
Suppose $\phi:\Bbb R^n\to\Bbb R$ is a solution of
$$-\Delta \phi=\exp \phi\tag{1}$$
And now let $t\in\Bbb R$ and define
$$\Omega(t)=\{\boldsymbol x\in\Bbb R^n:\phi(\boldsymbol x)>t\}\tag{2}$$
Now consider
$$F(t)=\int_{\Omega(t)}\exp \phi(\boldsymbol x)~\mathrm d^n\boldsymbol x\tag{3}$$
Which we could also write as
$$F(t)=-\int_{\Omega(t)}\Delta \phi(\boldsymbol x)\mathrm d^n\boldsymbol x \\ =-\int_{\Omega(t)}\nabla\cdot\nabla \phi(\boldsymbol x)\mathrm d^n \boldsymbol x $$
Apply divergence theorem:
$$ =-\int_{\partial\Omega(t)} \nabla \phi(\boldsymbol x)\cdot \boldsymbol n(\boldsymbol x)\mathrm d^{n-1} \boldsymbol x\tag{4}$$
Where $\boldsymbol n(\boldsymbol x)$ is the unit outward normal vector to $\partial \Omega(t)$ at the point $\boldsymbol x$.

Lemma. If $\Omega(t)=\{\boldsymbol x\in\Bbb R^n : \phi(\boldsymbol x)>t\}$ then the unit normal vector $\boldsymbol n(\boldsymbol x)$ to the $n-1$ dimensional surface $\partial\Omega(t)$ is independent of $t$ and given by
$$\boldsymbol n=-\frac{\nabla \phi}{|\nabla \phi|}$$

Proof.
Clearly
$$\partial\Omega(t)=\{\boldsymbol x\in\Bbb R^n :\phi(\boldsymbol x)=t\}$$
In other words it consists of the solutions to the equation
$$g(\boldsymbol x;t)=0$$
Where we regard $g(\boldsymbol x;t)=\phi(\boldsymbol x)-t$ as a function of $\boldsymbol x$ with $t$ just being a shape parameter. But, from equation 15 at this reference we know that the unit normal vector to this surface is exactly
$$\boldsymbol n(\boldsymbol x)=(\pm)\frac{\nabla g}{|\nabla g|}$$
The proof of this should be available in any introductory multivariable calculus text.
The choice of sign comes down to whether this normal vector points inwards or outwards to the surface. This is impossible to determine in general, and will just vary from example to example. But of course $\nabla g=\nabla \phi$, so going back to $(4)$, this means
$$F(t)=- \int_{\partial\Omega(t)}\nabla \phi(\boldsymbol x)\cdot (\pm)\frac{\nabla \phi(\boldsymbol x)}{|\nabla \phi(\boldsymbol x)|}\mathrm d^{n-1}\boldsymbol x \\ =\pm\int_{\partial\Omega(t)}|\nabla \phi(\boldsymbol x)|~\mathrm d^{n-1}\boldsymbol x$$
However, from $(3)$ we know that $F(t)$ must be non-negative, since $\exp \phi>0$ for $\phi\in\Bbb R$. And,  the integrand of the above equation is non-negative, so therefore we must choose the $\boldsymbol n=\color{red}{\boldsymbol{-}}\nabla\phi/|\nabla\phi|$. So
$$F(t)=\int_{\partial\Omega(t)}|\nabla \phi(\boldsymbol x)|~\mathrm d^{n-1}\boldsymbol x\tag{5}$$
This answers your first question.

Part 2.
This is the really hard one.
Recall the Reynolds transport theorem, essentially a generalization of the Leibniz Integral rule:
$$\frac{\mathrm d}{\mathrm dt}\int_{\Omega(t)} \mathbf A(\boldsymbol x,t)\mathrm d^n \boldsymbol x=\int_{\partial \Omega(t)}\big(\boldsymbol V(\boldsymbol x,t)\cdot \boldsymbol n(\boldsymbol x,t)\big)\mathbf A(\boldsymbol x,t)\mathrm d^{n-1}\boldsymbol x+\int_{\Omega(t)}\partial_t\mathbf A(\boldsymbol x,t)\mathrm d^n \boldsymbol x$$
Where $\mathbf A$ can be tensor valued. Here $\boldsymbol V(\boldsymbol x,t)$ is the "velocity of the area element". I still don't understand this very well myself, but you can regard it as the time derivative of the set $\partial\Omega (t)$. Recall that $\partial\Omega(t)$ is given by the solutions of the equation $\phi(\boldsymbol x)=t$. I don't know why this works, but try just differentiating w.r.t $t$:
$$\frac{\mathrm d}{\mathrm dt}\partial\Omega(t)~~\overset{\text{wtf?}}{=}~~\frac{\mathrm d}{\mathrm dt}\big(\phi(\boldsymbol x)=t\big) \\ \nabla\phi\cdot \frac{\mathrm d \boldsymbol x}{\mathrm dt}=1 \\ \nabla\phi\cdot \boldsymbol V=1$$
If we take $\boldsymbol V=\frac{\nabla\phi}{|\nabla\phi|^2}$ the above equation is satisfied. Now, we remark that
$$M(t)=\mu^n\big(\Omega(t)\big)=\int_{\Omega(t)}1~\mathrm d^n\boldsymbol x$$
So, applying the transport theorem, and taking $\boldsymbol V=\nabla\phi/|\nabla\phi|^2$ and $\boldsymbol n=-\nabla\phi/|\nabla\phi|$ from before, we get
$$M'(t)=\int_{\partial\Omega(t)}\bigg(\frac{\nabla \phi}{|\nabla\phi|^2}\cdot\frac{-\nabla\phi}{|\nabla\phi|}\bigg)(\boldsymbol x)~1~\mathrm d^{n-1}\boldsymbol x $$
$$M'(t)=\int_{\partial\Omega(t)}\frac{-1}{|\nabla\phi(\boldsymbol x)|}\mathrm d^{n-1}\boldsymbol x \tag{6}$$
As required.

Couldn't figure out the third one. Good luck.
