Justification of formal derivative there I'm having the following problem and so far I didn't find anything in the literature on this.
$\Phi\in C^1(\overline{\Omega}\times[0,1])$ with $\Phi'(x,0)=\Phi'(x,1)=\Phi(x,0)=0$ for every $x\in\Omega$, where $\Omega$ is a bounded domain and $'$ denotes the derivative w.r.t. the second variable. Furthermore $\Phi(x,s)$ is increasing fof fixed $x$ with $\Phi'(x,s)>0$.
Now assume, that there is $s\in L^\infty(\Omega,[0,1])$ with $x \mapsto \Phi(x,s(x))$ is in $ H^1(\Omega)$. Due to the degeneracy of $\Phi$ we don't get $H^1$ estimates on $s$. However, with appropriate truncations one obtains that $[\max(\min(1-\varepsilon,s),\varepsilon)]\in H^1(\Omega)$. I.e. somehow $s$  has a gradient when away from the set where $s=0$ or $s=1$. It is possible to show some identities for the limits $\varepsilon \to 0$. 
For any positive $\varepsilon$ one can show
$$ \mathbf{1}_{\{\varepsilon <s(x) <1-\varepsilon\}}(\nabla[\Phi(x,s(x))] -\nabla_x \Phi(x,s)) =\Phi'(x,s(x))\nabla [\max(\min(1-\varepsilon,s(x)),\varepsilon)] \tag{*}$$ for a.e. $x\in\Omega$,
where $\nabla_x$ is the gradient with respect to the first coordinate.
The goal is to let $\varepsilon$ pass to zero.
One obtains pointwise convergence on the set $\{x: 0<s(x)<1\}$ and the left hand side is integrable and dominates the right hand side. In particular, on all the sets $\{x: \varepsilon <s(x)<1-\varepsilon\}$ the identity $(*)$ holds. On the set $\{x:s(x)=0\}$ the identity is clear since $\Phi(x,0)=\Phi'(x,0)=0$ and since by Stampacchia's lemma $\nabla[\Phi(x,s(x))]=0$ a.e. on the set $\{x: \Phi(x,s)=\Phi(x,0)=0\}=\{x : s(x)=0\}$.
On the set where $\{x: s(x)=1\}$ one still obtains that the pointwise limit on the rhs is zero. However, on that set $\nabla_x \Phi(x,s)\neq 0$ unles $\Phi(x,1)$ is constant. Hence, for equality in the limit, we require that $\nabla [\Phi(x,s)]=\nabla_x \Phi(x,s)$ a.e. on the set $\{x:s(x)=1\}$. Which holds formally, but i don't see how to proof this rigorously.
Does anybody see an idea on how to proof this? Any comment is appreciated
Edit 1: I referred to the boundedness of the right hand side, since i want to show that the convergence holds in fact in $L^2(\Omega)$
Edit 2: So what one could ask essentially: When is it allowed to compare weak derivatives pointwise, i.e. is it possible to say that on the set $\{x:s(x)=1\}$ holds $\nabla \Phi(x,s)=\nabla\Phi(x,1)=\nabla_x\Phi(x,1)$. However, taking the union over all the level sets between zero and one we would never obtain a contribution of $\nabla s$.
Edit: Corrected the question.
Edit $L^2$ convergence in the case without $x$-dependence and also updated $(*)$ sorry the inconvenience.
However, consider $(*)$ without the $\nabla_x \Phi$ term. First of all, we obtain the obvious bound $|\nabla \Phi(s)| \geq |\Phi'(s(x)) \nabla[\max(\min(1-\varepsilon,s(x)),\varepsilon)]|$, basically due to $(*)$ and increasing the left hand side where the right hand side is zero. However, on the set $\{0<s(x)<1\}$ we trivially obtain pointwise convergence to the desired limit. On the sets where $s(x)=0$ and $s(x)=1$, we find $\Phi'(s(x)) \nabla[\max(\min(1-\varepsilon,s(x)),\varepsilon)]=0$. Furthermore, Stampacchia's lemma tells us, that a.e. on the sets where $\{\Phi(s)=\Phi(1)\}$ and $\{\Phi(s)=\Phi(0)\}$ we find $\nabla \Phi(s)=0$ a.e. Due to the monotonicity of $\Phi$. These sets coincide with $\{s=1\}$ and $\{s=0\}$ respectively. Hence, the pointwise convergence holds a.e. in $\Omega$ and with Lebesgue's theorem we obtain strong convergence.
This proof fails if $\Phi$ is $x$-dependent. Sorry for the inconvenience with $(*)$. I hope the notation for the sets was not soo sloppy.
 A: I think I can construct a counterexample.  Choose $\Omega = (0,1)$, and $\Phi(x,t) = \phi(t)$
smooth enough that $\phi(t)$ obeys the bound:
$$
\phi'(t) \leq C t.
$$
for sufficiently small $t$.
(And of course, such that $\phi(t)$ satisfies $\phi'(1) = 0$)  
Now choose a function $g_\alpha(x)$ with support in $(-1,1)$, and smooth outside of $0$, such that for 
$|x|<1/2$, $g_\alpha(x) = 1 - |x|^\alpha$.  Notice that $\alpha = 1/2$ is the
borderline case for $g_\alpha \in H^1$.  Indeed, for $\alpha > 1/2$, $g_\alpha$
satisfies the estimate:
$$
\int_{-1}^1 |\partial_x g_\alpha(y)|^2 \,dy 
\geq \int_{0}^{1/2} \alpha^2 y^{2\alpha - 2}\,dy
= \frac{\alpha^2}{2\alpha - 1} \left(\frac{1}{2}\right)^{2\alpha-1}. 
$$
Already we can see that there is no hope of an estimate of the form
$$
\|\nabla s(x)\|_{L^2} \lesssim \|\nabla (\Phi(x,s(x))\|_{L^2}.
$$
Indeed, let $\alpha_k = 1/2 + 2^{-k}$ and $s_k = 2^{-k} g_{\alpha_k}(x)$. 
Then by the above calculation,
$$
\|\nabla s_k(x)\|_{L^2} \gtrsim 2^k,
$$
but since $\nabla \Phi(x,s(x)) = \phi'(s(x)) \nabla s(x)$ (pointwise a.e), 
and $\phi'(s(x)) \leq 2^{-k}$, we have 
$$
\|\nabla (\Phi(x,s(x)))\|_{L^2} \sim 1.
$$
This is the main obstruction to uniform $L^2$ convergence, the following just constructs an explicit example using these ideas.
We are ready to construct the counterexample.  Set
$$
s(x) = \sum_{k=1}^\infty \frac{1}{k}g_{\alpha_k}(2^{k+2}(x - 2^{-k})) = \sum_{k=1}^\infty h_k(x).
$$
Which is a sequence of disjoint cusps $h_k(x)$ with support in 
$(2^{-k} - 2^{-k-2}, 2^{-k} + 2^{-k-2})$, and $\alpha_k$ will soon be chosen.
Indeed, if $\alpha_k > 1/2$ decays to $1/2$ rapidly enough, $s(x)\notin H^1$, since
$$
\int |\partial_x h_k(y)|^2 \,dy = 
\int_{2^{-k} - 2^{-k-2}}^{2^{-k} + 2^{-k-2}} \frac{1}{2^k} |\partial_x (g_{\alpha_k}(2^{k+2}(y - 2^{-k})))|^2 \, dy
\geq 2^{-k} \int_{-1/2}^{1/2} (\alpha_k)^2 |y|^{2\alpha_k - 1}\,dy.
$$
From this and the estimate above, $\alpha_k = \frac{1}{2} + \frac{1}{2^{k+1}}$
would work so that $\|h_k\|_{H^1} \sim_k 1$. (Meaning it's bounded above and below by a constant independent of $k$.)
On the other hand, $\Phi(x,s(x))\in H^1$ since 
$$
\int |\partial_x \Phi(y,s(y))|^2\,dy = 
\int |\phi'(s(y)) \partial_x s(y)|^2\,dy = 
\sum_k \int |\phi'(h_k(y)) \partial_x h_k(y)|^2\,dy
$$ 
Since $h_k(y) \leq 2^{-k}$, and $\phi'(t) \leq C t$, we can estimate
$$
\sum_k \int |\phi'(h_k(y)) \partial_x h_k(y)|^2\,dy 
\leq \sum_k \sup_k \|h_k\|_{H^1} 2^{-k} < \infty.
$$
Even if I made a mistake in the calculation, I think the general idea --that you make a sequence of smaller and smaller cusps, with unbounded $H^1$  norm-- is correct.
