# 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.

• hmm, no reaction so far? Is this to obvious or really hard? Is additional info needed? – Quickbeam2k1 Feb 11 '14 at 19:12

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.

• Hmm, just a first comment. If $\Phi$ is independent of $x$, then one can prove that $\nabla [\Phi(s(x)] =\lim_{\varepsilon\to 0} \Phi'(s(x)) \nabla[\max(\min(1-\varepsilon,s(x)),\varepsilon)]$ strongly in $L^2(\Omega)$. If you want, I can add some detail to this. Basically, one shows that $\nabla [\Phi(s(x)] \mathbf{1}_{\{\varepsilon<s(x)<1-\varepsilon\}}= \Phi'(s(x)) \nabla[\max(\min(1-\varepsilon,s(x)),\varepsilon)]$. Now the pointwise convergence follows from Stampacchia's lemma. – Quickbeam2k1 Feb 19 '14 at 22:13
• I agree you get pointwise convergence; even in this example if you cut off at some $\varepsilon>0$, you have pointwise convergence. However, $L^2$ convergence fails because of the degeneracy of $\Phi$. I may be wrong; perhaps you could provide more details? – felipeh Feb 19 '14 at 22:20
• I will, edit it in my question – Quickbeam2k1 Feb 19 '14 at 22:22
• done. sorry, I again found a mistake in $(*)$ :( – Quickbeam2k1 Feb 19 '14 at 22:35
• I edited my answer to make it a bit more clear that you can't expect quantitative control of the $L^2$ norm. Maybe this will help you understand the counterexample, I know it is quite messy. – felipeh Feb 19 '14 at 22:35