# Problematic terms in proof of Euler-Lagrange equation

In this question, for any function $f$, I will denote:

$$\operatorname{supt}(f)=\{f\neq0\}, \qquad \operatorname{supp}(f)=\overline{\operatorname{supt}(f)}.$$

The latter is called support of $f$, and $\mathcal{C}^1_c$ means continuously differentiable with compact support. Set:

$$I(v):=\int_\Omega L(x,\nabla v(x))dx,$$

where:

• $\Omega$ is an open subset of $\mathbb{R}^n$;
• $L$ is a nonnegative function satisfying Carathéodory conditions;
• For any $M\geq0$ there exists $\alpha_M\in L^1(\Omega)$ such that $L(x,\xi)\leq\alpha_M(x)$ whenever $|\xi|\leq M$;
• $L$ is convex in $\xi$.

Hypotheses:

• $u$ is a solution to the problem:

$$\min I(v)\qquad\text{where}\qquad v\in u^0+W^{1,1}_0(\Omega);$$

• $u^0$ is any boundary datum;

• $u$ is locally bounded, i.e. $L^\infty_{loc}(\Omega)$;
• $I(u)<\infty$;
• There exists $\overline v\in u^0+W^{1,1}_0(\Omega)$ such that $I(\overline v)<\infty$.

Known consequences of these hypotheses:

• $\langle\nabla_\xi L(\nabla u),\nabla(\overline v-u)\rangle$ is integrable and has nonnegative integral;
• $\langle\nabla_\xi L(\nabla u),\nabla u\rangle$ and $|\nabla_\xi L(\nabla u)|$ are both locally integrable ($L^1_{loc}(\Omega)$).

We then proceed to prove Euler-Lagrange, i.e. that, for $\eta\in\mathcal{C}^1_c(\Omega)$, we have:

$$\int_\Omega\langle\nabla_\xi L(\nabla u),\nabla\eta\rangle=0.$$

Since $u$ is locally bounded, there exists $R$ such that, for almost every $x\in\operatorname{supp}(\eta)$, we have $u(x)\in[-R,R]$. We set:

$$\overline v_t(x)= \begin{cases} \max\{t\eta(x)-R,u(x)\} & x\in\operatorname{supp}(\eta) \\ u(x) & \text{otherwise} \end{cases},$$

and:

$$\underline v_t(x)= \begin{cases} \min\{t\eta(x)+R,u(x)\} & x\in\operatorname{supp}(\eta) \\ u(x) & \text{otherwise} \end{cases},$$

which gives that:

\begin{align*} \nabla\overline v_t(x)={}& \begin{cases} \nabla u(x) & u(x)\geq t\eta(x)-R\text{ or }x\in\Omega\smallsetminus\operatorname{supp}(\eta) \\ t\nabla\eta(x) & x\in\operatorname{supp}(\eta)\text{ and }t\eta(x)-R\leq u(x) \end{cases}, \\ \nabla\underline v_t(x)={}& \begin{cases} \nabla u(x) & u(x)\leq t\eta(x)-R\text{ or }x\in\Omega\smallsetminus\operatorname{supp}(\eta) \\ t\nabla\eta(x) & x\in\operatorname{supp}(\eta)\text{ and }t\eta(x)-R\geq u(x) \end{cases}. \end{align*}

Or does it? What I know is that:

$$\nabla|f|= \begin{cases} \nabla f & f>0 \\ -\nabla f & f<0 \\ 0 & f=0 \end{cases}.$$

Since $\max\{f,g\}=\frac12(f+g+|f-g|)$, we would get:

$$\nabla\max\{f,g\}=\frac12(\nabla f+\nabla g+\nabla|f-g|)= \begin{cases} \nabla f & f>g \\ \nabla g & f<g \\ \frac12(\nabla f+\nabla g) & f=g \end{cases},$$

which, concentrating on the $\overline v_t$'s, should give:

$$\nabla\overline v_t(x)= \begin{cases} \nabla u(x) & u(x)>t\eta(x)-R\text{ or }x\in\Omega\smallsetminus\operatorname{supp}(\eta) \\ t\nabla\eta(x) & x\in\operatorname{supp}(\eta)\text{ and }t\eta(x)-R>u(x) \\ \frac12(t\nabla\eta(x)+\nabla u(x)) & t\eta(x)-R=u(x)\text{ and }x\in\operatorname{supp}(\eta) \end{cases}.$$

Am I doing something wrong? Why does my teacher not have that branch with the equality? Let us go on. By the first known consequence, since these $\overline v_t$'s have gradients different from $\nabla u$ only on a bounded subset, and hence $L(x,\nabla\overline v_t(x))$ is integrable, we have:

$$\int_\Omega\langle\nabla_\xi L(x,\nabla u(x)),\nabla(\overline v_t-u)\rangle\mathrm{d}x\geq0,$$

and that is finite. Now we consider $A_t=\{\eta(x)>\frac{u(x)+R}{t}\}\cap\operatorname{supp}(\eta^+)$. Clearly, $\chi_{A_t}$ converges pointwise to $\chi_{\operatorname{supt}(\eta^+)}$. And now comes the problem:

$$0\leq\frac1t\cdot\int_\Omega\langle\nabla_\xi L(x,\nabla u(x)),\nabla(\overline v_t-u)(x)\rangle\mathrm{d}x=\int_{A_t}\left\langle\nabla_\xi L(x,\nabla u(x)),\nabla\eta(x)-\frac1t\nabla u(x)\right\rangle\mathrm{d}x.$$

This equality is the problem. Now, outside the support, we clearly have $\nabla\overline v_t=\nabla u$, so the integral over there is zero. Over $A_t$ the gradient is $t\nabla\eta$, hence, dividing by $\frac1t$, we get that integral. But what about the points where $t\eta(x)-R=u(x)$? Over those points, we should get an integral where the second entry of the scalar product is $\frac12\nabla\eta(x)-\frac{1}{2t}\nabla u(x)$. Where did that term go? Can I safely neglect it, perhaps because those points are a null set? For the time being, I can see how the characteristic of the set of those points converges pointwise to zero outside $\operatorname{supp}(\eta)\smallsetminus\operatorname{supt}(\eta)$, and I can find a domination thanks to the second known consequence and the boundedness of $\nabla\eta$, so the limit passes under the integral, and if $\operatorname{supp}(\eta)\smallsetminus\operatorname{supt}(\eta)$ has measure zero, that term tends to zero, which is enough for me to neglect it in the rest of the proof. But at the moment, I only know that the part of that set where $\nabla\eta\neq0$ has measure zero, because in those points I can find a local chart that straightens the boundary of $\operatorname{supt}(\eta)$ locally, but I cannot deal with $\{\eta=\nabla\eta=0\}\cap\operatorname{supp}(\eta)$. How do I solve this problem?

I bet that the analogous problem arising from $\min\{f,g\}=\frac12(f+g-|f-g|)$ with the $\underline v_t$'s will be handled in much the same way. If there is any difference, please "warn" me.

If $f,g$ are Sobolev, $\nabla f=\nabla g$ a.e. on $\{f=g\}$.
$\{f=g\}=\{f-g=0\}$, $f-g$ is Sobolev, so by this we have $\nabla(f-g)=0$ a.e. on that set, QED.
In our case, this means $t\nabla\eta=\nabla u$ a.e. on $t\eta-R=u$, meaning we can eliminate the $t\eta-R=u$ branch and turn the $u>t\eta-R$ into a $\geq$ branch, and the problem vanishes.