# Euler-Lagrange equations for minimizer of energy with indicator function

I'm looking for a modern explanation/proof of the derivation of Euler-Lagrange (or first-order or the "first variation") conditions for $$\min_{u \in H^1_0(\Omega), u \geq 0} \int_\Omega |\nabla u|^2 + \int_\Omega \chi_{\{u > 0\}}\label{1}\tag{J}$$ (the constraint $$u \geq 0$$ could also be omitted if convenient). Here $$\chi_A$$ means the indicator function of the set $$A$$; so it is $$1$$ on $$A$$ and $$0$$ elsewhere.

This is (I believe) the Alt-Caffarelli functional analyised in their paper [1], which cites an old (and hard to decipher) German text by Friedrichs [2] for the derivation of the E-L equations. Is there a more modern textbook/paper or other that gives the proof of this functional (or related ones, involving the indicator function)?

References

[1] H. W. Alt; L. A. Caffarelli, Existence and regularity for a minimum problem with free boundary, J. Reine Angew. Math. 325, 105-144 (1981). MR618549, Zbl 0449.35105.

[2] K. O. Friedrichs, Über ein Minimumproblem für Potentialströmungen mit freiem Rande., Math. Ann. 109, 60-82 (1933). JFM 59.1447.01, Zbl 0008.16605.

• $\newcommand\supp{\operatorname{supp}}$I don't know how to derive the Euler-Lagrange equation, but, assuming no constraints on the sign of $u$ or its variation $v$, I get \begin{align*} \int_{\Omega} \chi_{u+tv} &= \begin{cases} V(\supp(u_+)\cup\supp(v_+)) &\text{ if }t > 0\\ V(\supp(u_+)) &\text{ if }t=0\\ V(\supp(u_+)\cup\supp(v_-)) &\text{ if }t < 0 \end{cases}. \end{align*} I do not see how to differentiate this at $t=0$. Commented Jul 21 at 15:24

Edit: An earlier version of this answer missed a factor of two, now corrected, thanks to Daniele Tampieri.

We seek the variation of the functional $$L[u]=\int_\Omega\left(|\nabla u|^2+1\right)\chi_{u>0}\,dx.$$ The indicator function restricts the integration to the volume $$V\subset\Omega$$ where $$u\geq 0$$.
It is essential that the indicator multiplies all terms, that was my initial mistake.

Introduce an infinitesimal variation $$u(x)\mapsto u(x)+\epsilon(x)$$, and compute the variation in $$L$$ to first order in $$\epsilon$$. Let $$S$$ be the surface boundary of $$V$$ on which $$u=0$$. The variation is $$L[u+\epsilon]=\int_\Omega\left(|\nabla (u+\epsilon)|^2+1\right)\chi_{u+\epsilon>0}\,dx$$ $$=L[u]+\int_V 2(\nabla u)\cdot(\nabla\epsilon) \,dx + \int_{S} \left(|\nabla u|^2+1\right)\frac{\epsilon}{|\nabla u|}\,ds+{\cal O}(\epsilon^2)$$ $$=L[u]-2\int_V \epsilon\nabla^2 u \,dx +\int_{S} \epsilon\left(2n\cdot\nabla u+\frac{|\nabla u|^2+1}{|\nabla u|}\right)\,ds+{\cal O}(\epsilon^2)$$ $$=L[u]-2\int_V \epsilon\nabla^2 u \,dx +\int_{S} \epsilon\left(-|\nabla u|+\frac{1}{|\nabla u|}\right)\,ds+{\cal O}(\epsilon^2).$$ On the second line I used the identity $$\int f(x)\frac{d}{du}\chi_{u>0}\,dx=\int_S f(x)|n\cdot\nabla u|^{-1}\,ds=\int_S f(x)|\nabla u|^{-1}\,ds,$$ with $$n$$ a unit vector normal to the surface $$S$$ and pointing outward. (Note that $$\nabla u$$ has only a normal component on $$S$$.) On the third line I carried out a partial integration of the volume integral, on the fourth line I used that $$n\cdot\nabla u=-|\nabla u|$$ on $$S$$.

The variation of $$L$$ vanishes for arbitrary $$\epsilon$$ if the integrand of the volume integral $$\int_V$$ and the integrand of the surface integral $$\int_{S}$$ each vanish identically, so $$\nabla^2 u=0\;\;\text{in}\;\;V,$$ $$|\nabla u|^2=1\;\;\text{on}\;\;S.$$

More generally, following the same steps to minimize $$L[u]=\int_V\left( f(x)|\nabla u|^2+h(u)+g(x)\right)\chi_{u>0}\,dx$$ gives the variation $$\delta L[u]=\int_V\epsilon\biggl(-2\nabla\cdot(f(x)\nabla u)+h'(u)\biggr)\,dx+\int_S\epsilon\left(2f(x)(n\cdot\nabla u)+\frac{f(x)|\nabla u|^2+h(u)+g(x)}{|\nabla u|}\right)\,ds$$ hence the Euler-Lagrange equations $$2\nabla\cdot(f(x)\nabla u)=h'(u)\;\;\text{in}\;\;V,$$ $$f(x)|\nabla u|^2=h(u=0)+g(x)\;\;\text{on}\;\;S.$$

• Thank you. How did you get the boundary term in the second line of your first displayed equation?
– BBB
Commented Jul 22 at 16:57
• boundary term: consider first a one-dimensional integration, with $u(x)>0$ for $x<0$ and $u(x)<0$ for $x>0$; then $$\int \frac{d}{du}\chi_{u>0}dx=\int \delta(u)dx=\int |u'(x)|^{-1}\delta(x)dx=|u'(0)|^{-1}.$$ I used that the derivative of the indicator function is a delta function, and $\delta(f(x))=|f'(x)|^{-1}\delta(x-x_0)$ if $f(x)$ vanishes at $x=x_0$. In the $d$-dimensional case a $d-1$ dimensional surface integral remains, $$\int \frac{d}{du}\chi_{u>0}dx=\int_S |n\cdot\nabla u|^{-1}ds=-\int_S (n\cdot\nabla u)^{-1}ds.$$ Commented Jul 22 at 17:05
• I notice a factor of two difference between the boundary term I have above, $2|\nabla u|^2=1$, and in the answer of Daniele Tampieri, $|\nabla u|=1$. I'm unsure why. The expression above is consistent with Eq. (2.8) of this source. Commented Jul 22 at 17:11
• @CarloBeenakker possibly the problem is caused by the fact that the regions $V=\Omega\cap\{u>0\}$ and its boundary $S$ are independent from the functional variation $\epsilon(x)$ even in the expression of $J(u+\epsilon)$ and this cannot be true, in this particular case. Commented Jul 22 at 18:01
• @DanieleTampieri You are right, I have fixed it. The factor of two is gone. Commented Jul 22 at 21:01

Edit. Corrected a wrong statement about the structure of the functional $$J$$, thanks to Carlo's answer, and added more notes.

The major difficulty of the derivation, as noted by Deane Yang in his comment, is the structure of the second term of the functional \ref{1}, called $$J_2$$ in the following, i.e. $$\DeclareMathOperator{\Dm}{d\!} \DeclareMathOperator{\Div}{\nabla\cdot} J_2(u)=\int_\Omega \chi_{\{u>0\}}\Dm x$$ Nevertheless we see that the following equation holds, $$J_2(u)=\int_\Omega \chi_{\{u>0\}}\Dm x=\int_\Omega H(u(x))\Dm x$$ where $$H(y)$$, $$y\in \Bbb R$$ is the standard Heaviside function, since we have that $$\chi_{\{u>0\}} = H(u(x))\quad \forall x\in \Omega,$$ as it's easy to prove by elementary set theoretic considerations. We can then try to use this relation and say that $$\begin{split} \left.\frac{\Dm J}{\Dm \epsilon}(u+\epsilon \varphi)\right|_{\epsilon=0} & = \left.\frac{\Dm}{\Dm \epsilon}\int_\Omega H\big(u(x)+\epsilon \varphi(x)\big)\Dm x \right|_{\epsilon=0} \\ & = -\int_\Omega H^\prime\big(u(x)+\epsilon\varphi(x)\big)\varphi(x)\Dm x \\ & = -\int_\Omega \varphi(x)\Dm \delta_{\partial\{u(x)>0\}} . \end{split}$$ where $$\delta_{\partial\{u>0\}}$$ is the Dirac measure supported on the boundary of subset of $$\Omega$$ where $$u>0$$. Here the derivative respect to the parameter $$\epsilon$$ as well as the gradient $$\nabla H(u)$$ (see below) must be intended in the sense of distributions.
Finally, let us approach the functional \ref{1}, or better its form considered by by Alt and Caffarelli (reference [1] of the question, p. 105): $$\begin{split} J(u)=J_1(u)+J_2(u) & = \int_{\Omega\color{red}{\cap\{u>0\}}} |\nabla u|^2 \Dm x+ \int_{\Omega\color{black}{\cap{\{u > 0\}}}} \Dm x \\ & = \int_\Omega \color{red}{\chi_{\{u>0\}}}|\nabla u|^2\Dm x + \int_\Omega \chi_{\{u>0\}}\Dm x \\ & = \int_\Omega \color{red}{H(u(x))}|\nabla u(x)|^2\Dm x + \int_\Omega H(u(x))\Dm x \end{split}$$ and applying the procedure developed above and the standard du Bois-Reymond's lemma we get $$\begin{split} \left.\frac{\Dm J}{\Dm \epsilon}(u+\epsilon \varphi)\right|_{\epsilon=0} & = \frac{\Dm}{\Dm \epsilon}\left[\int_\Omega {H(u+\epsilon \varphi)}|\nabla (u+\epsilon \varphi)|^2\Dm x + \int_\Omega H(u+\epsilon \varphi)\Dm x\right]_{\epsilon=0 }\\ & = \left[-\int_\Omega {H^\prime(u+\epsilon \varphi)}|\nabla (u+\epsilon \varphi)|^2\varphi\Dm x \right. \\ & \qquad +\left.2 \int_\Omega H(u+\epsilon \varphi)\big(\nabla u\cdot\nabla\varphi+\epsilon|\nabla\varphi|^2\big)\Dm x\right]_{\epsilon=0 } \\ &\qquad\qquad - \int_\Omega \varphi(x)\Dm \delta_{\partial\{u(x)>0\}} \\ & = -\int_\Omega |\nabla u(x)|^2\varphi(x)\Dm \delta_{\partial\{u(x)>0\}} \\ & \qquad-\int_\Omega \varphi(x)\Dm \delta_{\partial\{u(x)>0\}} \\ & \qquad\qquad + 2\int_\Omega H(u)\nabla u\cdot\nabla\varphi\Dm x \\ & = -\int_\Omega \big[|\nabla u(x)|^2+ 1\big] \varphi(x)\Dm \delta_{\partial\{u(x)>0\}} + 2\int_\Omega H(u)\nabla u\cdot\nabla\varphi\Dm x \\ & = -\int_\Omega \big[|\nabla u(x)|^2+ 1\big] \varphi(x)\Dm \delta_{\partial\{u(x)>0\}} - 2\int_\Omega \big[\Div \big(H(u)\nabla u\big)\big]\varphi\Dm x \\ & = -\int_\Omega \big[|\nabla u(x)|^2+ 1\big] \varphi(x)\Dm \delta_{\partial\{u(x)>0\}} \\ &\qquad - 2\int_\Omega \big(H(u)\Delta u\big)\varphi\Dm x \\ &\qquad\qquad - 2\int_\Omega \big(\nabla H(u)\cdot\nabla u\big)\varphi\Dm x \\ & = -\int_\Omega \big[|\nabla u(x)|^2+ 1\big] \varphi(x)\Dm \delta_{\partial\{u(x)>0\}} \\ &\qquad - 2\int_\Omega \big(H(u)\Delta u\big)\varphi\Dm x \\ &\qquad\qquad + 2\int_\Omega H^\prime(u)\lvert\nabla u\rvert^2\varphi\Dm x \\ & = -\int_\Omega \big[|\nabla u(x)|^2+ 1\big] \varphi(x)\Dm \delta_{\partial\{u(x)>0\}} \\ & \qquad +2 \int_\Omega |\nabla u(x)|^2\varphi(x)\Dm \delta_{\partial\{u(x)>0\}} \\ & \qquad\qquad - 2\int_\Omega \big(H(u)\Delta u\big)\varphi\Dm x\\ & = \int_\Omega \big[|\nabla u(x)|^2 - 1\big] \varphi(x)\Dm \delta_{\partial\{u(x)>0\}} - 2\int_{\Omega\cap\{u(x)>0\}} (\Delta u)\varphi\Dm x\\ \\ & \implies u=\min J \iff \begin{cases} \Delta u (x)= 0 & x\in \Omega\cap{\{u(x)>0\}} \\ \left.\begin{split} &u(x)=0 \\ &|\nabla u(x)| =1 \end{split}\right\} & x\in \Omega\cap{\partial\{u(x)>0\}} \\ \end{cases} \end{split}$$

Notes

• The result above coincides exactly with the one given by Alt and Caffarelli (reference [1] of the question, §0, p. 106 eq. (0.1)), when considering the functional $$f(x, u,\nabla u)= \lvert\nabla u\rvert^2 + Q^2$$ where $$Q$$ is a given function. In the case analysed here we have $$Q^2\equiv 1$$, but there absolutely no difference in the derivation respect to the general case.
• The only "modern" reference dealing with the work [1] of Alt and Caffarelli is the monograph [A1] written by their coAuthor Avner Friedman. Nevertheless he does not explicitly calculate the Euler-Lagrange equations of the functional involved as he use the direct method in the development of the analysis, as it is customary done for free boundary problems.

Reference

[A1] Avner Friedman, Variational principles and free-boundary problems, Pure and Applied Mathematics. A Wiley-Interscience Publication. New York: John Wiley & Sons, Inc. IX, 710 p. (1982). Zbl 0564.49002.

• Thank you for the answer and reference.
– BBB
Commented Jul 22 at 18:51
• @BBB you are welcome. I'm glad to have been of some help. I'm adding some more detail to it Commented Jul 22 at 18:54