What is the first variation of an indicator function?

Question

I read in a paper which wants to find the minimizer to the following energy function $$J(u) = \int_{\Omega} |\nabla u|^2 + \chi(\{u>0\})Q^2 dx $$ where $Q$ is a measurable function and $\chi$ is the indicator function.

I also read in another paper which wants to find the minimizer to $$J(u) = \int_{\Omega} |\nabla u|^2 + \mathcal L(\{u>0\}) dx $$ where $\mathcal L$ is the Lebesgue measure.

How can I compute the first variation of these functionals? I don't know how to deal with $\chi, \mathcal L$.

Answer 1

The term $\mathcal L(\{u>0\})$ (or its weighted analog $\int_{u>0} Q^2$) is not even continuous with respect to $u$: consider what happens for constant $u$. But one can write down some sort of variation when the gradient of $u$ does not vanish "too much".

Consider a point $x_0$ with $u(x_0)=0$ and $\nabla u(x_0)\ne 0$. Locally the function $u$ looks like $u(x) \approx (x-x_0)\cdot \nabla u(x_0)$. Within a ball of radius $r$, this linear function is positive on the set of measure $r^n|B^n|$ where $|B^n|$ is the measure of the unit $n$-dimensional ball. Adding a constant $c>0$ to the linear function increases the measure by about $c|\nabla u(x_0)|^{-1} r^{n-1} |B^{n-1}|$, because this increase comes from a cylinder of height $c|\nabla u(x_0)|^{-1}$.

From these local considerations, recognizing $r^{n-1} |B^{n-1}|$ as surface element of the (smooth, by the implicit function theorem) surface $\{u=0\}$, one concludes that the result of replacing $u$ by $u+\epsilon \phi$ is $$ \epsilon \int_{\{u=0\}} |\nabla u(x)|^{-1}\phi(x)\,dx $$ which means that the first variation of the functional $\mathcal L(\{u>0\})$ is $$ \phi \mapsto \int_{\{u=0\}} |\nabla u(x)|^{-1}\phi(x)\,dx $$ For the weighted case it is $$ \phi \mapsto \int_{\{u=0\}} |\nabla u(x)|^{-1}\phi(x)Q^2(x)\,dx $$

Answer 2

Agree with @bro for the discontinuity issue. However, it might be much clearer if one writes down the governing equation for $u$ when $J[u]$ is minimized, instead of simply going for the expression of the first variation. In fact, the discontinuity of $\chi$ on $\Omega$ would lead to a jump condition for $\nabla u$.

Notations

Suppose the original functional $$ J[u]=\int_{\Omega}\left(\left\|\nabla u\right\|^2+Q^2\cdot\mathbb{1}_{\left\{u>0\right\}}\right){\rm d}V $$ is equipped with a boundary condition $$ u|_{\partial\Omega}=g, $$ where $\Omega\subseteq\mathbb{R}^n$, and ${\rm d}V={\rm d}x_1{\rm d}x_2\cdots{\rm d}x_n$. Suppose, in a more general case, that $Q$ is a function of $\mathbf{x}\in\Omega$ and $u$, i.e., $Q=Q(\mathbf{x},u)$.

Now, consider a family of feasible functions $$ v:\Omega\times\mathbb{R}\to\mathbb{R},\quad\left(\mathbf{x},t\right)\mapsto v(\mathbf{x},t) $$ with $$ v|_{\partial\Omega}=g, $$ where $u=v(\mathbf{x},0)$ gives the minima of the functional $J[v]$, i.e., $$ \left(\frac{\rm d}{{\rm d}t}J[v(\cdot,t)]\right)\Bigg|_{t=0}=0. $$ Here $t$ is an auxiliary parameter, somehow playing the role of "time": as $t$ changes, the set $$ A_t:=\Omega\cap\left\{v(\cdot,t)>0\right\} $$ changes as well, making its boundary $\partial A_t$ deform continuously.

Finally, provided that $\nabla v(\cdot,0)$ may not be well defined on $\partial A_0$, let us define \begin{align} w(\cdot,t)&=v(\cdot,t)|_{A_t},\\ \omega(\cdot,t)&=v(\cdot,t)|_{B_t} \end{align} for further clarification, where $B_t=\Omega\setminus A_t$.

First variation: Formulism

With the notations above, our target functional reads $$ J[v]=\int_{B_t}\left\|\nabla\omega\right\|^2{\rm d}V+\int_{A_t}\left(\left\|\nabla w\right\|^2+Q^2(\cdot,w)\right){\rm d}V. $$ The first variation of $J$ is no more than $$ \frac{\rm d}{{\rm d}t}J[v(\cdot,t)]. $$ Note that in $J$, not only functions $\omega$ and $w$ depend on $t$, but also integral domains $A_t$ and $B_t$. Thus to take the derivative with respect to $t$, one must use Leibniz integral rule. With this rule, we have \begin{align} \frac{\rm d}{{\rm d}t}\int_{B_t}\left\|\nabla\omega\right\|^2{\rm d}V&=\int_{B_t}\frac{\partial}{\partial t}\left\|\nabla\omega\right\|^2{\rm d}V+\int_{\partial B_t}\left\|\nabla\omega\right\|^2\mathbf{w}\cdot{\rm d}\mathbf{S}\\ &=2\int_{B_t}\nabla\omega\cdot\nabla\dot{\omega}{\rm d}V+\int_{\partial B_t}\left\|\nabla\omega\right\|^2\mathbf{w}\cdot{\rm d}\mathbf{S}\\ &=2\int_{B_t}\left[\nabla\cdot\left(\dot{\omega}\nabla\omega\right)-\dot{\omega}\Delta w\right]{\rm d}V+\int_{\partial B_t}\left\|\nabla\omega\right\|^2\mathbf{w}\cdot{\rm d}\mathbf{S}\\ &=-2\int_{B_t}\dot{\omega}\Delta\omega{\rm d}V+2\int_{\partial B_t}\dot{\omega}\nabla\omega\cdot{\rm d}\mathbf{S}+\int_{\partial B_t}\left\|\nabla\omega\right\|^2\mathbf{w}\cdot{\rm d}\mathbf{S}, \end{align} where $\dot{\omega}$ stands for $\partial\omega/\partial t$, while $\mathbf{w}$ denotes the normal "velocity" of the moving part of $\partial B_t$ with respect to "time" $t$. We will come back to this $\mathbf{w}$ in the next section.

Similarly, we have \begin{align} &\frac{\rm d}{{\rm d}t}\int_{A_t}\left(\left\|\nabla w\right\|^2+Q^2(\cdot,w)\right){\rm d}V\\ &=\int_{A_t}\frac{\partial}{\partial t}\left(\left\|\nabla w\right\|^2+Q^2(\cdot,w)\right){\rm d}V+\int_{\partial A_t}\left(\left\|\nabla w\right\|^2+Q^2(\cdot,w)\right)\mathbf{w}\cdot{\rm d}\mathbf{S}\\ &=2\int_{A_t}\left(\nabla w\cdot\nabla\dot{w}+Q\frac{\partial Q}{\partial w}\dot{w}\right){\rm d}V+\int_{\partial A_t}\left(\left\|\nabla w\right\|^2+Q^2(\cdot,w)\right)\mathbf{w}\cdot{\rm d}\mathbf{S}\\ &=2\int_{A_t}\left(-\Delta w+Q\frac{\partial Q}{\partial w}\right)\dot{w}{\rm d}V+2\int_{\partial A_t}\dot{w}\nabla w\cdot{\rm d}\mathbf{S}+\int_{\partial A_t}\left(\left\|\nabla w\right\|^2+Q^2(\cdot,w)\right)\mathbf{w}\cdot{\rm d}\mathbf{S}. \end{align}

Depiction for $\mathbf{w}$

To figure out a depiction for the normal velocity of $\partial A_t$ (which is also the moving part of $\partial B_t$), $\mathbf{w}$, consider a particle $\mathbf{y}=\mathbf{y}(t)$ that tracks $\left\{w=0\right\}$, i.e., $\mathbf{y}(t)$ is such that $$ w(\mathbf{y}(t),t)=0. $$ This equality leads to $$ 0=\frac{\rm d}{{\rm d}t}w(\mathbf{y}(t),t)=\dot{\mathbf{y}}(t)\cdot\nabla w(\mathbf{y}(t),t)+\dot{w}(\mathbf{y}(t),t). $$ Suppose, in addition, that $\mathbf{y}(t)$ moves perpendicular to $\partial A_t$, for which $\dot{\mathbf{y}}(t)=\mathbf{w}(t)$. Using this assumption, the last equality reduces to $$ \dot{w}(\mathbf{y}(t),t)+\mathbf{w}(t)\cdot\nabla w(\mathbf{y}(t),t)=0. $$ Provided the arbitrariness of $\mathbf{y}(t)\in\partial A_t$, one eventually obtains $$ \dot{w}+\mathbf{w}\cdot\nabla w=0 $$ holds on $\partial A_t$. Similarly, $$ \dot{\omega}+\mathbf{w}\cdot\nabla\omega=0 $$ holds on the moving part of $\partial B_t$.

Surface integral of vector fields: Simplification

Let $\mathbf{n}$ be the outward unit normal of $A_t$.

Note that \begin{align} A_t&=\left\{w>0\right\},\\ \partial A_t&=\left\{w=0\right\}. \end{align} Thus $$ \mathbf{n}=-\frac{\nabla w}{\left\|\nabla w\right\|}. $$ Consequently, $$ \mathbf{w}\cdot{\rm d}\mathbf{S}_{\partial A_t}=\left(\mathbf{w}\cdot\mathbf{n}\right){\rm d}S_{\partial A_t}=-\left(\mathbf{w}\cdot\frac{\nabla w}{\left\|\nabla w\right\|}\right){\rm d}S_{\partial A_t}=\frac{\dot{w}}{\left\|\nabla w\right\|}{\rm d}S_{\partial A_t}. $$

Similarly, $$ \mathbf{n}=-\frac{\nabla\omega}{\left\|\nabla\omega\right\|}. $$ Yet since $\mathbf{n}$ is the outward unit normal of $A_t$, it is the inward unit normal of $B_t$. Consequently, $$ \mathbf{w}\cdot{\rm d}\mathbf{S}_{\partial B_t}=-\left(\mathbf{w}\cdot\mathbf{n}\right){\rm d}S_{\partial B_t}=\left(\mathbf{w}\cdot\frac{\nabla\omega}{\left\|\nabla\omega\right\|}\right){\rm d}S_{\partial B_t}=-\frac{\dot{\omega}}{\left\|\nabla\omega\right\|}{\rm d}S_{\partial B_t}. $$

Likewise, \begin{align} \nabla w\cdot{\rm d}\mathbf{S}_{\partial A_t}&=\left(\nabla w\cdot\mathbf{n}\right){\rm d}S_{\partial A_t},\\ \nabla\omega\cdot{\rm d}\mathbf{S}_{\partial B_t}&=-\left(\nabla\omega\cdot\mathbf{n}\right){\rm d}S_{\partial B_t}. \end{align}

First variation: Results

With all the results in the surface integral of vector fields, the first variation reads \begin{align} \frac{\rm d}{{\rm d}t}J[v]&=2\int_{A_t}\left(-\Delta w+Q\frac{\partial Q}{\partial w}\right)\dot{w}{\rm d}V+2\int_{B_t}\left(-\Delta\omega\right)\dot{\omega}{\rm d}V+\\ &\quad\quad\int_{\partial A_t}\left(\left\|\nabla w\right\|+\frac{Q^2}{\left\|\nabla w\right\|}+2\mathbf{n}\cdot\nabla w-\left\|\nabla\omega\right\|-2\mathbf{n}\cdot\nabla\omega\right)\dot{v}{\rm d}S. \end{align} Provided that \begin{align} \mathbf{n}\cdot\nabla w&=-\frac{\nabla w}{\left\|\nabla w\right\|}\cdot\nabla w =-\left\|\nabla w\right\|,\\ \mathbf{n}\cdot\nabla\omega&=-\frac{\nabla\omega}{\left\|\nabla\omega\right\|}\cdot\nabla\omega=-\left\|\nabla\omega\right\|, \end{align} the result is equivalent to \begin{align} \frac{\rm d}{{\rm d}t}J[v]&=2\int_{A_t}\left(-\Delta w+Q\frac{\partial Q}{\partial w}\right)\dot{w}{\rm d}V+2\int_{B_t}\left(-\Delta\omega\right)\dot{\omega}{\rm d}V+\\ &\quad\quad\int_{\partial A_t}\mathbf{n}\cdot\left[\left(1-\frac{Q^2}{\left\|\nabla w\right\|^2}\right)\nabla w-\nabla\omega\right]\dot{v}{\rm d}S. \end{align} Finally, due to the arbitrariness of $\dot{v}(\cdot,0)$, the governing equations are \begin{align} -\Delta w+Q\frac{\partial Q}{\partial w}&=0,&\text{in }A_0,\\ -\Delta\omega&=0,&\text{in }B_0,\\ \mathbf{n}\cdot\left[\left(1-\frac{Q^2}{\left\|\nabla w\right\|^2}\right)\nabla w-\nabla\omega\right]&=0,&\text{on }\partial A_0, \end{align} together with the continuity of $v(\cdot,0)$, i.e., \begin{align} w-\omega&=0,&\text{on }\partial A_0, \end{align} as well as the boundary condition for $v(\cdot,0)$, i.e., \begin{align} v(\cdot,0)&=g,&\text{on }\partial\Omega. \end{align}

As per the above governing equations, one can see that the characteristic-function $\chi$ in the functional $J$ eventually contributes to the jump condition for $\nabla v(\cdot,0)$ on the level set $\partial A_0$.