Shape derivative

Remarks

For $\norm{\theta}_W < 1$ the map $\mathrm{Id} + \theta$ is a bi-Lipschitz homeomorphism of $\R^d$, so $(\mathrm{Id} + \theta)(\Omega)$ is again open. Example: the volume $J(\Omega) = \abs{\Omega}$ of a bounded open set. By the change of variables formula $J((\mathrm{Id} + \theta)(\Omega)) = \int_\Omega \det(I + \nabla \theta) \, dx$. Since $\det(I + \nabla \theta) = 1 + \operatorname{div} \theta + O(\norm{\nabla\theta}^2)$, the shape derivative is $J'(\Omega)(\theta) = \int_\Omega \operatorname{div} \theta \, dx$. If $\Omega$ has a $C^1$ boundary with outward unit normal vector $n$, the divergence theorem gives $J'(\Omega)(\theta) = \int_{\partial \Omega} \theta \cdot n \, ds$. Non-example: $J(\Omega) = \abs{\abs{\Omega} - 1}$ at a set $\Omega$ with $\abs{\Omega} = 1$ is not shape differentiable. Take $\Omega$ bounded and a smooth function $\chi$ with compact support that equals $1$ on a neighbourhood of $\overline{\Omega}$. Along $\theta_s(x) = s \chi(x) x$ one gets $(\mathrm{Id} + \theta_s)(\Omega) = (1 + s) \Omega$ and $J = \abs{(1 + s)^d - 1}$, which has a corner at $s = 0$. For smooth $\Omega$ and smooth $J$, the Hadamard structure theorem gives $J'(\Omega)(\theta) = \int_{\partial \Omega} g \, (\theta \cdot n) \, ds$ for a function $g$ on $\partial \Omega$, the shape gradient.