Prove/disprove convexity of surface volume in terms of a deformation field

133 Views Asked by At

In the setting of convex optimization and mechanics, I am interested in the convexity of the potential energy $U$ of a pressure acting over some volume $V$ enclosed by a surface. Here pressure can be understood as the potential energy per unit volume, hence $U$ is given by $U=-\int_V p dV$. Assuming $p$ is constant, $U=-pV$, or if $p$ varies according to a law $p = C/V$, where $C$ is a constant, $U=-ClogV$. I would like to know if $U(V)$ of a deforming surface is convex in the deformation field $u$, defined below, or if it isn't, if convexity can be imposed with some restrictions.

The problem data:

A point $X$ in the reference configuration of a surface $S$ is given by a parametrization of its curvilinear coordinates:

$X=\phi_X(\xi^1, \xi^2), \quad (\xi^1, \xi^2) \in \mathbb{R}^2, \quad X \in \mathbb{R}^3$.

A covariant tangent basis is defined as $G_\alpha = \partial X /\partial \xi^\alpha $, thus the non-unit normal $G_3$ to $S$ is given by $G_3=G_1 \times G_2$ and the unit normal is $N =G_3/ |G_3|$. $S$ is described by the metric tensor with covariant components $G_{\alpha\beta} = G_\alpha \cdot G_\beta $.

$S$ is subject to some deformation, described by a parametrized displacement field $u(\xi^1,\xi^2)\in \mathbb{R}^3$, such that $x=X+u$. For simplicity, we assume that the boundary of $S$ is fixed, such that $u=0 \forall X \in \partial S.$ The deformed surface $s$ has analogous properties to $S$, that is the covariant basis is $g_\alpha = \partial x /\partial \xi^\alpha = G_\alpha + \partial u/\partial \xi^\alpha$, the non-unit normal $g_3$ and unit normal $n$.

Using Gauss theorem, we define the volume $V$ of $s$ by integrating a function of unit divergence: $V=\frac{1}{3}\int{x\cdot n} da$ , where the measure $da=\sqrt{det g_{\alpha\beta}} d\xi^1 d\xi^2 = |g_3| d\xi^1 d\xi^2 $. So

$ V = \frac{1}{3}\int x\cdot g_3 d\xi^1d\xi^2 = \frac{1}{3}\int (X+u)\cdot (g_1 \times g_2) d\xi^1d\xi^2$

Hence $V$ is the integral of some triple product, which doesn't immediately suggest convexity. However, perhaps there may be some hidden convexity since only displacements in the direction of $n$ affect V (note that the directional derivative of $V$ in a test direction $\delta v$ is $ DV(u)[\delta v] = \int n\cdot \delta v d\xi^1 d\xi^2$).

As a simple example, consider a sphere centered at the origin with radius r, and V = $4/3 \pi r^3$. Now $DV/dr = 4 \pi r^2$ which is obviously convex. The variation of V is $4/3 \pi (r+\delta r)^3$ gives a cubic function, which is locally convex for $V\geq$ or $ (r+\delta) r\geq 0$. Here a parametrization can be given by $X=[sin(\xi^1)cos(\xi^2), sin(\xi^1)sin(\xi^2), cos(\xi^1) ] , \quad \xi \in [0,\pi/2]\times[0,2\pi]$ and $G_3 = [sin^2(\xi^1)cos(\xi^2), sin^2(\xi^1)sin(\xi^2), sin(\xi^1)cos(\xi^1)]$. So perhaps one could show that $x\cdot g_3 $ is locally convex and then generalize this?

We can assume that the origin is inside the surface. Therefore, the integral V represents the sum of the volumes of pyramids of height $x$ and base $(g_1 \times g_2) d\xi^1d\xi^2$