Is a segment with density a stationary varifold?

Question

It easily seen that a segment with density 1 is a stationary point of the lenght functional. What about if we have a non costant density? My intuition says that it is, but: If $V$ is the varifold in $\mathbb{R}^2$ induced by the segment $[0,1]×{0}$ with density $f(x)$, if $X=(X_1,X_2)$ is a vector field, by the first variation formula we have $$ \delta V (X) = \int_0^1 div_{e_1} X(x,0) f(x) dx. $$ Integrating by parts, we obtain $$ \delta V (X) = X_1(1,0)f(1) - X(0,0)f(0) - \int_0^1 X_1(x,0) f'(x) dx $$ The presence of the integral term seems say that this varifold is not stationary. What is the truth?

Answer 1

First the rectifiable varifold $V$ associated to $M=(0,1)\times\{0\}\subset\mathbb{R^2}$ is stationary only when seen as varifold in an open domain $\Omega$ such that $(0,0),\,(1,0)\, \in \partial\Omega$ and admissible variations are made via vector-fields with compact support in $\Omega$. In fact in this case $$ \delta V(X, \Omega) = 0,\qquad\forall \, X \in C^1_c(\Omega, \mathbb{R}^2). $$ However when $(0,0),\,(1,0)\in \Omega$ we have $\delta V(\Omega,X) \neq 0$ for every $X=(X_1,X_2)\in\,C^1_c(\Omega,\mathbb{R}^2)$ such that $X_1(0)-X_1(1) \neq 0$. To get an intuition of this think that the vector-field $X$ represents an infinitesimal deformation field, that is it is associated with the derivative of the "length"/"mass" along the direction $$ M_t = \{x+\, tX(x)\, + o(t):\, x\in M\}. $$ In one case there is no-way to get $M_t$ with length smaller than $M$ in the other one can consider $M_t :=(t,1)\times\{0\}$ which can generated by a vector field $X=(X_1,0)$ where $X_1\equiv 1$ around $(0,0)$.

Similarly when $V$ is the rectifiable varifold concentrated on the segment $M$ and has a non-constant density $\theta:M\to\mathbb{R}$, the varifold $V$ is not stationary in $\Omega$ even in the case $(0,0),\,(1,0)\, \in \partial\Omega$. The point is that "infitesimal tangential changes" do not vary the legth/mass of $V$ when the density is constant, but this is not the case when $\theta$ is not constant