Plateau problem for fluxes of curves

Question

Consider $\mathbb{R}_t \times \mathbb{R}_x ^n$ , let $b_1(t,x)$ and $b_2 (t,x)$ be two velocity fields with all the regularity you want and consider the flow of the point $(0,x_0)$ for a time $T$. Now, I can view the trajectories of $(0,x_0)$ as curves in $\mathbb{R}^{n+1}$ like $ \gamma_1 =(t, \Phi_1 (t,x_0))$ and $\gamma _2= (t, \Phi_2 (t,x_0))$ with $\Phi_i$ the flux relative to $b_i$. The question is: if I consider $\tilde{\gamma} =\gamma_1 \cup \gamma_2 \cup I_T$ where $I_T$ is a segment in $\mathbb{R}^{n+1}$ joining $(T,\Phi_1 (T, x_0) )$ and $(T, \Phi_2 (T, x_0))$ , I can consider the Plateau problem associated to the rectifiable set $\tilde{\gamma}$, and by the general theory I'll have a minimal surface whose boundary is $\tilde{\gamma}$. The question is: given how my $\gamma$ is constructed, is there a way to estimate the area of such surface in terms of $b_1$ and $b_2$?

EDIT: Actually, what I want is to integrate such an estimate, starting from a certain mass distribution at time 0, so I would like something whixh takes that into account.

Answer 1

Almgren (https://mathscinet.ams.org/mathscinet-getitem?mr=855173) proves that the Plateau solution will satisfy the 2-dimensional Euclidean isoperimetric inequality. So you can bound $$ \textrm{Area(Plateau Solution)} \leq (4\pi)^{-1}(L(\gamma_1)+L(\gamma_2)+L(I_T))^2. $$ If you just want a coarse estimate you should be able to bound each term by the $L^\infty$ norm of $b_1,b_2$.

Of course, this inequality is unlikely to be close to optimal. (For example, if $b_1=b_2$ then the area of the Plateau solution is $=0$ while the RHS above is nonzero). If you want an estimate that works better in this case you might construct the explicit competitor $$ \Gamma : = \bigcup_{t=0}^T I_{\Phi_1(t,x_0),\Phi_2(t,x_1)} \times \{t\} $$ namely you union up all the lines between the points in the flow. You should be able to compute the area of this in terms of $b_1,b_2$ using the co-area formula.

EDIT: As requested here is a formula for the area of $\Gamma$. I did it via a parametrization instead of co-area but I think co-area would also work.

I parametrize $\Gamma$ by $F: [0,1]\times [0,T]\to\mathbb{R}^{n+1}$ $$ F(s,t) : = (s\Phi_1(t)+(1-s)\Phi_2(t),t) $$ (dropping the $x_0$ notation). Then we have $$ \partial_s F = \Phi_1-\Phi_2, \qquad \partial_t F = s\partial_t\Phi_1 +(1-s)\partial_t \Phi_2 + \mathbf{e}_{n+1}. $$ In particular, the induced metric is \begin{align*} g_{ss} & = |\Phi_1-\Phi_2|^2 \\ g_{st} & = \langle \Phi_1-\Phi_2,s\partial_t\Phi_1 +(1-s)\partial_t \Phi_2\rangle\\ g_{tt} & = 1 + |s\partial_t\Phi_1 +(1-s)\partial_t \Phi_2|^2. \end{align*} Hence $$ \det g = (1 + |s\partial_t\Phi_1 +(1-s)\partial_t \Phi_2|^2)|\Phi_1-\Phi_2|^2 - \langle \Phi_1-\Phi_2,s\partial_t\Phi_1 +(1-s)\partial_t \Phi_2\rangle^2. $$ This gives the explicit formula $$ \textrm{area}(\Gamma) = \int_{[0,1]\times [0,T]} \sqrt{(1 + |s\partial_t\Phi_1 +(1-s)\partial_t \Phi_2|^2)|\Phi_1-\Phi_2|^2 - \langle \Phi_1-\Phi_2,s\partial_t\Phi_1 +(1-s)\partial_t \Phi_2\rangle^2}. $$ If you want to write this in terms of $b_1$ and $b_2$ you can do the following (perhaps one can be more precise here to get a better estimate). First of all $$ |\partial_t \Phi_i| = |b_i| \leq \Vert b_i\Vert_{C^0} $$ Secondly, by the argument here we can bound $$ |\Phi_1-\Phi_2| \leq L^{-1}(e^{Lt}-1)|b_1-b_2|_{L^\infty}. $$ for $L$ the Lipschitz constant of one of the vector fields.

Using this above we should get $$ \textrm{area}(\Gamma) \leq (1 + (|b_1|_{C^0}+|b_2|_{C^0}) |b_1-b_2|_{C^0}^2 L^{-2}(e^{LT}-LT-1). $$

Note that I discarded the $g_{st}$ term, which could improve this estimate if you had some way of knowing it was big. But it does have the nice feature that it vanishes if $b_1=b_2$ and it's $O(T^2)$ for $T$ small.

I'm sure you can improve the estimate used above in the case of large $T$ (e.g. if $b_i$ are bounded, then the flows should diverge at most linearly).