Consider a family of smooth, boundary regions $U(\tau)\subset
\mathbb{R}^n$ that depend smoothly upon the parameter
$\tau\in\mathbb{R}$. Write $v$ for the velocity of the moving boundary
$dU(\tau)$ and $\nu $ for the outward pointing unit normal.If $f=f(x,\tau)$ is a smooth function, then
$\frac{d}{d\tau}\int_{U(\tau)} fdx=\int_{\partial U(\tau)}fv\cdot\nu
dS+\int_{U(\tau)}f_\tau dx$.
How is "velocity of the moving boundary" defined? Plus, can you explain the intuition of the formula in words ? thanks!
Best Answer
I interpret the velocity of the boundary here to be a vector field. Pick a point $x_0\in U(\tau_0)$, then consider the quantity $$\varphi(\tau)=\frac{U(\tau+\tau_0)-U(\tau_0)}{\tau}$$ $\varphi$ is smooth by assumption so we can take a limit and interpret this as a vector field defined on the boundary of $U(\tau)$. It carries with it information about how the point $x_0$ will evolve with $\tau$. As for the formula, I am not entirely sure what he means by $dU(\tau)$, I think it's supposed to be the manifolds map between tangent spaces. You can think of $U$ as a map between manifolds I suppose. Anyway, I think the intuition is that two contributions will cause a change in the value of the integral: the changes in $U$ and the changes in $f$. The first integral is integrating the flux of the vector field $f\vec{v}$ where $\vec{v}$ is the velocity of $U$ like we discussed. Hope this helps.