Suppose for each $t$, $S(t) \subset \mathbb{R}^n$ is a domain (hypersurface). We have a diffeomorphism $D^0_t:S(0) \to S(t)$ for each $t$ such that it solves the ODE $$\frac{d}{dt}D^0_t(\cdot) = V(D^0_t(\cdot),t)$$ $$D^0_0(\cdot) = \text{Identity}(\cdot)$$ where $V$ is discontinuous wrt. $t$. We also know that $D^0_t$ is $C^2$ in space.
Define the map $F_{-t}:H^1(S(t)) \to H^1(S(0))$ by $$(F_{-t} u)(x) = u(D^0_t(x)),$$ and define $F_t$ similarly. So $F_{-t}$ takes a function defined on $S(t)$ and turns into a function defined on $S(0)$, and $F_t$ does the opposite.
Suppose that for each $t$, $u(t):S(t) \to \mathbb{R}$ is a function defined on $S(t)$. Then $$(F_{-t}u(t))(\cdot) = u(t, D^0_t(\cdot))$$ and so $$\frac{d}{dt}(F_{-t}u(t))(\cdot) = \frac{d}{dt}u(t, D^0_t(\cdot)) = u_t|_{(t, D^0_t(\cdot))} + \nabla u|_{(t, D^0_t(\cdot))}\cdot V(D_t^0(\cdot),t).\tag{1}$$ Now suppose that $v \in C^1([0,T];H^1(S_0))$, so $$v(t):S(0) \to \mathbb{R}$$ for all $t$. Consider $u(t) := F_t v(t)$ for all $t$. Clearly $u(t):S(t) \to \mathbb{R}$. In this case then, $$\frac{d}{dt}(F_{-t}u(t))(\cdot) = \frac{d}{dt}(F_{-t}F_tv(t))(\cdot) = \frac{d}{dt}(v(t))(\cdot)$$ which is continuous wrt $t$ by definition. But this expression is also equal to (1), the right hand side of which is NOT continuous wrt. $t$ because of the $V$ present which is not continuous. So what is going on? Isn't this contradictory?
(Crossposted from https://math.stackexchange.com/questions/486343/a-contradiction-to-do-with-continuity-involves-chain-rule)
