Let $tau>0$, $dinmathbb N$ and $T_t$ be a $C^1$-diffeomorphism on $mathbb R^d$ for $tin(0,tau)$ with $T_0=operatorname{id}_{mathbb R^d}$. Assume $$T_t(x)=x+int_0^tv(s,T_s(x)):{rm d}s;;;text{for all }(t,x)in(0,tau)tag1$$ for some sufficiently nice $v:(0,tau)timesmathbb R^dtomathbb R^d$.

Let $mathcal M$ denote the set of embedded $C^1$-submanifolds of $mathbb R^d$ with boundary and $z:{partialOmega:Omegainmathcal M}tomathbb R$. We say that $z$ has a *material derivative* at $partialOmega$ in direction $v$ if $${rm d}y(Omega;v)(x):=left.frac{rm d}{{rm d}t}yleft(T_t(partialOmega)right)(T_t(x))right|_{t=0}tag2$$ exists for all $xinpartialOmega$.

If $M$ is an embedded $C^1$-submanifold of $mathbb R^d$ with or without boundary, let $operatorname P_M(x)$ denote the orthogonal projection of $mathbb R^d$ onto the tangent space $T_x:M$ for $xin M$ and if $f:Mtomathbb R$ is $C^1$-differentiable at $xin M$ define $${rm D}_Mf(x):=T_x(f)circoperatorname P_M(x)tag3.$$

Now assume $$z(partialOmega)=left.fright|_{partialOmega}tag4$$ for some $fin C^1(mathbb R^d)$. Why does it follow that $${rm d}z(partialOmega;v)(x)={rm D}_{partial M}f(x)v(0,x)tag5$$ for all $xinpartial M$?

Since, by assumption, $$yleft(T_t(partialOmega)right)(T_t(x))=f(T_t(x));;;text{for all }xinpartialOmegatag6,$$ shouldn’t we get $${rm d}z(partialOmega;v)(x)={rm D}f(x)v(0,x)tag7$$ for all $xinpartial M$?

The claim $(5)$ is made in Lemma 4.3.8-ii-II in this paper.