TITLE: How to find $v = \frac{u_t}{1-u_x}$ and $u_x = \frac{U_X}{1+U_X}$
QUESTION [1 upvotes]: I know that a material coordinate is denoted by $X$ and a spatial by $x = φ(X, t)$. With any function $f(x, t)$ we can associate a function $F(X, t)$:
$f(x, t) = f(φ(X, t), t) = F(X, t)$
I also know that in spatial coordinates the displacement is denoted by $u(x, t)$ and the velocity is
denoted $v(x, t)$.
Now I am kinda struggling how to find the following two formulas, because
I am not sure what I am missing to solve these formulas.
$v = \frac{u_t}{1-u_x}$ and $u_x = \frac{U_X}{1+U_X}$
Any help/tip/example would be grateful. Thanks in advance.
I was thinking of
$u(x,t) = u(φ (X,t), t) = U(X,t)$
$\frac{d}{dt} u(φ (X,t), t) = \frac{\partial u}{\partial t}(x,t) + u(x,t)$
But I do not know how to move on...

REPLY [1 votes]: Good start. This is a classical derivation in continuum mechanics, where we define coordinates in the reference configuration $(X,t)$ and in the deformed configuration $(x,t)$. The definition of velocity gives $v = \varphi_t$. With $x=\varphi(X,t)$, we have
$$
\partial_t U(X,t) = U_t = u_xv + u_t ,\qquad \partial_X U(X,t) = U_X = u_x\varphi_X \, .
$$
The definition of displacement gives $U = x-X$. Thus, we also have
$$
\partial_t U(X,t) = v , \qquad \partial_X U(X,t) = \varphi_X - 1 \, .
$$
Therefore, $v = \frac{u_t}{1-u_x}$ and $u_x = \frac{U_X}{1+U_X}$.
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