How to find $v = \frac{u_t}{1-u_x}$ and $u_x = \frac{U_X}{1+U_X}$

Question

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...

Answer 1

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 $u_x = \frac{U_X}{1+U_X}$.