In some articles that I read, I often encountered the same formulation for one-dimensional conservation laws of the form $$u_t+(F(u)u)_x=0$$ where $F(u)$ may also depend on $u_x$, etc., and $u(t=0,x)=u_0(x)$ is given. The unknown function is $u : \mathbb{R}_t \times \mathbb{R}_x \to \mathbb R$.
Let $X(t, \cdot) : \mathbb{R} \to \mathbb{R}$ defined by $X_t=F(t,u(t,X(t,x)))$, $X(0,x)=x$, which is the Lagrangian map of the field $F$, then solutions of the form $$u(t,X(t,x))=\frac{1}{X_x(t,x)}u_0(x)$$ may be sought.
Question: Is there an explanation of the last expression? Is it natural?
Because usually we want to write $u(t,x)=u_0(X(t,x))$ and then look for the equation satisfied by $X$. Here I find the expression for $u$ somehow unnatural and unexpected. The previous equations and notations arise in the article (1), §3.1.1 (p. 12 of the arXiv preprint).
(1) P. Germain, B. Harrop-Griffiths, J.L. Marzuola, "Existence and uniqueness of solutions for a quasilinear KdV equation with degenerate dispersion", Comm. Pure Appl. Math., 2019. doi:10.1002/cpa.21828 arXiv:1801.00420v1
If $\frac{\text d}{\text dt}(X_xu)$ would be zero, then we would have $X_xu=u_0(x)$. With the present notations, we find $$ \begin{aligned} \frac{\text d}{\text dt}(X_x u) &= X_{xt}u+X_x(u_t+u_XX_t) \\ &= b_xu+X_x(−(bu)_x+u_Xb)\\ &=X_x(1−X_x)(bu)_X \end{aligned} $$ which is nonzero in general. However, considering that $(bu)_x$ represents differentiation w.r.t. the second variable of $[bu](t,X(t,x))$, then we may rewrite the conservation law as $ u_t + (bu)_X = 0 $. Doing so, $\frac{\text d}{\text dt}(X_xu)$ vanishes, and the proposed solution is obtained.