I am reading the original paper (Ref.1) by Hunter and Saxton on solving the now famous PDE for $u(x,t)$ using method of characteristics: $$ \left(u_{t}+uu_{x}\right)_x=\tfrac12(u_x)^2,\qquad u(x,0)=F(x)\tag{1}$$
Their steps go like this: Suppose that $u(x, t)$ is a smooth solution of Eq.(1). We introduce a characteristic coordinate $\xi$, where $$x=X(\xi,t),\qquad U(\xi, t)=u[X(\xi,t),t],\tag{2}$$ and $X(\xi, t)$ satisfies $$X_t=U,\qquad X(\xi,0)=\xi\tag{3}$$
We also define $$V(\xi, t) =X_\xi(\xi, t)\tag{4}$$
It then follows from Eq.(1) that $V$ satisfies $$VV_{tt}=\tfrac12 (V_t)^2,\quad V(\xi, 0)=1,\quad V_t(\xi,0)=F'(\xi)\tag{5}.$$.
Questions (1) Where are the steps that I am missing in getting Eq.(5)?
Define $\omega=u_x$, then $\omega(x,t)$ satisfies $$ \omega_{t}+u\omega_{x}+\tfrac12\omega^2=0\tag{6}$$
Questions (2) Are they obtaining Eq.(6) from Eq.(5) by using method of characteristics?
Thanks a lot for the help!
Ref.1: J. K. HUNTER AND R. SAXTON, DYNAMICS OF DIRECTOR FIELDS, SIAM J. APPL. MATH. Vol. 51, No. 6, pp. 1498-1521, December 1991
I found a reference paper by Alejandro Sarria (arXiv:1307.4504) titled "Global estimates and blow-up criteria for the generalized Hunter-Saxton system". The 2-component generalization Hunter-Saxton equation was solved using method of characteristics.
I will just copy a few steps here in order to close this open problem.
First we rewrite (1) as: $$u_{xt}+u u_{xx}=(-\tfrac12)(u_{x})^2\tag{1B}.$$
From (2) and (3) above (in the question area), we obtain $$X_{t\xi}=U_\xi=u_x\,X_\xi\tag{2B}.$$ Replacing $X_\xi$ with $V$ (cf.(4) ) leads to $$V^{-1}V_{t}=u_x\,\tag{3B}.$$
On the other hand, the left hand side of (1B) is just: $$\frac{\mathrm{D}}{\mathrm{D}t}u_x[X(\xi,t),t]=\left(\frac{\partial}{\partial t}+X_t\frac{\partial}{\partial X}\right)u_x.\qquad \because X_t=U\tag{4B}$$
Thus we can rewrite (1B) as:
$$\left(V^{-1}V_{t}\right)_t=-\tfrac12\left(V^{-1}V_{t}\right)^2.\,\tag{5B}$$
or
$$\frac{VV_{tt}}{V^2}=\frac{(1-\tfrac12)(V_{t})^2}{V^2}.\,\tag{6B}$$
Eq.(6B) is just Eq.(5).