Given the system of two ODE's $\frac{{d{\mathbf{w}}}}{{d\xi }} = {\mathbf{r}}({\mathbf{w}}(\xi ))$, with $\lambda ({\mathbf{w}}(\xi )) = \xi $ and ${\mathbf{w}}(\lambda ({{\mathbf{u}}_L})) = {{\mathbf{u}}_L}$, ${\mathbf{w}}(\lambda ({{\mathbf{u}}_R})) = {{\mathbf{u}}_R}$
where $\xi=x/t$,
$x,\;t \in \mathbb{R}$
${{\mathbf{u}}_L},{{\mathbf{u}}_R} \in {\mathbb{R}^2}$,
${\mathbf{u}}:{\mathbb{R}^2} \to \mathbb{R}^2$,
${\mathbf{u}}(x,t) = {\mathbf{w}}(x/t) = {\mathbf{w}}(\xi )$
why does that imply that the function ${\mathbf{w(\xi)}}$ continuously connects the constants ${{\mathbf{u}}_R}$ and ${{\mathbf{u}}_L}$ (both given) for a fixed t? Looking at $\frac{{d{\mathbf{w}}}}{{d\xi }} = {\mathbf{r}}({\mathbf{w}}(\xi ))$ which gives the integral curves of the vector field ${\mathbf{r}}$, I integrate
$$\int_{{\mathbf{w}}(\xi = \lambda ({\mathbf{w}}(\xi ) = {{\mathbf{u}}_R}) = \lambda ({{\mathbf{u}}_R}))}^{{\mathbf{w}}(\xi = \lambda ({\mathbf{w}}(\xi ) = {{\mathbf{u}}_L}) = \lambda ({{\mathbf{u}}_L}))} {d{\mathbf{w}}(\xi ) = \int_{\xi = \lambda ({\mathbf{w}}(\xi ) = {{\mathbf{u}}_R}) = \lambda ({{\mathbf{u}}_R})}^{\xi = \lambda ({\mathbf{w}}(\xi ) = {{\mathbf{u}}_L}) = \lambda ({{\mathbf{u}}_L})} {{\mathbf{r}}({\mathbf{w}}(\xi ))} d\xi } $$
which gives
$${\mathbf{w}}(\lambda ({{\mathbf{u}}_L})) - {\mathbf{w}}(\lambda ({{\mathbf{u}}_R})) = {{\mathbf{u}}_L} - {{\mathbf{u}}_R} = \int_{\xi = \lambda ({{\mathbf{u}}_R})}^{\xi = \lambda ({{\mathbf{u}}_L})} {{\mathbf{r}}({\mathbf{w}}(\xi ))d\xi }. $$ Since the integral on the right hand side is a continuous function of ${\mathbf{w}}$ (${\mathbf{r}}$ is continuous), we can write ${{\mathbf{u}}_L} - {{\mathbf{u}}_R} = {\mathbf{f}}({\mathbf{w}}(\xi ))$, so that ${\mathbf{w}}(\xi )$ connects ${{\mathbf{u}}_L}$ and ${{\mathbf{u}}_R}$ continuously.
Is this explanation correct?
(This is from chapter 5.2, p.235 in the book 'Front Tracking for Hyperbolic Conservation Laws' by Risebro, Holden. Link to chapter 5: https://www.uio.no/studier/emner/matnat/math/MAT4380/v15/beskjeder/chapter5.pdf)