We needed to add the restriction $``z>0’’$ in Definition 4.9, that is,
$$ u(x+z,t)-u(x,t)\leq C\left( 1+\frac{1}{t} \right) z \quad \text{for a.e. x, z, t with $z>0$}. $$
In LectNote12, we proved that if $u\in C^1(\mathbb{R}\times [0,\infty))$ is a soltuion to
$$ \begin{equation} \begin{dcases} u_t(x,t)+b(x,t)u_x(x,t)=0, \quad (x,t)\in \mathbb{R}\times (0,\infty), \\ u(x,0)=u_0(x), \quad x\in \mathbb{R}, \end{dcases} \end{equation} $$
then $u$ should be
$$ \begin{equation} u(x,t)=u_0(\phi(0;x,t)), \end{equation} $$
where $(\phi(s;x,t))_{0\leq s\leq t}$ is the solution to
$$ \begin{equation} \begin{dcases} \frac{d}{ds}\phi(s;x,t)=b(\phi(s;x,t),s), \quad 0\leq s\leq t, \\ \phi(t;x,t)=x. \end{dcases} \end{equation} $$
However, we did not show that this is a solution. This is proved as follows.
Proof We first note that the identity
$$ \begin{equation} \phi(s;x,t)=\phi(s;\phi(\tau;x,t),\tau), \quad \tau \leq s \leq t \end{equation} $$
holds because both sides are the unique solution to the ODE
$$ \begin{dcases} \frac{d}{ds}\psi(s)=b(\psi(s),s), \quad \tau \leq s\leq t, \\ \psi(\tau)=\phi(\tau;x,t). \end{dcases} $$
Now differentiate (4) with respect to $\tau$ and use (3) to get
$$ \frac{\partial \phi(s;\phi(\tau;x,t),\tau)}{\partial \tau}+\frac{\partial \phi(s;\phi(\tau;x,t),\tau)}{\partial x}b(\phi(\tau;x,t),\tau)=0. $$
Set $\tau=t$ to obtain
$$ \begin{equation} \frac{\partial \phi(s;x,t)}{\partial t}+\frac{\partial \phi(s;x,t)}{\partial x}b(x,t)=0. \end{equation} $$
We may now use (5) to easily verify that (2) solves (1). QED