How do you solve $$ \begin{cases}u_t-\frac{x}{t^2+1}u_x=0, & x,t\in\mathbb{R}\\u(0,x)=u_0(x)\in C^1(\mathbb{R})\end{cases} $$ by the method of characteristics?
My approach is to consider the parametrisation $t=t(\tau,\xi), x=x(\tau,\xi), u=u(\tau,\xi)$. Then the ODE to solve are $$ \begin{align*} &\frac{dt}{d\tau}=1\\ &\frac{dx}{d\tau}=-\frac{x}{t^2+1}\\ &\frac{du}{d\tau}=0 \end{align*} $$ with initial conditions $$ \begin{align*} &t(0,\xi)=0\\ &x(0,\xi)=\xi\\ &u(0,\xi)=u_0(\xi). \end{align*} $$ Solving this, what I get is $$ t=\tau,\quad u=u_0(\xi),\quad x=-\frac{x}{t^2+1}t+\xi. $$
(Does this make sense? I have the feeling that it should be $x(t,\xi)=-\frac{\xi}{t^2+1}t+\xi$ instead.)
I have two questions:
(A) Isn't the right-hand side of $$ \frac{dx}{d\tau}=-\frac{x}{t^2+1} $$ globally Lipschitz continuous with respect to the second argument $x$, since $$ \left\lvert\frac{-x_1+x_2}{t^2+1}\right\rvert\leqslant \lvert x_2-x_1\rvert? $$ Doesn't this imply that this ODE has a unique global solution defined for all $(x,t)\in\mathbb{R}^2$ which then implies that the given PDE above has a unique global solution?
(2) How can I get an explicit expression of $u(x,t)$ depending on $u_0$?