How do we know the uniqueness of the transport equation? I'm learning PDE and have a basic question.
In Evans book on PDE, in order to solve $\partial_t u+\langle b, Du\rangle=0$ with the initial condition $u(0,\cdot)=g$, I first have to use the auxillary function $w(s)=u(t+s,x+sb)$ and realize that $u(t+s,x+sb)=u(t,x)$.  Now we can plug in $s=-t$ and use the initial condition to derive that the solution is of the form $u(x,t)=g(x-tb)$. But here, since we are using some "auxillary function," analytically speaking I don't know the uniquenss, right? Can somebody give me some insight on how to achieve uniqueness in the simplest way? Thanks!
 A: If $\partial_tu + \langle b,\nabla u\rangle = 0$ for some $b \in \Bbb R^n$, and we wish to solve for $u$ (defined on some subset of $\Bbb R^{n+1}$), we try to reconstruct the graph of $u$ inside $\Bbb R^{n+2}$, by rewriting the PDE as $$\langle (\partial_tu, \nabla u, -1),(1,b,0)\rangle = 0.$$Note that $(\partial_tu, \nabla u, -1)$ would be the normal direction to the graph of $u$. The flow of the vector field $V(t,x,u)=(1,b,0)$ (which will be tangent to the graph) is given by $$(t,x,u)=\Phi_V(s,(t_0,x_0,u_0))=(t_0+s,x_0+sb, u_0).$$So, saying that $u(0,x)=g(x)$ is saying that for $t_0=0$, we have that $u_0=g(x_0)$. Now, since $$x_0 = x-sb \quad\mbox{and}\quad s = t,\quad\mbox{and}\quad u=u_0$$we have $x_0 = x-tb$. Thus, the general solution is of the form $u(t,x)=g(x-tb)$. If you do not want to make reference to $g$, write $u(t,x) = u(0,x-tb)$, this is what the transport equation gave us.
In the same way when solving an ODE you get a constant of integration which depends on initial conditions, for PDEs you have not a constant, but a function which depends on initial conditions. So, changing the initial conditions changes this function as well and you don't quite have uniqueness.
