Method of Characteristics $au_x+bu_y+u_t=0 $ $au_x+bu_y+u_t=0$
$u(x,y,0)=g(x,y)$
solve $u(x,y,t)$
Our professor talked about solving this using Method of Characteristics. However, I am confused about this method. Since it's weekend, I think it might be faster to get respond here. In the lecture, he wrote down the followings:
Fix a point$(x,y,t)$ in $\mathbb{R}^3$.
$h(s)=u(x+as,y+bs,t+s)$,line $φ(s)=(x+as,y+bs,t+s)=(x,y,t)+s(a,b,1)$
$h'(s)=u_xa+u_yb+u_t=0$ for all $s$.
$h(-t)=u(x-at,y-bt,0)=g(x-at,y-bt)$ <----- u equal this value for all points on the line $(x+as,y+bs,t+s)$.
$h(0)=u(x,y,t)$
$u(x,y,t)=g(x-at,y-bt)$
The first question I have is that why we want to parametrize $x,y$ and $t$ this way. In addition, what is the characteristic system of this problem. If we have derived the formula already, why do we still need the characteristics system equations? Thank you!

I think it's easiest just to concisely re-explain the method, so that's what I'll do.
The idea: linear, first-order PDEs have preferred lines (generally curved) along which all the action happens. More specifically, because the differential bit takes the form of $\mathbf f \cdot \nabla u$ where in general $\mathbf f$ varies, it is actually always a directional derivative along the vector field $\mathbf f$.
Therefore, along a line given by $\dot{\mathbf x}(s) = \mathbf f$, we expect the PDE to reduce to an ODE involving $\mathrm d u(\mathbf x(s))/ \mathrm d s$. In fact, by the chain rule,
$\mathrm d u(\mathbf x(s))/ \mathrm d s = \dot{\mathbf x}(s) \cdot\nabla u = \mathbf f \cdot\nabla u$
which is exactly the term we said was in the PDE.
So $\mathbf f \cdot\nabla u = h(\mathbf x)$ is equivalent to 
$$\mathrm d u(\mathbf x(s))/ \mathrm d s = h(\mathbf x(s))$$
Therefore, by finding $\mathbf x(s)$ we can find the ODE $u$ satisfies, and find the initial conditions relevant to each line by saying that at $s=0$ we are on the space where the initial conditions are given.
Does this help? If you have a more specific question, ask away!
