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!

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$
$\dfrac{dx}{ds}=a$ , letting $x(0)=x_0$ , we have $x=as+x_0=at+x_0$
$\dfrac{dy}{ds}=b$ , letting $y(0)=y_0$ , we have $y=bs+y_0=bt+y_0$
$\dfrac{du}{dt}=0$ , letting $u(0)=f(x_0,y_0)$ , we have $u(x,y,t)=f(x_0,y_0)=f(x-at,y-bt)$
$u(x,y,0)=g(x,y)$ :
$f(x,y)=g(x,y)$
$\therefore u(x,y,t)=g(x-at,y-bt)$
