$$u_t+\text{yu}_x+\frac{1}{2}\text{(u(u-1)})_y\text{=0}$$ The initial condition is given as $$\text{u(x,y,0)=}u_0\text{(x,y)}$$
I know what $$\frac{\text{dt}}{\text{ds}}\text{ and}\frac{\text{dx}\backslash }{\text{ds}}\text{ is}$$. How do I know what $$\frac{\text{dy}}{\text{ds}}$$ is? I really need some help!
$u_t+yu_x+\dfrac{1}{2}(u(u-1))_y=0$
$u_t+yu_x+\dfrac{2u-1}{2}u_y=0$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dt}{ds}=1$ , letting $t(0)=1$ , we have $t=s$
$\dfrac{du}{ds}=0$ , letting $u(0)=u_0$ , we have $u=u_0$
$\dfrac{dy}{ds}=\dfrac{2u-1}{2}=\dfrac{2u_0-1}{2}$ , letting $y(0)=y_0$ , we have $y=\dfrac{(2u_0-1)s}{2}+y_0=\dfrac{(2u-1)t}{2}+y_0$
$\dfrac{dx}{ds}=y=\dfrac{(2u_0-1)s}{2}+y_0$ , letting $x(0)=f(y_0,u_0)$ , we have $x=\dfrac{(2u_0-1)s^2}{4}+y_0s+f(y_0,u_0)=yt-\dfrac{(2u-1)t^2}{4}+f\left(y-\dfrac{(2u-1)t}{2},u\right)$
$u(x,y,0)=u_0(x,y)$ :
$\therefore x=yt-\dfrac{(2u-1)t^2}{4}+u_0\left(y-\dfrac{(2u-1)t}{2},u\right)$