The question i just can't figure out one bit is:
Solve for the continious function $u(x,t)$: $$u_t+(1-2u)u_x=0 $$ $$-\infty < x < \infty, t>0$$ $$u(x,0)= \left\{ \begin{matrix} \frac{1}{4} & \mbox{for} & x < 0 \\ \frac{3}{4} & \mbox{for} & x>0\end{matrix}\right.$$
a.) solve for $u(x,t)$ (clarify in pictures how the solution evolves in time)
b.) Determine $u(x,t)$ if the IVs change to: $$u(x,0)= \left\{ \begin{matrix} \frac{1}{4} & \mbox{for} & x > 0 \\ \frac{3}{4} & \mbox{for} & x<0\end{matrix}\right.$$
I know you probably should start of like this:
$\frac{du}{dt} = \frac{\partial u}{\partial t} + \frac{dx}{dt}\frac{\partial u}{\partial x}$
If $ \frac{dx}{dt}=(1-2u)$ then $\frac{du}{dt} =0$
So $x = t - 2u(x_0,0)t + x_0$ with $x_0=x(t)$
Rewrite $x_0 = x - t + 2u(x_0,0)t$.
Use IVs with $x=x_0$:
$u(x_0,0)= \left\{ \begin{matrix} \frac{1}{4} & \mbox{for} & x_0 < 0 \\ \frac{3}{4} & \mbox{for} & x_0>0\end{matrix}\right.$
$u(x_0,0)= \left\{ \begin{matrix} \frac{1}{4} & \mbox{for} & x - t + 2 \cdot \frac{1}{4}t < 0 \\ \frac{3}{4} & \mbox{for} & x - t + 2\cdot \frac{3}{4}t>0\end{matrix}\right.$
$u(x,t)= \left\{ \begin{matrix} \frac{1}{4} & \mbox{for} & x < \frac{1}{2}t \\ \frac{3}{4} & \mbox{for} & x>-\frac{1}{2}t\end{matrix}\right.$
And from here on out i'm just completely blank.