Solve:
ut + ((x-1)u)x=0, x E(0,1) & t > 0
I want to try the solution u(x,t) = f(x + ct), but think this might be incorrect, any ideas?
If you directly substitute your solution form into the PDE you'll get
$\displaystyle
cf' +(x-1)f' + f = 0.
$
If you let $\displaystyle x = r - ct $ this becomes
$\displaystyle cf'(r) + (r - ct - 1)f'(r) + f(r) = 0$
Now differentiate wrt $\displaystyle t$ gives $\displaystyle f'(r) = 0$. So the only form is $\displaystyle f(r) = $constant.
The actual solution of the PDE is $\displaystyle u = e^{-t}f\left((x-1)e^{-t}\right)$.