Use the method of characteristics to solve the problem $$ u_t+ u u_x = 0 $$ with initial data $$ u(0,x) = \begin{cases} 1& x < 0 \\ 1- x & x \in [0,1] \\ 0 & x >1 . \end{cases} $$ What is the time interval where a Lipschitz solution exists ?
I have thus far the following: Consider the coefficients of each partial derivative, i.e. let: $$ \frac{dt}{ds} = 1 \qquad \frac{dx}{ds} = z \qquad \frac{dz}{ds} = 0$$ With the following conditions: $$ t(r,0) = 0 \qquad x(r,0) = r \qquad z(r,0) =\begin{cases} 1& r < 0 \\ 1- r & x \in [0,1] \\ 0 & r >1 . \end{cases}$$
First integration will give us:
$$t = s + c_1 \qquad z = c_2 \qquad \frac{dx}{ds} = c_2$$
Now since $t(r,0) = c_1 = 0$, we get that $t = s$. Hence $x = c_2(r)s + c_3(r) = c_2(r)t + c_3(r)$. So suppose the general form of this solution is $F(x-ut)$. Now the particular solution with regards to the initial condition, $u(x,0) = F(x)$, Hence the solution would be:
\begin{cases} 1 & \qquad x-ut < 0 \cr 1- (x-ut) & \qquad (x-ut) \in [0,1] \cr 0 & \qquad (x-ut) >1 . \end{cases} \begin{cases} 1 & \qquad x < t \cr 1- x+ ut & \qquad (x-ut) \in [0,1] \cr 0 & \qquad x >1. \end{cases}
What am doing right or wrong? Is the solution I have Lipschitz?
