Hello I have a question about sketching characteristics, I have 2 Burgers equations that are as follows: $$u_t+u\cdot u_x=0, \>\> u(x,0)=x^3$$ $$u_t+uu_x=0\>\> u(x,0)= \begin{cases} 1 & \text{ if } x\leq 0\\ 1-x & \text{ if } 0\leq x\leq 1\\ 0 & \text{ if } x\geq 1 \end{cases}$$ Now for the first I am asked to sketch the characteristics in general and sketch the profiles of $u(x,0)$ and $u(x,1)$, and for the second I am asked the sketch the characteristics in the $xt$-plane for $0\leq t\leq 1$ and then sketch $u(x,0), u(x,1/2), u(x,3/2),u(x,1)$.
I am beyond confused as to what to sketch, or even how to find the profiles.
I know the Burger's characteristics start off as vertical lines and then become less steeper until a point where the characteristics start sloping the other direction which causes shock (namely at $t=1$),and I believe the speed at which this happens is faster for the first PDE than the second, but this still doesn't help me sketch the characteristics. If anyone could help me with this I would appreciate it greatly. Thank you!
For the first, the solution is $$u=(x-tu)^3$$ which is an implicit form to express the solution $u(x,t)$. It could be made explicit in solving the cubic equation for $u$. But it is not necessary to answer for the wanted drawing.
To draw it on the $xt$-plane, for each value of $u$, draw : $$x=u\,t+u^{1/3}$$ that is a family of straight lines.
For the second, the solution is given in PDE with strange Auxiliary Conditions
To draw it on the xt-plane, for each value of $u$, draw the corresponding function $x(t)$ derived from the equation $u(x,t)$ explicitly found.
Especially observe what occurs then $t\to 1$.