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Consider the conservation law
$$ u_t + (u^4)_x = 0, $$
(a) Find the solution at $t=1$ with the following initial condition:
$$ u(x,0) = leftlbracebeginaligned &1 && x<0 \ &2 && 0leq x leq 2 \ &0 && x>2 endaligned right. . $$
(b) Solve the Riemann problem (You must consider both $u_l>u_r$ and $u_l<u_r$):
$$ u(x,0) = leftlbracebeginaligned &u_l && x<0 \ &u_r && x>0 endaligned right. . $$
(c) Find the Riemann solution at $x/t = 0$.

Try:

The characteristic are given by $x = 4 g(r)^3 t + r $ where $r$ is parameter. so we have

$$ x = begincases 4t+r, & r<0 \ 8t+r, & 0 leq r leq 2 \ r, & r > 2 endcases $$

We have two shocks formations at $x=0$ and $x=2$ for $t=0$. We first consider the shock at $x=0$, using R=H condition, we want

$$ xi_1'(t) = frac 2^4 - 1^4 2-1 = 15 implies xi_1(t) = 15t $$

and at $(x,t) = (2,0)$ we have

$$ xi_2'(t) = frac - 2^4 0-2 = 8 implies xi_2(t) = 8t+2$$

So we can write our solution for part a

$$ boxed u(x,t) = begincases 1, & x < 15 t \ 2, & 15t < x < 8t+2 \ 0, & x > 8t+2 endcases $$

IS this correct? I have a question as to what is it that they are asking in c)?

Here is a plot of the characteristic lines in the $x$-$t$ plane.

Since the flux $u mapsto u^4$ is convex, the classical theory for entropy solutions of conservation laws applies. Without entering into details, the rarefaction wave generated at $x=0$ and the shock wave generated at $x=2$ leads to the solution
$$
u(x,t) = leftlbrace
beginaligned
& 1 && x leq 4 t \
& sqrt[3]x/(4t) && 4 t leq x leq 32 t\
& 2 && 32 t leq x leq 2 + 8 t \
& 0 && x geq 2 + 8 t
endaligned
right.
$$
valid for small times $t < 1/12$. For larger times, one must compute the interaction of the rarefaction with the shock (see e.g. related posts on this site). Asking to find the Riemann solution at $x/t = 0$ is the same as asking to find the solution at $x = 0$ for nonzero time.