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I'm looking for a solution to the following first order problem $$ u_t+A(u)_x=0\quad\text{in }\mathbb{R}\times(0,+\infty) $$ with initial condition $$ u(x,0)=\begin{cases}1& x\leq0\\ 0& x>0.\end{cases} $$ Since the initial condition is decreasing, we may find a shock wave solution. I read that there is also the chance that we may find a solution with two discontinuity lines (I guess this depends mainly on the expression of $A$). For instance I read that this may happen when $A(u)=\frac{u^3}{3}$. Arguing by characteristics, we infer that the solution should only take values 0 and 1, but I cannot figure out geometrically how this double discontinuity could take place without breaking the Rankine–Hugoniot condition. I guess that I haven't figured out properly the characteristic lines.

Any hint is very appreciated.

Paul
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1 Answers1

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Consider a weak solution of the initial-value problem with two shocks: $$ u(x,t) = \left\lbrace \begin{aligned} &1 & & x<s_1 t\\ &u_m & & s_1 t < x< s_2 t \\ &0 & & s_2 t < x \\ \end{aligned} \right. $$ and the non-convex flux $A(u) = \frac13 u^3$. Now, we only need to adjust the jump value $u_m\in \Bbb R$ in such a way that the Rankine-Hugoniot conditions $$ s_1 = \underbrace{\frac{A({u_m}) - A(1)}{u_m - 1}}_{\frac13(u_m^2 + {u_m} + 1)} \qquad\quad <\qquad\quad s_2 = \underbrace{\frac{A(0) - A({u_m})}{0 - u_m}}_{\frac13 u_m^2} $$ are satisfied. Choosing $u_m< {-1}$ provides a weak solution to the Cauchy problem with two shock waves. Note that this solution isn't admissible, but that a semi-shock combined with a shock would be admissible (even though it has too many constant states...).

EditPiAf
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