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\begin{align} u_t+3u^2u_x &=0, \ x \in \mathbb{R} , \ t>0\tag 1 \end{align} $$u(x,0)=\begin{cases} 3 ,\quad x \leq 1 \quad \\ 2,\quad 1<x<2 \quad \\ 1 ,\quad x\geq 2 \end{cases}$$

The characteristics is: $$\frac{dx}{ds}=3u^2,\ \frac{dt}{ds}=1, \frac{du}{ds}=0$$

Completing now $\frac{du}{ds}=0$ with respect to $s$ yields $u(s)=c_1,\ c_1 \in \mathbb{R}$. Thus, we arrive at the relation $$\frac{dx}{3{c_1}^2}=\frac{dt}{1} \implies x=3{c_1}^2t+c_2, \ c_2 \in \mathbb{R}$$ The general solution of the M.D.E. is expressed in the form $$c_1=G(c_2)\Leftrightarrow u=G(x-3u^2t)$$

Applying now the initial conditions we have:

  • For $x\leq 1$ we have $u(x,0)=3$. Considering $x_1,t_1$ on the characteristic projection intersecting the $x$ axis at $(\bar{x_1},0)$ we have $$x_1-3u^2(x_1,t_1)t=\bar{x_1} \leq 1$$ and $u(x_1,t_1)=u(\bar{x_1},0)=3$

Therefore $$u(x,t)=3,\ x \leq 1$$

  • For $1<x< 2$, $u(x,0)=2$ holds. Consider $(x_2,t_2)$ on the characteristic projection intersecting the $x$ axis at $(\bar{x_2},0)$ we have $$x_2-3u^2(x_2,t_2)t=\bar{x_2} \in (1,2) $$ and $$u(x_2,t_2)=u(\bar{x_2},0)=2$$ So $1<x_2-t_2<2 \Leftrightarrow 1+t_2<x_2<2+t_2$

Therefore, $$u(x,t)=2,\ 1+t<x<2+t$$

  • For $x \geq 2$, $u(x,0)=1$. Consider $(x_3,t_3)$ on the characteristic projection intersecting the $x$ axis at $(\bar{x_3},0)$ we have $$x_3-3u^2(x_3,t_3)t=\bar{x_3} \geq 2 $$ and $$u(x_3,t_3)=u(\bar{x_3},0)=1$$

Therefore, $$u(x,t)=1,\ x\geq 2$$

Summarizing the results, we have the following:

$$u(x,t)=\begin{cases} 3 ,\quad x \leq 1 \quad \\ 2,\quad 1+t<x<2+t \quad \\ 1 ,\quad x\geq 2 \end{cases}$$

with characteristic projections

$$ x=27t+c_1 \ \text{with} \ u(\bar{x},0) \ \text{and} \ \bar{x}\leq 1$$ $$ x=12t+c_2 \ \text{with} \ u(\bar{x},0) \ \text{and} \ 1<\bar{x}< 2,$$

$$ x=3t+c_3 \ \text{with} \ u(\bar{x},0) \ \text{and} \ \bar{x} \geq 2$$

I have found three parallel projections. Is it possible? How could I found the breaking time?

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Athanasios Paraskevopoulos
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1 Answers1

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Here, shock waves are formed initially at the breaking time $t=0$, see footnote. To explain what happens, let us rewrite the PDE as $u_t + (u^3)_x =0$, and observe that we are dealing with Riemann problems for a scalar conservation law with nonconvex flux (see this post). The solution consists of two shock waves $$ u(x,t) = \left\lbrace \begin{aligned} &3, && x< 1+s_1 t\\ &2, && 1+s_1 t < x< 2+s_2 t \\ &1, && x> 2+s_2 t \end{aligned}\right. $$ with Rankine-Hugoniot speed \begin{aligned} &s_1 = \frac{3^3 - 2^3}{3-2} = 19,\\[5pt] &s_2 = \frac{2^3 - 1^3}{2-1} = 7 . \end{aligned} The characteristic curves $x = x_0+3u(x_0,0)^2 t$ must be stopped when they meet the trajectory of a shock wave $x=1+s_1t$ or $x=2+s_2t$.

It remains to discuss what happens when the two shock waves interact, i.e. $$ 1+s_1 t^* = 2+s_2 t^* $$ for some time $t^*>0$ to be determined, that is $t^* = \frac1{12}$. Here, a new shock wave is formed, with left state $u_l=3$ and right state $u_r=1$ at the time of interaction $t^*$ (I presume that this is the requested "breaking" time).

Note: Some authors denote the shock formation time breaking time. From there, the method of characteristics used in OP has limited validity, as characteristic curves must be stopped at discontinuities.

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