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Burgers' equation $$u_t+\Bigl(\frac{u^2}{2}\Bigr)_x=0$$ with initial conditions as below: $$\left\{\begin{aligned}&u(x,0)=2&&\text{if $0<x<1$}\\ &u(x,0)=0&&\text{otherwise} \end{aligned}\right.$$ So on the right hand side of $1$, it would have a shock wave of $x-1=t$. On the left hand side of $0$, it will have a rarefaction wave of $x=2t$.

At some $T_c$, the rarefaction wave would catch up with the shock wave. But the question asks what's the location of $x_s(t)$ after the two waves meet? And it asks to solve it in two ways:

  1. Solve an ODE using Rankine-Hugoniot jump condition
  2. For $t>T_c$, the exact solution is triangular-shaped. Use conservation law(?) to determine $x_s(t)$ based on the area of this triangle?

I am very lost on this problem... When $t<T_c$, I know the shock location is $(1+t,t)$, does it change after they meet?

pluton
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initial_D
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1 Answers1

11

Since the question is old, I'll give the complete solution, not just the $x_s$ part.

Before $T_c$

Draw the $xt$ coordinates, with $t$-axis pointing up. From the given $u(x,0)$, we find three families of characteristic lines:

  1. vertical lines $x=x_0$ with $x_0\le 0$
  2. slope $1/2$ lines $t=\frac12(x-x_0)$ with $0<x_0<1$
  3. vertical lines $x=x_0$ with $x_0\ge 1$

There is a gap between the families 1 and 2: it's a triangular region bounded by $x=0$ and $t=\frac12 x$.
This is the rarefaction wave: within this region, all characteristic lines go through $(0,0)$, the origin of the wave. Therefore, the line through $(x,t)$ has slope $t/x$, which yields $$ u(x,t)=x/t ,\qquad 0<x< 2t $$ This is family 4 of characteristics.

The families 2 and 3 appear to overlap. This means they are separated by shock wave, which originates at $x=1$ at time $t=0$ and moves to the right with velocity $\frac12(2+0)=1$ (the mean of velocities in front and behind the shock, as the jump condition says). Its trajectory is $x=1+t$.

The front edge of rarefaction wave $x=2t$ meets the shock wave $x=1+t$ when $t=1$. Thus, $T_c=1$. The meeting point is $(2,1)$.

After $T_c$

The characteristics of family 2 are no more; the shock is between families 3 and 4. The speed of shock wave as it passes through $(x,t)$ is the mean of velocities in front and behind the shock: $$ \frac{1}{2}\left( \frac{x}{t} +0 \right) = \frac{x}{2t} $$ Therefore, the trajectory of shock wave is described by the ODE $$ \frac{dx}{dt}=\frac{x}{2t} $$ Solve this separable ODE with the initial condition $x(1)=2$ to get $$ x_s(t) = 2 \sqrt{t} $$ This is the trajectory of shock for $t>1$.

Second method, conservation law

Introduce the quantity $P(t)=\int_{-\infty}^\infty u(x,t)\,dx$. It is actually independent of $t$ because $$\frac{dP}{dt} = \int_{-\infty}^\infty u_t \,dx = - \int_{-\infty}^\infty (u^2/2) _x \,dx = (u^2/2) \bigg|_{-\infty}^\infty =0$$ Since $P =2$ at $t=0$, it stays at $2$ for all times. At time $t>T_c$ the function $u$ is equal to $x/t$ for $0< x <x_s$, and is zero otherwise. Thus, its integral is $$ \frac{1}{t} \frac{x_s^2}{2} $$ Equating the above to $2$, we once again get $$ x_s(t) = 2 \sqrt{t} $$

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