# Ngô Quốc Anh

## March 27, 2010

### Wave equations: Shock propagation (shock fitting)

Filed under: PDEs — Ngô Quốc Anh @ 22:36

Let us consider the initial value problem

$\displaystyle u_t+uu_x=0, \quad x\in \mathbb R, t>0$

ubject to the initial condition

$\displaystyle {u_0}(x) = \left\{ \begin{gathered} 1, \quad x < 0, \hfill \\ - 1, \quad 0 < x < 1, \hfill \\ 0, \quad x > 1. \hfill \\ \end{gathered} \right.$

Note that the characteristics emanate with speed $u$ from the $x$ axis.

That means the characteristics starting from $x_0$ are given as the following

$\displaystyle \left\{ \begin{gathered} x-t=x_0, \quad x_0 < 0, \hfill \\ x+t=1-x_0, \quad 0 < x_0 < 1, \hfill \\ x=x_0, \quad x_0 > 1. \hfill \\ \end{gathered} \right.$

Actually, I have not shown this fact before, if I have time then I will post something regarding to the characteristics of nonlinear equations. Of course the shock must form $t=0$. The shock must start with discontinuity points. Clearly, $u_0$ is discontinuous at $x=0$ and $x=1$. However, at $x=1$, the characteristics do not intersect each other, therefore the shock must start at $x=0$. By the Rankine-Hugoniot condition, the speed of the shock is

$\displaystyle s=\frac{u_1+u_2}{2}$

where $u_1$ is the value of $u$ ahead of the shock and $u_2$ is the value of $u$ behind the shock. Clearly, a shock must from at $t=0$ with speed

$\displaystyle s=\frac{-1+1}{2}=0$

and propagate until time $t=1$. To continue the shock beyond $t=1$, we must know the solution ahead of the shock. Therefore, we introduce an expansion wave

$\displaystyle u=\frac{x-1}{t}$.

The shock beyond $t=1$ according to the jump condition has speed

$\displaystyle s=\frac{\frac{x-1}{t}+1}{2}$.

To find the shock path we note that $s=\frac{dx}{dt}$, so the above equation is a first order ordinary differential equation for the shock path. Obviously, this ODE is given as

$\displaystyle \frac{dx}{dt}-\frac{x}{2t}=\frac{1-\frac{1}{t}}{2}$,

which is also a linear equation. The initial condition is $x=0$ at $t=1$ where the shock starts beyond $t=1$. The solution is

$\displaystyle x=s(t)=t+1-2\sqrt{t},\quad 1 \leq t\leq 4$.

The above shock will propagate until $x=1$ and $t=4$. When $t>1$, the new jump condition is

$\displaystyle s=\frac{0+1}{2}=\frac{1}{2}$.

Therefore the shock is a straight line with speed $\frac{1}{2}$ for $t>4$ and its equation is

$\displaystyle x-1=\frac{t-4}{2}, \quad t>4$.

Source: J.D. Logan, An introduction to nonlinear partial differential equations, 2nd, 2008; Section 3.1.