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.

Leave a Comment »

No comments yet.

RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Create a free website or blog at

%d bloggers like this: