Ngô Quốc Anh

March 31, 2010

Wave equations: Shock Formation

Filed under: PDEs — Ngô Quốc Anh @ 0:01

So far once we have discontinuous initial data, we might have shocks. This also occurs even the given initial data is smooth. Let us still consider the following problem

\displaystyle u_t+c(u)u_x=0,\quad x\in \mathbb R, t>0

subject to the following initial condition

\displaystyle u(x,0)=u_0(x), \quad x\in \mathbb R

where c(u)>0, c'(u)>0 and u_0 \in C^1.

From a discussion of this entry we consider the case when u_0 is non-increasing as we need singularities (because c'>0). Moreover, the solution can be given implicitly as

\displaystyle\left\{ \begin{gathered} u(x,t) = {u_0}(\xi ), \hfill \\ x - \xi = c({u_0}(\xi ))t. \hfill \\ \end{gathered} \right.

Since u_0 is non-increasing, u'_0(x)<0 on \mathbb R. We now consider the characteristics which are straight lines, issuing from two points \xi_1 and \xi_2 on the x axis with \xi_1 < \xi_2 have speeds c(u_0(\xi_1)) and c(u_0(\xi_2)) respectively. Because u_0 is decreasing and c is increasing, it follows that

\displaystyle c(u_0(\xi_1))>c(u_0(\xi_2)).

In other words, the characteristic emanating from \xi_1 is faster than the one emanating from \xi_2.

Therefore the characteristics cross so a smooth solution cannot exist for all t>0.

We are now interested in calculating the breaking time. To this purpose, we calculate u_x along a characteristic which has equation


Let g(t)=u_x(x(t),t) denote the gradient of u along the characteristic x=x(t) given as above. Then

\displaystyle \frac{dg}{dt}=u_{tx}+c(u)u_{xx}.

By differentiating the PDE with respect to x we also have

\displaystyle u_{tx}+c(u)u_{xx}+c'(u)u_x^2=0.

Comparing gives

\displaystyle \frac{dg}{dt}=-c(u)g^2

along the characteristic. Solving this ODE gives us

\displaystyle g=\frac{g(0)}{1+g(0)c'(u_0(\xi))t}

where g(0) is the initial gradient at t=0. Thus

\displaystyle {u_x} = \frac{{{{u'}_0}(\xi )}}{{1 + {{u'}_0}(\xi )c'({u_0}(\xi ))t}}.

The fact that u'_0 and c' have different sign implies that u_0 will blow up at a finite time along the characteristic. What we need to do is to examine u_0 along all characteristics to find such \xi so that u_x first blows up.

A simple calculation show by denoting F(\xi)=c(u_0(\xi)) that the time of the first breaking is

\displaystyle t_b=-\frac{1}{F'(\xi_b)}

where \xi_b is such that |F'(\xi)| is maximum. An in-depth observation shows that if the initial data is not monotone, breaking will first occur on the characteristic \xi=\xi_b for which F'(\xi)<0 and |F'(\xi)| is maximum.

Source: J.D. Logan, An introduction to nonlinear partial differential equations, 2nd, 2008; Section 3.3.


  1. em thua thay theay co the post len may cai de thi cao hoc cua truong minh cac nam 2008 va 2009 khong ah ,em cam on thay

    Comment by nguyenvanphan — April 2, 2010 @ 14:18

    • Hi bạn, đề CH thì chắc là vẫn thế thôi, cứ ôn luyện đề thi của các năm trước nữa là ổn thôi mà. Chúc may mắn nhé.

      Comment by Ngô Quốc Anh — April 2, 2010 @ 14:25

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