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

x-\xi=c(u_0(\xi))t.

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.

2 Comments »

  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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